Geometry Transformation

A TASTE OF MATHEMATICS AIME–T–ON LES MATHEMATIQUES ´ Volume / Tome XII TRANSFORMATIONAL GEOMETRY Edward J. Barbeau University of TorontoThe ATOM series The booklets in the series, A Taste of Mathematics, are published by the Canadian Mathematical Society (CMS). They are designed as enrichment materials for high school students with an interest in and aptitude for mathematics. Some booklets in the series will also cover the materials useful for mathematical competitions at national and international levels. La collection ATOM Publi´es par la Soci´et´e math´ematique du Canada (SMC), les livrets de la collection Aime-t-on les math´ematiques (ATOM) sont destin´es au perfectionnement des ´etudiants du cycle secondaire qui manifestent un int´erˆet et des aptitudes pour les math´ematiques. Certains livrets de la collection ATOM servent ´egalement de mat´eriel de pr´eparation aux concours de math´ematiques sur l’´echiquier national et international. Editorial Board / Conseil de r´edaction Editor-in-Chief / R´edacteur-en-chef Bruce Shawyer Memorial University of Newfoundland / Universit´e Memorial de Terre-Neuve Associate Editors / R´edacteurs associ´es Edward J. Barbeau University of Toronto / Universit´e de Toronto Malgorzata Dubiel Simon Fraser University / Universit´e Simon Fraser Joseph Khoury University of Ottawa / Universit´e d’Ottawa Antony Thompson Dalhousie University / Universit´e Dalhousie Managing Editor / R´edacteur-g´erant Johan Rudnick CMS / SMCA TASTE OF MATHEMATICS AIME–T–ON LES MATHEMATIQUES ´ Volume / Tome XII TRANSFORMATIONAL GEOMETRY Edward J. Barbeau University of TorontoPublished by the Canadian Mathematical Society, Ottawa, Ontario and produced by the CMS ATOM Office, St. John’s, NL, Canada Publi´e par la Soci´et´e math´ematique du Canada, Ottawa (Ontario) et produit par le Bureau ATOM de la SMC, St. John’s, NL, Canada Printed in Canada by / imprim´e au Canada par Thistle Printing Limited ISBN 978-0-919558-23-6 All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information or retrieval system now known or to be invented, without permission in writing from the publisher: The Canadian Mathematical Society, 209-1725 St. Laurent Blvd., Ottawa, ON, CANADA K1G 3V4, except so far as may be allowed by law. Tous droits r´eserv´es. Aucune partie de cet ouvrage ne peut ˆetre reproduite ou utilis´ee par quelque proc´ed´e ou quelque fa¸con que ce soit, y compris les m´ethodes ´electroniques ou m´ecaniques, les enregistrements ou les syst`emes de mise en m´emoire et d’information, sans l’accord pr´ealable ´ecrit de l’´editeur, la Soci´et´e math´ematique du Canada, 209-1725 Boul. St. Laurent, Ottawa, ON, CANADA K1G 3V4, sauf dans les limites prescrites par la loi. c 2011 Canadian Mathematical Society / Soci´et´e math´ematique du Canadai Table of Contents The Author i Foreword ii Notation and Terminology iii 1 Getting into the transformational spirit 1 2 Isometries 9 3 Using isometries to solve problems 15 4 Dilations 29 5 Using dilations to solve problems 33 6 Inversion in a circle 49 7 Problems involving inversions 55 The Author Edward Barbeau is professor emeritus of mathematics at the University of Toronto. He was born in Toronto and received his Bachelor of Arts and Master of Arts degrees from the University of Toronto before going to the University of Newcastle-upon-Tyne in England to gain his PhD with a thesis on functional analysis written under the supervision of F.F. Bonsall. After being assistant professor at the University of Western Ontario in London, Ontario for two years and a NATO research fellow at Yale University in New Haven, Connecticut, UAS, for one year, he accepted an appointment at the University of Toronto, where he has remained. Dr. Barbeau is a life member of the Mathematical Association of America, the American Mathematical Society and the Canadian Mathematical Society, and has served all three societies on various committees, particularly having to do with mathematics education. He has published a number of books directed to students of mathematics and their teachers, including Polynomials (Springer), Power Play (MAA), Fallacies, Flaws and Flimflam (MAA) and After Math (Wall & Emerson, Toronto), has frequently given talks and workshops at professional meetings and in schools, has worked with high school students preparing for Olympiad competitions and has on five occasions accompanied the Canadian team to the International Mathematical Olympiad. He is currently associate editor in charge of the Fallacies, Flaws and Flimflam column in the College Mathematics Journal and education editor for the Notes of the Canadian Mathematical Society. He is a former chairman of the Education Committee of the Canadian Mathematical Society. His honours include the Fellowship of the Ontario Institute for Studies in Education, the David Hilbert Award from the World Federation of National Mathematics Competitions and the Adrien Pouliot Award from the Canadian Mathematical Society.ii Foreword This book is intended for secondary students with some experience in school geometry. It is assumed that they have had enough elementary Euclidean geometry to cover theorems about congruences of triangles, properties of isosceles and right triangles, basic area theorems for triangles and quadrilaterals, properties of circles and concyclic quadrilaterals. It is expected that the reader will have been introduced to the definitions of translations, rotations and reflections, but has not used them as a tool for solving geometric problems. The modern student who wishes to master a topic has an ample supply of material that can be accessed on the Internet. In particular, the sites http://www.theartofproblemsolving.com and http://www.cut-the-knot.org are well worth a visit. In addition, she can play with situations using dynamic software such as The Geometer’s Sketchpad (http://www.dynamicgeometry.com) and Cabri (http://www.cabri.com). Nevertheless, it seems to be useful to have something that can be put on a bookshelf and can be used as a reference as well as a supply of a few nice sample problems and their solutions. It should not be read like a novel; the reader should pause at each definition and result to see whether it is are understood and try to look at examples of whatever is being discussed. Before reading the solutions of the problems, the reader should think about them first and try to solve them first. Many of the solutions are attributed to secondary students who participated in correspondence programs. I am grateful to them for providing difference perspectives that I was unaware of when I posed the problems. I would also like to express my sincere gratitude to Bruce Shawyer of Memorial University in Newfoundland for electronically realizing the many diagrams that a geometry book requires. I conclude this foreword with a some notation, terminology and standard results for the convenience of the reader.iii NOTATION AND TERMINOLOGY Altitude of a triangle: The line segment that passes through a vertex of a triangle and is perpendicular to the opposite side. Area of a figure: The area of a geometric figure is indicated by square brackets: [· · · ] Centroid of a triangle: The point at which the medians intersect. Circumcircle of a polygon: A circle that passes through the vertices of a polygon. Its radius is called the circumradius. Collinear points: Points that lie on a straight line. Concylic quadrilateral: A quadrilateral through whose four vertices lie on a circle. Congruent geometric figures: Figures either of which can be obtained from the other by a rigid transformation that preserves angles and lengths. Congruence is indicated by the notation ≡. Congruence theorems for triangles: Two triangles are congruent if any one of the folling holds: (1) Two sides and the contained angle of one are correspondingly equal to two sides and the contained angle of the second (SAS); (2) Three sides of one are correspondingly equal to three sides of the second (SSS); (3) Two angles of one and the side connecting them are correspondingly equal to two angles of the second and the side connecting them (ASA). Distance between points: The distance between points A and B is denoted either by |AB| or AB, depending on context. Incentre of a triangle: The point at which the three angle bisectors of a triangle intersect; the centre of the incircle of the triangle. Incircle of a polygon: A circle that is tangent to each side of a polygon. It radius is called the inradius. Median of a triangle: The line segment that joins a vertex of a triangle to the midpoint of the opposite side. Orthocentre of a triangle: The point at which the three altitudes of the triangle intersect. Pedal point: The point at which the altitude of a triangle from a vertex intersects the opposite side. Produced: A segment or half line is produced when the full line containing it is drawn; the full line is called the production of the segment or half line. Similar geometric figures: Geometric figures of the same shape, in which one is a scaled version of the other. Figures whose linear dimensions are in a fixed proportion. Similarity is indicated by ∼.1 1 Getting into the transformational spirit The purpose of this book is not just to introduce a few techniques for solving geometry problems, but to encourage a different way of thinking about geometry among students who may have just been given a standard Euclidean approach. The best way to begin is with a few examples. Problem 1.1. In the diagram, AD = BC and ∠ABD + ∠BDC = 180◦ . Show that ∠BAD = ∠BCD. ...... ..... ...... ..... ...... ..... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ..... ...... ..... ...... ................................................................................................................................................................................................................ . B A D C .. .. .............. .... ... ....... o Figure 1.1a. Solution 1. This is quite a tough problem to tackle until we get the inspiration to start moving things around. Note that we have two equal segments in the situation as well as a pair of angles that are supplementary (add up to 180◦ ). Let us exploit this. Turn over the triangle BCD so that the positions of B and D are interchanged and C goes to C ′ . ......................................................................................................... .. ............................................................................................................................................................................ B A D C′ .. .. . ........ o Figure 1.1b. Then ∠DBC′ + ∠ABD = ∠BDC + ∠ABD = 180◦ . and DC′ = BC = AD, so that A, B, C ′ are collinear and triangle ADC′ is isosceles, whereupon ∠BAD = ∠DC′B = ∠BCD. Solution 2. Another possibility is to lay one equal side on top of the other. Suppose that we detach triangle BCD and lay point B on D and C on A, so that D falls on a position E, as in the diagram.2 . ....................................................................................................... ............ ............. .............. .............. .............. .............. .............. .............. ............... .............. .............. .............. ............. .. B A D E .. . ..... ................... ................... ................... ... o Figure 1.3b. Since ∠ABD + ∠DEA = ∠ABD + ∠BDC = 180◦ , it follows that ABDE is a concyclic quadrilateral. Also, ∠BAD and ∠DAE are subtended by equal chords at the circumference of the circumcircle of ABDE, so they are equal. Hence ∠BAD = ∠DAE = ∠BCD. Problem 1.2. Suppose that we have a circle of radius r, and construct as in the diagram, a rectangle whose sides OP and OQ lie along perpendicular diameters. What is the length of P Q? . . . .. .. .... .... ....... ....... .... .............. ............. ...... ......... ......... ....... ..... .... .. .. . . . ................................................................................................... . ........................... .......................................... .......................................... ........................................... .......................................... . . . ............................ .......................................... . O Q P Figure 1.2. Solution. The answer is immediate once we flip the rectangle over onto itself, so that its diagonal goes from the centre O to the vertex on the circumference of the circle. Problem 1.3. The unit square is partitioned into four triangles and one quadrilateral with areas a, b, c, d and e as indicated in the diagram below. . .................................................................................................................................................................................................. . ........................... .......................................... .......................................... ........................................... .......................................... . ........................................................................................................................................... a b c d e Figure 1.3. Each of the three partitioning lines join a vertex to the midpoint of one of the opposite sides. Determine the values of a, b, c, d, e. Solution. It is straightforward to determine that a = 1 4 , b + d = 1 2 , c + e = d + e = 1 4 , and so c = d; this can be done for example by noting how the square can be covered by four non-overlapping right triangles with arms of length 1 and 1 2 . However, this gives us only four independent equations for five variables, and it is not clear how we can find another condition to nail the values down.3 However, we note that if we rotate the triangle with area e through 90◦ clockwise, it falls on part of the triangle with area d, and one can see that by expanding the linear dimensions by a factor of 2, it will exactly cover the larger triangle. (Another way of looking at it is to note that one can cover the triangle of area d by four copies of the triangle with area e.) Thus, d = 4e and we find that (a, b, c, d, e) =  1 4 , 3 10 , 1 5 , 1 5 , 1 20 . Problem 1.4. Suppose that a unit square is partitioned into 9 polygons by various lines joining vertices to mid-points of sides as shown in the diagram. What is the area of the quadrilateral T UV W in the middle? . .................................................................................................................................................................................................. . ........................... .......................................... . .......................................... ........................................... .......................................... . . . . . . . . ....................................................................................................................................... ....................................................................................................................................... A B C D P Q R S W U V T Figure 1.4a. Solution. It seems evident that T UV W is in fact a square. Let us pursue this a little. Upon reflection, we realize that our intuition is fed by the symmetry of the situation. So let us try to capture this ingredient. If we rotate the square through an angle of 90◦ about its centre, then A → B, B → C, C → D, D → A, P → Q, Q → R, R → S, S → P (we will use “→” to mean “goes to”). The segment AP falls on BQ. Since the rotation is through a right angle, AP ⊥ BQ. Similarly, BQ ⊥ CR, CR ⊥ DS, so that T UV W is at least a rectangle. But T , the intersection of AP and BQ falls on U, the intersection of BQ and CR, and we find that T → U, U → V , V → W and W → T ; thus T U = UV = V W = W T . To answer the question posed, let triangle AT Q be rotated about the point Q, so that A falls on D and T on T ′ . Then T ′DW T is a square equal to T UV W. Similarly, we can rotate triangles BUR, CUS and DW P to form a cross consisting of five congruent squares, one of which is T UV W and all of which have combined area equal to that of the square. Accordingly, the area of T UV W is one-fifth of that of ABCD.