DOCSLIB.ORG

A Note on the Anticomplements of the Fermat Points

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
A Note on the Anticomplements of the Fermat Points Forum Geometricorum b Volume 9 (2009) 119–123. b b FORUM GEOM ISSN 1534-1178 A Note on the Anticomplements of the Fermat Points Cosmin Pohoata Abstract. We show that each of the anticomplements of the Fermat points is common to a triad of circles involving the triangle of reflection. We also gen- erate two new triangle centers as common points to two other triads of circles. Finally, we present several circles passing through these new centers and the anticomplements of the Fermat points. 1. Introduction The Fermat points F± are the common points of the lines joining the vertices of a triangle T to the apices of the equilateral triangles erected on the corresponding sides. They are also known as the isogonic centers (see [2, pp.107, 151]) and are among the basic triangle centers. In [4], they appear as the triangle centers X13 and X14. Not much, however, is known about their anticomplements, which are the points P± which divide F±G in the ratio F±G : GP± =1:2. Given triangle T with vertices A, B, C, (i) let A′, B′, C′ be the reflections of the vertices A, B, C in the respective opposite sides, and (ii) for ε = ±1, let Aε, Bε, Cε be the apices of the equilateral triangles erected on the sides BC, CA, AB of triangle ABC respectively, on opposite or the same sides of the vertices according as ε = 1 or −1 (see Figures 1A and 1B). ′ ′ ′ Theorem 1. For ε = ±1, the circumcircles of triangles A BεCε, B CεAε, C AεBε are concurrent at the anticomplement P−ε of the Fermat point F−ε. 2. Proof of Theorem 1 For ε = ±1, let Oa,ε be the center of the equilateral triangle AεBC; similarly for Ob,ε and Oc,ε. ′ (1) We first note that Oa,−ε is the center of the circle through A , Bε, and · π Cε. Rotating triangle Oa,εAB through B by an angle ε 3 , we obtain triangle Oa,−εCεB. Therefore, the triangles are congruent and Oa,−εCε = Oa,εA. Simi- ′ larly, Oa,−εBε = Oa,εA. Clearly, Oa,εA = Oa,−εA . It follows that Oa,−ε is the ′ center of the circle through A , Bε and Cε. Figures 1A and 1B illustrate the cases ε = +1 and ε = −1 respectively. (2) Let A1B1C1 be the anticomplementary triangle of ABC. Since AA1 and A+A− have a common midpoint, AA+A1A− is a parallelogram. The lines AA−ε and A1Aε are parallel. Since A1 is the anticomplement of A, it follows that the line A1Aε is the anticomplement of the line AA−ε. Similarly, B1Bε and C1Cε are the anticomplements of the lines BB−ε and CC−ε. Since AA−ε, BB−ε, CC−ε Publication Date: June 16, 2009. Communicating Editor: Paul Yiu. 120 C. Pohoata A B+ A Oa,− C− C+ Oa,− B C B C Oa,+ B− Oa,+ ′ A′ A Figure 1A Figure 1B concur at F−ε, it follows that A1Aε, B1Bε, C1Cε concur at the anticomplement of F−ε. This is the point P−ε. (3) A1B1C1 is also the ε-Fermat triangle of A−εB−εC−ε. (i) Triangles ABεCε and CBεA1 are congruent, since ABε = CBε, ACε = AB = 2π π CA1, and each of the angles BεACε and BεCA1 is min A + 3 , B + C + 3 . It follows that BεCε = BεA1. (ii) Triangles ABεCε and BA1Cε are also congruent for the same reason, and we have BεCε = A1Cε. ∠ π It follows that triangle A1BεCε is equilateral, and CεA1Bε = 3 . (4) Because P−ε is the second Fermat point of A1B1C1, we may assume ∠ π CεP−εBε = 3 . Therefore, P−ε lies on the circumcircle of A1BεCε, which is ′ the same as that of A BεCε. On the other hand, since the quadrilateral AA+A1A− is a parallelogram (the diagonals AA1 and A−A+ have a common midpoint D, the midpoint of segment BC), the anticomplement of the line AA− coincides with A1A+. It now follows that the lines A1A+, B1B+, C1C+ are concurrent at the anticomplement P− of the second Fermat points, and furthermore, ∠C+P−B+ = π 3 . Since ∠C+AB+ = 2π − (∠BAC+ + ∠CAB + ∠B+AC) 4π = − ∠CAB 3 π = + ∠ABC + ∠BCA 3 = ∠ABC+ + ∠ABC + ∠CBM = ∠C+BM, A note on the anticomplements of the Fermat points 121 B+ A− A C 1 B1 C+ Oa,+ F+ P− C− B C Oa B− ′ A A1 A+ Figure 2 it follows that the triangles C+AB+ and C+BM are congruent. Likewise, ∠C+AB+ = ∠MCB+, and so the triangles C+AB+ and MCB+ are congruent. Therefore, the ∠ π triangle C+MB+ is equilateral, and thus C+MB+ = 3 . Combining this with ◦ ∠C+P−B+ = 60 , yields that the quadrilateral MP−B+C+ is cyclic, and since ′ A MB+C+ is also cyclic, we conclude that the anticomplement P− of the sec- ′ ond Fermat point F− lies on the circumcircle of triangle A B+C+. Similarly, P− ′ ′ lies on the circumcircles of triangles B C+A+, and C A+B+, respectively. This completes the proof of Theorem 1. 3. Two new triangle centers By using the same method as in [6], we generate two other concurrent triads of circles. ′ ′ ′ ′ Theorem 2. For ε = ±1, the circumcircles of the triangles AεB C , BεC A , ′ ′ CεA B are concurrent. Proof. Consider the inversion Ψ with respect to the anticomplement of the second Fermat point. According to Theorem 1, the images of the circumcircles of triangles ′ ′ ′ ′ ′ ′ A B+C+, B C+A+, C A+B+ are three lines which bound a triangle A+B+C+, 122 C. Pohoata ′ ′ ′ where A+, B+, C+ are the images of A+, B+, and C+, respectively. Since the ′′ ′′ ′′ ′ ′ ′ ′ ′ ′ ′ images A , B , C of A , B , C under Ψ lie on the sidelines B+C+, C+A+, ′ ′ A+B+, respectively, by Miquel’s theorem, we conclude that the circumcircles of ′ ′′ ′′ ′ ′′ ′′ ′ ′′ triangles A+B C , B+C A , C+A B′′ are concurrent. Thus, the circumcircles ′ ′ ′ ′ ′ ′ of triangles A+B C , B+C A , C+A B are also concurrent (see Figure 3). Similarly, inverting with respect to the anticomplement of the first Fermat point, ′ ′ by Miquel’s theorem, one can deduce that the circumcircles of triangles A−B C , ′ ′ ′ ′ B−C A , C−A B are concurrent. B+ B′ A C+ C′ B C U+ A′ A+ Figure 3. Javier Francisco Garcia Capitan has kindly communicated that their points of concurrency do not appear in [4]. We will further denote these points by U+, and U−, respectively. We name these centers U+, U− the inversive associates of the anticomplements P+, P− of the Fermat points. 4. Circles around P± and their inversive associates We denote by O, H the circumcenter, and orthocenter of triangle ABC. Let J+, J− be respectively the inner and outer isodynamic points of the triangle. Though the last two are known in literature as the common two points of the Apollonius circles, L. Evans [1] gives a direct relation between them and the Napoleonic con- figuration, defining them as the perspectors of the triangle of reflections A′B′C′ A note on the anticomplements of the Fermat points 123 with each of the Fermat triangles A+B+C+, and A−B−C−. They appear as X15, X16 in [4]. Furthermore, let W+, W− be the Wernau points of triangle ABC. These points are known as the common points of the folloing triads of circles: AB+C+, BC+A+, CA+B+, and respectively AB−C−, BC−A−, CA−B− [3]. According to the above terminology, W+, W− are the inversive associates of the Fermat points F+, and F−. They appear as X1337 and X1338 in [4]. We conclude with a list of concyclic quadruples involving these triangle centers. The first one is an immediate consequence of the famous Lester circle theorem [5]. The other results have been verified with the aid of Mathematica. Theorem 3. The following quadruples of points are concyclic: (i) P+, P−, O, H; (ii) P+, P−, F+, J+; ′ (ii ) P+, P−, F−, J−; (iii) P+, U+, F+, O; ′ (iii ) P−, U−, F−, O; (iv) P+, U−, F+, W+; ′ (iv ) P−, U+, F−, W−; (v) U+, J+, W+, W−; ′ (v ) U−, J−, W+, W−. References [1] L. Evans, Some configurations of triangle centers, Forum Geom., 3 (2003) 49–56. [2] W. Gallatly, The Modern Geometry of the Triangle, 2nd ed., London: Hodgson, 1913. [3] D. Grinberg, Hyacinthos messages 6874, 6881, 6882, April, 2003. [4] C. Kimberling, Encyclopedia of Triangle Centers, available at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html. [5] J. Lester, Triangles III: Complex Triangle Functions, Aequationes Math., 53 (1997) 4–35. [6] C. Pohoata, On the Parry reflection point, Forum Geom., 8 (2008) 43–48. Cosmin Pohoata: 13 Pridvorului Street, Bucharest, Romania 010014 E-mail address: pohoata [email protected].
Load more

Recommended publications
  • Investigating Centers of Triangles: the Fermat Point
    Investigating Centers of Triangles: The Fermat Point A thesis submitted to the Miami University Honors Program in partial fulfillment of the requirements for University Honors with Distinction by Katherine Elizabeth Strauss May 2011 Oxford, Ohio ABSTRACT INVESTIGATING CENTERS OF TRIANGLES: THE FERMAT POINT By Katherine Elizabeth Strauss Somewhere along their journey through their math classes, many students develop a fear of mathematics. They begin to view their math courses as the study of tricks and often seemingly unsolvable puzzles. There is a demand for teachers to make mathematics more useful and believable by providing their students with problems applicable to life outside of the classroom with the intention of building upon the mathematics content taught in the classroom. This paper discusses how to integrate one specific problem, involving the Fermat Point, into a high school geometry curriculum. It also calls educators to integrate interesting and challenging problems into the mathematics classes they teach. In doing so, a teacher may show their students how to apply the mathematics skills taught in the classroom to solve problems that, at first, may not seem directly applicable to mathematics. The purpose of this paper is to inspire other educators to pursue similar problems and investigations in the classroom in order to help students view mathematics through a more useful lens. After a discussion of the Fermat Point, this paper takes the reader on a brief tour of other useful centers of a triangle to provide future researchers and educators a starting point in order to create relevant problems for their students. iii iv Acknowledgements First of all, thank you to my advisor, Dr.
    [Show full text]
  • On the Fermat Point of a Triangle Jakob Krarup, Kees Roos
    280 NAW 5/18 nr. 4 december 2017 On the Fermat point of a triangle Jakob Krarup, Kees Roos Jakob Krarup Kees Roos Department of Computer Science Faculty of EEMCS University of Copenhagen, Denmark Technical University Delft [email protected] [email protected] Research On the Fermat point of a triangle For a given triangle 9ABC, Pierre de Fermat posed around 1640 the problem of finding a lem has been solved but it is not known by 2 point P minimizing the sum sP of the Euclidean distances from P to the vertices A, B, C. whom. As will be made clear in the sec- Based on geometrical arguments this problem was first solved by Torricelli shortly after, tion ‘Duality of (F) and (D)’, Torricelli’s ap- by Simpson in 1750, and by several others. Steeped in modern optimization techniques, proach already reveals that the problems notably duality, however, Jakob Krarup and Kees Roos show that the problem admits a (F) and (D) are dual to each other. straightforward solution. Using Simpson’s construction they furthermore derive a formula Our first aim is to show that application expressing sP in terms of the given triangle. This formula appears to reveal a simple re- of duality in conic optimization yields the lationship between the area of 9ABC and the areas of the two equilateral triangles that solution of (F), straightforwardly. Devel- occur in the so-called Napoleon’s Theorem. oped in the 1990s, the field of conic optimi- zation (CO) is a generalization of the more We deal with two problems in planar geo- Problem (D) was posed in 1755 by well-known field of linear optimization (LO).
    [Show full text]
  • Mathematical Gems
    AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL 1 Mathematical Gems ROSS HONSBERGER MATHEMATICAL GEMS FROM ELEMENTARY COMBINATORICS, NUMBER THEORY, AND GEOMETRY By ROSS HONSBERGER THE DOLCIANI MATHEMATICAL EXPOSITIONS Published by THE MArrHEMATICAL ASSOCIATION OF AMERICA Committee on Publications EDWIN F. BECKENBACH, Chairman 10.1090/dol/001 The Dolciani Mathematical Expositions NUMBER ONE MATHEMATICAL GEMS FROM ELEMENTARY COMBINATORICS, NUMBER THEORY, AND GEOMETRY By ROSS HONSBERGER University of Waterloo Published and Distributed by THE MATHEMATICAL ASSOCIATION OF AMERICA © 1978 by The Mathematical Association of America (Incorporated) Library of Congress Catalog Card Number 73-89661 Complete Set ISBN 0-88385-300-0 Vol. 1 ISBN 0-88385-301-9 Printed in the United States of Arnerica Current printing (last digit): 10 9 8 7 6 5 4 3 2 1 FOREWORD The DOLCIANI MATHEMATICAL EXPOSITIONS serIes of the Mathematical Association of America came into being through a fortuitous conjunction of circumstances. Professor-Mary P. Dolciani, of Hunter College of the City Uni­ versity of New York, herself an exceptionally talented and en­ thusiastic teacher and writer, had been contemplating ways of furthering the ideal of excellence in mathematical exposition. At the same time, the Association had come into possession of the manuscript for the present volume, a collection of essays which seemed not to fit precisely into any of the existing Associa­ tion series, and yet which obviously merited publication because of its interesting content and lucid expository style. It was only natural, then, that Professor Dolciani should elect to implement her objective by establishing a revolving fund to initiate this series of MATHEMATICAL EXPOSITIONS.
    [Show full text]
  • IX Geometrical Olympiad in Honour of I.F.Sharygin Final Round. Ratmino, 2013, August 1 Solutions First Day. 8 Grade 8.1. (N
    IX Geometrical Olympiad in honour of I.F.Sharygin Final round. Ratmino, 2013, August 1 Solutions First day. 8 grade 8.1. (N. Moskvitin) Let ABCDE be a pentagon with right angles at vertices B and E and such that AB = AE and BC = CD = DE. The diagonals BD and CE meet at point F. Prove that FA = AB. First solution. The problem condition implies that the right-angled triangles ABC and AED are equal, thus the triangle ACD is isosceles (see fig. 8.1a). Then ∠BCD = ∠BCA + ∠ACD = = ∠EDA + ∠ADC = ∠CDE. Therefore, the isosceles triangles BCD and CDE are equal. Hence ∠CBD = ∠CDB = ∠ECD = ∠DEC. Since the triangle CFD is isosceles and BD = CE, we obtain that BF = FE. Therefore BFE 180◦ − 2 FCD 4ABF = 4AEF. Then AFB = ∠ = ∠ = 90◦ − ECD = 90◦ − DBC = ABF, ∠ 2 2 ∠ ∠ ∠ hence AB = AF, QED. BB BB CC CC PP FF FF AA AA AA DD AA DD EE EE Fig. 8.1а Fig. 8.1b Second solution. Let BC meet DE at point P (see fig. 8.1b). Notice that ∠CBD = ∠CDB = = ∠DBE, i.e., BD is the bisector of ∠CBE. Thus F is the incenter of 4PBE. Since the quadrilateral PBAE is cyclic and symmetrical, we obtain that A is the midpoint of arc BE of the circle (PBE). Therefore, by the trefoil theorem we get AF = AB, QED. Remark. The problem statement holds under the weakened condition of equality of side lengths. It is sufficient to say that AB = AE and BC = CD = DE. 8.2. (D. Shvetsov) Two circles with centers O1 and O2 meet at points A and B.
    [Show full text]
  • From Euler to Ffiam: Discovery and Dissection of a Geometric Gem
    From Euler to ffiam: Discovery and Dissection of a Geometric Gem Douglas R. Hofstadter Center for Research on Concepts & Cognition Indiana University • 510 North Fess Street Bloomington, Indiana 47408 December, 1992 ChapterO Bewitched ... by Circles, Triangles, and a Most Obscure Analogy Although many, perhaps even most, mathematics students and other lovers of mathematics sooner or later come across the famous Euler line, somehow I never did do so during my student days, either as a math major at Stanford in the early sixties or as a math graduate student at Berkeley for two years in the late sixties (after which I dropped out and switched to physics). Geometry was not by any means a high priority in the math curriculum at either institution. It passed me by almost entirely. Many, many years later, however, and quite on my own, I finally did become infatuated - nay, bewitched- by geometry. Just plane old Euclidean geometry, I mean. It all came from an attempt to prove a simple fact about circles that I vaguely remembered from a course on complex variables that I took some 30 years ago. From there on, the fascination just snowballed. I was caught completely off guard. Never would I have predicted that Doug Hofstadter, lover of number theory and logic, would one day go off on a wild Euclidean-geometry jag! But things are unpredictable, and that's what makes life interesting. I especially came to love triangles, circles, and their unexpectedly profound interrelations. I had never appreciated how intimately connected these two concepts are. Associated with any triangle are a plentitude of natural circles, and conversely, so many beautiful properties of circles cannot be defined except in terms of triangles.
    [Show full text]
  • Concurrency of Four Euler Lines
    Forum Geometricorum b Volume 1 (2001) 59–68. bbb FORUM GEOM ISSN 1534-1178 Concurrency of Four Euler Lines Antreas P. Hatzipolakis, Floor van Lamoen, Barry Wolk, and Paul Yiu Abstract. Using tripolar coordinates, we prove that if P is a point in the plane of triangle ABC such that the Euler lines of triangles PBC, AP C and ABP are concurrent, then their intersection lies on the Euler line of triangle ABC. The same is true for the Brocard axes and the lines joining the circumcenters to the respective incenters. We also prove that the locus of P for which the four Euler lines concur is the same as that for which the four Brocard axes concur. These results are extended to a family Ln of lines through the circumcenter. The locus of P for which the four Ln lines of ABC, PBC, AP C and ABP concur is always a curve through 15 finite real points, which we identify. 1. Four line concurrency Consider a triangle ABC with incenter I. It is well known [13] that the Euler lines of the triangles IBC, AIC and ABI concur at a point on the Euler line of ABC, the Schiffler point with homogeneous barycentric coordinates 1 a(s − a) b(s − b) c(s − c) : : . b + c c + a a + b There are other notable points which we can substitute for the incenter, so that a similar statement can be proven relatively easily. Specifically, we have the follow- ing interesting theorem. Theorem 1. Let P be a point in the plane of triangle ABC such that the Euler lines of the component triangles PBC, AP C and ABP are concurrent.
    [Show full text]
  • Equilateral Triangles Formed by the Centers of Erected Triangles
    International Journal of Computer Discovered Mathematics (IJCDM) ISSN 2367-7775 ©IJCDM Volume 6, 2021, pp. 43{67 Received 15 February 2021. Published on-line 15 April 2021 web: http://www.journal-1.eu/ ©The Author(s) This article is published with open access1. Equilateral Triangles formed by the Centers of Erected Triangles Stanley Rabinowitza and Ercole Suppab a 545 Elm St Unit 1, Milford, New Hampshire 03055, USA e-mail: [email protected] web: http://www.StanleyRabinowitz.com/ b Via B. Croce 54, 64100 Teramo, Italia e-mail: [email protected] Abstract. Related triangles are erected outward on the sides of a triangle. A triangle center is constructed in each of these triangles. We use a computer to find instances where these three centers form an equilateral triangle. Keywords. triangle geometry, erected triangles, equilateral triangles, computer- discovered mathematics, GeometricExplorer. Mathematics Subject Classification (2020). 51M04, 51-08. 1. Introduction The well-known result, known as Napoleon's Theorem, states that if equilateral triangles are erected outward on the sides of an arbitrary triangle ABC, then their centers (G, H, and I) form an equilateral triangle (Figure 1). Figure 1. Outer Napoleon Triangle 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. 43 44 Equilateral Triangles formed by the Centers of Erected Triangles Triangle GHI is called the outer Napoleon triangle of 4ABC. As is also well known, the equilateral triangles can be erected inward (Figure 2).
    [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]
  • The Fermat Point Aild the Steiner Problem J.N
    Virginia Journal of Science Volume 51, Number 1 Spring 2000 The Fermat Point aild The Steiner Problem J.N. Boyd and P.N. Raychowdhury, Department of Mathematical Sciences, Virginia Commonwealth University, Richmond, Virginia 23284-2014 ABSTRACT We revisit the convex coordinates of the Fermat point of a triangle. We have already computed these convex coordinates in a general setting. In this note, we obtain the coordinates in the context of the Steiner problem. Thereafter, we pursue calculations suggested by the problem. INTRODUCTION We came upon a lovely problem in Strang's outstanding calculus text (Strang, 1991 ). The problem, to which the author attaches the name of the great Swiss geometer J. Steiner, may be stated in the following manner: Find the point Fon the perimeter or in the interior of A Vi V2 V3 so that the sum of the distances from F to the three vertices of the triangle has its least possible value. The problem is to minimize d1 + d2 + d3 where d1, di, d3 are the distances from F to Vi, V2, V3, respectively. The geometry of the problem is suggested by Figure 1. We shall eventually show that the point F for which d 1 + d2 + d3 is a minimum is located in the interior of AVi ViVi in such a way that L ViFV2 = L V2FV3 = L VPi = 120° provided that the largest of the three angles of the triangle (L Vi, L Vi, L Vi) has a measure less than 120°. If one of these three angles equals or exceeds 120°, then the point Fis located at the vertex of that angle.
    [Show full text]
  • Heptagonal Triangles and Their Companions Paul Yiu Department of Mathematical Sciences, Florida Atlantic University
    Heptagonal Triangles and Their Companions Paul Yiu Department of Mathematical Sciences, Florida Atlantic University, [email protected] MAA Florida Sectional Meeting February 13–14, 2009 Florida Gulf Coast University 1 2 P. Yiu Abstract. A heptagonal triangle is a non-isosceles triangle formed by three vertices of a regular hep- π 2π 4π tagon. Its angles are 7 , 7 and 7 . As such, there is a unique choice of a companion heptagonal tri- angle formed by three of the remaining four ver- tices. Given a heptagonal triangle, we display a number of interesting companion pairs of heptag- onal triangles on its nine-point circle and Brocard circles. Among other results on the geometry of the heptagonal triangle, we prove that the circum- center and the Fermat points of a heptagonal tri- angle form an equilateral triangle. The proof is an interesting application of Lester’s theorem that the Fermat points, the circumcenter and the nine-point center of a triangle are concyclic. The heptagonal triangle 3 The heptagonal triangle T and its circumcircle C B a b c O A 4 P. Yiu The companion of T C B A′ O D = residual vertex A B′ C′ The heptagonal triangle 5 Relation between the sides of the heptagonal triangle b a b c a c a b c b a b b2 = a(c + a) c2 = a(a + b) 6 P. Yiu Archimedes’ construction (Arabic tradition) K H B A b a c C B AA K H F E The heptagonal triangle 7 Division of segment: a neusis construction Let BHPQ be a square, with one side BH sufficiently extended.
    [Show full text]
  • On Viviani's Theorem and Its Extensions
    On Viviani’s Theorem and its Extensions Elias Abboud Beit Berl College, Doar Beit Berl, 44905 Israel Email: eabboud @beitberl.ac.il April 12, 2009 Abstract Viviani’s theorem states that the sum of distances from any point in- side an equilateral triangle to its sides is constant. We consider extensions of the theorem and show that any convex polygon can be divided into par- allel segments such that the sum of the distances of the points to the sides on each segment is constant. A polygon possesses the CVS property if the sum of the distances from any inner point to its sides is constant. An amazing result, concerning the converse of Viviani’s theorem is deduced; Three non-collinear points which have equal sum of distances to the sides inside a convex polygon, is sufficient for possessing the CVS property. For concave polygons the situation is quite different, while for polyhedra analogous results are deduced. Key words: distance sum function, CVS property, isosum segment, isosum cross section. AMS subject classification : 51N20,51F99, 90C05. 1 Introduction Let be a polygon or polyhedron, consisting of both boundary and interior arXiv:0903.0753v3 [math.MG] 12 Apr 2009 points.P Define a distance sum function : R, where for each point P the value (P ) is defined as the sum ofV theP distances → from the point P to∈P the sides (faces)V of . We say thatP has the constant Viviani sum property, abbreviated by the ”CVS property”,P if and only if the function is constant. Viviani (1622-1703), who was a student andV assistant of Galileo, discovered the theorem which states that equilateral triangles have the CVS property.
    [Show full text]
  • Topics in Elementary Geometry
    Topics in Elementary Geometry Second Edition O. Bottema (deceased) Topics in Elementary Geometry Second Edition With a Foreword by Robin Hartshorne Translated from the Dutch by Reinie Erne´ 123 O. Bottema Translator: (deceased) Reinie Ern´e Leiden, The Netherlands [email protected] ISBN: 978-0-387-78130-3 e-ISBN: 978-0-387-78131-0 DOI: 10.1007/978-0-387-78131-0 Library of Congress Control Number: 2008931335 Mathematics Subject Classification (2000): 51-xx This current edition is a translation of the Second Dutch Edition of, Hoofdstukken uit de Elementaire Meetkunde, published by Epsilon-Uitgaven, 1987. c 2008 Springer Science+Business Media, LLC All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connec- tion with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper 987654321 springer.com At school I was good in mathematics; now I discovered that I found the so-called higher mathematics – differential and integral calculus – easier than the complicated (but elementary) plane geometry.
    [Show full text]