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Angle Chasing
Angle Chasing Ray Li June 12, 2017 1 Facts you should know 1. Let ABC be a triangle and extend BC past C to D: Show that \ACD = \BAC + \ABC: 2. Let ABC be a triangle with \C = 90: Show that the circumcenter is the midpoint of AB: 3. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove that H is the incenter of 4DEF . 4. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove (i) that A; E; F; H lie on a circle diameter AH and (ii) that B; E; F; C lie on a circle with diameter BC. 5. Let ABC be a triangle with circumcenter O and orthocenter H: Show that \BAH = \CAO: 6. Let ABC be a triangle with circumcenter O and orthocenter H and let AH and AO meet the circumcircle at D and E, respectively. Show (i) that H and D are symmetric with respect to BC; and (ii) that H and E are symmetric with respect to the midpoint BC: 7. Let ABC be a triangle with altitudes AD; BE; and CF: Let M be the midpoint of side BC. Show that ME and MF are tangent to the circumcircle of AEF: 8. Let ABC be a triangle with incenter I, A-excenter Ia, and D the midpoint of arc BC not containing A on the circumcircle. Show that DI = DIa = DB = DC: 9. Let ABC be a triangle with incenter I and D the midpoint of arc BC not containing A on the circumcircle. -
The Isogonal Tripolar Conic
Forum Geometricorum b Volume 1 (2001) 33–42. bbb FORUM GEOM The Isogonal Tripolar Conic Cyril F. Parry Abstract. In trilinear coordinates with respect to a given triangle ABC,we define the isogonal tripolar of a point P (p, q, r) to be the line p: pα+qβ+rγ = 0. We construct a unique conic Φ, called the isogonal tripolar conic, with respect to which p is the polar of P for all P . Although the conic is imaginary, it has a real center and real axes coinciding with the center and axes of the real orthic inconic. Since ABC is self-conjugate with respect to Φ, the imaginary conic is harmonically related to every circumconic and inconic of ABC. In particular, Φ is the reciprocal conic of the circumcircle and Steiner’s inscribed ellipse. We also construct an analogous isotomic tripolar conic Ψ by working with barycentric coordinates. 1. Trilinear coordinates For any point P in the plane ABC, we can locate the right projections of P on the sides of triangle ABC at P1, P2, P3 and measure the distances PP1, PP2 and PP3. If the distances are directed, i.e., measured positively in the direction of −→ −→ each vertex to the opposite side, we can identify the distances α =PP1, β =PP2, −→ γ =PP3 (Figure 1) such that aα + bβ + cγ =2 where a, b, c, are the side lengths and area of triangle ABC. This areal equation for all positions of P means that the ratio of the distances is sufficient to define the trilinear coordinates of P (α, β, γ) where α : β : γ = α : β : γ. -
Cyclic Quadrilaterals — the Big Picture Yufei Zhao [email protected]
Winter Camp 2009 Cyclic Quadrilaterals Yufei Zhao Cyclic Quadrilaterals | The Big Picture Yufei Zhao [email protected] An important skill of an olympiad geometer is being able to recognize known configurations. Indeed, many geometry problems are built on a few common themes. In this lecture, we will explore one such configuration. 1 What Do These Problems Have in Common? 1. (IMO 1985) A circle with center O passes through the vertices A and C of triangle ABC and intersects segments AB and BC again at distinct points K and N, respectively. The circumcircles of triangles ABC and KBN intersects at exactly two distinct points B and M. ◦ Prove that \OMB = 90 . B M N K O A C 2. (Russia 1995; Romanian TST 1996; Iran 1997) Consider a circle with diameter AB and center O, and let C and D be two points on this circle. The line CD meets the line AB at a point M satisfying MB < MA and MD < MC. Let K be the point of intersection (different from ◦ O) of the circumcircles of triangles AOC and DOB. Show that \MKO = 90 . C D K M A O B 3. (USA TST 2007) Triangle ABC is inscribed in circle !. The tangent lines to ! at B and C meet at T . Point S lies on ray BC such that AS ? AT . Points B1 and C1 lies on ray ST (with C1 in between B1 and S) such that B1T = BT = C1T . Prove that triangles ABC and AB1C1 are similar to each other. 1 Winter Camp 2009 Cyclic Quadrilaterals Yufei Zhao A B S C C1 B1 T Although these geometric configurations may seem very different at first sight, they are actually very related. -
The Role of Euclidean Geometry in High School
JOURNAL OF MATHEMATICAL BEHAVIOR 15, 221-237 (1996) The Role of Euclidean Geometry in High School HUNG-HSI Wu University of California When I was a high school student long ago, the need to study Euclidean geome- try was taken for granted. In fact, the novelty of learning to prove something was so overwhelming to me and some of my friends that, years later, we would look back at Euclidean geometry as the high point of our mathematics education. In recent years, the value of Euclid has depreciated considerably. A vigorous debate is now going on about how much, if any, of his work is still relevant. In this article, I would like to give my perspective on this subject, and in so doing, I will be taking a philosophical tour, so to speak, of a small portion of mathematics. This may not be so surprising because after all, in life as in mathematics, one does need a little philosophy for guidance from time to time. The bone of contention in the geometry curriculum is of course how many of the traditional “two-column proofs” should be retained. Let me first briefly indicate the range of available options in this regard: 1. A traditional text in use in a local high school in Berkeley not too long ago begins on page 1 with the undefined terms of the axioms, and formal proofs start on page 22. There is no motivation or explanation of the whys and hows of an axiomatic system. 2. A recently published text does experimental geometry (i.e., “hands-on ge- ometry” with no proofs) all the way until the last 130 pages of its 700 pages of exposition. -
Degree of Triangle Centers and a Generalization of the Euler Line
Beitr¨agezur Algebra und Geometrie Contributions to Algebra and Geometry Volume 51 (2010), No. 1, 63-89. Degree of Triangle Centers and a Generalization of the Euler Line Yoshio Agaoka Department of Mathematics, Graduate School of Science Hiroshima University, Higashi-Hiroshima 739–8521, Japan e-mail: [email protected] Abstract. We introduce a concept “degree of triangle centers”, and give a formula expressing the degree of triangle centers on generalized Euler lines. This generalizes the well known 2 : 1 point configuration on the Euler line. We also introduce a natural family of triangle centers based on the Ceva conjugate and the isotomic conjugate. This family contains many famous triangle centers, and we conjecture that the de- gree of triangle centers in this family always takes the form (−2)k for some k ∈ Z. MSC 2000: 51M05 (primary), 51A20 (secondary) Keywords: triangle center, degree of triangle center, Euler line, Nagel line, Ceva conjugate, isotomic conjugate Introduction In this paper we present a new method to study triangle centers in a systematic way. Concerning triangle centers, there already exist tremendous amount of stud- ies and data, among others Kimberling’s excellent book and homepage [32], [36], and also various related problems from elementary geometry are discussed in the surveys and books [4], [7], [9], [12], [23], [26], [41], [50], [51], [52]. In this paper we introduce a concept “degree of triangle centers”, and by using it, we clarify the mutual relation of centers on generalized Euler lines (Proposition 1, Theorem 2). Here the term “generalized Euler line” means a line connecting the centroid G and a given triangle center P , and on this line an infinite number of centers lie in a fixed order, which are successively constructed from the initial center P 0138-4821/93 $ 2.50 c 2010 Heldermann Verlag 64 Y. -
Chapter 10 Isotomic and Isogonal Conjugates
Chapter 10 Isotomic and isogonal conjugates 10.1 Isotomic conjugates The Gergonne and Nagel points are examples of isotomic conjugates. Two points P and Q (not on any of the side lines of the reference triangle) are said to be isotomic conjugates if their respective traces are symmetric with respect to the midpoints of the corresponding sides. Thus, BX = X′C, CY = Y ′A, AZ = Z′B. B X′ Z X Z′ P P • C A Y Y ′ We shall denote the isotomic conjugate of P by P •. If P =(x : y : z), then 1 1 1 P • = : : =(yz : zx : xy). x y z 320 Isotomic and isogonal conjugates 10.1.1 The Gergonne and Nagel points 1 1 1 Ge = s a : s b : s c , Na =(s a : s : s c). − − − − − − Ib A Ic Z′ Y Y Z I ′ Na Ge B C X X′ Ia 10.1 Isotomic conjugates 321 10.1.2 The isotomic conjugate of the orthocenter The isotomic conjugate of the orthocenter is the point 2 2 2 2 2 2 2 2 2 H• =(b + c a : c + a b : a + b c ). − − − Its traces are the pedals of the deLongchamps point Lo, the reflection of H in O. A Z′ L Y o O Y ′ H• Z H B C X X′ Exercise 1. Let XYZ be the cevian triangle of H•. Show that the lines joining X, Y , Z to the midpoints of the corresponding altitudes are concurrent. What is the common point? 1 2. Show that H• is the perspector of the triangle of reflections of the centroid G in the sidelines of the medial triangle. -
Prove It Classroom
How to Prove it ClassRoom SHAILESH SHIRALI Euler’s formula for the area of a pedal triangle Given a triangle ABC and a point P in the plane of ABC In this episode of “How (note that P does not have to lie within the triangle), the To Prove It”, we prove pedal triangle of P with respect to ABC is the triangle △ a beautiful and striking whose vertices are the feet of the perpendiculars drawn from formula first found by P to the sides of ABC. See Figure 1. The pedal triangle relates Leonhard Euler; it gives the area of the pedal triangle in a natural way to the parent triangle, and we may wonder of a point with reference whether there is a convenient formula giving the area of the to another triangle. pedal triangle in terms of the parameters of the parent triangle. The great 18th-century mathematician Euler found just such a formula (given in Box 1). It is a compact and pleasing result, and it expresses the area of the pedal triangle in terms of the radius R of the circumcircle of ABC and the △ distance between P and the centre O of the circumcircle. A O: circumcentre of ABC E △ P: arbitrary point Area ( DEF) 1 OP2 F △ = 1 P Area ( ABC) 4 − R2 O △ ( ) B D C Figure 1. Euler's formula for the area of the pedal triangle of an arbitrary point Keywords: Circle theorem, pedal triangle, power of a point, Euler, sine rule, extended sine rule, Wallace-Simson theorem Azim Premji University At Right Angles, July 2018 95 1 A E F P O B D C Figure 2. -
Let's Talk About Symmedians!
Let's Talk About Symmedians! Sammy Luo and Cosmin Pohoata Abstract We will introduce symmedians from scratch and prove an entire collection of interconnected results that characterize them. Symmedians represent a very important topic in Olympiad Geometry since they have a lot of interesting properties that can be exploited in problems. But first, what are they? Definition. In a triangle ABC, the reflection of the A-median in the A-internal angle bi- sector is called the A-symmedian of triangle ABC. Similarly, we can define the B-symmedian and the C-symmedian of the triangle. A I BC X M Figure 1: The A-symmedian AX Do we always have symmedians? Well, yes, only that we have some weird cases when for example ABC is isosceles. Then, if, say AB = AC, then the A-median and the A-internal angle bisector coincide; thus, the A-symmedian has to coincide with them. Now, symmedians are concurrent from the trigonometric form of Ceva's theorem, since we can just cancel out the sines. This concurrency point is called the symmedian point or the Lemoine point of triangle ABC, and it is usually denoted by K. //As a matter of fact, we have the more general result. Theorem -1. Let P be a point in the plane of triangle ABC. Then, the reflections of the lines AP; BP; CP in the angle bisectors of triangle ABC are concurrent. This concurrency point is called the isogonal conjugate of the point P with respect to the triangle ABC. We won't dwell much on this more general notion here; we just prove a very simple property that will lead us immediately to our first characterization of symmedians. -
Three Properties of the Symmedian Point / Darij Grinberg
Three properties of the symmedian point / Darij Grinberg 1. On isogonal conjugates The purpose of this note is to synthetically establish three results about the sym- median point of a triangle. Two of these don’tseem to have received synthetic proofs hitherto. Before formulating the results, we remind about some fundamentals which we will later use, starting with the notion of isogonal conjugates. B P Q A C Fig. 1 The de…nition of isogonal conjugates is based on the following theorem (Fig. 1): Theorem 1. Let ABC be a triangle and P a point in its plane. Then, the re‡ections of the lines AP; BP; CP in the angle bisectors of the angles CAB; ABC; BCA concur at one point. This point is called the isogonal conjugate of the point P with respect to triangle ABC: We denote this point by Q: Note that we work in the projective plane; this means that in Theorem 1, both the point P and the point of concurrence of the re‡ections of the lines AP; BP; CP in the angle bisectors of the angles CAB; ABC; BCA can be in…nite points. 1 We are not going to prove Theorem 1 here, since it is pretty well-known and was showed e. g. in [5], Remark to Corollary 5. Instead, we show a property of isogonal conjugates. At …rst, we meet a convention: Throughout the whole paper, we will make use of directed angles modulo 180: An introduction into this type of angles was given in [4] (in German). B XP ZP P ZQ XQ Q A C YP YQ Fig. -
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). -
The Secrets of Triangles: a Mathematical Journey
2 3 Published 2012 by Prometheus Books The Secrets of Triangles: A Mathematical Journey. Copyright © 2012 by Alfred S. Posamentier and Ingmar Lehmann. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, digital, electronic, mechanical, photocopying, recording, or otherwise, or conveyed via the Internet or a website without prior written permission of the publisher, except in the case of brief quotations embodied in critical articles and reviews. Cover image © Media Bakery/Glenn Mitsui Jacket design by Jacqueline Nasso Cooke Inquiries should be addressed to Prometheus Books 59 John Glenn Drive Amherst, New York 14228–2119 OICE: 716–691–0133 FAX: 716–691–0137 WWW.PROMETHE SBOOKS.COM 16 15 14 13 12 5 4 3 2 1 Library of Congress Cataloging-in-Publication Data Posamentier, Alfred S. The secrets of triangles : a mathematical journey / by Alfred S. Posamentier and Ingmar Lehmann. p. cm. Includes bibliographical references and index. ISBN 978–1–61614–587–3 (cloth : alk. paper) ISBN 978–1–61614–588–0 (ebook) 1. Trigonometry. 2. Triangle. I. Lehmann, Ingmar. II. Title. QA531.P67 2012 516'.154—dc23 2012013635 4 Printed in the nited States of America on acid-free paper 5 6 Acknowledgments Preface 1. Introduction to the Triangle 2. Concurrencies of a Triangle 3. Noteworthy Points in a Triangle 4. Concurrent Circles of a Triangle 5. Special Lines of a Triangle 6. seful Triangle Theorems 7. Areas of and within Triangles 8. Triangle Constructions 9. Inequalities in a Triangle 10. Triangles and Fractals Appendix Notes References Index 7 The authors wish to extend sincere thanks for proofreading and useful suggestions to Dr. -
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.