American Mathematical Monthly Geometry Problems 1894 –

Total Page:16

File Type:pdf, Size:1020Kb

American Mathematical Monthly Geometry Problems 1894 – YIU : Problems in Elementary Geometry 1 American Mathematical Monthly Geometry Problems 1894 – Vis171. (Marcus Baker) In a traingle ABC, the center of the circumscribed circle is O, the center of the inscribed circle is I, and the orthocenter is H. Knowing the sides of the triangle OIH, determine the sides of triangle ABC. R − 2 Solution by W.P. Casey: Let N be the nine-point center. From IN = 2 r and OI = R(R − 2r), R and r can be determined. The circumcircle and the incircle, and the nine-point circle, can all be constructed. Then, Casey wrote, “[i]t only remains to find a point C in the circumcircle of ABC so that the tangent CA CA to circle I may be bisected by the circle N in the point S, which is easily done”. [sic] The construction of ABC from OIH cannot be effected by ruler and compass in general. Casey continued to derive the cubic equation with roots cos α, cos β,cosγ, and coefficients in terms of R, r,and := OH,namely, r 4(R + r)2 − (2 +3R2) R2 − 2 x3 − (1 + )x2 + x − =0. R 8R2 8R2 Given triangle OIH,letN be the midpoint of OH. Construct the circle through N tangent to OI at O.ExtendIN to intersect this circle again at M. The diameter of the circumcircle is equal to the length of IM. From this, the circumcircle, the nine-point circle, and the incircle can be constructed. Now, it remains to select a point X on the nine point circle so that the perpendicular to OX is tangent to the incircle. This is in general not constructible. √ √ √ √ Here is an example: Let OI =2 2α, OH = 42α,andIH = 15α,whereα = 17. Then s − a, s − b, s − c are the roots of the equation x3 − 34x2 +172x − 2 · 172 =0. YIU : Problems in Elementary Geometry 2 The roots of this equation are inconstructible. Vis186. (Marcus Baker) (Malfatti circles) Inscribe in any plane triangle three circles each tangent to two sides and the other two circles. Three solutions were published. The second one, by U. Jesse Knisely, contains a construction: Thethirdsolution,byE.B.Seitz,1 gives the radii of the Malfatti circles as (1 + tan β )(1 + tan γ ) 4 4 · r x = α , 1+tan 4 2 and analogous expressions for the other two. Vis52. (James McLaughlin) (Inverse Malfatti problem) Three circles, radii p, q, r,are drawn in a triangle, each circle touching the other two and two sides of the triangle. Find the sides of the triangle. Solution by E.B.Seitz: write σ = p + q + r. √ √ √ √ √ √ √ √ √ pσ +(p( p + q + r) − pqr) σ − pqr( q + r − p) a = √ √ √ √ √ √ √ √ (q + r)2. (p q + q p + pqσ − r σ)(p r + r p + prσ − q σ) Vis289. (Christine Ladd) If R is the radius of the circumscribed circle of a triangle ABC, r the radius of the inscribed circle, p the radius of the circle inscribed in the orthic triangle, I the center of the inscribed circle of the triangle ABC,andQ the center of the circle inscribed 1The very first issue of the Monthly contains a biography of Seitz, written by Finkel. YIU : Problems in Elementary Geometry 3 in the triangle formed by joining the midpoints of the sides of ABC, show that 1 QI2 = . 2(7Rp − 6Rr +3r2 +2R2) This certainly is not right! The correct formula should be 1 QI2 = (2Rp − 6Rr +3r2 +2R2). 2 Q is the Spieker center X10 and 1 QI2 = (−a3 +2a2(b + c) − 9abc). 8s IG 2 Also, IX10 = 3 .Now, − 3 2 − − 3 − 2 − 2 ( a +2a (b + c) 9abc)= s1 +5s1s2 18s3 =2s(5r 16Rr + s ). 2 1 2 − 2 Therefore, QI = 4 (5r 16Rr + s ). YZ Now, for the orthic triangle, sin α =2R cos α. This means that YZ = R sin 2α.Thesemi- perimeter of the orthic triangle is therefore 1 abc 4·R s = R(sin 2α +sin2β +sin2γ)=2R sin α sin β sin γ = = = . 2 4R2 4R2 R Also, =2cosα cos β cos γ. YIU : Problems in Elementary Geometry 4 The inradius of the orthic triangle is p = =2R cos α cos β cos γ. s Now, (b2 + c2 − a2)(c2 + a2 − b2)(a2 + b2 − c2) cos α cos β cos γ = 8a2b2c2 − 6 4 − 2 2 − 3 − 2 ( s1 +6s1s2 8s1s2 8s1s3 +16s1s2s3 8s3 = 2 8s3 32r2s2[s2 − (2R + r)2] = 8(4Rrs)2 s2 − (2R + r)2 = . 8R2 s2−(2R+r)2 Therefore, p = 4R . s2 =(2R + r)2 +4Rp. Finally, 1 QI2 = (5r2 − 16Rr + s2) 4 1 = [5r2 − 16Rr +(2R + r)2 +4Rp] 4 1 = [6r2 − 12Rr +4R2 +4Rp] 4 1 = (2Rp − 6Rr +3r2 +2R2). 2 VIS 274. If the midpoints of three sides of a triangle are joined with the opposite vertices and R1,...,R6 are the radii of the circles circumscribed about the 6 triangles so formed, and r1, r2,...,r6 the radii of the circles inscribed in these triangles. prove that R1R3R5 = R2R4R6 and 1 1 1 1 1 1 + + = + + . r1 r3 r5 r2 r4 r6 VIS 282 (W.P.Casey) Let X, Y , Z be the traces of the incenter. Construct circles through I and with centers at X, Y , Z. The sum of the reciprocals of the radii of all the circles touching these circles is equal to four times the reciprocal of the inradius. YIU : Problems in Elementary Geometry 5 G1.941.S941. (B.F.Finkel) Show that the bisectors of the angles formed by producing the sides of an inscribed quadrilateral intersect each other at right angles. G2.941.S942;943. (From Bowser’s Trigonometry) Show that π 2 · 4 · 6 · 8 · 10 ··· 2 = , 2 1 · 3 · 5 · 7 · 9 ··· Wallis’ expression for π. P.H. Philbrick pointed out that this expression was not correct. It should be π 2 · 4 · 6 · 8 · 10 ···(2n) 2 1 = · . 2 1 · 3 · 5 · 7 · 9 ···(2n − 1) 2n +1 G3.941. (From Todhunter’s Trigonometry)IfA be the area of the circle inscribed in a triangle, A1, A2, A3 the ares of the escribed circles, show that 1 1 1 1 √ = √ + √ + √ . A A1 A2 A3 G4.941.S943. (From Todhunter’s Trigonometry) Three circles whose radii are a, b,andc touch each other externally; prove that the tangents at the points of contact meet in a point whose distance from any one of them is abc . a + b + c YIU : Problems in Elementary Geometry 6 G5.941.S943. (Adolf Bailoff) If from a variable point in the base of an isosceles triangle, perpendiculars are drawn to the sides, the sum of the perpendiculars is constant and equal to the perpendicular let fall from either extremity of the base to the opposite side. G6.941.S94:159–160. (Earl D. West) Having given the sides 6, 4, 5, and 3 respectively of a trapezium, inscribe in a circle, to find the diameter of the circle. [Note: trapezium means quadrilateral]. G7.941.S94:160. (William Hoover) Through each point of the straight line x = my + h is drawn a chord of the parabola y2 =4ax, which is bisected in the point. Prove that this chord touches the parabola (y − 2mn)2 =8a(x − h). G8.941.S944. (Adolf Bailoff) If the two exterior angles at the base of a triangle are equal, the triangle is isosceles. G9.941.S94:160–161. (J.C.Gregg) Two circles intersect in A and B. Through A two lines CAE and DAF are drawn, each passing through a center and terminated by the circumferences. Show that CA · AE = DA · AF . (Euclid) G10.941.S94:161. (Eric Doolittle) If MN be any plane, and A and B any point without the plane, to find a point P , in the plane, such that AP + PB shall be a minimum. G11.941. (Lecta Miller) A gentleman’s residence is at the center of his circular farm containing a = 900 acres. He gives to each of his m = 7 children an equal circular farm as large as can be made within the original farm; and he retains as large a circular farm of which his residence is the center, as can be made after the distribution. Required the area of the farms made. YIU : Problems in Elementary Geometry 7 G12.S94:198–199. (J.F.W.Scheffer) Let OA and OB represent two variable conjugate semi - diameters of the ellipse x2 y2 + =1. a2 b2 On the chord AB as a side describe an equilateral triangle ABC. Find the locus of C. G13.941. (Henry Heaton) Through two given points to pass four spherical surfaces tangent to two given spheres. G14.S94:232–233;268–269. (Henry Heaton) Through a given point to draw four circles tangent to two given circles. G15.S94:233–234. (Issac L. Beverage) A man starts from the center of a circular 10 acre field and walks due north a certain distance, then turns and walks south - west till he comes to the circumference, walking altogether 40 rods. How far did he walk before making the turn? G16.941. (H.C.Whitaker) Three lights of intensities 2, 4, and 5 are placed respectively at points the coordinates of which are (0, 3), (4, 5) and (9, 0). Find a point in the plane of the lights equally illuminated by all of them. G17.942:S94:269. (Robert J. Aley) Draw a circle bisecting the circumference of three given circles. G18.942.S94:271. (Henry Heaton) Through two given points to draw two circles tangnet to a given circle. G19.942:S94:315.
Recommended publications
  • 13-6 Success for English Language Learners
    LESSON Success for English Language Learners 13-6 The Law of Cosines Steps for Success Step I In order to create interest in the lesson opener, point out the following to students. • Trapeze, trapezium (a geometric figure as well as a bone in the wrist), trapezoid (also a geometric figure and a bone in the wrist), and trapezius (a muscle in the back) all come from the same root. They come from the Greek word trapeza, or “four-legged table.” Step II Teach the lesson. • Have students derive the remaining formulas in the Law of Cosines that were not derived in the text. • Technically speaking, a knot is not a nautical mile. A knot is a speed of one nautical mile per hour. • Heron’s Formula is also known as Hero’s Formula. A Heronian triangle is a triangle having rational side lengths and a rational area. Step III Ask English Language Learners to complete the worksheet for this lesson. • Point out that Example 1A in the student textbook is supported by Problem 1 on the worksheet. Remind students that, for example, side a is opposite angle A, not adjacent to it. • Point out that Example 3 in the student textbook is supported by Problem 2 on the worksheet. • Think and Discuss supports the problems on the worksheet. Making Connections • Students comfortableᎏᎏ with matrices may wish to verify the following equation: If ᭝ ϭ ͙s ͑ s Ϫ a ͒ ͑ s Ϫ b ͒ ͑ s Ϫ c ͒ is the area of the triangle under consideration, then 2 Ϫ1 11 a 2 2 2 2 Ϫ 2 ͑ 4 ͒ ϭ a b c 1 1 1 b .
    [Show full text]
  • Downloaded from Bookstore.Ams.Org 30-60-90 Triangle, 190, 233 36-72
    Index 30-60-90 triangle, 190, 233 intersects interior of a side, 144 36-72-72 triangle, 226 to the base of an isosceles triangle, 145 360 theorem, 96, 97 to the hypotenuse, 144 45-45-90 triangle, 190, 233 to the longest side, 144 60-60-60 triangle, 189 Amtrak model, 29 and (logical conjunction), 385 AA congruence theorem for asymptotic angle, 83 triangles, 353 acute, 88 AA similarity theorem, 216 included between two sides, 104 AAA congruence theorem in hyperbolic inscribed in a semicircle, 257 geometry, 338 inscribed in an arc, 257 AAA construction theorem, 191 obtuse, 88 AAASA congruence, 197, 354 of a polygon, 156 AAS congruence theorem, 119 of a triangle, 103 AASAS congruence, 179 of an asymptotic triangle, 351 ABCD property of rigid motions, 441 on a side of a line, 149 absolute value, 434 opposite a side, 104 acute angle, 88 proper, 84 acute triangle, 105 right, 88 adapted coordinate function, 72 straight, 84 adjacency lemma, 98 zero, 84 adjacent angles, 90, 91 angle addition theorem, 90 adjacent edges of a polygon, 156 angle bisector, 100, 147 adjacent interior angle, 113 angle bisector concurrence theorem, 268 admissible decomposition, 201 angle bisector proportion theorem, 219 algebraic number, 317 angle bisector theorem, 147 all-or-nothing theorem, 333 converse, 149 alternate interior angles, 150 angle construction theorem, 88 alternate interior angles postulate, 323 angle criterion for convexity, 160 alternate interior angles theorem, 150 angle measure, 54, 85 converse, 185, 323 between two lines, 357 altitude concurrence theorem,
    [Show full text]
  • An Innovative Analysis to Develop New Theorems on Irregular Polygon
    International Journal of Physics and Mathematical Sciences ISSN: 2277-2111 (Online) An Online International Journal Available at http://www.cibtech.org/jpms.htm 2013 Vol. 3 (1) January-March, pp.73-81/Kalaimaran Research Article AN INNOVATIVE ANALYSIS TO DEVELOP NEW THEOREMS ON IRREGULAR POLYGON *Kalaimaran Ara Construction & Civil Maintenance Unit, Central Food Technological Research Institute, Mysore-20, Karnataka, India *Author for Correspondence ABSTRACT The irregular Polygon is a four sided polygon of two dimensional geometrical figures. The triangle, square, rectangle, tetragon, pentagon, hexagon, heptagon, octagon, nonagon, dodecagon, parallelogram, rhombus, rhomboid, trapezium or trapezoidal, kite and dart are the members of the irregular polygon family. A polygon is a two dimensional example of the more general prototype in any number of dimensions. However the properties are varied from one to another. The author has attempted to develop two new theorems for the property of irregular polygon for a point anywhere inside of the polygon with necessary illustrations, appropriate examples and derivation of equations for better understanding. Key Words: Irregular Polygon, Triangle, Right-angled triangle, Perpendicular and Vertex INTRODUCTION Polygon (Weisstein, 2003) is a closed two dimensional figure formed by connecting three or more straight line segments, where each line segment end connects to only one end of two other line segments. Polygon is one of the most all-encompassing shapes in two- dimensional geometry. The sum of the interior angles is equal to 180 degree multiplied by number of sides minus two. The sum of the exterior angles is equal to 360 degree. From the simple triangle up through square, rectangle, tetragon, pentagon, hexagon, heptagon, octagon, nonagon, dodecagon (Weisstein, 2003,) and beyond is called n-gon.
    [Show full text]
  • The Brahmagupta Triangles Raymond A
    The Brahmagupta Triangles Raymond A. Beauregard and E. R. Suryanarayan Ray Beauregard ([email protected]) received his Ph.D. at the University of New Hampshire in 1968, then joined the University of Rhode Island, where he is a professor of mathematics. He has published many articles in ring theory and two textbooks. Linear Algebra (written with John Fraleigh) is currently in its third edition. Besides babysitting for his grandchild Elyse, he enjoys sailing the New England coast on his sloop, Aleph One, and playing the piano. E. R. Suryanarayan ([email protected]) taught at universities in India before receiving his Ph.D. (1961) at the University of Michigan, under Nathaniel Coburn. He has been at the University of Rhode Island since 1960, where is a professor of mathematics. An author of more than 20 research articles in applied mathematics, crystallography, and the history of mathematics, he lists as his main hobbies music, languages, and aerobic walking. The study of geometric objects has been a catalyst in the development of number theory. For example, the figurate numbers (triangular, square, pentagonal, . ) were a source of many early results in this field [41.Measuring the length of a diagonal of a rectangle led to the problem of approxin~atingfi for a natural number N. The study of triangles has been of particular significance. Heron of Alexandria (c. A.D. 75)-gave the well-known formula for the area A of a triangle in terms of its sides: A = Js(s - a)(s- b)(s- c),where s = (a + b + c)/2 is the semiperimeter of the triangle having sides a,b, c [41.He illustrated this with a triangle whose sides are 13,14,15 and whose area is 84.
    [Show full text]
  • Meeting in Mathematics
    226159-CP-1-2005-1-AT-COMENIUS-C21 527269-LLP-1-2012-1-AT-COMENIUS-CAM Pavel Boytchev Hannes Hohenwarter Evgenia Sendova Neli Dimitrova Emil Kostadinov Andreas Ulovec Vladimir Georgiev Arne Mogensen Henning Westphael Oleg Mushkarov MEETING IN MATHEMATICS 2nd edition • Universität Wien • Dipartimento di Matematica, Universita' di Pisà • VIA University College – Læreruddannelsen i Århus • Институт по математика и информатика , Българска академия на науките Authors Pavel Boytchev, Neli Dimitrova, Vladimir Georgiev, Hannes Hohenwarter, Emil Kostadinov, Arne Mogensen, Oleg Mushkarov, Evgenia Sendova, Andreas Ulovec, Henning Westphael Editors Evgenia Sendova Andreas Ulovec Project Evaluator Jarmila Novotná, Charles University, Prague, Czech Republic Reviewers Jarmila Novotná, Charles University, Prague, Czech Republic Nicholas Mousoulides, University of Nicosia, Cyprus Cover design by Pavel Boytchev Cartoons by Yovko Kolarov All Rights Reserved © 2013 No part of this work may be reproduced, stored in a retrieval system, or transmitted in any form or by means, electronic, mechanical, photocopying, microfilming, recording or otherwise, without written permission from the authors. For educational purposes only (i.e. for use in schools, teaching, teacher training etc.), you may use this work or parts of it under the “Attribution Non-Commercial Share Alike” license according to Creative Commons, as detailed in http://creativecommons.org/licenses/by-nc-sa/3.0/legalcode. This project has been funded with support from the European Commission. This publication reflects the views only of the authors, and the Commission cannot be held responsible for any use which may be made of the information contained therein. Published by Demetra Publishing House, Sofia, Bulgaria ISBN 978-954-9526-49-3 iii Contents PREFACE..................................................................................................................
    [Show full text]
  • The Malfatti Problem
    Forum Geometricorum b Volume 1 (2001) 43–50. bbb FORUM GEOM ISSN 1534-1178 The Malfatti Problem Oene Bottema Abstract. A solution is given of Steiner’s variation of the classical Malfatti problem in which the triangle is replaced by three circles mutually tangent to each other externally. The two circles tangent to the three given ones, presently known as Soddy’s circles, are encountered as well. In this well known problem, construction is sought for three circles C1, C2 and C3, tangent to each other pairwise, and of which C1 is tangent to the sides A1A2 and A1A3 of a given triangle A1A2A3, while C2 is tangent to A2A3 and A2A1 and C3 to A3A1 and A3A2. The problem was posed by Malfatti in 1803 and solved by him with the help of an algebraic analysis. Very well known is the extraordinarily elegant geometric solution that Steiner announced without proof in 1826. This solution, together with the proof Hart gave in 1857, one can find in various textbooks.1 Steiner has also considered extensions of the problem and given solutions. The first is the one where the lines A2A3, A3A1 and A1A2 are replaced by circles. Further generalizations concern the figures of three circles on a sphere, and of three conic sections on a quadric surface. In the nineteenth century many mathematicians have worked on this problem. Among these were Cayley (1852) 2, Schellbach (who in 1853 published a very nice goniometric solution), and Clebsch (who in 1857 extended Schellbach’s solution to three conic sections on a quadric surface, and for that he made use of elliptic functions).
    [Show full text]
  • Some Relations and Properties Concerning Tangential Polygons
    View metadata, citation and similar papers at core.ac.uk brought to you by CORE Mathematical Communications 4(1999), 197-206 197 Some relations and properties concerning tangential polygons Mirko Radic´∗ Abstract. The k-tangential polygon is defined, and some of its properties are proved. Key words: k-tangential polygon AMS subject classifications: 51E12 Received October 10, 1998 Accepted June 7, 1999 1. Preliminaries A polygon with the vertices A1, ..., An (in this order) will be denoted by A1...An. The lengths of the sides of the polygon A1...An will be denoted by |A1A2|, ..., |AnA1| or a1, ..., an. The interior angle at the vertex Ai will be denoted by αi or ∠Ai, i.e. ∠Ai = ∠An−1+iAiAi+1,i=1, ..., n (0 <αi <π). (1) Of course, indices are calculated modulo n. A polygon A = A1...An is a tangential polygon if there exists a circle C such that each side of A is on a tangent line of C. Definition 1. Let A = A1...An be a tangential polygon, and let k be a positive ≤bn−1 c ≤ n−1 ≤ n−2 integer such that k 2 , that is, k 2 if n is odd and k 2 if n is even. Then the polygon A will be called a k-tangential polygon if any two of its consecutive sides have only one point in common, and if there holds π β + ···+ β =(n − 2k) , (2) 1 n 2 where 2βi = ∠Ai, i =1, ..., n. Consequently, a tangential polygon A is k-tangential if n X ϕi =2kπ, i=1 where ϕi = ∠AiCAi+1 and C is the centre of the circle inscribed into the polygon A.
    [Show full text]
  • Triangle Centers Associated with the Malfatti Circles
    Forum Geometricorum b Volume 3 (2003) 83–93. bbb FORUM GEOM ISSN 1534-1178 Triangle Centers Associated with the Malfatti Circles Milorad R. Stevanovi´c Abstract. Various formulae for the radii of the Malfatti circles of a triangle are presented. We also express the radii of the excircles in terms of the radii of the Malfatti circles, and give the coordinates of some interesting triangle centers associated with the Malfatti circles. 1. The radii of the Malfatti circles The Malfatti circles of a triangle are the three circles inside the triangle, mutually tangent to each other, and each tangent to two sides of the triangle. See Figure 1. Given a triangle ABC, let a, b, c denote the lengths of the sides BC, CA, AB, s the semiperimeter, I the incenter, and r its inradius. The radii of the Malfatti circles of triangle ABC are given by C X3 r3 Y3 r3 O3 C3 X2 C2 r I 2 C Y1 1 O2 r1 O1 r2 r1 A Z1 Z2 B Figure 1 r r = (s − r − (IB + IC − IA)) , 1 2(s − a) r r = (s − r − (IC + IA − IB)) , 2 2(s − b) (1) r r = (s − r − (IA + IB − IC)) . 3 2(s − c) Publication Date: March 24, 2003. Communicating Editor: Paul Yiu. The author is grateful to the referee and the editor for their valuable comments and helps. 84 M. R. Stevanovi´c According to F.G.-M. [1, p.729], these results were given by Malfatti himself, and were published in [7] after his death.
    [Show full text]
  • Pythagorean Triples Before and After Pythagoras
    computation Review Pythagorean Triples before and after Pythagoras Ravi P. Agarwal Department of Mathematics, Texas A&M University-Kingsville, 700 University Blvd., Kingsville, TX 78363, USA; [email protected] Received: 5 June 2020; Accepted: 30 June 2020; Published: 7 July 2020 Abstract: Following the corrected chronology of ancient Hindu scientists/mathematicians, in this article, a sincere effortPythagorean is made to report Triples the origin Before of Pythagorean and After triples. Pythagoras We shall account for the development of these triples from the period of their origin and list some known astonishing directions. Although for researchers in this field, there is not much that is new in this article, we genuinely hope students and teachers of mathematicsRavi P Agarwal will enjoy this article and search for new directions/patterns. Department of Mathematics, Texas A&M University-Kingsville 700 University Blvd., Kingsville, TX, USA Keywords: Pythagorean triples; [email protected] and patterns; extensions; history; problems AMS Subject Classification: 01A16; 0A25; 0A32; 11-02; 11-03; 11D09 Abstract: Following the corrected chronology of ancient Hindu scientists/mathematicians by Lakshmikan- tham, et. al. [27], in this article a sincere effort has been made to report the origin of Pythagorean triples. We shall account the development of these triples from the period of its origin, and list some known aston- ishing directions. Although, for researchers in this field there is not much new in this article, we genuinely 1. Introductionhope
    [Show full text]
  • Arxiv:1712.00299V1 [Math.GT]
    POLYGONS WITH PRESCRIBED EDGE SLOPES: CONFIGURATION SPACE AND EXTREMAL POINTS OF PERIMETER JOSEPH GORDON, GAIANE PANINA, YANA TEPLITSKAYA Abstract. We describe the configuration space S of polygons with pre- scribed edge slopes, and study the perimeter as a Morse function on S. We characterize critical points of (these areP tangential polygons) and compute their Morse indices. ThisP setup is motivated by a number of re- sults about critical points and Morse indices of the oriented area function defined on the configuration space of polygons with prescribed edge lengths (flexible polygons). As a by-product, we present an independent computa- tion of the Morse index of the area function (obtained earlier by G. Panina and A. Zhukova). 1. Introduction Consider the space L of planar polygons with prescribed edge lengths1 and the oriented area as a Morse function defined on it. It is known that gener- ically: A L is a smooth closed manifold whose diffeomorphic type depends on • the edge lengths [1, 2]. The oriented area is a Morse function whose critical points are cyclic • configurations (thatA is, polygons with all the vertices lying on a cir- cle), whose Morse indices are known, see Theorem 2, [3, 5, 6]. The Morse index depends not only on the combinatorics of a cyclic poly- arXiv:1712.00299v1 [math.GT] 1 Dec 2017 gon, but also on some metric data. Direct computations of the Morse index proved to be quite involved, so the existing proof comes from bifurcation analysis combined with a number of combinatorial tricks. Bifurcations of are captured by cyclic polygons P whose dual poly- • gons P ∗ have zeroA perimeter [5]; see also Lemma 4.
    [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]
  • (Theorem 1) Is Proved
    Rad HAZU Volume 503 (2009), 41–54 ABOUT ONE RELATION CONCERNING TWO CIRCLES MIRKO RADIC´ AND ZORAN KALIMAN Abstract. This article can be considered as an appendix to the article [1]. Here the article [1] is extended to the cases when one circle is outside of the other and when circles are intersecting. 1. Preliminaries In [1] the following theorem (Theorem 1) is proved: Let C1 and C2 be any given two circles such that C1 is inside of the C2 and let A1 , A2 , A3 be any given three different points on C2 such that there are points T1 and T2 on C1 with properties |A1A2| = t1 +t2, |A2A3| = t2 +t3, (1) where t1 = |A1T1|, t2 = |T1A2|, t3 = |T2A3|.Then 2rR |A A | =(t +t ) , (2) 1 3 1 3 R2 − d2 where r = radius of C1 , R = radius of C2 , d = |IO|, I is the center of C1 , O is center of C2 .(SeeFigure1.) In short about the proof of this theorem. First the following lemma is proved. It t1 is given then t2 can be calculated using the expression √ t (R2 − d2) ± D (t ) = 1 1 (3a) 2 1,2 2 + 2 r t1 where = 2( 2 − 2)2 +( 2 + 2) 2 2 − 2 2 − ( 2 + 2 − 2)2 . D1 t1 R d r t1 4R d r t1 R d r (3b) The values (t2)1,2 given by (3) are solutions of the equation ( 2 + 2) 2 − ( 2 − 2)+ 2 2 − 2 2 +( 2 + 2 − 2)2 = . r t1 t2 2t1t2 R d r t1 4R d R d r 0 (4) Mathematics subject classification (2000): 51M04.
    [Show full text]