DOCSLIB.ORG

Angle Bisectors Bisect Arcs Below Was the Find Round of the 36Th

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Angle Bisectors Bisect Arcs Below Was the Find Round of the 36Th Volume 11, Number 2 April 2006 – May 2006 Olympiad Corner Angle Bisectors Bisect Arcs Below was the Find Round of the 36th Austrian Math Olympiad 2005. Kin Y. Li Part 1 (May 30, 2005) In general, angle bisectors of a triangle Solution. Let I be the incenter of ∆ABC. Problem 1. Show that an infinite do not bisect the sides opposite the By the theorem, we have 2IR = AR + BR number of multiples of 2005 exist, in angles. However, angle bisectors > AB and similarly 2IP > BC, 2IQ > CA. which each of the 10 digits 0,1,2,…,9 always bisect the arcs opposite the Also AI + BI > AB, BI + CI > BC and occurs the same number of times, not angles on the circumcircle of the CI + AI > CA. Adding all these counting leading zeros. triangle! In math competitions, this fact inequalities together, we get is very useful for problems concerning Problem 2. For how many integer 2(AP + BQ + CR) > 2(AB + BC + CA). angle bisectors or incenters of a triangle values of a with |a| ≤ 2005 does the system of equations x2 = y + a, y2 = x + a involving the circumcircle. Recall that Example 2. (1978 IMO) In ABC, AB = have integer solutions? the incenter of a triangle is the point AC. A circle is tangent internally to the where the three angle bisectors concur. circumcircle of ABC and also to the Problem 3. We are given real numbers sides AB, AC at P, Q, respectively. a, b and c and define s as the sum s = an n n Theorem. Suppose the angle bisector of Prove that the midpoint of segment PQ + bn + cn of their n-th powers for is the center of the incircle of ∆ABC. non-negative integers n. It is known that ∠ BAC intersect the circumcircle of ∆ABC at X ≠ A. Let I be a point on the A s1 = 2, s2 = 6 and s3 = 14 hold. Show that line segment AX. Then I is the incenter 2 | sn − sn−1 ⋅ sn+1 |= 8 of ∆ABC if and only if XI = XB = XC. A holds for all integers n > 1. Problem 4. We are given two P Q equilateral triangles ABC and PQR with I parallel sides, “one pointing up” and I “one pointing down.” The common area B C of the triangles’ interior is a hexagon. B C X Show that the lines joining opposite corners of this hexagon are concurrent. X Solution. Let I be the midpoint of line segment PQ and X be the intersection of (continued on page 4) Proof. Note ∠BAX =∠CAX =∠CBX. the angle bisector of ∠BAC with the arc So XB = XC. Then BC not containing A. Editors: ஻ Ի ஶ (CHEUNG Pak-Hong), Munsang College, HK I is the incenter of ∆ABC ଽ υ ࣻ (KO Tsz-Mei) By symmetry, AX is a diameter of the ∠CBI =∠ABI గ ႀ ᄸ (LEUNG Tat-Wing) ⇔ circumcircle of ∆ABC and X is the ∠IBX −∠CBX =∠BIX −∠BAX ୊ ፱ (LI Kin-Yin), Dept. of Math., HKUST ⇔ midpoint of the arc PXQ on the inside ؃ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU ∠IBX = ∠BIX ֔ ⇔ circle, which implies PX bisects .Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU ⇔ XI = XB = XC ∠QPB . Now ∠ABX = 90˚ = ∠PIX Acknowledgment: Thanks to Elina Chiu, Math. Dept., so that X, I, P, B are concyclic. Then HKUST for general assistance. Example 1. (1982 Australian Math Olympiad) Let ABC be a triangle, and On-line: ∠IBX =∠IPX =∠BPX =∠BIX. http://www.math.ust.hk/mathematical_excalibur/ let the internal bisector of the angle A The editors welcome contributions from all teachers and meet the circumcircle again at P. So XI = XB. By the theorem, I is the students. With your submission, please include your name, Define Q and R similarly. Prove that AP incenter of ∆ABC. address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, + BQ + CR > AB + BC + CA. are encouraged. The deadline for receiving material for the A Example 3. (2002 IMO) Let BC be a next issue is August 16, 2006. R diameter of the circle Γ with center O. For individual subscription for the next five issues for the Let A be a point on Γ such that 0˚ < 05-06 academic year, send us five stamped self-addressed Q envelopes. Send all correspondence to: I ∠AOB < 120˚. Let D be the midpoint Dr. Kin-Yin LI of the arc AB not containing C. The line Department of Mathematics B through O parallel to DA meets the line The Hong Kong University of Science and Technology C Clear Water Bay, Kowloon, Hong Kong AC at J. The perpendicular bisector of Fax: (852) 2358 1643 P OA meets Γ at E and at F. Prove that J is Email: [email protected] the incenter of the triangle CEF. Mathematical Excalibur, Vol. 11, No. 2, Apr. 06 - May 06 Page 2 A H, I on the circumcircle, we see that This implies J is the midpoint of KQ. P=BG∩CH, K=GJ∩HI and BI∩CJ= Hence the circle with center J and D F BD∩CE are collinear. Hence, BD∩CE is diameter KQ is tangent to circle O at K E on line PK, which is the same as line AP. and tangent to BC at Q. Since J is on the bisector of ∠BCA, this circle is J C O B Example 5. (2006 APMO) Let A, B be also tangent to AC. So this circle is O2. two distinct points on a given circle O and let P be the midpoint of line segment AB. Example 6. (1989 IMO) In an Let O1 be the circle tangent to the line AB acute-angled triangle ABC the internal at P and tangent to the circle O. Let ℓ be bisector of angle A meets the Solution. The condition ∠AOB < the tangent line, different from the line AB, circumcircle of the triangle again at A1. 120˚ ensures I is inside ∆CEF (when to O1 passing through A. Let C be the Points B1 and C1 are defined similarly. ∠ AOB increases to 120˚, I will intersection point, different from A, of ℓ Let A0 be the point of intersection of coincide with C). Now radius OA and and O. Let Q be the midpoint of the line the line AA1 with the external bisectors chord EF are perpendicular and bisect segment BC and O2 be the circle tangent of angles B and C. Points B0 and C0 are each other. So EOFA is a rhombus. to the line BC at Q and tangent to the line defined similarly. Prove that: Hence A is the midpoint of arc EAF. segment AC. Prove that the circle O2 is (i) the area of the triangle A0B0C0 is Then CA bisects ∠ECF. Since OA = tangent to the circle O. OC, ∠AOD = 1/2∠AOB = ∠OAC. twice the area of the hexagon Then DO is parallel to AJ. Hence N AC1BA1CB1, C ODAJ is a parallelogram. Then AJ = Z (ii) the area of the triangle A0B0C0 is at DO = EO = AE. By the theorem, J is least four times the area of the triangle L the incenter of ∆CEF. ABC. Q C Example 4. (1996 IMO) Let P be a J 0 C B point inside triangle ABC such that A P B 1 ∠APB −∠ACB = ∠APC −∠ABC. K A M I Let D, E be the incenters of triangles Solution. Let the perpendicular to AB B A APB, APC respectively. Show that AP, 1 1 through P intersect circle O at N and M B BD and CE meet at a point. 0 C A0 with N and C on the same side of line AB. A By symmetry, segment NP is a diameter Solution. (i) Let I be the incenter of of the circle of O1 and its midpoint L is the ∆ABC. Since internal angle bisector J I center of O1. Let line AL intersect circle O and external angle bisector are K again at Z. Let line ZQ intersect line CM perpendicular, we have ∠B0BA0 = 90˚. at J and circle O again at K. By the theorem, A I = A B. So A must H D E G 1 1 1 be the midpoint of the hypotenuse A0I Since AB and AC are tangent to circle O1, P of right triangle IBA0. So the area of AL bisects ∠ CAB so that Z is the B C ∆BIA0 is twice the area of ∆BIA1. midpoint of arc BC. Since Q is the F midpoint of segment BC, ∠ZQB = 90˚ = Cutting the hexagon AC1BA1CB1 into ∠LPA and ∠JQC = 90˚ =∠MPB. Next six triangles with common vertex I and Solution. Let lines AP, BP, CP applying a similar area fact like the last ∠ZBQ =∠ZBC =∠ZAC =∠LAP. intersect the circumcircle of ∆ABC statement to each of the six triangles, again at F, G, H respectively. Now So ∆ZQB, ∆LPA are similar. Since M is we get the conclusion of (i). the midpoint of arc AMB, ∠APB −∠ACB =∠FPG −∠AGB (ii) Using (i), we only need to show the =∠FA G. ∠JCQ =∠MCB =∠MCA =∠MBP. area of hexagon AC1BA1CB1 is at least Similarly, ∠APC − ∠ABC = ∠FA H. So ∆JQC, ∆MPB are similar. twice the area of ∆ABC. So AF bisects ∠HAG. Let K be the B incenter of ∆HAG. Then K is on AF By the intersecting chord theorem, AP·BP and lines HK, GK pass through the = NP·MP = 2LP·MP. Using the similar D H A2 A midpoints I, J of minor arcs AG, AH triangles above, we have respectively.
Load more

Recommended publications
  • Relationships Between Six Excircles
    Sangaku Journal of Mathematics (SJM) c SJM ISSN 2534-9562 Volume 3 (2019), pp.73-90 Received 20 August 2019. Published on-line 30 September 2019 web: http://www.sangaku-journal.eu/ c The Author(s) This article is published with open access1. Relationships Between Six Excircles Stanley Rabinowitz 545 Elm St Unit 1, Milford, New Hampshire 03055, USA e-mail: [email protected] web: http://www.StanleyRabinowitz.com/ Abstract. If P is a point inside 4ABC, then the cevians through P divide 4ABC into smaller triangles of various sizes. We give theorems about the rela- tionship between the radii of certain excircles of some of these triangles. Keywords. Euclidean geometry, triangle geometry, excircles, exradii, cevians. Mathematics Subject Classification (2010). 51M04. 1. Introduction Let P be any point inside a triangle ABC. The cevians through P divide 4ABC into six smaller triangles. In a previous paper [5], we found relationships between the radii of the circles inscribed in these triangles. For example, if P is at the orthocenter H, as shown in Figure 1, then we found that r1r3r5 = r2r4r6, where the ri are radii of the incircles as shown in the figure. arXiv:1910.00418v1 [math.HO] 28 Sep 2019 Figure 1. r1r3r5 = r2r4r6 In this paper, we will find similar results using excircles instead of incircles. When the cevians through a point P interior to a triangle ABC are drawn, many smaller 1This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.
    [Show full text]
  • Point of Concurrency the Three Perpendicular Bisectors of a Triangle Intersect at a Single Point
    3.1 Special Segments and Centers of Classifications of Triangles: Triangles By Side: 1. Equilateral: A triangle with three I CAN... congruent sides. Define and recognize perpendicular bisectors, angle bisectors, medians, 2. Isosceles: A triangle with at least and altitudes. two congruent sides. 3. Scalene: A triangle with three sides Define and recognize points of having different lengths. (no sides are concurrency. congruent) Jul 24­9:36 AM Jul 24­9:36 AM Classifications of Triangles: Special Segments and Centers in Triangles By angle A Perpendicular Bisector is a segment or line 1. Acute: A triangle with three acute that passes through the midpoint of a side and angles. is perpendicular to that side. 2. Obtuse: A triangle with one obtuse angle. 3. Right: A triangle with one right angle 4. Equiangular: A triangle with three congruent angles Jul 24­9:36 AM Jul 24­9:36 AM Point of Concurrency The three perpendicular bisectors of a triangle intersect at a single point. Two lines intersect at a point. The point of When three or more lines intersect at the concurrency of the same point, it is called a "Point of perpendicular Concurrency." bisectors is called the circumcenter. Jul 24­9:36 AM Jul 24­9:36 AM 1 Circumcenter Properties An angle bisector is a segment that divides 1. The circumcenter is an angle into two congruent angles. the center of the circumscribed circle. BD is an angle bisector. 2. The circumcenter is equidistant to each of the triangles vertices. m∠ABD= m∠DBC Jul 24­9:36 AM Jul 24­9:36 AM The three angle bisectors of a triangle Incenter properties intersect at a single point.
    [Show full text]
  • Figures Circumscribing Circles Tom M
    Figures Circumscribing Circles Tom M. Apostol and Mamikon A. Mnatsakanian 1. INTRODUCTION. The centroid of the boundary of an arbitrarytriangle need not be at the same point as the centroid of its interior. But we have discovered that the two centroids are always collinear with the center of the inscribed circle, at distances in the ratio 3 : 2 from the center. We thought this charming fact must surely be known, but could find no mention of it in the literature. This paper generalizes this elegant and surprising result to any polygon that circumscribes a circle (Theorem 6). A key ingredient of the proof is a link to Archimedes' striking discovery concerning the area of a circular disk [4, p. 91], which for our purposes we prefer to state as follows: Theorem 1 (Archimedes). The area of a circular disk is equal to the product of its semiperimeter and its radius. Expressed as a formula, this becomes A = Pr, (1) where A is the area, P is the perimeter, and r is the radius of the disk. First we extend (1) to a large class of plane figures circumscribing a circle that we call circumgons, defined in section 2. They include arbitrarytriangles, all regular polygons, some irregularpolygons, and other figures composed of line segments and circular arcs. Examples are shown in Figures 1 through 4. Section 3 treats circum- gonal rings, plane regions lying between two similar circumgons. These rings have a constant width that replaces the radius in the corresponding extension of (1). We also show that all rings of constant width are necessarily circumgonal rings.
    [Show full text]
  • Searching for the Center
    Searching For The Center Brief Overview: This is a three-lesson unit that discovers and applies points of concurrency of a triangle. The lessons are labs used to introduce the topics of incenter, circumcenter, centroid, circumscribed circles, and inscribed circles. The lesson is intended to provide practice and verification that the incenter must be constructed in order to find a point equidistant from the sides of any triangle, a circumcenter must be constructed in order to find a point equidistant from the vertices of a triangle, and a centroid must be constructed in order to distribute mass evenly. The labs provide a way to link this knowledge so that the students will be able to recall this information a month from now, 3 months from now, and so on. An application is included in each of the three labs in order to demonstrate why, in a real life situation, a person would want to create an incenter, a circumcenter and a centroid. NCTM Content Standard/National Science Education Standard: • Analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships. • Use visualization, spatial reasoning, and geometric modeling to solve problems. Grade/Level: These lessons were created as a linking/remembering device, especially for a co-taught classroom, but can be adapted or used for a regular ed, or even honors level in 9th through 12th Grade. With more modification, these lessons might be appropriate for middle school use as well. Duration/Length: Lesson #1 45 minutes Lesson #2 30 minutes Lesson #3 30 minutes Student Outcomes: Students will: • Define and differentiate between perpendicular bisector, angle bisector, segment, triangle, circle, radius, point, inscribed circle, circumscribed circle, incenter, circumcenter, and centroid.
    [Show full text]
  • G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter
    Regents Exam Questions Name: ________________________ G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter www.jmap.org G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter 1 Which geometric principle is used in the 3 In the diagram below, point B is the incenter of construction shown below? FEC, and EBR, CBD, and FB are drawn. If m∠FEC = 84 and m∠ECF = 28, determine and state m∠BRC . 1) The intersection of the angle bisectors of a 4 In the diagram below of isosceles triangle ABC, triangle is the center of the inscribed circle. AB ≅ CB and angle bisectors AD, BF, and CE are 2) The intersection of the angle bisectors of a drawn and intersect at X. triangle is the center of the circumscribed circle. 3) The intersection of the perpendicular bisectors of the sides of a triangle is the center of the inscribed circle. 4) The intersection of the perpendicular bisectors of the sides of a triangle is the center of the circumscribed circle. 2 In the diagram below of ABC, CD is the bisector of ∠BCA, AE is the bisector of ∠CAB, and BG is drawn. If m∠BAC = 50°, find m∠AXC . Which statement must be true? 1) DG = EG 2) AG = BG 3) ∠AEB ≅ ∠AEC 4) ∠DBG ≅ ∠EBG 1 Regents Exam Questions Name: ________________________ G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter www.jmap.org 5 The diagram below shows the construction of the 9 Triangle ABC is graphed on the set of axes below. center of the circle circumscribed about ABC. What are the coordinates of the point of intersection of the medians of ABC? 1) (−1,2) 2) (−3,2) 3) (0,2) This construction represents how to find the intersection of 4) (1,2) 1) the angle bisectors of ABC 10 The vertices of the triangle in the diagram below 2) the medians to the sides of ABC are A(7,9), B(3,3), and C(11,3).
    [Show full text]
  • Half-Anticevian Triangle of the Incenter
    International Journal of Computer Discovered Mathematics (IJCDM) ISSN 2367-7775 c IJCDM Volume 2, 2017, pp.55-71. Received 20 February 2017. Published on-line 28 February 2017 web: http://www.journal-1.eu/ c The Author(s) This article is published with open access1. Computer Discovered Mathematics: Half-Anticevian Triangle of the Incenter Sava Grozdeva, Hiroshi Okumurab and Deko Dekovc 2 a VUZF University of Finance, Business and Entrepreneurship, Gusla Street 1, 1618 Sofia, Bulgaria e-mail: [email protected] b Department of Mathematics, Yamato University, Osaka, Japan e-mail: [email protected] cZahari Knjazheski 81, 6000 Stara Zagora, Bulgaria e-mail: [email protected] web: http://www.ddekov.eu/ Abstract. By using the computer program ”Discoverer” we study the Half- Anticevian triangle of the Incenter. Keywords. Half-Anticevian triangle, triangle geometry, notable point, computer discovered mathematics, Euclidean geometry, “Discoverer”. Mathematics Subject Classification (2010). 51-04, 68T01, 68T99. 1. Introduction Let P be an arbitrary point in the plane of triangle ABC. Denote by P aP bP c the Anticevian triangle of point P . Let Ma be the midpoint of segment AP a and define Mb and Mc similarly. We call triangle MaMbMc the Half-Anticevian Triangle of Point P . Figure 1 illustrates the definition. In figure 1, P is an arbitrary point, P aP bP c is the Anticevian triangle, and MaMbMc is the Half-Anticevian Triangle of P . Note that the Half-Anticevian Triangle of the Centroid is the Medial Triangle. We encourage the students, teachers and researchers to investigate the Half- Anticevian triangles of other notable points, e.g.
    [Show full text]
  • Barycentric Coordinates in Olympiad Geometry
    Barycentric Coordinates in Olympiad Geometry Max Schindler∗ Evan Cheny July 13, 2012 I suppose it is tempting, if the only tool you have is a hammer, to treat everything as if it were a nail. Abstract In this paper we present a powerful computational approach to large class of olympiad geometry problems{ barycentric coordinates. We then extend this method using some of the techniques from vector computations to greatly extend the scope of this technique. Special thanks to Amir Hossein and the other olympiad moderators for helping to get this article featured: I certainly did not have such ambitious goals in mind when I first wrote this! ∗Mewto55555, Missouri. I can be contacted at igoroogenfl[email protected]. yv Enhance, SFBA. I can be reached at [email protected]. 1 Contents Title Page 1 1 Preliminaries 4 1.1 Advantages of barycentric coordinates . .4 1.2 Notations and Conventions . .5 1.3 How to Use this Article . .5 2 The Basics 6 2.1 The Coordinates . .6 2.2 Lines . .6 2.2.1 The Equation of a Line . .6 2.2.2 Ceva and Menelaus . .7 2.3 Special points in barycentric coordinates . .7 3 Standard Strategies 9 3.1 EFFT: Perpendicular Lines . .9 3.2 Distance Formula . 11 3.3 Circles . 11 3.3.1 Equation of the Circle . 11 4 Trickier Tactics 12 4.1 Areas and Lines . 12 4.2 Non-normalized Coordinates . 13 4.3 O, H, and Strong EFFT . 13 4.4 Conway's Formula . 14 4.5 A Few Final Lemmas . 15 5 Example Problems 16 5.1 USAMO 2001/2 .
    [Show full text]
  • Conic Construction of a Triangle from Its Incenter, Nine-Point Center, and a Vertex
    Conic Construction of a Triangle From Its Incenter, Nine-point Center, and a Vertex Paul Yiu Abstract. We construct a triangle given its incenter, nine-point center and a vertex by locating the circumcenter as an intersection of two rectangular hyper- bolas. Some special configurations leading to solutions constructible with ruler and compass are studied. The related problem of construction of a triangle given its circumcenter, incenter, and one vertex is revisited, and it is established that such a triangle exists if and only if the incenter lies inside the cardioid relative to the circumcircle. 1. Introduction In this note we solve the construction problem of a triangle ABC given its in- center I, its nine-point center N, and one vertex A. This is Problem 35 in Harold Connelly’s list of construction problems [1]. We shall show that the triangle is in general not constructible with ruler and compass, but can be easily obtained by in- tersecting conics. Some special configurations of I, N, A from which the triangle can be constructible with ruler and compass are studied in details in 2, 6 – 8. §§ 2. The case of collinear I, N, A In this case, the triangle ABC is isosceles, with AB = AC. The incircle and the nine-point circle are tangent to each other internally at the midpoint of the base BC. We set up a rectangular coordinate system with N at the origin, I at ( d, 0), and A at (a, 0) for a > 0, and seek the inradius r. The circumradius is − A r R = 2(d + r).
    [Show full text]
  • Lemmas in Euclidean Geometry1 Yufei Zhao [email protected]
    IMO Training 2007 Lemmas in Euclidean Geometry Yufei Zhao Lemmas in Euclidean Geometry1 Yufei Zhao [email protected] 1. Construction of the symmedian. Let ABC be a triangle and Γ its circumcircle. Let the tangent to Γ at B and C meet at D. Then AD coincides with a symmedian of △ABC. (The symmedian is the reflection of the median across the angle bisector, all through the same vertex.) A A A O B C M B B F C E M' C P D D Q D We give three proofs. The first proof is a straightforward computation using Sine Law. The second proof uses similar triangles. The third proof uses projective geometry. First proof. Let the reflection of AD across the angle bisector of ∠BAC meet BC at M ′. Then ′ ′ sin ∠BAM ′ ′ BM AM sin ∠ABC sin ∠BAM sin ∠ABD sin ∠CAD sin ∠ABD CD AD ′ = ′ sin ∠CAM ′ = ∠ ∠ ′ = ∠ ∠ = = 1 M C AM sin ∠ACB sin ACD sin CAM sin ACD sin BAD AD BD Therefore, AM ′ is the median, and thus AD is the symmedian. Second proof. Let O be the circumcenter of ABC and let ω be the circle centered at D with radius DB. Let lines AB and AC meet ω at P and Q, respectively. Since ∠PBQ = ∠BQC + ∠BAC = 1 ∠ ∠ ◦ 2 ( BDC + DOC) = 90 , we see that PQ is a diameter of ω and hence passes through D. Since ∠ABC = ∠AQP and ∠ACB = ∠AP Q, we see that triangles ABC and AQP are similar. If M is the midpoint of BC, noting that D is the midpoint of QP , the similarity implies that ∠BAM = ∠QAD, from which the result follows.
    [Show full text]
  • MYSTERIES of the EQUILATERAL TRIANGLE, First Published 2010
    MYSTERIES OF THE EQUILATERAL TRIANGLE Brian J. McCartin Applied Mathematics Kettering University HIKARI LT D HIKARI LTD Hikari Ltd is a publisher of international scientific journals and books. www.m-hikari.com Brian J. McCartin, MYSTERIES OF THE EQUILATERAL TRIANGLE, First published 2010. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission of the publisher Hikari Ltd. ISBN 978-954-91999-5-6 Copyright c 2010 by Brian J. McCartin Typeset using LATEX. Mathematics Subject Classification: 00A08, 00A09, 00A69, 01A05, 01A70, 51M04, 97U40 Keywords: equilateral triangle, history of mathematics, mathematical bi- ography, recreational mathematics, mathematics competitions, applied math- ematics Published by Hikari Ltd Dedicated to our beloved Beta Katzenteufel for completing our equilateral triangle. Euclid and the Equilateral Triangle (Elements: Book I, Proposition 1) Preface v PREFACE Welcome to Mysteries of the Equilateral Triangle (MOTET), my collection of equilateral triangular arcana. While at first sight this might seem an id- iosyncratic choice of subject matter for such a detailed and elaborate study, a moment’s reflection reveals the worthiness of its selection. Human beings, “being as they be”, tend to take for granted some of their greatest discoveries (witness the wheel, fire, language, music,...). In Mathe- matics, the once flourishing topic of Triangle Geometry has turned fallow and fallen out of vogue (although Phil Davis offers us hope that it may be resusci- tated by The Computer [70]). A regrettable casualty of this general decline in prominence has been the Equilateral Triangle. Yet, the facts remain that Mathematics resides at the very core of human civilization, Geometry lies at the structural heart of Mathematics and the Equilateral Triangle provides one of the marble pillars of Geometry.
    [Show full text]
  • Collinearity and Concurrence
    Collinearity and concurrence Po-Shen Loh 23 June 2008 1 Warm-up 1. Let I be the incenter of 4ABC. Let A0 be the midpoint of the arc BC of the circumcircle of 4ABC 0 which does not contain A. Prove that the lines IA , BC, and the angle bisector of ∠BAC are concur- rent. Hint: you shouldn’t need the Big Point Theorem1 for this one! Solution: Two of these lines are the angle bisector of ∠A, and of course that intersects with side BC. 2 Tools 2.1 Ceva and friends Ceva. Let ABC be a triangle, with A0 ∈ BC, B0 ∈ CA, and C0 ∈ AB. Then AA0, BB0, and CC0 concur if and only if: AC0 BA0 CB0 · · = 1. C0B A0C B0A Trig Ceva. Let ABC be a triangle, with A0 ∈ BC, B0 ∈ CA, and C0 ∈ AB. Then AA0, BB0, and CC0 concur if and only if: sin CAA0 sin ABB0 sin BCC0 ∠ · ∠ · ∠ = 1. sin ∠A0AB sin ∠B0BC sin ∠C0CA Menelaus. Let ABC be a triangle, and let D, E, and F line on the extended lines BC, CA, and AB. Then D, E, and F are collinear if and only if: AF BD CE · · = −1. FB DC EA Now try these problems. 1. (Gergonne point) Let ABC be a triangle, and let its incircle intersect sides BC, CA, and AB at A0,B0,C0 respectively. Prove that AA0, BB0,CC0 are concurrent. Solution: Ceva. Since incircle, we have BA0 = CA0, etc., so Ceva cancels trivially. 2. (Isogonal conjugate of Gergonne point) Let ABC be a triangle, and let D, E, F be the feet of the altitudes from A, B, C.
    [Show full text]
  • Advanced Euclidean Geometry What Is the Center of a Triangle?
    Advanced Euclidean Geometry What is the center of a triangle? But what if the triangle is not equilateral? ? Circumcenter Equally far from the vertices? P P I II Points are on the perpendicular A B bisector of a line ∆ I ≅ ∆ II (SAS) A B segment iff they PA = PB are equally far from the endpoints. P P ∆ I ≅ ∆ II (Hyp-Leg) I II AQ = QB A B A Q B Circumcenter Thm 4.1 : The perpendicular bisectors of the sides of a triangle are concurrent at a point called the circumcenter (O). A Draw two perpendicular bisectors of the sides. Label the point where they meet O (why must they meet?) Now, OA = OB, and OB = OC (why?) O so OA = OC and O is on the B perpendicular bisector of side AC. The circle with center O, radius OA passes through all the vertices and is C called the circumscribed circle of the triangle. Circumcenter (O) Examples: Orthocenter A The triangle formed by joining the midpoints of the sides of ∆ABC is called the medial triangle of ∆ABC. B The sides of the medial triangle are parallel to the original sides of the triangle. C A line drawn from a vertex to the opposite side of a triangle and perpendicular to it is an altitude. Note that in the medial triangle the perp. bisectors are altitudes. Thm 4.2: The altitudes of a triangle are concurrent at a point called the orthocenter (H). Orthocenter (H) Thm 4.2: The altitudes of a triangle are concurrent at a point called the orthocenter (H).
    [Show full text]