Hayk Sedrakyan Nairi Sedrakyan Methods of Proving

Problem Books in Mathematics

Hayk Sedrakyan Nairi Sedrakyan Geometric Inequalities Methods of Proving Problem Books in Mathematics

Series Editor: Peter Winkler Department of Mathematics Dartmouth College Hanover, NH 03755 USA

More information about this series at http://www.springer.com/series/714 Hayk Sedrakyan • Nairi Sedrakyan

Geometric Inequalities Methods of Proving

AC r1 + r2 > r3 Hayk Sedrakyan Nairi Sedrakyan University Pierre and Marie Curie Yerevan, Armenia Paris, France

ISSN 0941-3502 ISSN 2197-8506 (electronic) Problem Books in Mathematics ISBN 978-3-319-55079-4 ISBN 978-3-319-55080-0 (eBook) DOI 10.1007/978-3-319-55080-0

Library of Congress Control Number: 2017937367

Mathematics Subject Classiﬁcation (2010): 00A07

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This Springer imprint is published by Springer Nature The registered company is Springer International Publishing AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland To Margarita, a wonderful wife and a loving mother

To Ani, a wonderful daughter and a loving sister Preface

Geometric inequalities are one of the most interesting sections of elementary mathematics and have a wide range of applications in geometry and the other fields of mathematics, such as algebra and trigonometry. To prove geometric inequalities one often has to use, besides the geometric reasoning, algebraic transformations, trigonometric relations and inequalities, calculus and mathematical analysis. This book is the third book of the authors about inequalities. The first two books were dedicated to algebraic inequalities and were published in 2015 in South Korea. All these books reflect long years of experience of the authors in teaching. Most of the problems were created or proved by the authors during those classes. The authors have tried not to use, whenever possible, the concept of a derivative, therefore making the solutions of many problems understandable to students. The book contains more than 1000 problems. Approximately 800 problems in the book are with thorough solutions. Basically, these are non-standard problems. The majority of problems are for mathematics competitions and Olympiads. Many problems in the book and the majority of the solutions belong to the authors. Some of those problems were used by the authors to teach their students interested in mathematical Olympiads. In few cases, the solution was proposed by a student, so his/her name is indicated. Some problems of the authors, included in this book, were proposed in mathematical Olympiads (in different countries). Some problems were proposed in different mathematical journals, such as the American Mathe- matical Monthly (MAA), Crux Mathematicorum with Mathematical Mayhem (Canadian Mathematical Society), Mathematical Reflections (USA), and Kvant (Russia). The book is divided into eight chapters, each of the chapters consists of one, two, three, four or five paragraphs. The basis of the classification is usually the method of the solution.

vii viii Preface

The authors have tried to ﬁnd common approaches to different problems. The goal of the book is to teach the reader new and classical methods for proving geometric inequalities. The authors would like to express their gratitude to their family for the support.

Hayk Sedrakyan Nairi Sedrakyan Contents

1 Theorem on the Length of the Broken Line ...... 1 1.1 Triangle Inequality ...... 2 Solutions ...... 5 Problems for Self-Study ...... 20 1.2 Theorem on the Length of the Broken Line ...... 24 Solutions ...... 26 Problems for Self-Study ...... 36 2 Application of Projection Method ...... 39 2.1 Convex Polygon Lying Inside of Another Polygon ...... 40 Solutions ...... 41 Problems for Self-Study ...... 58 2.2 Sufﬁcient Conditions for Comparison of Lengths of Two Broken Lines on the Plane ...... 59 Solutions ...... 60 Problems for Self-Study ...... 65 2.3 Inscribed Polygons with the Least Perimeter ...... 66 Solutions ...... 67 Problems for Self-Study ...... 74 2.4 Method of Projections ...... 74 Solutions ...... 76 Problems for Self-Study ...... 88 3 Areas ...... 93 3.1 Inequalities with Areas ...... 93 Solutions ...... 98 Problems for Self-Study ...... 135

ix x Contents

4 Application of Vectors ...... 139 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities ...... 139 Solutions ...... 143 Problems for Self-Study ...... 163 5 Application of Trigonometric Inequalities ...... 167 5.1 Inequalities for the Angles of a Triangle ...... 168 Solutions ...... 170 Problems for Self-Study ...... 189 5.2 Inequalities for the Angles of Acute and Obtuse Triangles ...... 191 Solutions ...... 191 Problems for Self-Study ...... 198 5.3 Some Relations for a Triangle ...... 198 Solutions ...... 200 Problems for Self-Study ...... 205 5.4 Trigonometric Inequalities ...... 205 Solutions ...... 206 Problems for Self-Study ...... 217 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities ...... 218 Solutions ...... 223 Problems for Self-Study ...... 258 6 Inequalities for Radiuses ...... 261 6.1 Inequalities for Radiuses of Circles ...... 262 Solutions ...... 263 Problems for Self-Study ...... 273 6.2 Integer Lattice ...... 275 Solutions ...... 276 Problems for Self-Study ...... 283 7 Miscellaneous Inequalities ...... 285 7.1 Miscellaneous Inequalities ...... 286 7.2 Solutions ...... 300 7.3 Problems for Self-Study ...... 407 8 Some Applications of Geometric Inequalities ...... 413 8.1 Application of Geometric Inequalities for Solving Geometric problems ...... 413 Solutions ...... 416 Problems for Self-Study ...... 428 Contents xi

8.2 Using Geometric Inequalities for Proving Algebraic Inequalities ...... 430 Solutions ...... 432 Problems for Self-Study ...... 443

Basic Notations ...... 447

References ...... 449 Chapter 1 Theorem on the Length of the Broken Line

This chapter consists of two sections. Section 1.1 is devoted to the applications of one of the most important geometric inequalities, called the triangle inequality. As a generalization of triangle inequality, Section 1.2 is devoted to the theorem on the length of the broken line. Let us recall the triangle inequality: for any triangle, the sum of the lengths of any two sides is greater than or equal to the length of the remaining side. In other words, if A, B, C are arbitrary points, then AB AC þ BC. Moreover, the equality holds true, if and only if point C is on segment AB. The goal of Section 1.1 is to get the reader acquainted with the triangle inequality. This section consists of problems that can be proved by using, if necessary several times, the triangle inequality. In Section 1.1 selected problems are those that can be proved using the following techniques and statements: 1. For any convex quadrilateral ABCD it holds true AC þ BD > AB þ CD. 2. For any triangle with side lengths a, b, c and a median ma drawn to the side with < bþc length a it holds true ma 2 . 3. If M is an arbitrary point inside of triangle ABC, then MA þ MB < CA þ CB. 4. For any points A, B, C, D it holds true AC Á BD AB Á CD þ BC Á AD. In some problems, the triangle inequality is not always applied directly. In some cases, at ﬁrst one needs to do some geometric constructions or translations and only after that apply the triangle inequality. In Section 1.2, selected problems are those that can be proved using the theorem on the length of the broken line, that is: if A1, A2,...,An are arbitrary n points, then A1An A1A2 þ A2A3 þ ...þ An À 1An. Moreover, the equality holds true if and only if point Ai is on the segment Ai À 1Ai þ 1, for i ¼ 2, 3, . . . , n À 1. Some problems in this chapter were inspired by [7]. Nevertheless, even for these problems the authors have mostly provided their own solutions.

© Springer International Publishing AG 2017 1 H. Sedrakyan, N. Sedrakyan, Geometric Inequalities, Problem Books in Mathematics, DOI 10.1007/978-3-319-55080-0_1 2 1 Theorem on the Length of the Broken Line

1.1 Triangle Inequality

1.1.1. Consider a triangle ABC. Prove that (a) ∠C < ∠ A,ifAB < BC, (b) AB < BC,if∠C < ∠ A, (c) AB < BC þ AC, (d) the length of any side of the triangle is less than its semiperimeter. 1.1.2. Let A, B and C be arbitrary points. Prove that AC |AB À BC|. 1.1.3. Given a convex quadrilateral ABCD and a point M on a plane. Prove that (a) MA þ MB þ MC þ MD AC þ BD, (b) MA < MB þ MC þ MD,ifAB ¼ CD, (c) MA < MB þ MC þ MD,ifAC ¼ BD. 1.1.4. (a) Prove that in a convex quadrilateral the sum of its diagonals is greater than its semiperimeter and is smaller than its perimeter. (b) Let ABCD be a convex quadrilateral such that AB þ BD is not greater than AC þ CD. Prove that the length of side AB is less than the length of diagonal AC. (c) Prove that, if we connect the middle of all the adjacent sides of a convex polygon A1A2 ...An, then the perimeter of the obtained polygon is not less than the half of the perimeter of the polygon A1A2 ...An.

1.1.5. Let n>4 be an integer. For a convex n-gon A1A2...An consider the quadrilaterals AiAiþ1Aiþ2Aiþ3,i¼1,2,...,n, where Anþj¼Aj. Prove that from those quadrilaterals no more than n/2 have an inscirbed circle. Give an example of an octagon that has such 4 quadrilaterals. 1.1.6. Let ABCD be an inscribed quadrilateral. Prove that (a) |AB À CD| þ |AD À BC| 2| AC À BD|, (b) AB þ BD AC þ CD,if∠A ∠ D. (c) Prove that among all triangles inscribed in a given circle, the largest perimeter has the equilateral triangle. 1.1.7. Prove that bþcÀa < < bþc (a) 2 ma 2 , 3 < < (b) 4 ðÞa þ b þ c ma þ mb þ mc a þ b þ c, where ma, mb, mc are the lengths of the medians drawn from vertices A, B, C and a, b, c are, respectively, the lengths of sides BC, CA, AB of triangle ABC. 1.1.8. Let M be a point inside of triangle ABC. Prove that (a) MA þ MB < CA þ CB, (b) min(MA, MB, MC) þ MA þ MB þ MC < ma þ mb þ mc, (c) MA þ MB þ MC max (AB þ BC, AC þ BC, AC þ AB), pﬃﬃ 3 ; ; (d) MA þ MB þ MC 2 minðÞAB þ BC AC þ BC AC þ AB . 1.1 Triangle Inequality 3

1.1.9. (a) Let ABCD be a quadrilateral and E,F be the midpoints of sides AB, CD, BCþAD respectively. Prove that EF 2 . (b) Let M be the intersection point of medians AD and BE of triangle ABC. Prove that, if ∠AMB π/2, then AC þ BC > 3AB. (c) Let C1, A1 be points (different from the vertices) on sides AB, BC of triangle ABC.LetK be the midpoint of A1C1 and I be the in center of triangle ABC. Given that A1BC1I is an inscribed quadrilateral. Prove that AKC is an obtuse angle. 1.1.10. Let ABCD be a quadrilateral such that angles A and C are equal to 90. Prove that the perimeter of the inscribed quadrilateral in the quadrilateral ABCD is not smaller than 2AC. 1.1.11. Let M be a point inside of an equilateral triangle ABC. Prove that (a) MA þ MB > MC , (b) MA2 þ MB2 þ MC2 < 2 AB2. 1.1.12. Let E be a point on side AC of triangle ABC. Prove that (a) BE Á AC AE Á BC þ CE Á AB, (b) (EB À BA) Á AC (BC À AB) Á AE. 1.1.13. Let D be a point on side BC of triangle ABC such that AD > BC. Let point AE BD E on side AC be deﬁned by the following condition EC ¼ ADÀBC. Prove that AD > BE. 1.1.14. (a) Prove that for any distinct points A, B, C and D it holds true AC Á BD AB Á CD þ BC Á AD. (b) Let a square with the center O be externally constructed on the side AB of triangle ABC. Let M, N be the midpoints of sides BC, AC and the lengths of these sides be equal to a, b, respectively. Find the possible greatest value of the sum OM þ ON when angle ∠ACB changes. OAþOC (c) Given a rectangle ABCD on a plane. Find the smallest value of OBþOD, where O is an arbitrary point in that plane. (d) Prove that for any points A, B, C and D it holds true AB þ BC þ AC 2AD sin ∠ BDC. (e) Let D, E, F be points on sides BC, CA, AB of triangle ABC, respectively. Prove that 1 1 1 AB þ BC þ AC þ þ ðÞDE þ EF þ FD , AD BE CF R

where R is the circumradius of triangle ABC. (f) Given a triangle ABC and points D, E, F ,suchthat∠DBC ¼ ∠ ECA ¼ ∠ FAB > 0 and ∠DCB ¼ ∠ EAC ¼ ∠ FBA > 0. 4 1 Theorem on the Length of the Broken Line

Prove that AFþFBþBDþDCþCEþEA ADþBEþCF. (g) Given a point M and a parallelogram ABCD. Prove that MA Á MC þ MB Á MD AB Á AD. (h) Prove that for any distinct points A, B, C and D it holds true DA Á DB Á AB þ DB Á DC Á BC þ DC Á DA Á AC AB Á BC Á AC.

(i) Let ABCDA1B1C1D1 be a parallelepiped.

Prove that AB1 þ AD1 þ AC < AA1 þ AB þ AD þ AC1. (j) Let SABC be a tetrahedron. Prove that AB BC > AC : SAþSB þ SBþSC SAþSC (k) Let SABC be a tetrahedron. Prove that SA þ SB þ SC > 3 Á min (MA, MB, MC), where M is the intersection point of the medians of triangle ABC. 1.1.15. For any point T of a given triangle (the interior of the triangle included) denote by m(T) the smallest of segments TA, TB, TC. Find all points of triangle ABC, such that the value of m(T) is the greatest possible.

1.1.16. Let A1, A2,...,ÀÁA8 be the vertices of a parallelepiped and O be its center of 2 2 ::: 2 < ::: 2 symmetry. Prove that 4 OA1 þ OA2 þ þ OA8 ðÞOA1 þ OA2 þ þ OA8 . 1.1.17. Let G be the intersection point of the medians of triangle ABC. Prove that (a) if AB > AC, then AC þ BG < AB þ CG, < 1 (b) OG 3 ðÞOA þ OB þ OC , where O is an arbitrary.

1.1.18. Let O be a point in the hexagon A1A2A3A4A5A6, such that all its sides are visible under the angle of 60 . Prove that, if OA1 > OA3 > OA5 and OA2 > OA4 > OA6, then A1A2 þ A3A4 þ A5A6 < A2A3 þ A4A5 þ A6A1. 1.1.19. Given n distinct points. Prove that among those points there are points A, B, AB < 2 C, such that 1 AC 1 þ n, if (a) n ¼ 3, (b) n 5. 1.1.20. Given n (n 3) distinct points with the pairwise distances between them ::: λ > λnÀ1 λnÀ2 equal to a1, a2, , annðÞÀ1 . Given that n 0 and n þ n ¼ 1. Prove that there 2 ai 1 exist numbers i and j (i 6¼ j), such that 1 λ . For n ¼ 4, prove that the estimate qffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffi aj n pffiffiffiffi pffiffiffiffi 1 ¼ 3 9À 69þ 3 9þ 69¼ 1, 32471795:::: is not possible to make smaller. λ4 18 18 1.1.21. Given n (n 3) distinct points with the pairwise distances between them equal to a1, a2, :::, annðÞÀ1 , where a1 a2 ::: annðÞÀ1 : Given that δn > 0 and δn 2 2 nnðÞÀ1 1 δ 2 À > : ðÞ1 þ n 2 Prove that there exist numbers i and j(i 6¼ j), such that ai À 1 < δn: aj 1.1.22. Let M, N be distinct points on side BC of triangle ABC, such that ∠ ∠ ffiffiffiffiffiffiffiffiffiffiffiMNÁBCffiffiffiffiffiffiffiffiffiffiffi < sin ∠MAN < ffiffiffiffiffiffiffiffiffiffiffiMNÁBCffiffiffiffiffiffiffiffiffiffiffi : BAM ¼ CAN. Prove that p p 2 ∠ p p 2 ðÞBMÁCNþ BNÁCM sin BAC ðÞBMÁCNÀ BNÁCM 1.1 Triangle Inequality 5

Solutions

1.1.1. (a) Let D be a point on side BC, such that AB ¼ BD. Then, we have that ∠A > ∠ BAD ¼ ∠ BDA ¼ ∠ C þ ∠ DAC > ∠ C. Therefore ∠A > ∠ C. (b) Proof by contradiction argument. Assume that AB BC.IfAB ¼ BC, then ∠C ¼ ∠ A. This leads to a contradiction. If AB > BC, then according to problem 1.1.1а we have that ∠C > ∠ A. This leads to a contradiction. (c) Let D be a point on line AC, such that point C belongs to segment AD and CD ¼ BC. Hence, as ∠ABD ¼ ∠ B þ ∠ CBD > ∠ CBD ¼ ∠ BDC, then for triangle ABD, according to problem 1.1.1b, we obtain that AB < AD. Thus, it follows that AB < AC þ BC. bþcÀa aþbþc > (d) We have that p À a ¼ 2 , where p ¼ 2 . Therefore, p a. 1.1.2. Note that ÀAC AB À BC AC. Hence, |AB À BC| AC. 1.1.3. (a) As MA þ MC AC and MB þ MD BD, then MA þ MC þ MB þ MD AC þ BD. (b) We have that MA MB þ AB ¼ MB þ CD MB þ MC þ MD. Note that the equality MA ¼ MB þ MC þ MD cannot hold true. Otherwise, M belongs to segment CD, then MA < MB þ AB. (c) We have that MA MC þ AC ¼ MC þ BD MC þ MB þ MD. It is clear that the following equalities MA ¼ MC þ AC and BD ¼ MB þ MD simultaneously cannot hold true. Therefore, MA < MC þ MB þ MD. 1.1.4. (a) Let M be the intersection point of the diagonals of the given convex quadrilateral ABCD. Then, using the triangle inequality, we obtain that MA þ MB > AB, MB þ MC > BC, MC þ MD > CD, MA þ MD > AD. > 1 Summing up these inequalities, we deduce that AC þ BD 2 AB þ BCþð CD þ ADÞ. < < < 1 From AC AB þ BC and AC AD þ DC, it follows that AC 2 AB þ BCþð CD þ ADÞ. Therefore, AC þ BD < AB þ BC þ CD þ AD. Remark We have that AC þ BD ¼ (MA þ MB) þ (MC þ MD) > AB þ CD. (b) We have that AB þ BD AC þ CD and AB þ CD < AC þ BD. Therefore, 2AB þ CD þ BD < 2AC þ CD þ BD or AB < AC. (c) Let B1, B2,...,Bn be the midpoints of sides A1A2, A2A3,...,AnA1, respectively (Figure 1.1).

Figure 1.1 6 1 Theorem on the Length of the Broken Line

Then, for n 4, we have that

2B1B2 þ 2B2B3 þ ::: þ 2BnB1 ¼ A1A3 þ A2A4 þ A3A5 þ ::: þ AnA2 ¼ 1 ¼ ðÞðÞA A þ A A þðÞA A þ A A þ::: þ ðÞA A þ A A > 2 1 3 2 4 2 4 3 5 n 2 1 3 1 > ðÞðÞA A þ A A þ::: þ ðÞA A þ A A ¼A A þ A A þ ::: þ A A 2 1 2 3 4 n 1 2 3 1 2 2 3 n 1

(according to the remark in problem 1.1.4a). Hence, it follows that B1B2 þ B2B3 ::: > 1 ::: : þ þ BnB1 2 ðÞA1A2 þ A2A3 þ þ AnA1 For n ¼ 3 1 B B þ B B þ ::: þ B B ¼ ðÞA A þ A A þ ::: þ A A : 1 2 2 3 n 1 2 1 2 2 3 n 1

n 1.1.5. If it would be possible to cut off more than 2 circumscribed quadrilaterals, then among them there would be two neighboring quadrilaterals having two common sides. Let us denote these quadrilaterals by ABCD and BCDE (Figure 1.2). For any of them the sum of the opposite side is equal to AB þ CD ¼ BC þ AD, BC þ DE ¼ CD þ BE. Hence, we obtain that

AB þ DE ¼ AD þ BE: ð1:1Þ

The initial n-gon is convex; therefore its diagonals AD and BE intersect at some point P. According to the triangle inequality AD þ BE ¼ AP þ BP þ PD þ PE > AB þ DE. This leads to the contradiction with (1.1). In order to construct the required octagon, let us circumscribe an isosceles trapezoid A1A2A3A4 around the circle, such that its base is A1A4 and the base angle is equal to 45 . Then, construct it up to octagon A1A2 ...A8, as it is shown in Figure 1.3. In a similarÂÃ way, one can construct n-gon, such that it is possible to cut off by its n diagonals 2 circumscribed quadrilaterals. 1.1.6. (a) Let M be the intersection point of diagonals AC and BD of a quadrilateral ABCD. Note that ΔABM ~ ΔDCM. Therefore,

Figure 1.2 1.1 Triangle Inequality 7

Figure 1.3

CD CD AC À BD ¼ AM þ MC À BM À DM ¼ AM þ BM Á À BM À AM Á ¼ AB AB

AM À BM ¼ Á AB À CD AB À CD AB

(see problem 1.1.2). In a similar way, we obtain that |AC À BD| |AD À BC|, thus |AB À CD| þ |AD À BC| 2|AC À BD|. In the last inequality the equality holds true if and only if quadrilateral ABCD is a rectangle. ∠ ∠ CD CM DM (b) Note that MAD MDA. Therefore, MD MA.AsAB ¼ MB ¼ MA ¼ k 1 (see the proof of problem 1.1.6а), then AC þ CD À AB À BD ¼ (k À 1) (AB þ BM À AM) 0. (c) Let an irregular triangle ABC be inscribed in the given circle, such that ∠A ∠ B ∠ C. Note that ∠A > 60 > ∠ C. Let D be a point on arc ABC, such that ∠DAC ¼ 60.As∠A > ∠ A þ ∠ C À 60 ¼ ∠ ACD, then according to problem 1.1.6b (see the proof), AB þ BC þ AC < AD þ DC þ AC. If triangle ADC is equilateral, then this ends the proof. Otherwise, if triangle ADC is irregular, then repeating the above proof for triangle ADC, we obtain that its perimeter is smaller than the perimeter of the equilateral triangle inscribed in the given circle. Therefore, in this case also the perimeter of triangle ABC is smaller than the perimeter of the equilateral triangle inscribed into the given circle. Other proofs of this problem one can obtain using problems 5.1.6 and 8.2.1i. 1.1.7. (a) Consider Figure 1.4. < < bþc a > cþb We have that 2ma b þ c, thus ma 2 . Let c b, then ma þ 2 c 2 . > cþbÀa Hence, it follows that ma 2 . < bþc < aþc < aþb < (b) We have that ma 2 , mb 2 and mc 2 , thus ma þ mb þ mc a þ b þ c. 2 2 > 2 2 > Note that 3 ma þ 3 mc b. In a similar way, we obtain that 3 ma þ 3 mb c and 2 2 > 3 mc þ 3 mb a. Summing up these inequalities, we deduce that ma þ mbþ > 3 mc 4 ðÞa þ b þ c . 8 1 Theorem on the Length of the Broken Line

Figure 1.4

Figure 1.5

Figure 1.6

1.1.8. (a) Let N be the intersection point of lines AM and BC; then AM þ BM < AM þ MN þ BN ¼ AN þ NB < AC þ CN þ BN ¼ AC þ BC. Therefore, AM þ BM < AC þ BC.

(b) Let M be a point inside of triangle AGB1 (Figure 1.5), where G is the intersection point of the medians of triangle ABC. а b Using the result of problem 1.1.8 , we obtain that AM þ BM 2 þ mb and 2 2 AM þ MC 3 ma þ 3 mc. Therefore,

minðÞAM; BM; CM þAM þ BM þ CM 2AM þ BM þ CM b 2 2 m m 2 2 þ m þ m þ m < a þ c þ m þ m þ m ¼ m þ m þ m : 2 b 3 a 3 c 3 3 b 3 a 3 c b a c

We obtain that min(AM, BM, CM) þ AM þ BM þ CM < mb þ ma þ mc. Remark If the triangle is not obtuse, then taking the point M in the center of the circumcircle of triangle ABC, we obtain that ma þ mb þ mc > 4R, where R is the circumradius of triangle ABC. (c) Let us draw through point M segments parallel to AB, AC and BC (Figure 1.6). 1.1 Triangle Inequality 9

Let AB AC BC. Since triangles C2MC1, MB1B2, MA1A2 are similar to triangle ABC, then the smallest sides of these triangles are C1C2, MB1 and MA2, respectively. We have that

MA þ MB þ MC < ðÞþAB1 þ B1M ðÞþMA2 þ A2B ðÞMA1 þ A1C AB1 þ B1B2 þ A1A2 þ A2B þ CB2 þ A1C ¼ AC þ BC:

Hence, it follows that MA þ MB þ MC < AC þ BC. One can easily prove that the inequality holds true if M belongs to one of the sides of the triangle. (d) Let us consider two cases.

Case 1 Let there be a point M0 inside of the triangle, such that ∠AM0B ¼ ∠ BM0C ¼ ∠ AM0C ¼ 120 . Then, prove that AM þ BM þ CM AM0 þ BM0 þ CM0. Consider an equilateral triangle BCA1 constructed externally on the side BC of triangle ABC. We have that ∠BM0C þ ∠ BA1C ¼ 120 þ 60 ¼ 180 . Hence, M0BA1C is an inscribed quadrilateral. Thus, it follows that M0A1 ¼ M0B þ M0C (see the proof of problem 1.1.14а) and ∠AM0A1 ¼ ∠ AM0B þ ∠ BM0A1 ¼ 120 þ ∠ BCA1 ¼ 180 . Hence, AM0 þ BM0 þ CM0 ¼ AM0 þ M0A1 ¼ AA1. According to problem 1.1.14а, we have that MB þ MC MA1. Therefore, we deduce that MA þ MB þ MC MA þ MA1 AA1 ¼ AM0 þ BM0 þ CM0. pffiffi 3 ; Now, let us prove that AM0 þ BM0 þ CM0 2 minðAB þ BC ABþ AC; BC þ ACÞ.Letmax(∠A, ∠B, ∠C) ¼ ∠ C.Notethat∠AM0B > ∠ C. Hence, 60 ∠ C < 120 . Thus, it follows that 120 ∠ ACA1 < 180 . Therefore, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 AA1 ¼ AC þ BC À 2AC Á BC Á cosffiffiffi ∠ACA1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 3 AC2 þ BC2 þ AC Á BC ðÞAC þ BC : 2

Case 2 If there does not exist such point M0 (see Case 1), then one can easily prove that ∠C 120. Note that either ∠MCB < 120 or ∠MCA < 120. Without loss of generality one can assume that ∠MCB < 120 ; then according to problem 1.1.8а, MA þ MA1 AC þ CA1. Therefore, MA þ MB þ MC MA þ MA1 AC þ BC. This ends the proof. 1.1.9. (a) Let us denote by K the midpoint of diagonal BD. We have that AD BC EF EK þ KF ¼ 2 þ 2 , and the equality holds true if and only if K belongs to the segment EF, that is AD||BC. (b) Note that point M cannot be inside of the circle with diameter AB. Hence, AB MO 2 , where O is the midpoint of segment AB. According to problem AC BC > MO > 1.1.9a, we have that 4 þ 4 3 Á 2 . Thus, it follows that AC þ BC 3 Á AB. 10 1 Theorem on the Length of the Broken Line

Figure 1.7

Figure 1.8

(c) Let O be the midpoint of segment AC. Note that AC ¼ AC1 þ CA1. According to а К AC1þCA1 AC problem 1.1.9 , we have that O 2 ¼ 2 . Therefore, AKC is an obtuse angle. 1.1.10. Let a quadrilateral MNPK be inscribed in a quadrilateral ABCD (Figure 1.7). Let E and F be the midpoints of segments KM and NP, respectively. Since KM MNþKP а NP MNþNPþKPþKM AE ¼ 2 , EF 2 (see problem 1.1.9 ) and CF ¼ 2 , then 2 AE þ EF þ CF AF þ CF AC. Thus, it follows that MN þ NP þ KP þ KM 2AC. Remark If ∠A 90, ∠ C 90, then MN þ NP þ KP þ KM 2AC. 1.1.11. (a) Let us consider the Figure 1.8. Consider segments MA1, MB1, MC1, such that MA1||AB, MB1||BC and MC1||AC. One can easily prove that AC1MB1, BA1MC1 and CA1MB1 are isosceles trapezoids. Therefore, MA ¼ C1B1, MB ¼ A1C1 and MC ¼ A1B1. Hence, MA þ MB > MC .

(b) Let MA1 ¼ x, MB1 ¼ y, MC1 ¼ z, AB ¼ a (Figure 1.8). Therefore, x þ y þ z ¼ a and 1.1 Triangle Inequality 11

Figure 1.9

MA2 þ MB2 þ MC2 ¼ zaðÞþÀ x y2 þ xaðÞþÀ y z2 þ yaðÞþÀ z x2 ¼ ¼ a2 þ x2 þ y2 þ z2 À zx À xy À yz < a2 þ ðÞx þ y þ z 2 ¼ 2a2:

1.1.12. (a) If point E coincides with points A and C, then AB Á EC þ BC Á AE ¼ BE Á AC. If point E does not coincide with points A and C, then consider the following ﬁgures (Figure 1.9а, b), where MP ¼ AE Á EC, NP ¼ BE Á EC, MN ¼ AB Á EC, PK ¼ EB Á AE, MK ¼ BC Á AE. We have that MN þ MK > NK or AB Á EC þ BC Á AE > AC Á BE. (b) Note that the inequality of problem 1.1.12b is equivalent to the inequality of problem 1.1.12a. 1.1.13. We have that

AE Á AD ¼ BD Á EC þ AE Á BC: ð1:2Þ

From problem 1.1.12а, it follows that AC Á BE < AE Á BC þ AB Á EC. From the condition of the problem, it follows that point E does not coincide with points A and C. Let AD BE, then AC Á AD AC Á BE < AE Á BC þ AB Á EC. Using (1.2), we obtain that AC Á AD ¼ AD Á AE þ AD Á EC ¼ BD Á EC þ AE Á BC þ AD Á EC. There- fore, EC(BD þ AD À AB) < 0. This leads to a contradiction. Hence, AD > BE.

1 1.1.14. (a) Let us take a point A1 on a ray DA, such that DA1 ¼ DA. In a similar way, take points B1 and C1 on the rays DB and DC. One can easily prove that AB BC CA A1B1 ¼ DAÁDB, B1C1 ¼ DBÁDC and C1A1 ¼ DCÁDA. We have that A1B1 þ B1C1 A1C1. Thus, AB Á DC þ BC Á DA AC Á BD (see problem 4.1.9). (b) According to problem 1.1.14а, for points A, N, K, O we have that pffiffi NO Á AK AO Á NK þ AN Á OK or NO b þ 2 a, where K is the midpoint of 2 2 pffiffi МО a 2 side AB. In a similar way, we obtain that 2 þ 2 b. Hence, it follows 12 1 Theorem on the Length of the Broken Line

pﬃﬃ МО 2þ1 ∠ that þ NO 2 ðÞa þ b . The equality holds true if only ANO ¼ ∠ CMO ¼ 90, that is ∠C ¼ ∠ A þ ∠ B þ 90. Therefore ∠C ¼ 135. (c) According to problem 1.1.14a, for points A, B, C, O we have that AC Á BO AB Á CO þ BC Á AO . In a similar way, for points A, D, C, O we have that AC Á DO AD Á CO þ AO Á CD. Summing up these inequalities, we AOþCO AC obtain that BOþDO ABþAD. Taking O A, we obtain that the left-hand side of AC the last inequality is equal to ABþAD. Thus, the possible smallest value is equal AC to ABþAD. OAþOC ABþBC Remark One can prove that OBþOD BD . (d) If points B, C, D are on one line, then sin ∠ BDC ¼ 0; therefore AB þ BC þ AC 2AD sin ∠ BDC. If points B, C, D are not on one line and point O is the circumcenter of triangle BCD, then, according to problem 1.1.14а, for points A, B, O, C we have that

AB Á OC þ AC Á OB AO Á BC:

Hence,

AB Á OC þ AC Á OB þ BC Á OD BCðÞ AO þ OD : ð1:3Þ

We have that AO þ OD AD and OC ¼ OB ¼ OD. Thus from (1.3) we obtain that

BC AB þ AC þ BC AD Á ¼ 2AD sin ∠BDC: OD

EDþDFþFE ∠ BC (e) According to problem 1.1.14d, we obtain that AD 2 sin A ¼ R ;ina EDþDFþFE AC EDþDFþFE AB similar way we obtain that BEÀÁ R and CF R . Summing up 1 1 1 ABþBCþAC these inequalities, we deduce that AD þ BE þ CF ðÞED þ DF þ EF R . (f) Let BC ¼ mx, BD ¼ nx, DC ¼ kx, AC ¼ my, CE ¼ ny, AE ¼ ky, AB ¼ mz, AF ¼ nz, FB ¼ kz. According to problem 1.1.14а, for points A, B, D, C, we have that AB Á DC þ AC Á DB AD Á BC or kz þ ny AD; that is, FB þ CE AD. In a similar way we obtain that BD þ AE CF and AF þ DC BE. Summing up these inequalities, we deduce that AF þ FB þ BD þ DC þ CE þ EA AD þ BE þ CF. ! ! (g) Consider a point M0, such that MM0 ¼ AD. For points M, C, M0, D, according to problem 1.1.14a, we have that MC Á M0D þ CM0 Á MD CD Á MM0. As M0D ¼ AM, CM0 ¼ BM, MM0 ¼ AD, then MA Á MC þ MB Á MD AB Á AD. ! ! (h) Consider a point A0, such that BA0 ¼ AC. 1.1 Triangle Inequality 13

According to problem 1.1.14g, we have that AD Á DA0 þ BD Á DC AB Á AC; thus it follows that

AD Á BC Á DA0 þ BD Á DC Á BC AB Á AC Á BC: ð1:4Þ

According to problem 1.1.14a, we have that

BD Á CA0 þ CD Á BA0 BC Á DA0:

Thus, we deduce that

BD Á AD Á AB þ CD Á AD Á AC BC Á DA0 Á AD: ð1:5Þ

Summing up inequalities (1.4) and (1.5), we obtain that

DA Á DB Á AB þ DB Á DC Á BC þ DC Á DA Á AC AB Á BC Á AC:

(i) For points A, A1, B1, C1 according to problem 1.1.14a, we have that

AB1 Á A1C1 < AA1 Á B1C1 þ A1B1 Á AC1 ¼ AA1 Á AD þ AB Á AC1

(see the proof of problem 1.1.14а). Therefore,

AB1 Á AC < AA1 Á AD þ AB Á AC1: ð1:6Þ

In a similar way, for points A, D, D1, C1 and A, B, C, C1 we have that AD1 Á DC1 < AD Á D1C1 þ AC1 Á DD1 and AC Á BC1 < AB Á CC1 þ AC1 Á BC,or

AD1 Á AB1 < AD Á AB þ AC1 Á AA1, ð1:7Þ

and

AC Á AD1 < AB Á AA1 þ AC1 Á AD: ð1:8Þ

Let O, O1 be the intersection points of the diagonals of parallelograms ABCD and A1B1C1D1, respectively. Let line AC1 intersect with segments A1O and CO1 at points M and M1, respectively. Since AA1C1C is a parallelogram, then A1O1 ¼ OC and A1O1||OC, thus quadrilateral A1O1CO is also a parallelogram. Therefore, A1O|| O1C. We have that A1O1 ¼ O1C1 and A1M||O1M1. Hence MM1 ¼ M1C1. In a similar way, we obtain that AM ¼ MM1. Δ Δ A1M A1C1 AC Note that AOM C1A1M, thus MO ¼ AO ¼ AO ¼ 2. It follows that M is the intersection point of the medians of triangle A1BD. In a similar way, we obtain that M1 is the intersection point of the medians of triangle B1D1C. According to Stewart’s theorem, we obtain that 14 1 Theorem on the Length of the Broken Line 2 2 2 2 1 2 2 2 2 2 1 2 2 AB þ AD BD AM ¼ AA1 þ AO À A1O ¼ AA1 þ À À 3 3 9 3 3 2 4 2 A B2 þ A D2 BD2 1ÀÁ1ÀÁ À 1 1 À ¼ AA 2 þ AB2 þ AD2 À A B2 þ A D2 þ BD2 : 9 2 4 3 1 9 1 1

In a similar way, we obtain that

1 ÀÁ1 ÀÁ AM 2 ¼ AB 2 þ AD 2 þ AC2 À B D 2 þ D C2 þ CB 2 : 1 3 1 1 9 1 1 1 1

Therefore

1 ÀÁ AM 2 À AM2 ¼ AB 2 þ AD 2 þ AC2 À AB2 À AD2 À AA 2 1 3 1 1 1 as BD ¼ B1D1, A1D ¼ B1C, A1B ¼ D1C. 2 1 Note that AM1 ¼ 3 AC1, AM ¼ 3 AC1. Hence,

2 2 2 2 2 2 2 AB1 þ AD1 þ AC ¼ AB þ AD þ AA1 þ AC1 ; from this equality and inequalities (1.6), (1.7), (1.8) we obtain that

2 2 2 AB1 þ AD1 þ AC þ 2AB1 Á AC þ 2AD1 Á AB1 þ 2AC Á AD1 < 2 2 2 2 < AB þ AD þ AA1 þ AC1 þ 2AA1 Á AD þ 2AB Á AC1 þ 2AD Á ABþ þ2AC1 Á AA1 þ 2AB Á AA1 þ 2AC1 Á AD,

2 2 or (AB1 þ AC þ AD1) < (AA1 þ AB þ AD þ AC1) . Therefore,

AB1 þ AC þ AD1 < AA1 þ AB þ AD þ AC1:

(j) Note that

AB BC AB2 BC2 þ ¼ þ SA þ SB SB þ SC SA Á AB þ SB Á AB SB Á BC þ SC Á BC ðÞAB þ BC 2 : SA Á AB þ SC Á BC þ SBðÞ AB þ BC

It is sufﬁcient to prove that

ðÞAB þ BC 2ðÞSA þ SC > SA Á AB Á AC þ SC Á BC Á AC þ SB Á ACðÞ AB þ BC ðÞAB þ BC ðÞAB Á SA þ BC Á SC þ AB Á SC þ BC Á SA À SB Á AC > > ACðÞ SA Á AB þ SC Á BC , 1.1 Triangle Inequality 15

ðAB þ BC À ACÞðAB Á SA þ BC Á SCÞþ þðAB þ BCÞðAB Á SC þ BC Á SA À SB Á ACÞ > 0:

This holds true, as AB Á SC þ BC Á SA À SB Á AC 0 (see problem 1.1.14а) and AB þ BC > AC. (k) Let us consider two cases.

Case 1 Assume that there exists a point M0 inside of the triangle ABC, such that ∠AM0B ¼ ∠ BM0C ¼ ∠ AM0C ¼ 120 . We have that SA þ SB þ SC > M0A þ M0B þ M0C (see the solution of problem 1.1.8d). Let us prove that M0A þ M0B þ M0C 3 Á min (MA, MB, MC). ; ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiLet M0A ¼ x, M0B ¼ y, M0C ¼ z (x y z), then 3 Á minðÞ¼MA MB MC x2 þ y2 þ 4z2 þ 2xz þ 2yz À xy: Therefore, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x þ y þ z x2 þ y2 þ 4z2 þ 2xz þ 2yz À xy:

Case 2 Assume that there does not exist such a point M0. Let max (∠A, ∠B, ∠C) ¼ ∠ C. We have that (see the proof of problem 1.1.8d) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi SA þ SB þ SC > AC þ BC > 2AC2 þ 2BC2 À AB2 ¼ 3 Á minðÞMA; MB; MC :

See also problem 7.1.107d. 1.1.15. Let ∠A ∠ B ∠ C. If ∠A 90, then the required point T is the circumcenter O of the triangle. Let the radius of the circumcircle be equal to R; then m(O) ¼ R. If point T does not coincide with point O, then it is in one of the triangles AOB, TAþTB < OAþOB AOC, BOC. Let point T is in triangle AOB; then mTðÞ 2 2 ¼ R (see problem 1.1.8а). If ∠A > 90, then let us take on side BC points M and N, such that ∠BAM ¼ ∠ B and ∠CAM ¼ ∠ C. Then ∠BAM ¼ ∠ B < ∠ BAN ¼ ∠ A À ∠ C (Figure 1.10). Δ ATþBT AMþBM ∠ If T 2 ABM, then mTðÞ 2 2 ¼ AM AN ¼ mNðÞ,as AMN ¼ 2 ∠ B ∠ ANB ¼ 2 ∠ C and BN ¼ BM þ MN ¼ AM þ MN > AN.

Figure 1.10 16 1 Theorem on the Length of the Broken Line

0 If T 2 ΔAMN, then m(T) AT AA max (AM, AN) ¼ AN ¼ m(N), as max 0 0 (∠MA A, ∠NA A) 90. Δ ATþTC ANþCN If T 2 ANC, then mTðÞ 2 2 ¼ AN ¼ mNðÞ. Thus, the greatest value of m(T)is equal to m(N). From the proof, it follows that, if ∠B ¼ ∠ C, then points M and N are the required points. Otherwise, point N is the required point. ∠ > < BC Remark In the case, if A 90 , mTðÞCN 2 .

1.1.16. Let OA1 ¼ a, OA2 ¼ b, OA3 ¼ c, OA4 ¼ d and a ¼ max(a,b,c,d). According to problem 1.1.3b [it is true also, if M does not belong to the plane (ABC)], we have that b þ c þ d > a. Therefore

2 2 ðÞOAÀÁ1 þ ::: þ OA8 ¼ 4ðÞa þ b þ c þ d > > 2 2 2 2 > 4ÀÁa þ b þ c þ d þ abðÞþþ c þ d ab þ acÀÁþ ad > 2 2 2 2 2 2 2 2 2 ::: 2 : 4 a þ b þ c þ d þ a þ b þ c þ d ¼ 4 OA1 þ þ OA8

1 > 1 1.1.17. (a) Using the triangle inequality, we obtain that BG þ 2 CG 2 AB and 1 > 1 > CG þ 2 BG 2 AC. Summing up these inequalities, we deduce that 3(BG þ CG) AB þ AC. BG2ÀCG2 AB2ÀAC2 < AB2ÀAC2 < We have that BG À CG ¼ BGþCG ¼ 3ðÞBGþCG ABþAC . Hence, AC þ BG AB þ CG.

(b) Let A1 be the midpoint of side BC. According to problem 1.1.7a, we have that OBþOC OA1 2 . Point A2 is on the segment OA1 and A1A2 : A2O ¼ 1:2. AO 2 AO BO CO Then OG A2G þ A2O ¼ 3 þ 3 OA1 3 þ 3 þ 3 . It is clear that the equal- OBþOC ities OG ¼ A2G þ A2O and OA1 ¼ 2 do not hold true simultaneously. Thus, it < 1 follows that OG 3 ðÞAO þ BO þ CO . 0 0 0 1.1.18. Let A5 be a point on ray OA5, such that OA 5 ¼ OA3 and let A4 be a point on 0 0 0 > 0 the segment OA4, such that OA 4 ¼ OA6. Then, we have that A4A5 þ A4A5 A4A5 0 0 > 0 þA5A4; that is, A5A6 þ A4A5 A4A5 þ A5A6. Therefore,

A1A2 þ A3A4 þ A5A6 À A2A3 À A4A5 À A6A1 < 0 0 : : A1A2 þ A3A4 þ A5A6 À A2A3 À A4A5 À A6A1 ð1 9Þ

0 0 0 0 Let A be a point on the segment OA1, such that OA 1 ¼ OA3 ¼ OA 5 and let A be 1 0 2 a point on the segment OA2, such that OA 2 ¼ OA6. Then, we have that 0 0 < 0 0 0 < 0 A1A2 þ A1A2 A1A2 þ A2A1, thus A1A2 þ A1A6 A1A2 þ A1A6. Hence,

0 0 < A1A2 þ A3A4 þ A5A6 À A2A3 À A4A5 À A6A1 < 0 0 0 0 : ÀÁA1A2 þ A3A4 þ A5ÀÁA6 À A2A3 À A4A5ÀÁÀ A1A6 ¼ ð1 10Þ 0 0 0 0 : ¼ A1A2 À A2A3 þ A3A4 À A4A5 þ A5A6 À A1A6 ¼ 0 þ 0 þ 0 ¼ 0

From (1.9) and (1.10) it follows that A1A2 þ A3A4 þ A5A6< A2A3 þ A4A5 þ A6A1. 1.1 Triangle Inequality 17

1.1.19. Proof by contradiction argument. Let M and N be two points among those n, such that the distance MN ¼ d1 is the greatest. Consider plane Π, passing through the midpoint of segment MN and perpendicular to it. One of the half-spaces with boundary Π contains k points from the given n n points, where k 2. Let us denote these points by N, N1,...,Nk À 1 and MN ¼ d1 MN1 ¼ d2 .... MNk À 1 ¼ dk. According to our assumption, we have that 2 2 2 d 1 þ d , d 1 þ d , :::, d 1 þ d : 1 n 2 2 n 3 kÀ1 n k

Multiplying these inequalities, we obtain that 2 kÀ1 d 1 þ d : ð1:11Þ 1 n k ÀÁ 2 As MNk À 1 NNk À 1, then according to our assumption, MNkÀ1 1 þ n NNkÀ1. On the other hand, by the triangle inequality NNk À 1 MN À MNk À 1. Therefore, it follows that 2 2 2 þ d 1 þ d : ð1:12Þ n k n 1

From (1.11) and (1.12), we deduce that 2 2 k 2 þ 1 þ : ð1:13Þ n n ÀÁ 2 k 2 kðkÀ1Þ 4 ...> nÀ2 : Note that for n 6, we have that 1 þ n ¼ 1 þ k Á n þ 2 Á n2 þ 2 þ 2n 2 > nÀ2 From the last inequality and from 1.13, we deduce that 2 þ n 2 þ 2n .Fromthis < inequality it follows that n 6. This leads to a contradiction.ÀÁ ÀÁ 12 7 k 7 3 For n ¼ 5, from (1.13) we obtain that 5 5 5 . This leads to a contradiction. 5 5 For n ¼ 3, we have that MN 3 MN1 and MN1 3 NN1. Therefore, it follows 3 9 that MN MN1 þ NN1 5 MN þ 25 MN. This leads to a contradiction. pffiffi Remark 1. For n ¼ 4, there exist points A, B and C, such that 1 AB 1þ 5. pffiffi AC 2 1þ 5 Besides, in this inequality the estimate 2 is not possible to make smaller. 2. In a similar way, one can prove that among n points it is possible to choose points AB 1 k kÀ1 A, B, C, such that 1 λ , where λ0 is the positive root of λ þ λ ¼ 1 and ÂÃ AC 0 nþ1 k ¼ 2 . 18 1 Theorem on the Length of the Broken Line

It is known that the estimate 1 is not possible to make smaller for 3 n 7. λ0

1.1.20. The proof by contradiction method. Consider given points A1, A2,...,An, such that AiAj A1An ¼ d1, i, j 2 {1, ..., n} and max ðÞ¼A1Ai; AiAn di. i2fg2;:::;nÀ1

Without loss of generality one can assume that d2 d3 ... dn À 1. Then according to our assumption, we obtain that d1 > 1 , d2 > 1 , :::, dnÀ2 > 1 . d2 λn d3 λn dnÀ1 λn Multiplying these inequalities, we deduce that

λnÀ2 > : : n d1 dnÀ1 ð1 14Þ

Let A1An À 1 ¼ dn À 1. Then according to our assumption, we have that A1AnÀ1 1 > An 1An and according to the triangle inequality A A þ A A A A . λn À 1 n À 1 n À 1 n 1 n Therefore,

ðÞ1 þ λn dnÀ1 > d1: ð1:15Þ

λnÀ2 λnÀ1 > Using (1.14) and (1.15), we obtain that n þ n 1. This leads to a contradiction. Remark If n 3, λ > 0 and λnÀ2 þ λnÀ1 ¼ 1, then 1 < 1 þ 2. n n n λn n λ n ’ Indeed, if n nþ2, then using Bernoulli s inequality, we obtain that n nÀ2 n nÀ1 2nþ2 2nþ2 λnÀ2 þ λnÀ1 þ ¼ ÀÁnþ2 nþ2 n n n þ 2 n þ 2 2 nÀ2 1 þ 2 ðÞn À 2 < 1: 1 þ n n

This leads to a contradiction. qffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi pffiffiffiffi 3 One can easily prove that if 1 ¼ 3 9À 69þ 3 9þ 69, then 1 ¼ 1 þ 1 or λ0 18 18 λ0 λ0 λ3 λ2 λ λ 0 þ 0 ¼ 1. Therefore, 4 ¼ 0. Let us prove that for n ¼ 4, the estimate 1 is not possible to make smaller. λ4 Indeed, let A1, A2, A3, A4 be such points on one line, that A1A2 ¼ 1, A2A3 ¼ λ4, λ2 A3A4 ¼ 4, A2 belongs to the segment A1A3 and A3 belongs to segment A2A4. Then λ λ2 λ λ λ2 λ λ2 a1 ¼ 1 þ 4 þ 4, a2 ¼ 1 þ 4, a3 ¼ 4 þ 4, a4 ¼ 1, a5 ¼ 4, a6 ¼ 4. It is enough to note that a1 ¼ a2 ¼ a3 ¼ a4 ¼ a5 ¼ 1 . a2 a3 a4 a5 a6 λ4 1.1.21. Proof by contradiction argument. According to our assumption, for all i > j

ai we have that À 1 δn. Therefore, aj

nnðÞÀ1 2 2 À1 annðÞÀ1 ðÞ1 þ δn annðÞÀ1 ðÞ1 þ δn annðÞÀ1 ::: ðÞ1 þ δn a1, 2 2 À1 2 À2 hence

nnðÞÀ1 2 À1 annðÞÀ1 ðÞ1 þ δn a1: 2 1.1 Triangle Inequality 19

Let AB ¼ annðÞÀ1 , CD ¼ a1: Using the triangle inequality, we have that 2 AB AC þ BC AB. Then, without loss of generality one can assume that AC 2 and A 6¼ D. Note that À Á

minðÞAC; AD maxðÞÀAC; AD min AC; AD À 1 ¼ maxðÞAC; AD maxðÞAC; AD CD CD AB 2 < δ AC nnðÞÀ1 n, 2 À1 ðÞ1þδn 2

minðÞAC;AD < δ therefore maxðÞAC;AD À 1 n. This leads to a contradiction. 1.1.22. Let R and r be the circumradiuses of triangles ABC and MAN, respectively. sin ∠MAN MN 2R According to the law of sines, we have that sin ∠BAC ¼ 2r Á BC. BC2 < R < BC2 : Hence, it is enough to prove that pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 2 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 2 ðÞBMÁCNþ BNÁCM r ðÞBMÁCNÀ BNÁCM Let the circumcircle of triangle MAN intersect sides AB and AC of triangle ABC at points K and E, respectively (Figure 1.11). We have that ∠KEM ¼ ∠ BAM ¼ ∠ CAN ¼ ∠ EMC. Therefore, KE k BC. Thus, it follows that ΔKAE ΔBAC. pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Hence, KE ¼ r BC, AK ¼ pBMffiffiffiffiffiffiffiffiffiffiffiffiÁBN , AE ¼ pffiffiffiffiffiffiffiffiffiffiffiffiCMÁCN Using the triangle inequal- R R R R R : rÁðÞr À1 rÁðÞrÀ1 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi jjBMÁBNÀ CMÁCN ity, we obtain that |AK À AE| < KE < AK þ AE. Therefore, < pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi BC r < BMÁBNþ CMÁCN BC2 < R < BC2 : 1 À . Thus, pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 2 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 2 R BC ðÞBMÁCNþ BNÁCM r ðÞBMÁCNÀ BNÁCM

Figure 1.11 20 1 Theorem on the Length of the Broken Line

Problems for Self-Study

1.1.23. There are 50 correct clocks on the table. Prove that at some moment the sum of the distances from the center of the table to the end of the minute hands will be greater than the sum of the distances from the center of the table to the center of the clocks. 1.1.24. Let M be a point inside of the parallelogram ABCD. Prove that

AC þ BD MA þ MB þ MC þ MD < 2ðÞAB þ BC :

1.1.25. Prove that the sum of the lengths of the diagonals of a convex pentagon is greater than its perimeter and is smaller than twice the perimeter. 1.1.26. Prove that for any convex pentagon there are three diagonals that are sides of a triangle. 1.1.27. Prove that in a convex polygon there are no three sides that are greater than the largest diagonal of that polygon. 1.1.28. Prove that the arithmetic mean of the lengths of the sides of an arbitrary convex polygon is less than the arithmetic mean of the lengths of its diagonals. 1.1.29. Prove that 1 < 1 þ 1 , where h , h , h are the altitudes of some triangle. ha hb hc a b c 1.1.30. Prove the following inequalities. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (а) a2 þ b2 À ab þ b2 þ c2 À bc a2 þ c2 þ ac, where a, b, c > 0. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (b) c a2 þ b2 À ab þ a b2 þ c2 À bc b a2 þ c2 þ ac, where a, b, c > 0. 1.1.31. Let the bisectors (lines) of angles A, B, C intersect the circumcircle of triangle ABC at points A1, B1, C1, respectively. Prove that AA1 þ BB1 þ CC1 > AB þ BC þ AC. 1.1.32. Consider a triangle ABC, such that AB > AC and BM, CN are its medians. 1 < < 3 Prove that 2 ðÞAB À AC BM À CN 2 ðÞAB À AC . BMÀCN 3 ABþAC Hint Prove that ABÀAC ¼ 4 Á BMþCN. 1.1.33. Given that in triangle ABC side AC is the largest side. Prove that for any point M of plane AM þ CM is not less than BM. When does the equality hold true?

1.1.34. Given that points A1, A2,...,An are not on the same line. Let P and Q be distinct points, such that A1P þ A2P þ ...þ AnP ¼ A1Q þ A2Q þ ...þ AnQ ¼ S. Prove that A1K þ A2K þ ...þ AnK < S for some point K. 1.1.35. Given a tetrahedron ABCD. Prove that there exists a triangle with sides AB Á CD, AC Á BD and AD Á CB. 1.1.36. Let a, b, c be the side lengths of some triangle. Prove that 2 a2þ2bc b þ2ca c2þ2ab > 3. b2þc2 þ c2þa2 þ a2þb2 1.1 Triangle Inequality 21

1.1.37. Let ABCD be a tetrahedron, such that ∠CAB þ ∠ DAC þ ∠ BAD¼ ¼ ∠ ABC þ ∠ CBD þ ∠ DBA ¼ 180. Prove that CD AB. 1.1.38. Given that the acute angle of a parallelogram is equal to α, m and n are its > α m diagonals, m n. Prove that ctg 2 n. Hint Let ABCD be a parallelogram, such that ∠BAD ¼ α. Let O be the α < π circumradius of triangle ABD. Let A1 be the midpoint of arc BAD.As 2, point O is on the segment A1K, where K is the intersection point of diagonals AC 0 and BD. We have that A1K ¼ A1O þ OK ¼ AO þ OK AK. Let A be on the segment 0 ∠ 0 α α ∠ 0 m > A1K and A K ¼ AK, then BA K 2. Therefore ctg 2 ctg BA K ¼ n (AK BK, as the circle with diameter AC contains points B and D). 1.1.39. On sides AB and BC of triangle ABC are taken points D and F, respectively. Let E be the midpoint of segment DF. Prove that AD þ FC AE þ EC. 1.1.40. (a) A convex quadrilateral MNPQ is inside a convex quadrilateral ABCD. Line MP intersects the sides of quadrilateral ABCD at points K and L. Prove that the sum of the distances from one of these points to the vertices of the external quadrilateral is greater than the sum of the distances to the vertices of the inner quadrilateral. (b) Let four points be marked inside of a convex quadrilateral. Prove that on one of the sides of the quadrilateral there is a point, such that the sum of the distances from that point to the vertices of the quadrilateral is greater than the sum of the distances from that point to the marked points. 1.1.41. Let points H, I, K, M, O be the midpoints of sides AB, BC, CD, DE, EA of a convex pentagon, respectively. Prove that the length of a closed polyline HKOIMH is less than the length of polyline ACEBDA. BCþAD < ACþBD : Hint Prove that HK 2 2 1.1.42. Let on the side CD of a parallelogram ABCD be constructed an equilateral triangle CDE. Let X be an arbitrary point on a plane. Prove that XA þ XB þ AD XE. 0 ! ! Hint Let X be such a point that XX0 ¼ AD. According to problem 1.1.14а,we 0 0 0 0 0 0 have that CX þ X D EX . Thus, it follows that XA þ XB þ AD ¼ X C þ X D þ XX 0 0 X E þ XX XE.

1.1.43. Given circles ω(0, r) and ω1(0, r1). Let quadrilateral ABCD be inscribed in circle ω and the rays AB, BC, CD, DA intersect circle ω1 at points A1, B1, C1, D1, respectively (r1 > r). Prove that

r1 (a) A1B1 þ B1C1 þ C1D1 þ D1A1 r ðÞAB þ BC þ CD þ DA , 2 r1 (b) SA1B1C1D1 r2 SABCD. 22 1 Theorem on the Length of the Broken Line

Hint (a) See problem 1.1.14а. (b) Note that S Á S Á S Á S ¼ ðÞr 2 À r2 sin 2∠A Á sin 2∠B and D1AA1 BA1B1 CB1C1 DD1C1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 ACÁBD 2 ∠ ∠ 2 ∠ ∠ SABCD 2 2r sin A Á sin B 2r sin A Á sin B. 1.1.44. Given a quadrilateral ABCD, such that ∠A ∠ D and ∠B þ ∠ D 180. Prove that AC þ CD AB þ BD.

Hint Let C1 be the intersection point of line AC and a circle passing through points A, B, D.

Prove that C1 is on segment AC, and see problem 1.1.6b.

ÀÁ1.1.45. Prove that if a1, a2,...,ÀÁan (n 3) are positive numbers, such that 2 2 ::: 2 2 > 4 4 ::: 4 a1 þ a2 þ þ an ðÞn À 1 a1 þ a2 þ þ an , then one can construct a triangle with sides ai, aj, ak for any 1 i < j < k n. 2 Hint If m > 3 and a2 ::: a2 > m 1 a4 ::: a4 , then m 2 i1 þ þ im ðÞÀ i1 þ þ im ðÞÀ 2 a4 À 2a2 a2 þ ::: þ a2 þ ðÞm À 1 a4 þ ::: þ a4 À a2 þ ::: þ a2 im im i1 imÀ1 i1 imÀ1 i1 imÀ1 < 0:

Therefore, D > 0or 2 a2 þ ::: þ a2 > ðÞm À 1 ðÞm À 2 a4 þ ::: þ a4 i1 imÀ1 i1 imÀ1 : 2 2 ÀðÞm À 2 a2 þ ::: þ a2 ; a2 þ ::: þ a2 > ðÞm À 2 a4 þ ::: þ a4 i1 imÀ1 i1 imÀ1 i1 imÀ1

1.1.46. Let a, b, c be side lengths of some triangle. Prove that a b c bþcÀa þ cþaÀb þ aþbÀc 3. a 1 cþaÀb aþbÀc Hint We have that bþcÀa ¼ 2 bþcÀa þ bþcÀa . 1.1.47. Consider a triangle ABC. (a) Prove that for any point M on a plane AM sin ∠ A BM sin ∠ B þ CM sin ∠ C. (b) Let A1, B1, C1 be points on sides BC, AC, AB respectively, such that angles of triangle A1B1C1 are equal to α, β, γ. Prove that AA1 sin α þ BB1 sin β þ CC1 sin γ < BC sin α þ AC sin β þ AB sin γ. Hint See problem 1.1.14a.

1.1.48. Given n points A1, A2,...,An and a circle of radius 1. Prove that one can choose a point M on this circle, such that MA1 þ MA2 þ ...þ MAn n. 1.1.49. Let d be the smallest of the distances between the skew edges of tetrahedron ABCD and h be the smallest of its altitudes. Prove that 2d > h. 1.1 Triangle Inequality 23

Hint Let d ¼ MN, where M 2 BD and N 2 AC.AsBM þ DM BD, then without BD ⊥ loss of generality one can assume that BM 2 , MH (ABC) H 2 (ABC). Thus, it > h follows that d MH 2. 1.1.50. From two points A and B, the distance between them is equal to d km, are simultaneously observing during one second after the aircraft ﬂying in a straight line at aconstantspeed.FrompointA the observer reports that the plane has moved during that second by angle α, and from point B the observer reports that the plane has moved by angle β (α and β are acute angles). What could be the lowest speed of the plane? Hint Let during 1 second of observation the plane moved from point C to point D and M be the midpoint of the segment CD. Prove that AM CD ctg α and ÀÁ 2 2 CD β CD α β 2d BM 2 ctg 2. Then d ¼ AB 2 ctg 2 þ ctg 2 . Thus, CD α β. ctg 2þctg 2 1.1.51. Let O be the circumcenter of triangle ABC. On sides AB and BC are given points M and N, respectively, such that 2 ∠ MON ¼ ∠ AOC. Prove that NB þ BM þ MN AC. Hint See problem 1.1.14d. 1.1.52. The sum of the distances from point M to the two neighboring vertices of the square is equal to a. What is the largest value of the sum of the distances from point M to the other two vertices of the square? Hint See problem 1.1.14а. 1.1.53. Given that the perimeter of a convex quadrilateral is equal to 2004 and one of its diagonals is equal to 1001. Can the second diagonal be equal to 1? Can the second diagonal be equal to 2? Can the second diagonal be equal to 1001? 1.1.54. Let ABC be a triangle. Prove that 2 2 2 (a) maa þ mbb þ mcc mcab þ mabc þ mbac.

а a 2 c mb b mc Hint According to problem 1.1.14 , we have that 2 Á 3 ma 2 Á 3 þ 2 Á 3 . 2 3 (b) mamb þ mbmc þ mcma 2p À 4 ðÞab þ bc þ ac . а c a b Hint According to problem 1.1.14 , we have that mamb c Á 2 þ 2 Á 2.

a bÀc : Hint We have that ma À la 2 À ðÞp À b ¼ 2 1.1.55. Let A, X, D be points on a line, such that X is in between A and D. Let point B be such that ∠ABX 120 and point C be in between B and X. Prove that pffiffiffi 2AD 3ðÞAB þ BC þ CD .

2 2 2 3 2 Hint We have that AX AB þ BX þ AB Á BX 4 ðÞAB þ BX . 1.1.56. Let K and L be points on side AB of triangle ABC, such that AK ¼ KL ¼ LB. Prove that 24 1 Theorem on the Length of the Broken Line

(a) if AC < BC, then CK < CL. (b) ∠KCL > min (∠ACK, ∠BCL). Hint (b) Let AC BC, then CK < CB. Consider a parallelogram KCBD. 1.1.57. Let the diagonals of a convex quadrilateral ABCD intersect at point P. Prove that BC þ AD BD sin β þ AC sin γ, where β, γ > 0 and β þ γ ¼ ∠ BPC.

Hint Consider a point D1, such that quadrilateral BCD1D is a parallelogram. Then DD1¼BC. Note that BC þ AD AD1 BD sin β þ AC sin γ. 1.1.58. Prove that if a, b, c are the side lengths of some triangle, then

1 1 1 (a) aþb , bþc , aþc, a b c (b) 1þa , 1þb , 1þc, ðÞ1þa ðÞ1þb ðÞ1þb ðÞ1þc ðÞ1þa ðÞ1þc (c) aþbþ2ab , bþcþ2bc , aþcþ2ac , are also side lengths of some triangle. Hint 1 1 > 2 (a) aþb þ bþc aþbþc, a b > aþb (b) 1þa þ 1þb 1þaþb. 1.1.59. Let ABCD be a tetrahedron. Prove that AB Á CD þ BC Á AD S, where S is the total surface area of the tetrahedron.

Hint Prove that AB Á CD þ BC Á AD 2SABC þ 2SADC. Consider the layout of facets ABC, ADC and see problem 1.1.14а. 1.1.60. Prove that any polygon having a perimeter equal to 2a can be convered by a square whose diagonal is equal to a. Hint Inscribe the polygon into a square and use problem 1.1.10. 1.1.61. Given a triangle ABC and a point M. Let G be the intersection point of the medians of triangle ABC. Prove that MA þ MB þ MC þ 3MG 2(MA1 þ MB1 þ MC1), where points A1, B1, C1 are the midpoints of sides BC, AC, AB. Hint See problem 1.1.14f.

1.1.62. Let A1A2 ...An be a convex hexagon and M, N be given distinct points inside of it. Prove that max MA1; MA2; :::; MAn > 1: NA1 NA2 NAn Hint See problem 1.1.8a.

1.2 Theorem on the Length of the Broken Line

If A1, A2,...,An are distinct points, then A1An A1A2 þ A2A3 þ ...þ An À 1An. Note that the equality holds true only if simultaneously hold true the following conditions: point Ai is on the segment Ai À 1Ai þ 1 for all i ¼ 2, 3, . . . , n À 1. 1.2 Theorem on the Length of the Broken Line 25

1.2.1. Given two circles with radiuses R1 and R2, such that for the distance between their centers it holds true the following inequality d > R1 þ R2. Prove that d À R1 À R2 XY d þ R1 þ R2, where X and Y are arbitrary points of these two circles. 1.2.2. Prove that in any polygon there are at least two sides a and b, such that b < 1 a 2. 1.2.3. Given a convex hexagon ABCDEF, such that ∠A 90, ∠ D 90. Prove that BC þ CE þ EF þ FB 2AD. 1.2.4. Given the points A(a, 0), B(0, b), C(c, d) and O(0, 0) on a coordinate plane. Prove that AB þ BC þ CA 2CO. 1.2.5. Prove the following inequalities: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) a2 þ b2 þ ::: þ a2 þ b2 ðÞa þ ::: þ a 2 þ ðÞb þ ::: þ b 2. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 n nqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 n 1 n 2 2 2 2 a þ ka1b1 þ b þ ::: þ a þ kanbn þ b (b) q1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 n n 2 2 ðÞa1 þ ::: þ an þ kaðÞ1 þ ::: þ an ðÞþb1 þ ::: þ bn ðÞb1 þ ::: þ bn , where |k| 2. (c) ðÀa þ b þ cÞða À b þ cÞþða À b þ cÞða þ b À cÞþða þ b À cÞðÀa þ b þ cÞ pffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi abcð a þ b þ cÞ, where a, b, c > 0: 1.2.6. On the legs of a right-angled triangle with hypotenuse c and acute angle α are chosen points P and Q. Let PK and QH be perpendiculars to the hypotenuse drawn from points P and Q. Prove that KP þ PQ þ QH c sin 2α. 1.2.7. Prove that from all triangles inscribed in an acute triangle ABC, the smallest perimeter has the orthic triangle (the triangle whose vertices are the endpoints of the altitudes of triangle ABC). 1.2.8. Given a triangle ABC. Prove that ∠ACB 120 is a necessary and sufficient condition for the following inequality to hold true MA þ MB þ MC AC þ BC, for any point M of plane ABC.

A1A2 AnÀ1An A1An 1.2.9. (a) Prove that þ ::: þ , where M, A1,...,An are MA1ÁMA2 MAnÀ1ÁMAn MA1ÁMAn distinct points and n 3. When does the equality hold true? (b) Given a square ABCD inscribed in a circle. Let M be a point on the minor arc ^ AB: Prove that pffiffiffi 1. MC Á MD 3 3MA Á MB, ÀÁpffiffiffi 2. MC Á MD 3 þ 2 2 MA Á MB. 1.2.10. On sides AB and CD of a convex quadrilateral ABCD are constructed externally equilateral triangles ABE and CDF. Prove that for any points M and N it holds true AM þ BM þ MN þ CN þ DN EF. 26 1 Theorem on the Length of the Broken Line

1.2.11. Let ABCDEF be a convex hexagon such that AB ¼ BC ¼ CD, DE ¼ EF ¼ FA and ∠BCD ¼ ∠ EFA ¼ 60. Let G and H be two arbitrary points. Prove that AG þ GB þ GH þ DH þ HE CF. 1.2.12. (a) Let point M be the midpoint of side BC of a convex quadrilateral ABCD. ∠ 1 Given that AMD ¼ 120 . Prove that AB þ 2 BC þ CD AD.

(b) Let points A1, B1, C1, D1, E1, F1 be the midpoints of sides AB, BC, CD, DE, EF, FA of a convex hexagon ABCDEF, respectively. Given that all angles of pffiffi 2 3 hexagon A1B1C1D1E1F1 are equal. Prove that p 3 p1, where p and p1 are the perimeters of hexagons ABCDEF and A1B1C1D1E1F1, respectively. 1.2.13. Prove that any polygon with perimeter 2a can be covered by a circle whose diameter is equal to a. 1.2.14. Prove that the sum of the planar angles at a vertex of a tetrahedron: (a) is equal to 180, (b) is greater than 180, then any lateral edge of the tetrahedron is less than half- perimeter of its base. 1.2.15. Prove that any hexagonal cross-section plane of the unit cube has a perim- pffiffiffi eter greater than or equal to 3 2.

1.2.16. Given that a rectangle ABCD is inside of a convex polygon A1 ...An. Prove that

minðÞþAB; BC AB þ BC < A1A2 þ A2A3 þ ::: þ AnÀ1An:

1.2.17. Inside of the convex polygon with perimeter P are given two rectangles with the perimeter with sides a, b and c, d which do not have any common interior point. Prove that min(a, b) þ a þ b þ min (c, d) þ c þ d < P. 1.2.18. (а) Let points A and B be outside of circle ω. Through points A and B are drawn tangents AM and BN to circle ω. Prove that if segment AB intersects the circle ω, then AB > AM þ BN, and if it does not intersect, then AB AM þ BN. (b) Let points A, B, C be outside of circle ω, such that ω intersects segments AC, BC and does not intersect segment AB. Through points A, B, C are drawn tangents AM, BN, CK to circle ω. Prove that AM Á BC þ BN Á AC > CK Á AB.

Solutions

1.2.1. Let point X belong to a circle with center O1 and radius R1. Let point Y belong to a circle with center O2 and radius R2. For the broken line O1XYO2 we have that R1 þ XY þ R2 d. Thus, it follows that XY d À R1 À R2. 1.2 Theorem on the Length of the Broken Line 27

Similarly, for the broken line XO1O2Y we obtain that XY d þ R1 þ R2. This ends the proof.

1.2.2. Let a1 a2 ... an be lengths of the sides of a given polygon. We proceed with the proof by a contradiction argument. Assume that such two sides do not exist; then a2 2a1, a3 2a2,...,an 2an À 1. We have that a1 þ a2 þ ...þ an À 1 > an. Therefore, a1 > (...((an À an À 1) À an À 2) À ... À a1) (...(an À 1 À an À 2) À ... À > a1)(...(an À 2 À an À 3) À ... À a1) ... a1. Hence a1 a1. This leads to a contradiction. 1.2.3. Let points M, N, P be the midpoints of the segments BF, BE, CE, respectively. As ∠A 90, then point A is in the circle with diameter BF. BF CE Therefore, AM 2 . In a similar way, we obtain that DP 2 . Thus, it follows that 2AD 2AM þ 2MN þ 2NP þ 2DP BF þ EF þ BC þ CE. This ends the proof. ÀÁ ÀÁ 1.2.4. Let us consider points M a; b and N aþc ; d .As 2 2 qffiffiffiffiffiffiffiffiffi2 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 OM þ MN þ NC OC and OM ¼ a2þb ¼ AB, MN ¼ c þðÞbÀd ¼ BC and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 2 4 2 2 2 ðÞaÀc þd AC NC ¼ 4 ¼ 2 , then AB þ BC þ AC 2OC. See problem 1.1.14d. 1.2.5. (a) The proof follows straightforwardly from 1.2.5b for k ¼ 0. qffiffiffiffiffiffiffiffiffiffiffiffi k2 ::: ; ::: k ::: (b) Let us consider the points Ai 1 À 4 ðÞa1 þ þ ai b1 þ þ biþ 2ða1 þ

þaiÞÞ, i ¼ 1,...,n. Note that vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u0sffiffiffiffiffiffiffiffiffiffiffiffiffi 1 u 2 u k2 k 2 A A ¼ t@ 1 À a A þ b þ a i iþ1 4 iþ1 iþ1 2 iþ1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ::: ¼ aiþ1 þ kaiþ1biþ1 þ biþ1, i ¼ 0, , n À 1, where A0q(0;ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0) and 2 2 AoAn ¼ ðÞa1 þ ::: þ an þ kaðÞ1 þ ::: þ an ðÞþb1 þ ::: þ bn ðÞb1 þ ::: þ bn .As

A0A1 þ A1A2 þ ...þ An À 1An A0An, then qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 ka b b2 ::: a2 ka b b2 q1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ 1 1 þ 1 þ þ n þ n n þ n 2 2 ðÞa1 þ ::: þ an þ kaðÞ1 þ ::: þ an ðÞþb1 þ ::: þ bn ðÞb1 þ ::: þ bn :

nþk mþk mþn (c) Let Àa þ b þ c ¼ m, a À b þ c ¼ n, a þ b À c ¼ k, then a ¼ 2 , b ¼ 2 , c ¼ 2 . Therefore, if mn þ nk þ mk 0, then 28 1 Theorem on the Length of the Broken Line ! ffiffiffiffiffiffiffi ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffi p pffiffiffi 1 ÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 abcð a þ b þ cÞ¼ ðÞðn þ kÞm 2 þ ðÞn þ k mn þ nk þ mk þ 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ ðÞðm þ kÞn 2 þðm þ kÞ mn þ nk þ mk qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ ðÞðn þ mÞk 2 þðn þ mÞ mn þ nk þ mk qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4ðmn þ nk þ mkÞ2 þ 4ðm þ n þ kÞ2ðmn þ nk þ mkÞ 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4ðmn þ nk þ mkÞ2 þ 12ðmn þ nk þ mkÞ2 4 ¼ mn þ nk þ mk, according to problem 1.2.5а, and the following inequality (m þ n þ k)2 3mn þ 3nk þ 3mk for mn þ nk þ mk < 0. The proof is straightforward. 1.2.6. Let ABC be a right-angled triangle. Given that ∠C ¼ 90 and points M, N, R are the midpoints of segments PQ, QK, KH, respectively. Then CM þ MN þ NRCR h, where h is the altitude of triangle ABC drawn from vertex C. α α c α PQ PK QH Note that h ¼ c Á cos Á sin ¼ 2 sin 2 , CM ¼ 2 , MN ¼ 2 , NR ¼ 2 . Hence, it follows that KP þ PQ þ QH c Á sin 2α. 1.2.7. Let points M, N, P belong to sides AB, BC, AC, respectively. Let us draw from point P perpendiculars PE and PF to sides AB and BC. As points E and F belong to a circle whose diameter is BP, thus using the law of sines we obtain that EF ¼ BP Á sin ∠ B.IfQ, T are the midpoints of segments MP and NP, then MP MN NP EF EQ þ QT þ TF ¼ 2 þ 2 þ 2 . Thus, MP þ MN þ NP 2BP Á sin ∠ B. Let AA1, BB1, CC1 be the altitudes of triangle ABC; see Figure 1.12 and B1E1 ⊥ AB, B1F1 ⊥ BC. As points A, C1, A1, C are on a circle whose diameter is

Figure 1.12 1.2 Theorem on the Length of the Broken Line 29

AC, thus ∠BC1A1 ¼ ∠ C. In a similar way, we deduce that ∠AC1B1 ¼ ∠ C and ∠BA1C1 ¼ ∠ CA1B1 ¼ ∠ A. Let Q1, T1 be the midpoints of B1C1, A1B1, respectively. Then ∠E1Q1C1 ¼ 2 ∠ E1B1Q1 ¼ 2 ∠ B1C1C ¼ ∠ B1C1A1. Therefore E1Q1 k A1C1 and F1T1||C1A1. Hence 1 ∠ E1F1 ¼ E1Q1 þ Q1T1 þ T1F1 ¼ 2 ðÞB1C1 þ A1C1 þ A1B1 ¼BB1 sin B. Thus MP þ MN þ NP 2BP Á sin ∠ B 2BB1 Á sin ∠ B ¼ B1C1 þ A1C1 þ A1B1. Note that if points P and B1 do not coincide, then MP þ MN þ NP > B1C1 þ A1C1 þ A1B1. Therefore, MP þ MN þ NP ¼ B1C1 þ A1C1 þ A1B1 if and only if points M, N, P coincide with points C1, A1, B1, respectively. Second Proof Let points M, N, P belong to sides AB, BC, AC, respectively. According to problem 1.1.14d, for points M, N, P, B, we have that

MN þ NP þ MP 2BP sin ∠ sin B: ð1:16Þ

Let AA1, BB1, CC1 be the altitudes of triangle ABC.AsBP BB1, then from (1.16) we obtain that

MN þ NP þ MP 2BB1 sin ∠B: ð1:17Þ

Let the altitudes AA1, BB1, CC1 intersect at point H. Note that

2BB1 sin∠B ¼ BB1 sin∠B þ BB1 sin∠B ¼ BB1 sin∠B þ CC1 sin∠C ¼¼ BHsin∠B þ CHsin∠C þ B1H sin∠B þ C1Hsin∠C ¼ A1C1 þ A1B1 þþB1H cos∠C1B1H þ C1H cos∠B1C1H ¼ A1C1 þ A1B1 þ C1B1: ð1:18Þ

From (1.17) and (1.18), we obtain that

MN þ NP þ MP > A1C1 þ A1B1 þ C1B1:

Note that if points P and B1 do not coincide, then BP > BB1. Therefore, MN þ NP þ MP > A1C1 þ A1B1 þ C1B1. Hence, MP þ MN þ NP ¼ B1C1 þ A1C1 þ A1B1 if and only if points M, N, P coincide with points C1, A1, B1, respectively. This ends the proof. 1.2.8. Let the rotation by angle 180 À ∠ C and the rotation center C (Figure 1.13) point B moves to point B1, point M goes to point M1. We have that CB ¼ CB1. Therefore AC þ CB ¼ AB1. If ∠C 120 , then ∠MCM1 ¼ 180 À ∠ C 60 . Hence CM ¼ CM1, then CM MM1. Thus, it follows that 30 1 Theorem on the Length of the Broken Line

Figure 1.13

AM þ CM þ BM ¼ AM þ CM þ B1M1 AM þ MM1 þ B1M1 AB1 ¼ AC þ CB:

If for any point M of plane ABC it holds true that AM þ CM þ BM AC þ CB, then in the case when M coincides with A, we obtain that AB CB. On the other hand, if M coincides with B, then AB AC. Therefore, ∠C ∠ A, ∠C ∠ B. Now, assume that ∠C < 120.TheninsideangleCAB and outside triangle ABC one can ﬁnd a point B2,suchthat∠B2CB ¼ ∠ B2BC ¼ 60 .LetM0 be the second intersection point of line AB2 and circumcircle of triangle CBB2. Then AM0 þ BM0 þ CM0 ¼ AM0 þ M0B2 ¼ AB2 < AC þ CB2 ¼ AC þ CB. (See the proof of problem 1.2.9.). Hence, we have obtained that there exists a point M such that AM þ CM þ BM < AC þ CB. This leads to a contradiction. Thus ∠C 120 . Remark If max(∠A; ∠B; ∠C) < 120, then

AM þ BM þ CM AM0 þ BM0 þ CM0:

Indeed, according to problem 1.2.9a, we have that CB2 þ B2B BC or MCÁMB2 MBÁMB2 MCÁMB MB þ MC MB2. Therefore MA þ MB þ MC MA þ MB2 AB2 ¼ AM0 þ M0B2 ¼ AM0 þ BM0 þ CM0. 1 1.2.9. (а) Let us choose a point B on ray MA , such that MBi ¼ , i ¼ 1, . . . , n. i i MAi AiAj Note that BiBj ¼ . MAiÁMAj Indeed, if rays MAi and MAj are not on the same line, we obtain that triangle MAiAj is similar to triangle MBiBj. B B Hence, it follows that i j ¼ MBi ¼ 1 . AiAj MAi MAiÁMAj AiAj Therefore, BiBj ¼ . MAiÁMAj If rays MAi and MAj are on the same line, then

1 1 MAi Æ MAj AiAj BiBj ¼ MBi Æ MBj ¼ Æ ¼ ¼ : MAi MAj MAi Á MA MAi Á MAj 1.2 Theorem on the Length of the Broken Line 31

Then, the inequality we need to prove is equivalent to the following inequality B1B2 þ B2B3 þ ...þ Bn À 1Bn B1Bn. The equality holds true if B1, B2,...,Bn are on the same line (in the mentioned order). If M 2 B1Bn, then A1, A2,...,An 2 B1Bn. It is clear that points M, A1, A2,..., An are on the same line, in the following order: M, A1, A2,...,An or M, An, An À 1, ...,A1 or Ak, Ak À 1,...,A1, M, An, An À 1,...,Ak þ 1 or Ak þ 1,...,An, M, A1, ...,Ak, where k is an arbitrary positive integer not greater than n. If M2= B1Bn, then

∠A1 A2A3 þ ∠A1MA3 ¼ ∠A1A2M þ ∠A3A2M þ ∠A1MA3 ¼ ¼ ∠MB1B3 þ ∠B1B3M þ ∠B1MB3 ¼ 180 :

Therefore, points M, A1, A2, A3 are on the same circle. In a similar way, one obtains that points M, A2, A3, A4, ...,M, An À 2, An À 1, An are on the same circle. Then, points M, A1, A2,...,An are on the same circle (in the given order). 1 1 1 1 (b) 1. We have that MAÁMB ¼ MBÁMC þ MCÁMD þ MDÁMA а (see the proof of problem 1.2.9 ). qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ’ 1 3 1 Hence, using the Cauchy s inequality, we obtain that MAÁMB 3 2 2. pffiffiffi MAÁMBÁMC ÁMD Therefore, MC Á MD 3 3MA Á MB.

(c) 2. Note that ∠AMB ¼ 135, ∠CMD ¼ 45. Thus, it follows that MCÁMD ¼ SMCD ¼ MAÁMB SMAB ρðÞM;CD AB ρ ρðÞM;AB ¼ ρðÞM;AB þ 1 is the smallest if (M, AB) is the largest, i.e. M is the midpoint of arc AB. pffiffiffi MCÁMD 2π Therefore, MAÁMB ctg 8 ¼ 3 þ 2 2. 1.2.10. For points M, A, B, E and N, C, D, F using problem 1.2.9а we obtain that AM þ BM ME and CN þ DN NF. Thus, it follows that

AM þ MB þ MN þ CN þ DN ME þ MN þ NF EF:

This ends the proof.

0 0 1.2.11. Let points G and H be symmetric to points G and H with respect to line BE, respectively. As AB ¼ BD and AE ¼ ED, then points A and D are symmetric with 0 0 0 0 0 respect to the line BE. Using problem 1.2.10, we obtain that BG þ G D þ G H þ H 0 E þ H A CF. 0 0 0 0 0 0 Now, note that BG þ G D þ G H þ H E þ H A ¼ BG þ GA þ HG þ HE þ HD. Therefore, AG þ BG þ HG þ HE þ HD CF.

0 0 1.2.12. (a) Let points B and C be symmetric to points B and C with respect to lines 0 0 0 0 AM and DM, respectively. Note that BM ¼ MB , CM ¼ C M, ∠B MC ¼ 120 À (∠BMA þ ∠ CMD) ¼ 60. 0 0 0 BC Therefore B C ¼ B M ¼ BM ¼ 2 . 0 0 0 0 BC We have that AB þ B C þ C D AD or AB þ 2 þ CD AD. 32 1 Theorem on the Length of the Broken Line

(b) Note that ∠A1B1C1 ¼ ∠ B1C1D1 ¼ ...¼ ∠ F1A1B1 ¼ 120 . а AB BC CD : According to problem 1.2.12 , we have that 2 þ 2 þ 2 A1C1 BC CD DE ::: AF AB BC : In a similar way we obtain that 2 þ 2 þ 2 B1D1, , 2 þ 2 þ 2 F1B1 Summing up all these inequalities, we deduce that (see problem 1.2.5b)

3 p A1C1 þ B1D1 þ ...þ F1B1 ¼ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ A B 2 þ B C 2 þ A B Á B C þ B C 2 þ C D 2 þ B C Á C D 1 1pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 1 1 1 1 1 1 1 1 1 1 1 1 þ ::: þ A F 2 þ A B 2 þ A F Á A B ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 1 1 1 1 1 1 2 2 ¼ ðA1B1 þ B1C1 þ ...þ A1F1Þ þðB1C1 þ C1D1 þ ...þ A1B1Þ þðA1B1 þ B1C1 þ ...þ A1F1ÞðB1C1 þ C1D1 þ ...þ A1B1Þ pffiffiffi ¼ 3p1,

pﬃﬃ 2 3 : thus, p 3 p1 This ends the proof. 1.2.13. Points A, B are on the sides of a polygon and divide its perimeter to two equal parts. Note that for any point C, belonging to any side of the polygon, we have that ACþBC a OC OM þ MC 2 2, where O is the center of segment AB (Figure 1.14). a Thus, a circle with radius 2 and center O covers this polygon. This ends the proof.

1.2.14. (a) Given a tetrahedron SA1A2 ...An with a vertex S. Consider scanning its lateral surface, making the “cross section” along the edge SA1 (see Figure 1.15). Given that the sum of the planar angles at vertex S is equal to 180. Then 0 0 ∠A1SA1¼ 180 and 2SA1 < A1A2 þ A2A3 þ ...þ An À 1An þ AnA1 ¼ A1A2 þ A2A3 þ ...þ An À 1An þ AnA1. 0 (b) Let A be the intersection point of line A1S with a side of polygon A1A2 ...An A1, different from A1 (see the solution of problem 1.2.14а). Hence, if A 2 AkAk þ 1, 0 (An þ 1 A1), then A1A2 þ ...þ AkA > A1S þ SA, SA þ AAk þ 1 þ ... 0 0 þ AnA1 > SA1.

Therefore, A1A2 þ ...þ An À 1An þ AnA1 > 2SA1. 1.2.15. Consider the layout of the side surface of the cube (Figures 1.16 and 1.17). pffiffiffi Note that AA0 ¼ 3 2 and

Figure 1.14 1.2 Theorem on the Length of the Broken Line 33

Figure 1.15

Figure 1.16

Figure 1.17 34 1 Theorem on the Length of the Broken Line

AB þ BC þ CD þ DE þ EF þ FA ¼ AB þ BC þ CD þ DE þ EF þ FA0 AA0 pffiffiffi ¼ 3 2:

Second Proof We have that

ABpþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiBC þ CDqþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiDE þ EF pþ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiFA ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 a2 b2 b2 c2 c2 d2 d2 e2 e2 f 2 f 2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 þ 2 þ 1 þ 2 þ 1 þ 2 þ 1 þ 2 þ 1 þ 2 þ 1 þ 2 2 2 ðÞa1 þ b1 þ c1 þ d1 þ e1 þ f 1 þ ðÞa2 þ b2 þ c2 þ d2 þ e2 þ f 2 ¼ m,

(see problem 1.2.5а). Note that if a1 þqb1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ c1 þ d1 þ e1 þ f1 ¼ a,thena2 þ b2 qþ cffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 þ d2 þ e2 þ f2¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼6 À a, then m ¼ a2 þ ðÞ6 À a 2 ¼ 2a2 À 12a þ 36 ¼ 2ðÞa À 3 2 þ 18 pffiffiffi 3 2. pffiffiffi Therefore, it follows that AB þ BC þ CD þ DE þ EF þ FA 3 2. This ends the proof.

1.2.16. Let P ¼ A1A2 þ A2A3 þ ...þ An À 1An. Consider Figure 1.18. We have that P > P1 þ P2 þ P3 þ P4 and P1 AB Á sin α, P2 BC, P3 CD, P4 AD Á sin (90 À α). Thus, it follows that

P > AB Á sin α þ BC þ CD þ AD Á cos α minðÞAB; AD ðÞsin α þ cos α þBC þ CD minðÞAB; AD ðÞsin 2α þ cos 2α þBC þ CD ¼ minðÞþAB; AD BC þ CD:

Hence, we obtain that

A1A2 þ A2A3 þ ::: þ AnÀ1An > minðÞþAB; BC AB þ BC:

This ends the proof.

Figure 1.18 1.2 Theorem on the Length of the Broken Line 35

1.2.17. Lemma Let convex polygons B1B2 ...Bp and C1C2 ...Cm not have any common interior points. Then, there exists a line l passing through one of the sides of the polygons and separating these polygons. Indeed, let us choose a point O inside of the polygon B1B2 ...Bp, such that O is not on lines BiCj. Let k be the smallest positive integer, such that the image of the polygon B1B2 ...Bp by homothety with the center Oand ratio k has a common point with the polygon C1C2 ...Cm. Then, the following two cases are possible (Figure 1.19a, b). In both cases the proof of the lemma is straightforward. The proof of the problem follows from the lemma and problem 1.2.16. 1.2.18. (a) Let segment AB intersect circle ω, CD is tangent to the circle ω and CD|| AB (Figure 1.20a). We have that AC þ AB þ BD > CP þ PD ¼ ¼ AC þ AM þ BD þ BN. Therefore, AB > AM þ BN. If segment AB does not intersect circle ω (Figure 1.20b), then

AB AC þ CD þ BD ¼ AC þ CP þ PD þ DB ¼ AC þ CM þ BD þ DN ¼ AM þ BN

In this case, such placement of the points is also possible, as it is shown in Figure 1.21. Then, we have that AB < BK < BN < BN þ AM. (c) According to problem 1.2.18а, we have that

Figure 1.19 36 1 Theorem on the Length of the Broken Line

Figure 1.20

Figure 1.21

AM Á BC þ BN Á AC AMðÞþ BN þ CK BNðÞ AM þ CK > CKðÞ AM þ BN CK Á AB:

Therefore, AM Á BC þ BN Á AC > CK Á AB. This ends the proof.

Problems for Self-Study

1.2.19. Given two circles with radiuses R1, R2, such that the distance between their centers is equal to d and d þ R1 < R2. Prove that R2 À R1 À d XY d þ R1 þ R2, where X and Y are arbitrary points of these two circles. 1.2.20. Prove that in any quadrilateral there are at least two sides a and b, such that b < 1 a 1, 875. 1.2.21. Prove that any hexagonal cross section of a unit cube by a plane passing pﬃﬃ 3 2 through its center has an area not less than 4 . 1.2 Theorem on the Length of the Broken Line 37

1.2.22. In a quadrilateral ABCD angles A and C are not less than 90. Prove that the perimeter of the inscribed quadrilateral in the quadrilateral ABCD is not less than 2AC (see problem 2.3.6). 1.2.23. (a) In a tetrahedron SABC all planar angles at vertex S are equal to 60. Prove that AB þ BC þ AC SA þ SB þ SC. (b) In a convex hexagon ABCDEF any two of diagonals AD, BE and CF make an angle of 60. Prove that AB þ BC þ CD þ DE þ EF þ FA AD þ BE þ CF. (c) Among all convex quadrilaterals with given diagonals and given angles between them find the quadrilateral with the smallest perimeter. Hint (а) Let SA ¼ a, SB ¼ b, SC ¼ c; then sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 2 p 2 a 2 3 b 2 3 AB þ BC þ AC ¼ À b þ a þ À c þ b 2 2 2 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 2 c 2 3 þ À a þ c , 2 2

а and it is left topffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi use the inequality of problem 1.2.5 . 2 2 xþy : (b) Note that x þ y À xy 2 (c) Use the inequality of problem 1.2.5b.

1.2.24. Given a convex n-gon on a plane. Let ak be the length of its k-th side and dk be the length of its projection on a line containing that side. Prove that a1 þ a2 þ ::: þ an > 2. d1 d2 dn 1.2.25. Prove that for any polyhedron there are three edges, from which you can construct a triangle. 1.2.26. Given an inﬁnite set of points S on a plane, such that in any 1 Â 1 square there are ﬁnitely many number of points of the set S. Prove that there are two distinct points A and B from the set S, such that for any other point X belonging to the set S it holds true min(XA; XB) 0, 999AB. Hint Proceed the proof by contradiction argument. 1.2.27. Given a triangle ABC, such that ∠BAC 60.LetM be the midpoint of side BC and P be any point in plane ABC. Prove that PA þ PB þ PC 2AM. Hint Let the rotation by the angle 60 and center A, point P moves to point P1 and point C moves to point C1. Prove that PA þ PB þ PC ¼ PB þ PP1 þ P1C1 BC1 2AM. 1.2.28. On the sides of a unit square, as on hypotenuses, are externally constructed right-angled triangles. Let A, B, C, D be the vertices of the right angles and O1, O2, 38 1 Theorem on the Length of the Broken Line

O3, O4 be the incenters of these triangles, respectively. Prove that the area of quadrilateral: (a) ABCD is not greater than 2. (b) O1O2O3O4 is not greater than 1. Hint (a) Let M and N be the midpoints of the opposite sides of the square. Then AC AM þ MN þ CN ¼ 2. (b) Prove that if points O , O , O , O are on a circle, circumscribed around the p1 ffiffiffi 2 3 4 square, then O1O3 2.

1.2.29. Let SABC be a tetrahedron. Prove that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi H2 S þ S þ S ðÞAB þ BC þ AC 2 þ S2 , SAB SBC SAC 4 ABC where H is the length of the altitude of tetrahedron SABC, drawn from the vertex S. Hint See problem 1.2.5а. 1.2.30. Let P be the projection of point M on a plane containing points A, B, C. Prove that if from segments PA, PB, PC one can construct a triangle, then one can construct a triangle from segments MA, MB, MC too. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 HintqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiWe have that MA þ MB ¼ AP þ MP þ BP þ MP ðÞAP þ BP 2 þ 4MP2 (see problem 1.2.5a).

1.2.31. Let ABC be a triangle. Prove that 2pr ama pR. Hint See the solution of problem 1.2.7. 1.2.32. Let A , B , C be points on sides BC, AC, AB of triangle ABC, respectively. 1 1 1 pffiffiffi Prove that A1B1 þ B1C1 þ A1C1 3minðÞAA1; BB1; CC1 . Hint See the solutions of problems 1.2.7 and 1.2.3. Chapter 2 Application of Projection Method

This chapter consists of four sections and is devoted to the projection method. The projection method is one of the fundamental methods applied in order to deal with geometric inequalities. In this chapter we learn some techniques on how the projection method can be applied and to prepare a background for the application of projection method: in Section 2.1 we consider several problems and some properties of convex polygon lying inside of another polygon, in Section 2.2 we consider some problems with a sufficient condition for comparison of lengths of two broken lines on the plane; in Section 2.3 we deal with the inscribed polygons with the least perimeter and in Section 2.4 we consider problems that can be proved using these properties and the projection method. The main statement that we are going to use in order to apply the projection method is the following: if A0B0 is the orthogonal projection of segment AB onto a line l or onto a plane α, then AB A0B0. In Section 2.1 selected problems are those that deal with geometric inequalities related to two figures, such that one of them is inside of the other one. In this section, problems 2.1.1 and 2.1.10 are used as the main techniques of proofs. In Section 2.2 is provided a sufficiency condition of comparing the sum of the length of the segments belonging to two sets of segments on the plane. This beautiful condition is given in problem 2.2.18. Moreover, the following problems demonstrate the application of this condition. In Section 2.3 are given the solutions, for the triangle and quadrilateral, of the following problem: given a convex n-gon A1A2 ...An. Inscribe to A1A2 ...An a polygon B1B2 ...Bn, such that B1B2 ...Bn has the least perimeter (in Section 4.1 are provided some generalizations of this problem, see problem 4.1.18). The last section of this chapter (Section 2.4) is devoted to the application of projection method in different problems. Some problems in this chapter were inspired by [4, 8, 9, 12, 13, 15]. Nevertheless, even for these problems the authors have mostly provided their own solutions.

2.1 Convex Polygon Lying Inside of Another Polygon

2.1.1. Prove that, if a convex polygon lays inside of other polygon, then the perimeter of an internal polygon is less than the perimeter of the external polygon. 2.1.2. Prove that, if the sum of the plane angles at the top of a pyramid is greater than 180, then each lateral edge of a pyramid is smaller than the half-perimeter of its base. 2.1.3. Consider a convex quadrilateral with sides a , b , c , d laying inside of a unit 1 1 1 1 square. Prove that þ þ þ > 4. a b c d 2.1.4. Prove that, if a regular hexagon with side length b lays inside of an equilateral a triangle with side length a, then b < . 3 2.1.5. Prove that, if a rectangle is covered by a right-angled triangle, so that its smaller side is on a hypotenuse, then it is possible to cover this rectangle by the same right-angled triangle, such that its two sides are situated on the legs.

2.1.6. (a) Let S1 and S2 be squares with sides a and b such that they are inside of a unit square. Given that they have no common points. Prove that a þ b < 1. (b) Given that regular n-gons with sides b and c are located inside of a regular n-gon with side a and have no common points. Prove that b þ c < a.

2.1.7. If a regular polygon B1B2 ...Bn is inscribed into polygon A1A2 ...An (B1 2 A1A2, B2 2 A2A3, ..., Bn 2 AnA1), where ∠A1 ¼ ∠A2 ¼ ...¼ ∠An, then A1A2 ...An is also a regular polygon and one can place in polygon B1B2A3 ... AnA1 a polygon equal to B1B2 ...Bn, so that its sides are parallel to the sides of polygon A1A2 ...An. 0 0 0 2.1.8. Let two non-intersecting trianglespffiffiffi with altitudes h1 , h2 , hÀÁ3 and h1, h2, h3 lay ðÞþ; ; 0 ; 0 ; 0 inside of a unit square. Prove that 2 min h1 h2 h3 min h1 h2 h3 . a 2.1.9. (a) Prove that, if a regular 2n -gon with a side length 2 lays inside of a regular 2n-gon with a side length a and center O, then it covers point O. (b) A regular n-gon with a side length b is inside of a regular n-gon with a side a length a and does not contain its circumcenter. Prove that b < ÀÁ. 2 π 2cos 2n 2.1.10. Let a convex polygon M is inside of triangle ABC. Prove that polygon M can be covered by triangle ABC, so that one of the sides of polygon M lays on one of the sides of the triangle. 2.1.11. Consider two triangles, such that one triangle has sides a, b, c, the other has sides a0, b0, c0. Which relations between numbers a, b, c and a0, b0, c0 are necessary and sufficient in order the first triangle to be covered by the second one?

2.1.12. Prove that, if triangle A1B1C1 is covered by a triangle ABC, then

(a) A1B1 þ B1C1 þ A1C1 À maxðÞA1B1; B1C1; A1C1 AB þ BC þ AC À maxðÞAB; BC; AC , 2.1 Convex Polygon Lying Inside of Another Polygon 41

(b) T(A1B1C1) T(ABC), where for a triangle XYZ T(XYZ) ¼ min (MX þ MY þ MZ), and M is any point of a plane XYZ. 0 0 0 2.1.13. Let a triangleÀÁ altitudes h1, h2, h3 be inside of a triangle with altitudes h1, h2, 0 ; 0 ; 0 ðÞ; ; h3. Prove that min h1 h2 h3 min h1 h2 h3 . 2.1.14. Let a triangle be inside of the non-acute-angled triangle. Prove that the ﬁrst triangle can be in the second one in such a way that two of its vertices are on the largest side of the second triangle.

2.1.15. Let an equilateral triangle Ap1ffiffiffiB1C1 be inside of the non-acute-angled triangle ∠ π ABC ( C 2). Prove that AB 3A1B1.\ 2.1.16. Given on a plane an equilateral triangle XYZ with the side length equal to 1 and an equilateral triangle DEF,suchthat∠DEF ¼ 20. Prove that the area of a 1 convex figure containing triangles XYZ and DEF is greater than or equal to 2 cos 10 . 2.1.17. Let 1 Â 2 rectangle be inside of three mutually non-intersecting squares. Given that the side lengths of the squares are equal to a, b, and c. Prove that a þ b þ c 2. pffiffi 5À1 2.1.18. Given that a triangle ABC is covered by a unit square. Prove that r 4 , where r is the inradius of triangle ABC.

Solutions

2.1.1. Let us construct on the sides of the internal polygon, outside of it, half-strips, with parallel edges perpendicular to corresponding sides of the polygon (Figure 2.1). It is clear, that the perimeter of the internal polygon does not exceed that part of the perimeter of the external polygon which is inside of these strips. Hence, the perimeter of the internal polygon is less than the perimeter of the external polygon.

2.1.2. Given a pyramid SA1A2 ...An with a vertex S. Consider the layout of its lateral surface making the cut along the edge SA1 (Figure 2.2).

Figure 2.1 42 2 Application of Projection Method

Figure 2.2 A1'

Аn A1 S

Figure 2.3 K NB C

MA D L

Since according to the statement of the problem, the sum of the plane angles at the 0 ::: 0 vertex S is greater than 180 , then triangle SA1A1 lays inside of polygon A1A2 AnA1. þ 0 < þ þ :::þ 0 þ 0 According to problem 2.1.1 2SA1 A1A1 A1A2 A2A3 AnA1 A1A1. 2.1.3. According to problem 2.1.1, we have that a þ b þ c þ d < 4. If 1 þ 1 þ 1 þ 1 ÀÁa b c ÀÁd 4, thenÀÁ by summingÀÁ up these two inequalities, we deduce that þ 1 þ þ 1 þ þ 1 þ þ 1 < a a b b c c d d 8. This leads to a contradiction, as þ 1 > 1 þ 1 þ 1 þ 1 > x x 2(x 0). Therefore, a b c d 4. 2.1.4. It is sufficient to notice that the radius of the circle inscribed into a hexagon is less than the radius of the circle inscribed into the triangle. 2.1.5. Without loss of generality we can assume that rectangle is inscribed in a right-angled triangle (Figure 2.3). > SincepffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiAB AD, thenpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi we have that MN BC KC. Thus, it follows that NC ¼ MC2 À MN2 < MC2 À KC2 ¼ MK, where MNBA is a rectangle, such that triangle NCF can be placed inside of a triangle MKL similar to it. 2.1.6. (a) At first, let us prove the following lemma. Lemma If a square PQRS with a side length a is inscribed in a rectangle ABCD (Figure 2.4), then ABCD is also a square and the square with a side a can be placed inside of pentagon ABQRD, such that the sides of the square would be parallel to AB and AD. 2.1 Convex Polygon Lying Inside of Another Polygon 43

Figure 2.4 BQC

P1 Q1

P P2

R2 R

S2 S1 R1 AS D

Figure 2.5 a b

Indeed, consider the rotation of square PQRS around point O by an angle 90. Then point R will turn into point Q, hence line CD will turn into a line BC. Consequently, point O is equidistant from sides CD and BC. Similarly, we obtain that point O is equidistant from sides AB and BC, AD and AB. From above it follows that ABCD is a square with center O. Consider a square P1Q1R1S1 with center O, such that its sides are parallel to AB, AD, and P Q ¼ PQ (Figure 2.4). 1 1 ! Perform a translation by a vector Q1Q2 . Therefore, square P1Q1R1S1 turns into a square P0Q0R0S0 that is inside of pentagon ABQRD, as one can easily prove that Q1Q2 ¼ PP2 ¼ SS2. This ends the proof of the lemma. Using the lemma, we can place the square with a side a inside of a unit square, so that their sides are parallel and again the square with a side a would not have common points with a square with a side b (Figure 2.5). Using the lemma once again, we obtain that the sides of the squares with sides a and b, which do not have common points, are parallel to the sides of the unit square. It is not difﬁcult to prove that a þ b < 1.

(b) Let a regular n-gon B1B2 ...Bn with a side b is inside of a regular n-gon with a side length a. Let us draw lines passing through the vertices of B1B2 ...Bn and parallel to the sides of the n-gon with side a (Figure 2.6). 44 2 Application of Projection Method

Figure 2.6 а

cB2 B3 A2

B1 b

A1 Bn An

Figure 2.7 l Πc c dc db b

ΠΠb a

According to problem 2.1.7 A1A2 ...An is a regular n-gon. It is not difﬁcult to prove that a regular n-gon with a side c can have a common point only with one of the triangles B1A2B2, B2A3B3 ,. . .,BnA1B1. Then according to problem 2.1.7, without loss of generality, one can assume that the regular n-gon with a side b is inside of the regular n-gon with a side a, so that the sides of these polygons are parallel. Similar statement is true for a regular n-gon with a side c (Figure 2.7). Let l be a line passing along one of the sides of the polygons with sides b and c and separating these polygons (see the solution of problem 1.2.17). Let strips Πa, Πb, and Πc be those with minimal widths that contain these polygons having widths da, db, dc and their boundaries are parallel to line l. It is clear that d þ d < d and db ¼ b, dc ¼ c. Thus, b þ c < a. b c a da a da a ∠ ¼ ∠ ¼ ::: ¼ ∠ ¼ π À 2π 2.1.7. Note that B1A2B2 B2A3B3 BnA1B1 nÀÁ, (Figure 2.8). ∠ ¼ α ∠ ¼ 2π À α ∠ ¼ π À π À 2π þ 2π À α Let A2B1B2 , then A2B2B1 n , A3B2B3 n n ¼ α ∠ ¼ 2π À α Δ ¼ Δ ¼ , A3B3B2 n . Hence A2B2B1 A3B3B2. Consequently, B1A2 B2A3, A2B2 ¼ A3B3. Similarly, we obtain that A1B1 ¼ A2B2 ¼ ...¼ AnBn and B1A2 ¼ B2A3 ¼ ...¼ BnA1. Therefore A1A2 ¼ A2A3 ¼ ...¼ AnA1, this means that A1A2 ...An is a regular n-gon. Let O be the center of a regular n-gon A1A2 ...An. 2.1 Convex Polygon Lying Inside of Another Polygon 45

Figure 2.8 А3 B3¢ B2 B3

A2 B2¢

B1¢

A1 Bn An

Figure 2.9

P1 l

d2 P

Note that OB1 ¼ OB2 ¼ ...¼ OBn, or in the other words, O is the center of 0 0 ::: 0 regular n-gon B1B2 ...Bn. Choose points B1, B2, , Bn on segments OA1, 0 0 0 OA2,...,OAn correspondingly, so that OB 1 ¼ OB 2 ¼ ÁÁÁ ¼OB n ¼ OB1. Since 0 0 ::: 0 points B1, B2,...,Bn, B1, B2, , Bn lay on the same circumference, then from the ¼ 0 0 0 0 condition B1B2 ÀÁB2B3 it followsÀÁ that B1B2B2B3 is an equilateral trapezoid. ρ 0 ; ¼ ρ ; 0 0 Consequently,ÀÁÀÁB2 B1B2 ÀÁB2 B2B3 . Similarly,ÀÁ one canÀÁ prove that 0 0 0 0 0 0 0 0 ρ B1; B B ¼ ρ B ; B1B2 ¼ ρ B2; B B ¼ÁÁÁ¼ρ Bn; B B ¼ ρ B ; BnB1 . 1 2 2 2 3 n 1 ! 1 0 ⊥ 2 0 Let B2M B1B2 and M B1B2, then at translation by a vector B2M the image of 0 0 ::: 0 the regular n-gon B1B2 Bn will be in a polygon B1B2A3 ...AnA1. 2.1.8. Let straight line l contains one of the sides of the given triangles and separates these triangles (Figure 2.9). Consider bands Π1, Π2, Π containing these triangles and the square with minimal widths d1, d2, and pd,ffiffiffi their boundaries being parallel to line l.ÀÁ It is clear that þ 0 ; 0 ; 0 pd ffiffiffi d1 d2. Since 2 d ÀÁand d1 min (h1, h2, h3), d2 min h1 h2 h3 , then ðÞþ; ; 0 ; 0 ; 0 2 min h1 h2 h3 min h1 h2 h3 . 46 2 Application of Projection Method

Figure 2.10

a 2.1.9. (a) Suppose that the polygon with a side 2 does not contain point O, then the polygon symmetric to the given polygon with respect to point O is also inside of the a þ a < polygon with a side a. Then, according to problem 2.1.6(b), 2 2 a. This leads to a contradiction. (b) At ﬁrst, let us prove that the circumcircle of the regular n-gon with side b does not have points outside of the circumcircle of the regular n-gon with side a. Indeed, assume that these circles intersect. Let us choose point M on the larger circle, such that it is not a vertex of the regular n-gon with a side a and is inside of the smaller circle (Figure 2.10). Let given regular n-gons be A1A2 ...An (A1A2 ¼ a) and B1B2 ...Bn. Assume that point M is on the smaller arc A1A2. ∠ ¼ πðÞnÀ1 ... We have that A1MA2 n and that polygon A1A2 An is inside of angle ∠A1MA2. Consequently, polygon B1B2 ...Bn also is inside of that angle. πðÞnÀ1 Hence, ∠A1MA2 > max ∠BiMBj > . This leads to a contradiction. i, j n

Let us denote the radiuses of these circles by Ra and Rb and their centers by O and O1. From aforesaid, it follows that Ra Rb þ OO1, and since point O is not inside of polygon B1B2 ...Bn, then OO1 > rb, where rb is the radius of the circle inscribed in B1B2 ...Bn. a b a Thus, R > R þ r ,or > . We deduce that b < . a b b 2 sin π π b 2cos 2 π n 2 sin þ π 2n n 2tg n

2.1.10. It is enough to prove the problem for triangle A1B1C1 (Figure 2.11a, b), where A1B1 k AB, B1C1 k BC, A1C1 k AC. Consider the following two cases: Figure 2.11a, b. Let us consider ﬁrst the case of Figure 2.11a. Perform a rotation around vertex C2 by some angle, so that one of the sides with vertices C2, B2, A2 of polygon M becomes for the ﬁrst time parallel to lines A1B1, A1C1,orB1C1, respectively. Let u 2 [u1, u2], u1 and u2 being the values of u for aforesaid rotations (in positive and negative directions). Let us denote by Mu the image obtained by a rotation of 2.1 Convex Polygon Lying Inside of Another Polygon 47

Figure 2.11 B B1

C2 A2 u nf(u) j-u mf(u) Mu ag A1 B2 C1 AC a

B B1

u mf(u)

Mu ag A1 B2 C1 AC b

polygon M around C2 with subsequent similarity transformations with similitude centers A1 and B1. Note that ΔA2B2C2 is inscribed in triangle A1B1C1

It is clear that some of the sides of polygons Mu1 and Mu2 are on one of the sides of triangle A1B1C1. Let C2B2 ¼ mf(u), A2B2 ¼ nf(u). Let us prove that, on segment [u1, u2], function f(u) accepts its maximal value at point u1 or u2. Indeed, we have that

mfðÞ u sin u nfðÞ u sin u A C ¼ A B þ B C ¼ þ sin ðÞφ À u : 1 1 1 2 2 1 sin α sin γ

Thus, it follows that α γ α γ ðÞ¼ A1C1 sin sin ¼ A1C1 sin sin fu γ þ ðÞφ À α ðÞþ φ , m sin u sin n sin u sin a sin u 1 where a, φ, and φ1 are constants. Since, in case u 2 [u1, u2] f(u) > 0, or equivalently, sin(u þ φ1) > 0, then on segment [u1, u2], function sin(u þ φ1) accepts its minimal value at point u1 or u2. Hence, on segment [u1, u2], function f(u) accepts its minimal value at points u1 or u2. The proof is similar for the case of Figure 2.11b. This ends the proof. 48 2 Application of Projection Method

Figure 2.12 C'

a' b'

c B' c' A'

a' b'

ha'

c B' c' A'

Π a'

Figure 2.13

2.1.11. According to problem 2.1.10, if the first triangle is covered by the second triangle, then one can assume that one of the sides of the first triangle lays on the side of the second triangle. Let us find necessary and sufficient conditions for the first triangle to be covered by the second one in a way shown in Figure 2.12. 0 0 0 Let ha0 , hb0 , hc0 be the altitudes of the triangle with sides a , b , c drawn to the 0 0 0 Π 0 Π 0 Π 0 sides a , b , c , correspondingly, and a , b , c be the projections of the first triangle on the straight lines, containing heights ha0 , hb0 , hc0 (Figure 2.13). Let us prove that, or order that, one can place the triangle with sides a, b, c inside of the triangle with sides a0, b0, c0 in a way shown in Figure 2.12, it is necessary and sufficient that conditions

0 Π 0 0 Π 0 0 Π 0 ð : Þ ha a , hb b , hc c 2 1 hold true. The necessity of the conditions ha0 Πa0 follows from the fact that the triangle 0 0 0 0 2 0 with sides a, b, c is inside of a band with boundaries B C and l1, where A l1, 0 0 0 l1 B C , (see Figure 2.12). Similarly, one can prove the necessity of conditions 0 Π 0 0 Π 0 hb b and hc c . 2.1 Convex Polygon Lying Inside of Another Polygon 49

a' a bb'

β' βαα' B' B A A'

Figure 2.14

Figure 2.15 C

djl EF

0 Now, if condition (2.1) is satisfied, then it is clear that c c , otherwise Πa0>ha0 . Let us consider Figure 2.14. ! If the parallel displacement of point C by a vector BB0 has moved it in the shaded 0 < Π 0 half-band, then it means that hb b . This leads to a contradiction. This means that, if condition (2.1) is satisfied, then the first triangle can be placed inside of the second one. Note that, if point C and straight line l (Figure 2.15) are given, then the projection of segment CE (E 2 l) on line l1 is equal to CE| cos (φ À δ)|. 0 0 Π 0 ¼ Π 0 ¼ 1 ð Áj ðÞjþα þ β Áj ðα À α Þjþ Thus, we obtain that c hc, b 2 a sin b sin 0 0 0 0 α Þ Π 0 ¼ 1 ðÞÁj ðÞjþβ À β Áj ðÞjþβ þ α β c sin , a 2 a sin b sin c sin (see the proof of problem 2.2.1a). Then, condition (2.1) can be rewritten as:

1 0 0 0 0 0 ðÞÁj ðÞjþα þ β Áj ðÞjþα À α α hc hc, hb a sin b sin c sin , 2 ð2:2Þ 1 0 0 0 h 0 ðÞa Ájsin ðÞjþβ À β b Ájsin ðÞjþβ þ α c sin β : a 2

It is clear that for condition (2.2) one needs only the values of a, b, c, a0, b0, c0. Note that, for two triangles 18 variants of dispositions (similar to Figure 2.12) are possible and for the first triangle to be covered by the second triangle, it is 50 2 Application of Projection Method necessary and sufficient that condition (2.2) is satisfied at least for one of these 18 variants.

2.1.12. We have to prove that f(A1B1C1) f(ABC), if ΔABC covers ΔA1B1C1, where f(ABC) 2 R and that f(ABC) ¼ f(MNK), if ΔABC ¼ ΔMNK. According to problem 2.1.10, one can assume that one of the sides of triangle A1B1C1 lays on one of the sides of triangle ABC. Without loss of generality, we can assume that two sides of triangle A1B1C1 lay on two sides of triangle ABC, one of the two sides of triangle A1B1C1 coincides with one of the sides of triangle ABC. Indeed, in the case of Figure 2.16a, we have that

fAðÞ1B1C1 fAðÞ1B1A2 fABðÞ1A2 fABðÞ1C fðÞ ABC :

Therefore, f(A1B1C1) f(ABC), while in the case of Figure 2.16b, we have that f(A1B1C1) f(A1B1B2) f(A1BB2) f(A1BC) f(ABC). Thus, it follows that f(A1B1C1) f(ABC).

(a) Let triangles A1B1C1 and ABC are as it is shown in Figure 2.17.

Figure 2.16 CC

A2 B2 C1

BB1 A1 AB B1 A1 A a) б)

Figure 2.17 CºC1

b=b1 a1 a

a=a1 b1 b AºA1 c1 B1 B c 2.1 Convex Polygon Lying Inside of Another Polygon 51

Let us consider the following three cases: max(α1, β1, γ1) ¼ γ1, max(α1, β1, γ1) ¼ α1, and max(α1, β1, γ1) ¼ β1.

I. If max(α1, β1, γ1) ¼ γ1. Therefore, max(α, β, γ) ¼ γ. Note that ∠CB1B ¼ 180 À β1 > 90 , which means that a > a1. Hence, a þ b ¼ a þ b1 > a1 þ b1. II. If max(α1, β1, γ1) ¼ α1, then α ¼ α1 β1 > β. Consequently, max(α, β, γ) ¼ α or max(α, β, γ) ¼ γ. If max(α, β, γ) ¼ α, then b þ c ¼ c þ b1 > b1 þ c1. If max(α, β, γ) ¼ γ, then γ α ¼ α1 γ1. Therefore, ∠CB1B ¼ 180 À β1 > 90 . Consequently, a > a1, this means that a þ b ¼ a þ b1 > a1 þ b1 b1 þ c1.

III. Let max(α1, β1, γ1) ¼ β1. Consider the following three cases:

1. max(α, β, γ) ¼ β, then a þ c ¼ a þ BB1 þ c1 > a1 þ c1. 2. max(α, β, γ) ¼ α, then β1 α1. Consequently b1 a1, hence b þ c ¼ b1 þ c a1 þ c > a1 þ c1. 3. max(α, β, γ) ¼ γ. In this case, let us notice that if point E moves from point B1 to point B along segment B1B, then the value of ∠AEC À ∠ACE decreases from β1 À γ1 (β1 À γ1 0) to the value of β À γ (β À γ 0). Consequently, on segment B1B exists such point E that ∠AEC ¼ ∠ACE. Then a1 þ c1 AE þ CE ¼ AC þ CE ¼ b þ CE a þ b. Consequently, a1 þ c1 a þ b (∠CEB > 90 , if point E does not coincide with point B). This ends the proof of (a). (b) In order to end the proof, note that if max(α, β, γ) 120, then T(ABC) ¼ AB þ BC þ AC À max (AB, BC, AC) (see problem 1.2.8) and if max(α, β, γ) < 120, then T(ABC) ¼ BB0 (Figure 2.18). Indeed, since max(α, β, γ) < 120, then quadrilateral ABCB0 is convex and point B is outside of the circumcircle of the equilateral triangle ACB0. Let point М0 be such

Figure 2.18 B

M0 M A600 C 600 600

M¢

B¢ 52 2 Application of Projection Method that AМ0 ¼ AM and ∠MAM0 ¼ 60 (Figure 2.18). Then ΔAMC ¼ ΔAM0B0 and conse- 0 0 0 0 0 quently, AM þ BM þ CM ¼ MM þ BM þ M B BB ¼ BM0 þ M0B ¼ BM0 þ 0 AM0 þ CM0. This means that T(ABC) ¼ BB . Note that T(ABC) ¼ BM0 þ AM0 þ 0 0 CM0 > BM0 þ AM0 > AB and T(ABC) ¼ BB < AB þ AB ¼ AB þ AC. Similarly, one can prove that T(ABC) > BC, T(ABC) > AC, T(ABC) < BC þ AC, T(ABC) < AB þ BC. Without loss of generality, we can assume that side A1C1 is on side AC. Consider the following cases: I. If max(α, β, γ) < 120 , max(α1, β1, γ1) < 120 , then convex quadrilateral 0 A1B1C1B1 has a diameter T(A1B1C1) and is located inside of convex quadri- 0 lateral ABCB with a diameter T(ABC). Hence T(A1B1C1) T(ABC). II. If max(α, β, γ) 120 , max(α1, β1, γ1) < 120 , then according to problem 2.1.12a, we have that

TðÞ¼ ABC AB þ BC þ AC À maxðÞAB þ BC þ AC A1B1 þ B1C1 þ A1C1 À maxðÞA1B1 þ B1C1 þ A1C1 > TAðÞ1B1C1 :

Therefore, T(ABC) > T(A1B1C1). III. If max(α, β, γ) 120 , max(α1, β1, γ1) 120 , then according to problem 2.1.12a, we have that T(ABC) T(A1B1C1). IV. If max(α, β, γ) < 120 , max(α1, β1, γ1) 120 , then without loss of generality, one can assume that these triangles are ABC and AB1C, with vertex B1 being on side AB and ∠AB1C ¼ 120 . Then, the proof is similar to the proof of case I.

2.1.13. Let min(h1, h2, h3) ¼ h1 and Π is a band with a minimal width containing the 0 0 0 triangle with altitudes h1, h2, h3 and with boundaries perpendicular to the altitude with a length h1 (see Figure 2.19). ÀÁ ðÞ; ; 0 ; 0 ; 0 Π Then, min h1 h2 h3 h min h1 h2 h3 , where h is the width of the band . 2.1.14. According to problem 2.1.10, one can assume that one side of the second triangle is on one side of the first triangle and that the second triangle is in the first triangle. Without loss of a generality, we can assume that possible cases are those presented in Figure 2.20. In the case of Figure 2.20a the proof of the problem is obtained by rotating triangle AB1C around point A by angle ∠B1AB. In the case of Figure 2.20b,ifAB1 AC, then it is sufficient to consider the triangle, symmetric to triangle A1B1C with respect to the bisector of angle ∠CAB.

Figure 2.19

h1 2.1 Convex Polygon Lying Inside of Another Polygon 53

Figure 2.20 B B

B1 B1

A A≡A1 C1 C A1 C≡C1 ab

Figure 2.21 B

Figure 2.22 C

AA1 HMB1 B

> ∠ < ∠ ∠ > π Now, let AB1 AC, then AB1C ACB1. Consequently, BB1C 2. If ∠AB1A1 ∠BCB1, then the proof of the problem is obtained by rotating triangle A1B1C around point C by angle À ∠AB1A1. Let AB1 > AC and ∠AB1A1 > ∠BCB1, then ∠AB1C1 < ∠ABC. Since B1C < BC, A1B1 < AB, ∠AB1C1 < ∠ABC, then if triangle A1B1C1 is placed in such a way that points B1 and B coincide, while point A1 is on side AB, then vertex C1 of triangle A1B1C1 is inside of triangle ABC. This ends the proof. Remark If ABC is an acute triangle, then the statement of the problem is wrong (see the example in Figure 2.21).

2.1.15. According to problem 2.1.10, one can assume that points A1 and B1 lay on segment AB (Figure 2.22). 54 2 Application of Projection Method

∠ ∠ π Note that AC1B ACBpffiffiffi 2. Therefore, if M is the midpoint of segment AB, then AB 2C1M 2C1H ¼ 3A1B1.

Remark Similarly, one can prove that, if the triangle with altitudes h1, h2, h3 is inside of non-acute triangle ABC, then the inequality max(AB, BC, AC) 2 min (h1, h2, h3) holds true. Another proof of the problem can be obtained by using problem 2.1.13. 2.1.16. (Solution of M.D. Kovalev.) First we need to prove that the side of an equilateral triangle, containing ΔDEF, is not less than p2ffiffi cos 10 . According to 3 problem 2.1.10, it is sufficient to consider the following two cases (see Figure 2.23). pffiffi In case (a), we have that 3 a cos 10 . Consequently, a p2ffiffi cos 10 . 2 3 sin 100 p2ffiffi In case (b), we have that a D E ¼ ¼ cos 10 , where ∠FD E ¼ 60 . 1 sin 60 3 1

Consider now the minimal equilateral triangle X1Y1Z1 with sides parallel to the sides of triangle XYZ, which contains triangles XYZ and DEF (Figure 2.24).

aE a a a

1 F

200 DF D1 DE a b

Figure 2.23

Y1 Y1 Y1

YxYy a x aa Y х

XZXZ yz

XºX1 ZZ1 X1 Z1 X1 Z1 abc

Figure 2.24 2.1 Convex Polygon Lying Inside of Another Polygon 55

The area of a convex figure containing ΔXYZ and ΔDEF is not less than: pffiffiffi pffiffiffi pffiffiffi 3 1 3 3 in case (a): þ x ¼ þ ðÞa À 1 , p4 ffiffiffi 2 4 pffiffiffi 4 pffiffiffi 3 x þ y 3 3 in case (b): þ ¼ þ ðÞa À 1 , p4ffiffiffi 2 4 pffiffiffi4 pffiffiffi 3 x þ y þ z 3 3 in case (c): þ ¼ þ ðÞa À 1 . 4 2 4 4 pffiffi 3 1 It remains to note that 4 a 2 cos 10 . Remark Among all convex figures covering any triangle with the sides not exceeding 1, the least area has a triangle ABC, such that ∠A ¼ 60, AB ¼ 1, and the altitude 1 drawn to AB is equal to cos10 . The area of that triangle is equal to 2 cos 10 . 2.1.17. Let sides AB and BC of rectangle ABCD be equal to 1 and 2 respectively, and squares with sides a, b, c are not mutually intersecting and lay inside of rectangle ABCD. Note that, for each side of a rectangle one can find a “good” square, such that, while moving it in a direction perpendicular to the given side, it does not intersect with the other squares before intersecting with that side. According to the Dirichlet’s principle there is a “good” square simultaneously for two sides of the rectangle. If these are opposite sides, then this ends the proof (see Figure 2.25). In the case of Figure 2.25a, we have that a x þ y ¼ MN and b ND. Consequently, a þ b CD ¼ 1. We have that c 1. Thus, a þ b þ c 2. In the case of Figure 2.25b, according to problem 2.1.6a, we have that b þ c 2 À MN, a MN. Therefore, a þ b þ c 2 À MN þ MN ¼ 2. If these sides are adjacent, then it is possible to assume that the “good” square is located on one of these sides (see the proof of problem 2.1.6a). By repeating these reasonings for the new squares we eventually find that one of these squares lies in one of the corners of the rectangle (see Figure 2.26).

BCBC xМ yb 2-MN xN bcb

A DA M N D 2-MN ab

Figure 2.25 56 2 Application of Projection Method

Figure 2.26 BPC

AQD

babacba c

c abc

Figure 2.27

b xy c y c x а cb d a a ab

Figure 2.28

If the projections of any two squares on side CD do not intersect, then as it has been proven above, a þ b þ c 2. Otherwise, there exists a straight line parallel to AD crossing all three squares. If none of the squares with sides b and c crosses segment PQ (see Figure 2.26), then as we have already proven, a þ b þ c 2. Thus, possible positions of the squares are presented in ﬁg 2.27. From the proof of problem 2.1.6a, it follows that instead of Figure 2.27 one can consider the variants presented in Figure 2.28. In the case of Figure 2.28a, we have that a þ b þ c a þ b þ d 2 and, in the case of Figure 2.28b, we have that a þ b þ c a þ x þ y þ c ¼ (a þ x) þ (c þ y) 1 þ 1 ¼ 2.

2.1.18. At ﬁrst, note that if point B is on side AD of triangle ACD, then rABC rADC (by rXYZ we denote the radius of the incircle of triangle XYZ). Indeed, let O and O1 be the centers of the incircles of triangles ABC and ADC, respectively. Since, these points belong to bisector AA1 of triangle ADC, then ∠ ¼ 1 ∠ 1 ∠ ¼ ∠ 2 ACO 2 ACB 2 ACD ACO1. Therefore, AO AO1. Let E, E1 AC and OE ⊥ AC, O E ⊥ AC. Then, ΔAOE ΔAO E , thus rABC ¼ AO 1. 1 1 1 1 rADC AO1 2.1 Convex Polygon Lying Inside of Another Polygon 57

Let triangle ABC is covered by a unit square 1, and vertex B does not belong to any of the sides of the square. Then, if ray AB intersects one of the sides of the square at point D, we have that rABC rADC. Frompffiffi the aforesaid, it follows that, it is sufficient to prove the inequality 5À1 r 4 , for triangle ABC, such that all of its vertices are on the sides of the square. Let the vertices of the triangle are on the sides of square MNPQ and MN ¼ 1. Consider the following cases: (a) A, B 2 MN, C 2 PN, with B being on segment AN. pffiffi pffiffi pffiffi ¼ ¼ À 2 < 5À1 < 5À1 Since r rABC rACN rAPN rMPN 1 2 4 , then r 4 .

(b) A, B 2 MN, C 2 PQ with B being on segment AN. Let C0 be the midpoint of the segment PQ, then pffiffiffi À ¼ ¼ 1 1 ¼ ¼ 5 1 : r rABC rACN rMCN rMCN0 MC þ CN þ 1 MC0 þ C0N þ 1 4

(c) A 2 MQ, B 2 PQ, C 2 PN, and AM CN. Without loss of generality, one can assume that, the point A and M coincide. Indeed, it is sufficient to consider a unit square M1N1P1Q1, where M1 A, 2 k 2 N1 PN and M1N1 MN, Q M1Q1. pffiffi Let us prove that, if A M, B 2 PQ, C 2 PN, then r 5À1. 4 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi Denote BQ ¼ x, CN ¼ y, then we have to prove that 1 þ x2 þ 1 þ y2þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁpffiffiffi ðÞ1 À x 2 þ ðÞ1 À y 2 1 þ 5 ðÞ1 À xy , where x, y 2 [0; 1]. pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Firstly, we prove that 1 þ x2 Á 1 þ y2 þ ðÞ1 À x 2 þ ðÞ1 À y 2 pffiffiffi 5ðÞ1 À xy . pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi If xy 1, then 1 þ x2 Á 1 þ y2 þ ðÞ1 À x 2 þ ðÞ1 À y 2 1 þ xy 3 > pffiffi pffiffiffi 2 2 5 ðÞÀ 2 5 1 xy . < 1 If xy 2, then pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ x2 Á 1 þ y2 þ ðÞ1 À x 2 þ ðÞ1 À y 2 ¼ ðÞ1 À xy 2 þ ðÞx þ y 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ ðÞ1 À 2xy 2 þ ðÞ1 À x À y 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞ1 À xy þ 1 À 2xy 2 þ 12 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ðÞ1 À xy 2 þ 21ðÞÀ xy 1 À 2xy þ 1 À 2xy þ 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞ1 À xy 2 þ 21ðÞÀ xy ðÞþ1 À 2xy 1 À 2xy þ 1 pffiffiffi ¼ 5ðÞ1 À xy

(see problem 1.2.5a). 58 2 Application of Projection Method

Note that pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 þ þ 2 þ ðÞÀ 2 þ ðÞÀ 2 1 x 1 y 1 x 1 y pffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi À 1 þ x2 1 þ y2 þ ðÞ1 À x 2 þ ðÞ1 À y 2 ÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼À 1 þ x2 À 1 1 þ y2 À 1 þ 1 x2 y2 ¼ 1 À pffiffiffiffiffiffiffiffiffiffiffiffiffi Á pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À x Á y, 1 þ x2 þ 1 1 þ y2 þ 1 therefore pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 þ þ 2 þ ðÞÀ 2 þ ðÞÀ 2 þ 2 þ 2 1 qxffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 y 1 x 1 y 1 x 1 y þ ðÞ1 À 2 þ ðÞ1 À 2 þ 1 À ÀÁpffiffiffix y xy 1 þ 5 ðÞ1 À xy :

Problems for Self-Study

2.1.19. Let a triangle with sides a1, b1, c1 be inside of the triangle with sides a, b, c. 2 þ 2 þ 2 2 þ 2 þ 2 Is it true that a1 b1 c1 a b c ?

2.1.20. Given points A1, B1, and C1 on sides BC, CA, AB of triangle ABC, respectively, such that ∠AB1C1 þ ∠BC1A1 þ ∠CA1B1 ¼ 180 . Prove that triangle A1B1C1 can be placed inside of one of the triangles AB1C1, BC1A1, CA1B1.

2.1.21. Given a point O inside of triangle ABC and points A1, B1, C1 on sides BC, CA, AB, respectively, such that OB1 ⊥ AC, OA1 ⊥ BC, OC1 ⊥ AB. Prove that triangle A1B1C1 can be placed inside of one of triangles AB1C1, BC1A1, CA1B1. 2.1.22. Let a convex n-gon be inside of a unit square. Prove that one can ﬁnd three vertices A, B, C of this n-gon, so that the area of triangle ABC does not exceed 1 ¼ 1 ¼ 1 ¼ 8 (a) 2, for n 3, 4, (b) 4, for n 5, (c) 8, for n 6, (d)n2, for n 7. 2.1.23. Let a convex quadrilateral with the sum of the lengths of the diagonals equal to d0 be inside of a convex quadrilateral with the sum of the lengths of the diagonals equal to d. Prove that d0 < 2d.

2.1.24. Let a convex polygon with area S2 and perimeter P2 be inside of the convex polygon with area S and perimeter P . Prove that 2S1 > S2 . 1 1 P1 P2 2.1.25. Let a convex quadrilateral with the sum of the lengths of the pairwise distances of its vertices (i.e., the sum of all its sides and diagonals) equal to S2 be < 4S1 inside of a convex quadrilateral with the same sum equal to S1. Prove that S2 3 . 2.2 Sufﬁcient Conditions for Comparison of Lengths of Two Broken Lines on the Plane 59

2.1.26. Is it possibleqﬃﬃ to place inside of a unit square two regular triangles with sides 2 greater than 3, so that they do not intersect? 2.1.27. Prove that a regular n-gon with a side a can be placed inside of a regular n þ 1-gon with a side a.

2.1.28. Let a parallelogram with altitudes h1 and h2 be inscribed in a convex quadrilateral ABCD. Prove that quadrilateral ABCD can be placed inside of a rectangle with one of the sides equal to h1 þ h2. 2.1.29. (a) Prove that, if a convex polygon lays inside of a circle, then the perimeter of the polygon is less than the circumference of the circle. (b) Prove that, if a circle lays inside of a convex polygon, then the circumference of the circle is less than the perimeter of the polygon. Hint Inscribe regular polygons in the circle. 2.1.30. Let O be a given point in triangle ABC with the lengths of sides AB ¼ c, BC ¼ a, CA ¼ b, a b c. Prove that there exists a vertex of the triangle, such that the distance from point O does not exceed pbﬃﬃ. 2 Hint Consider cases when the point is either in triangle ADC or in ABD, where AD ⊥ BC(D 2 BC).

2.2 Sufﬁcient Conditions for Comparison of Lengths of Two Broken Lines on the Plane

2.2.1. (a) Given on a plane two sets of segments A1B1,...,AnBn and C1D1,..., CmDm, the sum of lengths of projections of the segments of the ﬁrst set on any straight line of this plane is greater than the sum of the lengths of projections of the segments of the second set on the same straight line. Prove that the sum of the lengths of the segments of the ﬁrst set is greater than the sum of the lengths of the segments of the second set.

(b) Given on a plane two sets of segments A1B1,...,AnBn and C1D1,...,CmDm. Let l1,...,ln be a straight line on that plane, such that AiBi ⊥ li, i ¼ 1, . . . , n. Given that for any i (i ¼ 1,...,n) the sum of the lengths of projections of the segments of the first set on any straight line li is greater than the sum of the lengths of projections of the segments of the second set on the same straight line. Prove that the sum of lengths of segments of the first set is greater than the sum of lengths of the segments of the second set. (c) Given on a plane two sets of segments A1B1,...,AnBn and C1D1,...,CmDm. Let l1,...,ln be such straight line on that plane that AiBi ⊥ li, i ¼ 1, . . . , n.Itis known that for any value of i (i ¼ 1, . . . , n) the sum of the lengths of projections of the segments of the first set on any straight line li is greater than the 60 2 Application of Projection Method

sum of the lengths of projections of the segments of the second set on the same straight line. Prove that the sum of the lengths of the segments of the ﬁrst set on any line on the plane is greater than the sum of the lengths of the segments of the second set on the same line. 2.2.2. Prove that for points A, B, C, D, E on a plane the following inequality AB þ CD þ DE þ EC AC þ AD þ AE þ BC þ BD þ BE holds true. 2.2.3. Four points on the straight line are denoted by letters A, B, C, D. Prove that AE þ ED þ |AB À CD| > BE þ CE holds true for any point E outside that line. ~ ~ ~ ~ ~ 2.2.4. Given on a plane vectors a, b, c, d, whose sum is equal to 0. Prove that ~a þ ~b þ ~c þ ~d ~a þ ~d þ ~b þ ~d þ ~c þ ~d .

2.2.5. Given on a plane two arbitrary triangles ABC and A1B1C1. Prove that

AA1 þ AB1 þ AC1 þ BA1 þ BB1 þ BC1 þ CA1 þ CB1 þ CC1 AB þ BC þ AC þ A1B1 þ B1C1 þ C1A1:

2.2.6. Given on a plane several segments, whose sum of lengths is equal to π. Prove that it is possible to choose a straight line, so that the sum of lengths of projections of the segments on this straight line is (a) less than 2, (b) more than 2. 2.2.7. Given on a plane several vectors, whose sum of lengths is equal to π. Prove that it is possible to choose several of these vectors, so that the length of their sum is greater than 1.

Solutions

2.2.1. (a) Given on a plane a segment AB and a straight line l. Consider projection of segment AB on line l and denote by l(AB) the length of that projection. If AB ⊥ l, then l(AB) ¼ 0. Similarly, for polygon ϕ denote by l(ϕ) the length of the projection of polygon ϕ on line l. It is not difﬁcult to prove that, if point C belongs to segment AB, then l(AB) ¼ l (AC) þ l(CB). On the other hand, if polygon A1A2 ...An is convex, then l(A1A2) þ l (A2A3) þ ...þ l(An À 1An) þ l(AnA1) ¼ 2l(A1A2 ... An) and if segments AB and CD are either on the same line or on the parallel lines, then CD Á l(AB) ¼ AB Á l(CD). (b) Let us ﬁrst prove the following lemma.

Lemma If on a plane are given such segments M1N1, M2N2,...,MpNp ( p 2), that no two of them are on the same or parallel lines, then there exists a convex and M N M N M1N1 M2N2 ::: p p M1N1 M2N2 ::: p p centrally symmetric polygon with sides 2 , 2 , , 2 , 2 , 2 , , 2 , MiNi so that the sides with length 2 are parallel to segments MiNi, for any 1 i p. We proceed the proof by mathematical induction. 2.2 Sufﬁcient Conditions for Comparison of Lengths of Two Broken Lines on the Plane 61

Figure 2.29 M2 N2

Figure 2.30 dd’

fk Nk+1

fk¢ A¢ О

A fk¢¢

Mk+1

Figure 2.31

For p ¼ 2 the proof is presented in Figure 2.29 Assume that the statement holds true for p ¼ k, prove that it holds true for p ¼ k þ 1. Given segments M1N1, M2N2,...,MkNk, Mk þ 1Nk þ 1. Consider a convex cen- ϕ M1N1 ::: MkNk M1N1 trally symmetric polygon k, the sides of which are equal to 2 , , 2 , 2 , ... MkNk , 2 and are parallel to segments M1N1, M2N2,...,MkNk. Consider a line d||Mk þ 1Nk þ 1 which has a common point with a polygon ϕk and one of the half-planes with a boundary d contains the ϕk polygon (Figure 2.30). It is clear that line d contains just one vertex of polygon ϕk, let us denote it by point A. Let lines d and d0 are symmetric with respect to point O (O is the center of symmetry 0 of polygon ϕk)andA is the symmetric point of point A with respect to point O.Take 0 0 on lines d and d points B and C, such that AB ¼ A C ¼ Mk þ 1Nk þ 1/2 (Figure 2.31). ϕ ϕ0 0 ϕ00 Note that polygon k þ 1, which is the sum of ﬁgures k, ABCA , and k, satisﬁes to the conditions of the lemma. This ends the proof of the lemma. If among segments A1B1,...,AnBn there are segments, for example, AiBi and AjBj, i 6¼ j on the same or parallel lines, then instead of them we consider the 62 2 Application of Projection Method

Figure 2.32 fA

M ¢ Bi

li EF

segment with length AiBi þ AjBj (parallel to segment AiBi). So, one obtains that segments A1B1,...,AnBn satisfy to the conditions of the lemma. Let ϕA be a convex polygon with a center of symmetry at point O, with the sides A1B1 ::: AnBn A1B1 ::: AnBn parallel to segments A1B1,...,AnBn and equal to 2 , , 2 , 2 , , 2 . The polygon (or segment) ϕC, for segments C1D1,...,CmDm with the same symmetry center, is deﬁned similarly. Let us prove that polygon ϕC is inside of polygon ϕA. Otherwise, there exists such a point M 2 ϕC, as it is shown in Figure 2.32. Let M0 be the symmetric point to point M with respect to point O. We have that

ðÞþ...þ ðÞ¼ðÞϕ 0 ðÞ¼ϕ ðÞþ... ðÞ: li A1B1 li AnBn 2li A 2MM 2li C li C1D1 li CmDm

Thus li(A1B1) þ ...þ li(AnBn) li(C1D1) þ ...þ li(CmDm). This leads to a contradiction. Thus, ϕC is inside of ϕA. Then, according to problem 2.1.1. A1B1 þ ...þ AnBn > C1D1 þ ...þ CmDm. Remark

1. Similarly, one can prove that if among segments A1B1,...,AnBn there are two located on non-parallel (crossing) lines and li(A1B1) þ ...þ li(AnBn) li(C1D1) þ ...þ li(CmDm), i ¼ 1, . . . , n, then A1B1 þ ...þ AnBn C1D1 þ ...þ CmDm. 2. If for any line l it holds true l(A1B1) þ ...þ l(AnBn) l(C1D1) þ ...þ l (CmDm), then A1B1 þ ...þ AnBn C1D1 þ ...þ CmDm.

(c) Since ϕC is inside of ϕA (see the solution of problem 2.2.1b), then l(ϕC) < l(ϕA), so 2l(ϕC) < 2l(ϕA). Therefore, l(A1B1) þ ...þ l(AnBn) > l(C1D1) þ ...þ l (CmDm).

Remark One can prove that, if there are given segments A1B1,...,AnBn on a plane and lines li ⊥ AiBi, i ¼ 1, . . . , n, then for any line l of that plane and for arbitrary mi > 0, i ¼ 1, . . . , n, holds true the following inequality 2.2 Sufﬁcient Conditions for Comparison of Lengths of Two Broken Lines on the Plane 63

m1lAðÞþ1B1 ...þ mnlAðÞnBn minðÞm1l1ðÞþA1B1 ...þ mnliðÞAnBn :

2.2.2. According to remark 2 (see the solution of problem 2.2.1b) it is enough to prove the inequality for the projections of points A, B, C, D, E on a line, i.e., it is enough to solve the problem for the case when points A, B, C, D, E are on the same line. As the inequality is symmetric with respect to points C, D, E, then we can take that E lays between points C and D, then DE þ EC ¼ CD. According to the triangle inequality AC þ AD CD, BC þ BD CD, AE þ BE AB. Summing up these three inequalities, we deduce that

AC þ AD þ AE þ BC þ BD þ BE AB þ CD þ DE þ EC:

2.2.3. Let AB > CD (in the case of AB < CD the proof is similar). One has to prove that AE þ ED þ AB > CD þ BE þ CE: ð2:3Þ

According to problem 2.2.1b, it is enough to prove inequality (2.3) for the projections of points A, B, C, D, E on a line li, where li is perpendicular to one of the segments AE, ED, AB (see problem 2.2.1b; must be used as follows: if li(A1B1) þ ...þ li(AnBn) li(C1D1) þ ...þ li(CmDm), i ¼ 1, . . . , n, and there exists a number j, such that lj(A1B1) þ ...þ lj(AnBn) > lj(C1D1) þ ...þ lj(CmDm), then A1B1 þ ...þ AnBn > C1D1 þ ...þ CmDm). If l ⊥ AE or AB, then we have an equality. While, if l ⊥ ED, then we have a strict inequality. In the case of AB ¼ CD, then we have to prove that AE þ ED > BE þ CE. Indeed, parallelogram EBNC (see Figure 2.33) is inside of parallelogram AEDN. Hence, 2BE þ 2CE < 2AE þ 2ED (see the solution of problem 2.1.1). Remark In the case of AB ¼ CD, for the set of segments AE, ED, AB and the set of segments CD, BE, CE, we have that l(AE) þ l(ED) þ l(AB) ¼ l(CD) þ l(BE) þ l (CE), at l ⊥ AE or l ⊥ ED or l ⊥ AB. On the other hand, AE þ ED þ AB > CD þ BE þ CE.

Figure 2.33 E

A BC D

N 64 2 Application of Projection Method

! ! ! 2.2.4. Consider points A, B, C, D, such that AB ¼ ~a, BC ¼ ~b, CD ¼ ~c, then ! DA ¼ ~d. We have to prove that

AB þ BC þ CD þ DA AC þ BD þ 2MN, ð2:4Þ where M and N are the midpoints of segments BD and AC. According to remark 2 (see the solution of problem 2.2.1b) it is sufﬁcient to prove inequality (2.4) for projections of points A, B, C, D on a line, i.e., it is enough to solve the problem for the case, when points A, B, C, D are on the same line. We can assume that A(0), B(b), C(c), D(d), and b, c, d 0, d b, then we must prove that |c À b| þ |c À d| c À 2b þ |d þ b À c|. If d þ b c, then |c À b| þ |c À d| |(c À b) À (c À d)| ¼ d À b ¼ c À 2b þ | d þ b À c|. If d þ b < c, then |c À b| þ |c À d| ¼ c À b þ c À d c À 2b þ |d þ b À c|. 2.2.5. According to remark 2 (see the solution of problem 2.2.1b) it is sufﬁcient to prove the inequality for the projections of points A, B, C, A1, B1, C1 on a line, i.e., it is enough to prove the problem for the case, when points A, B, C, A1, B1, C1 are on the same line. As the inequality is symmetric with respect to points A, B, C (A1, B1, C1), without loss of generality one can assume that point B lays between points A and C and point B1 lays between points A1 and C1, A1C1 AC. Then AB þ BC þ AC þ A1B1 þ B1C1 þ A1C1 ¼ 2AC þ 2A1C1. Since AC AB1 þ B1C, A1C1 A1B þ BC1, AC AA1 þ CA1, A1C1 AC AC1 þ CC1, then by summing up these inequalities we obtain that

2AC þ 2A1C1 AA1 þ AB1 þ AC1 þ BA1 þ BC1 þ CA1 þ CB1 þ CC1:

Therefore

AB þ BC þ AC þ A1B1 þ B1C1 þ A1C1 AA1 þ AB1 þ AC1 þ BA1 þ BB1 þ BC1 þ CA1 þ CB1 þ CC1:

2.2.6. Given segments A1B1,...,AnBn. We can assume that any two segments do not belong to the same or parallel straight lines. For n ¼ 1 the statement of the problem is evident, let n 2. Construct a polygon ϕA (see the solution of problem 2.2.1b).

(a) Let O be the center of symmetry of polygon ϕA, and d be the minimal of all distances between opposite sides (Figure 2.34).

Figure 2.34 l fA

O d 2.2 Sufﬁcient Conditions for Comparison of Lengths of Two Broken Lines on the Plane 65

Figure 2.35 fA

d1 O

d ϕ Since the circumference with a center O and radius 2 is inside of polygon A, the length of that circumference is less than the perimeter of polygon ϕA, i.e., πd < π (see problem 2.1.29b). Then the sum of the lengths of the projections of the segments A1B1,...,AnBn on line l is equal to 2d and 2d < 2 (Figure 2.34).

(b) Consider all diagonals of the ϕA polygon which pass through point O (Figure 2.35) and let d1 be the length of the largest of them.

ϕ d1 Since polygon A is inside of the circumference with a center O and radius 2 , the perimeter of polygon ϕA is less than the length of that circumference, i.e., πd1 > π (see problem 2.1.29a). Then the sum of the lengths of the projections of the segments A1B1,...,AnBn on line l1 (Figure 2.35) is equal to 2d1. Thus, it follows that the sum is larger than 2. ! ! ! ! ::: þ ::: þ ¼ π 2.2.7. Let be given vectors A1B1, , AnBn, such that A1B1 AnBn . Consider segments A1B1,...,AnBn, then according to problem 2.2.6b there exists a straight line l, such that l(A B ) þ ...þ l(A B ) > 2 (see notations in the 1 1 n n ! solution of problem 2.2.1). Let ~e be a unit vector on line l and A B ,~e 90, ! i i ¼ ~ > ¼ þ þ þ i 1,...,k and AiBi, e 90 , i k 1, . . . , n. Then, either l(A1B1) ... l > þ þ > þ þ > (AkBk) 1orl(Ak þ 1Bk þ 1) ... l(A nBn) 1. Let l(A1B1) ... l(AkBk) 1. ! ! We have that A1B þ ::: þ AkB lAðÞþ1B1 ::: þ lAðÞkBk > 1. Therefore, ! ! 1 k þ ::: þ > A1B1 AkBk 1.

Problems for Self-Study

2.2.8. Points C1, A1, B1 are taken correspondingly on sides AB, BC, AC of triangle ¼ λ ¼ λ ¼ λ 1 < λ < ABC, such that BA1 BC, CB1 CA, AC1 AB, where 2 1. Prove that for perimeter P of triangle ABC and for perimeter P1 of triangle A1B1C1 it holds true (2λ À 1)P < P1 < λP. 2.2.9. Point O is inside of triangle ABC with a perimeter P. Prove that P < þ þ < 2 AO BO CO P. 66 2 Application of Projection Method

2.2.10. Prove that, if the lengths of all sides and diagonals of a convex polygon is less than d, then its perimeter is less than πd. 2.2.11. Given several convex polygons and it is known that it is impossible to draw a straight line, so that it does not intersect any polygon and on both sides of it there is at least one polygon. Prove that all these polygons can be conﬁned within a polygon with the perimeter not exceeding the sum of their perimeters. 2.2.12. Given on a plane several vectors, whose sum of the length is equal to 1.

Prove that they can be broken into three groups (possiblyp empty),ﬃﬃ so that the sum 3 3 of the lengths of the vectors in these groups is more than 2π . Remark Add to these n vectors another 2n vectors obtained from the vectors of the given set by rotation by angle 120 clockwise or counterclockwise.

2.2.13. Given on a plane a convex n-gon. Let ak be the length of its k-th side and dk be the length of the projection of n-gon on a line containing that side (k ¼ 1, 2, . . . , n). Prove that a1 þ ::: þ an 4. d1 dn Remark First prove the statement of the problem for centrally symmetric polygons.

2.3 Inscribed Polygons with the Least Perimeter

2.3.1.ÀÁGiven convex polygon A1A2 ...An and points B1, B2,...,Bn on its sides B1 2A1A2; B2 2A2A3; :::; Bn 2AnA1; Bi Aj , so that for any points C1,...,Cn on the side of polygon A1 ...An (C1 2 A1A2,...,Cn 2 AnA1, Ci Aj) the inequality B1B2 þ B2B3 þ ...þ Bn À 1Bn þ BnB1 C1C2 þ C2C3 þ ...þ Cn À 1Cn þ CnC1 is true. Prove that

∠BnB1A1 ¼ ∠B2B1A2, ∠B1B2A2 ¼ ∠B3B2A3, :::, ∠BnÀ1BnAn ¼ ∠B1BnA1: ð2:5Þ

2.3.2.ÀÁGiven convex polygon A1A2 ...An and points B1, B2,...,Bn on its sides B1 2A1A2; B2 2A2A3; :::; Bn 2AnA1; Bi Aj , so that condition (2.5) of problem 2.3.1 is satisﬁed. Prove that for any points C1,...,Cn on the sides of polygon A1A2 ...An (C1 2 A1A2, ..., Cn 2 AnA1) the following inequality is true: B1B2 þ B2B3 þ ...þ Bn À 1Bn þ BnB1 C1C2 þ C2C3 þ ...þ Cn À 1Cn þ CnC1.

2.3.3. Prove that if A1A2A3 is not an acute-angled triangle, then on its sides do not exist such points B1, B2, B3 (B1 2 A1A2, B2 2 A2A3, B3 2 A3A1, Bi Aj) for which the condition (2.5) of problem 2.3.1 is satisﬁed.

2.3.4. Prove that if A1A2A3 is an acute-angled triangle, thenÁ on its sides exist such points B1, B2, B3 (B1 2 A1A2, B2 2 A2A3, B3 2A3A1, Bi Aj for which the condition (2.5) of problem 2.3.1 is satisﬁed. 2.3 Inscribed Polygons with the Least Perimeter 67

2.3.5. Given convex quadrilateral A1A2A3A4 and points B1, B2, B3, B4 on the sides of that quadrilateral (B1 2 A1A2, B2 2 A2A3, B3 2 A3A4, B4 2 A4A1, Bi Aj) so that the condition (2.5) of problem 2.3.1 is satisﬁed. Prove that quadrilateral A1A2A3A4 is inscribed and that maxð∠A A A ; ∠A A A ; ∠A A A ; ∠A A A ; ∠A A A ; ∠A A A ; ∠A A A ; ∠A A A Þ π 1 2 4 3 2 4 2 3 1 4 3 1 3 4 2 1 4 2 4 1 3 2 1 3 Â < : 2 ð2:6Þ

2.3.6. Prove that if a convex quadrilateral A1A2A3A4 is inscribed and for it condition (2.6) of problem 2.3.5 is satisﬁed, then on the sides of that quadrilateral exist such points B1, B2, B3, B4 (B1 2 A1A2, B2 2 A2A3, B3 2 A3A4, B4 2 A4A1, Bi Aj), for which the condition (2.5) of problem 2.3.1 is satisﬁed.

2.3.7. Given a regular tetrahedron A1A2A3A4 with edge 1 and points B1, B2, B3, B4 on its faces A2A3A4, A1A3A4, A1A2A4p,ffiffiffiffiffi and A1A2A3, respectively. Prove that B1B2 þ B2B3 þ B3B4 þ B4 B1 0, 4 10. 0 0 0 0 2.3.8. Given a cub ABCDA B C D with edge 1 and points B1, B2, B3, B4, B5, B6 on its 0 0 0 0 0 0 0 0 0 0 0 0 faces ABCD, AA B B, BB C C, A B C D , DD C C, pAAffiffiffi D D, respectively. Prove that B1B2 þ B2B3 þ B3B4 þ B4B5 þ B5B6 þ B6B1 2 3.

Solutions

2.3.1. Assume that condition (2.5) is not correct and let

∠BnB1A1 6¼ ∠B2B1A1: ð2:7Þ

0 Consider point B2 symmetric to B2 with respect to straight line A1A2 (Figure 2.36). 0 Denote the intersection point of straight lines B2Bn and A1A2 by B. Since ∠ ¼ ∠ 0 ¼ ∠ B2BM B2BM BnBA1, then according to (2.7) points B and B1 differ. 0 0 Take a point B1 on segment BB1 so that point B1 would lay on side A1A2. Let us 0 ::: consider points B1, B2, , Bn. According to problem 1.1.8a,

Figure 2.36 An

Bn A3 B2

MA2 BB1 A1

B'2 68 2 Application of Projection Method

Figure 2.37 ln Bn-1 An

A3 bn l1

B2 b1 C1 A1

C1'

b1 B1

C1'' l2

0 0 þ 0 < 0 þ 0 þ 0 < þ B2B1 B1Bn B2B1 B1Bn, thus B2B1 B1Bn B2B1 B1Bn, which means that

0 þ þ ::: þ þ 0 < þ þ þ : B1B2 B2B3 BnÀ1Bn BnB1 B1B2 B2B3 BnÀ1Bn BnB1

This leads to a contradiction.

2.3.2. Let ∠BnB1A1 ¼ ∠B2B1A2 ¼ β1, ∠B1B2A2 ¼ ∠B3B2A3 ¼ β2,...,∠Bn À 1BnAn ¼ ∠B1BnA1 ¼ βn. Draw through vertices A1, A2,...,An straight lines l1, l2,...,ln parallel to lines B1Bn,..,Bn À 1Bn, respectively (Figure 2.37). 0 00 0 00 Let C1 and C1 be projections of point C1 on lines l1 and l2, C2 and C 2 - 0 00 projections of point C2 on lines l2 and l3, etc., Cn and Cn - projections of point Cn on lines ln and l1. We have that

þ þ ::: þ þ 00 00 þ 00 0 þ ::: þ 00 0 þ 00 0 ¼ C1C2 C2C3 CnÀ1Cn CnC1 C1C2 C2C2 CnÀ1Cn CnC1 ¼ ðÞβ þ β þ::: þ ðÞβ þ β ¼ A2C1 cos 1 A2C2 cos 2 C1A1 cos 1 CnA1 cos n ¼ β þ β þ ::: þ β ¼ A1A2 cos 1 A2A3 cos 2 AnA1 cos n ¼ ðÞβ þ β þ::: þ ðÞβ þ β ¼ A2B1 cos 1 B1A1 cos 1 A1Bn cos n AnBn cos n

¼ B1B2 þ B2B3 þ ::: þ BnÀ1Bn þ BnB1: hence

C1C2 þ C2C3 þ ::: þ CnÀ1Cn þ CnC1 B1B2 þ B2B3 þ ::: þ BnÀ1Bn þ BnB1: 2.3 Inscribed Polygons with the Least Perimeter 69

Remark 1. The equality holds true if and only if

C1C2 k B1B2, C2C3 k B2B3, :::, CnÀ1Cn k BnÀ1Bn, CnC1 k BnB1:

2. For odd n the equality holds if C1 B1, C2 B2,...,Cn Bn. 2.3.3. Let ∠A1 90 and for points B1, B2, B3 condition (1) of problem 2.3.1 is satisﬁed (Figure 2.38). We have that

∠A2 þ ∠A3 ¼ 360 À ðÞ∠B2B3A3 þ ∠B2B1A2 ÀðÞ∠B3B2A3 þ ∠B1B2A2 ¼ ¼ 180 À ðÞ∠B1B3A1 þ ∠B3B1A1 þ180 À ðÞ∠B3B2A3 þ ∠B1B2A2 ¼ ∠A1 þ ∠B1B2B3 > ∠A1:

Hence ∠A2 þ ∠A3 > ∠A1, which is wrong.

2.3.4. Let A3B1, A1B2, A2B3 be the altitudes of acute triangle A1A2A3 (Figure 2.39).

Figure 2.38 A1

B3 B1

A3 B2 A2

Figure 2.39 A1

B1 B3

A3 B2 A2 70 2 Application of Projection Method

Figure 2.40 A2

B2 B1 O P A3 A1

Since points B1 and B3 lay on the circumference with a diameter A1H, we have ∠B3B1A1 ¼ ∠B3HA1. Similarly we get that ∠B2B1A2 ¼ ∠B2HA2 and since ∠B3HA1 ¼ ∠B2HA2, then ∠B3B1A1 ¼ ∠B2B1A2. The other two equalities are proved similarly. Remark According to problems 2.3.1 and 2.3.3, a triangle with a minimum perimeter can be inscribed in the given triangle A1A2A3 if it is an acute triangle (necessary condition). On the other hand, according to problems 2.3.2 and 2.3.4 this is also a sufﬁcient condition. 2.3.5. We have that (Figure 2.40)

∠A2 þ ∠A4 ¼ π À ∠B2B1A2 À ∠B1B2A2 þ π À ∠B4B3A4 À ∠B3B4A4 ¼ π À ∠B4B1A1 À ∠B1B4A1 þ π À ∠B3B2A3 À ∠B2B3A3 ¼ ∠A1 þ ∠A3,

thus quadrilateral A1A2A3A4 is inscribed. π ∠ π Let us assume, that one of these angles is not less than 2, let A1A3A4 2. Let ρ(M, l) be the distance from point M to straight line l. Note that ρ(A3, B1B2) ¼ ρ(A3, B2B3) ¼ ρ(A3, B3B4), similarly ρ(A1, B1B2) ¼ ρ(A1, B3B4). Hence line A1A3 is a locus of points equidistant from straight lines B1B2 and B3B4 (in the region of Π). Thus the bisectors of angles ∠B1B2B3 and ∠B4B3B2 intersect on line A1A3, let that point be O. Note that points O and A3 are on the different sides of line B2B3, ∠ ¼ π which means that point O belongs to ray A3A1 and since OB3A3 2, then ∠ < π A1A3A4 2, which is wrong.

2.3.6. Let the diagonals of inscribed quadrilateral A1A2A3A4 intersect at point O and points B1, B2, B3, B4 be the orthogonal projections of point O on the sides of quadrilateral A1A2A3A4 (Figure 2.41). We will prove that points B1, B2, B3, B4 satisfy condition (1) of problem 2.3.1. 2.3 Inscribed Polygons with the Least Perimeter 71

Figure 2.41 A2 B2 A3 B1

А1

Since points B1 and B4 are on the circle with diameter A1O, then ∠ ¼ ∠ ∠ ¼ π À ∠ ¼ π À ∠ ¼ B4B1A1 B4OA1. We have B4OA1 2 B4A1O 2 B2A2O ∠B2OA2. Since points B1 and B2 are on the circle with diameter A2O, then ∠B2OA2 ¼ ∠B2B1A2. Thus ∠B4B1A1 ¼ ∠B2B1A2. The other three inequalities one can prove similarly. Remark 1. According to problems 2.3.1 and 2.3.5, a quadrilateral with a minimum perimeter can be inscribed into the given quadrilateral A1A2A3A4 if the latter is inscribed and condition (2.6) of problem 2.3.5 is satisfied (necessary condition), and according to problems 2.3.2 and 2.3.6 these are also sufficient conditions. 2. If a quadrilateral with a minimum perimeter can be inscribed into the given quadrilateral A1A2A3A4 then the number of these would be infinite. According to remark 1 to problem 2.3.2, corresponding sides of all minimum possible perimeter quadrilaterals inscribed into A1A2A3A4 quadrilateral will be parallel to each other (Figure 2.42).

2.3.7. Let M be the midpoint of edge A1A3. Denote points symmetric to B1, B2, B3, 0 0 0 0 B4 with respect to plane A2MA4 by B1, B2, B3, B4, respectively, and the midpoints of 0 0 0 0 the segments B3B1, B2B2, B3B1, B4B4 by C1, C2, C3, C4,respectively. Note that points C1 and C3 are on facets A2A3A4 and A1A2A4 respectively, and points C2 and C4 are on segments MA4 and MA2, respectively According to problem 1.1.9a (see the solution) we have that þ þ þ þ ¼ B1B2 B2B3 þ B2B3 B1B2 þ B3B4 B4B1 þ B1B4 B3B4 2 2 2 2 ¼ B1B2 þ B2B3 þ B3B4 þ B4B1, thus C1C2 þ C2C3 þ C3C4 þ C4C1 B1B2 þ B2B3 þ B3B4 þ B4B1. 72 2 Application of Projection Method

Figure 2.42 A2

B2 A1

A3 B4

Let N be the midpoint of edge A2A4. Denote points symmetric to C1, C2, C3, C4 with respect to plane A1NA3 by 0 0 0 0 0 0 0 C1, C2, C3, C4, respectively, and the midpoints of segments C1C1, C2C4, C3C3, C4 0 C2 by D1, D2, D3, D4,respectively. Note that points D1, D2, D3, D4 are on segments NA3, MA4, NA1, MA2, respectively. According to problem 1.1.9a we get

D1D2 þ D2D3 þ D3D4 þ D4D1 þ 0 0 þ 0 0 þ 0 0 þ 0 0 C1C2 C4C1 þ C2C3 C3C4 þ C3C4 C2C3 þ C1C4 C1C2 2 2 2 2 ¼ C1C2 þ C2C3 þ C3C4 þ C4C1:

Thus D1D2 þ D2D3 þ D3D4 þ D4D1 C1C2 þ C2C3 þ C3C4 þ C4C1, which means that D1D2 þ D2D3 þ D3D4 þ D4D1 C1C2 þ C2C3 þ C3C4 þ C4C1 B1B2 þ B2B3 þ B3B4 þ B4B1, consequently,

B1B2 þ B2B3 þ B3B4 þ B4B1 D1D2 þ D2D3 þ D3D4 þ D4D1 ρðÞþNA3; MA4 ρðÞþMA4; NA1 ρðÞþNA1; MA2 ρðÞMA2; NA3 , where ρ(l1, l2) is the distance between straight lines l1 and l2. We shall prove that one can choose on segments MA4 and NA3 points K and E, respectively, so that KE ⊥ MA , KE ⊥ NA , then ρ(NA , MA ) ¼ KE (Figure 2.43). ! ! 4 ! 3 3 4 ~ A3E A4K Denote A3A ¼ ~a, A3A ¼ b, A3A2 ¼ ~c, ¼ λ, ¼ μ. 1 4 A3N A4M We have that ! ! ! ! EK ¼ EA 3 þ A3M þ MK ¼ ! ~a ! λ ~a ~a ¼ÀλA N þ þ ðÞ1 À μ MA4 ¼¼ À ~b þ~c þ þ ðÞ1 À μ ~b À ¼ 3 2 2 2 2 1 ¼ μ~a þÀðÞλ þ 2 À 2μ ~b À λ~c : 2 2.3 Inscribed Polygons with the Least Perimeter 73

Figure 2.43 A4

A 2 K b

A1 M c a

! ! Since jj~a ¼ ~b ¼ jj~c ¼ 1,~a~b ¼ ~b~c ¼ ~c~a ¼ 1 and EK ⊥ ~b þ~c , EK ⊥ 2 2~b À~a , then μ~a þ ðÞ2 À λ À 2μ ~b À λ~c ~b þ~c ¼ 0 and ðμ~a þ ðÞ2 À λ À 2μ ~b À λ~cÞ 2~b À~a ¼ 0, thus À3λ À 2μ þ 3 ¼ 0 and À2λ À 3μ þ 3 ¼ 0. Hence, we λ ¼ μ ¼ 3 obtain that 5. rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi ! ~ 2 Then ρðÞ¼NA3; MA4 EK ¼ EK ¼ 0, 1 3~a þ b À 3~c ¼ 0, 1 10. Simi- pffiffiffiffiffi ρðÞ¼; ρðÞ¼; ρðÞ¼; larly we get that MA4 NA1 pffiffiffiffiffiNA1 MA2 MA2 NA3 0, 1 10, thus B1B2 þ B2B3 þ B3B4 þ B4B1 0, 4 10. Another solution of this problem can be obtained by using a problem 4.1.24b.

Remark If points B1, B2, B3, B4 are on segments A3N, A4M, A1N, A2M, respectively,pffiffiffiffiffi A3B1 A4B2 A1B3 A2B4 3 and ¼ ¼ ¼ ¼ , then B1B2 þ B2B3 þ B3B4 þ B4B1 ¼ 0, 4 10. A3N A4M A1N A2M 5 2.3.8. Consider Cartesian coordinate system in the space and let B(0; 0; 0), A 0 (1; 0; 0), C(0; 1; 0), B (0; 0; 1), B1(x; y; 0). 0 0 Denote the points symmetric to B1, D1, D2, D3, D4 with respect to planes AA B B, 0 0 0 0 0 0 0 0 0 0 BB C C, A B C D , DD C C, AA D D by D1, D2, D3, D4, and D5, respectively. Then D1(x; Ày; 0), D2(Àx; Ày; 0), D3(Àx; Ày; 2), D4(Àx;2þ y; 2), D5(2 þ x;2þ y; 2). Note that

B1B2 þ B2B3 þ B3B4 þ B4B5 þ B5B6 þ B6B1 ¼

¼ D1B2 þ B2B3 þ B3B4 þ B4B5 þ B5B6 þ B6B1

D1B3 þ B3B4 þ B4B5 þ B5B6 þ B6B1 ¼ D2B3 þ B3B4 þ B4B5 þ B5B6 þ B6B1

D2B4 þ B4B5 þ B5B6 þ B6B1 ¼ D3B4 þ B4B5 þ B5B6 þ B6B1

D3B5 þ B5B6 þ B6B1 ¼ D4B5 þ B5B6 þ B6B1 D4B6 þ B6B1 ¼

¼ D5B6 þ B6B1 D5B1, 74 2 Application of Projection Method pffiffiffi consequently B1 B2 þ B2B3 þ B3B4 þ B4B5 þ B5B6 þ B6B1 D5 B1 ¼ 2 3. (See also problem 4.1.24c.)

Problems for Self-Study

2.3.9. Given an inscribed quadrilateral ABCD. Prove that the perimeter of the quadrilateral inscribed into ABCD is greater than or equal to 2AC sin ∠A. 2.3.10. Given straight lines l, m, and n. Find the triangle with the minimum possible perimeter and the vertices on lines l, m, and n. 2.3.11. Given an inscribed quadrilateral ABCD. Find on lines AB, BC, CD, DA, such points B1, B2, B3, B4, respectively, for which the sum B1B2 þ B2B3 þ B3B4 þ B4B1 would be minimal.

2.4 Method of Projections

2.4.1. Prove that radius of the incircle of the right-angled triangle is less than half of its leg. 2.4.2. Several circles, the sum of lengths of which is equal to 10, are located inside a unit square. Prove that one can ﬁnd a straight line crossing at least four of these circles. 2.4.3. Non-self-crossing broken line of length 1000 is located inside a unit square. Prove that that one can ﬁnd a straight line parallel to one of the sides of a square which would cross at least 500 segments of this broken line.

2.4.4. Given two straight lines a and b. A1B1, A2B2, A3B3 are perpendiculars drawn from three consecutive points A1, A2, A3 on line a to line b. Prove that A2B2 max (A1B1, A3B3). 2.4.5. Diagonals of a convex quadrilateral ABCD intersect at point O.LetP and Q be > ABþCD the centers of circumcircles of triangles AOB and COD. Prove that PQ 4 .

2.4.6. Points A1, B1, and C1 are taken on sides BC, CA, and AB of non-obtuse triangle ABC, respectively. Prove that 2(B1C1 cos α þ C1A1 cos β þ A1B1 cos γ) BC cos α þ AC cos β þ AB cos γ, where ∠A ¼ α, ∠B ¼ β, ∠C ¼ γ. Give an example of an obtuse triangle for which the inequality does not hold. 2.4.7. Let circles of unit radiuses have no common internal points and be inside of a band S. The band is formed by two parallel lines having a distance w. Let us call these circles a k-cloud, if each line that intersectspSffiffiffi,ÂÃ intersects also more than or þ k 2 equal to k circles. Prove that for k-cloud w 2 3 2 , where k N and [a] is the integer part of the number a. 2.4 Method of Projections 75

2.4.8. Let ON be the radius of the circle with a center at point O, which intersects chord AB at point M at a right angle. Let P be an arbitrary point on the larger arc AB, not coinciding with the point diametrically opposite to point N. Straight lines PM and PN deﬁne points Q and R on the circumference and on chord AB, respectively. Prove that RN > MQ. 2.4.9. In convex pentagon ABCDE,sideAB is perpendicular to side CD, and side BC is perpendicular to side DE. Prove that, if AB ¼ AE ¼ ED ¼ 1, then BC þ CD < 1. 2.4.10. (a) Opposite sides of convex hexagon ABCDEF are parallel (AB||DE, BC|| EF, CD||FA). Prove that RA þ RC þ RE P, where RA, RC, RE are the radiuses of the circumcircles of triangles FAB, BCD, and DEF, respectively, and p is the half- perimeter of hexagon ABCDEF.

(b) Let M be an arbitrary point inside triangle ABC, Ra, Rb, Rc the distances of point M from A, B, C, da, db, dc the distances from point M to lines BC, AC, and AB, respectively. Prove that Ra þ Rb þ Rc 2da þ 2db þ 2dc. (c) Let M be a point inside triangle ABC. Prove that one of the angles ∠MAB, ∠MBC, ∠MCA is less than or equal to 30.

2.4.11. Prove that for an acute triangle ha þ hb þ hc 4R þ r. 2.4.12. Given points U and V on sides AB and CD of square ABCD, respectively. Let straight lines DU and AV intersect at point P, and lines CU and BV at point Q. 1 Prove that PQ 2 AB. 2.4.13. Given 110 unit vectors on a plane, the sum of these being a zero vector. Prove that of these 110 vectors one can choose such 55 vectors, that the modulus of the sum of which will not be greater than 1. 2.4.14. Let the diagonals of convex quadrilateral ABCD intersect at point P. Points Q, R, S, and T are the feet of the perpendiculars drawn from point P to lines AB, BC, CD, and DA, respectively. Prove that

1 PQ þ PR þ PS þ PT ðÞAB þ BC þ CD þ DA : 2

2.4.15. Let ABCDEF be a convex hexagon. Given that AB ¼ CD ¼ EF. Prove that AD þ BE þ CF AB þ BC þ CD þ DE þ EF þ FA. For which hexagon does equality hold true? 2.4.16. The vertices of tetrahedron KLMN lay either (inside) on the facets or on the edges of another tetrahedron ABCD. Prove that the sum of the lengths of all edges of 4 tetrahedron KLMN is less than 3 of the sum of the lengths of all edges of tetrahedron ABCD.

2.4.17. Let M be a point inside a convex n-gon A1A2 ...An and p be half-perimeter ∠ ∠ ∠ A1 þ A2 þ ::: þ An of that n-gon. Prove that MA1 cos 2 MA2 cos 2 MAn cos 2 p.

2.4.18. (a) Points M, N, P are chosen on edges A1B1, C1C and AD of unit cube ABCDAp1ffiffiffiB1C1D1, respectively. Prove that the perimeter of triangle MNP is not less than 3 6=2. 76 2 Application of Projection Method

(b) Given a unit cube ABCDA1B1C1D1. Prove that the distance from the arbitrary point in the space to one of straight lines A B , C C, AD is not less than p1ffiffi. 1 1 1 2 (c) Given two spheres with diameters d and D insidepffiffiffi a unit cube so that they do not have common points. Prove that d þ D < 3 À 3.

Solutions

2.4.1. Consider the projection of a circle on the edge. 2.4.2. Let us consider the projection of these circles on side AB of the square. The 10 sum of the lengths of these projections is equal to π > 3. Thus, several segments Δ1, Δ2,...,Δk, with the sum of their lengths greater than 3, are located on the unit segment. We shall prove that there exists a point which belongs to at least four segments. Let different points A1, A2,...,Am be all endpoints of segments Δ1, Δ2,...,Δk (Figure 2.44). Replace segments Δ1, Δ2,...,Δk by segments of type AiAi þ 1, i ¼ 1, 2, . . . , m À 1, segment AiAi þ 1 being taken as many times as it belongs to segments Δ1, Δ2,...,Δk. It is clear that, by this replacement, the sum of the lengths of all segments will not change. Thus we can assume that any two segments either do not have any common internal point or do not coincide. Let M1, M2,...,Mn be the left endpoints of these segments. If each point Mi is a left endpoint of no more than three of those segments, then taking one of each segments with left endpoints M1, M2,...,Mn we obtain that the sum of the lengths of these segments is not greater than 1. Then the sum of the lengths of all segments will be not greater than 3. This means that there exists a point Mi which is a left endpoint of at least four segments. Thus there exists a point M which is an internal point of at least four segments. Then straight line l which passes through point M and is perpendicular to side AB of the square would intersect at least four of these circles. 2.4.3. Consider the projections of the segments of the broken line on the two perpendicular sides of the square. Let the lengths of the projections of the segments on one of the sides be a1, a2,...,an and those on the other side b1, b2,...,bn (Figure 2.45). Note that ai þ bi ci, where ci is the length i-th segment of the broken line. Therefore (a1 þ b1) þ ...þ (an þ bn) 1000. Hence, it follow that a1 þ ...þ an 500 or b1 þ b2 þ ...þ bn 500. To conclude the proof, see the solution of problem 2.4.2. 2.4.4. If straight lines a and b are in the same plane, then it is easy to see that min (A1B1, A3B3) A2B2 max (A1B1, A3B3).

Figure 2.44 AB

A1 A2 Am 2.4 Method of Projections 77

Figure 2.45

bi ci bi

Figure 2.46 A1¢

BA2¢

A3¢

If the straight lines a and b are not in the same plane then we shall consider the projections of points A1, A2, A3, B1, B2, B3 on the plane perpendicular to line b. Let 0 0 0 those points areÀÁA1, A2, A3, and B (Figure 2.46). ∠ 0 0 ; ∠ 0 0 0 0 > Since max BA 2A3 BA 2A1 90 , hence it follows that max(BA 1, BA 3) 0 0 0 0 BA 2 and we have that BA 1 ¼ B1A1, BA 2 ¼ B2A2, BA 3 ¼ B3A3. Therefore A2B2 < max (A1B1, A3B3).

Remark Given segments AB and CD, then inequality SABM max (SABC, SABD) holds true for any point M of segment CD. 2.4.5. Note that the lengths of the projections of segment PQ on lines AC and BD AC BD þ > AC þ BD are equal to 2 and 2 , respectively. Thus PQ PQ 2 2 , consequently, PQ > ACþBD > ABþCD 4 4 (see the remarks of problem 1.1.4a). β π 2.4.6. Let A2C2 be the projection of segment A1C1 on side AC. Since 2 and A1C1 A2C2, therefore we deduce that 2A1C1 cos β 2A2C2 cos β ¼ 2 cos β (AC À AC1 cos α À CA1 cos γ). Similarly we get that 2A1B1 cos γ 2 cos γ(AB À AB1 cos α À BA1 cos β) and 2B1C1 cos α 2 cos α(BC À BC1 cos β À CB1 cos γ). 78 2 Application of Projection Method

Summing up these three inequalities we obtain that

ðÞα þ β þ γ β þ γ þ αÀ 2 B1C1 cos C1A1 cos 1 A1B1 cos 2ACcos 2ABcos 2BCcos

À 2cosαÁ cosβðÞÀAC1 þ C1B 2cosβcosγðÞÀBA1 þ A1C 2cosγ cosαðÞAB1 þ B1C ¼ 2BCcosα þ 2ACcosβ þ 2ABcosγ À cosαðÞABcosβ þ ACcosγ À À cosβðÞABcosα þ BCcosγ ÀcosγðÞACcosα þ BCcosβ ¼ ¼ 2BCcosα þ 2ACcosβ þ 2ABcosγ À BCcosα À ACcosβ À ABcosγ ¼ ¼ BCcosα þ ACcosβ þ ABcosγ:

Hence, it follows that 2(B1C1 cos α þ C1A1 cos β1 þ A1B1 cos γ) BC cos α þ AC cos β þ AB cos γ. For α ¼ γ ¼ π, AC ¼ AB ¼ a, and C B k CB, B A k AB we have that C B ¼ a, 6 p1 ffiffi 3 3 1 1 1 1 1 1 3 ¼ 2 ¼ 7 A1B1 3 a, A1C1 3 a. Thus ðÞα þ β þ γ 2 B1C1pcosffiffiffi pffiffiffiC1A1 cospffiffiffi 1 A1B1 cos 3 3 À 7 3a ¼ a < ¼ BC cos α þ CA cos β þ AB cos γ: 3 2

2.4.7. Draw through the center of a certain circle of the k-cloud a straight line l, which is perpendicular to the boundaries of band S. Then that line l should intersect not less than k À 1 other circles. Line l divides the plane into two half-planes.ÂÃ It is k þ not difficult to understand that one of the half-planes contains not less than 2 1 centers of those circles (includingÂÃ the centers which are on line l). Let O1, O2,..., ¼ k þ On be these centers, where n 2 1 (Figure 2.47). 0 0 ::: 0 Let points O1, O2, , On be the projections of points O1, O2,...,On on line l. 0 Denote by di ¼ OiO , i ¼ 1, . . . , n. Note that for i ¼ 1, . . . , n À 1 we have that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffii qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 0 0 2 2 2 O O ¼ O O À ðÞd À d þ 4 À ðÞd À d þ 4 À 1 ¼ 3, i iþ1 i iþ1 i i 1 pi ffiffiffi i 1 pffiffiffiÂÃ ¼ 0 0 ðÞ¼À k since 0 di 1, i 1, . . . , n. Consequently O1On 3 n 1 3 2 , where 0 0 O1 and On are, respectively, the lowest and the highest points on line l.

Figure 2.47 l

On¢ On

O¢j Oj

O2¢ O2

O1¢ O1 2.4 Method of Projections 79

Figure 2.48 L

K ATMRB

Q N

Since band S should extend, at least by the radius of the circle, on each side of the 0 0 ends of segment O1On,p toffiffiffi containÂÃ the circles with centers O1 and On, then it should þ k have a width w 2 3 2 . 2.4.8. Consider the figure below (Figure 2.48). Segments MR and KM are projections of segments NR and QM on lines AB and ∠ ¼ π À ∠ ¼ π À ∠ ¼ ∠ KM, respectively. We have that QMK 2 KQM 2 MNR MRN. Therefore, it will be sufficient to prove that MR > MK. Note that _ _ _ _ ∠ ¼ BPþAN ¼ BPþBN ¼ ∠ BRP 2 2 MLP. This means that MLPR is an inscribe quadrilateral. Since ∠MLR ¼ ∠MPR ¼ ∠TLM, then MR ¼ MT > MK. 2.4.9. Let ∠CDB ¼ α, ∠CBD ¼ β, α β, and A0E0 is the projection of segment AE on line BD. We have that AE A0E0 ¼ BD þ sin α þ sin β, consequently

À α À β À α À 2α À α À β 1 sin sin < 1 sin < 1 sin < BD 1 sin sin α À β α À β α À β cos cos cos 2 2 2 α þ β α cos ðÞα þ β < cos 2 ¼ sin : α À β α À β sin α þ sin β cos cos 2 2

Hence

sin ðÞα þ β BD < : ð2:8Þ sin α þ sin β

þ ¼ BD sin α þ BD sin β Since BC CD sin ðÞαþβ sin ðÞαþβ , then according to (2.8) we get BC þ CD < 1. 2.4.10. (a) Consider Figure 2.49. MNþPK We have that BF MN and BF PK. Thus, we deduce that BF 2 or α ðÞa sin βþf sin γ þðÞc sin γþd sin β 2RA sin 2 . Therefore, it follows that 80 2 Application of Projection Method

Figure 2.49 MBbC P bg ac A a

D f a d gb NFeEK

1 sin β 1 sin γ R ða þ dÞ þ ðf þ cÞ : Similarly we get that A 4 sin α 4 sin α 1 sin α 1 sin β R ðf þ cÞ þ ðb þ eÞ , C 4 sin γ 4 sin γ 1 sin γ 1 sin α R ðb þ eÞ þ ða þ dÞ : E 4 sin β 4 sin β

þ 1 ðÞ> Summing up these inequalities and making use of inequality x x 2 x 0 , þ þ 1 ðÞþþ 1 ðÞþþ 1 ðÞ¼þ we obtain that RA RC RE 2 a d 2 b e 2 f c p. This ends the proof. Remark Equality holds if and only if hexagon ABCDEF is regular.

(b) Draw perpendiculars MB1, MD1, MF1 to lines BC, AC, AB, respectively, and construct parallelograms B1MF1A1, D1MB1C1, F1MD1E1. Since the radiuses of the circumcircles of triangles D1MF1, B1MF1 B1MD1 are Ra Rb Rc Ra ¼ Rb ¼ Rc ¼ equal to 2 , 2 , 2 , with 2 RE1 , 2 RA1 , 2 RC1 , then according to problem Ra þ Rb þ Rc ¼ þ þ 2.4.10a we have that 2 2 2 p da db dc. Thus

Ra þ Rb þ Rc 2da þ 2db þ 2dc:

(c) According to problem 2.4.10b MA þ MB þ MC 2MC1 þ 2MA1 þ 2MB1, where MA1 ⊥ BC, MB1 ⊥ AC, MC1 ⊥ AB (Figure 2.50). Thus, at least one of the following inequalities MA 2MC1, MB 2MA1, MC 2 MB1 is correct. Let MA 2MC1, consequently either ∠MAB 30 or ∠MAB 150 .If ∠MAB 150, then ∠MBC < 30. 2.4 Method of Projections 81

Figure 2.50 B A1

C1 M

AB1 C

Figure 2.51 B

C1 A1 kc ka O

AB1 C

2.4.11. Let O be the center of the circumcircle of acute triangle ABC and ka, kb, kc be the distances from point O to its sides (Figure 2.51).

Lemma Prove that ka þ kb þ kc ¼ R þ r. Let A1, B1, and C1 be the midpoints of sides BC, CA, and AB, respectively. By Ptolemy’s theorem akc þ cka ¼ bR akb þ bka ¼ cR, ckb þ bkc ¼ aR. On the other hand, aka þ bkb þ ckc ¼ 2S ¼ (a þ b þ c)r. By adding these equalities and reducing a þ b þ c, we obtain that ka þ kb þ kc ¼ R þ r. This ends the proof of the lemma. Projection of broken line AOA1 on straight line AA2 is equal to ha, where AA2 ⊥ BC, thus R þ ka ha. Similarly we get that R þ kb hb, R þ kc hc. By summing up these inequalities we get according to lemma that 4R þ r ¼ 3R þ ka þ kb þ kc ha þ hb þ hc (see problem 5.5.8b). 2.4.12. Let BU CV, then AU DV. Draw a median MN of the square (Figure 2.52) and P1P2||Q1Q2||AB. Projections of segments PP1 and QQ1 on side AB do not have common internal points, thus P P Q Q KN KM PQ AB À ðÞ¼PP þ QQ AB À 1 2 þ 1 2 AB À þ 1 1 2 2 2 2 1 ¼ AB: 2 82 2 Application of Projection Method

Figure 2.52

Figure 2.53 x a1 Sr-= a1 ++... an a n a2

2.4.13. First we will prove the following lemma. Lemma 2n unit vectors are drawn from point O on the plane. They are painted ~ ~ alternately into red and blue colors. Let S be the sum of n red vectors, r be the sum of n blue vectors. Prove that ~S À~r 2. Indeed let ~S À~r 6¼ ~0. Introduce a coordinate system with Ox axis along vector ~ S À~r. Since the sum of projections of vectors ~a1,~a2, :::,~an (Figure 2.53) on the Oy axis is equal to zero, then the length of vector ~S À~r is equal to the modulus of the difference of the sums of the lengths of the positive projections of these vectors on the Ox and that of the negative projections. Consequently, the length of the vector ~S À~r does not exceed either the sums of the lengths of the positive projections or sums of the lengths of the negative projections. It is clear that the sums of the lengths of the positive projections as well as sums of the lengths of the negative projections of vectors~a1,~a2, :::,~an on any axis does not exceed the diameter of the circle, i.e., does not exceed 2. (See also problem 7.1.72.) Consider unit vectors~e1, :::,~e110, the sum of which is a zero vector. According to the lemma they can be divided into two groups so that each group would contain ~ 55 vectors. The sum of the vectors of the ﬁrst group is S, the sum of the vectors of ~ ~À~ ~þ~ ¼ ~ ~À~ the second group is r, such that S r 2. Since S r 0, S r 2, then 2 ~S ¼ ~S À~r 2, consequently ~S 1. This ends the proof. 2.4 Method of Projections 83

Figure 2.54 B

A2 A1 AP C

D1 C1

2.4.14. We shall make use of the following fact: if XX1 is a bisector of angle X of triangle XYZ and ∠XX1Z 90 , then the midpoint of side YZ belongs to segment X1Z. Indeed, since ∠XX1Z 90 , then ∠Y ∠Z.consequently,

YX XY 1 ¼ 1: X1Z XZ

Let straight line A1C1 contain the bisector of angle APB and points A2, C2 be, respectively, the midpoints of sides AB, CD (see Figure 2.54). 0 0 Let also points A2, C2 be the projections of points A2, C2 on line A1C1. 0 0 0 0 ADþBC We have that A1C1 A2C2, thus it follows that A2C2 A2C2 2 (see ADþBC ABþCD problem 1.1.9a), consequently A1C1 2 . Similarly we get that B1D1 2 . Summing up these inequalities we obtain that ABþBCþCDþDA þ ¼ þ þ þ þ þ 2 A1C1 B1D1 A1P B1P C1P D1P PQ PR þ ABþBCþCDþDA þ þ þ PS PT, which means that 2 PQ PR PS PT. Remark The equality in the last inequality holds true if and only if quadrilateral ABCD is a rectangular. 2.4.15. Let points B0, C0 be projections of points B, C on line AD.ThenAD ¼ AB cos ∠BAD þ B0C0 þ CD cos ∠ADC AB cos ∠BAD þ BC þ CD cos ∠ADC. Similarly we obtain that BE AB cos ∠ABE þ AF þ EF cos ∠FEB and CF CD cos ∠FCD þ ED þ FE cos ∠CFE. Summing up these two inequalities we deduce that

AD þ BE þ CF ABðÞcos ∠BAD þ cos ∠ABE þCDðÞcos ∠ADC þ cos ∠FCD þ ∠BAD þ ∠ABE þ EFðÞcos ∠FEB þ cos ∠CFE þBC þ DE þ AF 2AB cos 2 ∠ADC þ ∠FCD ∠FEB þ ∠CFE þ 2CD cos þ 2EF cos þ BC þ DE þ AE 2 2 3AB þ BC þ DE þ AE ¼ AB þ BC þ CD þ DE þ EF þ FA: 84 2 Application of Projection Method

Figure 2.55 NC DP

MA FK

¼ ¼ ∠BADþ∠ABE þ ∠ADCþ∠FCD þ ∠FEBþ∠CFE ¼ π Since AB CD EF and 2 2 2 (see problem 5.1.4a). The equality holds if and only if ∠A ¼ ∠B ¼ ∠C ¼ ∠D ¼ ∠E ¼ ∠F. Remark We give an example of the centrally symmetric hexagon ABCDEF, for which AD þ BE þ CF > AB þ BC þ CD þ DE þ EF þ FA. Let MNPK be a unit square and let MA ¼ NB ¼ NC ¼ DP ¼ EK ¼ FK ¼ x (Figure 2.55). qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Then AD þ BE þ CF ¼ 3 1 þ ðÞ1 À 2x 2 and AB þ BC þ CD þ EF þ FA þ pffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi DE ¼ 2 2x þ 21ðÞþÀ 2x 2 x2 þ ðÞ1 À x 2. Note that at x ¼ 0 we have that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 1 þ ðÞ1 À 2x 2 ¼ 3 2 > 4 ¼ 2 2x þ 21ðÞþÀ 2x 2 x2 þ ðÞ1 À x 2. There- ÀÁ < < 1 fore, there exists such a number x 0 x 2 , such that

AD þ BE þ CF > AB þ BC þ CD þ DE þ EF þ FA:

2.4.16. (I. Ziganshin, Russia) Let KLM be the facet of tetrahedron KLMN which has the largest of the perimeters of all the facets of the tetrahedron, then

1À KL þ LM þ KM þ KN þ LN þ MN ¼ KL þ LM þ KM þ ðÞKN þ LN þ KL þ 2 Á þ ðÞKN þ MN þ KM þðÞLN þ MN þ LM 4ðÞKL þ LM þ KM ¼ 2ðÞKL þ LM þ KM : 2

Project tetrahedron ABCD on the plane of facet KLM. Denote the projections of vertices A, B, C, D of the tetrahedron by A1, B1, C1, D1, respectively. Let Γ be the convex envelope of points A1, B1, C1, D1 and PΓ be the perimeter of polygon Γ. Since triangle KLM is inside Γ, then according to problem 2.1.1 we have KL þ LM þ KM PΓ. Thus, it follows that 2.4 Method of Projections 85

Figure 2.56 B1 B1 C1

D1 A1

A1 C1 D1 ab

KL þ LM þ KM þ KN þ LN þ MN 2ðÞKL þ LM þ KM 2PΓ:

Now consider cases when Γ is a triangle or quadrilateral. In the ﬁrst case (Figure 2.56a) we have that

PΓ ¼ A1B1 þ B1C1 þ A1C1 ¼ 2 1 1 1 ¼ ðÞA B þ B C þ A C þA B þ B C þ A C < 3 1 1 1 1 1 1 3 1 1 3 1 1 3 1 1 2 1 < ðÞA B þ B C þ A C þðÞD A þ D B 3 1 1 1 1 1 1 3 1 1 1 1 1 1 þ ðÞD B þ D C þðÞD C þ D A ¼ 3 1 1 1 1 3 1 1 1 1 2 ¼ ðÞA B þ B C þ A C þ D A þ D B þ D C : 3 1 1 1 1 1 1 1 1 1 1 1 1

In the second case (Figure 2.56b) we have that

PΓ ¼ A1B1 þ B1C1 þ C1D1 þ D1A1 ¼ 2 1 ¼ ðÞA B þ B C þ C D þ D A þðÞA B þ B C þ C D þ D A < 3 1 1 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 2 2 < ðÞA B þ B C þ C D þ D A þðÞA C þ B D 3 1 1 1 1 1 1 1 1 3 1 1 1 1

See problem 1.1.4a. < 2 ðÞþ þ þ þ þ In both cases PΓ 3 A1B1 B1C1 A1C1 D1A1 D1B1 D1C1 . On the other hand, A1B1 AB, B1C1 BC, A1C1 AC, D1A1 DA, D1B1 DB, D1C1 þ þ þ þ þ < 4 ð þ þ þ DC. Therefore, KL LM KM KN LN MN 2PΓ 3 AB BC AC DA þ DB þ DCÞ. Remark Consider a regular pyramid DABC, where AB ¼ BC ¼ AC ¼ 1, DA ¼ DB ¼ DC ¼ n (n > 2). Choose on edges DA, DC, BA, BC points K, L, M, N, ¼ ¼ ¼ ¼ 1 respectively, such that DK DL BM BN n. Then we have that þ þ > > À 2 KM KD MB BD. Consequently KM n n. Similarly we deduce that > À 2 KN, ML, NL n n. Then 86 2 Application of Projection Method

ÀÁ KL þ KM þ KN þ ML þ NL þ MN 4 n À 2 > n : ð2:9Þ AB þ AC þ AD þ BC þ BD þ CD 3n þ 3

4ðÞnÀ2 !1 n ! 4 4 Since at n we have that 3nþ3 3, then it is clear that the value 3 in inequality 2.4.16 cannot be lowered.

2.4.17. Let ∠MAiAi þ 1 ¼ βi, ∠MAiAi À 1 ¼ γi, ∠Ai ¼ αi, i ¼ 1, . . . , n, An þ 1 Ai, A0 An, it is clear that αi ¼ βi þ γi. Note that

Xn Xn β þ γ β À γ ¼ ðÞβ þ γ ¼ i i i i 2p MAi cos i cos i 2MAi cos cos i¼1 i¼1 2 2 Xn αi 2MAi cos , i¼1 2

∠ ∠ A1 þ ::: þ An consequently, MA1 cos 2 MAn cos 2 p.

2.4.18. (a) Let M0, N0, and P0 be the midpoints of edges A1B1, C1C, and DApffiffi, respectively. From right-angled triangles AA P and A M P we have A P ¼ 5, pffiffi 1 0 1 0 0 1 0 2 M P ¼ 6, and cos ∠A M P ¼ p1ffiffi. Similarly, we get that the cosines of angles 0 0 2 1 0 0 6 ∠B M N , ∠C N M , ∠CN P , ∠DP N , and ∠AP M are equal to p1ffiffi. 1 0 0 1 0 0 0 0 0 0 0 0 6 0 Let point X be the projection of point X on line M0P0. Then, we have that

0 0 þ 0 0 þ 0 0 ¼ 0 þ þ 0 0 0 B1M M P P D B1M0 M0P0 P0D , M P 1 1 ¼ M0P0 þ pffiffiffi À ðÞB1M þ PD pffiffiffi : 6 6

Since MP M0P0, we get that

1 1 MP M0P0 þ pffiffiffi À ðÞB1M þ PD pffiffiffi : ð2:10Þ 6 6

In a similar way, we obtain that

1 1 MN M0N0 þ pffiffiffi À ðÞA1M þ CN pffiffiffi , ð2:11Þ 6 6 and

1 1 PN P0N0 þ pffiffiffi À ðÞAP þ NC1 pffiffiffi : ð2:12Þ 6 6 2.4 Method of Projections 87

Summing up (2.10), (2.11) and (2.12) we get pffiffiffi 3 6 MP þ MN þ PN M P þ M N þ P N ¼ : 0 0 0 0 0 0 2

(b) Let O be the center of the cube and M be an arbitrary point of the space.

Let’s consider projections of point M and straight lines A1B1, C1C, AD on plane M0N0P0 (see notations in the solution of problem 2.4.18a). 0 Denote the projection of point X on plane M0N0P0 by X . ⊥ ⊥ 0 0 0 0 \ 0 0 ¼ Since OM0 A1B1, we have OM0 A1B1. Let A1B1 A D A2, 0 0 \ 0 0 ¼ 0 0 \ 0 0 ¼ A1B1 C1C B2, A D C1C C2, then A2B2C2 is a regular triangle with center O. 0 þ 0 þ 0 We have that SM A2B2 SM B2C2 SM A2C2 SA2B2C2 , consequently

M0M0 Á A B M0N0 Á B C M0P0 Á A C 1 A B þ B C þ C A 2 2 2 þ 2 2 2 þ 2 2 2 pffiffiffi Á 2 2 2 2 2 2 , 2 2 2 2 2 thus maxðÞM0M0 ; M0N0 ; M0P0 p1ffiffi, where MP ⊥ AD, MM ⊥ A B , MN ⊥ C C, 2 2 2 2 2 2 1 1 2 1 and P2 2 AD, M2 2 A1B1, N2 2 C1C, which means that ÀÁ 0 0 0 0 0 0 1 maxðÞMM2; MN2; MP2 max M M 2; M N 2; M P2 pffiffiffi : 2

(c) Let O1 and O2 be the centers of these spheres and projections of segment O1O2 on the edges emerging from the same vertex are equal to a, b, and c. We have that a2 þ b2 þ c2 ¼ O O 2. Let a b, a c, in that case a Op1Offiffi 2. Note 1 2 3 d þ þ D (Figure 2.57) that 2 a 2 1. Consequently pffiffiffi þ þ 1 Op1Offiffi 2 þ d þ D > dpDffiffi þ d D. Hence, we deduce that d þ D < 3 À 3. 3 2 2 2 3 2

Figure 2.57

D О 2 2 а О1 d 2 88 2 Application of Projection Method

Problems for Self-Study

2.4.19. Prove that the distance from one of the vertices of the convex quadrilateral to the opposite diagonal does not exceed the half of that diagonal.

2.4.20. Let parallelogram P2 be inscribed in parallelogram P1 and parallelogram P3 be inscribed in parallelogram P3, such that the sides of P3 are parallel to the sides of P1. Prove that the length of at least one of the sides of P1 does not exceed the double length of the corresponding parallel side of P3. 2.4.21. Prove that inside a convex n-gon (n 7) one can ﬁnd a point, the sum of the distances from which to the vertices are greater than the perimeter. 2.4.22. Given a unit square and such a broken line, that inside it every line parallel to the side of the square intersects it in no more than one point. Prove that the length of the broken line is less than 2.

2.4.23. (a) Let us considerpffiffiffi on a plane a finite set of segments, the sum of lengths of which is less than 2. Prove that there exists an infinite net of unit squares, the sides of which do not intersect with any of these segments. Remark The statement of the problem holds true also if the number of segments is not finite. (b) A figure on the coordinate plane has an area S, where S > 1. Prove that it can be translated by a vector with the integer number of coordinates, so that the figure and its image do not intersect. 2.4.24. Let ABCDEF be a convex hexagon, such that AB|| ED, BC||EF, CD||AF, and AB þ DE ¼ BC þ EF ¼ CD þ AF. Prove that

AD þ BE þ CF AB þ BC þ CD þ DE þ EF þ AF:

2.4.25. Let ABCDEF be a convex hexagon. Prove that

BC Á DE BC Á AF AF Á DE AD þ BE þ CF AB þ CD þ EF þ þ þ : AF DE BC

Hint See the solutions of problems 2.4.15 and 5.1.22a. φ θ > φ þ θ ¼ π À α b sin θþc sin φ 2.4.26. Let , 0 and . Prove that ma 2 , where ma is the length of the median from vertex A of triangle ABC, AB ¼ c, AC ¼ b, α ¼ ∠A. pffiffiffiffiffiffiffiffiffiffiffiffiffi 2.4.27. Let ABCDA1B1C1D1 be a parallelepiped. Prove that V S1S2S3, where V is the volume of the parallelepiped and S1, S2, S3 are the areas of facets ABCD, AA1B1B, AA1D1D.

2.4.28. Given points A1,B1, and C1 on sides BC, CA, and AB of triangle ABC respectively. Prove that 2.4 Method of Projections 89

R sin α sin β sin γ 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , R ðÞα γ 2 þ ðÞα γ 2 À α α γ γ ðÞβ þ β sin 1 sin sin sin 1 2 sin sin 1 sin sin 1 cos 1 where ∠A ¼ α, ∠B ¼ β, ∠C ¼ γ, ∠B1A1C1 ¼ α1, ∠A1B1C1 ¼ β1, ∠A1C1B1 ¼ γ1, and R1, R are the radiuses of the circumcircles of triangles A1B1C1 and ABC, respectively.

Hint Let OA and OC be the centers of the circumcircles of triangles AC1B1 and CA1B1, then OAOC R sin β. It remains to prove that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin α 2 sin γ 2 sin α sin γ O O ¼ R 1 þ 1 À 2 1 1 cos ðÞβ þ β : A C 1 sin α sin γ sin α sin γ 1

2.4.29. Let p be the sum of the lengths of the edges of convex polyhedron and d be the maximal distance between its vertices. Prove that p > 3d. Hint Consider the projection of the polyhedron on straight line AB, where A and B are the vertices of the polyhedron and AB ¼ d. 2.4.30. The closed broken line passes on a surface of a unit cube and has common pointspffiffiffi with all its facets. Prove that the length of the broken line is not less than 3 2. Hint Consider the projections of the broken line on three mutually non-parallel edges of the cube. 2.4.31. A number of segments is located inside a unit cube, so that any plane parallel to one of the facets of the cube crosses not more than one of them. Prove that the sum of the lengths of these segments is not greater than 3. Hint Consider projections of these segments on three edges of a cube with common vertex. 2.4.32. A number of polygons is located inside a unit cube, so that any line parallel to one of the edges of the cube intersects with not more than one of them. Prove that the sum of the areas of these polygons is not greater than 3. Hint Consider projections of polygons on three facets of a cube with common vertex.

2.4.33. Prove that for any tetrahedron one can find two such planes that the ratiopffiffiffi of the areas of the projections of the tetrahedron on these planes is not less than 2. Hint For the tetrahedron SABC let MN be the common perpendicular of straight lines SA and BC. Consider planes α, β, γ, passing through line MN, where α ⊥ SA, β ⊥ BC, such that the angles between γ and straight lines SA, BC are equal. 90 2 Application of Projection Method

2.4.34. Given a triangular pyramid ABCD. R is the radius of its circumsphere, r is the radius of its insphere, a is the length of its longest edge, h is the length of the R > a least altitude (to one of its facets). Prove that r h. 2.4.35. Given 75 points inside a unit cube. Prove that the area of one of the triangles, with vertices belonging to these points, does not exceed 7/72. 2.4.36. The angles at base AD of trapeze ABCD satisfy condition ∠A < ∠D < 90. Prove that then AC > BD.

2.4.37. Given a plane n (n 2) unit vectorsÂÃ the sum of which is a zero vector. Prove n that of these n vectors one can choose 2 vectors, such that the sum of their modules is not greater than 1.

Hint Prove that if one has chosen k (k n À 2) vectors with a sum~S, where ~S 1, ~ ~ then from remaining vectors we can choose two vectors a and b, such that at least one of the inequalities ~S þ~a 1, ~S þ~b 1, ~S þ~a þ~b 1 holds true. (See also problem 2.4.13.)

2.4.38. Points A1, B1, C1 are taken on sides BC, CA, AB of triangle ABC, respectively. Given that AB1 þ AC1 ¼ BC1 þ BA1 ¼ CA1 þ CB1. þ þ 1 ðÞþ þ Prove that A1B1 B1C1 C1A1 2 AB BC AC . Hint See the solutions of problems 2.4.6 and 5.1.4a. 2.4.39. Let ABCDE be a convex pentagon. Given that AB ¼ BC ¼ CD ¼ DE ¼ EA and max(∠A, ∠B, ∠C, ∠D, ∠E) < 120. Prove that min(∠A, ∠B, ∠C, ∠D, ∠E) > 90. Hint Let ∠A 90, and M and N are the midpoints of segments AC and AD, respectively. Let X0 be the projection of point X on straight line MN. Then pffiffiffiffiffi ÀÁ pffiffi 0 0 > a þ a α þ a À α > a þ a þ 3 2a BE B E 2 2 cos 2 cos 30 2 2 1 2 , where AB ¼ a, ∠BMB0 ¼ α. This leads to a contradiction.

2.4.40. Let A1A2 ÁÁÁAn be a polygon, such that ∠A1 ¼ ∠A2 ¼ ÁÁÁ ¼ ∠An and A1A2 A2A3 ÁÁÁ An À 1An AnA1. Prove that A1A2 ÁÁÁAn is a regular polygon. 2.4.41. Given points D, E, F on sides BC, CA, AB of triangle ABC, respectively. It is ¼ ¼ ¼ ¼ ¼ knownpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi that DE EFpffiffiffi DF, pffiffiffiAB c, BC a, CA b. Prove that DE Á a2 þ b2 þ c2 þ 4 3S 2 2S, where S is the area of triangle ABC. Hint See problem 2.4.28.

2.4.42. Given points A1, B1, and C1 on sides BC, CA, and AB of triangle ABC, respectively. Let IA be the incenter of triangle AC1B1, and GA the centroid of triangle AC1B1. Points IB, GB and IC, GC are deﬁned similarly. Prove that IAIB þ IBIC þ ICIA GAGB þ GBGC þ GCGA. 2.4 Method of Projections 91

Hint Consider the projections of segments ICIA, IAIB, and IBIC on the straight lines AC, AB, and BC, respectively. Then, one can easily deduce that þ þ þ þ 1 ðÞþ IAIB IBIC ICIA A1B1 B1C1 A1C1. Prove that GCGA 3 A1C1 AC and GAGB þ GBGC þ GCGA A1B1 þ B1C1 þ A1C1. α r α ¼ α β γ 2.4.43. Prove the inequality cos R for a triangle, where max ( , , ). Hint Prove that cosβ þ cos γ 1. Chapter 3 Areas

This chapter is devoted to the inequalities related to areas and it consists of only one paragraph, that is Section 3.1. One of the methods for proving the inequalities related to areas (of some figures on the plane) is the following: if the figures with areas S1 , S2 ,...,Sk cover a figure with area S, then S1 þ S2 þ ...þ Sk S. Let us give a simple example. Prove that the area of parallelogram AMNK (M 2 AB, N 2 BC, K 2 AC) inscribed to triangle ABC is not greater than the half of the area of ABC. Consider parallelograms BNKE and NCKF; note that triangles AEK and KNF are, respectively, equal to triangles MBN and KNC. Moreover, triangles AEK and KNF cover parallelogram AMNK; therefore the sum of the areas of triangles MBN and KNCis not smaller than the area of parallelogram AMNK. This ends the proof of given example. One of the main methods of proving geometric inequalities related with areas is rewriting them as algebraic inequalities by introducing some notations. In order to compare areas of two figures, often one needs to consider consequent figures, such that the first figure and the last figure are the given figures. Afterwards, one needs to compare consequently areas of any two of the considered figures. Some problems in this chapter were inspired by [2, 4, 9, 11, 13–16]. Nevertheless, even for these problems the authors have mostly provided their own solutions.

3.1 Inequalities with Areas

3.1.1. (a) In a convex quadrilateral ABCD diagonals intersect at a point O and BC|| 1 AD. Prove that SOCD 4 SABCD. (b) Given points E and F on parallel sides BC and AD of the convex quadrilateral ABCD, respectively. Given also that AE, BF intersect at point P and segments 1 CF, ED at point Q. Prove that SPEQF 4 SABCD.

(c) In a convex quadrilateralpffiffiffiffiffiffiffiffiffiffiffiffi pABCDffiffiffiffiffiffiffiffiffiffi diagonalspffiffiffiffiffiffiffiffiffiffi AC, BD intersect at point O. Prove that SABCD SAOB þ SCOD. 3.1.2. Prove that the area of the polygon obtained by sequentially connecting the midpoints of a convex n-gon (n 5) will be greater than the half of the initial polygon. 3.1.3. Points K and M are taken on sides AB and CD of a convex quadrilateral ABCD, respectively. Let L be the point of intersection of segments AM and KD, N be the point of intersection of segments KC and BM. Prove that, if AK : ¼ ¼ < mn KB CM : MD m : n, then SKLMN m2þmnþn2 SABCD. 3.1.4. Let M and N be the midpoints of sides BC and CD of a convex quadrilateral 1 < < 1 ABCD, respectively. Prove that 4 SABCD SAMN 2 SABCD. 3.1.5. Bisectors ME , MF , MG , MH of triangles ABM, BMC, CMD, DMA are drawn from point M inside rectangle ABCDÀÁpffiffiffiwith area S. Prove that for area S0 of a quadrilateral EFGH, the inequality 2 À 1 S S0 0, 5S holds true. For which points does equality S0 ¼ 0,5S hold true? 3.1.6. Let M be a given point inside of an angle with a vertex O. Draw a segment AB passing through the point M and having the endpoints on the sides of the angle, such that

(a) the area SOAB is minimal. 1 þ 1 (b) MA MB is maximal.

3.1.7. (a) Given a convex quadrilateral ABCD with area S. Perpendiculars MA1, MB1, MC1, MD1 are drawn from the point of intersection of diagonals M to the sides S AB, BC, CD, DA, respectively. Prove that SA1B1C1D1 2. (b) Diagonal BD of the convex quadrilateral ABCD divides it into two isometric parts. Perpendiculars MA1, MB1, MC1, MD1 are drawn from the point M inside of the given quadrilateral to straight lines AB, BC, CD, and DA, respectively. þ < 1 Prove that SMA1D1 SMB1C1 2 SABCD. 3.1.8. Let M, N, and P be the feet of the perpendiculars drawn from the centroid of the triangle to the straight lines AB, BC and CA, respectively. Prove that 1 SMNP 4 SABC. 3.1.9. Points D, E and F are on sides BC, CA and AB of triangle ABC, respectively, such that they do not coincide with points A, B, C. Prove that, if quadrilateral AFDE 2 4SDEF EF is inscribed into a circle, then 2. SABC AD 3.1.10. Given D and E on sides AB and BC of triangle ABC, respectively. Points K and M divide segment DE into three equal segments (DK ¼ KM ¼ ME). The straight linespBKffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiand BM intersect side AC at points T and P, respectively. Prove that SBTP SABT Á SPBC. 2 2 2 3.1.11. Let P, Q, Rpffiffiffiffiffiffiffiffiffiffibe pointspffiffiffiffiffiffiffiffiffi in a trianglepffiffiffiffiffiffiffiffiffiffiABC, such that P [AB], Q [BC], R 3 3 3 [PQ]. Prove that SABC SAPRþ SQRC. 3.1 Inequalities with Areas 95

3.1.12. Let A1, B1, C1 be points on sides BC, CA, AB of a triangle ABC, respectively. Prove that (a) S Á S2 4S Á S Á S , ABC A1B1C1 AB1C1 BA1C1 CA1B1 ðÞ; ; (b) min SAB1C1 SBA1C1 SCA1B1 SA1B1C1 , (c) minðÞS ; S ; S 1 S , pffiffiffiffiffiffiffiffiffiffiffiffiAB1C1 pBAffiffiffiffiffiffiffiffiffiffiffiffi1C1 CA1pB1 ffiffiffiffiffiffiffiffiffiffiffiffi4 ABC pffiffiffiffiffiffiffiffiffiffi (d) S þ S þ S 3 S , AB1C1 BA1C1 CA1B1 2 ABC Á Á 1 ; AA1 BB1 CC1 (e) SA1B1C1 min 4SABC 2ðÞABþBCþAC , if segments AA1, BB1 and CC1 intersect at one point, Á 1 þ 1 þ 1 (f) SA1B1C1 S S S 3, rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiAB1C1 BA1C1 CA1B1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi S ÁS ÁS S ÁS ÁS AB1C1 BA1C1 CA1B1 AB1C1 BA1C1 CA1B1 (g) ðÞ; ; SA1B1C1 ðÞ; ; , max SAB1C1 SBA1C1 SCA1B1 min SAB1C1 SBA1C1 SCA1B1

if segments AA1, BB1 and CC1 intersect at one point. 3.1.13. Two points M and N are chosen on sides AB and BC of triangle ABC, respectively. Three parallel straight line passing through points M, B and N intersect side AC at points K, D and L, respectively. Prove that the area of the trapezoid (or parallelogram) KMNL is not greater than the area of one of the triangles ABD and DBC. 3.1.14. Two tangents PB and PC are drawn from point P to the circle, such that ∠BPC 90 . Point A is chosen on the smaller arc BC. Prove that the area of the triangle, cut from the angle BPC by the tangent to the circle drawn from point A, does not exceed the area of triangle ABC. 3.1.15. Let bisectors of the internal angles of triangle ABC intersect the circumcircle of triangle ABC. Given that theyÀÁ intersect the circle for the second time at points A1, SABC 2 ∠AÀ∠B 2 ∠BÀ∠C 2 ∠CÀ∠A B1, C1. Prove that min cos ; cos ; cos . SA1B1C1 2 2 2 3.1.16. Prove that the area (a) of the parallelogram placed inside the triangle does not exceed half the area of the triangle. (b) of the triangle placed inside the parallelogram does not exceed half the area of the parallelogram. 3.1.17. Prove that the convex polygon with the area S can be placed inside (a) a rectangle with the area not greater than 2S. (b) a triangle with the area not greater than 2S. 3.1.18. Prove that one can inscribe into a convex polygon with the area S

S (a) a parallelogram with the area not less than 2, 3S (b) a triangle with the area not less than 8 . 3S (c) a hexagon with the area not less than 4 . 96 3 Areas

3.1.19. Given a convex polygon in which one cannot place any triangle with area 1. Prove that this polygon can be placed into a triangle with area 4.

3.1.20. Given an acute triangle ABC.LetA1, B1, C1 be the points symmetric to points A, B, C with respect to straight lines BC, AC, AB, respectively. Prove that SA1B1C1 4SABC.

3.1.21. Given a triangle ABC. Segments BB1, CC1, AA1 are placed on rays AB, BC, CA starting at points B, C, A, respectively, such that BB1 ¼ AC, CC1 ¼ AB, ¼ þ þ AA1 BC. Prove that SAA1B SBB1C SCC1A 3SABC. 3.1.22. Three secants are drawn for each internal point X of triangle ABC parallel to its sides. As a result of this, one obtains three triangles each bounded by two secants and a side of the triangle. Let the areas of these triangles be S1, S2, S3. Prove that þ þ 1 S1 S2 S3 3 S. 3.1.23. In a convex hexagon ABCDEF the opposite sides are parallel to each other (AB||DE, BC||EF, CD||AF). Prove that 2SBDF SABCDEF. 3.1.24. Given points P, Q, R on sides AB, CD, EF of the centrally symmetric convex hexagon ABCDEF, respectively. Prove that 2SPQR SABCDEF. 3.1.25. The triangle is inscribed into a regular hexagon; one of the sides of the triangle passes through the center of symmetry of the hexagon. Prove that the area 1 of the triangle does not exceed 3 of the area of the hexagon. 3.1.26. Points D and E are chosen on the sides AB and AC of the triangle ABC, respectively. Let segments BE and CD intersect at point P. Prove that pffiffi (a) S 5 5À11 S , PDE ÀÁp2 ffiffiffi ABC (b) SPDE 5 2 À 7 SABC,ifSBCED ¼ 2SPBC. 3.1.27. The segment PQ passes through the centroid of triangle ABC and points P, 2 Q are on sides BC, AC, respectively. Prove that SMPQ 9 SABC, where M is the midpoint of side AB. 3.1.28. Prove that the area of any section of the cube by a plane passing through its center is not less than the area of the faces of the cube. 3.1.29. The part of a plane between two parallel straight lines is called a “strip”. Let several strips be given on a plane, and no two of them are parallel, i.e. no two boundary straight lines of different strips are parallel. How one should move the strips parallel to themselves to make the area of the polygon formed by the intersection of the strips the greatest possible? ÀÁ ω 1; 1 < 1 3.1.30. (a) Given a circle with a center A 2 2 and radius R, where R 2. Points M and N are on the positive semiaxes Oy and Ox so that segment MN is tangent to circle ω. Prove that 3.1 Inequalities with Areas 97 pffiffi 2 pffiffi 2 À < 2 (1) SMON 2 R ,if0 R 4 , pffiffi 1 À 2 2 < < 1 (2) SMON 4 R ,if 4 R 2. pffiffiffi (b) Prove that any section of the unit cube by a plane has an area not greater than 2.

3.1.31. (a) Given points B2,...,Bn and C2,...,Cn on the sides of triangle B1AC1 (B2,...,Bn 2 AB1, C2,...,Cn 2 AC1). The ray with a vertex at A intersects segments B1C1, B2C2,...,BnCn at points D1, D2,...,Dn, respectively. Prove that

AD1 þ AD2 þ ::: þ ADn maxðÞAB1 þ AB2 þ ::: þ ABn; AC1 þ AC2 þ ::: þ ACn :

(b) Prove that in a convex quadrilateral the length of any segment with ends on the sides of the quadrilateral, which passes through the point of intersection of diagonals do not exceed the length of one of the diagonals. (c) Prove that the area of any cross section of the tetrahedron by a plane does not exceed the area of one of its faces. 3.1.32. Given a convex quadrilateral ABCD. Prove that 2 2 2 (a) CD Á SABC þ BC Á SACD > AC Á SBCD,if∠A þ ∠ C > 180 , 2 2 2 (b) CD Á SABC þ BC Á SACD ¼ AC Á SBCD,if∠A þ ∠ C ¼ 180 , 2 2 2 (c) CD Á SABC þ BC Á SACD < AC Á SBCD,if∠A þ ∠ C < 180 . 3.1.33. In a convex hexagon ABCDEF opposite sides are parallel (AB||DE, BC||EF, CD||AF). Denote by A1, B1, C1, D1, E1, and F1 the midpoints of sides AB, BC, CD, DE, EF and FA, respectively. Prove that

(a) segments A1D1, B1E1 and C1F1 can be sides of a triangle. 1 < (b) 2 SBDF S1 SBDF, where S1 is the area of the triangle with sides having lengths A1D1, B1E1 and C1F1. 3.1.34. Given that the rectangle with sides a and b is inside the rectangle with sides c and d, such that max(a, b) > max (c, d). Prove that 2ab < cd. 3.1.35. O is an internal point of the convex quadrilateral ABCD with area S. Let points K, L, M and N be on segments AB, BC, CD and DA, respectively. Prove that if OKBL and OMDN are parallelograms, then pffiffiffi pffiffiffiffiffi pffiffiffiffiffi þ : (a) S pffiffiffi S1 pffiffiffiffiffiS2 pffiffiffiffiffi > þ : (b) 1, 25pffiffiffi S pffiffiffiffiffiT1 pffiffiffiffiffiT2 (c) C0 S T1 þ T2,

where S1, S2, T1, T2 are the areas of the quadrilaterals ONAK, OLCM, OKBL, OMDN, respectively and ÀÁ α þ π sin 2 4 C0 ¼ max : ½;π cos α 0 4 98 3 Areas

3.1.36. Let A1, B1, C1, D1 be the midpoints of sides BC, CD, DA, AB of a convex quadrilateral ABCD, respectively, S0 be the area of the quadrilateral, formed by the segments AA1, BB1, CC1, DD1, and S be the area of the quadrilateral ABCD. Prove 1 < 1 that 6 S S0 5 S. 3.1.37. Consider a triangle with the area S. Prove that if one places externally equilateral triangles on each side of the given triangle, then centers of these triangles are vertices of an equilateral triangle with the area not less than S. 3.1.38. Given on a plane mutually nonintersecting triangles all obtained from triangle ABC using a translation by a certain vector. All these triangles are contin- ued to become parallelograms, such that any of these parallelograms has a diagonal parallel to AB. Prove that S0 1, 5S, where S0 is the area of the union of these parallelograms and S is the sum of the areas of all triangles.

3.1.39. Given that in a convex hexagonpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiABCDEF triangles ACE and BDF are R Á similar. Prove that SABCDEF r SACE SBDF, where r and R are, respectively, inradius and circumradius of triangle ACE. 3.1.40. Given on a plane five points, so that the area of each of 10 triangles defined by these points is greater than 2. Provepffiffiffi that among those triangles exists a triangle, such that its area is greater than 1 þ 5.

3.1.41. The vertices of a convex hexagon A1A2A3A4A5A6 are on the sides of the unit square. Prove that the area of one of triangles A1A2A3, A2A3A4, A3A4A5, A4A5A6, 1 A5A6A1, A6A1A2 is not greater than 8. 3.1.42. Given on a plane a triangle ABC and a point P, such that ∠A ¼ 600, PA ¼ 1, ÀÁpffiffiffiffiffi pffiffi ¼ ¼ þ Á 3 PB 2, PC 3. Prove that SABC 13 73 8 . ∠ ¼ α π 3.1.43. Given on a plane a triangle ABC and a point P, such that A 2, PA ¼ r1, PB ¼ r2, PC ¼ r3,with0< r1 < r2 < r3, where r1, r2, r3, α are constants. Find the possible highest and lowest values of the area of triangle ABC. 3.1.44. Consider two parallelograms that intersect exactly in eight points. Prove that the common area of these parallelograms is greater than or equal to half of the area of one of them.

Solutions

3.1.1. (a) Consider the midpoints of segments AC, CD and BD and let those points be M, N and P, respectively. We have that MN||AD and NP||BC and also ¼ þ ¼ 1 þ 1 ¼ 1 þ 1 ¼ 1 SOCD SMCDP SMCN SPND 4 SACD 4 SBCD 4 SACD 4 SABC 4 SABCD. 1 Consequently, SOCD 4 SABCD. (b) Making use of problem 3.1.1a we get that 3.1 Inequalities with Areas 99

Figure 3.1 BC N M

K L

1 1 1 S ¼ S þ S S þ S ¼ S : PEQF PEF EQF 4 ABEF 4 FECD 4 ABCD

(c) We have that SAOB ¼ BO ¼ SBOC , consequently, SDOA OD SCOD pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi SABCD ¼ SAOB þ SBOC þ SCOD þ SDOA SAOB þ SCOD þ 2 SBOC Á SDOA ¼ pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi 2 ¼ ðÞSAOB þ SCOD , pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi we obtain that SABCD SAOB þ SCOD.

3.1.2. Let points B1, B2,...,Bn be the midpoints of sides A1A2, A2A3,...,AnA1 of the initial polygon A1A2 ...An, respectively. Denote by C1, C2,...,Cn the intersection points of the diagonalеs AnA2 and A1A3, A1A3 and A2A4,..., An À 1A1 and AnA2, respectively. We have that

1 S ::: ¼ S ::: À S À ::: À S ¼ S ::: À ðÞS þ ::: þ S ¼ B1B2 Bn A1A2 An B1A2B2 BnA1B1 A1A2 An 4 A1A2A3 AnA1A2 1 1 ¼ S ::: À ðÞ2S ::: À S À ::: À S À 2S ::: > S ::: : A1A2 An 4 A1A2 An C1A2C2 CnA1C1 C1C2 Cn 2 A1A2 An

3.1.3. We have that SMKC ¼ m (Figure 3.1). SMKD n Let SMKC ¼ mS1, SMKD ¼ nS1. Similarly we have SAKM ¼ mS2, SBKM ¼ nS2. Denote by ρ(X, l) the distance of the point X from the straight line l. Calculate 1 1 m n À Á S ¼ CD Á ρðÞ¼K; CD CD Á ρðÞþB; CD ρ A; CD ¼ CKD 2 2 m þ n m þ n 1 1 ¼ CM Á ρðÞþB; CD MD Á ρðÞ¼A; CD S þ S : 2 2 BCM AMD

Therefore, SKLMN ¼ SBNC þ SALD. Let SKNM ¼ x, SKLM ¼ y, then SNCM ¼ mS1 À x, SMLD ¼ nS1 À y, SAKL ¼ mS2 À y, SBKN ¼ nS2 À x. ðÞÀ ðÞÀ SBNC BN SBKN mS1 x nS2 x It is clear that ¼ ¼ , consequently, SBNC ¼ . SNCM NM SKNM x 100 3 Areas

ðÞÀ ðÞÀ ¼ nS1 y mS2 y ¼ þ Similarly, we deduce that SALD y . Since SKLMN SBNC SALD, ðÞÀ ðÞÀ ðÞÀ ðÞÀ then x þ y ¼ mS1 x nS2 x þ nS1 y mS2 y or 1 ¼ mS1nS2 þ mS1nS2 . Now x y ðÞS1þS2 ðÞmþn x ðÞS1þS2 ðÞmþn y we have to ﬁnd the maximal value of x þ y at x < min (mS1, nS2). mS1nS2 2 mS1nS2 2 π Let x y. Denote ¼ sin α, ¼ cos α, where α0 < α , ðÞS1þS2 ðÞmþn x ðÞS1þS2 ðÞmþn y 4 2 mS1nS2 sin α0 ¼ . ðÞS1þS2 ðÞmþn minðÞmS1;nS2 þ ¼ 4mS1nS2 < 4mS1nS2 Then x y 2 2 . ðÞS1þS2 ðÞmþn sin 2α ðÞS1þS2 ðÞmþn sin 2α0 2 nS2 If mS nS , then sin α0 ¼ and 1 2 ðÞS1þS2 ðÞmþn

4mS1nS2 ¼ mS1 ¼ mS1 ¼ ðÞS þ S ðÞm þ n sin 22α 1 À sin 2α nS2 1 2 0 0 1 À ðÞS1 þ S2 ðÞm þ n ðÞþ ðÞþ ðÞþ ðÞþ ðÞþ ðÞþ ¼ mm n S1 S1 S2 mm n S1 S1 S2 ¼ mn m n S1 S2 ¼ mS þ mS þ nS m2 m2 þ mn þ n2 1 2 1 mS þ S þ nS 1 n 1 1 mn ¼ S : m2 þ mn þ n2 ABCD

Thus, we obtain that mn S < S , ð3:1Þ KLMN m2 þ mn þ n2 ABCD

In the case of nS2 < mS1 the inequality (3.1) is obtained similarly.

3.1.4. We have that SABMND < SABCD,

1 1 1 S ¼ S þ S þ S ¼ S þ S þ S ¼ S þ S , ABMND AMN ABM AND AMN 2 ABC 2 ACD AMN 2 ABCD

< 1 ¼ þ ¼ þ consequently, SAMN 2 SABCD. Note that SABCD SABD SBCD SABD 4SCMN. ¼ þ > 1 Therefore, 4SAMN SABCD SABD,soSAMN 4 SABCD. 3.1.5. (Solution of G. Khotsanyan, 9th grade) First we shall prove two lemmas which are used at the proof of the inequality. Lemma 1 A straight line perpendicular to AB is drawn from point H of segment AB, where AH < BH. Points C and D are taken on that perpendicular so that CH < DH. Points E and F are the feet of the bisectors drawn from vertices C and D in triangles ABC and ABD. Prove that AE < AF.

Proof Indeed, according to thepffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi interior angle bisectorpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi theorem, for triangles ABC AE ¼ AC ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiCH2þAH2 AF ¼ AD ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiDH2þAH2 and ABD we get: EB BC , FB BD . CH2þpBHffiffiffiffiffiffiffiffiffiffiffiffi2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiDH2þBH2 ðÞ¼pffiffiffiffiffiffiffiffiffiffiffiffix2þAH2 ¼ þ AH2ÀBH2 Consider now a function fx 1 2þ 2 . Note that it grows in x2þBH2 x BH the interval x 2 [0 ; 1 ) (since AH2 À BH2 < 0). Consequently, from the condition 3.1 Inequalities with Areas 101

< < AE < AF CH DH, it follows that f(CH) f(DH). Hence, we deduce that EB FB or EBþAE < FBþAF > EB FB . From this inequality we obtain that AF AE. Lemma 2 Given that points E and F are the feet of the bisectors drawn from vertices C and D in triangles ABC and ABD, respectively, with CD||AB. Prove that, if segments CE and DF do not intersect, then CD > EF. Proof First introduce following notations (see also Figure 3.2) ∠ACB ¼ 2α, ∠ADB ¼ 2β, ∠BCD ¼ u, ∠ADC ¼ v. It is known that ∠ABC ¼ ∠BCD ¼ u, ∠BAD ¼ ∠ADC ¼ v. From triangles ABC and ABD we deduce that 2α þ u þ v þ ∠CAD ¼ 180 , 2β þ u þ v þ ∠CBD ¼ 180 . Hence, we have that 2α þ u þ v < 180 , 2β þ u þ v < 180 . Summing up these two inequalities, it follows that 2α þ 2β þ 2u þ 2v < 360 ,or∠ECD þ ∠FDC < 180 . If segments CE and DF do not intersect, then CD > EF. This ends the proof of the lemma. Now, we continue the proof of the problem. Let us introduce following notations (see Figure 3.3), AE ¼ a, AH ¼ b, DG ¼ c, BF ¼ d, AB ¼ CD ¼ e, BC ¼ AD ¼ f. We have that

S0 ¼ S À ðÞSAEH þ SBEF þ SCFG þ SDGH ¼ 1 ef 1 ¼ ef À ðÞab þ ðÞe À a d þ ðÞf À d ðÞþe À c ðÞf À b c ¼À ðÞc À a ðÞd À b : 2 2 2

Figure 3.2 CD uv abb a

vu AEF B

Figure 3.3 B d F f-d C

e-a e-c

EM G

A b H f-b D 102 3 Areas

Figure 3.4 BC1 C

B1 M D1

AA1 D

Thus,

S 1 S ¼ À ðÞc À a ðÞd À b : ð3:2Þ 0 2 2

If point M belongs to one of the medians of rectangle ABCD, then c ¼ a or d ¼ b, ¼ S i.e. S0 2. Now, assume that point M does not belong to any median and that point A is the vertex of the rectangle closest to M. Then, it is clear that M belongs to one of the four rectangles (obtained by dividing the given rectangle by medians) which contains vertex A (Figure 3.4). Let us now perform a translation of triangle MCD by a vector CB~ . Then side CD coincides with side BA and according to the first lemma c > a. Similarly, we obtain that d > b (Figure 3.4). < S From aforesaid and (3.2) we get that S0 2. From formula (3.2) it follows that S0 reaches its least value when the expression (c À a)(d À b) has a maximal value. We shall prove that both (c À a) and (d À b) reach their maximal values at point A. Compare the values of the (c À a) at points M, B1 and A. According to the first lemma the value of c at point B1 is greater than the value at point M, while the value of a is less. Consequently the value of the expression (c À a)atB1 is greater than at M. Denote the values of the c and a at points B1 and A by c1, a1 and c2, a2, respectively (Figure 3.5). Applying the second lemma to the case when bisectors drawn from the vertices B1 and A of triangles CB1D and CAD do not intersect, we obtain that a1 ¼ a1 À a2 > c1 À c2 or c2 À a2 > c1 À a1. If the bisectors intersect, then c2 > c1. Hence, it follows that c2 À a2 ¼ c2 > c1 À a1. We have proven that the value of (c À a) at point A is greater than the value at point B1, while the value at point B1 is greater than the value at point M (inside the rectangle). Similarly one can prove that the expression (d À b) reaches its maximal value at point A. Thus, S0 reaches its minimal value when the point M coincides with the vertex A, i.e. a ¼ 0, b ¼ 0. Therefore c ¼ pefffiffiffiffiffiffiffiffiffi, d ¼ pefffiffiffiffiffiffiffiffiffi. f þ e2þf 2 eþ e2þf 2 From formula (3.2) we deduce that 3.1 Inequalities with Areas 103

Figure 3.5 BC

c2 c1 a1

Figure 3.6 A E

A1 M

O FB

! S ef S0 1 À pffiffiffiffiffiffiffiffiffiffiffiffiffiffi : ð3:3Þ 2 ef þ ðÞe þ f e2 þ f 2 þ e2 þ f 2 pffiffiffiffiffi Since e þ f 2 ef , e2 þ f2 2ef, then qffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ef þ ðÞe þ f e2 þ f 2 þ e2 þ f 2 ef 3 þ 2 2 : ð3:4Þ

From inequalities (3.3) and (3.4) we get that ! S ef pffiffiffi S0 1 À ÀÁpffiffiffi ¼ 2 À 1 S: 2 ef 3 þ 2 2

3.1.6. Let us draw ME ⊥ OA, MF ⊥ OB (Figure 3.6). We have that Á ∠ Á ∠ 1 MA MB 1 2 sin A sin B 2 SMEF ¼ ME Á MF Á sin ∠EMF ¼ Á AB Á sin ∠O ¼ 2 AB2 2 sin ∠O Á MA MB 2 ¼ Á SABO Á sin ∠O: AB2 ð3:5Þ 104 3 Areas

(a) According to (3.5) would be minimal if the value of the expression MAÁMB SABO AB2 ðÞÀ 2 ¼ 1 À 1 Á MA MB would be maximal i.e. at ¼ . 4 4 AB2 MA MB

To construct segment AB one has to note that OA1 ¼ A1A, where MA1||OF. 1 Remark 1. SMEF 4 SABO. Indeed, according to (3.5), MAÁMB Á 1 . SMEF AB2 SABO 4 SABO 1 2. SOA1M 4 SOAB. (b) According to (3.5) we have that

1 1 AB S sin 2∠O 1 sin 2∠O þ ¼ ¼ ABO Á ¼ ON Á , MA MB MA Á MB AB SMEF 2 SMEF where ON ⊥ AB. If the straight line passing through point M perpendicular to line OM intersects the sides of the given angle at points A1 and B1, then segment A1B1 is what was required. Indeed, we have that

1 1 sin 2∠O sin 2∠O 1 1 þ ¼ ON Á OM Á ¼ þ : MA MB SMEF SMEF MA1 MB1

But if the straight line passing through point M perpendicular to line OM does not intersect the sides of the given angle, then the required segment cannot be constructed. Indeed, since N 2OK (Figure 3.7), where KM||OE, then ON does not have a maximal value.

Figure 3.7 3.1 Inequalities with Areas 105

Figure 3.8 D2

D3 DD'

D1 C1 C

MB1

AA1 BB3 B2

3.1.7. (a) We have that

1 S ¼ S þ S þ S þ S S A1B1C1D1 MA1B1 MB1C1 MD1C1 MA1D1 4 ABC 1 1 1 þ S þ S þ S ¼ 4 BCD 4 ACD 4 ABD 1 S ¼ S ¼ 2 ABCD 2

S (see the remark 1 of problem 3.1.6a). This means that SA1B1C1D1 2. (b) Draw from points C and M straight lines parallel to diagonal BD (Figure 3.8).

According to the remark 1 of problem 3.1.6a

1 1 1 S þ S S þ S 0 0 < S MA1D1 MB1C1 4 AB3D3 4 CB D 4 AB3D3 1 1 1 þ S ¼ S ¼ S : 4 B2B3D3D2 4 AB2D2 2 ABCD

Since from the statement of the problem, it follows that points C and A are equidistant from line BD, i.e. BD is the midline of triangle AB2D2. Remark 1. If ABCD is a square and point M coincides with C, then

1 S þ S ¼ S ¼ S : MA1D1 MB1C1 CBD 2 ABCD 106 3 Areas

Figure 3.9 B

B1 M G

APC1 C

Figure 3.10 C E

E' D

AFB

2. In general the inequality does not hold true: take in a square ABCD a point M close to point C and move vertices B and D along the sides in the direction of A. 2 2 1 SABCD SABCD 3. One can prove that, in general case, SMA D þ SMB C < max ; . 1 1 1 1 4 SBDC SABD

3.1.8. Draw through point G a straight line B1C1||BC (Figure 3.9). We have that S 1 S Á sin 2∠A (see the solution of problem 3.1.6a), and ÀÁ MGP 4 AB1C1 2 SAB1C1 AB1 4 SABC 2 ¼ ¼ . Therefore, SMGP Á sin ∠A. Similarly, we get that SABC AB 9 9 SABC Á 2∠ SABC Á 2∠ SMGN 9 sin B, SNGP 9 sin C.Thus S ÀÁS S ¼ S þ S þ S ABC Á sin 2∠A þ sin 2∠B þ sin 2∠C ABC : MNP MGP MGN NGP 9 4

(see problem 9.1). 0 0 3.1.9. Let DE k AB (Figure 3.10), then ∠ADE ¼ ∠DAF ¼ ∠DEF and 0 2 ∠ ¼ ∠ Δ Δ SDEF ¼ EF DAE DFE. Hence, it follows that AE D FDE. Thus 0 2 and we SAE D AD 3.1 Inequalities with Areas 107

Figure 3.11 B

E KM DE1

K1 AT P C

0 1 have to prove that SAE D 4 SABC. Note that, this follows from the solution of problem 3.1.6a (see remark 2). 3.1.10. Draw from the point A a straight line parallel to DE (Figure 3.11). Since DK ¼ KM ¼ ME, then AK1 ¼ K1M1 ¼ M1E1. Now by using Menelaus’ theorem for triangle AE1C and straight lines BT, BP, we obtain that

AK BE CT AM BE CP 1 Á 1 Á ¼ 1 ¼ 1 Á 1 Á : K1E1 BC TA M1E1 BC PA

Consequently,

1 S þ S S Á BPT BPC ¼ 2 Á BPC : 2 SABT SABT þ SBPT

Hence, it follows that (SBPT þ SBPC)(SABT þ SBPT) ¼ 4SABT Á SBPC. Á pUsingffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi furtherpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the Cauchy inequality,p weffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi deduce that 4SABT SBPC 2 SBPT Á SBPC Á 2 SABT Á SBPT. Thus SBPT SABT Á SBPC. þ SABT SBPC Remark From the last inequality it follows that SBPT 2 . Hence, we obtain AC that TP 3 . This problem was suggested on the 16th Russian Mathematical Olympiad. PB ¼ λ BQ ¼ μ 3.1.11. Let PA , QC , then using the Cauchy inequality, we deduce that rffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi S S S S 3 SAPR þ 3 QRC ¼ 3 SAPR Á PBQ þ 3 QRC Á PBQ ¼ S S S S S S ABCsffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiABC PBQ sABCffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPBQ ABC λ μ μS λ ¼ 3 SAPR Á 1 Á þ 3 QRC Á Á 1 S 1 þ λ 1 þ μ S 1 þ λ 1 þ μ PBQ PBQ ! 1 λSAPR þ μS 1 μ λ 1 QRC þ þ þ þ ¼ 1, 3 SPBQ 1 þ λ 1 þ μ 1 þ λ 1 þ μ

λ þ μ ¼ þ ¼ since SAPR SQRCpffiffiffiffiffiffiffiffiffiSPBRpffiffiffiffiffiffiffiffiffiffiSQBRpffiffiffiffiffiffiffiffiffiffiSPBQ. 3 3 3 Consequently SAPRþ SQRC SABC. 108 3 Areas

AC1 ¼ α BA1 ¼ β CB1 ¼ γ ¼ Á αðÞÀ γ ¼ 3.1.12. Let AB , BC , CA , then SAB1C1 SABC 1 , SBA1C1 Á ðÞÀ α β ¼ Á γðÞÀ β ¼ ð À αðÞÀÀ γ SABC 1 , SCA1B1 SABC 1 and SA1B1C1 SABC 1 1 βðÞÀ1 À α γðÞÞ1 À β . (a) We have to prove that (1 À α(1 À γ) À β(1 À α) À γ(1 À β))2 4αβγ(1 À α) (1 À β)(1 À γ), or

ðÞ1 À α À β À γ þ αγ þ βγ þ αβ 2 4αβγðÞ1 À α À β À γ þ αγ þ βγ þ αβ À αβγ , ðÞ1 À α À β À γ þ αγ þ βγ þ αβ À 2αβγ 2 0:

< ðÞ; ; (b) Assume that SA1B1C1 min SAB1C1 SBA1C1 SCA1B1 and let SAB1C1 SBA1C1

SCA1B1 . Then, taking into account the solution of the problem 3.1.12a, we get that

S Á S2 > S Á S2 4S Á S Á S ABC AB1C1 ABC A1B1C1 AB1C1 BA1C1 CA1B1 S2 Á 4S > S2 Á ðÞS þ S þ S þ S ¼S Á S2 : AB1C1 CA1B1 AB1C1 AB1C1 BA1C1 CA1B1 A1B1C1 ABC AB1C1

Consequently S Á S2 > S Á S2 . This leads to a contradiction. ABC AB1C1 ABC AB1C1 The Second Solution Consider the midpoints of sides BC, CA and AB. Let these points be A0, B0 and C0, respectively. If among points A1, B1, C1 there exist two points belonging to two of segments AC0, C0B, BA0, A0C, CB0, B0A with a common endpoint, then the statement is evident. Indeed, let C1 2 [C0B] and A1 2 [A0B], then M 2 BN, where M ¼ BB1 \ A1C1, N ¼ A0C0 \ BB1, consequently BM BN ¼ NB1 MB1. Therefore SA1C1B SA1B1C1 . Otherwise, one can assume that C1 2 [C0B], A1 2 [A0C], B1 2 [AB0], then ðÞ; ðÞ; ; SA1B1C1 min SA1B1C0 SA1B1B min SA0C0B1 SC0CB1 SA0B1B 1 minðÞS ; S ; S ¼S : A0C0B0 CC0B0 A0B0B 4 ABC

1 ðÞ; ; Consequently, 4 SABC min SAB1C1 SBA1C1 SCA1B1 .Thus, SA1B1C1 min ðÞ; ; SAB1C1 SBA1C1 SCA1B1 . Yet another solution can be obtained by using the solution of the problem 3.1.12e. (c) According to 3.1.12b we have

ðÞ; ; ¼ðÞ; ; ; min SAB1C1 SBA1C1 SCA1B1 min SAB1C1 SBA1C1 SCA1B1 SA1B1C1 1 S ðÞS þ S þ S þ S ¼ABC: 4 AB1C1 BA1C1 CA1B1 A1B1C1 4 3.1 Inequalities with Areas 109

(d) We have that pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ ¼ αðÞÀ γ þ βðÞÀ α þ γðÞÀ β SAB1C1 SBA1C1 SCA1B1 SABC 11 1 pffiffiffiffiffiffiffiffiffiffi α þ 1 À γ β þ 1 À α γ þ 1 À β Â S þ þ ABC 2 2 2 3pffiffiffiffiffiffiffiffiffiffi ¼ S : 2 ABC

(e) According to Ceva’s theorem αβγ ¼ (1 À α)(1 À β)(1 À γ). Hence

¼ ðÞÀ αðÞÀÀ γ βðÞÀÀ α γðÞÀ β SA1B1C1 SABC 1 1 1 1 ¼ SABCðÞðÞ1 À α ðÞ1 À β ðÞþ1 À γ αβγ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi S ¼ 2S αβγðÞ1 À α ðÞ1 À β ðÞ1 À γ ABC, ABC 4

ðÞÀ 1 since x 1 x 4. According to Stuart’s theorem we have ÀÁ 2 ¼ β 2 þ ðÞÀ β 2 À βðÞÀ β 2 ¼ βðÞÀ β 2 þ 2 À 2 AA1 b 1 c 1 a ÀÁ1 b c a þ β2b2 þ ðÞ1 À β 2c2 βðÞ1 À β b2 þ c2 À a2 þ 2βðÞ1 À β bc ¼ 4ðÞp À a pβðÞ1 À β , therefore pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Á Á Á ðÞþ þ αβγðÞÀ α ðÞÀ β ðÞÀ γ ¼ Á AA1 BB1 CC1 4SABC a b c 1 1 1 2SA1B1C1 Á Á ðÞþ þ AA1 BB1 CC1 a b c , hence SA1B1C1 2ðÞABþBCþAC . ðÞÀ αðÞÀÀ γ βðÞÀÀ α γðÞÀ β 1 þ 1 þ (f) We have to prove that 1 1 1 1 αðÞ1Àγ βðÞ1Àα 1 Þ γðÞ1Àβ 3, or 1 1 1 A ¼ ðÞðÞ1 À α ðÞ1 À β ðÞþ1 À γ αβγ þ þ 3: αðÞ1 À γ βðÞ1 À α γðÞ1 À β

Note that

ðÞ1 À α ðÞ1 À β βγ ðÞ1 À β ðÞ1 À γ A ¼ þ þ α 1 À γ β αγ ðÞÀ α ðÞÀ γ αβ À β þ þ 1 1 þ ¼ 1 1 À α γ 1 À β α β 1 À γ γ þβ À 1 þ À β þ þ γ À 1 þ À γ 1 À γ β 1 À α 1 À α α 1 À β α þ þ α À 1 þ À α ¼ þ γ 1 Àβ α 1 À β β 1 À γ γ 1 À α þ þ þ þ À 3 3, 1 À γ β 1 À α γ 110 3 Areas

þ 1 > since a a 2, where a 0. ¼ ð ; (g) Let SABC 1, then according to problems 3.1.12e and 3.1.12f max SAB1C1 ; Þ1 2 Á ðÞ; ; 1 2 SBA1C1 SCA1B1 4 and S A1B1C1 max SAB1C1 SBA1C1 SCA1B1 4 S A1B1C1 Á Á SAB1C1 SBA1C1 SCA1B1 . α À γ β À α γ À β Let (1 ) (1 ) (1 ) (see the solution of problempffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3.1.12a), then we À αðÞÀÀ γ βðÞÀÀ α γðÞÀ α αβðÞÀ γ ðÞÀ α have toq proveffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi that 1 1 1 1 1 1 or αβγ αβ Á αβγ γðÞÀ β 1 2 1Àβ, 1 4. But this is the statement of problem 3.1.12c. This ends the proof. BM ¼ λ BN ¼ μ ¼ ¼ 3.1.13. Let us denote that AB , BC , SABD S1, SDBC S2. Hence, it follows that

S ¼ S þ S À ðÞ1 À λ 2S À ðÞ1 À μ 2S À λμðÞ¼S þ S KMNL 1 ÀÁ2 1 2 ÀÁ1 2 2 2 2 2 ¼ S1 2λ À λ À λμ þ S2ðÞ2μ À μ À λμ 2λ þ 2μ À λ À μ À 2λμ maxðÞÂS1; S2

maxðÞS1; S2 , since 2λ À λ2 À λμ ¼ λ(1 À λ) þ λ(1 À μ) 0, 2μ À μ2 À λμ 0and2λ þ 2μ À λ2 À μ2 À 2λμ À 1 ¼À(λ þ μ À 1)2 0. 3.1.14. Let the tangents to a circle at point A intersect PB and PC at points M and N, respectively. Denote by MA ¼ a, NA ¼ b, ∠ABC ¼ β, ∠ACB ¼ α, then ∠MBA ¼ ∠MAB ¼ ∠ACB ¼ α, ∠NAC ¼ ∠NCA ¼ ∠ABC ¼ β. α Á β 1ðÞþ 2 sin 2 sin 2 a b ðÞα þ β SPMN ¼ 2 sin 2 2 ¼ 1 ¼ S 1 2 cos α Á cos β Á cos ðÞα þ β ABC 2a cos α Á 2b cos β Á sin ðÞα þ β 2 1 1 ¼ < pffiffi pffiffi ¼ 1, 2ðÞþα þ β ðÞÁα þ β ðÞα À β 2 2 cos cos cos 2 þ 2 2 2

2b cos β ¼ 2a cos α α À β < α þ β because sin β sin α and 0 45 . 3.1.15. Let ∠A ¼ α, ∠B ¼ β, ∠C ¼ γ and the radius of the circumcircle of triangle ∠ ¼ βþγ ∠ ¼ αþγ ABC is equal to R. Then, note that B1A1C1 2 , A1B1C1 2 and ∠ ¼ αþβ A1C1B1 2 . Therefore,

S 2R2 sin α sin β sin γ α β γ ABC ¼ ¼ 8 sin sin sin ¼ S α þ β β þ γ γ þ α 2 2 2 A1B1C1 2R2 sin sin sin 2 2 2 α À β α þ β γ α À β γ γ α À β ¼ 4 cos À cos sin ¼ 4 cos sin À 4sin 2 cos 2 , 2 2 2 2 2 2 2 ÀÁ αÀβ À γ 2 because cos 2 2 sin 2 0. 3.1 Inequalities with Areas 111

βÀγ γÀα SABC 2 SABC 2 Similarly, one can prove that S cos 2 and S cos 2 . A1B1ÀÁC1 A1B1C1 αÀβ βÀγ γÀα Thus, it follows that SABC min cos 2 ; cos 2 ; cos 2 . SA1B1C1 2 2 2 3.1.16. Consider the straight lines which pass through the vertices of a triangle and are parallel to one of the sides of the parallelogram (Figure 3.12). Note that it is sufﬁcient to solve the problem for the cases, when the triangle and the parallelogram have sides belonging to the same straight line (Figure 3.13a, b). 1 In a case of Figure 3.13a we have that SXYZT SAY1Z1T1 2 SABC (see the solution of problem 3.1.6a, remark 2), where Z1T1||AB. In the case of Figure 3.13b, we have 1 that SABC SXCT 2 SXYZT. 3.1.17. (a) Let points A and B be the most removed vertices of a polygon. Then it is clear that the polygon is inside the strip formed by the perpendiculars to the segment AB at points A and B. Consider the strip with a minimum width having boundaries parallel to the segment AB and which contains the given polygon (Figure 3.14). Then the intersection of these two strips contains the polygon and has area 2SABC þ 2SABD 2S.

(b) Let A, B, C be such vertices of the given polygon that SABC is maximal. It is clear that the given polygon is contained in the half-plane ΠA, the boundary of which passes through point A and is parallel to CB (Figure 3.15).

Figure 3.12

B YZ

Y1 YZZ1

AX T T1 C XA BT a b

Figure 3.13 112 3 Areas

The half-planes ΠB and ΠC are deﬁned similarly. The intersection of these three half-planes contain the given polygon and have the area equal to 4SABC. Thus, if S > S SABC 2, then this ends the proof. It remains to consider the case, whenSABC 2. Consider a triangle A1B1C1 with a minimum area having parallel sides to the sides of triangle ABC and which contains the given polygon (Figure 3.16).

Figure 3.14 C

Figure 3.15

PA A B

Figure 3.16 A1 D

A S2 B1 B S1 M

FS3 E C

C1 3.1 Inequalities with Areas 113

Figure 3.17 P F

l1(t) l2(t) AB C S1 S2 E

It is clear that AA , BB and CC intersect at point M. Let MC ¼ k, then S ¼ k 1 1 1 CC1 ABC (S1 þ S2 þ S3), where S1 ¼ SFAC, S2 ¼ SDAB, S3 ¼ SECB, and D, E, F are some vertices of the given polygon. > S > Because SABC 2, consequently k 1. We have that

¼ þ þ þ ¼ SA1B1C1 SABC SA1ACC1 SA1ABB1 SC1CBB1 2k þ 1 ¼ S þ ðÞS þ S þ S < ABC k 1 2 3 < SABC þ 3ðÞS1 þ S2 þ S3 < < 2ðÞSABC þ S1 þ S2 þ S3 2S:

In the absence of some of the points DEF the proof can be done similarly. 3.1.18. (a) Let points A, B be the most removed vertices of the given polygon and C be any point on segment AB (Figure 3.17). ¼ À ¼ AC Consider the function f(t) l1(t) l2(t) on the segment [0; 1], where t AB, while l1(t) and l2(t) are the lengths of the segments, cut off by the given polygon from the midpoint perpendiculars of segments AC and CB, respectively. It is not difﬁcult to understand that the graph of the function f(t) is a broken line and as l1(0) ¼ l2(1) ¼ 0, then we have that f(0) < 0 and f(1) > 0. Hence, it follows that there ¼ ¼ AC0 ¼ exists a certain point C0, such that l1(t0) l2(t0), where t0 AB (i.e. f(t0) 0). Consider the straight line passing through point C and perpendicular to segment AB. Assume that it divides the given polygon into two polygons with areas S1 and S2 (Figure 3.17). 1 Let us draw through points E and F support lines, then S1 l1(t) Á AC. Similarly, we obtain that S2 l2(t) Á CB. Consequently, l1(t) Á AC þ l2(t) Á CB S. For point C0 the quadrilateral EFPQ is a parallelogram, such that

1If the line l has at least one common point with a figure F and whole figure F is located on one side of l, then the line l is called a support line of the figure F. 114 3 Areas

AC þ C B AC Á l ðÞt C B Á l ðÞt S S ¼ 0 0 l ðÞ¼t 0 1 0 þ 0 2 0 : EFPQ 2 1 0 2 2 2 ÀÁ ÀÁ ¼ 1 1 Á AB þ 1 Á AB þ ¼ (b) At t ÀÁ2 we have thatÀÁl1 2 2 l2 2 2 ÀÁS. Note that SEFB SAPQ 3 Á 1 þ 3 Á 1 3 ; 3 8 AB l1 2 8 AB l2 2 4 S. Therefore,ÀÁÀÁ max SÀÁEFB SAPQ 8 S. ¼ 1 ¼ 3 Á 1 þ 1 3S (c) At t 2 we have that SAFPBQE 8 AB l1 2 l2 2 4 . 3.1.19. See the solution of problem 3.1.17b. 3.1.20. Let AB ¼ c, BC ¼ a, AC ¼ b, ∠A ¼ α, ∠B ¼ β, ∠C ¼ γ. We have that

1 1 1 S ¼ S þ S þ S þ S À ab sin 3γ À bc sin 3α À ac sin 3β ¼ A1B1C1 ABC ABC1 BCA1 ACB1 2 2 2 1 1 1 ¼ 4S À ab sin 3γ À bc sin 3α À ac sin 3β: ABC 2 2 2

Because sin3x ¼ sin x(2 cos 2x þ 1), then

1 1 1 ab sin 3γ þ bc sin 3α þ ac sin 3β ¼ S ðÞ2 cos 2γ þ 2 cos 2α þ 2 cos 2β þ 3 ¼ 2 2 2 ABC 2 2 2 ¼ SABCðÞ4cos γ þ 4cos α þ 4cos β À 3 0

(see problem 5.1.1). Consequently, SA1B1C1 4SABC. This ends the proof. 3.1.21. Let AB ¼ c, BC ¼ a, AC ¼ b, ∠A ¼ α, ∠B ¼ β, ∠C ¼ γ. We have that

1 1 1 S þ S þ S ¼ ac sin α þ ab sin β þ bc sin γ AA1B BB1C CC1A 2 2 2 α β γ ¼ sin þ sin þ sin SABC β γ α sinrffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin sin rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! sin α sin β sin γ S 2 Á þ 2 1 Á À 1 Â ABC sin β sin γ sin α sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffi ! sin α sin γ S 4 Á À 1 ¼ 3S , ABC sin γ sin α ABC pffiffiffiffiffi because x þ y 2 xy (x 0, y 0).

2 2 2 3.1.22. We have that S1 ¼ k , S2 ¼ n , S3 ¼ m (Figure 3.18). S ðÞkþmþn 2 S ðÞkþmþn 2 S ðÞkþmþn 2 Consequently 3.1 Inequalities with Areas 115

Figure 3.18 B

S3 S1 X mk S2 An C

Figure 3.19 DE

KMF N C

2 2 2 þ þ ¼ k þ n þ m Â S1 S2 S3 S kþmþn kþmþn kþmþn 2 k 1 2 n 1 2 m 1 S Á À þ Á À þ Á À 3 k þ m þ n 9 3 k þ m þ n 9 3 k þ m þ n 9 2 1 S ¼ S Á 1 À ¼ , 3 3 3

2 2 À 1 since x 3 x 9. 3.1.23. Let us draw through points B, D and F lines parallel to segments AF, EF and AB, respectively (Figure 3.19). Note that SBCD ¼ SBND, SDEF ¼ SDKF, SABF ¼ SBMF. Consequently 2SBDF ¼ SABCDEF þ SMNK SABCDEF. 3.1.24. Denote the points symmetrical to P, Q and R relative to the symmetry center 0 0 0 0 0 0 of hexagon ABCDEF by P , Q and R , respectively. It is clear that points P , Q and R 0 0 0 0 0 belong, respectively, to sides DE, AF and BC. Since Q R#R Q,2 PQ #P Q and RP # 0 ¼ 0 0 0 RP , we have 2SPQR SPR QP RQ SABCDEF (see the solution to problem 3.1.23). 3.1.25. Let ABCDEF be a regular hexagon with a symmetry center O and triangle MNP an inscribed triangle, with side MN passing through point O. One can assume

2AB # CD denotes that the segments AB and CD are parallel and equal. 116 3 Areas

that M 2 CD, N 2 AF and P 2 AB. We have that SMNP max (SMNA, SMNB), SMNA ¼ 1 1 ¼ 1 ¼ þ ¼ þ 2 SAMDN 2 SACDF 3 SABCDEF and SMNB SBON SBOM SBAO SBCO ¼ 1 1 3 SABCDEF. Consequently, SMNP 3 SABCDEF. EC ¼ λ BD ¼ μ ¼ ¼ 3.1.26. Let us introduce the following notations: AC , AB , SBPC x, SPDE y. Let SABC ¼ 1. We have that SBEC ¼ λ, SBCD ¼ μ, SADE ¼ (1 À λ)(1 À μ). As SPDB Á SPEC ¼ SPDE Á SPBC, then

ðÞλ À x ðÞ¼μ À x xy: ð3:6Þ

(a) According to (3.6) and

SADE ¼ ðÞ1 À λ ðÞ¼1 À μ 1 À λ À μ þ x À y, ð3:7Þ

λμðÞÀλ ðÞÀμ λμ we get that y ¼ 1 1 ¼ À λμ. λþμÀλμ λþμpÀλμffiffiffiffi pffiffiffiffiffi λμ Since λ þ μ 2 λμ, then y pffiffiffiffi À λμ. 2À λμ pffiffi pffiffi ðÞ¼ 3À 5 ¼ 5 5À11 ðÞ¼ t À 2 Note that max ft f 2 2 , where ft 2Àt t . ½0;1 pffiffiffiffiffi pffiffi 5 5À11 Therefore SPDE ¼ y f ðÞλμ max ftðÞ¼ , because 0 < λμ < 1. ½0;1 2

(b) Since SBCDE ¼ 2SBPC, then λ þ μ þ y À x ¼ 2x. We have that 3x À y ¼ λ þ μ and 2 ¼ λμ ¼ λ þ μ À λμ (3.6), consequently x 2 . According to (3.7) wep findffiffiffiffiffiffiffiffi that 2x , ¼ λ þ μ À λμ λ þ μ ¼ λμ þ λμ 2y 3pffiffiffiffiffiffiffiffi, consequently,pffiffiffiffiffi 2 , we deduce that λ þ μ ¼ λμ þ 2λμ 2 λμ. From the last inequality follows that ÀÁpffiffiffi 2 pffiffiffi λμ 2 À 2 ¼ 6 À 4 2. pffiffiffiffiffi pffiffiffi pffiffiffi ¼ λþμÀ3λμ ¼ 2λμÀ2λμ À λμ À We have thatpffiffiffiy 2 2 5 2 7, because 6 4 2. Thus SPDE ¼ y 5 2 À 7. QC ¼ λ CP ¼ μ 3.1.27. Denote AC and CB . Let point G be the centroid of triangle ABC.We S SCQG 2 SCGP 2 have that SCAM ¼ SBCM ¼ , where S ¼ SABC. Because ¼ λ, ¼ μ and 2 pffiffiffiffiffi SCAM 3 SCBM 3 SQCP ¼ λμ λμ ¼ λ þ μ λμ λμ λμ 4 S , we have 3 . Therefore 3 2 , hence 9. þ þ 7 λμ þ S ðÞþÀ λ We need to prove that SQCP SAQM SBMP 9 S,or S 2 1 S ðÞÀ μ 7 2 1 9 S. λμ þ 1 ðÞþÀ λ 1 ðÞ¼À μ À λμ 7 Indeed ÀÁ2 1 2 1 1 2 9. It remains to note that ¼ À þ þ À 7 ¼ 2 2 SQMP S SQCP SAQM SBMP S 9 S 9 S, consequently SQMP 9 S. 3.1.28. It is clear that the cross section of the cube by a plane passing through its center is a centrally symmetric convex polygon, which has an even number of sides. Thus, it is either a quadrilateral (Figure 3.20a), or a hexagon (Figure 3.20b). 2 In the first case (Figure 3.20a), we have that Sсеч ¼ bh a Á a ¼ a . 3.1 Inequalities with Areas 117

b D b B a c d g c h bd E

b A a F a b

Figure 3.20

Figure 3.21 BS2 C

A D

y i yi¢

In the second case (Figure 3.20b),pffiffiffi we have that Sсеч ¼ bd sin α þ bc sin β þ cd sin γ. Since 2b,2c,2d 2a and α þ β þ γ ¼ π, α β γ < π then according to the problem 5.2.3 in the case , , 2, we obtain that a2 ðÞα þ β þ γ > 2 Sсеч 2 sin sin sin a . γ π 2 a2 þ a2 ¼ 2 > Á 2 If 2, then DE 2 2 a , consequently Sсеч SABDE DE a a . 3.1.29. It is clear that one can leave two of the given strips as they are and move the remaining k strips. Let coordinate straight line be parallel to non of the boundary lines of these strips. Denote by xi (i ¼ 1, 2, . . . , k) coordinates of the intersection points of the line li with the coordinate line, where the straight line li is parallel to the bounding lines of the i-th strip and is equidistant from them. It is clear that there exist numbers ai and bi, such that ai xi bi [because the intersection of each of these k strips with parallelogram ABCD is not empty (Figure 3.21)], where ABCD is the intersection of two not moved strips. 118 3 Areas

b b x x x 0 a g a 180 -g

x b b=0 0 a a 180 -d d

a b

Figure 3.22

It is clear that to each ﬁgure which is the intersection of all these strips, corresponds a set of numbers x1, x2,...,xk. Denote the area of all these k þ 2 strips by S(x1, x2, ..., xk). If all these strips do not have common points we will set S(x1, x2, ..., xk) ¼ 0. We shall prove that there exists a constant number M, such that at any 0 2½; ¼ xi, xi ai bi , i 1, 2, . . . , k the following inequality holds true ÀÁ ÀÁ ðÞÀ; ; :::; 0 ; 0 ; :::; 0 j À 0 jþj À 0 jþ::: þj À 0 j : Sx1 x2 xk Sx1 x2 xk M x1 x1 x2 x2 xk xk ð3:8Þ ÀÁÀÁ ; :::; ; ; ; :::; À ; :::; ; 0; ; :::; þ Note that Sy1 yiÀ1 yi y iþ1 yk Sy1 yiÀ1 yi yiþ1 yk S1 0 À Á S2 (Figure 3.21), and S1, S2 yi yi d, where d is the longest diagonal of the þ 0 À Á parallelogram ABCD. Then S1 S2 2 yi yi d, and thus ÀÁ 0 0 0 SxðÞÀ1; x2; :::; xk Sx; x ; :::; x 1ÀÁ2 k ÀÁÀÁ 0 0 0 0 SxðÞÀ1; x2; :::; xk Sx; x2; :::; xk þ Sx; x2; :::; xk À Sx; x ; x3:::; xk þ ÀÁ1 ÀÁ1 1 2 0 0 0 0 0 0 þ ::: þ Sx; x ; :::; x À ; xk À Sx; x :::; x 1 2 k 1 1 2 k ÀÁ À 0 þ À 0 þ ::: þ À 0 ¼ j À 0 jþj À 0 jþ::: þj À 0 j 2d x1 x1 2d x2 x2 2d xk xk M x1 x1 x2 x2 xk xk , where M ¼ 2d. 0 0 ::: 0 Using the inequality (3.8) one can prove that there exist numbersÀÁx1, x2, , xk, 2 ¼ ðÞ; ; :::; 0; 0; :::; 0 such that for any xi [ai, bi], i 1, 2, . . . , k, Sx1 x2 xk Sx1 x2 xk . 0 0 ::: 0 We shall prove that all strips to which correspond numbers x1, x2, , xk have a common center of symmetry. Proof by contradiction argument. ÀÁ 0; 0; :::; 0 ¼ Let certain two parallel sides of the intersection with the area Sx1 x2 xk S be not equal. Let a > b (Figure 3.22). 3.1 Inequalities with Areas 119

Figure 3.23 y

E a K ON x

Let us shift the strip, one of the boarders of which contains a,byx (Figure 3.22). 0 Denote the area of intersection of new strips by S . Then for sufﬁciently enough small x we have that

2a þ xtgðÞβ þ tgα 2b À xtgðÞγ þ tgδ S0 À S ¼ x À x ¼ 2 2 2a À 2b þ xtgðÞβ þ tgα þ tgγ þ tgδ ¼ x > 0, 2

0 i.e. S > S. This leads to a contradiction. а ⊥ ⊥ 3.1.30. ( ) Let AE MN, OK MNp(Figureffiffi 3.23).pffiffi ∠ ¼ α ¼ 2 ¼ 2 α À α < π Denote AOK , then OA 2 and OK 2 cos R, where 0 4. ¼ OK ¼ OK Note that ON ðÞπÀα , OM ðÞπþα , consequently, cos 4 cos 4 pffiffi 2 2 α À 2 cos R S ¼ : MON cos 2α

ÀÁpffiffi pffiffi 2 pffiffi 2 cos αÀR 2 < 2 2 2 À (1) For 0 R 4 we have to prove that cos 2α 2 R ,or pffiffi pffiffiffiffiffiffiffiffiffiffi pffiffi pffiffiffiffiffiffiffiffiffiffi pffiffi ðÞcos αÀ cos 2α ðÞcos αÀ cos 2α 2 pffiffiffiffiffiffiffiffiffiffi R (α 6¼ 0). We need to prove that 2 pffiffiffiffiffiffiffiffiffiffi 2,or 2 1À cos 2α pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1À cos 2α 4 2 cos α À 1 cos 2α, i.e. (cosα À 1)2 0. (2) We shall prove that for 0 < R < 1 the following inequality holds true ÀÁ 2 pffiffi 2 2 cos αÀR ÀÁpffiffiffi 2 2 1 À 2 α À 1 cos 2α 4 R ,or 2R cos 2 0. pffiffi For 2 < R < 1 the equality may hold true, since p1ffiffi < p1ffiffi < 1, hence there 4 2 2 2 2R exists such an angle α, for which cos α ¼ p1ffiffi . 2 2R 120 3 Areas

B1 C1 B1 WC1 B1 C1 V A1 D1 P A1 A1 ND1 D1 S MN O K U BCBCBC b M O1 RT K ADADAQD a b c

Figure 3.24

(b) According to problem 3.1.29, if one shifts the plane parallel to itself so that it passes through the center of the cube, then the area of the cross section does not decrease. Thus, it is sufficient to make the proof for the case of dissection of the cube by a plane passing through its center. Consider the following cases (Figure 3.23a, b, c). pffiffiffi ¼ ðÞ¼; I. (Figure 3.24a) Ssec 2SANC1 2max SAD1C1 SADC1 2. ¼ ðÞ¼; ðÞ; II. (Figurepffiffiffi 3.24b) Ssec 2SNKP 2max SNKC1 SNKC max SANC1K SA1NCK 2. III. Let OO1 ⊥ ABC, where O is the center of the cube, O1K ⊥ QT and O1K ¼ R, ∠O1KO ¼ β (Figure 3.24c).

ÀÁpffiffi pffiffi 2 À 1À2 2ÀR < 2 ¼ 1 2STQD 2 If 0 R 4 , then Ssec cos β pffiffiffiffiffiffiR (see problem 3.1.30 a,1). 1þR2 4 pffiffi 2 pffiffi pffiffiffi pffiffi We need to prove that 1 À 2 2 À R pffiffiffiffiffiffiffiffi2R ,or2 2R À 2R2 p2ffiffiffiffiffiffiffiffiffiffi2R , 2 1þR2 þ 2 pffiffiffiffiffiffiffiffiffiffi 4 1 4R pffiffiffi pffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2À 2 À R pffiffiffiffiffiffiffiffiffiffi2 ,or 2 p1 ffiffiffiffiffiffiffiffiffiffi4R 1 R,or4 2R 1 þ 4R2 þ 1 þ 4R2. 2 2 1þ4R pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ4R pffiffiffi Indeed, 1 þ 4R2 þ 1 þ 4R2 > 4R þ 2R ¼ 6R > 4 2R. pffiffi À 1À2ðÞ1ÀR2 2 < 1 ¼ 1 2STQD 4 IV. If 4 R 2, then Ssec cos β pffiffiffiffiffiffiR (see problem 3.1.30a) 1þR2 4 pffiffiffiffiffiffiffiffiffiffi ÀÁ pffiffiffi 1 1þ4R2 1 þ 2 2 þ 2 þ (3) We need to prove that 2 2R R 2,or(1 x) (1 x) 8x, where 1 < ¼ 2 þ 3 À ¼ À 2 þ À 2 x 4R 1. Indeed, we have that (1 x) 8x (x 1)(x 4x 1) 0, À 2 þ À > 1 þ À > because x 1 0, x 4x 1 4 2 1 0. qffiffiffiffiffiffiffiffi pffiffi pffiffiffi pffiffiffi 1þR2 1 < 2 < 1 ¼ 4 < < V. If 2 R 2 , Ssec cos β R 2. Hence Ssec 2 3.1.31. (a) Consider Figure 3.25. 3.1 Inequalities with Areas 121

Figure 3.25 A

a-x x

BD C

We have that SABD þ SADC ¼ SABC, consequently AB Á AC sin α AD ¼ :ðÞ* AB sin x þ AC sin ðÞα À x

Taking into account (*), we obtain that AB1 Á AC1 ABn Á ACn AD1 þ ::: þ ADn ¼ þ ::: þ AB1 sin x þ AC1 sin ðÞα À x ABn sin x þ ACn sin ðÞα À x Â sin α, where ∠B1AC1 ¼ α, ∠B1AD1 ¼ x. For positive numbers a1,...,an, b1,...,bn one can prove by the method of mathematical induction that

a b a b ðÞa þ ::: þ a ðÞb þ ::: þ b 1 1 þ ::: þ n n 1 n 1 n : ð3:9Þ a1 þ b1 an þ bn a1 þ ::: þ an þ b1 þ ::: þ bn

Using further the inequality (3.9) for numbers ai ¼ ABi sin x and bi ¼ ACi sin (α À x)at0< x < α, we deduce that

ðÞAB1 þ ::: þ ABn ÁðÞAC1 þ ::: þ ACn AD1 þ ::: þ ADn sin α ðÞAB1 þ ::: þ ABn sin x þ ðÞAC1 þ ::: þ ACn sin ðÞα À x AB Á AC sin α ¼ ¼ AD, AB sin x þ AC sin ðÞα À x where AB ¼ AB1 þ ...þ ABn, AC ¼ AC1 þ ...þ ACn. One of angles ∠BDA and ∠CDA is not acute. Therefore AD < max (AB, AC), hence

AD1 þ ::: þ ADn AD < < maxðÞAB1 þ ::: þ ABn; AC1 þ ::: þ ACn :

0 0 0 (b) Let B , E , A be points symmetrical to B, E, A relative to point M (Figure 3.26). 122 3 Areas

Figure 3.26 C

BА'

EM E' F AB'

D B1

C1 M N1 K L 1 L1

D1 B

CN К A A1 a b

Figure 3.27

0 0 0 From problem 3.1.31a we have that ME þ MF max (MC þ MA , MB þ MD), or EF max (AC, BD). (c) Let the section of a tetrahedron ABCD be a quadrilateral KLMN (Figure 3.27a). We need to prove that SKLMN max (SAMC, SDKB). Let us project the tetrahedron on a plane perpendicular to KM (Figure 3.27b). ¼ 1 Á Since SKLMN 2 KM L1N1 andÀÁ from problem 3.1.31b we have that L1N1 max 1 Á ; 1 Á ðÞ; (A1C1, B1D1), then SKLMN max 2KM A1C1 2KM B1D1 max SAMC SBKD . It remains to note that SAMC max (SACD, SABC) and SBKD max (SABD, SCBD), (see remark to the solution of problem 2.4.4.). In the case of triangular section one can assume that it is passing, e.g., through vertex D, i.e., points L and M have coincided with point D.

3.1.32. Let points C1, B1, A1, D1 be H images of points C, B, A, D (Figure 3.28). 3.1 Inequalities with Areas 123

B1 D AC×CD

AB×CD H- image BD×AC CАC1 BC×CD A1

BC×AD BC×AC B

Figure 3.28

Figure 3.29 C2 CC1 D

Δ1 B1 D1 B2 Δ BE2 2

A1 E E1

A2 AF1 F

(a) If ∠A þ ∠C > 180 ,then∠B þ ∠D < 180 . Hence C1B1A1D1 is a convex quad- þ > ¼ 2 Á ¼ rilateral, thus SA1B1C1 SA1C1D1 SC1B1D1 . SA1B1C1 CD SABC, SA1C1D1 2 Á ¼ 2 Á 2 Á þ 2 Á > 2 Á BC SACD, SC1B1D1 AC SBCD. Consequently CD SABC BC SACD AC SBCD. ∠ þ ∠ ¼ þ ¼ 2 Á þ 2 Á (b) If A C 180 , then SA1B1C1 SA1C1D1 SC1B1D1 . Thus CD SABC BC S ¼ AC2 Á S . ACD BCD (c) If ∠A þ ∠C < 180 , then ∠B þ ∠D > 180 , hence point A1 is inside the triangle þ < C1B1D1, consequently SA1B1C1 SA1C1D1 SC1B1D1 . Therefore, it follows that 2 2 2 CD Á SABC þ BC Á SACD < AC Á SBCD.

3.1.33. Consider a hexagon A2B2C2C1E2F1, where CC2 ¼ AF1, AA2 ¼ CC1, C1E2|| A2B2||DE, C2B2||F1E2||BC (Figure 3.29). We have that the opposite sides of the hexagon A2B2C2C1E2F1 are parallel to ¼ AFþCD ¼ ¼ ¼ each other and C1C2 2 A2F1. Consequently, C1E2 B2A2, E2F1 B2C2. Let points E and E3 be symmetric to each other with respect to point E2, then ¼ ABþDE ¼ BCþEF ABCE3 is a parallelogram. Therefore, C1E2 2 , F1E2 2 . One can easily prove that B2C1||A1D1 and B2F1||B1E1, B2C1 ¼ A1D1, B2F1 ¼ B1E1. This ends the proof of the point (a) of the problem. 124 3 Areas

Figure 3.30 d

cc1 b a

Figure 3.31 NdCP

N1 P1 D

cO B M1 Q1

MA Q

1 1 ¼ ¼ ¼ ðÞþ Δ þ Δ We have that S1 SC1F1B2 2 SA2B2C2C1E2F1 2 SABCDEF S 1 S 2 (see the Δ ¼ Δ ¼ À solution of problem 3.1.23). Note that 1 2 and SABCDEF 2SBDF 4SΔ1 . Thus 1 ¼ À Δ ¼ À > S1 SBDF S 1 SBDF and SABCDEF 4S1 2SBDF. Finally we have S1 2 SBDF. 3.1.34. Let a b, c d, then b > d. Without loss of generality, one can assume that the rectangle with sides a and b is inscribed into the rectangle with sides c and d (Figure 3.30). Indeed, as c1 c and d1 d, then b > d d1. Hence, if we prove that 2ab < c1d1, then 2ab < cd. It is clear that after symmetry transformation with a center at point O (O is the center of symmetry of the rectangle with sides a and b) the rectangle with sides c and d transforms into itself. Hence, the centers of these rectangles coincide. Since the vertices of thepffiffiffiffiffiffiffiffiffi rectangle with sides a and b are on the circle with the center at a2þb2 O and a radius 2 , then it is clear that the positioning shown in Figure 3.31 is possible. To conclude the proof it remains to draw through the vertices A, B, C, D of the rectangle with the sides a and b a straight line, parallel to the sides of the rectangle with the sides c and d. We have that

¼ < þ þ þ ¼ þ þ þ ab SABCD SBCQ1 SADN1 SCDP1 SABM1 SBCN SADQ SCDP SABM ¼ cd À ab:

Thus 2ab < cd. 3.1.35. (a) If O is on AC, then ABCD, AKON, and OLCM are similar. Therefore pffiffiffiffi pffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffi pSffiffi1 þ pSffiffi2 ¼ AO þ CO ¼ 1. Consequently, S þ S ¼ S. But if O does not S S AC AC 1 2 belong to AC, then one can assume that O and D are on the same side of AC. 3.1 Inequalities with Areas 125

Denote the intersection points of a straight line passing through O with lines BA, OW ¼ AD, CD and BC, respectively, W, X, Y and Z. Assume that W X A. Then OX 1, OZ > and OY 1. Now we rotate the line around point O in such a way that it does not OW > pass through B until it reaches the position, such that Y Z C.ThenOX 1, while OZ ¼ OY 1. Therefore there exists an intermediate position of the line, such that OW ¼ OZ. Let P , P , Q and Q denote the areas of triangles WKO, OLZ, ONX and OX OY 1 2 1 2 pffiffiffi pffiffiffiffiffi pffiffiffiffiffi þ OMY, respectively.pffiffiffiffiffiffiffiffiffi The inequality S S1 S2 is equivalent to the inequality þ T1 T2 2 S1S2. pffiffiffiffiffiffiffiffiffiffiffi ÀÁ ÀÁ pffiffiffiffiffiffiffiffiffiffi 2 2 þ ¼ þ P1 ¼ OW ¼ OX ¼ Q1 þ ¼ T1 T2 2 P1P2 2 Q1Q2 and , then T1 T2 pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiP2 OZ pffiffiffiffiffiffiffiffiffiOY Q2 þ ¼ ðÞþ ðÞþ 2 P1P2 2 Q1Q2 2 P1 Q1 P2 Q2 2 S1S2 (IMO, 1995, proposed problems, Latvia). (b, c) Let O belong to triangle ACD and SABC ¼ a, SACD ¼ b and SOAC ¼ x.We = = = have that T1 2 Á T1 2 ¼ BK Á BL ¼ T1 2 , Therefore SAOB SBOC AB BC SABC

2S Á S T ¼ AOB BOC : 1 a

Á ¼ 2SAOD SCOD Similarly we obtain that T2 b . Consequently, pffiffiffiffiffi pffiffiffiffiffi SOAB þ SOBC SOAD þ SOCD a þ x b À x T1 þ T2 pffiffiffiffiffi þ pffiffiffiffiffi ¼ pffiffiffiffiffi þ pffiffiffiffiffi 2a 2b 2a 2b pffiffiffi pffiffiffi pffiffiffi pffiffiffi a þ b a À b ¼ pffiffiffi À pffiffiffiffiffiffiffiffi x: 2 2ab pffiffi pffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffi If a b, then T þ T apffiffi b a þ b ¼ S. 1 2 2 If a < b, then, since point O cannot be outside of parallelogram ABCE, then pffiffiffiffiffi pffiffiffiffiffi pffiffi pffiffi pffiffi pffiffi pffiffi pffiffi apþffiffi b paÀffiffiffiffiffiffib apþffiffi b x a. Therefore we deduce that T1 þ T2 À a ¼ À pffiffi pffiffi pffiffiffiffi 2 2ab 2 À pffiffiffi pa ffiffiffiffi b a ¼ bþ2pffiffiffiffiabÀa. 2b 2b pffiffiffi þ À ÂÃ b 2pffiffiffiab a ðÞαþπ= a ¼ 2α α2 ; π pffiffiffiffiffiffi2b ¼ sin 2 4 Denote tg , where 0 . Then þ α C0. b pffiffiffiffiffi pffiffiffiffiffi 4 pffiffiffiffi a b pffiffiffiffiffiffiffiffiffiffifficos pffiffiffi bþ2pffiffiffiffiabÀa π Consequently, T1 þ T2 C0 a þ b ¼ C0 S. When α ¼ , 2b pffiffiffiffiffi pffiffiffiffiffi pffiffiffi 4 sinðÞ 2αþπ=4 ¼ þ cos α 1, i.e. C0 1. Thus, in all cases T1 T2 C0 S. Note that if in a quadrilateral the conditions AB ¼ BC, AD ¼ CD are satisfied and SABC 2 π ABCO is a parallelogram, then using the condition ¼ tg α0, where 0 α0 pffiffiffiffiffi pffiffiffiffiffi SACDpffiffiffi 4 sinðÞ 2α0þπ=4 and C0 ¼ α we obtain that T1 þ T2 ¼ C0 S. cos 0 ÀÁ α π α þ π < 5 α We need to prove that, if 0 4, then sin 2 4 4 cos . This ends the proof of the point (b)ÂÃ of the problem. φ2 ; π φ ¼ 4 α < φ Indeed,ÀÁ let 0 4 and cos 5, then at 0 , we have that α þ π < ¼ 5 φ < 5 α φ α π φ ¼ 3 > sin 2 4 1 4 cos 4 cos . But if 4, then tg 4 126 3 Areas

Figure 3.32 AD1 B

O M1 C1 М3

K M2 D

B1 C pffiffiffi ÀÁ ÀÁpffiffi 2 À 1 ¼ tg π. Therefore, φ > π and sin 2α þ π sin 2φ þ π ¼ 2 Á 31 < pffiffi 8 8 4 4 2 25 2 Á 5 α 2 4 1, 25 cos . RemarkÀÁUsing the concept of a derivative it is possible to prove that tg 2α þ π tg α þ 2 ¼ 0, or tg3α þ 3tgα À 2 ¼ 0. Consequently, tg α ¼ ppffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 4 ppffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0ffiffiffi 0 0 0 3 3 2 þ 1À 2 À 1¼ 0, 59 ..., then C0 ¼ 1, 11.... 3.1.36. (Solution of O. Sarkisyan, 9th grade) Let the diagonals of the quadrilateral AO ¼ λ BO ¼ μ ABCD intersect at point O (Figure 3.32) and OC , OD . Let us express S0 through S, λ, μ. Since segment AA1, BB1, CC1 and DD1 medians of triangles ABC, BCD, ACD and ABD, then S S ¼ S þ S ¼ S þ S ¼ : AA1CC1 AA1C CC1A ABA1 CDC1 2 Similarly, S S ¼ S þ S ¼ : BB1DD1 BCB1 DAD1 2 Consequently, ¼ þ ¼ þ þ þ : S SAA1CC1 SBB1DD1 SABA1 SBCB1 SCDC1 SDAD1 ¼ þ þ þ Hence, we obtain that S0 SBA1M1 SCB1M2 SDC1M3 SAD1M4 . Let us now calculate SBA1M1 . From the similarity of triangles BMM1 and B1A1M1 we find that BM BM 1 ¼ : ð3:10Þ M1B1 A1B1

From the similarity of triangles AMO and AA1K, we have that 3.1 Inequalities with Areas 127

þ AK ¼ A1K ¼ BM MO ¼ 1 þ BM : AO MO 2MO 2 2MO

On the other hand, we have that

AK AO þ OK OC 1 ¼ ¼ 1 þ ¼ 1 þ : AO AO 2AO 2λ

þ 1 ¼ 1 þ BM ¼ λþ1 Therefore, 1 2λ 2 2MO, from which it follows that BM 2λþ1 BO. From (3.10) we deduce that λ þ BM1 ¼ BM ¼ 1 Á BO ¼ M1B1 A1B1 2λ þ 1 BD=2 ðÞλ þ μ ¼ 2 2 : ðÞ2λ þ 1 ðÞμ þ 1

Consequently,

ðÞλ þ μ BM1 ¼ 1 ¼ 2 1 : M B BB1 1 þ 1 1 4λμ þ 3μ þ 2λ þ 1 BM1

ðÞλþ μ SBM1A1 BM1ÁBA1 BM1 1 SABD AO Since ¼ Á ¼ ¼ λμþ μþ λþ and ¼ ¼ λ, then SBCB1 BB1 BC 2BB1 4 3 2 1 SBCD OC

ðÞλ þ 1 μ ðÞλ þ 1 μ S S ¼ S ¼ BCD BM1A1 4λμ þ 3μ þ 2λ þ 1 BCB1 4λμ þ 3μ þ 2λ þ 1 2 μ ¼ Á S ¼¼ f ðÞÁλ; μ S, 24ðÞλμ þ 3μ þ 2λ þ 1

y where fxðÞ¼; y ðÞþ þ þ . 24xy 3y 2x 1 ÀÁ Similarly, we obtain that S ¼ f μ; 1 S, S ¼ f 1; 1 S, CM2B1 λ DM3C1 λ μ ¼ 1; λ SAM4D1 f μ S. Therefore μ λ ¼ 1 þ 1 þ þ S0 λμ þ μ þ λ þ μ þ þ μλ þ λ þ λ þ μ þ λμ 2 4 3 2 1 4 3 2 4 3 2 ð : Þ λμ 3 11 þ Á S: 4λ þ 3λμ þ 2 þ μ

Denote 4λμ þ 3μ þ 2λ þ 1 ¼ a1,4μ þ 3 þ 2μλ þ λ ¼ a2,4þ 3λ þ 2μ þ λμ ¼ a3. À À À À À À þ λ ¼ 2a3 a2 5 μ ¼ 2a2 a1 5 λμ ¼ 2a1 a2 a3 5 Hence, it follows that 5 , 5 , 5 . Now we have to prove that 128 3 Areas

1 2a À a À 5 1 2a À a À 5 2a À a À a þ 5 2 < 2 1 þ þ 3 2 þ 1 2 3 ¼ A : 3 5a1 a2 5a3 5a4 5

It is not difﬁcult to prove that

(a) a1 þ a3 ¼ a2 þ a4, À ¼À1 ðÞðÞþþ À (b) a1a3 a2a4 2 5 a2 a4 a1a2 2a2a3 , (c) 3a1a3 À 2a2a4 > 0, (d) 3a2a4 À 2a1a3 > 0. À 2 Consider now the expression A 5. By reducing to a common denominator and regrouping the similar members, using the property (a) and presenting the expres- þ À À 2 ¼ sion a1a2a3a4 in the form a1a2a3(a1 a3 a2), we deduce that A 5 ðÞÀþ ðÞÀþ ðÞþÀ ðÞÀ 5a1a3 a2 a4 5a2a4 a1 a3 2a2a3 a1a3 a2a4 a1a2 a1a3 a2a4 . 5a1a2a3a4 Replacing in the obtained expression a2 þ a4 by a1 þ a3 we obtain that ðÞÀ ðÞðÞþþ À A À 2 ¼ a1a3 a2a4 5 a1 a3 a1a2 2a2a3 . According to the property (b), from the last 5 5a1a2a3a4 ðÞÀ 2 expression we obtain that A À 2 ¼À2 a1a3 a2a4 0. Thus A 2. Note that the 5 5a1a2a3a4 5 equality holds true if and only if a1a3 À a2a4 ¼ 0. However from the condition (a) it follows that 2(a1a3 À a2a4) ¼ (a2 À a1)(a2 þ a1 À a3 À a4). Since a1a3 À a2a4 ¼ 0, then either a2 À a1 ¼ 0ora2 þ a1 À a3 À a4 ¼ 0. If a2 ¼ a1, then we obtain that λμ þ μ þ λ þ ¼ μ þ þ λμ þ λ μ ¼ 2Àλ 4 3 2 1 4 3 2 ,or 2λÀ1. But if a2 þ a1 ¼ a3 þ a4, then accounting to (a) it follows that a1 ¼ a4, λμ þ μ þ λ þ ¼ λ þ λμ þ þ μ μ ¼ 2λþ1 i.e. 4 3 2 1 4 3 2 or λþ2 . Now, let us note that from a2 À a1 ¼ 0 (or a2 þ a1 À a3 À a4 ¼ 0) and the condition (a) we deduce that a1a3 À a2a4 ¼ 0. Therefore the last condition is equivalent to μ ¼ 2Àλ μ ¼ 2λþ1 either 2λÀ1 or λþ2 . ðÞÀ 2 Now, we need to prove that A > 1.Since A ¼ 2 À 2 a1a3 a2a4 , then it 3 5 5a1a2a3a4 ðÞÀ 2 13a a a a À6a2a2À6a2a2 remains to prove that 2 À 2 a1a3 a2a4 > 1,or 1 2 3 4 1 3 2 4 > 0, or 5 5a1a2a3a4 3 15a1a2a3a4 ðÞÀ ðÞÀ 3a1a3 2a2a4 3a2a4 2a1a3 > 0. 15a1a2a3a4 The last inequality is holds true due to conditions (c) and (d). 3.1.37. Note that (see Figure 3.33). pffiffiffi 2 2 ÀÁ2 2 2 c a 2 a þ b þ c 2 3 O O 2 ¼ þ À ac cos 60 þ β ¼ À S: 2 3 3 3 3 6 3 pffiffi 2 2 a2þb2þc2 2 3 Similarly we find that O1O3 ¼ O1O2 ¼ À S. pffiffi pffiffi 6 ÀÁ3 pffiffi ¼ 3 2 ¼ 3 c2 þ a2 À 2 À β þ 4 3 Thus SO1O2O3 4 O2O3 4 3 3 3ac cos 60 3 S S.

3.1.38. Let A1B1C1 and A2B2C2 be two nonintersecting triangles. Note that there exists a straight line l, containing one of the sides of one of these triangles, so that triangles A1B1C1 and A2B2C2 are on different sides of this line. Indeed, let O be any point inside triangle A1B1C1 and k be the minimal positive number at which 3.1 Inequalities with Areas 129

Figure 3.33 B O2 O3 b ca

A b C

А2

B¢ B¢ B2 A¢ A¢ B1 A1 B1 A1 C2 А2

C1 B2 C1 l

l C2 C¢ C¢ a b

Figure 3.34

0 0 0 0 0 0 triangles A B C and A2B2C2 have a common point, where A B C is the image of triangle A1B1C1 under similarity transformation with a center O and similarity coefﬁcient k. The existence of line l follows from Figure 3.34. Let A1B1C1 and A2B2C2 be two given triangles. Note that if line l contains sides A1C1 or B1C1, then parallelograms A1C1B1D1 and A2C2B2D2 do not have any common internal point, while if line l contains side A1B1, then hexagons A1M1N1B1K1P1 and A2M2N2B2K2P2 do not have common internal points [here MiNi and PiKi are midlines of triangles AiBiCi and AiBiDi (i ¼ 1, 2), respectively (Figure 3.35)]. From the aforesaid it follows that, if given triangles AiBiCi, i ¼ 1, 2, . . . , n do not have any common internal point, then any two of the hexagons AiMiNiBiKiPi, i ¼ 1, 2, . . . , n also do not have any common internal point. Pn Pn ¼ 3 ¼ 3 Therefore, S0 SAiMiNiBiKiP 2 SAiBiCi 2 S. i¼1 i¼1 130 3 Areas

Figure 3.35 D2

P2 K2

A2 B2 D1 M2 N2 P1 K1 A1 C2 B1 l

M1 N1

Figure 3.36 N

A1 C1

FE1 D

MEK

3.1.39. Let us draw through points A, C and E straight lines MN, NK and MK parallel to BF, BD and DF, respectively (Figure 3.36). Then ΔBFD ΔMNK, and let MN ¼ λBF. Note that

2 λ SBDF ¼ SMNK ¼ ¼ SBDF þ SMNBF þ SNKDB þ SMKDF: Thus, it follows that

ðÞλ þ 1 BF ðÞλ þ 1 BD ðÞλ þ 1 DF ÀÁ AA þ CC þ EE ¼ λ2 À 1 S : 2 1 2 1 2 1 BDF Therefore, we deduce that

BF Á AA BD Á CC DF Á EE S ¼ S þ 1 þ 1 þ 1 ¼ λS : ABCDEF BDF 2 2 2 BDF

We have that R > r1 (see the solution of problem 7.1.79), where r1 is the radius of the incircle of triangle BDF. Hence, we obtain that 3.1 Inequalities with Areas 131

Figure 3.37 A

BM E

D C

R R pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi SABCDEF SBDF ¼ SACE Á SBDF: r1 r Remark 1. If center O of the circumcircle of a triangle ACE is in a convex hexagon ¼ þ þ BFÁR þ BDÁR þ FDÁR ¼ ABCDEF, then SABCDEF SABOF SBCDO SDEFO 2 2 2 R SBDF. r1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2. If ABCDEF is a convex hexagon, then S R Á R1 Á S Á S , where r, ABCDEF r r1 ACE BDF

R and r1, R1 are the inradius and circumradius of triangles ACE and BDF, respectively. 3.1.40. Denote given five points by A, B, C, D, E. If the pentagon with vertices at A, B, C, D, E is not convex, then we can assumepffiffiffi that point D is inside triangle ABC, then SABC ¼ SABD þ SDBC þ SDAC > 6 > 5 þ 1. Consider now the case when pentagon ABCDE is convex. Let M be the point of intersection of BE with AC (Figure 3.37). pffiffi pffiffi pffiffi pffiffiffi ME 1þ 5 SACE ME 1þ 5 1þ 5 Suppose that , then ¼ , SACE SABC > 1 þ 5. BM pffiffi2 SABC BM 2 2 pffiffi ME 1þ 5 SBMD BM 5À1 In the case < , SMDE min (SCDE, SADE) > 2, ¼ > , SBMD > pffiffi pBMffiffiffi 2 SMDEpffiffiffi ME 2 pffiffiffi 5À1 ¼ À ¼ þ > À þ ¼ þ 2 SMDE 5 1. Therefore SBDE SBMD SMDE 5 1 2 5 1.

3.1.41. In the case when four vertices of the hexagon A1A2A3A4A5A6 are on two adjacent sides of the unit square ABCD, we have that if A , A 2 AB, A , A 2 BC, ÀÁ1 2 3 4 þ þ þ 4 ¼ 1 Á Á Á 1 A1A2 BA3 A2B A3A4 1 then SA1A2A3 SA2A3A4 4 A1A2 BA3 A2B A3A4 4 4 64. ðÞ; 1 Consequently, min SA1A2A3 SA2A3A4 8. It remains to consider the case when A1, A2 2 AB, A3 2 BC, A4, A5 2 CD, A6 2 AD. Let A4C A2B.IfDA5 AA1, then we proceed to the proof by contradiction 1 argument. Suppose that the areas of all these triangles are greater than 8. Then by ¼ ¼ 1 approaching point A3 to C, and point A6 to D, one can obtain SA3A4A5 SA4A5A6 8, 1 while other areas are greater than 8. Now by drawing together points A1 and A2,we ¼ ¼ 1 1 can reach SA6A1A2 SA1A2A3 8, while areas of other two triangles are greater than 8. Then we have that 132 3 Areas

1 þ A A 1 þ A A S ¼ S þ S ¼ 1 2 BA þ 4 5 CA ¼ A1A2A3A4A5A6 A1A2A3A6 A6A3A4A5 2 3 2 3 Á Á ¼ 1 þ A1A2 BA3 þ A4A5 CA3 ¼ 3: 2 2 2 4 On the other hand 1 A A þ A A 1 S ¼ S þ S þ S > þ 1 2 4 5 ¼ A1A2A3A4A5A6 A6A1A5 A2A3A4 A1A2A4A5 4 2 4 þ 1 1 þ 1 ¼¼ 1 þ 1 1 þ 1 ¼ 3: Á Á 2 8 BA3 A3C 4 8 BA3 A3C 4 2 Á ðÞBA3 þ A3C 4 This leads to a contradiction. If DA5 < AA1, then moving point A3 closer to C, and point A6 to A, one can obtain ¼ 1 ¼ 1 that SA6A1A2 8 , SA5A3A4 8, while the areas of remaining four triangles are greater 1 than 8. Now, let us move points A3 and A6 in a direction parallel to AB, so that ¼ ¼ 1 ¼ ¼ 1 > 1 > 1 ¼ SA6A1A2 SA6A1A5 8, SA2A3A4 SA3A4A5 8, SA1A2A3 8, SA6A5A4 8, PQ 1, QR < 1 (Figure 3.38). Note that A1A6||A2A5||A3A4. Let PO ¼ 1 and SV||A3A4. Denote A1A6 ¼ a, A3A4 ¼ b, ∠QA1A6 ¼ α, then α > 45 and RA3 > QA6. Therefore, a sin α þ b sin α < 1. We have that pffiffiffi 2 ¼ QO > QV ¼ QM þ MN þ NU þ UV ¼ a cos α sin α 1 1 b cos α sin α ¼ ÀÁ þ ÀÁ þ ÀÁ þ ÀÁ ¼ sin α þ 45 4a sin α þ 45 4b sin α þ 45 sin α þ 45 1 a þ b ¼ ÀÁ ðÞa þ b cos α sin α þ sin α þ 45 4ab 1 1 ÀÁ ðÞa þ b cos α sin α þ : sin α þ 45 a þ b

Figure 3.38 QA1 A2 R 450 a 1 4a M a N

A3 U A6 1 b V 4b a PA5 A4 SO 3.1 Inequalities with Areas 133 i Since the function fxðÞ¼x cos α sin α þ 1 decreases in 0; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 and x cos α sin α 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 1 a þ b < α < , then ðÞa þ b cos α sin α þ þ > α cos α sin αþ sin cos α sin α pffiffiffi a b sin α ¼ α þ α > 1 ðÞþ α α þ 1 sin cos sin . Consequently, 2 ðÞαþ a b cos sin aþb pffiffiffi sin 45 cos αþ sin α > ¼ 2. This leads to a contradiction. sin ðÞαþ45 This ends the proof. 1 Remark The number 8 cannot be made smaller, because if A1 A, A4 C, ¼ ¼ 1 ¼ ¼ ¼ ¼ ::: ¼ ¼ 1 A1A2 A1A6 2 A3A4 A4A5, then SA1A2A3 SA2A3A4 SA6A1A2 8. 0 0 ~ 0 ~ 0 3.1.42. Let us consider point B and P , such that BB ¼ PP ¼ AC~ . Let O be the 0 midpoint of segment BC and AB ¼ c, BC ¼ a, AC ¼ b, PB ¼ d. 0 0 According to problem 1.1.14a, for points P, B , P , C we have that bc d þ 6. 0 0 Since quadrilateral ABB C is a parallelogram, then O is the midpoint of segment AB . 0 2 ¼ 2þ2d2ÀAB02 2 ¼ 2 þ 2 À 2 2 ¼ 8þ18Àa2 Consequently, PO 4 , AB 2b 2c a , PO 4 . Hence, we obtain that b2 þ c2 À a2 ¼ d2 À 12, then d2 À 12 ¼ bc d þ 6. Thus pffiffiffiffi pffiffi pffiffi ÀÁpffiffiffiffiffi pffiffi 1þ 73 ¼ 3 3 ðÞþ þ 3 d 2 , therefore SABC 4 bc 4 d 6 13 73 8 . 3.1.43. First we prove that if given an angle ∠MAN ¼ α, then on sides AM, AN one can find points B0, C0, respectively, and inside the given angle a point P, such that PA0 ¼ r1, PB0 ¼ r2, PC ¼ r3 and ∠PB0A ¼ ∠PC0A. Indeed, let point K be inside angle MAN and ρ(K, AM) ¼ r2, ρ(K, AN) ¼ r3. Denote by P the intersection point of the circle with a center at A and radius r1 ρðÞ; with ray AK. Then AP ¼ r and P AM ¼ r2, and it remains to use the conditions 1 ρðÞP;AN r3 r2 > r1 and r3 > r1 to choose on rays AM and AN points B0 and C0, such that PB0 ¼ r2, PC0 ¼ r3 (Figure 3.39). Then ΔPEB0 and ΔPFC0 are similar. ∠ ¼ ∠ Hence PB0A PC0A. We need to prove that SABC SAB0C0 . Let ΔABC and point P satisfy the conditions of the problem. 0 0 ~ 0 ~ 0 ~ 0 Consider points B and P , such that BB ¼ PP ¼ AC , then (see the solution of 0 problem 3.1.42), if PB ¼ d, we obtain that

Figure 3.39 B0 M

r2 E P r1 r3

AF C0 N 134 3 Areas

bc dr1 þ r2r3 ð3:12Þ

2 ¼ 2 þ 2 À 2 þ α and d r2 r3 r1 2bc cos . Consequently,

2 2 þ 2À d r2 r3 À 2 þ ðÞþ α: r1 2 dr1 r2r3 cos pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi α þ 2 þ 2 þ α À 2 2α Hence d r1 cos r2 r3 2r2r3 cos r1sin . Therefore,

1 SABC ¼ bc sin α 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 r2 cos α þ r r2 þ r2 þ 2r r cos α À r2sin 2α þ r r sin α ¼ S , 2 1 1 2 3 2 3 1 2 3 AB0C0 since for triangle AB0C0 the inequality (3.12) turns into equality. Now we will find the minimal value of SABC. To do that, let us note that points A, B, C are on the circles with a center at point P and with radiuses r1, r2, r3, respectively (see Figure 3.40). 0 0 Note that triangle AB C also satisfies theÀÁ conditionsÀÁ of the problem and 0 0 Á 0 0 ¼ 1 Á Á Á 2α ¼ 1 2 À 2 2 À 2 2α SABC SAB C 4 AB AB AC AC sin 4 r2 r1 r3 r1 sin . Thus, it follows that ÀÁÀÁ ÀÁÀÁ 2 À 2 2 À 2 2α 1 2 2 2 2 2 1 r2 r1 r3 r1 sin SABC ¼ r À r r À r Á sin α : 2 1 3 1 0 0 4 SAB C 4SAB0Co

ðÞ2À 2 ðÞ2À 2 2α r2 r1 r3 r1 sin This means that the minimal value of SABC is equal to . 4SAB0Co This ends the proof.

Figure 3.40 C

a B A a B¢ P

C' 3.1 Inequalities with Areas 135

Figure 3.41

3.1.44. Consider the following two cases. (a) If any two of the constructed parallelograms (Figure 3.41а) do not have a common point, denote by S0 the intersection area of the parallelograms and ¼ > À let SA1B1C1D1 S1, then we have that S0 S1 S0.

> S1 : Thus, it follows that S0 2 (b) If two of the constructed parallelograms have common points (Figure 3.41b), then any two among the other four constructed parallelograms do not have a common point. > S ¼ Therefore S0 2 , where S SABCD.

Problems for Self-Study

3.1.45. Prove that in any convex polygon one can place a rectangle having the area 1 not less than 4 of the area of the given polygon. 3.1.46. There are 5 patches placed on the shirt with area 1, the area of each of them 1 being not less than 2. Prove that one can ﬁnd two patches so that the area of their 1 common parts is not less than 5.

3.1.47. Let T1 and T2 be two triangles with sides a1, b1, c1 and a2, b2, c2. Prove that there exists a triangle T with sides a, b, c, such that if S1, S2 and S are the areas of triangles T1, T2 and T, respectively, then qffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffi þ a2þa2 b2þb2 c2þc2 (a) S S1 S2, where a ¼ 1 2, b ¼ 1 2, c ¼ 1 2, p2 ffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 (b) S 4 S1 Á S2, where a ¼ a1 þ a2, b ¼ b1 þ b2, c ¼ c1 þ c2. 136 3 Areas

3.1.48. The area of a convex quadrilateral ABCD is equal to S. Prove that the area of the quadrilateral, with the vertices at the midpoints of segments AC, AD, BD and S BC, is less than 2. 3.1.49. Prove that in a triangle with area S one can inscribe a regular triangle, such S that its area is not greater than 4. 3.1.50. Prove that any acute trianglepffiffiffi with area 1 can be placed into right-angled triangle with area not greater than 3. 3.1.51. Given several squares the sum of area of which is equal to S. Prove that (a) one can place them without overlapping inside a square with area 2S, S (b) with these squares one can always cover the square with area 4. 3.1.52. Let AD be the altitude of the right-angled triangle ABC, ∠A ¼ 90 . The straight line passing through the centers of the incircles of triangles ABD and ACD intersects sides AB and AC, respectively, at points K and L. Prove that SABC 2SAKL. 3.1.53. Bisectors of angles A, B, C of an acute triangle ABC intersect its circumcircle at points A1, B1, C1, respectively. The straight line AA1 intersects the bisectors of the external angles B and Cof triangle ABC at point A0. Points B0, C0 are defined similarly. Prove that SA0B0C0 4SABC. 3.1.54. Let ABC be an acute triangle, points M, N and P be the feet of the perpendiculars drawn from the centroid of the triangle to sides AB, BC and CA, > 4 respectively. Prove that SMNP 27 SABC. 3.1.55. Let ABCD be a unit square. For any internal points M and N, such that line MN does not contain any of the vertices of the square, denote by S(M, N) the least of the areas of the triangles with the vertices from the set of the points {A, B, C, D, M, N}. Find the smallest number k, such that S(M, N) k for all such points M and N. Hint Let point N be inside of triangle CMD, then þ ¼ 1 ¼ þ þ SAMB SCMD 2, SCMD SCMN SMND SCND 3S(M, N). Therefore ðÞ; 1 ðÞ; 1 4SMN 2. Hence, it follows that SMN 8. 3.1.56. Points K, L, M and N are taken on sides AB, BC, CD and DA of a convex quadrilateral ABCD, respectively. Denote by S1, S2, S3 and S4 thepffiffiffiffiffi areaspffiffiffiffiffi of 3 þ 3 þ ptrianglesffiffiffiffiffi pffiffiffiffiffiAKNp, BKLffiffiffiffiffiffiffiffiffiffiffiffi, CLM and DMN, respectively. Prove that S1 S2 3 3 3 S3þ S4 2 SABCD. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 Hint Prove that S1 ¼ 3 AK Á AN Á SABD 1 AK þ AN þ SABD . SABCD AB AD SABCD 3 AB AD SABCD

3.1.57. Prove that if a convex pentagon A1A2A3A4A5 is inside of a parallelogram with area 1, then one of the triangles A1A2A3, A2A3A4, A3A4A5, A4A5A1, A5A1A2 has 1 an area that does not exceed 4. 3.1 Inequalities with Areas 137

Hint Draw a segment connecting the opposite sides of the parallelogram, then one Δ 1 can assume that A1A2A3 is inside the parallelogram with area 2. 1 Consequently, SA1A2A3 4. 3.1.58. (a) Let ABCDEF be a convex hexagon, such that AB||ED, BC||FE and CD|| AF. Prove that 1 6 S Á S Á S Á S Á S Á S S : ð3:13Þ ABC BCD CDE DEF EFA FAB 6 ABCDEF

Hint Prove that SABCDEF 2SABC þ 2SCDE þ 2SEFA.

Remark If BCEF is a rectangle, AB ¼ CD ¼ 3, AF ¼ DE ¼ 4, BF ¼ 5, BC ¼ 10, then (3.13) is wrong.

(b) Let ABCDEF be a convex hexagon. Prove that minðÞSABC; SBCD; SCDE; SDEF; SEFA; SFAB 1 6 SABCDEF . 3.1.59. Given points A, B, C, D inside a square with a side equal to 6, such that the distance between each two of them is not less than 5. Prove that points A, B, C, D form a convex quadrilateral with the area greater than 21. 3.1.60. Let ABCD be a convex quadrilateral and K be the intersection point of its diagonals. Let also the conditions KL||AB, LM||DC, MN||AB hold true for the points L 2 [AD], M 2 [AC], N 2 [BC]. Prove that SKLMN < 8 . SABCD 27 3.1.61. Let S be the area of a convex quadrilateral ABCD, a, b, c, d be the lengths of its consecutive sides, e, f be the lengths of the diagonals, and m, n the lengths of its medians connecting the midpoints of the opposite sides of the quadrilateral. Prove that: 1 (a) S 2 ef , (b) S mn, abþcd (c) S 2 , acþbd (d) S 2 , 1 ðÞþ ðÞþ (e) S 4 a c b d , 1 2 ¼ aþbþcþd (f) S 4 p, where p 2 , 1 ðÞþ 2 þ (g) S 4 a c bd . 3.1.62. Prove that none of the triangles inscribed into polygon M can have an area greater than the maximal area of the triangles, the vertices of which coincide with some of the vertices of M. 3.1.63. (a) Let M be a convex polygon and l be an arbitrary straight line. Prove that one can inscribe into M a triangle, оne side of which is parallel to l and the area of 3 which is not less than 8 of the area of M. 138 3 Areas

(b) Let M be a regular hexagon and l be a line parallel to one of its sides. Prove that one cannot inscribe into M a triangle, оne side of which is parallel to l and the 3 area of which is greater than 8 of the area of M. α > π 3.1.64. Solve problem 3.1.43 in the case of 2.

Hint Prove that there exists a triangle PB0C0 and a point A inside it, such that PA ¼ r1, PB0 ¼ r2, PC0 ¼ r3 and ∠BAC ¼ α, ∠AB0P ¼ ∠AC0P. Then SABC SAB0C0 . 3.1.65. A triangle with the area 1 is cut out of paper. Prove that one can flex it overpffiffiffi the segment of a line, so that area S of obtained figure would be less than 2 À 2.

Hint Let AB AC BC, if we ﬂex triangle ABC (SABC ¼ 1) over the bisector of ¼ BC angle C, then S BCþAC. On the other hand, if we ﬂex triangle ABC over a segment perpendicular to side BC, then one can prove that

2 S , þ tg ∠C 3 tg ∠B and that the equality can hold true. Then, prove that ! 1 2 pffiffiffi min ; < 2 À 2: 1 þ sin ∠B þ tg ∠C sin ðÞ∠Cþ∠B 3 tg ∠B

3.1.66. Given a triangle ABC. Prove that there exists a straight line l, such that if points A1, B1, C1 are symmetric to points A, B, C with respect to line ÀÁl,p thenffiffiffi the area of the common parts of triangles ABC and A1B1C1 is greater than 2 2 À 1 . Hint See the Hint of problem 3.1.64. 3.1.67. Let diagonals AD, BE and CF of the convex hexagon intersect at one point and A1 ¼ AD \ BF, D1 ¼ AD \ CE, B1 ¼ BE \ AC, E1 ¼ BE \ DF, C1 ¼ CF \ BD, F1 ¼ CF \ AE. Prove that 1 S S : A1B1C1D1E1F1 4 ABCDEF

Hint Let diagonals AD, BE, and CF intersect at a point O. Prove that SAOBÁSBOCÁSAOF 1 SA OB ¼ ðÞSAOF þ SBOC . 1 1 ðÞSAOBþSBOC ðÞSAOBþSAOF 8 3.1.68. The medians of triangle ABC intersect the circumcircle of the triangle for the second time at points A1, B1, C1. Prove that SA1B1C1 SABC. Remark Let point M be the centroid of triangle ABC. Then ðÞ2þ 2þ 2 2 ¼ a b c SMA1B1 2 2 SABC. 48mamb Chapter 4 Application of Vectors

This chapter is devoted to the application of vectors for proving geometric (or trigonometric) inequalities and consists of only one section, that is, Section 4.1. One of the methods of proving geometric inequalities is the use of vectors and the application of their properties. In particular, the following properties of the scalar product of two vectors is widely used. 2 ~ (a) ~a Á~a ¼ jj~a 0. Moreover, the equality holds true if and only if ~a ¼ 0.

(b) ~a Á~b jj~a Á ~b . Moreover, the equality holds true if and only if ~a and ~b are collinear vectors (parallel to one line or lying on one line). One of the main inequalities in Section 4.1 is the inequality of problem 4.1.8a. Very important inequality is also the inequality of problem 4.1.13. In Section 4.1 consider different types of problems and using vectors the authors apply some famous algebraic inequalities in order to prove geometric inequalities. Some problems in this chapter were inspired by [1, 7, 9, 10, 13]. Nevertheless, even for these problems the authors have mostly provided their own solutions.

4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities

4.1.1. Let R be the radius of the circumsphere of tetrahedron SABC. Prove that (a) SA2 + SB2 + SC2 +4R2 AB2 + BC2 + AC2, (b) SA2 + SB2 + SC2 + AB2 + BC2 + AC2 16R2, (c) xyAB2 + yzBC2 + xzAC2 + xtSA2 + ytSB2 + ztSC2 (x + y + z + t)2R2, where x, y, z, t are arbitrary numbers.

4.1.2. Let O be the circumcenter of regular tetrahedron SABC. Prove that φ þ ψ À2 φ ¼ ∠ ψ ¼ ∠ cos cos 3, where AOS, AOB. 4.1.3. Given four points A, B, C, and D. Prove that (а) AB2 + BC2 + CD2 + AD2 AC2 + BD2, (b) 4(AP2 + BQ2 + CR2 + DS2) 5(AB2 + BC2 + CD2 + AD2), where P, Q, R, S are the midpoints of segments BC, CD, DA, AB, respectively. 4.1.4. For arbitrary points A, B, C, D, E, and F prove the following inequalities: (a) 2(AB2 + BC2 + CD2 + DE2 + EF2 + FA2) AD2 + BE2 + CF2, 2 2 2 2 2 2 (b) 4(A1D1 + B1E1 + C1F1 ) 3((AB + DE) +(BC + EF) +(CD + AF) ), where A1, B1, C1, D1, E1, F1 are the midpoints of segments AB, BC, CD, DE, EF,andFA, respectively. 4.1.5. Given a trapezoid ABCD (AB k CD). Prove that, if AD > BC and AC > BD, then AB > CD.

4.1.6. Prove that for any tetrahedron A1A2A3A4 and for any point M inside of it 1 min cos ∠AiMAj À max cos ∠AiMAj. i

4.1.7. (a) Given 7 points on a unit sphere. Prove that amongpffiffiffi them one can find 2 points, such that the distance between them is less than 2. (b) Given 5 points on a unit sphere. Prove that among themp oneffiffiffi can find 2 points, such that the distance between them is not greater than 2. 4.1.8. For any points A, B, C, M and arbitrary numbers x, y, z prove that (a) (x + y + z)(xMA2 + yMB2 + zMC2) xyc2 + yza2 + xzb2, where AB ¼ c, BC ¼ a, AC ¼ b , (b) aMA2 + bMB2 + cMC2 abc, (c) MA Á MBc + MB Á MCa + MC Á MAb abc, (d) (x + y + z)(xMB2MC2 + yMA2MC2 + zMA2MB2) xyc2MC2 + yza2MA2 + xzb2MB2, (e) (a + b + c)(aMB2MC2 + bMA2MC2 + cMA2MB2) a2b2c2, (f) (MA Á MB + MB Á MC + MC Á MA)(MA + MB + MC) a2MA + b2MB + c2MC. 4.1.9. Prove the following inequality

A A A A A À A A A 1 2 þ 2 3 þ ::: þ n 1 n 1 n , MA1 Á MA2 MA2 Á MA3 MAnÀ1 Á MAn MA1 Á MAn where M, A1,...,An are arbitrary distinct points. 4.1.10. Let n-gon with area S be inscribed into a circle with radius R, such that on each side of the n-gon is marked one point. Prove that, if one connects consecu- 2S tively these marked points then the perimeter of the obtained n-gon is less than R . 4.1.11. Let M be a given point inside of triangle ABC and P, Q, R be points on straight lines AB, BC, AC, respectively, such that ∠PMA, ∠PMB, ∠QMB, ∠QMC, ∠ ∠ π RMA, RMC 2. Prove that 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 141 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi MA þ MB þ MC 2 pMAffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÁ MB Á cos ∠PMA Á cos ∠PMBþ þ2pMBffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÁ MC Á cos ∠QMB Á cos ∠QMCþ þ2 MA Á MC Á cos ∠RMA Á cos ∠RMC:

4.1.12. Let M be a given point inside of triangle ABC. Denote the distances from the point M to lines BC, CA, AB by da, db, dc and the distances from point M to vertices A, B, C by Ra, Rb, Rc, respectively. Prove that

(a) Ra + Rb + Rc 2da +2db +2dc, (b) RaRb + RbRc + RcRa 2Rada +2Rbdb +2Rcdc. 1 4.1.13. For any points A1, A2, ..., An, M and any numbers m1, ..., mn prove the following inequality m MA2 þ ::: þ m MA2 m GA2 þ ::: þ m GA2, where 1! 1 n! n 1 1 n n þ ::: þ ¼ ~ m1 +...+mn 0 and m1 GA 1 mn GA n 0.

4.1.14. Let points A1, A2,..., An be on the same circle, m1,...,mn > 0 and G be ! ! ~ such a point that m1 GA 1 þ ::: þ mn GA n ¼ 0. Given that straight lines GA1,..., GAn intersect the circle for the second time at points B1,...,Bn. Prove that

(a) m1GB1 +...+mnGBn m1GA1 +...+mnGAn, 2 þ ::: þ 2 2 þ ::: þ 2 (b) m1GB1 mnGBn m1GA1 mnGAn.

4.1.15. (а) Let M be a point inside of the convex n-gon with vertices A1, A2,...,An. Perpendiculars MB1, MB2,...,MBn are drawn from point M to lines A1A2, A2A3, ...,AnA1, respectively, and C1 2 A1A2, C2 2 A2A3, ..., Cn 2 AnA1 are arbitrary m1MB1 m2MB2 mnMBn points. Given positive numbers m1,...,mn, such that ¼ ¼ ::: ¼ . P P A1A2 A2A3 AnA1 2 2 Prove that mimjCiCj mimjBiBj . i

(b) Let M be a point inside of convex n-hedron. Perpendiculars MB1,...,MBn are draw from point M to planes Π1,..,Πn containing the faces of the polyhedrons with areas S1,...,Sn, respectively, and C1 2 Π1,...,Cn 2 Πn are arbitrary points. Given positive numbers m ,...,m such that m1MB1 ¼ ::: ¼ mnMBn. 1 n S1 Sn P P 2 2: Prove that mimjCiCj mimjBiBj i

4.1.16. (a) Given a triangle ABC and positive numbers x, y, z. Points A1, B1, and C1 are taken on straight lines BC, CA, and AB, respectively. Prove that ðÞþ þ xA B2 þ yB C2 þ zA C2 4 xy yz zx S2 , and that the equality holds true. 1 1 1 1 1 1 xAB2þyBC2þzAC2 ABC

1 The point Gis called center of mass for the system of points A1, A2, ..., An with masses m1, ¼ 2 þ ::: þ 2 m2,...,mn, and the expression IM m1MA1 mnMAn is called the moment of inertia of the system of points A1,...,An with masses m1,...,mn relative to the point M. 142 4 Application of Vectors

(b) Given a tetrahedron A1A2A3A4. Find on planes A2A3A4, A1A3A4, A1A2A4, 2 and A1A2A3 such points B1, B2, B3, and B4, respectively, that the sum B1B2 þ 2 þ 2 þ 2 þ 2 þ 2 B1B3 B1B4 B2B3 B2B4 B3B4 is minimal.

4.1.17. Given different points A1, A2,...,An and positive numbers m1, m2,...,mn. Pn Find a point M, such that the sum miMAi is minimal. i¼1 4.1.18. Given a convex polygon A1A2 ...An and positive numbers m1, m2,...,mn. Inscribe in a polygon A1A2 ...An a polygon C1C2 ...Cn (C1 2 A1A2, C2 2 A2A3, Pn ...,Cn 2 AnA1, Ci 6 Aj), so that the sum miCiCiþ1 is minimal, where Cn +1 C1. i¼1 4.1.19. Let centers O1 and O2 of the incircles of triangles A1B1C1 and A2B2C2, respectively, do not coincide. Prove that þ O1O2 < p1 p2 þ þ ðÞ; , A1A2 B1B2 C1C2 2max p1 p2 where p1 and p2 are perimeters of triangles A1B1C1 and A2B2C2, respectively. 4.1.20. (a) Let O be the center of a unit sphere tangent to all faces of trihedral angle with a vertex A. It is known that the measure of all linear angles of the trihedralpffiffiffi π π À 2 angle is not less than 2 and is not greater than arc cos 3. Prove that OA 3. (b) Is it possible that the distances of all vertices of the centihedron outscribed around the unit sphere from the center of the sphere is greater than 100? 4.1.21. Let ABCDA0B0C0D0 be a cube. Consider points K, L, M on edges AB, CC0, 0 pffiffiffi D0A , respectively. Prove that KL þ LM þ MK 1, 5 6AB. 4.1.22. The radius of the circumsphere of tetrahedron ABCD is equal to R, the lengths of the segments connecting vertices A, B, C, and D with centers of the opposite faces are equal to ma, mb, mc, and md, respectively. Prove that ma þ mbþ þ 16R mc md 3 . 4.1.23. Let ABCDbe a tetrahedron, such that AC ⊥ BC and AD ⊥ BD. Prove that the CD cosine of the angle between straight lines AC and BD is less than AB.

4.1.24. (a) Given points A1 and B1, A2 and B2, ..., An and Bn on faces Γ1, Γ2,...,Γn of a convex m-hedron (m n 3), respectively. Given that for any i 2 {1, 2, ..., n} vector m~e À m À ~e À is perpendicular to the plane containing face Γ , where m , i i i 1 i 1 ! i 1 AiAiþ1 m ,...,m are given positive numbers and~ei ¼ , i ¼ 1, 2, . . ., n, A A , 2 n AiAiþ1 n +1 1 m0 ¼ mn, ~e0 ¼ ~en. Prove that

m1B1B2 þ m2B2B3 þ ::: þ mnÀ1BnÀ1Bn þ mnBnB1 m1A1A2 þ m2A2A3 þ ::: þ mnÀ1AnÀ1An þ mnAnA1:

Given a tetrahedron C1C2C3C4, such that C1C2 ¼ C2C3 ¼ C3C4 ¼ C4C1. Find on faces C2C3C4, C1C3C4, C1C2C4, and C1C2C3 points B1, B2, B3, and B4, respectively, such that sum B1B2 + B2B3 + B3B4 + B4B1 is minimal. 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 143

(c) Given a cube ABCDA0B0C0D0. Find on faces ABCD, AA0BB0, BB0C0C, A0B0C0D0, 0 0 0 0 DD C C, and AA D D points B1, B2, B3,B4, B5, and B6, respectively, such that the sum B1B2 + B2B3 + B3B4 + B4B5 + B5B6 + B6B1 is minimal. 4.1.25. Given that circumcenter O of the tetrahedron is inside of that tetrahedron. Prove that (a) DA2 + DB2 + DC2 + AB2 + BC2 + AC2 > 12 Á OA2, (b) DA + DB + DC + AB + BC + AC > 6 Á OA.

Solutions

4.1.1. (a) The proof follows from problem 4.1.1c, if we take x ¼ y ¼ z ¼ 1, t ¼À1. (b) The proof follows from problem 4.1.1c, if we take x ¼ y ¼ z ¼ t ¼ 1. (c) Let O be the circumcenter of tetrahedron SABC. Then, we have that

2 þ 2 þ 2 þ 2 þ 2 þ 2 À ðÞþ þ þ 2 2 xyAB yzBC xzAC xtSAytSB ztSCx y z tR ! ! 2 ! ! 2 ! ! 2 ! ! 2 ¼ xy OB À OA þ yz OC À OB þ xz OC À OA þ xt OA À OS ! ! 2 ! ! 2 þyt OB À OS þþzt OC À OS À ðÞx þ y þ z þ t 2R2 ! ! ! ! 2 ¼À xOA þ yOB þ zOC þ tOS 0:

4.1.2. We have that ! ! ! ! 2 4R2 þ 6R2 cos φ þ 6R2 cos ψ ¼ OA þ OB þ OC þ OS 0 ðÞR ¼ OA :

Therefore, cos φ þ cos ψ À2. ! !3 ! 4.1.3. Let us denote AB ¼ ~rB, AC ¼~rC, AD ¼ ~rD. ~ 2 þ ðÞ~ À~ 2 þ ðÞ~ À~ 2þ ~2 ~2 þ ðÞ~ À~ 2 (a) We need to prove that rB rC rB rD rC rD rC rD rB , 2 or ðÞ~rB þ~rD À~rC 0. (b) We need to prove that ÀÁÀÁÀÁÀÁ ~ þ~ 2 ~ þ~ 2 ~ 2 ~ 2 4 rB rC þ rC rD À~r þ ~r À rD þ ~r À rB 2 2 B C 2 D 2 ~ 2 þ ðÞ~ À~ 2 þ ðÞ~ À~ 2 þ~2 5 rB rC rB rD rC rD ,

2 or ðÞ~rB þ~rD À~rC 0. 144 4 Application of Vectors

! ! ! ! ! 4.1.4. (a) Let us denote AB ¼ ~a, BC ¼ ~b, CD ¼ ~c, DE ¼ ~d, EF ¼ ~e, then one has to prove that 2 2 ~a2 þ~b2 þ~c2 þ ~d2 þ~e2 þ ~a þ~b þ~c þ ~d þ~e 2 2 2 ~a þ~b þ~c þ ~b þ~c þ ~d þ ~c þ ~d þ~e ,

But this inequality is equivalent to the following inequality 2 2 2 ðÞ~a þ~c þ~e 2 þ ~a þ ~d þ ~b þ~e þ ~a þ~b þ ~d þ~e 0:

! ! ! ! ! ! (b) Let AB ¼ ~a, BC ¼ ~b, CD ¼ ~c, DE ¼ ~d, EF ¼ ~e.Notethat2A D ¼ ! ! ! ! 1 1 þ , consequently, 4 2 ¼ 2 þ 2 þ 2 Á . Consider paral- BE AD A1D1 BE AD BE AD 2 ! ! ! 2 lelogram BADM.Wehavethat ~a À ~d ¼ EM2 ¼ BM À BE ¼ ! ! 2 ÀÁ 2 À 2 ¼ 2 þ 2 À ~À ~ AD BE ,hence4A1D1 2 AD BE a d . Then, we have that ÀÁ 2 þ 2 þ 2 ¼ 2 þ 2 þ 2À 4 A1D1 B1E1 C1F1 4CF 4BE 4AD 2 2 2 À ~a À ~d À ~a þ~b þ 2~c þ ~d þ~e À ~b À~e ¼ 2 2 2 ¼ 4 ~c þ ~d þ~e þ 4 ~a þ~b þ~c þ 4 ~b þ~c þ ~d À 2 2 2 À ~a À ~d À ~a þ~b þ ~d þ~e þ 2~c À ~b À~e ¼ 2 2 2 ¼ 3 ~a À ~d þ 3 ~a þ~b þ ~d þ~e þ 2~c þ 3 ~b À~e À 4ðÞ~a þ~c þ~e 2 2 2 2 3 ~a À ~d þ 3 ~a þ~b þ ~d þ~e þ 2~c þ 3 ~b À~e 3ðÞAB þ DE 2 þ 3ðÞBC þ EF 2 þ 3ðÞCD þ AF 2: ! ! ! ! 4.1.5. Let AD ¼ ~a, AC ¼ ~b, then DC ¼ ~b À~a, and AB ¼ k ~b À~a , where k > 0. 2 One has to prove that k > 1. We have that ~a2 > ðÞk À 1 ~b À k~a and 0 2 2 k À 1 B ðÞk À 1 ~b À k~aÞ ~b2 > k~b À ðÞk þ 1 ~a . Therefore, ¼ ~a2 À @ . þ ‘ 2 k 1 ~b2 À k~b À ðÞk þ 1 ~a > 0 Hence, k > 1. 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 145

! MAi 4.1.6. Let ~ei ¼ , i ¼ 1, 2, 3, 4, and ~e ¼ ~e1 þ~e2 þ~e3 þ~e4. Let us draw through MAi point M a plane Π perpendicular to vector ~e. It is clear that, if all points Ai were in the same half-space with boundary Π, then point M would have been inside the given tetrahedron. This means that there exists a vector~ek, such that~e Á~ek 0. Then we have that ðÞ~e À~ek ~ek À1. Hence, we deduce that min cos ∠AiMAj i

4.1.7. (а) Let points A1, A2, ...,A7 are on the unit sphere with center O.Letus pffiffiffi ! 2 ÀÁ 2 2 assume that AiAj 2 ðÞi 6¼ j , then AiAj ¼ AiA ¼ ~aj À~ai 2, where ! j OA i ¼ ~ai, i ¼ 1, . . ., 7. This means that for i 6¼ j we have that

~ai Á~aj 0: ð4:1Þ

Let ~e1 ¼ ~a1,~e2,~e3 be mutually perpendicular unit vectors and ~ ¼ ~ þ ~ þ ~ ¼ ¼ ¼ ¼ ~ Á~ ai xie1 yie2 zie3i, i 1, . . ., 7. As x1 1, y1 z1 0, and ai aj 0 (i 6¼ j), then x2,...,x7 0. ~ ðÞ; ::: ~ ðÞ; Let us consider vectors on a plane: b2 y2 z2 , , b7 y7 z7 . Note that there are ~ ~ ~ non-zero vectors among them. Indeed, if, for example, b2 ¼ b3 ¼ 0, then some two ~ of the vectors ~a1,~a2,~a3 coincide. This leads to a contradiction with (4.1). Let b3, :: ~ :, b7 be non-zero vectors. Then, some two of them form an acute angle, for example, ~ ~ ~ ~ ¼ þ þ ¼ þ~ ~ > b3 and b4. Then, we have that a3a4 x3x4 y3y4 z3z4 x3x4 b3b4 0, ~ ~ because x3x4 0 and b3b4 > 0. This leads to a contradiction with (4.1).

Remark Vectors ~a1ðÞ1; 0; 0 , ~a2ðÞÀ1; 0; 0 , ~a3ðÞ0; 1; 0 , ~a4ðÞ0; À1; 0 , ~a5ðÞ0; 0; 1 , ~a6 ðÞ0; 0; À1 satisfy condition (4.1). ÀÁ 2 (b) If ~aj À~ai > 2, (i 6¼ j), i, j, 2 {1, 2, 3, 4, 5}, then

~ai Á~aj < 0, ðÞi 6¼ j : ð4:2Þ

~ ~ ~ ~ Then, among vectors b2, b3, b4, b5 there is non-zero vector (see the proof of ~ ~ ! 4.1.7a), because otherwise, if b2 ¼ 0, then ~a2 ¼À1; 0; 0 and 0 ¼ ~a3 Á~a1þ

~a3~a2 < 0. This leads to a contradiction. 146 4 Application of Vectors

~ ~ ~ ~ Therefore, among vectors b2, b3, b4, b5 some two form an angle not greater than ~ ~ ~ ~ ~ ~ 90 . Let b2b3 0, then ~a2~a3 ¼ x2x3 þ b2b3 > 0, because x2, x3 < 0 and b2b3 0. This leads to a contradiction. pffiffi pffiffi pffiffi ~ ðÞ; ; ~ 2 2 ; ; À1 ~ À 2; À 6 ; À1 Remark Vectors a1 0 0 1 , a2 3 0 3 , a3 3 3 3 ,and pffiffi pffiffi ~ À 2; 6 ; À1 a4 3 3 3 satisfy condition (4.2). 4.1.8. (а) Note that ÀÁ ðÞþ þ 2 þ 2 þ 2 À 2 À 2 À 2 ¼ x y z xMA yMB zMC xyc yzaxzb ÀÁ! ! 2 ¼ ðÞx þ y þ z xMA2 þ yMB2 þ zMC2 À xy MB À MA À ! ! 2 ! ! 2 ! ! ! 2 Àyz MC À MB À xz MC À MA ¼ xMA þ yMB þ zMC 0:

(b) Using problem 4.1.8a for x ¼ a, y ¼ b, z ¼ c, we deduce that aMA2 + bMB2 + cMC2 abc. (c) For four points M, A, B, C we call their H-image the points M1, A1, B1, C1, where M1A1 ¼ MB Á MC, M1B1 ¼ MA Á MC, M1C1 ¼ MA Á MB, A1C1 ¼ MB Á AC, B1C1 ¼ MA Á BC, A1B1 ¼ MC Á AB (see Figure 4.1).

Let us rewrite inequality 4.1.8b for points M1, A1, B1, C1, this means that Á 2 þ Á 2 þ Á 2 Á Á B1C1 M1A1 A1C1 M1B1 B1A1 M1C1 B1C1 A1C1 A1B1,or

MB2 Á MC2 Á MA Á BC þ MA2 Á MC2 Á MB Á AC þ MA2 Á MB2 Á MC Á AB MA Á MB Á MC Á AB Á BC Á AC:

Therefore, MB Á MCa + MA Á MCb + MA Á MBc abc. If MA Á MB Á MC ¼ 0, then the proof of the inequality is straightforward. ðÞþ þ (d) LetÀÁM1, A1, B1, C1 be the H-image of points M, A, B, C. We have that x y z 2 þ 2 þ 2 2 þ 2 þ 2 xM1A1 yM1B1 zM1C1 xyB1A1 yzC1B1 xzA1C1 (see problem 4.1.8a). Therefore, (x + y + z)(xMB2MC2 + yMA2MC2 + zMA2MB2) xyc2MC2 + yza2MA2 + xzb2MB2.

A1 B MB×MC

MB×AC H- image MC×AB MCMC 1 MA×MB 1 MA×BC MA×MC A

Figure 4.1 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 147

(e) Using problems 4.1.8e and 4.1.8b, we obtain that for x ¼ a, y ¼ b, z ¼ c ÀÁ ðÞa þ b þ c aMB2MC2 þ bMA2MC2 þ cMA2MB2 abc2MC2 þ bca2MA2 þacb2MB2 a2b2c2:

Thus, (a + b + c)(aMB2MC2 + bMA2MC2 + cMA2MB2) a2b2c2. (f) If MA Á MB Á MC ¼ 0, then the proof of the inequality is straightforward. But, if MA Á MB Á MC 6¼ 0, then we obtain the required inequality from inequality ¼ 1 ¼ 1 ¼ 1 4.1.8a by taking x MA , y MB , z MC . ! ! ! Let us denote ¼~ and MAi ¼ ρ , ¼ 1, . . ., . Note that 4.1.9. MAi ri MA 2 i i n rffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffii 2 ÀÁ jj~r À~r ðÞ~riÀ~rj ! ! 2 ! ! AiAj ¼ i j ¼ ¼ ρ À ρ ¼ ρ À ρ Á ~2 ~2 . We have to prove MAi MAj ~ ~ r i r j i j i j ri rj !ρ À !ρ þ !ρ À !ρ þ ::: þ !ρ À !ρ !ρ À !ρ that 1 2 2 3 nÀ1 n 1 n . This can be obtained, if we use the following inequality jjþ~a1 jjþ~a2 ::: þ jj~anÀ1 jj~ þ~ þ ::: þ~ ~ ¼ !ρ À !ρ ¼ a1 a2 anÀ1 (see Section 1.2) for vectors ai iþ1 i, i 1, :::, n À 1. 4.1.10. Let us draw tangents to the circle at the vertices of the inscribed n-gon A1 ...An, and choose unit vectors ~e1, :::,~en on these tangents (Figure 4.2). Let B1 2 A1An, B2 2 A1A2,...,Bn 2 An À 1An be the marked points, then

B1B2 þ B2B3 þ ::: þ BnÀ1Bn þ BnB1 ! ! ! ! B1B2~e1 þ B2B3~e2 þ ::: þ BnÀ1Bn~enÀ1 þ BnB1~en ¼ ! ! ! ! ! ! ¼ B1A1 þ A1B2 ~e1 þ B2A2 þ A2B3 ~e2 þ ::: þ BnAn þ AnB1 ~en ¼ ! ! ! 2S ¼ A A ~e þ A A ~e þ ::: þ A A ~e ¼ , 1 2 1 2 3 2 n 1 n R ! ! ! ! because B2A2~e2 ¼ B2A2~e1, :::, AnB1~en ¼ AnB1~e1.

Figure 4.2 A e 2 2 e A3 1 e A1 B2 B3 3

B1 en An 148 4 Application of Vectors

∠ ∠ cos ðÞ∠PMAþ∠PMB þ1 ¼ 4.1.11. We have that cos PMA cos BMP 2 2 ∠PMAþ∠PMB 2 ∠AMB cos 2 cos 2 , consequently, it is sufﬁcient to prove that

MA þ MB þ MC pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ∠AMB pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ∠BMC pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ∠AMC 2 MA Á MB cos þ 2 MB Á MC cos þ 2 MA Á MC cos : 2 2 2 ð4:3Þ pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi Let A1, B1, C1 be such points that MA1 ¼ MA, MB1 ¼ MB, MC1 ¼ MC and ∠A MB ¼ ∠AMB , ∠B MC ¼ ∠BMC , ∠A MC ¼ π À ∠AMC. Then, from inequality 1 1 2 1 1 2 1 1 2 ! ! ! 2 MA1 À MB1 þ MC1 0 we obtain inequality (4.3).

Remark Inequality (4.3) coincides with the inequality of problem 5.1.22a by ¼ 2 ¼ 2 ¼ 2 ∠AMB ¼ α ∠BMC ¼ β ∠AMC ¼ γ replacing MA x , MB y , MC z , 2 , 2 , 2 . 4.1.12. (a) Let MP ⊥ AB, P 2 AB, MQ ⊥ BC, Q 2 BC, MR ⊥ AC, R 2 AC. Then MA cos ∠PMA ¼ MB cos ∠PMB ¼ dc, MB cos ∠QMB ¼ MC cos ∠QMC ¼ da, and MA cos ∠RMA ¼ MC cos ∠RMC ¼ db. Consequently, using problem 4.1.11, we deduce that Ra + Rb + Rc 2da +2db +2dc. (b) The required inequality for the H-image of points M, A, B, C can be obtained from 4.1.12a (see the proof of problem 4.1.8c)). 4.1.13. We have that Xn Xn ! ! 2 ¼ 2 ¼ þ IM miMAi mi MG GAi i¼1 !i¼1 Xn !2 !Xn ! Xn ¼ þ Á þ 2 ¼ mi MG 2MG mi GAi miGAi i¼1 i¼1 i¼1 2 ¼ m Á MG þ IG,

Pn where mi ¼ m. Therefore, IM IG. i¼1 2 2 4.1.14. (a) As mR ¼ I0 ¼ mOG + IG (see the proof of problem 4.1.13), then 2 2 m(R À OG ) ¼ IG,whereO is the center and R is the radius of the given circle. 2 2 Hence, point Gis inside of the given circle, thus GAi Á GBi ¼ R À OG . 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 149

We have that ! ! ! 2 Xn Xn Xn pffiffiffiffiffiffiffiffiffiffiffiffiffi 2Xn pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 Xn pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi miGBi miGAi ¼ miGBi miGAi mi GAi Á GBi ¼ i¼1 i¼1 i ¼1 ! i¼1 ! i¼1 ! ÀÁ Xn pffiffiffiffiffiffiffi Xn pffiffiffiffiffi Xn 2 2 2 2 2 2 ¼ m R À OG ¼ mIG ¼ mi ðÞmiGAi miGAi i¼1 i¼1 i¼1

Pn Pn (see the proof of problem 4.1.15a). Therefore, miGBi miGAi. i¼1 i¼1 Remark Inequality holds true, if instead of a circle one considers a sphere. (b) We have that

! ! ! ! ! Xn Xn Xn pffiffiffiffiffi Xn pffiffiffiffiffi Xn 2 2 2 ¼ ðÞ2 ðÞ2 Á ¼ miGBi miGAi miGBi miGAi miGAi GBi i¼1 i¼1 i¼1 i¼1 i¼1 ! Xn ÀÁ2 ÀÁÀÁ ¼ 2 À 2 ¼ 2 À 2 2 ¼ 2 mi R OG mR OG IG i¼1 ÀÁ ¼ 2 þ ::: þ 2 2: m1GA1 mnGAn

Pn Pn 2 2 Therefore, miGBi miGAi (see the proof of problem 4.1.14 a) i¼1 i¼1 Pn Pn < p p Remark If 0 p 2, then miGBi miGAi . i¼1 i¼1 Indeed, we have that ! ! ! 2 Xn Xn Xn pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 2 p p Á ¼ 2 À 2 ¼ miGBi miGAi mi GBi GAi m R OG i¼1 i¼1 i¼1 ÀÁ p ¼ 2 m1 2 þ ::: þ mn 2 : m m GA1 m GAn

p Because fxðÞ¼x2 at 0 < p 2 is concave function on the interval [0 ; + 1 )and Pn mi ¼ ’ m 1, then by using the Jensen s inequality we deduce that i¼1 m m m ÀÁ m ÀÁ f 1GA2 þ ::: þ nGA2 1 fGA2 þ ::: þ n fGA2 ,or m 1 m n m 1 m n p m m 2 m m 1GA2 þ ::: þ nGA2 1 GA p þ ::: þ n GA p, consequently, m 1 m n m 1 m n 150 4 Application of Vectors

! ! Xn Xn p p p 2 m1 2 þ ::: þ mn 2 miGBi miGAi m GA1 GAn i¼1 i¼1 m m ÀÁ p þ ::: þ p 2: m1GA1 mnGAn

Pn Pn p p Hence, it follows that miGBi miGAi . ¼ ¼ i 1! i 1 ! 4.1.15. (а) Note that m MB þ ::: þ m MB ¼ ~0. Indeed, let m MB ¼ kA A ,..., i 1 n n !1 1 !1 2 m MB ¼ kA A . Under rotations by 90 or –90 vectors m MB , :::, m MB trans- n n n 1 ! ! ! i 1 !n n form into vectors kA A , :::, kA A , and because kA A þ ::: þ kA A ¼ ~0, then ! !1 2 n 1 1 2 n 1 m MB þ ::: þ m MB ¼ ~0. 1 1 n n ! ! þ ::: þ ¼ ~ Let C0 be such a point that m1C0C1 mnC0Cn 0, then X X ! ! 2 X 2 ¼ À ¼ 2 mimjCiCj C0Cj C0Ci mimj mimjC0Cj < < < i j i Xj X i !j ! þ 2 À Á ¼ mimjC0Ci 2mimjC0Ci C0Cj i

Pn Pn 2 ¼ ðÞþ ::: þ 2 Similarly, we obtain that mimjBiBj m1 mn miMBi . i 2 þ ::: þ 2 ÀÁIt is known that for any a1,...,an and b1,...,bn 0 inequality a1 an 2 þ ::: þ 2 ðÞþ ::: þ 2 b1 bn a1b1 anbn holds true, the equality being reachedÀÁ only if a1 ¼ ::: ¼ an. Indeed, it is not difﬁcult to check that a2 þ ::: þ a2 ÀÁb1 bn P ÀÁ 1 n 2 þ ::: þ 2 À ðÞþ ::: þ 2 ¼ À 2 b1 bn a1b1 anbn aibj ajbi 0, where the equality i

Xn 2 X Xn 2 X A A A A þ pffiffiffiffiffi i iþ1 2 ¼ pi ffiffiffiffiffii 1 Á ðÞ2 Á ðÞþ ::: þ mimjCiCj miCoCi m1 mn mi mi i¼1 i

4.1.16. (a) Let m1, m2, m3 be positive numbers, such that x ¼ m1m2, y ¼ m2m3, z ¼ m1m3. Then,

2 þ 2 þ 2 ¼ 2 þ 2 þ 2 xA1B1 yB1C1 zC1A1 m1m2A1B1 m2m3B1C1 m1m3C1A1 ðÞþ þ 2 ðÞþ þ 2 4 m1 m2 m3 SABC ¼ 4 xy yz zx SABC 2 2 2 2 þ 2 þ 2 BC þ AC þ AB xAB yBC zAC m1 m2 m3

(see the proof of problem 4.1.15a). To prove that the equality holds true, it is sufﬁcient to prove that inside of triangle ABC there exists point M, such that Á Á Á m1 MA1 ¼ m2 MB1 ¼ m3 MC1 ⊥ ⊥ ⊥ 2 BC AC AB , where MA1 BC, MB1 AC, MC1 AB, and A1 BC, B1 2 AC, C1 2 AB. 152 4 Application of Vectors

Take a point M inside of triangle ABC, such that ctg ∠MAC ¼ ctg ∠A þ 1 m2AB and ctg ∠MCA ¼ ctg ∠C þ 1 m2BC. Then sin ∠A m3AC sin ∠C m1AC

MC MA Á sin ðÞ∠A À ∠MAC m Á AB 1 ¼ ¼ 2 and MB1 MA Á sin ∠MAC m3 Á AC Á ðÞ∠ À ∠ Á MA1 ¼ MC sin C ACM ¼ m2 BC : MB1 MC Á sin ∠ACM m1 Á AC m Á MA m Á MB m Á MC Therefore, 1 1 ¼ 2 1 ¼ 3 1 : BC AC AB

(b) Let us choose points C1, C2, and C3 on edges A1A4, A2A4, and A3A4, respectively, S2 S2 S2 A1C1 ¼ A1A2A3 A2C2 ¼ A1A2A3 A3C3 ¼ A1A2A3 so that 2 , 2 , 2 . Then, it is not difﬁcult to C1A4 S C2A4 S C3A4 S A2A3A4 A1A3A4 A1A2A4 construct common point M of planes A1C2A3, A2C1A3, and A1C3A2.

The point M is inside of tetrahedron A1A2A3A4. We need to prove that MB1 MB2 MB3 MB4 ¼ ¼ ¼ , where B1 2 (A2A3A4), B2 2 (A1A3A4), SA2A3A4 SA1A3A4 SA1A2A4 SA1A2A3 B3 2 (A1A2A4), B4 2 (A1A2A3) and MB1 ⊥ (A2A3A4), MB2 ⊥ (A1A3A4), MB3 ⊥ (A1A2A4) и MB4 ⊥ (A1A2A3). S2 V V S A1A2A3 ¼ A1C1 ¼ A1A2A3C1 ¼ A1A2A3M ¼ MB4 A1A2A3 Indeed, we have that 2 , conse- S C1A4 VA A A C VA A A M MB1 SA A A A2A3A4 2 3 4 1 2 3 4 2 3 4 quently, MB4 ¼ MB1 . Similarly, we obtain that MB4 ¼ MB2 and MB4 ¼ MB3 . SA1A2A3 SA2A3A4 SA1A2A3 SA1A3A4 SA1A2A3 SA1A2A4 Now, by using problem 4.1.15b for numbers m1 ¼ m2 ¼ m3 ¼ m4 ¼ 1, we obtain that the given sum is minimal at those points.

4.1.17. Lemma 1 Let M0 be such a point, that for any point M the following Pn Pn inequality holds true: miMAi miM0Ai.If i¼1 i¼1 ! Pn M0Ai ~ (a) M0 6 Ai, i ¼ 1, 2, . . ., n, then mi ¼ 0, (4.4) M0Ai i¼1 ! Pn M0Ai (b) M0 Ak, then mi jmk: (4.5) M0Ai i¼1, i6¼k

The Proof ! Pn ~ ¼ ~ 6¼ ~ ~ ¼ M0Ai (a) Let miei S 0, where ei M A . Take point Mt so that the equality ¼ 0 i !i 1 ¼ Á~ 6¼ ~ > M0Mt t S 0 holds true, where t 0. ! MtAi Let us denote ~eiðÞ¼t , then we have that MtAi 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 153 ! Xn Xn ! ! Xn Xn ! ~ ðÞ¼ ~ðÞ þ ~ðÞ miM0Ai miM0Ai ei t M0Mt mi ei t miMtAi ei t i¼1 i¼1 i¼1 i¼1 Xn ¼ ftðÞþ miMtAi: ð4:6Þ i¼1

Now, we need to prove that there exists t0 > 0, such that f(t0) > 0, then from (4.6) Pn Pn > it follows that miM0Ai miMt0 Ai. This leads to a contradiction the condition i¼1 i¼1 ~ ¼ ~ of the lemma. Hence S 0.qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ∠ jj~ðÞÀ~ ¼ ðÞ~ðÞÀ~ 2 ¼ MtAiM0 < ∠ Because ei t ei ei t ei 2 sin 2 MtAiM0, then there jj~S exists t > 0, such that jj~eiðÞÀt0 ~ei < , for all i ¼ 1, 2, . . ., n. Then, 0 2nmi

! Xn ! Xn Xn ! ftðÞ¼0 M0M mi~ei ðÞ¼t M M m~e þ m M M ðÞ~e ðÞÀt ~e t0 0 0 t0 i i i 0 t0 i 0 i i¼1 i¼1 i¼1

Xn ! t ~S2 À m M M jj~e ðÞÀt ~e 0 i 0 t0 i 0 i ¼ i 1 Xn ~ ~2 ~ S t0 ~2 > t0S À t0 S mi ¼ S > 0: i¼1 2nmi 2

Pn ! ~ ¼ ~ > ¼ ~ (b) Let S miei mk. Take point Mt so that the equality AkMt tS i¼1, i6¼k holds true, where t > 0. Then, we have that

Xn Xn ! miAkAi miAkAi~eiðÞ¼t i¼1, i6¼k i¼1, i6¼k ! Xn Xn ! Xn ¼ AkMt mi~ei þ miAkMt ðÞþ~eiðÞÀt ~ei miMtAi: i¼1, i6¼k i¼1, i6¼k i¼1, i6¼k

~ q Let us denote S ¼ mk þ q and choose t , such that jj~eiðÞÀt0 ~ei < , i ¼ 1, 0 ðÞnÀ1 mi 2, . . ., n, i 6¼ k. Then,

Xn Xn q Xn m A A > t ~S2 À t ~S m þ m M A i k i 0 0 i ðÞn À 1 m i t0 i i¼1 i¼1, i6¼k i i¼1, i6¼k Xn Xn ¼ þ ¼ : mkMt0 Ak miMt0 Ai miMt0 Ai i¼1, i6¼k i¼1 154 4 Application of Vectors

The obtained inequality contradicts to the conditions of the lemma. Conse-

Pn ~ quently, miei mk. This ends the proof of the lemma. i¼1, i6¼k Remark By using the Weierstrass theorem one can prove that there exists a point Pn M, such that the sum miMAi is minimal (see [2], problem 19). i¼1

Lemma 2 If for point M0 the condition (4.4) or (4.5) is fulﬁlled and points A1,..., An are not on the same line, then for any point M, different from M0, the inequality Pn Pn mi Á MAi > mi Á M0Ai holds true. i¼1 i¼1

Proof Indeed, if condition (4.4) is satisﬁed, then at M 6 M0, we have that

Xn Xn ! ! Xn Xn ! Xn mi Á MAi > mi Á MAi~ei ¼ MM0 mi~ei þ miÁM0Ai~ei ¼ mi Á M0Ai, i¼1 i¼1 i¼1 i¼1 i¼1 and if condition (4.5) is satisﬁed, then we have that

Xn Xn ! ! Xn Xn ! ! mi Á MAi > mi Á MAi~ei À MAk mi~ei ¼ mi MAi À MAk ~ei i¼1 i¼1, i6¼k i¼1, i6¼k i¼1, i6¼k Xn ¼ mi Á AkAi: i¼1

Remark 1. From lemmas 1 and 2 it follows that, if points A1, A2,...,An are not on the same line, then point M0 is unique. 2. If in a statement of lemma 2 the condition that points A1, A2,...,An are not on Pn Pn the same line was not given, then mi Á MAi mi Á M0Ai i¼1 i¼1 4.1.18. The proof of the problem follows from the following two lemmas.

Lemma 1 If B1B2 ...Bn is a polygon inscribed into a polygon A1A2 ...An, (Bi 6 Aj and Bi 2 AiAi +1, An +1 A1), so that for any polygon C1C2 ...Cn (Ci 6 Aj, Ci 2 AiAi +1) and any positive numbers mi the inequality Pn Pn miCiCiþ1 miBiBiþ1, where Cn +1 C1, Bn +1 B1, holds true, then the i¼1 i¼1 following conditions are satisﬁed: ! AiAiþ1 ðÞmiÀ1~eiÀ1 À mi~ei ¼0, i ¼ 1, :::, n, ð4:7Þ ! BiBiþ1 where ~ei ¼ , m ¼ m , ~eo ¼ ~en. BiBiþ1 0 n 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 155

Proof Suppose that, at i ¼ k condition (4.7) does not hold true, that means that ! ! 0 0 ! 0 B À B B B 6¼ ðÞ~ À ~ 2 ~0 ¼ k 1 k ~0 ¼ k kþ1 0 AkAkþ1 mkÀ1ekÀ1 mkek . Let Bk AkAkþ1, ekÀ1 0 , ek 0 . BkÀ1Bk BkBkþ1 0 If points Bk and Bk are different, then we have that ! 0 ! þ > ~ þ m B B ~e0 ¼ mkÀ1BkÀ1Bk mkBkBkþ1 mkÀ1BkÀ1Bke kÀ1 k k kþ1 k ! ! ! ! ¼ 0 ~0 þ 0 ~0 þ 0 ~0 þ 0 ~0 ¼ mkÀ1BkÀ1Bk ekÀ1 mkÀ1BkBkekÀ1 mkBkBk ek mkBkBkþ1ek ! ÀÁ ¼ 0 þ 0 þ 0 ~0 À ~0 : ð : Þ mkÀ1BkÀ1Bk mkBkBkþ1 BkBk mkÀ1ekÀ1 mkek 4 8 ! ! ÀÁÀÁ 0 ! 0 0 0 ! 0 0 Let B B ¼ λAkA þ , B B mkÀ1~e À À mk~e ¼ λAkA þ mkÀ1~e À À mk~e . k k k 1 k k! k 1 k k 1 k 1 k Let us choose λ, so that λ Á A A ðÞm À ~e À À m ~e > 0. We need to prove that k kþ1 k 1 k 1 k kÀÁ 0 0 ! 0 0 one can choose B , so that B 2AkAkþ1 and λ Á AkA þ mkÀ1~e À À mk~e > 0. k! k k 1 k 1 k ðÞ~ À ~ ¼6¼ 0 Indeed, let AkAkþ1 mkÀ1ekÀ1 mkek a 0. Let us choose Bk, so that

0 a ~e À~ei < ,ati ¼ k À 1;k (see the proof of lemma 1 of problem 4.1.17), i 4miÁAkAkþ1 then

! ÀÁ! λ Á ~0 À ~0 À λ Á ðÞ~ À ~ ¼ AkAkþ1 mkÀ1ekÀ1 mkek AkAkþ1 mkÀ1ekÀ1 mkek ÀÁÀÁÀÁ ! 0 0 ¼ λ Á A A m À ~e À~e À À m ~e À~e k kþ1 k 1 kÀ1 k 1 k k k ÀÁÀÁ λ ! 0 0 a λ A A m À ~e À~e À À m ~e À~e : k kþ1 k 1 kÀ1 k 1 k k k 2

Therefore,

ÀÁ λ ! 0 0 ! a λ Á A A m À ~e À m ~e λ Á A A ðÞm À ~e À À m ~e À k kþ 1 k 1 kÀ1 k k k kþ1 k 1 k 1 k k 2 a λ ¼ > 0: 2

From the last expression and (4.8) we deduce that

þ > 0 þ 0 : mkÀ1BkÀ1Bk mkBkBkþ1 mkÀ1BkÀ1Bk mkBkBkþ1

::: 0 ::: We have obtained that for polygon B1B2 BkÀ1BkBkþ1 Bn that the considered sum is less than for polygon B1B2 ...Bk ...Bn. This leads to a contradiction.

Remark If m1 ¼ m2 ¼ ...¼ mn, then it is possible to prove lemma 1 more simply than in the general case. On the other hand, Bi Aj is impossible.

Lemma 2 If for the inscribed polygon B1B2 ...Bn the condition (4.7) is satisﬁed, Pn ! ~ then for any inscribed polygon C1C2 ...Cn the sum miCiCiþ1ei is constant and i¼1 156 4 Application of Vectors

Pn Pn miCiCiþ1 miBiBiþ1. i¼1 i¼1 Proof We have that

Xn ! Xn ! ! ~ ¼ þ ~ miCiCiþ1ei mi CiAiþ1 Aiþ1Ciþ1 ei i¼1 i¼1 Xn ! Xn ! ¼ ~ þ ~ ¼ miCiAiþ1ei miAiþ1Ciþ1ei i¼1 i¼1 Xn ! Xn ! ¼ ~ þ ~ miCiAiþ1ei miÀ1AiCieiÀ1 i¼1 i¼1 Xn ! ! Xn ! ¼ À ~ þ ~ ¼ miÀ1 AiAiþ1 CiAiþ1 eiÀ1 miCiAiþ1ei i¼1 i¼1 Xn ! Xn ! ¼ ~ À ðÞ~ À ~ miÀ1AiAiþ1eiÀ1 CiAiþ1 miÀ1eiÀ1 miei i¼1 i¼1 Xn ! ¼ ~ miÀ1AiAiþ1eiÀ1, i¼1 note that the last one is a constant. Pn Pn ! Pn ! Pn ~ ¼ ~ ¼ Thus, miCiCiþ1 miCiCiþ1ei miBiBiþ1ei miBiBiþ1. i¼1 i¼1 i¼1 i¼1

Remark If n is odd, then there exists no more than one polygon B1B2 ...Bn, while for even n there can exist an inﬁnite number of polygons B1B2 ...Bn. 4.1.19. Let O be the incenter of triangle ABC. We need the following property of point O: ! ! ! BC Á OA þ AC Á OB þ AB Á OC ¼ ~0: ð4:9Þ

Let us consider points A1, B1, C1 (Figure 4.3). Since, the circle with diameter OA ¼ ∠ ¼ BCÁOA passes through points B1 and C1, then B1C1 OA sin A 2R , where R is the circumradius of triangle ABC. From the above said conditions OA ⊥ B1C1, OB ⊥ C A , OC ⊥ A B , it follows that, under the rotation by 90 vectors !1 1 ! 1 1 ! ! ! ! 2RB1C1 ,2RC1A1 ,2RA1B1 transform into vectors BC Á OA, AC Á OB, AB Á OC. Thus, we have that condition (4.9) is satisﬁed. According to that condition 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 157

Figure 4.3 B B 1 A1 C1

A B1 C

! ! ! ! Á þ Á þ Á ¼ Á B1C1 O1A1 A1C1 O1B1 A1B1 O1C1 B2C2 O2A2! ! þA2C2 Á O2B2 þ A2B2 Á O2C2 : ð4:10Þ ! ! ! ! ! ! Taking into account that O A ¼ O O þ O A þ A A , O B ¼ O O þ ! ! ! ! 2 !2 2!1 1 1 1 2 2 2 2 1 O1B1 þ B1B2 , O2C2 ¼ O2O1 þ O1C1 þ C1C2 ,from(4.10) it follows that ! ! ! p Á O O ¼ O A ðÞB C À B C þO B ðÞA C À A C 2 1 2 11! 2 2 1 1 1 1 2 2! 1 1 ! þO C ðÞA B À A B þþB C Á A A þ A C ÁB B ð4:11Þ 1 1 2 !2 1 1 2 2 1 2 2 2 1 2 þA2B2 Á C1C2 : Therefore,

Á À þ À þ À þ p2 O1O2 O1A1 B2C2 B1C1 O1B1 A2C2 A1C1 O1C1 A2B2 A1B1

þB2C2 Á A1A2 þ A2C2 Á B1B2 þ A2B2 Á C1C2: ð4:12Þ

þ ðÞO1A1 þ A2B2 þ O1B1 C1C2: ð4:13Þ

Let us come back again to Figure 4.3. 0 Let point B1 be symmetric to B1 with respect to point O. For medians AO and CO 0 0 < 0 of triangles AB1B1 and CB1B1 we have the inequalitiesÀÁAO (AB 1 + AB1)/2 and < 0 þ < 0 þ 0 = þ = CO (CB 1 + CB1)/2. Therefore, AO CO AB 1 B1C 2 AC 2. On the 0 0 þ 0 < þ other hand, point B1 is inside of triangle ABC, AB 1 B1C AB BC, and AB þ BC þ AC AO þ CO < : ð4:14Þ 2 Then, using inequalities (4.13) and (4.14) we deduce that 158 4 Application of Vectors p p p p Á O O 1 þ B C A A þ 1 þ A C B B þ 1 þ A B C C < 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2 1 2 p þ p < 1 2 ðÞA A þ B B þ C C : 2 1 2 1 2 1 2 þ O1O2 p1 p2 Thus, þ þ < . A1A2 B1B2 C1C2 2p2 It is clear that, from the very beginning we could assume that max( p1, p2) ¼ p2.

Remark If A1 6 A2, B1 6 B2, then (4.13) takes the following form p2 Á O1O2 (O1A1 + A1B1 + O1B1)C1C2. Thus, p2 Á O1O2 < (C1A1 + A1B1 + C1B1) ¼ Á C1C2 p1 C1C2. Hence, it follows that O1O2 < min p1; p2 . C1C2 p2 p1 4.1.20. (a) Let P, Q, and R be the tangential points of the unit sphere with the faces ! ! ! ~ of trihedral angle. Then, ~p ¼ OP, q ¼ OQ, and ~r ¼ OR are unit vectors and 2 ~c~ π 2 ~c~ π 2 ~c~ π arccos3 p, q 2, arccos3 q, r 2, arccos3 p, r 2 . On the other ! ~ ¼ ~~ ¼ ~~ ¼ ~~ ¼ hand, for vector a OA, we havepffiffiffi that ap aq ar 1. We need to prove that jj~a 3. c~ c ~c c~ Let max ~p, q ; ~p,~r ; q,~r ¼ ~p, q ¼ φ, then there exist angles φ1 and ! ! α φ φ φ α < π ~ ¼ ; ; ~ ¼ φ; φ; , such that 2 1 , 2, and p 1 0 0 , q cos sin 0 , ! ~r ¼ cos φ cos α; sin φ cosα; sin α (see Figure 4.4). 1 p1 ffiffiffi If α ¼ π, then a ¼ 3. 2ÀÁ! If ~a ¼ x, y, z , then we find that x ¼ 1, x cos φ + y sin φ ¼ 1 and x cos φ1 α φ α α ¼ ¼ ¼ φ ¼ ÀÁcos + y sin 1 cosÀÁ+ z sin ÀÁ1. Consequently, x 1, y tg 2, z φ φ φ cos À cos α cos φ À = sin α cos . 2 1 2 ÀÁ2 ÀÁ φ À α φ À φ 2 2α 1þ3 cos φ Then, we have to prove that cos 2 cos cos 1 2 sin 2 .

Figure 4.4 z

О a y p j1 q

x 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 159

Figure 4.5

Figure 4.6 B¢ C¢

L ¢ ¢ A M0 MD e L 2 0 e 3 BC e 1

K0 K AD

ÀÁ ÀÁ φ φ À φ ðÞ¼ φ À α 2 WeÀÁÀÁ have that cos 2 cosÂÃ1 2 1. Therefore, ft cos 2 t cos φ φ max f cos ; f ðÞ1 ,att2 cos ; 1 . 2 ÀÁ 2ÀÁ ÀÁ φ þ φ þ φ þ φ Note that f cos ¼ 1 cos 4 sin α 4 1 3 cos 4 sin α 4 1 3 cos sin 2α. 2 2 2 ÀÁ2 2 2 ðÞ¼ φ À α 2 1þ3 cos φ 2α It remains to prove that f 1 cos 2 cos 2 sin ,or ðÞþ φ 2α À φ α À φ 31 cos cos 4 cos2 cos 2 cos 0. ~c~ ¼ α ðÞφ À φ We have that cos q, r cos cos 1 . Thus, it follows that pffiffi φ cos α cos π cos α cos cos α cos ðÞφ À φ 2. Hence cos α 2 2. We have 4 hip2ffiffi 1 3 3 ¼ α2 ; 2 2 ðÞ¼ ðÞþ φ 2 À φ À that t cos 0 3 . We deduce that gt 31 cos t 4 cos 2 t pffiffi φ 2 2 ; ðÞ 2 cos max g 3 g 0 . pffiffi pffiffi ¼À φ 2 2 ¼ 8 ðÞÀþ φ 8 2 φ À Note that g(0) 2 cos 0, and g 3 3 1 cos 3 cos 2 pffiffi hipffiffi φ ¼ 4 2φ À 8 2 φ þ ¼ φ 2 2; 2 cos 3 cos 2 3 cos 2 2 0, because t cos2 2 1 . Therefore, 160 4 Application of Vectors

pffiffi pffiffi pffiffi φðÞ¼4 2 À 8 2 À φ 2 ; φðÞ φ 2 ¼ t 3 t 3 t 2 max 2 1 0, because 2 0 and pffiffi φðÞ¼10 À 8 2 < 1 3 3 0. (b) Yes, it is possible (see Figure 4.5). First one has to construct “such” tetrahedron and then add another 96 faces. One should take AB ¼ CD, AD ¼ DB ¼ BC ¼ AC, where AB is large enough. 4.1.21. Let points K , L , M be the midpoints of edges AB, CC0, and A0D0, 0 0 0 ! ! ! K0M0 M0L0 L0M0 respectively. Let us denote ~e1 ¼ ,~e2 ¼ ,~e3 ¼ (see Figure 4.6). K0M0 M0L0 L0M0 Note that ! ! ! ! ! !~ þ ~ ¼ þ 0 þ 0 ~ þ KM~e1 þ ML e2 LK e3 KA AA A M e1 ! ! ! 0 0 0 0 ! ! ! þ MD þ D C þ C L ~e2þ LC þ CB þ BK ~e3¼ ! ! 0 0 0 ! ! ! ¼ AA ~e1þD C ~e2þCB~e3 þ KA~e1 þ BK~e3 þ ! ! ! 0 0 0 ! þ A M~e1 þ MD ~e2 þ C L~e2 þ LC~e3 ¼ ! ! 0 0 0 ! ! ! ¼ AA ~e1þD C ~e2 þ CB~e3þ BA~e1þBK ðÞ~e3À~e1 ! ! ! 0 0 0 0 þþA D ~e2 þ A MðÞþ~e1 À~e2 C C~e3 ! 0 þ C LðÞ~e2 À~e3 : ð4:15Þ

! ! ! Because BK ðÞ¼~e À~e 0, A0MðÞ¼~e À~e 0, C0LðÞ¼~e À~e 0 (see the proof 3 1 1 2 2 3! ! ! of problem 2.4.18а), then from (4.15) it follows that, the sum KM~e1 þ ML~e2 þ LK ~e3 is constant, this means that ! ! ! ! ! !~ þ ~ þ ~ ¼ ~ þ ~ þ ~ KM e1 MLe2 LK e3 K0M0e1 M0L0e2 L0K0e3

¼ K0M0 þ M0L0 þ L0K0: Thus, rffiffiffi ! ! ! 3 KM þ ML þ LK KM~e þ ML~e þ LK~e ¼ K M þ M L þ L K ¼ 3 AB 1 2 3 0 0 0 0 0 0 2 pffiffiffi ¼ 1, 5 6AB: 4.1.22. Let G be the center of mass of tetrahedron ABCD, and O be the circumcenter. ¼ 4 ¼ 4 ¼ 4 ¼ 4 þ þ Then, ma 3 AG, mb 3 BG, mc 3 CG, md 3 DG. We have that ma mb mc 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 161 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þm ¼ 4 ðÞAG þ BG þ CG þ DG 8 AG2 þ BG2 þ CG2 þ DG2 , because R2 ¼ d 3 3 ! ! 2 ! ! AO2 ¼ AG þ GO ¼ AG2 þ GO2 þ 2 Á GO Á AG . We obtain that

!! ! ! ! 4R2 ¼ AG2 þ BG2 þ CG2 þ DG2 þ 4GO2 þ 2GO AG þ BG þ CG þ DG ¼ ¼ AG2 þ BG2 þ CG2 þ DG2 þ 4GO2 AG2 þ BG2 þ CG2 þ DG2: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ þ 8 2 þ 2 þ 2 þ 2 16 Thus, ma mb mc md 3 AG BG CG DG 3 R (see problem 4.1.13). ! ! ! 4.1.23. Let CA ¼ a Á~i, CB ¼ b Á~j, CD ¼ x Á~i þ y Á~j þ z Á~k, where ~i,~j,~k are ! ! BD coordinate vectors and z 6¼ 0. We have that cos φ ¼ CA Á ACÁBD x ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, where φ is the angle between lines AC and BD,as x2 þ ðÞy À b 2 þ z2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2þ 2þ 2 ! ! CD ¼ px ffiffiffiffiffiffiffiffiffiy z and 0 ¼ AD Á BD ¼ xxðÞþÀ a yyðÞþÀ b z2. We have to prove AB a2þb2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ that x2 a2 þ b2 < ðÞx2 þ y2 þ z2 x2 þ y2 þ z2 þ b2 À 2by ,or x2 a2 þ b2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ < ðÞax þ by ax À by þ b2 , b2(x2 + y2) < b2(ax + by), 0 < z2.

4.1.24. (a) From the statement of the problem we have that ! ðÞmi~ei À miÀ1~eiÀ1 AiBi ¼ 0, i ¼ 1, 2, . . ., n. Consequently,

m1A1A2 þ m2A2A3 þ ::: þ mnÀ1AnÀ1An þ mnAnA1 ¼ ! ! ! ! ¼ m1A1A2~e1 þ m2A2A3~e2 þ ::: þ mnÀ1AnÀ1An~enÀ1 þ mnAnA1~en ¼ ! ! ! ! ! ! ¼ m1 A1B1 þ B1B2 þ B2A2 ~e1 þ m2 A2B2 þ B2B3 þ B3A3 ~e2 þþ::: ! ! ! ! ! ! þ þ þ ~ þþ þ þ ~ ¼ mnÀ1 AnÀ1BnÀ1 BnÀ1Bn BnAn enÀ1 mn AnBn BnB1 B1A1 en ! ! ! ¼ ðÞþ~ À ~ ðÞþ~ À ~ ::: þ ðÞ~ À ~ A1B1 m1e1 mnen A2B2 m2e2 m1e1 AnBn mnen mnÀ1enÀ1 ! ! ! ! þþ ~ þþ ~ þ ::: þ ~ þ ~ ¼ m1B1B2e1 m2B2B3e2 mnÀ1BnÀ1BnenÀ1 mnBnB1en ! ! ! ! ¼ ~ þ ~ þ ::: þ ~ þ ~ m1B1B2e1 m2B2B3e2 mnÀ1BnÀ1BnenÀ1 mnBnB1en

m1B1B2 þ m2B2B3 þ ::: þ mnÀ1BnÀ1Bn þ mnBnB1:

Hence, we deduce that m1B1B2 + m2B2B3 +...+mn À 1Bn À 1Bn + mnBnB1 m1A1A2 + m2A2A3 +...+mn À 1An À 1An + mnAnA1.

(b) Let point M be the midpoint of edge C1C3 and point N be the midpoint of edge C2C4. Let segment A1A4 be the common perpendicular of lines C3N and C2M, that is, A1 2 C3N, A4 2 C2M, A1A4 ⊥ C2M, A1A4 ⊥ C3N. From the statement of the problem it follows that points C1 and C3 are symmetric with respect to plane 162 4 Application of Vectors

C2C4M. Therefore, if point A3, symmetric to A1 with respect to plane C2C4M, then A3 2 C1N, A3A4 ⊥ C2M, A3A4 ⊥ C1N.

Similarly, we obtain that, if point A2 is symmetric to A4 with respect to plane C1C3N, then A2 2 C4M and A2A3 ⊥ C1N, A2A3 ⊥ C4M, A1A2 ⊥ C4M, A A ⊥ C N. 1 2 3! AiAiþ1 Let~ei ¼ , i ¼ 1, 2, 3, 4; A A . Because~e3⊥C2M and~e4⊥C2M, we have AiAiþ1 5 1 that ~e4 À~e3⊥C2M. Also, A4A1 ¼ A4A3, A1A3 ⊥ C2C4M, C1C3 ⊥ C2C4M. Consequently, A1A3||C1C3 and ~e4 À~e3⊥A1A3. Hence ~e4 À~e3⊥C1C3. We have obtained that ~e4 À~e3⊥C2M, ~e4 À~e3⊥C1C3. Therefore, ~e4 À~e3⊥C1C2C3. The proof that other three conditions of problem 4.1.24a hold also true can be done similarly. Thus, B1B2 + B2B3 + B3B4 + B4B1 A1A2 + A2A3 + A3A4 + A4A1.To complete the solution we have to notice that points A1, A2, A3, A4 are on segments C3N, C4M, C1N, C2M, respectively. Indeed, to prove, for example, that point A1 is on segment C3N one has to consider projections of points A4, A1, C3, N on a plane passing through point C2 and perpendicular to line C2M. 0 0 0 0 0 (c) Let points A1, A2, A3, A4, A5, A6 belong to the segments AC, A B, B C, A C , CD , DA0, respectively, and

0 0 0 0 0 0 0 0 AA1 : AC ¼ BA2 : BA ¼ B A3 : B C ¼ C A4 : C A ¼ D A5 : D C 0 ¼ DA6 : DA ¼ 1 : 3:

Now, it is not difﬁcult to check that the conditions of problem 4.1.24a are satisﬁed, where m1 ¼ m2 ¼ m3 ¼ m4 ¼ m5 ¼ m6 ¼ 1. Thus, it follows that

B1B2 þ B2B3 þ B3B4 þ B4B5 þ B5B6 þ B6B1 A1A2 þ A2A3 þ A3A4 þ A4A5 þ A5A6 þ A6A1:

4.1.25. (a) Let us consider midpoints M, N, P, K, F, E of edges AD, BD, CD, AB, AC, BC, respectively. As OM ⊥ AD, OK ⊥ AB, OF ⊥ AC, then AO > AM, AO > AK, and AO > AF. According to problem 7.1.38a tetrahedron AMKF does not contain point O. Then, we obtain that point O is inside of the polyhedron with faces MPF, MFK, FKE, PNE, MNK, FPE, MPN, NKE. For point X 6 O draw a plane α passing through ! point O and perpendicular to vector OX. It is obvious, that at least one of the points M, N, P, K, F, E and point X are on the different sides of plane α. If that is the point ! ! M, then OM Á OX < 0. ! ! ! ! Let ~e ¼ OA ,~e ¼ OB ,~e ¼ OC ,~e ¼ OD and ~e ¼ ~e þ~e þ~e þ~e .We 1 OA 2 OB 3 OC 4 OD 1 2 3 4 need to prove that 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 163

~e < 2: ð4:16Þ

Indeed, it is sufﬁcient to prove inequality (4.16) in case ~e 6¼ ~0. ! ! ! Let OX ¼ ~e, then as it was mentioned OM Á OX < 0. Consequently, ðÞ~ þ~ ~ < ~ ~ ~ ~ ~ ~ ~ þ~ þð e1 e4 e 0. We have that e 2e e and e3e e . Thus, e1 e2 2 ~e3 þ~e4Þ~e < 2 ~e . Hence ~e < 2 ~e , then ~e < 2. Note that

DA2 þDB2 þDC2 þ AB2 þ BC2 þ AC2 ¼ 2 2 2 2 2 2 2 ¼ AO ðÞ~e1 À~e4 þ ðÞ~e2 À~e4 þ ðÞ~e3 À~e4 þ ðÞ~e1 À~e2 þ ðÞ~e2 À~e3 þ ðÞ~e1 À~e3 ¼ ¼ AO2ðÞ16 Àj~e2 > 12 Á AO2:

Thus, DA2 + DB2 + DC2 + AB2 + BC2 + AC2 > 12 Á AO2. (b) Note that

2OA Á DA þ 2OA Á DB þ 2OA Á DC þ 2OA Á AB þ 2OA Á BC þ 2OA Á AC > > DA2 þ DB2 þ DC2 þ AB2 þ BC2 þ AC2 > 12 Á AO2

(see problem 4.1.25a). Therefore, DA + DB + DC + AB + BC + AC > 6 Á OA.

Remark Given a polyhedron with n vertices and a point M inside of it. Let ~ei be a unit vector directed from point M to the i-th vertex of the polyhedron, then

~e1 þ~e2 þ ::: þ~en < n À 2.

Problems for Self-Study

4.1.26. Let ABCD be a quadrilateral circumscribed around a circle. Given that the opposite sides AB and CD, BC and AD are on the lines intersecting at points M and N. Prove that cos ∠A + cos ∠B + cos ∠C + cos ∠D + cos ∠M + cos ∠N 2.

4.1.27. Let ABC be an arbitrary triangle. Prove thatÀÁ for any equilateral triangle A B C the inequality A A2 þ B B2 þ C C2 1 AB2 þ BC2 þ CA2 À p2ﬃﬃ S 1 1 1 1 1 1 6 3 ABC holds true.

4.1.28. Given points A1,...,An and positive numbers m1,...,mn. For any point X let us denote by f(X) the expression m1A1X +...+mnAnX. Given that point M is BM ¼ α α À α on segment AB, such that AB . Prove that f(M) f(A)+(1 )f(B).

4.1.29. Given that pointsA1, A2,...,An are on the same sphere, m1,...,mn > 0 and ! ! ~ G is such a point that m1GA1 þ ::: þ mnGAn ¼ 0. Let straight lines GA1,...,GAn intersect this sphere (for the second time) at points B1, B2,...,Bn. Prove that P þ ::: þ P P þ ::: þ P m1GB1 mnGBn m1GA1 mnGAn , where 0 p 2. 164 4 Application of Vectors

4.1.30. Let O be the incenter of triangle ABC. Prove that ab + bc + ac (AO + BO + CO)2. Hint Prove that aAO2 + bBO2 + cCO2 ¼ abc (see problem 4.1.8b). 4.1.31. Let the medians of faces ABD, ACD, BCD drawn from vertex D of tetrahedron ABCDform equal angles with edges AB, AC, BC, respectively. Prove that the area of each of these faces is not greater than the sum of the areas of the other two faces. ! ! ! DA þ DB ; ¼ α 6¼ Hint Let 2 AB 90 , then

! ! ! ¼ 1 Á DA þ DB jjα ¼ 1 2 À 2 α SADB 2 AB 2 ctg 4 DA DB ctg . For α ¼ 90 see problem 1.1.14a (not only α ¼ 90). 4.1.32. Let in a tetrahedron ABCD the angles ADB, ADC, BDC be obtuse and the lengths of edges AD, BD, CD be equal. Prove that ABC is an acute triangle. ! ! Hint Prove that CA Á CB > 0. 4.1.33. Prove that the difference of the squares of the lengths of the adjacent sides of a parallelogram is less than the product of its diagonals. 4.1.34. Let n points be inside of the unit sphere. Prove that the sum of the squares of distances between all possible pairs of these points does not exceed n2. 4.1.35. (a) Consider a convex hexagon, such that the length of each of its sides is more than 1. Will there always be in it a diagonal with the length more than 2? (b) Consider a convex hexagon ABCDEF, such that the lengths of the diagonals AD, BE, CFare greater than 2. Will it always have a side with the length more than 1? Hint See problem 4.1.4a. 4.1.36. Let M be the intersection point of the diagonals of an inscribed quadrilateral, N be the intersection point of its midlines and O be its circumcenter. Prove that OM ON (the midline is a segment connecting the midpoints of the opposite sides). ! ¼ Hint Let ABCD be an inscribed quadrilateral, then ON ! ! ! ! ! ! ! ! ! 2 1 þ þ þ 2 OA þ OB þOCþOD 4 OA OB OC OD , Prove that OM 4 .

4.1.37. Prove that for any triangle ABC the following inequalities hold true: 2 þ 2 þ 2 2; α þ β þ γ À3 (a) a b c 9R cos 2 cos 2 cos 2 2, (b) a3 þb3 þc3 þ3abc a2b þ b2a þ a2c þ ac2 þ b2c þ bc2; cos α þ cos β þ cos γ 3 2, 2 2 þ 2 þ 2; α þ β À γ 3 (c) c a b R cos 2 cos 2 cos 2 2. 4.1 Application of Vectors for Proving Geometric and Trigonometric Inequalities 165

4.1.38. Let α1, α2,...,α6 be (the linear angles of) the dihedral angles of a tetrahedron. Prove that cosα1 + cos α2 +...+cosα6 2. ¼ ¼ ¼ 1 4.1.39. Use problem 4.1.8a for x y z 3 to prove problem 4.1.4b. ! ! ! ! ! Hint Consider points X, Y, Z, and M, such that XY ¼ BC þ CD À EF À FA , ! ! ! ! ! ! ! ! YZ ¼ FA þ AB À DE À CD, and XM ¼ BC À EF . 4.1.40. The rectangular projection of a triangular pyramid to some plane has the greatest possible area. Prove that this plane is parallel either to one of the sides of a pyramid or to two skew edges of the pyramid. 4.1.41. (a) Given that in a tetrahedron ABCD the sum of the cosines of all plane angles at vertex D does not exceed –1. Prove that for any point M, other than D, holds true MA + MB + MC + MD > DA + DB + DC. (b) Given that in a tetrahedron ABCD the sum of the cosines of all plane angles at vertex D does not exceed –1. Prove that inside this tetrahedron one can ﬁnd a point M0, such that ∠AM0B ¼ ∠CM0D, ∠AM0C ¼ ∠BM0D, ∠AM0D ¼ ∠BM0C.

Moreover, for any point M, other than M0, it holds true MA + MB + MC + MD > M0A + M0B + M0C + M0D. Hint See the proof of problem 4.1.17.

4.1.42. Let triangles A1A2A3 and B1B2B3 with orthocenters H1 and H2, respectively, be inscribed in a circle of radius R. Prove that H1H2 < 4R + A1B1. Hint Let G and G be the centroids of those triangles A A A and B B B , 1 2 ! ! 1 2 !3 !1 2 3 ¼ ¼ 1 þ þ : respectively. Prove that H1H2 3G1G2 and G1G2 3 A1B1 A2B2 A3B3 Chapter 5 Application of Trigonometric Inequalities

This chapter consists of five sections, that is, Sections 5.1, 5.2, 5.3, 5.4, and 5.5. Many problems in geometry can be solved by applying trigonometry. In particular, many problems related to geometric inequalities can be deduced to trigonometric inequalities. Section 5.1 mainly provides trigonometric inequalities concerning to angles of a triangle. Let us emphasize few methods of proving such inequalities: note that, if α β γ πÀα πÀβ πÀγ , , are the angles of some triangle, then 2 , 2 , 2 are the angles of some acute triangle. Therefore, if some inequality holds true for angles α, β, γ of some πÀα πÀβ πÀγ triangle, then substituting these angles by 2 , 2 , 2 one can obtain a “new” inequality for angles α, β, γ. Moreover, if some inequality holds true for the angles of any acute triangle, then from this inequality one can obtain a “new” inequality for the angles of any triangle. In Section 5.1 many problems are proved using the maximal (minimal) values of a quadratic polynomial. One of the crucial methods explained in this section is the method based on the following statement: if the quadratic coefficient of a quadratic function is positive (negative), then on any segment that function accepts its maximal (minimal) value in one of the endpoints of the considered segment. Section 5.2 selects such inequalities, concerning the angles of a triangle, that hold true either only for acute triangles or only for obtuse triangles. One of the most important methods of proving geometric inequalities is based on modifications of trigonometric expressions. Therefore, Section 5.3 is devoted to some important mathematical identities related to triangles. Section 5.4 considers some trigonometric inequalities that are later on applied in Section 5.5 in order to prove geometric inequalities. Summarizing the above mentioned, Sections 5.1, 5.2, 5.3, and 5.4 provides methods related to trigonometry in order to prove geometric inequalities. These types of geometric inequalities are considered in Section 5.5. Some problems in this chapter were inspired by [1, 2]. Nevertheless, even for these problems the authors have mostly provided their own solutions.

© Springer International Publishing AG 2017 167 H. Sedrakyan, N. Sedrakyan, Geometric Inequalities, Problem Books in Mathematics, DOI 10.1007/978-3-319-55080-0_5 168 5 Application of Trigonometric Inequalities

5.1 Inequalities for the Angles of a Triangle

Let α, β, and γ be the angles of some triangle. In the problems of this section it is required to prove the following inequalities. 2α 2β 2γ 3 5.1.1. cos þ cos þ cos 4. α β γ 1 5.1.2. cos cos cos 8. α β γ 3 5.1.3. cos2 þ cos2 À cos2 2. < α β γ 3 5.1.4. (a) 1 cos þ cos þ cos 2, ffiffiffiÀÁÀÁ p α β γ 2 α β γ α β γ (b) 3 cos 2 þ cos 2 þ cos 2 4 cos 2 þ cos 2 þ cos 2 þ 2 cos 2 cos 2 cos 2. < α β γ 3 5.1.5. 1 sin 2 þ sin 2 þ sin 2 2. pffiffi α β γ 3 3 5.1.6. sin þ sin þ sin 2 . pffiffi α β γ 3 3 5.1.7. cos 2 þ cos 2 þ cos 2 2 . pffiffiffi 5.1.8. (a) ctgα þ ctgβ þ ctgγ 3, ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 2β 2γ 1þ8cos 2α 1þ8cos 1þ8cos (b) sin α þ sin β þ sin γ 6. ffiffiffi α β γ p 5.1.9. (a) tg 2 þ tg 2 þ tg 2 3, ÀÁ β γ β γ (b) tg α tg tg tg α tg tg 1. 2 2 2 2 þ 2 þ ÀÁ2 3 β γ β γ (c) ctg 2α þ ctg 2 þ ctg 2 ctg α þ ctg þ ctg ðÞctgα þ ctgβ þ ctgγ . 2 2 2 ffiffiffi2 2 2 α β γ p 5.1.10. ctg 2 þ ctg 2 þ ctg 2 3 3. α β γ 1 5.1.11. sin 2 sin 2 sin 2 8. pffiffi α β γ 3 3 5.1.12. sin Á sin Á sin 8 . pffiffi α β γ 3 3 5.1.13. cos 2 cos 2 cos 2 8 . α β β γ γ α 3 5.1.14. cos cos þ cos cos þ cos cos 4. 5.1.15. sin2α þ sin 2β þ sin 2γ sin α þ sin β þ sin γ. α β γ α β γ 5.1.16. sin þ sin þ sin cos 2 þ cos 2 þ cos 2. α β γ α β γ 5.1.17. ctg þ ctg þ ctg tg 2 þ tg 2 þ tg 2. α β γ α β γ 5.1.18. cos þ cos þ cos sin 2 þ sin 2 þ sin 2. 2α 2β 2γ 2α 2β 2γ 5.1.19. ctg þ ctg þ ctg tg 2 þ tg 2 þ tg 2. α β γ α β γ 5.1.20. (a) cos Á cos Á cos sin 2 sin 2 sin 2, α β γ α β γ (b) sin Á sin Á sin cos 2 cos 2 cos 2, α β γ 2α 2β 2γ (c) cos cos cos 8sin 2 sin 2 sin 2, 5.1 Inequalities for the Angles of a Triangle 169

pffiffiffi α β γ 1 α β γ α β γ (d) sin 2 þ sin 2 þ sin 2 2 À sin 2 sin 2 sin 2 þ 3 cos 2 cos 2 cos 2. 2 αÀβ 2 βÀγ 2 γÀα α β γ (e) cos 4 þ cos 4 þ cos 4 2ðÞ cos þ cos þ cos , α β β γ γ α (f) cos À þ cos À þ cos À p2ffiffi ðÞsin α þ sin β þ sin γ , 2 2 2 3 pffiffiffi (g) cos α þ cos β þ cos γ þ ctgα þ ctgβ þ ctgγ 1, 5 þ 3, (h) (1 À cos α)(1 À cos β)(1 À cos γ) cos α cos β cos γ (4 À 2 cos α À 2 cos β À 2 cos γ), pffiffiffi (i) sin 2α þ sin 2β þ sin 2γ 2 3ðÞcos α cos β þ cos β cos γ þ cos γ cos α : qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi jjαÀβ jjβÀγ jjγÀα 71þ17 17 5.1.21. sin 2 þ sin 2 þ sin 2 32 . 5.1.22. (a) 2xy cos α þ 2yzcosβ þ 2xz cos γ x2 þ y2 þ z2, where α þ β þ γ ¼ π and α, β, γ, x, y, z are arbitrary numbers.

cos α cos β cos γ (b) α þ β þ γ ctg α1 þ ctg β þ ctg γ , where α þ β þ γ ¼ π and sin 1 sin 1 sin 1 1 1 α1, β1, γ1 are angles of some triangle. 2 2 2 (c) a ctgα1 þ b ctgβ1 þ c ctgγ1 4S, where a, b, c, S are respectively, the sides and the area of some triangle, and α1, β1, γ1 the angles of another triangle. (d) (b2 þ c2)(1 À cos φ) þ a2 cos φ 4S| sin φ|, where φ is any angle and S is the areaÀÁ of the triangle withÀÁ sides a, b, c. ÀÁ 2 2 2 2 2 2 2 2 2 2 2 2 (e) a b1 þ c1 À a1 þ b a1 þ c1 À b1 þ c a1 þ b1 À c1 16SS1, where a, b, c, S are respectively, the sides and the area of some triangle, and a1, b1, c1, S1 of another triangle. (f) (xa2 þ yb2 þ zc2)2 16S2(xy þ yz þ zx), where x, y, z are arbitrary numbers and S is the area of a triangle with sides a, b, c. 2 2 2 (g) k tgα1 þ l tgβ1 þ m tgγ1 S, where k, l, m are the distances from the circumcenter of triangle ABC to lines BC, AC, AB, respectively, S is the area of triangle ABC, and α1, β1, γ1 are the angles of some other acute triangle. ÀÁ 2α 2β 2γ α β γ 2 5.1.23. cos 2 þ cos 2 þ cos 2 sin 2 þ sin 2 þ sin 2 . α β γ πþα πþβ πþγ 5.1.24. sin 2 þ sin 2 þ sin 2 cos 8 þ cos 8 þ cos 8 . 2α 2β γ > 3 5.1.25. cos þ cos þ cos 4. 2α 2β 2β 2γ 2γ 2α 3 α β γ 5.1.26. cos cos þ cos cos þ cos cos 2 cos cos cos . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2þy2þ2xy cos α y2þz2þ2yz cos β z2þx2þ2zx cos γ 5.1.27. sin α þ sin β þ sin γ 2x þ 2y þ 2z, where x 0, y 0, z 0. 5.1.28. 3(cosα þ cos β þ cos γ) 2(sinα sin β þ sin β sin γ þ sin γ sin α). α β γ 3π α β γ 5.1.29. sin 3 Á sin 3 Á sin 3 8sin 9 Á sin 2 Á sin 2 Á sin 2. 2α 2β 2γ γ 4 5.1.30. tg 2 þ tg 2 þ tg 2 2, where 2arctg3. α β γ α β γ α α β β γ γ 1 1 1 5.1.31. sin sin 1 þ sin sin 1 þ sin sin 1 144 sin 2 sin 2 sin 2 sin 2 sin 2 sin 2 , where α, β, γ and α1, β1, γ1 are angles of some triangles. 170 5 Application of Trigonometric Inequalities

5.1.32. Prove that

α α β β γ γ sin sin 1 þ sin sin 1 þ sin sin 1 α β γ α β γ 2 þ ðÞcos þ cos þ cos À 1 ðÞcos 1 þ cos 1 þ cos 1 À 1 , where α, β, γ and α1, β1, γ1 are the angles of some triangles.

Solutions

5.1.1. We have that 1 þ cos 2α 1 þ cos 2β cos 2α þ cos 2β þ cos 2γ ¼ þ þ cos 2γ ¼ 2 2 ¼ 1 þ cos ðÞα þ β cos ðÞþα À β cos 2γ ¼ 3 sin 2ðÞα À β ¼ 1 À cos γ cos ðÞþα À β cos 2γ ¼ þ 4 4 α β 2 3 þ cos γ À cos ðÞÀ : 2 4 This ends the proof. 5.1.2. Since (see the proof of the problem 5.1.1)

cos 2α þ cos 2β þ cos 2γ ¼ 3 ¼ 1 À cos γðÞcos ðÞÀα À β cos γ ¼1 À 2 cos α cos β cos γ , 4 1 then cos α Á cos β Á cos γ . 8 This ends the proof. 5.1.3. Let us note that 3 1 À ðÞcos 2α þ cos 2β À cos2γ ¼2cos 2γ þ 2 cos γ cos ðÞþα À β ¼ 2 2 1 ¼ ðÞ2 cos γ þ cosðÞα À β 2 þ sin 2ðÞα À β 0: 2 This ends the proof. 5.1.4. (a) We have that γ α À β γ cos α þ cosβ þ cos γ ¼ 2 sin cos þ 1 À 2sin 2 ¼ 2 2 2 α β γ ¼ 1 þ 4 sin sin sin > 1: 2 2 2 α β γ πÀα πÀβ πÀγ If , , are the angles of a triangle, then 2 , 2 , 2 are also angles of some acute triangle. Therefore, using the inequality of the problem 5.1.2, we deduce that πÀα πÀβ πÀγ 1 α β γ 1 α β γ cos 2 cos 2 cos 2 8 or sin 2 sin 2 sin 2 8, hence cos þ cos þ cos 1 3 1 þ 4 Á 8 ¼ 2. 5.1 Inequalities for the Angles of a Triangle 171

(b) We have that α β γ α þ β β þ γ γ þ α cos þ cos þ cos ¼ 4 cos cos cos , 2 2 2 4 4 4 α β γ β þ γ α þ γ α þ β cos cos cos ¼ sin sin sin ¼ 2 2 2 2 2 2 α þ β β þ γ α þ γ α þ β β þ γ γ þ α ¼ 8 cos cos cos Á sin sin sin : 4 4 4 4 4 4 We have to prove that

pffiffiffi α þ β β þ γ γ þ α α þ β β þ γ γ þ α 3 cos cos cos 1 þ sin sin sin , 4 4 4 4 4 4 or pffiffiffi 3 α β γ 1 α β γ cos þ cos þ cos 1 þ sin þ sin þ sin À 1 , 4 2 2 2 4 2 2 2 α π β π γ π 3 cos þ þ cos þ þ cos þ , 2 6 2 6 2 6 2

(see the problem 5.1.4a). This ends the proof. < πÀα πÀβ πÀγ 3 5.1.5. From the problem 5.1.4a, we have that 1 cos 2 þ cos 2 þ cos 2 2. < α β γ 3 Hence, 1 sin 2 þ sin 2 þ sin 2 2. This ends the proof. π xþy xÀy xþy 5.1.6. If x, y 2 [0, ], then sin x þ sin y ¼ 2 sin 2 cos 2 2 sin 2 . Consequently,

π α β γ π= π α þ β γ þ þ þ þ 3 π sin α þ sin β þ sin γ þ sin 2 sin þ 2 sin 3 4 sin 2 2 ¼ 4 sin : 3 2 2 2 3 pffiffi α β γ π 3 3 Hence, we obtain that sin þ sin þ sin 3 sin 3 ¼ 2 . This ends the proof. π β π γ 5.1.7. Using the inequality of the problem 5.1.6 for angles πÀα , À , À , we deduce pffiffi 2 2 2 α β γ 3 3 that cos 2 þ cos 2 þ cos 2 2 . This ends the proof. 5.1.8. (a) Note that sin γ cos γ 2 sin γ cos γ ctgα þ ctgβ þ ctgγ ¼ þ ¼ þ sin α sin β sin γ cosðÞþα À β cos γ sin γ γ 0 1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 u 2 sin γ cos γ γ 1 À tg 1 1 γ u 1 γ pffiffiffi 2 @ A t : γ þ γ ¼ 2tg þ γ ¼ γ þ 3tg γ 3tg ¼ 3 1 þ cos sin 2 2tg 2 tg 2 tg 2 2 2 2 172 5 Application of Trigonometric Inequalities

(b) We have that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 8cos 2α 1 þ 8cos 2β 1 þ 8cos 2γ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ ¼ 1 þ 9ctg 2α þ 1 þ 9ctg 2β þ 1 þ 9ctg 2γ sinα sinβ sinγ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 þ 9ðÞctgα þ ctgβ þ ctgγ 2 6 (see the problem 1.2.5a and 5.1.8a). This ends the proof. π β π γ 5.1.9. (a) Using the inequality of the problem 5.1.8a for angles πÀα , À , À ,we pffiffiffi 2 2 2pffiffiffi πÀα πÀβ πÀγ α β γ obtain that ctg 2 þ ctg 2 þ ctg 2 3. Therefore, tg 2 þ tg 2 þ tg 2 3. ÀÁ β γ β β γ γ (b) We have that α , consequently α α 1. ÀÁtg 2 þ 2 ¼ ctg 2 tg 2 tg 2 þ tg 2 tg 2 þ tg 2 tg 2 ¼ α β β γ α γ α β α γ β γ 2 Hence 1 3 tg 2tg 2tg 2tg 2 þ tg 2tg 2tg 2tg 2 þ tg 2tg 2tg 2tg 2 , since (x þ y þ z) 3(xy þ yz þ zx). ÀÁ α β γ 2α 2β 2γ α (c) One needs to prove that tg 2 Á tg 2 Á tg 2 ctg 2 þ ctg 2 þ ctg 2 ctg þ ctgβ þ ctgγ, or equivalently α β α γ α α β β γ β 1 À tg Á tg À tg Á tg Á ctg þ 1 À tg Á tg À tg Á tg Á ctg þ 2 2 2 2 2 2 2 2 2 2 α β γ α γ β γ γ tg tg tg 1 2 1 2 1 2 þ 1 À tg Á tg À tg Á tg Á ctg α À þ À þ γ À , 2 2 2 2 2 2 β 2 2 2tg 2tg 2tg 2 2 2 1 α β γ α β γ or tg Á tg Á tg tg þ tg þ tg . 3 2 2 2 2 2 2 The last inequality follows from problem 5.1.9b. This ends the proof. ÀÃ ; π sin ðÞxþy 2 sin ðÞxþy xþy 5.1.10. If x, y2 0 2 , then ctgx þ ctgy ¼ sin x sin y 1À cos ðÞxþy ¼ 2ctg 2 . Hence, it follows that γ π α þ β þ γ π α β γ π α þ β þ þ π ctg þ ctg þ ctg þ ctg 2ctg þ 2ctg 2 6 4ctg 4 12 ¼ 4ctg : 2 2 2 6 4 2 2 6 ffiffiffi α β γ π p Therefore, ctg 2 þ ctg 2 þ ctg 2 3ctg 6 ¼ 3 3. This ends the proof.

πÀα πÀβ πÀγ 5.1.11. Using the inequality of the problem 5.1.2 for angles 2 , 2 , 2 , we obtain α β γ 1 that sin 2 sin 2 sin 2 8. This ends the proof. 5.1.12. Note that 1 1 À cos ðÞx þ y sin x Á sin y ¼ ðÞcos ðÞÀx À y cos ðÞx þ y ¼ 2 2 x þ y ¼ sin 2 : 2 5.1 Inequalities for the Angles of a Triangle 173

Therefore, we obtain that

π α β γþπ π α þ β γ þ þ þ 3 π sin α sin β sin γ sin sin 2 Á sin 2 3 sin 4 2 2 ¼ sin 4 : 3 2 2 2 3 pffiffi α β γ 3π 3 3 This means that sin sin sin sin 3 ¼ 8 . This ends the proof. π β π γ 5.1.13. Using the inequality of the problem 5.1.12 for angles πÀα , À , À ,we pffiffi 2 2 2 α β γ 3 3 obtain that cos2 cos 2 cos 2 8 . This ends the proof. α < π 5.1.14. Let 2, then 3 cos αcosβ þ cosβ cos γ þ cos γ cos α À ¼ 4 β þ γ β À γ 1 3 ¼ 2 cos α cos cos þ ðÞcos ðÞþβ À γ cos ðÞβ þ γ À 2 2 2 4 α 1 3 1 α 2 α 2 cos α sin þ ðÞÀ1 À cos α ¼À 2 sin À 1 4 sin þ 3 0: 2 2 4 4 2 2

This ends the proof. 5.1.15. We have that sin2α þ sin 2β ¼ 2 sin(α þ β) cos(α À β) 2 sin γ. Similarly, sin2β þ sin 2γ 2 sin α and sin2α þ sin 2γ 2 sin β. Summing up these three inequalities, we deduce that sin2α þ sin 2β þ sin 2γ sin α þ sin β þ sin γ. This ends the proof.

πÀα πÀβ πÀγ 5.1.16. Using the inequality of the problem 5.1.15 for angles 2 , 2 , 2 ,we α β γ α β γ deduce that sin þ sin þ sin cos2 þ cos 2 þ cos 2. This ends the proof. 5.1.17. We have that

sin ðÞα þ β 2 sin γ 2 sin γ γ ctgα þ ctgβ ¼ ¼ ¼ 2tg : sin α sin β cos ðÞÀα À β cos ðÞα þ β 1 þ cos γ 2

β γ α γ α β Similarly, we obtain that ctg þ ctg 2tg 2 and ctg þ ctg 2tg 2. α β γ α β γ Therefore, ctg þ ctg þ ctg tg 2 þ tg 2 þ tg 2. This ends the proof. α β γ αÀβ γ 5.1.18. Note that cos þ cos ¼ 2 sin 2 cos 2 2 sin 2. Thus, it follows that cosα þ cos β cosβ þ cos γ cos γ þ cosα cos α þ cosβ þ cos γ ¼ þ þ 2 2 2 γ α β sin þ sin þ sin : 2 2 2 This ends the proof. 174 5 Application of Trigonometric Inequalities

2α 2β 1 α β 2 2γ 5.1.19. We have that ctg þ ctg 2 ðÞctg þ ctg 2tg 2 (see the proof of 2α 2β 2γ 2α 2β 2γ the problem 5.1.17). Hence ctg þ ctg þ ctg tg 2 þ tg 2 þ tg 2. This ends the proof. 5.1.20. (a) If the triangle is not acute angled, then inequality is correct, as α β γ cos α cos β cos γ 0 < sin sin sin q. ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÂÃ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 2 2 ; π p 1þ cos ðÞxþy xþy If x, y2 0 2 , then cos x cos y 2 ¼ cos 2 . But if the triangle is an acute triangle, then pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cos α cos β cos γ ¼ cos α cos β cos β cos γ cos γ cos α α þ β β þ γ α þ γ γ α β cos cos cos ¼ sin sin sin : 2 2 2 2 2 2

(b) We have that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin α sin β sin γ ¼ sin γ sin β sin α sin γ sin α sin β α β γ cos cos cos 2 2 2 (see the proof of the problem 5.1.12). (c) Let α β γ, then we need to prove thatðÞ cos ðÞþα À β cos γ cos γ 4 ÀÁÀÁ αÀβ γ 2 2γ αÀβ γ 2 cos 2 À sin 2 sin 2, f cos 2 0, where ftðÞ¼ðÞ2 cos À 1 t þ 4 γ γ γ sin 3 t À cos 2 cos γ À 2sin 4 . 2 2 2 ÀÁ αÀβ αÀβ ; Since 0 cos 2 1, then we have that f cos ÀÁ2 maxðÞf ðÞ0 f ðÞ1 . αÀβ We need to prove that, if f(0) 0, f(1) 0, then f cos 2 0. 2 γ γ 4γ 2γ ÀÁWe have that f ðÞ¼À0 cos 2 cos À 2sin 2 0, and f ðÞ¼À1 sin 2 γ 2 2 sin 2 À 1 0. α β γ (d) Let . We need to prove that ÀÁ ÀÁ β γ ‘ α β γ γ γ sin α sin sin 1 cos À cos π sin cos π ,or ÀÁ2 þ 2 þ 2 2 þ 2 2 þ 6 þ 2 2 À 6 αÀβ f cos 4 ÀÁ0, where ÀÁ ÀÁ γ π γ γ γ γ γ ft 2 cos π t2 2 sin À t 1 sin cos π sin cos π : ðÞ¼ 2 þ 6 À ÀÁ4 Á þ 2 À 2 À 2 þ 6 þ 2 2 À 6 α β α β As 0 cos À 1, then f cos À maxðÞf ðÞ0 ; f ðÞ1 . 4 4 ÀÁ αÀβ We need to prove that, if f(0) 0 and f(1) 0, then we have that f cos 4 0. We have that 1 γ γ π γ γ π f ðÞ¼0 À sin À cos þ þ sin cos À ¼ 2 2 2 6 2 2 6 pffiffiffi pffiffiffi 1 π γ 3 3 γ ¼ sin γ À À sin þ À cos 0, 2 6 2 2 2 2 π < γ π γ π since À6 À 6 2 6. 5.1 Inequalities for the Angles of a Triangle 175

Because γ γ π 1 γ π π À γ f ðÞ¼1 sin cos À À 1 þ þ cos þ À 2 sin 2 2 6 2 2 6 4 1 γ π π À γ π À γ π À γ π þ cos þ À 2 sin ¼ 2 sin sin þ À 1 0, 2 2 6 4 4 4 3 we have that f(1) 0. (e) Let γ β α. One needs to prove that 3 1 α À β β À γ γ À α þ cos þ cos þ cos cos α þ cos β þ cos γ: 4 4 2 2 2 γ 1 β À α 1 β À α π À 3γ 3 2 sin À cos À cos cos þ cos γ À 0: 2 4 2 2 4 4 4

γ Note that γ π. Therefore, 2 sin À 1 3. ÀÁ3 2 4 4 γ 1 2 1 πÀ3γ γ γ 1 If fxðÞ¼2 2 sin 2 À 4 x À 2 x cos 4 þ cos À 2 sin 2 À 2, then one needs to prove that ÀÁ βÀα βÀα f cos 4 0, as 0 cos 4 1 and for x 2 [0, 1]. We have that f(x) max( f(0), f(1)), then it is sufﬁcient to prove that f(0) 0 and f(1) 0. Indeed, we have that

γ 1 1 γ γ 1 1 1 f ðÞ¼0 cos γ À 2 sin À ¼ À 2sin2 À 2sin2 À 2 Á À 2 Á < 0: 2 2 2 2 2 2 2 4 γ 1 π À 3γ f ðÞ¼1 2 sin À 1 À cos þ cos γ: 2 2 4

Let γ ¼ π À 2φ, then ÀÁÀÁ φ 1 3φ φ 1 φ φ 2 φ f ðÞ¼1 2 cos À 1 À 2 sin 2 À cos 2 ¼À2 sin 2 2 sin 2 À 1 4 sin 2 þ 3 0, < φ π : as 0 3

Remark For α þ β þ γ ¼ π the inequality of problem 5.1.20e may not hold true. α π β 5π γ π For example, if ¼ 2 , ¼ 2 , ¼À2 . (f) Let α β γ. One needs to prove that

α À β α À β π À 3γ 4 γ α À β 2 cos þ 2 cos cos pffiffiffi cos cos þ pffiffiffi sin γ, 2 4 4 3 2 2 3 176 5 Application of Trigonometric Inequalities or 4 γ α À β π À 3γ α À β 2 4 γ 2 pffiffiffi cos À 1 cos 2 À 2 cos cos þ pffiffiffi sin γ À pffiffiffi cos þ 1 0: 3 2 4 4 4 3 3 2 γ π γ γ Let fxðÞ¼2 p4ffiffi cos À 1 x2 À 2 cos À3 x þ p2ffiffi sin γ À p4ffiffi cos þ 1. 3 2 4 3 3 2 Note that pffiffi γ π α β α β π 2 p4ffiffi cos À 1 2 p4ffiffi cos À 1 ¼ 2 and 2 cos À 1, as 0 À . 3 2 3 6 2 4 4 4 One needs to prove that α À β f cos 0: ð5:1Þ 4

We have that pffiffiffi α À β 2 f cos max f ðÞ1 ; f : 4 2 pffiffi 2 Let us prove that f 2 0 and f(1) 0, then (5.1) holds true. We have that pffiffiffi 2 2 pffiffiffi π À 3γ 2 π pffiffiffi π À 3γ f ¼ pffiffiffi sin γ À 2 cos pffiffiffi sin À 2 cos ¼ 2 3 4 3 3 4 pffiffiffi π À 3γ pffiffiffi π pffiffiffi π ¼ 1 À 2 cos < 1 À 2 cos < 1 À 2 cos ¼ 0, 4 4 4 pffiffi 2 < therefore f 2 0. Now, let us prove that f(1) 0. π γ γ We need to prove that 1 þ 2 cos À3 p4ffiffi cos þ p2ffiffi sin γ. 4 3 2 3 Consider the following function ÂÃ π γ γ π gðÞ¼γ 1 þ 2 cos À3 À p4ffiffi cos À p2ffiffi sin γ on 0; . 4 3 2 3 3 < γ π For 0 3, we have that

0 3 π À 3γ 2 γ 2 2 γ 2 g ðÞ¼Àγ sin þ pffiffiffi sin À pffiffiffi cos γ pffiffiffi sin À pffiffiffi cos γ 2 4 3 2 3 3 2 3 2 π 2 π pffiffiffi sin À pffiffiffi cos ¼ 0: 3 6 3 3 ÀÁ γ π Hence, gðÞg 3 ¼ 0. Hence, it follows that f(1) 0. 5.1 Inequalities for the Angles of a Triangle 177

α β γ γ γ π (g) Let max( , , ) ¼ , then 3. We need to prove that γ α À β sin γ 3 pffiffiffi 2 sin cos þ cos γ þ þ ctgγ þ 3: 2 2 2 αÀβ 2 γ 2 cos 2 À sin 2 Consider the following function γ γ iγ γ sin γ ; : fxðÞ¼ 2 sin x þ cos þ 2 2γ þ ctg on sin 1 2 x À sin 2 2

We have that ÀÁ ÀÁ γ 2 γ γ 2 γ γ x2 À sin2 À 2x cos 1 À sin2 À 2 sin cos 0 γ 2 γ 2 f ðÞ¼x 2 sin Á ÀÁ2 < 2 sin Á ÀÁ2 2 ¼ 2 2 2γ 2 2 2 2γ 2 x À sin 2 x À sin 2 γ γ γ cos 3 À 2 sin cos 3 À 1 ¼ sin γ Á ÀÁ2 2 sin γ Á ÀÁ2 0: 2 2γ 2 2 2γ 2 x À sin 2 x À sin 2

ÀÃ γ; 0 < Thus, it follows thatÀÃ for x2 sin 2 1 , we have that f (x) 0. Therefore, f(x)isa γ; decreasing function on sin 2 1 . α À β 3 pffiffiffi Note that, it is sufficient to prove that f cos þ 3. i2 2 α À β γ α À β As cos 2 sin ; 1 , then f cos f ðÞ1 . 2 2 2 3 pffiffiffi α À β 3 pffiffiffi Let us prove that f ðÞ1 þ 3, then f cos þ 3. 2 2 2 γ γ γ sin γ We have that f ðÞ¼1 2 sin þ cos þ 2 γ þ ctg . 2 1 À sin 2 Let γ ¼ π À 2φ, then 0 < φ < π and one needs to prove that 3 pffiffiffi φ φ φ φ 3 2 cos À cos 2 þ 2ctg À ctg 2 2 þ 3,or 2φ pffiffiffi 2 3 þ tg φ 1 : φ À 3 2 cos À ÀÁ2ffiffiffitg ÀÁ2 p 2 2 3 À tgφ 4tgφ cos φ À 1 , ÀÁpffiffiffi 2 ÀÁ φ φ 2 φ φ φ 1 2 3 cos À sin 4 cos sin cos À 2 , ÀÁÀÁpffiffiffi 2 3cos 2φ sin2φ 2 2 sin 2φ cos φ 1 2 3 cos φ sin φ , ðÞÀ À 2 þ 1 pffiffiffi 2 ðÞ2 cos φ À 1 2ðÞ2 cos φ þ 1 2 sin 2φðÞ2 cos φ À 1 2 3 cos φ þ sin φ : 2 178 5 Application of Trigonometric Inequalities

It is sufficient to prove that 1 pffiffiffi 2 ðÞ2 cos φ þ 1 2 sin 2φ 3 cos φ þ sin φ : 2 Àpffiffiffi φ 2 1 φ φ The last inequality holds true, as (2 cos þ 1) 4 and 2 sin 2 3 cos þ φ 2 1 sin Þ 2 Á 1 Á 4 ¼ 2. (h) Let α β γ, then we need to prove that

1ÀðÞcosαþ cosβ þ cosγ þcosαcosβ þðÞcosαþ cosβ cosγ 5cosαcosβcosγ À2cosαcosβcosγðÞcosαþ cosβ þ cosγ , αÀβ γ 1 αÀβ γ 1À2cos sin À cosγ þ ðÞcosðÞÀαÀβ cosγ þ2cos sin cosγ 2 2 2 2 2 5 αÀβ γ cosγðÞcosðÞÀαÀβ cosγ ÀðÞcosðÞÀαÀβ cosγ cosγ 2cos sin þ cosγ , 2 2 2 αÀβ γ 1 αÀβ αÀβ γ 1À2cos sin À cosγ þ 2cos2 À1À cosγ þ2cos sin cosγ 2 2 2 2 2 2 5 αÀβ αÀβ αÀβ γ cosγ 2cos2 À1À cosγ À 2cos2 À1À cosγ cosγ 2cos sin þ cosγ : 2 2 2 2 2

Let us consider the following function γ ÀÁγÀÁ fxðÞ¼4 cos γ sin x3 þ 1 À 5 cos γ þ 2cos 2γ x2 À 2 sin 1 þ cos 2γ x 2 2 Àcos 3γ þ 1, 5cos 2γ þ cos γ þ 0, 5 on [0; 1]. ÀÁ αÀβ We need to prove that f cos 2 0. < γ π We have that 0 3 , therefore γ ÀÁγÀÁ f 0ðÞ¼x 12 cos γ sin x2 þ 21À 5 cos γ þ 2cos 2γ x À 2 sin 1 þ cos 2γ , γÀÁ2 2 f 0ðÞ¼À0 2 sin 1 þ cos 2γ < 0, γÀÁ2 f 0ðÞ¼1 2 sin 6 cos γ À 1 À cos 2γ þ 2 À 10 cos γ þ 4cos 2γ 2 6 cos γ À 1 À cos 2γ þ 2 À 10 cos γ þ 4cos 2γ ¼ ¼ ðÞcos γ À 1 ðÞ3 cos γ À 1 < 0:

As 0 x 1, then f0(x) max( f0(0), f0(1)) < 0. αÀβ As 0 cos 2 1, then α À β γ ÀÁγÀÁ f cos f ðÞ¼1 4 cos γ sin þ 1 À 5cosγ þ 2cos 2γ À 2 sin 1 þ cos 2γ À 2 2 ÀÁ2 3γ 2γ γ γ 2 γ ; 2 : Àcos þ 1, 5cos þ cos þ 0, 5 ¼ 21ðÞÀ cos sin 2 À 0 5 0 5.1 Inequalities for the Angles of a Triangle 179

α β γ γ π α β π (i) Let .If ¼ 3, then ¼ ¼ 3 and pffiffiffi 3 3 pffiffiffi sin 2α þ sin 2β þ sin 2γ ¼ ¼ 2 3ðÞcos α cos β þ cos β cos γ þ cos γ cos α : 2

γ < π If 3, then pffiffiffi sin 2α þ sin 2β þ sin 2γ À 2 3ðÞcos α cos β þ cos β cos γ þ cos γ cos α ¼ pffiffiffi ¼ 2 sin ðÞα þ β cos ðÞþα À β sin 2γ À 3ðÞcos ðÞþα À β cos ðÞα þ β À pffiffiffi α þ β α À β pffiffiffi À4 3 cos γ cos cos ¼ 2 sin γ À 3 cos ðÞÀα À β 2 2 pffiffiffi γ α À β pffiffiffi pffiffiffi α À β À4 3 cos γ sin cos þ sin 2γ þ 3 cos γ ¼ 2 2 sin γ À 3 cos 2 À 2 2 2 pffiffiffi γ α À β pffiffiffi pffiffiffi À4 3 cos γ sin cos þ sin 2γ þ 3 cos γ À 2 sin γ þ 3: 2 2

Let us consider the following function pffiffiffi pffiffiffi γ pffiffiffi pffiffiffi fxðÞ¼2 2 sin γ À 3 x2 À 4 3 cos γ sin x þ sin 2γ þ 3 cos γ À 2 sin γ þ 3 2 on [0; 1]. ÀÁpffiffiffi We have that 2 2 sin γ À 3 < 0, and pffiffiffi pffiffiffi pffiffiffi γ γ γ f ðÞ¼0 sin 2γ þ 3 cos γ À 2 sin γ þ 3 ¼ 2 3cos 2 À 8 cos sin 3 ¼ 2 2 2 γ pffiffiffi γ 3 3 γ ¼ 2 cos 3 cos À þ À 4sin 3 > 0, 2 2 2 2 2 ÀÁpffiffiffi pffiffiffi γ pffiffiffi pffiffiffi f ðÞ¼1 2 2 sin γ À 3 À 4 3 cos γ sin þ sin 2γ þ 3 cos γ À 2 sin γ þ 3 ¼ 2 pffiffiffi pffiffiffi γ ¼ 2 sin γ þ sin 2γ À 3ðÞÀ1 À cos γ 4 3 cos γ sin ¼ 2 γγ pffiffiffi pffiffiffi γ ¼ 2 sin 4cos 3 À 2 3 cos γ À 3 sin : 2 2 2 pffiffiffi pffiffiffi ÂÃ 3x x ; π : Consider a function gxðÞ¼4cos 2 À 2 3 cos x À 3 sin 2 on 0 3 < < π As 0 x 3 , we have that pffiffiffi x x pffiffiffi x 3 g0ðÞ¼x cos 2 sin 2 3 À 3 cos À < 0, 2 2 2 2 pffiffiffi pffiffi x < 1 < x < 3 as sin 2 2 ,0 2 3 À 3 cos 2 2 . 180 5 Application of Trigonometric Inequalities ÀÁ < < π > π Then, for 0 x 3 , it follows that gxðÞ g 3 ¼ 0. Thus, it follows that γ γ > : f ðÞ¼1 2 sin 2 gðÞ 0 Note that, for 0 < x 1, f(x) min( f(0), f(1)) > 0. Hence, we obtain that pffiffiffi sin 2αþ sin 2β þ sin 2γ À 2 3ðÞcos α cos β þ cos β cos γ þ cos γ cos α ¼ α À β α À β ¼ f cos > 0, as 0 < cos 1: 2 2

This ends the proof. 5.1.21. Let α β γ. Let us denote α À β ¼ 2y, β À γ ¼ 2x. We have that x 0, π α β γ γ π π y 0 and 2 ¼ 2 þ 2 þ 2 ¼ 32 þ 2x þ y. Therefore, 2x þ y 2 and x 4. Then jjα À β jjβ À γ jjγ À α π π sin þ sin þ sin sin À 2x þ sin x þ sin À x ¼ 2 2 2 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ sin x þ cos x þ cos 2x ¼ 1 þ sin 2x þ 1 À sin22x: pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi 2 Consider the functionpffiffiffiffiffiffiftðÞ¼ 1 þ t þ 1 À t on the interval [0; 1]. Since 2 f 0 t p1ffiffiffiffiffiffi pffiffiffiffiffiffiffit p1ÀffiffiffiffiffiffiffitÀ2t pffiffiffiffiffiffiffi4t þptÀffiffiffiffiffiffi1 . ðÞ¼2 1 t À 2 ¼ 2 ¼À 2 þ 1Àt 2 1Àt 2 1Àt ðÞq1ffiffiffiffiffiffiffiffiffiffiffiÀtþ2t qffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi pffiffiffiffi pffiffiffiffi 17À1 7þ 17 23þ 17 Consequently, max ftðÞ¼f 8 ¼ 8 þ 32 . ½0;1 qffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi pffiffiffiffi pffiffiffiffi jjαÀβ jjβÀγ jjγÀα 7þ 17 23þ 17 71þ17 17 Thus, sin 2 þ sin 2 þ sin 2 8 þ 32 ¼ 32 . pffiffiffiffi β 17À1 Remark The given estimate is exactqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi since at arcsin 8 , we have that pffiffiffiffi jjαÀβ jjβÀγ jjγÀα 71þ17 17 sin 2 þ sin 2 þ sin 2 32 . Note that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi jjα À β jjβ À γ jjγ À α 71 þ 17 17 sin þ sin þ sin < : 2 2 2 32

This ends the proof. 5.1.22. (a) We have that

x2 þ y2 þ z2 À 2xy cos α À 2yzcosβ À 2zx cos γ ¼ ¼ ðÞx À ðÞy cos α þ z cos γ 2 þ ðÞy sin α À z sin γ 2 0

1 1 1 (b) Let x, y, z > 0 and 2xy ¼ α ,2yz ¼ β ,2xz ¼ γ , this means that sin 1 sin 1 sin 1 β γ α pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 1 x ¼ α β γ , y ¼ α β γ , z ¼ α β γ . Then from the 2 sin 1 sin 1 sin 1 2 sin 1 sin 1 sin 1 2 sin 1 sin 1 sin 1 problem 5.1.22a, it follows that 5.1 Inequalities for the Angles of a Triangle 181

cos α cos β cos γ þ þ ¼ 2xy cos α þ 2yz cos β þ 2xz cos γ x2 þ y2 þ z2 ¼ sin α sin β sin γ 1 1 1 β γ α 1 sin 1 sin 1 sin 1 ¼ α γ þ α β þ γ β ¼ 2 sin 1 sin 1 sin 1 sin 1 sin 1 sin 1 1 ¼ ðÞðÞctg α þ ctg γ þ ðÞctg α þ ctg β þ ðÞctg γ þ ctg β 2 1 1 1 1 1 1 α β γ : ¼ ctg 1 þ ctg 1 þ ctg 1

1 1 1 1 ÀÁ S ¼ R2 sin 2u þ R2 sin 2v þ R2 sin 2w ¼ a2ctgu þ b2ctgv þ c2ctgw : 2 2 2 4 Therefore,

2 α 2 β 2 γ a ctg 1 þ b ctg 1 þ c ctg 1 À 4S ¼ ¼ a2ðÞctg α À ctgu þb2ðÞþctg β À ctgv c2ðÞctg γ À ctgw ¼ 1 1 1 sin ðÞu À α sin u sin ðÞv À β sin v sin ðÞw À γ sin w ¼ 4R2 1 þ 1 þ 1 ¼ sin α sin β sin γ 1 1 1 cosðÞ 2u À α cosðÞ 2v À β cos2ðÞw À γ ¼ 2R2 ctg α À 1 þ ctg β À 1 þ ctg γ À 1 ¼ 1 sin α 1 sin β 1 sin γ 1 1 1 α β γ 2 α β γ cos cos cos ¼ 2R ctg 1 þ ctg 1 þ ctg 1 À α À β À γ 0 sin 1 sin 1 cos 1

(see the problem 5.1.22b), since α ¼ 2u À α1, β ¼ 2v À β1, γ ¼ 2w À γ1 and α þ β þ γ ¼ π. (d) It is sufﬁcient to prove the inequality for 0 φ 2π. For φ ¼ 0orπ the inequality is obvious. Using the inequality 5.1.22c, for the angles given below, we obtain the required inequality: < φ < π γ β π φ α φ If 0 , then we take 1 ¼ 1 ¼ 2 À 2 , 1 ¼ . π < φ < π γ β φÀπ α π φ If 2 , then we take 1 ¼ 1 ¼ 2 and 1 ¼ 2 À . ÀÁ 2 2 2 2 2 α 2 α (e) Note that a b1 þ c1 À a1 ¼ a 2b1c1 cos 1 ¼ 4S1a ctg 1. Therefore, ÀÁÀÁÀÁ 2 2 2 2 2 2 2 2 2 2 2 2 a b1 þÀÁc1 À a1 þ b a1 þ c1 À b1 þ c a1 þ b1 À c1 ¼ 2 α 2 β 2 γ ¼ 4S1 a ctg 1 þ b ctg 1 þ c ctg 1 16SS1,

(see the problem 5.1.22c). 182 5 Application of Trigonometric Inequalities

(f) If xy þ yz þ zx ¼ 0, then the proof is obvious. If xy þ yz þ zx 6¼ 0, then without loss of generality one can assume that xy þ yz þ zx ¼ 1 and x 0, y 0. 2 2 2 2 2 We need to prove that (xa þ yb þ zc ) 16S . ÀÃ α β γ α β ; π γ ; π Let x ¼ ctg 1, y ¼ ctg 1, z ¼ ctg 1, where 1, 1 2 0 2 , 1 2ðÞ0 . Then, we have that ctgα1ctgβ1 þ ctgβ1ctgγ1 þ ctgγ1ctgα1 ¼ 1. Therefore α β 1Àctg 1ctg 1 ctg γ ¼ α β or ctgγ1 ¼ ctg(π À α1 À β1). Thus, it follows that α1, β1, γ1 are 1 ctg 1þctg 1 the angles of some triangle. According to problem 5.1.22c, we obtain that xa2 þ yb2 þ zc2 4S. Hence, we deduce that ÀÁ 2 xa2 þ yb2 þ zc2 16S2

(g) Let points A1, B1, and C1 be the midpoints of sides BC, AC, and AB, respectively. Using the problem 4.1.8а for x ¼ tgα1, y ¼ tgβ1, z ¼ tgγ1, we obtain that α β 2 β γ 2 α γ 2 2 2 2 tg 1tg 1A1B1þtg 1tg 1B1C1þtg 1tg 1A1C1 tg α1 Á OA þ tg β Á OB þ tg γ Á OC α β γ , 1 1 1 1 1 tg 1þtg 1þtg 1 2 α 2 β 2 γ γ c2 α a2 β b2 or k tg 1 þ l tg 1 þ m tg 1 ctg 1 Á 4 þ ctg 1 Á 4 þ ctg 1 Á 4 ,as tgα1 þ tgβ1 þ tgγ1 ¼ tgα1tgβ1tgγ1. À 1 2 α 2 β According to problem 5.1.22c, it follows that 4 a ctg 1 þ b ctg 1þ 2 γ c ctg 1ÞS. 2 2 2 Therefore, k tgα1 þ l tgβ1 þ m tgγ1 S. This ends the proof. α β α β π α β π 5.1.23.ÀÁIn any triangleÀÁ there are angles and , such that either 2 , 2 6 or 2 , 2 6, α β then 1 À 2 sin 2 1 À 2 sin 2 0. Thus, it follows that α β γ α β γ 2 cos 2 þ cos 2 þ cos 2 À sin þ sin þ sin ¼ 2 2 2 2 2 2 α β γ α β ¼ cos α þ cos β þ cos γ À 2 sin sin À 2 sin sin þ sin ¼ 2 2 2 2 2 α β γ α β γ α β ¼ 1 þ 4 sin sin sin À 2 sin sin À 2 sin sin þ sin ¼ 2 2 2 2 2 2 2 2 γα β α β γ ¼ 1 þ sin 1 À 2 sin 1 À 2 sin À 2 sin sin À sin ¼ 2 2 2 2 2 2 α À β γ α β ¼ 1 À cos þ sin 1 À 2 sin 1 À 2 sin 0, 2 2 2 2

(see the proof of the problem 5.1.4a). Hence, we obtain that 5.1 Inequalities for the Angles of a Triangle 183

α β γ α β γ 2 cos 2 þ cos 2 þ cos 2 sin þ sin þ sin : 2 2 2 2 2 2

This ends the proof. 5.1.24. We have that (see the problem 5.1.15)

sin 2α þ sin 2β þ sin 2γ sin α þ sin β þ sin γ: ð5:2Þ

πÀα πÀβ πÀγ Using (5.2) for angles 2 , 2 , 2 , we obtain that α β γ sin α þ sin β þ sin γ cos þ cos þ cos : ð5:3Þ 2 2 2 In the same way from (5.3), we deduce that α β γ π α π β π γ cos þ cos þ cos cos À þ cos À þ cos À : ð5:4Þ 2 2 2 4 4 4 4 4 4

From the last inequality, in the same way, it follows that π α π β π γ π þ α π þ β π þ γ cos À þ cos À þ cos À cos þ cos þ cos : 4 4 4 4 4 4 8 8 8 ð5:5Þ From inequalities (5.2)–(5.5), we deduce that π þ α π þ β π þ γ sin 2α þ sin 2β þ sin 2γ cos þ cos þ cos : 8 8 8 This ends the proof. γ < 2α 2β γ > 2α 2β 2γ 3 5.1.25. If 90 , then cos þ cos þ cos cos þ cos þ cos 4, (see the problem 5.1.1). In the case γ 90 , we have that

cos 2α þ cos 2β þ cos γ ¼ 1 þ cos ðÞα þ β cos ðÞα À β 3 1 2 3 þ cos γ 1 þ cos 2ðÞþα þ β cos γ ¼¼ þ cos γ þ : 4 2 4

Here the equality cannot hold true. Otherwise, we obtain that α þ β ¼ 90 and γ ¼ 120 . This leads to a contradiction. This ends the proof. 184 5 Application of Trigonometric Inequalities

5.1.26. We have that

cos 2α cos 2β þ cos 2β cos 2γ þ cos 2γ cos 2α ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ðÞcos 2α cos 2β 2 þ ðÞcos 2β cos 2γ 2 þ ðÞcos 2γ cos 2α 2þ

þ2cos 2α cos 2β cos 2γðÞcos 2α þ cos 2β þ cos 2γ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3 cos 2α cos 2β cos 2γðÞcos 2α þ cos 2β þ cos 2γ cos α cos β cos γ: 2 Here, we have used the inequality x2 þ y2 þ z2 xy þ yz þ zx and the problem 5.1.1. This ends the proof. 5.1.27. Note that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 x2 þ y2 þ 2xy cos α ðÞx sin γ þ y sin β þ ðÞx cos γ À y cos β sin γ sin β ¼ x þ y : sin α sin α sin α sin α pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y2 z2 2yz cos β γ Similarly, we get the inequalities: þ þ y sin α þ z sin and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin β sin β sin β 2 2 γ z þx þ2zx cos sin β sin α sin γ z sin γ þ x sin γ. By summing up these inequalities, we get pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 α y2 þ z2 þ 2yz cos β z2 þ x2 þ 2zx cos γ x þ y þ 2xy cos þ þ α sin β sin γ sin sin γ sin α sin β sin α sin γ sin β x þ þ y þ þ z þ 2x þ 2y þ 2z, sin α sin γ sin α sin β sin β sin γ

> 1 as for a 0, we have that a þ a 2. This ends the proof. 5.1.28. Let α β γ, then we have that

2ðÞ sinαsinβ þ sinβ sinγ þ sinγ sinα À3ðÞ cosα þ cosβ þ cosγ ¼ β þ γ γ À β γ À β β þ γ ¼ 4sinαsin cos þ cosðÞÀγ À β cosðÞÀγ þ β 3cosα À 6cos cos ¼ 2 2 2 2 γ À β γ À β α α ¼ 2cos 2 þ 2cos 2sinαcos À 3sin À 2cosα À 1: 2 2 2 2 γ β γ γ β Since 0 À < < π, then 0 < cos À 1. 2 2 2 2 ÀÁ 2 α α α α Consider the function fxðÞ¼2x þ 2x 2 sin cos 2 À 3 sin 2 À 2 cos À 1in the interval [0; 1]. Note that max fxðÞ¼maxðÞf ðÞ0 ; f ðÞ1 . We need to prove that, if ÀÁ½0;1 γÀβ f(0) 0 and f(1) 0, then f cos 2 0. ÀÁ α < < α π Indeed, we have that f(0) ¼À1 À 2 cos 0, 0 3 . 5.1 Inequalities for the Angles of a Triangle 185

α 3α α α 2 α f ðÞ¼1 1 À 2 cos α þ 2 sin þ sin À 3 sin ¼À 2 sin À 1 2 sin þ 1 0: 2 2 2 2 2 This ends the proof. < < π 5.1.29. We need to prove that, if 0 x, y 3, then sin 2x sin 2y sin2ðÞx þ y : ð5:6Þ sin 3x sin 3y 2 3xþ3y sin 2 α β α < β < 2π Let x y, denote x À y ¼ , x þ y ¼ , then 0 3 . Therefore, α > β cos 2αÀ cos 2β 1À cos 2β α β cos 2 cos 2. We have to prove that cos 3αÀ cos 3β 1À cos 3β, or (cos2 À cos 2 ) (1 À cos 3β) (cos3α À cos 3β)(1 À cos 2β). Note that

ðÞcos 2α À cos 2β ðÞÀ1 À cos 3β ðÞcos 3α À cos 3β ðÞ¼1 À cos 2β 3α 3β ¼ 4 sin2βsin2 À sin2αsin2 ¼ 2 2 α β α β α β 3α 3β ¼ 8 sin sin 4 cos cos þ 1 cos À cos sin β sin þ sin α sin 0: 2 2 2 2 2 2 2 2 Using the inequality (5.6), we obtain that α β α β γ π 2 γ π sin 2 sin 2 sin 2 sin 2 sin þ sin2 þ 6 6 6 18 6 6 6 18 Á γ π γ π α β α β 2 sin 3 sin 3 sin 3 sin 3 sin23 þ sin 3 þ 6 6 6 18 12 36 12 12 α β γ π 4 π sin þ þ þ sin4 12 12 12 36 9 : α β γ π ¼ π sin4 þ þ þ 6 sin43 12 12 12 36 2 α β γ 3π α β γ Therefore, sin 3 sin 3 sin 3 8sin 9 sin 2 sin 2 sin 2. This ends the proof. ÀÁ 2α 2β 2γ α β γ 2 5.1.30. We have that tg 2 þ tg 2 þ tg 2 ¼ tg 2 þ tg 2 þ tg 2 À 2, (see the proof of the problem 5.2.1b. Note that α þ γ α þ γ α γ 2 sin 2 sin tg þ tg ¼ 2 2 ¼ 2 2 γ À α γ þ α 2φ À ðÞγ þ α À 2φ γ þ α cos þ cos cos þ cos 2 2 2 2 γ þ α À 2φ 2φ ¼ tg þ tg , 2 2 αÀγ 2φÀðÞγþαÀ2φ γÀα φ 4 φ > π since 2 2 2 , where ¼ arctg3, 4.Thus 186 5 Application of Trigonometric Inequalities

α β γ β γ þ α À 2φ 4 tg þ tg þ tg tg þ tg þ ¼ 2 2 2 2 2 3 π φ 2 sin À 4 2 cos φ 4 ¼ 2 þ þ ¼ 2: β γ þ α À 2φ π 3 1 þ sin φ 3 cos À þ cos À φ 2 2 2

2α 2β 2γ Therefore, tg 2 þ tg 2 þ tg 2 2. This ends the proof. 5.1.31. We have that

α α β β γ γ sin sin 1 þ sin sin 1 þ sin sin 1 α β γ α β γ ¼ sin sin sin sin 1 sin 1 sin 1 02 2 2 2 2 2 1 α α β β γ γ cos cos 1 cos cos 1 cos cos 1 B 2 2 2 2 2 2 C ¼ 4@ þ α α γ γ þ A β β γ γ 1 1 β β α α sin sin 1 sin sin 1 sin sin sin sin sin sin 1 sin sin 1 2 2 2 2 2 2 2 2 2 2 2 2 0 α α β β B cos cos 1 cos cos 1 B22 2 2 16@ þ α γ α γ þ β þ γ β þ γ þ 1 þ 1 1 À cos 1 À cos 1 1 1 À cos 1 À cos 2 2 2 2 1 γ γ cos cos 1 C 2 2 C þ A ¼ β þ α β þ α 1 À cos 1 À cos 1 1 2 2 π þ α π þ α π þ β π þ β π þ γ π þ γ ¼ 16 tg tg 1 þ tg tg 1 þ tg tg 1 4 4 4 4 4 4 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3 π þ α π þ α π þ β π þ β π þ γ π þ γ p p 48 tg tg 1 tg tg 1 tg tg 1 48 3 3 3 Á 3 3¼ 144 4 4 4 4 4 4

π α π β π γ π α π β π γ þ þ þ π þ 1 þ 1 þ 1 π Since 4 þ 4 þ 4 ¼ , 4 þ 4 þ 4 ¼ , (see the proof of the problems 5.2.1a and 5.2.9). This ends the proof. 5.1.32. Without loss of generality, we can assume that α β γ. The expression cosα1 þ cos β1 þ cos γ1 is symmetric with respect to α1, β1, γ1, and the value of the α α0 β β0 γ γ0 α0 β0 γ0 expressionÈÉ sin sin 1 þ sin sin 1 þ sin sin 1 is maximal, when 1 1 1, α0 ; β0 ; γ0 α ; β ; γ where 1 1 1 ¼ fg1 1 1 . Therefore, without loss of generality, we can α β γ assume that 1 1 1. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi α α β β 2α 2β 2α 2β Since sin sin 1 þ sin sin 1 ðÞsin þ sin ðÞsin 1 þ sin 1 , then it is sufficient to prove that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin2α þ sin2β sin2α þ sin2β þ sin γ sin γ 1 1 1 ð5:7Þ α β γ α β γ : 2 þ ðÞcos þ cos þ cos À 1 ðÞcos 1 þ cos 1 þ cos 1 À 1 5.1 Inequalities for the Angles of a Triangle 187

Note that

cos 2α þ cos 2β sin2α þ sin2β ¼ 1 À ¼ 1 þ cos γ cos ðÞα À β 2 α À β ¼ 2 cos γcos 2 þ 1 À cos γ and 2

γ α À β cos α þ cos β ¼ 2 sin cos : 2 2

We have to prove that α À β 2 cos γcos 2 þ 1 À cos γ ðÞsin2α þ sin2β 2 1 1 γ α À β 2 þ 2 sin ðÞcos α þ cos β þ cos γ À 1 cos þ 2 1 1 1 2 γ α β γ γ γ 2: þðÞcos À 1 ðÞcos 1 þ cos 1 þ cos 1 À 1 ÀÀsin sin 1Þ

The last inequality can be rewritten as: γ α À β 2 cos γðÞsin2α þ sin2β À4sin2 ðÞcos α þ cos β þ cos γ À 1 2 cos 2 þ 1 1 2 1 1 1 2 α À β þB cos þ C 0: 2 ð5:8Þ

We need to prove that

2α 2β α β γ 2: : sin 1 þ sin 1 ðÞcos 1 þ cos 1 þ cos 1 À 1 ð5 9Þ

Indeed, we have that

2α 2β α β γ 2 γ α β sin 1 þ sin 1 À ðÞcos 1 þ cos 1 þ cos 1 À 1 ¼ 1 þ cos 1 cos ðÞÀ1 À 1 γ α β 2 γ α1 À β À 2 sin 1 cos 1À 1 þ cos γ À 1 ¼ 2 cos γ À 4sin2 1 cos 2 1 þ 2 2 1 1 2 2 γ α À β þ 41ðÞÀ cos γ sin 1 cos 1 1 þ cos γ ðÞ1 À cos γ 0: 1 2 2 1 1 π γ γ α β γ γ 2 1 γ 1 1À 1 As 1 3, then 2 cos 1 1 4sin 2 and 4ðÞ 1 À cos 1 sin 2 cos 2 0, cosγ1(1 À cos γ1) 0. γ 2γ According to (5.9) and 2 cos 4sin 2, we obtain that γ 2α 2β 2γ α β γ 2 A ¼ 2 cos sin 1þð sin 1ÞÀ4sin 2 ðÞcos 1 þ cos 1 þ cos 1 À 1 0. 188 5 Application of Trigonometric Inequalities

2 γ αþβ αÀβ Let f(x) ¼ Ax þÀÁBxÀÁþ C.Sincesin ¼ cos cos 1andA 0, then γ 2 2 2 max fxðÞ¼max f sin ; f ðÞ1 ; thus, if we prove that the inequality (5.7) γ; 2 ½sin 2 1 αÀβ γ αÀβ holds true for cos 2 ¼ sin 2 and cos 2 ¼ 1, then it holds true (in the general case). αÀβ γ If cos 2 ¼ sin 2, then (5.7) has the following form: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2γ γ α β γ γ 1 À cos 1 þ cos 1 cos ðÞ1 À 1 þ sin sin 1 2 or pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γ γ α β γ γ : sin Á 1 þ cos 1 cos ðÞ1 À 1 þ sin sin 1 2 pffiffi Since sin γ 3, cos(α À β ) 1, then it is sufficient to prove that pffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2ffiffiffi 1 1 3 1 þ cos γ þ 3 sin γ 4, but the last inequality holds true, as pffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 pffiffiffi 1 pffiffiffi γ γ 3 < 3 1 þ cos 1 þ 3 sin 1 6 þ 2 4. αÀβ If cos 2 ¼ 1, then (5.7) has the form: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi π À γ π À γ sin2 þ sin2 ðÞsin2α þ sin2β þ sin γ sin γ 2 2 1 1 1 π À γ π À γ 2 þ cos þ cos þ cos γ À 1 ðÞcos α þ cos β þ cos γ À 1 : 2 2 1 1 1 ð5:10Þ

Similarly, to the proof given above in this case too it is sufficient to prove the α β π 1À 1 γ γ inequality (5.10) for cos 2 ¼ 1. Then under the condition , 1 3 one has to prove the inequality pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γ γ γ γ ðÞ1 þ cos ðÞ1 þ cos 1 þ sin sin 1 γ γ 2 þ 2 sin þ cos γ À 1 2 sin 1 þ cos γ À 1 , 2 2 1 or γ γ γ γ γ γ 2 cos cos 1 þ sin γ sin γ 2 þ 4 sin sin 1 1 À sin 1 1 À sin : ð5:11Þ 2 2 1 2 2 2 2

Indeed, we have that γ γ γ γ γ γ γ À γ D ¼ 2 cos cos 1 þ sin γ sin γ À 4 sin sin 1 1 À sin 1 1 À sin ¼ cos 1 þ 2 2 1 2 2 2 2 2 γ þ γ γ γ γ γ γ γ γ γ þ cos 1 þ 4 sin sin 1 cos cos 1 À 1 þ sin 1 þ sin À sin 1 sin ¼ 2 2 2 2 2 2 2 2 2 γ À γ γ þ γ γ γ γ γ γ þ γ γ À γ ¼ cos 1 þ cos 1 þ 4 sin sin 1 cos þ 1 À 1 þ 2 sin 1 cos 1 : 2 2 2 2 2 2 4 4

ÀÁγ γ γ γ γ γ γ γ γ γ 1 þ 1 À 1 À 1 þ 1 If cos 2 þ 2 À 1 þ 2 sin 4 cos 4 0, then D cos 2 þ cos 2 2. 5.1 Inequalities for the Angles of a Triangle 189 ÀÁ γ γ γ γ γ γ 1 þ 1 À 1 > If cos 2 þ 2 À 1 þ 2 sin 4 cos 4 0, then γ À γ γ þ γ γ γ γ þ γ γ þ γ γ À γ D ¼ cos 1 þ cos 1 þ 4sin sin 1 cos 1 À 1 þ 2sin 1 cos 1 2 2 2 2 2 4 4 γ þ γ γ γ γ þ γ γ þ γ 1 þ cos 1 þ 4sin sin 1 cos 1 À 1 þ 2sin 1 2 2 2 2 4 γ þ γ γ þ γ γ þ γ γ þ γ 1 þ cos 1 þ 21À cos 1 cos 1 À 1 þ 2sin 1 ¼ D : 2 2 2 4 0 ÀÁ γ γ γ 2 γ γ 2 0 2γ 2 0 0 γ þ 1 Then D0 ¼ 2cos 2 þ sin 0 À 4sin 2 1 À sin 2 , where 0 ¼ 2 . γ ÀÁγ 2 0 0 2 As 2 À D0 ¼ 2sin 2 1 À 2 sin 2 0, then D0 2. Hence, D 2. This ends the proof of (5.11).

Problems for Self-Study

Prove the inequalities 5.1.33–5.1.52. qffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi α β γ 4 3 5.1.33. sin þ sin þ sin 3 4.

sin αþ sin βþ sin γ 5.1.34. sin α sin β sin γ 4. pffiffi α β γ 3 3 5.1.35. À2 sin 3 þ sin 3 þ sin 3 2 . pffiffiffi 5.1.36. sin2α þ sin2β þ sin2γ 2 3 sin α sin β sin γ. pffiffiffi 5.1.37. 3sin2α þ 3sin2β À sin2γ 2 3 sin α sin β sin γ. 5.1.38. 2 sin α sin β þ 2 sin β sin γ þ 2 sin γ sin α À sin2α À sin2β À sin2γ pffiffiffi 2 3 sin α sin β sin γ: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi α β γ α β γ 5.1.39. sin þ sin þ sin 3 6 sin 2 sin 2 sin 2. 5.1.40. (sin2β þ sin2γ)(1 À cos φ) þ sin2α cos φ 2 sin α sin β sin γ sin φ, where φ is any angle. 5.1.41. sin2α þ sin2β þ sin2γ < 2(sinα sin β þ sin β sin γ þ sin γ sin α). ÀÁ ÀÁ ÀÁ α β β γ γ α > α β γ 5.1.42. sin 2 þ þ sin 2 þ þ sin 2 þ sin þ sin þ sin .

1 1 1 γ 5.1.43. (a) α β þ β γ þ sin α sin 12, sin 2 sin 2 sin 2 sin 2 2 2 pffiffiffi 1 1 1 γ (b) cos α þ β þ cos 2 3, 2 cos 2 2 1 1 1 γ (c) sin α þ β þ sin 6, 2 sin 2 2 1 1 1 9 (d) sin α þ sin β þ sin γ α β γ. 4 cos 2 cos 2 cos 2 α β γ α β γ 5.1.44. ctg Á ctg Á ctg tg 2 Á tg 2 Á tg 2. 190 5 Application of Trigonometric Inequalities

pﬃﬃ α β γ 2 3 5.1.45. sin Á sin Á sin 2 9 . 5.1.46. ctg2β ctgαctgγ, if 2sin2β ¼ sin2α þ sin2γ.

5.1.47. (a) cos α cos β cos ðÞþα À β cos β cos γ cos ðÞþβ À γ cos γ cos α cos ðÞγ À α 1 À cos α cos β cos γ, pffiffi β γ β γ (b) sin α sin sin cos α cos cos 3 6, 4 4 4 þ 4 4ÀÁ4 8 β γ β β γ γ (c) 6À3k sin α sin sin k sin α sin sin sin sin sin α , where k 2, 4 2 þ 2 þ 2ÀÁÀ 2 2 þ 2 2 þ 2 2 3 α β γ α β β γ γ α 6À3k < (d) sin 2 þ sin 2 þ sin 2 À k sin 2 sin 2 þ sin 2 sin 2 þ sin 2 sin 2 4 , where 0 k 2, 7 ÀÁ α β γ α β β γ γ α α β γ (e) sin 2 þ sin 2 þ sin 2 þ k sin 2 sin 2 þ sin 2 sin 2 þ sin 2 sin 2 À sin 2 sin 2 sin 2 3 5k 4 2 þ 8 , where k À5, (f) 7 k α β γ α β β γ γ α where þ sin þ sin þ sin þ k sin sin þ sin sin þ sin sin À 5 2 2 2 2 2 2 2 2 2 2 4 α β γ À þ 2k sin sin sin 5 2 2 2 4 k À5. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi α β γ 15 α β β γ γ α 5.1.48. (a) sin þ sin þ sin 4 þ cos ðÞþÀ cos ðÞþÀ cos ðÞÀ , (b) À1 cos ðÞα À β cos ðÞβ À γ cos ðÞγ À α 1, 8 pffiffi pffiffi 3 3 α β β γ γ α 3 3 (c) À 8 sin ðÞÀ sin ðÞÀ sin ðÞÀ 8 . β γ α β β γ γ α 1 β γ Hint (c) If sin( À ) 0, then sin ðÞÀ sin ðÞÀ sin ðÞÀ 2 sin ðÞÀ ðÞ1 À cos ðÞβ À γ , while at sin(β À γ) < 0 we have sin ðÞα À β sin ðÞβ À γ γ α sin ðÞγÀβ β γ sin ðÞÀ 2 ðÞ1 þ cos ðÞÀ . 5.1.49. (a) cos(α À β) cos(β À γ) cos(γ À α) 8 cos α cos βcosγ, (b) cos2(α À β) þ cos2(β À γ) þ cos2(γ À α) 24 cos α cos βcosγ. ÀÁ 2 αÀβ 2 βÀγ 2 γÀα α β γ 3 5.1.50. cos 2 cos 2 cos 2 8 sin 2 sin 2 sin 2 . Hint See the problems 5.5.1b and 5.3.4. ÀÁ α β γ 2 αÀβ ; 2 βÀγ ; 2 γÀα 5.1.51. 8 sin 2 sin 2 sin 2 min cos 2 cos 2 cos 2 . 5.1.52. sin2α þ sin2β þ sin2γ 2 þ (cosα þ cos β þ cos γ À 1)2. 5.1.53. Find the smallest value of the expression

cosðÞþ 2α þ β cosðÞþ 2β þ γ þ cosðÞÀ 2γ þ α cos 2α À cos 2β À cos 2γ þ cos α þ cos β þ cos γ: 5.2 Inequalities for the Angles of Acute and Obtuse Triangles 191

5.2 Inequalities for the Angles of Acute and Obtuse Triangles

Let α, β, and γ be the angles of some acute triangle. Prove the inequalities of the problems 5.2.1–5.2.8. pffiffiffi 5.2.1. (a) tgα þ tgβ þ tgγ 3 3, 2α 2β 2γ < (b) tg 2 þ tg 2 þ tg 2 2. α β γ α β γ 5.2.2. (a) tg þ tg þ tg ctg 2 þ ctg 2 þ ctg 2, α β γ α β γ (b) tg Á tg Á tg ctg 2 ctg 2 ctg 2. 5.2.3. (a) sinα þ sin β þ sin γ > 2, α β γ > 1 (b) cos 2 cos 2 cos 2 2. 5.2.4. cos(α À β) cos(β À γ) cos(γ À α) 8 cos α cos β cos γ. 5.2.5. (4 cos α þ 1)2 þ (4 cos β þ 1)2 þ (4 cos γ þ 1)2 27. 2α 2β 2γ 2α 2β 2γ 5.2.6. tg þ tg þ tg ctg 2 þ ctg 2 þ ctg 2. 5.2.7. cos2α þ cos2β þ cos2γ 4cos2αcos2β þ 4cos2βcos2γ þ 4cos2γcos2α. 2 5.2.8. (a) sin α þ sin β þ sin γ p2ffiffi ðÞcos α þ cos β þ cos γ , pffiffiffi 3 (b) sin α þ sin β þ sin γ 2 3ðÞcos α cos β þ cos β cos γ þ cos γ cos α , (c) sin2α þ sin2β þ sin2γ (cosα þ cos β þ cos γ)2. Let α, β, and γ be the angles of some obtuse triangle. Prove the inequalities. 5.2.9. tgα þ tgβ þ tgγ < 0. α β γ > γ > π 5.2.10. cos2 þ cos 2 À cos 2 1, if 2. pffiffiffi 5.2.11. (a) sin α þ sin β þ sin γ < 1 þ 2, pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffi (b) sin α þ sin β þ sin γ < 1 þ 4 8, α β γ < 1 (c) sin sin sin 2. pffiffi α β γ < 1þ 2 5.2.12. (a) cos 2 cos 2 cos 2 4 , pffiffi α β γ < 2À1 (b) sin 2 sin 2 sin 2 4 . 5.2.13. 1 þ cos α cos β cos γ > 2 sin α sin β sin γ.

Solutions

π α π β π γ α À 1 β À 1 γ À 1 α β γ π 5.2.1. (a) Note that, if ¼ 2 , ¼ 2 , ¼ 2 , then 1 þ 1 þ 1 ¼ and α β γ > α β γ 1, 1, 1 0. Hence, 1, 1, 1 are angles offfiffiffi some triangle. By such replacement β γ p α1 1 1 we obtain that ctg 2 þ ctg 2 þ ctg 2 3 3 (see the problem 5.1.10). 192 5 Application of Trigonometric Inequalities

π π Remark For α ¼ 2 , β ¼ γ ¼ , we have that tgα þ tgβ þ tgγ ¼Àp1ﬃﬃ. 3 6 3 α β β γ α γ (b) Note that tg 2 tg 2 þ tg 2 tg 2 þ tg 2 tg 2 ¼ 1. Indeed, α β γ β γ α β γ β γ β γ tg tg þ tg þ tg tg ¼ tg tg þ 1 À tg tg þ tg tg ¼ 2 2 2 2 2 2 2 2 2 2 2 2 α α β γ β γ ¼ tg ctg 1 À tg tg þ tg tg ¼ 1: 2 2 2 2 2 2

Then, α β γ tg 2 þ tg 2 þ tg 2 ¼ 2 2 2 ÀÁ β γ 2 α β β γ γ α ¼ tg α þ tg þ tg À 2 tg tg þ tg tg þ tg tg : 2 2 2 2 2 2 2 2 2

α β γ < We need to prove that tg 2 þ tg 2 þ tg 2 2. We have that α þ β 2 sin α β γ 2 γ tg þ tg þ tg ¼ þ tg < 2 2 2 α À β α þ β 2 cos þ cos 2 2 α þ β 2 sin γ ÀÁ2 < þ tg , 90 À α þ β À 90 α þ β 2 cos þ cos 2 2 ÀÁ α β α β α β α β since 90 À þ > À > À90 . This means that cos 90 À þ < cos À . 2 2 2 2 α β γ < αþβÀ90 γ < Thus tg 2 þ tg 2 þ tg 2 tg 45þ tg 2 þ tg 2 tg 45 þ tg 45 ¼ 2, since α β γ α β γ α β γ tg þ À90 þ tg ¼ tg þ À90 þ Á 1 À tg þ À90 tg . 2 2 2 ÀÁ2 2 2α 2β 2γ α β γ 2 < Therefore, tg 2 þ tg 2 þ tg 2 ¼ tg 2 þ tg 2 þ tg 2 À 2 2. < < π sin ðÞxþy 2 sin ðÞxþy xþy 5.2.2. (a) If 0 x, y 2, then tgx þ tgy ¼ cos x cos y 1þ cos ðÞxþy ¼ 2tg 2 . Therefore, 1 α þ β β þ γ α þ γ γ α β tgα þ tgβ þ tgγ 2tg þ 2tg þ 2tg ¼ ctg þ ctg þ ctg : 2 2 2 2 2 2 2 γ > π α β γ < < α β γ Remark If 2, then tg þ tg þ tg 0 ctg 2 þ ctg 2 þ ctg 2, (see the problem 5.2.9). (b) As tgα þ tgβ þ tgγ ¼ tgαtgβtgγ (see the proof of the problem 5.2.9) and 5.2 Inequalities for the Angles of Acute and Obtuse Triangles 193

α β γ π À α π À β π À γ π À α π À β π À γ ctg þ ctg þ ctg ¼ tg þ tg þ tg ¼ tg tg tg ¼ 2 2 2 2 2 2 2 2 2 α β γ ¼ ctg ctg ctg , 2 2 2

α β γ α β γ then according to the problem 5.2.2a, we have that tg tg tg ctg 2 ctg 2 ctg 2. γ > π α β γ < < α β γ Remark If 2, then tg tg tg 0 ctg 2 ctg 2 ctg 2, 5.2.3. (a) Note that

sin α þ sin β þ sin γ > sin2α þ sin2β þ sin2γ ¼ ¼ 2 À cos ðÞα þ β cos ðÞÀα À β cos 2γ ¼ 2 þ 2 cos α cos β cos γ > 2:

pffiffi γ 2π α β π α β γ 3 < : Remark For ¼ 3 , ¼ ¼ 6, we have that sin þ sin þ sin ¼ 1 þ 2 2 α β γ γ αÀβ γ γ (b) We have that sin þ sin þ sin ¼ 2 cos 2 cos 2 þ 2 sin 2 cos 2 ¼ 4 α β γ : cos 2 cos 2 cos 2 α β γ > 1 Therefore, according to the problem 5.2.3a, we deuce that cos 2 cos 2 cos 2 2. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5.2.4. We have that sin α cos ðÞ¼β À γ 1 ðÞsin 2γ þ sin 2β sin 2β sin 2γ. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 sin 2β sin 2γ Therefore, cos ðÞβ À γ . Similarly, we deduce that cos ðÞα À β pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin α sin 2α sin 2β γ α sin 2α sin 2γ sin γ and cos ðÞÀ sin β . By multiplying last three inequalities, we obtain that cos(α À β) cos(β À γ) cos(γ À α) 8 cos α cos β cos γ. Remark The inequality holds true for any triangle. α β < π γ γ π α γ π β α β β γ Indeed, let 2 .If 2 þ or 2 þ , then cos( À ) cos( À ) γ α α β γ π α < γ < π β < γ α < cos( À ) 0 8 cos cos cos .If2 þ 2 þ , then 0 À cos( À ) Àcos γ and 0 < cos(α À β) cos(β À γ) 8 cos α cos β cos γ. 5.2.5. Note that cos2α þ cos2β ¼ 1 þ cos(α þ β) cos(α À β) ¼ 1 À cos γ cos(α À β) 1 À cos γ. Thus, it follows that

1ÀÁÀÁÀÁÀÁ cos 2α þ cos 2β þ cos 2γ ¼ cos 2α þ cos 2β þ cos 2β þ cos 2γ þ cos 2γ þ cos 2α 2 1 ðÞ3 À cosα À cosβcosγ : 2

Therefore,

ðÞ4 cos α þ 1 2 þ ðÞ4 cos β þ 1 2 þ ðÞ4 cos γ þ 1 2 27: 194 5 Application of Trigonometric Inequalities

2 2α 2β ðÞtgαþtgβ 2 αþβ 2γ 5.2.6. Note that tg þ tg 2 2tg 2 ¼ 2ctg 2 (see the proof of the problem 5.2.2a). Thus, tg 2α þ tg 2β þ tg 2γ ¼ 1ÀÁ1ÀÁ1ÀÁγ α β ¼ tg 2α þ tg 2β þ tg 2β þ tg 2γ þ tg 2α þ tg 2γ ctg 2 þ ctg 2 þ ctg 2 : 2 2 2 2 2 2 Remark For angles β ¼ α, γ ¼ π À 2α, where α is a sufﬁciently small number, the inequality does not hold true. 5.2.7. We have that

4cos 2αcos 2β þ 4cos 2βcos 2γ þ 4cos 2γcos 2α À cos 2α À cos 2β À cos 2γ ¼ ¼ ðÞcosðÞÀα À β cosγ 2 þ 4cos 2γðÞ1 À cosγ cosðÞα À β À1 þ cosγ cosðÞÀα À β cos 2γ ¼ ¼ cos 2ðÞÀα À β ðÞ4cos 3γ þ cosγ cosðÞþα À β 4cos 2γ À 1:

Consider a quadratic trinomial f(x) ¼ x2 À (4cos3γ þ cos γ)x þ 4cos2γ À 1. Note that f(cosγ) ¼À4cos4γ þ 4cos2γ À 1 ¼À(2cos2γ À 1)2 0, f(1) ¼À4cos3 γ À cos γ þ 4cos2γ ¼Àcos γ(2 cos γ À 1)2 0. Therefore, max fxðÞ¼maxðÞf ðÞcos γ ; f ðÞ1 0. ½cos γ;1 As 0 |α À β| γ, then cosγ cos(α À β) 1, this means that

4cos 2αcos 2β þ 4cos 2βcos 2γ þ 4cos 2γcos 2α À cos 2α À cos 2β À cos 2γ ¼ ¼ f ðÞcos ðÞα À β 0: α β π γ 2π Remark For ¼ ¼ 6 , ¼ 3 the inequality does not hold true. 5.2.8. (a) Note that

2 pffiffiffiðÞcos α þ cos β þ cos γ 2 À sin α À sin β À sin γ ¼ 3 2 γ α À β 2 γ α À β 8 γ α À β ¼ pffiffiffi 2 sin cos þ cos γ À 2cos cos À sin γ ¼ pffiffiffisin2 cos 2 þ 2 2 2 2 2 2 3 3 4 γ γ α À β 2 þ2 pffiffiffi sin cos γ À cos cos þ pffiffifficos 2γ À sin γ: 3 2 2 2 3

Consider a quadratic trinomial 8 γ 4 γ γ 2 fxðÞ¼pffiffiffi sin2 x2 þ 2 pffiffiffi sin cos γ À cos x þ pffiffiffi cos 2γ À sin γ: 3 2 3 2 2 3 γ π Let 3, then 2 γ 2 γ 2 3 γ γ f ðÞ¼1 pffiffiffi 2 sin þ cos γ À 2 cos À sin γ pffiffiffi Á 2 sin þ cos γ À 2cos À sin γ, 3 2 2 3 2 2 2 5.2 Inequalities for the Angles of Acute and Obtuse Triangles 195 ÀÁ γ γ 3 γ 1 2 3 since 2 sin 2 þ cos ¼ 2 À 2 sin 2 À 2 2. Thus, it follows that pffiffiffi γ γ γ π π f ðÞ1 3 2 sin þ cos γ À 2 cos À sin γ ¼ 2 2 sin À þ sin À γ ¼ 2 2 2 6 3 γ π γ π ¼ 4 sin À 1 À cos À 0: Therefore; f ðÞ1 0: 2 6 2 6 ÀÁ γ 2 We need to prove that f cos ¼ p2ffiffi ðÞsin γ þ cos γ À ðÞsin γ þ cos γ À1 0. 2 3 pffiffiffi ÀÁpffiffiffi Indeed, we have that 1 ¼ sin2γ þ cos 2γ sin γ þ cos γ 2 sin γ þ π 2. ÀÁ ffiffiffipffiffi 4 γ p Thus, f cos ¼ðÞsinγþcosγ p2ffiffiðÞÀsinγþcosγ 1 À1 2 2pffiffi2À1 À1<0. 2 3 3 αÀβ γ γ αÀβ Since 0 2 2, then we have that cos 2 cos 2 1, this means that 2 pffiffiffiðÞcos α þ cos β þ cos γ À 1 2 À sin α À sin β À sin γ ¼ 3 α À β γ ¼ f cos himax fxðÞ¼max f cos ; f ðÞ1 0: 2 γ 2 cos ; 1 2 Therefore,

2 sin α þ sin β þ sin γ pffiffiffi ðÞcos α þ cos β þ cos γ 2: 3

Remark For angles β ¼ α, γ ¼ π À 2α, where α is a sufficiently small number the inequality does not hold true. 2 1 2 2 2 (b) We have ðÞx þ y þ z À 3xy À 3yz À 3xz ¼ 2 ðÞx À y þ ðÞy À z þ ðÞx À z 0. Therefore, (x þ y þ z)2 3xy þ 3yz þ 3xz. Thus, it follows that 2 sin α þ sin β þ sin γ pffiffiffiðÞcos α þ cos β þ cos γ 2 3 pffiffiffi 2 3ðÞcos α cos β þ cos β cos γ þ cos γ cos α (see the problem 5.2.8а). This means that pffiffiffi sin α þ sin β þ sin γ 2 3ðÞcos α cos β þ cos β cos γ þ cos γ cos α :

α β γ γ γ π γ αÀβ (c) Let max( , , ) ¼ , then 3 and cos 2 cos 2 1. 196 5 Application of Trigonometric Inequalities

We have to prove that

α À β ðÞ4 cos γ À 2 cos 2 2 γ α À β À 4 sin cos γ cos þ 1 À cos γ þ sin2γ À cos 2γ 0: ð5:12Þ 2 2 ÀÁÀÁ γ Since 4 cos γ À 2 0, then min fxðÞ¼min f cos ; f ðÞ1 , where fxðÞ¼ γ; 2 ½cos 2 1 γ 2 γ γ γ 2γ 2γ: ðÞ4 cos À 2 x À 4 sin 2 cosÀÁx þ 1 À cos þ sin ÀÀÁcos γ 2 γ γ γ We have that f ðÞ¼1 2 sin 2 À 1 cos 0 and f cos 2 ¼ 1 À sin 2 0. Thus, the inequality (5.12) holds true. 5.2.9. We have that

tgα þ tgβ þ tgγ ¼ tg ðÞα þ β ðÞþ1 À tgαtgβ tgγ ¼ π π ¼ÀtgγðÞþ1 À tgαtgβ tgγ ¼ tgαtgβtgγ < 00< α; β < ; γ > : 2 2

5.2.10. We have that

cos 2α þ cos 2β À cos 2γ ¼ ¼ 1 þ 2 cos ðÞα þ β cos ðÞÀα À β 2cos 2γ ¼ 1 À 4 sin α sin β cos γ > 1:

γ > π α β < π 5.2.11. (a) Let 2, then þ 2. Consequently, sin α þ sin β þ sin γ ¼ α þ β α À β α þ β π pffiffiffi ¼ 2 sin cos þ sin γ 2 sin þ sin γ < 2 sin þ 1 ¼ 1 þ 2: 2 2 2 4

pffiffi pffiffiffi α β γ π α β γ 3 3 > Remark For ¼ ¼ ¼ 3, we have that sin þ sin þ sin ¼ 2 1 þ 2. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γ > π 2 2 (b) Let 2. Note that x þ y 2ðÞx þ y . Therefore, ffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi p 2 pffiffiffiffiffiffiffiffiffiffi 2 pffiffiffiffiffiffiffiffiffiffi sin α þ sin β þ sin γ 2 sin α þ 2ðÞsin β þ sin γ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi α þ β pffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 4 sin þ sinγ < 2 2 þ 1 ¼ 1 þ 4 8: 2 5.2 Inequalities for the Angles of Acute and Obtuse Triangles 197

γ > π (c) Let 2, then 1 sin α Á sin β Á sin γ ¼ ðÞcos ðÞÀα À β cos ðÞα þ β sin γ 2 1 sin γ sin 2γ sin γ 1 ðÞ1 þ cos γ sin γ ¼ þ < < : 2 2 4 2 2

α β γ < 1 Hence, sin sin sin 2. γ > π 5.2.12. (a) Let 2, then α β γ 1 α À β α þ β γ 1 γ γ cos cos cos ¼ cos þ cos cos 1 þ cos cos ¼ 2 2 2 2 2 2 2 2 2 2 pffiffiffi pffiffiffi 1 γ 1 2 1 1 þ 2 ¼ cos þ sin γ < þ ¼ : 2 2 4 4 4 4

pffiffi pffiffi α β γ π α β γ 3 3 > 1þ 2 Remark For ¼ ¼ ¼ 3, we have that cos 2 Á cos 2 Á cos 2 ¼ 8 4 . (b) Let γ > 90 , then

α β γ α γ þ β À 90 45 45 sin sin sin < sin sin sin 45 sin sin sin 45 ¼ 2 2 2 2 ffiffiffi 2 2 2 p 1 À cos 45 2 À 1 β γ 1 γ À β β þ γ ¼ sin 45 ¼ , since sin sin ¼ cos À cos , 2 4 2 2 2 2 2 ÀÁ γþβÀ90 1 α βþγ sin 2 sin 45 ¼ 2 cos 2 À cos 2 ÀÁÀÁ ÀÁ α γþβÀ90 1 α 245 1 sin 2 sin 2 ¼ 2 cos 45 À À cos 45 , sin 2 ¼ 2 1 À cos 45 , α < α < γÀβ < π γÀβ < α cos(45 À ) 1, and 0 2 2 2. Therefore, cos 2 cos 2. γ > π 5.2.13. Let 2, then

1 þ cos α cos β cos γ À 2 sin α sin β sin γ ¼ 1ÀÁ ¼ 2 þ cos ðÞα À β ðÞcos γ À 2 sin γ Àcos 2γ À 2 sin γ cos γ 2 1ÀÁ 2 þ cos γ À 2 sin γ À cos 2γ À 2 sin γ cos γ , 2 since cosγ À 2 sin γ < 0. Consider a quadratic trinomial f(x) ¼Àx2 þ (1 À 2 sin γ)x þ 2 À 2 sin γ. Note that f(0) ¼ 2 À 2 sin γ > 0 and f(À1) ¼ 0. Hence,f(x) > 0, for all À1 < x < 0 . Thus 2 þ cos γ À 2 sin γ À cos2γ À 2 sin γ cos γ ¼ f(cosγ) > 0. Therefore, 1 þ cos α cos β cos γ > 2 sin α sin β sin γ. α β γ π α β γ < α β γ Remark For ¼ ¼ ¼ 3, we have that 1 þ cos cos cos 2 sin sin sin . 198 5 Application of Trigonometric Inequalities

Problems for Self-Study

Let α, β, and γ be the angles of some acute triangle. Prove the following inequalities. 5.2.14. sin2α > sin 2β > sin 2γ,ifα < β < γ. 5.2.15. cos2α þ cos 2β À cos 2γ < 1. 5.2.16. 2 cos(α À β) cos(β À γ) cos(γ À α) 1 þ 8 cos α cos β cos γ. < 2α 2β 2γ 3 1 α β γ 5.2.17. 2 sin þ sin þ sin 2 þ 2 ðÞcos þ cos þ cos . 5.2.18. (a) sinα þ sin β þ sin γ > cos α þ cos β þ cos γ, pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi (b) sin α þ sin β þ sin γ > 2. pffiffiffiffiffi 5.2.19. tg nα þ tg nβ þ tg nγ 3 3n, where n 2 N.

tg 5αþtg 5βþtg 5γ 5.2.20. tgαþtgβþtgγ 9. β γ 5.2.21. 1 tg 2α þ tg 2 þ tg 2 < 2. 2 2 ÀÁ2 ÀÁ ÀÁ α β γ π α π β π γ 5.2.22. tg þ tg þ tg ctg 8 þ 8 þ ctg 8 þ 8 þ ctg 8 þ 8 . 3 3 3 pffiffiffi sin αþsin βþsin γ > 5.2.23. sin α sin βðÞsin αþ sin βþ sin γ 2. Hint Prove that sin 3α þ sin 3β þ sin 3γ > > ðÞsin α þ sin β sin α sin β þ sin γðÞsin2α þ sin2β sin α sin βðÞsinα þ sin β þ 2sinγ :

5.2.24. ðÞcos α cos β þ cos α cos γ þ cos β cos γ 2 cos 2αsin2γ þ cos 2γsin2β þ cos 2βsin2α: Let α, β, and γ be the angles of some obtuse triangle. Prove the following inequalities. 5.2.25. sin2α þ sin2β þ sin2γ < 2. 5.2.26. cosα cos β cos γ > À 1.

5.3 Some Relations for a Triangle

Let α, β, γ be the angles of triangle ABC. Prove the following relations. α β γ p 5.3.1. (a) sin þ sin þ sin ¼ R, 5.3 Some Relations for a Triangle 199

α β γ pr (b) sin sin sin ¼ 2R2, α β γ 2pr (c) sin 2 þ sin 2 þ sin 2 ¼ R2 . α β γ p 5.3.2. ctg 2 þ ctg 2 þ ctg 2 ¼ r. α β γ p 5.3.3. cos 2 cos 2 cos 2 ¼ 4R. α β γ r 5.3.4. sin 2 sin 2 sin 2 ¼ 4R. α β γ r 5.3.5. tg 2 tg 2 tg 2 ¼ p. α β γ Rþr 5.3.6. cos þ cos þ cos ¼ R . α β γ a2þb2þc2 p2Àr2À4rR 5.3.7. ctg þ ctg þ ctg ¼ 4S ¼ 2pr . 2α 2β 2γ p2Àr2À4rR 5.3.8. (a) sin þ sin þ sin ¼ 2R2 , 2α 2β 2γ 6R2Àp2þ4Rrþr2 (b) cos þ cos þ cos ¼ 2R2 . α β γ 3R2Àp2þr2þ4rR 5.3.9. (a) cos 2 þ cos 2 þ cos 2 ¼ R2 , α β γ p2ÀðÞ2Rþr 2 (b) cos cos cos ¼ 4R2 . α β β γ γ α p2þ4Rrþr2 5.3.10. (a) sin sin þ sin sin þ sin sin ¼ 4R2 , α β β γ γ α r2þp2À4R2 (b) cos cos þ cos cos þ cos cos ¼ 4R2 , 2 2 2 (c) 1 þ 1 þ 1 ¼ p þr À4R . cos α cos β cos γ p2ÀðÞ2Rþr 2

αÀβ βÀγ γÀα p2þ2Rrþr2 5.3.11. cos 2 cos 2 cos 2 ¼ 8R2 . 5.3.12. tgα þ tgβ þ tgγ ¼ 4S ¼ 2pr . a2þb2þc2À8R2 p2ÀðÞ2Rþr 2

2 2 5.3.13. tgαtgβ þ tgβtgγ þ tgγtgα ¼ p Àr À4Rr . p2ÀðÞ2Rþr 2 α β γ 4Rþr 5.3.14. tg 2 þ tg 2 þ tg 2 ¼ p . 5.3.15. sin 3α cosβ cos γ þ sin 3β cos α cos γ þ sin 3γ cos α cos β ¼ pr ¼ 2R2 À p2 þ ðÞ2R þ r 2 : 4R4 β À γ α À γ α À β 5.3.16. sin2α cos 2 þ sin2β cos 2 þ sin2γ cos 2 ¼ 2 2 2 ðÞ2R þ r p2 þ 4Rr2 þ r3 ¼ : 8R3 200 5 Application of Trigonometric Inequalities

Solutions

α β γ a b c 5.3.1. (a) Taking into account the relation sin þ sin þ sin ¼ 2R þ 2R þ 2R,we α β γ p obtain that sin þ sin þ sin ¼ R. α β γ abc abc α β γ pr (b) Since sin sin sin ¼ 8R3 and 4R ¼ S ¼ pr, then sin sin sin ¼ 2R2.

sin 2α þ sin 2β þ sin 2γ ¼ ¼ 2 sin ðÞα þ β cos ðÞþα À β sin 2γ ¼ 2 sin γðÞcos ðÞÀα À β cos ðÞα þ β ¼ 4 sin α sin β sin γ,

α β γ pr 2pr then sin 2 þ sin 2 þ sin 2 ¼ 42R2 ¼ R2 (see the problem 5.3.1). α pÀa β pÀb γ pÀc 5.3.2. Taking into account that ctg 2 ¼ r , ctg 2 ¼ r , ctg 2 ¼ r and p À a þ α β γ p p À b þ p À c ¼ p, we deduce that ctg 2 þ ctg 2 þ ctg 2 ¼ r. 5.3.3. Note that

α þ β α À β γ γ sin α þ sin β þ sin γ ¼ 2 sin cos þ 2 sin cos ¼ 2 2 2 2 γ α À β α þ β α β γ ¼ 2 cos cos þ cos ¼ 4 cos cos cos : 2 2 2 2 2 2

Therefore,

α β γ 1 p cos cos cos ¼ ðÞsin α þ sin β þ sin γ ¼, 2 2 2 4 4R

(see the problem 5.3.1a). 5.3.4. We have that

α β γ sin α sin β sin γ pr 2p r sin sin sin ¼ ¼ : ¼ , 2 2 2 α β γ 2 R 4R 8 cos 2 cos 2 cos 2 2R

(see the problem 5.3.1b and 5.3.3). 5.3 Some Relations for a Triangle 201

5.3.5. Since

β γ α β γ sin α sin sin r p r tg tg tg ¼ 2 2 2 ¼ : ¼ 2 2 2 α β γ 4R 4R p cos 2 cos 2 cos 2

(problems 5.3.3 and 5.3.4). 5.3.6. Note that

α þ β α À β γ α À β γ cosα þ cosβ þ cosγ ¼ 2cos cos þ cosγ ¼ 2sin cos þ 1 À 2sin2 ¼ 2 2 2 2 2 γ α À β α þ β α β γ r R þ r ¼ 2sin cos À cos þ 1 ¼ 1 þ 4sin sin sin ¼ 1 þ ¼ , 2 2 2 2 2 2 R R

(see the problem 5.3.4). 5.3.7. We have that

cosα cosβ cosγ b2 þ c2 À a2 a2 þ c2 À b2 a2 þ b2 À c2 ctgα þ ctgβ þ tgγ ¼ þ þ ¼ þ þ ¼ sinα sinβ sinγ 2bcsinα 2acsinβ 2absinγ b2 þ c2 À a2 a2 þ c2 À b2 a2 þ b2 À c2 a2 þ b2 þ c2 ¼ þ þ ¼ : 4S 4S 4S 4S pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Taking into account that pr ¼ ppðÞÀ a ðÞp À b ðÞp À c and abc ¼ 4SR, we obtain that p2r2 ¼ p( p À a)( p À b)( p À c). Thus, it follows that pr2 ¼ p3 À p2 (a þ b þ c) þ p(ab þ bc þ ac) À 4SR ¼ p3 À 2p3 þ p(ab þ bc þ ac) À 4pRr, this means that ab þ bc þ ac ¼ p2 þ r2 þ 4Rr and a2 þ b2 þ c2 ¼ (a þ b þ c)2 À 2(ab þ bc þ ac) ¼ 2p2 À 2r2 À 8Rr. Thus, a2 þ b2 þ c2 2p2 À 2r2 À 8Rr p2 À r2 À 4Rr ¼ ¼ : 4S 4pr 2pr

5.3.8. (a) We have that a2 þ b2 þ c2 p2 À r2 À 4Rr sin2α þ sin2β þ sin2γ ¼ ¼ , 4R2 2R2

(see the proof of the problem 5.3.7). (b) Note that ÀÁ6R2 À p2 þ r2 þ 4Rr cos 2α þ cos 2β þ cos 2γ ¼ 3 À sin2α þ sin2β þ sin2γ ¼ , 2R2 (see the problem 5.3.8a). 202 5 Application of Trigonometric Inequalities

5.3.9. (a) We have that

cos 2α þ cos 2β þ cos 2γ ¼ cos 2α þ cos 2β þ cos 2γ À ðÞsin2α þ sin2β þ sin2γ ¼ 6R2 À p2 þ r2 þ 4Rr p2 À r2 À 4Rr 3R2 À p2 þ r2 þ 4Rr ¼ À ¼ , 2R2 2R2 R2

(see the problems 5.3.8a and b). (b) Note that

cos 2α þ cos 2β þ cos 2γ ¼ 2 cos ðÞα þ β cos ðÞþα À β 2cos 2γ À 1 ¼À1 À 4 cos α cos β cos γ:

Hence, 1 p2 À ðÞ2R þ r 2 cos α cos β cos γ ¼À ðÞ1 þ cos 2α þ cos 2β þ cos 2γ ¼, 4 4R2 (see the problem 5.3.9a). 5.3.10. (a) We have that

sin α sin β þ sin β sin γ þ sin γ sin α 1 ¼ ðÞsin α þ sin β þ sin γ 2 À sin2α À sin2β À sin2γ ¼ 2 1 p2 p2 À r2 À 4rR p2 þ r2 þ 4Rr ¼ À ¼ 2 R2 2R2 4R2 (see the problems 5.3.1a and 5.3.8a). (b) We have that cos α cos β þ cos β cos γ þ cos γ cos α ¼ 1 ¼À ðÞcos α þ cos β þ cos γ 2 À cos 2α À cos 2β À cos 2γ ¼ 2 ! 1 ðÞR þ r 2 6R2 À p2 þ 4Rr þ r2 p2 À 4R2 þ r2 ¼ À ¼ , 2 R2 2R2 4R2

(see the problems 5.3.6 and 5.3.8b). (c) Note that

1 1 1 cos α cos β þ cos β cos γ þ cos α cos γ þ þ ¼ ¼ cos α cos β cos γ cos α cos β cos γ r2 þ p2 À 4R2 p2 À ðÞ2R þ r 2 r2 þ p2 À 4R2 ¼ : ¼ , 4R2 4R2 p2 À ðÞ2R þ r 2

(see the problems 5.3.10b and 5.3.9b). 5.3 Some Relations for a Triangle 203

5.3.11. Note that α À β β À γ γ À α 1 α À γ α þ γ À 2β γ À α cos cos cos ¼ cos þ cos cos ¼ 2 2 2 2 2 2 2 1 α À γ 1 α þ γ À 2β γ À α 1 ¼ cos 2 þ cos cos ¼ ðÞ1 þ cosðÞþγ À α cosðÞþβ À γ cosðÞα À β ¼ 2 2 2 2 2 4 1ÀÁÁ ¼ 1 þ cosγ cosαþ cosβcosγ þ cosαcosβ þ sinαsinβ þ sinβsinγ þ sinγ sinα ¼ 4 1 r2 þ p2 À 4R2 p2 þ 4Rr þ r2 p2 þ r2 þ 2Rr ¼ 1 þ þ ¼ , 4 4R2 4R2 8R2

(see the problems 5.3.10a and 5.3.10b). 5.3.12. We have that

tgα þ tgβ þ tgγ ¼ tg ðÞα þ β ðÞþ1 À tgαtgβ tgγ ¼ÀtgγðÞþ1 À tgαtgβ tgγ ¼ sin α sin β sin γ pr p2 À ðÞ2R þ r 2 2pr ¼ tgαtgβtgγ ¼ ¼ : ¼ , cos α cos β cos γ 2R2 4R2 p2 À ðÞ2R þ r 2

(see the problems 5.3.1b and 5.3.9b). Since a2 þ b2 þ c2 À 8R2 ¼ 2p2 À 2r2 À 8Rr À 8R2 ¼ 2( p2 À (2R þ r)2), (see the proof of the problem 5.3.7) and S ¼ pr, then

2pr 4S tgα þ tgβ þ tgγ ¼ ¼ : p2 À ðÞ2R þ r 2 a2 þ b2 þ c2 À 8R2

5.3.13. Note that

ctgα þ ctgβ þ ctgγ p2 À r2 À 4rR tgαtgβ þ tgβtgγ þ tgαtgγ ¼ ¼ tgαtgβtgγ ¼ ctgαctgβctgγ 2pr p2 À r2 À 4rR p2 À r2 À 4rR 2pr p2 À r2 À 4rR ¼ ðÞtgα þ tgβ þ tgγ ¼ ¼ , 2pr 2pr p2 À ðÞ2R þ r 2 p2 À ðÞ2R þ r 2

(see the problem 5.3.7 and the proof of the problem 5.3.12). 5.3.14. Note that

α β γ r r r tg þ tg þ tg ¼ þ þ 2 2 2 p À a p À b p À c ab þ bc þ ac À p2 r2 þ 4Rr r þ 4R ¼ r ¼ r ¼ , ðÞp À a ðÞp À b ðÞp À c pr2 p

(see the proof of the problem 5.3.7). 204 5 Application of Trigonometric Inequalities

5.3.15. We have that sin 3α cos β cos γ þ sin 3βcosαcos γ þ sin 3γcosα cos β ¼ ¼ sin 3αðÞcos ðÞþβ þ γ sin β sin γ þsin 3βðÞcosðÞþα þ γ sin α sin γ þ þsin 3γðÞcos ðÞþα þ β sin αsin β ¼ 1 À cos 2α sin 2α 1 À cos 2β sin 2β 1 À cos 2γ sin 2γ ¼À À À þ 2 2 2 2 2 2 þ sin α sin β sin γðÞsin2α þ sin2β þ sin2γ ¼ 1 1 ¼À ðÞsin 2α þ sin 2β þ sin 2γ þðÞsin 4α þ sin 4β þ sin 4γ þ 4 8 þ sin α sin β sin γðÞsin2α þ sin2β þ sin2γ ¼ 1 ¼Àsin αsin β sin γ þ ðÞ2 sinðÞ 2α þ 2β cosðÞþ 2α À 2β 2 sin 2γ cos 2γ þ 8 þ sin α sin β sin γðÞsin2α þ sin2β þ sin2γ ¼ 1 ¼Àsin αsin β sin γ þ ðÞÀ sin 2γ cosðÞþ 2α À 2β sin 2γ cosðÞ 2α þ 2β þ 4 þ sin α sin β sin γðÞsin2α þ sin2β þ sin2γ ¼ 1 ÀÁ ¼Àsin αsin β sin γ À sin 2α sin 2β sin 2γ þ sin α sin β sin γ sin2α þ sin2β þ sin2γ ¼ 2 ¼ sin α sin β sin γðÞsin2α þ sin2β þ sin2γ À 1 À 4cosα cos β cos γ ¼ ¼ sin α sin β sin γðÞsin2α þ sin2β þ sin2γ þ cos 2α þ cos 2β þ cos 2γ ¼ pr 6R2 p2 4Rr r2 α β γ 2α 2β 2γ À þ þ ¼ sin sin sin ðÞcos þ cos þ cos ¼2 2 ¼ 2R 2R pr ¼ 2R2 À p2 þ ðÞ2R þ r 2 , 4R4

(see the proof of the problems 5.3.1c, 5.3.9b and the problems 5.3.1b, 5.3.8b). 5.3.16. We have that β À γ α À γ α À β sin2α sin2αcos 2 þ sin2βcos 2 þ sin2γcos 2 ¼ ðÞ1 þ cos ðÞβ À γ þ 2 2 2 2 sin2β sin2γ þ ðÞ1 þ cos ðÞα À γ þðÞ1 þ cos ðÞα À β ¼ 2 2 1ÀÁðÞ1 À cos 2α cos β cos γ ðÞ1 À cos 2β cos α cos γ ¼ sin2α þ sin2β þ sin2γ þ þ þ 2 2 2 ðÞ1 À cos 2γ cos α cos β sin α sin β sin γ þ þ ðÞsin α þ sin β þ sin γ ¼ 2 2 sin2α þ sin2β þ sin2γ cos α cos β þ cos β cos γ þ cos γ cos α ¼ þ À 2 2 cosα cos β cos γ sin α sin β sin γ À ðÞcosα þ cos β þ cos γ þðÞsin α þ sin β þ sin γ ¼ 2 2 p2 À r2 À 4rR r2 þ p2 À 4R2 p2 À ðÞ2R þ r 2 R þ r pr p ¼ þ À Á þ Á ¼ 4R2 8R2 8R2 R 4R2 R ðÞ2R þ r p2 þ 4Rr2 þ r3 ¼ 8R3 (see the problems 5.3.1a, 5.3.1b, 5.3.6, 5.3.8a, 5.3.9b, 5.3.10b). 5.4 Trigonometric Inequalities 205

Problems for Self-Study

Let α, β, γ be the angles of triangle ABC. Prove the following relations. 2 1 1 1 4Rþr α γ 5.3.17. cos 2 þ 2β þ cos 2 ¼ 1 þ p . 2 cos 2 2 1 1 1 32R2þ8Rr α γ 5.3.18. 2α 2β þ 2β 2γ þ cos 2 cos 2 ¼ p2 . cos 2cos 2 cos 2cos 2 2 2

5.4 Trigonometric Inequalities

α β γ αþβ βþγ γþα < α 5.4.1. Prove that ctg þ tg þ ctg ctg 2 þ ctg 2 þ ctg 2 , where 0 , β γ < π , 2. 5.4.2. Prove that α β β γ γ α 3 (a) sin cos þ sin cos þ sin cos 2, 2α β 2β γ 2γ α < 3 (b) sin cos þ sin cos þ sin cos 2. pffiffiffi 5.4.3. Prove that cos α þ cos β þ cos γ 5, if sinα þ sin β þ sin γ 2. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 5.4.4. Prove that a sin α þ b sin β þ c sin γ 2 a2 þ b2 þ c2 , if cos2α þ cos2 β þ cos2γ ¼ 1. α β γ α β γ 1 1 1 > < α α β 5.4.5. Prove that ctg 2 þ ctg 2 þ ctg 2 ctg 2 þ ctg 2 þ ctg 2,if0 1 , 0 < γ1 γ β, β1 6¼ β and α1 þ β1 þ γ1 ¼ α þ β þ γ ¼ π. α α ::: α < π < α < π 5.4.6. Prove that 1 þ 2 þ þ n 2 ðÞn À 1 ,if0 i 2, i ¼ 1, 2, . . . , n and 2 2 2 cos α1 þ cos α2 þ ...þ cos αn > 1. φ φ ::: φ nπ φ φ φ > 5.4.7. Prove that cos 1 cos 2 Á Á cos n cos n,if 1, 2,..., n 0, and φ1 þ ...þ φn ¼ π 5.4.8. Prove that π (a) tgα þ tgβ þ tgγ pffiffi 2 ,if0 α, β, γ < , 3 cos α cos β cos γ 2 pffiffi α β γ δ 3 3 α β γ δ < π (b) tg þ tg þ tg þ tg 4 cos α cos β cos γ cos δ,if0 , , , 2. 5.4.9. Prove that sin2α þ sin 2β þ sin 2γ þ sin 2δ 16 sin α sin β sin γ sin δ,if α, β, γ, δ > 0 and α þ β þ γ þ δ ¼ π. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5.4.10. Prove that ab sin α þ cd sin β 2 ðÞp À a ðÞp À b ðÞp À c ðÞp À d ,ifa, b, c, aþbþcþd 2 2 d, p À a, p À b, p À c, p À d are positive numbers, where p ¼ 2 and a þ b À 2ab cos α ¼ c2 þ d2 À 2cd cos β. 206 5 Application of Trigonometric Inequalities

5.4.11. Prove that (a) 1 þ 4 cos α cos β cos γ cos α þ cos β þ cos γ,ifα, β, γ > 0 and α þ β þ γ ¼ π, α β γ 17 2 α β γ < α β γ < π (b) cos þ cos þ cos 12 þ 3 cos cos cos ,if0 , , 2 and α þ β þ γ ¼ π, α β γ α β γ 17 2 α β γ < α (c) 1 þ 4 sin 2 sin 2 sin 2 sin 2 þ sin 2 þ sin 2 12 þ 3 sin 2 sin 2 sin 2,if0 , β, γ and α þ β þ γ ¼ π, α β β γ γ α 1 α β γ α β γ > (d) sin 2 sin 2 þ sin 2 sin 2 þ sin 2 sin 2 2 þ 2 sin 2 sin 2 sin 2, where , , 0 and α þ β þ γ ¼ π. 5.4.12. Prove that cosα þ cos β 1, if α, β > 0 and 2α þ β π,2β þ α π. sin ðÞβþλα sin β < α β α β < π λ 5.4.13. Prove that sin ðÞαþλβ sin α,if0 , þ , and 0 1.

5.4.14. Prove that, if α1 þ ...þ αn βi þ ...þ βn π, α1,...,αn > 0 and β β α α β ,...,β 0, i ¼ 1, . . . , n, then cos 1 þ ::: þ cos n cos 1 þ ::: þ cos n. 1 n sin α1 sin αn sin α1 sin αn α α ::: α α α π 5.4.15. Prove that sin 1 þ sin 2 þ þ sin n n sin n,if0 i , i ¼ 1, . . . , n, and α1 þ ...þ αn ¼ α. α α ::: α π α α ::: α 1þ 2þ þ n α < 5.4.16. Prove that tg 1 þ tg 2 þ þ tg n ntg n ,if0 i 2, i ¼ 1, 2, . . . , n. 5.4.17. Prove that 0 α À sin α À sin β À sin γ þ sin(α þ β) þ sin(α þ γ) π,if α þ β þ γ π and α, β, γ > 0. ÀÁ φ π; βþγ 3φ α φ β φ γ φ 5.4.18. Prove that min 6 3 , if sin ¼ sin( À ) sin( À ) sin( À ) and 0 < φ < γ β α, α þ β þ γ ¼ π. 5.4.19. Prove that

x1x2 cos α1 þ x2x3 cos α2 þ ::: þ xnÀ1xn cos αnÀ1 þ xnx1 cos αn π ÀÁ cos x2 þ x2 þ ::: þ x2 , ð5:13Þ n 1 2 n where α1 þ α2 þ ...þ αn ¼ π, for (a) n ¼ 3, (b) n ¼ 4, and (c) n ¼ 6.

Solutions ÀÁ ; π 5.4.1. Note that if x, y2 0 2 , then

2 sin ðÞx þ y 2 sin ðÞx þ y x þ y ctgx þ ctgy ¼ ¼ 2ctg , cos ðÞÀx À y cos ðÞx þ y 1 À cos ðÞx þ y 2 thus 5.4 Trigonometric Inequalities 207

1 1 1 ctgα þ ctgβ þ ctgγ ¼ ðÞþctgα þ ctgβ ðÞþctgβ þ ctgγ ðÞctgγ þ ctgα 2 2 2 α þ β β þ γ γ þ α ctg þ ctg þ ctg : 2 2 2

a2þb2 5.4.2 (a) Since ab 2 , then

sin2α þ cos 2β sin2β þ cos 2γ sin2γ þ cos 2α sinαcosβ þ sinβ cosγ þ sinγ cosα þ þ ¼ 2 2 2 ðÞsin2α þ cos 2α þðÞsin2β þ cos 2β þðÞsin2γ þ cos 2γ 3 ¼ ¼ : 2 2

Remark Similarly, one can prove that

ðÞcos α þ cos β þ cos γ 2 þ ðÞsin α þ sin β þ sin γ 2 ¼ ¼ 3 þ 2 cos ðÞþα À β 2 cos ðÞþβ À γ 2 cos ðÞγ À α 9

α β γ 2 α β γ 2 and since (sin þ sin þ sin ) 4, then (cospffiffiffi þ cos þ cos ) 5. Consequently, cos α þ cos β þ cos γ 5. 5.4.4. Note that sin2α þ sin2β þ sin2γ ¼ 1 À cos2α þ 1 À cos2β þ 1 À cos2γ ¼ 2, consequently, if a2 þ b2 þ c2 6¼ 0. Then,

a sin α b sin β c sin γ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁþ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁþ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ¼ 2 a2 þ b2 þ c2 2 a2 þ b2 þ c2 2 a2 þ b2 þ c2 a sin α b sin β c sin γ ¼ qÀÁffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi þ qÀÁffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁpffiffiffi a2 þ b2 þ c2 2 a2 þ b2 þ c2 2 a2 þ b2 þ c2 2 1 a2 sin2α 1 b2 sin2β þ þ þ 2 a2 þ b2 þ c2 2 2 a2 þ b2 þ c2 2 1 c2 sin2γ þ þ ¼ 1, 2 a2 þ b2 þ c2 2 208 5 Application of Trigonometric Inequalities

(see the proof of the problem 5.4.2a), this means that a sin α b sin β c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ þ þ sin γ 2 a2 þ b2 þ c2 . If a2 b2 c2 0, then a b c 0 and a sin α b sin β c sin γ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁþ þ ¼ ¼ ¼ ¼ þ þ ¼ 0 ¼ 2 a2 þ b2 þ c2 .

β γ β γ γ α1 1 1 α1 1þ 1À 5.4.5. We need to prove that ctg 2 þ ctg 2 þ ctg 2 ctg 2 þ ctg 2 þ γ α β γ ctg 2 ctg 2 þ ctg 2 þ ctg 2. As α1 α and γ1 γ, then β1 β (β1 6¼ β). Hence, β1 > β. We have to prove that

β γ β þ γ À γ γ ctg 1 þ ctg 1 ctg 1 1 þ ctg , ð5:14Þ 2 2 2 2 this means that

β γ β γ sin 1þ 1 sin 1þ 1 2 2 : β γ β γ β γ γ β γ 1À 1 1þ 1 1þ 1À2 1þ 1 cos 2 À cos 2 cos 2 À cos 2

π β γ β γ γ β γ π > 1À 1 1þ 1À2 > 1À 1 > The last inequality holds true, as 2 2 2 À 2 À2. Let us prove the second inequality:

α β þ γ À γ α β ctg 1 þ ctg 1 1 ctg þ ctg , ð5:15Þ 2 2 2 2 that is π γ π γ sin À sin À 2 2 : α γ β γ π γ β α π γ 1þ À 1À 1 À À À cos 2 À cos 2 cos 2 À cos 2

π β γ α γ β α α γ β γ π > 1þ 1À 1À À 1þ À 1À 1 > This holds true, as 2 2 2 2 À2. Note that in (5.14) and (5.15) the equality cannot simultaneously hold true. Otherwise, γ1 ¼ γ and α1 ¼ α, then β1 ¼ β. This leads to a contradiction. Thus, α β γ α β γ 1 1 1 > ctg 2 þ ctg 2 þ ctg 2 ctg 2 þ ctg 2 þ ctg 2. 5.4.6. Consider the triangle in Figure 5.1. a2þa2þ:::þa2 2α 2α ::: 2α 1 2 n > We have that cos 1 þ cos 2 þ þ cos n ¼ 4 1, from which it follows that 2 2 ::: 2 > : : a1 þ a2 þ þ an 4 ð5 16Þ

Now we shall arrange triangles in a way shown in Figure 5.2.

Figure 5.1 11

ai ai ai 5.4 Trigonometric Inequalities 209

Figure 5.2

an-1 a2

p-2an-1 p-2a2

p-2an p-2a1 a1

Figure 5.3 an-1

b2 an

bn-2 a2 b1

аa1

π α π α π α > π α α ::: α < π nÀ1 Note that À 2 1 þ À 2 2 þ ...þ À 2 n ,or 1 þ 2 þ þ n 2 . Indeed, if π À 2α1 þ π À 2α2 þ ...þ π À 2αn π, then (see Figure 5.3) 2 2 2 2 2 2 ::: 2 ::: 2 2 a a1 þ b1 a1 þ a2 þ b2 a1 þ þ anÀ2 þ bnÀ2 2 ::: 2 2 2 2 2 ::: 2 a1 þ þ anÀ2 þ anÀ1 þ an. Hence, we obtain that 4 a1 þ a2 þ þ an.This leads to a contradiction with (5.16). α α ::: α π α α ::: α π nÀ1 2α Remark When 1 0, 2 ¼ ¼ n 2 and 1 þ 2 þ þ n 2 , cos 1 þ 2 2 cos α2 þ ...þ cos αn 1. < φ π 5.4.7. Let 0 i 2, i ¼ 1, . . . , n. φ φ φ φ φ ::: φ nπ If 1 ¼ 2 ¼ ...¼ n(1), then cos 1 cos 2 Á Á cos n ¼ cos n. If the condi- φ < π < φ tion (5.16) is not satisﬁed, then there exist such i, j, that i j, but then n ÀÁ 2π ÀÁÀÁcos φ þ φ þ cos φ þ φ À cos φ þ φ þ cos φ À φ i j i j n cosφ cos φ ¼ i j j i < ¼ i j 2 2 π π ¼ cos φ þ φ À cos , i j n n 210 5 Application of Trigonometric Inequalities

π < φ φ < φ φ 2π < φ φ < π since À2 i À j i þ j À n j À i 2. φ φ π φ φ π Thus, it follows that, if i þ j is replaced by n and i þ j À n, then the product cosφ1 Á ...Á cos φn increases and if we repeat this action not more than n À 1 times, φ ::: φ < nπ we deduce that cos 1 Á Á cos n cos n. φ > π φ ::: If for some i we have that i 2, then at n 2, we obtain cos 1 Á Á φ < nπ cos n 0 cos n. 5.4.8. (a) Since A ¼ cos α cos β cos γðÞtgα þ tgβ þ tgγ ¼ ¼ sin ðÞα þ β cos γ þ cos α cos β sin γ, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁÀÁ α α 2 2 2 2 then using the inequality 1b1 þ 2b2 a1 þ a2 b1 þ b2 , we obtain that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðÞsin2ðÞþα þ β cos 2αcos 2β ðÞcos 2γ þ sin2γ ¼ À 1 ¼ sin2ðÞþα þ β ðÞcos ðÞþα þ β cos ðÞα À β 2 4 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4 3 1 2 2 sin2ðÞþα þ β ðÞjjcos ðÞα þ β þ 1 2 ¼ À jjcos ðÞα þ β À pffiffiffi: 4 3 4 3 3

Thus, cos α cos β cos γðÞtgα þ tgβ þ tgγ p2ffiffi. 3 Hence, it follows that tgα þ tgβ þ tgγ p2ffiffi Á 1 . 3 cos α cos β cos γ (b) We have that

B ¼ cosαcosβ cosγ cosδðÞtgα þ tgβ þ tgγ þ tgδ ¼ ¼ sinðÞα þ β cosγ cosδ þ sinðÞγ þ δ cosαcosβ ÀÁ ÀÁ α β cosγþcosδ 2 γ δ cosαþcosβ 2 sinðÞþ 2 þ sinðÞþ 2 γ þ δ α þ β sinðÞα þ β cos 2 þ sinðÞγ þ δ cos 2 ¼ 2 2 α þ β γ þ δ α þ β þ γ þ δ ¼ 2cos cos sin ¼ 2 2 2 α þ β þ γ þ δ α þ β À γ À δ α þ β þ γ þ δ ¼ cos þ cos sin 2 2 2 α þ β þ γ þ δ α þ β þ γ þ δ α þ β þ γ þ δ α þ β þ γ þ δ 1 þ cos sin 2cos 2 sin ¼ 2 2 4 2 α þ β þ γ þ δ α þ β þ γ þ δ ¼ 4cos 3 sin : 4 4 ÂÁ 3 αþβþγþδ ; π We have obtainedÀÁ that B 4cos t sin t, where t ¼ 4 2 0 2 . aþb 2 Since ab 2 , then 5.4 Trigonometric Inequalities 211

cos 2t cos 2t cos 2t ÀÁ cos 6tsin2t ¼ 27 1 À cos 2t 3 3 3 2 2 2 4 cos tþcos tþcos tþ1Àcos 2t 27 27 3 3 3 ¼ : 4 44

pffiffi pffiffi pffiffi Consequently, 4 3 3 3 3, thus α β γ δ 3 3 . B 42 ¼ 4 tg þ tg þ tg þ tg 4 cos α cos β cos γ cos δ 5.4.9. We have that sin 2α þ sin 2β þ sin 2γ þ sin 2δ ¼ sin α sin β sin γsinδ 4 sin α sin β sin ðÞÀγ þ δ sinðÞþ 2γ þ 2δ 4 sin γ sin δ sin ðÞÀα þ β sinðÞ 2α þ 2β ¼ ¼ sin α sin β sin γsinδ 4 sin α sin β sin ðÞþγ þ δ 4 sin γ sin δ sin ðÞα þ β ¼ sin α sin β sin γ sin δ ¼ 4ðÞctgα þ ctgβ þ ctgγ þ ctgδ ,

(see the proof of the problem 5.3.1c). < π xþy Note that, if 0 x, y and x þ y , then ctgx þ ctgy 2ctg 2 . Indeed, we have that

sin ðÞx þ y 2 sin ðÞx þ y 2 sin ðÞx þ y x þ y ctgx þ ctgy ¼ ¼ ¼ 2ctg : sin x sin y cos ðÞÀx À y cos ðÞx þ y 1 À cos ðÞx þ y 2

Therefore, α þ β γ þ δ α þ β þ γ þ δ ctgα þ ctgβ þ ctgγ þ ctgδ 2ctg þ 2ctg 4ctg ¼ 4: 2 2 4

sin 2αþ sin 2βþ sin 2γþ sin 2δ α β γ Thus, sin α sin β sin γ sin δ 16, this means that sin2 þ sin 2 þ sin 2 þ sin 2δ 16 sin α sin β sin γ sin δ. 5.4.10. We have that (ab sin α þ cd sin β)2 þ (ab cos α À cd cos β)2 ¼ a2b2 þ c2d2 α β α β a2þb2Àc2Àd2 À 2abcd cos( þ ), since ab cos À cd cos ¼ 2 , then we deduce that 2 α β 2 2 2 2 2 α β a2þb2Àc2Àd2 ðÞabsin þ cd sin ¼ a b þ c d À 2abcd cosðÞÀþ 2 ¼ 2 2 a2þb2Àc2Àd2 α β ¼ ðÞab þ cd À 2 À 2abcdðÞ1 þ cosðÞþ ¼ a2 þ b2 À c2 À d2 a2 þ b2 À c2 À d2 α þ β ¼ ab þ cd À ab þ cd þ À 4abcdcos 2 ¼ 2 2 2 ðÞc þ d 2 À ðÞa À b 2 ðÞa þ b 2 À ðÞc À d 2 α þ β ¼ À 4abcdcos 2 ¼ 2 2 2 α þ β ¼ 4ðÞp À a ðÞp À b ðÞp À c ðÞÀp À d 4abcdcos 2 4ðÞp À a ðÞp À b ðÞp À c ðÞp À d : 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Therefore, ab sin α þ cd sin β 2 ðÞp À a ðÞp À b ðÞp À c ðÞp À d . 212 5 Application of Trigonometric Inequalities

α 5.4.11. (a) According to the problem 5.1.20, it follows that 1 þ 4sin2 β γ α β γ α β γ sin 2 sin 2 1 þ 4cos cos cos and it remains to note that cos þ cos þ cos α β γ ¼ 1 þ 4sin2 sin 2 sin 2 (see the proof of the problem 5.3.6). (b) Using the problems 5.3.6 and 5.3.9b, we need to prove the inequality: 2 Rþr 17 2 p2ÀðÞ2Rþr 2 3 2 2 R 12 þ 3 4R2 , that is p 2 R þ 10Rr þ r . 2 2 2 According to the problem 5.5.23, we have that p 2R þ 8RrÀÁþ 3r . Note that 2 2 3 2 2 2 2 3 2 2 2R þ 8Rr þ 3r 2 R þ 10Rr þ r , since 2R þ 8Rr þ 3r À 2R þ 10Rr þ r 1 2 2 3 2 2 ¼ 2 ðÞR À 2r 0. Consequently, p 2 R þ 10Rr þ r . πÀα πÀβ πÀγ π (c) Since 2 þ 2 þ 2 ¼ , then according to the problems 5.4.11a and b, we obtain that π À α π À β π À γ π À α π À β π À γ 1 þ 4 cos cos cos cos þ cos þ cos 2 2 2 2 2 2 17 2 π À α π À β π À γ þ cos cos cos , 12 3 2 2 2

α β γ α β γ 17 2 α β γ or 1 þ 4 sin 2 sin 2 sin 2 sin 2 þ sin 2 þ sin 2 12 þ 3 sin 2 sin 2 sin 2. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁ α β β α α β 1 β α β α (d) We have that 2 sin 2 sin 2 ¼ tg 2 sin tg 2 sin 2 tg 2 sin þ sin tg2 , this means that

α β β γ α γ 2 sin sin þ 2 sin sin þ 2 sin sin 2 2 2 2 2 2 1 α 1 β 1 γ tg ðÞþsin β þ sin γ tg ðÞþsin α þ sin γ tg ðÞsin α þ sin β ¼ 2 2 2 2 2 2 α β À γ β α À γ γ α À β ¼ sin cos þ sin cos þ sin cos ¼ 2 2 2 2 2 2 α β γ ¼ cosα þ cos β þ cos γ ¼ 1 þ 4 sin sin sin , 2 2 2

(see the proof of the problem 5.3.6). Another proof of the problem can be obtained by the inequality of the problem πÀα πÀβ πÀγ 5.2.8c for the triangle with angles 2 , 2 , 2 . 5.4.12. Since α < α þ β π À α, then sin(α þ β) sin α, analogously sin(α þ β) sin β. < α β < π α β α β β α It is clear that 0 , 2, consequently, sin( þ ) ¼ sin cos þ sin cos sin(α þ β) cos β þ sin(α þ β) cos α. Thus, it follows that cosα þ cos β 1. α λβ > π β λα α λβ > π α λβ β λα > π 5.4.13. If þ 2 2, then þ 2 þ 2 2. Therefore, þ þ þ , thus sin α sin ðÞαþλβ sin ðÞβþλα sin β sin β 1 sin ðÞβþλα or sin ðÞαþλβ sin α. 5.4 Trigonometric Inequalities 213 ÀÁ ÀÁ α λβ π β λα α λβ α λβ β λα < β α < π If þ 2 2, then cos þ 2 cos þ 2 ,as þ 2 þ 2 þ . λα λβ 1 1 α λβ π We have that sin 2 sin 2 and sin β sin α, consequently, at þ 2 2 the following inequality holds true: ÀÁ ÀÁ λβ λβ λα cos β þ λα cos α þ 2 sin 2 2 2 sin 2 2 : sin β sin α

sin ðÞÀβþλα sin β sin ðÞÀαþλβ sin α sin ðÞβþλα sin β Hence, we deduce that sin β sin α .Thus sin ðÞαþλβ sin α. 5.4.14. We proceed to prove by mathematical induction. β α For n ¼ 1we have that 0 < α β π, consequently, cos 1 cos 1. 1 1 sin α1 sin α1 At n ¼ 2 we have that α1, α2 > 0, β1, β2 0, and α1 þ α2 β1 þ β2 π. We have to prove that

cos β cos β cos α cos α 1 þ 2 1 þ 2 : sin α1 sin α2 sin α1 sin α2

Let α and α be constant numbers. Consider the expression cos x1 þ cos x2, where 1 2 sin α1 sin α2 α1 > 0, α2 > 0, α1 þ α2 x1 þ x2 π, and x1, x2 0 and let this expression accept its maximal value at x1 ¼ β1 and x2 ¼ β2. Let α1 α2, then one can assume that β1 β2. Otherwise, we have that β β β < β β β α α cos 1 cos 2 cos 1 cos 2. Therefore, (cos 1 À cos 2)(sin 2 À sin 1) 0or sin α þ sin α β β 1 2 cos 2 þ cos 1. sin α1 sin α2 If β1 ¼ 0, then

cos β cos β cos α cos α 1 cos ðÞα þ α cos α cos α 1 þ 2 À 1 À 2 þ 1 2 À 1 À 2 ¼ sin α1 sin α2 sin α1 sin α2 sin α1 sin α2 sin α1 sin α2 α α α 1 1 α 1 α α α 2 sin sin þ 2 sin sin ðÞ1 þ 2 ¼ tg 1 À 2 2 ¼À 2 0: α α1 2 sin 2 cos sin α 2 2

β < β β β cos 1 If 0 1 2, then with the decrease of the value of 1, the expression sin α β β β 1 þ cos 2 increases. Thus, β þ β ¼ α þ α . We need to prove that sin 1 ¼ sin 2. sin α2 1 2 1 2 sin α1 sin α2 β β Indeed, note that the function fxðÞ¼cos ðÞ1þx þ cos ðÞ2Àx accepts its maximal sin α1 sin α2 value on the interval [Àβ1; β2] at the point x ¼ 0. Therefore, according to the β β Fermat’s theorem f 0ðÞ¼À0 sin 1 þ sin 2 ¼ 0. sin α1 sin α2 Lemma

α ::: α β ::: β > α ::: α β ::: β If 1, , n, 1, , n 0, 1 þ þ n ¼ 1 þ þ n sin β sin β π, and 1 ¼ ::: ¼ n ¼ λ, ð5:17Þ sin α1 sin αn then λ ¼ 1. 214 5 Application of Trigonometric Inequalities

We proceed the proof by contradiction argument. Let λ 6¼ 1, then one can assume that λ < 1. Let α1 α2 ... αn, then it follows from (5.17) that β β β β < α ::: β < α π 1 2 ... n and 1 1, , nÀ1 nÀ1 2. Note that

β β λ α β α β > λ α α α α sin ðÞ¼1 þ 2 ðÞsin 1 cos 2 þ sin 2 cos 1 ðÞsin 1 cos 2 þ sin 2 cos 1 ¼ ¼ λ sin ðÞα1 þ α2 :

Analogously, we obtain that

β β β β β β β β β > sin ðÞðÞþ1 þ 2 3 ¼sin ðÞ1 þ 2 cos 3 þ sin 3 cos ðÞ1 þ 2 > λ α α β α β β > ðÞsin ðÞ1 þ 2 cos 3 þ sin 3 cos ðÞ1 þ 2

> λðÞsin ðÞα1 þ α2 cos α3 þ sin α3 cos ðÞα1 þ α2 ¼λ sin ðÞα1 þ α2 þ α3 , β β > λ α α etc., sin( 1 þ ... þ n À 1) sin( 1 þ ... þ n À 1). β sin n Hence, sin ðÞφ À β > sin ðÞφ À αn , where φ ¼ α þ ...þ α ¼ β þ ... n sin αn 1 n 1 þ βn. Thus, sinαn sin(φ À βn) > sin βn sin(φ À αn)or sinαn sin φ cos βn À sin αn sin βn cos φ > sin βn sin φ cos αn À sin βn sin αn cos φ, sinφ sin(αn À βn) > 0. Since 0 < φ π, then φ 6¼ π and sinφ > 0, then αn À βn > 0. This means that αn > βn. We have obtained that α1 > β1,...,αn À 1 > βn À 1, αn > βn, then α1 þ ...þ αn > β1 þ ...þ βn. This leads to a contradiction. Therefore, λ ¼ 1. This ends the proof of the lemma. Since sinα1 ¼ sin β1 and sinα2 ¼ sin β2, then α1 ¼ β1 and α2 ¼ β2,orα1 ¼ β1 and π α β α α α π α α π α β À 2 ¼ 2 ( 1 þ 2 ¼ 1 þ À 2, 2 ¼ 2, this means that 2 ¼ 2). cos β cos β cos α cos α Consequently, 1 þ 2 ¼ 1 þ 2. sin α1 sin α2 sin α1 sin α2 Let now n 3 and the inequality holds true for n À 1, we need to prove that it holds true also for n. Let 0 < α1 α2 ... αn be constant numbers. Consider the expression cos x1 þ ::: þ cos xn, where x , x ,...,x 0 and α þ ...þ α x þ sin α1 sin αn 1 2 n 1 n 1 ...þ xn π and let this expression reach its greatest value at x1 ¼ β1,...,xn ¼ βn. Then, β1 β2 ... βn (see the case of n ¼ 2). One can consider β1 > 0, otherwise for n À 1 we have that

cos β cos β cos β cos ðÞα þ α cos α 2 þ 3 þ ::: þ n 1 2 þ ::: þ n : ð5:18Þ sin ðÞα1 þ α2 sin α3 sin αn sin ðÞα1 þ α2 sin αn

Note that

1 cos β cos α cos α cos β cos ðÞα þ α þ 2 À 1 À 2 2 À 1 2 : ð5:19Þ sin α1 sin α2 sin α1 sin α2 sin ðÞα1 þ α2 sin ðÞα1 þ α2 5.4 Trigonometric Inequalities 215 Indeed, since cos β 1 À 1 1 À 1 ,(α þ (α þ α ) 2 sin α2 sin ðÞα1þα2 sin α2 sin ðÞα1þα2 2 1 2

α1 þ α2 þ α3 π, sin (α1 þ α2) sin α2); thus it is sufﬁcient to prove that 1 cosα 1 cosα 1 cos α α ÀÁ À 1 À 2 À ðÞ1 þ 2 α1 α2 α1 α2 þ À 0, or tg 2 þ tg 2 À tg 2 þ 2 0, ÀÁsinα1 sin α2 sinðÞα1 þ α2 α α α α À tg 1 þ tg 2 tg 1 tg 2 2 2 2 2 0. α1 α2 1 À tg 2 tg 2 α α π < 1 2 < The last inequality holds true, since 0 2 , 2 4. By summing up the inequalities (5.18) and (5.19), we deduce the inequality at n. Thus, we have that 0 < β1 β2 ... βn. If the value of β1 decreases, then the β β cos 1 ::: cos n α α β β expression sin α þ þ sin α increases. Thus, 1 þ ...þ n ¼ 1 þ ...þ n. Then β 1 β n sin 1 ¼ ::: ¼ sin n (see the case of n ¼ 2). Consequently, according to the lemma, sin α1 sin αn α1 ¼ β1,...,αn ¼ βn. This ends the proof of the lemma.

5.4.15. Since at 0 α1, α2 π we have that sin α1 þ sin α2 ¼ α α α α α α 1þ 2 1À 2 1þ 2 ’ 2 sin 2 cos 2 2 sin 2 , then according to the Jensen s inequality, we α α ::: α α deduce that sin 1 þ sin 2 þ þ sin n n sin n. α α < π 5.4.16. As 0 1, 2 2, we have that

sin ðÞα þ α 2 sin ðÞα þ α tg α þ tg α ¼ 1 2 1 2 1 2 1 α α α α α α 2 ðÞcos ðÞþ1 À 2 cos ðÞ1 þ 2 1 þ cos ðÞ1 þ 2 α þ α ¼ 2tg 1 2 2

then according to the Jensen’s inequality, we deduce that tg α1 þ tg α2þ α α ::: α ::: α 1þ 2þ þ n þ tg n ntg n . 5.4.17. Note that, if x, y 0 and x þ y π, then sin(x þ y) ¼ | sin(x þ y)| | sin x|| cos y| þ | sin y|| cos x| sin x þ sin y, this means that

α À sin α À sin β À sin γ þ sin ðÞþα þ β sin ðÞα þ γ α À sin α À sin β À sin γ þ sin α þ sin β þ sin α þ sin γ ¼ α þ sin α ¼ α þ sin ðÞπ À α α þ π À α ¼ π:

Here, we use the inequality sinx x at x 0 (see the remark of the proof of the problem 7.1.87). The proof of the second part is done as follows:

α À sin α þ sin ðÞÀα þ β sin β þ sin ðÞÀα þ γ sin γ sin ðÞÀα þ β sin β þ sin ðÞÀα þ γ sin γ ¼ αα α ¼ 2 sin cos þ β þ cos þ γ α22 α 2 α 2 sin cos π À þ γ þ cos þ γ ¼ 0: 2 2 2 216 5 Application of Trigonometric Inequalities

5.4.18. We have that sin3φ ¼ sin(α À φ) sin(β À φ) sin(γ À φ). Therefore,

4sin 3φ þ sin 3φ ¼ sin ðÞþφ þ 2α sin ðÞφ þ 2β sin ðÞφ þ 2γ , 3 sin φ À sin ðÞφ þ 2α sin ðÞφ þ β þ γ , 2 sin φ À 2 sin α cos ðÞφ þ α sin ðÞφ þ β þ γ , β þ γ β þ γ α α À2 sin cos φ þ 2 sin cos cos ðÞφ þ α , 2 2 2 2 β þ γ 3α α À 2 cos φ þ sin þ φ À sin φ þ , 2 2 2 ð5:20Þ α α 3α α sin φ À þ sin φ þ sin þ φ À sin φ À , 2 2 2 2 α α sin φ cos sin α cos φ þ , 2 2 α α sin φ 2 sin cos φ þ , 2 2 2 sin φ sin ðÞα þ φ :

φ 1 φ < π φ π From (ÀÁ5.20) we obtain that sin 2, and since 3 , we deduce that 6. π; βþγ π If min ÀÁ6 3 ¼ 6, then the inequality holds true. π; βþγ βþγ β γ π φ < If min 6 3 ¼ 3 , then þ 2 and we obtain from (5.20) that sin2 φ α φ β γ φ φ β γ φ φ βþγ 2 sin sin ( þ ) ¼ sin ( þ À ). Thus, 2 þ À ,or 3 . 5.4.19. Note that

ðÞa cos α þ b cos β 2 ðÞa cos α þ b cos β 2 þ ðÞa sin α À b sin β 2 ¼ a2 þ b2 þ 2ab cos ðÞα þ β : ð5:21Þ

(a) We have that

x1x2 cosα1 þ x2x3 cosα2 þ x3x1 cosα3 ¼ x2ðÞx1 cosα1 þ x3 cosα2 þx3x1 cosα3 x2 þ ðÞx cosα þ x cosα 2 2 1 1 3 2 þ x x cosα 2 3 1 3 x2 þ x2 þ x2 þ 2x x cosðÞα þ α 2 1 3 1 3 1 2 þ x x cosα ¼ 2 3 1 3 πÀÁ ¼ cos x2 þ x2 þ x2 , 3 2 1 3 ÀÁ α α α π 2 2 2 consequently, x1x2 cos 1 þ x2x3 cos 2 þ x3x1 cos 3 cos 3 x2 þ x1 þ x3 , (see also the proof of the problem 5.1.22а). 5.4 Trigonometric Inequalities 217

(b, c) We need to prove that, if the inequality (5.20) holds true for n ¼ k, then it holds true also at n ¼ 2k,(k 2 N, k 2).

Indeed, let α1 þ α2 þ ...þ α2k ¼ π, using the inequalities (5.21) and the Cauchy-Bunyakovsky inequality, and also (5.20) for n ¼ k, we deduce that

x1x2 cosα1 þ x2x3 cosα2 þ ::: þ x2kÀ1x2k cosα2kÀ1 þ x2kx1 cosα2k x ðÞx cosα þ x cosα þ::: þ x ðÞx cosα þ x cosα 2 1 1 3 2 2k 2kÀ1 2kÀ1 1 2k

x2 Á jjx1 cosα1 þ x3 cosα2 þ ::: þ jjx2k Á jjx2kÀ1 cosα2kÀ1 þ x1 cosα2k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 α α ::: 2 2 α α x2 Á x1 þ x3 þ 2x1x3 cosðÞ1 þ 2 þ þ jjx2k Á x2kÀ1 þ x1 þ 2x1x2kÀ1 cosðÞ2kÀ1 þ 2k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁÀÁ 2 ::: 2 2 2 ::: 2 α α ::: α α x2 þ þ x2k 2x1 þ 2x3 þ þ 2x2kÀ1 þ 2x1x3 cosðÞþ1 þ 2 þ 2x1x2kÀ1 cosðÞ2kÀ1 þ 2k rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁÀÁπ 2 x2 þ ::: þ x2 x2 þ x2 þ ::: þ x2 1 þ cos 2 2k 1 3 2kÀ1 k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi π ÀÁÀÁ ¼ 2cos x2 þ ::: þ x2 x2 þ x2 þ ::: þ x2 2k 2 2k 1 3 2kÀ1 π ÀÁ cos x2 þ x2 þ ::: þ x2 : 2k 1 2 2k

Since the inequality (5.20) holds true for n ¼ 2 and n ¼ 3, then according to the proved statement it holds true also for n ¼ 4 and n ¼ 6.

Problems for Self-Study

Prove the inequalities. ÀÁ 2 2 α β γ 2 1 2 1 1 < α β 5.4.20. ðÞtg þ tg þ tg 1 þ cos α þ 1 þ cos β þ 1 þ cos γ ,if0 , , γ < π α β γ π 2 and þ þ ¼ . P ÀÁ α α n 5.4.21. cos i À j À2. 1i 2α β β α < α < β α β < π 5.4.22. sin sin 2 sin þ 2 sin sin 2 sin þ 2 ,if0 and þ . sin α sin β sin β sin γ sin γ sin α γ α α β γ > α β γ π 5.4.23. sin2 þ sin2 þ 2β 9, where , , 0 and þ þ . 2 2 sin 2 sin α sin β 4 sin α sin β Hint Prove that γ À 1. 2 2γ sin 2 sin nÀ1 α ::: α ðÞnÀ1 2 1 α < π 5.4.24. tg 1 þ þ tg n nÀ2 cos α Á:::Á cos α , where n 2, 0 i 2, i ¼ n 2 1 n 1, . . . , n. 218 5 Application of Trigonometric Inequalities

α ::: α 1þ þ n α α α α Hint Let n ¼ ,if 1 2, then

ðÞtg α1 þ ::: þ tg αn cos α1 Á ::: Á cos αn ¼ sin ðÞα1 þ α2 cos α3 Á ::: Á cos αn þ 1 þ ðÞcos ðÞþα þ α cos ðÞα À α ðÞtg α þ ::: þ tg α cos α Á ::: Á cos α 2 1 2 2 1 3 n 3 n 1 sin ðÞα þ ðÞα þ α À α cos α Á ::: Á cos α þ ð cos ðÞα þ ðÞα þ α À α þ 1 2 3 n 2 1 2 þ cos ðÞα À ðÞα1 þ α2 À α ÞðÞtg α3 þ ::: þ tg αn Âcos α3 Á ::: Á cos αn ¼

¼ ðÞtgα þ tg ðÞþα1 þ α2 À α tg α3 þ ::: þ tg αn

Â cos α cos ðÞα1 þ α2 À α cos α3 Á ::: Á cos αn,

π < α α α α α α α α < π since À2 1 À 2 À ðÞ1 þ 2 À 2 À 1 2. Consequently, we obtain that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi nÀ1 ÀÁ n cos 2α nÀ1 2 cosα1 Á ::: Á cosαnðÞtgα1 þ ::: þ tgαn cos α Á ntgα ¼ nnðÞÀ 1 2 sin α sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi nÀ1 nÀ1 2 2 n nÀ1 cos αþ:::þcos αþsin2α ðÞn À 1 2 2 nÀ1 nÀ1 : nnðÞÀ 1 n ¼ nÀ2 n 2

5.4.25. Let π > α1, α2,...,α6 > 0 and α1 þ α2 þ ...þ α6 ¼ 2π. Is the inequality cosα1 þ ...þ cos α6 3 correct? α α α π α α π α π Hint Take 1 ¼ 2 ¼ 3 ¼ 6, 4 ¼ 5 ¼ 4, and 6 ¼ . sin α sin β sin γ sin δ 1 α β γ δ α β γ 5.4.26. 1À sin α sin β sin γ sin δ 12 ðÞsin 2 þ sin 2 þ sin 2 þ sin 2 , where , , , δ > 0 and α þ β þ γ þ δ ¼ π. n sin 2π α ::: α n α ::: α α α > 5.4.27. sin 2 1 þ þ sin 2 n π n sin 1 Á Á sin n, where 1 Á ...Á n 0 ðÞsin n and α1 þ α2 þ ...þ αn ¼ π. 5.4.28. sin2α þ sin 2β þ sin 2γ þ sin 2δ 4k þ 1(sin α sin β sin γ sin δ)k, where k 2 N, α, β, γ, δ > 0, and α þ β þ γ þ δ ¼ π. α β γ α β β γ γ α α β 5.4.29. sin 2 þ sin 2 þ sin 2 2 sin 2 sin 2 þ 2 sin 2 sin 2 þ 2 sin 2 sin 2, where , , γ > 0 and α þ β þ γ ¼ π. À β α γ β α γ α β 5.4.30. p2ffiffi ðÞsin α þ sin β þ sin γ cos À þ cos À þ cos À p2ffiffi cos þ cos þ 3 2 2 2 3 2 2 γ α β γ > α β γ π cos 2Þ,if , , 0 and þ þ ¼ .

5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities

Prove the inequalities for the elements of triangle ABC (5.5.1–5.5.11). 5.5.1. (a) R 2r, 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 219

2 βÀγ 2r (b) cos 2 R . pffiffiffi 5.5.2. (a) a2 þ b2 þ c2 4 3S, pffiffiffi (b) 3a2 þ 3b2 À c2 4 3S, a2 bc (c) S 3 þ 6 . 5.5.3. a2 þ b2 þ c2 < 8R2, if triangle ABC is obtuse. 5.5.4. (a) a2 þ b2 þ c2 > 8R2, if triangle ABC is acute, (b) b þ c > 2R þ 2r, if triangle ABC is acute. 5.5.5. (a) a2 þ b2 þ c2 8R2 þ 4r2,

2 2 2 2 2Rr2 (b) a þ b þ c 8R þ RÀr, (c) a2 þ b2 þ c2 24Rr À 12r2, (d) b2 þ c2 þ R2 a2. 5.5.6. p2 4R2 þ 4Rr þ 3r2. 5.5.7. (a) p > 2R þ r, if triangle ABC is acute, 2 2 2 2 > 2 2 (b) laa þ lbb lcc , if triangle ABC is acute. 5.5.8. (a) p < 2R þ r, if triangle ABC is obtuse.

(b) ma þ mb þ mc min( p þ 2R,4R þ r). pffiffiffi 5.5.9. (a) a2 þ b2 þ c2 À ðÞa À b 2 À ðÞb À c 2 À ðÞa À c 2 4 3S. pffiffiffi (b) a2 þ b2 þ c2 À 1 ðÞjjþa À b jjþb À c jjc À a 2 4 3S, 2 pffiffiffiffiffiffiffiffiffi (c) aa0 þ bb0 þ cc0 4 3SS0, where S0 is the area of triangle with sides a0, b0, c0, (d) For any tetrahedron ABCD proves that pffiffiffi AC Á BD þ AB Á CD þ BC Á AD 3S, where S is the surface area of tetrahedron ABCD.

(e) Given that any edge of tetrahedron ABCD has an angle not smaller than 120 . Prove that AC Á BD þ BC Á AD þ AB Á CD > 2S, where S is the surface area of tetrahedron ABCD. (f) For a convex hexagon ABCDEF with area S prove that pffiffiffi ACðÞ BD þ BF À DF þCEðÞ BD þ DF À BF þAEðÞ BF þ DF À BD 2 3S: pffiffiffi (g) a2 þ b2 þ c2 4 3S þ 3 ðÞa À b 2 þ ðÞb À c 2 þ ðÞa À c 2 :

5.5.10. (a) a3 þ b3 þ c3 þ 3abc a2b þ b2a þ a2c þ c2a þ c2b þ b2c, 2 2 (b) p 16Rr À 5r , (c) p a p b p b p c p c p a 1 1 1 9. ðÞðÞÀ ðÞþÀ ðÞÀ ðÞþÀ ðÞÀ ðÞÀ a2 þ b2 þ c2 4 220 5 Application of Trigonometric Inequalities

2 2 (d) a2b b c2 c2a2 9R2, if triangle ABC is acute. c2 þ a2 þ b2 pffiffiffi 5.5.11. lalb þ lblc þ lcla 3 3S. 5.5.12. Point H is the orthocenter of the acute triangle ABC, and AD, BE, and CF are altitudes of that triangle. Prove that (a) HA2 þ HB2 þ HC2 4(HE2 þ HD2 þ HF2), pffiffiffi (b) AB þ BC þ AC 2 3ðÞHD þ HE þ HF , (c) (HD þ HE þ HF)2 AF2 þ BD2 þ CE2. 5.5.13. Given a tetrahedron ABCD, such that the edges AD, BD, CD are mutually perpendicular and AD ¼ a, BD ¼ b, and CD ¼ c. Prove that the sum of the distances from A, B, and C to the straight line l, passing through D and intersecting the face qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ ABC, does not exceed 2 a2 þ b2 þ c2 . For which location of line l does the equality hold true? 5.5.14. Let a circle with the center O be inscribed in triangle ABC. Given that the tangents to the circle are drawn from the intersection points of the circle with rays OA, OB, and OC. Prove that these tangents define a triangle, such that its perimeter does not exceed the perimeter of the given triangle.

5.5.15. The central angle AOB is less than 90 . Find on the smaller arc AB a point M, SMA2B2 such that the ratio has the minimal value. Here, MA1 ⊥ OA, A1 2 OA, SAA1A2 þSBB1B2 MB1 ⊥ OB, B1 2 OB, A2 is the intersection point of segments MA1 and AB, B2 is the intersection point of segments MB1 and AB. 5.5.16. Given a point M inside of triangle ABC. Prove that SABC > ðÞABþBCþAC 2 SAMC ,if∠B max(∠A, ∠C). ðÞAMþMCþAC 2 5.5.17. Prove that if the sides of the inscribed hexagon ABCDEF are, such that AB ¼ BC, CD ¼ DE, EF ¼ FA, then the area of triangle ACE does not exceed the area of triangle BDF.

5.5.18. The incircle of triangle ABC touches the sides of the triangle at points A1, B1, 2 2 2 p2 r and C1. Prove that A1B1 þ B1C1 þ A1C1 3 and 2pR A1B1 þ B1C1 þ A1C1, where p is the semiperimeter, and R and r are the circumradius and inradius of triangle ABC, respectively. 5.5.19. Let ABC be an acute triangle with the circumcenter O and the circumradius R. Let AO intersect the circumcircle of triangle OBC at point D, BO intersect the circumcircle of triangle OCA at point E, and CO intersect the circumcircle of triangle OAB at point F. Prove that (a) OD Á OE Á OF 8R3, AD BE CF (b) OD þ OE þ OF 4, 5. 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 221

5.5.20. Let M be a point inside a convex n-gon A1A2 ...An. Denote the distances MA1, MA2,...,MAn of the point from the vertices of the polygon by R1, R2,..., Rn, and the distances MP1, MP2,...,MPn from the straight lines A1A2, A2A3,..., ::: nπ ::: AnA1 by d1, d2,...,dn. Prove that R1 Á R2 Á Á Rncos n d1 Á d2 Á Á dn.

5.5.21. Let the rectangle P1 be inscribed in the rectangle P2 with sides c and α c d (c d), where d is smaller than the greatest side of P1. Prove that sin 2 d, where α is the angle between two straight lines containing any two of the sides of the rectangles P1 and P2. 5.5.22. Prove that of all quadrilaterals with the given side lengths a, b, c, d the greatest area has the one that can be inscribed in a circle. 5.5.23. Prove the inequality for the acute triangle it holds true p2 2R2 þ 8Rr þ 3r2. 5.5.24. Let a, b, c be the sides of triangle ABC, and S be its area. Prove that (a) x a4 þ y b4 þ z c4 8S2, yþz xþz xþy pffiffiffi x 2 y 2 z 2 > (b) yþz a þ xþz b þ xþy c 2 3S,ifx þ y, x þ z, y þ z 0.

5.5.25. Given points A1, A2,...,An on a unit circle, dividing the circle into n arcs. Let P be the midpoint of the greatest of these n arcs. Prove that

(a) PA1 Á ...Á PAn 2, for n ¼ 2, 3, 4, (b) There exist points A1, A2,...,An, such that PA1 Á PA2 Á ...Á PAn < 2, for n 5.

5.5.26. (a) Given points A1, A2,...,An on a unit circle. Prove that on this circle exists a point P, such that PA1 Á PA2 ...Á PAn 2.

(b) Given points A1, A2,...,An on a plane and a circle with a center O. Prove that on that circle exists a point P, such that PA1 Á PA2 Á ::: Á PAn 1 ::: . 2nÀ1 ðÞOP þ OA1 ðÞÁOP þ OA2 Á ðÞOP þ OAn 5.5.27. Let point P be inside of the convex quadrilateral ABCD. Prove that at least ∠ ∠ ∠ ∠ π one of angles PAB, PBC, PCD, PDA is not greater than 4. 5.5.28. (a) Let P be the intersection point of the cevians CD, BE of the triangle ABC (D 2 AB, E 2 AC) and BD CD. Prove that AD þ DP > AE þ EP. (b) Let P be the intersection point of the cevians CD, BE of the isosceles triangle ABC (AB ¼ AC) and AD < AE. Prove that AD þ DP > AE þ EP. 5.5.29. Let O be the intersection point of the diagonals AC and BD of the convex quadrilateral ABCD. Let circles S1, S2, S3, S4 with centers O1, O2, O3, O4 be the incircles of triangles AOB, BOC, COD, DOA, respectively. Prove that (a) The sum of the diameters of the circles S , S , S , S is less than or equal to ÀÁpffiffiffi 1 2 3 4 2 À 2 ðÞAC þ BD ,

(b) O1O2 þ O2O3 þ O3O4 þ O4O1 < AC þ BD. 5.5.30. Let D be a point inside of angle ACB, such that ∠DAC ¼ ∠ DCB and ∠DBC ¼ ∠ DCA. Prove that CD R, where R is the circumradius of triangle ABC. 222 5 Application of Trigonometric Inequalities

5.5.31. Let M be a point inside of the convex n-gon A1A2 ...An. Denoted by MAi ¼ Ri, ρ(M, AiAi þ 1) ¼ di, i ¼ 1, 2, ...n, where An þ 1 A1. 1 Prove that, R1 þ R2 þ ...þ Rn π ... . cos nðÞd1þd2þ þdn 5.5.32. Prove that

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 ðÞR1 þ R3 þ ::: þ R2kÀ1 ðÞR2 þ R4 þ ::: þ R2k π ðÞd1 þ d2 þ ::: þ d2k , cos 2k

(see the notations of the problem 5.5.31).

5.5.33. Prove the following inequalities for a convex 2n-gon A1A2 ...A2n. π ÀÁ (a) 4sin2 A A2 þ A A2 þ ::: þ A A2 ðÞA A þ A A 2þ 2n 1 nþ1 2 nþ2 n 2n 1 2 nþ1 nþ2 þðÞA A þ A A 2 þ ::: þ ðÞA A þ A A 2, 2π 3ÀÁnþ2 nþ3 n nþ1 2n 1 (b) 4tg 2 B B2 þ B B2 þ ::: þ B B2 ðÞA A þ A A 2þ 2n 1 nþ1 2 nþ2 n 2n 1 2 nþ1 nþ2 2 2 þðÞA2A3 þ Anþ2Anþ3 þ ::: þ ðÞAnAnþ1 þ A2nA1 , where points B1B2 ...B2n are the midpoints of segments A1A2, A2A3,..., A2nA1 respectively. 5.5.34. Given two circles with radiuses r and R. Let the distance between the centers of these circles be equal to d,(d > 0). Given that any of those circles touches three sides of a convex quadrilateral with perimeter p (Figure 5.14). 5.5.35. Given two circles on a plane, such that any of those circles touches three sides of a convex quadrilateral ABCD (see Figure 5.14). For which quadrilateral S ABCD expression p2 is the greatest possible, where S is the area of quadrilateral ABCD and p is its perimeter. 5.5.36. Let two circles have the radiuses r and R, and any of those two circles touches three sides of a convex quadrilateral ABCD (see Figure 5.14). Prove that S Rpþffiffiffiffir p2, where S is the area of quadrilateral ABCD and p is its perimeter. 32 Rr 5.5.37. Given two circles with radiuses r and R not having any common interior points. Given also that any of those circles touches three sides of a convex quadrilateral with perimeter p. Prove that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 9ðÞþR þ r 81ðÞR þ r 2 À 144ðÞR þ r Rr p : 2 π : 5.5.38. Given that all angles of triangle ABC are greater 4 Let C1, A1, B1 be points on sides AB, BC, AC of triangle ABC, respectively. Consider the circumcircle of triangle ABC and for any points M, N, belonging to the inner part of this circle, denote the length of the chord passing through points M, N by X(M, N). Prove that

(a) A1B1 þ B1C1 þ A1C1 > X(A1, B1), (b) max(AB1 þ B1C1 þ AC1, A1B þ BC1 þ A1C1, A1B1 þ B1C þ A1C) > X(A1, B1). 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 223

5.5.39. Let r be the inradius of triangle ABC and let M, M0 be given points inside of triangle ABC, such that ∠MAB ¼ ∠ M0AC and ∠MBA ¼ ∠ M0BC. 0 0 0 0 Denote the distances da, db, dc and da, db, dc from pointsM and M to lines BC, 0 0 0 : AC, AB, respectively. Find the greatest possible value of the product dadbdcdadbdc

5.5.40. Inside of triangle ABC is given a point M. Let da, db, dc be the distances from point M to lines BC, CA, AB and Ra, Rb, Rc be the distances from point M to vertices A, B, C, respectively. Prove that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; ; 2 2 2 dadbdc : maxðÞRa Rb Rc da þ db þ dc þ maxðÞda; db; dc

5.5.41. Let the area of a convex pentagonABCDE be equal to S and the circumradiuses of triangles ABC, BCD, CDE, DEA, EAB be equal to R1, R2, R3, R4, 4 4 4 4 4 4 2 R . Prove that R þ R þ R þ R þ R S : 5 1 2 3 4 5 5sin2108

Solutions

Rþr α 5.5.1. (a) Using the problems 5.3.6 and 5.1.4a, we deduce that R ¼ cos þ β γ 3 cos þ cos 2. Therefore, R 2r. β γ (b) According to the problem 5.3.4, we need to prove that cos 2 À ÀÁ 2 α β γ βÀγ α 2 8 sin 2 sin 2 sin 2, or cos 2 À 2 sin 2 0. pffiffiffi ÀÁÀÁ 5.5.2. (a) Note thata2 þ b2 þ c2 À 4 3S ¼ 2ðÞa À b 2 þ 4ab 1 À cos γ À π 0: pffiffiffi 3 Thus, it follows that a2 þ b2 þ c2 4 3S.. pffiffiffi ÀÁÀÁ 2 2 2 2 γ π : (b) Note that 3a þ 3b À c À 4 3S ¼ 2ðÞa À b þ 4ab 1 À sin À 6 0 pffiffiffi Therefore, 3a2 þ 3b2 À c2 4 3S.

2 ÀÁÀÁ a2 bc ðÞbÀc 5bc α 3 : (c) Note that 3 þ 6 À S ¼ 3 þ 6 1 À sin þ arccos5 0 a2 bc Therefore, 3 þ 6 S. 5.5.3. According to the problem 5.2.9, it follows that tgα tgβ tgγ 4S þ þ ¼ a2þb2þc2À8R2 < 0 (see the problem 5.3.12). Thus, we obtain that a2 þ b2 þ c2 < 8R2. 5.5.4. (a) According to the problem 5.3.12, it follows that 4S a2þb2þc2À8R2 ¼ tgα þ tgβ þ tgγ > 0. Consequently, a2 þ b2 þ c2 > 8R2. (b) Note that b þ c À 2R À 2r ¼ a þ 2ðÞÀp À a 2RÀ 2r ¼ α α α 1 r α α α ¼ 2Rsinα þ 2rctg À 2R À 2r ¼ 2R cos À sin α À sin cos À sin ¼ 2 2 2 sin R 2 2 2 2 pffiffiffi α π β À γ pffiffiffi α π β À γ pffiffiffi ¼ 2 2Rcos þ 2cos À 2sin þ > 0, as 2cos > 2: 2 4 2 2 4 2 224 5 Application of Trigonometric Inequalities

Remark If ABC is a right angle triangle, then a2 þ b2 þ c2 ¼ 2c2 ¼ 8R2 5.5.5. If triangle ABC is non-acute, then a2 þ b2 þ c2 8R2 (see the problem 5.5.3). 2 2 2 < 2 2 2 2 2 < 2 2Rr2 Thus, a þ b þ c 8R þ 4r and a þ b þ c 8R þ RÀr. If ABC is an acute triangle, then: (a) According to the problem 5.2.2a, we have that 4pr ¼ tgα þ tgβ þ tgγ a2þb2þc2À8R2 α β γ p 2 2 ctg 2 þ ctg 2 þ ctg 2 ¼ r (see the problems 5.3.12 and 5.3.2). Therefore, a þ b þ c2 8R2 þ 4r2. (b) According to the problems 5.3.12 and 5.3.4, we need to prove that 4S 2r2 tgαþtgβþtgγ α β γ. 1À4 sin 2 sin 2 sin 2 As S ¼ pr ¼ Rr(sinα þ sin β þ sin γ) and tgα þ tgβ þ tgγ ¼ tgαtgβtgγ (see the proof of the problem 5.3.12), then it remains to prove that ÀÁ β γ 2 sin α sin sin cos α cos β cos γ 2 2 2 see the proof of the problem 5:3:3 : 1 α β γ ðÞ 4 À sin 2 sin 2 sin 2

We have to prove that α β γ 2 α β γ 1 sin sin sin þ cos α cos β cos γ sin sin sin À cos α cos β cos γ 0, 2 2 2 2 2 2 4 or ÀÁ α β γ 2 γ α À β γ α À β γ γ cos À À sin sin2 þ 2 cos 2 À cos 2 cos À sin sin cosγÀ 2 2 2 2 2 2 2 2 αÀ β γ À cos 2 À cos 2 cosγ 0: 2 2 ð5:22Þ α β γ αÀβ Let and x ¼ cos 2 , we need to prove that for 0 x 1, we have the inequality:ÀÁ ÀÁÀÁ ÀÁ γ 2 2γ 2 2γ γ γ γ 2 2γ fxðÞ¼ x À sin 2 sin 2 þ 2 x À cos 2 x À sin 2 sin 2 cos À x À cos 2 cos γ 0. ÀÁ 0 γ γ 2 2γ γ γ 2γ Indeed, we have that f ðÞ¼x 6 cos sin 2 x þ 2 sin 2 À cos À 2 cos sin 2 3γ 2γ γ γ 0 0 x À 2sin 2 À 2cos 2 sin 2 cos . Note that, if f (x1) ¼ f (x2) ¼ 0. Then, according to the Vieta’s theorem x1x2 < 0. We need to prove that f0(1) < 0. Indeed, we have that γ γ γ γ γ γ f 0ðÞ¼1 6cosγ sin þ 2 sin2 À cosγ À 2cosγsin2 À 2sin 3 À 2cos 2 sin cosγ ¼ 2 2 2 2 2 2 γ γ γ γ γ γ 1 ¼ 21À sin 2sin 3 sin À 1 þ 2sin2 þ sin À 1 < 0, since 0 < sin , ÀÁ2 2 2 2 2 2 2 3γ γ < 2γ γ 2sin 2 sin 2 À 1 0 and 2sin 2 þ sin 2 À 1 0. 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 225

0 < We have obtainedÀÁ that, for x 2 [0; 1], f (x)ÀÁ0. Therefore, for 0 x 1 4γ γ 2 αÀβ fxðÞf ðÞ¼1 sin 2 2 sin 2 À 1 0. Hence, f cos 2 0, this means that (5.22) holds true. (c) We have that a2 þ b2 þ c2 ¼ 2p2 À 2r2 À 8Rr (see the proof of problem 5.3.7), then from the problem 5.5.10b we obtain that a2 þ b2 þ c2 24Rr À 12r2. (d) Note that b2 þ c2 þ R2 À a2 ¼ R2((2 cos α À cos(β À γ))2 þ sin2(β À γ)) 0. Thus, it follows that b2 þ c2 þ R2 a2. 5.5.6. Since a2 þ b2 þ c2 ¼ 2p2 À 2r2 À 8Rr (see the proof of the problem 5.3.7), then according to the problem 5.5.5, we have that 2p2 À 2r2 À 8Rr 8R2 þ 4r2. Hence, p2 4R2 þ 4Rr þ 3r2. 5.5.7. (a) According to the problem 5.3.12, it follows that 2pr ¼ tgα þ tgβþ p2ÀðÞ2Rþr 2 tgγ > 0. Consequently, p > 2R þ r. α; β; γ γ < π γ γ φ Second Solution Let maxðÞ¼2, we have that a sin 2 þ b sin 2 ¼ c cos (see Figure 5.4). ÀÁ γ γ γ γ Note that φ ¼ β À 90 À < . Therefore, ðÞa þ b sin > c Á cos , thus γ γ 2 2 2 2 cðÞsin 2þ cos 2 p > γ . 2 sin 2 c γ γ 1 γ p À 2R À r ¼ p À À ðÞp À c tg ¼ p 1 À tg À c À tg > sin γ 2 2 sin γ 2 γ γ γ γ γ c sin þ cos cos À sin c 1 À 2sin2 > 2 2 2 2 2 : γ γ À γ ¼ 0 2 sin cos 2 sin 2 2 ÀÁ γ (b) We have that l2a2 þ l2b2 h2a2 þ h2b2 ¼ 2h2c2 > l2c2, since hc ¼ sin þ β pﬃﬃ a b a b c c lc 2 > 2 > γ β > γþβ > 2 (135 2 þ 2 45 ). 5.5.8. (a) According to the problems 5.3.12 and 5.2.9, it follows that 2pr ¼ p2ÀðÞ2Rþr 2 tgα þ tgβ þ tgγ < 0, consequently, p < 2R þ r. γ > γ > 0 < Second Solution Let max(a, b, c) ¼ c, 90 , then 2 45 . Thus, p À c r, consequently, p < c þ r < 2R þ r.

Figure 5.4 j

g/2 r

g/2 b 226 5 Application of Trigonometric Inequalities

Figure 5.5 C

R AB

C1 RR O

(b) If triangle ABC is not obtuse, then according to the Carnot’s theorem ka þ kb þ kc ¼ R þ r (see the proof of problem 2.4.11). Then, it follows that ma R þ ka, mb R þ kb, mc R þ kc.

Therefore, ma þ mb þ mc 4R þ r min( p þ 2R,4R þ r). If triangle ABC is obtuse (see Figure 5.5). c < < b c According to the problem 1.1.8a, we have that mc þ 2 2R, ma 2 þ 2, mb < a c 2 þ 2 (see problem 1.1.7a). Therefore, ma þ mb þ mc < p þ 2R < 4R þ r (see problem 5.5.8a). 5.5.9. (a) We have that ÀÁ a2 þ b2 þ c2 À ðÞa À b 2 À ðÞb À c 2 À ðÞa À c 2 ¼ 2ðÞÀab þ bc þ ac a2 þ b2 þ c2 ¼ ¼ 2ðÞp2 þ r2 þ 4Rr ÀðÞ2p2 À 2r2 À 8Rr ¼4r2 þ 16Rr,

(see the proof of the problem 5.3.7). pffiffiffi According to the problems 5.1.9 and 5.3.14, we have that 4Rþr 3. Hence, ppffiffiffi pffiffiffi a2 þ b2 þ c2 À ðÞa À b 2 À ðÞb À c 2 À ðÞa À c 2 ¼ 4rðÞ4R þ r 4 3pr ¼ 4 3S. α β γ γ Second Solution Let max( , , ) ¼ . pffiffi γ 2 2 2 γ 2 2 ab γ 3 If 120 , then c ¼ a þ b À 2ab cos a þ b þ ab, S ¼ 2 sin 4 ab. Therefore,

2ab þ 2bc þ 2ac À a2 À b2 À c2 ¼ 2ab þ 2caðÞÀþ b a2 À b2 À c2 > pffiffiffi > 2ab þ 2c2 À a2 À b2 À c2 3ab 4 3S, pffiffiffi thus 2ab þ 2bc þ 2ac À a2 À b2 À c2 > 4 3S. If γ < 120 , then inside of triangle ABC exists a point M, such that ∠AMB ¼ ∠ ∠ BMC ¼ CMA ¼ 120 . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 LetpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiMA ¼ x, MB ¼ y, pMCffiffi ¼ z, then a ¼ y þ yz þ z , b ¼ x þ xz þ z , 2 2 3 c ¼ x þ y þ xy S ¼ 4 ðÞxy þ yz þ zx . Consequently, 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 227 pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ab þ 2bc þ 2ac À a2 À b2 À c2 À 4 3S ¼ 2 x2 þ xy þ y2 y2 þ yz þ z2þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ x2 þ xy þ y2 x2 þ xz þ z2 þ y2 þ yz þ z2 x2 þ xz þ z2 À ðÞx þ y þ z 2

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁ pffiffi 2 ÀÁ pffiffi 2 x 2 3 z 2 3 2 y þ 2 þ 2 x y þ 2 þ 2 z þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁ pffiffi 2 ÀÁ pffiffi 2 y 2 3y z 2 3 þ x þ 2 þ 2 x þ 2 þ 2 z þ ! rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁ pffiffi 2 ÀÁ pffiffi 2 y 2 3y x 2 3 2 þ z þ 2 þ 2 z þ 2 þ 2 x À ðÞx þ y þ z x z 3xz y z 3yz y x 2 y þ y þ þ þ x þ x þ þ þ z þ z þ þ 2 2 4 2 2 4 2 2 3xy þ À ðÞx þ y þ z 2 ¼ 0, 4 pffiffiffi thus 2ab þ 2bc þ 2ac À a2 À b2 À c2 4 3S. Remark The inequality pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 xy y2 y2 yz z2 y2 yz z2 z2 zx x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ þ þ þ þ þ þ þ þ z2 þ zx þ x2 x2 þ xy þ y2 ðÞx þ y þ z 2 holds true for arbitrary values of x, y, z.

2 Third Solution Weffiffiffi need to prove that 2ab þ 2bc þ 2ac 2ðÞOA þ OB þ OC 2 p a2 þ b þ c2 þ ÀÁ4 3S, whereÀÁO is the incenter of triangle ABC. 1 1 1 2 2 2 2 We havethat a þ b þ c a Á OA þ b Á OB þ c Á OC ðÞOA þ OB þ OC and a Á OA2 þ b Á OB2 þ c Á OC2 ¼ abc. Consequently, ab þ bc þ ac (OA þ OB þ OC)2, (see the proof of the problem 4.1.8b). We need to prove that for any point M in plane pABCffiffiffi the following inequality holds true: 2ðÞMA þ MB þ MC 2 a2 þ b2 þ c2 þ 4 3S. Let ∠BMC ¼ α1, ∠AMC ¼ β1, ∠AMB ¼ γ1, then we have that pffiffiffi pffiffiffi pffiffiffi γ 4 3ðÞSAMB þ SBMC þ SCMA ¼2 3 Á MA Á MB sin 1 þ 2 3 Á MB pffiffiffi α γ : Á MC sin 1 þ 2 3 Á MA Á MC sin 1 ð5 23Þ

Note that sin(x À 30 ) 1, consequently pffiffiffi 2 3 sin x 2 cos x þ 4 ð5:24Þ 228 5 Application of Trigonometric Inequalities

According to (5.23) and (5.24), we deduce that pffiffiffi γ α 4 3S 2 Á MA Á MB cos 1 þ 4 Á MA Á MB þ 2 Á MB Á MC cos 1þ β 2 þ4 Á MB Á MC þ 2 Á MA Á MC cos 1 þ 4 Á MA Á MC ¼ 2ðÞMA þ MB þ MC Àa2 À b2 À c2, hence pffiffiffi 2ðÞMA þ MB þ MC 2 a2 þ b2 þ c2 þ 4 3S: ð5:25Þ

Remark The equality in (5.25) holds true, only if ∠BMB ¼ ∠ CMA ¼ ∠ BMC ¼ 120 . Hence, if point M0 is such that ∠BM0A ¼ ∠ CM0A ¼ ∠ BM0C ¼ 120 and M=6 M0, then MA þ MB þ MC > M0A þ M0B þ M0C. (b) Let a b c and a ¼ x þ y, b ¼ y þ z, c ¼ x þ z, then y x z 0. One needs to 2 2 2 1 2 provepffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi that ðÞx þ y þ ðÞy þ z þ ðÞx þ z À 2 ððÞþpx Àffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz ðÞþy À z ðÞÞy À x 4 3xyzðÞ x þ y þ z ,orxxðÞþþ y þ z 3xyz 2 xxðÞþ y þ z 3yz. pffiffiffiffiffi Obviously the last inequality holds true, as u þ v 2 uv. (c) We have that ÀÁ 0 0 0 0 aa0 þ bb þ cc0 2 sin α sin α þ sin β sin β þ sin γ sin γ pffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi SS0 sin α sin β sin γ sin α0 sin β0 sin γ0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 sin α sin β sin γ sin α0 sin β0 sin γ0 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin α sin β sin γ sin α0 sin β0 sin γ0 6 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 sin α sin β sin γ sin α0 sin β0 sin γ0 6 pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffi¼ 4 3, pffiffi 2 6 3 3 8 ffiffiffiffiffiffiffiffiffi 0 p (see problem 5.1.12). Thus, it follows that aa0 þ bb þ cc0 4 3SS0 . Remark It also holds true for the following inequality pffiffiffiffi ffiffiffiffi ffiffiffiffi ffiffiffiffi ffiffiffiffi pffiffiffiffi 0 p 2 0 p p 2 0 p 2 aaÀ b0 À c0 þ bbÀ a0 À c0 þ ccÀ a0 À b0 pffiffiffiffiffiffiffiffiffi 4 3SS0 : 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 229

(d) Let BC ¼ a, AB ¼ c, AC ¼ b, AD ¼ x, BD ¼ y, CD ¼ z. Then, according to problem 5.5.9а, we have that

2ab þ 2bc þ 2ac À a2 À b2 À c2 þ 2bx þ 2bz þ 2xz À b2 À x2 À z2 þ 2ayþ pffiffiffi þ2az þ 2yz À y2 À z2 À a2 þ 2cx þ 2cy þ 2xy À c2 À x2 À y2 4 3S:

In order to prove this inequality, it is sufﬁcient to prove that

2ax þ 2by þ 2cz ab þ bc þ ac þ bx þ bz þ xz þ ay þ az þ yz þ cx þ cy þ xyÀ Àa2 À b2 À c2 À x2 À y2 À z2:

This is obvious, as it is equivalent to the following inequality

ðÞa þ x 2 þ ðÞb þ y 2 þ ðÞc þ z 2 ðÞa þ x ðÞþb þ y ðÞb þ y ðÞc þ z þ ðÞc þ z ðÞa þ x :

(e) Let max(AB, BC, AC, AD, BD, CD) ¼ AB, then ∠ADB 120 , ∠ ACB 120 .

Note that ∠DAC < 120 . Indeed, if ∠DAC 120 , then we deduce that ∠DAB þ ∠ BAC > ∠ DAC 120 , but ∠DAB < 60 , ∠ BAC < 60 . This leads to a contradiction. Without loss of generality one can assume that ∠ADC 120 . Let AB ¼ c, BC ¼ a, AC ¼ b, AD ¼ x, BD ¼ y, CD ¼ z. We have that c2 a2 þ b2 þ ab and b2 x2 þ z2 þ xz, therefore pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi AC Á BD þ BC Á AD þ AB Á CD ¼ ax þ by þ cz ax þ y x2 þ z2 þ xz þ z a2 þ b2 þ ab pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 3 3 3 3 3 3 3 ax þ ðÞx þ z y þ ðÞa þ b z > ax þ xy þ zy þ az þ bz > 2 2 2 2 2 2 2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 3 3 3 3 3 3 3 3 > axðÞþþ z xy þ zy þ bz > ab þ xy þ zy þ bz 2 2 2 2 2 2 2 2 2ðÞþABC 2ðÞþABD 2ðÞþBDC 2ðÞ¼ADC 2S:

Hence, we obtain that AC Á BD þ BC Á AD þ AB Á CD > 2S. (f) Let max(∠EAC, ∠ACE, ∠AEC) ¼ ∠ EAC. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi ∠ 2 2 3 : If EAC 120 ,then CE AC þ AE þ AC Á AE 2 ðÞAC þ AE Thus, it follows that

ACðÞ BD þ BF À DF þCEðÞ BD þ DF À BF þAEðÞ BF þ DF À BD > pffiffiffi pffiffiffi 3 3 > ACðÞ BD þ BF À DF þðÞAC þ AE ðÞBD þ DF À BF þ 2 2 pffiffiffi 3 pffiffiffi pffiffiffi pffiffiffi pffiffiffi þ AEðÞ BF þ DF À BD ¼3AC Á BD þ 3AE Á DF 2 3S þ 2 3S 2 ABCD ADEF pffiffiffi ¼ 2 3S: 230 5 Application of Trigonometric Inequalities

Hence, we obtainpffiffiffi that ACðÞ BD þ BF À DF þCEðÞ BD þ DF À BF þ AEðÞ BF þ DF À BD > 2 3S: If ∠EAC < 120 , inside of triangle ACE exists a point T, such that ∠ATE ¼ ∠ ATC ¼ ∠ CTE ¼ 120 . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiLet AT ¼ x, CT ¼ yp, ETffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi¼ z, BF ¼ m þ n, BDpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi¼ n þ k, DF ¼ m þ k, then AC ¼ 2 2 2 2 2 2 x þ y þ xy, CE ¼ y þ z þ yz, AE ¼ x þ z þ xz and S ¼ SABTF þ SBCDT þ SDEFT. Therefore pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 3S 3xðm þ nÞþ 3yðn þ kÞþ 3zðm þ kÞ¼ 3mðx þ zÞþ 3nðx þ yÞ pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 3kðy þ zÞ2m x2 þ z2 þ xz þ 2n x2 þ y2 þ xy þ 2k y2 þ z2 þ yz ¼ ¼ ACðBD þ BF À DFÞþCEðBD þ DF À BFÞþAEðBF þ DF À BDÞ:

Thus, it follows that pffiffiffi ACðÞ BD þ BF À DF þCEðÞ BD þ DF À BF þAEðÞ BF þ DF À BD 2 3S:

See also the proof of problem 7.1.115b. (g) Let a ¼ m þ n, b ¼ n þ k, c ¼ m þ k, then m > 0, n > 0, k > 0. One needs to prove that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2mn þ 2nk þ 2mk À m2 À n2 À k2 3mnkðÞ m þ n þ k : ð5:26Þ

At first, we need the following lemma. Lemma If m > 0, n > 0, k > 0, 2mn þ 2nk þ 2mk À m2 À n2 À k2 > 0, then for any numbers x, y, z it holds true (2mn þ 2nk þ 2mk À m2 À n2 À k2)(myz þ nxz þ kxy) mnk(x þ y þ z)2. In order to prove inequality (5.26), consider the case 2mn þ 2nk þ 2mk À m2 À n2 À k2 0. Hence (5.26) holds true. In the second case 2mn þ 2nk þ 2mk À m2 À n2 À k2 > 0, according to the lemma 2 2 2 9mnk : for x ¼ y ¼ z ¼ 1, we havepffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi that 2mn þ 2nk þ 2mk À m À n À k mþnþk 9mnk 2 2 Notep thatffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffimþnþk 3mnkðÞ m þ n þ k , therefore 2mn þ 2nk þ 2mk À m À n Àk2 3mnkðÞ m þ n þ k : Now, let us prove the lemma. Let max(m, n, k) ¼ m and A ¼ 2mn þ 2nk þ 2mk À m2 À n2 À k2. Thus, it follows that

A À 4mk ¼ 2nk þ 2mn À m2 À n2 À k2 À 2mk ¼ÀðÞm þ k À n 2 < 0:

One needs to prove that

mnkz2 þ ðÞ2mnkðÞÀ x þ y AmyðÞþ nx z þ mnkðÞ x þ y 2 À Akxy 0, 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 231

in order to prove this it is enough to prove the following inequality: D ¼ (2mnk(x þ y) À A(my þ nx))2 À 4mnk(mnk(x þ y)2 À Akxy) 0or

n2ðÞA À 4mk x2 þ 2ðÞAmn þ 2mnkðÞ k À m À n xy þ m2ðÞA À 4nk y2 0: ð5:27Þ

Note that A À 4mk < 0 and D ¼ 4ðÞAmn þ 2mnkðÞ k À m À n 2 À 4ðÞA À 4mk ðÞA À 4nk m2n2 y2 ¼ ¼ 16m2n2k2y2 A þ ðÞk À m À n 2 À 4mn ¼ 0:

Therefore, the inequality (5.27) holds true. See also the proof of problem 8.2.14. 5.5.10. (a) Note that

a3 þ b3 þ c3 þ 3abc À a2b À b2a À b2c À c2b À a2c À c2a ¼ 8S2 ¼ abc À ðÞa þ b À c ðÞa À b þ c ðÞ¼b þ c À a 4RS À p ¼ 4SRðÞÀ 2r 0,

(see the problem 5.5.1a). (b) Let a ¼ x þ y, b ¼ y þ z, c ¼ z þ x, then x ¼ p À b > 0, y ¼ p À c > 0, z ¼ p À a > 0 . Therefore,

1ÀÁ p2 À ðÞ¼16Rr À 5r2 p3 À 16Rpr þ 5r2p ¼ p 1 ¼ ðÞx þ y þ z 3 À 4ðÞx þ y ðÞy þ z ðÞþx þ z 5xyz ¼ p 1ÀÁ ¼ x3 þ y3 þ z3 þ 3xyz À x2y À y2x À x2z À z2x À y2z À z2y ¼ p 1 ¼ ðÞxyz À ðÞx þ y À z ðÞx À y þ z ðÞy þ z À x , p

(see the proof of the problem 5.5.10a). If x, y, z are the lengths of the sides of a certain triangle, then according to 5.5.10a, p2 16Rr À 5r2. If x, y, z are not the lengths of the sides of a triangle, then xyz > 0 (x þ y À z) (x À y þ z)(y þ z À x). Hence, p2 > 16Rr À 5r2. (c) We have that abc ¼ 4pRr and ab þ bc þ ac ¼ p2 þ r2 þ 4Rr (see the proof of problem 5.3.7), thus 232 5 Application of Trigonometric Inequalities 1 1 1 ðÞðÞp À a ðÞþp À b ðÞp À b ðÞþp À c ðÞp À c ðÞp À a þ þ ¼ a2 b2 c2 ðÞab þ bc þ ac À p2 ðÞab þ bc þ ac 2 À 4abcp ¼ a2b2c2 2 ðÞr þ 4R ðÞp2 þ r2 þ 4Rr À 16p2R2r ¼ : 16p2R2r

We have to prove that (4R þ r)(( p2 þ r2 þ 4rR)2 À 16p2R2r) 36p2R2r or ÀÁ ðÞ4R þ r p4 À 2r 34R2 À r2 p2 þ r2ðÞ4R þ r 3 0: ð5:28Þ

Let f(x) ¼ (4R þ r)x2 À 2r(34R2 À r2)x þ r2(4R þ r)3. rðÞ34R2Àr2 вер < According to the problems 5.5.1a and 5.5.10b, we have that x ¼ 4Rþr rÁ34R2 < 2 2 2 2 3 2 4R ¼ 8, 5Rr 16Rr À 5r p . Therefore, f( p ) f(16Rr À 5r ) ¼ 4r (R À 2r) 0, thus (5.28) holds true. This proof was provided by D. Harutyunyan, ninth grade. (d) Note that 1 a2b2 b2c2 c2a2 sin2αsin2β sin2βsin2γ sin2γsin2α þ þ ¼ 4 þ þ R2 c2 a2 2 sin2γ sin2α sin2β b ! 1 1 1 ¼ 4 þ þ ¼ ðÞctgβ þ ctgα 2 ðÞctgγ þ ctgβ 2 ðÞctgα þ ctgγ 2 ¼ 4ðÞctgαctgβ þ ctgβctgγ þ ctgγctgα Â ! 1 1 1 Â þ þ 9, ðÞctgβ þ ctgα 2 ðÞctgγ þ ctgβ 2 ðÞctgα þ ctgγ 2 according to problem 5.5.10c.

lb a la b 5.5.11. Using the law of sines, we obtain that sin γ ¼ β α and sin γ ¼ α β . sin ðÞ2þ sin ðÞ2þ lalb 2 sin γ 2 sin γ Therefore, S ¼ β α α β ¼ γÀα γÀβ. sin ðÞ2þ sin ðÞ2þ cos 2 cos 2 We have to prove that

α À β α À γ β À γ 2 sin γ cos þ 2 sin β cos þ 2 sin α cos 2 2 2 pffiffiffi α À β β À γ γ À α 3 3 cos cos cos : 2 2 2

Note that 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 233

α À β α À γ β À γ 2 sin γ cos þ 2 sin β cos þ 2 sin α cos ¼ 2 2 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi α À β γ À α β À γ 4 sin2γcos 2 þ sin2βcos 2 þ sin2αcos 2 þ 2 2 2 α À β α À γ γ À β γ À α þ8 sin γ sin β cos cos þ 8 sin β sin α cos cos þ 2 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 2 β À γ β À α ðÞ2R þ r p2 þ 4Rr2 þ r3 þ8 sin α sin γ cos cos þ , 2 2 2R3 α À β β À γ γ À α þ8 cos cos cos ðÞsin α sin β þ sin β sin γ þ sin γ sin α ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 2 2 ðÞ2R þ r p2 þ 4Rr2 þ r3 p2 þ 2Rr þ r2 p2 þ 4Rr þ r2 þ 8 ¼ 2R3 8R2 4R2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ p4 þ 4R2 þ 8Rr þ 2r2 p2 þ ðÞ4Rr þ r2 2 ¼ 2R2

(see the problems 5.3.16, 5.3.11,pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi and 5.3.10a). 4 2 2 2 2 2 ffiffiffi p þðÞ4R þ8Rrþ2r p þðÞ4Rrþr p p2þ2Rrþr2 We need to prove that 2R2 3 3 8R2 . Denote m ¼ p2 þ 2Rr þ r2, then according to the problems 5.5.6 and 5.5.10b, we 2 2 2 have that m1 ¼ 18Rr À 4r m 4R þ 6Rr þ 4r ¼ m2. Note that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p4 4R2 8Rr 2r2 p2 4Rr r2 2 p2 4Rr r2 2 4p2R2 þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ þ þ ðÞþ ¼ ðÞþ þ þ ¼ ¼ ðÞm þ 2Rr 2 þ 4R2ðÞm À 2Rr À r2 :

2 2 2 27 2 It remains to prove that ðÞm þ 2Rr þ 4R ðÞm À 2Rr À r 16 m .Thatis 11 2 3 À16 m þ 4RRðÞþ r m À 8R r 0. 11 2 3 Consider a function fmðÞ¼À1 16 m þ 4RRðÞþ r m À 8R r on a segment [m1, m2]. Since min fmðÞ¼minðÞfmðÞ1 ; fmðÞ2 , then it remains to verify that f(m1) ½m1;m2 0 and f(m2) 0. Indeed, we have that 11 f ðm Þ¼ð4R2 þ 6Rr þ 4r2Þ 4RðR þ rÞÀ ð4R2 þ 6Rr þ 4r2Þ À 8R3r ¼ 2 16 1 ¼ ð2R2 þ 3Rr þ 2r2Þð10R2 À Rr À 22r2ÞÀ8R3r 4 1 ð2R2 þ 3Rr þ 2r2Þð10R2 À Rr À 11RrÞÀ8R3r ¼ 4 1 1 ¼ ð2R2 þ 3Rr þ 2r2Þð5R À 6rÞR À 8R3r ð7Rr þ 2r2Þð5R À 6rÞR À 8R3r ¼ 2 2 Rr Rr ¼ ð19R2 À 32Rr À 12r2Þ ð6Rr À 12r2Þ0: Therefore, f ðm Þ0: 2 2 2 234 5 Application of Trigonometric Inequalities

We have that, 11ÀÁ fmðÞ¼ðÞ18Rr À 4r2 4RRðÞÀþ r 9Rr À 2r2 À 8R3r ¼ 1 8 ðÞ9R À 2r r ÀÁ ¼ 32R2 À 67Rr þ 22r2 À 8R3r ÀÁ4 2 2 3 ¼ 2RrÀÁ32R À 67Rr þ 22r À 8R r ¼ ¼ 2Rr 28R2 À 67Rr þ 22r2 ¼ 2RrðÞ R À 2r ðÞ28R À 11r 0:

Hence, f(m1) 0. AF b cos α 5.5.12. Note that HA ¼ ¼ ¼ 2R cos α and HE ¼ HA sin(90 À γ) ¼ cosðÞ 90 Àβ sin β 2R cos α cos γ. In a similar way, we deduce that HB ¼ 2R cos β, HC ¼ 2R cos γ, HD ¼ 2R cos β cos γ, HF ¼ 2R cos α cos β. (a) According to the problem 5.2.7, we have that

4R2 cos 2α þ 4R2 cos 2β þ 4R2 cos 2γ 16R2 cos 2αcos 2β þ 16R2 cos 2β cos 2γ þ 16R2 cos 2γ cos 2α:

Therefore, HA2 þ HB2 þ HC2 4(HE2 þ HD2 þ HF2). (b) According to the problem 5.2.8b, we have that 2R sin α þ 2R sin β þ 2R sin γ pffiffiffi 2 3 ðÞ2R cos α cos β þ 2R cos β cos γ þ 2R cos γ cos α . Thus, AB þ BC þ pffiffiffi AC 2 3ðÞHD þ HE þ HF . (c) According to the problem 5.5.12b, we have that 1 ðÞHD þ HE þ HF 2 ðÞAF þ BD þ CE þ BF þ CD þ AE 2 12 1 ðÞAF þ BD þ CE 2 þ ðÞBF þ CD þ AE 2 6 1ÀÁÀÁÀÁ 3 AF2 þ BD2 þ CE2 þ 3 BF2 þ CD2 þ AE2 ¼ AF2 þ BD2 þ CE2, 6 since AF2 À BF2 þ BD2 À CD2 þ CE2 À AE2 ¼ AH2 À BH2 þ BH2 À CH2 þ CH2 À AH2 ¼ 0. 5.5.13. Let the straight line l intersect lines DA, DB, and DC and form angles α, β, γ, respectively. Then, cos2α cos2β cos2γ 1 (see the proof of the problem 7.1.10). þ þ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ We have to prove that a sin α þ b sin β þ c sin γ 2 a2 þ b2 þ c2 (see the problem 5.4.4). sin α sin β sin γ The equality holds true, if and only if a ¼ b ¼ c , this means that pffiffiffi pffiffiffi pffiffiffi 2a 2b 2c sin α ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , sinβ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , sinγ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , a2 þ b2 þ c2 a2 þ b2 þ c2 a2 þ b2 þ c2 0 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 B b2 þ c2 À a2 a2 þ c2 À b2 a2 þ b2 À c2C @cos α ¼ , cosβ ¼ , cosγ ¼ A: a2 þ b2 þ c2 a2 þ b2 þ c2 a2 þ b2 þ c2 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 235

Figure 5.6 A

A1 A2 M

a B2 b

OB1 B

Remark If max(a, b, c) ¼ a and a2 > b2 þ c2, then the equality cannot hold true. 5.5.14.ÀÁLet ∠A ¼ α, ∠ B ¼ β, ∠ÀÁC ¼ γ and r be the inradius of triangle ABC. Then, α β γ αþβ βþγ γþα 2r ctg 2 þ ctg 2 þ ctg 2 and 2r ctg 4 þ ctg 4 þ ctg 4 are the perimeters of the given and obtained triangles, respectively. According to the problem 5.4.1, we have that α β γ α þ β β þ γ γ þ α 2r ctg þ ctg þ ctg 2r ctg þ ctg þ ctg : 2 2 2 4 4 4

5.5.15. Consider Figure 5.6. Let OA ¼ OB ¼ OM ¼ R, ∠AOM ¼ α, ∠ BOM ¼ β. Note that

R2 S ¼ S þ S ¼ ðÞsin 2α þ sin 2β ¼ OA1MB1 MOA1 MOB1 4 R2 R2 ¼ sin ðÞα þ β cos ðÞα À β sin ðÞ¼α þ β S : 2 2 AOB α β Therefore, SMA2B2 SAA1A2 þ SBB1B2 , and the equality holds true for ¼ . S Thus, MA2B2 1 and the greatest value is reached, when M is the midpoint SAA1A2 þSBB1B2 of the arc AB

5.5.16. Let ∠A ¼ α, ∠ B ¼ β, ∠ C ¼ γ and ∠MAC ¼ α1, ∠ MCA ¼ γ1, ∠ AMC ¼ β1 (see Figure 5.7). Therefore, α1 < α β, γ1 < γ β, and β1 > β, According to the problem 5.4.5, it α β γ α β γ follows that ctg 1 þ ctg 1 þ ctg 1 > ctg þ ctg þ ctg . 2 2 2 2ÀÁ2 2 2 2 β γ Note that ðÞABþBCþAC ¼ 4p ¼ 4p ¼ 4 ctg α þ ctg þ ctg , (see the problem SABC pr r 2 2 2 5.3.2). 2 α β γ Similarly we obtain that ðÞAMþMCþAC ¼ 4 ctg 1 þ ctg 1 þ ctg 1 . SAMC 2 2 2 2 2 Thus, ðÞAMþMCþAC > ðÞABþBCþAC . SAMC SABC 236 5 Application of Trigonometric Inequalities

Figure 5.7 B

b M

aa1 g1 g AC

5.5.17. If point O is the circumcenter with the radius R of the given hexagon ABCDEF and ∠COD ¼ ∠ DOE ¼ α, ∠AOB ¼ ∠ BOC ¼ β, ∠FOA ¼ ∠ EOF ¼ γ. Then, α þ β þ γ ¼ 180 and using the problem 5.1.15, we obtain that

1 1 S ¼ R2ðÞsin 2α þ sin 2β þ sin 2γ R2ðÞsin α þ sin β þ sin γ ¼ ACE 2 2 1 ¼ R2ðÞsin ðÞþβ þ γ sin ðÞþα þ γ sin ðÞα þ β ¼S : 2 BDF

5.5.18. Let ∠A1OB1 ¼ 2γ, ∠A1OC1 ¼ 2β, and ∠B1OC1 ¼ 2α, where O is the incen- < α β γ < π α β γ π ter of triangle ABC. Then, we have that 0 , , 2, þ þ ¼ and α β γ α β γ A1B1 þ B1C1 þ A1C1 ¼ 2r(sin þ sin þ sin ), p ¼ r(tgpffiffiffiþ tg þ tg ). Therefore, A1B1 þ B1C1 þ A1C1 ¼ 2rðÞsin α þ sin β þ sin γ 3 3r r(tgα þ tgβ þ tgγ) ¼ p (see the problems 5.1.6 and 5.2.1). We have that

2pr 2r2ðÞtgα þ tgβ þ tgγ ¼ ¼ 8rtgðÞα þ tgβ þ tgγ cos α cos β cos γ ¼ R rtgðÞα þ tgβ 2 sin ðÞπ À 2γ ¼ 8r sin α sin β sin γ ¼ 2rðÞsin 2α þ sin 2β þ sin 2γ :

2pr α β γ α β γ Consequently, R ¼ 2rðÞsin 2 þ sin 2 þ sin 2 2r ðÞsin þ sin þ sin ¼ A1B1 þ B1C1 þ A1C1 (see the problem 5.1.15). ∠ π β, ∠ ∠ π α ∠ 5.5.19. Note that COD ¼ À 2 ODC ¼ OBC ¼ 2 À . Hence, OCD ¼ π β γ 2 þ À . π β γ R sin ðÞ2þ À R cos ðÞβÀγ Using the law of sines, we obtain that OD ¼ π α ¼ cos α . Similarly, sin ðÞ2À R cos ðÞγÀα R cos ðÞαÀβ we deduce that OE ¼ cos β and OF ¼ cos γ .

3 cos ðαÀβÞcos ðβÀγÞcos ðγÀαÞ 3 (a) We have thatOD Á OE Á OF ¼ R cos αcos βcos γ 8R , (see the problem 5.2.4). (b) Since AD AOþOD R cos α sin 2α sin 2α OD ¼ OD ¼ 1 þ OD ¼ 1 þ cos ðÞβÀγ ¼ 1 þ 2 sin α cos ðÞβÀγ ¼ 1 þ sin 2βþ sin 2γ. 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 237

BE sin 2β CF sin 2γ Similarly, OE ¼ 1 þ sin 2αþ sin 2γ and OF ¼ 1 þ sin 2αþ sin 2β. Thus, we have to prove sin 2α sin 2β sin 2γ that A ¼ sin 2βþ sin 2γ þ sin 2αþ sin 2γ þ sin 2αþ sin 2β 1, 5. Denote by x ¼ sin 2β þ sin 2γ, y ¼ sin 2α þ sin 2γ, z ¼ sin 2α þ sin 2β. It is clear that x, y, z > 0 and sin 2α ¼ yþzÀx, sin 2β ¼ xþzÀy, sin 2γ ¼ xþyÀz. Thus, ÀÁ2 2 2 1 x y 1 x z 1 y z > A ¼ 2 y þ x þ 2 z þ x þ 2 z þ y À 1, 5 1, 5, as for a 0 we have that 1 a þ 2 2. φ 1 ∠ 5.5.20. Denote by i ¼ 2 AiMAiþ1, i ¼ 1, . . . , n, where An þ 1 A1. We have that

2 ∠ ∠ di ¼ MAi Á MAiþ1 cos AiMPi cos Aiþ1MPi ¼ cosðÞ∠A MP þ cos∠A MP þcosðÞ∠A MP À cos∠A MP ¼ MA Á MA i i iþ1 i i i iþ1 i i iþ1 2 1 þ cosðÞ∠AiMPi þ ∠Aiþ1MPi MAi Á MAiþ1 2 Á ∠AiMPi þ ∠Aiþ1MPi ¼ MA Á MA cos 2 i iþ1 2 ∠A MP þ ∠A MP π MA Á MA cos 2φ , sinceφ i i iþ1 i < : i iþ1 i i 2 2

Thus, we obtain that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d1d2 ...d MA1 Á MA2 Á MA2 Á MA3 Á ...Á MA Á MA1 Á cosφ Á cosφ Á ...Á cosφ ¼ n π n 1 2 n ... φ ... φ nn ... : : : ¼ R1 Á R2 Á Á Rncos 1 Á Á cos n cos R1 Á R2 Á Á Rn ðsee the problem 5 4 7Þ

5.5.21. Let a and b(a b) be the sides of the rectangle P1. We can assume that α is the angle between the lines containing the greatest sides of the rectangles P1 and P2. We have that d ¼ b cos α þ a sin α c ¼ b sin α þ a cos α and b > b cos α þ a sin α. α π > sin α Therefore, 4 and b a 1À cos α. α c α b sin αþa cos α cos 2αðÞb sin αÀa cos α Thus, it follows that sin 2 À d ¼ sin 2 À b cos αþa sin α ¼ b cos αþa sin α a cos 2α b cos αþa sin α 0. α c Hence, we obtain that sin 2 d. 5.5.22. Let in a quadrilateral AB ¼ a, BC ¼ b, CD ¼ c, DA ¼ d and ∠ABC ¼ α, ∠ β aþbþcþd ADC ¼ , p ¼ 2 . According to the problem 5.4.10, we have that

ab cd pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi S sin α þ sin β ðÞp À a ðÞp À b ðÞp À c ðÞp À d : ABCD 2 2

< a2þb2Àc2Àd2 < α π Note that À1 2abþ2cd 1. Thus, there exists an angle 2 (0; ), such that α a2þb2Àc2Àd2 cos ¼ 2abþ2cd . 238 5 Application of Trigonometric Inequalities

Let us construct a trianglerABCffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiwith AB ¼ a, BC ¼ b and ∠B ¼ α. Then, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 2 2 α cdðÞ a þb þabðÞ c þd AC ¼ a þ b À 2ab cos ¼ abþcd . It is not difficult to verify that AC < c þ d, AC þ c > d, AC þ d > c . Hence, there exists a quadrilateral ABCD, so that AB ¼ a, BC ¼ b, CD ¼ c, AD ¼ d and ∠ABC ¼ α.Let∠ADC ¼ β, then AC2 ¼ a2 þ b2 À 2ab cos α ¼ c2 þ d2 À 2cd cos β ¼ c2 þ d2 þ 2cd cos α. Hence, α þ β ¼ π and

1 SABCD ¼ ðÞab þ cd sin α ¼ 2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u ÀÁ! u 2 1 t a2 þ b2 À c2 À d2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ðÞab þ cd 1 À ¼ ðÞp À a ðÞp À b ðÞp À c ðÞp À d : 2 2ðÞab þ cd

This ends the proof. 5.5.23. According to the problem 5.2.8c, we have that sin2α þ sin2β þ sin2 γ (cosα þ cos β þ cos γ)2. p2Àr2À4Rr ðÞRþr 2 Using the problems 5.3.8a and 5.3.6, we obtain that 2R2 R2 . Thus, p2 2R2 þ 8Rr þ 3r2. nþkÀm kþmÀn 5.5.24. Denote by y þ z ¼ m, x þ z ¼ n, x þ y ¼ k, then x ¼ 2 , y ¼ 2 , mþnÀk z ¼ 2 . (a) One needs to prove that

n þ k k þ m m þ n a4 þ b4 þ c4 À a4 À b4 À c4 16S2: m n k

Note that n þ k k þ m m þ n n m k m a4 þ b4 þ c4 À a4 À b4 À c4 ¼¼ a4 þ b4 þ a4 þ c4 þ m n k m n m k rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k m n m k m k n þ b4 þ c4 À a4 À b4 À c4 2 a4 Á b4 þ 2 a4 Á c4 þ 2 b4 Á c4À n k m n m k n k Àa4 À b4 À c4 ¼ 2a2b2 þ 2a2c2 þ 2b2c2 À a4 À b4 À c4 ¼ 16S2:

We have to prove that

n þ k k þ m m þ n pffiffiffi a2 þ b2 þ c2 À a2 À b2 À c2 4 3S: m n k 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 239

Figure 5.8 A3 · 2a2

A2 · 2a1 · An A1 ·

2an

Note that

n þ k k þ m m þ n a2 þ b2 þ c2 À a2 À b2 À c2 ¼ m n k n m k m k m ¼ a2 þ b2 þ a2 þ c2 þ b2 þ c2 À a2 À b2 À c2 m n m k n k rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n m k m k n 2 a2 Á b2 þ 2 a2 Á c2 þ 2 b2 Á c2 À a2 À b2 À c2 ¼ m n m k n k pffiffiffi ¼ a2 þ b2 þ c2 À ðÞa À b 2 À ðÞb À c 2 À ðÞa À c 2 4 3S

(see problem 5.5.9).

5.5.25. (a) Let A1A2 ¼ 2α1, A2A3 ¼ 2α2, ..., AnA1 ¼ 2αn (see Figure 5.8), where α α π α α α α 1 þ ...þ n ¼ and let max( 1, 2, ..., ÀÁn) ¼ n. α 2 π 2 π > α For n ¼ 2,ÀÁ we have that PA1 Á PA2 ¼ 2 sin 2 and 2 2. Therefore, π 2 α π PA1 Á PA2 2 sin 4 ¼ 2. It is clear that, the equality holdsÀÁ true, only if 2 ¼ 2 . α α 2 3 α 3 α α For n ¼ 3, we have that PA1 Á PA2 Á PA3 ¼ 8sinÀÁ2 sin 2 þ 2 , where 2 1. α α α π α α α 2 3þ2 2 α 3 > 3 3þ2 2 Note that PA1 Á PA2 Á PA3 8sin 6 sin 2 þ 2 , because 2 2 6 . α α π π 3þ2 2 α α Denote by 6 ¼ , then it is not difﬁcult to prove that 4 6. Hence, 2 PA1 Á PA2 Á PA3 8sin α sin 3α 4 sin α sin 3α ¼ 2(cos2α(1 À 2 cos 2α) þ 1) 2. Obviously, PA1 Á PA2 Á PA3 ¼ 2, if and only if α3 ¼ α2 ¼ α1. For, n ¼ 4 we need the following lemma. Lemma Given points M, A, B, C, D on a unit circle (see Figure 5.9), so that ABc ¼ CD.d Then, MA Á MD < MB Á MC. Indeed, let ABd ¼ CDd ¼ 2α, BCc ¼ 2β, and MAd ¼ 2γ. Then, we have that

MA Á MD ¼ 4 sin γ sinðÞ¼ 2α þ β þ γ 2ðÞ cosðÞÀ 2α þ β cosðÞ 2α þ β þ 2γ < < 2ðÞ cos β À cosðÞ 2α þ β þ 2γ :

As β < 2α þ β < π, thus MA Á MD < 2(cosβ À cos(2α þ β þ 2γ)) ¼ 4 sin(α þ γ) sin(α þ β þ γ) ¼ MB Á MC. 240 5 Application of Trigonometric Inequalities

Figure 5.9 D · 2a

C · 2b · M B · 2g 2a · A

2a2 A2 · A2 2a2 · A3 2a3 A3 2a1 A¢ 2a3 ® 2 · 2a3 · A4 · A4º A3¢ A1 · A1 · a4 · 2a4 P a4 · P

Figure 5.10

2 2a2 a2 2a A2 A2 A · · A¢¢ 3¢¢ A a -a 2 x · 3 A 4 2 · A3 A¢ 2¢ x 3 a -a 4 2 A¢ ® A 2 3¢ ® A · A4 A 4 · 4 A A 1x 1 · A1 x · A4¢¢ · A · 1¢¢ P P P

Figure 5.11

α α α π Now we need to prove the problem for n ¼ 4. Let 1 3.If 4 2, then ÀÁÀÁpﬃﬃ α α α 4 2 4 α 4 α α 4 > 2 PA1 Á PA2 Á PA3 Á PA4 ¼ 16sin 2 sin 1 þ 2 sin 1 þ 2 þ 2 16 2 ¼ 4. α < π If 4 2, then the proof consists in moving points A1, A2, A3, A4 along the circumference in such a way that PA1 Á PA2 Á PA3 Á PA4 decrease. If α4 α2 þ 2α3, then 3α4 ¼ α4 þ α4 þ α4 α4 þ α1 þ α2 þ 2α3 > π. Using the lemma we obtain that (Figure 5.10) 0 0 0 0 PA1 Á PA2 Á PA3 Á PA4 > PA1 Á PA 2 Á PA 3 Á PA4 > PA1 Á PA 2 Á PA4 > 2(PA 3 > 1 and for n ¼ 3 the statement of the problem holds true). Now, let α4 < α2 þ 2α3. Choose x, such that 2α4 À 2x ¼ 2α1 À (α4 À α2) þ 2x, x 3α4Àα2À2α1 ¼ 4 (see Figure 5.11). 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 241

0 Using the lemma twice, we obtain that PA1 Á PA2 Á PA3 Á PA4 (PA1 Á PA 2) Á 0 00 00 00 00 (PA 3 Á PA4) (PA 1 Á PA 2) Á (PA 3 Á PA 4). 00 00 00 00 00 00 α 00 00 α π α α < π Let A1A2 ¼ A2A3 ¼ A1A4 ¼ 2 ,asA3A4 2 , then 2 8 and 6 2 . π α < π 00 00 00 00 Therefore, 4 3. We have to prove that PA1 Á PA2 Á PA3 Á PA ¼ 2α 3α 5α 16sin 2 sin 2 sin 2 2, or

2ðÞ cos α À cos 2α ðÞcos 2α À cos 3α 1, 2 cos α cos 2α À 2cos 22α À 2 cos α cos 3α þ 2 cos 2α cos 3α 1, cos α þ cos 3α À 1 À cos 4α À cos 2α À cos 4α þ cos α þ cos 5α 1, 2 cos α þ cos 3α þ cos 5α 2 þ cos 2α þ 2 cos 4α, 2 cos α þ 2 cos α cos 4α cos 2αðÞ4 cos 2α þ 1 , 4 cos αcos 22α cos 2αðÞ4 cos 2α þ 1 :

π α < 2π α As 2 2 3 , then cos2 0. We have to prove that 4 cos α cos 2α 4 cos 2α þ 1; this means that 8cos3 α À 8cos2α À 4 cos α þ 3 0, (2 cos α À 1)(4cos2α À 2 cos α À 3) 0. The last α > 1 2α < α < α inequality holds true since cos 2 and 4cos 4 cos 2 cos þ 3. The equality holds when the quadrilateral A1A2A3A4 is a square.

(b) Take the points, so that A1A2A3An is a square and PA4, ÁÁÁ, PAn À 1 < 1. This π < π leads to a contradiction, since PAn ¼ 2 sin 8 2 sin 6 ¼ 1. Then, we have that PA1 Á PA2 Á ...Á PAn < PA1 Á PA2 Á PA3 Á PAn ¼ 2.

5.5.26. (a) Let points A1, A2,...,An and M be represented in the complex plane by the numbers α1, α2,...αn and z, respectively. Then, MA1 Á MA2,...,MAn ¼ | n n À 1 (z À α1) Á ... Á (z À αn)| and let f(z) ¼ (z À α1) Á ...Á (z À αn) ¼ z þ cn À 1z π n 2 ji þ ...þ c1z þ c0, where |c0| ¼ |α1 Á ... Á αn| ¼ 1 and w ¼ c0. Denote by wj ¼ e n , xj ¼ wwj, j ¼ 0, . . . , n À 1, then |xj| ¼ |w| Á |wj| ¼ 1 and

jjþfxðÞ 0 jjþfxðÞ1 ::: þ jjfxðÞnÀ1 jjfxðÞþ0 fxðÞþ1 ::: þ fxðÞnÀ1 ¼

Xn XnÀ1 XnÀ1 n n nÀ1 nÀ1 ::: ¼ w wj þ cnÀ1w wj þ þ c1w wj þ nc0 j¼0 j¼0 j¼0 n ¼ jjnw þ 0 þ ::: þ 0 þ nc0 ¼ 2ncjj0 ¼ 2n:

Thus, exists a number i, such that |f(xi)| 2.

Second Solution Lemma Let Àm be the smallest and M the greatest values of the trigonometric λ λ μ ... λ μ polynomial of theq n-ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffith order g(t) ¼ 0 þ 1 cos t þ 1 sin t þ þ n cos nt þ n sin nt, λ2 μ2 then M þ m 2 n þ n. 242 5 Application of Trigonometric Inequalities qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi λ2 μ2 φ λ λ μ Indeed, let gtðÞ¼ftðÞþ n þ n sin ðÞnt þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi,wheref(t) ¼ 0 þ 1qcosffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit þ 1 sin t ... λ μ λ2 μ2 φ μ λ2 μ2 φ þ þ n À 1 cos(n À 1)t þ n À 1 sin(n À 1)t, þ cos ¼ n, þ sin ÀÁ n ÀÁn n n π 0 3π ¼ λn, tk ¼ À φ þ 2πk =n, k ¼ 1, ..., n and t ¼ À φ þ 2πk =n. Since gtðÞk q2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁ ÀÁ qk ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ¼ ftðÞþ λ2 þ μ2 M and gt0 ¼ ft0 À λ2 þ μ2 Àm, k ¼ 1, ..., n,then k qn ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin k k nqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin Pn Pn ÀÁ 1 λ2 μ2 1 0 λ2 μ2 n ftðÞþk n þ n M and n ftk þ m n þ n.Thus k¼1 k¼1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 Xn 1 Xn ÀÁ ftðÞþ2 λ2 þ μ2 M þ m þ ft0 : ð5:29Þ n k n n n k k¼1 k¼1

Pn Pn ÀÁ 1 λ 1 0 We need to prove that n ftðÞ¼k 0 ¼ n ftk , then from (5.29) it follows qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k¼1 k¼1 λ2 μ2 that M þ m 2 n þ n. ðÞnþ1 α β nα sin 2 sin ðÞþ 2 It is known that sin β þ sin ðÞþβ þ α ...þ sin ðÞ¼β þ nα α and sin 2 ðÞnþ1 α β nα sin 2 cos ðÞþ 2 cos β þ cos ðÞþβ þ α ...þ cos ðÞ¼β þ nα α . sin 2 Hence, 0π 1 À φ m þ 2πm cosmt þ cosmt þ ...þ cosmt ¼ cos@ 2 Aþ 1 2 n n 0π 1 0π 1 φ π φ π À m þ 2 m 2πm À m þ 2 m 2πm þcos@ 2 þ A þ...þ cos@ 2 þ ðÞn À 1 A ¼ n n n n 0π 1 φ π 2πm À m þ 2 m 2πm sin n sin@ 2 þ ðÞn À 1 A 2n n 2n ... : ¼ πm ¼ 0, for m ¼ 1, ,n À 1 sin n

Pn Pn Pn 0 0 Similarly, we need to prove that sin mtk ¼ 0, sin mt k ¼ 0, cos mt k ¼ 0, k¼1 k¼1 k¼1 Pn Pn ÀÁ ... 1 λ 1 0 for m ¼ 1, , n À 1. Therefore, n ftðÞk ¼ 0 ¼ n ftk . k¼1 k¼1 Let Ai(cos2αi, sin2αi), i ¼ 1, ..., n and P(cos2t, sin2t). Then, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi PA ¼ ðÞcos 2t À sin 2α 2 þ ðÞsin 2t À cos 2α 2 ¼ 2 À 2 cosðÞ 2t À 2α i i i i

¼ 2 sin ðÞt À αi :

n À 1 Thus, PA1 Á PA2 ÁÁÁPAn ¼ 2|2 sin(t À α1) sin(t À α2) ÁÁÁsin(t À αn)|. 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 243

n À 1 We need to prove, by mathematical induction, that 2 sin(t À α1) sin (t À α2) ÁÁÁsin(t À αn) is a trigonometric polynomial of the n-th order, where λ2 μ2 n þ n ¼ 1. Indeed, for n ¼1 we have that2 sin(t À α1) ¼Àsin α1 cos t þ cos α1 sin t. λ2 μ2 α 2 α 2 Thus, 1 þ 1 ¼ÀðÞsin 1 þ ðÞcos 1 ¼ 1. kÀ1 Let for n ¼ k, we have that 2 sin tÀð α1Þ sin ðÞÁÁÁt À α2 sin ðÞ¼t À αk kPÀ1 λ μ λ μ λ2 μ2 ðÞi cos it þ i sin it þðÞk cos kt þ k sin kt , where k þ k ¼ 1. Therefore, i¼0

2k sin ðÞt À α sin ðÞÁÁÁt À α sin ðÞ¼t À α 1 2 !kþ1 XkÀ1 λ μ α α ¼ ðÞi cos it þ i sin it ðÞ2 cos kþ1 sin t À 2 sin kþ1 cos t þ i¼0 λ μ α α þðÞk cos kt þ k sin kt ðÞ2 cos kþ1 sin t À 2 sin kþ1 cos t ¼ λ α μ α ¼ÀðÞk sin kþ1 À k cos kþ1 cos ðÞk þ 1 t μ α λ α þÀðÞk sin kþ1 þ k cos kþ1 sin ðÞk þ 1 tþ Xk þ ðÞai cos it þ bi sin it : i¼0

λ2 μ2 It remains to prove that kþ1 þ kþ1 ¼ 1; this means that

ðÞÀλ sin α À μ cos α 2 þÀðÞμ sin α þ λ cos α 2 ¼ ÀÁk kþ1 k kþ1 k kþ1 k kþ1 λ2 μ2 2α 2α : ¼ k þ k ðÞsin kþ1 þ cos kþ1 ¼1

Let ÀÁ nÀ1 max 2 sin ðÞt À α1 sin ðÞÁÁÁt À α2 sin ðÞt À αn ¼ ½0;2π nÀ1 ¼ 2 sin ðÞt0 À α1 sin ðÞÁÁÁt0 À α2 sin ðÞ¼t0 À αn M

and ÀÁ nÀ1 min 2 sin ðÞt À α1 sin ðÞÁÁÁt À α2 sin ðÞt À αn ¼ ½0;2π nÀ1 ¼ 2 sin ðÞt1 À α1 sin ðÞÁÁÁt1 À α2 sin ðÞ¼Àt1 À αn m:

Consequently, according to the lemma, M 1orm 1. Thus, |M| 1or|m| 1. Hence, one of points P(cos2t0, sin2t0) and P(cos2t1, sin2t1) satisﬁes the condition PA1 Á PA2 Á ...Á PAn 2.

Remark One can prove that, if points A1, A2,...,An (n 3) are not the vertices of the regular n-gon, then exists a point Pon the circle, so that PA1 Á PA2 Á ...Á PAn > 2. 244 5 Application of Trigonometric Inequalities

(b) The proof by mathematical induction. For n ¼ 1, the inequality obviously holds true. Let us consider two cases:

(1) At least one of points Ai coincides with point O. Let An 0, then by mathematical induction there exists a point on the circle (with the center O), so that

PA1 Á PA2 Á ::: Á PAnÀ1 Á PAn ¼ PA1 Á PA2 Á ::: Á PAnÀ1Á ðÞÁOP þ OA ::: Á ðÞOP þ OA PO 1 nÀ1 PO ¼ 2nÀ2 ðÞÁOP þ OA ::: Á ðÞOP þ OA ðÞOP þ OA ¼ 1 nÀ1 n 2nÀ2 ::: > ðÞÁOP þ OA1 Á ðÞOP þ OAnÀ1 ðÞOP þ OAn : 2nÀ1

(2) None of points Ai coincides with point O.

Let Bi be the intersection point of ray OAi with the given circle and let point Ci be diametrically opposite to point Bi. We need to prove that PAi PBi , i ¼ 1, 2, . . . n, where P is a point on a OPþOAi OPþOBi given circle. If P coincides with point Bi or Ci, then the proof is obvious. Let P not coincide with any of points B and C . Then we have that PAi ¼ i i OPþOAi PAi sin ∠PCiAi PBi PBi ¼ sin ∠PiCiAi ¼ ¼ . AiCi sin ∠AiPCi BiCi OPþOBi Let us consider a homothety with a center O, such that the image of the given circle is a unit circle. Let M0 be the image of point M by this homothety. Then, 0 0 0 according to the problem 5.5.26a there exists a point P , such that P B 1 Á ...Á 0 0 0 0 0 0 PBi P B i P B i PB1 ::: P B n 2, as ¼ 0 0 ¼ . Therefore, it follows that Á Á OPþOBi OP þOB i 2 OPþOB1 PBn 2 1 PA1 ::: PAn PB1 ::: PBn 1 n ¼ nÀ1. Thus, Á Á Á Á nÀ1 or PA1 OPþOBn 2 2 OPþOA1 OPþOAn OPþOB1 OPþOBn 2 ::: 1 ::: . Á Á PAn 2nÀ1 ðÞÁOP þ OA1 Á ðÞOP þ OAn This ends the proof.

5.5.27. Denote ∠PAB ¼ α1, ∠ PBC ¼ β1, ∠ PCD ¼ γ1, ∠ PDA ¼ δ1 and ∠PAD ¼ α2, ∠ PBA ¼ β2, ∠ PCB ¼ γ2, ∠ PDC ¼ δ2, we have to prove that α ; β ; γ ; δ π minðÞ1 1 1 1 4. α ; β ; γ ; δ > π PB Let minðÞ1 1 1 1 4, then according to the law of sines 1 ¼ PA Á β γ PC PD PA sin α1 sin 1 sin 1 sin δ1 Á Á ¼ β Á γ Á δ Á α . Therefore, PB PC PD sin 2 sin 2 sin 2 sin 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi α β γ δ p α β γ δ α β γ δ sin 1 sin 1 sin 1 sin 1 ¼ sin 1 sin 1 sin 1 sin 1 sin 2 sin 2 sin 2 sin 2 α þ β γ þ δ α þ β γ þ δ sin 1 1 sin 1 1 sin 2 2 sin 2 2 2 2 2 2 α þ β þ γ þ δ α þ β þ γ þ δ sin2 1 1 1 1 sin2 2 2 2 2 4 4 α þ β þ γ þ δ þ α þ β þ γ þ δ π sin4 1 1 1 1 2 2 2 2 ¼ sin4 , 8 4

π xþy ; π 2 xþy since for x, y, 2 [0; ], we have that 2 2½0 and 0 sin x sin y sin 2 . 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 245 ÀÁ α β γ δ π 4 α ; β ; γ ; δ > π We ﬁnd that sin 1 sin 1 sin 1 sin 1 sin 4 and minðÞ1 1 1 1 4. α ; β ; γ ; δ > 3π Thus, maxðÞ1 1 1 1 4 . Let point P be inside triangle OCB, where O is the intersection point of the diagonals AC and BD, then α1 þ β1 < π, β1 þ γ1 < π. α ; β ; γ ; δ δ > 3π Therefore, maxðÞ¼1 1 1 1 1 4 . α ; β ; γ ; δ α ; β ; γ ; δ > π Note that maxðÞ2 2 2 2 minðÞ1 1 1 1 4. To prove this, one has to consider a segment with the length max(PA, PB, PC, PD) and use that in a triangle where the longer side is opposite to the larger angle. Let E be the intersection point of the straight lines PC and AB. Through points 0 β0 ∠ 0 β P and C draw a circle touching the half-line EB at point B , then 1 ¼ PB C 1 ∠ 0 γ0 ∠ 0 β0 and PCB ¼ 2 ¼ PB A ¼ 2. α δ < π δ δ < π β ; γ α ; β ; γ ; δ > π Since 2 þ 1 , 1 þ 2 , then maxðÞ¼2 2 maxðÞ2 2 2 2 4. δ α δ < π π > α β Hence, 1 þ 2 þ 2 . Otherwise, we have that 2 1 þ 1þð γ α δ δ β ; γ > 3π π π π 1ÞþðÞþ2 þ 1 þ 2 maxðÞ2 2 4 þ þ 4 ¼ 2 . This leads to a contradiction. Thus, the half-lines DC and AP intersect (at point F). Let us draw through points P and A a circle touching the half-line FC at point D0, δ0 ∠ 0 ∠ δ α0 ∠ 0 ∠ 0 δ0 then 1 ¼ PD A PDA ¼ 1 and 2 ¼ PAD ¼ PD C ¼ 2. Note that point D0 is not between points C and F. Otherwise, the circumcircle of 0 γ > ∠ > triangle APD contains point D and hence also point C, but then 1 ACP δ > 3π α β γ δ > π π 3π 3π 1 4 . This leads to a contradiction, as ð 1 þ 1 þ 1 þ 1 4 þ 4 þ 4 þ 4 Þ. Since the obtained quadrilateralÀÁAB0CD0 containsÀÁ segments PA, PB0, PC PD0, α0 ; β0 ; γ0 ; δ0 α ; β0 ; γ ; δ0 then one canÀÁ prove that max 2 2 2 2 ÀÁmin 1 1 1 1 . Then, we have π α β0 γ δ0 α0 β0 > 3π π π π π π that 2 ¼ 1 þ 1 þ 1 þ 1 þ 2 2 þ 2 2 4 þ 4 þ 4 þ 4 þ 24 ¼ 2 , but this α ; β ; γ ; δ π is not possible. Thus, ðÞ1 1 1 1 4. Second Solution Let AB ¼ a, BC ¼ b, CD ¼ c, DA ¼ d, PA ¼ x, PB ¼ y, PC ¼ z, and PD ¼ t. Then, we have that

1 1 1 1 S ¼ ax sin α þ by sin β þ cz sin γ þ dt sin δ ð5:30Þ ABCD 2 1 2 1 2 1 2 1

2 2 2 2 2 According to the law of cosines, we obtain that y ¼ a þ x À 2ax cos α1, z ¼ y 2 2 2 2 2 2 2 þ b À 2yb cos β1, t ¼ c þ z À 2cz cos γ1, and x ¼ d þ t À 2dt cos δ1. Therefore,

2 2 2 2 α β γ δ : a þ b þ c þ d ¼ 2ax cos 1 þ 2by cos 1 þ 2cz cos 1 þ 2dt cos 1 ð5 31Þ

ab cd a2þb2 c2þd2 Note that SABCD ¼ SABC þ SACD 2 þ 2 4 þ 4 . From the last inequalities, using (5.30) and (5.31), we deduce that ax sin α1 þ by sin β1 þ cz sin γ1 þ dt sin δ1 ax cos α1 þ by cos β1 þ cz cos γ1 þ dt cos δ1. α α α β γ δ α α α π Hence, there exists such , that 2 { 1, 1, 1, 1} and sin cos , then 4 (see also the problem 7.1.71c). 246 5 Application of Trigonometric Inequalities

Figure 5.12 A

E D M P K N

5.5.28. (a) At ﬁrst, note that it is sufﬁcient to prove the problem for BD ¼ CD. Indeed, if BD > CD, then consider on segment BD a point B0, such that B0D ¼ CD. Let line B0P intersect segment EC at point E0. Since B0D ¼ CD, then for triangle B0AC, we have that

AD þ DP > AE0 þ E0P ¼¼ AE þ EE0 þ E0P > AE þ EP:

Thus, AD þ DP > AE þ EP. Now, we need to prove the problem for BD ¼ CD. Let ∠DBP ¼ α, ∠PBC ¼ β, ∠DCA ¼ γ. Let us draw through point P a segment MK, parallel to line BC (Figure 5.12). Let N be a point on ray EC, such that EN ¼ EP. We have that ∠ ∠ ∠EBCþ∠ECB αþγ β > β ∠ EPN ¼ ENP ¼ 2 ¼ 2 þ ¼ EPK. Thus, point N is on ray KC. Let ∠PMN ¼ x, ∠ PNM ¼ y. We have that ∠AMN þ ∠ MNA ¼ ∠ ABC þ ∠ αþγ BCA, consequently, x þ y ¼ 2 . Note that, AM ¼ AD þ DM ¼ AD þ DP and > αþγ AN ¼ AE þ EN ¼ AE þ EP and we have to prove that AM AN; this means that 2 β ∠ > ∠ α β > α þ þ y ¼ MNA MAN ¼ þ þ x or y 2. α We proceed the proof by contradiction argument. Let y 2. Let ∠BAP ¼ u, ∠CAP ¼ v, then according to the Ceva’s theorem, we have that for trianglesÀÁABC and AMN: sinβ sin u sin γ ¼ sin α sin v sin(α þ β) and αþγ β α β sin x sin β sin γ sin x sin u sin 2 þ ¼ sin y sin ðÞþ sin v. Therefore, sin y ¼ αþγ β α. sin ðÞ2 þ sin α γ sin þ Ày α γ α γ β γ sin x ðÞ2 þ þ < α < π sin sin Since sin y ¼ sin y ¼ sin 2 ctgy À cos 2 and 0 y 2 2. αþγ β α ¼ sin ðÞ2 þ sin α γ α γ α γ α γ γ γ þ þ þ α þ 2 sin ctgy À cos sin ctg À cos ¼ sin α. Thus, 2 sin β cos 2 ÀÁ2 2 ÀÁ2 2 ÀÁ2 sin 2 2 αþγ β α β γ β γ α sin 2 þ cos 2 or sin À 2 sin þ 2 þ . π < β γ < β α γ ∠ABCþ∠BCA < π This leads to a contradiction, as À2 À 2 þ þ 2 ¼ 2 2. (b) Let A0 be a point on side AC, such that AA0 ¼ AD and let D0 be the intersection point of segments CD and BA0, then BD0 ¼ D0C. Hence, according to the problem 5.5.28a, we obtain for triangle A0BC that A0D0 þ D0P > A0E þ EP. Thus, DD0 þ D0P > A0E þ EP or AD þ DP > AE þ EP, since AD ¼ AA0 and DD0 ¼ A0D0. 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 247

5.5.29. (a) Let ∠AOB ¼ α, then according to the law of cosines, we have that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi AB ¼ ðÞAO þ BO 2 À 2AO Á BOðÞ1 þ cos α rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 α ðÞAO þ BO 2 À ðÞAO þ BO 2ðÞ1 þ cos α ¼ ðÞAO þ BO sin : 2 2

α Similarly, we obtain that CD ðÞCO þ OD sin 2. Since, α α 2r þ 2r ¼ ðÞAO þ BO À AB tg þ ðÞCO þ DO À CD tg ¼ 1 3 2 2 α α α ¼ ðÞAC þ BD À ðÞAB þ CD tg ðÞÁAC þ BD 1 À sin tg 2 2 2 ÀÁ α α and similarly we deduceÀÁÀÁ that 2r2 þ 2r4ÀÁ ðÞAC þ BD 1 À cos 2 ctg 2. Then, α α α α 2r1 þ 2r2 þ 2r3 þ 2r4 1 À sin 2 tg 2 þ 1 À cos 2 ctg 2 ðÞAC þ BD , where r1, r2, r3, r4 are the radiuses of the circles S1, S2, S3, S4, respectively.ÀÁ α α ÀÁIn order to end theffiffiffi proof, it remains to prove that 1 À sin 2 tg 2 þ α α p 1 À cos 2 ctg 2 2 À 2. ÀÁÀÁ α α > α α α α Let t ¼ sin 2 þ cosffiffiffi2, then ÀÁt 0ffiffiffi andÀÁ 1 Àffiffiffi sin 2 tg 2 þ 1 À cos 2 ctg 2 ¼ 3 2 p p p t À3tþ2 ¼ t þtÀ2 2 À 2, i.e., t À 2 t þ 2 2 À 1 0. The last inequality t2À1 tþ1 pffiffiffi t2À1 α α 1 α 1 holds true, as 2 ¼ sin 2 cos 2 ¼ 2 sin 2, thus t 2. (b) According to the problem 1.1.4a, we have that

2r1 þ 2r3 2r2 þ 2r4 O1O2 þ O2O3 þ O3O4 þ O4O1 < 2ðÞO1O3 þ O2O4 ¼α þ α sin cos 0 1 α α 2 2 1 À sin 1 À cos B 2 2C 2 ðÞAC þ BD @ α þ α A ¼ ðÞAC þ BD < AC þ BD, cos sin t þ 1 2 2

α α t2À1 > > as sin 2 cos 2 ¼ 2 0, this means that t 1. This ends the proof. ∠ α ∠ β ∠ α β 5.5.30. Letpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi us denote AC ¼ b, CB ¼ a, DAC ¼ , DBC ¼ , then ACB ¼ þ a2þb2À2ab cos ðÞαþβ and R ¼ 2 sin ðÞαþβ . Using the law of sines for triangles ΔADC and ΔBDC, we obtain that 2 ab sin αÁ sin β CD ¼ sin2ðÞαþβ . We have that a2 þ b2 À 2ab cos ðÞα þ β 2abðÞ1 À cos ðÞα þ β ¼ ¼ 2abðÞ2 sin α Á sin β þ 1 À cos ðÞα À β 2ab Á 2 sin α Á sin β ¼ 4ab sin α Á sin β:

Thus, a2 þ b2 À 2ab cos(α þ β) 4ab sin α Á sin β or R2 CD2. Hence, we deduce that R CD. This ends the proof. 248 5 Application of Trigonometric Inequalities

φ 1 ∠ ... 2 5.5.31. Let us denote by i ¼ 2 AiMAiþ1, i ¼ 1, , n.Wehavethat, di ¼ MAi Á MAiþ1 cos ∠AiMPi Á cos ∠Aiþ1MPi,wherePi 2 AiAi þ 1 and MPi ⊥ AiAi þ 1. Hence, 1 þ cosðÞ∠A MP þ ∠A MP d2 ¼ R R cos∠A MP Á cos∠A MP R R Á i i iþ1 i ¼ i i iþ1 i i iþ1 i i iþ1 2 ∠A MP þ∠A MP ¼ R R cos 2 i i iþ1 i R R cos 2φ , i iþ1 2 i iþ1 i

∠ ∠ φ AiMPiþ Aiþ1MPi < π as i 2 2. Consequently pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi d1 þ d2 þ ...þ dn R1R2 cos φ þ R2R3 cos φ þ ... pffiffiffiffiffiffiffiffiffiffi 1 2 φ : : þ RnR1 cos n ð5 32Þ

It is clear that φ1 þ φ2 þ ... þ φn ¼ π and if we need to prove that pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi φ φ ... φ R1R2 cos 1 þ R2R3 cos 2 þ þ RnR1 cos n π cos ðÞR þ R þ ...þ R , ð5:33Þ n 1 2 n then from (5.32) and (5.33) we obtain that

... 1 R1 þ R2 þ þ Rn π ... : cos nðÞd1 þ d2 þ þ dn

... Consider on the coordinate plane points B1(a1,0),pBffiffiffiffiffi2(a2, b2), ,pBnffiffiffiffiffi(an, bn), ... Bn þ 1(Àpa1ffiffiffiffiffi, 0) (see Figure 5.13), such that OB1 ¼ R1, OB2 ¼ R2, , ∠ φ ∠ φ ... ∠ φ OBn ¼ Rn, B1OB2 ¼ 1, B2OB3 ¼ 2, , BnOBn þ 1 ¼ n. pffiffiffiffiffiffiffiffiffiffiffiffiffi OB2þOB2 ÀB B2 φ i iþ1 i iþ1 According to the law of cosines RiRiþ1 cos i ¼ 2 .

Figure 5.13 y

B2 Bn j2

jn j1

Bn+1 O B1 x 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 249 ÀÁ 2 2 ... 2 Therefore, the inequality (5.33) can beÀÁ rewritten, as 2 OB1 þ OB2 þ þ OBn 2 2 ... 2 π 2 ... 2 ÀB1B2 À B2B3 À À BnBnþ1 2 cos n OB1 þ þ OBn ,or

a1a2 þ a2a3 þ ...þ anÀ1an À ana1 þ b2b3 þ b3b4 þ ...þ bnÀ1bn πÀÁ ð5:34Þ cos a2 þ ...þ a2 þ b2 þ ...þ b2 : n 1 n 2 n If we prove that for any numbers π ÀÁ x x þ x x þ ...þ x x À x x cos x2 þ x2 þ ...þ x2 , ð5:35Þ 1 2 2 3 nÀ1 n n 1 n 1 2 n then we can use the inequality (5.35) for x1 ¼ a1, ..., xn ¼ an and x1 ¼ 0, x2 ¼ b2, ..., xn ¼ bn, in order to prove the inequality (5.34). Thus, we have to prove that for n 3 and any numbers x1, x2x3, ..., xn it holds true π ÀÁ A ¼ 2 cos x2 þ x2 þ ...þ x2 À 2x x À 2x x À , ..., À 2x x þ 2x x 0: n 1 2 n 1 2 2 3 nÀ1 n n 1 ð5:36Þ One can easily verify that 1 2π π π 2 A ¼ x sin À x sin þ x sin þ π 2π 1 n 2 n n n sin sin n n 1 3π 2π π 2 þ x sin À x sin þ x sin þ ... 2π 3π 2 n 3 n n n sin sin n n 1 ðÞn À 1 π ðÞn À 2 π π 2 þ x sin À x sin þ x sin : ðÞn À 2 π ðÞn À 1 π nÀ2 n nÀ1 n n n sin sin n n ð5:37Þ

Therefore,0 the inequality (5.36) holds true. To prove (5.37), it remains1 to note that B C πB 1 1 1 C sin2 @ þ þ ...þ A ¼ n π 2π 2π 3π ðÞn À 2 π ðÞn À 1 π sin sin sin sin sin sin n n n n n n 0 1 2π π 3π 2π ðÞn À 1 π ðÞn À 2 π Bsin À sin À sin À C πB n n n n n n C ¼ sin @ þ þ ...þ A ¼ n π 2π 2π 3π ðÞn À 2 π ðÞn À 1 π sin sin sin sin sin sin n n n n n n π π 2π 2π 3π ðÞn À 2 π ðÞn À 1 π ¼ sin ctg À ctg þ ctg À ctg þ ...þ ctg À ctg ¼ n n n n n n n π π ðÞn À 1 π π ¼ sin ctg À ctg ¼ 2cos : n n n n This ends the proof. 250 5 Application of Trigonometric Inequalities

5.5.32. We have that (a cos α þ b cos β)2 (a cos α þ b cos β)2 þ (a sin α À b sin β)2 ¼ a2 þ b2 þ 2ab cos(α þ β) (see the solution of the problem 5.5.31), thus pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi ::: φ φ ::: φ φ d1 þ d2 þ þ d2k R1R2 cos 1 þ R2R3 cos 2 þ þ R2kÀ1R2k cos 2kÀ1 þ R2kR1 cos 2k pffiffiffiffiffiÀÁpffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiÀÁpffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi φ φ ::: φ φ R2 R1 cos 1 þ R3 cos 2 þ þ R2k R2kÀ1 cos 2kÀ1 þ R1 cos 2k rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁpffiffiffiffiffi pffiffiffiffiffi ÀÁpffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi ::: φ φ 2 ::: φ φ 2 ðÞR2 þ R4 þ þ R2k R1 cos 1 þ R3 cos 2 þ þ R2kÀ1 cos 2kÀ1 þ R1 cos 2k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi À pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Á ::: φ φ ::: φ φ ðÞR2 þ R4 þ þ R2k R1 þ R3 þ 2 R1R3 cosðÞþ1 þ 2 þ R2kÀ1 þ R1 þ 2 R2kÀ1R1 cosðÞ2kÀ1 þ 2k rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi π ðÞR þ R þ ::: þ R 2ðÞR þ R þ ::: þ R þ2ðÞR þ R þ ::: þ R cos 2 4 2k 1 3 2kÀ1 1 3 2kÀ1 k π pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2cos ðÞR þ R þ ::: þ R ðÞR þ R þ ::: þ R : 2k 2 4 2k 1 3 2kÀ1

This ends the proof. ! ! ~ ::: ! ~ ~2 5.5.33.(a) Let A1Anþ1 ¼ d1, A2Anþ2 ¼ d2, , AnA2n ¼ dn, di ¼ Ri, i ¼ 1, . . . , n, d d d ~ ~ φ ::: ~ ~ φ ~ ~ φ d1, d2 ¼ 1, , dnÀ1, dn ¼ nÀ1 and dn,À d1 ¼ n, note that

φ1 þ φ2 þ ...þ φn ¼ π. ! ! ! ! We have that ~d þ A A À d þ A A ¼ ~0. Thus, it follows that ~d À d 1 nþ1 nþ2 2 2 1 1 2 2 ! ! ~ ! ¼ A1A2 À Anþ1Anþ2 A1A2 þ Anþ1Anþ2 ; therefore, d1 À d2 ðA1A2 þ Anþ1 pffiffiffiffiffiffiffiffiffiffi 2 φ 2 Anþ2Þ or R1 À 2 R1R2 cos 1 þ R2 ðÞA1A2 þ Anþ1Anþ2 . Similarly, we obtain that pffiffiffiffiffiffiffiffiffiffi φ 2 R2 À 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR2R3 cos 2 þ R3 ðÞA2A3 þ Anþ2Anþ3 , and so on recurrently, p 2 R 1 À 2 R 1R cos φ þ R ðÞA 1A þ A2 1A2 . nÀ nÀ n nÀ1 n nÀ n nÀ n ~ ! ! ! ~ ~ ! ! ! As d1 þ Anþ1An þ dn þ A2nA1 ¼ 0, then d1 þ dn ¼ A1A2n À Anþ1An An pffiffiffiffiffiffiffiffiffiffi π φ 2 Anþ1 þ A2nA1; therefore. R1 þ 2 R1Rn cos ðÞþÀ n Rn ðÞAnAnþ1 þ A2nA1 . Summing up these inequalities, we deduce that ÀÁpffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi ::: φ φ ::: φ 2ðÞR1 þ R2 þ þ Rn À2 R1R2 cos 1 þ R2R3 cos 2 þ þ RnR1 cos n 2 2 2 ðÞA1A2 þ Anþ1Anþ2 þ ðÞA2A3 þ Anþ2Anþ3 þ ðÞAnAnþ1 þ A2nA1 : ð5:38Þ We have pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi φ φ ::: φ R1R2 cos 1 þ R2R3 cos 2 þ þ RnR1 cos n π cos ðÞR þ R þ ::: þ R ð5:39Þ n 1 2 n

(see the proof of problem 5.5.31); thus from (5.38) and (5.39) it follows that π 21À cos ðÞR þ R þ ::: þ R n 1 2 n 2 2 2 ðÞA1A2 þ Anþ1Anþ2 þ ðÞA2A3 þ Anþ2Anþ3 þ ::: þ ðÞAnAnþ1 þ A2nA1 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 251

or ÀÁ π 2 4sin2 A A2 þ A A2 þ ::: þ A A2 ðÞA A þ A A 2þ 2n 1 nþ1 2 nþ2 n 2n 1 2 nþ1 nþ2 2 2 þðÞA2A3 þ Anþ2Anþ3 þ ::: þ ðÞAnAnþ1 þ A2nA1 :

Remark This estimate is an optimal one, as for a regular 2n-gon the equality holds true. pffiffiffiffiffiffiffiffiffiffi ! ~ ! 2 φ (a) We have that 2B1Bnþ1 ¼ d1 þ d2 , therefore 4B1Bnþ1 ¼ R1 þ 2 R1R2 cos 1þ R2. pffiffiffiffiffiffiffiffiffiffi Similarly, we obtain that 4B B2 ¼ R þ 2 R R cos φ þ R , and so on 2 pnþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 2 2 3 2 3 recurrently, 4B B2 ¼ R þ 2 R R cos φ þ R . nÀ12nÀ1 nÀ1 nÀ1 n nÀ1 n pffiffiffiffiffiffiffiffiffiffi ! ~ ! 2 φ We have that 2BnB2n ¼ dn À d1 , hence 4BnB2n ¼ Rn þ 2 RnR1 cos n þ R1. Summing up these n inequalities, we obtain that ÀÁ 4 B B2 B B2 ::: B B2 2 R R ::: R ÀÁ1pffiffiffiffiffiffiffiffiffiffinþ1 þ 2 nþ2 þ þpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin 2n ¼ ðÞ1 þ p2 þffiffiffiffiffiffiffiffiffiffiþ n þ φ ::: φ φ : þ2 R1R2 cos 1 þ þ RnÀ1Rn cos nÀ1 þ RnR1 cos n

From the proof of problem 5.5.33а, it follows that ÀÁπ 2 2 ::: 2 ::: 4 B1Bnþ1 þ B2Bnþ2 þ þ BnB2n 21þ cos ðÞR1 þ R2 þ þ Rn π n ctg 2 ðÞA A þ A A 2 þ ðÞA A þ A A 2 þ ::: þ ðÞA A þ A A 2 : 2n 1 2 nþ1 nþ2 2 3 nþ2 nþ3 n nþ1 2n 1

We deduce that π ÀÁ 4tg 2 B B2 þ B B2 þ ::: þ B B2 2n 1 nþ1 2 nþ2 n 2n 2 2 2 ðÞA1A2 þ Anþ1Anþ2 þ ðÞA2A3 þ Anþ2Anþ3 þ ::: þ ðÞAnAnþ1 þ A2nA1 :

5.5.34. Let any of the given circles touch three sides of a convex quadrilateral ABCD. Let us ﬁnd out for which quadrilateral ABCD its perimeter is the smallest. Consider another two circles, such that the ﬁrst circle touches lines BC, AD, AB and the second circle touches lines BC, AD, CD (Figure 5.14).

Figure 5.14 252 5 Application of Trigonometric Inequalities

Let O1O2D0 ¼ 2α. One can prove that MA0 ¼ AB, ND0 ¼ CD and p ¼ 2AB þ 2CD þ 2A0D0. Therefore, p ¼ 2MN. Note that p is the smallest, if MN is the smallest. As MN ¼ O3O4 sin 2α, then MN is the smallest, if O3O4 is the smallest. Note that O3O4 is the smallest, if the circles with centers O3 and O4 touch the circles with centers O1 and O2, respectively. Therefore, the perimeter of quadrilateral ABCD is the smallest, if AB ⊥ O1O2 and CD ⊥ O1O2. Hence, we obtain that it is sufﬁcient to prove the following inequality: p 2d ðÞRÀr 2 þ4r þ 4R þ 3 d , for quadrilateral ABCD, such that sides AB and CD are perpendicular to line O1O2. ðÞRÀr 2 : Note that, if R ¼ r, then p ¼ 2d þ 8R ¼ 2d þ 4r þ 4R þ 3 d > α α α RÀr : If R r, then AB ¼ 2rtg , CD ¼ 2Rctg , A0D0 ¼ ðÞR À r tg 2 , d ¼ cos 2α Thus, we need to prove that

R À r 3 2rtgα þ 2Rctgα þ ðÞR À r tg 2α þ 2r þ 2R þ ðÞR À r cos 2α, ð5:40Þ cos 2α 2 < α < π : where 0 4 The inequality (5.40) can be rewritten as 2R 2r ðÞcos α À sin α 2 À ðÞcos α À sin α ðÞR À r þ sin α cos α cos 2α ð5:41Þ 3 ÀÁ þ ðÞR À r cos 2α À sin2α : 2

Note that α α > 2R 2r > 2RÀ2r : cos À sin 0 and sin α À cos α sin α Therefore, in order to prove (5.41) it is sufficient to prove that 2 > 1 3 α α α 2 > 1 α: sin α cos αþ sin α þ 2 ðÞcos þ sin , or cos ðÞ3cos À 1 2 sin α > α 2α > 1 α: The last inequality holds true, as cos sin andðÞ 3cos À 1 2 sin Remark One can prove that pffiffiffi (a) p 2 6jjR À r þ 4R þ r: ðÞRÀr 2 : (b) If d ¼ R þ r, then p 6R þ 6r þ 3 Rþr 5.5.35. Consider Figure 5.14. S ⊥ Let us prove that the value of p2 is the greatest possible one, if AB O1O2 and CD ⊥ O1O2. Indeed, we have that

S AB Á r þ CD Á R þ A D ðÞR þ r ðÞAB þ CD þ A D ðÞÀR þ r AB Á R À CD Á r ¼ 0 0 ¼ 0 0 ¼ p2 p2 p2 R þ r MA Á R þ ND Á r R þ r 2rtgα Á R þ 2Rctgα Á r R þ r 4Rr ¼ À 0 0 À ¼ À : 2p p2 2p p2 2p p2 sin2α ð5:42Þ 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 253

Let quadrilateral A1B1C1D1 be such that any of the given two circles touches its three sides and A1B1 ⊥ O1O2 and C1D1 ⊥ O1O2. Let the area of quadrilateral A1B1C1D1 be equal to S1 and its perimeter be equal to p1. We have that

S R þ r 4Rr 1 ¼ À : ð5:43Þ 2 2 α p1 2p1 p1 sin 2

From (5.42) and (5.43), it follows that it is sufﬁcient to prove the following inequality,

R þ r 4Rr R þ r 4Rr À À : ð5:44Þ 2 α 2 α 2p p sin 2 2p1 p1 sin 2

According to the proof of problem 5.5.34, we have that p p1. Therefore, in order to prove (5.44), it is sufﬁcient to prove that, 4Rr 1 1 R þ r : : α þ ð5 45Þ sin 2 p1 p 2

As p p1, then in order to prove (5.45), it is sufﬁcient to prove that 4Rr 2 Rþr 16Rr α Á ,orp ðÞR þ r α. sin 2 p1 2 1 sin 2 If R ¼ r, then p1 ¼ 8R þ 2d. Therefore, it follows that p18R. If R > r, then we have that 0<α<π/4, and one needs to prove that,

ðÞ2rtgα þ 2Rctgα þ ðÞR À r tg 2α sin 2αðÞR þ r 8Rr, R2 À r2 ðÞ2rsin2 þ 2Rcos 2α ðÞþR þ r tg 2α sin 2α 4Rr, 2 R2 À r2 ÀÁ 2ðÞR þ r 2cos 2α þ tg 2α sin 2α À 2 R2 À r2 sin2α 4Rr: 2

The last inequality holds true as 2(R þ r)2cos2α > (R þ r)2 4Rr and tg2α sin 2α 4sin2α. 5.5.36. According to the proof of problem 5.5.35 (see 5.42), we have that

S R þ r 1 4Rr 1 ðÞR þ r 2 Á À Á sin 2α: p2 2 p sin 2α p2 64Rr

Let the distance between the centers of those circles be equal to d. If 0 < d R þ r, then we have that

R À r R À r cos 2α ¼ : d R þ r 254 5 Application of Trigonometric Inequalities

Thus, it follows that pffiffiffiffiffi 2 Rr sin 2α : R þ r

Hence, we deduce that S Rpþffiffiffiffir p2. 32 Rr If d > R þ r, then according to problem 7.1.109b, we have that

p2 p2 R þ r S < pffiffiffiffiffi p2: 18 16 32 Rr

Therefore, we obtain that

R þ r S < pffiffiffiffiffi p2: 32 Rr

This ends the proof. 5.5.37. According to the proof of problem 5.5.34, it is sufficient to prove the given inequality for quadrilateral ABCD, such that AB ⊥ O1O2 and CD ⊥ O1O2. а S 1 According to problem 7.1.109 , we have that p2 18. Moreover, according to the S Rþr 4Rr 1 proof of problem 5.5.35, we obtain that p2 ¼ 2p À p2 sin 2α 18. On the other hand, we have that cos 2α ¼ RÀr RÀr. pffiffiffiffi d Rþr α 2 Rr We deduce that sin 2 Rþr . Therefore, we obtain that pffiffiffiffiffi R þ r 2 RrðÞR þ r 1 À : 2p p2 18

Thus, it follows that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 9ðÞþR þ r 81ðÞR þ r 2 À 144ðÞR þ r Rr p , 2

(see problem 7.1.109a). 5.5.38. (а) According to the proof of problem 1.2.7, we have that

A1B1 þ B1C1 þ A1C1 a cos α þ b cos β þ c cos γ ¼ RðÞsin 2α þ sin 2β þ sin 2γ : ð5:46Þ

α β γ > π α β γ < π As , , 4 , then , , 2. According to problem 4.1.2.1f, we obtain that 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 255

sin 2α þ sin 2β þ sin 2γ ¼ sin ðÞþπ À 2α sin ðÞþπ À 2β sin ðÞπ À 2γ > > 2 π α 2 π β 2 π γ : πðÞþÀ 2 πðÞþÀ 2 πðÞ¼À 2 2 ð5:47Þ

Hence, from (5.46)and(5.47), we deduce that A1B1 þ B1C1 þ A1C1 > 2R X(A1,B1). Therefore, A1B1 þ B1C1 þ A1C1 > X(A1, B1). > (b) Note that PAC1B1 þ PBA1C1 þ PCA1B1 ¼ PABC þ PA1B1C1 , and PABC 4R, (see problem 5.2.3а). According to the proof of problem 5.5.39а, we have that > > PA1B1C1 2R. Therefore, PAC1B1 þ PBA1C1 þ PCA1B1 6R, hence ; ; > > ; maxðÞPAC1B1 PBA1C1 PCA1B1 2R XAðÞ1 B1 . ; ; > ; : Thus, it follows that maxðÞPAC1B1 PBA1C1 PCA1B1 XAðÞ1 B1 This ends the proof. 5.5.39. At ﬁrst, let us prove the following lemma. ρðÞX;QR ∠ Lemma If inside of angle PQR is chosen a point X, then ρðÞX;QP ¼ sin PQR Á ctg ∠PQX À cos ∠PQR:

ρðÞX;QR QX sin ∠RQX sin ðÞ∠PQRÀ∠PQX Proof of the lemma Indeed, we have that ρðÞX;QP ¼ QX sin ∠PQX ¼ sin ∠PQX ∠ ∠ ∠ , ρðÞX;QR ∠ ∠ ¼ sin PQR Á ctg PQX À cos PQR therefore ρðÞX;QP ¼ sin PQR Á ctg PQX À cos ∠PQR: This ends the proof of the lemma. According to the lemma, we have that

0 dc ∠ ∠ ∠ ∠ ∠ 0 ∠ db : ¼ sin Actg MAC À cos A ¼ sin Actg M AB À cos A ¼ 0 db dc

0 0 : 0 0 Thus, it follows that dcdc ¼ dbdb Similarly, we deduce that dcdc ¼ dada, 0 0 : therefore dada ¼ dbdb 0 da db Hence ¼ 0 , and according to the lemma, we obtain that sin ∠ Cctg ∠ db da MCA À cos ∠ C ¼ sin ∠ Cctg ∠ M0CB À cos ∠ C. Therefore, ∠MCA ¼ ∠ M0CB. Without loss of generality one can assume that ∠C ∠ B ∠ A. 0 2 Let us prove that dcdc r , where r is the inradius of triangle ABC. Let ∠A ¼ α, ∠ B ¼ β, ∠ MAB ¼ φ, ∠ MBA ¼ ψ. Note that

AB AB AB d ¼ , d0 ¼ , r ¼ c ctgφ þ ctgψ c ctg ðÞþα À φ ctg ðÞβ À ψ α β ctg 2 þ ctg 2, then by Cauchy-Bunyakovsky inequality, it follows that 256 5 Application of Trigonometric Inequalities

AB2 d d0 ¼ c c ðÞctgφ þ ctgψ ðÞctg ðÞþα À φ ctg ðÞβ À ψ 2 AB : ÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ctgφctgðÞα À φ þ ctgψctgðÞβ À ψ As cos α ctgφctgðÞ¼α À φ ctgφctgðÞÀα À φ 1 þ 1 ¼ þ 1 ¼ sin φ sin ðÞα À φ 2 cos α 2 cos α α ¼ þ 1 þ 1 ¼ ctg 2 , cos ðÞÀα À 2φ cos α 1 À cos α 2 then

AB2 d d0 ÀÁ¼ r2: c c α β 2 ctg 2 þ ctg 2 ÀÁ 0 0 0 0 3 6 Then dadbdcdadbdc ¼ dcdc r , and if point M is the incenter of triangle 0 0 0 6: ABC, we obtain that dadbdcdadbdc ¼ r 0 0 0 6 Therefore, the greatest possible value of expression dadbdcdadbdc is equal to r .

2 2 2 02 0 ðÞR ÀOM ðÞR ÀOM Remark One can prove that dada ¼ 4R2 , where O is the circumcenter of triangle ABC and R is its circumradius. 5.5.40. Let R be a positive root of ÀÁ 3 2 2 2 : x À da þ db þ dc x À 2dadbdc ¼ 0, ð5 48Þ

α β γ α da β db γ dc and , , be acute angles, such that cos ¼ R , cos ¼ R , cos ¼ R . Then, from (5.48) it follows that

cos 2α þ cos 2β þ cos 2γ þ 2 cos α cos β cos γ ¼ 1:

From the last equation, we deduce that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cos γ ¼Àcos α cos β þ cos 2αcos 2β À cos 2α À cos 2β þ 1 ¼ ¼Àcos α cos β þ sin α sin β ¼ cos ðÞπ À ðÞα þ β :

Thus, it follows that α þ β þ γ ¼ π. Obviously, one of the following inequalities holds true α ∠ A, β ∠ B, γ ∠ C.Ifα ∠ A, then 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 257 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 α 2 2 ∠ db þ dc þ 2dbdc cos db þ dc þ 2dbdc cos A R ¼ ¼ R sin α sin ∠A a

maxðÞRa; Rb; Rc :

Similarly, if β ∠ B or if γ ∠ C, then R max(Ra, Rb, Rc). α β γ α α π α 1 Let min( , , ) ¼ , we have that 3. Therefore cos 2 and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi maxðÞR ; R ; R R ¼ R cos 2α þ cos 2β þ cos 2γ þ 2 cos α cos β cos γ a b c rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d d d 2α 2β 2γ β γ 2 2 2 a b c : R cos þ cos þ cos þ cos cos ¼ da þ db þ dc þ maxðÞda; db; dc qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 dadbdc Hence, maxðÞRa; Rb; Rc d þ d þ d þ : a b c maxðÞda;db;dc This ends the proof. 5.5.41. We are going to use the following lemmas.

Lemma 1 Let the area S of a convex n-gon A1A2 ...Ansatisfy to the following inequality 4S AnA2 Á R1 þ A1A3 Á R2 þ ...þ An À 1A1 Á Rn, where Riis the circumradius of triangle Ai À 1AiAi þ 1, i ¼ 1, 2, . . . , n, A0 An, An þ 1 A1.

Let Mi be the midpoints of AiAi þ 1, for i ¼ 1, 2, . . . , n. For any i consider a quadrilateral, created by segments AiMi and AiMi À 1, and also by perpendiculars to these segments drawn from points Mi and Mi À 1, respectively. Let us prove that, these n quadrilateral cover given n-gon. Indeed, let P be a point inside of n-gon. Let PAk be the smallest of the distances и PA1, PA2,...,PAn. We have that PAk PAk þ 1 PAk PAk À 1, therefore point P is inside of n-gon and any of the two semiplanes containing Ak are bounded by the midpoints of perpendiculars to AkAk þ 1 and AkAk À 1, this means that in k-th quadrilateral. In order to end the proof, it is sufﬁcient to note that the area of i-th 1 AiÀ1Aiþ1 : quadrilateral is not greater than 2 Á 2 Á Ri According to the assumptions of our problem, it follows that

2 ∠ 2 ∠ ::: 2 ∠ : 4S 2R1 sin A1 þ 2R2 sin A2 þ þ 2R5 sin A5

By Cauchy-Bunyakovsky inequality, we obtain that qÀÁffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 ::: 4 2∠ ::: 2∠ 2S R1 þ þ R5 ðÞsin A1 þ þ sin A5 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 2 0 4 ::: 4 5sin 108 R1 þ þ R5 ,

4 4 4 4 4 4 2 therefore R þ R þ R þ R þ R S : 1 2 3 4 5 5sin2108 In this inequality we have used the following lemma. 2 Lemma 2 If α1, α2,...,α5are the angles of a convex pentagon, then sin α1 þ ... 2 2 þ sin α5 5sin 108 . 258 5 Application of Trigonometric Inequalities

The given sum does not depend on the order of the angles; thus without loss of generality one can assume that α1 α2 ... α5. If α1 ¼ 108 , then α2 ¼ ...¼ α5 ¼ 108 , and the inequality becomes an equality. If α1 < 108 , then α5 > 108 . Note that α1 þ α5 < 270 . Otherwise, if α1 þ α5 270 , then α2 þ α3 þ α4 270 . Therefore α2 90 , and α1 90 . Hence, we deduce that α5 180 . This leads to a contradiction. Hence, we obtain that ÀÁ 2 2 α α 2α 2α sin 108 þ sin 1 þÀÁ5 À 108 À sinÀÁ1 À sin 5 ¼ ¼ 2 cos ðÞα1 þ α5 sin α1 À 108 sin α5 À 108 > 0:

This means that substituting α1 by 108 and α5 by α1 þ α5 À 108 makes the sum of the squares of sines greater. Repeating these steps several times, one can make all the angles to be equal to 108 . This ends the proof.

Problems for Self-Study

Prove the inequalities for the elements of triangle ABC (5.5.33–5.5.37). r α β γ 17 r 5.5.42. 1 þ R sin 2 þ sin 2 þ sin 2 12 þ 6R. α β β γ γ α 1 r 5.5.43. sin 2 Á sin 2 þ sin 2 Á sin 2 þ sin 2 Á sin 2 2 þ 2R. 2 2 2 27 2 5.5.44. (a) ma þ mb þ mc 4 R , pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 (b) 3 3rRðÞþ r ma þ mb þ mc R, ppffiffiffiffiffiffiffiffiffiffiffiffi 2 (c) ma þ mb þ mc 3 3S, pffiffiffi a2þb2þc2 (d) 8ðÞmamb þ mbmc þ mcma 3 2 þ 6 3S , (e) 2R(m þ m þ m ) p2 þ r2 þ 4Rr, pffiffiffi a b c (f) 4 3S 9abc aqþbffiffiffiffiþc 2r (g) ha la Á R , 2 2r2 2ðÞrþR (h) 10r À R ha þ hb þ hc R , 2 (i) lalb þ lblc þ lcla 13Rr þ r . 5.5.45. a cos α þ b cos β þ c cos γ p. 2 2 5.5.46. 13, 5Rr lalblc 16Rr . 5.5.47. Let ABC be an arbitrary triangle and I be its incenter. Denote the intersection points of the straight lines AI, BI, CI by A0, B0, C0 with the circumcircle of the 4 0 0 0 triangle ABC. Prove that 5 ðÞR À 2r IA þ IB þ IC À IA À IB À IC 2ðÞR À 2r . 5.5 Using Trigonometric Inequalities for Proving Geometric Inequalities 259

! ! ! 5.5.48. Let point O be inside of triangle ABC, such that OK þ OM þ ON ¼ ~0, where K, M, N are the feet of the perpendiculars drawn from point O to sides AB, BC, AC. Prove that OKþOMþON p1 ffiffi. ABþBCþCA 2 3 5.5.49. Let I be the incenter of triangle ABC, and R and r be, respectively, the circumradius and the inradius of triangle ABC. Prove that (a) R3 IA Á IB Á IC, pffiffiffiffiffiffiffiffiffiffi (b) 3 3 4Rr2 IA þ IB þ IC 2R þ 2r. Hint Prove that IA Á IB Á IC ¼ 4Rr2. 5.5.50. Prove that for the triangle with angles α, β, γ and circumradius R the α β γ 9R2 inequality tg 2 þ tg 2 þ tg 2 4S holds true, where S is the area of the triangle. β < π γ < π < bþc 5.5.51. Let ABC be a triangle, such that 2 , 2. Prove that R þ r 2 , if and α < π only if 2.

5.5.52. Let O be the intersection point of bisectors AA1, BB1, CC1 of the triangle ABC. Prove that 8 AO Á BO Á CO 4R, where R and r are, respectively, the A1O B1O C1O r circumradius and the inradius of triangle ABC.

5.5.53. Prove that 36r1r2 a1a2 þ b1b2 þ c1c2 8R1R2 þ 4r1r2, where Ri and ri are, respectively, the circumradius and the inradius of the triangles with sides ai, bi, ci (i ¼ 1, 2). Hint See the problems 5.1.31 and 5.1.32. 5.5.54. Let triangle ABC be cut of the paper. Bend triangle ABC around the segment passing through A, so that the area of obtained ﬁgure is minimal, if: (a) ∠C ∠ B 3 ∠ C, (b) ∠A ¼ 40 , ∠B ¼ 125 . Hint Prove that triangle ABC should be bent (a) around bisector AD, (b) around segment AX, where X 2 [BC] and ∠XAC ¼ ∠ C. 5.5.55. Let D, E, F be points on sides BC, CA, AB of triangle ABC, respectively. Prove that

(a) PDEF min(PAEF, PBDF, PCDE), (b) rDEF min(rAEF, rBDF, rCDE),

where PXYZ and rXYZ are the perimeter and the inradius of triangle XYZ, respectively. Chapter 6 Inequalities for Radiuses

This chapter consists of Sections 6.1 and 6.2. In Section 6.1, selected problems related to the inequalities with radiuses of circles are provided. Perhaps, the most well known among them is the following one. Prove that R 2r, where R and r are the circumradius and inradius of triangle ABC, respectively. One of the proofs of this problem is obtained by Euler’s formula, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d ¼ R2 À 2Rr, where d is the distance between the circumcenter and incenter of triangle ABC. One of the methods to prove that R 2r and many other inequalities is the following statement: if triangle ABC is circumscribed to a circle with radius r and points M, N, K are three points not on the same line, belonging to lines AB, BC, AC, then the circumradius of triangle MNK is not less than r. In the proofs of many problems of this paragraph is used the following statement, if a circle does not have any points outside of triangle ABC, then the radius of that circle is not greater than the inradius of triangle ABC. Most of the problems provided in Section 6.1 can be proved using trigonometry. In Section 6.2, problems related to polygons, such that all of their vertices are chosen at the nodes of the integer lattice are considered. Among those problems, we would like to emphasize the following one, as it has a wide range of applications: if the vertices of a triangle are at the nodes of the integer lattice, then the area of the triangle is not less than 0.5. Some problems in this chapter were inspired by [4, 7, 15]. Nevertheless, even for these problems the authors have mostly provided their own solutions.

6.1 Inequalities for Radiuses of Circles

6.1.1. Prove that for the triangle with sides a, b, and c, the inequality 2uvR (u + v)2 r holds true, where 2u ¼ min (a + b À c, b + c À a, c + a À b) and 2v ¼ max (a + b À c, b + c À a, c + a À b). 6.1.2 One of the inner angles of the triangle is equal to α. Prove that

2R ÀÁ1 (a) α α , r sin 2 1 À sin 2 (b) R 2r. 6.1.3. Consider three circles pairwise not having any common internal points intersecting pairwise, such that their centers are on one straight line. Prove that, if the fourth circle is tangent to all three circles its radius cannot be less than the radiuses of these three circles.

6.1.4. (a) Circle S1 touches sides AC, AB of triangle ABC, circle S2 touches sides BC, AB, and S1, S2 touch each other externally. Prove that the sum of the radiuses of these circles is greater than the radius of the incircle S of triangle ABC.

(b) Given a triangle ABC. Circles with radiuses r1, r2, and r3 are such that each touches the other two circles and two sides of the triangle. Prove that r1 + r2 > r3. 6.1.5. Let two circles touch each other externally and touch the circle with the radius R internally on diametrically opposite points. Given a circle with radius R1 R touching all three circles, prove that R1 3.

6.1.6. Let M be the midpoints of side BC of triangle ABC, r1 and r2 be the radiuses of the incircles of triangles ABM and ACM, respectively. Prove that r1 < 2r2. 6.1.7. A rectangle is cut into rectangles. Prove that the sum of the circumradiuses of all these rectangles is not less than the circumradius of the initial rectangle. 6.1.8. (a) Prove that, if the polygon circumscribed around a circle with radius r is broken into convex circumscribed polygons, then the sum r1 + ... + rn of the inradiuses of these polygons is not less than r.

(b) Prove that, if a circle with radius r0 is inside of the polygon circumscribed around a circle with radius r, then r0 r. 6.1.9. Let point M belong to the diameter AB ¼ 2R of some half-circle. From point M are drawn segments MC and MD, such that α ¼ ∠AMC ¼ ∠BMD 90 and points C, D are on that half-circle. Given that the circle with radius r1 touches the half-circle and segments CM, AM. Given also that the circle with radius r2 touches the half-circle and segments DM, MB. Prove that ÀÁpffiffiffi (a) r1 þ r2 2 2 À 1 R, α 2 (b) r1 þ r2 2 sin α . 1 þ sin 2R 6.1 Inequalities for Radiuses of Circles 263

6.1.10. A circle is inscribed into triangle A1B1C1. Let the vertices of triangle Ai +1Bi +1Ci +1 be the intersection points of the bisectors of triangle AiBiCi with the given circle, i ¼ 1, 2, ...,andri be the inradius of triangle AiBiCi.Provethat

(a) rn +1 rn, for n ¼ 1, 2, ..., (b) 2rn rn À 1 + rn +1, for n ¼ 3, 4, .... 6.1.11. Given an acute triangle ABC and a point M on the line AB. Prove that R1 + R2 > R, where R1, R2, R are the circumradiuses of triangles ACM, BCM, ABC, respectively.

6.1.12. (a) Let ABCD be a convex quadrilateral, RA, RB, RC, and RD the circumradiuses of triangles DAB, ABC, BCD, and CDA, respectively. Prove that RA + RC > RB + RD, if and only if ∠A + ∠C > ∠B + ∠D.

(b) Let ABCD be a convex quadrilateral, rA, rB, rC, and rD the inradiuses of the triangle DAB, ABC, BCD, and CDA, respectively. Prove that rA + rC > rB + rD,if and only if ∠A + ∠C < ∠B + ∠D.

6.1.13. Let points C1, B1, and A1 be chosen, respectively, on sides AB, AC, and BC of the equilateral triangle ABC with side length 2. What is the greatest possible value of the sum of the inradiuses of triangles AB1C1, A1BC1, A1B1C ?

6.1.14. Let the inradius and the circumradius of the regular n-gon be equal to rn and 2 rn rnþ1 Rn, respectively. Prove that . Rn Rnþ1

6.1.15. Let r be the inradius of triangle ABC, rA be the radius of its excircle, touching sides AB, AC,(rB and rC are deﬁned similarly). r2 Prove that (a) rA + rB + rC r, (b) rArB þ rBrC þ rCrA 3 . 6.1.16. Prove that R 3r, where R is the radius of the circumsphere of the tetrahedron ABCD, and r is the radius of its insphere.

Solutions

6.1.1. Let a b c,then2u ¼ a + b À c and 2v ¼ b + c À a.Wehavetoprove a þ b À c b þ c À a abc S that 2 Á R b2r.Wehave R ¼ , r ¼ ,and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 2 4S p S ¼ ppðÞÀ a ðÞp À b ðÞp À c . Thus, we have to prove that ac b(a + c À b), or equivalently, (b À c)(a À b) 0. The last inequality obviously holds true. This ends the proof. β γ 6.1.2. (a) We have a ¼ rctg þ rctg ¼ 2R sin α, consequently 2 2 264 6 Inequalities for Radiuses β γ β γ sin þ 2R ctg þ ctg 2 2 1 ¼ 2 2 ¼ ¼ ¼ r sin α β γ α β γ sin α sin sin 2 sin sin sin 2 2 2 2 2 1 1 1 ¼ ¼ αα : α β À γ β þ γ α β þ γ sin cos À cos sin 1 À cos sin 1 À sin 2 2 2 2 2 2 2 ÀÁ < α α 1 (b) Note that 0 sin 2 1 À sin 2 4. According to problem 6.1.2a, we obtain 2R that r 4, or R 2r. This ends the proof.

6.1.3. Denote the centers and the radiuses of these circles (see Figure 6.1)byO, O1, O2, O3 and r, r1, r2, r3, respectively. We have that OO1 + OO3 O1O3 and r + r1 OO1, r + r3 OO3, O1O3 r1 +2r2 + r3. Therefore, r + r1 + r + r3 r1 +2r2 + r3,orr r2.

∠ > < O1O2 r1þr2 6.1.4. (a) Since O1OO2 90 , we have that OM 2 ¼ 2 , where M is the r1þr2 < midpoint of segment O1O2 (see Figure 6.2). Hence, r OM þ 2 r1 þ r2.

Remark If circles S1 and S2 do not touch each other, then 2r O1O2 + r1 + r2. The equality holds true, only if these circles coincide.

(b) According to problems 6.1.4a and 6.1.8b, we have that r1 + r2 > r > r3.

Figure 6.1

O1 O2 O3

Figure 6.2 C

O1 M r

AB 6.1 Inequalities for Radiuses of Circles 265

Figure 6.3

R1 R-x+R1 xR-R1

R-x x

6.1.5. We have that x < R (see Figure 6.3). > R Proof by contradiction argument. Assume that R1 3. By Stuart’s theorem, we have that

2 2 2 ðÞR À R1 R ¼ ðÞR1 þ x x þ ðÞR þ R1 À x ðÞÀR À x ðÞR À x xR:

Therefore, 4R3 R 2 4R 2 > ðÞR À R 2R > þ x x þ À x ðÞÀR À x ðÞR À x xR: 9 1 3 3

Hence, we deduce that 0 > (R À 2x)2. This leads to contradiction.

6.1.6. Let r be the inradius of triangle ABC. It is clear that r1 < r (see problem 6.1.8b). By homothety with center B and ratio 2, the image of triangle ABM < r < < contains triangle ABC. Therefore, r 2r1. Thus, 2 r1 r. r < < < Similarly, we can prove that 2 r2 r. Hence, r1 2r2. 6.1.7. We have that (see Figure 6.4),

2R ¼ AB ¼ AA1 þ A1A2 þ ...þ AkB 2R1 þ 2R2 þ ...þ 2Rkþ1 2R1 þ ...þ 2Rn:

Therefore, R R1 + ... + Rn. 6.1.8. (a) Let the semiperimeters of the polygons circumscribed around the circle with radiuses r1, r2, ..., rn, r are equal to p1, p2, ..., pn, p, respectively. We have that S ¼ S1 + ... + Sn and pi < p (see problem 2.1.1). Consequently, S1 Sn S1þ...þSn S r1 þ ...þ rn ¼ þ ...þ > ¼ ¼ r. p1 pn p p 266 6 Inequalities for Radiuses

Figure 6.4 B

Ak 2Rk+1 2R2

A3 A2 A1 2R3

2R1 А

(b) Let the sides of the polygons circumscribed around the circle with radius r be ... a1r0 a2r0 ... anr0 equal to a1, a2, , an, and the area to S. Then, S 2 þ 2 þ þ 2 . 2S Hence, r0 ¼ r and the equality holds true, if and only if the circle with a1þ...þan radius r0 is the incircle of the given polygon.

6.1.9. Let O1, O2, and O be circles with radiuses r1, r2, and R, respectively. Consider a point symmetric to point O1 with respect to the straight line AB. 0 ∠ 0 ∠ ∠AMC Let that point is O1. Since O1MA ¼ O1MA ¼ 2 and ∠ ∠BMD , ∠ ∠ ∠ 0 ∠ O2MB ¼ 2 AMC ¼ BMD, we have that O1MA ¼ O2MB.This 0 means that points O1, M, O2 are on the same line. According to the triangle 0 0 inequality, we have that O1O2 O1O þ OO2. (a) Note that O0 O ¼ O O ¼ R À r , OO ¼ R À r and ∠O MA ¼ ∠O MB 45, 1 1 pffiffiffi 1 2 2 p1ffiffiffi 2 r1 r2 thus O1M ¼ 2r1 and O2M ¼ 2r2. sin ∠O1MA sin ∠O2MB pffiffiffi pffiffiffi To end the proof we have to use the obtained inequalities: 2r1 þ 2r2 O1 M þ O M ¼ O0 M þ O M ¼ O0 O O0 O þ OO ¼ R À r þ R À r : 2 1 2 1 ÀÁ2pffiffiffi 1 2 1 2 Therefore, r þ r p2ffiffiR ¼ 2 2 À 1 R. 1 2 2þ1 (b) We have that

r1 r2 0 0 0 α þ α ¼ O1M þ O2M ¼ O1M þ O2M ¼ O1O2 O1O þ OO2 ¼ sin sin 2 2 ¼ O1O þ O2O ¼ R À r1 þ R À r2:

α 2 Thus, it follows that r1 þ r2 2 sin α . sin 2þ1R 6.1.10. Let us deduce (recall) the formula expressing the inradius of the triangle through its angles and the circumradius. Let α, β, γ be the angles of triangle ABC and I, R be the incenter and the circumradius, respectively (see Figure 6.5). 6.1 Inequalities for Radiuses of Circles 267

Figure 6.5 B

a/2 g/2 AC

We have that γ α AC sin α 2R sin β γ α r ¼ AI sin ¼ 2 sin ¼ sin sin ¼ 2 α þ γ 2 β 2 2 sin cos 2 2 α β γ α À β α þ β γ ¼ 4R sin sin sin ¼ 2R cos À cos sin ¼ 2 2 2 2 2 2 α þ γ À β β þ γ À α α þ β þ γ α þ β À γ ¼ R sin þ sin À sin þ sin ¼ 2 2 2 2 ¼ RðÞcos α þ cos β þ cos γ À 1 :

Without loss of generality, one can take R ¼ 1. If ρ is the inradius of the triangle αþβ βþγ γþα with the angles 2 , 2 , 2 , then α þ β β þ γ γ þ α ρ À r¼ cos þ cos þ cos À cosα À cosβ À cosγ ¼ 2 2 2 α þ β 1 β þ γ 1 ¼ cos À ðÞcosα þ cosβ þ cos À ðÞcosβ þ cosγ þ 2 2 2 2 ð6:1Þ γ þ α 1 þ cos À ðÞcosγ þ cosα ¼ 2 2 α þ β α À β β þ γ β À γ α þ γ α À γ ¼ 2cos sin 2 þ 2cos sin 2 þ 2cos sin 2 : 2 4 2 4 2 4

Let us now continue the proof of our problem. The part (a) follows immediately from (6.1), since all summands in it are nonnegative and from triangle AiBiCi with the angles αi, βi, γi by a transition α β β γ γ α iþ i iþ i iþ i described in the problem we get a triangle with the angles equal to 2 , 2 , 2 . (b) Since for any initial triangle the second triangle is acute-angled, we consider an acute triangle with angles α, β, γ. Hence, we get a triangle with the angles βþγ π α γþα π β αþβ π γ 2 ¼ 2 À 2 , 2 ¼ 2 À 2 , 2 ¼ 2 À 2 . This triangle, in turn, will transform into a triangle with the angles equal to π βþγ π α π γþα π β π αþβ π γ 2 À 4 ¼ 4 þ 4 , 2 À 4 ¼ 4 þ 4 , 2 À 4 ¼ 4 þ 4. 268 6 Inequalities for Radiuses

Let r2, r3, r4 be the corresponding inradiuses of these triangles and α β γ < π for simplicity let us assume that 2. According to (6.1), we have that β γ β γ α γ β α r r 2 sin α sin 2 À 2 sin sin 2 À 2 sin sin 2 À , and r r 3 À ÀÁ2 ¼ 2 4 þ ÀÁ2 4 þ 2 ÀÁ4 4 À 3 ¼ π α 2 βÀγ π β 2 γÀα π γ 2 βÀα 2 sin 4 À 4 sin 8 þ 2 sin 4 À 4 sin 8 þ 2 sin 4 À 4 sin 8 . The last equality can be obtained from the previous one substituting α, β, γ by βþγ γþα αþβ 2 , 2 , 2 and by a simple transformation. We have that β À γ α β À γ 1 π α 2r À r À r ¼ ðÞÀr À r ðÞ¼r À r 8sin 2 sin cos 2 À sin À þ 3 2 4 3 2 4 3 8 2 8 4 4 4 γ À α β γ À α 1 π β þ 8sin 2 sin cos 2 À sin À þ 8 2 8 4 4 4 β À α γ β À α 1 π γ þ 8sin 2 sin cos 2 À sin À : 8 2 8 4 4 4

α π If 6, then each summand in the right-hand side is nonnegative, α β À γ 1 π α α π 1 π α sin cos 2 À sin À > sin cos 2 À sin À 2 8 4 4 4 2 16 4 4 4 π π 1 π π 1 π π sin cos 2 À sin À ¼ 2sin À sin > 0 12 16 4 4 24 4 12 24

Thus, it follows that, 2r3 À r2 À r4 0. α < π π < β γ < π β α > π γ α > π γ β < π Let 6, then 3 2, and as À 6, À 6,0 À 6,we obtain that γ β α β γ π α 2 À 2 À 1 2r3 À r2 À r4 > 8sin sin cos À sin À þ 8 2 8 4 4 4 π β γ À α 1 π β þ8sin 2 sin cos 2 À sin À þ 482 8 4 4 4 π γ β À α 1 π γ þ8sin 2 sin cos 2 À sin À > 48 2 8 4 44 γ À β 1 π π π γ À α 1 π > 8sin 2 À sin þ 8sin 2 sin cos 2 À sin þ 8 4 4 48 6 8 4 6 π π β À α 1 π π 1 π þ8sin 2 sin cos 2 À sin > 8sin 2 À sin þ 48 6 8 4 6 48 4 4 π 1 γ À α β À α 1 þ8sin 2 cos 2 þ cos 2 À ¼ 48 02 8 8 4 1 γ À α β À α pffiffiffi π B1 þ cos 1 þ cos 1 2C ¼ 8sin 2 @ 4 þ 4 À À A ¼ 48 4 4 4 8 pffiffiffi π 2 À 2 1 γ À α 1 β À α ¼ 8sin 2 þ cos þ cos > 0, 48 8 4 4 4 4

Therefore, 2r3 À r2 À r4 > 0. This ends the proof. 6.1 Inequalities for Radiuses of Circles 269

Remark If α ¼ 0, β ¼ 0, γ ¼ π, then

α β À γ β γ À α γ β À α 2 sin sin 2 þ 2 sin sin 2 þ 2 sin sin 2 < 2 4 2 4 2 4 π α β À γ π β γ À α < 2 sin À sin 2 þ 2 sin À sin 2 4 4 8 4 4 8 π γ β À α þ 2 sin À sin 2 : 4 4 8

AC BC 6.1.11. Note that R1 2 and R2 2 . ACþBC > AOþBO Therefore, R1 þ R2 2 2 ¼ R, where O is the circumcenter of triangle ABC (see problem 1.1.8a). 6.1.12. (a) Let O be the intersection point of segments AC and BD. It is clear that one of the angles ∠AOB and ∠BOC is not less than 90, and we may consider that ∠BOC 90 (otherwise A and C can be interchanged). Let us denote ∠DAO ¼ α, ∠CBO ¼ β, ∠BCO ¼ γ and ∠ADO ¼ δ, then the angles α, β, γ, δ are acute. By the law of sines we have that, 2RD sin α ¼ CD ¼ 2RC sin β and 2RA sin δ ¼ AB ¼ 2RB sin γ. Therefore, sin γ sin α RA þ RC ¼ RB sin δ þ RD sin β. If ∠A + ∠C > ∠B + ∠D, we have that ∠A + ∠C > 180. This means that point A is inside of the circumcircle of triangle BCD. Hence, it follows that δ < γ and β < α > sin δ sin β . Thus, RA þ RC RB sin δ þ RD sin β ¼ RB þ RD. If ∠A + ∠C < ∠B + ∠D, then ∠B + ∠D > ∠A + ∠C. Hence, RB + RD > RA + RC or RA + RC < RB + RD. Therefore, if RA + RC > RB + RD, then ∠A + ∠C < ∠B + ∠D does not hold true. For ∠A + ∠C ¼ ∠B + ∠D, we have that ABCD is an inscribed quadrilateral, but then RA ¼ RB ¼ RC ¼ RD. Thus, RA + RC ¼ RB + RD. This leads to contradiction. Thus, it follows that ∠A + ∠C > ∠B + ∠D. (b) We need to prove that, if ∠B + ∠D > 180 , then rA + rC > rB + rD.

Denote by OA, OB, OC, and OD the incenters of triangles DAB, ABC, BCD, and CDA, respectively. At ﬁrst, let us prove the following statements.

1. OAOBOCOD is a convex quadrilateral. Note that

1 1 ∠ABO ¼ ∠ABD < ∠ABC ¼ ∠ABO and ∠BAO < ∠BAO : A 2 2 B B A

Therefore, segments AOB and BOA intersect. Denote their intersection point by M (Figure 6.6) and the intersection point of segments BOC and COB by N. 270 6 Inequalities for Radiuses

Figure 6.6 B

OB N M

O OA C AC

We have that

1 ∠AO B ¼ 90 þ ∠ACB: ð6:2Þ B 2

Therefore, ∠AOBB > 90 . Similarly, ∠BOBC > 90 and ∠AOBC > 90 , this means that point OB is inside triangle BMN. Thus, it is also inside triangle OABOC. Exactly in the same way, we can prove that point OD is inside triangle OADOC. Since the quadrilateral OABOCD is convex, we obtain that the quadrilateral OAOBOCOD is convex too. 2. If ∠B + ∠D ¼ 180 , then ∠OAOBOC ¼ 90 . Indeed, since ∠B + ∠D ¼ 180, then one can circumscribe a circle around the quadrilateral ABCD. Therefore, ∠ADB ¼ ∠ACB. Taking this into consideration and by (6.2) we deduce that ∠AOAB ¼ ∠AOBB. This means that, points A, B, OB,andOA ∠ ∠ 1 ∠ are on the same circle. Thus, BOBOA ¼ 180 À BAOA ¼ 180 À 2 BAD. Simi- ∠ 1 ∠ larly, we obtain that BOBOC ¼ 180 À 2 BCD. Hence, it follows that ∠ ∠ 1 ∠ ∠ ∠ BOBOA þ BOBOC ¼ 360 À 2 ðÞ¼A þ C 270 .Thus, OAOBOC ¼ 90 . Corollary If ∠B + ∠D ¼ 180 , then OAOBOCOD is a rectangle. 3. If ∠B + ∠D > 180 , then ∠OAOBOC > 90 . Since ∠B + ∠D > 180, then point D is inside the circumcircle of the triangle ABC. Let the straight line BD intersects that circle at point D0 (Figure 6.7). 0 0 0 0 Denote the incenters of triangles ABD and CBD by OA and OC, respectively. As ∠ 0 > ∠ 0 BAO A BAOA, then point OA is on segment OAB. 6.1 Inequalities for Radiuses of Circles 271

Figure 6.7 B

OA C

AO¢A O¢C

D¢

0 Similarly, point OC is on segment OCB. By the statement 2, we have that ∠ 0 0 ∠ > ∠ 0 0 OAOBOC ¼ 90 . Therefore, OAOBOC OAOBOC ¼ 90 . Corollary

If ∠B þ ∠D > 180 , then OBOD < OAOC: ð6:3Þ

Indeed, we have that ∠OAOBOC > 90 and ∠OAODOC > 90 . Therefore, points OB and OD are inside a circle with diameter OAOC. Thus, OBOD < OAOC. Let the incircle of triangle DAB touches side BD at point A1 (points B1, C1 and D1 are deﬁned similarly).

4. Prove that A1C1 ¼ B1D1. We have that,

AD þ BD À AB BD þ CD À BC A1C1 ¼ jjDA1 À DC1 ¼ À ¼ 2 2 ð6:4Þ AD þ BC À AB À CD ¼ ¼ B1D1: 2

Let us now continue the proof of the problem. Note that,

2 2 2 2 2 2: : OAOC ¼ A1C1 þ ðÞrA þ rC and OBOD ¼ B1D1 þ ðÞrB þ rD ð6 5Þ

If ∠B + ∠D > 180, then from (6.3), (6.4), and (6.5) we deduce that rA + rC > rB + rD. If ∠B + ∠D ¼ 180, then using the corollaries of the statements 2, 3, and 4, we obtain that rA + rC ¼ rB + rD. If ∠B + ∠D < 180 , then ∠A + ∠C > 180 . Therefore, rB + rD > rA + rC. 272 6 Inequalities for Radiuses

Let rA + rC > rB + rD,thenaswehavealreadyproven,∠A + ∠C > 180 or ∠A + ∠C ¼ 180 are impossible. This means that ∠A + ∠C < 180.Hence,we obtain that ∠A + ∠C < ∠B + ∠D. Let ∠A + ∠C < ∠B + ∠D, then ∠A + ∠C < 180 . Therefore, rA + rC > rB + rD.

6.1.13. Let r1, r2, r3 be the inradiuses of triangles AB1C1, BA1C1, CA1B1, respectively. Note that π π π ∠AC1B1 ∠AB1C1 6 ¼ 2r1ctg þ 2r2ctg þ 2r3ctg þ r1 ctg þ ctg þ 6 6 6 2 2 ∠BC A ∠BA C ∠CA B ∠CB A þ r ctg 1 1 þ ctg 1 1 þ r ctg 1 1 þ ctg 1 1 : 2 2 2 3 2 2 ð6:6Þ

x þ y ÀÃ According to inequality ctgx þ ctgy 2ctg , where x, y2 0; π (see the 2 2 π π proof of problem 6.1.10). From (6.6) we deduce that 6 2r1ctg 6 þ 2r2ctg 6 þ 2r3 π π π π π ctg þ 2r1ctg þ 2r2ctg þ 2r3ctg 4ctg ðÞr1 þ r2 þ r3 . Thus, it follows that 6 6 pffiffi 6 6 6 3 r1 þ r2 þ r3 2 . If A1, B1, C1 are the midpoints of sides BC, AC, AB, respectively, then pffiffi 3 r1 þ r2 þ r3 ¼ . 2 pffiffi 3 Thus, the greatest value of the sum of radiuses r1, r2, r3 is equal to 2 . π π π 6.1.14. We have that rn ¼ cos . We have to prove that cos cos 2 , for n ¼ 3, Rn n pffiffiffi n nþ1 ... 2 π 2 π π π 4, 5, , or sin nþ1 2sin 2n, sin nþ1 2 sin 2n. sin α sin ðÞα À β α À β α If 0 < β < α < π, then ¼ þ cos ðÞα À β < þ 1 ¼ . 2 sin β tgβ β β pffiffiffi π π π π 4 2 2 Also, for n 3, nþ1 4. Hence, sin π π . Therefore, sin . 4 n þ 1 n þ 1 sin π n þ 1 nþ1 π nþ1 pffiffi π π > π Then, for n 4 we have that sin 2 2 ,as(4À )n (4 À )4 . π nþ1 sin > π pffiffi 2n > 2 2n π nþ1 While, for n ¼ 3, we have that sin π . π 4 ffiffi sin ¼ sin π p 2n sin ¼ 2 ffiffiffi 6 π p π Thus, for n 3, we obtain that sin nþ1 2 sin 2n. ∠ 6.1.15. (a) We have that sin A ¼ rÀrA, according to problem 5.1.5, it follows 2 rþrA ∠ ∠ ∠ sin A þ sin B þ sin C 3. Therefore, rÀrA þ rÀrB þ rÀrC 3. Hence, r þ 2 2 2 2 rþrA rþrB rÀÁþrC 2 rþrA r þ r 9. Note that 9 r þ r þ r rþrA þ rþrB þ rþrC 9 Á rþrB rþrC 4 rþrA rþrB rþrC r r r 4 3rþrAþrBþrC r . Thus rA + rB + rC r. 6.1 Inequalities for Radiuses of Circles 273

∠A 1À sin 2 2 πÀ∠A (b) rA ¼ r ∠A ¼ r Á tg 4 . Therefore, 1þ sin 2 rArBþ rBrC þ rCrA ¼ π À ∠A π À ∠B π À ∠B π À ∠C π À ∠C π À ∠A ¼ r2 tg 2 tg 2 þ tg 2 tg 2 þ tg 2 tg 2 4 4 4 4 4 4 r2 π À ∠A π À ∠B π À ∠B π À ∠C π À ∠C π À ∠A 2 tg tg þ tg tg þ tg tg ¼ 3 4 4 4 4 4 4 r2 π À ∠B π À ∠A π À ∠C π À ∠A π À ∠C ¼ tg tg þ 1 À tg tg þ 3 4 4 4 4 4 Á 2 2 r þtg πÀ∠C tg πÀ∠A ¼ : 4 4 3

6.1.16. Let the center O of the insphere of the tetrahedron ABCD be inside of 1 ρ ; 1 ρ ; it. Then, V ¼ 3 SBDC Á ðÞA BDC 3 SBDC Á ðÞR þ ðÞO BDC . We need to obtain similar inequalities for other three faces of the tetrahedron. 1 Summing up all these four inequalities, we deduce that 4V 3 SR þ V, and as 1 V ¼ 3 SR, then it follows that R 3r. Now, let points O and A are not at the same side of the plane BCD. Then, ρ(A, BCD) R. Let the plane passing through the center of the insphere of the tetrahedron ABCD and parallel to the plane BCD intersects the edges AB, AC, AD at points B0, C0, and D0, respectively. 0 0 0 0 Denote the circumcenters of triangles B C D and BCD by O and O1, respec- ρ ; 0 0 tively. As O0D0 > 2r and ðÞÀA BDC r ¼ O D > 2r , then 1 > r þ 2r r þ 2r. ρðÞA;BDC O1D O1D ρðÞA;BDC O1D R R Thus, it follows that R > 3r. This ends the proof.

Problems for Self-Study

6.1.17. Prove that for any triangle ABC ρ ρ ρ 3 ρ ρ ρ (a) r a þ b þ c 4 R, where a, b, c are the radiuses of three circles, inscribed into the segments cut of the circumcircle K of triangle ABC by the sides of triangle BC, CA, AB (circles touching the circumcircle and one of the sides of the triangle at the midpoint (see Figure 6.8)). 4 16 2 8 64 3 (b) 4r pa, pb, pc 3 ðÞR þ r , 3 r papb þ pbpc þ pcpa 3 Rr, 27 r papbpc 32 2 27 Rr , where pa, pb, pc are the radiuses of three circles touching two sides of the triangle (not their extensions) and the circumcircle of the triangle (see Figure 6.9). 274 6 Inequalities for Radiuses

Figure 6.8 A

Figure 6.9 A

Hint

bcr (b) Prove that pa ¼ ppðÞÀa . 6.1.18. Let a quadrilateral be inscribed into a unit square, such that its sides are the hypotenuses of the triangles. Given that in each triangle is inscribed a circle. Prove pffiffiffi that the sum of the radiuses of these circles does not exceed 2 À 2 and is equal to pffiffiffi 2 À 2, only if the sides of the inscribed quadrilateral are parallel to the diagonals of the square. Hint See problem 1.1.10. 6.1.19. Prove that the inradius of any face of the tetrahedron is greater than the radius of its insphere. 6.1.20. Prove that for any triangle ABC 6.2 Integer Lattice 275

(a) 1 ¼ 1 þ 1 þ 1 2, r ra pffiffiffiffiffiffiffiffirb rc pffiffiffiffiffiffiffiffiR pffiffiffiffiffiffiffiffi (b) 9r rarb þ rbrc þ rcra r þ 4R, 2 2 2 2 (c) ra þ rb þ rc 27r , (d) 4R < ra + rb + rc 4, 5R.

180 6.1.21. Prove that cos n Á R r, where R is the circumradius and r is the inradius of any convex n-gon. 6.1.22. Given a point B on segment AC. Points M and N are such that AM ¼ MB, BN ¼ NC and MN||AC. Let r, r1 and r2 be, respectively, the inradiuses of triangles MBN, AMB, and BNC. Prove that, if r1 < r2, then r1 < r < r2.

Hint Prove that 2SMBN ¼ SAMB + SBNC and 2PMBN ¼ PAMB + PBNC.

6.2 Integer Lattice

6.2.1. An integer lattice is the set of all points (knots) that have integer coordinates (in a rectangular coordinate system). Prove that, if the vertices of a parallelogram coincide with the lattice points of an integer lattice and inside of the parallelogram or on its border there are other lattice points, then the area of such parallelogram is greater than 1. 6.2.2. The vertices of a convex pentagon are in the integer lattice. Prove that, the area of the pentagon is not less than 2.5. 6.2.3. (a) The vertices of the quadrilateral ABCD are in the integer lattice. Given that the angles A and C of the quadrilateral are equal, and the angles B and D are not equal. Prove that |AB Á BC À CD Á DA| 1. (b) Vertices of the inscribed quadrilateral ABCD are in the integer lattice. Given that ABCD is neither a trapezoid nor a rectangle. Prove that |AC Á AD À BC Á BD| 1. 6.2.4. Given on a plane a circle with the radius r and with the center at the origin of the rectangular coordinate system. Let δ(r) be the distance from the nearest point δ < p2ffiffi with the integer coordinates to that circle. Prove that ðÞr r. (The distance between the point on a plane and a circle is defined as follows: draw a line through the given point and the center of the circle, then find the distance between the point and the nearest intersection point of the drawn line with the circle). 6.2.5. Given a figure, with the area smaller than 1, on a grid plane having a grid size (length) equal to 1. Prove that, it is possible to translate this figure on the grid plane, such that none of the grid points (knots) is inside of it. 6.2.6. (a) Vertices of an acute triangle ABC are in the integer lattice. Given that pffiffiffiffiffi minðÞAB; BC; AC 65. Prove that SABC 30. 276 6 Inequalities for Radiuses

(b) Vertices of an acute triangle ABC are in the integer lattice. Given that min(AB, BC, AC) 2000. Prove that inside of that triangle there is a lattice point, such that either the difference or the sum of its coordinates is divisible by 2000. 6.2.7. Given on a coordinate plane a square S with dimensions n Â n. Prove that, for any disposition of the square it covers not more than (n +1)2 integer points (both x and y coordinates are integer numbers), where n 2 N. 6.2.8. Prove that any convex integer polygon (a polygon with the integer coordinates of all vertices) with area S can be placed into a integer parallelogram with area 4S. 6.2.9. Let each point of a plane with the integer coordinates be the center of a disk 1 with the radius 1000. Prove that (a) there exists an equilateral triangle, such that all the three vertices are in different disks, (b) the side of any equilateral triangle, such that all vertices are in different disks, is greater than 100. 6.2.10. Consider on a plane nonself-intersecting polygons, such that their all sides are expressed by integers, and angles are either right or are equal to 270. Let S be the area of one of these polygons. Prove that 2n À 1 S n2, if all polygons have the same given perimeter equal to 4n. 6.2.11. Let the vertices of a triangle be in the knots of the integer lattice. Given that inside of the triangle there are n lattice points, where n 2 N. What is the greatest number of the lattice points that can be on the boundaries of the triangle (vertices included)?

Solutions

6.2.1. We need to prove that, if the vertices of the triangle coincide with the lattice 1 points of the integer lattice, then the area of the triangle is not smaller than 2. Indeed, any triangle can be built up to a rectangle by adding rectangular triangles (Figure 6.10). As the area of the obtained rectangle is integer, and the area of every right n 1 triangle has a form 2, where n 2 N, then SABC 2. Let the vertices of the parallelogram ABCD coincide with the lattice points and the lattice point M be inside or on the boundary of the parallelogram ABCD. As SABCD ¼ SAMB + SBMC + SCMD + SDMA and three of the numbers SAMB, SBMC, 1 SCMD, SDMA are not smaller than 2, then SABCD 1.5. b Remark One can prove that SABC ¼ c þ 2 À 1, where c is the number of the lattice points inside of triangle ABC, аnd b is the number of the lattice points on the boundary of triangle ABC. 6.2 Integer Lattice 277

Figure 6.10 B y

C C

B A

Figure 6.11 B M C A

D E

(Hint. Check that the proved statement holds true for a rectangle, with the side passing along the lines of the grid and for right triangle, with the legs passing along the lines of the grid). 6.2.2. It is not difficult to note that, if the vertices of the pentagon are in the lattice points, then for some two vertices A and B, the corresponding coordinates have the same parity. Hence, the midpoint of segment AB is a lattice point. This means that, if the vertices of convex pentagon ABCDE are in the lattice points, then there exists a lattice point M, inside or on the boundary of pentagon ABCDE (see Figure 6.11). In the first case, we have that 1 : SABCDE ¼ SAMB þ SBMC þ SCMD þ SDME þ SMEA 5 Á 2 ¼ 2 5 (see the proof of problem 6.2.1), and in the second case, we obtain that SABCDE 2. Since the pentagon MBCDE (see Figure 6.11) satisfies the conditions of the problem, then, as we have proven, SMBCDE 2. Thus, SABCDE ¼ SAME + SMBCDE 0.5 + 2 ¼ 2.5. 6.2.3. (a) Consider a parallelogram ABA0D (Figure 6.12). Since points A, B, and D are the lattice points, then point A0 is also a lattice point. It is clear that points C and A0 differ (∠B 6¼ ∠D). We have that ∠BCD ¼ ∠BA0D. Hence, points B, C, A0, and D are on one circle. According to problem 1.2.9a, we have that BA0 Á CD ¼ CB Á A0D + CA0 Á BD or BC Á A0D ¼ BA0 Á CD + CA0 Á BD. Thus, | AB Á BC À CD Á AD| ¼ CA0 Á BD 1, as CA0 1 and BD 1. 278 6 Inequalities for Radiuses

Figure 6.12 C A¢

B D

(b) Note that ∠CBD ¼ ∠CAD and let ∠CAD ¼ α. We have that, jAC Á ADÀ 2 BC Á BDj¼ sin α jjSACD À SBCD 2jjSACD À SBCD .IfSACD ¼ SBCD, then 1 AB k CD. This leads to contradiction. Hence, jjSACD À SBCD 2 (see the proof of problem 6.2.1). Therefore, |AC Á AD À BC Á BD| 1.

6.2.4. It is clear that δðÞr 1, and if 1 < p2ffiffi. Then, this ends the proof. 2 2 rqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Let r 16. Note that the lattice point Ar½; r2 À ½r 2 þ 1 is outside that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 circle. Indeed, ½r 2 þ r2 À ½r 2 þ 1 > ½r 2 þ r2 À ½r 2 ¼ r2. Hence, OA > r, where O is the origin of the coordinate system. Obviously, δ(r) OA À r. We need to estimate the value of OA À r. We have that, OA2Àr2 < OA2Àr2 OA À r ¼ OAþr 2r . Note that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 OA2 À r2¼ 1 þ 2 r2 À ½r 2 þ r2 À ½r 2 À r2 À ½r 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffi 1 þ 2 r2 À ½r 2 1 þ 2 r2 À ½r 2 ¼ 1 þ 2 fgr ðÞr þ ½r < 1 þ 2 2r < 1 þ 3 r:

ffiffi p pffiffi Hence, OA À r < 1þ3 r. Since for r 16 we have that 1 < r, then pffiffi 2r 1 3 r < þ < p2ffiffi < p2ffiffi δ < p2ffiffi OA À r 2r r. Therefore OA À r r, and hence ðÞr r. 6.2.5. Let the points of this figure are red. We attach the figure to the transparent grid paper, cut the paper into the cells and put them in a pile, translating them parallelly, but without turning, such that one cell remains in its place. The red points cannot cover completely the cell since the area of the red parts is smaller than that of the cell. Now, let us recall how the figure was located on the grid paper, and shift parallelly only the grid paper such that one of its lattice points gets into a not red point from a pile. Therefore, we obtain the required translation (location) of the figure. 6.2 Integer Lattice 279

B «mid.vert» B «mid.vert» N

m m ¢ b a h a A h-1 M b AM «mid.horiz» n¢ n «mid.horiz» n K mC C «mid.vert» ab

Figure 6.13

6.2.6. Consider three vertical lines of the integer lattice passing through the vertices A, B, C. One of these lines is between the other two. A vertex of the triangle is called a “mid-vertical,” if it belongs to this line. Similarly, we deﬁne the “mid-horizontal” vertex of triangle AB C. Since AB C is an acute triangle, then none of its vertices can be simultaneously “mid-vertical” and “mid-horizontal,” Consider two cases (see Figure 6.13a, b). (a) If in the case I (Figure 6.13a), h 6. Since h2 + m2 65, then it follows that m, n 6. Hence α 45, β 45. Therefore, ∠A 90. This leads to a contradiction. Hence, h 7, and since m + n 9, then SABC 31.5. In case II (Figure 6.13b), one can assume that a b.Ifa 7, then ab an bm 7Á7 7Á1 7Á1 : SABC ¼ 2 þ 2 þ 2 2 þ 2 þ 2 ¼ 31 5. 2 2 Let a 6 and b0 is the minimal positive integer, such that a þ b0 65. It is clear that a b0 b. Let n0 be a minimal nonnegative integer, such that b2 þ n2 65. The vertex C is on the marked part of the plane (Figure 6.14), hence 0 0 ÀÁ ab bbðÞoÀa ano; ab bbðÞoÀaþ1 aÁ1 bboþano ; bboþaþb S min 2 þ 2 þ 2 2 þ 2 þ 2 ¼ min 2 2 . Hence, we deduce that at a 3, bo 8 and SABC 32. ÀÁ 65; 60 For a ¼ 4, we obtain that bo ¼ 7, no ¼ 4, thus SABC min 2ÀÁ2 ¼ 30. 69; 61 : For a ¼ 5, we obtain that bo ¼ 7, no ¼ 4, so that SABC min 2 2 ¼ 30 5. 42 7Á2 6Á1 > For a ¼ 6, b 7, we have that SABC 2 þ 2 þ 2 30. 36 6Á4 For a ¼ 6, b ¼ 6, we have that SABC 2 þ 2 ¼ 30. (b) In the case of Figure 6.13b, we have that a2 + b2 20002. Since a b, then (a + b)2 > b(a + b) 20002. Hence, a + b 2001. This means that on the broken 280 6 Inequalities for Radiuses

Figure 6.14 B

b0-a A a a

line AMB there are not less than 2000 lattice points (apart from A and B), then the sum of the coordinates of these lattice points accepts the values S, S +1, ..., S + 1999. Hence, one of them is divisible by 2000 (for other disposition of the broken line AMB it can be necessary to consider the difference of the coordinates). In the case of Figure 6.13a, either α < 45,orβ < 45. Let α < 45, then h > m and h + m 2001 (see the proof in the case of 0 hÀ1 > 0 = Figure 6.13b). Since m ¼ h m m À 1 and m 2 N, then inside of the triangle there are at least h + m À 2 lattice points (on the broken line AMN), for h + m 2002 the proof of the problem is straightforward. While, for h + m ¼ 2001, if m > 1, then h 1999 and h2 + m2 < h(h + m) 1999 Á 2001 < 20002. This leads to contradiction. 0 hÀ1 > If h + m ¼ 2001 and m ¼ 1, then h ¼ 2000, n 1999 and n ¼ h n 1. Thus, the number of lattice points located on segment AM is equal to 1999. Let us consider one more lattice point on segment MK (n0 > 1), then the difference of the coordinates of one of these lattice points (their total number is 2000) is divisible by 2000. This ends the proof. 6.2.7. Let M be the set of all lattice points covered by square S, and H be the convex envelope of the set M. Since H is contained in S, then its area cannot exceed the area 2 of square S. Therefore, SH n . By Pick’s theorem (see the remark to the proof of problem 6.2.1) SH ¼ c þ b 2 b 2 2 À 1 n and c þ 2 n þ 1. According to problem 2.1.1, the perimeter H cannot exceed the perimeter of S. Thus, it follows that PH 4n. It is clear that, the distance between no two lattice points can be less than 1. Hence, the number of the lattice points on the boundaryÀÁH cannot be more than 4n.Thus,we b b 2 2 obtain that b 4n. Therefore, c þ b ¼ c þ 2 þ 2 n þ 1 þ 2n ¼ ðÞn þ 1 . 6.2 Integer Lattice 281

C2 CC1

D2 DD1

Figure 6.15

6.2.8. Let A and B be the vertices of the given polygon, such that the distance between them is the greatest, and C,D be the most distant vertices of the polygon from the straight line AB and are located at different sides of this line. Construct parallelograms ABC1C, ABCC2, ABD1D, and ABDD2 (Figure 6.15). We need to prove that C1C2D2D1 is the required parallelogram. Indeed, since C1C2 ¼ 2AB ¼ D1D2 and C1C2 k AB k D1D2, then quadrilateral C1C2D2D1 is a parallelogram. It is clear that points C1, D1, D2, C2 are integer lattice points (as A, B, C, D are lattice points).

We have that SC1C2D2D1 ¼ 4ðÞSABC þ SABD 4S. Finally, the given polygon is located in the union of the trapezoids ABC1C2 and ABD1D2. Otherwise, if the vertex E of the polygon is in the strip with boundaries AB and C1C2, but not inside the trapezoid ABC1C2 (Figure 6.15), then SACE > SABC and ρ(E, AB) > ρ(C, AB). This leads to contradiction. pffiffiffi < 1 6.2.9. (a) Note that there exist such natural numbers m and n, that 3n À m 1000. Indeed, (2;1) is a solution of the equation

x2 À 3y2 ¼ 1: ð6:7Þ

If (x0, y0) is a solution of this equation, then (2x0 +3y0; x0 +2y0) is also a solution. > Consequently, ffiffiffi if the Eq. 6.7 has a solution (m, n), where m, n 2 Nm, n 1000,ÀÁ thenffiffiffi p p 3n À m ¼ pffiffi 1 < 1 < 1 and triangle with vertices (0; 0); (2n; 0); n; 3n 3nþm m 1000 satisfies the conditions of the problem. (b) Proof by contradiction argument. Let A, B, C be the centers of three circles containing the vertices of equilateral triangle with the side length no greater than 100. ÀÁ 2 2 4 4 < Then, AB À BC ¼ jjAB À BC Á jjAB þ BC 1000 Á 2 Á 100 þ 1000 1 (see problem 1.2.1). Therefore, AB ¼ BC. Similarly, we obtain that AB ¼ AC. This leads pffiffi 3 2 to contradiction, as SABC ¼ 4 AB must be rational (see the proof of problem 6.2.1). 6.2.10. From the condition of the problem, it follows that, one can assume that the sides of the polygon are on the lines of the integer lattice, and vertices are in the 282 6 Inequalities for Radiuses

b lattice points. It is known that S ¼ c þ 2 À 1, where c is the number of the lattice points inside the polygon and b the number of the lattice points on its boundaries. Since, c 0, b ¼ 4n, then S 2n À 1. Let us consider the smallest rectangle containing the polygon with area S, the sides (passing) on the lines of the integer lattice . Let k and m be the sides of the constructedÀÁ rectangle. It is clear that 2k +2m 4n, consequently, mþk 2 2 2 S mk 2 n . Thus, it follows that S n . This ends the proof. 6.2.11. We need to prove that the maximal number of the lattice points that can be on the boundaries of the triangle (vertices included) is equal to 9 for n ¼ 1 and to 2n + 6, for n 2 (see Figure 6.16) Let A, B, C be the lattice points and the number of the lattice points inside triangle ABC is equal to n, while those on sides AB, BC, and CA (and different from the vertices of triangle ABC ) are k, l, and p, respectively, where k l p. Note that if X, Y, Z are the lattice points, then point T, where XY~ ¼ ZT~ , is also a lattice point. Let l 1 and the lattice points A1, A2, ..., Al are on side BC, then from the ... aforesaid it follows that BA1 ¼ A1A2 ¼ ¼ Al À 1Al ¼ AlC. Similarly, if points B1, B2, ..., Bp are on side CA, then CB1 ¼ B1B2 ¼ ... ¼ Bp À 1Bp ¼ BpA (B0 C). Let CADB be a parallelogram, then we deduce that (see Figure 6.17)

A B

12…n BCA C n=1 n ³ 2

Figure 6.16

Figure 6.17 D

A Bp Al

B2 B1 C 6.2 Integer Lattice 283

Figure 6.18 B

E F

U M V

A C

lp 2n þ k: ð6:8Þ

For k ¼ 0, we have that k + l + p ¼ l + p ¼ lp +1À (l À 1)( p À 1) lp +1 2n +1. Hence, k + l + p +3< 2n +6. For k 1, we have that (l À k)( p À k) 0, consequently

lp þ k2 2n k þ l þ p þ 3 þ k þ 3 þ 2k þ 4: ð6:9Þ k k

If n ¼ 1, then from (6.8), we obtain that k2 lp 2+k. Therefore, k ¼ 1ork ¼ 2, then from (6.9), we deduce that l + p + k +3 9. If n > 1, then from (6.8) we obtain that k(k À 1) 2n.Thusk n, we deduce that 2n l þ p þ k þ 3 k þ 2k þ 4 2n þ 6. Case l ¼ 0 remains to be considered. In this case, we have that k ¼ 0. Let EF be the midline of triangle ABC (see Figure 6.18), and point M is the closest lattice point to side AC. Then, point M belongs to the trapezoid AEFC.Otherwise,pointB0,whereB0 2 BM and BM ¼ MB0 is a lattice point and is closer to AC than point M.LetUV k AC and CB1 ¼ a, then the number of the lattice points on segment UV is not greater than n,and UV < (n +1)a. Thus, it follows that AC ¼ 2EF 2UV < (2n +2)a. On the other hand, AC ¼ ( p +1)a, this means that p < 2n + 1. Therefore, l + p + k +3¼ p +3< 2n +6.

Problems for Self-Study

6.2.12. What is the least perimeter of a convex 32-gon, such that its all vertices are in the knots of a unit grid paper? 6.2.13. A nonempty set of the lattice points is marked on the integer lattice. Given also the ﬁnal set of nonzero vectors with the integer coordinates. Given that, if one takes the origin of all given vectors at any of the marked lattice points, then among their endpoints there are more marked lattice points than unmarked ones. Prove that the number of the marked lattice points is inﬁnite. 284 6 Inequalities for Radiuses

6.2.14. Prove that, if points A, B, C coincide with the knots of the integer lattice and ÀÁpffiffiffi 2 ÀÁpffiffiffi 2 AC À 2AB þ AC À 2BC > 0, then (AB + BC)2 À 4AB Á BC sin ∠ABC 1. Hint For points A(x; y), B(0; 0) and C(u; v), we have that AB Á BC sin ∠ABC ¼ |uy À vx|. 6.2.15. Vertices of the convex 2n-gon are in the knots of a unit integer lattice. Prove nnðÞÀ1 n3 that the area of 2n-gon is not smaller than (a) 2 , (b) 100. Hint Firstly, prove the statement of the problem for centrally symmetric polygons. 6.2.16. A triangle is placed on a coordinate plane, so that its translations by vectors with integer coordinates do not overlap. Find the greatest possible area of such triangle. Hint See problem 3.1.38. 6.2.17. The number of the lattice points inside of the convex figure with area S and semiperimeter p is n. Prove that n > S À p. Chapter 7 Miscellaneous Inequalities

In this chapter we consider problems that can be proved either by the methods described in the previous chapters or by some other methods introduced in this chapter. For example, complex numbers, the method of coordinates and application of geometric transformations are used in order to prove some inequalities. In Section 7.1, we consider some problems related to combinatorial geometry and provide many inequalities related to the parts of a triangle. Perhaps, the most important advice to deal with such problems is the following one: in order to prove inequalities related to the parts of a triangle, in some cases one needs to make a substitution a ¼ m þ n, b ¼ m þ k, c ¼ n þ k, where m > 0, n > 0, k > 0. Some problems related to the parts of a triangle can be proved considering the triangle made of the medians of the given triangle. Every inequality related to this new triangle holds true also for the given triangle. In this section we have selected the most useful inequalities of different themes related to geometric inequalities. This is the reason why Section 7.1 is much longer than the other sections. In order to prove any new inequality that is not included in Section 7.1, one can ﬁnd a similar inequality in Section 7.1 and use its method of proof. In this chapter many inequalities related to different spaces are included, such that in their proofs the following properties of a polyhedral angle are used: 1. The sum of all planar angles of a convex polyhedral angle is less than 360. 2. The sum of all planar angles of a convex polyhedral angle is greater than its any planar angle multiplied by 2. Some problems in this chapter were inspired by [1–6, 9, 11–16]. Nevertheless, even for these problems the authors have mostly provided their own solutions.

7.1 Miscellaneous Inequalities

7.1.1. Given a point M inside of triangle ABC. Straight lines AM, BM and CM intersect the sides of triangle ABC at points A1, B1 and C1, respectively. Prove that

(a) MA1 þ MB1 þ MC1 max (AA1, BB1, CC1), (b) MA1 þ MB1 þ MC1 < MA þ MB þ MC, 2 2 2 < 2 2 2 (c) MA1 þ MB1 þ MC1 MA þ MB þ MC , (d)AM þ BM þ CM 6, A1M B1M C1M (e) AM Á BM Á CM 8, A1M B1M C1M (f) AM Á BM þÁBM Á CM þ CM AM 12 A1M B1M B1M C1M C1M A1M (g) MA Á MA1 þ MB Á MB1 þ MC Á MC1 2MB1 Á MA1 þ 2MC1 Á MB1 þ 2MA1 Á MC1, (h) A1M þ B1M þ C1M 3, AM BM CM 2 ÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (i) MA þ MB þ MC 2 MA Á MB þ MB Á MC þ MC Á MA , 1 1 ÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 p1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 (j) MAþMBþMCþ2ðÞMA1þMB1þMC1 2 MAÁMBþ MBÁMCqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ MC Á MA , 2 2 2 3 2 2 2 (k)MA þMB þMC 2ðÞMA1 ÁMB1þMB1 ÁMC1þMC1ÁMA1 þ6 MA1MB1MC1, (l) MAqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÁ MB þ MBÁ MC þ MC Á MA MA1 Á MB1 þ MB1 Á MC1 þ MC1 Á MA1þ 3 2 2 2 9 MA1MB1MC1 2 2 2 2 2 2 2 2 2 2 2 (m) 2ða1MA þ b1MB þ c1MCÞ a ðb1 þ c1 À a1Þþb ða1 þ c1 À b1Þþ c ða1þ 2 2 b1 À c1Þþ16SS1, where SABC ¼ S and S1 is the area of the triangle with vertices A1, B1, C1 and sides a1, b1, c1. 7.1.2. (a) Prove that the inradius r of the right triangle is less than one-fourth of the hypotenuse. (b) Prove that the inradius r of an isosceles triangle is less than one-third of the lateral side. 7.1.3. Prove that for non-obtuse triangle

min(ha, hb, hc) r þ R max (ha, hb, hc).

7.1.4. Given a triangle ABC and bisectors AA1, BB1, CC1 of the internal angles, prove that

1 (a) A1B1 þ B1C1 þ A1C1 2 ðÞAB þ BC þ CA , 2 2 2 1 2 (b) A1B1 þ B1C1 þ A1C1 12 ðÞAB þ BC þ CA . 7.1.5. Let D, E, F be points on sides BC, CA, AB of triangle ABC, respectively and points P, Q, R are (the second) intersection points of AD, BE, CF, respectively, with AD BE CF the circumcircle of triangle ABC. Prove that PD þ QE þ RF 9 and ﬁnd when does the equality hold true. 7.1.6. Given that the side lengths of the convex hexagon ABCDEF satisfy the BC DE FA 3 conditions AB ¼ BC, CD ¼ DE, EF ¼ FA, prove that BE þ DA þ FC 2. 7.1 Miscellaneous Inequalities 287

7.1.7. Let a, b and c, d, respectively be the sides of two triangles such that a < c d < b. Prove that the first rectangle can be placed inside the second one, if and only if (b2 À a2)2 (bd À ac)2 þ (bc À ad)2. w 2 2 u 2 2 v 2 2 2 7.1.8. For triangle ABC, prove that uþv a b þ vþw b c þ uþw a c 8S , where u, v, w > 0. 7.1.9. Given two similarly oriented triangles ABC and O on a plane, such that ∠A ¼ ∠A1, ∠B ¼ ∠B1, prove that AA1 Á BC BB1 Á CA þ CC1 Á AB. 7.1.10. Place in a cube a circle of the greatest possible radius. 7.1.11. (a) Given three points inside of a unit cube, prove that one can find twopffiffiffi points among these points, such that the distance between them is less than 2. (b) Given eight points inside of a unit cube, prove that one can find two points among these points, such that the distance between them is less than 1. (c) Is it possible to place seven points inside of a cube with an edge 2000, so that the distance between any two of them is not greater than 2001? 7.1.12. Let us fix the position of unit square S on a plane. What is the maximal number of non-intersecting pairs of unit squares that can be arranged in a plane, so that they touch the given square, but do not intersect with it? (Two squares are considered intersecting, if their common part is a polygon). > 5 7.1.13. Prove that for an acute-angled triangle, mambmc 8 abc holds true.

7.1.14. Given a point M inside of triangle ABC and A1, B1, C1 as the midpoints of sides BC, AC, AB respectively, prove that

(a) max(MA, MB, MC) 2min(MA1, MB1, MC1), (b) min(MA, MB, MC) 2 max (MA1, MB1, MC1), (c) min(∠A, ∠B, ∠C) < ∠MAB þ ∠MBC þ ∠MCA < π À min (∠A, ∠B, ∠C), (d) for a point M inside of the rectangle ABCD, ∠MAB þ ∠MBC þ ∠MCD þ ∠MDA > ∠ ; ∠ π minðÞBAC DAC þ2.

7.1.15. Given a point M inside of triangle ABC and A1, B1, C1 as the feet of 2 2 2 2 2 2 the altitudes of ABCp,ffiffiffi prove that MA þ MB þ MC MA1 þ MB1 þ MC1,if maxðÞAB; BC; AC 2minðÞAB; BC; AC .

7.1.16. Given a point M inside of triangle ABC and A1, B1, C1 as the feet of the perpendiculars drawn from point M to lines BC, AC, AB respectively, prove that pffiffiffi < ∠ , ∠ , ∠ 0 (a) MA þ MB þ MC 2 2ðÞMA1 þ MB1 þ MC1 ,if A B C 45 , < 1 1 ; 1 1 ; 1 1 (b) MA þ MB þ MC max sin α þ sin β sin β þ sin γ sin γ þ sin α ðþMA1 MB1 þ MC1Þ, where α, β and γ are the angles of triangle ABC. 288 7 Miscellaneous Inequalities

7.1.17. Prove that the sum of the distances of the points inside of the tetrahedron from its vertices is less than the perimeter of the tetrahedron. 7.1.18. (a) Prove that, among 21 distances between the pairs of seven different points on a plane, no number can occur more than 12 times. (b) What is the maximal number of occurrence of the same number among 15 distances of the pairs of six different points on a plane? 7.1.19. Given that the sides of the articulated quadrilateral ABCD are equal (consequently) to a, b, c, d, what is the greatest possible value of the sum of the midlines of such quadrilateral? (The midline is a segment connecting the midpoints of the opposite sides). 7.1.20. (a) Let ABCD be a convex quadrilateral, such that its sides and diagonals do not exceed 1. Prove that the perimeter of the quadrilateral ABCD does not exceed π 2 þ 4 sin 12. (b) Let points A, B, C and D be in a space such that no more than one of the distances AB, AC, AD, BC, BD, CD is greater than 1. Prove that the sum of these six pffiffiffi distances is smaller than or equal to 5 þ 3. (c) Given that the lengths of five edges of a tetrahedron are smaller than or equal to 1 1, prove that its volume is smaller than or equal to 8.

7.1.21. Given positive numbers m1, m2,...,mn and points A1, A2,...,An (n 2) on Pn a plane. For any line l on the plane let us define ρðÞ¼l miρðÞAi; l . Let M be the i¼1 set of lines AiAj (i 6¼ j). Prove that if line l does not belong to M, then min ρðÞl0 < ρðÞl . l02M 7.1.22. Let the quadrilateral ABCD be circumscribed around a circle and M, N be the intersection points of lines AB and CD, AC and BD, respectively. Prove that, if ∠ 1 ∠ max(AB, BC, CD, AD) ¼ AD, then 90 AND 90 þ 2 AMD. 7.1.23. Given a tetrahedron ABCD inside of a unit cube, prove that AB Á CD Á d < 2, where d is the distance between the straight lines AB and CD. 7.1.24. Given a tetrahedron ABCD inside of a parallelepiped, prove that the volume V of ABCD is less than 3, where V is the volume of the parallelepiped. 7.1.25. Let the straight line intersect sides AB and BC of triangle ABC at points M and K respectively. Given that the area of triangle MBK is equal to the area of the MBþBK > 1 quadrilateral AMKC, prove that AMþCAþKC 3. 7.1.26. Let the triangular section of a cube touch the insphere of the cube. Prove that the area of this section is less than the half of the area of the face of the cube. 7.1.27. (a) Prove that if the length of each of the bisectors of the triangle is greater than 1, then its area is greater than p1ffiffi. 3 7.1 Miscellaneous Inequalities 289

(b) Prove that if the length of each of the bisectors of the triangle is less than 1, then its area is less than p1ffiffi. 3 (c) Let points A1, B1, C1 be, respectively, on sides BC, AC and AB of the triangle ABC. Given that AA1 1, BB1 1, CC1 1, prove that the area of triangle ABC is smaller than or equal to p1ffiffi. 3 7.1.28. Find the greatest value of the number k, such that the inequality a2 þ b2 þ c2 > k(a þ b þ c)2 holds true for any obtuse triangle. 7.1.29. For triangle ABC prove the following inequalities: 2 2 2 < 1 (a) a þ b þ c þ 4abc 2,ifa þ b þ c ¼ 1, (b) a(2a2 À b2 À c2) þ b(2b2 À c2 À a2) þ c(2c2 À a2 À b2) 0, 2 2 2 (c) a b(a À b) þ b c(b À c) þ c a(c À a) 0, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi ÀÁ (d) aðÞþa þ c À b bðÞþa þ b À c cðÞb þ c À a ðÞa þ b þ c a2 þ b2 þ c2 , pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi (e) a þ b À c þ b þ c À a þ a þ c À b a þ b þ c, 2 2 2 (f) aÀÁþ b þ c þ 2abc < 2, where a þ b þ c ¼ 2, a b c a c b (g) 2 b þ c þ a c þ b þ a þ 3, a b c a c b < (h) b þ c þ a À c À b À a 1, (i) a2pq þ b2qr þ c2pr 0, where p þ q þ r ¼ 0, (j) a2ðÞþ2b þ 2c À a b2ðÞþ2c þ 2a À b c2ðÞ2a þ 2b À c 9abc, (k) abþbcþca > 1, ðÞaþbþc 2 4 (l) ðÞaþb ðÞbþc ðÞaþc > 1, ðÞaþbþc 3 4 (m) a3 þ b3 þ c3 < (a þ b þ c)(ab þ bc þ ac), (n) 3 a b c < 2, 2 bþc þ cþapþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþb pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (o) a þ b þ c b2 þ c2 À a2 þ a2 þ c2 À b2 þ a2 þ b2 À c2, where ABC isq notffiffiffiffiffiffiffiffiffiffiffi an obtuseqffiffiffiffiffiffiffiffiffiffiffi triangle,qffiffiffiffiffiffiffiffiffiffiffi (p) abc þ abc þ abc a þ b þ c, aþbÀc aÀb þc bþcÀa aÀb bÀc cÀa < 1 (q) aþb þ bþc þ cþa 8. 7.1.30. Let A, B, C be given points on a plane. Prove that for any point X of that plane, m(ABC) m(ABX) þ m(AXC) þ m(XBC), where for points P, Q, R on a plane we denote by m(PQR) the smallest of the lengths of the altitudes of triangle PQR (if points P, Q, R belong to the same line, then m(PQR) ¼ 0). 7.1.31. Given a triangle ABC and its incenter I and A0, B0, C0 be the intersection points of the bisectors of angles CAB, ABC, BCA with sides BC, CA, 1 < AI BI CI 8 ABrespectively, prove that 4 AA0 Á BB0 Á CC0 27. 7.1.32. Given a plane π, a point P on that plane and a point Q outside plane π, find π QPþPR all points R in plane , so that the ratio QR has the greatest value. 7.1.33. Given on a plane a finite number of strips with the sum of widths equal to 11, and a circle with a radius 1, prove that each of these strips can be displaced parallelly to itself so that together they cover the circle. 290 7 Miscellaneous Inequalities

7.1.34. Givenp thatffiffi the right triangle ABC can be covered by two unit circles, prove 3 3 that SABC 2 . 7.1.35. Prove that (a) in a trihedral angle, each plane angle is less than the sum of two other flat angles, (b) in a convex polyhedral angle, the sum of all face angles is less than 360, (c) in a polyhedral angle, each face angle is less than the sum of all other face angles, (d) ∠AMB þ ∠BMC þ ∠CMD > 180, if point M is inside of the tetrahedron ABCD, (e) ∠AMB þ ∠BMC þ ∠AMC þ ∠AMD þ ∠BMD þ ∠CMD > 3π, (f) the sum of all dihedral angles of the trihedral angle is greater than π, (g) the sum of all dihedral angles of a tetrahedron is between 2π and 3π, 3π < ∠ ∠ ∠ < π (h) 4 A1AC1 þ BAC1 þ DAC1 ,ifABCDA1B1C1D1 is a rectangular parallelepiped. 7.1.36. (а) Prove that if the cube with an edge b does not have points outside the < a cube with an edge a and does not contain its center of symmetry, then b 2. (b) Given a rectangular coordinate system in a space and a cube with an edge a. Given also that all coordinates of any point belonging to this cube are non-negative, prove that, there is a point belonging to the cube, such that all its coordinates are not less than a, (c) Cubes with edges a and b do not have common points and are inside of the cube with an edge c. Prove that a þ b < c. 7.1.37. For triangle ABC, prove the following inequalities:

p ha hb hc (a) R a þ b þ c , ma mb mc p (b) a þ b þ c 2r, (c) p ra þ rb þ rc p , R a b c pffiffi2r (d) ma þ mb þ mc 3 3, a b c 2pffiffiffi (e) a þ b þ c 2 3, ma mb mc (f) ma þ mb þ mc 1 þ R, ha hb hcÀÁr 4Rþr 3 (g) mambmc 3 , (h) l2 þ l2 þ l2 p2, a b c pffiffiffi 2 2 2 2 2 2 (i) ma þ mb þ mc À ðÞma À mb À ðÞmb À mc À ðÞma À mc 3 3S, (j) ra þ rb þ rc 3, hahb hc ÀÁ (k) 3 a þ b þ c 4 ra þ rb þ rc , ra rb rc a b c 2 2 2 27R2 (l) ra þ rb þ rc 4 , (m) ha þ hb þ hc 9r, 2 2 2 27 2 (n) ma þ mb þ mc 4 R , 7.1 Miscellaneous Inequalities 291

(o) 1 þ 1 þ 1 2, ma mb mc R (p) r þ r þ r 41 R, a b c pffiffiffiÀÁ2 (q) 1 þ 1 þ 1 < 2 1 þ 1 þ 1 ,ifABC is an acute triangle, la lb lc a b c (r) la hc,ifa b c, (s) bc l pbcffiffiffiffiffiffi, 2R a 2 2Rr a2þb2þc2 (t) ma þ mb þ mc 2R , (u) ma þ mb þ mc 1, mbmc mamc mamb r 2 RÀ2r 2 2 2 (v) p 2r ðÞa À b þ ðÞb À c þ ðÞc À a , (w) mamb þ mbmc þ mcma lara þ lbrb þ lcrc, < 4 (x) la þ lb 3 ðÞa þ b , 2 2 2 2 2 (y) 4acmamc 16S þ (b À a )(c À b ), pffiffi (z) 1 þ 1 þ 1 3, mamb mbmc mcma S (aa) 1 þ 1 p2ffiffi , mab mba 3S (bb) S 2abmamb , a2þb2þc2 (cc) la þ lb þ lc > 1, ma mb mc (dd) ha þ hb þ hc 2R þ 5r, > (ee) ha þ hb þ hc 2Rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ 4r,ifABC is an acute triangle, 3 p (ff) la þ lb þ lc 2 ab þ bc þ ac. 7.1.38. For any tetrahedron ABCD and any point M inside of it, prove that

(a) MA Á MB Á MC Á MD 81MA1 Á MB1 Á MC1 Á MD1, where A1, B1, C1, D1 are the intersection points of the straight lines AM, BM, CM, DM with the opposite faces of the tetrahedron ABCD. (b) at least one of segments AM, BM, CM is less than the corresponding segment AD, BD, CD, (c) DM < max (AD, BD, CD), (d) AM þ BM þ CM þ DM < AB þ BC þ CD þ AD þ BD þ AC, (e) AM þ BM þ CMþ DM << max (AB þ AC þ AD, BA þ BC þ BD, CA þ CB þ CD, DA þ DB þ DC).

7.1.39. Given points A, B, A0, B0 on a line l, and a point P outside that line. Given also that ∠APB ¼ ∠A0PB0 and ∠PA0B0 ¼ ∠PB0A0, prove that

(a) AB A0 B0, (b) |PA À PB| AB À A0 B0. 7.1.40. Given a triangle ABC, such that AB > AC. The bisector of angle ∠A intersects side BC at point D. Let P be a point on segment AD, and Q,R the intersection points of BP and CP with sides AC and AB, respectively. Prove that, PB À PC > RB À QC > 0. 7.1.41. Let P be any point inside of the equilateral triangle ABC. Prove that (a) |∠PAB À ∠PAC| |∠PBC À ∠PCB|, 292 7 Miscellaneous Inequalities

jj∠PAB À ∠PAC jj∠PAB À ∠PAC jj∠PAB À ∠PAC arcsin 2 sin À (b) 2 2 jj∠PBC À ∠PCB :

0 7.1.42. Consider two triangles ABC and A0B0C0, such that ∠A 90 and ∠A 90. 0 0 1 1 1 Let ha and ha be the altitudes drawn to sides a and a .Provethat, 0 0 þ 0 , haha bb cc where a, b, c and a0, b0, c0 are the sides of triangles ABC and A0B0C0, respectively. 7.1.43. Given a square ABCD and P and Q are points on segments BC and CD respectively. Let E and F be the intersection points of line PQ with the extension of π ∠ ∠ < 5π segments AB and AD respectively. Prove that PAQ þ ECF 4 . 7.1.44. Let the straight line l be perpendicular to plane P. Three spheres touch each other, so that each of them touches plane P and the straight line l. Given that the radius of the largest spherepffiffiffi is equal to 1, the radius of the smallest sphere is equal to r, prove that r 3 À 2 2. 7.1.45. Given an equilateral triangle DAC and a point B on ray CD such that ∠BAC ¼ 70. Let point E be on segment AB, so that ∠ECA ¼ 55 and point K be the midpoint of segment ED. Without using a computer, calculator or protractor, show that 60 > ∠AKC > 57, 5. 7.1.46. (a) Let M be the centroid of triangle ABC.LetAB touch the circumcircle of triangle AMC. Prove that sin ∠CAM þ sin ∠CBM p2ffiffi. 3 (b) Let M be the centroid of triangle ABC. Prove that sin ∠CAMþ sin ∠CBM p2ffiffi. 3 7.1.47. Given on a plane points M, N and triangle ABC, prove that AMÁAN BMÁBN CMÁCN ABÁAC þ BAÁBC þ CAÁCB 1.

7.1.48. Given distinct points A1, A2, :::, An on a plane and arbitrary points B1, B2,...,Bn À 1 (n 2), prove that ::: ::: ::: B1A1ÁB2A1Á ÁBnÀ1A1 þ B1A2ÁB2A2Á ÁBnÀ1A2 þ ::: þ B1AnÁB2AnÁ ÁBnÀ1An 1. A2A1ÁA3A1Á:::ÁAnA1 A1A2ÁA3A2Á:::ÁAnA2 A1AnÁA2AnÁ:::ÁAnÀ1An

7.1.49. Given distinct points A1, A2, :::, An on a plane and arbitrary points B1,..., B1A1Á:::ÁBnÀ2A1 B1AnÀ1ÁB2AnÀ1Á:::ÁBnÀ2AnÀ1 B1AnÁB2AnÁ:::ÁBnÀ2An Bn 2 (n 3), prove that þ :::þ . À A2A1Á:::ÁAnA1 A1AnÀ1ÁA2AnÀ1Á:::ÁAnAnÀ1 A1AnÁA2AnÁ:::ÁAnÀ1An 7.1.50. Let M be an arbitrary point inside of triangle ABC with the semiperimeter p. Prove that, AM sin ∠BMC þ BM sin ∠AMC þ CM sin ∠AMB p and that the equality holds true, if and only if M coincides with its incenter.

7.1.51. Let equilateral triangles MNP and M1N1P1 be inscribed in triangle ABC such that M, M1 2 BC, N, N1 2 AB, P, P1 2 AC,(M ≢ M1, N ≢ N1, P≢ P1). Prove that one can construct a triangle with segments MM1, NN1 and PP1. 7.1.52. Given a unit square ABCD and points P and Q on its sides BC and CD ∠ respectively so that PAQ 45 . Let E, F be the intersection pointspffiffiffi of the straight line PQ with lines AB, AD respectively. Prove that, AE þ AF 2 2.

7.1.53. Let a convex polygon A1A2 ...An be inscribed in a circle with the radius R. Let A be a point on that circle, distinct from the vertices of the polygon. Denote by 7.1 Miscellaneous Inequalities 293

ai ¼ AAi and let bi be the distance from point A to line AiAiþ1, i ¼ 1, . . . , n, where a2 a2 a2 A A . Prove that, 1 þ 2 þ ::: þ n 2nR. n þ 1 1 b1 b2 bn 7.1.54. (a) Let points M and N be chosen on side AC of triangle Q0 such that ∠ABM ¼ ∠MBN ¼ ∠NBC. Prove that AN Á CM < 4 Á AM Á NC. (b) Let points M and N be chosen on sides AB and AC of triangle ABC respectively. Prove that BN þ MN þ CM 2BC, if max(∠A, ∠B, ∠C) 120 and ∠A ∠B, ∠C. 7.1.55. Let O be the intersection point of the diagonals of the convex quadrilateral ABCD. Given that AB ¼ a, BC ¼ b, CD ¼ c, DA ¼ d, and ra, rb, rc, rd are the inradiuses of triangles AOB, BOC, COD, DOA respectively, prove that (a) a þ c > b þ d and 1 þ 1 > 1 þ 1 ,ifa þ c > b þ d, where S ¼ S , S ¼ Sa Sc Sb Sd ra rc rb rd a AOB b SBOC, Sc ¼ SCOD, Sd ¼ SDOA, (b) ∠IdIaIb þ ∠IdIcIb > π,ifa þ c > b þ d, where Ia, Ib, Ic, Id are the incenters of triangles AOB, BOC, COD, DOA, respectively.

ÀÁ7.1.56. Given a convex polygon A1A2 ...An and points B1, B2,...,Bn on its sides B1 2A1A2; B2 2A2A3; :::; Bn 2AnA1; Bi ≢ Aj . Given also points C1, C2, :::, Cn on segments A1B1, A2B2,...,AnBn respectively so that ∠BnB1B2 ¼ ∠CnC1C2 ¼ β1, ∠B1B2B3 ¼ ∠C1C2C3 ¼ β2,..., ∠Bn À 1BnB1 ¼ ∠Cn À 1CnC1 ¼ βn, prove that B1C1 sin β1 þ B2C2 sin β2 þ ...þ Bn À 1Cn À 1 sin βn À 1 > BnCn sin βn. 7.1.57. Let the real number λ > À 1. Let a, b, c be the sides of a triangle. Prove that ðÞpþλa ðÞpþλb ðÞpþλc λ 3 aþbþc ðÞpÀa ðÞpÀb ðÞpÀc ðÞ2 þ 3 , where p ¼ 2 . 7.1.58. Let a, b, c be the sides of a triangle ABC, and P be any point inside of it. Let also AP intersect the circumcircle of triangle BPC (for the second time) at point A0. Points B0 and C0 are defined similarly. Prove that pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi AB0 þ B0C þ CA0 þ A0B þ BC0 þ C0A 2 ab þ 2 bc þ 2 ac. 7.1.59. Let O be the incenter of a triangle ABC and D, E be the midpoints of segments AB, AC, respectively. Let K, L be the intersection points of segments BO and CO with DE respectively. Prove that AO þ BO þ CO > BC þ KL. 7.1.60. For a non-obtuse triangle, prove that

h2 h2 h2 (a) a þ b þ c 9, ap2 ffiffiffi b2 c2 4 (b) 3maxðÞha; hb; hc p. 7.1.61. Let M be the midpoint of segment AB of triangle ABC and line l contains the bisector ofp angleffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiACB. Prove that the distance from point M to line l does not aþb aþbþc exceed 2 À ppðÞÀ c , where p ¼ 2 . pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7.1.62. Given a triangle, prove that mc 3p À ppðÞÀ a À ppðÞÀ b . 7.1.63. Let the bisectors of the internal angles A and C of triangle ABC intersect with the median BM at points E and F respectively. Prove that AE < CF if BC > AB. 294 7 Miscellaneous Inequalities

7.1.64. Let E and F be the intersection points of the altitudes of the internal angles A and C of triangle ABC with the median BM, respectively. Prove that AE < CF if BC > AB. 7.1.65. Consider a triangle ABC, such that BC > AB. Let points E and F be chosen on the median BM such that ∠BAE ¼ ∠BCF. Prove that (a) AE < CF, (b) BE < BF, if points E and B are different. 7.1.66. Given a square ABCD. Let E be a point on ray AB before point B and F be a point on ray AD before point D such that EF ¼ 2AB. Let P and Q be the intersection points of EF with BC and R respectively. Prove that (a) the APQ is an acute triangle, (b) ∠PAQ 450. 7.1.67. Let the circumradius of triangle ABC be equal to 1, O be its center, P be a point inside of the triangle such that OP ¼ x. Prove that AP Á BP Á CP (1 þ x)2(1 À x) and that equality holds true only if P O. 7.1.68. Let A, B, C be points on a circle. Let P, Q, R be the midpoints of the arcs BC, CA, AB respectively. Given that segments AP, BQ, CR intersect sides BC, CA, AB at AL BM CN points L, M, N respectively, prove that PL þ QM þ RN 9. For which triangle ABC does the equality hold true?

7.1.69. (a) Let A1A2 ...An be a regular n-gon, inscribed in a unit circle with the Pn center O. Given a point M on ray OA outside of the n-gon, prove that 1 > n . 1 MAi MO i¼1

(b) Let A1A2 ...An be a regular polygon and let M be a point inside of it. Prove that

2π π sin ∠A MA þ sin ∠A MA þ ::: þ sin ∠A MA > sin þ ðÞn À 2 sin : 1 2 2 3 n 1 n n

7.1.70. Let point M be chosen inside of angle BAC and the straight line l drawn through it be such that the foot of the perpendicular drawn from point A to line l is symmetric to point M with respect to the midpoint of segment UV, where UV is segment of the line cut by the sides of the angle. Let the segment PQ pass through point M, where points P and Q are on the sides of angle BAC. Prove that PQ UV. ÀÁ αβγ π 3 7.1.71. (a) Given a point M inside of triangle ABC, prove that 6 , where α ¼ ∠MAB, β ¼ ∠MBC, γ ¼ ∠MCA. ::: α ∠ (b) Given a point M inside of the convex polygon A1A2 An, let i ¼MAiAi þ 1, n α α ::: α πðÞnÀ2 i ¼ 1, 2, . . . , n, An þ 1 A1. Does the inequality 1 2 Á Á n 2n hold true for n 4? (c) Given a point M inside of the convex polygon A1A2:::An, let αi ¼ ∠MAiAi þ 1, α ; α ; :::; α πðÞnÀ2 i ¼ 1, 2, . . . , n, An þ 1 A1. Prove that minðÞ1 2 n 2n . 7.1 Miscellaneous Inequalities 295

7.1.72. Let 2n-gon A1A2 ...A2n be inscribed in a unit circle with the center O. Prove that À À ! ! ::: ! 1 ∠ ∠ ::: ∠ A1A2 þ A3A4 þ þ A2nÀ1A2n 2 sin 2 A1OA2 þ A3OA4 þ þ A2n À1OA2nÞ. 7.1.73. Let O be the intersection point of the diagonals of the inscribed quadrilateral AB CD BC AD OA OC OB OD ABCD. Prove that CD þ AB þ AD þ BC OC þ OA þ OD þ OB. 7.1.74. Let the incircle of triangle ABC touch its sides BC and AC at points A0 and B0 respectively. Given that point L is the midpoint of segment A0B0,prove that angle ALB is obtuse. 7.1.75. Given an acute triangle with the smallest side cand angle γ opposite to it. Given that it is possible to paint the triangle in two colors, so that the distance between any two points having the same color is not greater than c, prove that γ 36.

7.1.76. Given a point M inside of triangle ABC and da, db, dc denote the distances from point M to lines BC, CA, AB, and Ra, Rb, Rc denote the distances from point M to vertices A, B, C respectively. Prove that

(a) daRa þ dbRb þ dcRc 2(dadb þ dbdc þ dcda), (b) RaRb þ RbRc þ RcRa 2ðÞdaRa þ dbRb þ dcRc , (c) Ra þ Rb þ Rc 2(da þdb þ dc), (d) 1 þ 1 þ 1 2 1 þ 1 þ 1 , dadb dbdc dcda daRa dbRb dcRc (e) 1 þ 1 þ 1 2 1 þ 1 þ 1 , daRa dbRb dcRc RaRb RbRc RcRa (f) 1 þ 1 þ 1 2 1 þ 1 þ 1 . da db dc Ra Rb Rc 7.1.77. Let r be the radius of the insphere of the tetrahedron ABCD. Prove that < ABÁCD r 2ðÞABþCD . 7.1.78. Let ABCD be a tetrahedron such that ∠ADB ¼ ∠ADC ¼ ∠BDC ¼ 90. Given that r and R are the radiuses of the insphere and circumsphere of the tetrahedron ABCD respectively, prove that (a) DA þ BD þ DC 2R þ 6r, 2 (b) ðÞR þ r ðÞR À 3r OO1, where O and O1 are the centers of the spheres with the radiuses r and R, respectively. 7.1.79. Let M be an arbitrary point in the plane of triangle ABC. Given that R and r are the circumradius and the inradius of triangle ABC respectively and point M be at a distance d from the circumcenter of triangle ABC, prove that MA Á MB Á MC 2rjR2 À d2j.

7.1.80. Given the circles ω(O, r) and ω1(O, r1), where r1 > r and let n-gon A1A2:::An be inscribed in the circle ω and rays A1A2, A2A3, :::, AnÀ1An, AnA1 intersect the circumference of ω1 at points B1, B2,...,BnÀ1, Bn, respectively. Prove that 296 7 Miscellaneous Inequalities

::: r1 ::: (a) B1B2 þ B2B3 þ þ BnÀ1Bn þ BnB1 r ðÞA1A2 þ A2A3 þ þ AnÀ1An þ AnA1 , r2 ::: 1 ::: (b) SB1B2 Bn r2 SA1A2 An , for n ¼ 3, 4.

7.1.81. Consider on a plane circles C1(O1, r1),...,Cn(On, rn). Given that any straightP line on the plane intersects not more than two of these circles, prove that riþrj nnðÞÀ1 sin π, where n 3. OiOj 2 n 1i 1, 5 ∠CAK, (b) the factor 1.5 in (a) is the greatest possible.

7.1.83. Let B1, B2,...,Bn be n unit spheres, where n 3, arranged so that each sphere touches externally other two spheres and let C1, C2,...,Cn be n tangential points of these spheres. Let P be a point outside of all these spheres. Denote by ti the length of the tangent drawn from point P to the Bi sphere (1 i n). Prove that the product of ti is not greater than the product of the distances PCi. 7.1.84. Prove that among all convex n-gons located inside of the given circle, the inscribed regular n-gon has the greatest (a) area, and (b) perimeter. 7.1.85. Prove that among all convex n-gons containing the given circle, the circumscribed regular n-gon has the smallest (a) area, and (b) perimeter. 7.1.86. Prove that, among all convex n-gons with the given perimeter, the regular n-gon has the greatest area. π π 7.1.87. Let M be a convex n-gon. Prove that 2ntg2n Á H P 2n sin 2n Á D, where H, P, D are respectively, the width,1 perimeter and diameter of the n-gon. ~ 7.1.88. Let to any non-vanishing vector Vi on the plane, i ¼ 1, 2, . . . , n, be brought ~ into correspondence a strip p i, with the boundaries perpendicular to vector Vi, and ~ the width Wi is less than Vi . Prove that for any point O of that plane there exist numbers a1,...,an 2 {0; 1} such that point M does not belong to any of these ! ~ ~ ~ strips, where OM ¼ a1V1 þ a2V2 þ ::: þ anVn.

7.1.89. Let A1(x1, y1ÀÁ), A2(x2, y2), . . . , An(xn, yn) be different points on the coordinate ::: ::: x1þx2þ þxn ; y1þy2þ þyn plane, n 2 and M n n is their center of mass. Denote by C the center of a circle with the minimum radius r, containing all points A1, A2,...,An. Denote by d the distance between points M and C. Prove that d nÀ2 r n . 7.1.90. Let a cube be sectioned by a plane, such that this plane passes through one of the vertices of the cube and the section is a pentagon. Prove that the ratio of some of the sides of the pentagon is greater than (a) 1,7; (b) 1,83; and (c) 1,84.

1H ¼ min l(M), D ¼ max l(M) (See notations in the solution of the problem 4.1b.) 7.1 Miscellaneous Inequalities 297

7.1.91. Prove that if each of the plane angles at the vertex of the pyramid is not less than 60, then the sum of the lengths of all its edges is not greater than the perimeter of the base.

7.1.92. Given points A1, B1, C1 respectively on the edges SA, SB and SC of the tetrahedron SABC with a volume V, and points A2, B2, C2 on sides B1C1, C1A1, A1B1 of triangle A1B1C1 respectively. Let V1, V2ffiffiffiffiffiffi, V3 beffiffiffiffiffiffi the volumesffiffiffiffiffiffi pffiffiffiffi of tetrahedrons p3 p3 p3 3 AA1B2C2, BB1C2A2, CC1A2B2, prove that, V1þ V2þ V3 V. 7.1.93. Let R be the circumradius of the quadrilateral ABCD and S be its area. Prove that (a) (AB þ BC)(CD þ AD) 4R Á AC, (b) ðÞAB Á CD þ BC Á AD 2ðÞAB Á BC þ CD Á AD R2(AB þ BC)(BC þ CD)(CD þ AD)(AD þ AB). (c) AB þ BC þ CD þ DA < AC þ BD þ 2R, pffiffiffi 3 (d) 2RS ðÞabcd 4. 7.1.94. Let in the acute triangle ABC, bisector AD, the median BM and the altitude CH be concurrent. Prove that angle BAC is greater than 45. 7.1.95. (a) Let O, A, B, C, D be distinct points. Prove that ∠AOB Á ∠COD þ ∠BOC Á ∠AOD 0, 5 ∠AOC Á ∠BOD. (b) Let to each two points A, B on the plane, by a certain rule, is brought into correspondences the number kAB k0, such that three conditions are satisfied: 1. kABk¼0, if and only if points A and B coincide, 2. kABk¼kBAk, for any points A and B, 3. kACkkABkþkBCk, for any points A, B and C. Prove that, for any points A, B, C, D on the plane

kABkÁkCDkþkBCkÁkADk 0, 5 kACkÁkBDk : ð1Þ

7.1.96. Let O be the intersection point of segments AA1, BB1, CC1 (A1 2 BC, B1 2 AC, C1 2 AB), where O is the circumcenter of triangle ABC. Prove that,

(a) OA1 þ OB1 þ OC1 1, 5R, (b) 1 þ 1 þ 1 6. OA1 OB1 OC1 R

7.1.97. Given a point M inside of a triangle with sides a, b, c, where a b, c, and da, db, dc denote the distances of point M from the lines containing the sides of the triangle. Prove that qffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi ffiffiffiffiffi ffiffiffiffiffi pffiffi p p p 3 3 (a) da þ db þ dc a, pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi2 (b) da þ db þ dc 1, 92a, if triangle with sides a, b, c is obtuse. 298 7 Miscellaneous Inequalities

7.1.98. Let the convex pentagon ABCDE be inscribed in a unit circle and AB ¼ a, BC ¼ b, CD ¼ c, DE ¼ d, AE ¼ 2. Prove that a2 þ b2 þ c2 þ d2 þ abc þ bcd < 4.

7.1.99. (a) Vectors ~a1,~a2, :::,~an are on a plane such that their lengths are smaller than or equal to 1. Prove that oneffiffiffi can choose the signs in the sum p ~c ¼Æ~a1Æ~a2 Æ ::: Æ~an so that ~c 2. (b) Let ~a ,~a , :::,~a be vectors with the lengths smaller than or equal 1. Prove that 1 2 n pffiffiffi ~ ~ ~ ::: ~ ~ one can choose the signs in the sum c ¼Æa1 Æ a2 Æ Æ an so that c 7.

(c) Vectors ~a1,~a2, :::,~an are on a plane such that ~a1 þ ~a2 þ ::: þ ~an 1. Prove that there exist positive integers 1 i1 < i2 < ...< ik n, such that pffiffi ~ ~ ::: ~ 2 ai1 þ ai2 þ þ aik 8 . 7.1.100. Given a right-angled triangle ABC, with ∠ACB ¼ 90 . Let A1 2 BC, B1 2 AC, C1 2 AB, and triangle A1B1C1 be right-angled. What is the smallest possible value of the length of the hypotenuse of triangle A1B1C1? 7.1.101. Given an arbitrary point O on the base ABC of the triangular pyramid SABC, such that all plane angles at vertex S are not greater than 60. Prove that at least one of angles SAO, SBO and SCO is not less than 60. 7.1.102. Let ABC be a triangle with the semiperimeter p and inradius r. Half-circles with diameters BC, CA, AB are drawn outside of triangle ABC. Given that the circle pffiffi p < p 3 touching all these three half circles has a radius t, prove that 2 t 2 þ 1 À 2 r.

7.1.103. Given distinct points A1, A2,...,An on a plane, where n 2 and D ¼ max AiAj, d ¼ min AiAj. Prove that 1i 3 ðÞn À 1 d, p2 ffiffiffi (e) D 3d, for n ¼ 6. 7.1.104. Given a convex n-gon. The length of any segment, connecting the vertex of that polygon with a certain point of its boundary and dividing its area into two isometric parts is smaller than or equal to 1. Prove that the area of the n-gon pffiffi 3 (a) is smaller than or equal to 3 ,ifn ¼ 3, 2 (b) is less than 3,ifn ¼ 4, n π (c) is less than 4 sin n,ifn 5. 7.1.105. In a given triangle each median is greater than any of its altitudes. Prove that the smallest angle of that triangle is less than 5300. 7.1 Miscellaneous Inequalities 299

7.1.106. Given on a plane an acute triangle ABC and the straight line l. Let u, v, w be, respectively, the distances from points A, B and C to line l. Prove that u2tgα þ v2tgβ þ w2tgγ S, where ∠A ¼ α, ∠B ¼ β, ∠C ¼ γ and S is the area of triangle ABC. 7.1.107. Given a point M and a triangle ABC with sides a, b, c. Prove that (a) MA þ MB þ MC pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3abc , a2b2þb2c2þc2a2 (b) MA þ MB þ MC 6r. 7.1.108. In a non-isosceles triangle ABC, let O, I, H be the circumcenter, incenter and orthocenter, respectively. Prove that ∠OIH > 135. 7.1.109. (a) Let the inscribed circle of triangle ABC touch sides AB, BC, AC at points C1, A1, B1 respectively. Prove that segments AA1, BB1, CC1 intersect at one point (denote it by M). Prove that AM p2ffiffi AB : 3 1 (b) Let ABCDEF be a convex hexagon with area S. Prove that pffiffiffi ACðÞ BD þ BF À DF þCEðÞ BD þ DF À BF þAEðÞ BF þ DF À BD 2 3S. (c) Three circles externally touch each other at points X, Y, Z.

Let us multiply the radiuses of these circles by p2ffiffi, keeping the centers fixed, and 3 consider these new circles. Prove that any point of triangle XYZ is covered at least by one of the new circles. (d) Given triangle ABC with sides a, b, c and point M, prove that

MA þ MB MB þ MC MC þ MA pffiffiffi R þ þ 2 3 Á : p À c p À a p À b r

7.1.110. (a) Let A1A2 ...An be a convex polygon. Point P is chosen inside of this polygon, such that its projections P1, P2,...,Pn on lines A1A2,...,AnA1 respectively lie on the sides of the polygon. Prove that for arbitrary points X1, X2,...,Xn on sides A1A2,...,AnA1 respectively, X X X X max 1 2 ; :::; n 1 1: P1P2 PnP1

(b) Let A1A2 ...An be a cyclic convex polygon whose circumcenter is strictly in its interior. Let B1, B2,...,Bn be arbitrary points on sides A1A2,...,AnA1 respectively other than the vertices. Prove that

B B B B B B 1 2 þ 2 3 þ ::: þ n 1 > 1: A1A3 A2A4 AnA2 300 7 Miscellaneous Inequalities

7.1.111. Let MNP be an equilateral triangle inscribed in triangle ABC such that M 2 BC, N 2 AC, P 2 AB. Point I is not inside of triangle MNP, where I is the incenter of triangle ABC. Prove that ∠A > 120. 7.1.112. Consider triangle ABC with inradius r. Let M and M0 be two points inside 0 0 of triangle, such that ∠MAB ¼ ∠M AC and ∠MBA ¼ ∠M BC. Denote by da, db, dc 0 0 0 0 and da, db, dc the distances from M and M to sides BC, CA, AB respectively. Prove 0 0 0 6 that dadbdcdadbdc r . 7.1.113. (a) Given convex hexagon ABCDEF with AB k DE, BC k EF, CD k FA. The distance between lines AB and DE is equal to the distance between lines BC and EF, and to the distance between lines CD and FA. Prove that the sum AD þ BE þ CF does not exceed the perimeter of hexagon ABCDEF.

(b) Let ABCDEF be a cyclic hexagon, such that AC ¼ CE ¼ EA . Given that diag- DE CD < onals AD, BE, CF intersect at one point, prove that EF À BC 1.

7.2 Solutions

1 ∠ MA1 2MA1ÁBC sin AA1C SMBC 7.1.1. (a) Note that ¼ 1 ∠ ¼ . In the same way, one can prove AA1 2AA1ÁBC sin AA1C SABC that MB1 ¼ SMAC and MC1 ¼ SMAB. Consequently, MA1 þ MB1 þ MC1 ¼ 1, or 1 ¼ MA1 þ BB1 SABC CC1 SABC AA1 BB1 CC1 AA1 MB1 þ MC1 MA1 þ MB1 þ MC1 . Hence, we deduce BB1 CC1 maxðÞAA1; BB1; CC1 maxðÞAA1; BB1; CC1 maxðÞAA1; BB1; CC1 that MA1 þ MB1 þ MC1 max (AA1, BB1, CC1).

Remark 1 max(AA1, BB1, CC1) < max (AB, BC, AC), consequently, the inequality MA1 þ MB1 þ MC1 < max (AB, BC, AC) holds true.

Remark 2 MA1 þ MB1 þ MC1 min (AA1, BB1, CC1).

(b), (c) Let max(AA1, BB1, CC1) ¼ AA1, then according to the problem 7.1.1a MA MB1 þ MC1. Note that MA1 max (MB, MC). Therefore, > MA þ MB þ MC MA þ max (MB,ÀÁMC) MB1 þ MC1 þ MA1, 2 2 2 2 2 2 2 MA þ MB þ MC > MA þ max MB ; MC ðÞMC1 þ MB1 þ 2 > 2 2 2: MA1 MA1 þ MB1 þ MC1 (d) We have that MA1 þ MB1 þ MC1 ¼ 1 (see the proof of the problem 7.1.1a). Also AA1 BB1 CC1 AA BB CC MA MB MC 1 þ 1 þ 1 1 þ 1 þ 1 ¼ MA MB MC AA BB CC 11 1 1 1 1 AA MB BB MA AA MC CC MA BB MC CC MB ¼ 3 þ 1 1 þ 1 1 þ 1 1 þ 1 1 þ 1 1 þ 1 1 BB1MA1 AA1MB1 CC1MA1 AA1 MC1 CC1MB1 BB1 MC1 AA BB CC MA MB MC 3 þ 2 þ 2 þ 2 ¼ 9, hence 1 þ 1 þ 1 9: Thus, þ þ 6 MA1 MB1 MC1 MA1 MB1 MC1 7.2 Solutions 301

(e) We have that MA1 þ SMABþSMCB, MB ¼ SMABþSMCB, MC ¼ SMCBþSMAC (see the proof of AA1 SMAC MB1 SMAC MC1 SMAB the problem 7.1.1а). Hence, it follows that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi MA MB MC 2 S Á S 2 S Á S 2 S Á S Á Á MAB MAC Á MAB MCB Á MCB MAC ¼ 8: MA1 MB1 MC1 SMBC SMAC SMAB

(f) According to the Cauchy’s inequality and the problem 7.1.1e, we have that

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 pffiffiffiffiffi AM BM BM CM CM AM 3 AM BM CM 3 Á þ Á þ Á 3 Á Á 3 82¼ 12: A1M B1M B1M C1M C1M A1M A1M B1M C1M

(g) We have that

MA MB MC 1 ¼ 1 þ 1 þ 1 ¼ MA þ MA1 MB þ MB1 MC þ MC1 MA2 MB2 MC2 ¼ 1 þ 1 þ 1 MA1ðÞMA þ MA1 MB1ðÞMB þ MB1 MC1ðÞMC þ MC1 ðÞMA þ MB þ MC 2 1 1 1 : MA1ðÞþMA þ MA1 MB1ðÞþMB þ MB1 MC1ðÞMC þ MC1

Therefore, MA Á MA1 þ MB Á MB1 þ MC Á MC1 2MA1 Á MB1 þ 2MB1 Á MC1 þ 2MC1Á MA1. (h) We have that,

A M B M C M S S S 3 1 þ 1 þ 1 ¼ MBC þ MAC þ MAB AM BM CM SMAB þ SMAC SMAB þ SMBC SMAC þ SMBC 2

(see the proof of the problem 5.5.19b) (i) We have that,

MA MB MC 1 ¼ 1 þ 1 þ 1 MA þÀÁMA1 MB þ MB1 MC þ MC1 pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi 2 MA1 þ MB1 þ MC1 , MA þ MB þ MC þ MA1 þ MB1 þ MC1 consequently pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi MA þ MB þ MC 2 MA1 Á MB1 þ MB1 Á MC1 þ MC1 Á MA1 : 302 7 Miscellaneous Inequalities

(j) We have that

ÀÁpffiffiffiffiffiffiffi pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi 2 MA MB MC MA þ MB þ MC 2 ¼ þ þ : MA þ MA1 MB þ MB1 MC þ MC1 MA þ MB þ MC þ MA1 þ MB1 þ MC1

Thus, it follows that

MA þ MB þ MC þ 2ðÞMA1 þ MB1 þ MC1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 MA Á MB þ MB Á MC þ MC Á MA :

(k), (l) Denote by SMAB ¼ SC, SMBC ¼ SA, SMAC ¼ SB. We have that, MA ¼ SBþSC SAþSC SAþSB Á MA1, MB ¼ Á MB1, MC ¼ Á MC1, consequently SA SB SC S2 S2 S2 S2 MA2 þ MB2 þ MC2 ¼ BMA2 þ AMB2 þ CMA2 þ AMC2 þ S2 1 S2 1 S2 1 S2 1 A B A C S2 S2 S S S S S S þ CMB2 þ BMC2 þ 2 B C MA2 þ A C MB2 þ A B MC2 S2 1 S2 1 S2 1 S2 1 S2 1 B C A B qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiC 3 2 2 2: 2ðÞMA1 Á MB1 þ MB1 Á MC1 þ MC1 Á MA1 þ6 MA1 Á MB1 Á MC1

We have that,

MAÁ MB þ MB Á MC þ MC Á MA ¼ MA1 Á MB1 þ MB1 Á MC1 þ MC1 Á MA1þ S2 S S S2 þ C MA Á MB þ CMA Á MB þ CMA Á MB þ A MB Á MC þ S S 1 1 S 1 1 S 1 1 S S 1 1 A B B A B C S S S2 S S þ AMB Á MC þ AMB Á MC þ B MC Á MA þ BMC Á MA þ BMC Á MA S 1 1 S 1 1 S S 1 1 S 1 1 S 1 1 B C A C qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA C 9 6 6 6 MA1 Á MB1 þ MB1 Á MC1 þ MC1 Á MA1 þ 9 MA Á MB Á MC ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 1 3 2 2 2: ¼ MA1 Á MB1 þ MB1 Á MC1 þ MC1 Á MA1 þ 9 MA1 Á MB1 Á MC1

(m) Let ∠BMC¼α2, ∠AMC¼β2 and ∠AMB¼γ2. We have that

γ α β 2S 2SMAB þ 2SMBC þ 2SMAC MA Á MB sin 2 þ MC Á MB sin 2 þ MA Á MC sin 2 ð7:2Þ 7.2 Solutions 303

(point M can be inside or outside triangle ABC). Since, sin x sin y 1 þ cos x cos y, then from (7.2) we deduce that

γ γ α α β β 1 þ cos 1 cos 2 1 þ cos 1 cos 2 1 þ cos 1 cos 2 4S 2MA Á MB γ þ 2MC Á MB α þ 2MA Á MC β ¼ sin 1 sin 1 sin 1 ÀÁ 2MA Á MB 2MC Á MB 2MA Á MC 2 2 2 γ ¼ γ þ α þ β þ MA þ MB À c ctg 1þ ÀÁsin 1 sin 1 sin ÀÁ1 2 2 2 α 2 2 2 β þ MB þ MC À a ctg 1 þþ MA þ MC À b ctg 1, where α1, β1, γ1 are the angles of the triangle with sides a1, b1, c1. Consequently,

2MA MB 2MC MB 2 α 2 β 2 γ Á Á 4S þ a ctg 1 þ b ctg 1 þ c ctg 1 γ þ α þ sin 1 sin 1 2MA Á MC 2 γ β 2 α γ þ β þ MA ðÞctg 1 þ ctg 1 þMB ðÞctg 1 þ ctg 1 sin 1 2 α β 2MA Á MB 2MC Á MB 2MA Á MC þ MC ðÞctg 1 þ ctg 1 ¼γ þ α þ β sin 1 sin 1 sin 1 2 α 2 β 2 γ 2 MA sin 1 MB sin 1 MC sin 1 ðÞa1MA þ b1MB þ c1MC : þ β γ þ α γ þ α β ¼ sin 1 sin 1 sin 1 sin 1 sin 1 sin 1 2S1

Thus, it follows that ÀÁ 2 α 2 β 2 γ 2: 8SS1 þ 2S1 a ctg 1 þ b ctg 1 þ c ctg 1 ðÞa1MA þ b1MB þ c1MC

To complete the proof, it remains to note that ÀÁ 2 α 2 β 2 γ 2 α 2 β 4S1 a ctg 1 þ b ctg 1 þ c ctg 1 ¼ a Á 2b1c1 cos 1 þ b Á 2a1c1 cos 1þ c2 Á 2a b cos γ ¼ ÀÁ1 1 1 ÀÁÀÁ 2 2 2 2 2 2 2 2 2 2 2 2 : ¼ a b1 þ c1 À a1 þ b a1 þ c1 À b1 þ c a1 þ b1 À c1

Remark If point M0 is, such that ∠BM0C ¼ 180 À α1, ∠AM0C ¼ 180 À β2, ∠AM0B ¼ 180 À γ1, then a1MA þ b1MB þ c1MC > a1M0A þ b1M0B þ c1M0C, where M ≢ M0 (see also problem 4.1.17). 7.1.2. (a) Let us complete triangle up to a rectangular (Figure 7.1). < < BD Then, EF BD or 4r EF BD. Hence, we obtain that r 4 . : AC > BO (b) Let AB ¼ BC. Denote AB 2 ¼ k, then k 1. We have that OH ¼ k (see Figure 7.2).

2 2 2 Hence, AB2 ¼ AB þ ðÞr þ rk 2. Therefore, AB2 ¼ r2 ðÞkþ1 k > 9r2,ask2(k þ 1)2 k2 k2À1 À 9(k2 À 1) ¼ (k2 À 3)2 þ 2k2(k À 1) > 0. 304 7 Miscellaneous Inequalities

Figure 7.1

O2 F

EO1 AD

Figure 7.2 B

AHC

7.1.3. By Carnot’s theorem, we have that

ak bk ck ak þ bk þ ck r þ R ¼ k þ k þ k ¼ a þ b þ c a b c ¼ a b c a b c minðÞa; b; c 2S ¼ ¼ maxðÞh ; h ; h , minðÞa; b; c a b c ak bk ck ak þ bk þ ck 2S r þ R ¼ k þ k þ k ¼ a þ b þ c a b c ¼ a b c a b c maxðÞa; b; c maxðÞa; b; c

¼ minðÞha; hb; hc

(see the proof of problem 2.4.11). 7.1.4. (a) Denote BC ¼ a, CA ¼ b, AB ¼ c and ∠C ¼ γ. By triangle angle bisector ab ab proportionality theorem, one can easily obtain that CA1 ¼ bþc, CB1 ¼ aþc. Note that A1B1 can be found by using the cosine law for triangles A1CB1 and ABC, 7.2 Solutions 305

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ðÞab 2 A B ¼ ab þ ab À 2 cos γ ¼ 1 1 bþc aþc ðÞa þ c ðÞb þ c sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ðÞab 2 a2 þ b2 À c2 ¼ ab þ ab À 2 ¼ bþc aþc ðÞa þ c ðÞb þ c 2ab vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u ÀÁ ÀÁ ! u2 2 2 2 2 t 2 ab a þ b þ 2caðÞþ b þ c ab À a þ b À c ðÞab þ caðÞþ b þ c ¼ ab ðÞaþc ðÞbþc ab sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c2ðÞa þ c ðÞÀb þ c caðÞÀ b 2ðÞa þ b þ c abc2 ¼ ab ðÞaþc ðÞbþc ab ðÞa þ c ðÞb þ c sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi a þ b 2 þ c abc abc ab þ c 2 a þ b þ 2c pffiffiffiffiffi pffiffiffiffiffi ¼ ¼ : 2 ac Á 2 bc 4 4 4 8

bþcþ2a aþcþ2b Similarly, we deduce that B1C1 8 and A1C1 8 . Therefore, 4aþ4bþ4c 1 A1B1 þ B1C1 þ A1C1 8 ¼ 2 ðÞAB þ BC þ AC . (b) We have that pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi c ab b ac a bc ðÞA B 2 þ ðÞB C 2 þ ðÞA C 2 þ þ 1 1 1 1 1 1 4 4 4 caðÞþ b baðÞþ c abðÞþ c 1 þ þ ðÞa þ b þ c 2 8 8 8 12

(see the proof of problem 7.1.4a). AD 7.1.5. We need to prove that DP is the smallest, if AD is the bisector of angle A. Indeed, we have that

AD AD2 AD2 AD AD sin β sin γ ¼ ¼ ¼ Á ¼ Á ¼ DP DP Á AD BD Á DC BD DC sin ∠BAD sin ∠CAD β γ β γ β γ 2 2 sin sin 2 sin sin sin sin AD1 ¼ ∠ ∠ ¼ a ¼ , cos ðÞBAD À CAD Àcos a 1 À cos a sin 2 D1B Á D1C 2 where ∠BAD1 ¼ ∠CAD1, D1 2 BC. 2 2 ac ba 2 bca AD1 It is known that BD1 ¼ , CD1 ¼ and AD ¼ bc À 2. Hence, ¼ bþc bþc 1 ðÞbþc D1BÁD1C 2 2 2 2 2 2 ðÞbþc Àa . Therefore, AD ðÞbþc Àa . Similarly, we obtain that BE ðÞaþc Àb and a2 DP a2 EQ b2 2 CF ðÞaþb Àc2 RF c2 . By summing up these three inequalities, we deduce that 306 7 Miscellaneous Inequalities

AD BE CF ðÞb þ c 2 ðÞa þ c 2 ðÞa þ b 2 1 b þ c a þ b a þ c 2 þ þ þ þ À 3 þ þ À 3 ¼ DP EQ RF a2 b2 c2 3 a c b 1 b a a c c b 2 1 ¼ þ þ þ þ þ À 3 ðÞ2 þ 2 þ 2 2 À 3 ¼ 9, 3 a b c a b c 3

2 2 2 1 2 1 > AD BE CF as x þ y þ z 3 ðÞx þ y þ z and x þ x 2(x 0). Thus, DP þ EQ þ RF 9. The equality holds true only for an equilateral triangle ABC, with AD, BE, CF being the bisectors of angles A, B, C respectively. 7.1.6. Let AC ¼ a, CE ¼ b, AE ¼ c. According to problem 1.1.14a, we get that BC a AB Á CE þ BC Á AE AC Á BE. This means that BE bþc. Similarly, we deduce that DE b FA c DA aþc and FC aþb. By summing up these three inequalities, we obtain that BC DE FA a b c 3 BE þ DA þ FC bþc þ aþc þ aþb 2 (see the proof of problem 5.5.19b). Therefore, BC DE FA 3 BE þ DA þ FC 2. 7.1.7. Let the rectangle with sides a and b be inside of the rectangle with sides c and d. Consider a rectangle with sides c1 and d1, such that its sides are parallel to the sides of the rectangle with sides c and d. Moreover, the rectangle with sides a and b is inscribed in it (see Figure 7.3). It is clear that c1 c and d1 d. Hence, one can assume that the centers of these rectangles coincide (prove that the center O of the rectangle with sides a and b is the center of rectanglepffiffiffiffiffiffiffiffiffi with sides c1 and d1). Let us consider a circle with the center a2þb2 O and the radius 2 (see Figure 7.4).

Figure 7.3 d

d1 b cO

O1 c1 a

Figure 7.4

Ad 7.2 Solutions 307

Figure 7.5 b

a cos αα+ b sin a α b cos αα+ a sin

We have that a AB (see the proof of problem 3.1.34), which means that

0 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 0 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 d a2 þ b2 c2 c a2 þ b2 d2 a2 @ À À A þ @ À À A , 2 4 4 2 4 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁÀÁÀÁ 2 2 cd À a2 þ b2 À a2 þ b2 c2 þ d2 þ c2d2 4a2b2,

(cd À 2ab)2 (a2 þ b2)2 À (a2 þ b2)(c2 þ d2) þ c2d2,(b2 À a2)2 (bd À ac)2 þ (bc À ad)2. 2 2 2 2 2 Let a < c d < b and (b À a ) (bd À ac) ÀÁþ (bc À ad) . Denote by α α > α ; π bd À ac ¼ R cos , bc À ad ¼ R sin , where R 0, 2 0 2 . Then, from the condition (bd À ac)2 þ (bc À ad)2 (b2 À a2)2, we deduce that R b2 À a2. Note that, α α α α d ¼ RbðÞcos þa sin , c ¼ RaðÞcos þb sin . Hence, d b cos α þ a sin α, b2Àa2 b2Àa2 c a cos α þ b sin α. Thus, the rectangle with sides b cos α þ a sin α, a cos α þ b sin α can be placed inside of the rectangle with sides c and d. Then it can also contain the rectangle with sides a and b (see Figure 7.5). Second Solution (G. Nersisyan). Without loss of generality, one can assume that two adjacent vertices of the rectangle with sides a and b are on the sides of the second rectangle. In that case, the following conditions must be fulﬁlled. a cos α þ b sin α c, ð7:3Þ b cos α þ a sin α d:

Note that sin α 0 and cos α 0. Consider the following inequalities: ax þ by c, ð7:4Þ bx þ ay d, c c < 1, > 1, ð7:5Þ b a d d < 1, > 1: ð7:6Þ b a The set of the inequalities in (7.3) has a solution, if and only if the solution of equations ax þ by ¼ c, bx þ ay ¼ d is outside the circle (see Figure 7.6). Hence, (b2 À a2)2 (bc À ad)2 þ (bd À ac)2. This ends the proof. 308 7 Miscellaneous Inequalities

Figure 7.6 y

d/a

bx+ay=d

c/a x 1 ax+by=c

yþzÀx 7.1.8. Let a b c. Denote u þ v ¼ x, v þ w ¼ y, and u þ w ¼ z, Then, w ¼ 2 , xþzÀy xþyÀz u ¼ 2 , v ¼ 2 . If ab ac þ bc, we have that w u v 4w u v a2b2 þ b2c2 þ a2c2 > 4S2 þ þ ¼ u þ v v þ w u þ w u þ v v þ w u þ w y x 2z x 1 z y ¼ 4S2 2 þ þ þ þ þ À 3 4S2ðÞ2 þ 2 þ 1 À 3 ¼8S2: x 2y x 2z 2 y z

w 2 2 u 2 2 v 2 2 > 2 Thus, uþv a b þ vþw b c þ uþw a c 8S . If ab < ac þ bc, we have that

w u v a2b2 þ b2c2 þ a2c2 ¼ u þv v þ w u þ w ya2b2 xb2c2 za2b2 xa2c2 zb2c2 ya2c2 1ÀÁ ¼ þ þ þ þ þ À a2b2 þ b2c2 þ a2c2 2x 2y 2x 2z 2y 2z 2 1ÀÁ ab2c þ a2bc þ abc2 À a2b2 þ b2c2 þ a2c2 : 2

We need to prove that 2ab2c þ 2a2bc þ 2abc2 16S2 þ a2b2 þ b2c2 þ a2c2, that is

2ab2c þ 2a2bc þ 2abc2 ðÞa þ b þ c ðÞa þ b À c ðÞa þ c À b ðÞþb þ c À a a2b2 þ b2c2 þ a2c2, 2ab2c þ 2a2bc þ 2abc2 ðÞa þ b 2 À c2 c2 À ðÞa À b 2 þ a2b2 þ b2c2 þ a2c2, ÀÁ 2 2abcðÞ a þ b þ c c2 ðÞa þ b 2 þ ðÞa À b 2 À c4 À a2 À b2 þ a2b2 þ b2c2 þ a2c2, 7.2 Solutions 309

a4 þ b4 þ c4 þ 2abcðÞÀ a þ b þ c 3a2b2 À 3b2c2 À 3a2c2 0, ÀÁ 2 c4 À ðÞac þ bc À ab 2 þ a2 À b2 À 2c2ðÞa À b 2 0, ÀÁÀÁ c2 þ ac þ bc À ab c2 À ac À bc þ ab þ ðÞa À b 2 ðÞa þ b 2 À 2c2 0, ÀÁ ðÞc À a ðÞc À b c2 þ ac þ bc À ab þ ðÞa À b 2 ðÞa þ b 2 À 2c2 0:

The obtained inequality holds true, since (c À a)(c À b) 0, c2 þ ac þ bc > ac þ bc > ab,(a þ b)2 4b2 > 2c2.

7.1.9. Let complex numbers a, b, c and a1, b1, c1 correspond to the vertices of triangles ABC and A1B1C1 in a complex plane. According to the assumptions of the problem, triangles ABC and A1B1C1 are similar and have the same orientation, hence

b À a b À a ∗ 1 1 ¼ : ðÞ c1 À a1 c À a

We need to prove that there exist complex numbers z and v such that 8 <> vzðÞ¼À a a1 À a,7ðÞ:7 vzðÞ¼À b b À b,7ðÞ:8 :> 1 vzðÞ¼À c c1 À c: ðÞ7:9

a1Àb1 From (7.7) and (7.8), it follows that v ¼ bÀa þ 1, while from (7.7) and (7.9) we a1Àc1 a1Àb1 a1Àc1 obtain that v ¼ cÀa þ 1. From (*), we have that bÀa þ 1 ¼ cÀa þ 1. If v ¼ 0, then from (7.7) and (7.8), we have that a1 ¼ a, b1 ¼ b, but from the assumptions of the problem, we have that c1 ¼ c. Thus, it follows that

AA1 Á BC ¼ 0 ¼ BB1 Á CA þ CC1 Á AB:

a1Àa If v 6¼ 0, then z ¼ v þ a. Let the complex number z corresponds to point M. Then |a1 À a| ¼ |v||z À a|, |b1 À b| ¼ |v||z À b|, |c1 À c| ¼ |v||z À c|, that is AA1 ¼ |v| Á MA, BB1 ¼ |v| Á MB and CC1 ¼ |v| Á MC. Using the inequality of the problem 1.1.14a, for points M, A, B, C, we obtain that MA Á BC MB Á AC þ MC Á AB. Hence, |v|MA Á BC |v|MB Á AC þ |v|MC Á AB,orAA1 Á BC BB1 Á AC þ CC1 Á AB.

7.1.10. Let the length of the edge of the cube ABCDA1B1C1D1 be equal to a. Consider the midpoints P, Q, R, S, T, U of the edges AB, BC, CC1, C1D1, D1A1, A1A, respectively. Since segments RU, SP and QT pass through the center O of the cube and QP k RU k ST, then points P, Q, R, S, T, U belong to the same plane.

We have that,pffiffi QR ¼ RS ¼ ST ¼ TU ¼ UP ¼ PQ ¼ OQ ¼ OR ¼ OS ¼ OT ¼ OU ¼ OP ¼ a 2, thus the hexagon PQRSTU is regular. The inradius of the hexagon pffiffi 2pffiffi pffiffi is equal to a 2 Á 3 ¼ 6 a. We need to prove that if a circle with the radius r can be 2 2 4 pffiffi > 6 placed inside of the cube, then it is impossible that r 4 a. 310 7 Miscellaneous Inequalities

Figure 7.7 P

O1 P1 M a h

pﬃﬃ > 6 Let O1 be the center of that circle and the radius r 4 a. Consider a plane Π containing that circle. Denote by α Πd, AA B B , ¼ 1 1 β Πd, BB C C , γ Πd, ABCD . Note that cos2α cos2β cos2γ 1. ¼ 1 1 ¼ þ þ ¼

Indeed, let B(0; 0; 0), A(a;0;0),C(0; a;0),B1(0; 0; a),~n⊥Π,~nuðÞ; v; w and ~n ¼ 1, 2 2 2 2 2 2 2 2 2 then cos α þ cos β þ cos γ ¼ ðÞ~e1 Á~n þ ðÞ~e2 Á~n þ ðÞ~e3 Á~n ¼ u þ v þ w ¼

1, where ~e1ðÞ0; 1; 0 ,~e2ðÞ1; 0; 0 ,~e3ðÞ0; 0; 1 . Since Πd, DD1 C1C ¼ α, then one can Π a assume that the distance of O1 from the plane 1 AA1Bq1Bffiffi is not greater than 2. ρ ;Π ðÞO1 1 h a a affiffi 2 Therefore, sin α ¼ ¼ < p ¼ (see Figure 7.7). ρðÞO1;Π1\Π O1M 2O1M 2r 6 3 qffiffi 2 4 a Thus, sin α < 2. Hence, cos α > p1ffiffi. Similarly, we obtain that cos β > p1ffiffi and 3 3 3 cos γ > p1ffiffi. Then, cos2α þ cos2β þ cos2γ > 1. This leads to a contradiction. 3

7.1.11. (a) Let points Ai(xi; yi; zi), i ¼ 1, 2, 3 be given inside of the cube with vertices (0; 0; 0), (0; 1; 0), (0; 0; 1), (1; 0; 0), . . . , (1; 1; 1), then xi, yi, zi 2 (0; 1), i ¼ 1, 2, 3. We have that,

2 2 2 A1A2 þ A2A3 þ A3A1 ÀÁ 2 ÀÁ 2 2 < : 2max xi À xj þ 2max yi À yj þ 2max zi À zj 6 pffiffiffi Hence, it follows that minðÞA1A2; A2A3; A3A1 < 2. (b) Let d be the smallest number, such that eight cubes with edges d, placed in each of the vertices of the cube, contain all given points (see Figure 7.8). 1 Then, it is clear that d 2. Thus, each of these cubes contains only one of these points;ffiffiffi otherwisepffiffi the distance between two points in the same cube is less than p 3 3d 2 . In this case the problem would have been proven. Since d is minimal, then at least one of the points is on the surface ofone ofthe cubes. Thus, some two points M and N from eight given points are in a rectangular parallelepiped with dimensions qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d Â d Â (1 À d). Then, MN d2 þ d2 þ ðÞ1 À d 2 ¼ 1 À dðÞ2 À 3d < 1.

Figure 7.8

Figure 7.9 128

64 506

2000 64 506

Figure 7.10

7.1.12. Figure 7.10 gives an example with eight adjoining squares. Denote by A and B the centers of two squares touching the fixed square S with the center O. Let OA ¼ a, OB ¼ b and AB ¼ c. We have that pffiffiffi pffiffi OA ¼ a ONþ KA 1 þ 1 ¼ 1 and a, b 2. Also, a ¼ OA OM þ MA 2 pffiffi pffiffiffi 2 2 2 2 þ 2 ¼ 2 (see Figure 7.11). Applying the law of cosines to triangle OAB, we obtain that c2 ¼ a2 þ b2 α α a2þb2Àc2 α ∠ À 2ab cos . Hence, cos ¼ 2ab , where ¼ AOB. α a2þb2À1 Since c 1, then cos 2ab . 2 2 We need to prove that a þb À1 3, or equivalently, 4a2 À 6ab þ 4b2 À 4 0. 2ab 4 ÂÃpffiffiffi Indeed, consider a function f(x) 4x2 6xb 4b2 4on the interval 1; 2 .We ¼ÀÁÀ ÀÁpþffiffiffi À ; have that faðÞmaxpffiffi fxðÞ¼max f ðÞ1 f 2 . ½1; 2 312 7 Miscellaneous Inequalities

Figure 7.11

M А N K O

pffiffiffi Note that, f(1) ¼ 4b2 À 6b¼ 4b(b À 1, 5) < 0, as 1 b 2 < 1, 5 and ÀÁpffiffiffi pffiffiffi pffiffi ÀÁpffiffiffi pffiffi pffiffiffi 2 2 2 < f 2 ¼ 4b À 6 2b þ 4 ¼ 4 b À 2 b À 2 0, as 2 1 b 2. Thus, f(a) ¼ 4a2 À 6ab þ 4b2 À 4 0. α α 3 α WeÀÁ have obtained that angle is, such that cos 4 ¼ cos 0, where α ; π α 3α α 9 < 1 0 2 0 2 . We have that cos 3 0 ¼ 4cos 0 À 3 cos 0 ¼À16 À2. Hence, α > 2π α α > 2π 3 0 3 . Thus, 0 9 , it follows that the 360 -layout of segments connecting the centers of the adjoining squares with point O cannot contain such nine angles α 2π greater than 9 ¼ 40 . Therefore, the greatest number of the adjoining squares is equal to eight. 7.1.13. Since the triangle is acute, then a2, b2, c2 are the sides of some triangle and consequently, there exist numbers m, n, k > 0 such that a2 ¼ m þ n, b2 ¼ n þ k, c2 ¼ k þ m. It remains to prove that (4m þ n þ k)(4n þ k þ m)(4k þ m þ n) > 25 (m þ n)(n þ k)(k þ m), or 12mnk þ m3 þ n3 þ k3 > m2k þ k2m þ n2m þ m2n þ k2 n þ n2k. We need to prove that 3mnk þ m3 þ n3 þ k3 m2k þ k2m þ n2m þ m2n þ k2n þ n 2k,ormnk (m þ n À k)(k þ m À n)(n þ k À m). If m, n .,k are not the sides of some triangle, then mnk > 0 (m þ n À k) (k þ m À n)(n þ k À m). If m, n .,k are the sides of some triangle, then

ðÞm þ n À k ðÞk þ m À n ðÞ¼n þ k À m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ðÞm þ n À k ðÞk þ m À n Á ðÞk þ m À n ðÞn þ k À m Á ðÞm þ n À k ðÞn þ k À m m þ n À k þ k þ m À n k þ m À n þ n þ k À m m þ n À k þ n þ k À m Á Á ¼ mnk: 2 2 2

7.1.14. (a), (b) Let max(∠AMB, ∠BMC, ∠AMC) ¼ ∠BMC and min (∠AMB, ∠BMC, ∠AMC) ¼ ∠AMB.As∠AMB þ ∠BMC þ ∠AMC ¼ 360, then ∠BMC 120 , ∠AMB 120 . Consider parallelograms CMBA2 and AMBC2 (see Figure 7.12). Note that ∠MBA2 60 .Thus,2MA1 ¼ MA2 max (MB, BA2) ¼ max (MB, MC). Hence, 2 min (MA1, MB1, MC1) 2MA1 max (MB, MC) max (MA, MB, MC). Therefore, 2 min (MA1, MB1, MC1) max (MA, MB, MC). As ∠AMB 120 , then ∠MBC2 60 . Therefore, we deduce that 2MC1 ¼ MC2 min (MB, BC2) ¼ min (MB, MA). Hence, it follows that 2 max (MA1, MB1, MC1) 2MC1 min (MB, MA) min (MA, MB, MC). Thus, 2 max (MA1, MB1, MC1) min (MA, MB, MC). 7.2 Solutions 313

Figure 7.12 B

C2 C1 A1 A2

(c) Suppose that ∠MAB þ ∠MBC þ ∠MCA min (∠A, ∠B, ∠C), then ∠MAB < ∠B À ∠MBC ¼ ∠MBA, hence MB < MA. Similarly, we obtain that ∠MBC < ∠C À ∠MCA ¼ ∠MCB. Thus, MC < MB, and ∠MCA < ∠A À ∠MAB ¼ ∠MAC. Hence, MA < MC. Thus, we have obtained that MA < MC < MB < MA. This leads to a contradiction. Hence, ∠MAB þ ∠MBC þ ∠MCA > min (∠A, ∠B, ∠C). Therefore, it follows that ∠A À ∠MAB þ ∠B À ∠MBC þ ∠C À ∠MCA > min (∠A,∠B,∠C). Hence, ∠MAB þ ∠MBC þ ∠MCA < π À min (∠A,∠B,∠C). (d) Let us draw the diagonals AC and BD dividing the quadrilateral into four triangles AOB, BOC, COD, DOA.PointM is in one of these triangles. (If point M is on the common border of two triangles, then one can consider it being in both triangles.) If point M is in triangle COD, then according to problem 7.1.14c, we have that ∠MBC þ ∠MCD þ∠MDB > min (∠BDC,∠DBC) ¼ φ, where φ ¼ min (∠BAC, ∠DAC). Consequently, we deduce that

∠MAB þ ∠MBC þ ∠MCD þ ∠MDA ∠OAB þ ∠MBC þ ∠MCD þ ∠MDB þ ∠ODA ¼ π π ¼ þ ∠MBC þ ∠MCD þ ∠MDB > þ φ: 2 2

If point M is in triangle AOD, then by making a change of variables A ! D, B ! A, C ! B, D ! C, we obtain that M 2 ΔCOD. This ends the proof as min (∠BAC,∠DAC) ¼ min (∠ABD,∠CBD). The proof for the other cases can be done similarly. 2 2 7.1.15. Let max(AB, BC, AC) ¼ AC, then AB2 þ BC2 pACffiffi þ pACffiffi ¼ AC2. 2 2 Thus, ABC is a non-obtuse triangle. We need to prove that the inequality holds true for the vertices of triangle ABC. 2 2 2 2 2γ 2 2γ For vertex C, we have to prove that a þ b hc þ a cos þ b cos ; this is γ 2 γ 2 2 2 2 2 ðÞa sin þ ðÞb sin hc,orha þ hb hc (let min(AB, BC, AC) ¼ AB). Indeed, we have that h2 h2 4S2 4S2 4S2 4S2 h2. a þ b ¼ a2 þ b2 2c2 þ 2c2 c It remains to note that, if M(x; y) satisfies the condition 2 2 2 2 2 2 MA þ MB þ MC MA1 þ MB1 þ MC1, then it is equivalent to that the numbers x and y satisfy the inequality a1x þ b1y þ c1 0 for some constant numbers a1, b1, c1. Thus, the inequality holds true for any points inside triangle ABC. 314 7 Miscellaneous Inequalities

7.1.16. (a) We have that

MB1 þ MC1 ¼ MAðÞsin ∠MAC þ sin ∠MAB α ∠MAC À ∠MAB ¼ 2MA Á sin cos > 2 2 pffiffiffi α α 2 > 2MA Á sin cos ¼ MA Á sin α MA, 2 2 2 pffiffi since 90 α 45. We have obtained that MB þ MC > 2 MA. Similarly, we pffiffi 1 p1ffiffi 2 > 2 > 2 deduce that MB1 þ MA1 2 MC and MA1 þ MC1 2 MB. Summing up these three inequalities, we deduce that pffiffiffi MA þ MB þ MC < 2 2ðÞMA1 þ MB1 þ MC1 :

(b) Note that, if M A, then AA AA 1 1 MA þ MB þ MC ¼ AB þ AC ¼ 1 þ 1 ¼ MA þ sin β sin γ 1 sin β sin γ 1 1 1 1 1 1 max þ ; þ ; þ Á MA ¼ sin α sin β sin β sin γ sin γ sin α 1 1 1 1 1 1 1 ¼ max þ ; þ ; þ Á ðÞMA þ MB þ MC : sin α sin β sin β sin γ sin γ sin α 1 1 1 ð7:10Þ

Similarly, one can prove the inequality (7.10) for vertices B and C. If points M and M satisfy the condition (7.10) and M 2 [M M ], with MM1 ¼ λ 1 2 1 2 M1M2 (0 < λ < 1), then we have that

! ! ! AM ¼ AM ¼ λAM2 þ ðÞ1 À λ AM1 λAM2 þ ðÞ1 À λ AM1 ð7:11Þ and ρ(M, BC) ¼ ρ(M2, BC) Á λ þ (1 À λ) Á ρ(M1, BC). As, at least for one of the vertices of ABC the condition (7.11) is an inequality, then

λ λ AM þ BMþ CM ðÞAM2 þ BM2 þ CM2 þ ðÞ1 À ðÞAM1 þ BM1 þ CM1 À 1 1 ; 1 1 ; 1 1 ρ ; λ λ ρ ; max α þ β β þ γ γ þ α ðÞÁM2 BC þ ðÞÁ1 À ðÞþM1 BC sin sin sin sin sin sin Á ρ ; λ λ ρ ; ρ ; λ λ ρ ; þ ðÞÁM2AB þ ðÞÁ1 À ðÞþM1 AB ðÞÁM2 AC þ ðÞÁ1 À ðÞM1 AC ¼ 1 1 1 1 1 1 ¼ max þ ; þ ; þ Â ðÞρðÞM;BC þ ρðÞM;AB þ ρðÞM;AC : sinα sinβ sinβ sinγ sinγ sinα

We have obtained that the inequality holds true for point M. This ends the proof. 7.2 Solutions 315

7.1.17. Given a point M inside of the tetrahedron ABCD, we have to prove that MA þ MB þ MC þ MD < AB þ BC þ CD þ AC þ AD þ BD ¼ p. Denote by f(M) ¼ MA þ MB þ MC þ MD. Let us prove that set of points M, such that f(M) < p, is a convex ﬁgure. Indeed, let f(M ) < p, f(M ) < p, M 2 [M M ] and MM1 ¼ λ. Then, f(M) < λf 1 2 1 2 M1M2 (M2) þ (1 À λ)f(M1) < λp þ (1 À λ)p ¼ p (see the proof of problem 7.1.16b). Note that, f(A) ¼ AB þ AC þ AD < p and similarly f(B) < p, f(C) < p, f(D) < p. Thus, for any point M of the tetrahedron f(M) < p. See also the proof of problem 7.1.38e.

7.1.18. (a) Let the distance a between pairs of points A1, A2,...,A7 on the plane occur at most k times. Denote by ni the number of the segments with lengths a, drawn from point Ai. Without loss of generality, one can assume that n1 n2 ... n7. Note that n1 þ n2 9; otherwise on the circles with centers A1 and A2 and radius a will be located at least n1 þ (n2 À 2) 8 of seven points. This leads to a contradiction. It is clear that the distance between any two of points A1, A2, A3, A4 is not equal to a. Let AiAj 6¼ a (i 6¼ j). In that case, on the circles with centers Ai, Aj and radius a are located at least ni þ nj À 2 points. Since, Ai, Aj are not on those circles, ni þ nj À 2 þ 2 7. Hence, n3 þ n4 ni þ nj 7. Thus, we deduce that n4 3. Con- sequently, n7 n6 n5 3. 1 1 We have that k ¼ 2 ðÞðÞþn1 þ n2 ðÞþn3 þ n4 n5 þ n6 þ n7 2 ð9 þ 7 þ 3þ 3 þ 3Þ. Hence it follows that k 12. In a regular hexagon with a side a, the distance a for the vertices and center occurs 12 times. Thus, k ¼ 12. (b) In a similar way, as for the proof of problem 7.1.18a, one can obtain that 1 k 2 ðÞ8 þ 6 þ 3 þ 3 ¼10. However, it is easy to check that the cases n1 ¼ 5, n2 ¼ n3 ¼ ...¼ n6 ¼ 3 and n1 ¼ n2 ¼ 4, n3 ¼ ...¼ n6 ¼ 3 are not possible. (In the last case for points Ai, the only point not located on the circle with a radius a and center A1, ni 2). Answer 9. The example is presented in Figure 7.13. 7.1.19. Let in a quadrilateral ABCD we have that AB ¼ a, BC ¼ b, CD ¼ c, DA ¼ d, the midline KM ¼ m and the distance between the midpoints of the diagonals PQ ¼ x.AsPKQM is a parallelogram with sides a/2 and c/2 (K is on BC, M is on 2 2 2 2 2 a2 c2 2 AD), then m ¼ KM ¼ 2PK þ 2KQ À PQ ¼ 2 þ 2 À x .

Figure 7.13 316 7 Miscellaneous Inequalities

Figure 7.14 M

D DЈ CЈ C AB

2 b2 d2 2 Similarly, if n is the second midline, then n ¼ 2 þ 2 À x . Thus, each midline accepts its greatest value when x has the smallest value. Therefore, in that case the sum of the midlines also accepts the greatest value. 1 However, by the triangle inequality, PQ ¼ x 2 c À a . Similarly, x 1 2 d À b . For the simplicity, assume that |d À b| |c À a|. In this case, the greatest value of 1 the sum of the midlines is reached if, x ¼ 2 d À b , this means that the required quadrilateral is a trapezoid with bases b,d and lateral sides a,c (or a parallelogram). bþd qTheffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sum of midlines of such trapezoid is equal to 2 þ 1 2 2 2 2 2a þ 2c À ðÞd À b . 7.1.20. (a) Let us consider two cases. 1. If max (AB, BC, CD, DA, AC, BD) ¼ max (AB, BC, CD, DA). Let max (AB, BC, CD, DA) ¼ AB, then one can assume that points C and D are inside of the figure AMB (Figure 7.14). In the case, if ray BD intersects the arc BM, we have that the quadrilateral ABCD is inside of the quadrilateral ABC0D0. Hence, it follows that AB þ BC þ CD þ DA AB þ BC0 þ C0D0 þ D0A (see the problem 2.1.1). ∠ 0 0 α ∠ 0 β α β π 0 0 Let AB ¼ R, C AD ¼ 2 and C AB ¼ 2 , then 2 þ 2 3 and C D þ 0 α β αþβ π C B ¼ 2R(sin þ sin )4R sin 2 4R sin 12. Therefore, AB þ BC þ CD þ αþβ π DA 2R þ 4R sin 2 2 þ 4 sin 12. Similarly, one can consider the case when ray AC interests the arc AM. It remains to consider the case presented in Figure 7.15. ∠ 0 π ∠ 0 π ∠ 0 0; ∠ 0 0 2π Since BAD 3 and ABC 3, then minðÞAD C BC D 3 . ∠ 0 0 2π 0 0 0 Let AD C 3 .WehavethatAB þ BC þ CD þ DA AB þ BC þ C D þ D0A.IfweprovethatAD0 þ D0C0 AM þ MC0, then

AB þ BC þ CD þ DA AB þ BC0 þ C0D0 þ D0A AB þ BC0 þ C0M þ MA π π 2R þ 4R sin 2 þ 4 sin : 12 12

∠ 0 5π ∠ 0 0 2π ∠ 0 ∠ 0 0 Indeed, since AD M ¼ 6 and AD C 3 , then AD K C D M. 7.2 Solutions 317

Figure 7.15 T M

DЈ CЈ DC K

Figure 7.16 CЈ A

KNDЈ MT

AЈ

∠ < π < ∠ 0 ∠ 0 ∠ 0 < π While AMK 6 C MT,as AMC ¼ AC M 2. We have that ∠AD0K > ∠AMK. Let A0 be the point symmetric to A with respect to line KT (Figure 7.16). Then from the conditions ∠AD0K ∠C0D0T and ∠AMK < ∠C0MT it follows that, points N, D0, M are on line KT, as it is shown in Figure 7.15. According to problem 2.1.1, we have that A0C0 þ A0D0 þ C0D0 A0C0 þ A0M þ MC0. Hence, we obtain that AD0 þ C0D0 AM þ MC0. 2. If max(AB, BC, CD, DA, AC, BD) > max (AB, BC, CD, DA). Let max(AB, BC, CD, DA, AC, BD) ¼ AC and l be the perpendicular bisector of segment AC. One can assume that points B and D are in the same half-plane with the boundary l (see Figure 7.17a). Indeed, otherwise we have for the quadrilateral ABCD0 that AD0 ¼ CD, CD0 ¼ AD and BD0 < BN þ ND0 ¼ BD. Now, let l || AC, l || AC and l || BD (see Figure 7.17b) and points B0 and D0 1 2 3 0 0 !0 !0 0 0 0 0 are, such that B 2 l1, D 2 l2, BB ¼ DD and max(AB , B C, CD , D A) ¼ AC. 0 Since max(CD0, CB0) > AC, then we can assume that D 2 [DD0]. Now it is clear that the quadrilateral AB0CD0 is convex and that max(AB0, B0C, CD0, D0A, AC, B0D0) ¼ max (AB0, B0C, CD0, D0A) ¼ AC 1. Moreover, AB þ BC þ 0 0 CD þ DA AB0 þ B0C þ CD0 þ D0A. According to the case (a), it followsAB þ B 0 0 π π Cþ CD þ D A 2 þ 4sin12.Hence,AB þ BC þ CD þ DA 2þ 4sin12. This solution was proposed by D. Harutyunyan and A. Kalantaryan, Ninth grade. (b) Let, apart from BD, all other distances are not greater than 1. According to pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi problem 4.1.3, if AC ¼ c, then BD 4 À c2. Hence AC þ BD c þ 4 À c2. Let c ¼ 2 sin α, where 0 α π. Then AC þ BD 2 sin α þ 2 cos α ¼ pffiffiffi ÀÁpffiffiffi ÀÁ6 pffiffiffi 2 2 sin α þ π 2 2 sin π þ π ¼ 1 þ 3. 4 6 4 pffiffiffi Thus, we obtain that AB þ BC þ AD þ CD þ AC þ BD 5 þ 3. 318 7 Miscellaneous Inequalities

l l3 l B B0 B¢ Bl1

A C AC

l2 D¢ D D0 D¢ D

a b

Figure 7.17

Draw the altitudes SE and CF of triangles ASB and ACB, respectively.pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Since AF þ FB a, then AF a or FB a. Let AF a, then CF ¼ AC2 À AF2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 qffiffiffiffiffiffiffiffiffiffiffiffi 2 a2 a2 1 À AF 1 À 4 . Similarly, we deduce that SE 1 À 4 . 1 We have that V ¼ 3 SABCSH, where SH is the altitude of the tetrahedron SABC. 1 ABÁCF a 1 a2 1 3 Hence, V ¼ 3 Á 2 Á SH 6 CF Á SE 6 a 1 À 4 8,asa À 4a þ 3 ¼ (a À 1) (a2 þ a À 3) 0(a 1, a2 þ a À 3 2 À 3 < 0). ⊥ 1 Note that if SB ¼ SA ¼ AB ¼ AC ¼ BC ¼ 1 and (SAB) (ABC), then V ¼ 8. 7.1.21. If line l does not contain any of the given points, then one can move it parallel to itself, so that the ρ(l) does not increase and it approaches to one of the given points. Draw a straight line through one of the given points and start turning it around that point. Let φ be the turning angle, 0 φ < π. Now, if φk, k ¼ 1, . . . , m (m n À 1), be the values of φ, at which the line passes (sequentially) through the remaining points of our set, then ρ(l) can be expressed as ρðÞ¼l Pm φ φ > ak sin ðÞÀ k , ak 0. k¼1 φ φ But on each of the intervals [ k, k þ 1] the considered function, being a sum of convex (upward) functions, is itself a convex function, consequently, cannot reach the minimal value inside that interval. 7.1.22. From triangles AND, BNC, ABN, CND, according to the law of cosines, we obtain that

AD2 þ BC2 À AB2 À CD2 ¼ AN2 þ ND2 À 2AN Á ND Á cos αþ þ BN2 þ CN2 À 2BN Á CN Á cos α À AN2 À BN2 À 2AN Á BN Á cos α À CN2 À ND2À À 2CN Á ND Á cos α ¼À2AC Á BD Á cos α, where α ¼ ∠AND. 7.2 Solutions 319

Figure 7.18 M

BB1 C

B2 C2 А1 N C1

A2 D2

AD1 D

Since AD þ BC ¼ AB þ CD, then

À2AC Á BD Á cos α ¼ ¼ AD2 þ BC2 À AB2 À CD2 ¼ ¼ 2ðÞÁAD À AB ðÞAD À CD 0:

Therefore, ∠AND ¼ α 90. We need to prove that the center O of the incircle of the quadrilateral ABCD is _ _ inside of triangle AND. Note that DD BB , hence D1 D2C1 A1 B2B1. _ _ 1 1 Thus, A2 D2C2 A2 B2C2 (see Figure 7.18). _ _ Similarly, we deduce that B2 A2D2 B2 C2D2. Consequently, point O is inside ∠ ∠ 1 ∠ of triangle AND. Hence, AND AOD ¼ 90 þ 2 AMD. 7.1.23. Denote the midpoints of segments AB and CD by M and N, respectively. We need to prove that AB2 þ CD2 þ 2MN2 < 6. Let A0, B0, C0, D0, M0, N0 be the points, symmetric to points A, B, C, D, M, N with respect to the center of the cube. Consider the parallelograms D0CDC0, ABA0B0and MNM0N0. Since in a parallelogram the sum of the squares of the adjacent sides does not exceed the square of the largest diagonal, 0 0 0 0 then CD2 þ NN 2 < 3andAB2 þ MM 2 < 3. Therefore, CD2 þ AB2 þ NN 2 þ MM 2 < 6. Thus, CD2 þ AB2 þ 2MN2 þ 2M0N2 < 6. Hence, CD2 þ AB2 þ 2MN2 < 6. Since, for x, y, z 0, we have that ÀÁffiffiffi ffiffi ÀÁffiffiffi ÀÁffiffi ÀÁffiffi ffiffiffi pffiffiffiffiffiffiffi 1 p pffiffiffi p p pffiffiffi 2 pffiffiffi p 2 p p 2 x þ y þ z À 3 3 xyz ¼ 3 xþ 3 yþ 3 z 3 xÀ 3 y þ 3 yÀ 3 z þ 3 zÀ 3 x 0: 2 Thus, we deduce that pffiffiffiffiffiffiffi x þ y þ z 3 3 xyz: ð7:12Þ 320 7 Miscellaneous Inequalities

2 2 2 pAccordingffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi to the inequality (7.12), it follows CD þ AB þ 2MN 3 2 2 2: 3 CD Á AB Á 2MN pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Hence, we obtain that 6 > 3 3 CD2 Á AB2 Á 2MN2 3 3 2AB2 Á CD2d2. Therefore, AB Á CD Á d < 2. This ends the proof. 7.1.24. Given a tetrahedron ABCD inside of the parallelepiped with the volume V. Draw through point A a line l not parallel to plane BCD. Let A1 and A2 be the intersection points of the straight line l with faces of the parallelepiped. Then, < ; VABCD maxðÞVA1BCD VA2BCD . ; Let maxðÞVA1BCD VA2BCD ¼VA1BCD. Let us draw through point A1 a line intersecting the edges of the parallelepiped at points A3 and A4. Therefore, ; VA1BCD maxðÞVA3BCD VA4BCD ¼VA3BCD. If point A3 belongs to the edge A5A6 ; of the parallelepiped, then VA3BCD maxðÞVA5BCD VA6BCD . Thus, we have proven < that there exists a vertex A0 of the parallelepiped, such that VABCD VA0BCD. Similarly, one can prove that there exist vertices B0, C0, D0 of the parallelepiped, such that

< : VABCD VA0BCD VA0B0CD VA0B0C0D VA0B0C0D0

V V It remains to note that VA0B0C0D0 ¼ 3 or 6.

7.1.25. Let A1 and C1 be the midpoints of sides BC and AB, respectively. Note that < 1 point K belongs to segment A1C; otherwise we deduce that SBMK SABA1 ¼ 2 SABC. BC This leads to a contradiction. Consequently, BK 2 . AB Similarly, one can prove that BM 2 . Hence, it follows that

AB þ BC ABþBC MB þ BK 2 2 > 2 1 : AB BC ABþBC ¼ AM þ CA þ KC 2 þ CA þ 2 2 þ AB þ BC 3

7.1.26. Let ABC be a triangular section of the cube with the edge 2 and the center D (1;1;1) (see Figure 7.19).

Figure 7.19 z

ОB Ay

x 7.2 Solutions 321

Denote by OA ¼ a, OB ¼ b, OC ¼ c, then the equation of plane ABC is x y z а ρ 2 2 2 2 a þ b þ c ¼ 1, nd (D, ABC) ¼ 1. It is known that SABC ¼ SAOB þ SBOC þ SAOC jj1þ1þ1À1 and ρðÞ¼D; ABC pa ffiffiffiffiffiffiffiffiffiffiffiffiffib c . 1 þ 1 þ 1 a2 b2 c2 We have to prove that, if 0 < a, b, c 2 and 2(a þ b þ c) À 2(ab þ bc þ ca) þ abc ¼ 0, then a2b2 þ b2c2 þ c2a2 < 16. Note that

a2b2 þ b2c2 þ c2a2 ¼ ÀÁ ab bc ca 2 2abc a b c a b c abc 2 2abc a b c ¼ ðÞÀÁþ þ À ðÞ¼þ þ þ þ þ 2 À ðÞ¼þ þ abc 2: ¼ a þ b þ c À 2

abc < Hence, we have to prove that a þ b þ c À 2 4. abc abc > We have that a þ b þ c À 2 3a À 2 0 (let min(a, b, c) ¼ a). Since 8 À 2(a þ b þ c) ¼ (2 À a)(2 À b)(2 À c) 0, then a þ b þ c 4. Hence, abc < a þ b þ c À 2 4.ÀÁ Note that the area of the total surface of the tetrahedron 1 abc ab bc ac OABC is equal to 2 a þ b þ c À 2 þ 2 þ 2 þ 2 ¼ a þ b þ c 4. Thus, we have obtained that the area of total surface of the tetrahedron OABC is not greater than the area of the face of the cube. 7.1.27. (а) Let α 60, then according to problem 3.1.6a α 2 1 2 1 SABC l tg pffiffiffi l > pffiffiffi : a 2 3 a 3

(b) Let minðÞ¼a; b; c c < p2ffiffi, then S ¼ chc clc < p1ffiffi. 3 ABC 2 2 3 If minðÞa; b; c p2ffiffi and max(a, b, c) ¼ a, then α 60. Let point B on side BC 3 1 be such that ∠BAB1 ¼ 60 . Then ρ(B, AB1) ¼ AB Á sin 60 1. Since lb ρ(B, AB1) 1, then lb 1. This leads to a contradiction. (c) See the proof of problem 7.1.27b. ÀÁpffiffiffi 2 pffiffiffi 7.1.28. Answer 2 2 À 1 . Let a ¼ b and c ¼ 2a þ 1, where n 2 N and ÀÁpffiffiffi n > 1 2 2 2 > 2 2 À 2 a n. Then for n !1, from the inequality a þ b þ c k(a þ b þ c) , ÀÁpffiffiffi 2 we deduce that k 2 2 À 1 . ÀÁpffiffiffi 2 We need to prove that a2 þ b2 þ c2 > 2 2 À 1 ðÞa þ b þ c 2. ÀÁpffiffiffi 2 2 2 > Let max(ffiffiffi a, b, c) ¼ c, a þ b ¼ 1 and k0 ¼ 2 2 À 1 , then c 1 and p 2 2 2 2 a þ b 2. Consider a function f(x) ¼ apffiffiþ b þ x À k0(a þ b þ x) on the interval (1, þ1). Note that x ¼ k0 ðÞa þ b 2k0 < 1. Hence, it follows that b 1Àk0 1Àk0 ffiffiffi 2 2 p fcðÞ> f ðÞ¼1 2 À k0ðÞa þ b þ 1 2 À k0 2 þ 1 ¼ 0:

ÀÁpffiffiffi 2 Thus, a2 þ b2 þ c2 > 2 2 À 1 ðÞa þ b þ c 2. 322 7 Miscellaneous Inequalities

7.1.29. (a) From the condition a þ b þ c ¼ 1, according to the triangle inequality, < 1 1 > 1 we deduce that a, b, c 2. Denote by a ¼ 2 À k, then k 0 and c ¼ 2 þ k À b .We < 1 < 1 have to prove that 2k(k À b)(1 À 2b) 0. Note that c ¼ 2 þ k À b 2, consequently k À b < 0. (b) Note that ÀÁÀÁÀÁ a 2a2 À b2 À c2 þ b 2b2 À a2 À c2 þ c 2c2 À a2 À b2 ¼ ÀÁ ÀÁÁ À ÀÁÁ À ¼ aa2 À b2 þ aaðÞþ2 À c2 bb2 À a2 þ b b2 À c2 þ cc2 À a2 þ c c2 À b2 ¼ ¼ ðÞa þ b ðÞa À b 2 þ ðÞa þ c ðÞa À c 2 þ ðÞb þ c ðÞb À c 2 0 Hence, a(2a2 À b2 À c2) þ b(2b2 À a2 À c2) þ c(2c2 À a2 À b2) 0. Remark The inequality holds true, only if a, b, c 0. (c)Denote by a ¼ m þ n, b ¼ n þ k, c ¼ m þ k. Then, m ¼ p À b, n ¼ p À c, k ¼ p À a, hence m, n, k > 0. We have that

a2baðÞþÀ b b2cbðÞþÀ c c2acðÞÀ a ¼ km3 þ mn3 þ nk3 À k2mn À m2nk À n2mk ¼ ¼ kmðÞ m À n 2 þ mnðÞ n À k 2 þ nkðÞ k À m 2 0

Consequently, a2b(a À b) þ b2c(b À c) þ c2a(c À a) 0. (d) Note that pffiffiffi pffiffiffi pffiffiffi aðÞþa þ c À b bðÞþa þ b À c cðÞ¼b þ c À a pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ aaðÞþ c À b Á a þ c À b þ baðÞþ b À c Á a þ b À c þ cbðÞþ c À a Á b þ c À a ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! ¼ aaðÞþ c À b ; baðÞþ b À c ; cbðÞþ c À a Á a þ c À b; aþ b À c; b þ c À a

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! p p p aaðÞþ c À b ; baðÞþ b À c ; cbðÞþ c À a Á a þ c À b; a þ b À c; b þ c À a ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ aaðÞþþ c À b baðÞþþ b À c cbðÞþ c À a Á a þ c À b þ a þ b À c þ b þ c À a ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ ¼ ðÞa þ b þ c a2 þ b2 þ c2 : qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi aþbÀcþbþcÀa (e) Note that a þ b À c þ b þ c À a 2 2 ¼ 2 b. Similarly, we pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi obtain that a þ b À c þ a þ c À b 2 a and b þ c À aþ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ c À b 2 c. Summing up these three inequalities, we get a þ b À cþ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi b þ c À a þ a þ c À b ÀÁa þ bÀÁþ c.ÀÁ a 2 b 2 c 2 a b c < 1 (f) According to problem 7.1.29a, 2 þ 2 þ 2 þ 42 Á 2 Á 2 2, it follows that a2 þ b2 þ c2 þ 2abc < 2. (g) Let max(a, b, c) ¼ b, we have to prove that f(b) ¼ (2a À c)b2 þ (2c2 À a2 À 3ac) b þ ac(2a À c) 0. 7.2 Solutions 323

Note that f(a) ¼ a(a À c)2 0, f(c) ¼ c(a À c)2 0 and f(a þ c) ¼ a3 þ c(a À c)2 > 0. If 2a À c 0, f(b) min ( f(c), f(a þ c)) 0, as c b < a þ c. a2 3ac 2c2 > ϐ þ À If 2a À c 0 and c a, we have that b ep ¼ 22ðÞaÀc c. Hence, f(b) f(c) 0. a2 3ac 2c2 > ϐ þ À < If a À c 0, we have that b ep ¼ 22ðÞaÀc a. Hence, f(b) f(a) 0. Second Solution Using the notations a ¼ m þ n, b ¼ n þ k, c ¼ m þ k, the given inequality can be rewritten in the form m3 þ n3 þ k3 þ m2n þ k2m þ n2k 2(m2k þ n2m þ k2n). The last inequality holds true, since m3 þ k2m 2m2k, n3 þ m2n 2n2m, k3 þ n2k 2k2n.

a b c a c b jjaÀb jjbÀc jjcÀa < < (h) Note that b þ c þ a À c À b À a ¼ c Á a Á b 1, since ja À bj c, jb À cj < a, jc À aj < b. (i) Note that one of the numbers pq, qr, pr is non-negative. Let pq 0, then ÀÁ a2pq þ b2qr þ c2rp ¼ a2pq À b2q þ c2p ðÞ¼p þ q ¼ÀðÞb þ c À a ðÞb þ c þ a pq À ðÞbq À cp 2 0:

(j) Note that

2 2 2 a ðÞþ2ÀÁb þ 2c À a b ðÞþ2c ÀÁþ 2a À b c ðÞ¼2a þÀÁ2b À c ¼ b 2a2 þ 2c2 À b2 þ a 2b2 þ 2c2 À a2 þ c 2a2 þ 2b2 À c2 ¼ 4m2a þ 4m2b þ 4m2c ¼ ÀÁ ÀÁ ÀÁ a b c 2 2 2 2 2 2 ¼ 9 3ma aþ 3mb b þ 3mc c 9abc

(see problem 4.1.8b). (k) Note that

ðÞa þ b þ c 2 ¼ a2 þ b2 þ c2 þ 2ab þ 2bc þ 2ac < < abðÞþþ c baðÞþþ c caðÞþþ b 2ab þ 2bc þ 2ac ¼ 4ab þ 4bc þ 4ac

Hence abþbcþac > 1. ðÞaþbþc 2 4 (l) Note that

ðÞa þ b þ c 3 ¼ ðÞa þ b 3 þ 3ðÞa þ b 2c þ 3ðÞa þ b c2 þ c3 < < 3 2 2 ðÞa þ b þ 3ðÞa þ b c þ3ðÞa þ b c þ ðÞa þ b c2 ¼ ðÞa þ b ðÞa þ b 2 þ 3ðÞa þ b c þ 4c2 ¼ ¼ ðÞa þ b ðÞa À b 2 þ 4ab þ 3ðÞa þ b c þ 4c2 : 324 7 Miscellaneous Inequalities

Since |a À b| < c and |a À b| < a þ b, then (a À b)2 < c(a þ b). Thus, it follows that ðÞa þ b þ c 3 < ðÞa þ b ðÞa À b 2 þ 4ab þ 3ðÞa þ b c þ 4c2 < < ðÞa þ b ðÞcaðÞþþ b 4ab þ 3ðÞa þ b c þ 4c2 ¼4ðÞa þ b ðÞa þ c ðÞb þ c :

Hence ðÞaþb ðÞaþc ðÞbþc > 1. ðÞaþbþc 3 4 (m) a3 þ b3 þ c3 ¼ a2 Á a þ b2 Á b þ c2 Á c < a2ðÞþb þ c b2ðÞþa þ c c2ðÞb þ a < <(a þ b þ c)(ab þ bc þ ac). 2 a b c a2 b2 c2 ðÞaþbþc 3 (n) bþc þ aþc þ aþb ¼ abþac þ abþcb þ acþbc 2ðÞabþbcþac 2. Let maxðÞ¼a; b; c c, then a b c a b c c þ þ þ þ ¼ 1 þ < 2: b þ c a þ c a þ b b þ a a þ b a þ b a þ b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 À (o) We have that b2 þ c2 À a2 þ a2 þ c2 À b2 2 b2 þ c2 À a2 þ c2þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 2 2 2 2 2 a À b Þ¼4c . Consequentlypffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib þ c Àpaffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ a þ c À b 2c. 2 2 2 2 2 2 pSimilarly,ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi wep getffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi that b þ c À a þ a þ b À c 2b and a2 þ b2 À c2 þ a2 þ c2 À b2 2a. pSummingffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi upp theffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi last threepffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi inequalities, we deduce that b2 þ c2 À a2 þ a2 þ c2 À b2 þ a2 þ b2 À c2 a þ b þ c.

abc abc abc 2 2 2 (p) We need to prove that aþbÀc þ aÀbþc þ bþcÀa a þ b þ c . Indeed, without loss of generality one can assume that a b c. Then, we have that

abc abc abc À a2 þ À b2 þ À c2 ¼ b þ c À a a þ c À b a þ b À c aaðÞÀ b ðÞa À c bbðÞÀ a ðÞb À c ccðÞÀ a ðÞc À b ¼ þ þ b þ c À a a þ c À b a þ b À c aaðÞÀ b ðÞa À c bbðÞÀ a ðÞb À c baðÞÀ b ðÞb À c bbðÞÀ a ðÞb À c þ þ b þ c À a a þ c À b b þ c À a a þ c À b baðÞÀ b ðÞb À c bbðÞÀ a ðÞb À c þ ¼ 0: a þ c À b a þ c À b

Note that 7.2 Solutions 325

qffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffi 2 abc abc abc abc þ abc þ abc ¼ þ þ þ aþb0Àc aÀbþc bþcÀa a þ b À c a À b þ c b1þ c À a B 1 1 1 C þ2abc@qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA c2 a b 2 b2 c a 2 a2 b c 2 À ðÞÀ À ðÞÀ À ðÞÀ 1 1 1 a2 þ b2 þ c2 þ 2abc þ þ ¼ ðÞa þ b þ c 2, a b c qffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffi abc abc abc Thus aþbÀc þ aÀbþc þ bþcÀa a þ b þ c. (q) We have that

a À b b À c c À a a À b b À а а À c c À a þ þ ¼ þ þ þ ¼ a þ b b þ c c þ a a þ b b þ c b þ c c þ a

ðÞa À b ðÞc À a ðÞа À c ðÞa À b ðÞa À b ðÞc À a ðÞc À b ¼ þ ¼ < ðÞa þ b ðÞb þ c ðÞb þ c ðÞc þ a ðÞa þ b ðÞb þ c ðÞc þ a c Á b Á a abc 1 < pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ : ðÞa þ b ðÞb þ c ðÞc þ a 2 ab Á 2 bc Á 2 ca 8

7.1.30. At ﬁrst, let us prove the following lemmas. Lemma 1 If max(AB, BC, AC, AX, BX, CX) ¼ max (AB, BC, AC), then m(ABC) m (ABX) þ m(ACX) þ m(BCX). aÁmðÞ ABX aÁmðÞ ACX Indeed, we have that SABC SABX þ SBCX þ SACX 2 þ 2 þ aÁmðÞ BCX 2SABC 2 , where a ¼ max (AB, BC, AC). Hence, mðÞ¼ ABC a mðÞþ ABX mðÞþ ACX mðÞ BCX . Lemma 2 If point C is inside of triangle ABX, then m(ABC) m(ABX). Consider two cases. (a) max(AB, BC, AC, AX, BX, CX) ¼ max (AB, BC, AC) ¼ a. Then from the inequal- aÁmðÞ ABC aÁmðÞ ABX ity SABC SABX , we obtain that 2 SABX 2 Thus m(ABC) m(ABX). (b) maxðÞAB; BC; AC; AX; BX; CX ¼AX. Then the perpendicular drawn from vertex B to the straight line AX intersects ray AC. Hence, m(ABC) m(ABX). If the convex envelope of points A, B, C, X is a quadrilateral, then by lemma 2, we have that m(ABO) þ m(ACO) þ m(BCO) m(ABX) þ m(ACX) þ m(BCX), while by using lemma 1, we deduce that m(ABO) þ m(ACO) þ m(BCO) m(ABC), where O is the intersection point of the diagonals of the quadrilateral with vertices A, B, C, X. If the convex envelope of points A, B, C, X is a triangle ABC, then from lemma 1, the proof of the problem follows straight away. 326 7 Miscellaneous Inequalities

If the convex envelope of points A, B, C, X is a triangle ABX, then m(ABC) m (ABX) m(ABX) þ m(ACX) þ m(BCX). If the convex envelope of points A, B, C, X is a segment, then m(ABC) ¼ 0 ¼ m (ABX) þ m(ACX) þ m(BCX). 0 bc 7.1.31. Using the property of the bisector of the triangle, we obtain that AB ¼ aþc BI aþc BI aþc AI bþc and IB0 ¼ b . Consequently BB0 ¼ aþbþc. Similarly, we get that AA0 ¼ aþbþc and CI aþb CC0 ¼ aþbþc. Then,

AI BI CI ðÞa þ c ðÞb þ c ðÞa þ b 1 Á Á ¼ > : AA0 BB0 CC0 ðÞa þ b þ c 3 4

(see the problem 7.1.29l). We have that ! 3 AI BI CI aþc þ bþc þ aþb 8 Á Á aþbþc aþbþc aþbþc ¼ : AA0 BB0 CC0 3 27 ÀÁ > xþyþz 3 See the proof of problem 7.1.23, for x, y, z 0, we have that xyz 3 .

7.1.32. Draw through point Q a perpendicular QQ0 to plane π and consider a circle ω(Q0, Q0R) on plane π and let R0 be a point, such that for any point R of the circumference ω we have PR PR0 (see Figure 7.20). Then, QPþPR QPþPR0. QR QR0 Consider Figure 7.21, where PM0 ¼ PR0. ∠ QPþPR0 QM0 sin QR0M0 1 QPþPR Then ¼ ¼ α α. Hence the value of is the greatest if QR0 QR0 sin 2 sin 2 QR ∠ α QPþPR QR0Q0 ¼ 90 À 2. When points Q0 and P coincide, the ratio QR reaches its greatest value at any point R belonging to the circumference of ω0(Q0, QQ0).

7.1.33. Let us take on the plane a point O and draw lines l1,...,ln parallel to the borders of the given strips. Let us choose a coordinate system with the point of origin at O. Consider those lines of the set l1,...,ln having points in the quarter I. Without loss of generality, one can assume that the strips corresponding 11 to these lines are such that the sum of their widths is not less than 2 . Now, consider line l that is a bisector of the second quarter (Figure 7.22) and angles α1, α2,...,αk, formed by the considered lines with line l.

Figure 7.20 Q

w R p

R0 Q0 P 7.2 Solutions 327

Figure 7.21 Q

R0 Q0 P

a 2

Figure 7.22 y lk

a1 a2 ak

Figure 7.23 C

P1 P2 Pk A a1 a2 ak B l

Let us arrange on the plane the strips Π1, Π2,...,Πk corresponding to lines l1, l2,...,lk in a way shown on Figure 7.23. Then, it is clear that triangle ABC is Π Π 11 α π π α π entirely covered by the strips 1,..., k, AB 2 and 1 4 , À k 4. Hence, 328 7 Miscellaneous Inequalities

AB AB α π α pffiffi for the inradius r of triangle ABC, the condition r ¼ 1 À k π 11 > ctg 2 þctg 2 2ctg 8 4 ðÞ2À1 1 holds true. Therefore, r > 1. ! To complete the proof one has to make a parallel translation by a vector OO1, where O is the incenter of triangle ABC, and O1 is the center of the given circle. pffiffi ∠ AB AB < 3 3 7.1.34. Let C ¼ 90 .IfAB 2, then SABC ¼ 2 Á hc hc mc ¼ 2 1 2 . Let ω1 and ω2 be given circles. If AB > 2, then one can assume that A, C 2 ω1.

Let us choose a point C1 on segment CB, such that AC1 ¼ 2. Then,pffiffiffiffiffiffiffiffi it is clear that B, ω ðÞ2þx 4Àx2 C1 2 2. Hence, it follows that BC1 2, and SABC 2 , where CC1 ¼ x. Note that pffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞ2 þ x 4 À x2 ¼ 2 ffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p p 2 2 3 3 2 þ x 2 þ x 2 þ x 3 3 2þx þ 2þx 2þx þ 2 À x ¼ Á Á ðÞ2 À x 3 3 Á 3 2 3 3 3 2 2 2 ffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi p 4 p 3 3 2þx þ 2þx þ 2þx þ 2 À x 3 3 3 3 3 ¼ : 2 4 2

pﬃﬃ 3 3 Hence, SABC 2 . 7.1.35. (a) Let us display on edges of the trihedral angle equal segments SA, SB, SC, where S is the vertex of the trihedral angle. Let point O be the foot of the perpendicular drawn from point S to plane ABC, Then, OA ¼ OB ¼ OC. Take on segment SC1 point O1, so that O1C1 ¼ C1O, where C1 is the midpoint of segment AB. Then, we have that ∠AOB ¼ ∠AO1B ¼ ∠ASB þ ∠O1AS þ ∠O1BS > ∠ASB. Consequently, ∠ASB þ ∠BSC þ ∠ASC < ∠AOB þ ∠BOC þ ∠AOC 360, thus

∠ASB þ ∠BSC þ ∠ASC < 360 : ð7:13Þ

Let ASC be the largest plane angle in the trihedral angle SABC. Using (7.13) we obtain that for the trihedral angle SAB0C, where B0 is the point, symmetric to point B with respect to point S 180 À ∠ASB þ 180 À ∠BSC þ ∠ASC < 360,hence∠ASC < ∠ASB þ ∠BSC. (b) Let us produce a plane crossing all sides of the given polyhedral angle with a vertex S. Let the resulting section be a convex n-gon A1A2,...,An. According to problem 7.1.35a, we have that

180 ðÞ¼n À 2 ∠AnA1A2 þ ∠A1A2A3 þ ::: þ ∠AnÀ1AnA1 <

< ðÞ∠AnA1S þ ∠A2A1S þðÞ∠A1A2S þ ∠A3A2S þ::: þ ðÞ∠AnÀ1AnS þ ∠A1AnS ¼ ¼ ðÞ∠AnA1S þ ∠A1AnS þ::: þ ðÞ∠A2A1S þ ∠A1A2S ¼180 n À ∠A1SA2 À ::: À ∠AnSA1:

Therefore, ∠A1SA2 þ ∠A2SA3 þ ...þ ∠AnSA1 < 360 . 7.2 Solutions 329

Figure 7.24 D

AB D1

(c) Let AnSA1 be the largest plane angle in the n-hedral angle SA1A2 ...An. According to the problem 7.1.35a, we have that ∠A1SA2 þ ∠A2SA3 þ ...þ ∠An À1SAn> ∠A1SA3þ ∠A3SA4 þ ...þ ∠An À 1SAn ∠A1SA4 þ ∠A4SA5 þ ...þ ∠An À 1SAn ... ∠A1SAn, hence ∠A1SA2 þ ∠A2SA3 þ ...þ ∠An À 1SAn > ∠A1SAn. (d) We have to prove that ∠AMB þ ∠BMC > ∠CMD1 (see Figure 7.24). We have that

∠AMB þ ∠BMC ¼ ∠AMB þ ∠BMD2 þ ∠D2MC > ∠AMD2 þ ∠D2MC ¼ ¼ ∠AMD1 þ ∠D1MD2 þ ∠D2MC > ∠AMD1 þ ∠CMD1 > ∠CMD1:

Hence, ∠AMB þ ∠BMC > ∠CMD1. (e) According to the solution of problem 7.1.35d, we have that ∠AMB þ ∠BMC >>∠AMD1 þ ∠CMD1 ¼ π À ∠AMD þ π À ∠CMD. Consequently, ∠AMB þ ∠BMC þ ∠AMD þ ∠CMD > 2π. Similarly, one can prove the inequalities ∠AMC þ ∠BMD þ ∠AMB þ ∠CMD > 2π, ∠AMD þ ∠BMC þ ∠BMD þ ∠AMC > 2π. Summing up the last three inequalities we obtain the required inequality. (f) Let O be any point inside of the trihedral angle with the dihedral angles α, β, γ. Draw through point O rays OA, OB, OC perpendicular to the faces of the trihedral angle. According to problem 7.1.35b, we have that π À α þ π À β þ π À γ < 2π, consequently α þ β þ γ > π. (g) Let O be any point inside of the tetrahedron with the dihedral angles α, β, γ, α1, β1, γ1. Let rays OA, OB, OC, OD be perpendicular to the faces of the tetrahedron. Then, point O is inside of the tetrahedronа ABCD, since one can choose ! ! ! ! ~ points A1, B1, C1, D1 on those rays, such that OA 1 þ OB 1 þ OC1 þ OD1 ¼ 0. According to problem 7.1.35e, it follows that π À α þ π À β þ π À γ þ π À α1 þ π À β1 þ π À γ1 > 3π. Therefore, α þ β þ γ þ α1 þ β1 þ γ1 < 3π. 330 7 Miscellaneous Inequalities

Figure 7.25

b1 a1 g1

According to problem 7.1.35f, it follows that α þ β þ γ > π, α þ β1 þ γ1 > π, β þ α1 þ γ1 > π, γ þ α1 þ β1 > π. Summing up all these inequalities, we deduce that α þ β þ γ þ α1 þ β1 þ γ1 > 2π (see Figure 7.25).

(h) Let O and O1 be the midpoints of segments AC1 and BD respectively. Then, ∠BOC1 ¼ 2 ∠BAC1, ∠DOC1 ¼ 2 ∠DAC1.

∠BOD ¼ 2 ∠BOO1 ¼ 2 ∠AOO1 ¼ 2 ∠A1AC1. According to problem 7.1.35b ∠BOC1 þ ∠DOC1 þ ∠BOD < 2π. Consequently ∠BAC1 þ ∠A1AC1 þ ∠DAC1 < π. However, according to problem 7.1.35a

∠BAC1 þ ∠A1AC1 þ ∠DAC1 ¼ 1 1 1 ¼ ðÞ∠BAC þ ∠A AC þðÞ∠A AC þ ∠DAC þðÞ∠BAC þ ∠DAC > 2 1 1 1 2 1 1 1 2 1 1 π π π 3π > þ þ ¼ : 4 4 4 4

7.1.36. (a) Denote the cubes with edges a and b by Φa and Φb respectively, and the 0 cube, symmetric to the cube Φb with respect to the center of the cube Φa by Φb . It is not difﬁcult to prove that the cube Φb and the center of the cube Φa are on different sides of plane α, containing one of the faces of the cube Φb. Consider two following cases:

(a) α is parallel to one of the edges of the cube Φa, (b) α is not parallel to any of the edges of the cubeΦa. 0 Note that the cubes Φb and Φb do not have any common points. Indeed, if points 0 0 M 2 Φb and M 2 Φb , then point M , symmetric to point M with respect to the center of the cube Φa, also belongs to Φb. Consequently, the center of the cube Φa, that is 0 the midpoint of segment MM , belongs to the cube Φb. This leads to a contradiction. In the case (a) the proof is obtained if one considers the projection of the cubes 0 Φb and Φb on the face of the cube Φa, that is perpendicular to plane α. 7.2 Solutions 331

It is clear that these projections do not have any common points and are rectangles with sides b and b(sin φ þ cos φ), where φ is the angle between the planes containing one of the faces of the cubes Φa and Φb. Since sin φ þ cos φ sin2φ þ cos2φ ¼ 1, then these projections contain squares < < a with sides b. Then, according to the problem 2.1.6b, we have that 2b a.Thus,b 2. Let us continue now the case (b). Let the cube Φb be inside of a triangular pyramid ABCD (see the proof of problem 7.1.36c), where A is the vertex of the cube Φa, and the edges of the cube Φa, drawn from vertex A, belong to the rays AB, AC, AD, with points B, C, D belonging to plane α. Let us consider now a plane (B0C0D0), parallel to the plane (BCD) and containing 0 0 0 one of the faces of the cube Φb, with B 2 AB, C 2 AC, D 2 AD. Using problem 2.1.10, we obtain that the cube Φb can be placed inside of the pyramid ABCD,so that one of its edges belongs to, let us say, segment B0C0. Now we will consider the sections of the tetrahedron ABCD, obtained by the 0 intersections with planes β and β , containing the faces of the cube Φb and perpendicular to the straight line B0C0. Since AD ⊥ (ABC), then AD ⊥ B0C0.We have that B0C0 ⊥ β, B0C0 ⊥ β0, thus AD k β, AD k β0,orAD belongs to one of planes β and β0. Note that these sections are right-angled triangles, and using problem 2.1.5, we obtain that the cube Φb is in a triangular pyramid ABCD, with the face ABC containing one of the faces of the cube Φb. Repeating this reasoning once again, we obtain that the cube Φb is inside of the triangular pyramid ABCD and point A is one of the vertices of the cube Φb.Itis < a Φ clear that b 2, since the center of the cube a is not inside of the pyramid ABCD.

(b) Denote the cube with the edge a by Φa. Let b, c, d be the smallest non-negative numbers, such that by the parallel translation, given by the relations x0 ¼ x À b, 0 0 0 y ¼ y À c, z ¼ z À d, the cube Φa transforms into the cube Φa , having vertices M1, M2, M3 belonging to the coordinate planes Oxy, Oyz and Oxz, respectively.

Now, consider a cube Φ2a with sides 2a, with one of the vertices at the point of the origin and three edges on the positive semiaxes Ox, Oy, Oz. Note that the cube 0 Φa does not have points outside the cube Φ2a. 0 Indeed, if M(x, y, z) 2 Φa and M2= Φ2a, then without loss of generality, we can > > assume that x 2a. Consequentlypffiffiffi MM2 2a. However, on the other hand, M2, 0 0 M 2 Φa . Therefore, MM2 3a. This leads to a contradiction. Thus, the cube Φa does not have points outside the cube Φ2a. 0 0 0 If M0 (a, a, a)2= Φa , then since M0 is the center of the cube Φ2a and according to а < 2a problem 7.1.36 , it follows that a 2 . This leads to a contradiction. 0 0 Thus, M0 2 Φa . Hence all coordinates of point M0(a þ b, a þ c, a þ d) are not less than a and it belongs to the cube Φa.

(c) Denote the cubes with edges a, b and c by Φa, Φb and Φc respectively. Since the cubes Φa and Φb do not have common points, then there exists a plane α, such that these cubes are on the different sides of that plane. Since the cubes Φa and Φb are on different sides of plane α, then consequently the cube Φc has vertices on the both sides of plane α. Let A be the most distant vertex of Φc from plane α and is in the half-space with a boundary α containing the cube Φa. 332 7 Miscellaneous Inequalities

Vertex B of the cube Φc is defined similarly. Note that the ray with a vertex at A, containing one of the edges of the cube Φc is either parallel to plane α or intersects it. If the chosen coordinate system with the origin at point A is such that the positive semiaxes Ox, Oy, Oz contain the edges of the cube Φc, then according to problem 7.1.36b, there exists a point M0(x0, y0, z0), belonging to the cube Φa and x0 a, y0 a, z0 a. In that case, if the cube with the edge a, with one of the vertices at point A and its three edges are on the positive semiaxes Ox, Oy, Oz, then it is in the half-space with a boundary α, containing point A. From the aforesaid, it follows that one can assume that each of the cubes Φa and Φb, that do not have common points, has three edges on the edges of the cube Φc. The following cases are possible. (a) AB is the edge of Φ , then a þ b < c. c pffiffiffi pffiffiffi pffiffiffi (b) AB is a diagonal of one of the faces of the cube Φc, then 2a þ 2b < 2c. Therefore, a þ b < c. pffiffiffi pffiffiffi pffiffiffi (c) AB is a diagonal of the cube Φc, then 3a þ 3b < 3c. Consequently, a þ b < c. Thus, in all cases, we deduce that a þ b < c. 7.1.37. (a) We have that

h h h ah bh ch ðÞab 2 þ ðÞbc 2 þ ðÞac 2 a þ b þ c ¼ a þ b þ c ¼ 2S Á a b c a2 b2 c2 ðÞabc 2 ðÞab ðÞþbc ðÞab ðÞþac ðÞbc ðÞac 4Sp p 2S Á ¼ ¼ : ðÞabc 2 abc R

ma mb mc maha mbhb mchc (b) Since a þ b þ c ¼ 2S þ 2S þ 2S , then we have to prove that 2 maha þ mbhb þ mchc p . α β γ γ γ > π Let max( , , ) ¼ .If 2, then c b c a m h þ m h þ m h < þ Á b þ þ Á a a a b b c c 2 2 2 2 c cb ca ab a2 þ b2 þ h < þ þ þ ¼ p2: 2 c 2 2 2 2 γ π If 2, then we need to prove that maha p( p À a), mbhb p( p À b) and mchc p( p À c). We have that c b c a m h þ m h þ m h < þ Á b þ þ Á a a a b b c c 2 2 2 2 c cb ca ab a2 þ b2 þ h < þ þ þ ¼ p2: 2 c 2 2 2 2 7.2 Solutions 333

γ π If 2, then we need to prove that maha p( p À a), mbhb p( p À b) and mchc p( p À c). We have that S2 p2ðÞp À a 2 À m2h2 ¼ p2ðÞp À a 2 À ðÞb À c 2 þ ðÞb þ c 2 À a2 Á ¼ a a a2 ! ðÞb À c 2 ðÞp À b ðÞp À c ¼ ppðÞÂÀ a ppðÞÀÀ a ðÞp À b ðÞÀp À c 4ppðÞÀ a ¼ a2 a2 ! ðÞa þ c À b ðÞa þ b À c ðÞb À c 2 ¼ ppðÞÀ a ppðÞÀ a 1 À À ðÞp À b ðÞp À c ¼ a2 a2 ðÞb À c 2 ppðÞÀ a ðÞb À c 2 ÀÁ ¼ ppðÞÀ a ðÞppðÞÀÀ a ðÞp À b ðÞp À c ¼b2 þ c2 À a2 ¼ a2 2a2 ppðÞÀ a ðÞb À c 2 ¼ bccosα 0: a2

Hence maha p( p À a). Similarly, we obtain that mbhb p( p À b) and mchc p( p À c). Summing up these three inequalities, we obtain that 2 maha þ mbhb þ mchc p . S S S (c) It is known that ra ¼ pÀa , rb ¼ pÀb , rc ¼ pÀc. Hence, r r r a þ b þ c ¼ a b c S S S p ðÞpÀb ðÞpÀc ðÞpÀa ðÞpÀc ðÞpÀb ðÞpÀa ¼ þ þ ¼ þ þ apðÞÀa bpðÞÀb cpðÞÀc S a b c p 1 1 1 p p p ðÞpÀbþpÀc 2 þ ðÞpÀaþpÀc 2 þ ðÞpÀbþpÀc 2 ¼ Á ¼ : S 4a 4b 4c S 2 2r

ra rb rc S S S p 4Sp Since a þ b þ c ¼ apðÞÀa þ bpðÞÀb þ cpðÞÀc and R ¼ abc, then we have to prove bc ac ab that pÀa þ pÀb þ pÀc 4p. Note that

bc ac ab ðÞpÀaþpÀc ðÞpÀaþpÀb ðÞpÀbþpÀc ðÞpÀaþpÀb þ þ ¼ þ þ pÀa pÀb pÀc pÀa pÀb ðÞpÀbþpÀc ðÞpÀaþpÀc ðÞpÀa ðÞpÀb ðÞpÀb ðÞpÀc ðÞpÀc ðÞpÀa þ ¼ þ þ þ3p sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipÀc spffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀc pÀa pÀb ðÞpÀa ðÞpÀb ðÞpÀb ðÞpÀc ðÞpÀa ðÞpÀb ðÞpÀc ðÞpÀa Á þ Á þ pÀc pÀa pÀc pÀb sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞpÀb ðÞpÀc ðÞpÀc ðÞpÀa þ Á þ3p¼4p, pÀa pÀb since x2 þ y2 þ z2 xy þ yz þ zx. 334 7 Miscellaneous Inequalities

(d) We have that m m m 2 a þ b þ c ¼ a b c 2 2 2 2 2 2 m m c þ m m a þ m m b m 2 m 2 m 2 9 3 a 3 b 3 b 3 c 3 a 3 c ¼ a þ b þ c þ Á a2 b2 c2 2 abc m 2 m 2 m 2 9 a þ b þ c þ a2 b2 c2 2

(see problem 4.1.8c). 2 2 2 2 2 2 2 2 2 Since ma mb mc 1 2 b a 2 b c 2 a c 3 9, then a2 þ b2 þ c2 ¼ 4 a2 þ b2 þ c2 þ b2 þ c2 þ a2 À 4 ÀÁ pffiffi ma mb mc 2 9 9 27 ma mb mc 3 3 a þ b þ c 4 þ 2 ¼ 4 . Thus, a þ b þ c 2 . 3 3 3 (e) Since the triangle with sides ma, mb, mc has medians with the lengths 4 a, 4 b, 4 c (see Figure 7.26). pffiffi 3a 3b 3c Then, according to problem 7.1.37d, it follows that 4 þ 4 þ 4 3 3. Hence, pffiffiffi ma mb mc 2 a þ b þ c 2 3. ma mb mc α β γ γ γ π (f) Let max( , , ) ¼ .If 2, then we have that ma R þ ka, mb R þ kb, mc R þ kc (See the proof of problem 2.4.11). Hence, m m m 1 1 1 k k k a þ b þ c R þ þ þ a þ b þ c ¼ ha hb hc ha hb hc a b c aka bkb ckc aka bkb ckc a b c Rp þ þ R R þ þ þ 2 þ 2 þ 2 ¼ þ 2 2 2 ¼ þ 1: 2S 2S 2S aha bhb chc S S r 2 2 2

γ > π < bþc < aþc < c If 2, then ma 2 , mb 2 and mc 2. Consequently,

Figure 7.26

a/2

ma 2 mc ma 3 3 2 3 mb 3 7.2 Solutions 335

m m m b þ c a þ c c b a c a þ b þ c < þ þ ¼ þ þ ¼ ha hb hc 2ha 2hb 2hc 2ha 2hb 2r ab1 c 2R c ¼ þ ¼ þ : 2S 2r c 2r

2R c < R > We need to prove that c þ 2r 1 þ r ,or(2R À c)(c À 2r) 0. This leads to a contradiction since 2R > c > ha > 2r.

(g) According to problem 5.5.8b, we have that ma þ mb þ mc min ( p þ 2R,4R þ r) 4R þ r. ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p3 Thus, it follows that ma þ mb þ mc 4R þ r. Therefore 3 mambmc maþ mb þ mc 4R þ r. Hence, we deduce that 4R þ r 3 m m m : a b c 3

(h) It is known that l2 ¼ 4bc ppðÞÀ a . Hence l2 ppðÞÀ a . Similarly, we get that a ðÞbþc 2 a 2 2 lb ppðÞÀ b , lc ppðÞÀ c and summing up these three inequalities, we obtain 2 2 2 2 that la þ lb þ lc p . 2 2 2 S (i) Since the triangle with sides 3 ma, 3 mb, 3 mc has an area 3 (see the proof of problem 7.1.37e), then according to problem 5.5.9, it follows that 2 2 2 2 2 2 2 2 2 2 2 2 m þ m þ m À m À m À m À m À 3 a 3 b 3 c 3 a 3 b 3 b 3 c 2 2 2 pffiffiffi S m À m 4 3 Á : 3 a 3 c 3

pffiffiffi 2 2 2 2 2 2 Hence, ma þ mb þ mc À ðÞma À mb À ðÞmb À mc À ðÞma À mc 3 3S. 2 2 2 Remark Using the fact that the triangle with sides 3 ma, 3 mb, 3 mc has the medians a b c S 2 , 2 , 2 and the area 3, one can propose «new» problems arising from this problem. For example, according to problem 4.1.8c, we have that mambc þ mbmca 9 þmcmab 4 abc, then the inequality abmc þ bcma þ acmb 4mambmc holds true. (j) We have that r r r S S S 1 a b c a þ b þ c ¼ þ þ ¼ þ þ ¼ ha hb hc ðÞp À a ha ðÞp À b hb ðÞp À c hc 2 p À a p À b p À c 1 p À b þ p À c p À a þ p À c p À a þ p À b ¼ þ þ ¼ 2 p À a p À b p À c 1 p À b p À a 1 p À b p À c 1 p À a p À c ¼ þ þ þ þ þ 1 þ 1 þ 1 ¼ 3: 2 p À a p À b 2 p À c p À b 2 p À c p À a 336 7 Miscellaneous Inequalities

Figure 7.27

b/2 OA ra A1 g/2

ÀÁ β γ α 2 (k) Note that BC ¼ CA1 þ A1B¼ra tg 2 þ tg 2 ¼ ra cos β γ (see Figure 7.27). cos 2 cos 2 α β a 2 b 2 Hence, ¼ cos β γ. Similarly, we get that ¼ cos α γ, ra cos cos rb cos 2 cos 2 γ 2 2 c 2 ¼ cos β α. rc cos 2 cos 2 We have to prove that α β γ α β β γ α γ 3 cos 2 þ cos 2 þ cos 2 4 cos 2 cos 2 þ cos 2 cos 2 þ cos 2 cos 2 , 2 2 2 2 2 2 2 2 2 or 3 cos α þ cos β þ cos γ 3 þ 2 2 ðÞ1 þ cos α ðÞþ1 þ cos β ðÞ1 þ cos β ðÞþ1 þ cos γ ðÞ1 þ cos α ðÞ1 þ cos γ , thismeansthat3 cos α þ cos β þ cos γ þ 2(cos α cos β þ cos β cos γ þ cos γ cos α). The last inequality is obtained according to problems 5.1.4a and 5.1.14. β γ a cos 2 cos 2 (l) We have thatra ¼ α α β γ (see the proof of problem 7.1.37k). cos 2 ¼ 4R sin 2 cos 2 cos 2 β α γ γ α β Similarly, rb ¼ 4R sin 2 cos 2 cos 2 and rc ¼ 4R sin 2 cos 2 cos 2. Therefore,

2 2 2 ra þ rb þ rc ¼ α β γ β α γ γ α β ¼ 16R2 sin 2 cos 2 cos 2 þ sin 2 cos 2 cos 2 þ sin 2 cos 2 cos 2 ¼ 2 2 2 2 2 2 2 2 2 ¼ 2R2ðþðÞ1 À cos α ðÞ1 þ cos β ðÞþ1 þ cos γ ðÞ1 À cos β ðÞ1 þ cos α ðÞ1 þ cos γ þðÞ1 À cos γ ðÞ1 þ cos α ðÞÞ¼1 þ cos β 2R2ð3 þ cos α þ cos β þ cos γÀ À cos α cos β À cos β cos γ À cos γ cos α À 3 cos α cos β cos!γÞ¼ R þ r r2 þ p2 À 4R2 p2 À ðÞ2R þ r 2 ¼ 2R2 3 þ À À 3 Á ¼ ðÞ4R þ r 2 À 2p2 R 4R2 4R2

(see problems 5.3.6, 5.3.10b and 5.3.9b). 7.2 Solutions 337

2 2 2 2 2 2 According to problem 5.5.6, p 4R þ 4Rr þ 3r . Hence, ra þ rb þ rc 8 2 2 27 2 2 2 2 27 2 R À 5r 4 R (see problem 5.5.1a), thus ra þ rb þ rc 4 R . 1 1 1 ha hb hb hc (m) We have that ðÞha þ hb þ hc þ þ ¼ 3 þ þ þ þ þ ha hb hc hb hc hc hb ha þ hc 9 and 1 þ 1 þ 1 ¼ a þ b þ c ¼ 1 ¼ 1. Hence, h þ h þ hc ha ha hb hc 2S 2S 2S S=p r a b hc 9r. ÀÁ 2 2 2 3 2 2 2 2 2α 2β 2γ (n) We have that ma þ mb þ mc ¼ 4 a þ b þ c ¼ 3R ðÞsin þ sin þ sin 27 2 4 R (see problem 5.1.1). (o) Since (x þ y þ z)2 3x2 þ 3y2 þ 3z2, according to problem 7.1.37n m þ m þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ a b 2 2 2 1 1 1 9 mc 3 m þ m þ m 4, 5R. Consequently, þ þ a b c ma mb mc maþmbþmc 9 ¼ 2. Hence, 1 þ 1 þ 1 2. 4, 5R R ma mb mc R (p) We need to prove that ra þ rb þ rc ¼ 4R þ r. Indeed, ra þ rb þ rc ¼ S S S SpðÞðÞÀ a ðÞþp À b ðÞp À b ðÞþp À c ðÞp À c ðÞp À a ¼ þ þ ¼ ¼ p À a p À b p À c ðÞp À a ðÞp À b ðÞp À c SabðÞþ bc þ ac À p2 SrðÞ2 þ 4rR ¼ ¼ , ðÞp À a ðÞp À b ðÞp À c ðÞp À a ðÞp À b ðÞp À c

(see the proof of problem 5.3.7). SrðÞ rþ4R S2ðÞrþ4R Since ra þ rb þ rc ¼ ðÞpÀa ðÞpÀb ðÞpÀc ¼ ppðÞÀa ðÞpÀb ðÞpÀc ¼ r þ 4R, then according to problem 5.5.1a, ra þ rb þ rc 4, 5R. α pffiffi 1 1 α 1 α 2bc cos 2 2bc (q) Since S ¼ bc sin α ¼ bla sin þ cla sin , then la ¼ > . 2 ÀÁ2 2 2 2 bþc bþÀÁc Consequently, 1 < p1ffiffi 1 þ 1 . Similarly, we obtain that 1 < p1ffiffi 1 þ 1 and ÀÁla 2 b c lb 2 a c 1 < p1ffiffi 1 þ 1 . Summing up these three inequalities, we deduce that lc 2 a bpffiffiffiÀÁ 1 þ 1 þ 1 < 2 1 þ 1 þ 1 . la lb lc a b c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (r) We have that l ppðÞÀ a (see the proof of problem 7.1.37h). Hence, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia 2S 2 ppðÞÀa ðÞpÀb ðÞpÀc 2laðÞpÀb hc ¼ c ¼ c c la. (s) Let point A1 be on segment BC and straight line AA1 intersect the circumcircle of triangle ABC (for the second time) at point A2. Then, AA1 Á A1A2 ¼ BA1 Á A1C and A1A2 ¼ AA2 À AA1 2R À AA1. Consequently, BA1 Á A1C AA1(2R À AA1). Thus,

2 AA þ BA Á A C ∗ AA 1 1 1 ðÞ 1 2R

2 bc One can easily prove that, if AA1 ¼ la, AA1 þ BA1 Á A1C ¼ bc, then la 2R. 338 7 Miscellaneous Inequalities

Figure 7.28 A

BA1 C

AЈ

pffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi We have that l 2 bc pp a ,2Rr abc S abc to prove the inequal- aq¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffibþc ðÞÀ ¼ 2S Á P ¼ aþbþc pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ity l pbcffiffiffiffiffiffi ¼ 1 bc Á 2p, we have to prove that 4 apðÞÀ a ðÞb þ c 2, a 2 2Rr 2 a or (b þ c À 2a)2 0. (t) According to (*) (see the proof of problem 7.1.37s), when AA1 ¼ ma, we obtain 2 2 2 2 2 a 2 2 2 a b c 2 2 2 maþ 4 maþmbþmc þ 4 þ 4 þ 4 a þb þc that ma 2R . Hence, ma þ mb þ mc 2R ¼ 2R . 0 (u) Let GA1 ¼ A1A (see Figure 7.28), then using the inequality in problem 7.1.37t 4 2 2 2 0 a b c 9ðÞmaþmbþmc for triangle BGA , we deduce that þ þ 2m Á2m Á2m . 2 2 2 3 a 3 b 3 c 2SBGA0

1 ma mb mc P 1 Note that S 0 ¼ SABC, consequently þ þ ¼ . BGA 3 mbmc mamc mamb SABC r

m2þm2þm2 Remark We have that, ma þ mb þ mc ¼ a b c mambþmbmcþmcma ¼ 1 þ mbmc mamc mamb mambmc mambmc ma 1 þ 1 2 (see problem 7.1.37o). Hence, ma þ mb þ mc 2. mb mc R mbmc mamc mamb R (v) We shall prove that if a, b, c 0, then (a þ b)(b þ c)(a þ c)(a þ b þ c)2 24abc (a2 þ b2 þ c2). 1 Indeed, without loss of generality, we can assume that a þ b þ c ¼ 1 and c 3. Then, we have to prove that (1 À a)(1 À b)(1 À c) 24abc(1 À 2ab À 2bc À 2ac), or

ab þ ðÞa þ b c À abc 24abcðÞ1 À 2ab À 2caðÞþ b : ð7:14Þ Let f(x) ¼ 48cx2 þ (1 À 25c þ 48c2(1 À c))x þ c(1 À c), then (7.14) is equivalent to the inequality f(ab) 0. ðÞaþb 2 ðÞ1Àc 2 Note that 0 ab 4 ¼ 4 ¼ x0. If 1 À 25c þ 48c2(1 À c) 0orf(x) does not have roots, then it is clear that f(ab) 0. 2 Let 1 À 25c þ 48c (1 À c) < 0 and x1 and x2 be its roots. Then by Vie`tta’s theorem x1 > 0, x2 > 0 (in the case of c ¼ 1, a ¼ b ¼ 0 and then f(ab) ¼ 0). Note 1 2 2 that fxðÞ¼0 4 ðÞ3c À 1 ðÞ2c À 1 0. Consequently, x0 x1 x2 or x0 x2 x1 . 7.2 Solutions 339

4 2 ðÞ1Àc 1Àc In the case of x0 x2 x1, we have that x0 x1x2. Hence, 16 48 . 3 1 3 ÀÁTherefore,ðÞ 1 À c 3. This leads to a contradiction asðÞ 1 À c 2 3 8 < 1 3 ¼ 27 3. Thus, ab x0 x1 x2. It follows that f(ab) 0. For numbers p À a > 0, p À b > 0, p À c > 0, we have that

2 ðÞðÞþp À a ðÞp À b ðÞðÞþp À b ðÞp À c ðÞðÞþp À a ðÞp À c ÂðÞðÞþp À a ðÞþp À b ðÞp À c 24ðÞp À a ðÞp À b ðÞÂp À c ðÞp À a 2 þ ðÞp À b 2 þ ðÞp À c 2 :

Consequently, abcp2 24ðÞp À a ðÞp À b ðÞp À c ðÞp À a 2 þ ðÞp À b 2 þ ðÞp À c 2 , S2 4RSp2 24 Á ðÞp À a 2 þ ðÞp À b 2 þ ðÞp À c 2 , p Rp2 2r Á 3 ðÞp À a 2 þ ðÞp À b 2 þ ðÞp À c 2 , Rp2 2rpðÞÀ a þ p À b þ p À c 2 þ ðÞðÞÀp À a ðÞp À b 2þ þðÞðÞÀp À b ðÞp À c 2 þ ðÞðÞÀp À a ðÞp À c 2 :

RÀ2r 2 2 2 2 Hence, we obtain that 2r Á p ðÞa À b þ ðÞb À c þ ðÞc À a . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi (w) We have that la ppðÞÀ a ¼ rbrc (see the proof of problem 7.1.37h). Note 2 2 2b2þ2c2Àa2 ðÞbþc Àa2 that ma ¼ 4 4 ¼ ppðÞ¼À a rbrc. Hence, pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi m m þ m m þ m m r r Á r r þ r r Á r r þ r r Á r r ¼ a b pffiffiffiffiffiffiffiffib c pcffiffiffiffiffiffiffiffia pb ffiffiffiffiffiffiffiffic c a a c b a a b b c ¼ rc Á rarb þ ra rbrc þ rb rarc rclc þ rala þ rblb:

(x) See the proof of problem 7.1.37q, 2bc 2ac b2 a2 l þ l < þ ¼ 2b þ 2a À 2 þ a b b þ c a þ c b þ c a þ c 2ðÞb þ a 2 2ðÞb þ a 2 4 2ðÞÀb þ a < 2ðÞÀb þ a ¼ ðÞa þ b : a þ b þ 2c 3ðÞa þ b 3 (y) We have that qÀÁffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 2 2 2 2 2 2 4acmamc ¼ ac 2b þ 2c À a 2a þ 2b À c ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁÀÁ ¼ ac 22b2 À a2 À c2 a2 þ b2 þ c2 þ 9a2c2

and 16S2 þ (b2 À a2)(c2 À b2) ¼ (2b2 À a2 À c2)(a2 þ c2 À b2) þ 3a2c2. 2 2 2 2 2 2 pDenoteffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix ¼ 2b À a À c , y ¼ a þ c , z ¼ b , then we have to prove that ac 2xyðÞþþ z 9a2c2 xyðÞþÀ z 3a2c2 (if x(y À z) þ 3a2c2 < 0, then the 340 7 Miscellaneous Inequalities

inequality is evident) or 2a2c2x(y þ z) x2(y À z)2 þ 6x(y À z)a2c2,2a2c2x (4z À 2y) x2(y À z)2. Since 2z À y ¼ x, then 4a2c2x2 x2(y À z)2, x2(2ac À y þ z)(2ac þ y À z) 0, x 2(b þ a À c)(b þ c À a)(a þ c À b)(a þ b þ c) 0. pffiffi (z) According to problem 5.1.6, we have that a þ b þ c 3 3. Therefore, pffiffiffi 2R 2R 2R 2 abc a þ b þ c 3 3 4S . Then, according to the remark of the problem 7.1.37i, it follows from the last inequality that

2 2 2 pffiffiffi 2 2 2 pffiffiffi ma Á mb Á mc 1 1 1 3 m þ m þ m 3 33 3 3 : Thus, þ þ : 3 a 3 b 3 c S m m m m m m S 4 a b b c c a 3

(aa) See the second proof of problem 7.1.46b. (ab) See the second proof of problem 7.1.46b. (ac) Let a b c, then max(ma, mb, mc) ¼ ma, the median drawn from vertex A intersects the bisector drawn from vertex B at point E, аnd I is the incenter of triangle ABC. Then, la þ lb þ lc > laþlbþlc > BEþEIþIA BEþEA > 1. ma mb mc ma ma ma (ad) We have that

ha þ hb þ hc ¼ 2RðÞsin α sin β þ sin β sin γ þ sin α sin γ ¼ p2 þ 4Rr þ r2 4R2 þ 8Rr þ 4r2 ¼ 2R Á 2R þ 5r 4R2 2R (see problems 5.3.10а, 5.5.6 and 5.5.1a). p2þ4Rrþr2 (ae) See the proof of problem 7.1.37ad. We have that ha þ hb þ hc ¼ 2R > ðÞ2Rþr 2þ4Rrþr2 > 2R 2R þ 4r (see problem 5.5.7a). (af) We have that (see the proof of problem 7.1.37q) pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi 2 bc α pffiffiffiffiffi 2 ac β pffiffiffiffiffi 2 ab γ la þ lb þ lc ¼ bc Á cos þ ac Á cos þ ab Á cos sbffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ c 2 a þ c 2 a þ b 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2bc 2ac 2ab ab þ bc þ ac ðÞþ1 þ cosα ðÞþ1 þ cosβ ðÞ1 þ cosγ ¼ ðÞb þ c 2 ðÞa þ c 2 ðÞa þ b 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 b2 c2 ¼ ab þ bc þ ac Á 3 À À À ðÞb þ c 2 ðÞa þ c 2 ðÞa þ c 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 a b c 2 3pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ab þ bc þ ac 3 À þ þ ab þ bc þ ac 3 b þ c a þ c a þ c 2

(see the proof of problem 5.5.19b). 7.1.38. (a) Denote by x ¼ MA , x ¼ MB , x ¼ MC and x ¼ MD . The ratio of the A MA1 B MB1 C MC1 D MD1 altitudes AP and MU of the tetrahedrons ABCD and MBCD is equal to 1 þ xA :1. Hence, VMBCD ¼ 1 . Similarly, VMACD ¼ 1 , VMABD ¼ 1 and VMABC ¼ 1 . Apart VABCD 1þxA VABCD 1þxB VABCD 1þxC VABCD 1þxD from that, since V þ V þ V þ V ¼ V , then 1 þ 1 þ MBCD MACD MABD MABC ABCD 1þxA 1þxB 1 1 1þxA 1þxA 1þxA þ ¼ 1. Thus, it follows that xA ¼ þ þ 1þxC 1þxD 1þxB 1þxC 1þxD 7.2 Solutions 341 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 ffiffiffiffiffiffiffi 3 ðÞ1þxA p 3 ,as x þ y þ z 3 3 xyz, for x, y, z 0. Similarly, xB qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðÞ1þxB ðÞ1þxC ðÞ1þxD qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3 3 3 ðÞ1þxB 3 ðÞ1þxC 3 ðÞ1þxD 3 , xC 3 , xD 3 . ðÞ1þxA ðÞ1þxC ðÞ1þxD ðÞ1þxA ðÞ1þxB ðÞ1þxD ðÞ1þxA ðÞ1þxB ðÞ1þxC Multiplying these four inequalities, we deduce that xAxBxCxD 81. Hence, it follows that MA Á MB Á MC Á MD 81MA1 Á MB1 Á MC1 Á MD1. (b) Consider a plane α, passing through point M and perpendicular to segment DM. Then it is clear that plane α intersects at least one of the edges DA, DB, DC, otherwise we obtain that point M is outside of the tetrahedron ABCD.LetDB be ∠ > π ∠ > ∠ that edge, then we have that BMD 2. Consequently, BMD MDB. Thus, it follows that BD > BM. (c) Let the straight line DM and the face ABC intersect at point D1. Let lines AD1 and BC intersect at point D2. Then, we have that DM < DD1 < max (AD, DD2) and DD2 < max (BD, CD). Thus, it follows that DM < max (AD, BD, CD). (d) According to the problem 7.1.38a, without loss of generality, we can assume DM 3D1M. Consider a plane passing through point M and parallel to plane ABC. Let A2, B2, C2 be the intersection points of that plane with the edges AD, BD, CD, respectively. Let also max(A2D, B2D, C2D) ¼ A2D, then according to the triangle inequality and the problem 7.1.38c), we deduce that

AM þ BM þ CM þ DM < AA2 þ A2M þ BB2 þ B2M þ CC2 þ C2M þ A2D BD CD BD CD AD þ þ þ A M þ B M þ C M < AD þ þ þ A B þ B C þ A C < 4 4 2 2 2 4 4 2 2 2 2 2 2 BD CD < AD þ þ þ AB þ BC þ AC 4 4

(see problem 1.1.8c). Consequently, AM þ BM þ CM þ DM < AB þ BC þ CD þ AD þ BD þ AC (see problem 7.1.38e). (e) We need to prove that, if for any distinct points M1 and M2 point M is the inner point of segment M1M2, then f(M) < max ( f(M1), f(M2)), where f(X) ¼ AX þ BX þ CX þ DX. Indeed, let M1M ¼ M1M0 ¼ α, where M is the point of segment AM (if M A, M1M2 M1A 0 1 1 then we take M0 on segment AM2). Then, we obtain that AM AM0 þ M0M ¼ (1 À α)AM1 þ αAM2, the equality holds true, if and only if A is on the straight line M1M2. Then at least one of the vertices of the tetrahedron ABCD does not belong to the straight line M1M2. Thus it follows that

fMðÞ< ðÞ1 À α fMðÞþ1 αfMðÞ2 maxðÞfMðÞ1 ; fMðÞ2 : ð7:15Þ 342 7 Miscellaneous Inequalities

Figure 7.29 P0 P l1

AA1 B1 B l

a-j jaj

ll AHBAB H а) b)

Figure 7.30

According to the inequality (7.4) (see the notations of problem 7.1.38c, it follows f(M) < max ( f(D), f(D1)) < max ( f(D), f(A), f(D2)) < max ( f(D), f(A), f(B), f(C)), thus f(M) < max ( f(D), f(A), f(B), f(C)). This ends the proof.

7.1.39. (a) Draw through point P a line l1 parallel of the straight line l (Figure 7.29). Take on the straight line l a point P , such that P A ¼ P B. Then the straight line 1 0 0 0 _ ∠ AB > ∠ l1 touches the circumcircle of triangle AP0B. We have that AP0B ¼ 2 APB. Consequently, one can take points A1 and B1 on segment AB such that ∠A1P0B1 ¼ ∠APB and A1P0 ¼ B1P0. Hence, AB A1B1 ¼ A0B0. (b) Let PH ⊥ l and PA PB (Figure 7.30). Consider Figure 7.30a (in the case of Figure 7.30b, the proof is similar). Denote by ∠APB ¼ α, PH ¼ h, ∠BPH ¼ φ. We have that ∠APH ¼ α À φ φ ¼ ∠BPH. h h α φ φ α Then,ÀÁ we have to prove that cos ðÞαÀφ À cos φ htgðÞðÞþÀ tg À2htg2, that 2 sin α À φ sin α 2 sin α cos α 2 sin α is, 2 2 2 2 À 2,or cos ðÞα À φ cos φ cos ðÞα À φ cos φ cos α α α α 2 sin À φ cos cos 2 À cos ðÞα À φ cos φ, α2 α2 2α α π α sin À φ cos sin 2 À φ ¼ sin À φ cos À þ φ 2 2 2 2 2 2 α φ < π α < π α φ π The last inequality holds true since 0 2 À 2,0 2 2 À 2 þ 2.

Remark If PA 6¼ PB, then |PA À PB| > AB À A0 B0. 7.1.40. Consider points C0 and Q0 symmetric to points C and Q with respect to the straight line AD (Figure 7.31). 7.2 Solutions 343

Figure 7.31 A

QЈ Q R

PCЈ

CD B

Note that ΔPQC ¼ ΔPQ0C0, hence ∠PC0B ¼ π À ∠PCQ > ∠PBC0. Thus, it follows that PB > PC0 ¼ PC. Since ∠BPR ¼ ∠CPQ ¼ ∠C0PQ0, then ∠RPQ0 ¼ ∠BPC0. 0 > 0 0 According to problem 7.1.39b, we have that PB À PC BC À B0C0, where B0 0 ∠ 0 ∠ 0 ∠ 0 ∠ 0 , C0 2AB and B0PC 0 ¼ BPC , PB0C0 ¼ PC 0B0. Moreover, according to 0 0 0 > 0 0 problem 7.1.39a, it follows that Q R B0C0. Hence PB À PC BC À Q R.This means that PB À PC > BR À QC since BC0 À Q0R ¼ BR À Q0C0 ¼ BR À QC. We have to prove that BR > QC. We have that ρ(P, AC) ¼ ρ(P, AB) BR SPBR PRÁPB CPþPR Hence ¼ ¼ . Since AB > AC, then h < h . Note that hc ¼ Á QC SPQC PQÁPC c b PR ρ ; PBþPQ ρ ; CPþPR < PBþPQ ðÞP AB and hb ¼ PQ Á ðÞP AC . Thus, PR PQ , consequently CP Á PQ < PR Á PB. Therefore, BR > QC. 7.1.41. (a) If P belongs to the altitude AH, then |∠PAB À ∠PAC| ¼ 0 ¼ |∠PBC À ∠PCB|. Otherwise consider points P0 and P symmetric to each other with respect to the straight line AH (Figure 7.32). Note that |∠PAB À ∠PAC| ¼ ∠PAP0 and |∠PBC À ∠PCB| ¼ ∠PCP0 ¼ ∠PRP0. Hence, we have to prove that RE AE. _ ∠ 0 ∠ ∠ 0 ∠ ∠ 0 BRC Since BP C þ BRC ¼ 180 , BN C þ BSC ¼ 180 and BP C ¼ 2 > ∠BN0C, then ∠BRC < ∠BSC. Thus, it follows that RE > SE. We need to prove that SE AE. This means that SN0 N0A ¼ NN0. _ _ _ Let ∠NBN0 ¼ α. Since α 60 and SBN0 ¼ 180 À α, N0QN ¼ 2α, then SBN0 _ N0QN. Consequently, SN0 NN0. (a) Let ∠PBC ¼ β ∠PCB ¼ γ, then PC PB. Therefore P is inside of the triangle ABH, where AH is the altitude of triangle ABC. Consequently, ∠PAC ¼ 30 þ α, ∠PAB ¼ 30 À α, where 0 α 30. We have to prove that β À γ arcsin (2 sin α) À α 2α, or sin(β À γ þ α) 2 sin α sin 3α. Note that sin3α À sin α ¼ 2 sin α cos 2α sin α, since 0 2α 60. We have to prove that sin(β À γ) cos α þ cos (β À γ) sin α 2 sin α,or α sin ðÞβÀγ tg 2À cos ðÞβÀγ . 344 7 Miscellaneous Inequalities

Figure 7.32

According to the law of cosines, it follows that α β PC PA PB sinðÞ 30 þ sinðÞ 60 À sin γ 1 ¼ Á Á ¼ Á Á . PA PB PC sinðÞ 60 Àγ sinðÞ 30 Àα sin β sinðÞ 30 Àα sinðÞ 60 Àβ sin γ Consequently ¼ sinðÞ 30 þα sinðÞ 60 Àγ sinβ pffiffi pffiffi 1À 3tgα ðÞ3 cos βÀ sin β sin γ Thus, pffiffi ¼ pffiffi . Hence, we deduce that tgα ¼ 1þ 3tgα ðÞ3 cos γÀ sin γ sin β sin ðÞβÀγ pffiffi . 3 sin ðÞÀβþγ 2 sin β sin γ ffiffiffi sin ðÞβÀγ sin ðÞβÀγ p pffiffi β γ We have to prove that β γ β γ 2À cos ðÞβÀγ or 2 À cos ðÞÀ 3 3 sin ðÞÀpffiffiþ 2 sin sin β γ β γ 3 β γ 1 β γ β γ 0 sin ðÞÀþ 2 sin sin ,1 2 sin ðÞþþ 2 cos ðÞþ ,1 sin ( þ þ 30 ). 0 0 haha haha γ γ0 β β0 γ β0 γ 7.1.42. We have that bb0 þ cc0 ¼ sin sin þ sin sin sin cos þ cos β0 γ β0 γ0 β0 β γ 1 1 1 sin ¼ sin ðÞþ 1, as 90 À and 90 À . Thus, 0 0 þ 0. haha bb cc 7.1.43. Denote by ∠PAB ¼ α, ∠QAD ¼ β, ∠BCE ¼ γ, ∠DCF ¼ δ and BP ¼ b, DQ ¼ c, DF ¼ a, BE ¼ d (see Figure 7.33). ∠ ∠ π α β γ δ π π γ δ α β Note that PAQ þ ECF ¼ 2 À ðÞþþ ðÞþþ 2 ¼ þ þ À À , γ δ α β < π hence we have to prove that 0 þ À À 4. 7.2 Solutions 345

Figure 7.33 E d P g C B b

d Q

a c

b ADa F ÀÁ γ δ α β ; π π < γ δ α β < π Since þ , þ 2 0 2 , then À2 þ À À 2. Thus, it is sufﬁcient to prove that 0 tg(γ þ δ À α À β) < 1. Let AB ¼ 1, we have that

tg ðÞÀγ þ δ tg ðÞα þ β aþd À bþc tg ðÞγ þ δ À α À β ¼ ¼ 1Àad 1Àbc : γ δ α β aþd bþc 1 þ tg ðÞþ tg ðÞþ 1 þ 1Àad Á 1Àbc

Since ΔDQF ΔBEP, then ad ¼ bc. Similarly, we have that ΔDQF ΔPQC and ΔBEP ΔPQC. Thus, one can cðÞ1Àb bðÞ1Àc easily deduce that a ¼ 1Àc and d ¼ 1Àb . We have that tg ðÞγ þ δ À α À β ¼ðÞaþdÀbÀc ðÞ1Àbc and a þ d À b À c ¼ ðÞ1Àbc 2þðÞaþd ðÞbþc ðÞcÀb 2 ðÞ1Àc ðÞ1Àb 0. Therefore, it follows that tg(γ þ δ À α À β) 0. γ δ α β < aþdÀbÀc < 2 > Note that tg ðÞþ À À ðÞaþd ðÞbþc , since 1 À bc 1 and (1 À bc) 0. aþdÀbÀc < aþd 1 If b þ c 1, then ðÞaþd ðÞbþc ðÞaþd ðÞbþc ¼ bþc 1. Thus, we obtain that tg (γ þ δ À α À β) < 1. < cðÞ1Àb bðÞ1Àc < c b < c b If b þ c 1, then a þ d ¼ 1Àc þ 1Àb 1Àc þ 1Àb 1ÀcÀb þ 1ÀcÀb, hence < γ δ α β < aþdÀbÀc < a þ d À (a þ d)(c þ b) c þ b. Consequently, tg ðÞþ À À ðÞaþd ðÞbþc 1. Thus, tg(γ þ δ À α À β) < 1. This ends the proof. 7.1.44. Denote the radius of the middle sphere by R. Let l \ P ¼ O, and A, B, C be the points at which the spheres with radiuses 1, R, r touch plane P (Figure 7.34), respectively. pffiffi pffiffiffi pffiffiffiffiffi We have that OA ¼ 1, OB ¼ R, OC ¼ r and AC ¼ 2 r, AB ¼ 2 R, BC ¼ 2 Rr. According to the triangle inequality, it follows that OC AC OA, thus pffiffi pffiffiffi þ 1 r þ 2 r. Hence, r 3 À 2 2p. ffiffiffi pffiffiffi Note that the equality r ¼ 3 À 2 2 is possible. Indeed, at r ¼ 3 À 2 2 we have that OC þ AC ¼ OA, thus C lies on segment OA (Figure 7.35). 346 7 Miscellaneous Inequalities

Figure 7.34 A

CO B

Figure 7.35 A1-r C r O

2 Rr 2 R R

Figure 7.36 C

DA K BE

Using Stewart’s theorem, we get that 4Rr ¼ 4Rr þ R2(1 À r) À r(1 À r)or pffiffiffi pffiffiffi ÀÁpffiffiffi 2 pffiffiffi R ¼ 2 À 1. Since 1 > R ¼ 2 À 1 > 2 À 1 ¼ r and R þ 2 R > rþ pffiffi pffiffiffiffiffi 2 r ¼ 1, then triangle AOB exists, which means that BC ¼ 2 Rr and thus there exist three spheres satisfying the conditions of the problem. 7.1.45. Note that (see Figure 7.36) ∠CEA ¼ 180 À 70 À 55 ¼ 55. Hence, EA ¼ CA ¼ AD. Consequently ∠KAD ¼ 5, ∠KAC ¼ 65. In triangle ACE ,we have that CE > AC; thus CE > CD. Let ∠DCN ¼ ∠ECN (Figure 7.37). Then, DN DC < ∠ < ∠ ∠ > ∠ NE ¼ CE 1. Hence KCE 2, 5 ¼ NCE and KCA ECA ¼ 55 . Consequently, ∠AKC ¼ 115 À ∠KCA > 57, 5 and ∠AKC ¼ 115 À ∠KCA < 60. S 7.1.46. (a) Let SABC ¼ S, then SAMC ¼ SBMA ¼ SCMB ¼ 3. We have that ∠MAB ¼ ∠MCA; consequently ΔAMK ΔACK, whereqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiK is the 2 intersection point of lines CM and AB. Thus, m c m b, since m 2b þ2c2Àa2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ¼ c a ¼ 4 pffiffi pffiffi 2a2þ2b2Àc2 2 2 2 3b 3a and mc ¼ 4 , we obtain that 2c ¼ a þ b and ma ¼ 2 , mb ¼ 2 . 7.2 Solutions 347

Figure 7.37 C

D N E

Note that

2S 2S S S sin ∠CAM þ sin ∠CBM ¼ AMC þ CMB ¼ þ ¼ AMÁ AC BM Á BC mab mba 1 a b 1 a b 1 a2 þ b2 ¼ þ sin γ ¼ pffiffiffi þ sin γ ¼ pffiffiffi Á sin γ: 2 ma mb 3 b a 3 ab

Since 2c2 ¼ a2 þ b2, then according to the law of cosines, we deduce that 2 2 a þb ¼ 4 cos γ. Thus, sin ∠CAM þ sin ∠CBM ¼ p2ffiffi sin 2γ. Hence, sin ∠CAMþ ab 3 sin ∠CBM p2ffiffi. 3 (b) Consider two circles passing through points M and C and touching the straight line AB at points B1 and A1, where point B1 belongs to ray KB, and K is the point of intersection of lines CM and AB. 2 2 CM Since KB1 ¼ KM Á KC ¼ KA1 and MK ¼ 2, then point M is the orthocenter of triangle CA1B1. We have that ∠CBM ∠CB1M and ∠CAM ∠CA1M. If ∠CB1M > 90 , then let us consider points E and N on segment A1M, such that 1 ME ¼ EN ¼ 8 MA1. Then, NP k B1C, where P is the intersection point of lines B1M and A1C. We have that ∠MPN ¼ ∠MB1C > 90 ; consequently point P is inside of the < 1 circle with a diameter MN. Hence EP 2 MN. According to the law of sines, EP sin ∠A1PE EP MN 1 sin ∠MA1C ¼ < ¼ . Hence, it follows that sin ∠CBM þ sin A1E A1E 2A1E 7 ∠CAM 1þ sin ∠CA M < 8 < p2ffiffi, since ∠CAM ∠CA M < 90. 1 7 3 1 It remains to consider the case when ∠CB1M 90 and ∠CA1M 90 . According to problem 7.1.46a, it follows that sin ∠CAM þ sin ∠CBM sin ∠CA M þ sin ∠CB M p2ffiffi. 1 1 3 Second Solution We have that sin ∠CAM þ sin ∠CBM ¼ S þ S . mab mba abmamb We have to prove that S pffiffi pffiffi . 3 b Á m þ 3 a Á m ffiffi 2 a ffiffi 2 b x2 þ y2 p p Since xy , then 3 b Á m þ 3 a Á m 2 2 a 2 b 1 3 1 3 a2 þ b2 þ c2 b2 þ m2 þ a2 þ m2 ¼ : Hence, it is sufficient to 2 4 a 2 4 b 2 prove that S 2abmamb ,or(a2 b2 c2) sin γ 4m m . a2þb2þc2 þ þ a b 348 7 Miscellaneous Inequalities

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁÀÁ 2 2 2 2 2 2 2 2 We have that 4mamb ¼ 22c À a À b c þ a þ b þ 9a b . Denote by c2 ¼ x, ab cos γ ¼ y. Then we have to prove that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γ 9y2 2ðÞx þ y sin 2ðÞx À 2y ðÞþ2x þ 2y cos 2γ,or 2γ 2 γ 4cos 2γx2 À 4xyðÞþ1 þ 2sin 2γ 9À8cos Àsin 2 y2 0, cos 2γ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 γ 9À8cos 2γÀsin 22γ 2 cos x À cos γ y 0, since ÀÁÀÁÀÁ 2 9 À 8cos 2γ À sin 22γ ¼ 9 À 81À sin 2γ À 4sin 2γ 1 À sin 2γ ¼ 1 þ 2sin 2γ :

7.1.47. Let the numbers a, b, c and m, n correspond in the complex plane to the vertices of triangle ABC and points M, N respectively. Then, we have to prove that jjmÀa jjnÀa jjmÀb jjnÀb jjmÀc jjnÀc jjbÀa jjcÀa þ jjaÀb jjcÀb þ jjaÀc jjbÀc 1. Note that

ðÞm À a ðÞn À a ðÞm À b ðÞn À b ðÞm À c ðÞn À c þ þ ¼ ðÞb À a ðÞc À a ðÞa À b ðÞc À b ðÞa À c ðÞb À c mnðÞ c À b þ a À c þ b À a ÀmacðÞðÞþÀ b baðÞþÀ c cbðÞÀ a À ¼ ðÞa À b ðÞb À c ðÞc À a ÀnacðÞðÞþÀ b baðÞþÀ c cbðÞÀ a þa2ðÞþc À b b2ðÞþa À c c2ðÞb À a ¼ ðÞa À b ðÞb À c ðÞc À a ÀÁ a2ðÞþc À b b2ðÞþa À c c2ðÞb À a a2 À b2 c À abðÞÀ a À b c2ðÞa À b ¼ ¼ ¼ ðÞa À b ðÞb À c ðÞc À a ðÞa À b ðÞb À c ðÞc À a ac þ bc À ab À c2 ¼ ¼ 1: ðÞb À c ðÞc À a

Consequently, according to the triangle inequality,

ðÞm À a ðÞn À a ðÞm À b ðÞn À b ðÞm À c ðÞn À c 1 ¼ þ þ ðÞb À a ðÞc À a ðÞa À b ðÞc À b ðÞa À c ðÞb À c

ðÞm À a ðÞn À a ðÞm À b ðÞn À b ðÞm À c ðÞn À c þ þ ðÞb À a ðÞc À a ðÞa À b ðÞc À b ðÞa À c ðÞb À c

ðÞm À a ðÞn À a ðÞm À b ðÞn À b ðÞm À c ðÞn À c þ þ ¼ ðÞb À a ðÞc À a ðÞa À b ðÞc À b ðÞa À c ðÞb À c jjm À a jjn À a jjm À b jjn À b jjm À c jjn À c ¼ þ þ , jjb À a jjc À a jja À b jjc À b jja À c jjb À c thus 7.2 Solutions 349

jjm À a jjn À a jjm À b jjn À b jjm À c jjn À c þ þ 1: jjb À a jjc À a jja À b jjc À b jja À c jjb À c

Remark If points M and N coincide, then we obtain the inequality of problem 4.1.8b.

7.1.48. Let the numbers a1, a2,...,an and b1, b2,...,bn À 1 correspond in the complex plane to points A1, A2,...,An and B1, B2,...,Bn À 1 respectively. By mathematical induction we need to prove that

ðÞb À a ðÞÁb À a ::: Á ðÞb À a ðÞb À a ðÞÁb À a ::: Á ðÞb À a 1 1 2 1 nÀ1 1 þ 1 2 2 2 nÀ1 2 þ ::: ðÞa2 À a1 ðÞÁa3 À a1 ::: Á ðÞan À a1 ðÞa1 À a2 ðÞÁa3 À a2 ::: Á ðÞan À a2 ðÞb À a ðÞÁb À a ::: Á ðÞb À a þ 1 n 2 n nÀ1 n ¼ 1 ðÞa1 À an ðÞÁa2 À an ::: Á ðÞanÀ1 À an

Indeed, for n ¼ 2, we have that b1Àa1 þ b1Àa2 ¼ 1. a2Àa1 a1Àa2 Let for n ¼ k the statement holds true. Let us prove that the statement holds true also for n ¼ k þ 1. ::: Consider the expression pzðÞ¼ðÞÁb1Àa1 ÁðÞbkÀ1Àa1 ðÞzÀa1 þ :::þ ðÞÁa2Àa1 :::ÁðÞakÀa1 ðÞakþ1Àa1 ::: ðÞÁb1Àakþ1 ÁðÞbkÀ1Àakþ1 ðÞzÀakþ1 . ðÞÁa1Àakþ1 :::ÁðÞakÀ1Àakþ1 ðÞakÀakþ1 It has a form of Az þ B, for the value z ¼ a1. We have that ðÞÁb1Àa2 :::ÁðÞbkÀ1Àa2 ðÞÁb1Àakþ1 :::ÁðÞbkÀ1Àakþ1 paðÞ¼1 þ ::: þ ¼ 1, since for numbers a2, ðÞÁa3Àa2 :::ÁðÞakþ1Àa2 ðÞÁa2Àakþ1 :::ÁðÞakÀakþ1 a3,...,ak þ 1 and b1,...,bk À 1 the statement holds true (n ¼ k). Similarly, we obtain that p(a2) ¼ 1, Thus Aa1 þ B ¼ 1 and Aa2 þ B ¼ 1. Hence, A ¼ 0 and B ¼ 1, then p(bk) ¼ 1. For the numbers z1,...,zn one can easily prove that |z1| þ ...þ |zn| | z1 þ ... þ zn| (see the proof of problem 7.1.47); hence from the equality (*), we obtain the given inequality.

7.1.49. Let numbers a1, a2,...,an and b1, b2,...,bn À 2 correspond in the complex plane to points A1,...,An and B1,...,Bn À 2 respectively. By mathematical induction we need to prove that

ðÞb À a ðÞÁb À a ::: Á ðÞb À a ðÞb À a ðÞÁb À a ::: Á ðÞb À a ∗ 1 1 2 1 nÀ2 1 þ ::: þ 1 n 2 n nÀ2 n ¼ 0 ðÞ ðÞa2 À a1 ðÞÁa3 À a1 ::: Á ðÞan À a1 ðÞa1 À an ðÞÁa2 À an ::: Á ðÞanÀ1 À an

Indeed, for n ¼ 3, we have that

b À a b À a b À a 1 1 þ 1 2 þ 1 3 ¼ ðÞa2 À a1 ðÞa3 À a1 ðÞa1 À a2 ðÞa3 À a2 ðÞa1 À a3 ðÞa2 À a3 b ðÞa À a þ a À a þ a À a Àa ðÞÀa À a a ðÞÀa À a a ðÞa À a ¼ 1 3 2 1 3 2 1 1 3 2 2 1 3 3 2 1 ¼ 0: ðÞa1 À a2 ðÞa2 À a3 ðÞa3 À a1

Let for n ¼ k (*) holds true. 350 7 Miscellaneous Inequalities

We need to prove that (*) holds true also for n ¼ k þ 1. Consider the expression

ðÞÁb À a ::: Á ðÞb À a ðÞz À a pzðÞ¼ 1 1 kÀ2 1 1 þ ::: ðÞÁa2 À a1 ::: Á ðÞakþ1 À a1 ðÞÁb À a ::: Á ðÞb À a ðÞz À a þ 1 kþ1 kÀ2 kþ1 kþ1 ðÞÁa1 À akþ1 ::: Á ðÞak À akþ1

We have that p(z) ¼ Az þ B and

ðÞÁb1 À a1 ::: Á ðÞbkÀ2 À a1 ðÞÁb1 À ak ::: Á ðÞbkÀ2 À ak paðÞ¼kþ1 þ ::: þ ¼ 0 ðÞn ¼ k : ðÞÁa2 À a1 ::: Á ðÞak À a1 ðÞÁa1 À ak ::: Á ðÞakÀ1 À ak

Similarly, we obtain that p(ak) ¼ 0. Hence, it follows that A ¼ 0, B ¼ 0. Thus, it follows that p(bk À 1) ¼ 0. Therefore, we have that

::: ::: B1An Á B2An Á Á BnÀ2An ðÞb1 À an ðÞÁb2 À an Á ðÞbnÀ2 À an ¼ ¼ A A Á A A Á ::: Á A A ðÞa À a ðÞÁa À a ::: Á ðÞa À a 1 n 2 n nÀ1 n 1 n 2 n nÀ1 n ::: ::: ðÞÁb1 À a1 Á ðÞbnÀ2 À a1 ðÞÁb1 À anÀ1 Á ðÞbnÀ2 À anÀ1 ¼ þ ::: þ ðÞÁa À a ::: Á ðÞa À a ðÞÁa À a ::: Á ðÞa À a ðÞa À a 2 1 n 1 1 nÀ1 nÀ2 nÀ1 n nÀ1 ::: ::: ðÞÁb1 À a1 Á ðÞbnÀ2 À a1 ðÞÁb1 À anÀ1 Á ðÞbnÀ2 À anÀ1 þ ::: þ ¼ ðÞÁa2 À a1 ::: Á ðÞan À a1 ðÞÁa1 À anÀ1 ::: Á ðÞan À anÀ1 B A Á ::: Á B A B A Á ::: Á B A ¼ 1 1 nÀ2 1 þ ::: þ 1 nÀ1 nÀ2 nÀ1 : A2A1 Á ::: Á AnA1 A1AnÀ1 Á ::: Á AnAnÀ1

This ends the proof.

7.1.50. Let M1, A1, B1, C1 be such points that their H images are points M, A, B, C (Figure 7.38). (see the proof of problem 4.1.8c).

H - image M M1

A1 C1 A C

Figure 7.38 7.2 Solutions 351

We have that

MA ¼ M1B1 Á M1C1, MB ¼ M1A1 Á M1C1, MC ¼ M1A1 Á M1B1, AB ¼ M1C1 Á A1B1,

BC ¼ B1C1 Á M1 A1, AC ¼ M1B1 Á A1C1

∠ ∠ ∠ ∠ ∠ (such points exist, sinceqffiffiffiffiffiffiffiffiffiffiffiA1MB1 ¼ qAMBffiffiffiffiffiffiffiffiffiffiffi, B1M1C1q¼ffiffiffiffiffiffiffiffiffiffiffiBMC, A1M1C1 ¼ ∠ MBÁMC MAÁMC MAÁMB AMC and A1M1 ¼ MA , B1M1 ¼ MB , C1M1 ¼ MC ). We have to prove that

M1A1 ÁM1C1 sin∠A1M1C1 þM1B1 ÁM1C1 sin∠B1M1C1 þM1A1 ÁM1B1 sin∠A1M1B1 1 ðÞM A ÁB C þM B ÁA C þM C ÁA B : 2 1 1 1 1 1 1 1 1 1 1 1 1

Note that

M1A1 Á M1C1 sin∠A1M1C1 þ M1B1 Á M1C1 sin∠B1M1C1 þ M1A1 Á M1B1 sin∠A1M1B1 ¼ A M Á B C B M Á A C C M Á A B ¼ 2S ¼ S þ S þ S 1 1 1 1 þ 1 1 1 1 þ 1 1 1 1 , A1B1C1 A1B1M1C1 A1B1C1M1 A1M1B1C1 2 2 2 where the equality holds true, if A1M1 ⊥ B1C1, B1M1 ⊥ A1C1, C1M1 ⊥ A1B1. Then, ∠MAB ¼ ∠M1B1A1 ¼ ∠M1 C1A1 ¼ ∠MAC. Similarly, we obtain that ∠MBA ¼ ∠MBC and ∠MCB ¼ ∠MCA. This means that point M coincides with the incenter of triangle ABC. This ends the proof. 7.1.51. There are two possible cases (Figure 7.39a, b). ! ! ! ! ! ! Let us construct vectors OX ¼ MM1, OY ¼ NN 1 and OZ ¼ PP 1 (Figure 7.40). We need to prove that triangle XYZ is equilateral. Denote by R60 ~a the image of the vector ~a obtained by rotation by an angle of 60. Note that

N b M1 B N1 b M N1 M1 N

M aga g

APP1 CA PP1 C а) b)

Figure 7.39 352 7 Miscellaneous Inequalities

Figure 7.40 X Y

1800-g

b 0 -

0 OZOa Z 180 -b g 180 1800-a

Y X ab

60 ! 60 ! ! 60 ! ! R XY ¼ R OY À OX ¼¼R NN1 À MM1 ¼ 60 ! ! 60 ! 60 ! ! ! ¼ R NM À N1M1 ¼¼R NM À R N1M1 ¼ PM À P1M1 ¼ ! ! ! ! ! ¼ PP 1 À MM1 ¼ OZ À OX ¼ XZ ,

! ! hence R60 XY ¼ XZ . Consequently, triangle XYZ is equilateral. In case (a), according to problem 1.1.11а, one can construct a triangle using segments OX ¼ MM1, OY ¼ NN1 and OZ ¼ PP1. In case (b), according to problem 1.2.9, one cannot construct a triangle using segments OX, OY and OZ only if points O, X, Y, Z are on the same circumference, and in the last case, we obtain that α ¼ β ¼ γ ¼ 60 . Let K and K1 be the centers of d circumcircles of triangles MNP and M1N1P1 respectively. Since β ¼ 60 , NKM ¼ 120 and NK ¼ KM, then point K is on the bisector of the angle B. Similarly, we need to prove that point K is on the bisector of angle A. Hence point K is the incenter of triangle ABC. The same is true also for point K1, because points K and K1 coincide. But then we have the result from the case (a). This leads to a contradiction. Remark See problem 7.1.9. 7.1.52. According to problem 7.1.43, it follows that ∠ECF 135. Therefore γ δ 45 (see the solution of this problem). We have that AE AF þ pffiffiffi þ γ δ γþδ π ¼ 2 þ tg þ tg 2 þ 2tg 2 2 þ 2tg 8 ¼ 2 2 (see the proof of problem 5.2.2а).

2 ai 1 ai ai 7.1.53. Note that ¼ ai Á b ¼ ¼ 2R Á (an 1 ¼ a1), i ¼ 1, 2, . . . , n. bi i sin ∠Aiþ1AiA aiþ1 þ ai a2 a2 Hence, 1 þ ::: þ n ¼ 2R a1 þ a2 þ ::: þ anÀ1 þ an . According to Cauchy’s inequal- b1 bn a2 a3 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffian a1 a2 a2 ity, a1 þ a2 þ ::: þ anÀ1 þ an n n a1 Á a2 Á ::: Á anÀ1 Á an¼ n.Thus, 1 þ 2 þ :::þ a2 a3 an a1 a2 a3 an a1 b1 b2 a2 n 2nR. bn φ 7.1.54. (a) Let ∠ABM ¼ ∠MBN ¼ ∠NBC ¼ φ. We have that AM ¼ SABM ¼ AB sin . MC SMBC BC sin 2φ CN BC sin φ 4AMÁNC 1 > Similarly, we deduce that AN ¼ AB sin 2φ. Thus, ANÁCM ¼ cos 2φ 1. Hence, 4AM Á NC > AN Á CM. 7.2 Solutions 353

(b) Denote the points symmetric to points B and C with respect to the straight lines AC and AB by B1 and C1 respectively. We have that BN ¼ B1N and CM ¼ C1M. Consequently BN þ MN þ CM ¼ B1N þ MN þ C1M B1C1.

We need to prove that B1C1 2BC, then BN þ MN þ CM 2BC. Let AB ¼ c, AC ¼ b, BC ¼ a, ∠A ¼ α, ∠B ¼ β, ∠C ¼ γ, and R be the circumradius of triangle ABC. Then cos ∠B1AC1 ¼ cos 3α and AB1 ¼ c, AC1 ¼ b. According to the law of cosines, we have that b2 þ c2 À a2 ¼ 2bc cos α;then (b2 þ c2 À a2)sin2α ¼ bc sin α sin 2α. Therefore, 4(b2 þ c2 À a2)sin2α ¼ 2bc cos α À 2bc cos 3α, Thus ÀÁ 2 2 2 2α 2 2 2 : 4 b þ c À a sin À 3a ¼ B1C1 À 4a ð7 16Þ pffiffi pffiffiffi α ; β; γ 3 Since a ¼ 2R sin and maxðÞ¼b c 2RmaxðÞ sin sin 2R 2 ¼ R 3,we 2 2 2 2 have that b þ c À a À 3R 0. Then, from (7.2), we deduce that B1C1 2a. 7.1.55. Denote by OA ¼ x, OB ¼ y, OC ¼ z, OD ¼ t and ∠AOB ¼ α. Note that

S x S S y S a ¼ ¼ d , b ¼ ¼ a ð7:17Þ Sb z Sc Sc t Sd According to (7.3), we deduce that

SaSc ¼ SbSd ð7:18Þ and 1 1 1 1 1 x þ y þ a z þ t þ c y þ z þ b x þ t þ d þ À À ¼ þ À À ¼ r r r r 2 S S S S a c b d a c b d 1 x z y t z x t y a c b d ¼ À þ À þ À þ À þ þ À À ¼ 2 S S S S S S S S S S S S a b a d c d c b a c b d 1 a c b d ¼ 0 þ 0 þ 0 þ 0 þ þ À À 2 Sa Sc Sb Sd

Hence, we obtain that 1 1 1 1 1 a c b d þ À À ¼ þ À À ð7:19Þ ra rc rb rd 2 Sa Sc Sb Sd

We need to prove that a c 2 b d 2 ðÞa þ c 2 ðÞb þ d 2 þ À þ ¼ À ð7:20Þ Sa Sc Sb Sd SaSc SbSd

Then from (7.4), (7.5), and (7.6), we obtain the proof of the case (a). By using the law of cosines, we deduce that 354 7 Miscellaneous Inequalities

2 2 2 a ¼ x þ y À 4Sactgα, 2 2 2 b ¼ y þ z þ 4Sbctgα, ð7:21Þ 2 2 2 c ¼ z þ t À 4Scctgα, 2 2 2 d ¼ x þ t þ 4Sdctgα

Consequently,

2 2 2 2 ðÞ4Sb þ 4Sd þ 4Sa þ 4Sc ctgα ¼ b þ d À a À c ð7:22Þ

We have that 2 2 S S a þ c À S S b þ d ¼ a c Sa Sc b d Sb Sd a2 c2 b2 d2 ¼ Á Sc þ Á Sa À Á Sd À Sb þ 2ac À 2bd ¼ Sa Sc Sb Sd x2 þ y2 À 4S ctgα z2 þ t2 À 4S ctgα y2 þ z2 þ 4S ctgα ¼ a S þ c S À b S À S c S a S d a c b x2 þ t2 þ 4S ctgα x2S2 z2S2 y2S2 t2S2 À d S þ 2ac À 2bd ¼ c À d þ c À b þ S b S S S S S S S S d a c d b a c b d 2 2 2 2 2 2 2 2 z Sa x Sb t Sa y Sd þ À þ À À ðÞ4Sc þ 4Sa þ 4Sd þ 4Sb ctgα þ 2ac À 2bd ¼ SaSc SbSd SaSc SbSd ¼ 0 þ 0 þ 0 þ 0 þ a2 þ c2 À b2 À d2 þ 2ac À 2bd ¼ ðÞa þ c 2 À ðÞb þ d 2 2 2 a c b d 2 2 We obtain that SaSc þ À SbSd þ ¼ ðÞa þ c À ðÞb þ d . Sa Sc Sb Sd 2 2 2 2 According to (7.4), we have that a þ c À b þ d ¼ ðÞaþc À ðÞbþd . Sa Sc Sb Sd SaSc SbSd

(b) Denote by OA1 ¼ x1, OB1 ¼ y1, OC1 ¼ z1 and OD1 ¼ t1 (Figure 7.41). α ∠ ∠ π α ∠ ∠ Since 2 ¼ IaOB1 ¼ IcOD1 and 2 À 2 ¼ IbOC1 ¼ IdOA1, then ΔIaOB1 ΔIcOD1 ΔIbOC1 ΔIdOA1. Thus,

Figure 7.41 bC B Ib B1 rb ra C1 aA1 Orc Ic c Ia rd D1 AId

D 7.2 Solutions 355

t1 rc : x1y1 ¼ rard, y1z1 ¼ rarb, ¼ ð7 23Þ y1 ra

> xþyþzþtÀbÀd xþyþzþtÀaÀc We have that a þ c b þ d and x1 þ z1 ¼ 2 , y1 þ t1 ¼ 2 . Hence, x1 þ z1 > y1 þ t1. Consequently, according to (7.12) and 1 þ 1 > 1 þ 1 (see problem 7.1.55a), ra rcrb rd rard rarb > rc 1 1 > 1 1 we obtain that þ y1 þ y1 ¼ rcy1 þ rcy1 þ . Thus, qffiffiffiffiffiffiffiffiffi y1 y1 ra rc ra rb rd y < rarbrd. 1 rc OIbÁOId rb rd rarbrd > > Note that ¼ Á ¼ 2 1. Hence OIb Á OId OIa Á OIc. OIaÁOIc y1 t1 rcy1 If we draw a circle through points Ia, Ib, Id, then according to the last inequality, point Ic is inside of that circle. Thus ∠IdIaIb þ ∠IdIcIb > π.

7.1.56. Since ∠D1C1D2 ¼ ∠D1B1D2, then points D1, C1, B1 , and D2 (Figure 7.42) are located on the same circumference. Hence, ∠C1D1B1 ¼ ∠C1D2B1 ¼ ∠C2D2B2. Similarly, we ﬁnd that ∠C2D2B2 ¼ ∠C2D3B2 and so on. Denote by φ ¼ ∠C1D1B1 ¼ ∠C1D2B1 ¼ ∠C2D2B2 ¼ ∠C2D3B2 ¼ ... ¼ ∠CnD1Bn. D1D2 B1C1 According to the law of sines, we have that β ¼ 2R1 ¼ φ, where R1 is the sin 1 sin circumradius of the quadrilateral D1C1B1D2. Consequently, B1C1 sin β1 ¼ D1D2 sin φ. Similarly, we obtain that B2C2 sin β2 ¼ D3D2 sin φ and, BnCn sin βn ¼ DnD1 sin φ. Thus, we have that

A3 C3 B3

B2 D3 C2

C1 D1 Dn

A1 Bn Cn An

Figure 7.42 356 7 Miscellaneous Inequalities

β ::: β φ φ ::: φ > B1C1 sin 1 þ þ BnÀ1CnÀ1 sin nÀ1 ¼ D1D2 sin þ D2D3 sin þ þ DnÀ1Dn sin > φ β : D1Dn sin ¼ BnCn sin n

β β > β Hence, B1C1 sin 1 þ ...þ Bn À 1Cn À 1 sin n À 1 BnCn sin n. Remark See problems 7.1.9 and 7.1.51.

7.1.57. Note that, if a1, a2, a3, b1, b2, b3 > 0, then according to the Cauchy’s inequality rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a a a b b b 3 1 Á 2 Á 3 þ 3 1 Á 2 Á 3 a1 þ b1 a2 þ b2 a3 þ b3 a1 þ b1 a2 þ b2 a3 þ b3 a a a b ba ba 1 þ 2 þ 3 1 þ 2 þ 3 a þ b a þ b a þ b a þ b a þ b a þ b 1 1 2 2 3 3 þ 1 1 2 2 3 3 ¼ 1: 3 3 Hence, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi 3 p3 3 ðÞa1 þ b1 ðÞa2 þ b2 ðÞa3 þ b3 a1a2a3þ b1b2b3: ð7:24Þ According to the inequality (7.24), we have that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 p þ λa p þ λb p þ λc ¼ p À a p À b p À c sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 a b c ¼ 1 þ ðÞλ þ 1 1 þ ðÞλ þ 1 1 þ ðÞλ þ 1 p À a p À b p À c rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi abc 1 þ ðÞλ þ 1 3 ¼ p a p b p c sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðÞÀ ðÞÀ ðÞÀ

3 ðÞp À b þ p À c ðÞp À a þ p À c ðÞp À a þ p À b ¼ 1 þ ðÞλ þ 1 ðÞp À a ðÞp À b ðÞp À c sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2 ðÞp À b ðÞp À c Á 2 ðÞp À a ðÞp À c Á 2 ðÞp À a ðÞp À b 1 þ ðÞλ þ 1 ¼ 2λ þ 3: ðÞp À a ðÞp À b ðÞp À c

ðÞpþλa ðÞpþλb ðÞpþλc λ 3 Thus, ðÞpÀa ðÞpÀb ðÞpÀc ðÞ2 þ 3 .

7.1.58. Denote by ∠PBC ¼ α1, ∠APC ¼ β1, ∠APB ¼ γ1. Let R1 be the radius of the circumference passing through points P, B, A0 and C. According to the law of sines, γ β 0 0 π γ π β sin 1 sin 1 we have that BA þ A C ¼ 2R1ðÞsin ðÞþÀ 1 sin ðÞÀ 1 ¼a sin α þ a sin α . γ 1 1 0 0 sin 1 sin α1 Similarly, we deuce that AB þ B C ¼ b sin β þ b sin β and β 1 1 0 0 sin α1 sin 1 A C þ C B ¼ c γ þ c γ . Hence, it follows that sin 1 sin 1 7.2 Solutions 357

Figure 7.43 A

LKE O

0 0 0 0 0 0 ABþ B C þ CA þ A B þ BCþ C A ¼ sin γ sin α sin β sin α sin γ sin β ¼ a 1 þ c 1 þ a 1 þ b 1 þ b 1 þ c 1 sin α sin γ sin α sin β sin β sin γ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 γ α β α γ β sin 1 sin 1 sin 1 sin 1 sin 1 sin 1 2 a α Á c γ þ 2 a α Á b β þ 2 b β Á c γ ¼ sin 1 sin 1 sin 1 sin 1 sin 1 sin 1 pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ 2 ac þ 2 ab þ 2 bc: pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi Thus, AB0 þ B0C þ CA0 þ A0B þ BC0 þ C0A 2 ac þ 2 ab þ 2 bc. 7.1.59. Since DE k BC and ∠DBK ¼ ∠KBC, then ∠DKB ¼ ∠KBC ¼ ∠DBK (see Figure 7.43). Hence DK ¼ BD ¼ AD. Consequently ∠AKB ¼ 90. Similarly, we obtain that ∠ALC ¼ 90. Consequently, points K and L are on the circumference with a ∠ ∠ ∠A diameter AO. Thus, AO LK . (Note that LOK ¼ BOC ¼ 90 þ 2 ; this means that AO > LK.) According to the triangle inequality, BO þ CO > BC; hence AO þ BO þ CO > LK þ BC.

7.1.60. (a) For a non-obtuse triangle, we have that a ¼ ha(ctgβ þ ctgγ), b ¼ hb(ctgα þ ctgγ), c ¼ hc(ctgα þ ctgβ), where ctgα, ctgβ, ctgγ 0. Note that ctgαctgβ þ ctgβctgγ þ ctgγctgα ¼ 1. Indeed, we have that γ α β 1ÀctgαÁctgβ α β β γ γ α ctg ¼Àctg ðÞ¼þ ctgαþctgβ ,orctg ctg þ ctg ctg þ ctg ctg ¼ 1. Denote by ctgα ¼ x, ctgβ ¼ y, ctgγ ¼ z, then we need to prove that, if x, y, z 0 and x þ y > 0,y þ z > 0, x þ z > 0, then ! 1 1 1 9 ðÞxy þ yz þ zx þ þ : ð7:25Þ ðÞx þ y 2 ðÞy þ z 2 ðÞx þ z 2 4

The last inequality is equivalent to the following inequality.

2xyzðÞ xðÞ x À y ðÞþx À z yyðÞÀ z ðÞþy À x zzðÞÀ x ðÞz À y þxyðÞ x2 þ xy þ y2 ðÞx À y 2þ 2 þyzðÞ y2 þ yz þ z2 ðÞy À z 2 þ xzðÞ x2 þ xz þ z2 ðÞx À z 2 þ 3xyðÞ x2 À y2 þ 2 2 þ3yzðÞ y2 À z2 þ 3xzðÞ x2 À z2 0:

It remains to prove that x(x À y)(x À z) þ y(y À z)(y À x) þ z(z À x)(z À y) 0. 358 7 Miscellaneous Inequalities

Indeed, let min(x, y, z) ¼ z 0, then x(x À y)(x À z) þ y(y À z)(y À x) þ z(z À x) (z À y) ¼ (x À y)2(x þ y À z) þ z(z À x)(z À y) 0, since (x À y)2 0, x þ y À z 0, z 0, (z À x)(z À y) 0. As

h2 h2 h2 1 1 1 a þ b þ c ¼ þ þ ¼ a2 b2 c2 ðÞctgα þ ctgβ 2 ðÞctgβ þ ctgγ 2 ðÞctgγ þ ctgα 2

1 1 ¼ ðÞctgαctgβ þ ctgβctgγ þ ctgγctgα þ þ ðÞctgα þ ctgβ 2 ðÞctgβ þ ctgγ 2 ! ! 1 1 1 1 9 þ ¼ ðÞxy þ yz þ xz þ þ : ðÞctgγ þ ctgα 2 ðÞx þ y 2 ðÞy þ z 2 ðÞx þ z 2 4

h2 h2 h2 Therefore, a b c 9. a2 þ b2 þ c2 4 Remark Another proof of the inequality (7.25) can be obtained from problem 5.5.10c. pffiffiffi (b) Let a b c, then we have to prove that 2 3S pa,or3(p À a)(a2 À (b À c)2) pa2,3(b þ c À a)(a2 À (b À c)2) (a þ b þ c)a2. 2 2 2 2 Let a ¼ 1, c ¼ b þ x, thenp fromffiffiffi conditions a þ b c , we obtain that 1 x þ 2bx x2 þ 2x. Hence, x 2 À 1. We have to prove that 3(2b þ x À 1)(1 À x2) 1 þ 2b þ x, b(4 À 6x2) 4 À 2x À 3x2 þ 3x3. We have that b(4 À 6x2) 4 À 6x2 4 À 2x À 3x2 þ 3x3. 7.1.61.pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiLet a < b. (If a ¼ b, then this distance is equal to zero, and aþb À ppðÞÀ c > aþb À pþpÀc ¼ 0.) 2 2 2 γ γ þÀa sin ρ ; 2 ðÞ2 We have that ðÞ¼M l b sin 2 (see Figure 7.44).

Figure 7.44 y C

g g 2 2 ab

B х M l A 7.2 Solutions 359 qffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ ; bÀa 1À cos γ bÀa ðÞpÀa ðÞpÀb Therefore, ðÞ¼M l 2 2 ¼¼ 2 ab . Note that rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞb À a ðÞp À a ðÞp À b 1 p2 À paðÞþþ b ab ¼ ðÞa þ b 2 À 4ab ¼ 2vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiab 2 ab u ! u 2 1t ðÞa þ b ðÞÀp2 þ paðÞþ b ¼ ðÞa þ b 2 þ 4paðÞÀþ b 4p2 À þ 4ab 2 ab qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞa þ b 2 þ 4paðÞÀþ b 4p2 À 4ðÞa þ b paðÞÀþ b p2 ¼ 2 a þ b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ À paðÞÀþ b p2 ¼ À ppðÞÀ c : 2 2

pffiffiffiffi γ 2 ab Remark The equality holds true when cos 2 ¼ aþb . qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2a2þ2b2Àc2 ðÞðÞÀpÀa ðÞpÀb 7.1.62. We have that mc ¼ 4 ¼ ppðÞþÀ c 4 and we have rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiqffiffiffiffiffiffi qffiffiffiffiffiffi 2 pffiffiffi to prove that A ¼ pÀc þ 1 pÀa À pÀb þ pÀa þ pÀb 3. qffiffiffiffiffiffip 4 p qp ffiffiffiffiffiffi p p qffiffiffiffiffiffi qffiffiffiffiffiffi pÀa pÀb pÀa pÀb Denote by p ¼ x þ t, p ¼ x À t ; then 2x ¼ p þ p rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiqffiffiffiffi pffiffiffi pffiffi pÀa pÀb 2c < < < 2 2 p þ p ¼ p 2. Consequently, 0 x 2 and

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 A ¼ 1 À ðÞx þ t 2 À ðÞx À t 2 þ ðÞx þ t 2 À ðÞx À t 2 þ 2x ¼ 4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ¼ 1 À 2x2 À 21ðÞÀ 2x2 t2 þ 2x 1 À 2x2 þ 2x 3,

ÀÁpffiffiffi 2 since 1 À 2x2 3 À 2x . ÀÁpffiffiffi 2 This means that 3x À 1 0. BE AB < BC BC BF BE < BF 7.1.63. Figure 7.45 is correct since EM ¼ AM AM ¼ CM ¼ FM or EM þ 1 FM þ 1, FM < EM.

Figure 7.45 B

I E F

AMC 360 7 Miscellaneous Inequalities

Consider triangle AIC. According to Menelaus’s theorem, it follows that AE IF CM AE IE ∠ ∠C ∠ IE Á FC Á MA ¼ 1, or CF ¼ IF. We have that EFI ¼ 2 þ FBC and ∠ ∠A ∠ ∠ < ∠Bþ∠C < ∠Bþ∠A < ∠ IEF ¼ 2 þ ABE. Consequently, EFI 2 2 IEF. Thus, EI < IF, hence AE < CF. This ends the proof. 7.1.64. From the statement of the problem it follows that triangle ABC is either acute or right. Let triangle ABC be acute and point N be the midpoint of segment BC > AB ∠ > ∠ ∠ AB, then MN ¼ 2 2 ¼ NB, ABM BMN ¼ MBC. ∠ ∠ < ∠B < ∠ Figure 7.46 is correct since ABH ¼ 90 À A 2 ABM. We have to prove that HF > HE (see the proof of problem 7.1.63), or ∠HEF ¼ 90 À ∠B þ ∠ABM > 90 À ∠B þ ∠MBC ¼ ∠HFE is correct. If triangle ABC is right-angled, then if ∠B ¼ 90, we have that AE ¼ AB < BC ¼ CF, while if ∠A ¼ 90, we deduce that CF ¼ CM ¼ AM > AE, since ∠AEM ¼ 90 À ∠MBC > ∠C þ ∠MBC ¼ ∠AFE. 7.1.65. Consider three cases (see Figure 7.47, I, II, III). (I) For triangle AKC and line BM, according to Menelaus’s theorem, we obtain AE FK CM AE KE < ∠ ∠ ∠ < that EK Á FC Á MA ¼ 1, CF ¼ KF 1, since EFK ¼ FCB þ MBC ∠EAB þ ∠MBA ¼ ∠KEF. (see the proof of problem 7.1.64). (II) Since ∠A > ∠C, then ∠EAC > ∠ECA. Hence, CF > EA.

Figure 7.46 B

E NHF

AMC

B B B E K E=F F FК E

AMCAMCAMC I II III

Figure 7.47 7.2 Solutions 361

AE FK CM AE KE < (III) We have that EK Á FC Á MA ¼ 1. Consequently, CF ¼ EF 1, since ∠EFK < ∠KEF. sin ∠BAE sin ∠BCF < sin ∠BCF ∠ < (b) We have that BE ¼ AE sin ∠ABE ¼ AE sin ∠ABE AE sin ∠FBC, since FBC ∠ < ∠ < sin ∠BCF < sin ∠BCF ABE 180 À FBC. Hence, BE AE sin ∠FBC CF sin ∠FBC ¼ BF (see problem 7.1.65a). Thus, BE < BF. 7.1.66. Denote by AB ¼ a, ∠EFA ¼ α, ∠BAP ¼ β, ∠DAQ ¼ γ, then we have to prove that

∠EPA > 90 , ð7:26Þ

∠FQA > 90, and β þ γ 45. (а) To prove (7.26), we have to prove that AB Á BE > BP2. This means that point P is inside of the circle with a diameter AE.

2 BP2 ðÞðÞ2a sin αÀa ctgα cos α sin 2αÀ1 < We have that ABÁBE ¼ aÁðÞ2a sin αÀa ¼ 1 þ sin 2α 1. Similarly, we need to prove that ∠FQA > 90. (b) We have to prove that tg(β þ γ) 1, that is tgβ þ tgγ þ tgβtgγ 1. β BP α α γ α α We have that tg ¼ AB ¼ ðÞ2 sin À 1 ctg and tg ¼ (2 cos À 1)tg . γ β γ β cos 22α Therefore, tg þ tg þ tg tg ¼ 1 À sin α cos α 1. 7.1.67. Let point P be inside triangle ABO (Figure 7.48). _ _ Note that if P ≢ O, then P1B< B1P2, for which it is sufﬁcient to draw a diameter _ _ parallel to BB1. Similarly, P1A< A1P2, consequently ∠PBA1 > ∠BA1P. Hence, it follows that PA1 > PB. We have that (1 À x)(1 þ x) ¼ PA Á PA1 > PA Á PB and PC PO þ OC ¼ 1 þ x. Thus, PA Á PB Á PC < (1 þ x)2(1 À x). If P O, then PA Á PB Á PB ¼ 1 and x ¼ 0. Hence, AP Á BP Á CP ¼ (1 þ x)2(1 À x). AL AL2 AL2 ABÁACÀBLÁLC Δ Δ 7.1.68. We have that LP ¼ LPÁAL ¼ BLÁLC ¼ BLÁLC , since BAL APC. Thus, AL AB 2 AC ¼ ALþLP,orAL ¼ AB Á AC À AL Á LP ¼ AB Á AC À BL Á LC. Consequently, AL ¼ ABÁAC À 1 4ABÁAC À 1 ¼ 4ABÁAC À 1. LP BLÁLC ðÞBLþLC 2 BC2

Figure 7.48 B A1 P1 P

AC O

P2 B1 362 7 Miscellaneous Inequalities

Similarly, we deduce that BM 4ABÁBC 1 and CN 4ACÁBC 1. QM AC2 À RN AB2 À Summing up the above inequalities, we obtain that AL BM CN AB Á AC AB Á BC AC Á BC þ þ 4 þ þ À 3 LP QM RN BC2 AC2 AB2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3 AB Á AC AB Á BC AC Á BC 12 Á Á À 3 ¼ 9: BC2 AC2 AB2

AL BM CN Hence LP þ QM þ RN 9. Note that the equality holds true, only if BL ¼ LC, AM ¼ CM, AN ¼ BN and ABÁAC ABÁBC ACÁBC. This means that, triangle ABC is equilateral. BC2 ¼ AC2 ¼ AB2 See also problem 7.1.5. 2π 2π ::: 2πðÞnÀ1 2πðÞnÀ1 7.1.69. (a) Let z1 ¼ 1, z2 ¼ cos n þ i sin n , , zn ¼ cos n þ i sin n correspond in the complex plane to the vertices of the regular n-gon A1A2 ...An. Then, n n the numbers z1,...,zn are the roots of the polynomial z À 1so that z À 1 ¼ (z À z1). . .(z À zn). For points M(z) we have that z 2 R and z > 1; consequently,

MA1 Á MA2 Á ::: Á MAn ¼ jjz À z1 jjÁz À z2 ::: Á jj¼z À zn jjðÞz À z1 ðÞÁz À z2 ::: Á ðÞz À zn ¼ ¼ jjzn À 1 ¼ zn À 1 ¼ MOn À 1:

Pn qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi According to Cauchy’s inequality, we have that 1 n n 1 Á 1 Á ::: Á 1 . MAi MA1 MA2 MAn i¼1 Pn Pn 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 ffiffiffiffiffiffiffiffiffiffiffin > n 1 > n Thus n pn ¼ pn n , consequently . MAi MA1Á:::ÁMAn MO À1 MO MAi MO i¼1 i¼1

∠ α > [AiAiþ1 π , , , :::, , (b) We have that AiMAiþ1 ¼ i 2 ¼ n i ¼ 1 2 n Anþ1 A1 and α1 þ α2 þ ...þ αn ¼ 2π. π < α < π ::: α α α π Let us prove that, if n i , i ¼ 1, 2, , nand 1 þ 2 þ ...þ n ¼ 2 , then α α ::: α > 2π π : sin 1 þ sin 2 þ þ sin n sin n þ ðÞn À 2 sin n α > π β > π α β < π Note that, if n , n and þ 2 , then π π sin α þ sin β > sin þ sin α þ β À : n n α α α α α α ::: Let n ¼ max ( 1, 2, ..., nÀÁ). Thus, it follows that sin 1 þ sin 2 þ þ α > π α α ::: α π : sin nÀ1 ðÞn À 2 sin n þ sin 1ÀÁþ 2 þ þ nÀ1 À ðÞn À 2 n α α ::: α π α It is sufﬁcient to note that sin 1 þ 2 þ þ nÀ1 À ðÞn À 2 n þ sin n sin 2π π 2π : α α ::: n þ sin ¼ sin n Hence, we obtain that sin 1 þ sin 2 þ þ α > 2π π : sin n sin n þ ðÞn À 2 sin n This ends the proof. ÀÁ ∠ α ∠ β ∠ φ α β φ ; π 7.1.70. Denote NAV ¼ , NAB ¼ , PMU ¼ . It is clear that , , , 2 0 2 (Figure 7.49a, b). 7.2 Solutions 363

B B P UP P1 U 1

P M M

j N j N

b Q1 b a a QV AVQAQ1 l l a b

Figure 7.49

Consider Figure 7.49a.IfQQ1 PP1, then PQ P1Q1 UV. Hence, PQ UV. Applying the law of sines to triangles VQQ1 and UPP1 , we obtain that

QQ VQ sin α P U sin β VQ sin α cos ðÞβ À φ VM sin α cos ðÞβ À φ 1 ¼ 1 : 1 ¼ 1 Á ¼ Á ¼ PP1 cos ðÞα þ φ cos ðÞβ À φ P1U cos ðÞα þ φ sin β MU cos ðÞα þ φ sin β UN sin α cos ðÞβ À φ ANtgβ sin α cos ðÞβ À φ cos ðÞÁβ À φ cos α ¼ Á ¼ Á ¼ ¼ VN cos ðÞα þ φ sin β ANtgα cos ðÞα þ φ sin β cos β Á cos ðÞα þ φ cos α cos β cos φ þ sin β sin φ cos α ¼ > 1, cos α cos β cos φ À sin α sin φ cos β since cos(α þ φ) > 0 and sin φ sin (α þ β) > 0. Let us consider Figure 7.49b and prove that QQ1 PP1, then PQ P1Q1 UV. Indeed, in the same way as in the ﬁrst case, we deduce that

QQ VQ sin α P U sin β cos α cos ðÞβ þ φ 1 ¼ 1 : 1 ¼ Á ¼ PP1 cos ðÞα À φ cos ðÞβ þ φ cos ðÞα À φ cos β cos α cos β cos φ À sin β sin φ cos α ¼ < 1, cos α cos β cos φ þ sin α sin φ cos β as sin φ sin (α þ β) > 0 and cos(α À φ) > 0. 7.1.71. (a) First we need to prove that there exists a triangle ABC and a point M inside of it, such that αβγ has the greatest value. Consider the set X of all possible products αβγ. Since, α, β, γ < π, then αβγ < π3. We have obtained that the set X is bounded from above; hence it has a supremum. Let sup X ¼ a.Ifa 2 X, then this ends the proof of the statement. Suppose that a 2 X. Then there exists a sequence of triangles AnBnCn and points Δ α β γ α ∠ β Mn (Mn 2 AnBnCn) such that lim ðÞ¼n n n a, where n ¼ MnAnBn, n ¼ n!1 ∠MnBnCn,γn ¼ ∠MnCnAn. 364 7 Miscellaneous Inequalities

α0 ∠ β0 ∠ γ0 ∠ α β γ ÀÁLetÀÁn ¼ÀÁMnAnCn, n ¼ MnBnA, n ¼ MnCnBn. Since the ðÞn , ðÞn , ðÞn , α0 β0 γ0 n , n , n sequences are bounded, then without loss of Generality, one can α α β β γ γ assume that they are convergent. Let lim n ¼ , lim n ¼ , lim n ¼ , n!1 n!1 n!1 lim α0 ¼ α0, lim β0 ¼ β0 lim γ0 ¼ γ0. Then, we have that α þ α0 þ β þ β0 þ γ þ γ0 n!1 n ÀÁn!1 n n!1 n α α0 β β0 γ γ0 π α β γ αβγ ¼ lim n þ þ n þ þ n þ ¼ and a ¼ lim ðÞ¼n n n , hence n!1 n n n n!1 α, β, γ > 0. According to the law of sines we have that,

M A M B M C sinβ0 sinγ0 sinα0 1 ¼ n n Á n n Á n n ¼ n Á n Á n Á Consequently; MnBn MnCn MnAn sinαn sinβ sinγ n n ÀÁ α β γ α β γ α0 β0 γ0 α0 β0 γ0: sin sin sin ¼ lim ðÞsin n sin n sin n ¼ lim sin sin sin ¼ sin sin sin n!1 n!1 n n n

Hence,

sin α sin β sin γ ¼ sin α0 sin β0 sin γ0 ð7:27Þ

Thus α0, β0, γ0 > 0. Now, consider triangle ABC and point M inside of it, such that ∠MAB ¼ α, ∠MBA ¼ β0, ∠MAC ¼ α0, ∠MBC ¼ β, then ∠ACB ¼ γ þ γ0. 00 Let ∠MCB ¼ γ , then ÀÁ sin α sin β sin γ þ γ0 À γ00 ¼ sin α0 sin β0 sin γ00 ð7:28Þ

0 00 sin γ sin ðÞγþγ Àγ From (7.27) and (7.28), we obtain that sin γ0 ¼ sin γ00 , 00 Thus sin(γ þ γ0)ctgγ0 À cos (γ þ γ0) ¼ sin (γ þ γ0)ctgγ À cos (γ þ γ0) 00 Hence γ0 ¼ γ . We have obtained that a 2 X. This leads to a contradiction. This ends the proof. Thus, there exists a triangle ABC and a point M inside of it, such that the product αβγ has the greatest value. We need to prove that α0 ¼ β0 ¼ γ0. Indeed, let for example α0 6¼ β0, then let us draw through points B and M a circle touching ray KA at point A0, where K is the intersection point of the straight lines 0 0 BM and AC. Since, α 6¼ β , then points A and A0 are different. Thus α < α0, where α0 ¼ ∠MA0B, hence a ¼ αβγ < α0βγ. This leads to a contradiction. Now by using the following lemma, we can prove that α ¼ β ¼ γ. Lemma If point M given inside of triangle ABC is, such that α ¼ ∠MAB 6¼ ∠MBC ¼ β and α þ β þ 2 ∠MBA < π, then there exists such a point D on the straight line BC, that point M is again inside of triangle ADC and α1β1 > αβ, where α1 ¼ ∠MAD, β1 ¼ ∠MDC.

Let D be any point of line BC (M 2 △ ADC). Suppose that α1β1 αβ. Then α ¼ β. This leads to a contradiction. 7.2 Solutions 365

Indeed, let ∠DAB ¼ x,∠DMB ¼ y. Note that x ! 0 , y ! 0, Therefore, lim x ¼ lim x Á sin x Á sin y ¼ lim sin x. x!0 y x!0 sin x sin y y x!0 sin y Let ∠ABM ¼ φ. According to the law of sines,

sin x sin x sin β sin ðÞβ þ φ ¼ Á Á sin y sin ðÞβ þ φ sin y sin β BD MD sin ðÞβ þ φ MD sin ðÞβ þ φ ¼ Á Á ¼ Á : AD BD sin β AD sin β

Consequently

x sin x MB sin ðÞβ þ φ sin α sin ðÞβ þ φ lim ¼ lim ¼ Á ¼ : ð7:29Þ x!0 y x!0 sin y AB sin β sin ðÞα þ φ sin β

If point D is inside of segment BC (M 2 ΔADC), then α1β1 ¼ (α À x)(β þ y) αβ. α x α x Consequently, β ; hence lim β lim , thus þy y x!0 þy x!0 y α x lim : ð7:30Þ β n!0 y

If point D is outside of segment BC(M 2 ΔADC), then α1β1 ¼ (α þ x) β αβ x αþx ( À y) . Consequently, y β , thus x α lim ð7:31Þ x!0 y β

α x sin α sin ðÞβþφ From (7.29), (7.30), and (7.31), we obtain that β ¼ lim ¼ α φ β. x!0 y sin ðÞþ sin α sin ðÞαþφ β sin ðÞβþφ Thus sin α ¼ sin β . Let α > β, then since α þ φ þ β þ φ < π, we obtain

sin ðÞβ þ φ < sin ðÞα þ φ ð7:32Þ ÀÁ α β < α β β < ββ< π We have that sin( À ) À and tg 2 ; consequently

sin α sin ðÞα À β α À β α ¼ þ cos ðÞα À β < þ 1 ¼ ð7:33Þ sin β tgβ β β

α sin ðÞαþφ > β sin ðÞβþφ According to (7.32) and (7.33), we deduce that sin α sin β . This leads to a contradiction and ends the proof of the lemma. α0 β0 γ0 φ α β γ 3α 3φ Since ¼ ¼ ¼ , ¼ ¼ , then accordingÀÁ to (7.27) sin ¼ sin , conse- α β γ φ π αβγ π 3 quently, ¼ ¼ ¼ ¼ 6 and a ¼ ¼ 6 . ÀÁ αβγ α β γ π 4 Remark If given a point M is inside of triangle ABC, then ðÞþ þ 3 6 , where α ¼ ∠MAB, β ¼ ∠MBC, γ ¼ ∠MCA. 366 7 Miscellaneous Inequalities ÀÁ α2βγ π 4 Hint Similar to the solution of problem 7.1.71a, one can prove that 6 . (b) For any n 4 we need to give an example of an n- gon and a point M inside of it, n α α ::: α > πðÞnÀ2 such that 1 2 Á Á n 2n .

Consider triangle ABC and pointpffiffi M inside of it suchp thatffiffi ∠ ∠ ∠ ∠ 2π ∠ ∠ π 2π MCA ¼ MCB ¼ MBC ¼ MBA ¼ 8 , then MAC ¼ MAB ¼ 2 À 4 . Since ∠MAB 6¼ ∠MBC, ∠MAB þ ∠MBC þ 2 ∠MBA < π, then from the lemma above it follows that there exists a point D on line BC such that point M is inside of pffiffi pffiffi γ β > π 2π 2π γ ∠ β ∠ triangle ADC and 1 1 2 À 4 Á 8 , where 1 ¼ MAD, 1 ¼ MDC.

Let point D1 be on side AC and close to point A and point A2, located on the half- line MD1, close to D1, so that the quadrilateral A1A2A3A4 is convex. Here A1 C, A A, A D and apart from that the inequality α γ β ¼ 3 4pffiffi pffiffi pffiffi pffiffi p2ffiffi 1 1 α α α > π π 2π π 2π 2π 2π3 α > 2π 2 3 4 À 2 À 4 2 À 4 8 ¼ 64 holds true. Since 1 8 , then pffiffi pffiffi ÀÁ α α α α > 2π3 2π π 4 1 2 3 4 64 Á 8 ¼ 4 . Let n 5 and A1A2 ...An À 1 be a regular (n À 1)-gon with the center M. Take 0 0 point An on side A1An À 1 and point An on the half-line MA n outside of the polygon A1A2 ...An À 1, so that the polygon A1A2 ...An À 1An is convex. Then, nÀ1 α α ::: α > πðÞnÀ3 α < π πðÞnÀ3 1 2 Á Á nÀ1 2ðÞnÀ1 . Evidently, for any given angle À 2ðÞnÀ1 , one 0 0 α > α can take point An close to point A1 and point An close to point An, such that n . nÀ1 πðÞnþ1 πðÞnÀ3 < α α ::: α Thus, point An can be chosen, such that 2ðÞnÀ1 2ðÞnÀ1 1 2 Á Á n.We need to prove that for n 5 the following inequality holds true nÀ1 n πðÞnþ1 πðÞnÀ3 > πðÞnÀ2 2ðÞnÀ1 2ðÞnÀ1 2n ,or

n n2 À 3n n À 3 > ð7:34Þ n2 À 3n þ 2 n þ 1 n n ’ n2À3n 2 According to Bernoulli s inequality, we have that n2À3nþ2 ¼ 1 À n2À3nþ2 2n > 4 1 À n2À3nþ2 1 À nþ1 at n 7, Hence for n 7, (7.34) holds true. Let us check that (7.34) holds trueÀÁ for n ¼ 5 and n ¼ 6. 5 5 > 1 > For n ¼ 5 , we have to check thatÀÁ6 3, or 3125 2592. 9 6 > 3 > For n ¼ 6, we have to check that 10 7, or 3720087 3000000. (c) At ﬁrst, let us prove the following lemma.

Lemma If given α1,...,αn, β1,...,βn > 0, such that α1 þ ...þ αn þ β β π α β < πα β < π β β 1 þ ...þ n ¼ (n À 2) and i þ i i þ i þ 1 , i ¼ 1, 2, . . . , n, n þ 1 ¼ 1. Moreover, if

α ... α β ... β : sin 1 Á Á sin n ¼ sin 1 Á Á sin n, ð7 35Þ 7.2 Solutions 367

then there exists a convex polygon A1 ...Anand point M inside of it, so that ∠MAiAi þ 1 ¼ αi, ∠MAiAi À 1 ¼ βi, i ¼ 1, 2, . . . , n, where An þ 1 A1, A0 An. Indeed, consider triangles ΔMAiAi þ 1i ¼ 1, 2, . . . , n À 1, such that ∠MAiAi þ 1 ¼ α ∠ β i, i ¼ 1,...,n À 1, MAiAi À 1 ¼ ii ¼ 2, 3, . . . , n,withpointsAi À 1 and Ai þ 1 being on different sides of the straight line MAi, i ¼ 2,...,n À 1. ∠ π π α β α β π Note that A1MAn ¼ 2 À ((n À 1) À ( 1 þ 2) À ... À ( n À 1 þ n)) ¼ À (αn þ β1) and ∠MA1An þ ∠MAnA1 ¼ αn þ β1. Let ∠MA1An ¼ β , ∠MAnA1 ¼ α. According to the law of sines, it follows that 1 ¼ MA1 Á MA2 Á :::Á MA2 MA3 β β β β MAn sin 2 sin 3 ::: sin n sin MA ¼ sin α Á sin α Á Á sin α Á sin α. From the last relation and (7.35) we obtain 1 β 1 β 2 nÀ1 that sin ¼ sin 1 and as α þ β ¼ α þ β, then sin(α þ β)ctgα À cos (α þ β) ¼ sin sin α sin αn n 1 (α þ β)ctgαn À cos (α þ β). Therefore, α ¼ αn and β ¼ β1. Hence, the polygon A1A2 ...An is convex and point M is inside of the polygon A1A2 ...An, since there exist close enough points to Ai on the AiM (i ¼ 1, 2, ..., n), such that these points are inside of the polygon. This ends the proof of the lemma. α ; :::; α > πðÞnÀ2 ε > Let minðÞ1 n 2n , then there exists a number 0, such that

πðÞn À 2 α þ ε, i ¼ 1, :::, n: ð7:36Þ i 2n

Consider the set X of all possible products α1 Á ...Á αn, such that the condition (7.36) holds true. Let sup X ¼ a. In the case a 2 X, let α1 Á ...Á αn ¼ a. Then as it was shown in (a), β β φ α πðÞnÀ2 ε ::: 1 ¼ ...¼ n ¼ . We have that i 2n þ , i ¼ 1, , n and

n sin φ ¼ sinÀÁα1::: sin αn ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi α ::: α φ ::: φ n α ::: α nφ sin 1þ þ sin nþ sin þ þ sin ¼ sin 1 sin nsin 2n α þ ::: þ α þ nφ πðÞn À 2 sin n 1 n ¼ sin n , 2n 2n

φ α < π φ πðÞnÀ2 and since þ i , we have 2n . If max(MA1, ..., MAn) ¼ MAi, then from φ α πðÞnÀ2 α triangle MAiAi þ 1, we obtain that i, consequently, 2n i. This leads to a contradiction. If a2 X, then we can deduce that there exist (see the proof of problem 7.1.71a) numbers α1,...,αn > 0, β1,...,βn 0 such that α1 þ ...þ αn þ β1 þ ...þ βn ¼ π(n À 2), α1 Á ...Á αn ¼ a, and αi þ βi þ 1 π, i ¼ 1, . . . , n (βn þ 1 ¼ β1), αi þ βi π, i ¼ 1, . . . , n. Apart from that, we also have that sin α1 Á ...Á α β β α πðÞnÀ2 ε sin n ¼ sin 1 Á ...Á sin n, i 2n þ , i ¼ 1, . . . , n . Let αi 6¼ π, i ¼ 1, . . . , n, then from the condition sin α1 Á ...Á sin αn ¼ sin β1 Á ...Á sin βn, we deduce that β1,...,βn > 0.

It is clear that there exists a number i, such that αi þ βi ¼ π or αi þ βi þ 1 ¼ π. Otherwise, according to the lemma, we deduce that a 2 X. 368 7 Miscellaneous Inequalities

If αi þ βi ¼ π for some i, then consider the set α1, β1, α2, β2,...,αi À 1, βi À 1, αi þ 1, βi þ 1,,...,αn, βn.Ifαi þ βi þ 1 ¼ π, then consider the set of angles α1, β1,...,αi À 1, βi À 1, αi þ 1, βi, αi þ 2, βi þ 2,...,αn, βn. Excluding all such pairs, α β ::: α β we obtain a set of angles 1, 1, , k, k satisfying the conditions of the lemma with n > k 3, because the number of those pairs is not greater than n À 3. By mathematical induction, we obtain that (problem 2.4.10c is the case of n ¼ 3),ifk < n, α ; :::; α πðÞkÀ2 < πðÞnÀ2 α ::: α α ; :::; α then minðÞ1 k 2k 2n ,as 1, , k 2fg1 n . Then, α ; :::; α α ; :::; α < πðÞnÀ2 minðÞ1 n minðÞ1 k 2n . This leads to a contradiction. Let α1 ¼ π, then β1 ¼ 0 and α2 þ ...þ αn À 1 þ αn þ β2 þ ...þ βn À 1 þ βn ¼ π(n À 3). Therefore, it follows that n > 3. 0 ::: 0 Hence there exists a convex polygon A1 An and a point M inside of it, such that α0 π ε α0 ::: α0 πðÞnÀ2 ε ε π ε > 1 À , 2, , n 2n þ ( can be chosen initially, such that À πðÞnÀ2 ε α0 ∠ 0 0 0 0 2n þ ), where i ¼ MA iAiþ1, i ¼ 1, . . . , n, Anþ1 A1. 0 ::: 0 IfÀÁ point M is inside of the polygon A2 An, then either α0 ; :::; α0 πðÞnÀ3 < πðÞnÀ2 ∠ 0 0 min 2 nÀ1 2ðÞnÀ1 2n . This leads to a contradiction, or MA nA2 πðÞnÀ3 < πðÞnÀ2 ∠ 0 0 < πðÞnÀ2 ε α0 < πðÞnÀ2 ε 2ðÞnÀ1 2n . Then, MA nA1 2n þ which means that n 2n þ .This leads also to a contradiction. 0 0 0 α0 < ε If point M is in triangle AnA1A2, then n . α0 πðÞnÀ2 ε This leads to a contradiction, as n 2n þ . This ends the proof.

! ! ::: ! ! ! ::: 7.1.72. We have that, A1A2 þ A3A4 þ þ A2nÀ1A2n ¼ÀA2A3 À A4A5 À À ! ! ! ! A2nA1 j¼ A2A3 þ A4A5 þ ::: þ A2nA1 and 1 1 sin ðÞ∠A1OA2 þ ::: þ ∠A2nÀ1OA2n ¼ sin 180 À ðÞ∠A2OA3 þ ::: þ ∠A2nOA1 ¼ 2 2 1 ¼ sin ðÞ∠A OA þ ::: þ ∠A OA : 2 2 3 2n 1

Then, one can assume that ∠A1OA2 þ ...þ ∠A2n À 1OA2n π. We shall prove the inequality by induction. At n ¼ 1 , we get an evident equality. Note that on the arc A A A exist such points A0 and A0 , that 2nÀ3 2nÀ2 2n 2nÀ3 2nÀ2 0 !0 ! ! A2nÀ3A 2nÀ2 ¼ A2nÀ3A2nÀ2 þ A2nÀ1A2n (see Figure 7.50).

Figure 7.50 A2n-2

A2n-1 a b b ab+ B

A2n-3 A2n 7.2 Solutions 369

Let A be a point on the arc A2n À 2A2n À 1A2n such that A2n À 2A ¼ A2n À 1A2n. Then, ∠ < ∠ 0 0 < A2n À 3A2n À 2B A2n À 3A2n À 2A. Therefore, A2nÀ3A2nÀ2 ¼ A2nÀ3B A2nÀ3A ; ∠ 0 0 < ∠ ∠ > thus A2nÀ3OA 2nÀ2 A2nÀ3OA2nÀ2 þ A2nÀ1OA2n. For n 1 , we deduce that

! ! ! ! ! A A þ ::: þ A A þ A A ¼ A A þ ::: þ A0 A0 1 2 2nÀ3 2nÀ2 2nÀ1 2n 1 2 2nÀ3 2nÀ2 ÀÁ 1 0 0 sin ∠A1OA2 þ ::: þ ∠A OA 2nÀ2 < 2 2nÀ3 1 < sin ðÞ∠A OA þ ::: þ ∠A OA þ ∠A OA : 2 1 2 2nÀ3 2nÀ2 2nÀ1 2n

Δ Δ Δ Δ AB AO BC CO 7.1.73. Note that AOB COD and AOD BOC; Thus CD ¼ OD and AD ¼ OD, BO Á OD ¼ OC Á OA. 2 We have to prove that AO OD CO OD OA OC OCÁOA OD or OD OD þ AO þ OD þ OC OC þ OA þ OD2 þ AOÁOC Á (OA þ OC)(OD2 þ AO Á OC) OA2 Á OD2 þ OD2 Á OC2 þ OA2 Á OC2 þ OD4. Indeed, we have that ÀÁ 2 2 ODÁOAþODÁOCþOD2þOAÁOC ðÞOD Á OA þ OD Á OC OD þ OA Á OC 2 ¼ 2 ODÁOAþODÁOCþOD2þOAÁOC ¼ 4 Á 4 OD2 Á OA2 þ OD2 Á OC2 þ OA2 Á OC2 þ OD4 4 Á ¼ 4 ¼ OD2 Á OA2 þ OD2 Á OC2 þ OA2 Á OC2 þ OD4:

7.1.74. Let O be the incenter of triangle ABC and lines AO,BOÀÁintersect line A0B0 at 00 00 00 points A and B respectively. Note that ∠AA B0 ¼ 180 À ∠A þ 90 þ ∠C ¼ ∠B; 00 2 00 2 2 consequently the quadrilateral OA0A B is inscribed, thus ∠AA B ¼ ∠OA0B ¼ 90. 00 Similarly, one can prove that ∠AB B ¼ 90. ∠ 00 00 ∠ ∠C ⊥ 00 00 00 00 Since A OB ¼ AOB ¼ 90 þ 2 and OL A B , then L is on segment A B . Thus, L is inside of the circle with a diameter AB, hence ∠ALB > 90. 7.1.75. Consider triangle ABC with the lengths of sides AB ¼ c, AC ¼ b, BC ¼ a, with a b c. Denote by α, β and γ the angles at vertices A, B and C, respectively. Let B1 be a point on side AC, such that ∠B1BC ¼ γ. From the assumptions of the problem, it follows that some two of points B, B1, C have the same colour. If B and C have the same colour, then a c. Consequently, a ¼ b ¼ c and γ ¼ 60 > 36. If B and B1 or C and B1 are of the same colour, then BB1 ¼ B1C c.IfB1 coincides with point A, then β ¼ γ and 2γ > 90, γ > 45. If points B1 and A are different, then in triangle BAB1, c ¼ AB BB1 holds true. Consequently α ¼ ∠BAB1 ∠BB1A ¼ 2γ. Then, it follows that 180 ¼ α þ β þ γ 2α þ γ 5γ. Thus, we obtain that γ 36. 370 7 Miscellaneous Inequalities

Figure 7.51 B А0

AB0 C

7.1.76. Let us introduce the following deﬁnition. A set of the positive numbers (a, b, c,Ra, Rb, Rc,da, db, dc) is called a describer,if there exists a triangle ABC and a point M inside of it, such that BC ¼ a, AC ¼ b, AB ¼ c, MA ¼ Ra, MB ¼ Rb, MC ¼ Rc,MA1 ¼ da, MA1 ¼ da, MB1 ¼ db, MC1 ¼ dc, where MA1 ⊥ BC, MB1 ⊥ AC, MC1 ⊥ AB (Figure 7.51). At ﬁrst, let us prove the following lemmas.

Lemma 1 If (a, b, c,Ra, Rb, Rc,da, db, dc)isaÁ describer, then prove that aRa bRb cRc dbdc dadc dadb ( , , , da, db, dc, , , is also a describer , where R is the 2R 2R 2R Ra Rb Rc circumradius of the triangle with sides a, b, c.

Proof of Lemma 1 Note that point M is inside of triangle A1B1C1 (Figure 7.51) and that one can describe a circle around the quadrilateral AC1MB1 . According to ∠ aRa the law of sines, we have that B1C1 ¼ Ra sin A ¼ 2R . We have that ∠MC1A2 ¼ ∠MAB1. MA2 db Hence, ΔMC1A2 ΔMAB1, we deduce that ¼ (MA2 ⊥ B1C1). Hence, dc Ra dbdc MA2 ¼ . This ends the proof of the lemma 1. Ra

Lemma 2 If (a, b, c,Ra, Rb, Rc,da, db, dc)isadescriber, then prove that (aRa, bRb, cRc, RbRc, RaRc, RaRb, daRa, dbRb, dcRc) is also a describer.

Proof of Lemma 2 Multiply the sides of triangles MAB, MBC and MAC by Rc, Ra and Rb, respectively and combine these three triangles into one (Figure 7.52). This ends the proof of lemma 2.

(a) Let ha, hb, hc be the altitudes drawn from vertices A, B, C of triangle ABC, respectively. We have that da þ db þ dc ¼ SMBC þ SMAC þ SMAB ¼ 1. On the other ha hb hc SABC SABC SABC hand, h R þ d , h R þ d , h R þ d . Hence, da þ db þ dc 1. a a a b b b c c c Raþda Rbþdb Rcþdc a2 a2 a2 2 Using the inequality 1 þ 2 þ 3 ðÞa1þa2þa3 , where b , b , b > 0, we obtain b1 b2 b3 b1þb2þb3 1 2 3

2 2 2 2 da db dc ðÞda þ db þ dc : 1 2 þ 2 þ 2 2 2 2 Rada þ da Rbdb þ db Rcdc þ dc Rada þ da þ Rbdb þ db þ Rcdc þ dc 7.2 Solutions 371

Figure 7.52 b1

bк b2

Hence, Rada þ Rbdb þ Rcdc 2(dadb þ dbdc þ dcda). Now, if we write the inequality (a) for the describer (aRa, bRb, cRc, d , d , d , dbdc, dadc, dadb ), we obtain 2R 2R 2R a b c Ra Rb Rc the inequality (b). So, the inequality (b) follows from (a) and lemma 1. Inequality (c) follows from (b) and lemma 2, inequality (d) follows from (c) and lemma 1, inequality (e) follows from (d) and lemma 1, inequality (f) follows from (e) and lemma . Thus, we have proven that a ) b ) c ) d ) e ) f. Note that, as (a) follows from (f) and lemma 1, we obtain that a , b , c , d , e , f. 7.1.77. Let PQRS be the section of the tetrahedron ABCD passing through the center O of the sphere with the radius r and parallel to lines AB and CD. Since the circle with the radius r is in the parallelogram PQRS, then if PQ ¼ x Á CD (PQ k CD, QR k AB,0< x < 1), we obtain that QR ¼ (1 À x)AB and PQ > 2r, RQ > 2r. > 2r > 2r > 2r 2r < ABÁCD Therefore, x CD,1À x AB. Thus, 1 CD þ AB. Hence, r 2ðÞABþCD . ÀÁ a; b; c 7.1.78. (a) Let DA ¼ a, DB ¼ b, DC ¼ c. Note that point O1 2 2 2 is equidistant from points D(0; 0; 0), A(a; 0; 0), B(0; b; 0), C(0; 0; c), consequently R ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ ÀÁ ÀÁ a 2 b 2 c 2. We have that V abc and V 1 S r 2 þ 2 þ 2pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ABCD ¼ 6 ABCD ¼ 3 n Á ¼ 1 ab bc ac 1 2 2 2 2 2 2 3 r 2 þ 2 þ 2 þ 2 a b þ b c þ a c Thus r ¼ pabcffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi abþbcþacþ a2b2þb2c2þa2c2 1 1 1 Let a ¼ x , b ¼ y , c ¼ z, then we have to prove that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 1 1 1 1 6 Á 1 Á 1 Á 1 þ þ þ þ þ qx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiy z x y z x2 y2 z2 1 1 1 1 1 1 1 1 1 1 1 1 x Á y þ y Á z þ x Á z þ x2 Á y2 þ y2 Á z2 þ x2 Á z2, 372 7 Miscellaneous Inequalities

or 1 1 1 1 1 1 2 x Á y þ y Á z þ x Á z 6 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , 1 1 1 1 1 1 x þ y þ z þ x2 þ y2 þ z2 x þ y þ z þ x2 þ y2 þ z2

or pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x þ y þ z xy þ yz þ xz þ x2y2 þ y2z2 þ z2x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : 3 x þ y þ z þ x2 þ y2 þ z2

Without loss of generality, one can assume that x þ y þ z ¼ 1 and min(x, y, z) ¼ z, then z 1. Let xy þ yz þ xz ¼ p, xyz ¼ q. 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 2 It remains to prove that 3 þ 3 1 À 2p p þ p À 2q. We have that (x y z)2 3(xy yz xz). Therefore, p 1. Thus we have to ÀÁffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ þ þ þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi3 ÀÁ 1 1p 2 2 2À8p 2 p 1 prove that 3 þ 3 1 À 2p À p p À 2q,or 9 þ 2q þ 3 1 À 2p 3 À p 0. 1 4 It is sufficient to prove that 9 þ q À 9 p 0. Indeed, we have that 1 4 1 4 1 4 4 þ q À p ¼ þ xyz À ðÞzxðÞþþ y xy ¼þ xy z À À zðÞ1 À z 9 9 9 9 9 9 9 1 x þ y 2 4 4 1 1 À z 2 9z À 4 4 þ z À À zðÞ¼1 À z þ Á À zðÞ¼1 À z 9 2 9 9 9 2 9 9 1 ÀÁzðÞ3z À 1 2 ¼ 9z3 À 6z2 þ z ¼ 0, 36 36 1 4 hence þ q À p 0. 9 9 a b c (b) Note that O(r, r, r) and O ; ; ; , consequently 1 2 2 2 ÀÁÀÁÀÁ 2 2 2 a 2 b 2 c 2 ðÞR þ r ðÞÀR À 3r OO1 ¼ R À 2Rr À 3r À r À 2 À r À 2 À r À 2 ¼ ¼ raðÞÀþ b þ c 2Rr À 6r2 ¼ raðÞþ b þ c À 2R À 6r 0

(see problem 7.1.78а).

7.1.79. Let A1 and A2 be the intersection points, respectively, of the straight lines AM and AО with circumcircle of triangle ΔABC. Here О is the center of the circumcircle of triangle ABC. Let us choose on the half-lines AB, AA1, AA2, AC 0 0 0 0 0 0 1 0 1 and AM, respectively, points B , A , A , C and M , such that AB ¼ , AA1 ¼ , 1 2 AB AA1 0 1 0 1 0 1 AA2 ¼ , AC ¼ , and AM ¼ . AA2 AC AM 7.2 Solutions 373

0 0 0 0 0 0 MAÁBC Note that points A1, A2 are on the straight line B C and B C ¼ ABÁACÁMA, M0C0 ¼ MCÁAB , B0M0 ¼ BMÁAC , and A0 M0 ¼ A1M , ∠AA 0A 0 ¼ ∠AA A ¼ 90 AMÁACÁAB ABÁAMÁAC 1 AMÁAA1 2 1 1 2 2 (see the proof of problem 1.2.9а). Thus, it follows that SM ¼ AM Á 2 2 AC Á AB Á SB0M0C0 , where SM is the area of the triangle with sides MA Á BC, MB Á AC, MC Á AB. 0 0 0 ⊥ 0 0 0 Δ 0 0 Δ Let H 2 B C and M H B C , then M H||AA2. Thus, A1M H AA1A2. 0 0 0 M H A1 M 0 A1M Therefore, ¼ . Hence, M H ¼ . Then, we deduce that S 0 0 0 ¼ AA1 AA2 2AMÁR B M C 1 0 0 0 BC A1M BCÁA1MÁMA 2 2 2 B C M H and S AM AC AB S 0 0 0 2 Á ¼ 2ÁABÁAC Á 2ÁAM ÁR ¼ 4RÁAB ÁACÁAM2 M ¼ Á Á Á B M C ¼ 2 2 SABC Á A1M Á MA ¼ SABC Á R À d . Let points A0, B0, C0 be the feet of the perpendiculars drawn from point M to 1 SABC 2 2 lines BC, CA, AB, respectively. Then, SA0B0C0 ¼ 4R2 Á SM ¼ 4R2 Á R À d (see the proof of lemma 1 of problem 7.1.76). If points A0, B0, C0 are on the same straight line, then one can prove that d ¼ R. Hence, it follows that MA Á MB Á MC 0 ¼ 2r|R2 À d2|. Let R0 be the circumradius of triangle A0B0C0, then R0 r (see Figure 7.53). A0B0ÁB0C0ÁA0C0 SABC 2 2 Thus, we have that ¼ SA B C ¼ 2 Á R À d . Hence, 4R0 0 0 0 4R

MA Á BC MB Á AC MC Á AB SABC 2 2 Á Á ¼ R À d Á 4R0: 2R 2R 2R 4R2

2 2 2 2 Therefore, it follows that MA Á MB Á MC ¼ 2R0|R À d | 2r|R À d |.

7.1.80. (a) According to problem 1.1.14а, we obtain for points O, A2, B1, B2, that r1 Á B1A2 þ r Á B1B2 r1 Á A2B2. Similarly, r1 Á B2A3 þ r Á B2B3 r1 Á A3B3,..., r1 Á BnA1 þ r Á BnB1 r1 Á A1B1. Summing up these inequalities we obtain that

rBðÞ1B2 þ B2B3 þ ::: þ BnÀ1Bn þ BnB1 r1ðÞA1B1 À B1A2 þr1ðÞA2B2 À B2A3 þ::: þ r1ðÞAnBn À A1Bn :

aij

Oi OOj

Ci aji Cj

Figure 7.53 374 7 Miscellaneous Inequalities

::: r1 ::: Therefore, B1B2 þ B2B3 þ þ BnÀ1Bn þ BnB1 r ðA1A2 þ A2A3 þ þ AnÀ1An þ AnA1Þ. ::: 2 2 (b) Note that A1B1 Á B1A2 ¼ A2B2 Á B2A3 ¼ ¼ AnBn Á BnA1 ¼ r1 À r and S1 ¼ 1 ∠ 1 ∠ , ::: SA1B1Bn ¼ 2 A1Bn Á A1B1 sin A1, S2 ¼ 2 A2B1 Á A2B2 sin A2 ¼ SA2B1B2 , 1 ∠ Sn ¼ SAnBnÀ1Bn ¼ 2 AnBnÀ1 Á AnBn sin An. Multiplying these inequalities, we deduce that ÀÁ 1 n S S Á ::: Á S ¼ r2 À r2 sin ∠A Á sin ∠A Á ::: Á sin ∠A Á 1 2 n 2n 1 1 2 n

For n ¼ 3orn ¼ 4, we need to prove that

2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi nr n S ::: Á sin ∠A Á ::: Á sin ∠A : ð7:37Þ A1A2 An 2 1 n

For n ¼ 3, we have that

A A Á A A Á A A S ¼ 1 2 2 3 3 1 ¼ 2r2 sin ∠A Á sin ∠A Á ∠: sin A A1A2A3 4r 1 2 3 3r2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 sin ∠A sin ∠A sin ∠A 2 1 2 3

(see problem 5.1.12). For n ¼ 4, we have that

1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi S A A Á A A ¼ 2r2 sin ∠A sin ∠A 2r2 sin ∠A sin ∠A ¼ A1A2A3A4 2 1 3 2 4 2 1 2 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4 ¼ 2r sin ∠A1 sin ∠A2 sin ∠A3∠: sin A4:

For n 5 the inequality (7.37) may be wrong. ::: We have that SB1B2:::Bn ¼ SA1A2:::An þ S1 þ S2 þ þ Sn. According to the arithmetic mean-geometric mean of the inequality, we deduce that ÀÁ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n nr1 À r n S ::: S ::: þ n S Á S Á ::: Á S ¼ S ::: þ Á sin∠A Á ::: Á sin∠A : B1B2 Bn A1A2 An 1 2 n A1A2 An 2 1 n

According to (7.37), we have that (for n ¼ 3orn ¼ 4) SB1B2:::Bn r2Àr2 r2 ::: 1 ::: 1 ::: SA1A2 An þ r2 SA1A2 An ¼ r2 Á SA1A2 An . r2 ::: 1 ::: Remark For n 5 we do not know, whether the inequality SB1B2 Bn r2 SA1A2 An holds true or not.

7.1.81. Note that at any i 6¼ j the circles Ci and Cj do not have common points. Indeed, if the circles Ci and Cj have a common point A, then let us draw through point A a line, intersecting the circle Ck, where k 6¼ i, k 6¼ j, but this is in a contradiction with the assumptions of the problem. 7.2 Solutions 375

Figure 7.54 Bm Ak Bk

Om Cr Ok

tm Am

Figure 7.55 Ai+1

Ai w

Consider the smallest convex polygon, containing the centers of the circles Ci (i ¼ 1, 2, ..., n). Denote by β1, β2,...,βk the external angles (Figure 7.54) of the polygon, and by Cij the union of two arcs of the circles Ci such that if point M 2 Cij, then the tangent to the circle Ci drawn from point M intersects (has a common point with) the circle Cj. Note that from the assumptions of the problem follows that Cij \ Cil ¼ ∅,ifj 6¼ l. α PDenoteÀÁ by 2 ij the radian measure of the arcs Cij, then α α π β β ::: β π 2 ij þ ji 2 n À ðÞ1 þ 2 þ þ k ¼2 ðÞn À 1 . Let OiOj ¼ dij. 1i 1, 5α. α α β < π α β α α β > Let ! 0, þ 2 2 and sin( þ 2 ) ¼ 4 sin ( , 0). α β αþ2β αþ2β 4 sin α β Then, þ 2 ! 0 and α ¼ sin ðÞαþ2β Á α ! 4, α ! 1, 5. (To construct ΔABC, ﬁrst construct ΔCAL).

7.1.83. Note that ΔOmAmCr ¼ ΔOkAkCr (see Figure 7.56). 2 Hence, AmCr ¼ CrAk ¼ xr. We have that tmÀÁ¼ PCr Á ðÞPCr À xr and 2 2 2 2 2 2 4 tk ¼ PCr Á ðÞPCr þ xr . Therefore, tmtk ¼ PCr Á PCr À xr PCr . Thus, 2 PCr tmtk. Multiplying all these inequalities, we obtain that PC1 Á PC2 Á ...Á PCn t1 Á t2 Á ...Á tn.

7.1.84. Let the convex polygon A1A2 ...An be inside of the circle ω. Take any point M inside of the polygon A1A2 ...An. Let ray MAi intersects the circle ω at point Bi (i ¼ 1, 2, ..., n). It is clear that the polygon A1A2 ...An is inside of the polygon

B1B2 ...Bn, thus SA1A2:::An SB1B2:::Bn and PA1A2:::An PB1B2:::Bn (see problem 2.1.1). Let point O be the center of the circle ω and ∠B2OB1 ¼ α1, ∠B3OB2 ¼ α2,..., ∠BnOBn À 1 ¼ αn À 1, ∠B1OBn ¼ αn. If point O is inside or is on one of the sides of 1 2 ::: α the polygon B1B2 ...Bn, then weÀÁ obtain that SB1B2 Bn ¼ 2 R sin 1þð α α α ::: α ::: 1 ::: n sin 2 þ þ sin nÞ, PB1B2 Bn ¼ 2R sin 2 þ þ sin 2 , where R is the radius

Figure 7.56 B1 C1 B2

a2 C2 a1 a3

a4 B3

B4 7.2 Solutions 377

ω of the circle . According to problem 5.4.15, it follows that SB1B2:::Bn 1 2 2π π ::: 2 R n Á sin n and PB1B2 Bn 2Rn sin n. If point O is outside of the polygon B1B2 ...Bn and max(α1, ..., αn À 1, αn) ¼ αn, then αn ¼ α1 þ ...þ αn À 1 < π. Thus, we deduce that

1 2 S ::: < R ðÞsin α þ sin α þ ::: þ sin α B1B2 Bn 2 1 2 n 1 α þ ::: þ α 1 2π R2n sin 1 n R2n sin , 2 n 2 n

α ::: α π π 1þ þ n < 2 Since n n 2 (n 4). < 1 2 α α < 2 < 1 2 2π For n ¼ 3, we have that SB1B2B3 2 R ðÞsin 1 þ sin 2 R 2 R Á 3 sin 3 . In the same way, we obtain that α1 αn α1 þ ::: þ αn π P ::: ¼ 2R sin þ ::: þ sin 2Rn sin < 2Rn sin : B1B2 Bn 2 2 2n n

1 2 2π π ::: ::: We have obtained that SA1A2 An 2 R n Á sin n and PA1A2 An 2Rn sin n. To end the proof it remains to note that the area and the perimeter of the regular n-gon, 1 2 2π π inscribed into the circle with the radius R, are equal to 2 R n Á sin n and 2Rn sin n respectively.

7.1.85. Let the convex polygon A1A2 ...An contains the circle ω. Let the half-plane Π ω i contains the circle , has a boundary li||AiAi þ 1 (An þ 1 A1), i ¼ 1, 2, . . . , n, line li is tangent to the circle ω, and Πi is in the half-plane with a boundary AiAi þ 1, containing the polygon (see Figure 7.57). If Π1 \ Π2 \ ...\ Πn ¼ B1B2 ...Bn, then obviously B1B2 ...Bn is circumscribed around the circle ω and B1B2 ...Bn is in the polygon A1 ...An. Thus,

SA1A2:::An SB1B2:::Bn and PA1A2:::An PB1B2:::Bn (see problem 2.1.1). We deduceÀÁ that π ::: α α ::: α α ; (see Figure 7.58) PB1B2 Bn ¼ rtgðÞ1 þ tg 2 þ þ tg 2n , where i 2 0 2 , α1 þ α2 þ ...þ α2n ¼ 2π and r is the radius of the circle ω. According to problem 2π ::: α α ::: α 5.4.16, it follows that PB1B2 Bn ¼ rtgðÞ1 þ tg 2 þ þ tg 2n 2nr Á tg 2n ¼ π 1 2 π ::: ::: 2nr Á tg n. Similarly, we obtain that SB1B2 Bn ¼ 2 PB1B2 Bn Á r nr tg n. 2 π π ::: ::: We have obtained that SA1A2 An nr tg n and PA1A2 An 2nrtgn. To end the proof it remains to note that the area and the perimeter of the regular n-gon, 2 π π circumscribed around the circle with the radius R, are equal to nr tg n and 2nrtgn respectively.

Figure 7.57 378 7 Miscellaneous Inequalities

Figure 7.58 A3 A2

Figure 7.59 y

B C

a x 0A

Figure 7.60 A2 An A1 DAk

w CAk+1

7.1.86. This proof is based on the fact of the existence, among all possible n-gon with a given perimeter, of a n-gon with a maximal area. First of all, note that the n-gon with a given perimeter that has a maximal area must be convex. Otherwise, one can construct a n-gon with the same perimeter, but with the greater area (see Figure 7.59). Now, let us do the second remark about the properties of the required n-gon. If (n 4) any four consecutive vertices lie on one circle, then the n-gon is an inscribed polygon. Indeed, it is sufﬁcient to change only vertices A2 and A3 and make use of the fact, that the articulated quadrilateral A1A2A3A4 (see Figure 7.60) has a maximal area, when the quadrilateral A1A2A3A4 is inscribed (see problem 5.5.22). 7.2 Solutions 379

We need to prove that all sides of the required n-gon are equal. Otherwise, without loss of generality, one can assume that A1A2 ¼ a þ x, A2A3 ¼ a À x, x 6¼ 0. Then, if A A 2b, then by Heron’s formula, we have that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 3 ¼ ÀÁqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 2 2 < 2 2 2 SA1A2A3 ¼ ðÞa þ b ðÞa À b b À x a À b Á b . This leads to a contradic- 0 0 tion, as S ::: < S 0 ::: , where A A ¼ A A ¼ a. This ends the proof. A1A2 An A1A2A3 An 1 2 2 3 7.1.87. Consider instead of the given n-gon M a centrally symmetric convex 2m-gon and denote it by M0, such that the projections of M and M0 on any straight line are the same. Such 2m-gon can be constructed (see the proof of problem 2.2.1b). Since H ¼ min l(M) ¼ min l(M0) and D ¼ max l(M) ¼ l(M0), and polygons M,M0 have equal perimeters, then it is sufficient to prove the problem for the polygon M0. The polygon M0 contains a circle with the diameter H and it is contained in the circle with the diameter D. Applying to it the problems 7.1.84b and 7.1.85b, we deduce that π π 2mtg Á H P 2m sin Á D: 2m 2m

Since 2m 2n, then m n and to complete the proof, we have to prove the π π π π sin 2m 2m tg 2m 2m following inequalities π π and π π . sin 2n 2n tg 2n 2n These inequalities hold true (see the solution of problem 7.1.71a). Moreover, if < β α < π 0 2, then tgα α β cos ðÞαÀβ α β 1 αÀβ α tgβ ¼ tg ðÞÀ cos α sin β þ 1 tg ðÞÁÀ sin β þ 1 β þ 1 ¼ β.

Remark In the proof of the problem, we have used the following inequalities. If < α < π α < α < α < < 0 2, then sin tg . It follows from the inequality SAOB SsectAOB SAOC (see Figure 7.61).

7.1.88. Let us choose in the strip pi a point Mi, equidistant from the boundaries of ~ ! the strip pi, where i ¼ 1,...,nÀÁand deﬁne bi ¼ Vi Á OMi. ~ ~ ~ 2 Consider the numbers a1V1 þ a2V2 þ ::: þ anVn À 2a1b1 À ::: À 2anbn, n where ai 2 {0, 1}, i ¼ 1, 2, . . . , n, in total 2 numbers. There are maximal numbers

Figure 7.61 QD R C

B T

SA 380 7 Miscellaneous Inequalities

among these numbers, let these maximums are reached at the values a1,...,an, then ÀÁ ~ ~ ~ 2 a1V1 þ a2V2 þ ::: þ anVn À 2a1b1 À ::: À 2anbn ÀÁ ~ ~ ~ ~ ~ 2 a1V1 þ ::: þ aiÀ1ViÀ1 þ ðÞ1 À ai Vi þ aiþ1Viþ1 þ ::: þ anVn À

À2a1b1 À ::: À 2aiÀ1biÀ1 À 21ðÞÀ ai bi À 2aiþ1biþ1 À ::: À 2anbn, where i ¼ 1, 2, . . .ÀÁ , 2n. ÀÁ ~ ::: ~ 2 ~ ::: ~ ::: ~ 2 Consequently a1V1 þ ÀÁÀÁþ anVn À a1V1 þ þ ðÞ1 À ai Vi þ þ anVn ~ ~ ::: ~ ~ 2biðÞ2ai ÀÀÁÀÁ1 . ViðÞ2ai À 1 2 a1V1 þ þ anVn À ðÞ2ai À 1 Vi 2biðÞ2ai À 1 . ~ ::: ~ ~ ~ 2 22ðÞai À 1 a1V1 þ þ anVn V i À bi Vi . ~ ! ~ ! ~ 1 ~ 2 > jjVi ÁWi ! ~ > Hence, OM Á Vi À OMi Á Vi 2 Vi 2 , this means that MMi Á Vi ~ jjVi 2 Á Wi. Therefore, point M is not inside of the strip pi, where i ¼ 1, 2, . . . n.

7.1.89. Lemma. Let n 2 and A1, A2,...,An are different points on the plane and ω(C, r) is the circle with the smallest radius containing points A1,...,An. Then, on the circle ω there are either two points Ai and Aj, such that AiAj ¼ 2r, or three points Ai, Aj, Ak, so that AiAjAk is a non-obtuse triangle. Indeed, the case CAi < r, i ¼ 1, 2, . . . , n leads to a contradiction since the circle ω0(C, r0), where r0 ¼ max CAi, contains all points Ai. However, r0 < r. 1in The case CA1 ¼ r, CAi < r, i ¼ 2, 3, . . . , n leads also to a contradiction, since the rþr1 circle ω0(C0, r0), (where r0 ¼ , r1 ¼ max CAi, and C0 is a point of segment CA1, 2 2in such that C0A1 ¼ r0), contains all points Ai. rþr1 < rÀr1 As r0 ¼ 2 r and C0Ai CC0 þ CAi 2 þ r1 ¼ r0, i ¼ 2, . . . n, the last statement holds true. The case on Figure 7.62, where CA1 ¼ r,...,CAk ¼ r and CAi < r, i ¼ k þ 1, . . . n, k 2, leads also to a contradiction, since the circle ω0(C0, r0) (where C0 is a point rÀr1 of segment CD, such that C0C < , where r1 ¼ max CAi, and ÀÁ2 kþ1in ; :::; ; rþr1 < r0 ¼ max C0A1 C0Ak 2 ), contains all points Ai and r0 r. Indeed, as ∠CDAi 90 , i ¼ 1, . . . , k, then ∠CC0Ai > ∠CDAi 90 , i ¼ 1,...,k. Hence, C0Ai < r, i ¼ 1, . . . , k and C0Ai CC0þ < rÀr1 rþr1 < CAi 2 þ r1 ¼ 2 r, i ¼ k þ 1, . . . n. This ends the proof of the lemma. Without lossqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi of generality, one can assume that C(0; 0) and r ¼ 1. Then, we have ::: 2 ::: 2 to prove that ðÞx1 þ x2 þ þ xn þ ðÞy1 þ þ yn n À 2.

If A1, A2 are such that A1A2 ¼ 2, then x1 þ x2 ¼ y1 þ y2 ¼ 0. Hence, 7.2 Solutions 381

Figure 7.62 N

QD R C M PE B

FA qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ::: 2 ::: 2 ::: 2 ::: 2 ðÞx1 þ þ xn þ ðÞy1 þ þ yn ¼ ðÞx3 þ þ xn þ ðÞy3 þ þ yn qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ::: 2 2 : ðÞn À 2 x3 þ y3 þ þ xn þ yn ðÞn À 2 ðÞn À 2 ¼ n À 2 2 2 2 2 2 2 Δ If points A1, A2, A3 are, such that x1 þ y1 ¼ x2 þ y2 ¼ x3 þ y3 ¼ 1 and A1A2A3 2 2 is non-obtuse, then (x1 þ x2 þ x3) þ (y1 þ y2 þ y3) 1, since according to 2 2 2 problem 5.5.4a, we obtain that (x1 À x2) þ (y1 À y2) þ ...þ (y2 À y3) 8 and 2 2 2 2 (x1 þ x2 þ x3) þ (y1 þ y2 þ y3) ¼ 9 À (x1 À x2) À ...À (y2 À y3) . Consequently, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ::: 2 ::: 2 ðÞx1 þ þ xn þ ðÞy1 þ þ yn rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ::: 2 2 2 ::: 2 : ðÞn À 2 ðÞx1 þ x2 þ x3 þ x4 þ þ xn þ ðÞy1 þ y2 þ y3 þ y4 þ þ yn n À 2

7.1.90. (a) We shall prove that for any pentagonalpffiffi section of the cube the ratio of 2 some of its two sides is greater than 1 þ 2 . We proceed the proof by contradiction argument. a Let AB ¼ a, AE ¼ b, BC ¼ b1, CD ¼ c1, PQ ¼ c, DE ¼ a1. We have that λ, pffiffi a1 b λ, where λ ¼ 1 þ 2. b1 2 Note that ΔBPC ΔETA and ΔDRE ΔASB (see Figure 7.63). Therefore, c ¼ b λ¸ c ¼ a λ. PC b1 DR a1 c c Hence, we deduce that PC ÀÁλ, DR λ. ÀÁ 1 1 Thus, it follows that CQ c 1 À λ , DQ c 1 À λ . Frompffiffiffi triangleÀÁCQD, according to the Pythagorean theorem, it follows that 1 c1 2c 1 À λ . a > c pffiffi 1 λ Thus, we obtain that c c 1 ¼ . 1 1 2ðÞ1Àλ 382 7 Miscellaneous Inequalities

Figure 7.63 N N¢ QDD¢ S1

CE¢ M¢ M F1 PE B B¢

Figure 7.64 C

g g a b 2 2 B A

Hence, we obtain that a > λ. This leads to a contradiction (see our assumption). c1 pffiffi To end the proof of the problem it remains to note that 1 þ 2 > 1, 7. pffiffi 2 The estimate 1 þ 2 is exact, since one can take a section, with the sides 2 pffiffi c c 2cðÞλÀ1 c “slightly” different from the numbers c, c, λ, λ, λ ¼ λ. (b) We proceed the proof by contradiction argument. Let a b (see the notations of the part [a]). Note that ΔBMN ¼ ΔAES (see Figure 7.64). pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiThus MK þ ES ¼ MK þ MN ¼ KN c, this means that a À c þ 2 2 > b À c c. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Hence, we obtain that 2 a2 À c2 > c, consequently pffiffiffi a 5 > ð7:38Þ c 2 pffiffiffi a Let λ ¼ 1, 83. Since λ,thena c1λ 2cðÞλ À 1 (see the proof of the part [a]). c1 pffiffiffi λ Similarly, we deduce that b 2cðÞÀ 1 . It is notp difficultffiffiffi to note that, if we take a section ABCDE, such that AB ¼ AE ¼ 2cðÞλ À 1 ,then, CD ¼ pffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 2c 2 À p c1. ?From the last inequality and (7.38), it follows that 2λ2À4λþ1 7.2 Solutions 383

pffiffiffiffiffi a a 10 > > λ, c pffiffiffi 1 2c 2 À pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 42À pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 2λ2À4λþ1 2λ2À4λþ1 as ! 1 1 1 4λ 2 À pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 7, 32 2 À pffiffiffiffiffiffiffiffiffiffiffiffiffiffi < 7, 32 2 À ¼ λ2 λ 0, 3778 0, 615 2 À 4 þ 1 pffiffiffiffiffi ¼ 7, 32ðÞ 2 À 1; 626 < 3 < 10:

Thus, we obtain that a > λ. This leads to a contradiction (see our assumption). c1 (c) Let a b (see the notations of the part [a]). At first, let us prove the following lemmas. Âpffiffiffi 4 2 λ LemmaÁ 1 Prove that the equation 2x À 6x À 2x þ 1 ¼ 0 has one root 0in 2, 1 , with λ0 2 (1, 84; 1, 85). 4 2 2 Indeed, it is sufficient to note that the functionÀÁpffiffiffif(x) ¼ 2x À 6x À 2x þ 1 ¼ 2(x À 2)2 þ 2(x À 0, 5)2 À 7, 5 is increasing in 2; 1 and that f(1, 84) ¼À 0, 06902528, f(1, 85) ¼ 0, 1920125.

Lemma 2 Prove that a ¼ b, if and only if a1 ¼ b1. Indeed, wep haveffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi that PB þ BFpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi¼ c ¼ RE þ ES (seepffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Figure 7.64).p Onffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the other b1 2 2 2 2 a1 2 2 2 2 hand, PB ¼ b b À c , BF ¼ a À c , RE ¼ a a À c , ES ¼ b À c Consequently pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b 1 À 1 a2 À c2 ¼ 1 À 1 b2 À c2 ð7:39Þ a b

a1 b1 If a ¼ b, then from (7.39) it follows that 1 À a ¼ 1 À b , this means that a1 ¼ b1. However, if a1 ¼ b1, but a 6¼ b, then a > b. On the other hand, from (7.39), it a1 < b1 > follows that 1 À a 1 À b . Hence, we deduce that b a. This leads to the contradiction and ends the proof of lemma 2.

Remark If a > b, then a1 > b1. Indeed, from (7.39), we obtain that 1 À a1 < 1 À b1. Thus, a1 > a > 1. Hence, a b b1 b a1 > b1. a Lemma 3 Prove that, if a ¼ b ¼ c , then ¼ λ0 (see the notations of lemma 1). 1 1 1 a1 Indeed, if a1 ¼ b1 ¼ c1, then according to lemma 2 we have that a ¼ b, and according to the problem 7.1.90, it follows that λ ¼ a > 1, 7. On the other hand, ÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a1 pffiffiffi ÀÁ 1 2 2 1 a λ þ 1 a À c ¼ c. From triangle CQD, we obtain that 2c 1 À λ ¼ c1 ¼ λ. 1 2 2 2 1þðÞ1þλ Consequently, 2 λ 1 a . Hence, we deduce that 2λ4 6λ2 ðÞÀ ¼ c2 ¼ 1 2 À ðÞ1þλ À 2λ þ 1 ¼ 0, and according to lemma 1, we have that λ ¼ λ0. 384 7 Miscellaneous Inequalities

a Lemma 4 Prove that, if a ¼ b, then λ0 (see the notations of lemma 1). minðÞb1;c1 a a Let us introduce the following notations, a1 ¼ b1 ¼ , c1 ¼ . λ λ1 If we prove that λ λ0 or λ1 λ0, then lemma 4 would be proven. Let

1 < λ < λ0, λ1 < λ0, ð7:40Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi p ÀÁ p λ then we have that a2 À c2 1 þ 1 ¼ c and a ¼ 2c ðÞÀ1 . λ λ1 λ 1 2 λ2 1ÀÁþðÞ1þλ Consequently, 2 1 ¼ 2 . Now from the last equality and (7.40), it follows 1À 1 λ2 2 2 1 1 1þ 1þλ λ2 > λ2 1ÀÁþðÞ1þλ > 0 that 2 0 2 1 ¼ 2 2 1À 1 λ2 1 1Àλ2 0 2 2 λ2 1 > 1 λ4 λ2 λ > Hence 2 0 1 À λ2 1 þ 1 þ λ , or equivalently, 2 0 À 6 0 À 2 0 þ 1 0. 0 0 This leads to a contradiction and ends the proof of the lemma 4. According to the remark of lemma 2, we have that min (a, b, a1, b1, c1) ¼ min (b1, c1). Let λ ¼ a . minðÞb1;c1 If a ¼ b, then according to lemma 4, it follows that λ λ0. If a > b, then ∠PMA ¼ ∠FAB > ∠SAE ¼ ∠ANP. Thus, PM < PN (see Figure 7.65). Let us draw from point C a line parallel to the straight line F1S1 and consider the 0 0 0 0 0 0 0 0 pentagonal section AB CD E , with the sides equal to a , b1, c1, a1, b , where 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > AB ¼ a , B C ¼ b1, CD ¼ c1, D E ¼ a1, E A ¼ b . Note that a ¼ b , b1 b1, 0 0 a a0 a0 0 c > c1, a > a , consequently, λ ¼ > > ¼ λ . 1 1 minðÞb1;c1 minðÞb1;c1 0 ; 0 minðÞb1 c1 0 According to lemma 4, we have that λ λ0. Consequently, λ > λ0. Thus, we have proven that in all cases λ λ0 > 1, 84 (see lemma 1). This ends the proof of the part [c]. From lemma 3 it follows that the number λ0 ¼ 1, 84267036 . . . is an exact estimate. This ends the proof. 7.1.91. Note that, if in triangle ABC, AC ¼ b, BC ¼ a, AB ¼ c and ∠C ¼ γ, then c γ ðÞa þ b sin 2 (see Figure 7.66).

Figure 7.65 B

C A

D D¢ 7.2 Solutions 385

ab ab cb e e f

k ak df cd c d

a b c

Figure 7.66

γ aþb Therefore, if 60 , then c 2 . For the pyramid SA1A2 ...An we have that

2A1A2 SA1 þ SA2, :::,2AnA1 SAn þ SA1:

Summing up all obtained inequalities, we obtain that

A1A2 þ A2A3 þ ::: þ AnA1 SA1 þ SA2 þ ::: þ SAn

This ends the proof. V V 7.1.92. Note that V1 ¼ A1B2 Á A1C2 , AA1B1C1 ¼ AA1 , ASB1C1 ¼ SB1 Á SC1, consequently VAA1B1C1 A1C1 A1B1 VASB1C1 SA V SB SC V1 ¼ A1B2 Á A1C2 Á AA1 Á SB1 Á SC1. Thus, it follows that V A1C1 A1B1 SA SB SC rffiffiffiffiffiffi AA1 A1B2 SB1 A1C2 SC1 SA1 A1B2 SB1 A1C2ÁSC1 3 V þ Á þ Á 1 À þ Á þ 1 SA A1C1 SB A1B1 SC ¼ SA A1C1 SB A1B1ÁSC : V 3 3

Similarly, one can prove the following inequalities rffiffiffiffiffiffi rffiffiffiffiffiffi SB1 B1A2 SA1 B1C2 SC1 SC1 C1B2 SB1 C1A2 SA1 3 V 1 À þ Á þ Á 3 V 1 À þ Á þ Á 2 SB B1C1 SA A1B1 SC and 3 SC A1C1 SB C1B1 SA : V 3 V 3 qffiffiffiffi qffiffiffiffi qffiffiffiffi Summing up the last three inequalities, we obtain that 3 V1þ 3 V2þ 3 V3 1, or ffiffiffiffiffiffi ffiffiffiffiffiffi ffiffiffiffiffiffi pffiffiffiffi V V V p3 p3 p3 3 equivalently, V1þ V2þ V3 V. This ends the proof. 7.1.93. (a) Let DD0||AC (see Figure 7.67). Then according to Ptolemy’s theorem, we obtain that AB Á CD0 þ BC Á AD0 ¼ AC Á BD0, AB Á CD þ BC Á AD ¼ AC Á BD. Note that CD ¼ AD0, CD0 ¼ AD. Hence, it follows that (AB þ BC)(CD þ AD) ¼ AC(BD þ BD0) 4R Á AC. 386 7 Miscellaneous Inequalities

Figure 7.67 a

db Mdc yz da yxz a

Figure 7.68 C

A1 B1

M E

AC1 B

(b) Let AB ¼ a, BC ¼ b, CD ¼ c, AD ¼ d, AC ¼ e, BD ¼ f (see Figure 7.68). We have that

R2ðÞa þ b ðÞb þ c ðÞc þ d ðÞ¼d þ a 2 ¼ ðÞðÞþac þ bd ad þ bc ðÞðÞþac þ bd cd þ ab R ¼ pffiffiffiffiffi ¼ ðÞef þ ek ðÞef þ fk R2 ¼ R2efðÞ f þ k ðÞe þ k 4R2efk ef e2f 2 Á fk ¼ ¼ ðÞac þ bd 2ðÞab þ cd , as 2R e,2R f. (c) If we prove that e þ f þ k > a þ b þ c þ d (see the notations of the proof of problem 7.1.93b), then e þ f þ 2R e þ f þ k > a þ b þ c þ d. Note that ÀÁ ðÞe þ f þ k 2 À ðÞa þ b þ c þ d 2 ¼ e2 þ f 2 þ k2 À a2 þ b2 þ c2 þ d2 ¼ ðÞab þ cd ðÞac þ bd ðÞab þ cd ðÞad þ bc ðÞad þ bc ðÞac þ bd ¼ þ þ À a2 À b2 À c2 À d2 ¼ ad þ bc ac þ bd ab þ cd abcdðÞ a þ b þ c À d ðÞa þ b À c þ d ðÞÀa À b þ c þ d ðÞa þ bþ c þ d ¼ > 0 ðÞad þ bc ðÞac þ bd ðÞab þ cd holds true. Therefore, we obtain that e þ f þ k > a þ b þ c þ d. 7.2 Solutions 387

1 k 2 (d) We have that S ¼ 2 efqÁ 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR, consequentlyðÞ 4RS ¼ ef Á fk Á ek ¼ ðÞac þ bd ðÞad þ bc ðÞab þ cd 8 ðÞabcd 3.

pffiffiffi 3 Hence, it follows that 2RS ðÞabcd 4. This ends the proof. ’ AH BD 7.1.94. According to Ceva s theorem, we have that HB Á DC ¼ 1. AH AB AH BC AB Consequently, HB Á AC ¼ 1, or AC Á HB Á BC ¼ 1. α 1 sin γ Hence, cos Á cos β Á sin α ¼ 1. Thus, tgα ¼ sin α þ cos α Á tgβ. If α 45, then β > 45. Consequently, tgβ > 1. Hence we deduce that tgα ¼ sin α þ cos αtgβ > sin α þ cos α > sin2α þ cos2α ¼ 1, thus α > 45.This leads to a contradiction, with the assumption that α 45. Therefore, it follows that α > 45. This ends the proof. 7.1.95. (a) At first, we need to prove that, if a, b, c, d 0, then 2ac þ 2bd min (a þ b, c þ d) min (a þ d, b þ c). Let min(a, b, c, d) ¼ d. Note that bd d2, a(cÀd)d(cÀb). Therefore, 2ac þ 2bd (c þ d)(a þ d) min (a þ b, c þ d) Á min (a þ d, b þ c). Thus, it follows that

∠AOB Á ∠COD þ ∠BOC Á ∠AOD 0, 5minðÞ∠AOB þ ∠BOC; ∠COD þ ∠AOD ÁÁ minðÞ∠AOB þ ∠AOD; ∠BOC þ ∠COD 0, 5 Á ∠AOC Á ∠BOD

(see problem 7.1.35a). (b) We have that kAB kÁkCD kþkBC kÁkAD k0, 5 min (kAB kþkBCk, kCD kþkADk) Á min (kAB kþkADk, kBC kþkCDk) (see the proof of the problem 7.1.95a). According to the condition (3) of the problem we have that kAB kþ k AD kkBDk and kBC kþkCD kkBDk, consequently min(kAB kþ k ADk, kBC kþkCDk) kBDk. Similarly, we obtain that min(kAB kþkBCk, kCD kþkADk) kACk. Hence, kAB kÁkCD kþkBC kÁkAD k0, 5 k AC kÁkBDk. (B) Let point O does not belong to plane α. For any points A and B on plane α , define kAB k¼∠AOB. Then, it is not difficult to verify that the conditions (1), (2), and (3) of the problem are satisfied (see problem 7.1.35a).

Let C0 > 0 be such that the inequality kAB kÁkCD kþ k BC kÁkAD kC0 k AC kÁkBDk holds true for any points A, B, C and D on 0 plane α. Let O be the projection of point O on plane α and AnBnCnDn be a square 0 0 with the center O and O An ¼ n. 388 7 Miscellaneous Inequalities

We have that kAnBn kÁkCnDn kþkBnCn kÁkAnDn kC0 k AnCn kÁ k BnDnk. Since lim k AnBn k¼ lim k BnCn k¼ lim k CnDn k¼ lim k AnDn k¼ n!1 n!1 n!1 n!1 π π π π π and lim k AnCn k¼ lim k BnDn k¼ π, then obtain that Á þ Á C0π Á π. 2 n!1 n!1 2 2 2 2 1 Therefore, it follows that C0 2. This leads to a contradiction. 7.1.96. (a) Note that OA1 þ OB1 þ OC1 ¼ SBOC þ SAOC þ SAOB ¼ 1. Consequently 2 ¼ AA1 BB1 CC1 SABC SABC SABC R R R 9R2 þ þ 2 ; hence OA1 þ OB1 þ OC1 1, 5R. RþOA1 RþOB1 RþOC1 3R þROAðÞ1þOB1þOC1 (b) We have that 1 1 1 1 9 . ¼ 1þ R þ 1þ R þ 1þ R 3þ R þ R þ R OA1 OB1 OC1 OA1 OB1 OC1 Therefore, we obtain that 1 þ 1 þ 1 6. OA1 OB1 OC1 R

db y da x dc z 7.1.97. Note that 2S ¼ x y z, 2S ¼ x y z, 2S ¼ x y z (see Figure 7.69). b þ þ a þ þ c þ þ Therefore, it follows that rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi 1 x 1 y 1 z d þ d þ d ¼ 2S Á þ Á þ Á a b c a x þ y þ z b x þ y þ z c x þ y þ z sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 1 1 1 x y z 2S þ þ þ þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia b c x þ y þ z x þ y þ z x þ y þ z 2S ¼ ðÞab þ bc þ ac ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiabc ab þ bc þ ac ¼ : 2R ffiffiffiffiffi ffiffiffiffiffi ffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p p p a2þb2þc2 Consequently, da þ db þ dc 2R . If α > 90, then b2 c2 < a2. Hence, ab bc ac ab c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ þÀÁffiffiffi þ þ ¼ ðÞþþ 2 2 p bc a 2 b2 þ c2 þ b þc < 2 þ 1 a2 < 1, 92a2. 2 q2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi Therefore, d þ d þ d < 1, 92a2 1, 92a. a b c 2R pffiffiffi If α 90, then 60 α 90. Consequently, a 3R and since a2þb2þc2 2α 2β 2γ 2R ¼ 2RðÞsin þ sin þ sin , then according to the inequality 5.1.1, we have that

Figure 7.69 C

O b M b M0 b AB 7.2 Solutions 389

qffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi pffiffiffi pffiffi 2α 2β 2γ 9 3 3 3 3 da þ db þ dc 2RðÞsin þsin þsin 2R ¼ 2 Á 3R 2 Áa. α π β γ π Remark For the triangle with angles ¼ 2 , ¼ ¼ 4 and aq pointffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiM on the altitude pffiffiffi ffiffiffiffiffi ffiffiffiffiffi ffiffiffiffiffi ÀÁpffiffiffi AM p p p 1 AH, where MH ¼ 2 2, we have that da þ db þ dc ¼ 2 þ 2 a. 7.1.98. Since AE ¼ 2, then AE is a diameter of the circle. Consequently, 4 ¼ AE2 ¼ AC2 þ CE2 ¼ a2 þ b2 þ 2ab cos α þ c2 þ d2 þ 2cd sin α, where ∠AEC ¼ α. Hence, 4 ¼ a2 þ b2 þ c2 þ d2 þ ab Á CE þ cd Á AC. Since ∠ABC, ∠CDE > 90, then CE > CD ¼ c, AC > BC ¼ b. Therefore, 4 > a2 þ b2 þ c2 þ d2 þ abc þ bcd. 7.1.99. (a) The proof is shown by mathematical induction. Sinceffiffiffi 2 2 2 2 p ðÞ~a1 þ~a2 þ ðÞ~a1 À~a2 ¼ 2~a1 þ 2~a2 4, then minðÞjj~a1 þ~a2 ; jj~a1 À~a2 2. Hence, the statement holds true for n ¼ 2. For n 3, we have that, if ~a1,~a2,~a3 are non-vanishing vectors, then two of the vectors Æ~a1, Æ~a2, Æ~a3 form an angle not exceeding 60 . Thus, the difference of those two vectors has a length not exceeding 1. If even one of ~a1,~a2,~a3 is a non-vanishing vector, then the last statement is also correct. Consequently, by replacing these two vectors by their difference we obtain n À 1 vector, satisfying the conditions of the problem. This ends the proof. (b) At first, we need to prove that, if there are 15 non-vanishing vectors in the space, then among them one can find two vectors, such that the angle between them is less than 60. Move the starting points of all 15 vectors to the same point and consider a unit sphere with the center at that point. Consider 15 cones, such that their altitudes contain the given vectors and the bases of the cones lie on the surface of the sphere and form “caps.” Let the angles of the axial sections of these cones be equal to 60, then the area of one “cap” is equal pffiffi ÀÁpffiffiffi ÀÁpffiffiffi to 2πRH ¼ 2π 1 À 3 ¼ π 2 À 3 . We need to prove that 15π 2 À 3 > 4π pffiffiffi 2 or 26 > 15 3, 676 > 675. It follows that, among these “caps” there are two overlapping, then the angle between corresponding vector is less than 60. ~ ~ ~ ::: ~ ::: Let n 7. Consider all possible vectorsÀÁc1 ¼ a1 þ a2 þ þ an, , 2 2 n 2 2 ~c2n ¼À~a1 À ::: À~an, then ~c1 þ ::: þ~c 2n ¼ 2 ~a1 þ ::: þ~an . Consequently, n 2 2 2 there exists a number i, such that 2 ~c ~c ::: ~c n ffiffiffi ffiffiffi jji jj1 þ þ jj2 ¼ n 2 2 n p p 2 jj~a1 þ ::: þ jj~an n2 . Hence, jj~ci n 7. We proceed the proof of the problem by mathematical induction. At n 7, this ends the proof of the statement. If among the vectors~a1, :::,~an is a zero-vector, then we obtain that the statement needs to be proven for n À 1 vectors. This leads to a contradiction. If there is no zero-vector among~a1, :::,~an vectors, then since~a1, :::,~a8 vectors are non-vanishing, then among 16 vectors ~a1, :::,~a8, À~a1, :::, À~a8 there are two vectors, forming angle not exceeding 60. Therefore, the length of the difference 390 7 Miscellaneous Inequalities of these vectors does not exceed 1. Consequently, replacing these two vectors by their difference we obtain n À 1 vectors, satisfying the conditions of the problem, such that the statement holds true. ~ ; ::: ~ ; (c) Let us obtain the vectors a1ðÞx1 y1 , , anðÞxn yn .

Let M1 ¼ {(x, y)| x 0, y 0}, M2 ¼ {(x, y)| x 0, y 0}, M3 ¼ {(x, y)| x 0, y 0}, M4 ¼ {(x, y)| x 0, y 0}. As (xi, yi) 2 M1 [ M2 [ M3 [ Mk, then there are numbers i ,...,i , such that x ; y M j 1,...,k and ~a ::: ~a 1. 1 k ij ij 2 p ¼ jji1 þ þ jjik 4 Consequently, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁÀÁ 2 2 1 jj~ai þ ::: þ~ai ¼ ðÞxi þ ::: þ xi þ y þ ::: þ y pffiffiffi jjxi þ ::: þ xi þ y þ ::: þ y ¼ 1 k 1 k i1 ik 2 1 k i1 ik ÀÁ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 2 2 2 2 ¼ pffiffiffi jjþxi y þ ::: þ jjþxi y pffiffiffi xi þ y þ ::: þ xi þ y ¼ 2 1 i1 k ik 2 1 i1 k ik pffiffiffi 1 2 ¼ pffiffiffiðÞjjþ~ai ::: þ jj~ai : 2 1 k 8 pffiffi Remark If the vectors ~a , :::,~a are given in the space, then the number 2 must be 1pffiffi n 8 3 replaced by the number 24 . 7.1.100. (1) Let ∠A1C1B1 ¼ 90 . Note that points C and C1 lie on the circle with the AB β β ∠ diameter A1B1. Consequently, A1B1 CC1 hc ¼ 2 sin 2 , where ¼ ABC. (2) Let ∠A1B1C1 ¼ 90 and a circle with the diameter A1B1 intersects segment A1C1 at point E (see Figure 7.70). Take a point M on the arc B1E, such that ∠A1MC1 ¼ 180 À β. This is possible since ∠A1B1C1 ¼ 90 < 180 À β < 180 ¼ ∠A1EC1.As∠A1MC1 þ ∠A1BC1 ¼ 180 , then points A1, B, C1, M lie on the same circle with a radius R. By the law of sines, we have that

A1C1 ¼ 2R sin β BM sin β ð7:41Þ Note that ∠B1MC1 ¼ 360 À ∠B1MA1 À ∠A1MC1 ¼ 90 þ β ¼ 180 À ∠BAC. Hence, points A, B1, M, C1 lie on the same circle.

Figure 7.70 A

1200

S CЈ OЈ BЈ 7.2 Solutions 391

Figure 7.71 T

TЈ

Since ∠AMC ¼ ∠AMB1 þ ∠CMB1 ¼ ∠AC1B1 þ ∠CA1B1 ¼ 90 þ β, then M is on the arc with the endpoints A and C (Figure 7.71). Thus, BM þ MO BO ¼ BM0 þ M0O. Consequently, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi AB2 AB BM BM ¼ BO À OM ¼ AB2 þ tg 2β À tgβ ð7:42Þ 0 0 4 2

Accordingpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi to (7.2) and (7.3), we deduce that A1C1 AB 2β β β 2 4 þ tg À tg sin . If ∠B A C 90, similarly we obtain that B C AB pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 1 ¼ 1 1 2 4 þ ctg 2β À ctgβ cos β. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Let β 45, we need to prove that 4 þ tg 2β À tgβ sin β pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 þ ctg 2β À ctgβ cos β sin 2β. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi The last inequality holds true, as 4 þ ctg 2β ctgβ þ 2 sin β or 4cos2 β β 4 cos . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi We have to prove that 4 þ tg 2β À tgβ tgβ 4 þ ctg 2β À ctgβ,or pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 31ðÞÀtg 2β 4 þ ctg 2β þ ctgβ tgβ 4 þ tg 2β þ tgβ,1À tgβ pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4þtg 2βþ 4tg 2βþ1 Therefore, we needp toffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi prove that 4 þ tg 2β þpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4tg 2β þ 1 3 þ 3tgβ. Note that, this holds true, as 4 þ tg 2β < 2 þ tgβ and 4tg 2β þ 1 < 2tgβ þ 1. ⊥ ⊥ ⊥ One can easily prove that,p if ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiM0A1 BC, M0C1 AB, M0B1 AC, then ∠ AB 2β β β A1B1C1 ¼ 90 and A1C1 ¼ 2 4 þ tg À tg sin . Hence, the smallest possiblepffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi value for the hypotenuse of triangle A1B1C1 is equal to AB 2β β β 2 4 þ tg À tg sin . 7.1.101. We need to prove that, if ∠SAC < 60, ∠SAB < 60, then ∠SAO < 60. Consider a cone with the altitude AS (see Figure 7.72). Let the angle of the axial section of the cone be equal to 120 and lines AB, AC, AO intersect the plane of the 392 7 Miscellaneous Inequalities

i i

3 k 3

2 α2 2 k

α1 αk 1 1 O

Figure 7.72 base of the cone at points B0, C0, O0, respectively. Then, points B0 and C0 belong to the base of the cone, and consequently O0 also belongs to the base of the cone, then ∠SAO ¼ ∠SAO0 < 60. We proceed the proof by contradiction argument. Assume that, either ∠SAC or ∠SAB 60. Let ∠SAC 60, consequently ∠SCA < 60. Hence, ∠SCB 60 and ∠SBC < 60, ∠SBA 60 and ∠SAB < 60. Then, we obtain that SC > SA, SB > SC and SA > SB. Therefore, SC > SA > SB > SC. This leads to a contradiction. 7.1.102. Let O be the center of the circle with the radius t and points F,E be the midpoints of sides AB,AC, respectively. Without loss of generality, one can assume = > AC AB > BC > p that O2 EF. Consequently, OE þ OF EF or t À 2 þ t À 2 2 . Hence, t 2. Let m, n, k, x > 0. Note that the only solution of the equation pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mnkðÞ m þ n þ k ¼ mnxðÞ m þ n þ x þ mkxðÞ m þ k þ x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ nkxðÞ n þ k þ x ð7:43Þ

mnk is the number x0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi. mn þ nk þ mk þ 2 mnkðÞ m þ n þ k It is clear that the equation (7.4) has not more than one solution and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁpffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mn kðÞþm þ n mnðÞ m þ k þ n k mnx0ðÞm þ n þ x0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ mn þ nk þ mk þ 2 mnkðÞ m þ n þ k ÀÁpffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi mn kðÞþm þ n mn mnðÞ m þ k þ n k ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mn þ nk þ mk þ 2 mnkðÞ m þ k þ n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mnkðÞþ m þ n mn mnkðÞ m þ k þ n ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , mn þ nk þ mk þ 2 mnkðÞ m þ k þ n thus 7.2 Solutions 393 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mnxðÞ m þ n þ x þ mkxðÞ m þ k þ x þ nkxðÞ n þ k þ x ¼ 0 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 mnkðÞ2m þ 2n þ 2k þðÞmn þ nk þ mk mnkðÞ m þ k þ n ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mn þ nk þ mk þ 2 mnkðÞ m þ k þ n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ mnkðÞ m þ n þ k :

Let D be the midpoint of side BC. Denote by DE ¼ m þ n, FD ¼ m þ k, FE ¼ n þ k, then OD ¼ t À (n þ k) ¼ x þ m, where x ¼ t À (m þ n þ k) > 0, OE ¼ x þ n, OF ¼ x þ k. One can easily prove that point O is inside of triangle DEF(see the proof of problem 1.1.4a). Therefore, SDEF ¼ SDOE þ SDOF þ SEOF; hence x ¼ mnkpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi. If we take into account that mn þ nk þ mk mnþnkþmkþ2 mnkðÞ mþnþk qffiffiffiffiffiffiffiffiffiffiffi pffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁpffiffiffi 3 mnkðÞ m þ n þ k , then it follows that x 2 À 3 mnk ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mþnþk ÀÁpffiffiffi p ÀÁpffiffiffi pffiffi mnkðÞ mþnþk r p 3 2 À 3 mþnþk ¼ 2 À 3 Á 2. Hence, t À 2 1 À 2 r.

7.1.103. (a) Let T be an equilateral triangle containing all points A1,...,An, such that each side of the triangle contains at least one of these points. Denote by T0 another equilateral triangle with the same properties as of T (Figure 7.73), sides of T and T0 are parallel. Note that, the sum of the distances between the parallel sides of these triangles (see the proof of problem 7.1.76a) is equal to the sum of the lengths of the altitudes of these triangles. On the other hand, it does not exceed 3D. Hence, it follows that the altitude of at least one of the triangles does not exceed 3D, and the side of that pffiffiffi 2 triangle does not exceed 3D. (b) Here, we use that the length of the side of the equilateral triangle T is a continuous function of its direction. Consequently, the difference of the sides of triangles T and T0 by the rotation of the side of triangle T by angle π changes the sign. Hence, there exists a direction of the side of triangle T, such that triangles T and T0 have equal sides.

Figure 7.73 ii+1 EB D α i αi

O M

A C 394 7 Miscellaneous Inequalities

Intersection of such triangles is a centrally-symmetric “hexagon,” the distances between its parallel sides do not exceed D. Consequently, that hexagon can be 2π placed inside of the centrally-symmetric hexagon with angles 3 , with the distances between the parallel sides equal to D. Then, it is a regular hexagon and the length of its side is equal to pDffiffi. 3 (c) This result follows from (b). We provide also another proof. Let n 3. Consider the circles ω A ; pDffiffi , i ¼ 1, 2, . . . , n. Note that any three of these circles have a i i 3 common point.

Indeed, if points Ai, Aj, Ak lie on the same straight line, such that point Aj belongs to segment AiAk, then the midpoint of that segment belongs to the circles with the centers A , A , A ,asD < pDﬃﬃ. i j k 2 3 If points Ai, Aj, Ak are the vertices of some acute triangle, then the center O of the circumcircle of that triangle belongs to the circles with the centers Ai, Aj, Ak, as one ∠ π can assume that AiAjAk 3. Hence, we obtain that

AiAk D OAi ¼ OAj ¼ OAk ¼ π 2 sin ∠A A A 2 sin ¼ pDﬃﬃ : i j k 3 3

∠ π If AiAjAk 2, then the required point is the midpoint of segment AiAk. Since any three of the circles ω1, ω2,...,ωn have a common point, then according to Helly’s theorem, all these circles have a common point. Let O be that point, then the circle ω O; pDffiffi contains all these points A , A ,...,A . 3 1 2 n (d) According to problem (c), there exists a circle ω O; pDffiffi containing all points ÀÁ 3 ω ; d A1, A2,...,An. Then the circles i Ai 2 i ¼ 1, 2, . . . n do not have common internal points and are contained in the circle ω0 O; pDffiffi þ d . Consequently, the 3 2 0 sum of the areas of the circles ω1,...,ωn is less than the area of the circle ω . pffiffi 2 2 pffiffiffi Therefore n Á π Á d < π pDffiffi þ d , hence D > 3 ðÞn À 1 d. 4 3 2 2 (e) According to the problem (c), there exists a circle ω O; pDffiffi containing all i 3 points A1, A2,...,A6. If point O coincides with point Ai, then for j 6¼ i, we have that d A A ¼ A O pDffiffi. j i j 3

If O does not coincide with any of points A1, A2,...,A6, thenÀÁ there exist points π A and A , such that ∠A OA . Consequently, d A A max A O; A O pDffiffi. i j i j 3 i j i j 3 2 7.1.104. (a) Let G be the centroid of triangle ABC. We have that AG, BG, GG 3, since each median divides the triangle into two triangles of the equal area. Note that 7.2 Solutions 395 maxðÞ∠AGB; ∠BGC; ∠AGC 2π. Let ∠AGB 2π, then S ¼ 3S ¼ 3 pffiffi pffiffi 3 ABC AGB 3 Á 1 AG Á BG sin ∠AGB 3 Á 2 Á 2 Á 3 ¼ 3. For the equilateral triangle with the 2 2 3 3 2 3 pffiffi 3 median equal to 1, we have that the area is equal to 3 . (b) Let O be the intersection point of the diagonals of the convex quadrilateral ABCD and BO OD, CO AO. BO AO If AB k DC, then OD ¼ CO, consequently BO ¼ OD and AO ¼ OC. Therefore, the quadrilateral ABCD is a parallelogram. 1 1 < 2 Then, AC 1, BD 1, consequently SABCD 2 AC Á BD 2 3. If AB=k DC, then let M be the intersection point of rays BA and CD, with BC ¼ b, ∠ABC ¼ β and ∠DCB ¼ γ. < 1 b2 We have that S ¼ SABCD SAMC ¼ 2 Á ctgβþctgα, consequently

b2 SðÞ ctgα þ ctgβ < : ð7:44Þ 2

S Let A1 and D1 be points on sides AB and CD, respectively, such that SBA1C ¼ 2 and S S. Then, we have that CA 1 and BD 1, hence BD1C ¼ 2 ÀÁ1 1 2 β 2 2 CA2 ¼ b2 þ S À 2S cos ¼ b À Sctgβ þ S 1. Similarly, we obtain that 1 ÀÁb2sin 2β sin β b b2 S2 b Sctgγ 2 1. Consequently, according to (7.5), it follows that b2 þ À b 2S2 1 S 2 2S2 1 b 2 2S2 9b2 2 þ 2b À ðÞctgβ þ ctgγ > þ 2b À ¼ þ 3S: b2 2 b b2 2 2 b2 8

< 2 Hence, we deduce that S 3. ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 For the quadrilateral ABCD, where AB ¼ BC ¼ CD ¼ p pffiffi and pffiffiffi 38À14 7 ∠B ¼ ∠C ¼ β, with cos β ¼ 3 À 7, one canffiffiffiffiffiffiffiffiffiffiffiffiffiffi easily verify that the assumptions p pffiffi 20þ14 7 ::: of the problem are satisfied, and that S ¼ 12 ¼ 0, 6293764 . (c) Let us call the median the segment connecting the vertex of n-gon with the point on its boundary, dividing its area S into two equal parts. Let us draw the medians of the polygon and note that any two of them either coincide or intersect. Otherwise, they divide n-gon into three parts, such that two of S them have the area 2. Let n-gon have k medians (3 k n). Let us draw through the center O of the circle ω diameters parallel to these medians (Figure 7.74) and enumerate these diameters (in the clockwise direction) by the numbers 1, 2, . . . k. Two medians are called adjacent, if they are parallel to the diameters with numbers i and i þ 1(1 i k À 1) or k and 1. 396 7 Miscellaneous Inequalities

Figure 7.74 B AMCЈ C AЈ 0 0 A0 M0 C0 O ω

Figure 7.75

! ! Note that, if the medians AB and CD are adjacent, then points B and D (A and C) belong to the same side of the n-gon. Otherwise, there is one more diameter between the diameters with numbers i and i þ 1 (Figure 7.75). ! ! Therefore the adjacent medians AB and CD form a “butterfly.” This means a figure composed of triangles AMC and BMD, where M is the intersection point of the medians. Note that SAMC ¼ SBMD, then the area of the “butterfly” is equal to SAMC þ SBMD. Since (AM þ MC) þ (BM þ MD) ¼ AB þ CD 2, then AM þ MC 1or BM þ MD 1. Let AM þ MC 1, then SAMC þ SBMD ¼ 2SAMC ¼ AM Á MC Á sin αi 2 α ðÞAMþMC α sin i 4 sin i 4 . One can prove by mathematical induction, that the “butterflies” formed by the medians 1 and 2, 2 and 3, . . . , i À 1 and i cover the shaded part of the n-gon (Figure 7.73). Consequently, all “butterflies” cover all of the n-gon. Hence, it α α ::: α π 1 α α ::: α 1 1þ 2þ þ k k follows that S 4 ðÞsin 1 þ sin 2 þ þ sin k 4 sin k ¼ 4 sin k n π n π 4 sin n. This means that S 4 sin n (see problems 5.4.15 and 8.2.1e). Note that n π S ¼ 4 sin n,ifk ¼ n, all medians are concurrent and are divided by the intersection point into two equal parts. This leads to a contradiction. Hence, we obtain that < n π S 4 sin n. Remark If each chord dividing the area of a planar convex figure into two equal π parts has a length not exceeding 1, then the area of that figure does not exceed 4. 7.2 Solutions 397

Figure 7.76

7.1.105. Let in triangle ABC we have that AB BC AC and point M is the > 2SABC 4SABM ∠ midpoint of segment AC. Since BM AB ¼ AB ¼ 2BM sin ABM, then ∠ < 1 ∠ < ∠ > sin ABM 2. Consequently, ABM 30 or ABM 150 . If ∠ABM < 30, then consider a parallelogram ABCD. In triangle BCD we have that BC AB ¼ CD, consequently ∠CBM ∠CDB ¼ ∠ABM. Hence, ∠ABC 2 ∠ABM < 60. This leads to a contradiction. We have obtained that ∠ABM > 150. Consider the circumcircle of triangle ABM. Take on that circle points A0 and M0, such that ∠A0BM0 ¼ 150 and AM|| A0M0. Let A0M0 ¼ M0C0 (see Figure 7.76), since AM||A0M0 and A0M0 ¼ M0C0, then 0 > 0 we have that A0M MC 0. > 0 > 0 ∠ < ∠ 0 Therefore CM ¼ AM A0M MC 0, consequently ACB AC 0B ¼ ∠A0C0B. Let B0 lie on the arc A0BM0 and C0B0 be a tangent to the circle ω.Then ∠A0C0B ∠A0C0B0, hence ∠ACB < ∠A0C0B0. If we prove that ∠A0C0B0 < 0 5 30 , then this ends theffiffiffi proof of the problem.ÀÁqffiffi p Note that B C ¼ 2A M and cos 30 þ ∠A C B ¼ 2. Let us prove that, 0 0 0 q0 ffiffi 0 0 0 3 if 0 < α < 90 and cos α ¼ 2, then α < 35300. 3 ÀÁ α 1 а α 7 We have that cos 2 ¼ 3, cos 180 À 4 ¼ 9. Now, if we prove that > 7 α < α < cos 38 9, then as 45 and cos(180 À 4 ) cos 38 , it follows that 180 À 4α > 38. This means that α < 35300. φ 7 0 < φ < φ 329 < 324 4 Let cos ¼ 9 and 0 ÀÁ90 , then cos 3 ¼À729 À729 ¼À9. φ < φ > 4 4 > Since 60 and cos 180 À 3 9, then if we prove that 9 sin 24 ¼ cos 66 , we deduce that 180 À 3φ < 66,orφ > 38. Hence, 7 φ < 9 ¼ cos cos 38 . < 4 β 4 < β < It remains to prove that sin 24 9. Let sin ¼ 9 and 0ffiffiffiffiffiffiffiffiffiffiffiffiffi 90 . Since p pffiffi β < < β 10þ2 5 < 716 30 , then it is sufficient to prove that sin72ffiffiffi sin 3 ÀÁ,or 4 ffiffiffi 729. 2 p p Note that 162ffiffiffiffiffiffiffiffiffiffiffiffiffi< 415 Á 83, hence 162 5 < 415, or 81 10 þ 2 5 < 1225. p pffiffi 10þ2 5 < 35 < 716 Therefore, 4 36 729. This ends the proof. 7.1.106. Without loss of generality, one can assume that AH ¼ 1 and the equation of line l has the form x cos φ þ y sin φ À ρ ¼ 0. 398 7 Miscellaneous Inequalities

Note that, if B(Àctgβ, 0), C(ctgγ, 0), A(0; 1), then u ¼jsin φ À ρj, v ¼jÀcos φctgβ À ρj, w ¼jcosφctgγ À ρj, а 2S ¼ ctgβ þ ctgα. We have to prove that (sin φ À ρ)2tgα þ (cos φ Á ctgβ þ ρ)2tgβ þ (cos φ Á ctgγ À ρ)2tgγ ctgα þ ctgβ,or

ðÞtgβ þ tgγ À tg ðÞβ þ γ ρ2 þ 2 sin φtgðÞβ þ γ ρ þ cos 2φðÞÀctgβ þ ctgγ tgβ þ tgγ Àsin 2φtgðÞβ þ γ : tgβtgγ

Since tgβ, tgγ > 0 and tgβ Á tgγ > 1, then we have to prove that tg2βtg2γ Á ρ2 À 2ρ sin φtgβtgγ þ sin2φ 0, or (ρtgβtgγ À sin φ)2 0, which is evident. Note that the equality holds true, if and only if the straight line l passes through the orthocenter of triangle ABC. 7.1.107. (a) Let max(∠A, ∠B, ∠C) ¼ ∠A ¼ α. If α 120, then according to problem 1.2.8, we have that MA þ MB þ MC b þ c. We need to prove that b þ c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3abc or a2b2þb2c2þa2c2 ÀÁ a2 ðÞb þ c 2 b2 þ c2 À 9b2c2 þ ðÞb þ c 2b2c2 0 ð7:45Þ

Since 0 a2 (b þ c)2, then it is sufficient to prove the inequality (7.6) for a2 ¼ 0 and a2 ¼ (b þ c)2. If a2 ¼ 0, then the proof is evident in (7.6), and if a2 ¼ (b þ c)2, then the inequality (7.6) is obtained from the inequalities b2 þ c2 2bc and (b þ c)2 4bc. If α < 120, then there exists a point T inside of triangle ABC, such that ∠ATB ¼ ∠BTC ¼ ∠ATC ¼ 120 (seeqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the proof of problem 1.2.8), and MA þ MB þ MC TA þ TB þ TC ¼ b2 þ c2 À 2bc cosðÞ 60 þ α . We need to prove that TA þ TB þ TC pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3abc or a2b2þb2c2þc2a2

1 1 1 9 þ þ À 0: ð7:46Þ b2 c2 b2 þ c2 À 2bc cos α b2 þ c2 À 2bc cosðÞ 60 þ α

x 1 9 Consider the function fxðÞ¼ þ À on [2q; 1), where q2 xÀ2q cos α xÀ2q cosðÞ 60 þα 0 ðÞxÀ2q cos αÀq ðÞxÀ2q cos αþq 9 q ¼ bc. Since f ðÞ¼x 2 þ 2 and x 2q q þ q2ðÞxÀ2q cos α ðÞxÀ2q cosðÞ 60 þα 2q cos α (60 α < 120), then for x 2q, we have that f0(x) > 0. Hence, it follows that f(x) f(2q). 1 9 We need to prove that f(2q) 0or4þ . 1À cos α 1À cosðÞ 60 þα The last inequality can be reduced to the inequality. (sin(30 þ α) À 1)(4 sin (30 þ α) À 1) 0. Note that, this inequality holds true, as sin(30 þ α) 1 and 30 þ α 2 [90; 150). Hence, 4 sin (30 þ α) > 2. Since b2 þ c2 2bc ¼ 2q, then f(b2 þ c2) f(2q) 0. Hence, f(b2 þ c2) 0. Thus, it follows the inequality (7.10). 7.2 Solutions 399

(b) According to problem 7.1.107a, we have that MA þ MB þ MC pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3abc . a2b2þb2c2þc2a2 We need to prove that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3abc 6r,or a2b2þb2c2þc2a2

1 1 1 1 þ þ , a2 b2 c2 4r2 h2 h2 h2 1 a þ b þ c : 2 2 2 2 2 2 2 a ha b hb c hc 4r S2 h2 þ h2 þ h2 ¼ p2: a b c r2

The last inequality holds true, as 2 2 2 2 2 2 2 ha þ hb þ hc la þ lb þ lc p (see the problem 7.1.37h). This ends the proof. ∠ > ∠ > ∠ ∠ > 1 ∠ : 7.1.108. Assume that A B C. Let us prove that OIH 180 À 2 A ∠ < 1 ∠ > ∠ > As A 90 , then 180 À 2 A 135 and therefore OIH 135 . Let us draw altitudes AA1 and CC1 in triangle ABC (Figure 7.77). As ∠OAB ¼ 90 À ∠C > 90 À ∠B ¼ ∠BAA1 and ∠OCA ¼ 90 À ∠B > 90 À ∠A ¼ ∠ACC1, then point O is inside of triangle A1HC. Note that ∠BAA1 ¼ ∠OAC ¼ 90 À ∠B and ∠ACC1 ¼ ∠OCB ¼ 90 À ∠A. Hence, ∠HAI ¼ ∠IAO and ∠HCI ¼ ∠ICO. Let lines AI and CI intersect segment HO at points E and F, respectively. Let C2 be the midpoint of side AB and K be intersection point of lines OC2 and AI.

Figure 7.77 400 7 Miscellaneous Inequalities

HE AH < HC HC HF As EO ¼ AO AO ¼ CO ¼ FO, and lines CI, OC2 intersect on the circumcircle of triangle ABC, then point I is on segment EK. < HC HF < HG Note that HC 2 Á CO, therefore CO ¼ FO 2 ¼ GO. Thus, it follows that point G is on segment FO. ∠ > ∠ ∠ 1 ∠ > , , ∠ (a) We have that GIH FIE ¼ AIC ¼ 90 þ 2 B 112 5 PNA ¼ [X1B1À[B1X [X1B1 < ∠ > 2 2 30 , B 45 . ∠ ∠ 1 ∠ ∠ ∠ : (b) If B 60 ,then AIC ¼ 90 þ 2 B 180 À B ¼ AHC Note that point I is inside of the circle passing through points A, H, C. Then, it follows that ∠AIH > ∠ACH ¼ 90 À ∠A and ∠OIH ¼ ∠OIA þ ∠AIH > ∠OKA þ ∠AIH 1 ∠ ∠ > 1 ∠ : ¼ 90 þ 2 A þ AIH 180 À 2 A ∠ > 1 ∠ : ∠ < ∠ We deduce that OIH 180 À 2 A If B 60 ,then AIC ¼ 90 þ 1 ∠ > ∠ ∠ : 2 B 2 B ¼ AOC Hence, point I is inside of the circle passing through points ∠ < 1 ∠ : A, O, C. Note that IOA 2 C Then,

1 ∠OIA ¼ 180 À ∠IAO À ∠IOA ¼ 180 À ðÞÀ∠B À ∠C ∠IOA > 2 1 1 1 > 180 À ðÞÀ∠B À ∠C ∠C ¼ 180 À ∠B: 2 2 2

∠ > 1 ∠ ∠ > ∠ > 1 ∠ > 1∠ : Thus, OIA 180 À 2 B and OIH OIA 180 À 2 B 180 À 2 A ∠ > 1 ∠ : We obtain that, OIH 180 À 2 A Remark This estimate is an exact one, as for ∠A ¼ 90, ∠B ! 90 we have that ∠ 1 ∠ ∠ , ∠ ∠ IHO ¼ 2 ðÞ!B À C 45 IOH ! 0 and thus OIH ! 135 .

7.1.109. (a) Note that AC1 ¼ AB1, BC1 ¼ BA1, CA1 ¼ CB1, therefore AC1 Á BA1 Á CB1 ¼ 1: According to Ceva’s theorem, it follows that segments AA , BC1 CA1 AB1 1 BB1 and CC1 intersect at one point. Point M is called the Gergonne Point. Let AC1 ¼ AB1 ¼ m, BC1 ¼ BA1 ¼ n, CA1 ¼ CB1 ¼ k. According to Stewart’s theorem, from triangle ABC, we obtain that 2 2 2 ðÞmþn kþðÞmþk nÀnkðÞ nþk : AA1 ¼ nþk Note that AC1 þ AB1 ¼ SMAC þ SMAB ¼ SMACþSMAB ¼ SMAC ¼ MA ,asSMAC ¼ BC1 CB1 SMBC SMBC SMA1CþSMA1B SMA1C MA1 SMA1C SMAB ¼ MA : SMA1B MA1 MA m m mkþmn We obtain that ¼ þ , therefore MA ¼ AA1: Hence, we deduce MA1 n k mkþmnþnk 3 that MA2 ¼ m ðÞkþn ðÞmkþmnþ4nk : ðÞmkþmnþnk 2 2 4 2 One needs to prove that MA 3 AB1 ,or mkðÞþn ðÞmkþmnþ4nk 4 : ðÞmkþmnþnk 2 3 The last inequality holds true, as it is equivalent to (mn þ mk À 2nk)2 0. 7.2 Solutions 401

Figure 7.78

(b) Let point M be the Gergonne Point of triangle BDF, then from problem 7.1.109а, it follows BM p1ffiffi ðÞBF þ BD À DF , DM p1ffiffi ðÞDF þ BD À BF 3 3 и FM p1ffiffi ðÞBF þ DF À BD : 3 Note that

AC Á BM CE Á DM S ¼ S þ S þ S þ ABCM CDEM AFEM 2 2 AE Á FM 1 þ pffiffiffiACðÞ BF þ BD À DF 2 2 3 1 1 þ pffiffiffiCEðÞ DF þ BD À BF þpffiffiffiAEðÞ BF þ DF À BD : 2 3 2 3

Thus, it followsp thatffiffiffi ACðÞ BF þ BD À DF þCEðÞ DF þ BD À BF þ AEðÞ BF þ DF À BD 2 3S: (c) Let A, B, C be the centers of those circles (Figure 7.78). ABþACÀBC As AY ¼ AZ, BX ¼ BZ, CX ¼ CY, then we obtain that AY ¼ AZ ¼ 2 , BX BCþABÀAC BCþACÀAB : ¼ BZ ¼ 2 , CX ¼ CY ¼ 2 Thus, it follows that the incircle of triangle ABC touches sides BC, AC, AB at points X, Y, Z, respectively. Let point M be the Gergonne Point of triangle ABC, then from problem 7.1.109а, it follows that the circles with centers A, B, C and radiuses p2ffiffi AY, p2ffiffi BZ, p2ffiffi CX cover 3 3 3 triangles MZY, MXZ, MXY, respectively. Therefore, these circles cover triangle XYZ too (point M is inside of triangle XYZ). (d) Let N be the Gergonne Point of triangle ABC, then from problem 7.1.115а,it follows that NA p2ffiffi ðÞp À a , NB p2ffiffi ðÞp À b , NC p2ffiffi ðÞp À c : 3 3 3

p2ffiffi ðÞÁpÀa МA p2ffiffi ðÞÁpÀb МB а 3 3 According to problem 7.1.47 , we have that bc þ ac þ p2ffiffi ðÞÁpÀc МC 3 MAÁNA þ MBÁNB þ MCÁNC 1: Therefore, MAþMB þ MBþMC þ MCþMA abpffiffiffi bc ac ab pÀc pÀa pÀb R : 2 3 Á r 402 7 Miscellaneous Inequalities

7.1.110. (a) Let Pn þ 1 P1, Xn þ 1 X1, An þ 1 A1. Note that, if point M is inside of polygon X1X2 ...Xn, then we have that

ðÞ∠X1A2X2 þ ∠X1MX2 þ::: þ ðÞ∠XnA1X1 þ ∠XnMX1 ¼

¼ ðÞ∠X1A2X2 þ ::: þ ∠XnA1X1 þðÞ∠X1MX2 þ ::: þ ∠XnMX1 ¼ ðÞn À 2 π þ 2π ¼ nπ:

∠ ∠ πn π: Therefore, the exists a number i, such that XiAiþ1Xiþ1 þ XiMXiþ1 n ¼ Hence, point M is inside of circle passing through points Xi, Ai þ 1, Xi þ 1. Thus, it follows that there exists a number i, such that point P is inside of circle passing through points Xi, Ai þ 1, Xi þ 1. Then, we have that

XiXiþ1 PAiþ1 , 2 sin ∠XiAiþ1Xiþ1 and

PiPiþ1 PAiþ1 ¼ , ∠XiAiþ1Xiþ1 ¼ ∠PiAiþ1Piþ1: 2 sin ∠PiAiþ1Piþ1

Therefore, PiPi þ 1 XiXi þno1. Hence, we obtain that max X1X2 ; X2X3 ; :::; XnX1 1: P1P2 P2P3 PnP1

0 0 (b) Let ∠A1 ¼ α1, ∠A2 ¼ α2,...,∠An ¼ αn and segment B1 B2 be the projection of segment B1B2 on the line perpendicular to the bisector of angle A1A2A3. We have that, α α B B B 0B 0 ¼ B 0A þ A B 0 ¼ B A cos 90 À 2 þ A B cos 90 À 2 ¼ 1 2 1 2 1 2 2 2 1 2 2 2 2 2 α B A þ A B ¼ ðÞB A þ A B sin 2 > 1 2 2 2 sinα 1 2 2 2 2 2 2 B A þ A B ¼ 1 2 2 2 A A : 4R 1 3

Therefore B1B2 > B1A2þA2B2 , where R is the circumradius of polygon A A ...A . A1A3 4R 1 2 n Similarly, we deduce that B2B3 > B2A3þA3B3 , :::, BnB1 > BnA1þA1B1 : A2A4 4R AnA2 4R Summing up these inequalities, we obtain that

B B B B B B B A þ A B B A þ A B B A þ A B 1 2 þ 2 3 þ ::: þ n 1 > 1 2 2 2 þ 2 3 3 3 þ ::: þ n 1 1 1 ¼ A1A3 A2A4 AnA2 4R 4R 4R A A þ A A þ ::: þ A A ¼ 1 2 2 3 n 1 : 4R

::: Hence, we obtain that B1B2 þ B2B3 þ ::: þ BnB1 > A1A2þA2A3þ þAnA1 : A1A3 A2A4 AnA2 4R 7.2 Solutions 403

Figure 7.79

Figure 7.80

According to problem 1.1.7b and problem 1.1.8b, one can prove that A A þ A A þ ...þ A A > 4R, therefore B1B2 þ B2B3 þ ::: þ BnB1 > 1: 1 2 2 3 n 1 A1A3 A2A4 AnA2 7.1.111. Let points I and N be on the different sides of line PM. Without loss of generality, one can assume that point N is on segment B1C, where B1 is the point at which the incircle of triangle ABC touches side AC. The following cases (Figures 7.79 and 7.80) are possible.

(a) Point P is on segment C1B (Figure 7.79), where C1 is the point at which the incircle of triangle ABC touches side AB.

∠ [XYÀ[X1Y1 , Note that 60 ¼ MPN ¼ 2 therefore [XY 120 . ∠ [X1B1À[B1X [X1B1 < , < We have that PNA ¼ 2 2 30 as [Y1BY 180 . Similarly, we deduce that ∠APN < 30. Thus, it follows that ∠A ¼ 180 À ∠PNA À ∠APN > 120.

(b) Point P is on segment AC1 (Figure 7.80). We have that ∠ANP < 30, note that ∠NPI > 60 and ∠IPB > ∠NPI > 60. Thus, it follows that ∠BPN > 120. Therefore ∠APN < 60, hence ∠A ¼ 180 À ∠APN À ∠ANP > 90. Let point K is on segment AB1 and the incircle of triangle ABC touches segment ∠ 1 ∠ > , ∠ > ∠ AB1. Note that IAC1 ¼ 2 BAC 45 thus IAC1 AIC1. Therefore r ¼ IC1 > APþPKþAK > > AC1 ¼ 2 PK, then r PK. Let us draw, through point I , a line parallel to line PM, and note that PM > 2r. Using the triangle inequality, we obtain that 404 7 Miscellaneous Inequalities

KN PN À PK ¼ PM À PK > 2r À PK > PK. Hence, KN > PK. We deduce that, ∠KPN > ∠PNK. Thus, it follows that

∠KPB ¼ 2∠KPI > 120 þ 2∠KPN:

Therefore, ∠APN < 60 À ∠KPN < 60 À ∠PNK. Hence, we obtain that ∠A ¼ 180 À ∠PNA À ∠APN > 120. This ends the proof. 7.1.112. Let a triangle with area S and perimeter P contain these circles with radius R. At ﬁrst, let us prove the following lemmas.

Lemma 1 There exists a triangle with area S1and perimeter P1, such that any side of this triangle touches at least one of the circles and S S1, P P1. Indeed, for this one needs to translate (moving parallel) the lines containing the sides of triangle ABC with area S and perimeter P, such that any of them touches at least one of the circles (Figure 7.81). Note that triangle A1B1C1 is the required triangle, as S1 ¼ SA1B1C1 S and P1 P.

Lemma 2 There exists a triangle with area S2and perimeter P2, such that one of its sides touches both circles and any of the other two sides touches at least one of the circles. Moreover, S2 S1 and P2 P1. Indeed, according to lemma 1 we have that any of the sides of triangle A1B1C1 touches at least one of the circles. We proceed the proof by contradiction arguement. Assume that none of the sides of triangle A1B1C1 touches simultaneously both circles. Let centers O1 and O2 of those circles be connected by a segment. Now, let us rotate this ﬁxed ﬁgure such that one of the sides, for example A1C1 touches both circles (Figure 7.82). ∠ 0 ∠ 0 < π ∠ ∠ , We have that O1O2O2 ¼ O1O2 O2 2 O1NC1 ¼ O1O2C2 then ∠ 0 > ∠ 0 > π : 0 > O2 O2M O2 O2N 2 Therefore, O2 K O2M ¼ R. This means that the 0 circle with center O2 is inside of triangle A1B1C1. In order to end the proof of lemma 1, let us note that one can translate (moving parallel) side B1C1 and decrease the area and the perimeter of the triangle.

Figure 7.81 7.2 Solutions 405

Figure 7.82

Figure 7.83

Let S2 ¼ (A2B2C2), P2 ¼ A2B2 þ B2C2 þ A2C2, then S1 S2 and P1 P2.

Lemma 3 There exists an equilateral triangle with area S3and perimeter P3, such that its base touches both circles and any of the other two sides touches at least one of the circles. Moreover, S2 S3 and P2 P3. Indeed, let line l be the common tangent of those circles (Figure 7.83). 0 0 0 Denote by A1 , B2 and C2 , respectively, the symmetric points of points A1, B2 and C2 with respect to line l respectively. Let points A0, B0, C0 be the midpoints of 0 0 0 0 0 segments C2 A1, B2B2 and A1 C2,respectively. As C2 A1 ¼ A1 C2, then A0C0 ¼ A1C2. We have that S2 ¼ (A0B0C0) ¼ S0 and p2 p0, where p0 is the perimeter of triangle A0B0C0. Moreover, it is obvious that the circles are in triangle A0B0C0 (Figure 7.83). Consider Figure 7.84. Let area of equilateral triangle A3B0C3 is equal to S3, аnd its perimeter is equal to p3. Then, we have that S2 ¼ S0 S3 and p2 p0 p3. Therefore, S2 S3 and p2 p3. This ends the proof of lemma 1. Lemma 4 Let an isosceles triangle, with angles 45 ,45 ,90 and area S4, be such that its bases touches both circles and any of the other two sides touches at least one of the circles. Then, S3 S4. Let ∠A3C3B0 ¼ 2α, note that A3C3 ¼ 2R(1 þ ctgα), B0H ¼ R(1 þ ctg(45 À α)), where B0H is the altitude of triangle A3C3B0. 406 7 Miscellaneous Inequalities

Figure 7.84

2R2ðÞ1þctgα ctgα : 2 2α Therefore, S3 ¼ ctgαÀ1 Thus, it follows that the equation 2R ctg þ 2 2 2 (2R À S3)ctgα þ S3 ¼ 0 must have a solution. We deduce that D ¼ (2R À S3) 2 > π 2 2 À 8ffiffiffiR S3 0. On the otherffiffiffi hand, S3 2 R . Hence, we obtain that S3 6R þ p p 2 4 2R2: If S ¼ 6R2 þ 4 2R2,thenD ¼ 0. Thus, it follows that ctgα ¼ S3À2R pffiffiffi 3 4R2 : α π : ¼ 1 þ 2 Therefore, ¼ 8 We have that S S1 S2 S3 S4,hence S S4. This means that the smallest possible value ofp areaffiffiffi of a triangle containing two tangent circles with radius R is equal to 6R2 þ 4 2R2: α α 1 4Rctg2α We have that p3 ¼ 2RðÞþ1 þ ctg 2RðÞÁ1 þ ctg cos 2α ¼ ctgαÀ1 , therefore the 2 equation 4Rctg α À p3ctgα þ p3 ¼ 0 must have a solution. Thus, it follows that 2 D ¼ p3 À 16p3R 0. We deduce that, p3 16R. α p3 : Moreover, p3 ¼ 16R,ifctg ¼ 8R ¼ 2 We obtain that, the smallest possible value of perimeter of a triangle containing two tangent circles with radiuses R is equal to 16R. 7.1.113. (a) At first, let us prove the following lemma. Lemma Given a convex hexagon ABCDEF, such that ABjjDE, BCjjEF, CD||FA. Given that the distance between lines AB,DE is equal to the distance between lines BC,EF and is equal to the distance between lines CD, FA. Prove that AD2 ¼ (AB þ DE)(FA þ CD). Note that point D is equidistant from lines AB and FA. Then, we have that ∠BAD ¼ ∠FAD. As AB||DE,CD||FA, then ∠BAD ¼ ∠FAD ¼ ∠ADE ¼ ∠ADC. Similarly, we obtain that ∠ABE ¼ ∠CBE ¼ ∠FEB ¼ ∠DEB and ∠BCF ¼ ∠DCF ¼ ∠CFA ¼ ∠CFE. Let ∠BAD ¼ α, ∠ABE ¼ β and ∠BCF ¼ γ. As the sum of all interior angles of a convex hexagon is equal to 720, then it follows that α þ β þ γ ¼ 180. Let us consider parallelogram ADEK. We have that BK ¼ BA þ AK ¼ BA þ DE, KE ¼ AD, ∠AKE ¼ ∠BAD ¼ α, ∠KBE ¼ ∠ABE ¼ β, ∠BEK ¼ γ. 7.3 Problems for Self-Study 407

Then, the triangles with sides AD, AB þ DE, BE and FA þ CD, AD, CF are sim- 2 ilar. Тhus, it follows that AD ¼ (AB þ DE)(FA þ CD).pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi According to the lemma, it follows that AD ¼ ðÞAB þ DE ðÞFA þ CD ABþDEþFAþCD 2 . ABþBCþEFþED BCþCDþEFþFA Similarly, we deduce that BE 2 , CF 2 . Summing up the last three inequalities, we obtain that

AD þ BE þ CF AB þ BC þ CD þ DF þ EF þ FA:

(b) Let diagonals AD, BE and CF intersect at point M. Note that ∠MFE ¼ ∠CFE ¼ ∠CAE ¼ 60. Similarly, we deduce that ∠MDE ¼ 60, ∠CDM ¼ 60, ∠CBM ¼ 60. Let ∠BMC ¼ α, ∠CMD ¼ β, ∠DME ¼ γ, then ∠FME ¼ ∠BMC ¼ α. DE ME EF According to the law of sines, we have that ¼ ¼ : sin γ sin 60 sin α DE sin γ : Thus, it follows that EF ¼ sin α CD sin β : Similarly, we deduce that BC ¼ sin α As α þ β þ γ ¼ π, then by triangle inequality, we obtain that β γ < α DE CD < : jsin À sin j sin . Therefore, it follows that EF À BC 1 Remark Let ABCDEF be a cyclic hexagon. Given that diagonals AD, BE and CF DEÁAE CDÁAC < : intersect at one point. Prove that EF À BC CE

7.3 Problems for Self-Study

7.1.114. Prove the following inequalities for triangle ABC. 1 1 1 4 (a) 2 þ 2 þ 2 2, ðÞma þ mb À mc ðÞma À mb þ mc ðÞÀma þ mb þ mc 3R m2 m2 m2 9 (b) a þ b þ c , a2 b2 c2 4 9 m m þ m m þ m m 5 (c) < a b b c c a < , 20 ab þ bc þ ac 4 a2 b2 c2 pffiffiffi (d) þ þ 3 3R, Àa þ b þ c a À b þ c a þ b À c r r r R (e) a þ b þ c À 1. rb þ rc rc þ ra ra þ rb r 7.1.115. The parallelogram ABCD lies in the base of the pyramid SABCD. A plane passing through vertex A and the midpoint K of the edge SC intersects the edges 1 V1 3 SB and SD at points M and N. Prove that, 3 V 8, where V, V1 are the volumes of the pyramids SABCD, SAMKN, respectively. 408 7 Miscellaneous Inequalities

7.1.116. Let the lateral edge of a regular triangular pyramid has a length a. Prove a3 that the volume of such a pyramid does not exceed 6 . 7.1.117. Prove that the ratio of the volume V of the regular n-gon pyramid to the volume V1 of its insphere, satisﬁes the following inequality:

V tg ðÞπ=n : 2V1 π=n

pffiffi r2 3À 8 7.1.118. Prove that the inequality 2 2 5 holds true for the right-angled maþmb triangle, where ma, mb are the lengths of the medians drawn to the legs, r is the circumradius of the triangle. rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 R 1 1 1 1 7.1.119. Prove the inequality 2 þ þ for triangle ABC, where S is the 2S 3 ha hb hc area of the triangle, R is the circumradius, ha, hb, hc are its altitudes.

7.1.120. Given a point M inside of angle AOB. Let M1 and M2 be its projections on 1 2 ∠ the sides of the angle. Prove that SOM1MM2 2 OM sin AOB. 7.1.121. Let a, b, c be the sides of triangle ABC, r, and R be its inradius and circumradius, respectively. Prove that 1 1 1 1 1 2 1 1 1 1 þ þ þ þ : 2Rr 3 a b c a2 b2 c2 4r2

7.1.122. Given a point O inside of triangle ABC, x, y, z are the distances of the points O to the sides of the triangle. Theq circumradiusffiffiffi of triangle ABC is R. ffiffiffi ffiffiffi ffiffi p p p R Prove that x þ y þ z 3 2. ÀÁÀÁ 2 2 2 2 2 2 2 7.1.123. Prove that ma þ mb þ mc ha þ hb þ hc 27S , where ma, mb, mc are the medians, ha, hb, hc are the altitudes, and S is the area of triangle ABC. φ 7.1.124. Thepffiffi median BM of triangle ABC forms an angle with side BC . Prove that ∠ 2 2À3 cos φ ctg A sin φ . 7.1.125. Let a radius drawn in the half-circle with the diameter AB divide it into two sectors. Given that circles are inscribed into each of these sectors and M, N arepffiffi the 8À 2 tangential points of these circles with the diameter AB, prove that MN 8 R, where AB ¼ 2R. β γ 7.1.126. Given a triangle ABC, prove that α þ þ 3, where h , h , h are the ha hb hc 2 a b c altitudes of ABC, α, β, γ are the distances of the feet of the bisectors of angles A, B, C from the sides of the triangle. 7.3 Problems for Self-Study 409

7.1.127. Lateral edges of a triangular pyramid are mutually perpendicular.Prove 9h2 that S1 þ S2 þ S3 2 , where S1, S2, S3 are the areas of lateral facets and h is the altitude of the pyramid.

7.1.128. Given triangle P1P2P3 and a point P inside of the triangle. The straight lines P1P, P2P, P3P intersect the opposite sides at points Q1, Q2, Q3 respectively. Prove that at least one of the numbers P1P , P2P , P3P is less than or is equal to PQ1 PQ2 PQ3 2 and at least one of these numbers is greater than or equal to 2. 2 2 2 2 > 7.1.129. Prove that r þ ra þ rb þ rc 4S, where r, ra, rb, rc are, respectively, the radiuses of the incircle and excircles of triangle ABC, S is the area of the triangle. 7.1.130. For triangle ABC prove that rffiffiffiffiffiffi 1 1 1 9 R þ þ : α β γ π 2r

7.1.131. In a rectangular sheet of paper, n rectangular holes were cut, and sides of the holes are parallel to the edges of a sheet. What is the least guaranteed number of the rectangular parts, such that this sheet with the holes is possible to cut? (Show that, in all cases it is possible to cut the sheet into the number of parts found by you, but in some cases, it is impossible to cut it into smaller number of parts). Hint The proof by mathematical induction, if n rectangular holes were cut in a polygon, such that all internal angles are equal to either 90 or 270, with sides of the holes parallel to sides of that polygon, then that sheet with the holes can be cut into 3n þ k þ 1 rectangular parts, where k is the number of internal angles equal to 270. Separately, consider the case, if n ¼ 0. To prove these statements one has to consider the rectangular, with the left vertical side being the leftmost and that angle of 270from the angles of the polygon, is the one with the leftmost vertex. 7.1.132. Triangle ABC is cut into finite number of triangles, such that in each triangle there is an angle greater than 120. Prove that max(∠A, ∠B, ∠C) > 120. 7.1.133. Is it possible to cut any convex polygon into a finite number of non-convex quadrilaterals? 7.1.134. What is the minimal number of non-overlapping tetrahedrons that a cube can be divided into? 7.1.135. Given a point M inside of the square ABCD, prove that ∠ ∠ ∠ ∠ > 3π MAB þ MBC þ MCD þ MDA 4 . Hint See problem 7.1.14d. 7.1.136. Given on a plane n vectors the length of each being equal to 1. The sum of all these vectors is a zero-vector. Prove that these vectors can be enumerated in such a way, that for all k ¼ 1, 2, . . . , n the followingpffiffi condition holds true: the sum of 5 initial k vectors has a length not greater than 2 . 410 7 Miscellaneous Inequalities

7.1.137. Given several acute triangles. Let a new triangle be constructed, from the sides of the given triangles, by the following rule: the smallest side of it is equal to the sum of the smallest sides of the triangles, the length of the middle side is equal to the sum of the middle sides of the triangles, and the largest side is equal to the sum of the largest sides of the triangles. Provepffiffiffi that the cosine of the largest angle of the obtained triangle is greater than 1 À 2.

Hint Consider the triangles with sides ai bi cffiffiffii, i ¼ 1, 2, . . . , n. Then, we have 2 p to prove that ðÞb1 À a1 þ ::: þ bn À an þ 2 2ðÞa1 þ ::: þ an ðÞb1 þ ::: þ bn 2 ðÞc1 þ ::: þ cn . rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁpffiffiffiÀÁÀÁ 2 2 2 2 Prove that ðÞbi À ai bj À aj þ 2 aibj þ ajbi bi þ ai bj þ aj .

7.1.138. Given a point M inside of triangle ABC (see the notations of problem 7.1.76), prove that

(a) ha þ hb þ hc 9, da db dc (b) hahbhc 27dadbdc, (c) (ha À da)(hb À db)(hc À dc) 8dadbdc, (d) aRa bdb þ cdc, (e) aRa cdb þ bdc, 2 2 (f) aRaRbRc bdbRb þ cdcRc , R (g) RaRbRc 2r ðÞda þ db ðÞdb þ dc ðÞdc þ da , 2R (h) RaRb þ RbRc þ RcRa r ðÞdadb þ dbdc þ dcda , 2 2 2 2 (i) a þ b þ c 4RRðÞaþRbþRc , dbdc dcda dadb RaRbRc (j) aRa þ bRb þ cRc 2(ada þ bdb þ cdc), (k) aRada þ bRbdb þ cRcdc ÀÁ2(adbdc þ bdadc þ cdadb), 2 2 2 (l) ðÞa þ b þ c RaRbRc 2 adaRa þ bdbRb þ cdcRc , (m) R2R2R2 ðÞd R þ d R ðÞd R þ d R ðÞd R þ d R , paffiffiffiffiffib c pffiffiffiffiffi a pa ffiffiffiffiffi b pb ffiffiffiÀÁpb ffiffiffiffiffib pc ffiffiffiffiffic pc ffiffiffiffiffic a a (n) Ra þ Rb þ RÀÁc 2 da þ db þ dc , 2 2 2 > 2 2 2 (o) Ra þ Rb þ Rc 2 da þ db þ dc , (p) Ra Rb Rc 1, a2 þ b2 þ c2 R 2 2 2 Ra Rb Rc (q) bc þ ca þ ab 1, (r) RR0 RaRbRc, where R0 is the circumradius of the triangle with sides aRa, bRb, cRc. Hint See the proof of problems 7.1.47, 7.1.76, 7.1.79, and 7.1.97. 2 1 7.1.139. Prove that, for triangle ABC, the inequality ma 4 ðÞ4b þ 4c À 5a a holds true, where BC ¼ a, AC ¼ b, AB ¼ c, ma is the median drawn to side a. For which triangle the equality holds true? 1 7.1.140. Prove the inequality r 16 ðÞh1 þ h2 þ h3 þ h4 for a triangular pyramid, where r is the inradius and hi are the altitudes of the pyramid (i ¼ 1, 2, 3, 4). 7.3 Problems for Self-Study 411

7.1.141. The angle is divided by the rays drawn from its vertex into 2n þ 1 equal angles. These angles cut on some straight line 2n þ 1 segments. Prove that the sum of the lengths of the ﬁrst, third, ...,(2n þ 1)-th segments is greater than the sum of the remaining segments. Hint Draw a perpendicular from the vertex of the angle to the line. 7.1.142. For triangle ABC (see the notations of problem 7.1.76), prove that α β γ d þ d 2 d þ d 2 d þ d 2 R2sin 2 þ R2sin 2 þ R2sin 2 a b þ b c þ c a : a 2 b 2 c 2 2 2 2

dbþdc α Hint Prove that Ra sin α cos 2. 7.1.143. For an acute triangle ABC (see the notations of problem 7.1.76), prove that 2 2 2 2 2α 2 2β 2 2γ da þ db þ dc Rasin 2 þ Rbsin 2 þ Rc sin 2.

2 2 2 2 α dbþdc 2 sin xþsin ðÞÀx 2 2α Hint Prove that 2 ¼ Ra 2 Rasin 2.

7.1.144. Given a regular n-gon A1A2 ...An inscribed into a unit circle and a point π 2 M on the minor arc A1An, prove that 2ctg MA1 þ MA2 þ ::: þ MAn π . 2n sin 2n

Hint Let M0 be the midpoint of the minor arc A1An and M is on the minor arc A1M0, then according to problem 1.1.6b, we have that A1Ak þ 1 þ A1An À k Ak þ 1M þ An À kM Ak þ 1M0 þ An À kM0. Consequently, it follows that

A1A2 þ A1A3 þ ::: þ A1An MA1 þ MA2 þ ::: þ MAn M0A1 þ M0A2 þ ::: þ M0An:

7.1.145. A quadrilateral has a circumcribedpffiffiffi circle with radius R and an inscribed circle with radius r. Prove that R 2r. pffiffiffiffiffiffiffiffiffiffi 2 abcd 1 Hint Prove that r ¼ ðÞaþc ðÞbþd 4 abcd and see problem 7.1.93d). Chapter 8 Some Applications of Geometric Inequalities

This chapter consists of two sections, that is, Sections 8.1 and 8.2. Section 8.1 is devoted to geometric problems that can be solved applying geometric inequalities. In the proofs of many problems of this chapter are used the following statements about a triangle. (a) Greater side of a triangle is opposite the greater angle. (b) Greater angle of a triangle is opposite the greater side. In some problems of this paragraph, one needs to understand when does the equality holds true in the corresponding geometric inequality. In many proofs of Section 8.1 is used the following statement: if points M and N are on the same side of segment AB and point M is inside of triangle ANB, then ∠AMB < ∠ ANB. In Section 8.2 is shown how one can prove algebraic inequalities using geometric inequalities. In this paragraph are selected such inequalities that can be proved by comparing areas (volumes) of two geometric ﬁgures. In order to prove some inequalities in Section 8.2 is used very useful substitution x ¼ tgα, y ¼ tgβ, z ¼ tgγ, where α, β, γ are the angles of some triangle (see problem 8.2.9). On the other hand, in order to deal with many geometric inequalities and to transform them into algebraic inequalities, often is used the method of coordinates. Some problems in this chapter were inspired by [1, 13]. Nevertheless, even for these problems, the authors have mostly provided their own solutions.

8.1 Application of Geometric Inequalities for Solving Geometric problems

8.1.1. Given a point M inside of square ABCD, such that ∠MBC ¼ ∠ MCB ¼ 15. Prove that triangle AMD is equilateral.

© Springer International Publishing AG 2017 413 H. Sedrakyan, N. Sedrakyan, Geometric Inequalities, Problem Books in Mathematics, DOI 10.1007/978-3-319-55080-0_8 414 8 Some Applications of Geometric Inequalities

∠ ¼ 8.1.2. Given a point D on side AC ofp triangleffiffiffi ABC, such that ABD 90 , ∠DBC ¼ 30. Find segment AD,ifAB ¼ 3, CD ¼ 1. 8.1.3. Given points E and F on sides AB and BC of triangle ABC, respectively, such that AE ¼ EF ¼ BF. Find angle AEC,if∠A ¼ 75, ∠B ¼ 10. 8.1.4. Given that in a convex quadrilateral AB ¼ BC and ∠ABD ¼ 65, ∠CBD ¼ 35, ∠ADC ¼ 130. Find the unknown angles of quadrilateral ABCD. 8.1.5. Given a triangle ABC, such that ∠B ¼ ∠ A +30 and AB ¼ 2BC. Find the angles of triangle ABC. 8.1.6. Given a point M inside of triangle ABC, so that ∠MAB ¼ ∠ MCB ¼ 20, ∠MAC ¼ 80, ∠MCA ¼ 30. Find the value of angle ∠AMB. 8.1.7. Given a point M inside of triangle ABC, so that ∠MAB ¼ ∠ MBC ¼ 20, ∠MBA ¼ 10, ∠MAC ¼ 80. Find the value of angle ∠AMC. 8.1.8. The circumcenter of triangle ABC is inside of the incircle of that triangle. Find the angles of triangle ABC,ifla ¼ R. 8.1.9. In triangle ABC, the bisectors of angles A and C intersect with the median BM at points E and F, respectively. Given that AE ¼ CF. Prove that AB ¼ BC. 8.1.10. Prove that a convex quadrilateral ABCD is circumscribed, if and only if quadrilateral MNPQ is inscribed, where M, N and P, Q are the tangential points of the incircles of triangles ABC and ACD with sides AB, BC and CD, AD, respectively. 8.1.11. Let in a convex quadrilateral ABCD diagonals AC and BD be perpendicular. Given that the opposite sides AB and DC are not parallel to each other. Perpendic- ular bisectors of sides AB and DC intersect at a point P inside of ABCD. Prove that a circle can be circumscribed around quadrilateral ABCD, if and only if the areas of triangles ABP and CDP are equal. 8.1.12. Given a point D inside of the acute triangle ABC, such that DA Á DB Á AB + DB Á DC Á BC + DC Á DA Á CA ¼ AB Á BC Á CA. Find the locus of point D. 8.1.13. A convex polygon is broken by nonintersecting diagonals into triangles. Prove that the sum of inradiuses of all these triangles does not depend on the way of breaking, if and only if the convex polygon is inscribed. (The diagonals may have common endpoints).

8.1.14. Given points A1, B1, and C1 on the sides BC, CA, and AB of triangle ABC, respectively, such that ∠AB1C1 + ∠ BC1A1 + ∠ CA1B1 ¼ 180 . Given also that the (a) areas, (b) perimeters, (c) inradiuses of triangles AB1C1, BC1A1, and CA1B1 are equal. Prove that A1 , B1 and C1 are the midpoints of sides BC, CA, and AB. 8.1 Application of Geometric Inequalities for Solving Geometric problems 415

8.1.15. Given points D and E on sides AC and BC of triangle ABC, respectively, such ∠CDE ¼ ∠CED that ∠BDE ∠AED. Is it true that ABC is an isosceles triangle, if AE and BD are the (a) altitudes, (b) medians, (c) bisectors of ABC? 8.1.16. (a) Prove that if for the circumscribed quadrilateral ABCD the condition p2 ¼ AC2 + BD2 holds true, where p is the semiperimeter of the quadrilateral, then it is a rhombus. (b) Let quadrilateral ABCD be inscribed in circle ω. Let E be the intersection point of rays AB, DC, and F be the intersection point of rays BC, AD. Given that AC ⊥ EF. Prove that segment AC is a diameter of the circle ω.

8.1.17. In triangle ABC the bisectors of angles ∠ABC and ∠BCA intersect sides CA and AB at points D and E, respectively. Let (a) AE ¼ BD and AD ¼ CE, AD ¼ BD ∠ (b) CE AE and A 36 . Find the angles of triangle ABC. 8.1.18. Let two isosceles triangles ABC and ADC be constructed on the base AC in different half-planes, such that ∠ADC ¼ 3 ∠ ACB. Let AE be the bisector of triangle ABC and F be the intersection point of segments DE and AC. Prove that CEF is an isosceles triangle. 8.1.19. Given a convex hexagon, such that for its each pair of opposite sides the following condition holds true: the ratio of the distance betweenp theﬃﬃ midpoints of 3 these sides to the sum of the lengths of these sides is equal to 2 . Prove that all angles of the hexagon are equal. (A convex hexagon ABCDEF has three pairs of opposite sides: AB and DE, BC and EF, CD and FA). 8.1.20. Prove that one can circumscribe a circle around the convex quadrilateral ABCD, if and only if the common tangent of the incircles of triangles ABD and ACD, different from AD, is parallel to BC. α ha 1 8.1.21. (a) For triangle ABC, prove that a 2 ctg 2. (b) Let circle ω be inscribed in quadrilateral ABCD and I be the center of ω. Given that (AI + DI)2 +(BI + CI)2 ¼ (AB + CD)2. Prove that ABCD is an isosceles trapezoid or a square. (c) Let point I be inside of a convex quadrilateral ABCD. Given that ∠BIC ¼ ∠ BAI + ∠ CDI, ∠ CID ¼ ∠ CBI + ∠ DAI, and ∠AIB + ∠ CID ¼ 180. Prove that one can inscribe a circle in quadrilateral ABCD. 416 8 Some Applications of Geometric Inequalities

8.1.22. (a) Let ABCDEF be a convex hexagon, such that AD ¼ BC + EF, ¼ ¼ AB ¼ CD ¼ EF BE AF + CD, CF AB + DE.ProvethatDE AF BC. (b) Let ABCDEF be a convex hexagon, such that the sum of the distances of each interior point to the six sides (AB, BC, CD, DE, EF, and FA) is equal to the sum of the distances between the midpoints of AB and DE, BC and EF, CD and FA. Prove that ABCDEF is a cyclic hexagon. 8.1.23. Let equilateral triangles ACB0 and BDC0 are drawn on the diagonals of a convex quadrilateral ABCD, such that points B, B0 are on the same side of AC, and points C, C0 are on the same side of BD. Find ∠BAD + ∠ CDA,ifB0C0 ¼ AB + CD.

Solutions

8.1.1. Note that ΔMAB ¼ ΔMCD, consequently, MA ¼ MD. If MA > AD, then ∠MAD ¼ ∠ MDA > ∠ AMD. Therefore ∠MAD > 60, thus ∠MAB < 30 and MA > AB. Hence ∠ABM > 75. This leads to a contradiction. Similarly, we can prove that if MA < AD, then ∠ABM < 75. We deduce that MA ¼ AD, which means that triangle AMD is equilateral. 8.1.2. If BD > 1 ¼ CD, then ∠DCB > ∠ DBC ¼ 30. Let point E be on the segment BD and BE ¼ 1, thus ∠BAD > ∠ BAE ¼ 30. We obtain that ∠A + ∠ C > 60.This leads to a contradiction. Similarly, one can prove that if BD < 1, then ∠A + ∠ C < 60. Hence, BD ¼ 1, therefore AD ¼ 2. One can easily verify that there exists a triangle satisfying these conditions. 8.1.3. Note that ∠EFC ¼ 20. If EC > AE ¼ EF, then ∠ECF < ∠ EFC ¼ 20 and ∠ECA < ∠ EAC ¼ 75. Consequently, ∠ACB ¼ ∠ ECF + ∠ ECA < 95. This leads to a contradiction. Similarly, one can prove that if EC < AE, then ∠ACB > 95. Since ∠ACB ¼ 95, hence EC ¼ AE. Thus ∠AEC ¼ 30. One can easily verify that there exists a triangle satisfying these conditions. 8.1.4. If BD > AB ¼ BC, then ∠BDA < ∠ BAD. Consequently, ∠BDA < 57.5 and ∠BDC < ∠ BCD. Hence ∠BDC < 72.5, therefore ∠ADC < 57.5 + 72.5 ¼ 130. This leads to a contradiction. Similarly, one can prove that if BD < AB, then ∠ADC > 130. On the other hand, since ∠ADC ¼ 130, then we have that BD ¼ AB. Consequently, ∠A ¼ 57.5, ∠C ¼ 72.5. This ends the proof. 8.1.5. Let BC ¼ a, thus AB ¼ 2a > BC. Therefore, one can take on side AB a point D, such that ∠DCA ¼ ∠ A ¼ α. Thus AD ¼ DC ¼ b and ∠CDB ¼ 2α. 8.1 Application of Geometric Inequalities for Solving Geometric problems 417

If b > a, then b > a > 2a À b. Therefore for triangle BDC, we have the inequality α +30 > 2α > 150 À 3α. This leads to a contradiction. If b < a, then b < a < 2a À b. Hence α +30 < 2α < 150 À 3α. This leads to a contradiction. If b ¼ a, then triangle BDC is equilateral. Therefore ∠A ¼ 30, ∠B ¼ 60, ∠C ¼ 90. This ends the proof. 8.1.6. Let N be a point symmetric to point M with respect to line AC. We have that ∠NAC ¼ ∠ MAC ¼ 80. Consequently, N lies on line AB. Since ∠NCA ¼ ∠ MCA ¼ 30 and MC ¼ NC, then triangle MNC is equilateral. If BM < MN ¼ MC, then ∠NBM > ∠ MNB ¼ 10 and ∠MBC > ∠ MCB ¼ 20. Therefore, ∠ABC > 30. This leads to a contradiction. Similarly, one can prove that if BM > MN ¼ MC, then ∠ABC < 30. This leads to a contradiction. Consequently, BM ¼ MN. Hence ∠AMB ¼ 150. 8.1.7. Let N be a point symmetric to point M with respect to line AC, and lines AC, MN intersect at point K, which means MK ⊥ AC and MK ¼ NK. Note that N lies on line AB. > ∠ > ∠ ¼ ¼ MN ¼ BM > MC If BM MC, then MCB MBC 20 . Since MK 2 2 2 , then ∠MCK > 30. Consequently, ∠ACB > 50. This leads to a contradiction. Similarly, one can prove that if BM < MC, then ∠ACB < 50. This leads to a contradiction. Hence BM ¼ MC, ∠MCB ¼ 20, and ∠AMC ¼ 70. This ends the proof.

8.1.8. Let O and O1 be the incenter and circumcenter of triangle ABC, respectively. By the triangle inequality, R ¼ AO AO1 + O1O (la À r)+r ¼ R. Hence, la ¼ AO1 + r, O1O ¼ r, and O1 2 [AO], consequently, ∠B ¼ ∠ C and ∠A ¼ 90. Therefore ∠B ¼ ∠ C ¼ 45. This ends the proof. 8.1.9. If BC > AB, then AE < CF (see problem 7.1.63). Similarly, if BC < AB, then AE > CF. Hence BC ¼ AB. This ends the proof. 8.1.10. Let quadrilateral ABCD be circumscribed, thus AB + CD ¼ AD + BC.We ¼ AB þ AC À BC ¼ AD þ AC À CD ¼ have that AM 2 2 AQ. Similarly, one can prove that CN ¼ CP. Consequently,

ÀÁ ÀÁ ∠QMN þ ∠QPN ¼ 180 À ∠AMQ À ∠BMN þ 180 À ∠CPN À ∠DPQ ∠A þ ∠B ∠C þ ∠D ¼ þ ¼ 180 : 2 2

Hence quadrilateral MNPQ is inscribed. If AB + CD > AD + BC, then AM > AQ and CP > CN. Therefore, ∠AQM > ∠ ∠ > ∠ ∠ < À ∠A ∠ < À ∠C AMQ and CNP CPN. Hence AMQ 90 2 and CPN 90 2 . Consequently, 418 8 Some Applications of Geometric Inequalities ∠B ∠D ∠QMN þ ∠QPN ¼ 90 À ∠AMQ þ þ 90 À ∠CPN þ 2 2 ∠A ∠B ∠C ∠D > þ þ þ ¼ 180 : 2 2 2 2

Similarly, one can prove that, if AB + CD < AD + BC, then ∠QMN + ∠ QPN < 180. Thus, if ∠QMN + ∠ QPN ¼ 180, then AB + CD ¼ AD + BC. This means that, if quadrilateral MNPQ is inscribed, then quadrilateral ABCD is circumscribed. This ends the proof. 8.1.11. Let quadrilateral ABCD be inscribed, thus point P is the center of the circumcircle of ABCD. Consequently, PA ¼ PB ¼ PC ¼ PD and ∠APB + ∠ CPD ¼ 2 ∠ BCA +2 ∠ ¼ ¼ 1 Á ∠ ¼ 1 Á CBD 180 . Hence, it follows that SABP 2 PA PB sin APB 2 PC PD sin ∠CPD ¼ SCDP. Let in the convex quadrilateral ABCD diagonals AC and BD are perpendicular. Let the perpendicular bisectors to sides AB and DC intersect at point P inside ABCD. We need to prove that if PA > PD, then SABP > SCDP. Let E 2 AC, F 2 BD, and PE k BD, PF k AC, thus AE > EC, BF > FD. Let C1 be a point symmetric to point A with respect to point E, and D1 be a point symmetric to point B with respect to point F. Thus, it follows that PC1 ¼ PA ¼ PB ¼ PD1 and segment PM intersects line CD, where M is the midpoint of segment C1D1. Therefore PM > PH, where PH ⊥ CD, H 2 CD. > ¼ ¼ 1 Á > 1 Á We have that C1D1 CD, consequently, SABP SC1D1P 2 C1D1 PM 2 CD PH ¼ SCDP. It is clear that if PA < PD, then SABP < SCDP. Thus, if SABP ¼ SCDP, then PA ¼ PD. Therefore, quadrilateral ABCD is inscribed. This ends the proof. 8.1.12. According to problem 4.1.8c, we have that DA Á DB Á AB + DB Á DC Á BC + DC Á DA Á CA AB Á BC Á CA, the equality holds true, if

! ! ! ~ A1B1 Á D1C1 þ B1C1 Á D1A1 þ A1C1 Á D1B1 ¼ 0, ð8:1Þ where D1, A1, B1, C1, are H images of points D, A, B, C (see the proof of problems 4.1.8a,c). If D1 is the incenter of triangle A1B1C1, then condition (8.1) is satisﬁed (see the proof of problem 4.1.19). Let O be any point, then condition (8.1) can be rewritten ! ! ! ! ! ! ~ as A1B1 Á OC1 À OD1 þ B1C1 Á OA1 À OD1 þ A1C1 Á OB1 À OD1 ¼ 0. ! ! ! ! Á þ Á þ Á A1B1 OC1 B1C1 OA1 A1C1 OB1 Therefore, OD1 ¼ . Thus, condition (8.1) is holds A1B1þB1C1þA1C1 true only for one point D1 (the incenter of triangle A1B1C1). 8.1 Application of Geometric Inequalities for Solving Geometric problems 419

We have that ∠D1A1C1 ¼ ∠ D1A1B1 and ∠D1A1C1 ¼ ∠ DCA, ∠ D1A1B1 ¼ ∠ DBA. Hence ∠DCA ¼ ∠ DBA. Similarly, we obtain that ∠DBC ¼ ∠ DAC and ∠DAB ¼ ∠ DCB. ∠ þ ∠ þ ∠ ¼ 1 ðÞ∠ þ ∠ þ ∠ ¼ Consequently, DAC DAB DBA 2 A B C 90 . Hence, BD ⊥ AC. In the same way, we deduce that AD ⊥ BC and CD ⊥ AB. Hence, point D is the orthocenter of triangle ABC. This ends the proof.

8.1.13. Let the convex polygon A1A2 ...An be inscribed. Let A1A2 ...An be broken by nonintersecting diagonals into k triangles. Therefore, we have that ∠A1 + ∠ A2 + ... + ∠ An ¼ 180 k and ∠A1 + ∠ A2 + ... + ∠ An ¼ 180 (n À 2). Thus, it follows that k ¼ n À 2. Let the inradiuses of these triangles be equal to r1 , r2 , ... , rn À 2, and R be the circumradius of polygon A1A2 ...An. Using problem 5.3.6 for all these n À 2 triangles, one can easily obtain that _ r1 rnÀ2 1 n À 2 þ þ ...þ ¼ cos α1 þ cos α2 þ ...þ cos αn, where αi ¼ AiA iþ1, R R _ 2 i ¼ 1, ... , n and An +1 A1, and the arc AiA iþ1 does not contain the vertices of polygon A1A2 ...An other than Ai and Ai +1. Consequently, r1 + ... + rn À 2 does not depend on the way of breaking. Let the convex polygon A1A2 ...An (n 4) be such that for any way of breaking it by nonintersecting diagonals into triangles, the sum of the inradiuses of these triangles is the same. Then, note that quadrilateral A1A2A3A4 also satisﬁes that condition and using problem 6.1.12b we deduce that quadrilateral A1A2A3A4 is inscribed. Similarly, one can prove that for n 5 quadrilateral AiAi +1Ai +2Ai +3, i ¼ 2, ..., n À 3, is inscribed, consequently, polygon A1A2 ...An is inscribed. This ends the proof.

8.1.14. Let A0 , B0 , C0 be the midpoints of sides BC, CA, AB, respectively, and let points A1, A0 be different. Let point A1 belong to segment BA0.IfB1 belongs to segment CB0,thenC1 belongs to segment BC0. Otherwise, we have that ∠C ¼ ∠ AB0C0 ∠ AB1C1, ∠A ¼ ∠ BC0A0 ∠ BC1A1,and∠B ¼ ∠ B0A0C > ∠ CA1B1. Thus, 180 ¼ ∠ C + ∠ A + ∠ B > ∠ AB1C1 + ∠ BC1A1 + ∠ CA1B1. This leads to a contradiction. We obtain that ΔBC1A1 ΔBC0A0 ¼ ΔAC0B0 ΔAC1B1, thus conditions (a), (b), and (c) are wrong (see problem 2.1.1 and the proof of problem 6.1.8b). If B1 belongs to segment AB0, and C1 belongs to segment AC0, then ΔAB1C1 ΔAB0C0 ¼ ΔCA0B0 ΔCA1B1. This leads to a contradiction. If B1 belongs to segment AB0, and C1 belongs to segment BC0, then ΔBC1A1 ΔBC0A0 ¼ ΔCA0B0 ΔCA1B1. This leads to a contradiction. Similarly, one can prove that if A1 belongs to segment CA0, then this case also leads to a contradiction. Hence points A1, B1 and C1 are the midpoints of sides BC, CA, and AB. ∠CDE 8.1.15. Let ¼ k, ∠BDE ¼ y, ∠AED ¼ x, thus ∠CDE ¼ ky and ∠CED ¼ kx. ∠BDE 420 8 Some Applications of Geometric Inequalities

Figure 8.1 C

kx ky E D y x M

A B

þ ¼ π ¼ þ ¼ (a) We have that kx x 2 ky y. Consequently, x y. Since points D and ¼ π À ∠ ¼ π À ∠ E are on the circle with a diameter AB, then x 2 A, y 2 B. Hence, ∠A ¼ ∠ B, therefore AC ¼ BC. (b) We have that ∠A ¼ ky, ∠B ¼ kx, ∠MAB ¼ x, ∠ MBA ¼ y, x ¼ y. Let x y. > > > If x y, then AC BC and MB MA (see Figure 8.1). 4 2AB2 þ 2BC2 À AC2 2AC2 þ 2AB2 À BC2 Thus, MB2 À MA2 ¼ À < 0. 9 4 4 This leads to a contradiction. Hence, x ¼ y, then ∠A ¼ ∠ B. Therefore, AC ¼ BC. (c) Note that ∠A ¼ 2(ky À x) and ∠B ¼ 2(kx À y), ∠C ¼ π À kx À ky. Consequently, 2(ky À x)+2(kx À y)+π À kx À ky ¼ π, kx + ky ¼ 2(x + y), k ¼ 2. Let N be the intersection point of the bisectors of triangle CDE, thus CN and CM are bisectors of angle ACB. Consequently, C, N, M belong to the same line. ⊥ ⊥ π À ¼ π À ¼ ∠ ¼ ∠ Since MN DE, then CM DE. Hence 2 2y 2 2x, y x, thus A B. This ends the proof.

8.1.16. (a) Let A1, B1, C1, D1 be the midpoints of the sides AB, BC, CD, DA, respectively. Then, since quadrilateral A1B1C1D1 is a parallelogram, we have that AC2 þ BD2 2A B2 þ 2B C2 ¼ A C2 þ B D2. Therefore, we obtain that ¼ 1 1 1 1 1 1 1 1 2 2 þ 2 A1C1 B1D1. p p According to problem 1.1.9a, we have that A1C1 2 , B1D1 2 . Consequently, 2 2 2 AC + BD p . The equality holds true, if AB k B1D1 k CD and BC k A1C1 k AD, which means that ABCD is a parallelogram and as AB + CD ¼ BC + AD, then AB ¼ BC ¼ CD ¼ AD. This means that ABCD is a rhombus. (b) Note that ∠A + ∠ D < 180, ∠A + ∠ B < 180, consequently 2 ∠ A + ∠ B + ∠ D < 360. Hence, it follows that ∠A < 90. Segment AC intersects segment EF, consequently AEF is an acute triangle. 8.1 Application of Geometric Inequalities for Solving Geometric problems 421

Figure 8.2 E

B B0

A C 0 C К

D0 D

Let C0 be the orthocenter of that triangle. If points C0 and C do not coincide, then ∠B < 90 and ∠D < 90 or ∠B > 90 and ∠D > 90. Hence, it follows that 180 ¼ ∠ B + ∠ D < 180 (see Figure 8.2) or 180 ¼ ∠ B + ∠ D > 180. Both these inequalities are wrong. Thus, points C0 and C coincide. Hence, it follows that ∠B ¼ 90, which means that AC is the diameter of circle ω. 8.1.17. (a) (Proof of Hayk Sedrakyan, ninth grade) Let BC ¼ a, AB ¼ c, AC ¼ b ∠A ¼ 2α, ∠ B ¼ 2β, ∠ C ¼ 2γ. bc bc We have that BD ¼ AE ¼ and EC ¼ AD ¼ . On the other hand, a þ b a þ c BD2 ¼ AB Á BC À AD Á DC, EC2 ¼ BC Á AC À AE Á EB. b2c ab2 c2b c2a Therefore, ¼ a À , ¼ a À . ðÞa þ b 2 ðÞa þ c 2 ðÞa þ c 2 ðÞa þ b 2 x2y Let b ¼ xa and c ¼ ya, then from the last equalities, we deduce that ðÞx þ 1 2 x2 xy2 y2 ¼ 1 À and ¼ 1 À .Asb + c > a,thenx + y > 1. ðÞy þ 1 2 ðÞy þ 1 2 ðÞx þ 1 2 x2y xy2 y2 x2 Note that À ¼ À , consequently, (y À x) ðÞx þ 1 2 ðÞy þ 1 2 ðÞx þ 1 2 ðÞy þ 1 2 (x2y2 À xy) ¼ (y À x)(y3 + xy2 + x2y + y3 +2(y2 + xy + x2)+y + x). Hence y ¼ x or x2y2 ¼ y3 + xy2 + x2y + x3 +2y2 +3xy +2x2 + x + y. Otherwise, x2y2 > xy2 + yx2 +3xy or xy > x + y +3> 4. We have obtained that bc > 4a2. By the law of sines, it follows that sin2β sin 2γ > 4sin22α. Note that BD AE sin 2α sin γ ¼ , hence ¼ . Therefore, sin2β sin 2γ > 4 sin β sin γ or AD EC sin β sin 2α cosβ cos γ > 1. This leads to a contradiction. 422 8 Some Applications of Geometric Inequalities

We have that x ¼ y, which means that b ¼ c. Thus, AD ¼ AE ¼ BD, consequently, β ¼ γ ¼ 2α. Hence, ∠A ¼ 36, ∠ B ¼ 72, ∠ C ¼ 72. AD CE (b) See the notations of problem 8.1.17a. We have that 2α 36 and ¼ . BD AE sin β sin 2α Hence ¼ , sinβ sin γ ¼ sin22α. sin 2α sin γ Let β ¼ φ + x, γ ¼ φ À x, where Àφ < x < φ, β + γ 72, φ 36. We have that sin(φ + x) sin (φ À x) ¼ sin24φ, consequently, cos2x ¼ 2sin24φ + cos 2φ. Note that 2sin24φ + cos 2φ 1, since φ 4φ 180 À φ, sin4φ sin φ. Therefore cos2x 1. Consequently, x ¼ 0 and 4φ ¼ 180 À φ. Hence ∠A ¼ 36, ∠B ¼ 72, ∠C ¼ 72. This ends the proof. ∠ ¼ γ ∠ ¼ À γ ∠ ¼ À γ ∠ ¼ À 8.1.18. Let ACB , thus DAE 90 , DCE 90 2, AEC 180 3γ > 2 .IfDE AD, then from triangles ADE and DEC, we obtain that ∠ < ∠ ¼ À γ ∠ < ∠ ¼ À γ AED DAE 90 , DEC DCE 90 2. Therefore, it follows ∠ < À 3γ that AEC 180 2 . This leads to a contradiction. Similarly, we obtain that, if DE < AD, then this also leads to a contradiction. ¼ ∠ ¼ À γ Hence DE DC. Thus, it follows that DEC 90 2. ∠ ¼ À γ ¼ Consequently, EFC 90 2, therefore CE CF. This ends the proof. 8.1.19. Let M be the intersection point of diagonals AD, CF (see Figure 8.3), and points P and Q be the midpointspffiffiffi of sidespffiffiffiCD and AF, respectively. pffiffiffi 3 3 3 As MP þ MQ PQ ¼ CD þ AF, then either MP CD or pffiffiffi 2 2 2 3 MQ AF. 2 pffiffiffi 3 Let MP CD, we need to prove that ∠CMD 60. Indeed, if pffiffiffi 2 3 MP ¼ CD, then using the law of cosines and the formula for the median, we 2 obtain that

MC2 þ MD2 À CD2 CD2 CD2 1 cos ∠CMD ¼ ¼ ¼ : 2 Á MC Á MD 2 Á MC Á MD M2C þ MD2 2

Figure 8.3 CPD

B ME

AQ F 8.1 Application of Geometric Inequalities for Solving Geometric problems 423 pffiffiffi 3 Consequently, ∠CMD 60.IfMP > CD let us take a point M on segment pffiffiffi 2 0 3 PM, such that M P ¼ CD. Thus, as we have proven above, ∠CM D 60. 0 2 0 We have that ∠CMD < ∠ CM0D, hence ∠CMD < 60 . Similarly, we obtain that ∠BNC 60 and ∠BKA 60, where N ¼ CF \ BE, K ¼ AD \ BE. Since ∠CMD + ∠ BNC + ∠ BKA ¼ 180, then ∠CMD ¼

∠ BNC ¼ ∠ BKA ¼ 60. Note that, the equality ∠CMD ¼ 60 is possible if MP pffiffiffi pffiffiffi 3 3 þMQ ¼ CD þ AF and MC ¼ MD. Thus, we deduce that ∠MCD ¼ 2 2 pffiffiffi 3 ∠ DMC ¼ 60, MQ ¼ AF, hence ∠MAF ¼ ∠ MFA ¼ 60. 2 Similarly, we obtain that ∠NBC ¼ ∠ NCB ¼ ∠ NEF ¼ ∠ NFE ¼ 60 and ∠KAB ¼ ∠ KBA ¼ ∠ KED ¼ ∠ KDE ¼ 60,hence∠A ¼ ∠ B ¼ ... ¼ ∠ F ¼ 120. See also problem 4.1.4b.

8.1.20. Let us denote the incenters of triangles ABD and ACD by O1 and O2, respectively, and the common tangent of these incircles, different from AD,by l (see Figure 8.4). Note that l and AD are symmetric to each other with respect to O1O2, thus l and BC are parallel, if and only if ∠CBD À ∠ ADB ¼ 2(∠O2O1D À ∠ O1DA). This means that

∠CBD ¼ 2∠O2O1D: ð8:2Þ

C l

Figure 8.4 424 8 Some Applications of Geometric Inequalities

If ABCD is an inscribed quadrilateral, then ∠ABD ¼ ∠ ACD. Thus, it follows ∠ ¼ þ 1 ∠ ¼ þ 1 ∠ ¼ ∠ that AO1D 90 2 ABD 90 2 ACD AO2D. Consequently, points A, O1, O2, D belong to the same circle. Therefore, we have obtained that ∠CBD ¼ ∠ CAD ¼ 2 ∠ O2AD ¼ 2 ∠ O2O1D, which means that l k BC. Let l k BC, prove that quadrilateral ABCD is inscribed. Denote by ∠CAD ¼ α, ∠CAB ¼ α1, ∠ABD ¼ β, ∠DBC ¼ β1, ∠ACB ¼ γ, ∠ACD ¼ γ1, ∠CDB ¼ δ, ∠BDA ¼ δ1. According to the law of sines, we have that γ δ α β ¼ AB Á BC Á CD Á DA ¼ sin Á sin Á sin Á sin : 1 α β γ δ BC CD DA AB sin 1 sin 1 sin 1 sin 1

Consequently, α β γ δ ¼ α β γ δ : ð : Þ sin sin sin sin sin 1 sin 1 sin 1 sin 1 8 3 β 1 We have that l k BC, thus according to (8.2) we obtain that ∠O2O1D ¼ and γ 2 ∠AO O ¼ (the proof is similar to the proof of (8.2)). 2 1 2 α α β On the other hand, ∠O AD ¼ , ∠O AO ¼ 1, ∠AO D ¼ 90 þ , 2 2 2 1 2 1 2 γ δ δ ∠AO D ¼ 90 þ 1, ∠O DA ¼ 1, ∠O DO ¼ . Now, by (8.3), for the quadri- 2 2 1 2 2 1 2 lateral AO1O2D we obtain that α β γ δ α β γ δ sin cos sin sin ¼ sin 1 sin 1 cos 1 sin 1 : ð8:4Þ 2 2 2 2 2 2 2 2

Dividing equality (8.3) by equality (8.4), we deduce that α β γ δ α β γ δ cos sin cos cos ¼ cos 1 cos 1 sin 1 cos 1 : ð8:5Þ 2 2 2 2 2 2 2 2

Now, let us prove that β ¼ γ1. Indeed, let β 6¼ γ1, thus without loss of generality one can assume that β < γ1. In that case point C is inside of the circumcircle of triangle ABD. Therefore, α > β1 and γ > δ1, thus from (8.4) we deduce that β δ α γ α γ δ β < 1 1 1 1 < cos 2 sin 2 sin 2 cos 2 , while from (8.5) we have that cos 2 sin 2 cos 2 sin 2. δÀβ α Àγ < 1 1 Adding these two last inequalities, we obtain that sin 2 sin 2 . This leads to a contradiction, as δ À β ¼ α1 À γ1 Therefore, β ¼ γ1. Thus, ABCD is an inscribed quadrilateral. This ends the proof. (c) We are going to use the following lemma. 8.1 Application of Geometric Inequalities for Solving Geometric problems 425

Figure 8.5

Lemma Let point P be inside of a convex quadrilateral ABCD (see Figure 8.5). Given that ∠PAD + ∠ PDA + ∠ PBC + ∠ PCB 180 and ∠BPC ¼ ∠ BAP + ∠ CDP. Prove that AD + BC AB + CD. Proof of the Lemma Indeed, let ∠BAP ¼ x, then ∠CDP ¼ ∠ BPC À x. We have that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jjþ ðÞ∠ À ðÞþ ∠ 2 þ ðÞ∠ 2 ¼ AP cospxffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiDP cos BPC x AP DPpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficos BPC DP sin BPC ¼ AP2 þ DP2 þ 2AP Á DP cos ∠BPC AP2 þ DP2 À 2AP Á DP cos ∠APD ¼ AD, therefore

jjAP cos x þ DP cos ðÞ∠BPC À x AD: ð8:6Þ

According to (8.6), we obtain that AD DN À AM and BC BM À CN. Hence, we deduce that AD + BC AB + CD. According to the lemma, we have that AD + BC AB + CD and AB + CD AD + BC. Therefore AB + CD ¼ AD + BC. Remark One can prove that point I is the incenter of quadrilateral ABCD. 8.1.22. (a) We need to prove that for any hexagon ABCDEF, the inequality (AB + DE)2 +(AF + CD)2 +(BC + EF)2 AD2 + BE2 + CF2 holds true and that the equal- ! ! ! ! ! ! ! ! ity holds true, if and only if AB "" ED, CD "" AF , BC "" FE , and AB þ CD ! þ EF ¼ ~0. 426 8 Some Applications of Geometric Inequalities

! ! ! ! ! Indeed, let AB ¼ ~a, BC ¼ ~b, CD ¼ ~c, DE ¼ ~d, EF ¼ ~e. Thus, we have that

AB þ DE ¼ jj~a þ ~d ~a À ~d , BC þ EF ¼ ~b þ jj~e ~b À~e , ð : Þ 8 7 CD þ AF ¼ jj~c þ ~a þ~b þ~c þ ~d þ~e ~a þ~b þ~c þ~c þ ~d þ~e :

Therefore, we deduce that

ðÞþ 2 þ ðÞþ 2 þ ðÞþ 2 AB DE AFCD BC EF 2 2 2 ~a À ~d þ ~b À~e þ ~a þ~b þ~c þ~c þ ~d þ~e ¼ 2 2 2 ¼ ~a þ~b þ~c þ ~b þ~c þ ~d þ ~c þ ~d þ~e þ ðÞ~a þ~c þ~e 2 2 2 2 ~a þ~b þ~c þ ~b þ~c þ ~d þ ~c þ ~d þ~e ¼ AD2 þ BE2 þ CF2, hence

ðÞAB þ DE 2 þ ðÞAF þ CD 2 þ ðÞBC þ EF 2 AD2 þ BE2 þ CF2: ð8:8Þ

Note that the equality in (8.8) holds true, if and only if it holds true in the inequalities of (8.7) and~a þ~c þ~e ¼ ~0, which means that~a "# ~d,~b "# ~e,~c "" ~a þ~b þ~c þ ~d þ~e and ~a þ~c þ~e ¼ ~0. From the assumptions of the problem, it follows that (AB + DE)2 +(AF + CD)2 +(BC + EF)2 ¼ AD2 + BE2 + CF2. Therefore, AB k ED, BC k EF, CD k AF and ! ! ! AB þ CD þ EF ¼ ~0. Consider now parallelogram ABKF, then CDEK is also a parallelogram, consequently, K 2 BE and K 2 CF. FK KE EF AB CD EF Thus ΔEKF ΔBKC, hence ¼ ¼ ,or ¼ ¼ . KC BK BC DE AF BC This ends the proof. Second Solution Let AD ¼ p, BE ¼ q, CF ¼ r, and BE \ CF ¼ M.Asq + r ¼ BM + ME + CM + MF ¼ BM + CM + ME + MF > BC + EF ¼ AD ¼ p, then q + r > p. Similarly, we obtain that p + q > r, p + r > q. Thus, it follows that one can construct a triangle with sides p, q, r. Note that α α1 (see Figures 8.6 and 8.7). Indeed, if α < α1, then

cos α > cos α1: ð8:9Þ

Let us consider parallelogram BCKE (Figure 8.6). We have that FK2 ¼ r2 + q2 2 2 2 À 2rq cos α1 > r + q À 2rq cos α ¼ p , therefore FK > p. On the other hand FK FE + EK ¼ FE + BC ¼ AD ¼ p. This leads to a contradiction. Similarly, we obtain that β β1 and γ γ1.Asα + β + γ ¼ 180 ¼ α1 + β1 + γ1, then α ¼ α1 , β ¼ β1 , γ ¼ γ1. 8.1 Application of Geometric Inequalities for Solving Geometric problems 427

Figure 8.6

Figure 8.7

We have that α ¼ α1, therefore FK ¼ p ¼ FE + EK, which means that FE||BC. As β ¼ β1, then AD||FE (Figure 8.6). Similarly, we obtain that CD||BE||AF and AB||CF||DE, then quadrilaterals ABMF EF ME MF and DCME are parallelograms. As ΔFME ΔBMC, then ¼ ¼ ,or BC BM CM EF ¼ CD ¼ AB : BC AF DE (b) Let diagonals AD and CF, CF and BE, BE and AD intersect at points M, N, P, respectively (Figure 8.8). Let point O intersects the bisector of triangle MNP,if points M, N, P coincide, then point O coincides with point M.

Let A1, B1, C1, D1, E1, F1 be the midpoints of sides AB, BC, CD, DE, EF, FA of a hexagon and da, db, dc, dd, de, df be the distances from point O to lines AB, BC, CD, DE, EF, FA, respectively. Note that A0D0 da + dd, B0E0 db + de, C0F0 dc + df, therefore

A0D0 þ B0E0 þ C0F0 da þ db þ dc þ dd þ de þ df : ð8:10Þ

On the other hand, we have that ∠A1A0D0 90 and ∠D1D0A0 90 , as in any triangle the bisector is between the altitude and the median (if the bisector, altitude, and median are drawn from the same vertex of a triangle). Therefore, the projection of segment A1D1 on line A0D0 contains segment A0D0, which means that A1D1 A0D0. Similarly, we obtain that B1E1 B0E0, C1F1 C0F0. 428 8 Some Applications of Geometric Inequalities

Figure 8.8

Hence, from the last three inequalities and (8.10), we obtain that

A1D1 þ B1E1 þ C1F1 da þ db þ dc þ dd þ de þ df : ð8:11Þ

From the assumptions of the problem, it follows that (8.11) is an equality. Note that, it holds true if points A1, B1, C1, D1, E1, F1 coincide with points A0, B0, C0, D0, E0, F0, respectively. Then, we obtain that lines B0C0, C0F0, A0D0 are the mid-perpendiculars of segments BC and EF, CD and AF, DE and AB, respectively. Then, point O is equidistant from points A, B, C, D, E, F. This means that ABCDEF is a cyclic hexagon. 8.1.23. Let us externally construct an equilateral triangle BCF on side BC of a convex quadrilateral ABCD. Note that ∠FBC ¼ 60 ¼ ∠ C0BD, therefore ∠FBC0 ¼ ∠ CBD. On the other hand, we have that BF ¼ BC and BC0 ¼ BD, then ΔBFC0 ¼ ΔBCD. Thus, it follows that FC0 ¼ CD and ∠BFC0 ¼ ∠ BCD. Similarly, we obtain that B0F ¼ AB and ∠B0FC ¼ ∠ ABC (Figure 8.9). We have that B0C0 ¼ AB + CD, therefore B0C0 ¼ B0F + FC0. Hence, it follows that point F is on segment B0C0. We deduce that ∠B0FC + ∠ BFC0 ¼ 180 + ∠ BFC ¼ 240. Thus, we obtain that ∠BCD + ∠ ABC ¼ ∠ BFC0 + ∠ B0FC ¼ 240 and ∠BAD + ∠ CDA ¼ 120.

Problems for Self-Study

8.1.24. Let O be the intersection point of diagonals AC and BD of a convex quadrilateral ABCD. Given that BO ¼ OD and AB + BC ¼ AD + DC. Prove that either BD ⊥ AC or ABCD is a parallelogram. 8.1.25. Prove that, if in triangle ABC holds true p ¼ 2R + r, then triangle ABC is right angled. 8.1 Application of Geometric Inequalities for Solving Geometric problems 429

Figure 8.9

Hint See problems 5.5.7a and 5.5.8a. ÀÁpffiffiffi 8.1.26. Find the angles of triangle ABC,ifla ¼ R 1 þ 2 r. 8.1.27. Let ABCD be a convex quadrilateral. Prove that if MNPQ is an inscribed quadrilateral, then RSTU is also an inscribed quadrilateral, where M, N and P, Q are tangential points of incircles of triangles ABC and ACD with sides AB, BC and CD, AD, respectively, R, S and T, U are the tangential points of the incircles of triangles ABD and BCD with sides AD, AB and BC, CD, respectively. Hint See problem 8.1.10.

8.1.28. Prove that if in triangle ABC holds true la ¼ lb, then a ¼ b. 8.1.29. The altitudes drawn from the vertices A and C of triangle ABC intersect the median BM at points E and F, respectively. Given that AE ¼ CF. Prove that ABC is an isosceles triangle. 8.1.30. Prove that, if three bisectors of one triangle are equal, respectively to three bisectors of another triangle, then these triangles are congruent.

8.1.31. The base A1A2 ...An of pyramid PA1A2 ...An is a regular polygon. Prove that if ∠PA1A2 ¼ ∠ PA2A3 ¼ ... ∠ PAn À 1An ¼ ∠ PAnA1, then the pyramid is regular. 8.1.32. Let O be the intersection point of diagonals AC and BD of a convex quadrilateral ABCD. Given that the perimeters of all triangles AOB, BOC, COD, DOA are equal. Prove that ABCD is a rhombus. 8.1.33. Let O be the intersection point of the diagonals of a convex quadrilateral ABCD. Prove that if the inradiuses of triangles AOB, BOC, COD, and DOA are equal to each other, then ABCD is a rhombus. 8.1.34. Given points D, E on sides AB, AC of triangle ABC, respectively. Prove that if DP ¼ PE and AB ¼ AC, where P is the intersection point of segments BE and CD, then BP ¼ PC. Hint If BP > PC, then consider a point B0 on ray PC, such that PB0 ¼ PB. Thus, ∠PCB > ∠ PBC and ∠PCE > ∠ PB0E ¼ ∠ PBD. 430 8 Some Applications of Geometric Inequalities

8.2 Using Geometric Inequalities for Proving Algebraic Inequalities

8.2.1. Prove the following inequalities (a) |sinx| |x|, π (b) tgx > x, where 0 < x < , 2 π (c) α À sin α < β À sin β, where 0 < α < β < , π 2 (d) tgα À α < tgβ À β, where 0 < α < β < , 2 sin α sin β π (e) > , where 0 < α < β < , α β 2 2 π (f) sin x > x, where 0 < x < , π 2 tgα tgβ π (g) < , where 0 < α < β < , α β pffiffiffi 2 3 3 (h) sin α þ sin β þ sin γ , where α, β, γ > 0 and α + β + γ ¼ π. 2 8.2.2. Prove the inequalities π þ þ > þ þ < < (a) 2 2 sin x cos y 2 sin y cos z sin 2x sin 2y sin 2z, where 0 x y < z < π, 2 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (b) 0:785n2 À n < n2 À 1 þ n2 À 22 þ ...þ n2 À ðÞn À 1 2 < 0:79n2, where n 2 N. 1 À h Xn 1 þ h 8.2.3. Prove that the inequality < x2iðÞx2iþ1 À x2iÀ1 < holds true 2 i¼1 2 for real numbers 0 ¼ x1 < x2 < ... < x2n < x2n +1¼ 1, such that xi +1À xi h, 1 i 2n.

8.2.4. Let the real numbers a1, ..., an, an +1 and b1, ..., bn be such that 0 bk 1, k ¼ 1, ..., n, and a1 a2 ... an +1¼ 0. Let [b1 + ... + bn] ¼ p. Prove that a1b1 + a2b2 + ... + anbn a1 + a2 + ... + ap + ap +1. 8.2.5. Prove the inequalities (a) a2 + b2 2ab, where a, b > 0, 3 3 3 > (b) a + b + c 3abc, wheresa,ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib, c 0, 2 pffiffiffiffiffi a þ b a2 þ b2 (c) ab , where a, b > 0. 1 þ 1 2 2 a b 8.2 Using Geometric Inequalities for Proving Algebraic Inequalities 431 pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 8.2.6. Prove that 6 3abðÞþ c a2 þ c2 þ a2 þ b2 þ b þ c , where a, b, c > 0. 8.2.7. Prove the inequalities (a) ((x + y)(y + z)(x + z))2 xyz(2x + y + z)(2y + x + z)(2z + x + y), where x, y, z > 0, a b c a þ b þ c (b) 1 þ 1 þ 1 þ 21þ pffiffiffiffiffiffiffi , where a, b, c > 0, b c a 3 abc 1 1 1 9 (c) þ þ , where x, y, z > 0 and x + y + z ¼ 1. ðÞxy þ z 2 ðÞxz þ y 2 ðÞyz þ x 2 16xyz qffiffiffiffiffiffi qffiffiffiffiffiffi qffiffiffiffiffiffi 2a þ 2b þ 2c > 8.2.8. Prove the inequality aþb bþc cþa 3, where a, b, c 0. ðÞþ ðÞþ 8.2.9. Find thepffiffiffiffiffiffiffi maximal constant number C, such that the inequality x y y z ðÞz þ x C xyz holds true for all x, y, z > 0 and x + y + z ¼ 1. 8.2.10. Prove that a ðÞþþ b ðÞþþ c ðÞþ > bþc y z cþa x z aþb x y 3, where a + b, b + c, c + a 0, x, y, z > 0, and xy + yz + zx ¼ 3.

8.2.11.pProveffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ þ 2 þ 2 þ þ 2 þ 2 þ þ 2 xy pxffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixy y yz y yz z zx x xz z ðÞx2 þ xy þ y2 ðÞy2 þ yz þ z2 ðÞx2 þ xz þ z2 , where x, y, z > 0.

8.2.12.pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiProve that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ þ 2 þ 2 þ þ 2 þ 2 þ þ 2 xpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixy y y yz z x xz z 5x2 þ 5y2 þ 5z2 þ 4xy þ 4yz þ 4xz, where x, y, z > 0. 8.2.13. Prove that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi 2 þ þ 2 Á 2 þ þ 2 Á 2 þ þ 2 3 ðÞþ þ 3 > x xy y y yz z z zx x 8 x y z , where x 0, y > 0, z > 0. 8.2.14. Prove that αβγðÞx þ y þ z 2 αyz þ βzx þ γxy , where α > 0, β > 0, 2αβ þ 2βγ þ 2γα À α2 À β2 À γ2 γ > 0 and 2αβ +2βγ +2γα À α2 À β2 À γ2 > 0.

8.2.15.pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiProve that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2ðÞx3 þ y3 þ z3 þ xyz x2 þ y2 þ z2 þ yz, where y > 0, z > 0 and max (x, y, z) ¼ x. 432 8 Some Applications of Geometric Inequalities

Solutions

8.2.1. (a) As sin(Àx) ¼Àsin x and |sinx| 1, then it is sufﬁcient to prove the inequality for 0 < x < 1. We have that (see Figure 8.10) SΔAOB ¼ 1 2 ¼ sin x ¼ OA2 Á ¼ x < < 2 OA sin x 2 , S sec AOB 2 x 2, since SΔAOB SsecAOB. Then, sinx x, hence at 0 < x < 1, we have that |sinx| < |x|. (b) We have that (see Figure 8.10)

¼ OAÁOC ¼ 1 > > SAOC 2 2 tgx, since SΔAOC SsecAOB. Therefore, tgx x.

αÀ sin α < βÀ sin β α À α < β À β Therefore, 2 2 , hence sin sin .

Figure 8.10

Figure 8.11 8.2 Using Geometric Inequalities for Proving Algebraic Inequalities 433

tgαÀα tgβÀβ (d) We have that S _ < S _ (see Figure 8.11), consequently < . Hence B0BA C0CA 2 2 tgα À α < tgβ À β. (e) We have that

β ðÞβ À α α þ ðÞβ À α α sin ¼ sin cos cos sin ¼ β β sin α α 1 sin α 1 sin α ¼ sin ðÞβ À α þ α cos ðÞβ À α Á < ðÞÁðÞþβ À α α ¼ α tgα β α β α

(see problems 8.2.1a and b). (f) We have that

π π π sin sin À x cos x þ sin x cos À x π 2 ¼ 2 π 2 ¼ 2 2 sin x π x π 2 sin x π 2 ¼ sin À x þ x cos À x Á < À x þ x Á , x 2 tgx 2 π x 2 π

> 2 hence sin x π x. tgβ cos ðÞβ À α 1 (g) We have that ¼ tg ðÞβ À α þ 1 > tg ðÞÁβ À α þ 1 > tgα cos β sin α sin α β À α tgα tgβ þ 1, hence < . α α β pffiffiffi 3 3 (h) Let α β γ.Ifα ¼ γ, then sin α þ sin β þ sin γ ¼ . 2 π If α 6¼ γ, then α < < γ. Let the triangle with angles α, β, γ is inscribed in the 3 circle with the radius 0.5 (see Figure 8.12). β π β As ∠DD0C ¼ ∠DAC ¼ α þ and ∠CDD0 ¼ þ , then ∠DD0C < ∠ CDD0. 2 3 2 π ÀÁ < 0 α þ γ < þ α þ γ À π Therefore CD CD , which means that sin sin sin sin 3 . 3π π π In the same way, we obtain that sin β þ sin α þ γ À sin þ sin , 3 3 3 consequently, π π π sin α þ sin γ þ sin β < sin þ sin β þ sin α þ γ À 3 sin : 3 3 3 pffiffiffi 3 3 Hence, sin α þ sin γ þ sin β < . 2 þ þ < π ¼ 8.2.2. (a) Note that S1 S2 S3 4 (see Figure 8.13) and S1 sin x cos x, S2 ¼ cos y(siny À sin x), S3 ¼ cos z(sinz À sin y). 434 8 Some Applications of Geometric Inequalities

Figure 8.12 D D¢

b b 2 2

B¢ B b b

p g a 3 CA

Figure 8.13 D(cosz;sinz)

S3 C(cosy;siny) S2 B(cosx;sinx) yS1 xz A(1;0)

π > þ ðÞþÀ ðÞÀ Therefore, 4 sin x cos x cos y sin y sin x cos z sin z sin y , hence π þ þ > þ þ 2 2 sin x cos y 2 sin y cos z sin 2x sin 2y sin 2z. pffiffiffiffiffiffiffiffiffiffiffiffiffi πn2 (b) Note that S þ S þ ...þ S À < (see Figure 8.14a), S ¼ n2 À 1 , pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 2 nq1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4 1 2 2 2 2 S2 ¼ n À 2 , ...,,SnÀ1 ¼ n À ðÞn À 1 .

pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 À þ 2 À 2 þ ...þ 2 À ðÞÀ 2 < π 2 < : 2 Therefore, n 1 n 2 n n 1 4 n 0 79n . 8.2 Using Geometric Inequalities for Proving Algebraic Inequalities 435

Figure 8.14

Figure 8.15

π 2 Note that S0 þ S1 þ ::: þ SnÀ1 > n (see Figure 8.14b) and S0 ¼ n, pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 2 S1 ¼ n À 1 S2 ¼ n À 2 , ..., SnÀ1 ¼ n À ðÞn À 1 , consequently, pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 À þ 2 À 2 þ ...þ 2 À ðÞÀ 2 > π 2 À > : 2 À n 1 n 2 n n 1 4 n n 0 785n n. 8.2.3. Note that (see Figure 8.15)

Xn 2 1 ðÞx2 À x1 x2iðÞ¼x2iþ1 À x2iÀ1 S1 þ S2 þ ...þ Sn < þ þ ... i¼1 2 2 2 ðÞx2n À x2nÀ1 1 h 1 h þ þ ðÞx À x þ ...þ x À x À < þ 2 2 2 2 1 2n 2n 1 2 2 436 8 Some Applications of Geometric Inequalities

We have that

2 2 1 ðÞx À x ðÞx þ À x S þ S þ ...þ S > À 1 0 À ...À 2n 1 2n 1 2 n 2 2 2 1 h 1 h À ðÞx À x þ ...þ x þ À x > À : 2 2 1 0 2n 1 2n 2 2

8.2.4. Note that the sum a1b1 + a2b2 + ... + anbn is equal to the area of the shaded figure (see Figure 8.16), while the sum a1 + ... + ap +1 is equal to the area of the second figure, containing the first figure. Therefore, a1b1 + a2b2 + ... + anbn a1 + a2 + ... + ap + ap +1. This ends the proof. þ ¼ a2 þ b2 8.2.5. (a) See Figure 8.17. We have that ab S1 S2 2 2 . (b) See Figure 8.18.

¼ ¼ ¼ Let V1 VABB2C2D2 , V2 VADB3C3D3 , V3 VAA1B4C4D4 . ¼ þ þ ¼ a3 þ b3 þ c3 We have that abc V V1 V2 V3 3 3 3 . x þ y þ z 3 Remark If x, y, z > 0, then xyz . pffiffiffi p3ffiffiffi pffiffi It is sufficient to take a ¼ 3 x, b ¼ 3 y, c ¼ 3 z .

Figure 8.16

Figure 8.17 8.2 Using Geometric Inequalities for Proving Algebraic Inequalities 437

B4 c C4

C3 B1 C1

A1 D1 D4

c D 3 B2 C2 b

ab BD2 C a

A b D

Figure 8.18 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 pffiffiffiffiffi a þ b a2 þ b2 (c) If a ¼ b, then ¼ ab ¼ ¼ . 1 þ 1 2 2 a b If a 6¼ b, then without loss of generality, one can assume that a > b. a À b pffiffiffiffiffi Let AC ¼ , CB ¼ ab, ∠C ¼ 90 (see Figure 8.19). 2 a þ b ab 2 Thus, AB ¼ and HB ¼ ¼ , consequently, HB < CB < AB, that is, aþb 1 1 2 2 þ pffiffiffiffiffi a b 2 < < aþb ab 2 . 1 þ 1 a b ¼ aÀb ¼ aþb ∠ ¼ Let AC 2q,ffiffiffiffiffiffiffiffiffiBC 2 , C 90 (see Figure 8.20). qffiffiffiffiffiffiffiffiffi ¼ a2þb2 < aþb < a2þb2 Thus AB 2 , and BC AB, which means that 2 2 . 8.2.6. We have that (see Figure 8.21.) 438 8 Some Applications of Geometric Inequalities

Figure 8.19 A

ab- 2

C ab B

Figure 8.20 A

Figure 8.21 B

ma n

A b D c C p

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi abðÞþ c 1 S ¼ ¼ ðÞm þ n þ p ðÞm þ n À p ðÞm À n þ p ðÞn þ p À m ABC 2 16 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m þ n þ p 3 1 ðÞm þ n þ p ¼ pffiffiffi ðÞm þ n þ p 2 ¼ 16 3 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3 1 2 ¼ Á 2 þ 2 þ 2 þ 2 þ þ pffiffiffi3 a b a c b c , 4 3

(see the remark of problem 8.2.5b). 8.2 Using Geometric Inequalities for Proving Algebraic Inequalities 439

Therefore, we obtain that pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 6 3abðÞþ c a2 þ b2 þ a2 þ c2 þ b þ c . Note that the equality holds pffiffiffi pffiffiffi true, if and only if 3b ¼ 3c ¼ a. 8.2.7. (Proof of Hayk Sedrakyan, ninth grade) (a) Without loss of generality, one can assume that xyz ¼ 1. Let x + y ¼ a, y + z ¼ b, z + x ¼ c, thus we have to prove that (a + b)(b + c)(a + c) a2b2c2. Since a, b, c are the sides of some triangle, its area denoted by S,wehavethatS2 a þ b þ c a þ b À c a À b þ c Àa þ b þ c a þ b þ c ¼ Á Á Á ¼ and a2b2c2 ¼ 16S2R2. 2 2 2 2 2 Therefore, we have to prove that 2(sinα + sin β + sin γ) (sinα + sin β)(sinα + sin γ)(sinβ + sin γ). According to the remarks of problems 8.2.5b and 8.2.1h, it follows that 3 ðÞα þ β ðÞα þ γ ðÞβ þ γ 2ðÞ sin αþ sin βþ sin γ ¼ sin sin sin sin sin sin 3 8 ¼ ðÞsin α þ sin β þ sin γ ÁðÞsin α þ sin β þ sin γ 2 27 8 27 ðÞsin α þ sin β þ sin γ Á ¼ 2ðÞ sin α þ sin β þ sin γ : 27 4

(b) Without loss of generality, one can assume that abc ¼ 1. ¼ ¼ ¼ Let a +pbffiffiffi x, b + c y, a + c z, then x, y, z are the sides of some triangle of the area S ¼ p (see the proof of problem 8.2.7a). We have to prove that xyz 2+x + y + z or 2RS 1+p. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 Á 2 ¼ Accordingrffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi to problems 8.2.1h and 5.3.1, we have that 2RS 4R S pffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 3 2 Á 4 3 48p2 27p2þ63p þ ¼ xþyþz 3 xyz ¼ 4R p 4 27 p 27 27 p 1,asp 2 2 3, 1 xþyÀz Á yþzÀx Á xþzÀy xyz 2 2 2 8 (see the proof of problem 7.1.13). (c) We have to prove that

1 þ 1 þ 1 9 ðÞ1 À x 2ðÞ1 À y 2 ðÞ1 À z 2ðÞ1 À x 2 ðÞ1 À y 2ðÞ1 À z 2 16xyz or 16ðÞx þ y þ z xyzðÞ x þ z 2 þ ðÞy þ z 2 þ ðÞx þ y 2 9ðÞx þ z 2ðÞy þ z 2ðÞx þ y 2: 440 8 Some Applications of Geometric Inequalities

Figure 8.22 B

2x R

O G

AMC

Let x + y ¼ a, y + z ¼ b, x + z ¼ c, where xyz ¼ 1. Therefore, S2 ¼ p, thus we have to prove that 8(a + b + c)(a2 + b2 + c2) 9a2b2c2, or equivalently, a2 + b2 + c2 9R2 (see problem 5.1.1). According to Stuart’s theorem, we have that (see Figure 8.22) 2 ¼ 1 Á 2 þ 2 Á 2 À b2 À 2 2c2þ2a2Àb2 ¼ 2 À a2þb2þc2 2 GO 3 R 3 R 4 9 4 R 9 0. Hence, a + b2 + c2 9R2. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁ abþbcþac ¼ α abþbcþac ¼ β abþbcþac ¼ γ α β γ 2 ; π 8.2.8. Let a tg , b tg , c tg , where , , 0 2 . Note that tgα + tgβ + tgγ ¼ tgα Á tgβ Á tgγ. tgαþtgβ ¼À γ α β ¼ π À γ α β Therefore, it follows that 1ÀtgαÁtgβ tg ,ortg( + ) tg( ), hence + + γ ¼ π. We have that rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u u 2a 2b 2c u 2 u 2 u 2 þ þ ¼ t þ u þ t ¼ a þ b b þ c c þ a tgα t tgβ tgγ 1 þ 1 þ 1 þ tgβ tgγ tgα 1 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin 2α Á sin β þ sin 2β Á sin γ þ sin 2γ Á sin α sin α sin β sin γ 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞsin 2α þ sin 2β þ sin 2γ ðÞsin 2α þ sin 2β þ sin 2γ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin α sin β sin γ ¼ 4ðÞ sin 2α þ sin 2β þ sin 2γ 3

(see problemsqffiffiffiffiffiffi 5.3.19bqffiffiffiffiffiffi and 5.1.1).qffiffiffiffiffiffi 2a þ 2b þ 2c Hence, aþb bþc cþa 3. This ends the proof. 8.2 Using Geometric Inequalities for Proving Algebraic Inequalities 441

pffiffi ¼ ¼ ¼ 1 8 3 8.2.9.Forx y z 3, we have that C 9 . We need to prove that for x > 0, y > 0, z > 0, the following inequality holds true pffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8 3 ðÞx þ y ðÞy þ z ðÞz þ x xyzðÞ x þ y þ z 3: ð8:12Þ 9

WithoutÀÁ loss of generality, one can assume that x ¼ tgα, y ¼ tgβ, z ¼ tgγ, where α β γ 2 ; π α β γ , , 0 2 and , , are the angles of some triangle. ÀÁ ¼ α ¼ β ¼ γ α β γ 2 ; π Indeed, denote by x tg , y tg , z tg , where , , 0 2 and we can assume that x + y + z ¼ xyz, then α, β, γ are the angles of some triangle. Then, inequality (8.12) is equivalent to the following inequality sin α Á sin β pffiffi Á γ 3 3 sin 8 (see problem 5.1.12). If x, y, z > 0 and x + y + z ¼ 1, then from inequality (8.12), we obtain that pffiffi pffiffiffiffiffiffiffi ðÞx þ y ðÞy þ z ðÞz þ x 8 3 xyz. This means that the greatest value of C is equal pffiffi 9 8 3 to 9 . This ends the proof. > ¼ 1 Remark Similarly, one can prove that if x, y, z 0, x + y + z 1 and k 2, then (x + y)(y + z)(z + x) 8 Á 27k À 1(xyz)k. ¼ ¼ ¼ ¼ nþkÀm ¼ mþkÀn 8.2.10. Denote by b + c m, c + a n, a + b k, then a 2 , b 2 , ¼ nþmÀk > c 2 , note that for m, n, k, x, y, z 0. We need to prove that

n m k m k n ðÞþy þ z ðÞþx þ z ðÞþy þ z ðÞþx þ y ðÞþx þ z ðÞx þ y m n m k n k 6 þ 2x þ 2y þ 2z:

As n m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞþy þ z ðÞx þ z 2 ðÞy þ z ðÞx þ z , m n k m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞþy þ z ðÞx þ y 2 ðÞy þ z ðÞx þ y , m k k n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞþx þ z ðÞx þ y 2 ðÞx þ z ðÞx þ y , n k then it if sufficient to prove that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞx þ y ðÞy þ z þ ðÞx þ y ðÞx þ z þ ðÞx þ z ðÞy þ z 3 þ x þ y þ z or 442 8 Some Applications of Geometric Inequalities pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 3 þ y2 þ 3 þ z2 þ 3 3 þ x þ y þ z: ð8:13Þ pffiffiffi pffiffiffi pffiffiffi ÀÁ ¼ α ¼ β ¼ γ α β γ 2 ; π Denote by x 3ctg , y 3ctg , z 3ctg , where , , 0 2 . From the condition xy + yz + zx ¼ 3, we deduce that α + β + γ ¼ π. Then, inequality (8.13) is equivalent to the following inequality 1 þ pffiffiffi pffiffiffi sin α 1 þ 1 þ α þ β þ γ α þ β þ γ : sin β sin y 3 ctg ctg ctg ,ortg 2 tg 2 tg 2 3 The last inequality holds true, according to problem 5.1.9a. 8.2.11. Let us consider points M, A, B, C on plane, such that MA ¼ x, MB ¼ y, MC ¼ z and ∠AMB ¼ ∠ BMC ¼ ∠ CMA ¼ 120. According to problem 4.1.8c (orp 1.1.14i),ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi we havep thatffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixyAB + yzBC Á Á ¼ 2 þ þ 2 ¼ 2 þ þ 2 + zxACpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiAB BC AC. Note that AB x pxyffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiy , BC ypffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiyz z , AC ¼ 2 þ þ 2 2 þ þ 2 þ 2 þ þ 2þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix xz z ,pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi therefore xy x xy y yz y yz z zx x2 þ xz þ z2 ðÞx2 þ xy þ y2 ðÞy2 þ yz þ z2 ðÞx2 þ xz þ z2 : 8.2.12. We need to prove that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 x2 þ xy þ y2 Á y2 þ yz þ z2 þ 2 y2 þ yz þ z2 Á z2 þ zx þ x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ2 x2 þ xy þ y2 Á z2 þ zx þ x2 3x2 þ 3y2 þ 3z2 þ 3xy þ 3yz þ 3zx, or equivalently pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 x2pþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixy þ y2 Á y2pþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiyz þ z2 þ 2 y2 þ yz þ z2 Á z2 þ zx þ x2 þ 2 x2 þ xy þ y2 Á z2 þ zx þ x2 À pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ÀÁpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 À x2 þ xy þ y2 þ y2 þ yz þ z2 þ z2 þ zx þ x2 ðÞx þ y þ z 2:

Let us consider points M, A, B, C on plane, such that MA ¼ x, MB ¼ y, MC ¼ z, and ∠AMB ¼ ∠ BMC ¼ ∠ CMA ¼ 120. À 2 À 2 À 2 2 Hence, wep needffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi to prove that 2pcaffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi+2ab +2bc a pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib c (MA + MB + MC) , where a ¼ y2 þ yz þ z2, b ¼ z2 þ zx þ x2, c ¼ x2 þ xy þ y2: Note that 4( p À a)( p À b)+4(p À b)( p À c)+4(p À c)( p À a) ¼ 2ac +2ab +2bc À a2 À b2 À 2 ¼ a þ b þ c : c , where p 2 It is sufficient to prove that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðÞp À a ðÞþp À b ðÞp À b ðÞþp À c ðÞp À c ðÞp À a MA þ MB þ MC:

The last inequality holds true according to problem 7.1.107c. This ends the proof. 8.2.13. Let us consider equilateral triangle ABC and point P inside of it, such that PA1 ¼ x, PB1 ¼ y, PC1 ¼ z (Figure 1.7). Thus, it follows that PA ¼ 8.2 Using Geometric Inequalities for Proving Algebraic Inequalities 443 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ þ 2 ¼ 2 þ þ 2 ¼ 2 þ þ 2 y yz z , PB z zx pxffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, PC x pxyffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiy , and accordingpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi to prob- 2 þ þ 2 Á 2 þ þ 2 Á 2 þ þ 2 lempffiffi 7.1.67b, we have that x xy y y yz z z zx x 3 ðÞþ þ 3: 8 x y z This ends the proof. ÀÁpffiffiffi pffiffiffi pffiffiffi 8.2.14. We have that 2αβ þ 2βγ þ 2γα À α2 À β2 À γ2 ¼ α þ β þ γ Â ÀÁpffiffiffi pffiffiffi pffiffiffi ÀÁpffiffiffi pffiffiffi pffiffiffi ÀÁpffiffiffi pffiffiffi pffiffiffi À α þ β þ γ α À β þ γ α þ β À γ > 0, therefore pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi β þ γ > α, α þ γ > β, α þ β > γ: pffiffiffi pffiffiffi pffiffiffi Consider points M, A, B, C on plane, such that BC ¼ α, AC ¼ β, AB ¼ γ, and MA ¼ MB ¼ MC, From problem 4.1.8а, it follows that (x + y + z)2MA2 αyz + βzx + γxy. pffiffi pffiffi pffiffi αÁ βÁ γ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ðÞ¼αβ þ βγ þ γα À α2 À β2 À γ2: We have that MA 4ðÞABC and 4 ABC 2 2 2 αβγðÞþ þ 2 Thus, it follows that αyz þ βzx þ γxy x y z : 2αβ þ 2βγ þ 2γαÀ α2 À β2 À γ2 This ends the proof. 2 2 2 8.2.15. Without loss of generality, one can assumeÀÁ that x + y + z + yz ¼ 1. ¼ β ¼ γ β γ 2 ; π : Denote by y cos , z cos , where , 0 2 2 2 2 2 2 2 Note that 4x x + y + z + yz ¼ 1 and y + z < 1.ÀÁ 1 2β < À 2γ ¼ 2 π À γ : Therefore, x 2 and cos 1 cos cos 2 β þ γ > π : Thus, it follows that 2 Let α ¼ π À β À γ, then cos2α + cos2β + cos2γ + 2 cos α cos β cos γ ¼ 1 (see the proof of problem 5.1.2). We deduce that x2 + y2 + z2 +2xyz x2 + y2 + z2 + yz, and x2 + y2 + z2 + yz ¼ 1 ¼ cos2α + cos2β + cos2γ + 2 cos α cos β cos γ. Hence, we obtain that x cos α. This ends the proof.

Problems for Self-Study

8.2.16. Prove the inequalities (a) x3 + y3 + z3 +3xyz x2y + y2x + y2z + z2y + z2x + x2z, where x, y, z > 0, 4 4 4 4/3 2 2 2 2 2 2 (b) a + b + c +3(abc) 2(a b + b c + a c ), where a, b, c > 0, pffiffiffiffiffiffiffi ÀÁpffiffiffi pffiffiffi 2 ÀÁpffiffiffi pffiffiffi 2 pffiffiffi pffiffiffi a þ b þ c À 3 1 À þ À þ ðÞÀ 2 (c) 3 abc 3 a b b c c a , where a, b, c > 0, ðÞÀ þ þ ðÞþÀ þ ðÞÀ þ ðÞþþ À ðÞÀþ À ðÞþ þ (d) paffiffiffiffiffiffiffibÀÁpffiffiffic apffiffiffib pcffiffiffi a b c a b c a b c a b c abc a þ b þ c , where a, b, c > 0. Hint (b) In inequality 8.2.10a, take x ¼ a4/3, y ¼ b4/3, z ¼ c4/3 and use inequality 8.2.5c. 444 8 Some Applications of Geometric Inequalities

pffiffiffiffiffiffiffi qffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi a þ b þ c > > 8.2.17. Prove the inequality b þ c c þ a a þ b 2, where a, b, c 0. 8.2.18. Prove the inequality (a2 + b2 + c2)(a + b À c)(b + c À a)(c + a À b) abc(ab + bc + ac), where a, b, c > 0. Hint See the proof of problem 7.1.29p.

8.2.19. Prove the inequality 1 þ 1 þ 1 9 , where x, y, x2 þ xy þ y2 y2 þ yz þ z2 z2 þ zx þ x2 ðÞx þ y þ z 2 z > 0. Hint See the proof of problem 7.1.107. 8.2.20. Prove that if x, y, z > 0 and x2 + y2 + z2 +2xyz ¼ 1, then 1 (a) xyz 8, þ þ 3 (b) x y z 2, þ þ 3 2 þ 2 þ 2 (c) xy yz xz 4 x y z , þ þ 1 þ (d) xy yz xz 2 2xyz. ÀÁ ¼ α ¼ β ¼ γ α β γ 2 ; π α β γ ¼ π Hint Let x tg , y tg , z tg , where , , 0 2 , then + + . 8.2.21. Prove the inequalitypffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi xy þ yz þ xz 3 þ 1 þ x2 þ 1 þ y2 þ 1 þ z2, where x, y, z > 0 and x + y + z ¼ xyz. ÀÁ ¼ α ¼ β ¼ γ α β γ 2 ; π α β γ ¼ π Hint Let x tg , y tg , z tg , where , , 0 2 , then + + . 8.2.22. Prove that if x, y, z > 0 and xy + yz + zx +2xyz ¼ 1, then 1 (a) xyz 8, þ þ 3 (b) x y z 2, 1 þ 1 þ 1 ðÞþ þ (c) x y z 4 x y z , 2 1 þ 1 þ 1 À ðÞþ þ ðÞ2zÀ1 ¼ (d) x y z 4 x y z zðÞ2zþ1 , where z max (x, y, z). ÀÁ ¼ 2α ¼ 2β ¼ 2γ α β γ 2 ; π Hint Let xy cos , yz cos , xz cos , where , , 0 2 . Then α + β + γ ¼ π. 8.2.23. Prove that, if x, y, z > 0 and xyz ¼ x + y + z + 2, then (a) xypffiffiffi+ yzp+ffiffiffizx p2(ffiffix + yp+ffiffiffiffiffiffiffiz), þ þ 3 (b) x y z 2 xyz. ÀÁ 1 ¼ 2α 1 ¼ 2β 1 ¼ 2γ α β γ 2 ; π α β Hint Let xy cos , yz cos , xz cos , where , , 0 2 , then + + γ ¼ π. 8.2.24. Prove the inequality 64(x + y + z)3xyz 27(x + y)2(y + z)2(x + z)2, where x, y, z > 0. Hint Let x + y ¼ a, y + z ¼ b, x + z ¼ c. 8.2 Using Geometric Inequalities for Proving Algebraic Inequalities 445

8.2.25. Prove the inequality (3 À 2a)(3 À 2b)(3 À 2c) a2b2c2, where a, b, c > 0 and a + b + c ¼ 3. 8.2.26. Prove the inequality

φ þ φ þ ::: þ φ þ φ x1x2 cos 1 x2x3 cos 2 xnÀ1xn cos nÀ1 xnx1 cos n πÀÁ cos x2 þ x2 þ ::: þ x2 , n 1 2 n where n 2 and φ1 + φ2 +...+φn ¼ π. Hint See the proof of problem 5.5.31. 8.2.27. Prove the inequalities pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) a2 þ ab þ b2 þ b2 þ bc þ c2 þ c2 þ ac þ a2 3 ab þ bc þ ac, where > ap, bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, c 0, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi (b) a2 þ ab þ b2 þ b2 þ bc þ c2 þ c2 þ ac þ a2 3ðÞa þ b þ c , where a, b, c > 0. 8.2.28. Prove the inequality qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi 3 3 1 þ ðÞþ þ 3 > ¼ 3 abc 6 a b c abc, where a, b, c 0 and ab + bc + ca 1. ÀÁ ¼ α ¼ β ¼ γ α β γ 2 ; π Hint Denote by a ctg , b ctg , c ctg , where , , 0 2 . 8.2.29. Prove that ðÞxy þ yz þ zx 1 þ 1 þ 1 9, where x, y, z > 0. ðÞxþy 2 ðÞyþz 2 ðÞxþz 2 4 Hint See problem 5.5.10c. Basic Notations

ρ(X, l) The distance from the point X to the line l. ρ(X, Π) The distance from the point X to the plane Π. AB # CD The segments AB and CD are parallel and equal. supX Supremum of the set X. M N The points M and N coincide. M=6 N The points M and N do not coincide. (ABC) The plane containing points A , B , C.

Φ Φ1 All points of the ﬁgure Φ are inside the ﬁgure Φ1.

Φ Φ1 The ﬁgure Φ is covered by the ﬁgure Φ1. AB \ CD ¼ X The lines AB and CD intersect at point X. _ ABC The arc ABC. [AB] The segment AB. S A A A A1A2...An The area of the polygon 1 2 ... n. P A A A A1A2...An The perimeter of the polygon 1 2 ... n. ω(O, R) A circle with a center O and radius R. R The set of real numbers. [a] The whole part of the number a. {a} The fractional part of the number a.

Notation for the Elements of Triangle ABC

a , b , c The lengths of the sides BC , AC , AB. p The half-perimeter of the triangle. SABC or S The area of the triangle. α , β , γ The values of the angles at the vertices A , B , C. ma , mb , mc The lengths of the medians drawn from the vertices A , B , C. ha , hb , hc The lengths of the altitudes drawn from the vertices A , B , C. la , lb , lc The lengths of the bisectors drawn from the vertices A , B , C. r and R The inradius and circumradius of the triangle. ra , rb , rc The radiuses of the excircles. References

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