Geometry with Trigonometry, Second Edition [2ed.] 0128050667, 978-0-12-805066-8

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Geometry with Trigonometry

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Geometry with Trigonometry 2nd Edition

Patrick D. Barry Professor Emeritus of Mathematics, School of Mathematical Sciences, University College, Cork.

“All things stand by proportion.” George Puttenham (1529–1590)

“Mathematics possesses not only truth, but supreme beauty – a beauty cold and austere like that of sculpture, and capable of stern perfection, such as only great art can show.” Bertrand Russell in The Principles of Mathematics (1872–1970)

Woodhead Publishing is an imprint of Elsevier 80 High Street, Sawston, Cambridge, CB22 3HJ, UK 225 Wyman Street, Waltham, MA 02451, USA Langford Lane, Kidlington, OX5 1GB, UK Copyright © 2016 Elsevier Ltd. All rights reserved. First Published by Horwood Publishing Limited, 2001 Reprinted by Woodhead Publishing Limited, 2014 © P. D. Barry, 2001, 2014. 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 storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein). Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein.

ISBN: 978-0-12-805066-8 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress For information on all Woodhead Publishing Publications visit our website at http://store.elsevier.com/

To my late wife Fran, my children Conor, Una and Brian, and my several grandchildren

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Contents About the author . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xvii Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xix 1 Preliminaries 1.1 Historical note . . . . . . . . . . . . . . . . . . . 1.2 Note on deductive reasoning . . . . . . . . . . . . 1.2.1 Definitions . . . . . . . . . . . . . . . . . . . . . 1.2.2 Proof . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Euclid’s The Elements . . . . . . . . . . . . . . . 1.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Definitions . . . . . . . . . . . . . . . . . . . . . 1.3.3 Postulates and common notions . . . . . . . . . . 1.3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.5 Congruence . . . . . . . . . . . . . . . . . . . . 1.3.6 Quantities or magnitudes . . . . . . . . . . . . . 1.4 Our approach . . . . . . . . . . . . . . . . . . . 1.4.1 Type of course . . . . . . . . . . . . . . . . . . . 1.4.2 Need for preparation . . . . . . . . . . . . . . . . 1.5 Revision of geometrical concepts . . . . . . . . . 1.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.2 The basic shapes . . . . . . . . . . . . . . . . . . 1.5.3 Distance; degree-measure of an angle . . . . . . . 1.5.4 Our treatment of congruence . . . . . . . . . . . 1.5.5 Parallel lines . . . . . . . . . . . . . . . . . . . . 1.6 Pre-requisites . . . . . . . . . . . . . . . . . . . 1.6.1 Set notation . . . . . . . . . . . . . . . . . . . . 1.6.2 Classical algebra . . . . . . . . . . . . . . . . . . 1.6.3 Other algebra . . . . . . . . . . . . . . . . . . . 1.6.4 Distinctive property of real numbers among fields

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1 1 3 3 3 4 4 4 6 7 7 8 9 9 9 10 10 10 15 17 18 19 19 20 20 20

2 Basic shapes of geometry 2.1 Lines, segments and half-lines . . . . . . . . . . . . 2.1.1 Plane, points, lines . . . . . . . . . . . . . . . . . . 2.1.2 Natural order on a line . . . . . . . . . . . . . . . . 2.1.3 Reciprocal orders . . . . . . . . . . . . . . . . . . 2.1.4 Segments . . . . . . . . . . . . . . . . . . . . . . . 2.1.5 Half-lines . . . . . . . . . . . . . . . . . . . . . . 2.2 Open and closed half-planes . . . . . . . . . . . . . 2.2.1 Convex sets . . . . . . . . . . . . . . . . . . . . . 2.2.2 Open half-planes . . . . . . . . . . . . . . . . . . . 2.2.3 Closed half-planes . . . . . . . . . . . . . . . . . . 2.3 Angle-supports, interior and exterior regions, angles

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Contents 2.3.1 2.3.2 2.3.3 2.4 2.4.1 2.4.2 2.4.3 2.4.4

Angle-supports, interior regions . . Exterior regions . . . . . . . . . . Angles . . . . . . . . . . . . . . . Triangles and convex quadrilaterals Terminology . . . . . . . . . . . . Triangles . . . . . . . . . . . . . . Pasch’s property, 1882 . . . . . . . Convex quadrilaterals . . . . . . .

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3 Distance; degree-measure of an angle 3.1 Distance . . . . . . . . . . . . . . . 3.1.1 Axiom for distance . . . . . . . . . 3.1.2 Derived properties of distance . . . . 3.2 Mid-points . . . . . . . . . . . . . . 3.2.1 . . . . . . . . . . . . . . . . . . . . 3.3 A ratio result . . . . . . . . . . . . . 3.3.1 . . . . . . . . . . . . . . . . . . . . 3.4 The cross-bar theorem . . . . . . . . 3.4.1 . . . . . . . . . . . . . . . . . . . . 3.5 Degree-measure of angles . . . . . . 3.5.1 Axiom for degree-measure . . . . . 3.5.2 Derived properties of degree-measure 3.6 Mid-line of an angle-support . . . . 3.6.1 Right-angles . . . . . . . . . . . . . 3.6.2 Perpendicular lines . . . . . . . . . 3.6.3 Mid-lines . . . . . . . . . . . . . . . 3.7 Degree-measure of reflex angles . . . 3.7.1 . . . . . . . . . . . . . . . . . . . .

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4 Congruence of triangles; parallel lines 4.1 Principles of congruence . . . . . . . . . . . . 4.1.1 Congruence of triangles . . . . . . . . . . . . . 4.2 Alternate angles, parallel lines . . . . . . . . . . 4.2.1 Alternate angles . . . . . . . . . . . . . . . . . 4.2.2 Parallel lines . . . . . . . . . . . . . . . . . . . 4.3 Properties of triangles and half-planes . . . . . 4.3.1 Side-angle relationships; the triangle inequality 4.3.2 Properties of parallelism . . . . . . . . . . . . . 4.3.3 Dropping a perpendicular . . . . . . . . . . . . 4.3.4 Projection and axial symmetry . . . . . . . . .

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Contents 5 The parallel axiom; Euclidean geometry 5.1 The parallel axiom . . . . . . . . . . . . . . . . . . . . . 5.1.1 Uniqueness of a parallel line . . . . . . . . . . . . . . . 5.2 Parallelograms . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Parallelograms and rectangles . . . . . . . . . . . . . . . 5.2.2 Sum of measures of wedge-angles of a triangle . . . . . . 5.3 Ratio results for triangles . . . . . . . . . . . . . . . . . 5.3.1 Lines parallel to one side-line of a triangle . . . . . . . . 5.3.2 Similar triangles . . . . . . . . . . . . . . . . . . . . . . 5.4 Pythagoras’ theorem, c. 550B.C. . . . . . . . . . . . . . 5.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Mid-lines and triangles . . . . . . . . . . . . . . . . . . 5.5.1 Harmonic ranges . . . . . . . . . . . . . . . . . . . . . . 5.6 Area of triangles, and convex quadrilaterals and polygons 5.6.1 Area of a triangle . . . . . . . . . . . . . . . . . . . . . 5.6.2 Area of a convex quadrilateral . . . . . . . . . . . . . . . 5.6.3 Area of a convex polygon . . . . . . . . . . . . . . . . .

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6 Cartesian coordinates; applications 6.1 Frame of reference, Cartesian coordinates 6.1.1 . . . . . . . . . . . . . . . . . . . . . . . 6.2 Algebraic note on linear equations . . . . 6.2.1 . . . . . . . . . . . . . . . . . . . . . . . 6.3 Cartesian equation of a line . . . . . . . . 6.3.1 . . . . . . . . . . . . . . . . . . . . . . . 6.4 Parametric equations of a line . . . . . . . 6.4.1 . . . . . . . . . . . . . . . . . . . . . . . 6.5 Perpendicularity and parallelism of lines . 6.5.1 . . . . . . . . . . . . . . . . . . . . . . . 6.6 Projection and axial symmetry . . . . . . 6.6.1 . . . . . . . . . . . . . . . . . . . . . . . 6.6.2 Formula for area of a triangle . . . . . . . 6.6.3 Inequalities for closed half-planes . . . . . 6.7 Coordinate treatment of harmonic ranges . 6.7.1 New parametrisation of a line . . . . . . . 6.7.2 Interchange of pairs of points . . . . . . . 6.7.3 Distances from mid-point . . . . . . . . . 6.7.4 Distances from end-point . . . . . . . . . 6.7.5 Construction for a harmonic range . . . .

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7 Circles; their basic properties 103 7.1 Intersection of a line and a circle . . . . . . . . . . . . . . . . . . . 103 7.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 7.2 Properties of circles . . . . . . . . . . . . . . . . . . . . . . . . . . 105

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8 Translations; axial symmetries; isometries 8.1 Translations and axial symmetries . . . 8.1.1 . . . . . . . . . . . . . . . . . . . . . 8.2 Isometries . . . . . . . . . . . . . . . 8.2.1 . . . . . . . . . . . . . . . . . . . . . 8.2.2 . . . . . . . . . . . . . . . . . . . . . 8.3 Translation of frame of reference . . .

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9 Trigonometry; cosine and sine; addition formulae 9.1 Indicator of an angle . . . . . . . . . . . . . . 9.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Cosine and sine of an angle . . . . . . . . . . 9.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Polar coordinates . . . . . . . . . . . . . . . 9.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Angles in standard position . . . . . . . . . . 9.3.1 Angles in standard position . . . . . . . . . . 9.3.2 Addition of angles . . . . . . . . . . . . . . . 9.3.3 Modified addition of angles . . . . . . . . . . 9.3.4 Subtraction of angles . . . . . . . . . . . . . 9.3.5 Integer multiples of an angle . . . . . . . . . 9.3.6 Standard multiples of a right-angle . . . . . . 9.4 Half angles . . . . . . . . . . . . . . . . . . . 9.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 The cosine and sine rules . . . . . . . . . . . 9.5.1 The cosine rule . . . . . . . . . . . . . . . . . 9.5.2 The sine rule . . . . . . . . . . . . . . . . . .

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Contents 9.5.3 . . . . . . . . . . . . . . . . . . . . . . . . 9.5.4 The Steiner-Lehmus theorem, 1842 . . . . . 9.6 Cosine and sine of angles equal in magnitude 9.6.1 . . . . . . . . . . . . . . . . . . . . . . . .

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10 Complex coordinates; sensed angles; angles between lines 10.1 Complex coordinates . . . . . . . . . . . . . . . . . 10.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Complex-valued distance . . . . . . . . . . . . . . . 10.2.1 Complex-valued distance . . . . . . . . . . . . . . . 10.2.2 A complex-valued trigonometric function . . . . . . . 10.3 Rotations and axial symmetries . . . . . . . . . . . . 10.3.1 Rotations . . . . . . . . . . . . . . . . . . . . . . . . 10.3.2 Formula for an axial symmetry . . . . . . . . . . . . 10.4 Sensed angles . . . . . . . . . . . . . . . . . . . . . 10.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Sensed-area . . . . . . . . . . . . . . . . . . . . . . 10.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.2 Sensed-area of a triangle . . . . . . . . . . . . . . . . 10.5.3 A basic feature of sensed-area . . . . . . . . . . . . . 10.5.4 An identity for sensed-area . . . . . . . . . . . . . . 10.6 Isometries as compositions . . . . . . . . . . . . . . 10.6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 Orientation of a triple of non-collinear points . . . . . 10.7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.8 Sensed angles of triangles, the sine rule . . . . . . . . 10.8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.9 Some results on circles . . . . . . . . . . . . . . . . 10.9.1 A necessary condition to lie on a circle . . . . . . . . 10.9.2 A sufficient condition to lie on a circle . . . . . . . . 10.9.3 Complex cross-ratio . . . . . . . . . . . . . . . . . . 10.9.4 Ptolemy’s theorem, c. 200A.D. . . . . . . . . . . . . 10.10 Angles between lines . . . . . . . . . . . . . . . . . 10.10.1 Motivation . . . . . . . . . . . . . . . . . . . . . . 10.10.2 Duo-sectors . . . . . . . . . . . . . . . . . . . . . . 10.10.3 Duo-angles . . . . . . . . . . . . . . . . . . . . . . 10.10.4 Duo-angles in standard position . . . . . . . . . . . 10.10.5 Addition of duo-angles in standard position . . . . . 10.10.6 Addition formulae for tangents of duo-angles . . . . 10.10.7 Associativity of addition of duo-angles . . . . . . . 10.10.8 Group properties of duo-angles; sensed duo-angles . 10.10.9 An application . . . . . . . . . . . . . . . . . . . . 10.11 A case of Pascal’s theorem, 1640 . . . . . . . . . . . 10.11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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145 . 145 . 145 . 148 . 148 . 148 . 150 . 150 . 151 . 153 . 153 . 156 . 156 . 156 . 157 . 157 . 158 . 158 . 160 . 160 . 162 . 162 . 164 . 164 . 165 . 165 . 166 . 168 . 168 . 168 . 169 . 169 . 171 . 173 . 174 . 175 . 176 . 176 . 176

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Contents 10.11.2 10.11.3

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11 Vector and complex-number methods 11.1 Equipollence . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Sum of couples, multiplication of a couple by a scalar . . . . . 11.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.2 Vector space over R . . . . . . . . . . . . . . . . . . . . . . . 11.3 Scalar or dot products . . . . . . . . . . . . . . . . . . . . . . 11.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Components of a vector . . . . . . . . . . . . . . . . . . . . . 11.4.1 Components . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4.2 Areal coordinates . . . . . . . . . . . . . . . . . . . . . . . . 11.4.3 Cartesian coordinates from areal coordinates . . . . . . . . . . 11.4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Vector methods in geometry . . . . . . . . . . . . . . . . . . . 11.5.1 Menelaus’ theorem, c. 100A.D. . . . . . . . . . . . . . . . . . 11.5.2 Ceva’s theorem and converse, 1678 A.D. . . . . . . . . . . . . 11.5.3 Desargues’ perspective theorem, 1648 A.D. . . . . . . . . . . 11.5.4 Pappus’ theorem, c. 300A.D. . . . . . . . . . . . . . . . . . . 11.5.5 Centroid of a triangle . . . . . . . . . . . . . . . . . . . . . . 11.5.6 Orthocentre of a triangle . . . . . . . . . . . . . . . . . . . . . 11.5.7 Incentre of a triangle . . . . . . . . . . . . . . . . . . . . . . . 11.6 Mobile coordinates . . . . . . . . . . . . . . . . . . . . . . . 11.6.1 Grassmann’s supplement of a vector . . . . . . . . . . . . . . 11.6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6.3 Handling a triangle . . . . . . . . . . . . . . . . . . . . . . . 11.6.4 Circumcentre of a triangle . . . . . . . . . . . . . . . . . . . . 11.6.5 Other distinguished points for a triangle . . . . . . . . . . . . 11.6.6 Euler line of a triangle . . . . . . . . . . . . . . . . . . . . . . 11.6.7 Similar triangles . . . . . . . . . . . . . . . . . . . . . . . . . 11.6.8 Centroids of similar triangles erected on the sides of a triangle . 11.6.9 Circumcentres of similar triangles on sides of triangle . . . . . 11.6.10 Triangle with vertices the mid-points of sides of given triangle 11.6.11 The nine-point circle . . . . . . . . . . . . . . . . . . . . . . 11.6.12 Parametric equations of lines. . . . . . . . . . . . . . . . . . 11.7 Some well-known theorems . . . . . . . . . . . . . . . . . . . 11.7.1 Feuerbach’s theorem, 1822 . . . . . . . . . . . . . . . . . . . 11.7.2 The Wallace-Simson line, 1797 . . . . . . . . . . . . . . . . . 11.7.3 The incentre on the Euler line of a triangle . . . . . . . . . . . 11.7.4 Miquel’s theorem, 1838 . . . . . . . . . . . . . . . . . . . . . 11.8 Isogonal conjugates . . . . . . . . . . . . . . . . . . . . . . .

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185 . 185 . 185 . 187 . 187 . 188 . 190 . 190 . 192 . 192 . 193 . 193 . 193 . 194 . 195 . 195 . 196 . 199 . 201 . 203 . 203 . 205 . 206 . 206 . 207 . 207 . 209 . 211 . 215 . 215 . 217 . 218 . 219 . 222 . 225 . 226 . 226 . 229 . 231 . 232 . 234

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11.8.1 Isogonal conjugates . . . . . . . . . . . . . . . . . . . . . . . . . . 234 11.8.2 Concurrency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 11.8.3 Symmedians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 12 Trigonometric functions in calculus 12.1 Repeated bisection of an angle . . . . . 12.1.1 . . . . . . . . . . . . . . . . . . . . . . 12.2 Circular functions . . . . . . . . . . . . 12.2.1 . . . . . . . . . . . . . . . . . . . . . . 12.2.2 Definition of π . . . . . . . . . . . . . . 12.3 Derivatives of cosine and sine functions . 12.3.1 . . . . . . . . . . . . . . . . . . . . . . 12.4 Parametric equations for a circle . . . . 12.4.1 Area of a disk . . . . . . . . . . . . . . 12.4.2 Length of an arc of a circle . . . . . . . 12.4.3 Radian measure . . . . . . . . . . . . . 12.5 Extension of domains of cosine and sine 12.5.1 . . . . . . . . . . . . . . . . . . . . . . 12.5.2 . . . . . . . . . . . . . . . . . . . . . .

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Appendix; List of axioms

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Bibliography

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Index

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About the author Paddy Barry was born in Co. Westmeath in 1934 and his family moved to Co. Cork five years later. After his secondary school education at Patrician Academy, Mallow he studied at University College, Cork from 1952 to 1957, obtaining a degree of BSc in Mathematics and Mathematical Physics in 1955 and the degree of MSc in 1957. He then did research in complex analysis under the supervision of Professor W. K. Hayman, FRS, at Imperial College of Science and Technology, London (1957–1959) for which he was awarded the degree of PhD in 1960. He took first place in the Entrance Scholarships Examination to University College, Cork in 1952 and was awarded a Travelling Studentship in Mathematical Science at the National University of Ireland in 1957. He was appointed Instructor in Mathematics at Stanford University, California in 1959–1960. Returning to University College, Cork in 1960 he made his career there, becoming Professor of Mathematics and Head of Department in 1964. He has been Professor Emeritus of Mathematics there from 1999 to date. He participated in the general administration of the College, being a member of the Governing Body on a number occasions, and was the first modern Vice-President of the College from 1975 to 1977. He was also a member of the Senate of the National University of Ireland from 1977 to 1982. University College, Cork was upgraded to National University of Ireland, Cork in 1997 and both titles are now used.

His mathematical interests expanded in line with his extensive teaching experience. As examiner for matriculation for many years he had to keep in contact with the detail of secondary school mathematics and the present book arose from that context, as it seeks to give a thorough account of the geometry and trigonometry that is done, necessarily incompletely, at school.

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Preface I have for a long time held the view that whereas university courses in algebra, number systems and analysis admirably consolidate the corresponding school material, this is not the case for geometry and trigonometry. These latter topics form an important core component of mathematics, as they underpin analysis in its manifold aspects and applications in classical applied mathematics and sundry types of science and engineering, and motivate other types of geometry, and topology. Yet they are not well treated as university topics, being either neglected or spread over a number of courses, so that typically a student picks up a knowledge of these incidentally and relies mainly on the earlier intuitive treatment at school. Clearly the treatment of geometry has seriously declined over the last fifty years, in terms of both quantity and quality. Lecturers and authors are faced with the question of what, if anything, should be done to try to restore it to a position of some substance. Bemoaning its fate is not enough, and surely authors especially should ponder what kinds of approach are likely to prove productive. Pure or synthetic geometry was the first mathematical topic in the field and for a very long time the best established. It was natural for authors to cover as much ground as was feasible, and ultimately there was a large bulk of basic and further geometry. That was understandable in its time but perhaps a different overall strategy is now needed. Synthetic geometry seems very difficult. In it we do not have the great benefit of symbolic manipulations. It is very taxing to justify diagrams and to make sure of covering all cases. From the very richness of its results, it is difficult to plan a productive approach to a new problem. In the proofs that have come down to us, extra points and segments frequently need to be added to the configuration. It is true that, as in any approach, there are some results which are handled very effectively and elegantly by synthetic methods, but that is certainly not the whole story. On the other hand, what is undeniable is that synthetic geometry really deals with geometry, and it forces attention to, and clarity in, geometrical concepts. It encourages the careful layout of sequential proof. Above all, it has a great advantage in its intuitive visualisation and concreteness. The plan of this book is to have a basic layer of synthetic geometry, essentially five chapters in all, because of its advantages, and thereafter to diversify as much as possible to other techniques and approaches because of its difficulties. More than that, we assume strong axioms (on distance and angle-measure) so as to have an efficient approach from the start. The other approaches that we have in mind are the use of coordinates, trigonometry, position-vectors and complex numbers. Our emphasis is on clarity of concepts, proof and systematic and complete development of material. The synthetic geometry that we need is what is sufficient to start coordinate geometry and trigonometry, and that takes us as far as the ratio results for triangles and Pythagoras’ theorem. In all, a considerable portion of traditional ground involving straight-lines and circles is covered. The overall approach is innovative as is the detail on trigonometry in Chapter 9 and on what are termed ‘mobile coordinates’ in Chapter 11. Some

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new concepts and substantial new notation have been introduced. There is enough for a two-semester course; a one-semester one could be made from Chapters 2-9, with Chapter 7 trimmed back. My object has been to give an account at once accessible and unobtrusively rigorous. Preparation has been in the nature of unfinished business, stemming from my great difficulties when young in understanding the then textbooks in geometry. I hold that the reasoning in geometry should be as convincing as that in other parts of mathematics. It is too much to hope that there are no errors, mathematical or typographical. I should be grateful to be told of any at the email address [email protected].

Acknowledgements I am grateful to students in a succession of classes who responded to this material in its nascent stages, to departmental colleagues, especially Finbarr Holland and Des MacHale, who attended presentations of it, and to American colleagues the late David Rosen of Swarthmore and John Elliott of Fort Kent, Maine, who read an earlier approach of mine to geometry. I am especially grateful to Dr. P.A.J. Cronin, a colleague in our Department of Ancient Classics, for preparing the translations from Greek and Latin in the Glossary on pp. xix-xx. Paddy Barry National University of Ireland, Cork, November, 2000.

PREFATORY NOTE TO THE REVISED EDITION Several changes have taken place since the preparation of this material in the years leading up to 2001. The original publisher HORWOOD was acquired in 2010 by WOODHEAD which in turn was acquired in 2013 by ELSEVIER. However WOODHEAD is retaining its imprint. The reason for a revised edition was that the stock for the original edition was exhausted. There is not a lot of alteration in this edition, some correction of errors and the strengthening of some proofs apart from Chapter 11 in which there is considerable alteration of detail. There is a great alteration in the software available and required in mathematical word processing, particularly on those who have relied heavily on pstricks for preparing diagrams, and I owe a huge debt in this regard to my colleagues Paul Keegan, Marc van Dongen, Peter Flynn, Stephen Wills, Ben McKay and Joseph Manning. I also thank Ulrike Fischer, who is a troubleshooter for TeX, and without whose involvement this project would not have been completed. Paddy Barry National University of Ireland, Cork, February, 2015.

Glossary Greek and Latin roots of mathematical words acute < L acutus, sharp-pointed (perf.partic. of acuere, to sharpen). addition < L additio, an adding to (addere, to add). angle < L angulus, corner < Gk agkylos, bent. area < L area, a vacant space. arithmetic < Gk arithmetike (sc. tekhne), the art of counting (arithmein, to count; arithmos, number). axiom < Gk axioma, self-evident principle (axioun, to consider worthy; axios, worthy). calculate < L calculatus, reckoned (perf. partic. of calculare < calculus, pebble). centre < Gk kentron, sharp point (kentein, to spike). chord < Gk khorde, string of gut. circle < L circulus, ring-shaped figure (related to Gk kyklos, ring; kirkos or krikos, ring). congruent < L congruens (gen. congruentis), agreeing with (pres. partic. of congruere, to agree with). curve < L curvus (curvare, to bend). decimal < L (Med) decimalis, of tenths (decima (sc. pars), tenth part; decem, ten). degree < OF degre < L degredi, descend (de, down; gradi, to step). diagonal < L diagonalis, diagonal < Gk diagonios, from angle to angle (dia, through ; gonia, angle). diagram < Gk diagramma, plan, figure indicated by lines (dia, through; gramma, a thing which is drawn; graphein, to draw). diameter < Gk diametros, diametrical (dia, through ; metron, measure). distance < L distantia, remoteness (distare, to stand apart). divide < L dividere, to separate. equal < L equalis, equal (aequare, to make equal; aequus, equal).

example < L exemplum, sample < eximere, to take out. exponent < L exponens (gen. exponentis), setting forth (pres. partic. of exponere, to set forth; ex, out; ponere to place). factor < L factor, maker, doer (facere, to make). focus < L focus, hearth. fraction < L fractio, a breaking into pieces (frangere, to break). geometry < Gk geometria, measuring of land (ge, land; metrein, to measure). graph < Gk graphos, drawing, picture (graphein, to draw). hypotenuse < Gk hypoteinousa (sc. gramme), the line extending underneath (pres. partic. of hypoteinein, to extend under; gramme, line). hypothesis < Gk hypothesis, supposition, assumption (hypotithenai, to place beneath). inclination < L inclinatio, a leaning to one side (inclinare, to cause to lean). induction < L inductio, a leading into (inducere, to lead in). isosceles < Gk isoskeles, having equal legs (isos, equal; skelos, leg). line < L linea, a linen thread (linum, flax < Gk. linon). logic < Gk logike (sc. techne), the art of reasoning (logos, reason; logikos, endowed with reason; techne, art). magnitude < L magnitudo, size, greatness (magnus, great). mathematics < Gk mathematika, things that require mathematical or scientific reasoning (mathema, lesson; mathemat/ikos, mathematical or scientific; manthanein, to learn). measure < F mesure < L mensura, measure (metiri, to measure). minus < L minus, less. multiply < L multiplicare, multiply (multus, much; plicare, to lay together). negative < L negativus, denying (negare, to deny). number < F nombre < L numerus, number.

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oblong < L oblongus, longish. obtuse < L obtusus, blunted (perf. partic. of obtundere, to blunt). orthogonal < Gk orthogonios, rectangular (orthos, right; gonia, angle). parallel < Gk parallelos, beside one another (para, beside; allelous, one another). perimeter < Gk perimetron, circumference (peri around; metron, measure). perpendicular < L perpendiculum, plumb-line (perpendere, to weigh precisely). plane < L planum, level ground (planus, level). point < L punctum, small hole (pungere, to pierce). polygon < Gk polygonon, thing with many angles (polys, many; gonia, angle). positive < L positivus, settled (ponere, to place). postulate < L postulare, to ask for. power < OF poeir < L posse, to be able. product < L productus, brought forth (perf. partic. of producere, to bring forth).

rectangle < L rectiangulum, rightangled (rectus, right; angulus, angle). rhombus < Gk rhombos, a device whirled round (rhembein, to whirl round). science < L scientia, knowledge (scire, to know). secant < L secans (gen. secantis), cutting (secare, to cut). square < OF esquarre < L quattuor, four. subtract < L subtractus, withdrawn (perf. partic. of subtrahere, to withdraw). sum < L summa, top. tangent < L tangens (sc. linea) (gen. tangentis), touching line (tangere, to touch; linea, line). technical < Gk tekhnikos, (tekhne, art).

artistic, skilful

theorem < Gk theorema, thing observed, deduced principle (theorein, to observe). total < L totus, whole, all. trapezium < Gk trapezion, small table (trapeza, table). triangle < L triangulum, triangle (tris, three; angulus, angle).

proportion < L portio, comparative relationship (pro, according to; portio, part).

trigonometry < Gk trigonometria, measurement of triangles (trigonon, triangle; metrein, to measure).

quadrangle < L quadrangulum, thing with four angles (quattuor, four; angulus, angle).

vector < L vector, bearer (vehere, to bear).

quotient < L quotiens, how often.

volume < OF volum < L volumen, roll, book (volvere, to roll).

radius < L radius, rod, spoke of wheel.

vertex < L vertex, summit.

1

Preliminaries 1.1 HISTORICAL NOTE This is one in a long line of textbooks on geometry. While all civilisations seem to have had some mathematical concepts, the most significant very old ones historically were the linked ones of Sumer, Akkad and Babylon, largely in the same region in what is now southern Iraq, and the separate one of Egypt. These are the ones which have left substantial traces of their mathematics, which was largely arithmetic, and geometrical shapes and measurement. The outstanding contribution to mathematics was in Greece about 600 B.C.– 200 B.C. The earlier mathematics conveyed techniques by means of examples, but the Greeks stated general properties of the mathematics they were doing, and organised proof of later properties from ones taken as basic. There was astonishing progress in three centuries and the fruit of that was written up in Euclid’s The Elements, c. 300B.C. He worked in Alexandria in Egypt, which country had come into the Greek sphere of influence in the previous century. Euclid’s The Elements is one of the most famous books in the world, certainly the most famous on mathematics. But it was influential widely outside mathematics too, as it was greatly admired for its logical development. It is the oldest writing on geometry of which we have copies by descent, and it lasted as a textbook until after 1890, although it must be admitted that in lots of places and for long periods not very many people were studying mathematics. It should probably be in the Guinness Book of Records as the longest lasting textbook in history. The Elements shaped the treatment of geometry for 2,000 years. Its style would be unfamiliar to us today, as apart from using letters to identify points and hence line-segments, angles and other figures in diagrams, it consisted totally of words. Thus it did not use symbols as we do. It had algebra different from ours in that it said things in words written out in full. Full symbolic algebra as we know it was not perfected until about 1600 A.D. in France, by Vieta and later Descartes. Another very significant feature of The Elements was that it did not have numbers ready-made, and used distance or length, angle-measure and area as separate quantities, although links Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50001-X © 2016 Elsevier Ltd. All rights reserved.

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between them were worked out. Among prominent countries, The Elements lasted longest in its original style in the U.K., until about 1890. They had started chipping away at it in France in the 16th century, beginning with Petrus Ramus 1515-1572. There is a very readable account of the changes which were made in France in Cajori [3, pages 275 – 289]. These changes mainly involved disimprovements logically; authors brought in concepts which are visually obvious, but they did not provide an account of the properties of these concepts. Authors in France, and subsequently elsewhere, started using our algebra to handle the quantities and this was a major source of advance. One very prominent textbook of this type was Elements of Geometry by Legendre, (first edition 1794), which was very influential on the continent of Europe and in the U.S.A. All in all, these developments in France shook things up considerably, and that was probably necessary before a big change could be made. Although The Elements was admired widely and for a long time for its logic, there were in fact logical gaps in it. This was known to the leading mathematicians for quite a while, but it was not until the period 1880-1900 that this geometry was put on what is now accepted as an adequate logical foundation. Another famous book Foundations of Geometry by Hilbert (1899) was the most prominent in doing this. The logical completion made the material very long and difficult, and this type of treatment has not filtered down to school-level at all, or even to university undergraduate level except for advanced specialised options. Another sea-change was started in 1932 by G.D. Birkhoff; instead of building up the real number system via geometrical quantities, he assumed a knowledge of numbers and used that from the start in geometry; this appeared in his ‘ruler postulate’ and ‘protractor postulate’. His approach allowed for a much shorter, easier and more efficient treatment of geometry. In the 1960’s there was the world-wide shake-up of the ‘New Mathematics’, and since then there are several quite different approaches to geometry available. In this Chapter 1 we do our best to provide a helpful introduction and context, and suggest a re-familiarisation with the geometrical knowledge already acquired. Logically organised geometry dates from c. 600-300 B.C. in Greece; by c. 350B.C. there was already a history of geometry by Eudemus. From the same period and earlier, date positive integers and the treatment of positive fractions via ratios. The major mathematical topics date from different periods: geometry as just indicated; full algebra from c. 1600 A.D.; full coordinate geometry from c. 1630A.D.; full numbers (negative, rational, decimals) from c. 1600 A.D.; complex numbers from c. 1800 A.D.; calculus from c. 1675 A.D.; trigonometry from c. 200 B.C., although circles of fixed length of radius were used until c. 1700 A.D. when ratios were introduced. There is an account of the history of geometry of moderate length by H. Eves in [2, pages 165-192] It should be clear from this history that the Greek contribution to geometry greatly influenced all later mathematics. It was transmitted to us via the Latin language, and we have included a Glossary on pp. xix-xx showing the Greek or Latin roots of

Sec. 1.2

Note on deductive reasoning

3

mathematical words.

1.2 NOTE ON DEDUCTIVE REASONING The basic idea of a logical system is that we list up-front the terms and properties that we start with, and thereafter proceed by way of definitions and proofs. There are two main aspects to this.

1.2.1 Definitions The first aspect concerns specifying what we are dealing with. A definition identifies a new concept in terms of accepted or known concepts. In practice a definition of a word, symbol or phrase E is a statement that E is to be used as a substitute for F, the latter being a phrase consisting of words and possibly symbols or a compound symbol. We accept ordinary words of the English language in definitions and what is at issue is the meaning of technical mathematical words or phrases. In attempting a definition, there is no progress if the technical words or symbols in F are not all understood at the time of the definition. The disconcerting feature of this situation is that in any one presentation of a topic there must be a first definition and of its nature that must be in terms of accepted concepts. Thus we must have terms which are accepted without definition, that is there must be undefined or primitive terms. This might seem to leave us in a hopeless position but it does not, as we are able to assume properties of the primitive terms and work with those. There is nothing absolute about this process, as a term which is taken as primitive in one presentation of a topic can very well be a defined term in another presentation of that topic, and vice versa. We need some primitive terms to get an approach under way.

1.2.2 Proof The second aspect concerns the properties of the concepts that we are dealing with. A proof is a finite sequence of statements the first of which is called the hypothesis, and the last of which is called the conclusion. In this sequence, each statement after the hypothesis must follow logically from one or more statements that have been previously accepted. Logically there would be a vicious circle if the conclusion were used to help establish any statement in the proof. There is also a disconcerting feature of this, as in any presentation of a topic there must be a first proof. That first proof must be based on some statements which are not proved (at least the hypothesis), which are in fact properties that are accepted without proof. Thus any presentation of a topic must contain unproved statements; these are called axioms or postulates and these names are used interchangeably. Again there is nothing absolute about this, as properties which are taken as axiomatic in one presentation of a topic may be proved in another presentation, and

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vice versa. But we must have some axioms to get an approach under way.

1.3 EUCLID’S The Elements 1.3.1 The Elements involved the earliest surviving deductive system of reasoning, having axioms or postulates and common notions, and proceeding by way of careful statements of results and proofs. Up to c. 1800, geometry was regarded as the part of mathematics which was best-founded logically. But its position was overstated, and its foundations not completed until c. 1880-1900. Meanwhile the foundations of algebra and calculus were properly laid in the 19th century. From c. 1800 on, some editions used algebraic notation in places to help understanding.

1.3.2 Definitions The Greeks did not appreciate the need for primitive terms, and The Elements started with an attempt to define a list of basic terms. DEFINITIONS 1. A POINT is that which has no parts, or which has no magnitude. 2. A LINE is length without breadth. 3. The EXTREMITIES of a line are points. 4. A STRAIGHT LINE is that which lies evenly between its extreme points. 5. A SUPERFICIES is that which has only length and breadth. 6. The EXTREMITIES of a superficies are lines. 7. A PLANE SUPERFICIES is that in which any two points being taken, the straight line between them lies wholly in that superficies. 8. A PLANE ANGLE is the inclination of two lines to one another in a plane, which meet together, but are not in the same direction. 9. A PLANE RECTILINEAL ANGLE is the inclination of two straight lines to one another, which meet together, but are not in the same straight line. 10. When a straight line standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a RIGHT ANGLE; and the straight line which stands on the other is called a PERPENDICULAR to it.

Sec. 1.3

Euclid’s The Elements

5

11. An OBTUSE ANGLE is that which is greater than a right angle. 12. An ACUTE ANGLE is that which is less than a right angle. 13. A TERM or BOUNDARY is the extremity of anything. 14. A FIGURE is that which is enclosed by one or more boundaries. 15. A CIRCLE is a plane figure contained by one line, which is called the CIRCUMFERENCE, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another. 16. And this point is called the CENTRE of the circle. 17. A DIAMETER of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. 18. A SEMICIRCLE is the figure contained by a diameter and the part of the circumference cut off by the diameter. 19. A SEGMENT of a circle is the figure contained by a straight line and the circumference which cuts it off. 20. RECTILINEAL FIGURES are those which are contained by straight lines. 21. TRILATERAL FIGURES, or TRIANGLES, by three straight lines. 22. QUADRILATERAL FIGURES by four straight lines. 23. MULTILATERAL FIGURES, or POLYGONS, by more than four straight lines. 24. Of three-sided figures, an EQUILATERAL TRIANGLE is that which has three equal sides. 25. An ISOSCELES TRIANGLE is that which has two sides equal. 26. A SCALENE TRIANGLE is that which has three unequal sides. 27. A RIGHT-ANGLED TRIANGLE is that which has a right angle. 28. An OBTUSE-ANGLED TRIANGLE is that which has an obtuse angle. 29. An ACUTE-ANGLED TRIANGLE is that which has three acute angles. 30. Of four-sided figures, a SQUARE is that which has all its sides equal, and all its angles right angles. 31. An OBLONG is that which has all its angles right angles, but not all its sides equal.

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Ch. 1

32. A RHOMBUS is that which has all its sides equal, but its angles are not right angles. 33. A RHOMBOID is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles. 34. All other four-sided figures besides these are called TRAPEZIUMS. 35. PARALLEL STRAIGHT LINES are such as are in the same plane, and which being produced ever so far both ways do not meet. Although in The Elements these definitions were initially given, some of these were treated just like motivations (for instance, for a point where no use was made of the fact that ‘it has no parts’) whereas some were genuine definitions (like that of a circle, where the defining property was used). Our definitions will differ in some respects from these.

1.3.3 Postulates and common notions The Greeks understood the need for axioms, and these were laid out carefully in The Elements. The Elements has two lists, a first one of POSTULATES and a second one of COMMON NOTIONS. It is supposed by some writers that Euclid intended his list of POSTULATES to deal with concepts which are mathematical or geometrical, and the second list to deal with concepts which applied to science generally. POSTULATES Let it be granted, 1. That a straight line may be drawn from any one point to any other point. 2. That a terminated straight line may be produced to any length in a straight line. 3. And that a circle may be described from any centre, at any distance from that centre. 4. All right angles are equal to one another. 5. If a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these straight lines, being continually produced, shall at length meet on that side on which are the angles which are less than two right angles.

COMMON NOTIONS 1. Things which are equal to the same thing are equal to one another.

Sec. 1.3

Euclid’s The Elements

7

2. If equals be added to equals the wholes are equals. 3. If equals be taken from equals the remainders are equal. 4. Magnitudes which coincide with one another, [that is, which exactly fill the same space] are equal to one another. 5. The whole is greater than its part.

1.3.4 The Elements attempted to be a logically complete deductive system. There were earlier Elements but these have not survived, presumably because they were outclassed by Euclid’s. The Elements are charming to read, proceed very carefully by moderate steps and within their own terms impart a great sense of conviction. The first proposition is to describe an equilateral triangle on [A, B]. With centre A a circle is described passing through B, and with centre B a second circle is described passing through A. If C is a point at which these two circles cut one another, then we take the triangle with vertices A, B,C. It is ironical that, with The Elements being so admired for their logical proceeding, there should be a gap in the very first proposition. The postulates and common notions did not make any provisions which would ensure that the two circles in the proof would have a point in common. This may seem a curious choice as a first proposition, dealing with a very special figure. But in fact it is used immediately in Proposition 2, from a given point to lay off a given distance. The main logical lack in The Elements was that not enough assumed properties were listed, and this fact was concealed through the use of diagrams.

1.3.5 Congruence Two types of procedure in The Elements call for special comment. The first is the method of superposition by which one figure was envisaged as being moved and placed exactly on a second figure. The second is the process of construction by which figures were not dealt with until it was shown by construction that there actually was such a type of figure. In the physical constructions, what were allowed to be used were straight edges and compasses. The notion of superposition is basic to Euclid’s treatment of figures. It is visualised that one figure is moved physically and placed on another, fitting perfectly. We use the term congruent figures when this happens. Common Notion 4 is to be used in this connection. This physical idea is clearly extraneous to the logical set-up of primitive and defined terms, assumed and proved properties, and is not a formal part of modern treatments of geometry. However it can be used in motivation, and properties motivated by it can be assumed in axioms.

8

Preliminaries

Ch. 1

1.3.6 Quantities or magnitudes The Elements spoke of one segment (then called a line) being equal to or greater than another, one region being equal to or greater than another, and one angle being equal to or greater than another, and this indicates that they associated a magnitude with each segment (which we call its length), a magnitude with each region (which we call its area), and a magnitude with each angle (which we call its measure). They did not define these magnitudes or give a way of calculating them, but they gave sufficient properties for them to be handled as the theory was developed. In the case of each of them the five common notions were supposed to apply. Thus in The Elements, the quantities for which the Common Notions are intended are distance or equivalently length of a segment, measure of an angle and area of a region. These are not taken to be known, either by assumption or definition, but congruent segments are taken to have equal lengths, congruent angles are taken to have equal measures, and congruent triangles are taken to have equal areas. Thus equality of lengths of segments, equality of measures of angles, and equality of areas of triangles are what is started with. Treatment of area is more complicated than the other two, and triangles equal in area are not confined to congruent triangles. Addition and subtraction of lengths are to be handled using Common Notions 1, 2, 3 and 4; so are addition and subtraction of angle measures; so are addition and subtraction of area. Taking a unit length, there is a long build-up to the length of any segment. They reached the stage where if some segment were to be chosen to have length 1 the length of any segment which they encountered could be found, but this was not actually done in The Elements. Taking a right-angle as a basic unit, there was a long build-up to handling any angle. They reached the stage where if a right-angle was taken to have measure 90◦ , the measure of any angle which they encountered could be found, but this was not actually done. There is a long build up to the area of figures generally. The regions which they considered were those which could be built up from triangles, and they reached the stage where if some included region were chosen to have area 1 the area of any included region could be found. This is not actually completed in The Elements but the materials are there to do it with. All this shows that The Elements although very painstaking, thorough and exact were also rather abstract. It should be remembered that the Greeks did not have algebra as we have, and used geometry to do a lot of what we do by algebra. In particular, considering the area of a rectangle was their way of handling multiplication of quantities. Traditionally in arithmetic the area of a rectangle was dealt with as the product of the length and the breadth, that is by multiplication of two numbers. However, reconciling the geometrical treatment of area with the arithmetical does not seem to have been handled very explicitly in books, not even when from 1600 A.D. onwards real numbers were being detached from the ‘quantities’ of Euclid.

Sec. 1.4

Our approach

9

1.4 OUR APPROACH 1.4.1 Type of course Very scholarly courses in geometry assume as little as possible, and as a result are long and difficult. Shorter and easier courses have more or stronger assumptions, and correspondingly less to prove. What is difficult in a thorough course of geometry is not the detail of proof usually included, but rather is, first of all, locational viz. to prove that points are where diagrams suggest they are, that is to verify the diagrams, and secondly to be sure of covering all cases. In particular, the type of approach which assumes that distance, angle-measure and area are different ‘quantities’ leads to a very long and difficult treatment of geometry. To make things much easier and shorter, we shall suppose that we know what real numbers are, and deal with distance/length and angle-measure as basic concepts given in terms of real numbers, and develop their properties. Moreover, we shall define area in terms of lengths. What we provide, in fact, is a combination of Euclid’s original course and a modification of an alternative treatment due to the American mathematician G.D. Birkhoff in 1932.

1.4.2 Need for preparation What this course aims to do is to revise and extend the geometry and trigonometry that has been done at school. It gives a careful, thorough and logical account of familiar geometry and trigonometry. At school, a complete, logically adequate treatment of geometry is out of the question. It would be too difficult and too long, unattractive and not conducive to learning geometry; it would tend rather to put pupils off. Thus this is not a first course in geometry. It is aimed at third level students, who should have encountered the basic concepts at secondary, or even primary, school. It starts geometry and trigonometry from scratch, and thus is self-contained to that extent. But it is demanding because of a sustained commitment to deductive reasoning. In preparation the reader is strongly urged to start by revising the geometry and trigonometry which was done at school, at least browsing through the material. It would also be a good idea to read in some other books some descriptive material on geometry, such as the small amount in Ledermann and Vajda [10, pages 1 – 26], or the large amount in Wheeler and Wheeler [13, Chapters 11 – 15]. Similarly trigonometry and vectors can be revised from McGregor, Nimmo and Stothers [11, pages 99 – 123, 279 – 331]. It would moreover be helpful to practise geometry by computer, e.g. by using software systems such as The Geometer’s Sketchpad or Cabri-Géomètre. Material which can be found in elementary books should be gone over, and also a look forwards could be had to the results in this book.

10

Preliminaries

Ch. 1

1.5 REVISION OF GEOMETRICAL CONCEPTS 1.5.1 As part of the preliminary programme, we now include a review of the basic concepts of geometry. Geometry should be thought of as arising from an initial experimental and observational stage, where the figures are looked at and there is a great emphasis on a visual approach.

1.5.2 The basic shapes 1. The plane Π is a set, the elements of which are called points. Certain subsets of Π are called lines. By observation, given any distinct points A, B ∈ Π, there is a unique line to which A and B both belong. It is denoted by AB. 2. Given distinct points A and B, the set of points consisting of A and B themselves and all the points of the line AB which are between A and B is called a segment, and denoted by [A, B].

B A Figure 1.1. A line AB. The arrows indicate that the line is to be continued unendingly.

B A Figure 1.2. A segment [A, B].

NOTE. Note that the modern mathematical terminology differs significantly from that in The Elements. What was called a ‘line’ is now called a segment, and we have added the new concept of ‘line’. This is confusing, but the practice is well established. In ordinary English and in subjects cognate to mathematics, ‘line’ has its old meaning. 3. The set consisting of the point A itself and all the points of the line AB which are on the same side of A as B is, is called a half-line, and denoted by [A, B . If A is between B and C, then the half-lines [A, B and [A,C are said to be opposite.

Sec. 1.5

Revision of geometrical concepts

B

11

A

C

B

A Figure 1.3. A half-line [A, B .

Opposite half-lines.

4. If the points B,C are distinct from A, then the pair of half-lines {[A, B , [A,C } is called an angle-support and denoted by |BAC; if [A, B and [A,C are opposite half-lines, then |BAC is called a straight angle-support. In each case A is called its vertex, [A, B and [A,C its arms.

C A

B

A

C

B Figure 1.4. An angle-support.

A straight angle-support.

5. The set of all the points on, or to one side of, a line AB is called a closed half-plane, with edge AB.

B A Figure 1.5. A closed half-plane shaded.

6. If the points A, B,C are not collinear, then the set of points which are in both the closed half-plane with edge AB, containing C, and the closed half-plane with edge AC, containing B, is called the interior region of |BAC and denoted by I R(|BAC); also (Π \ I R(|BAC)) ∪ |BAC is called the exterior region of |BAC and denoted by E R(|BAC). When C ∈ [A, B the interior and exterior regions of |BAC are taken to be [A, B and Π, respectively.

12

Preliminaries

C

Ch. 1

C B

B

A Figure 1.6. An interior region.

A The corresponding exterior region.

7. If |BAC is a non-straight angle-support, then the couples (|BAC, I R(|BAC)), (|BAC, E R(|BAC)), are called the wedge-angle and reflexangle, respectively, with support |BAC; this wedge-angle is denoted by ∠BAC. Thus a wedge-angle is a pair of arms in association with an interior region, while a reflexangle is a pair of arms combined with an exterior region. If |BAC is a straight angle-support, and H1 , H2 are the closed half-planes with edge the line AB, then the couples (|BAC, H1 ), (|BAC, H2 ), are called the straightangles with support |BAC. If C ∈ [A, B then the wedge-angle ∠BAC = ∠BAB is called a null-angle, and the reflex-angle with support |BAB is called a full-angle.

Figure 1.7. A wedge-angle.

A reflex-angle.

A straight-angle.

NOTE. The reason we call |BAC an angle-support and not an angle is that it supports two angles. If we were confining ourselves to pure geometry, and not concerned to go forward to coordinate geometry and trigonometry, we could confine ourselves to wedge and straight angles. Even more if we were to confine ourselves to the angles

Sec. 1.5

Revision of geometrical concepts

13

in triangles, we could take |BAC = [A, B ∪ [A,C . However when A is between B and C, this would result in a straight-angle being a line, and it would not have a unique vertex. In the early part of our course, we can confine our attention to wedge and straight angles. C C A

B

B

A

Figure 1.8. Supports bearing two angles each.

8. If A is between B and C and D ∈ BC, the wedge-angles ∠BAD, ∠CAD are called supplementary. If A, B,C are not collinear, and A is between B and B1 , and A is between C and C1 , then the wedge-angles ∠BAC, ∠B1 AC1 are called opposite angles at a vertex. D C

C

A

B1

B

B A C1

Figure 1.9. Supplementary angles.

Opposite angles at a vertex.

9. If A, B,C are non-collinear points and [A, D is in the interior region of | BAC, then [A, D is said to be between [A, B and [A,C .

D

C

B A Figure 1.10. [A, D between [A, B and [A,C .

10. If A, B,C are non-collinear points, let H1 be the closed half-plane with edge BC, containing A, H3 be the closed half-plane with edge CA, containing B, H5 be the closed half-plane with edge AB, containing C. Then the intersection H1 ∩ H3 ∩

14

Preliminaries

Ch. 1

H5 is called a triangle. The points A, B,C are called its vertices and the segments [B,C], [C, A], [A, B] its sides. If a vertex is not the end-point of a side ( e.g. the vertex A and the side [B,C]), then the vertex and side are said to be opposite each other. We denote the triangle with vertices A, B,C by [A, B,C]. If at least two sides of a triangle have equal lengths, then the triangle is called isosceles.

C

A B

A Figure 1.11. A triangle [A, B,C].

B

D

C

An isosceles triangle.

11. Let A, B, C, D be points no three of which are collinear, and such that [A,C] ∩ [B, D] = 0. / For this let H1 be the closed half-plane with edge AB, containing C, H3 be the closed half-plane with edge BC, containing D, H5 be the closed half-plane with edge CD, containing A, H7 be the closed half-plane with edge DA, containing B. Then the intersection H1 ∩ H3 ∩ H5 ∩ H7 is called a convex quadrilateral. The points A, B, C, D are called its vertices, the segments [A, B], [B,C], [C, D], [D, A] its sides, and the segments [A,C], [B, D] its diagonals. Two vertices which have a side in common are said to be adjacent, and two vertices which have a diagonal in common are said to be opposite. Thus A and B are adjacent as they both belong to [A, B] which is a side; A and C are opposite as they both belong to [A,C] which is a diagonal. Two sides which have a vertex in common are said to be adjacent, and two sides which do not have a vertex in common are said to be opposite. Thus the sides [A, B], [D, A] are adjacent as the vertex A belongs to both, while the sides [A, B], [C, D] are opposite as none of the vertices belongs to both of them. The wedge angles ∠DAB, ∠ABC, ∠BCD, ∠CDA are called the angles of the convex quadrilateral; two of these angles are said to be adjacent or opposite according as their two vertices are adjacent or opposite vertices of the convex quadrilateral. We denote the convex quadrilateral with vertices A, B, C, D, with A and C opposite, by [A, B,C, D].

Sec. 1.5

Revision of geometrical concepts D

15

C

B A Figure 1.12. A convex quadrilateral.

1.5.3 Distance; degree-measure of an angle 1. With each pair (A, B) of points we associate a non-negative real number |A, B|, called the distance from A to B or the length of the segment [A, B]. In all cases |B, A| = |A, B|. By observation, given any non-negative real number k, and any half- line [A, B there is a unique point P ∈ [A, B such that |A, P| = k.

k

y

P

x

A

P

x

R

Q +y

Addition of distances.

Figure 1.13. Laying off a distance k.

By observation, if Q ∈ [P, R] then |P, Q| + |Q, R| = |P, R|. In all cases |A, A| = 0, while |A, B| > 0 if A = B. 2. Given distinct points A and B, choose the point C ∈ [A, B so that |A, C| = 12 |A, B|. Then C is between A and B and 1 1 |C, B| = |A, B| − |A, C| = |A, B| − |A, B| = |A, B| = |A, C|. 2 2 The point C which is on the line AB and equidistant from A and B, is called the mid-point of A and B. It is also called the mid-point of the segment [A, B].

+

+

B C

A Figure 1.14. Mid-point of A and B.

16

Preliminaries

Ch. 1

3. With each wedge-angle ∠BAC we associate a non-negative number, called its degree-measure, denoted by |∠BAC|◦ , and for each straight-angle α we take |α |◦ = 180.

C

D

D x

y x+y

y 180 A

x B

A

B

C Figure 1.15. Addition of angle-measures.

C

H1 k

B

A Figure 1.16. Laying off an angle.

By observation, we note that if A, B,C are non-collinear and [A, D is between [A, B and [A,C , then |∠BAD|◦ + |∠CAD|◦ = |∠BAC|◦ , while if [A, B and [A,C are opposite and D ∈ AB, then |∠BAD|◦ + |∠CAD|◦ = 180. By observation, given any number k with 0 ≤ k < 180 and any half-line [A, B , on each side of the line AB there is a unique wedge-angle ∠BAC with |∠BAC|◦ = k. In all cases |∠BAB|◦ = 0, so that the degree-measure of each null angle is 0, while if ∠BAC is not null then |∠BAC|◦ > 0. It follows from the foregoing, that if ∠BAD is any wedge-angle then |∠BAD|◦ < 180, and that if ∠BAD, ∠CAD are supplementary angles, then |∠CAD|◦ = 180 − |∠BAD|◦ . 4. Given points B and C distinct from A such that C ∈ [A, B , we can choose a point D such that |∠BAD|◦ is equal to half the degree-measure of the wedge or straight angle with support |BAC. Then for all points P = A on the line AD we have |∠BAP|◦ = |∠PAC|◦ . We call AP the mid-line or bisector of the support |BAC.

Sec. 1.5

Revision of geometrical concepts

D

C

D P

P B

A

17

A

C

B

Figure 1.17. Mid-line of an angle-support.

5. Any angle ∠BAC such that 0 < |∠BAC|◦ < 90 is called acute, such that |∠BAC|◦ = 90 is called right, and such that 90 < |∠BAC|◦ < 180 is called obtuse. If ∠BAC is a right-angle, then the lines AB and AC are said to be perpendicular to each other, written AB ⊥ AC.

C

x

90

|

|

B A

y

|

B

A

|

y

|

|

z

|

|

C

|

x

|

|

A

B Figure 1.18. Perpendicular lines.

z C

|

Figure 1.19. Congruent triangles.

1.5.4 Our treatment of congruence If [A, B,C], [A , B ,C ] are triangles such that |B, C| = |B , C |, |C, A| = |C , A |, |A, B| = |A , B |, |∠BAC|◦ = |∠B A C |◦ , |∠CBA|◦ = |∠C B A |◦ , |∠ACB|◦ = |∠A C B |◦ , then we say by way of definition that the triangle [A, B,C] is congruent to the triangle [A , B ,C ]. Behind this concept is the physical idea that [A, B,C] can be placed on [A , B ,C ], fitting it exactly. By observation if [A, B,C], [A , B ,C ] are such that |C, A| = |C , A |, |A, B| = |A , B |, |∠BAC|◦ = |∠B A C |◦ , then [A, B,C] is congruent to [A , B ,C ]. This is known as the SAS (side, angle, side) condition for congruence of triangles. Similarly by observation if [A, B,C], [A , B ,C ] are such that |B, C| = |B , C |, |∠CBA|◦ = |∠C B A |◦ , |∠BCA|◦ = |∠B C A |◦ ,

18

Preliminaries

Ch. 1

then [A, B,C] is congruent to [A , B ,C ]. This is known as the ASA (angle, side, angle) condition for congruence of triangles. It can be proved that if T and T are triangles with vertices {A, B,C}, {A , B ,C }, respectively, for which |B, C| = |B , C |, |C, A| = |C , A |, |A, B| = |A , B |, then T is congruent to T . This is known as the SSS(side-side-side) principle of congruence.

1.5.5 Parallel lines 1. Distinct lines l, m are said to be parallel if l ∩ m = 0; / this is written as l m. We also take l l. m

l

Figure 1.20. Parallel lines.

By observation, given any line l and any point P there cannot be more than one line m through P which is parallel to l.

Q R

Q

◦ P ◦ A

R B

Figure 1.21. Alternate angles for a transversal.

S ◦ T P ◦ A

B

Corresponding angles.

It can be shown that two lines are parallel if and only if alternate angles made by a transversal, as indicated, are equal in magnitude, or equivalently, if and only if corresponding angles made by a transversal are equal in magnitude. 2. A convex quadrilateral in which opposite side-lines are parallel to each other is called a parallelogram. A parallelogram in which adjacent side-lines are perpendicular to each other is called a rectangle.

Sec. 1.6

Pre-requisites

19

D

A

A T

T B

D

C

B

C A rectangle.

Figure 1.22. A parallelogram.

1.6 PRE-REQUISITES Although this book re-starts geometry and trigonometry from the beginning, it does not take mathematics from a start. Consequently there is material from other parts of mathematics which is assumed known. This also should be revised at the start, or at the appropriate time when it is needed. At the beginning, we presuppose a moderate knowledge of set theory, sufficient to deal with sets, relations and functions, in particular order and equivalence relations. From Chapter 3 on we assume a knowledge of the real number system, and the elementary algebra involved. Later requirements come in gradually.

1.6.1 Set notation For set notation we refer to Smith [12, pages 1 – 38]. We mention that we use the word function where it uses map. We should also like to emphasise the difference between a set {a, b} and a couple or ordered pair (a, b). In a set, the order of the elements does not matter, so that {a, b} = {b, a} in all cases, and {a, b} = {c, d} if and only if either a=c

and

b=d

a=d

and

b = c.

or In a couple (a, b) it matters which is first and which is second. Thus (a, b) = (b, a) unless a = b, and (a, b) = (c, d) if and only if a = c and

b = d.

20

Preliminaries

Ch. 1

1.6.2 Classical algebra We need a knowledge of the real number system and the complex number system, and the classical algebra involving these, up to dealing with quadratic equations and two simultaneous linear equations in two unknowns. For this very elementary material there is an ample provision of textbooks entitled College Algebra by international publishers.

1.6.3 Other algebra For matrices and determinants we refer to Smith [12, pages 95 – 124] and McGregor, Nimmo and Stothers [11, pages 243 – 278], and for the little that we use on group theory to Smith [12, pages 125 – 152].

1.6.4 Distinctive property of real numbers among fields For the properties that distinguish the field of real numbers from other ordered fields, we refer to Smith [12, pages 153 – 196].

2

Basic shapes of geometry COMMENT. Geometry deals with our intuitions as to the physical space in which we exist, with the properties of the shapes and sizes of bodies as mathematically abstracted. It differs from set theory in that in geometry there are distinguished or special subsets, and relations involving them. To start with we presuppose a moderate knowledge of set theory, sufficient to deal with sets, relations and functions. From Chapter 3 on we assume a knowledge of the real number system, and the elementary algebra involved. In this first chapter we introduce the plane, points, lines, natural orders on lines, and open half-planes as primitive concepts, and in terms of these develop other special types of geometrical sets.

2.1 LINES, SEGMENTS AND HALF-LINES 2.1.1 Plane, points, lines Primitive Terms . Assuming the terminology of sets, the plane, denoted by Π, is a universal set the elements of which are called points. Certain subsets of Π are called (straight) lines. We denote by Λ the set of all these lines. AXIOM A1 . Each line is a proper non-empty subset of Π. For each set {A, B} of two distinct points in Π, there is a unique line in Λ to which A and B both belong. | We denote by AB the unique line to which distinct points A and B belong, so that A ∈ AB and B ∈ AB. It is an immediate consequence of Axiom A1 that AB = BA; that if C and D are distinct points and both belong to the line AB, then AB = CD; and that if A, B are distinct points, both on the line l and both on the line m, then l = m. Furthermore if l, m are any two lines in Λ, then either l ∩ m = 0, / Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50002-1 © 2016 Elsevier Ltd. All rights reserved.

22

Basic shapes of geometry

Ch. 2

or l ∩ m is a singleton, or l = m and in this last case

l ∩ m = l = m.

Moreover the plane Π is not a line, as each line is a proper subset of Π. If three or more points lie on one line we say that these points are collinear. If one point lies on three or more lines we say that these lines are concurrent.

2.1.2 Natural order on a line COMMENT. The two intuitive senses of motion along a line give us the original examples of linear (total) orders, and we refer to these as the two natural orders on that line. On a diagram the sense of a double arrow gives one natural order on l, while the opposite sense would yield the other natural order on l. We now take natural order as a primitive term, and go on to define segments and half-lines in terms of this and our existing terms.

B B A A Figure 2.1. A line AB. The arrows indicate that the line is to be continued unendingly.

Figure 2.2. The double arrow indicates a sense along the line AB.

Primitive Term . On each line l ∈ Λ there is a binary relation ≤l , which we refer to as a natural order on l. We read A ≤l B as ‘A precedes or coincides with B on l ’. AXIOM A2 . Each natural order ≤l has the properties:(i) A ≤l A for all points A ∈ l; (ii) if A ≤l B and B ≤l C then A ≤l C; (iii) if A ≤l B and B ≤l A, then A = B; (iv) for any points A, B ∈ l, either A ≤l B or B ≤l A. | COMMENT. We refer to (i), (ii), (iii) in A2 as the reflexive, transitive and antisymmetric properties, respectively, of a binary relation; property (iii) can be reworded as, if A ≤l B and A = B then B ≤l A. A binary relation with these three properties is commonly called a partial order. A binary relation with all four properties (i), (ii), (iii) and (iv) in A2 is commonly called a linear order or a total order.

Sec. 2.1

Lines, segments and half-lines

23

2.1.3 Reciprocal orders If A ≤l B we also write B ≥l A and read this as ‘B succeeds or coincides with A on l’. Then ≥l is also a total order on l, i.e. ≥l satisfies A2 (i), (ii), (iii) and (iv), as can readily be checked as follows. First, on interchanging A and A in A2 (i), we have A ≥l A for all A ∈ l. Secondly, suppose that A ≥l B and B ≥l C; then C ≤l B and B ≤l A, so by A2 (ii) C ≤l A; hence A ≥l C. Thirdly, suppose that A ≥l B and B ≥l A; then B ≤l A and A ≤l B so by A2 (iii) A = B. Finally, let A, B be any points on l; by A2 (iv), either A ≤l B or B ≤l A and so either B ≥l A or A ≥l B. We say that ≥l is reciprocal to ≤l . If now we start with ≥l and let l be its reciprocal we have A l B if B ≥l A; then we have A l B if and only if A ≤l B. Thus l coincides with ≤l , and so the reciprocal of ≥l is ≤l . The upshot of this is that ≤l and ≥l are a pair of total orders on l, each the reciprocal of the other. There is no natural way of singling out one of ≤l , ≥l over the other, and the notation is equally interchangeable as we could have started with ≥l . Having this pair is a nuisance but it is unavoidable, and we try to minimise the nuisance as follows. Given distinct points A and B, let l = AB. Then exactly one of A ≤l B, A ≥l B holds; for by A2 (iv) either A ≤l B or A ≥l B, and by A2 (iii) both cannot hold as that would imply that A = B. Thus we can choose the natural order on l in which A precedes B, by taking ≤l when A ≤l B, and by taking ≥l when A ≥l B; we will use the notation ≤l for this natural order. Let A and B be distinct points in Π, let l = AB and A ≤l B. Let C be a point of l, distinct from A and B. Then exactly one of (a) C ≤l A ≤l B,

(b) A ≤l C ≤l B,

(c) A ≤l B ≤l C,

holds. Proof . If C ≤l A then clearly (a) holds. If C ≤l A is false, then by A2 (iv) A ≤l C; by A2 (iv) we have moreover either C ≤l B or B ≤l C, and these yield, respectively, (b) and (c). Thus at least one of (a), (b), (c) holds. On the other hand, if (a) and (b) hold, we have A = C by A2 (iii) and this contradicts our assumptions. Similarly if (b) and (c) hold we would have B = C. Finally if (a) and (c) hold, from (a) we have C ≤l B by A2 (ii) and then B = C.

2.1.4 Segments Definition . For any points A, B ∈ Π, we define the segments [A, B] and [B, A] as follows. Let l be a line such that A, B ∈ l and ≤l , ≥l a pair of reciprocal natural orders on l. Then if (2.1.1) A ≤l B so that B ≥l A, we define [A, B] ={P ∈ l : A ≤l P ≤l B} = {P ∈ l : A ≤l P and P ≤l B}, [B, A] ={P ∈ l : B ≥l P ≥l A},

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Ch. 2

A ≥l B,

(2.1.2)

while if B ≤l A

so that

we define [B, A] ={P ∈ l : B ≤l P ≤l A}, [A, B] ={P ∈ l : A ≥l P ≥l B}. We should use a more complete notation such as [A, B]≤l ,≥l , [B, A]≤l ,≥l , but make do with the less precise one. Note that (2.1.2) comes from (2.1.1) on interchanging A and B, or on interchanging ≤l and ≥l . When A = B, by A1 l = AB; by A2 (iv) at least one of (2.1.1) and (2.1.2) holds, and by A2 (iii) only one of (2.1.1) and (2.1.2) holds. When A = B, l can be any line B through A, and we find that {P ∈ l : A ≤l P ≤l A} = {A}, {P ∈ A l : A ≥l P ≥l A} = {A}, for the singleton {A}. To see this we Figure 2.3. A segment [A, B]. note that A ≤l A ≤l A by A2 (i), while if A ≤l P ≤l A then P = A by A2 (iii). The same argument holds for ≥l . Thus [A, A] = {A}. Segments have the following properties:(i) If A = B, then [A, B] ⊂ AB. (ii) A, B ∈ [A, B] for all A, B ∈ Π. (iii) [A, B] = [B, A] for all A, B ∈ Π. (iv) If C, D ∈ [A, B] then [C, D] ⊂ [A, B]. (v) If A, B,C are distinct points on a line l, then precisely one of A ∈ [B,C], B ∈ [C, A], C ∈ [A, B], holds.

Proof . In each case we suppose that A ≤l B so that we have (2.1.1) above; otherwise replace ≤l by ≥l throughout to cover (2.1.2). (i) By A1 , l = AB so [A, B] is a set of points on AB. (ii) By A2 (i) A ≤l A ≤l B and A ≤l B ≤l B. (iii) As A ≤l B, then B ≥l A so [B, A] = {P ∈ l : B ≥l P ≥l A}. Now if P ∈ [A, B], then A ≤l P and P ≤l B. It follows that B ≥l P and P ≥l A. Thus P ∈ [B, A] and so [A, B] ⊂ [B, A]. By a similar argument [B, A] ⊂ [A, B] and so [A, B] = [B, A].

Sec. 2.1

Lines, segments and half-lines

25

(iv) Let C, D ∈ [A, B] so that A ≤l C ≤l B and A ≤l D ≤l B. By A2 (iv) either C ≤l D or D ≤l C. If C ≤l D and P ∈ [C, D], then C ≤l P ≤l D. Thus A ≤l C, C ≤l P so by A2 (ii), A ≤l P. Also P ≤l D, D ≤l B so by A2 (ii) P ≤l B. Thus P ∈ [A, B]. If D ≤l C, we interchange C and D in the last paragraph. (v) This follows immediately from 2.1.3.

2.1.5 Half-lines Definition . Given a line l ∈ Λ, a point A ∈ l and a natural order ≤l on l, then the set ρ (l, A, ≤l ) = {P ∈ l : A ≤l P}, is called a half-line or ray of l, with initial point A. Given distinct points A, B let ≤l be the natural order on l = AB for which A ≤l B; then we also use the notation [A, B for ρ (l, A, ≤l ).

C

B

A B

A Figure 2.4. A half-line [A, B .

Opposite half-lines.

As ≥l is also a natural order on l,

ρ (l, A, ≥l ) = {P ∈ l : A ≥l P} = {P ∈ l : P ≤l A} is also a half-line of l, with initial point A. We say that ρ (l, A, ≤l ) and ρ (l, A, ≥l ) are opposite half-lines. Half-lines have the following properties:(i) In all cases ρ (l, A, ≤l ) ⊂ l. (ii) In all cases A ∈ ρ (l, A, ≤l ). (iii) If B,C ∈ ρ (l, A, ≤l ), then [B,C] ⊂ ρ (l, A, ≤l ).

Proof . (i) By the definition of ρ (l, A, ≤l ), we have P ∈ l for all P ∈ ρ (l, A, ≤l ) and so ρ (l, A, ≤l ) ⊂ l. (ii) By A2 (i) A ≤l A, so A ∈ ρ (l, A, ≤l ). (iii) As B,C ∈ ρ (l, A, ≤l ) we have A ≤l B and A ≤l C. Since B,C ∈ l, by A2 (iv) either B ≤l C or C ≤l B. When B ≤l C, we have B ≤l P for all P ∈ [B,C]; with A ≤l B this gives A ≤l P by A2 (ii), and so P ∈ ρ (l, A, ≤l ). When C ≤l B, we have a similar proof.

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2.2 OPEN AND CLOSED HALF-PLANES 2.2.1 Convex sets Definition . A set E is said to be convex if for every P, Q ∈ E , [P, Q] ⊂ E holds. NOTE. By 2.1.4(iv) every segment is a convex set; by 2.1.5(iii) so is every halfline. In preparation for the next subsection, we note that by A1 , for each line l ∈ Λ we have Π \ l = 0. /

2.2.2 Open half-planes Primitive Term . Corresponding to each line l ∈ Λ, there is a pair {G1 , G2 } of nonempty sets called open half-planes with common edge l. AXIOM A3 . Open half-planes G1 , G2 with common edge l have the properties:(i) Π \ l = G1 ∪ G2 ; (ii) G1 and G2 are both convex sets; (iii) if P ∈ G1 and Q ∈ G2 , then [P, Q] ∩ l = 0. / | We note the following immediately. Open half-planes {G1 , G2 } with common edge l have the properties:/ l ∩ G2 = 0. / (i) l ∩ G1 = 0, / (ii) G1 ∩ G2 = 0. (iii) If P ∈ G1 and [P, Q] ∩ l = 0/ where Q ∈ l, then Q ∈ G2 . (iv) Each line l determines a unique pair of open half-planes.

Proof . (i) By A3 (i), l ∩ (G1 ∪ G2 ) = 0/ and as G1 ⊂ G1 ∪ G2 it follows that l ∩ G1 = 0. / The other assertion is proved similarly. / there is some point R in both G1 and G2 . By A3 (iii) with P = (ii) If G1 ∩ G2 = 0, R, Q = R, we have that [R, R] ∩ l = 0. / But R is the only point in [R, R] so R ∈ l. This / contradicts the fact that l ∩ G1 = 0. / (iii) For otherwise by A3 (i), Q ∈ G1 and then by A3 (ii) [P, Q] ⊂ G1 . As l ∩ G1 = 0, it follows that [P, Q] ∩ l = 0/ which contradicts the assumptions. (iv) Suppose that Π \ l = G1 ∪ G2 = G1 ∪ G2 , where {G1 , G2 } and {G1 , G2 } are both sets of open half-planes with common edge l. Then G1 ⊂ G1 ∪ G2 = G1 ∪ G2 so at one of the following holds

Sec. 2.2 (a) G1 ⊂ G1

Open and closed half-planes or

(b) G1 ⊂ G2

or

27

(c) G1 ∩ G1 = 0, / G1 ∩ G2 = 0. /

In (c) we have P ∈ G1 , P ∈ G1 and Q ∈ G1 , Q ∈ G2 for some P and Q. But then we have [P, Q] ⊂ G1 , by A3 (ii) applied to G1 , and [P, Q] ∩ l = 0, / by A3 (iii) applied to / Thus (c) cannot happen. {G1 , G2 }. This gives a contradiction as l ∩ G1 = 0. By similar reasoning, we must have either (d) G1 ⊂ G1

or

(e) G1 ⊂ G2 .

Now (a) and (d) give G1 = G1 and it follows that G2 = G2 as (G1 ∪ G2 ) \ G1 = G2 , (G1 ∪ G2 ) \ G1 = G2 . Similarly (b) and (e) give G1 = G2 and it follows that G2 = G1 . Finally, we cannot have (a) and (e) as that would imply G1 ⊂ G2 . Neither can we have (b) and (d). TERMINOLOGY. If two points are both in G1 or both in G2 they are said to be on the one side of the line l, while if one of the points is in G1 and the other is in G2 they are said to be on different sides of l.

B A Figure 2.5. A closed half-plane shaded.

2.2.3 Closed half-planes Definition . If G1 , G2 are open half-planes with common edge l, we call H1 = G1 ∪ l, H2 = G2 ∪ l, closed half-planes with common edge l. Closed half-planes H1 , H2 with common edge l have the properties:(i) H1 ∪ H2 = Π. (ii) H1 ∩ H2 = l. (iii) Each of H1 , H2 is a convex set. (iv) If A ∈ l and B = A is in H1 , then [A, B ⊂ H1 .

Proof . (i) By A3 (i), Π = G1 ∪ G2 ∪ l = (G1 ∪ l) ∪ (G2 ∪ l) = H1 ∪ H2 . (ii) For (G1 ∪ l) ∩ (G2 ∪ l) = (G1 ∩ G2 ) ∪ (G1 ∩ l) ∪ (G2 ∩ l) ∪ (l ∩ l) = l ∩ l = l. (iii) We prove that H1 is convex; proof for H2 is similar. Let A, B ∈ H1 ; we wish to show that [A, B] ⊂ H1 . CASE (a). Let A, B ∈ G1 . Then the conclusion follows from A3 (ii).

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CASE (b). Let A, B ∈ l. Then [A, B] ⊂ l ⊂ H1 , so the result follows. CASE (c). Let one of A, B be on l and the other in G1 , say A ∈ l, B ∈ G1 . Suppose that [A, B] is not a subset of H1 . Then there is some point C ∈ [A, B] such that C ∈ G2 . Note that C = A,C = B as A, B ∈ G2 ,C ∈ G2 . Now B ∈ G1 ,C ∈ G2 so by A3 (iii) there is some point D of [B,C] on l, so that D ∈ [B,C], D ∈ l. Then A, B,C are collinear and distinct, and C ∈ [A, B] so by 2.1.4 we cannot have A ∈ [B,C]. Hence A = D. But A ∈ l, D ∈ l so by A1 , AD = l. However AB = BC and D ∈ BC, so D ∈ AB. Then AB = AD = l, so B ∈ l. This gives a contradiction. Thus the original supposition is untenable so [A, B] ⊂ H1 , and this proves (iii). (iv) CASE (a). Let B ∈ l. Then [A, B ⊂ l ⊂ H1 , which gives the desired conclusion. CASE (b). Let B ∈ G1 . Suppose that [A, B is not a subset of H1 . Then there is some point C ∈ [A, B such that C ∈ G2 . Clearly C = A,C = B. Now A, B,C are distinct collinear points, so by 2.1.4 precisely one of A ∈ [B,C], B ∈ [C, A], C ∈ [A, B], holds. We cannot have A ∈ [B,C] as that would put B,C in different half-lines with initial-point A, whereas they are both in [A, B. This leaves us with two subcases. Subcase 1. Let C ∈ [A, B]. We recall that A, B ∈ H1 so by part (iii) of the present result [A, B] ⊂ H1 . As C ∈ [A, B],C ∈ G2 , we have a contradiction. Subcase 2. Let B ∈ [A,C]. We recall that A ∈ H2 ,C ∈ H2 so by part (iii) of the present result, [A,C] ⊂ H2 . Then B ∈ H2 , B ∈ G1 which gives a contradiction. Thus the original supposition is untenable, and this proves (iv). NOTE. The terms ‘open’ and ‘closed’ are standard in analysis and point-set topology. What is significant is that an open half-plane contains none of the points of the edge, while a closed half-plane contains all of the points of the edge.

2.3 ANGLE-SUPPORTS, INTERIOR AND EXTERIOR REGIONS, ANGLES 2.3.1 Angle-supports, interior regions

C B

C

A

A B

Figure 2.6. An angle-support.

A straight angle-support.

Sec. 2.3

Angle-supports, interior and exterior regions, angles

29

Definition . We call a pair {[A, B , [A,C } of co-initial half-lines an angle-support. For this we use the notation |BAC. When A ∈ [B,C], this is called a straight anglesupport. We call the half-lines [A, B and [A,C the arms, and the point A the vertex, of |BAC. Note that we are assuming B = A and C = A from the definition of half-lines. In all cases we have |BAC = |CAB. COMMENT. The reason that we do not call |BAC an angle is that there are two angles associated with this configuration.

C

C B

B

A Figure 2.7. An interior region.

A The corresponding exterior region.

Definition. Consider an angle-support |BAC which is not straight. When A, B,C are not collinear, let H1 be the closed half-plane with edge AB in which C lies, and H3 the closed half-plane with edge AC in which B lies. Then H1 ∩ H3 is called the interior region of |BAC, and we denote it by I R(|BAC). When A, B,C are collinear we have [A, B = [A,C and we define I R(|BAC) = [A, B . Interior regions have the following properties:(i) [A, B and [A,C are both subsets of I R(|BAC). (ii) If P, Q ∈ I R(|BAC) then [P, Q] ⊂ I R(|BAC), so that an interior region is a convex set. (iii) If P ∈ I R(|BAC) and P = A, then [A, P ⊂ I R(|BAC).

Proof . (i) When A, B,C are non-collinear, by 2.1.5 [A, B ⊂ AB ⊂ H1 and by 2.2.3 [A, B ⊂ H3 so [A, B ⊂ H1 ∩ H3 . Similarly for [A,C . When [A, B = [A,C the result is trivial. (ii) When A, B,C are non-collinear, we have that [P, Q] is a subset of H1 by 2.2.3. It is a subset of H3 similarly, and so is a subset of the intersection of these closed half-planes. When [A, B = [A,C , the result follows from 2.1.5. (iii) When A, B,C are non-collinear, by 2.2.3 [A, P is a subset of each of H1 and H3 , and so of their intersection. When [A, B = [A,C we have I R(|BAC) = [A , B and [A, P = [A, B .

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2.3.2 Exterior regions Definition . If |BAC is an angle-support which is not straight and I R(|BAC) is its interior region, then {Π \ I R(|BAC)} ∪ [A, B ∪ [A,C is called the exterior region of |BAC, and denoted by E R(|BAC). Thus the interior and exterior regions have in common only the arms.

2.3.3 Angles

Figure 2.8. A wedge-angle.

A reflex-angle.

A straight-angle.

Definition . Let |BAC be an angle-support which is not straight, with interior re-  pair |BAC, I R(|BAC) gion I R(|BAC) and exterior region ER(|BAC). Then the  is called a wedge-angle, and the pair |BAC, E R(|BAC) is called a reflex-angle. If |BAC is a straight-angle support and H1 , H2 are the closed half-planes with common edge AB, then each of the pairs (|BAC, H1 ), (|BAC, H2 ) is called a straight-angle In each case the point A is called the vertex of the angle, the half-lines [A, B and [A,C are called the arms of the angle, and |BAC is called the support of the angle. We denote a wedge-angle with support |BAC by ∠BAC. The wedge-angle ∠BAB is said to be a null-angle. A reflex-angle with support |BAB is called a full-angle.

2.4 TRIANGLES AND CONVEX QUADRILATERALS 2.4.1 Terminology COMMENT. The terminology which we have used hitherto is established, apart from ‘angle-support’ and ‘wedge-angle’ which we have coined. Now we are reaching terminology which is of long standing but is used in slightly varying senses. In Euclidean geometry it is generally accepted that the concept of triangle is associated with: (i) a set {A, B,C} of three points which are not collinear; (ii) a union of segments [B,C] ∪ [C, A] ∪ [A, B], where the points A, B,C are as in (i); (iii) an intersection of half-planes H1 ∩ H3 ∩ H5 , where A, B,C are as in (i), H1 is the closed half-plane with edge BC in which A lies, H3 is the closed half-plane

Sec. 2.4

Triangles and convex quadrilaterals

31

with edge CA in which B lies, and H5 is the closed half-plane with edge AB in which C lies. However in some courses the actual definition of a triangle is taken to be (i), in other courses it is taken to be (ii), and in other courses it is taken to be (iii), with (ii) and (iii) very common. In yet other courses a combination of (i) and (ii) is taken. Having to make a choice for the sake of precision, we opt for (iii); then for us (i) will be the set of vertices of our triangle, and (ii) will be the perimeter of our triangle, with the individual segments being the sides. We shall then be able to refer naturally to the area of a triangle and the length of its perimeter. Consideration similar to (i), (ii) and (iii) for a triangle surround each of the terms quadrilateral, parallelogram, rectangle and square, and we adopt our terminology consistently.

2.4.2 Triangles NOTE. Let A, B,C be points which do not lie on one line. Then by A1 , A, B,C are distinct points, and A ∈ BC, B ∈ CA, C ∈ AB. In fact these lines are not concurrent; for BC and CA cannot have a point P in common other than P = C, while C ∈ AB. Definition . For non-collinear points A, B,C let H1 be the closed half-plane with edge BC in which A lies, H3 the closed half-plane with edge CA in which B lies, and H5 the closed half-plane with edge AB in which C lies. Then the intersection H1 ∩ H3 ∩ H5 is called a triangle, and is denoted by [A, B,C].

D

C

C

B B A Figure 2.9. A triangle [A, B,C].

A Figure 2.10. A convex quadrilateral.

The points A, B,C are called its vertices; the segments [B,C], [C, A], [A, B] are called its sides; the lines BC,CA, AB are called its side-lines. The union [B,C]∪[C, A]∪[A, B] of its sides is called its perimeter. A side and a vertex not contained in it are said to be opposite; thus A is opposite [B,C] but is not opposite [C, A] or [A, B]. Triangles have the following properties:(i) [A, B,C] is independent of the order of the points A, B,C. (ii) Each of the vertices A, B,C is an element of [A, B,C]. (iii) If P, Q ∈ [A, B,C], then [P, Q] ⊂ [A, B,C] so that a triangle is a convex set.

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(iv) Each of the sides [B,C], [C, A], [A, B] is a subset of [A, B,C].

Proof . (i) As ∩ is commutative, H1 ∩ H3 ∩ H5 is independent of the order of H1 , H3 , H5 . (ii) The vertex A is in H1 by definition. It is also in the edge of each of H3 and H5 , so by 2.2.3 it is in each of these closed half-planes. The vertices B and C are treated similarly. (iii) By definition of an intersection, P and Q are in each of H1 , H3 , H5 . By 2.2.3, [P, Q] is a subset of each of these closed half-planes, and so it is a subset of their intersection. (iv) This follows from parts (ii) and (iii) of the present result.

2.4.3 Pasch’s property, 1882 Pasch’s property. If a line cuts one side of a triangle, not at a vertex, then it will either pass through the opposite vertex, or cut one of the other two sides. Proof . Let [A, B,C] be the triangle and l a line which cuts the side [A, B] at a point which is not a vertex. If C ∈ l we have the first conclusion. Otherwise suppose that l does not cut [B,C]. Then A and B are on different sides of l, but B and C are on the same side of l. It follows that A and C are on different sides if l, so by A3 (iii) a point of [A,C] lies on l.

2.4.4 Convex quadrilaterals Definition . Let A, B,C, D be four points in Π, no three of which are collinear, and such that [A,C] ∩ [B, D] = 0. / Let H1 be the closed half-plane with edge AB in which D lies, H3 the closed half-plane with edge BC in which A lies, H5 the closed halfplane with edge CD in which B lies, and H7 the closed half-plane with edge DA in which C lies. Then the intersection H1 ∩ H3 ∩ H5 ∩ H7 of these four half-planes is called a convex quadrilateral, and we denote it by [A, B,C, D]. As each of the closed half-planes is a convex set this intersection is also a convex set. Each of the four points A, B,C, D is called a vertex; the segments [A, B], [B,C], [C, D], [D, A] are called the sides, and AB, BC,CD, DA are called the side-lines ; the union of the sides [A, B]∪[B,C]∪[C, D]∪[D, A] is called the perimeter. The segments [A,C], [B, D] are called the diagonals, and AC, BD the diagonal lines. Vertices which are the end-points of a side are called adjacent while vertices which are the endpoints of a diagonal are called opposite; thus A and B are adjacent as [A, B] is a side, and A and C are opposite as [A,C] is a diagonal. Sides which have a vertex in common are said to be adjacent while sides which do not have a vertex in common are said to be opposite; thus the sides [A, B], [A, D] are adjacent as the vertex A is in both, while the sides [A, B], [C, D] are opposite as neither C nor D is in AB and so neither of them could be A or B.

Sec. 2.4

Triangles and convex quadrilaterals

33

If we write D / \ A C \ / B then two vertices in [A, B,C, D] will be adjacent if the letters for them in this diagram are linked.

Exercises 2.1 Let P be a fixed point in Π. Identify the union of all lines l ∈ Λ such that P ∈ l. 2.2 Prove that segments have the following properties:(i) If C ∈ [A, B], then [A,C] ∪ [C, B] = [A, B] and [A,C] ∩ [C, B] = {C}. (ii) If C ∈ [A, B] and B ∈ [A,C] then B = C. (iii) If C ∈ [A, B] and D ∈ [A,C], then C ∈ [D, B]. (iv) If B = A, B ∈ [A,C] and B ∈ [A, D], then either C ∈ [B, D] or D ∈ [B,C]. 2.3 Prove that half-lines have the following properties:(i) If B ∈ ρ (l, A, ≤l ), then ρ (l, B, ≤l ) ⊂ ρ (l, A, ≤l ). (ii) If B ∈ ρ (l, A, ≤l ), then ρ (l, A, ≤l ) = [A, B] ∪ ρ (l, B, ≤l ) and [A, B] ∩ ρ (l, B, ≤l ) = {B}. (iii) Let B ∈ ρ (l, A, ≤l ), A = B and A ∈ [B,C]. Then C ∈ ρ (l, A, ≤l ) only if C = A. (iv) Let B ∈ ρ (l, A, ≤l ) and A = B. Then C ∈ ρ (l, A, ≤l ) if and only if either B ∈ [A,C] or C ∈ [A, B]. (v) In all cases

ρ (l, A, ≤l ) ∪ ρ (l, A, ≥l ) = l and ρ (l, A, ≤l ) ∩ ρ (l, A, ≥l ) = {A}. (vi) Let B ∈ ρ (l, A, ≤l ) and A = B. Then C ∈ ρ (l, A, ≥l ) if and only if A ∈ [B,C]. (vii) Let B ∈ ρ (l, A, ≤l ) and A = B. Then

ρ (l, A, ≤l ) ∪ ρ (l, B, ≥l ) = l, ρ (l, A, ≤l ) ∩ ρ (l, B, ≥l ) = [A, B], ρ (l, A, ≥l ) ∩ ρ (l, B, ≤l ) = 0, / ρ (l, A, ≥l ) ∪ ρ (l, B, ≤l ) ∪ [A, B] = l. (viii) If A = B, A = C and C ∈ [A, B, then [A, B = [A,C .

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Basic shapes of geometry

Ch. 2

2.4 If [A, B], [C, D] are both segments of a line l such that [A, B] ∩ [C, D] = 0, / show that [A, B] ∩ [C, D] and [A, B] ∪ [C, D] are both segments. 2.5 Show that if A = B and C, D are both in AB\[A, B], then either [A, B]∩[C, D] = 0/ or [A, B] ⊂ [C, D]. 2.6 Let ≤E be a total order on the set E and f : E → F a 1:1 onto function. If for a, b ∈ F, a ≤F b when f −1 (a) ≤E f −1 (b), show that ≤F is a total order on F. 2.7 Use Ex.2.6 to show that if F is an infinite set and there is a total order on F, then there are infinitely many total orders on F. 2.8 Show that interior regions have the following properties:(i) If P ∈ I R(|BAC) and P = A, then AP ∩ I R(|BAC) = [A, P . (ii) If A, B,C are non-collinear and U ∈ [A, B , V ∈ [A,C but neither U nor V is A, then UV ∩ I R(|BAC) = [U,V ]. (iii) If A,U,V are distinct collinear points, and U and V are both in I R(|BAC), then V ∈ [A,U . 2.9 Show that an exterior region has the following properties:(i) The arms [A, B and [A,C are both subsets of E R(|BAC). (ii) If P ∈ E R(|BAC) and P = A, then [A, P ⊂ E R(|BAC). 2.10 Show that convex quadrilaterals [A, B,C, D] have the following properties:(i) Each of 2[A, D,C, B],

3[C, B, A, D],

4[C, D, A, B],

5[B, A, D,C],

6[B,C, D, A],

7[D, A, B,C],

8[D,C, B, A], is equal to 1[A, B,C, D]. (ii) Each of the vertices A, B,C, D is an element of [A, B,C, D]. (iii) The lines AC and BD cannot have distinct points E and F in common. (iv) Each side and each diagonal is a subset of [A, B,C, D]. (v) Any pair of opposite sides are disjoint. 2.11 For non-collinear points A, B, D let C be a point which is in the interior region I R(|BAD) but not in [A, B , [A, D or [A, B, D]. Show that then A, B,C, D are the vertices of a convex quadrilateral.

3

Distance; degree-measure of an angle COMMENT. In this chapter we introduce distance as a primitive concept, relate it to the properties of segments, and define the notion of the mid-point of two points. We also introduce as a primitive concept the notion of the degree-measure of a wedgeangle and of a straight-angle, relate it to the properties of interior-regions and halfplanes, and define the notion of the mid-line of an angle-support.

3.1 DISTANCE 3.1.1 Axiom for distance Notation. We denote by R the set of real numbers. Primitive Term . There is a function | | : Π × Π → R called distance. We read |A, B| as the distance from A to B. We also refer to |A, B| as the length of the segment [A, B]. AXIOM A4 . Distance has the following properties:(i) |A, B| ≥ 0 for all A, B ∈ Π; (ii) |A, B| = |B, A| for all A, B ∈ Π; (iii) If Q ∈ [P, R], then |P, Q| + |Q, R| = |P, R|; (iv) Given any k ≥ 0 in R, any line l ∈ Λ, any point A ∈ l and either natural order ≤l on l, there is a unique point B ∈ l such that A ≤l B and |A, B| = k, and a unique point C ∈ l such that C ≤l A and |A, C| = k. | COMMENT. Note that A4 (iv) states that we can lay off a distance k, uniquely, on l on either side of A. The fact that different letters A, B,C are used is not to be taken as a claim that A, B,C are distinct in all cases. Axiom A4 (iv) implies that each line l Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50003-3 © 2016 Elsevier Ltd. All rights reserved.

36

Distance; degree-measure of an angle

Ch. 3

contains infinitely many points and this supersedes the specification in A1 that l = 0; / nevertheless it was convenient to stipulate the latter to avoid a trivial situation. In A4 (iii) addition + of real numbers is involved.

y x

P

x

k

R

Q +y

B

A

Figure 3.1. Addition of distances.

Laying off a distance k.

3.1.2 Derived properties of distance Distance has the following properties:(i) For all A ∈ Π |A, A| = 0, and we have |A, B| > 0 if A = B. (ii) If P ∈ [A, B], then |A, P| ≤ |A, B|. If additionally P = B, then |A, P| < |A, B|. (iii) If A = C and B lies on the line AC but outside the segment [A,C], then |A, B| + |B, C| > |A, C|. (iv) If C ∈ [A, B is such that |A, B| < |A, C|, then B ∈ [A,C].

Proof . (i) By A4 (iii) with P = Q = A and any R ∈ Π, we have |A, A| + |A, R| = |A, R|, i.e. x + y = y where x = |A, A| and y = |A, R|. It follows that x = 0. Next with A = B let l = AB and ≤l be the natural order on l for which A ≤l B. Then we have A ≤l B, A ≤l A, |A, A| = 0, so that if we also had |A, B| = 0, then by the uniqueness part of A4 (iv) with k = 0, we would have A = B and so have a contradiction. To avoid this we must have |A, B| > 0. (ii) As P ∈ [A, B], by A4 (iii) we have |A, P|+ |P, B| = |A, B|. But by A4 (i) |P, B| ≥ 0 and so |A, P| ≤ |A, B|. If P = B, then by (i) of the present theorem |P, B| > 0 and so |A, P| < |A, B|. (iii) As B ∈ [A,C] we have B = A, B = C and so by 2.1.4 we have either A ∈ [B,C] or C ∈ [A, B]. In the first of these |B, A| + |A, C| = |B, C| by A4 (iii) and as |A, B| = |B, A| > 0 this gives |A, C| < |B, C| < |A, B| + |B, C|. In the second case we have |A, C| + |C, B| = |A, B| by A4 (iii) and as |C, B| = |B, C| > 0, then |A, C| < |A, B| < |A, B| + |B, C|.

Sec. 3.1

Distance

37

(iv) We have A = B by definition of [A, B , and A = C as 0 < |A, B| < |A, C| so that 0 < |A, C|. We also have B = C as |A, B| < |A, C| combined with B = C would give |A, B| < |A, B|, whereas once (A, B) is known |A, B| is uniquely determined. We cannot have A ∈ [B,C] as C ∈ [A, B . Then by 2.1.4 either B ∈ [C, A] or C ∈ [A, B]. But by (ii) of the present result, if C ∈ [A, B] we would have |A, C| ≤ |A, B|. As this is ruled out by assumption, we must have B ∈ [A,C]. Segments and half-lines have the following further properties:(i) Let l ∈ Λ be a line, A ∈ l and ≤l a natural order on l. Then there are points B and C on l such that A ≤l B and B = A, and such that C ≤l A and C = A. (ii) If A = B, there are points X ∈ [A, B] such that X = A and X = B. (iii) If [A, B = [C, D then A = C.

B

B X

A C

A Figure 3.2.

Proof. (i) By A4 (iv) with any k > 0, there is some B ∈ l such that A ≤l B and |A, B| = k. As |A, B| > 0, we have A = B. Proof for the existence of C is similar. (ii) Let ≤l be the natural order on l = AB for which A ≤l B. As A = B we have |A, B| > 0 and then with any k such that 0 < k < |A, B|, there is a point X ∈ l such that A ≤l X and |A, X| = k. As |A, X| = 0, we have A = X ; as |A, X| < |A, B| then X = B. As X ∈ [A, B and |A, X| < |A, B|, we have X ∈ [A, B]. (iii) With the notation of (ii), P ∈ [A, B if and only if A ≤l P. Now C ∈ [C, D = [A, B so A ≤l C; similarly A ≤l D. CASE 1. Let C ≤l D, so that [C, D = {Q ∈ l : C ≤l Q}. As A ∈ [A, B = [C, D we have C ≤l A and this combined with A ≤l C implies A = C. CASE 2. Let D ≤l C. Then [C, D = {Q ∈ l : Q ≤l C}. By (i) of the present result there is an X ∈ l such that X ≤l A and X = A. Then X ≤l A, A ≤l C so X ≤l C and thus X ∈ [C, D . However X ∈ [A, B as otherwise we would have A ≤l X which combined with X ≤l A implies X = A and involves a contradiction. Then X ∈ [C, D , X ∈ [A, B which contradicts the fact that [A, B = [C, D and so this case cannot occur.

38

Distance; degree-measure of an angle

D

B C

A =C

DX

Ch. 3

B

A

Figure 3.3.

3.2 MID-POINTS 3.2.1 If A = B there is a unique point X on l = AB such that |A, X| = |X, B|. In this in fact X ∈ [A, B] and X = A, X = B. Proof . Existence . Let ≤l be the natural order on l for which A ≤l B. With k = 12 |A, B|, by A4 (iv) there is a point X on l such that A ≤l X and |A, X| = 12 |A, B|. Clearly X ∈ [A, B . As |A, B| > 0 we have |A, X| < |A, B|; by 3.1.2 this implies that X ∈ [A, B], X = B. By A4 (iii) |A, X| + |X, B| = |A, B| and so |X, B| = |A, B| − 12 |A, B| = 12 |A, B|. Thus |A, X| = |X, B| as required. We have already seen that X ∈ [A, B] and X = B; as |A, X| > 0 we also have X = A. Uniqueness . Suppose now that Y ∈ l and |A, Y| = |Y, B|. Then Y cannot be A or B, as e.g. Y = A implies that |A, A| = |A, B|, i.e. 0 = |A, B|. Thus by 2.1.4 one of Y ∈ [A, B], A ∈ [Y, B], B ∈ [A,Y ],

+

+

holds. The second of these would imply |Y, A| + |A, B| = |Y, B| and so |Y, A| < |Y, B| as |A, B| > 0. The third of these would imply |A, B| + |B, Y| = |A, Y| and so |B, Y| < |A, Y|. As these contradict our assumptions, we must have Y ∈ [A, B]. Then |A, Y| + |Y, B| = |A, B| and as |A, Y| = |Y, B| this implies that |A, Y| = 21 |A, B|. Then A ≤l X , A ≤l Y and |A, X| = |A, Y| so by the uniqueness in A4 (iv) we must have X = Y . Definition . Given any points B A, B ∈ Π, we define the midpoint of A and B as follows: if A = B then the mid-point is A; when A = B the mid-point is the X unique point X on the line AB such that |A, X| = |X, B|, which A has just been guaranteed. We deFigure 3.4. Mid-point of A and B. note the mid-point of A and B by mp(A, B). Mid-points have the following properties:(i) For all A, B ∈ Π, mp(A, B) = mp(B, A).

Sec. 3.3

A ratio result

39

(ii) For all A, B ∈ Π, mp(A, B) ∈ [A, B]. (iii) In all cases mp(A, A) = A, and mp(A, B) = A, mp(A, B) = B when A = B. (iv) Given any points P and Q in Π, there is a unique point R ∈ Π such that Q = mp(P, R).

Proof . (i) When A = B this follows from the definition and A4 (ii); when A = B it is immediate. (ii) When A = B this follows from the preparatory result. When A = B it amounts to A ∈ {A}. (iii) This follows from the definition and preparatory result. (iv) Existence . If Q = P we take R = P and then mp(P, R) = mp(P, P) = P = Q. Suppose then that P = Q, let l = PQ and let ≤l be the natural order on l under which P ≤l Q. Take R on l so that P ≤l R and |P, R| = 2|P, Q|. Then P precedes both Q and R on l, while |P, Q| = 12 |P, R|. By our initial specification of X in the preparatory result we see that Q = mp(P, R). Uniqueness . Suppose that also Q = mp(P, S). We again first take Q = P. Now in the preparatory result we had X = A, so that cannot be the situation here as Q = P; thus we must have S = P and so S = R. Next suppose that Q = P; then we cannot have S = P, as we had X = A. Then Q ∈ PS, so by A1 S ∈ PQ. In fact Q ∈ [P, S] so as P ≤l Q we must have P ≤l S; moreover |P, R| = |P, S| as each is twice the distance |P, Q|. By the uniqueness in A4 (iv) we must then have R = S.

3.3 A RATIO RESULT 3.3.1 Let A, B,C be distinct collinear points, and write |A, C| |A, C| = r, = s. |A, B| |C, B| Then if C ∈ [A, B] we have

s r , r= . 1−r 1+s Proof . Let |A, C| = x, |C, B| = y. As C ∈ [A, B] we have |A, B| = x + y. Then s=

x =r x+y Hence

y 1 = −1 x r

so that

and so

x+y 1 = . x r x |A, C| r = = . y |C, B| 1 − r

40

Distance; degree-measure of an angle In turn s=

r 1−r

s − sr = r,

and so

giving s = r(1 + s)

Ch. 3

r=

and thus

s . 1+s

3.4 THE CROSS-BAR THEOREM 3.4.1 The cross-bar theorem. Let A, B,C be non-collinear points, X = A any point on [A, B and Y = A any point on [A,C . If D = A is any point in the interior region / I R(|BAC), then [A, D ∩ [X ,Y ] = 0.

Proof . If D is on [A, B or [A,C the result is clear, so we turn to other cases. By 3.1.2 there is a point E = A such that A ∈ [E, X ]. Thus X and E are on different sides of the line AC.

C Y D X E

A

B

Figure 3.5. The Cross-Bar Theorem.

Then by 2.2.3(iv) every point of [Y, E (other than Y ) is on one side of AC, while every point of [A, D (other than A) is on a different side of AC; thus [A, D does not meet [Y, E]. Moreover the other points of the line AD are on one side of the line AB, while the points of [E,Y (other than E) are on the other side of AB. On combining these two, we see that the line AD does not meet the side [E,Y ] of the triangle [E, X ,Y ]. As AD does meet the side [E, X ] of that triangle, we see by 2.4.3 that AD must meet the third side [X ,Y ] of that triangle at some point F. As F ∈ [X ,Y ] ⊂ I R(|BAC), F must be on the part of AD in I R(|BAC), that is F ∈ [A, D .

3.5 DEGREE-MEASURE OF ANGLES 3.5.1 Axiom for degree-measure Primitive Term . There is a function | |◦ on the set of all wedge-angles and straightangles, into R. Thus with each angle α , either a wedge-angle α = ∠BAC or a straightangle with support |BAC, there is associated a unique real number |α |◦ , called its degree-measure.

Sec. 3.5

Degree-measure of angles

41

AXIOM A5 . Degree-measure | |◦ of angles has the following properties:(i) In all cases |α |◦ ≥ 0; (ii) If α is a straight-angle, then |α |◦ = 180; (iii) If ∠BAC is a wedge-angle and the point D = A lies in the interior region I R(|BAC), then |∠BAD|◦ + |∠DAC|◦ = |∠BAC|◦ , while if |BAC is a straight angle-support and D ∈ AB, then |∠BAD|◦ + |∠DAC|◦ = 180; (iv) If B = A, if H1 is a closed half-plane with edge AB and if the half-lines [A,C and [A, D in H1 are such that |∠BAC|◦ = |∠BAD|◦ , then [A, D = [A,C ; (v) If B = A, if H1 is a closed half-plane with edge AB and if 0 < k < 180, then there is a half-line [A,C in H1 such that |∠BAC|◦ = k. | C

D

D y x+y A

x y 180 A

x B

B

C Figure 3.6. Addition of angle-measures.

COMMENT. The properties and proofs for degree-measure are quite like those for distance, with the role of interior regions analogous to that of segments. We note that A5 (i) is like A4 (i), A5 (iii) is like A4 (iii), A5 (iv) is like the uniqueness part of A4 (iv) and A5 (v) is like the existence part of A4 (iv). Wedge-angles ∠BAD and ∠DAC such as those in the second part of A5 (iii) are said to be supplementary

3.5.2 Derived properties of degree-measure Definition . For a wedge-angle ∠BAC, if we take a point B1 = A so that A ∈ [B, B1 ] and a point C1 = A so that A ∈ [C,C1 ], then ∠B1 AC1 is called the vertically opposite angle of ∠BAC.

42

Distance; degree-measure of an angle

C

C

H1 k

Ch. 3

B

B1

A

B

A

C1

Figure 3.7. Laying off an angle.

Figure 3.8. Opposite angles at a vertex.

Degree-measure has the properties:(i) The null-angle ∠BAB has degree-measure 0. (ii) For any non-null wedge-angle ∠BAC, we have 0 < |∠BAC|◦ < 180. (iii) If ∠B1 AC1 is the angle vertically opposite to ∠BAC, then |∠B1 AC1 |◦ = |∠BAC|◦ , so that vertically opposite angles have equal degree-measures.

Proof . (i) Let C be a point not on AB. Then by A5 (iii) with D = B, |∠BAB|◦ + |∠BAC|◦ = |∠BAC|◦ . It follows that |∠BAB|◦ = 0. (ii) Given any non-null wedge-angle ∠BAC, let H1 be the closed half-plane with edge AB in which C lies. If we had |∠BAC|◦ = 0, then we would have |∠BAC|◦ = |∠BAB|◦ and so by A5 (iv) we would have [A,C = [A, B . This would imply that ∠BAC is null, contrary to assumption. Then by A5 (i) |∠BAC|◦ > 0. Now choose the point E = A so that A ∈ [B, E]. Then by A5 (iii), as we have supplementary angles, |∠BAC|◦ + |∠CAE|◦ = 180. But [A, E = [A,C as ∠BAC is a wedge-angle, so ∠CAE is not a null-angle. By the last paragraph we then have |∠CAE|◦ > 0 and it follows that |∠BAC|◦ < 180. (iii) As |BAB1 , |CAC1 are straight we have |∠BAC|◦ + |∠CAB1|◦ = 180, |∠CAB1 |◦ + |∠B1 AC1 |◦ = 180, there being two pairs of supplementary angles. It follows that |∠BAC|◦ + |∠CAB1|◦ = |∠CAB1 |◦ + |∠B1 AC1 |◦ , from which we conclude by subtraction that |∠BAC|◦ = |∠B1 AC1 |◦ .

Sec. 3.5

Degree-measure of angles

43

Degree-measure has the further properties:(i) If ∠BAC is a wedge-angle and D = A is in I R(|BAC), then |∠BAD|◦ ≤ |∠BAC|◦ . If, further, D ∈ [A,C then |∠BAD|◦ < |∠BAC|◦ . (ii) For non-collinear points A, B,C let H1 be the closed half-plane with edge AB in which C lies. If D = A is in H1 and |∠BAD|◦ ≤ |∠BAC|◦ , then D ∈ I R(|BAC).

H4 H3 C

D

H1

D

C B

A Figure 3.9.

H1 B

E A

Proof . (i) As D ∈ I R(|BAC), by A5 (iii) |∠BAD|◦ + |∠DAC|◦ = |∠BAC|◦ . By A5 (i), |∠DAC|◦ ≥ 0 so |∠BAD|◦ ≤ |∠BAC|◦ . If D ∈ [A,C then ∠DAC is not a null-angle, so |∠DAC|◦ > 0 and hence |∠BAD|◦ < |∠BAC|◦ . (ii) Let E = A be such that A ∈ [B, E]. Let H3 , H4 be the closed half-planes with common edge AC, with B ∈ H3 and E ∈ H4 . Then H1 = H1 ∩ Π = H1 ∩ (H3 ∪ H4 ) = (H1 ∩ H3 ) ∪ (H1 ∩ H4 ) = I R(|BAC) ∪ I R(|EAC). As D ∈ H1 , then either D ∈ I R(|BAC) or D ∈ I R(|EAC). Now suppose that D ∈ I R(|BAC), so that D ∈ I R(|EAC) and D ∈ [A,C . By A5 (iii), |∠EAD|◦ + |∠DAC|◦ = |∠EAC|◦ . Hence by A5 (iii), as we have supplementary pairs of angles, 180 − |∠BAD|◦ + |∠DAC|◦ = 180 − |∠BAC|◦ . From this |∠BAC|◦ + |∠DAC|◦ = |∠BAD|◦ , and as |∠DAC|◦ > 0, we have |∠BAC|◦ < |∠BAD|◦ . This gives a contradiction with our hypothesis.

44

Distance; degree-measure of an angle

Ch. 3

3.6 MID-LINE OF AN ANGLE-SUPPORT 3.6.1 Right-angles Definition . Given any point P = A of a line AB, by A5 (v) there is a half-line [P, Q such that |∠APQ|◦ = 90. Then ∠APQ is called a right-angle. If R = P is such that P ∈ [A, R] then ∠RPQ is also a right-angle. For |APR is a straight angle-support, so having supplementary angles, |∠APQ|◦ + |∠QPR|◦ = 180. As |∠APQ|◦ = 90 it follows that |∠RPQ|◦ = 180 − 90 = 90.

3.6.2 Perpendicular lines Definition . If l, m are lines in Λ, we say that l is perpendicularto m, written l ⊥ m, if l meets m at some point P and if A = P is on l, and Q = P is on m, then ∠APQ is a right-angle. COMMENT. In 3.6.1, we say that a perpendicular PQ has been erected to the line AB at the point P on it. Perpendicularity has the following properties:(i) If l ⊥ m, then m ⊥ l. (ii) If l ⊥ m, then l = m and l ∩ m = 0. /

Proof . These follow immediately from the definition of perpendicularity.

P

C Q

90 P

P

A

B

A

Figure 3.10. Perpendicular lines.

Mid-line of an angle-support.

Sec. 3.6

Mid-line of an angle-support

45

3.6.3 Mid-lines Given any angle-support |BAC such that C ∈ [A, B , there is a unique line l such that A ∈ l and for all P = A but P ∈ l, |∠BAP|◦ = |∠PAC|◦ . Proof . Existence . This was already shown in 3.6.1 in the case when |BAC is straight, so we may assume that A, B,C are non-collinear. By A5 (v) and 3.5.2, as 0 < |∠BAC|◦ < 180 and so 0 < 12 |∠BAC|◦ < 90, there is a half-line [A, P with P on the same side of AB as C is, such that |∠BAP|◦ = 12 |∠BAC|◦ . Then [A, P ⊂ I R(|BAC) by 3.5.2, so by A5 (iii) |∠BAP|◦ + |∠PAC|◦ = |∠BAC|◦ . H4 H3

H4 H3 H1

C A

P

C B1

B

H1

P A

B

B1 Figure 3.11.

It follows that 1 1 |∠PAC|◦ = |∠BAC|◦ − |∠BAC|◦ = |∠BAC|◦ 2 2 and so |∠BAP|◦ = |∠PAC|◦ . If P = A is such that A ∈ [P, P ], then by A5 (iii) |∠BAP |◦ = 180 − |∠BAP|◦ = 180 − |∠PAC|◦ = |∠P AC|◦ .

Uniqueness . When |BAC is straight, by A5 (iii) 2|∠BAP|◦ = 180 so |∠BAP|◦ = 90. By A5 (iv) this determines l uniquely. For the remainder we suppose then that we have a wedgeangle ∠BAC. Let H1 , H2 be the closed half-planes with common edge AB, with C ∈ H1 , and H3 , H4 be the closed half-planes with common edge AC, with B ∈ H3 . Let B1 = A be such that A ∈ [B, B1 ]. Now if l contains a point Q = A in H4 it will also contain a point R = A of H3 , so we may assume that l contains a point P = A of H3 . As H1 =H1 ∩ Π = H1 ∩ (H3 ∪ H4 ) = (H1 ∩ H3 ) ∪ (H1 ∩ H4 ) =I R(|BAC) ∪ I R(|B1 AC), we then have P ∈ I R(|BAC) or P ∈ I R(|B1 AC). We get a contradiction if l is either AB or AC. For if l = AB, then we have |∠BAP|◦ = 0, |∠PAC|◦ > 0. Similarly if l = AC.

46

Distance; degree-measure of an angle

Ch. 3

We also get a contradiction if l contains a point P = A in I R(|B1 AC) which is not on AC. For then by 3.5.2 |∠B1 AP|◦ < |∠B1 AC|◦ , so that 180 − |∠BAP|◦ < 180 − |∠BAC|◦ and so |∠BAC|◦ < |∠BAP|◦ . It follows from 3.5.2 that [A,C ⊂ I R(|BAP) and so |∠BAC|◦ + |∠CAP|◦ = |∠BAP|◦. But |∠BAC|◦ > 0 and so |∠CAP|◦ < |∠BAP|◦ , which gives a contradiction. Thus l must contain a point P = A in I R(|BAC). As then |∠BAP|◦ + |∠PAC|◦ = |∠BAC|◦ and |∠BAP|◦ = |∠PAC|◦ , we must have |∠BAP|◦ = 21 |∠BAC|◦ which determines [A, P uniquely. Definition . We define the mid-line or bisector of the angle-support |BAC as follows:- if C ∈ [A, B then it is the line AB, and otherwise it is the unique line l just noted. We use the notation ml(|BAC) for this.

3.7 DEGREE-MEASURE OF REFLEX ANGLES 3.7.1 Definition . Let α be a reflex angle with support |BAC. We first suppose that C ∈ AB, and as in 3.5.2 let ∠B1 AC1 be the vertically opposite angle of the wedge-angle ∠BAC. Then B1 ∈ AC, C1 ∈ AB and ∠B1 AC is the vertically opposite angle for ∠BAC1. By 3.5.2 we note that 180 + |∠B1 AC|◦ = |∠BAC1|◦ + 180, and we define the degree-measure of α to be the common value of these: |α |◦ = 180 + |∠B1 AC|◦ = |∠BAC1 |◦ + 180. Secondly, if C ∈ [A, B so that α is a full-angle, we define |α |◦ = 360. Then for each reflex-angle α , |α |◦ is defined; by 3.5.2 it satisfies 180 < |α |◦ < 360 unless α is a full-angle in which case |α |◦ = 360. C1

C1 A

B

B1

D A

B

B1 C

C

Figure 3.12. Measure of a reflex angle.

Let α be a non-full reflex-angle with support |BAC and take B1 = A, C1 = A so that A ∈ [B, B1 ], A ∈ [C,C1 ]. Let [A, D ⊂ I R(|B1 AC1 ) but D ∈ [A,C1 , D ∈ [A, B1 . Then |∠BAD|◦ + |∠DAC|◦ = |α |◦ .

Sec. 3.7

Degree-measure of reflex angles

47

Proof . As [A, D ⊂ I R(|B1 AC1 ), D is in the closed half-plane with edge AB in which C1 lies, and also in the closed half-plane with edge AC in which B1 lies. By 3.5.2, |∠B1 AD|◦ < |∠B1 AC1 |◦ so by A5 (iii) |∠BAC1 |◦ < |∠BAD|◦ . By 3.5.2 then [A,C1 ⊂ I R(|BAD), and by similar reasoning [A, B1 ⊂ I R(|DAC). Then |∠BAD|◦ + |∠DAC|◦ = |∠BAD|◦ + (|∠DAB1 |◦ + |∠B1 AC|◦ ) = (|∠BAD|◦ + |∠DAB1|◦ ) + |∠B1 AC|◦ = 180 + |∠B1 AC|◦ = |α |◦ . COMMENT. We could use this last result to employ the measures of reflex-angles to a significant extent, but in fact do not do so until our full treatment of them in Chapter 9.

Exercises 3.1 If B ∈ [A,C], then |A, B| ≤ |A, P| ≤ |A, C| for all P ∈ [B,C]. 3.2 Let A, B,C be points of a line l, and M = mp(A, B). If C is A or B, or if C ∈ l \ [A, B], then |C, A| + |C, B| = 2|C, M|. 3.3 Let A, B,C be distinct points and D = mp(B,C), E = mp(C, A), F = mp(A, B). Prove that D, E, F are distinct. If A ∈ BC, show that neither E nor F belongs to BC. 3.4 If A = B, show that {P ∈ AB : |B, A| + |A, P| = |B, P|} is the half-line of AB with initial-point A which does not contain B, while [A, B = {P ∈ AB : |A, P| + |P, B| = |A, B| or |A, B| + |B, P| = |A, P|}. 3.5 Find analogues of 3.3.1 when A ∈ [B,C] and when B ∈ [C, A]. 3.6 Show that if A, B,C, D are distinct collinear points such that C ∈ [A, B], B ∈ [A, D], and |A, C| |A, D| = , |C, B| |D, B| then

1 1 2 + = . |A, C| |A, D| |A, B|

3.7 Show that if A, B,C are non-collinear points, and P = A is a point of I R(|BAC), then I R(|BAP) ∪ I R(|PAC) = I R(|BAC), I R(|BAP) ∩ I R(|PAC) = [A, P .

48

Distance; degree-measure of an angle

3.8 Show that if d is any positive real number and | d| | is also a distance function.

Ch. 3

| is a distance function, then

3.9 If α is the reflex angle with support |BAC and β is the reflex angle with support |BAF, show that if [A, F ⊂ I R(|BAC) then |α |◦ + |∠CAF|◦ = |β |◦ . 3.10 Prove that if l = ml(|BAC) and m = ml(|BAC1 ) where A ∈ [C,C1 ], then l ⊥ m. 3.11 Suppose that B,C, B1 and C1 are points distinct from A and that A ∈ [B, B1 ], A ∈ [C,C1 ]. Show that then ml(|B1 AC1 ) = ml(|BAC).

4

Congruence of triangles; parallel lines COMMENT. In this chapter we deal with the notion of congruence of triangles, and make a start on the concept of parallelism of lines. As we have distance and anglemeasure, we do not need special concepts of congruence of segments and congruence of angles, and we are able to define congruence of triangles instead of having it as a primitive term as in the traditional treatment. As a consequence there is a great gain in effectiveness and brevity.

4.1 PRINCIPLES OF CONGRUENCE 4.1.1 Congruence of triangles A x | | |

y

B | |

z C

|

y

| | | |

x

| |

z

C

A

B Figure 4.1. Congruent triangles.

Definition . Let T be a triangle with the vertices {A, B,C} and T a triangle with vertices {A , B ,C }. We say that T is congruent to T in the correspondence A → A , B → B , C → C , if |B, C| = |B , C |,

|C, A| = |C , A |,

|A, B| = |A , B |,

|∠BAC|◦ = |∠B A C |◦ , |∠CBA|◦ = |∠C B A |◦ , |∠ACB|◦ = |∠A C B |◦ . We denote this by T

≡

(A,B,C)→(A ,B ,C )

T .

Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50004-5 © 2016 Elsevier Ltd. All rights reserved.

50

Congruence of triangles; parallel lines

Ch. 4

We say that T is congruent to T , written T ≡ T , if T is congruent to T in at least one of the correspondences (A, B,C) →

(A , B ,C ), (A, B,C) → (A ,C , B ), (A, B,C) → (B ,C , A ),

(A, B,C) →

(B , A ,C ), (A, B,C) → (C , A , B ), (A, B,C) → (C , B , A ).

COMMENT. Originally, behind the concept of congruence lay the idea that T can be placed on T , fitting it exactly. AXIOM A6 . If triangles T and T , with vertices {A, B,C} and {A , B ,C }, respectively, are such that |C, A| = |C , A |, |A, B| = |A , B |, |∠BAC|◦ = |∠B A C |◦ , then T

≡

(A,B,C)→(A ,B C )

T . |

COMMENT. This is known as the SAS (side, angle, side) principle of congruence for triangles. Triangles have the following properties:If in a triangle [A, B,C] :(i) |A, B| = |A, C| then |∠ABC|◦ = |∠ACB|◦ ; (ii) |A, B| = |A, C| and D is the mid-point of B and C, then AD ⊥ BC; (iii) B = C, D is the mid-point of B and C, and A = D is such that AD ⊥ BC, then |A, B| = |A, C|; (iv) |BAC is not straight, E ∈ [A, B , F ∈ [A,C are such that |A, E| = |A, F| > 0, and G = mp(E, F), then AG = ml(|BAC).

Proof . (i) Note that for the triangle T with vertices {A, B,C}, under the correspondence (A, B,C) → (A,C, B), |A, B| = |A, C|, |A, C| = |A, B|, |∠BAC|◦ = |∠CAB|◦ , so by the SAS principle T

≡

(A,B,C)→(A,C,B)

T . In particular |∠ABC|◦ = |∠ACB|◦ .

Sec. 4.1

Principles of congruence

(ii) Note that if T1 , T2 are the triangles with vertices {A, B, D}, {A,C, D}, respectively, then

51

A

|A, B| = |A, C|, |B, D| = |C, D|, |∠ABD|◦ = |∠ACD|◦ , by (i), so by the SAS principle, ≡ T1 (A,B,D)→(A,C,D) T2 . In particular |∠ADB|◦ = |∠ADC|◦ . B D C As D ∈ [B,C], the sum of the Figure 4.2. An isosceles triangle. degree-measures of these angles is 180 and so they must be right-angles. (iii) As AD ⊥ BC we know that A ∈ BC. If T1 , T2 are the triangles with vertices {A, B, D}, {A,C, D}, respectively, then |B, D| = |C, D|, |A, D| = |A, D|, |∠BDA|◦ = |∠CDA|◦ , ≡ so by the SAS principle, T1 (A,B,D)→(A,C,D) T2 . In particular |A, B| = |A, C|. (iv) As in (ii), the triangles [A, E, G], [A, F, G] are congruent, and so |∠EAG|◦ = |∠FAG|◦.

Definition . A triangle is said to be isosceles if at least two of its sides have equal lengths. If T, T are triangles with vertices {A, B,C}, {A , B ,C }, respectively, for which |B, C| = |B , C |, |∠CBA|◦ = |∠C B A |◦ , |∠BCA|◦ = |∠B C A |◦ , ≡ then T (A,B,C)→(A ,B ,C ) T . Proof . Suppose that |C , A | = |C, A|. Choose the point D on the half-line [C , A such that |C , D | = |C, A|. Then if T is the triangle with vertices {B ,C , D }, under the correspondence (B,C, A) → (B ,C , D ) we have

|B, C| = |B , C |, |C, A| = |C , D |, |∠BCA|◦ = |∠B C D |◦ . A

B C

C B

D Figure 4.3.

A

52

Congruence of triangles; parallel lines

Then by the SAS principle, T

≡

(B,C,A)→(B ,C ,D )

Ch. 4

T . In particular

|∠C B D |◦ = |∠CBA|◦ = |∠C B A |◦ . Then we have different wedge-angles ∠C B A , ∠C B D , laid off on the same side of B C and having the same degree-measure. This gives a contradiction. Thus |C , A | = |C, A|, and as we also have |C , B | = |C, B|, |∠B C A |◦ = |∠BCA|◦ , ≡ by the SAS principle we have T (B,C,A)→(B ,C ,A ) T . This is known as the ASA (angle, side,angle) principle of congruence. If T and T are triangles with vertices {A, B,C}, {A , B ,C }, respectively, for which |B, C| = |B , C |, |C, A| = |C , A |, |A, B| = |A , B |,

then T

≡

(A,B,C)→(A ,B ,C )

T . A

A

B

C

E D

B E

C

D

Figure 4.4. The SSS principle of congruence.

Proof . Choose D on the opposite side of BC from A, so that |∠CBD|◦ = |∠C B A |◦ and |B, D| = |B , A |. Let T be the triangle with vertices {B,C, D}. Then as |B, C| = |B , C |, by the SAS principle ≡ T (B,D,C)→(B ,A ,C ) T .

A

B C

E

Now |B, A| = |B , A | = |B, D| so D we have an isosceles triangle and |∠BAD|◦ = |∠BDA|◦ . Similarly |∠CAD|◦ = |∠CDA|◦ . Note that A and D are on different sides of BC, so a point E of [A, D] is on BC. CASE 1. Let E ∈ [B,C]. Then [A, D ⊂ I R(|BAC) and [D, A ∈ I R(|BDC). It follows that |∠BAC|◦ = |∠BAD|◦ + |∠DAC|◦ = |∠BDA|◦ + |∠ADC|◦ = |∠BDC|◦ .

Sec. 4.2

Alternate angles, parallel lines

53

CASE 2. Let B ∈ [E,C]. Then [A, B ⊂ I R(|DAC) and [D, B ∈ I R(|ADC). It follows that |∠BAC|◦ = |∠DAC|◦ − |∠DAB|◦ = |∠ADC|◦ − |∠ADB|◦ = |∠BDC|◦ . CASE 3. Let C ∈ [B, E]. Then [A,C ⊂ I R(|BAD) and [D,C ∈ I R(|BDA). It follows that |∠BAC|◦ = |∠BAD|◦ − |∠DAC|◦ = |∠BDA|◦ − |∠ADC|◦ = |∠BDC|◦ . Now combining the cases, by the SAS principle, as |A, B| = |D, B|, |A, C| = |D, C|, |∠BAC|◦ = |∠BDC|◦ , ≡ ≡ we have T (A,B,C)→(D,B,C) T . But T (D,B,C)→(A ,B ,C ) T so ≡ T (A,B,C)→(A ,B ,C ) T .

This is known as the SSS(side, side,side) principle of congruence for triangles.

4.2 ALTERNATE ANGLES, PARALLEL LINES 4.2.1 Alternate angles Let A, B,C be non-collinear points, and take D = C so that C ∈ [A, D]. Then |∠BCD|◦ > |∠CBA|◦ . Proof . Let E = mp(B,C) and choose F so that E = mp(A, F). Then if T1 , T2 are the triangles with vertices {A, B, E}, {F,C, E}, respectively, by the SAS principle of ≡ congruence T1 (A,B,E)→(F,C,E) T2 . In particular, |∠EBA|◦ = |∠ECF|◦ , i.e. |∠CBA|◦ = |∠BCF|◦ . But [C, F ⊂ I R(|BCD) as E, and so F, is in the closed half-plane with edge AC in which B lies, and D and F are on the opposite side of BC from A. Also F ∈ AD as F ∈ AD would imply that E = C. Then by 3.5.2 |∠BCF|◦ < |∠BCD|◦ .

COROLLARY 1 . In the theorem let G = C be such that C ∈ [B, G]. Then |∠ACG|◦ > |∠ABC|◦ . Proof . This follows immediately as ∠ACG and ∠BCD are vertically opposite angles.

54

Congruence of triangles; parallel lines

Ch. 4

COMMENT. If D and H are on opposite sides A of BC, then ∠CBH and H ∠BCD are known as alC G ternate angles . This last result implies that if alE B ternate angles ∠CBH and ∠BCD are equal in meaD F sure, then CD and BH cannot meet at some point Figure 4.5. Result on alternate angles. A. COROLLARY 2 . Given any line l and any point P ∈ l, there is a line m which contains P and is such that l ∩ m = 0. / Proof . Take any points A, B ∈ l and lay off an angle ∠APQ on the opposite side of AP from B, so that |∠APQ|◦ = |∠PAB|◦ . Then by the recent comment the line PQ does not meet l. In this ∠APQ and ∠PAB are alternate angles which are equal in measure.

4.2.2 Parallel lines Definition . If l and m are lines in Λ, we say that l is parallel to m, written l m, if l = m or l ∩ m = 0. / Parallelism has the following properties:(i) l l for all l ∈ Λ; (ii) If l m then m l; (iii) Given any line l ∈ Λ and any point P ∈ Π, there is at least one line m which contains P and is such that l m. (iv) If the lines l and m are both perpendicular to the line n, then l and m are parallel to each other. P Q

m

n m P

Q

B

l B

A l

A

Figure 4.6. Parallel lines.

Sec. 4.3

Properties of triangles and half-planes

55

Proof . (i) and (ii) follow immediately from the definition, while (iii) follows from 4.2.1. (iv) As perpendicular lines form right-angles with each other at some point, l must meet n at some point A, and m must meet n at some point P such that if B is any other point of l and Q is any point of m on the other side of n from B, then |∠PAB|◦ = 90, |∠APQ|◦ = 90. Then, as these are alternate angles equal in measure, by 4.2.1 Corollary 1, l m.

4.3 PROPERTIES OF TRIANGLES AND HALF-PLANES 4.3.1 Side-angle relationships; the triangle inequality If A, B,C are non-collinear points and |A, B| > |B, C|, then |∠ACB|◦ > |∠BAC|◦ , so that in a triangle a greater angle is opposite a longer side. B A D

B

C

C D A Figure 4.7. Angle opposite longer side.

Figure 4.8. The triangle inequality.

Proof . Choose D ∈ [B, A so that |B, D| = |B, C|. Then D ∈ [B, A] as |B, D| < |B, A|. Now |∠ACB|◦ > |∠DCB|◦ as [C, D ⊂ I R(|BCA), and |∠DCB|◦ = |∠BDC|◦ by 4.1.1. But |∠BDC|◦ > |∠DAC|◦ by 4.2.1 Corollary 1, so |∠ACB|◦ > |∠DCB|◦ = |∠BDC|◦ > |∠DAC|◦ . Hence |∠ACB|◦ > |∠DAC|◦ and ∠DAC = ∠BAC as D ∈ [B, A].

COROLLARY . If A, B,C are non-collinear points and |∠ACB|◦ > |∠BAC|◦ , then |A, B| > |B, C|, so that in a triangle a longer side is opposite a greater angle. Proof . For if |A, B| ≤ |B, C|, we have |∠ACB|◦ ≤ |∠BAC|◦ by 4.1.1(i) and this result. The triangle inequality. If A, B,C are non-collinear points, then |C, A| < |A, B| + |B, C|. Proof . Take a point D so that B ∈ [A, D] and |B, D| = |B, C|. As [C, B ⊂ I R(|ACD) we have |∠DCA|◦ > |∠DCB|◦ . But |∠DCB|◦ = |∠CDB|◦

56

Congruence of triangles; parallel lines

Ch. 4

by 4.1.1, so by our last result, |A, D| > |A, C|. However |A, D| = |A, B| + |B, D| as B ∈ [A, D], and the result follows. A m

P

l

B

E

F

G1 l

C

D

Figure 4.10.

Figure 4.9.

4.3.2 Properties of parallelism Let l ∈ Λ be a line, G1 an open half-plane with edge l and P a point of G1 . If m is a line such that P ∈ m and l m, then m ⊂ G1 . Proof . As P ∈ l, P ∈ m we have l = m. Then as l m we have l ∩ m = 0. / Thus there cannot be a point of m on l. Neither can there be a point Q of m in G2 , the other open half-plane with edge l. For then we would have [P, Q] ∩ l = 0/ and so a point R of m would be on l, as [P, Q] ⊂ PQ = m. Let AB,CD be distinct lines and l distinct from and parallel to both. If l meets [A,C] in a point E, then l meets [B, D] in a point F. Proof . By the Pasch property applied to [A, B,C] as l does not meet [A, B] it meets [B,C] at some point G. Then by the Pasch property applied to [B,C, D], as l does not meet [C, D] it meets [B, D] in some point F.

4.3.3 Dropping a perpendicular P P l S

B Q

A R

l S T

U

Figure 4.11. Dropping a perpendicular.

Given any line l ∈ Λ and any point P ∈ l, there is a unique line m such that P ∈ m and l ⊥ m.

Sec. 4.3

Properties of triangles and half-planes

57

Proof . Existence . Let A, B be distinct points of l. Take a point Q on the opposite side of l from P and such that |∠BAQ|◦ = |∠BAP|◦ . Also take R ∈ [A, Q so that |A, R| = |A, P|. As P and R are on opposite sides of l, [P, R] meets l in a point S. We first suppose that A ∈ PR so that A = S. Then [A, P, S] and [A, R, S] are congruent by the SAS-principle, so in particular |∠ASP|◦ = |∠ASR|◦ . As S ∈ [P, R] it follows that these are right-angles and so PR ⊥ l. In the second case suppose that A ∈ PR so that A = S. Then S ∈ [P, R] and by construction |∠BSR|◦ = |∠BSP|◦ . Again these are right-angles so PR ⊥ l.

Uniqueness . Suppose that there are distinct points S, T ∈ l such that PS ⊥ l, PT ⊥ l. Choose U = T so that T ∈ [S,U ]. Then |∠UTP|◦ = |∠USP|◦ = 90 and this contradicts 4.2.1. COMMENT. We refer to this last as dropping a perpendicular from the point P to the line l. Let A, B,C be non-collinear points such that AB ⊥ AC and let D be the foot of the perpendicular from A to BC. Then D ∈ [B,C], D = B, D = C.

Proof . By 4.2.1, in a right-angled triangle each of the other two angles have degree-measure less than 90. By 4.3.1 it then follows that the side opposite the right-angle is longer than each of the other sides. It follows that |B, D| < |A, B| < |B, C|. By a similar argument |C, D| < |B, C|.

A

B

D

C Figure 4.12.

We cannot then have B ∈ [C, D] as that would imply |C, B| ≤ |C, D|, and similarly we cannot have C ∈ [B, D] with as that would imply |B, C| ≤ |B, D|. Hence D ∈ [B,C], D = B, D = C.

4.3.4 Projection and axial symmetry Definition . For any line l ∈ Λ we define a function πl : Π → l by specifying that for all P ∈ Π, πl (P) is the foot of the perpendicular from P to l. We refer to πl as projection to the line l.

58

Congruence of triangles; parallel lines

P

Ch. 4

P l

l

πl (P) sl (P) Figure 4.13. Projection to the line l.

Axial symmetry in the line l.

Definition . For any line l ∈ Λ we define a function sl : Π → Π by specifying that for all P ∈ Π, sl (P) is the point Q such that πl (P) = mp(P, Q). We refer to sl as axial symmetry in the line l. Let H1 , H2 be closed half-planes with common edge l, let P1 ∈ H1 \ l and P2 = sl (P1 ). Then, for all P ∈ H1 , |P, P1 | ≤ |P, P2 |. Proof . If P ∈ l, then |P, P1 | = |P, P2 |, by 4.1.1 when P ∈ P1 P2 , and as P = mp(P1 , P2 ) otherwise. When P ∈ G1 = H1 \ l we suppose first that P ∈ P1 P2 . Then [P, P2 ] meets l in a point Q and we have |P, P2 | = |P, Q| + |Q, P2 | = |P, Q| + |Q, P1 |. Now we cannot have Q ∈ [P, P1 ] as [P, P1 ] ⊂ G1 and Q ∈ l. Thus either Q ∈ PP1 or Q ∈ PP1 \ [P, P1 ]. We then have |P, Q| + |Q, P1 | > |P, P1 | by 4.3.1 and 3.1.2. For the case when P ∈ P1 P2 , we denote by R the point of intersection of P1 P2 and l, so that R = mp(P1 , P2 ).

P P1 H1 R

Q l

H2

P2 Figure 4.14. Distance and half-planes.

Now P ∈ [R, P1 so either P ∈ [R, P1 ] or P1 ∈ [R, P]. In the first of these cases we have |P1 , P| < |P1 , R| = |R, P2 | < |P, P2 |, as R ∈ [P, P2 ]. In the second case we have |P, P1 | < |P, R| < |P, P2 | as R ∈ [P, P2 ].

Exercises 4.1 If D = A is in [A, B,C] but not in [B,C], then |B, D| + |D, C| < |B, A| + |A, C| and |∠BDC|◦ > |∠BAC|◦ .

Sec. 4.3

Properties of triangles and half-planes

59

4.2 There is an AAS-principle of congruence that if |∠BAC|◦ = |∠EDF|◦ , |∠CBA|◦ = |∠FED|◦ , |B, C| = |E, F|, then the triangles [A, B,C], [D, E, F] are congruent. [Hint. Suppose that |∠BCA|◦ < |∠EFD|◦ ; lay off an angle ∠BCG equal in magnitude to ∠EFD and with G on the same side of BC as A is; then [C, G meets [A, B] at a point H; also [H, B,C] ≡ [D, E, F] and in particular |∠BHC|◦ = |∠EDF|◦ = |∠BAC|◦ ; deduce a contradiction and then apply the ASA-principle.] 4.3 There is an ASS-principle of congruence for right-angled triangles, that if BC ⊥ BA, EF ⊥ ED, |C, A| = |F, D|, |A, B| = |D, E|, then [A, B,C] ≡ [D, E, F]. [Hint. Take C so that E ∈ [F,C ] and |E, C | = |B, C|.] 4.4 If P ∈ ml(|BAC) and Q = πAB (P), R = πAC (P), then |P, Q| = |P, R|. Conversely, if P ∈ I R(|BAC) and |P, Q| = |P, R| where Q = πAB (P), R = πAC (P), then P ∈ ml(|BAC). 4.5 In triangles [A, B,C], [D, E, F] let |A, B| = |D, E|, |A, C| = |D, F|, |∠BAC|◦ > |∠EDF|◦ . Then |B, C| > |E, F|. [Hint. Lay off the angle ∠BAG with |∠BAG|◦ = |∠EDF|◦ and with G on the same side of AB as C is. If G ∈ BC proceed; if G ∈ BC, let K = mp(G,C) and show that [A, K meets [B,C] in a point H.] 4.6 If AB AC, then AB = AC. 4.7 Let H1 be a closed half-plane with edge l, let P ∈ H1 and let O = πl (P). Then if m is any line in Λ such that O ∈ m, we must have πm (P) ∈ H1 . ≡ 4.8 With the notation of 4.1.1 show that in all cases T (A,B,C)→(A,B,C) T and if ≡ ≡ T (A,B,C)→(A T . Show also that if ,B ,C ) T then T (A ,B ,C )→(A,B,C) ≡ ≡ ≡ T (A,B,C)→(A ,B ,C ) T and T (A ,B ,C )→(A ,B ,C ) T , then T (A,B,C)→(A ,B ,C ) T . Thus congruence of triangles is reflexive, symmetric and transitive, and thus is an equivalence relation.

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5

The parallel axiom; Euclidean geometry COMMENT. The effect of introducing any axiom is to narrow things down, and depending on the final axiom still to be taken, we can obtain two quite distinct wellknown types of geometry. By introducing our final axiom, we confine ourselves to the familiar school geometry, which is known as Euclidean geometry.

5.1 THE PARALLEL AXIOM 5.1.1 Uniqueness of a parallel line We saw in 4.2 that given any line l and any point P ∈ l there is at least one line m such that P ∈ m and l m. We now assume that there is only one such line ever. AXIOM A7 . Given any line l ∈ Λ and any point P ∈ l, there is at most one line m such that P ∈ m and l m. | COMMENT. By 4.2 and A7 , given any line l ∈ Λ and any point P ∈ Π, there is a unique line m through P which is parallel to l. Let l ∈ Λ, P ∈ Π and n ∈ Λ be such that l = n, P ∈ n and l n. Let A and B be any distinct points of l and R a point of n such that R and B are on opposite sides of AP. Then |∠APR|◦ = |∠PAB|◦ , so that for parallel lines alternate angles must have equal degree-measures. Proof . Let m be the line PQ in 4.2.1 such that |∠APQ|◦ = |∠BAP|◦ . Then l m. As m and n both contain P and l is parallel to both of them, by A7 we have m = n, so that R ∈ [P, Q and so |∠APR|◦ = |∠APQ|◦ . Thus |∠APR|◦ = |∠APQ|◦ = |∠PAB|◦ . Let l, n be distinct parallel lines, A, B ∈ l and P, T ∈ n be such that B and T are on the one side of AP, and S = P be such that P ∈ [A, S]. Then the angles ∠BAP, ∠T PS have equal degree-measures. Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50005-7 © 2016 Elsevier Ltd. All rights reserved.

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Ch. 5

Proof . Choose R = P so that P ∈ [T, R]. Then R ∈ n and B and R are on opposite sides of AP, so that ∠BAP, ∠APR are alternate angles and so have equal degreemeasures. But ∠APR and ∠T PS are vertically opposite angles and so have equal degree-measures. Hence |∠BAP|◦ = |∠TPS|◦ . n Q R

Q

◦P ◦

R B

Figure 5.1. Alternate angles.

◦ A

A l

n

S ◦ T P B

l Corresponding angles.

We call such angles ∠BAP, ∠T PS corresponding angles for a transversal. If lines l, m, n are such that l m and m n, then l n. Proof . If l = n, the result is trivial as l l, so suppose l = n. If l is not parallel to n, then l and n will meet at some point P, and then we will have distinct lines l and n, both containing P and both parallel to m, which gives a contradiction by A7 . Thus parallelism is a transitive relation. Combined with the properties in 4.2.2 this makes it an equivalence relation. If lines are such that l ⊥ n and l m, then m ⊥ n. Proof . As l is perpendicular to n they must meet at some point A. As l m, we cannot have m

n n, as by transitivity that would A imply l n. Thus m meets n in B ◦ Q some point P, and if we choose l ◦ P B on l, Q on m on opposite sides of n, then we have |∠APQ|◦ = |∠PAB|◦ as these are alternate m angles for parallel lines. Hence Figure 5.2. ◦ |∠APQ| = 90 and m ⊥ n.

5.2 PARALLELOGRAMS 5.2.1 Parallelograms and rectangles Definition . Let points A, B,C, D be such that no three of them are collinear and AB

CD, AD BC. Let H1 be the closed half-plane with edge AB in which C lies; as CD AB then, by 4.3.2, D ∈ H1 . Similarly let H3 be the closed half-plane with edge BC in which A lies; as AD BC, then D ∈ H3 . Thus D ∈ H1 ∩ H3 = I R(|ABC) and so by the cross-bar theorem [A,C] meets [B, D in some point T , which is unique

Sec. 5.2

Parallelograms

63

as AC = BD would imply B ∈ AC. Similarly C ∈ I R(|BAD) so T is on [B, D]. Thus [A,C] ∩ [B, D] = 0/ so as in 2.4.4 a convex quadrilateral [A, B,C, D] can be defined, and in this case it is called a parallelogram. The terminology of 2.4.4 then applies. D

A

A

D T

T B

C

B

Figure 5.3. A parallelogram.

C A rectangle.

Definition . If [A, B,C, D] is a parallelogram in which AB ⊥ AD, then, as AB

CD, by 5.1.1 we have AD ⊥ CD. Thus if two adjacent side-lines of a parallelogram are perpendicular, each pair of adjacent side-lines are perpendicular; we call such a parallelogram a rectangle. Parallelograms have the following properties:(i) Opposite sides of a parallelogram have equal lengths. (ii) The point of intersection of the diagonals of a parallelogram is the mid-point of each diagonal.

Proof . (i) With the notation above for a parallelogram, the triangles with vertices {A, B, D} and {C, D, B} are congruent in the correspondence (A, B, D) → (C, D, B) by the ASA principle. First note that |B, D| = |D, B|. Secondly note that AB CD and A and C are on opposite sides of BD so that ∠ABD and ∠CDB are alternate angles, and hence |∠ABD|◦ = |∠CDB|◦ . Finally AD BC, and A and C are on opposite sides of BD, so that ∠ADB and ∠CBD are alternate angles and hence |∠ADB|◦ = |∠CBD|◦ . It follows that |A, B| = |C, D|, |A, D| = |B, C|. (ii) Let T be the point of intersection of the diagonals. Then the triangles [A, B, T ], [C, D, T ] are congruent by the ASA principle, as |A, B| = |C, D|, |∠ABT|◦ = |∠CDT|◦ , |∠BAT|◦ = |∠DCT|◦ . It follows that |A, T| = |C, T|, |B, T| = |D, T|.

5.2.2 Sum of measures of wedge-angles of a triangle If A, B,C are non-collinear points, then |∠CAB|◦ + |∠ABC|◦ + |∠BCA|◦ = 180.

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The parallel axiom; Euclidean geometry

Ch. 5

Thus the sum of the degree-measures of the wedge-angles of a triangle is equal to 180. Proof . Let l be the line through A which is parallel to BC. If m is the line through B which is Q l AR parallel to AC, then we cannot x y have l m as we would then have l m, m AC which would imm ply l AC; we would then have BC l, l AC and so BC AC; thus as BC ∩ AC = 0/ we would y x have BC = AC; this would make B C A, B,C collinear and contradict Figure 5.4. Angles of a triangle. our assumption. Thus m meets l at some point, Q say. Then [A,C, B, Q] is a parallelogram and [A, B], [Q,C] meet at a point T . Now Q is on the opposite side of AB from C, so that ∠CBA and ∠BAQ are alternate angles and so |∠CBA|◦ = |∠BAQ|◦ . Moreover [A, B ⊂ I R(|CAQ) and so |∠CAB|◦ + |∠BAQ|◦ = |∠CAQ|◦ . Choose R = A so that A ∈ [Q, R]. Then R ∈ l and R is on the opposite side of AC from Q. But BQ AC so B and Q are on the same side of AC, and hence B and R are on opposite sides of AC. Then ∠BCA and ∠CAR are alternate angles, so |∠BCA|◦ = |∠CAR|◦ . Thus (|∠CAB|◦ + |∠CBA|◦) + |∠BCA|◦ = (|∠CAB|◦ + |∠BAQ|◦ ) + |∠BCA|◦ = |∠CAQ|◦ + |∠CAR|◦ = 180. COROLLARY. If the points A, B,C are non-collinear, and D = C is chosen so that C ∈ [B, D], then |∠ACD|◦ = |∠BAC|◦ + |∠CBA|◦. Thus the degree-measure of an exterior wedge-angle of a triangle is equal to the sum of the degree-measures of the two remote wedge-angles of the triangle. Proof . For each of these is equal to 180 − |∠ACB|◦ , as C ∈ [B, D].

5.3 RATIO RESULTS FOR TRIANGLES 5.3.1 Lines parallel to one side-line of a triangle Let A, B,C be non-collinear points, and with l = AB, m = AC, let ≤l , ≤m be natural orders such that A ≤l B, A ≤m C. Let D1 , D2 , D3 be points of AB such that A ≤l D1 ≤l D2 ≤l D3 ≤l B and |D1 , D2 | = |D2 , D3 |, so that D2 is the mid-point of D1 and D3 . Then the lines through D1 , D2 and D3 which are all parallel to BC, will meet AC in points E1 , E2 , E3 , respectively, such that A ≤m E1 ≤m E2 ≤m E3 ≤m C and |E1 , E2 | = |E2 , E3 |.

Sec. 5.3

Ratio results for triangles

65

1

2

|

|

|

|

|

Proof . By Pasch’s property in 2.4.3 applied to the triangle A [A, B,C], the lines through [D1 , D2 , D3 ] which are parallel to BC will meet [A,C] in points E1 , E2 , E3 , respectively. By Pasch’s property applied to E1 D1 [A, D3 , E3 ], since D2 ∈ [A, D3 ] y and the lines through D2 and x D2 E2 D3 parallel to BC are parallel to F y each other, E2 ∈ [A, E3 ]. x D3 E3 By Pasch’s property applied to G [A, D2 , E2 ], since D1 ∈ [A, D2 ] and the lines through D1 and B C D2 parallel to BC are parallel to each other, E1 ∈ [A, E2 ]. It reFigure 5.5. Transversals to parallel lines. mains to show that E2 is equidistant from E1 and E3 . By Pasch’s property applied to [A, D2 , E2 ], since E1 ∈ [A, E2 ] the line through E1 which is parallel to AB = AD2 will meet [D2 , E2 ] in a point F. By Pasch’s property applied to [A, D3 , E3 ], since E2 ∈ [A, E3 ] the line through E2 which is parallel to AB = AD3 will meet [D3 , E3 ] in a point G. Let T1 , T2 be the triangles with vertices {E1 , F, E2 }, {E2 , G, E3 }, respectively. Our objective is to show that ≡ T1 (E ,F,E )→(E T. ,G,E ) 2 2

3

Now D1 E1 D2 F, D1 D2 E1 F, so [D1 , D2 , F, E1 ] is a parallelogram, and so by 5.2.1 |D1 , D2 | = |E1 , F|. Similarly D2 E2 D3 G, D2 D3 E2 G so [D2 , D3 , G, E2 ] is a parallelogram, and so |D2 , D3 | = |E2 , G|. But |D1 , D2 | = |D2 , D3 | and hence |E1 , F| = |E2 , G|. Let H1 be the closed half-plane with edge AC in which B lies. Then [A, B] ⊂ H1 , so D2 , D3 ∈ H1 . Then [D2 , E2 ], [D3 , E3 ] ⊂ H1 , so F, G ∈ H1 . Then F and G are on the one side of the line AC, and as D2 E2 D3 E3 and E2 ∈ [E3 , E1 ], the angles ∠FE2 E1 , ∠GE3 E2 are corresponding angles for parallel lines and so have equal degree-measures. Thus |∠FE2 E1 |◦ = |∠GE3 E2 |◦ . By transitivity E1 F E2 G as both are parallel to AB, F and G are on the one side of AC, and E2 ∈ [E1 , E3 ], so the angles ∠FE1 E2 and ∠GE2 E3 are corresponding angles for parallel lines and so have equal degree-measures. Thus |∠FE1 E2 |◦ = |∠GE2 E3 |◦ . As |∠FE2 E1 |◦ = |∠GE3 E2 |◦ , |∠FE1 E2 |◦ = |∠GE2 E3 |◦ , by 5.2.2 |∠E1 FE2 |◦ = |∠E2 GE3 |◦ . Thus |E1 , F| = |E2 , G|, |∠FE1 E2 |◦ = |∠GE2 E3 |◦ , |∠E1 FE2 |◦ = |∠E2 GE3 |◦ , so by the ASA principle, the triangles T1 , T2 are congruent in the correspondence (E1 , F, E2 ) → (E2 , G, E3 ). It follows that |E1 , E2 | = |E2 , E3 |.

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The parallel axiom; Euclidean geometry

Ch. 5

Let A, B,C be non-collinear points and let P ∈ [A, B and Q ∈ [A,C be such that PQ BC. Then |A, P| |A, Q| = . |A, B| |A, C| A A B1 B2 P B4 B5 B6 B7 B8 B9

C1 C2 Q C4 C5 C6 C7 C8 C9

B

Pu P

Qu Q

Pv

Qv

B

C

C Figure 5.6.

Proof . We assume first that P ∈ [A, B]. Within this first case, we suppose initially that |A, P| s = , |A, B| t where s and t are positive whole numbers with s < t, so that s/t is an arbitrary rational number between 0 and 1. For 0 ≤ j ≤ t let B j be the point on [A, B such that |A, Bj | j = , |A, B| t so that B0 = A, Bt = B and Bs = P. If AB = l and ≤l is the natural order for which A ≤l B, then A ≤l B j−1 ≤l B j ≤l B j+1 ≤l B and |Bj−1 , Bj | = |Bj , Bj+1 |. If AC = m and ≤m is the natural order for which A ≤m C, then by the last result applied with (D1 , D2 , D3 ) = (B j−1 , B j , B j+1 ), for 1 ≤ j ≤ t − 1 the line through B j which is parallel to BC will meet AC in a point C j such that A ≤m C j−1 ≤m C j ≤m C j+1 ≤m C and |Cj−1 , Cj | = |Cj , Cj+1 |. It follows that, for 0 ≤ j ≤ t, |A, Cj | = j|A, C1 | and so as Ct = C, |A, Cj | j|A, C1 | j = = . |A, C| t|A, C1 | t In particular, as Cs = Q, it follows that |A, Q| s |A, P| = = . |A, C| t |A, B|

Sec. 5.3

Ratio results for triangles

67

Still within the first case, now suppose that |A, Q| |A, P| = x, = y, |A, B| |A, C| where x is an irrational number with 0 < x < 1. If u is any positive rational number less than x, and Pu is a point chosen on [A, B] so that |A, Pu | = u, |A, B| then the line through Pu which is parallel to BC will meet [A,C] in a point Qu such that |A, Qu | = u. |A, C| Similarly if v is any rational number such that x < v < 1, and Pv is a point chosen on [A, B] so that |A, Pv | = v, |A, B| then the line through Pv which is parallel to BC will meet [A,C] in a point Qv such that |A, Qv | = v. |A, C| As |A, Pu | < |A, P| < |A, Pv | we have P ∈ [Pu , Pv ]. It follows by 4.3.2 that Q ∈ [Qu , Qv ] and so u < y < v. Thus for all rational u and v such that u < x < v we have u < y < v. It follows that x = y. This completes the first case. For the second case note that if P ∈ [A, B] we have B ∈ [A, P]. Then by the first case |A, B| |A, C| = , |A, P| |A, Q| so the reciprocals of these are equal.

5.3.2 Similar triangles Let A, B,C and A , B ,C be two sets of non-collinear points such that |∠BAC|◦ = |∠B A C |◦ , |∠CBA|◦ = |∠C B A |◦ , |∠ACB|◦ = |∠A C B |◦ . Then

|B , C | |C , A | |A , B | = = . |B, C| |C, A| |A, B|

Thus if the degree-measures of the angles of one triangle are equal, respectively, to the degree-measures of the angles of a second triangle, then the ratios of the lengths of corresponding sides of the two triangles are equal.

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The parallel axiom; Euclidean geometry

Ch. 5

A B

C

A

C

C B

B Figure 5.7. Similar triangles.

textProof. Choose B ∈ [A, B and C ∈ [A,C so that |A, B | = |A , B |, |A, C | = |A , C |. Then as |∠B AC |◦ = |∠BAC|◦ = |∠B A C |◦ , by the SAS principle we see that the triangles [A, B ,C ], [A , B ,C ] are congruent in the correspondence (A, B ,C ) → (A , B ,C ). In particular |∠AB C |◦ = |∠A B C |◦ and so |∠AB C |◦ = |∠ABC|◦ . These are corresponding angles in the sense of 5.1.1, so B C BC and then by 5.3.1 |A, B | |A, C | = , |A, B| |A, C| so |A , B | |A , C | = . |A, B| |A, C| By a similar argument on taking a triangle [B, E, F] which is congruent to [B ,C , A ], we have |B , C | |B , A | = . |B, C| |B, A| COMMENT. Triangles like these, which have the degree-measures of corresponding angles equal and so the ratios of the lengths of corresponding sides are equal, are said to be similar in the correspondence (A, B,C) → (A , B ,C ). Let A, B,C and A , B ,C be two sets of non-collinear points such that |A , B | |A , C | = , |∠B A C |◦ = |∠BAC|◦ . |A, B| |A, C| Then the triangles are similar. Proof . Choose B ∈ [A, B , C ∈ [A,C so that |A, B | = |A , B |, |A, C | = |A , C |. Then as |∠B A C |◦ = |∠BAC|◦ = |∠B AC |◦ , by the SAS principle we see that the triangles [A , B ,C ], [A, B ,C ] are congruent. We note that |A, B | |A, C | = . |A, B| |A, C|

Sec. 5.4

Pythagoras’ theorem, c. 550B.C.

69

Now the line through B which is parallel to BC will meet [A,C in a point D such that |A, B | |A, D| = . |A, B| |A, C| Hence

|A, C | |A, D| = , |A, C| |A, C|

from which it follows that |A, D| = |A, C |. As C , D ∈ [A,C we then have D = C and so B C BC. Thus the degree-measures of the angles of [A, B,C] are equal to those of the corresponding angles of [A, B ,C ] and so in turn to those of the corresponding angles in [A , B ,C ].

5.4 PYTHAGORAS’ THEOREM, c. 550B.C. 5.4.1 Pythagoras’ theorem. Let A, B,C be non-collinear points such that AB ⊥ AC. Then |B, C|2 = |C, A|2 + |A, B|2 .

Proof . Let D be the foot of the perpendicular from A to BC; then by 4.3.3 D is between B and C. The triangles [D, B, A], [A, B,C] are similar as |∠ADB|◦ = |∠CAB|◦ = 90, |∠DBA|◦ = |∠ABC|◦ , and then by 5.2.2 |∠BAD|◦ = |∠BCA|◦ . Then by the last result |A, B| |B, D| = , |B, C| |A, B| so that |A, B|2 = |B, D||B, C|. By a similar argument applied to the triangles [D,C, A], [A, B,C] we get that |A, C|2 = |D, C||B, C|. Then by addition, as D ∈ [B,C], |A, B|2 + |A, C|2 = (|B, D| + |D, C|)|B, C| = |B, C|2 . C

E

A

B B

D

A

C

Figure 5.8. Pythagoras’ theorem.

Figure 5.9. Impossible figure for converse.

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The parallel axiom; Euclidean geometry

Ch. 5

CONVERSE of Pythagoras’ theorem. Let A, B,C be non-collinear points such that |B, C|2 = |C, A|2 + |A, B|2 . Then ∠BAC is a right-angle. Proof . Choose the point E so that |A, C| = |A, E|, E is on the same side of AB as C is, and ∠BAE is a right-angle. By Pythagoras’ theorem, |B, E|2 = |A, E|2 + |A, B|2 = |C, A|2 + |A, B|2 = |B, C|2 . Thus |B, E| = |B, C|, and the lengths of the sides of the triangle [B, A,C] are equal to those of [B, A, E]. By the SSS principle, [B, A,C] ≡ [B, A, E]. In particular |∠BAC|◦ = |∠BAE|◦ and this latter is a right-angle by construction. In fact E = C. NOTE. In a right-angled triangle, the side opposite the right-angle is known as the hypotenuse.

5.5 MID-LINES AND TRIANGLES 5.5.1 Harmonic ranges Let A, B,C be non-collinear points such that |A, B| > |A, C|. Take D = A so that A ∈ [B, D]. Then the mid-lines of |BAC and |CAD meet BC at points E, F, respectively, such that {E, F} divide [B,C] internally and externally in the same ratio. D

H A

G

K B

E

F

C Figure 5.10.

Proof . By the cross-bar theorem the mid-line of |BAC meets [B,C] in a point E. Let G be a point of the mid-line of |CAD, on the same side of AB as C is. We cannot have AG BC as that would imply |∠BCA|◦ = |∠CAG|◦ = |∠GAD|◦ = |∠CBA|◦ , and this in turn would imply that |A, B| = |A, C|, contrary to hypothesis. Then AG meets BC in some point F. Take H ∈ [A, D so that |A, H| = |A, C|. Then |∠AHC|◦ = |∠ACH|◦ . We have that |∠BAC|◦ = |∠AHC|◦ + |∠ACH|◦ , |∠BAC|◦ = |∠BAE|◦ + |∠EAC|◦,

|∠AHC|◦ = |∠ACH|◦ , |∠BAE|◦ = |∠EAC|◦ .

Sec. 5.5

Mid-lines and triangles

71

It follows that |∠EAC|◦ = |∠ACH|◦ , and as E, H are on opposite sides of AC this implies that AE HC. It then follows that |B, A| |B, A| |B, E| = = . |E, C| |A, H| |A, C| Next choose K ∈ [A, B so that |A, K| = |A, C|. Then |A, K| < |A, B| so K ∈ [A, B]. Now |∠HAC|◦ = |∠AKC|◦ + |∠ACK|◦ ,

|∠AKC|◦ = |∠ACK|◦ ,

|∠HAC|◦ = |∠HAG|◦ + |∠GAC|◦ ,

|∠HAG|◦ = |∠GAC|◦ .

It follows that |∠GAC|◦ = |∠ACK|◦ . But H, K are on opposite sides of AC, H, G are on the same side, and so G, K are on opposite sides. This implies that AG KC. Now AG meets BC at F, and K ∈ [A, B] so C ∈ [B, F]. It follows that |B, A| |B, A| |B, F| = = . |F, C| |A, K| |A, C| On combining the two results, we then have |B, E| |B, F| = . |E, C| |F, C| NOTE. We also refer to the mid-line of |CAD above as the external bisector of |BAC. When {E, F} divide [B,C] internally and externally in the same ratio, we say that (B,C, E, F) form a harmonic range. Let (A, B,C, D) be a harmonic range and S ∈ AB. Let the line through C, parallel to SD, meet SA at G and SB at H. Then C is the mid-point of G and H. S

G D

B A

C H

Proof . We are given

so

Figure 5.11.

|A, C| |A, D| = , |C, B| |D, B| |A, C| |C, B| = . |A, D| |D, B|

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The parallel axiom; Euclidean geometry

Ch. 5

As GC SD the triangles [A, D, S] and [A,C, G] are similar, so |A, C| |G, C| = . |A, D| |S, D| In the similar triangles [B,C, H] and [B, D, S], |B, C| |C, H| = . |B, D| |S, D| Then |G, C| |C, H| = . |S, D| |S, D| It follows that |G, C| = |C, H|. S G A

K B

C

H

D

Figure 5.12.

Let (A, B,C, D) be a harmonic range, S ∈ AB and K = S be such that S ∈ [A, K]. Suppose that CS ⊥ DS. Then CS and DS are the mid-lines of |ASB and |BSK.

Proof . Let the line through C, parallel to DS meet SA at G and SB at H. Then C is the mid-point of G and H. Also CS ⊥ SD, SD GH so SC ⊥ GH. It follows that the triangles [G,C, S] and [H,C, S] are congruent by the SAS-principle. In particular |∠GSC|◦ = |∠HSC|◦ and so SC is the mid-line of |ASB. But also |∠CGS|◦ = |∠CHS|◦ and in fact the triangle [S, G, H] is isosceles. Now ∠CGS and ∠DSK are corresponding angles and ∠CHS and ∠DSH are alternate angles. It follows that |∠DSK|◦ = |∠DSH|◦ and so the mid-line of |BSK is SD.

5.6 AREA OF TRIANGLES, AND CONVEX QUADRILATERALS AND POLYGONS 5.6.1 Area of a triangle Let A, B,C be non-collinear points, and D ∈ BC, E ∈ CA, F ∈ AB points such that AD ⊥ BC, BE ⊥ CA, CF ⊥ AB. Then |A, D||B, C| = |B, E||C, A| = |C, F||A, B|.

Sec. 5.6

Area of triangles, and convex quadrilaterals and polygons A

73

A

F

F E

B

E

B

C

D

C

Figure 5.13.

Proof . The triangles [A, B, E] and [A,C, F] are similar in the correspondence (A, B, E) → (A,C, F), as ∠BAE = ∠CAF is in both, |∠AEB|◦ = |∠AFC|◦ = 90, and then by 5.2.2 |∠ABE|◦ = |∠ACF|◦ . By 5.3.2 |A, B| |B, E| = . |C, F| |C, A| On cross multiplication, |B, E||C, A| = |C, F||A, B|. By a similar argument, we can show that |A, D||B, C| is equal to these. Definition . With the notation of the last result, the area of the triangle [A, B,C], denoted by Δ[A, B,C], is the common value of: 1 1 1 2 |A, D||B, C|, 2 |B, E||C, A|, 2 |C, F||A, B|.

Area of triangles has the following properties:(i) If P ∈ [B,C] is distinct from B and C, then Δ[A, B, P] + Δ[A, P,C] = Δ[A, B,C]. (ii) If [A, B,C, D] is a convex quadrilateral, then Δ[A, B, D] + Δ[C, B, D] = Δ[B,C, A] + Δ[D,C, A]. (iii) If two triangles are congruent then their areas are equal.

74

The parallel axiom; Euclidean geometry

Ch. 5

A

D A T

B

P

D

B

C

C

Figure 5.14.

Proof . (i) For D is the foot of the perpendicular from the vertex A to the opposite side-line in each of the triangles [A, B, P] and [A, P,C], so with p1 = |A, D| we have Δ[A, B, P] = 12 p1 |B, P|, Δ[A, P,C] = 12 p1 |P, C|, and the sum of these is 1 1 2 p1 (|B, P| + |P, C|) = 2 p1 |B, C|,

as P ∈ [B,C]. (ii) As in 5.2.1 denote by T the point which [A,C] and [B, D] have in common. Then by (i) above, Δ[A, B, D] + Δ[C, B, D] = (Δ[A, B, T ] + Δ[A, D, T ]) + (Δ[C, B, T ] + Δ[C, D, T ]) , Δ[A, B,C] + Δ[A, D,C] = (Δ[A, B, T ] + Δ[C, B, T ]) + (Δ[A, D, T ] + Δ[C, D, T ]) , and these are clearly equal. (iii) Let T be a triangle with vertices {A, B,C} and T a triangle with vertices ≡ {A , B ,C } such that T (A,B,C)→(A ,B ,C ) T . Suppose that D and D are the feet of the perpendiculars from A and A onto BC and B C , respectively. There are three cases to be considered. A

A

B=D

C

B

D Figure 5.15.

A

C

B

C

D

Sec. 5.6

Area of triangles, and convex quadrilaterals and polygons

75

First suppose that D = B. Then |∠ABC|◦ = 90 = |∠A B C |◦ and so D = B . Then 1 2 |A, D||B,C|

= 12 |A, B||B,C| = 12 |A , B ||B ,C | = 12 |A , D ||B ,C |,

so the areas are equal. There is a similar proof when D = C. Secondly suppose that D lies in [B,C] but is not B or C. Then the angles ∠CBA = ∠DBA and ∠BCA = ∠DCA are both acute. It follows that the angles ∠C B A and ∠B C A are both acute and so D lies in [B ,C ] but is not B or C . Now the triangles with vertices {A, B, D} and {A , B , D } have two angles pairwise equal in magnitude and so the third angles ∠DAB and ∠D A B are also equal. Then by the ASA principle of congruence these two triangles are congruent in the correspondence (A, B, D) → (A , B , D ). In particular |A, D| = |A , D | and so 1 1 2 |A, D||B,C| = 2 |A , D ||B ,C |. Thus the areas are equal. Thirdly suppose that C lies in [B, D] but is not D. Then the angles ∠CBA = ∠DBA and ∠DCA are acute and the angle ∠BCA is obtuse. It follows that ∠C B A is acute and ∠B C A is obtuse and so C lies in [B , D ] but is not D . Now the triangles with vertices {A, B, D} and {A , B , D } are congruent as in the second case and so the equality of areas follows. There is a similar proof when B lies in [D,C] but is not D, on interchanging the roles of B and C in this last argument.

5.6.2 Area of a convex quadrilateral NOTE. In the quadrilateral in Figure 5.14 we have Δ[A, B, D] + Δ[C, B, D] = {Δ[A, B, T ] + Δ[A, T, D]} + {Δ[C, B, T ] + Δ[C, T, D]} = {Δ[A, B, T ] + Δ[C, B, T ]} + {Δ[A, T, D] + Δ[C, T, D]} = Δ[B, A,C] + Δ[D, A,C]. Definition. We define the area of the convex quadrilateral [A, B,C, D] as in Figure 5.14 to be Δ[A, B, D] + Δ[C, B, D] = Δ[B, A,C] + Δ[D, A,C], and denote it by Δ[A, B,C, D]. If [A, B,C, D] is a rectangle, then Δ[A, B,C, D] = |A, B||B, C|, that is the area is equal to the product of the lengths of two adjacent sides. Proof . For Δ[A, B, D] = 12 |A, B||A, D|, Δ[C, B, D] = 12 |D, C||B, C|. As by 5.2.1 |D, C| = |A, B| and |B, C| = |A, D|, the result follows by addition.

5.6.3 Area of a convex polygon As a preparation for generalisation we first show that in 2.4.4 for the convex quadrilateral [A, B,C, D] we have C ∈ H1 , D ∈ H3 , A ∈ H5 , and B ∈ H7 , so that each of the vertices is in each of these closed half-planes.

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The parallel axiom; Euclidean geometry

Ch. 5

We cannot have A = T as then we would have A ∈ BD; nor can T be any other point of the line AB, for then we would have AT = AB and so C ∈ AB. Thus T ∈ AB and similarly T is not on any of the other side-lines. As B, D ∈ H1 and T ∈ [B, D] we have T ∈ H1 . Then A ∈ AB, T ∈ H1 and T = A so by 2.2.3(iv) [A, T ⊂ H1 . Now A, T,C are distinct and collinear and T ∈ [A,C] so by 2.1.4(v) we cannot have A ∈ [T,C]. Thus C ∈ [A, T and so C ∈ H1 . By a similar argument the other cases follow. We now establish a converse of this, starting with A, B,C, D as in 2.4.4, with no three of them collinear and H1 , H3 , H5 , H7 as defined there. We now suppose that C ∈ H1 , D ∈ H3 , A ∈ H5 , B ∈ H7 and wish to prove that [A,C] ∩ [B, D] = 0. / To start with we note that H1 ∩ H3 = I R(|ABC) and [A,C] is a cross-bar of this. Now we are given that D ∈ H1 , D ∈ H3 so D is in this interior region and by the cross-bar theorem in 3.4.1 [B, D meets [A,C] in a point T . Similarly H1 ∩ H7 = I R(|BAD) and [B, D] is a cross-bar of this. Thus C is in this interior region and by the cross-bar theorem [A,C meets [B, D] in some point. But the lines AC and BD can meet only once as otherwise we would have AC = BD which is ruled out. Thus we have T ∈ [B, D] as well.

Definition . For an integer n ≥ 3 let P1 , P2 , . . . , Pn be n points such that no three of them are collinear. For each integer j such that 1 ≤ j ≤ n − 1 let H2 j−1 , H2 j be the closed half-planes with common edge the line Pj Pj+1 , and H2n−1 , H2n the closed half-planes with edge Pn P1 . Suppose that all the points Pk lie in H2 j−1 in each case.  Then the intersection nj=1 H2 j−1 is called a convex polygon. The intersection of the corresponding open half-planes is called the interior of the convex polygon. The notation for convex quadrangles is extended to convex polygons in a straightforward way. Consider a convex polygonal region with sides [P1 , P2 ], [P2 , P3 ], . . . , [Pn , P1 ]. Let a point U interior to the polygon be joined by segments to the vertices. Then n−1

n−1

j=1

j=2

∑ Δ[U, Pj , Pj+1] + Δ[U, Pn, P1] = ∑ Δ[P1, Pj , Pj+1].

(5.6.1)

Proof . CASE 1. We first take the case of a triangle so that n = 3. Now [P1 ,U will meet [P2 , P3 ] in a point V . Then by 5.6.1(i) Δ[U, P1 , P2 ] + Δ[U, P2 , P3 ] + Δ[U, P3 , P1 ] = Δ[U, P1 , P2 ] + {Δ[U, P2 ,V ] + Δ[U,V, P3 ]} + Δ[U, P3 , P1 ] = {Δ[U, P1 , P2 ] + Δ[U, P2 ,V ]} + {Δ[U,V, P3 ] + Δ[U, P3 , P1 ]} = Δ[P1 , P2 ,V ] + Δ[V, P3 , P1 ] = Δ[P1 , P2 , P3 ].

Sec. 5.6

Area of triangles, and convex quadrilaterals and polygons P1

77

P1 P4

U

U

V

P2

P3

P2

P3 P6

P1

P1 P4

P2

U

P5

U P3

P2

P3

P4

Figure 5.16.

CASE 2. Secondly we take the case of a convex quadrilateral so that n = 4. Suppose first that U ∈ [P1 , P3 ]. Then by 5.6.1(i) used twice, {Δ[U, P1 , P2 ] + Δ[U, P2 , P3 ]} + {Δ[U, P3 , P4 ] + Δ[U, P4 , P1 ]} = Δ[P1 , P2 , P3 ] + Δ[P1, P3 , P4 ]. Suppose next that U ∈ [P1 , P3 ]. Then U is interior to [P1 , P2 , P3 ] or [P1 , P3 , P4 ], say U ∈ [P1 , P3 , P4 ]. We need to show that we have a convex quadrilateral [P1 , P2 , P3 ,U ]. Now U ∈ I R(|P1 P2 P3 ) for which [P1 , P3 ] is a cross-bar. Thus [P2 ,U meets [P1 , P3 ] in a point T . Now we cannot have U ∈ [P2 , T ] as that would make U ∈ [P1 , P2 , P3 ], nor can we have P2 ∈ [T,U ] as that would put U outside H1 and H3 , so we must have T ∈ [P2 ,U ]. Then by 5.6.1(ii) Δ[U, P1 , P2 ] + Δ[U, P2 , P3 ] = Δ[P1 , P2 , P3 ] + Δ[U, P1 , P3 ], so Δ[U, P1 , P2 ] + Δ[U, P2 , P3 ] + Δ[U, P3 , P4 ] + Δ[U, P4 , P1 ] = Δ[P1 , P2 , P3 ] + {Δ[U, P1 , P3 ] + Δ[U, P3 , P4 ] + Δ[U, P4 , P1 ]} = Δ[P1 , P2 , P3 ] + Δ[P1, P3 , P4 ] by CASE 1. CASE 3. We now suppose that the result holds, for some n ≥ 4, for any convex polygonal region with n sides. Then for that n consider any convex polygonal region with n + 1 sides, [P1 , P2 ], [P2 , P3 ], . . . , [Pn , Pn+1 ], [Pn+1 , P1 ]. As n + 1 ≥ 5, [P1 , P2 , P3 ] and [P1 , Pn , Pn+1 ] have only P1 in common, so U cannot be in both. For suppose that U is an interior point which is common to both. Then

78

The parallel axiom; Euclidean geometry

Ch. 5

[P1 ,U meets [P2 , P3 ] at a point V = P2 so that U ∈ [P1 ,V ]. Also [P1 ,U meets [Pn−1 , Pn ] at a point W = Pn so that U ∈ [P1 ,W ]. If we write l = P1U and have P1 ≤l U , then U ≤l V, U ≤l W and as V ≤l W or W ≤l V we have V ∈ [U,W ] or W ∈ [U,V ]. One possibility is that W = V . We cannot have either Pn−1 or Pn on P2 P3 as that would make three points Pk collinear. Then we have that V = W ∈ [Pn−1 , Pn ] and V ∈ P2 P3 so that Pn−1 and Pn are on different sides of P2 P3 , that is one is in H3 and the other is not, whereas all the vertices Pk are in H3 . This contradiction implies that V = W . A second possibility is that V ∈ [U,W ], V = W . Then W ∈ H3 whereas Pn−1 , Pn ∈ H3 and as W ∈ [Pn−1 , Pn ] ⊂ H3 we have a contradiction. A third and final possibility is that W ∈ [U,V ], W = V . Then V ∈ H2n−3 whereas P2 , P3 ∈ H2n−3 and as V ∈ [P2 , P3 ] ⊂ H2n−3 we have a contradiction again. With this preparation suppose now that U ∈ [P1 , P2 , P3 ]. By 5.6.1(ii) Δ[U, P1 , P2 ] + Δ[U, P2 , P3 ] = Δ[P1 , P2 , P3 ] + Δ[U, P1 , P3 ]. Hence as U is interior to the polygon with n sides [P1 , P3 ], [P3 , P4 ], . . ., [Pn , Pn+1 ], [Pn+1 , P1 ], n

∑ Δ[U, Pj , Pj+1] + Δ[U, Pn+1, P1]

j=1

n

= Δ[P1 , P2 , P3 ] + Δ[U, P1 , P3 ] + ∑ Δ[U, Pj , Pj+1 ] + Δ[U, Pn+1 , P1 ] j=3

n

= Δ[P1 , P2 , P3 ] + ∑ Δ[P1 , Pj , Pj+1 ] = j=3

n

∑ Δ[P1, Pj , Pj+1].

j=2

If instead U ∈ [P1 , Pn , Pn+1 ] we get the same conclusion by similar reasoning. The result now follows by induction on n. Definition . The area of the polygonal region in the present section is defined to be the sum of the areas of the triangles involved, as in (5.6.1).

Exercises 5.1 Opposite wedge-angles in a parallelogram have equal degree-measures. 5.2 If two adjacent sides of a rectangle have equal lengths, then all the sides have equal lengths. Such a rectangle is called a square. 5.3 If the diagonals of a parallelogram have equal lengths, it must be a rectangle. 5.4 If the diagonal lines of a rectangle are perpendicular, it must be a square. 5.5 Let A, B,C be non-collinear points and let P ∈ [A, B and Q ∈ [A,C be such that |A, P| |A, Q| = . |A, B| |A, C|

Sec. 5.6

Area of triangles, and convex quadrilaterals and polygons

79

Then PQ BC. 5.6 Let AB ⊥ AC and let D = mp(B,C). Prove that |D, A| = |D, B| = |D, C|. 5.7 Let A, B,C be non-collinear points and for A ∈ [B, P and A ∈ [C, Q let PQ BC. Show that then |A, P| |A, Q| = . |A, B| |A, C| 5.8 Suppose that A, B,C are non-collinear points with |A, B| > |A, C| and let D = πBC (A). Prove that then |A, B|2 − |A, C|2 = |B, D|2 − |C, D|2 . 5.9 Suppose that A, B,C are non-collinear points and D is the mid-point of B and C. Prove that then |A, B|2 + |A, C|2 = 2|B, D|2 + 2|A, D|2 . [Hint. Consider the foot of the perpendicular from A to BC.] 5.10 Show that the AAS-principle of congruence in Ex.4.2 can be deduced from 5.2.2 and the ASA-principle. 5.11 Show that the AAS-principle of congruence for right-angled triangles in Ex.4.3 can be deduced from Pythagoras’ theorem and the SSS-principle. 5.12 For C ∈ AB, suppose that m is the line through C which is parallel to AB. Prove that for any point D ∈ AB the line AD meets m in a unique point E. When, additionally, D ∈ I R(|BAC) then E is on [A, D and is also on m∩I R(|BAC). 5.13 In a triangle [A, B,C], let |A, B| > |A, C|. Let D ∈ [A, B be such that |A, D| = |A, C|. Prove that then 2|∠BCD|◦ = |∠ACB|◦ − |∠CBA|◦.

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6

Cartesian coordinates; applications COMMENT. Hitherto we have confined ourselves to synthetic or pure geometrical arguments aided by a little algebra, and traditionally this is continued with. This is a difficult process because of the scarcity of manipulations, operations and transformations to aid us. The main difficulties in synthetic proofs are locational, to show that points are where the diagrams suggest they should be, and in making sure that all possible cases are covered. For ease and efficiency we now introduce coordinates, and hence thoroughgoing algebraic methods. These not only enable us to deal with the concepts already introduced but also to elaborate on them in an advantageous way. In Chapter 6 we do the basic coordinate geometry of lines, segments, half-lines and half-planes. The only use we make of angles here is to deal with perpendicularity.

6.1 FRAME OF REFERENCE, CARTESIAN COORDINATES 6.1.1 Definition . A couple or ordered pair F = ([O, I , [O, J ) of half-lines such that OI ⊥ OJ, will be called a frame of reference for Π. With it, as standard notation, we shall associate the pair of closed half-planes H1 , H2 , with common edge OI, and with J ∈ H1 , and the pair of closed half-planes H3 , H4 , with common edge OJ, and with I ∈ H3 . We refer to Q1 = H1 ∩ H3 , Q2 = H1 ∩ H4 , Q3 = H2 ∩ H4 and Q4 = H2 ∩ H3 , respectively, as the first, second, third and fourth quadrants of F . We refer to OI and OJ as the axes and to O as the origin. Given any point Z in Π, (rectangular) Cartesian coordinates for Z are defined as follows. Let U be the foot of the perpendicular from Z to OI and V the foot of the Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50006-9 © 2016 Elsevier Ltd. All rights reserved.

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Cartesian coordinates; applications

Ch. 6

perpendicular from Z to OJ. We let  x=



|O, U| if Z ∈ H3 , −|O, U| if Z ∈ H4 ,

and

y=

|O, V| −|O, V|

if Z ∈ H1 , if Z ∈ H2 .

H4 H3 H4 H3 Q2

V

Q1

J

J H1

O

Z

I

Q3

O

I

H1 U H2

H2 Q4 Figure 6.2. Cartesian coordinates.

Figure 6.1. Frame of reference.

Then the ordered pair (x, y) are called Cartesian coordinates for Z, relative to F . We denote this in symbols by Z ≡F (x, y), but when F is fixed and can be understood, we relax this notation to Z ≡ (x, y). Cartesian coordinates have the following properties:(i) If Z ∈ Q1 , then x ≥ 0, y ≥ 0; if Z ∈ Q2 , then x ≤ 0, y ≥ 0; if Z ∈ Q3 , then x ≤ 0, y ≤ 0; if Z ∈ Q4 , then x ≥ 0, y ≤ 0. (ii) If Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ) and U1 = πOI (Z1 ), V1 = πOJ (Z1 ), U2 = πOI (Z2 ), V2 = πOJ (Z2 ), then |U1 , U2 | = ±(x2 − x1 ), |V1 , V2 | = ±(y2 − y1 ). (iii) If Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ), then |Z1 , Z2 | =



(x2 − x1 )2 + (y2 − y1 )2 .

(iv) If Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ) and Z3 ≡ (x3 , y3 ) where x3 = 12 (x1 + x2 ), y3 = 12 (y1 + y2 ), then Z3 = mp(Z1 , Z2 ). (v) Let ≤l be the natural order on l = OI under which O ≤l I. If x1 < x2 , U1 ≡ (x1 , 0) and U2 ≡ (x2 , 0), then U1 ≤l U2 .

Sec. 6.1

Frame of reference, Cartesian coordinates

83

Proof . (i) This is clear from the definition of coordinates. (ii) For if Z1 , Z2 ∈ H3 we have |O, U2 | = x1 , |O, U2 | = x2 , and so as U1 ,U2 ∈ [O, I ,

|U1 , U2 | = ±(x2 − x1 )

according as U1 ∈ [O,U2 ] or U2 ∈ [O,U1 ]. Similarly if Z1 , Z2 ∈ H4 , we have |O, U1 | = −x1 , |O, U2 | = −x2 and |U1 , U2 | = ±[−x2 − (−x1 )] according as U1 ∈ [O,U2 ] or U2 ∈ [O,U1 ]. Finally if Z1 ∈ H3 , Z2 ∈ H4 then |O, U1 | = x1 , |O, U2 | = −x2 and O ∈ [U1 ,U2 ] so that

|U1 , U2 | = x1 + (−x2 );

similarly if Z1 ∈ H4 , Z2 ∈ H3 . That |V1 , V2 | = ±(y2 − y1 ) can be shown in the same way. (iii) Now the lines through Z1 parallel to OI and through Z2 parallel to OJ are perpendicular to each other, and so meet in a unique point Z4 . Clearly πOI (Z4 ) = πOI (Z2 ) = U2 so Z2 and Z4 have the same first coordinate, x2 ; πOJ (Z4 ) = πOJ (Z1 ) = V1 so Z1 and Z4 have the same second coordinate, y1 . Thus Z4 has coordinates (x2 , y1 ). If the points Z1 , Z2 , Z4 are not collinear, then by Pythagoras’ theorem |Z1 , Z2 |2 = |Z1 , Z4 |2 + |Z4 , Z2 |2 ; if they are collinear we must have Z1 = Z4 or Z2 = Z4 and this identity is trivially true. But |Z1 , Z4 | = |U1 , U2 | as [Z1 , Z4 ,U2 ,U1 ] is a rectangle, or else Z1 = Z4 and U1 = U2 , or Z1 = U1 , Z4 = U2 . Similarly |Z2 , Z4 | = |V1 , V2 |. Thus we have the distance formula |Z1 , Z2 |2 = |U1 , U2 |2 + |V1 , V2 |2 = (x2 − x1 )2 + (y2 − y1 )2 , which expresses the distance |Z1 , Z2 | in terms of the coordinates of Z1 and Z2 . (iv) If Z1 = Z2 , then x2 = x1 , y2 = y1 so that x3 = x1 , y3 = y1 . Thus Z3 = Z1 = mp(Z1 , Z1 ), as required. Suppose then that Z1 = Z2 . Note that  |Z1 , Z3 |2 =

x1 + x2 − x1 2

2

 +

y1 + y2 − y1 2

2

 =

x2 − x1 2

2

 +

y2 − y1 2

2

84

Cartesian coordinates; applications

Ch. 6

and so |Z1 , Z3 | = 12 |Z1 , Z2 |. Similarly  |Z3 , Z2 |2 =

x1 + x2 − x2 2

2

 +

y1 + y2 − y2 2

2

 =

x1 − x2 2

2

 +

y1 − y2 2

2

and so |Z3 , Z2 | = 21 |Z1 , Z2 |. Then |Z1 , Z3 | + |Z3 , Z2 | = |Z1 , Z2 |. It follows by 3.1.2 and 4.3.1 that Z3 ∈ [Z1 , Z2 ] ⊂ Z1 Z2 . As |Z1 , Z3 | = |Z3 , Z2 | it then follows that Z3 = mp(Z1 , Z2 ).

Z4

H4 H3 V1

U1 O Z1 O

J U2 Z2

O

U2

I

U1

H1 H2

U1 U2

U1 U2 O

V2

Figure 6.3. The distance formula.

Order of points on the x-axis.

(v) By 2.1.4 at least one of (a) O ∈ [U1 ,U2 ], (b) U1 ∈ [O,U2 ], (c) U2 ∈ [O,U1 ], holds. In (a), U1 and U2 are in different half-lines with end-point O. We cannot have U1 ∈ [O, I as then we would have x1 ≥ 0, x2 ≤ 0, a contradiction. Thus U2 ∈ [O, I so that U1 ≤l O, O ≤l U2 and thus U1 ≤l U2 . In (b) we cannot have U1 ≤l O. For then we would have U2 ≤l O and |O, U1 | = −x1 , |O, U2 | = −x2 . As U1 ∈ [O,U2 ] we have |O, U1 | ≤ |O, U2 | which yields −x1 ≤ −x2 and so x1 ≥ x2 , a contradiction. Hence O ≤l U1 and so as U1 ∈ [O,U2 ], U1 ≤l U2 . In (c) we cannot have O ≤l U1 . For then we would have O ≤l U2 and so |O, U1 | = x1 , |O, U2 | = x2 . As U2 ∈ [O,U1 ] we have |O, U2 | ≤ |O, U1 |, so that x2 ≤ x1 , a contradiction. Hence U1 ≤l O so U1 ≤l U2 .

Sec. 6.2

Algebraic note on linear equations

85

6.2 ALGEBRAIC NOTE ON LINEAR EQUATIONS 6.2.1 It is convenient to note here some results on solutions of two simultaneous linear equations in two unknowns. (a) If a1,1 a2,2 − a1,2 a2,1 = 0,

(6.2.1)

then the pair of simultaneous equations

a1,1 x + a1,2 y = k1 , a2,1 x + a2,2 y = k2 ,

(6.2.2)

has precisely one solution pair (x, y), and that is given by (x, y) =

a2,2 k1 − a1,2 k2 a1,1 k2 − a2,1 k1 , a1,1 a2,2 − a1,2 a2,1 a1,1 a2,2 − a1,2 a2,1

.

(6.2.3)

(b) If (a1,1 , a1,2 ) = (0, 0)

and

(a2,1 , a2,2 ) = (0, 0),

(6.2.4)

and a1,1 a2,2 − a1,2 a2,1 = 0,

(6.2.5)

then there is some j = 0 such that a2,1 = ja1,1 , a2,2 = ja1,2 .

(6.2.6)

(c) If (6.2.4) holds, then for the system (6.2.2) of simultaneous equations to have either no, or more than one, solution pair (x, y) it is necessary and sufficient that (6.2.5) hold. Note in particular that when (6.2.4) holds, for the pair of homogeneous linear equations a1,1 x + a1,2 y = 0, a2,1 x + a2,2 y = 0,

(6.2.7)

to have a solution (x, y) other than the obvious one (0,0), it is necessary and sufficient that (6.2.5) hold.

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Cartesian coordinates; applications

Ch. 6

6.3 CARTESIAN EQUATION OF A LINE 6.3.1 Given any line l ∈ Λ, there are numbers a, b and c, with the case a = b = 0 excluded, such that Z ≡ (x, y) ∈ l if and only if ax + by + c = 0.

Proof . Take any point Z2 ≡ (x2 , y2 ) ∈ l and let Z3 ≡ (x3 , y3 ) = sl (Z2 ). Then Z2 = Z3 . Now l is the perpendicular bisector of [Z2 , Z3 ], so by 4.1.1 Z ∈ l if and only if |Z, Z2 | = |Z, Z3 |. As these are both non-negative, this is the case if and only if |Z, Z2 |2 = |Z, Z3 |2 . By 6.1.1 this happens if and only if (x − x2 )2 + (y − y2 )2 = (x − x3 )2 + (y − y3 )2 . This simplifies to 2(x3 − x2 )x + 2(y3 − y2 )y + x22 + y22 − x23 − y23 = 0. On writing a = 2(x3 − x2 ), b = 2(y3 − y2 ), c = x22 + y22 − x23 − y23 , we see that Z ≡ (x, y) ∈ l if and only ax + by + c = 0. Now a = b = 0 corresponds to x2 = x3 , y2 = y3 , which is ruled out as Z2 = Z3 . COROLLARY. Let Z0 ≡ (x0 , y0 ), Z1 ≡ (x1 , y1 ) be distinct points and Z ≡ (x, y). Then Z ∈ Z0 Z1 if and only if −(y1 − y0 )(x − x0 ) + (x1 − x0 )(y − y0 ) = 0.

Proof . By the theorem, there exist numbers a, b, c, with the case a = b = 0 excluded, such that Z ∈ Z0 Z1 if and only if ax + by + c = 0. As Z0 , Z1 ∈ Z0 Z1 we then have ax0 + by0 + c = 0, ax1 + by1 + c = 0. We subdivide into two cases as follows. CASE 1. Let x0 = x1 . We rewrite our equations as ax1 + c = −by1 , ax0 + c = −by0 , and regard these as equations in the unknowns a and c. As x1 − x0 = 0, we note that by 6.2.1 we must have a=

−b(x1 y0 − x0 y1 ) −b(y1 − y0 ) , c= . x1 − x0 x1 − x0

Sec. 6.3

Cartesian equation of a line

87

Note that b = 0, as b = 0 would imply a = 0 here. On inserting these values for a and c above we see that Z ∈ l if and only if −b(x1 y0 − x0 y1 ) −b(y1 − y0 ) x + by + = 0, x1 − x0 x1 − x0 and so as b/(x1 − x0 ) = 0, if and only if −(y1 − y0 )x + (x1 − x0 )y − x1 y0 + x0 y1 = 0. This is equivalent to the stated equation. CASE 2. Let y0 = y1 . We rewrite our equations as by1 + c = −ax1 , by0 + c = −ax0 , and note that, as y1 − y0 = 0, by 6.2.1 we must have b=

−a(x1 − x0 ) −a(y1 x0 − y0 x1 ) , c= . y1 − y0 y1 − y0

Note that a = 0, as a = 0 would imply b = 0 here. On inserting these values for b and c above we see that Z ∈ l if and only if ax +

−a(x1 − x0 ) −a(y1 x0 − y0 x1 ) y+ = 0, y1 − y0 y1 − y0

and so as −a/(y1 − y0 ) = 0, if and only if −(y1 − y0 )x + (x1 − x0 )y − x1 y0 + x0 y1 = 0. This is equivalent to the stated equation. Now either CASE 1 or CASE 2 (or both) must hold, as otherwise we have x0 = x1 , y0 = y1 and so Z0 = Z1 , contrary to what is given.

Definition . If l ∈ Λ and l = {Z ≡ (x, y) : ax + by + c = 0}, we call ax + by + c = 0 a Cartesian equation of l relative to F , and we write l ≡F ax + by + c = 0. When F can be understood we relax this to l ≡ ax + by + c = 0. Let l ∈ Λ be a line, with Cartesian equation (i) ax + by + c = 0. Then l also has (ii) a1 x + b1 y + c1 = 0, as an equation if and only if (iii) a1 = ja, b1 = jb, c1 = jc,

88

Cartesian coordinates; applications

Ch. 6

for some j = 0. Proof . Necessity . Suppose first that l can be expressed in each of the forms (i) and (ii) above. We subdivide into four cases as follows. CASE 1. Suppose that a = 0, b = 0 and c = 0. Then we note from (i) that the points A ≡ (−c/a, 0) and B ≡ (0, −c/b) are in l, and are in fact the only points of l in either OI or OJ, as A is the only point with y = 0 and B is the only point with x = 0. We now note that none of a1 , b1 , c1 can be equal to 0. For if a1 = 0, by (ii) we would have y = −c1 /b1 for all points Z in l; this would make l parallel to OI and give a contradiction. Similarly b1 = 0 would imply that x = −c1 /a1 for all points Z in l, making l parallel to OJ and again giving a contradiction. Moreover if c1 = 0, by (ii) we would have that O ∈ l, again a contradiction. We note from (ii) that the points A1 ≡ (−c1 /a1 , 0), B1 ≡ (0, −c1 /b1 ) are in l and are in fact the only points of l in either OI or OJ. Thus we must have A1 = A, B1 = B and so c1 c c c1 − =− , − =− . a1 a b1 b Thus a1 b1 c1 = = , a b c and if we denote the common value of these by j, we have j = 0 and (iii). CASE 2. Suppose that a = 0. Then b = 0 and by (i) for every Z ∈ l we have y = −c/b, so that l contains B and is parallel to OI; when c = 0, l has no point in common with OI, and when c = 0, l coincides with OI. Now we must have a1 = 0, as otherwise l would meet OI in the unique point A1 , and that would give a contradiction. Then b1 = 0 and for every Z ∈ l we have y = −c1 /b1 , so that l contains B1 and is parallel to OI. Thus we must have −

c1 c =− . b1 b

When c = 0, this implies that c1 = 0, so that if we take j = b1 /b, we have satisfied (iii). When c = 0, we must have that b1 /b = c1 /c, and if we take j to be the common value of these we have (iii) again. CASE 3. Suppose that b = 0. This is treated similarly to CASE 2. CASE 4. Finally suppose that a = 0, b = 0 and c = 0. Then by (i) we see that O ∈ l and then by (ii) we must have c1 = 0. We see from (i) that C ≡ (1, −a/b) is in l, and on using this information in (ii) we find that a1 + b1 (−a/b) = 0. This implies that a1 /a = b1 /b, and if we take j to be the common value of these, we must have (iii). This establishes the necessity of (iii).

Sufficiency Suppose now that (iii) holds. Then a1 x + b1 y + c1 = j(ax + by + c) and as j = 0 we have a1 x + b1 y + c1 = 0 if and only if ax + by + c = 0.

Sec. 6.4

Parametric equations of a line

89

6.4 PARAMETRIC EQUATIONS OF A LINE 6.4.1 Let l be a line with Cartesian equation ax + by + c = 0. (i) If Z0 ≡ (x0 , y0 ) is in l, then l = {Z ≡ (x, y) : x = x0 + bt, y = y0 − at, (t ∈ R)}. (ii) If Z1 ≡ (x1 , y1 ) = (x0 + b, y0 − a) and ≤l is the natural order on l for which Z0 ≤l Z1 , then for Z2 ≡ (x0 + bt2 , y0 − at2 ), Z3 ≡ (x0 + bt3 , y0 − at3 ) we have t2 ≤ t3 if and only if Z2 ≤l Z3 . (iii) If Z1 ≡ (x1 , y1 ) = (x0 + b, y0 − a), then [Z0 , Z1 ] = {Z ≡ (x, y) : x = x0 + bt, y = y0 − at, (0 ≤ t ≤ 1)}. (iv) With Z1 as in (ii), [Z0 , Z1 = {Z ≡ (x, y) : x = x0 + bt, y = y0 − at, (t ≥ 0)}.

Proof . (i) If Z ∈ l then ax + by + c = 0, ax0 + by0 + c = 0, so that b(y − y0 ) = −a(x − x0 ).

(6.4.1)

When b = 0, let us define t by t = (x − x0 )/b; then by (6.4.1) we must have, y − y0 = −at. Thus x = x0 + bt, y = y0 − at, (6.4.2) for some t ∈ R. When b = 0 then a = 0, and by (6.4.1) we must have x = x0 . If we define t by t = (y − y0 )/(−a), then we have (6.4.2) for some t ∈ R. Conversely suppose that (6.4.2) holds for any t ∈ R. Then ax + by + c = a(x0 + bt) + b(y0 − at) + c = ax0 + by0 + c = 0. (ii) We first suppose that l is not perpendicular to m = OI, so that b = 0. We recall that Z0 , Z1 are distinct points on l for which Z0 ≤l Z1 . Let ≤m be the natural order on m for which O ≤m I. Let U0 = πm (Z0 ), U1 = πm (Z1 ) so that U0 ≡ (x0 , 0), U1 ≡ (x0 + b, 0). Z Z0 1 Z1 Z0

U0 U1 Figure 6.4. Direct correspondence.

U1 U0 Indirect correspondence.

90

Cartesian coordinates; applications

Ch. 6

If b > 0, then x0 < x0 + b and so by 6.1.1 U0 ≤m U1 . In this case we say that the correspondence between ≤l and ≤m is direct. If b < 0 then x0 + b < x0 and so U1 ≤m U0 . In this case we say that the correspondence between ≤l and ≤m is indirect. In what follows we assume that b > 0 so that the correspondence between ≤l and ≤m is direct. The other case can be covered by replacing ≤m by ≥m in the following. Z0 Z2 Z3

U2 U3 U0

Z3

Z0 Z2

U3

Suppose now that Z2 ≤l Z3 ; we wish to show that U2 ≤m U3 where U2 = πm (Z2 ), U3 = πm (Z3 ). We subdivide into three cases.

U0U2

Z1 Z2

U1U2

Z3

U3

Z3

U3

Figure 6.5.

CASE 1. Suppose that Z2 ≤l Z0 . Then Z2 ≤l Z0 ≤l Z1 so that Z0 ∈ [Z2 , Z1 ]. Then by 4.3.2, U0 ∈ [U2 ,U1 ]. As U0 ≤m U1 , we then have U2 ≤m U0 . There are now two possibilities, that Z3 ≤l Z0 or that Z0 ≤l Z3 . In the first of these subcases, Z3 ∈ [Z2 , Z0 ] so U3 ∈ [U2 ,U0 ]. As U2 ≤m U0 we then have U2 ≤m U3 . In the second of these subcases we have Z0 ∈ [Z2 , Z3 ] so U0 ∈ [U2 ,U3 ]. As U2 ≤m U0 we have U0 ≤m U3 so U2 ≤m U3 . CASE 2. Suppose that Z0 ≤l Z2 ≤l Z1 . Then Z2 ∈ [Z0 , Z1 ] so U2 ∈ [U0 ,U1 ]. As U0 ≤m U1 then U0 ≤m U2 ≤m U1 . Now Z2 ∈ [Z0 , Z3 ] so U2 ∈ [U0 ,U3 ]. As U0 ≤m U2 it follows that U2 ≤m U3 . CASE 3. Suppose that Z1 ≤l Z2 . Then Z1 ∈ [Z0 , Z2 ] so that U1 ∈ [U0 ,U2 ]. As U0 ≤m U1 we then have U1 ≤m U2 . Then Z2 ∈ [Z1 , Z3 ] so U2 ∈ [U1 ,U3 ]. As U1 ≤m U2 we have U2 ≤m U3 . Now continuing with all three cases, we note that U2 ≡ (x0 + bt2 , 0), U3 ≡ (x0 + bt3 , 0) and as U2 ≤m U3 by 6.1.1 we have x0 + bt2 ≤ x0 + bt3 . As b > 0 this implies that t2 ≤ t3 . We also have that t2 ≤ t3 implies Z2 ≤l Z3 . For otherwise Z3 ≤l Z2 and so by the above t3 ≤ t2 , which gives a contradiction unless Z2 = Z3 . When l is perpendicular to OI we use πOJ instead of πm . By a similar argument we reach the same conclusion. (iii) This follows directly from (ii) of the present theorem. It can also be proved as follows. Note that in (6.4.2) t = 0 gives Z0 and t = 1 gives Z1 . Then for Z ≡ (x, y)

Sec. 6.4

Parametric equations of a line

91

with x and y as in (6.4.2), by 6.1.1 we have    |Z0 , Z| = (x − x0 )2 + (y − y0 )2 = (bt)2 + (−at)2 = |t| b2 + a2 ,    |Z1 , Z| = (bt − t)2 + (a − at)2 = (t − 1)2 (b2 + a2 ) = |t − 1| b2 + a2 ,   |Z0 , Z1 | = b2 + (−a)2 = b2 + a2 . Thus when t < 0,

  |Z0 , Z| = (−t) b2 + a2 , |Z1 , Z| = (1 − t) b2 + a2 ,

and so |Z, Z0 | + |Z0 , Z1 | = |Z, Z1 |; thus by 3.1.2 and (i) above, Z0 ∈ [Z, Z1 ], Z0 = Z, Z = Z1 . When 0 ≤ t ≤ 1,   |Z0 , Z| = t b2 + a2 , |Z, Z1 | = (1 − t) b2 + a2 , and so |Z0 , Z| + |Z, Z1| = |Z0 , Z1 | ; thus Z ∈ [Z0 , Z1 ]. When t > 1,   |Z0 , Z| = t b2 + a2 , |Z1 , Z| = (t − 1) b2 + a2 , and so |Z0 , Z1 | + |Z1 , Z| = |Z0 , Z|; thus Z1 ∈ [Z0 , Z] and Z = Z0 , Z = Z1 . These combined show that the values of t for which 0 ≤ t ≤ 1 are those for which Z ∈ [Z0 , Z1 ]. (iv) This follows directly from (ii) of the present theorem. It can also be proved as follows. As in the proof of (iii) above, we see that the values of t for which t ≥ 0 are those for which Z ∈ [Z0 , Z1 . COROLLARY. Let Z0 ≡ (x0 , y0 ) and Z1 ≡ (x1 , y1 ) be distinct points. Then the following hold:(i) Z0 Z1 = {Z ≡ (x, y) : x = x0 + t(x1 − x0 ), y = y0 + t(y1 − y0 ), t ∈ R}. (ii) Let ≤l be the natural order on l = Z0 Z1 for which Z0 ≤l Z1 . Let Z2 ≡ (x0 + t2 (x1 − x0 ), y0 + t2 (y1 − y0 )) , Z3 ≡ (x0 + t3 (x1 − x0 ), y0 + t3 (y1 − y0 )) . Then we have t2 ≤ t3 if and only if Z2 ≤l Z3 . (iii) [Z0 , Z1 ] = {Z ≡ (x, y) : x = x0 + t(x1 − x0 ), y = y0 + t(y1 − y0 ), 0 ≤ t ≤ 1}.

92

Cartesian coordinates; applications

Ch. 6

(iv) [Z0 , Z1 = {Z ≡ (x, y) : x = x0 + t(x1 − x0 ), y = y0 + t(y1 − y0 ), t ≥ 0}.

Proof . By 6.3.1, in the above we can take a = −(y1 − y0 ), b = x1 − x0 and the conclusions follow immediately. NOTE. We refer to x = x0 + bt, y = y0 − at, (t ∈ R) in 6.4.1 as parametric equations of the line l, and t as the parameter of the point Z ≡ (x, y).

6.5 PERPENDICULARITY AND PARALLELISM OF LINES 6.5.1 Let l ≡ ax + by + c = 0, m ≡ a1 x + b1 y + c1 = 0. (i) Then l ⊥ m if and only if (ii) Also l m if and only if

aa1 + bb1 = 0.

(6.5.1)

ab1 − a1 b = 0.

(6.5.2)

Proof . (i) Suppose that l ⊥ m. Then l meets m in a unique point which we denote by Z0 . By 6.4.1 Z1 ≡ (x0 + b, y0 − a) is a point of l and similarly Z2 ≡ (x0 + b1 , y0 − a1 ) is a point of m. Now by Pythagoras’ theorem |Z0 , Z1 |2 + |Z0 , Z2 |2 = |Z1 , Z2 |2 and so by 6.1.1 [b2 + (−a)2 ] + [b21 + (−a1 )2 ] = (b − b1 )2 + (a1 − a)2 . This simplifies to (6.5.1). Conversely suppose that (6.5.1) holds. Then we cannot have (6.5.2) as well. For if we did, on multiplying (6.5.1) by a and (6.5.2) by b we would find that a2 a1 + abb1 = 0, −b2 a1 + abb1 = 0, so that (a2 + b2 )a1 = 0, and hence as (a, b) = (0, 0), a1 = 0. Similarly aba1 + b2 b1 = 0, −aba1 + a2 b1 = 0, so that b1 = 0 as well, giving a contradiction. We now search for a point of intersection of l and m, and so consider solving for (x, y) the simultaneous equations ax + by = −c, a1 x + b1 y = −c1 .

Sec. 6.6

Projection and axial symmetry

93

As ab1 − a1 b = 0, by 6.2.1 these will have a unique solution, yielding a point which we shall denote by Z0 ≡ (x0 , y0 ). Then by 6.4.1 l = {Z ≡ (x, y) : x = x0 + bt, y = y0 − at, t ∈ R}, m = {Z ≡ (x, y) : x = x0 + b1t, y = y0 − a1t, t ∈ R}. We choose Z1 ∈ l, Z2 ∈ m as above, and from (6.5.1) find that |Z0 , Z1 |2 + |Z0 , Z2 |2 = |Z1 , Z2 |2 . By 6.4.1 we can conclude that l ⊥ m. (ii) By 6.2.1 the equations ax + by + c = 0, a1 x + b1 y + c1 = 0 have either no solution or more than one if and only if (6.5.2) holds. Alternatively, by (i) above we have l m if and only if there is some (a2 , b2 ) = (0, 0) such that aa2 + bb2 = 0, a1 a2 + b1 b2 = 0. But the equations au + bv = 0, a1 u + b1 v = 0, have a solution (u, v) other than (0, 0) if and only if ab1 − a1 b = 0. Thus (6.5.2) is a condition for l and m to be parallel. COROLLARY . (i) The lines Z1 Z2 and Z3 Z4 are perpendicular if and only if (y2 − y1 )(y4 − y3 ) + (x2 − x1 )(x4 − x3 ) = 0. (ii) These lines are parallel if and only if −(y2 − y1 )(x4 − x3 ) + (y4 − y3 )(x2 − x1 ) = 0.

6.6 PROJECTION AND AXIAL SYMMETRY 6.6.1 Let l ≡ ax + by + c = 0 and Z0 ≡ (x0 , y0 ). Then (i) |Z0 , πl (Z0 )| =

|ax0 + by0 + c| √ . a2 + b2

(ii)  πl (Z0 ) ≡ x0 −

a a2 + b2

(ax0 + by0 + c), y0 −

b a2 + b2

 (ax0 + by0 + c) .

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Cartesian coordinates; applications

Ch. 6

(iii)  sl (Z0 ) ≡ x0 −

2a a2 + b2

(ax0 + by0 + c), y0 −

2b a2 + b2

 (ax0 + by0 + c) .

Proof . Let m be the line such that l ⊥ m and Z0 ∈ m. Then as l ⊥ m, by 6.5.1 we will have m ≡ −bx + ay + c1 = 0 for some c1 , and as Z0 ∈ m we have c1 = bx0 − ay0 . To find the coordinates (x, y) of πl (Z0 ) we need to solve simultaneously the equations ax + by = −c, −bx + ay = −bx0 + ay0 . As for (i) we shall then go on to apply 6.1.1 it is (x − x0 )2 and (y − y0 )2 that we shall actually use, and it is easier to work directly with these. We rewrite the equations as a(x − x0 ) + b(y − y0 ) = −(ax0 + by0 + c), −b(x − x0 ) + a(y − y0 ) = 0. Now on squaring each of these and adding, we find that (a2 + b2 )[(x − x0 )2 + (y − y0 )2 ] = (ax0 + by0 + c)2 . The conclusion (i) now readily follows. For (ii) we solve these equations, obtaining x − x0 = −

a a2 + b2

(ax0 + by0 + c), y − y0 = −

b a2 + b2

(ax0 + by0 + c)).

For (iii) we recall that if sl (Z0 ) ≡ (x1 , y1 ) and πl (Z0 ) ≡ (x2 , y2 ), then as mp(Z0 , sl (Z0 )) = πl (Z0 ) we have x1 + x0 = 2x2 , y1 + y0 = 2y2 . Now x2 and y2 are given by (ii) of the present theorem, and the result follows.

6.6.2 Formula for area of a triangle Let Z1 ≡F (x1 , y1 ), Z2 ≡F (x2 , y2 ) and Z3 ≡F (x3 , y3 ) be non- collinear points. Then the area Δ[Z1 , Z2 , Z3 ] is equal to |δF (Z1 , Z2 , Z3 )| where

δF (Z1 , Z2 , Z3 ) = 12 [x1 (y2 − y3 ) − y1 (x2 − x3 ) + x2 y3 − x3 y2 ] ⎛ ⎞ x1 y1 1 = 12 det ⎝ x2 y2 1 ⎠ . x3 y3 1 Proof . By 6.3.1 Z2 Z3 ≡ −(y3 − y2 )(x − x2 ) + (x3 − x2 )(y − y2 ) = 0, so by 6.6.1 |Z1 , πZ2 Z3 (Z1 )| is equal to |−(y3 −y2 )(x1 −x2 )+(x3 −x2 )(y1 −y2 )| √ . 2 2

Z1

(y3 −y2 ) +(x3 −x2 )

= But Δ[Z1 , Z2 , Z3 ] 1 and 2 |Z2 , Z3 ||Z1 , πZ2 Z3 (Z1 )|, the denominator above is equal to |Z2 , Z3 |. Hence the area is equal to half the numerator.

Z2

πZ2 Z3 (Z1 )

Z3

Figure 6.6. Area of a triangle.

Sec. 6.7

Coordinate treatment of harmonic ranges

95

6.6.3 Inequalities for closed half-planes Let l ≡ ax + by + c = 0. Then the sets {Z ≡ (x, y) : ax + by + c ≤ 0},

(6.6.1)

{Z ≡ (x, y) : ax + by + c ≥ 0},

(6.6.2)

are the closed half-planes with common edge l. Proof . Let Z1 ≡ (x1 , y1 ) be a point not in l, and let sl (Z1 ) = Z2 ≡ (x2 , y2 ). Let Z ≡ (x, y). Then as in 6.3.1, Z ∈ l if and only if |Z, Z1 |2 = |Z, Z2 |2 , and this occurs when (x − x1 )2 + (y − y1 )2 = (x − x2 )2 + (y − y2 )2 , which simplifies to 2(x2 − x1 )x + 2(y2 − y1 )y + x21 + y21 − x22 − y22 = 0. This is an equation for l and so by 6.3.1 there is some j = 0 such that   ax + by + c = j 2(x2 − x1 )x + 2(y2 − y1 )y + x21 + y21 − x22 − y22 . By 4.3.4 the sets   Z ≡ (x, y) : 2(x2 − x1 )x + 2(y2 − y1 )y + x21 + y21 − x22 − y22 ≤ 0 ,   Z ≡ (x, y) : 2(x2 − x1 )x + 2(y2 − y1 )y + x21 + y21 − x22 − y22 ≥ 0 ,

(6.6.3) (6.6.4)

are the closed half-planes with edge l, as they correspond to |Z, Z1 | ≤ |Z, Z2 | and |Z, Z1 | ≥ |Z, Z2 |, respectively. But when j > 0, (6.6.1) and (6.6.3) coincide as do (6.6.2) and (6.6.4), while when j < 0, (6.6.1) and (6.6.4) coincide as do (6.6.2) and (6.6.3).

6.7 COORDINATE TREATMENT OF HARMONIC RANGES 6.7.1 New parametrisation of a line As in 6.4.1, if Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ), Z ≡ (x, y) where x = x1 + t(x2 − x1 ), y = y1 + t(y2 − y1 ), then Z ∈ Z1 Z2 and |Z1 , Z|2 = [t(y2 − y1 )]2 + [t(y2 − y1 )]2 = t 2 |Z1 , Z2 |2 , |Z, Z2 |2 = [(1 − t)(x2 − x1 )]2 + [(1 − t)(y2 − y1 )]2 = (1 − t)2 |Z1 , Z2 |2 ,   |Z1 , Z|  t  = . |Z, Z2 |  1 − t  Accordingly, if we write

t 1−t

= λ where λ = 0 and so have t = |Z1 , Z| = |λ |. |Z, Z2 |

λ 1+λ ,

we have

96

Cartesian coordinates; applications

Ch. 6

Thus Z divides (Z1 , Z2 ) in the ratio |λ | : 1. Changing our notation slightly, if we denote by Z3 ≡ (x3 , y3 ) the point with

λ λ 1 (x2 − x1 ) = x1 + x2 , 1+λ 1+λ 1+λ λ λ 1 (y2 − y1 ) = y1 + y2 , y3 = y1 + 1+λ 1+λ 1+λ x3 = x1 +

then Z3 divides (Z1 , Z2 ) in the ratio |λ | : 1. Consequently if we denote by Z4 ≡ (x4 , y4 ) the point with x4 =

λ λ 1 1 x1 + x2 , y4 = y1 + y2 , 1+λ 1+λ 1+λ 1+λ

where λ = −λ , so that x4 =

λ λ 1 1 x1 − , y4 = y1 − y2 , 1−λ 1−λ 1−λ 1−λ

then Z4 also divides (Z1 , Z2 ) in the ratio |−λ | : 1 = |λ | : 1. t s Now λ = 1−t and if we write −λ = 1−s we have Z4 in the original format, x4 = x1 + s(x2 − x1 ), y4 = y1 + s(y2 − y1 ). Then

s t =− 1−t 1−s

so that s= Thus s − 12 = Hence



s − 12

1 2t

t − 12

−

1 2t

t − 12

.

1 t − 12 t + 14 1 1 = 14 . = 2 2 t − 12 t − 12

  1 1  t − 2 = 4 , (s − 12 )(t − 12 ) = 14 .

(6.7.1)

Then we have three possibilities, (a) (b) (c)

 1 1   t −  < , s − 1  > 1 , 2 2 2   2   s − 1  < 1 , t − 1  > 1 , 2 2  2   2  s − 1  = 1 , t − 1  = 1 . 2 2 2 2

In (a) we have − 12 < t − 12 < 12 and either s− 12 < − 12 or s− 12 > 12 . Hence 0 < t < 1 and either s < 0 or s > 1. It follows that Z3 ∈ [Z1 , Z2 ], Z4 ∈ [Z1 , Z2 ]. The situation in (b) is like that in (a) with the roles of t and s, and so of Z3 and Z4 interchanged.

Sec. 6.7

Coordinate treatment of harmonic ranges

97

In (c) − 12 = t − 12 or t − 12 = 12 , so either t = 0 or t = 1. Similarly either s = 0 or s = 1. We rule out the case of t = 1 as then λ would be undefined, and we rule out the case of s = 1 as then −λ would be undefined. What remains is t = s = 0 and we excluded this by taking λ = 0; it would imply that Z3 = Z4 = Z1 . Thus just one of Z3 , Z4 is in the segment [Z1 , Z2 ] and the other is on the line Z1 Z2 but outside this segment. Hence Z3 and Z4 divide {Z1 , Z2 } internally and externally in the same ratio. We recall that we then call (Z1 , Z2 , Z3 , Z4 ) a harmonic range. We note above that there can be no solution for s if t = 12 ; thus there is no corresponding Z4 when Z3 is the mid-point Z0 of Z1 and Z2 . Similarly there can be no solution for t if s = 12 ; thus there is no corresponding Z3 when Z4 is Z0 .

6.7.2 Interchange of pairs of points If the points Z3 and Z4 divide {Z1 , Z2 } internally and externally in the same ratio, then it turns out that the points Z1 and Z2 also divide {Z3 , Z4 } internally and externally in the same ratio. Proof . For we had

λ λ 1 1 x1 + x2 , y3 = y1 + y2 , 1+λ 1+λ 1+λ 1+λ λ λ 1 1 x4 = x1 − x2 , y4 = y1 − y2 . 1−λ 1−λ 1−λ 1−λ x3 =

Then (1 + λ )x3 = x1 + λ x2 , (1 − λ )x4 = x1 − λ x2 . By addition and subtraction, we find that 1+λ 1−λ x3 + x4 , 2 2 λ −1 1+λ x3 + x4 , x2 = 2λ 2λ x1 =

and by a similar argument, 1+λ 1−λ y3 + y4 , 2 2 λ −1 1+λ y3 + y4 . y2 = 2λ 2λ y1 =

If we define μ by

1+λ 1 = , 1+μ 2

so that

μ=

1−λ , 1+λ

μ 1−λ = , 1+μ 2

98

Cartesian coordinates; applications

then x1 =

μ μ 1 1 x3 + x4 , y1 = y3 + y4 . 1+μ 1+μ 1+μ 1+μ

If we define μ by

1+λ 1 = , 1 + μ 2λ

so that

μ = then x2 =

Ch. 6

λ −1 , 1+λ

μ λ −1 = , 1+μ 2λ

μ μ 1 1 x3 + x4 , y2 = y3 + y4 . 1+μ 1+μ 1+μ 1 + μ

As μ = −μ , this shows that Z1 and Z2 divide {Z3 , Z4 } internally and externally in the same ratio.

6.7.3 Distances from mid-point Let Z0 be the mid-point of distinct points Z1 and Z2 . Then points Z3 , Z4 ∈ Z1 Z2 divide {Z1 , Z2 } internally and externally in the same ratio if and only if Z3 and Z4 are on the one side of Z0 on the line Z1 Z2 and |Z0 , Z3 ||Z0 , Z4 | = 14 |Z1 , Z2 |2 .

Proof . We have Z0 ≡ (x0 , y0 ) where x0 = 12 (x1 + x2 ), y0 = 12 (y1 + y2 ). Then

and so

x3 − x0 = (t − 12 )(x2 − x1 ),

y3 − y0 = (t − 12 )(y2 − y1 ),

x4 − x0 = (s − 12 )(x2 − x1 ),

y4 − y0 = (s − 12 )(y2 − y1 ),

   |Z0 , Z3 ||Z0 , Z4 | = | t − 12 s − 12 ||Z1 , Z2 |2 .

By (6.7.1) if and 1 , Z2 } internally and externally in thesame  {Z  1ratio   Z31,Z4 divide 1 1 1 1 . This is equivalent to having | s − t − = t − | = only if s − 2 4 2 2 4 and   12  1 1 1 s − 2 t − 2 > 0. The latter is equivalent to having either s − 2 > 0 and t − 2 > 0, or s − 12 < 0 and t − 12 < 0, so that Z3 and Z4 are on the one side of Z0 on the line Z1 Z2 .

6.7.4 Distances from end-point Let {Z3 , Z4 } divide {Z1 , Z2 } internally and externally in the same ratio with Z2 ∈ [Z1 , Z4 ]. Then

1 1 1 1 + = . 2 |Z1 , Z3 | |Z1 , Z4 | |Z1 , Z2 |

Proof . We have as before x3 = x1 +

λ λ (x2 − x1 ), y3 = y1 + (y2 − y1 ), 1+λ 1+λ

Sec. 6.7

Coordinate treatment of harmonic ranges x4 = x1 +

99

λ λ (x2 − x1 ), y4 = y1 + (y2 − y1 ), λ −1 λ −1

Now λ /(λ − 1) > 1 and so λ > 1. Hence 12 < λ /(1 + λ ) < 1, and so Z3 ∈ [Z1 , Z2 ]. Thus Z2 , Z3 and Z4 are on the one side of Z1 on the line Z1 Z2 . Then |Z1 , Z4 | λ λ |Z1 , Z3 | = , = , |Z1 , Z2 | λ + 1 |Z1 , Z2 | λ − 1 so that

|Z1 , Z2 | λ + 1 |Z1 , Z2 | λ − 1 , , = = |Z1 , Z3 | λ |Z1 , Z4 | λ

and so

|Z1 , Z2 | |Z1 , Z2 | λ + 1 λ − 1 + = + = 2. |Z1 , Z3 | |Z1 , Z4 | λ λ

Hence 1 2



1 1 + |Z1 , Z3 | |Z1, , Z4 |

=

1 . |Z1 , Z2 |

This is expressed by saying that |Z1 , Z2 | is the harmonic mean of |Z1 , Z3 | and |Z1 , Z4 |.

6.7.5 Construction for a harmonic range W3 l Z1

W2 Z3

Z4

Z2 W1 Figure 6.7.

Let Z1 , Z2 , Z3 be distinct collinear points with Z3 not the mid-point of Z1 and Z2 . Take any points W1 and W2 , not on Z1 Z2 , so that Z2 is the mid-point of W1 and W2 . Let l be the line through Z1 which is parallel to W1W2 and let W3 be the point in which W1 Z3 meets l, with Z4 the point in which W2W3 meets Z1 Z2 . Then (Z1 , Z2 , Z3 , Z4 ) is a harmonic range. Proof. Without loss of generality we may take the x-axis to be the line Z1 Z2 and so take coordinates Z1 ≡ (x1 , 0), Z2 ≡ (x2 , 0), Z3 ≡ (x3 , 0), Z4 ≡ (x4 , 0), and W1 ≡ (u1 , v1 ), W2 ≡ (2x2 − u1 , −v1 ). The lines l and W1 Z3 have equations (u1 − x2 )y = v1 (x − x1 ),

(u1 − x3 )y = v1 (x − x3 ),

100

Cartesian coordinates; applications

Ch. 6

respectively, and so W3 has coordinates x3 (u1 − x2 ) − x1 (u1 − x3 ) , x3 − x2

u3 =

v3 = v1

x3 − x1 . x3 − x2

On forming the equation of W2W3 and finding where it meets Z1 Z2 we obtain x4 =

−x3 (x1 + x2 ) + 2x1 x2 , x1 + x2 − 2x3

from which it follows that (x1 − x2 )(x3 − x1 ) x4 − x1 = , x1 + x2 − 2x3 From these we see that

x2 − x4 =

(x1 − x2 )(x3 − x2 ) . x1 + x2 − 2x3

x3 − x1 x4 − x1 =− . x2 − x4 x2 − x3

Exercises 6.1 Suppose that Z1 , Z2 , Z3 are non-collinear points and Z5 = mp{Z3 , Z1 }, Z6 = mp{Z1 , Z2 }. Show that if |Z2 , Z5 | = |Z3 , Z6 |, then |Z3 , Z1 | = |Z1 , Z2 |. (Hint. Select a frame of reference to simplify the calculations). 6.2 Let l1 , l2 be distinct intersecting lines and Z0 a point not on either of them. Show that there are unique points Z1 ∈ l1 , Z2 ∈ l2 such that Z0 is the mid-point of Z1 and Z2 . 6.3 Suppose that Z1 , Z2 , Z3 are non-collinear points. Show that the points Z ≡ (x, y), the perpendicular distances from which to the lines Z1 Z2 , Z1 Z3 are equal, are those the coordinates of which satisfy −(y2 − y1 )(x − x1 ) + (x2 − x1 )(y − y1 )  (x2 − x1 )2 + (y2 − y1 )2 −(y3 − y1 )(x − x1 ) + (x3 − x1 )(y − y1 )  ± = 0. (x3 − x1 )2 + (y3 − y1 )2 Show that if x = (1 − t)x2 + tx3 , y = (1 − t)y2 + ty3 then Z lies on the line with equation −(y2 − y1 )(x − x1 ) + (x2 − x1 )(y − y1 )  (x2 − x1 )2 + (y2 − y1 )2 −(y3 − y1 )(x − x1 ) + (x3 − x1 )(y − y1 )  + = 0, (x3 − x1 )2 + (y3 − y1 )2 if and only |Z1 , Z2 |2

. |Z1 , Z2 |2 + |Z2 , Z3 |2 Deduce that this latter line is the mid-line of |Z2 Z1 Z3 . t=

Sec. 6.7

Coordinate treatment of harmonic ranges

101

6.4 If the fixed triangle [Z1 , Z2 , Z3 ] is isosceles, with |Z1 , Z2 | = |Z1 , Z3 |, and Z is a variable point on the side [Z2 , Z3 ], show that the sum of the perpendicular distances from Z to the lines Z1 Z2 and Z1 Z3 is constant.[Hint. Select a frame of reference to simplify the calculations.] 6.5 Let [A, B,C, D] be a parallelogram, E = mp{C, D}, F = mp{A, B}, and let AE and CF meet BD at G and H, respectively. Prove that AE CF and |D, G| = |G, H| = |H, B|.

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7

Circles; their basic properties Hitherto our sets have involved lines and half-planes, and specific subsets of these. Now we introduce circles and study their relationships to lines. We do not do this just to admire the circles, and to behold their striking properties of symmetry. They are the means by which we control angles, and simplify our work on them.

7.1 INTERSECTION OF A LINE AND A CIRCLE 7.1.1 Definition . If O is any point of the plane Π and k is any positive real number, we call the set C (O; k) of all points X in Π which are at a distance k from O, i.e. C (O; k) = {X ∈ Π : |O, X| = k}, the circle with centre O and length of radius k. If X ∈ C (O; k) the segment [O, X ] is called a radius of the circle. Any point U such that |O, U| < k is said to be an interior point for this circle. Any point V such that |O, V| > k is said to be an exterior point for this circle. For every circle C (O; k) and line l, one of the following holds:(i) l ∩ C (O; k) = {P} for some point P, in which case every point of l \ {P} is exterior to the circle. (ii) l ∩ C (O; k) = {P, Q} for some points P and Q, with P = Q, in which case every point of [P, Q] \ {P, Q} is interior to the circle, and every point of PQ \ [P, Q] is exterior to the circle. (iii) l ∩ C (O; k) = 0, / in which case every point of l is exterior to the circle.

Proof . Let M = πl (O), and let m be the line which contains M and is perpendicular to l, so that O ∈ m. Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50007-0 © 2016 Elsevier Ltd. All rights reserved.

104

Circles; their basic properties Q

P=M O

Ch. 7

l

M

O

P (i) Suppose that |O, M| = k, so that M is a point of the circle. M We write P = M. Then P ∈ l, P ∈ C (O; k) and OP ⊥ l. Thus if V is l any point of l, other than P, by O 4.3.1 we have |O, V| > |O, P| = k. Hence V is exterior to the cirFigure 7.1. cle, and so there is no point common to l and the circle except P. (ii) Suppose that |O, M| < k, so that M is interior to the circle. Then k2 − |O, M|2 > 0 so that its square rootcan be extracted as a positive real number. By A4 (iv) choose P ∈ l so that |M, P| = k2 − |O, M|2 . There is also a point Q ∈ l on the other side of M from P and such that |M, Q| = |M, P|. Clearly M is the mid-point of P and Q. When M = O, this gives |O, P| = k so that P ∈ C (O; k). By 2.1.3 any point X = P of the half-line [O, P must satisfy either X ∈ [O, P] or P ∈ [O, X ]. If X ∈ [O, P] then by 3.1.2 |O, X| < |O, P| = k, so that X is interior to the circle. On the other hand if P ∈ [O, X ] then |O, X| > |O, P| = k, and so X is an exterior point for the circle. Moreover Q is also on the circle and similar results hold when X ∈ [O, Q . When M = O, we have MP = l, MO = m, so that MP ⊥ MO and then by Pythagoras’ theorem |O, P|2 = |O, M|2 + |M, P|2 = |O, M|2 + [k2 − |O, M|2 ] = k2 ; thus again |O, P| = k, so that P is on the circle. By 2.1.3 any point X = P of the half-line [M, P must satisfy either X ∈ [M, P] or P ∈ [M, X ]. If X ∈ [M, P], then by 3.1.2 |M, X| < |M, P|; when X = M, clearly X an interior point; when X = M, by Pythagoras’ theorem this gives |O, X|2 = |O, M|2 + |M, X|2 < |O, M|2 + |M, P|2 = k2 , so that |O, X| < k and so again X is interior to the circle. If on the other hand P ∈ [M, X ], while still X = P, then by 3.1.2 |M, X| > |M, P|; by Pythagoras’ theorem we have |O, X|2 = |O, M|2 + |M, X|2 > |O, M|2 + |M, P|2 = k2 , so that |O, X| > k and so X is exterior to the circle. Thus the points of [M, P] \ {P} are interior to the circle, and the points of ([M, P ) \ [M, P] are exterior to the circle.

Sec. 7.2

Properties of circles

105

Similar results hold when X ∈ [M, Q , that is the points of [M, Q] \ {Q} are interior to the circle while the points of ([M, Q ) \ [M, Q] are exterior to the circle. But M ∈ [P, Q] so that [P, M] ∪ [M, Q] = [P, Q], [M, P ∪ [M, Q = PQ and so we can take these results together. Thus the points of [P, Q], other than P and Q, are interior to the circle, and the points of PQ \ [P, Q] are exterior to the circle, leaving just the points P and Q of the line l = PQ in the circle. (iii) Suppose that |O, M| > k, so that M is exterior to the circle. Then M ∈ l and OM ⊥ l. If X ∈ l, X = M, then by 4.3.1, |O, X| > |O, M| > k, so that X is exterior to the circle.

Definition . If l is a line such that l ∩ C (O; k) = {P} for a point P, then l is called a tangent to C (O; k) at P, and P is called the point of contact. If l ∩ C (O; k) = {P, Q} for distinct points P and Q, then l is called a secant for C (O; k) and the segment [P, Q] is called a chord of the circle; when O ∈ l = PQ, the chord [P, Q] is called a diameter of the circle; in that case O = mp(P, Q). If l ∩ C (O; k) = 0, / then l is called a non-secant line for the circle. NOTE. By the above every point of a tangent to a circle, other than the point of contact, is an exterior point. If [P, Q] is a chord, every point of the chord other than its end-points P and Q is an interior point, while every point of PQ \ [P, Q] is exterior. Every point of a non-secant line is an exterior point.

7.2 PROPERTIES OF CIRCLES 7.2.1 Circles have the following properties:(i) If [Q, S] is a diameter of the circle C (O; k) and P any point of the circle other than Q and S, then PQ ⊥ PS. (ii) If points P, Q, S are such that PQ ⊥ PS, then P is on a circle with diameter [Q, S]. (iii) If P is any point of the circle C (O; k), [Q, S] is any diameter and U = πQS (P), then U ∈ [Q, S] and |Q, U| ≤ 2k. (iv) If Q is a point of a circle with centre O and l is the tangent to the circle at Q, then every point of the circle lies in the closed half-plane with edge l in which O lies.

106

Circles; their basic properties

P S

O

M Q S

O

Ch. 7

P

P Q

S

O

U

Q

Figure 7.2.

Proof. (i) By 5.2.2 |∠OSP|◦ + |∠SPO|◦ + |∠POS|◦ = 180, |∠OQP|◦ + |∠QPO|◦ + |∠POQ|◦ = 180. But by 4.1.1, and so

|∠OSP|◦ = |∠SPO|◦ , |∠OQP|◦ = |∠QPO|◦ , 2|∠SPO|◦ + 2|∠QPO|◦ + |∠POS|◦ + |∠POQ|◦ = 360.

Now O ∈ [Q, S] so |∠POS|◦ + |∠POQ|◦ = 180, and as [P, O ⊂ I R(|QPS) we have |∠SPO|◦ + |∠OPQ|◦ = |∠QPS|◦ . Thus |∠QPS|◦ = 90. (ii) Let O be the mid-point of Q and S and through O draw the line parallel to PQ. It will meet [P, S] in a point M. Then by 5.3.1 M is the mid-point of P and S. But PQ ⊥ PS and PQ MO so by 5.1.1 MO ⊥ PS. Then [O, P, M] ≡ [O, S, M] by the SAS principle of congruence. It follows that |O, P| = |O, S|. (iii) If P ∈ QR, then by (i) of the present theorem and 4.3.3 U ∈ [Q, S]. If P ∈ QS then U is either Q or S and so U ∈ [Q, S]. Then by 3.1.2 |Q, U| ≤ |Q, S|. But as O = mp(Q, S), by 3.2.1 |Q, O| = 12 |Q, S|, and so |Q, S| = 2k. (iv) Let [Q, S] be the diameter containing Q and H1 the closed half-plane with edge l which contains O. Then by 2.2.3 every point of [Q, O lies in H1 . If P is any point of the circle and U = πQO (P) then by (iii) above U ∈ [Q, S] ⊂ [Q, O so U ∈ H1 . But l ⊥ QS, U P ⊥ QS so U P l. Then by 4.3.2 P ∈ H1 .

7.2.2 Equation of a circle Let Z0 ≡ (x0 , y0 ) and k > 0. Then Z ≡ (x, y) is on C (Z0 ; k) if and only if (x − x0 )2 + (y − y0 )2 = k2 .

Proof . This is immediate by the distance formula in 6.1.1.

7.2.3 Circle through three points Given any three non-collinear points A, B and C, there is a unique circle which passes through them.

Sec. 7.3

Formula for mid-line of an angle-support

107

Proof . Let l and m be the perpendicular bisectors of [B,C] and [C, A], respectively. Then if we had l m we would have l m, m ⊥ CA and so l ⊥ CA by 5.1.1; this would yield BC ⊥ l, CA ⊥ l and so BC CA by 4.2.2(iv). This would make the points A, B,C collinear and so give a contradiction. Thus l must meet m in a unique point, D say. Then by 4.1.1(iii) D is equidistant from B and C as it is on l, and it is equidistant from C and A as it is on m. Thus the circle with centre D and length of radius |D, A| passes through A, B and C. Conversely, suppose that a circle passes through A, B and C. Then by 4.1.1(ii) its centre must be on l and on m and so it must be D. The length of radius then must be |D, A|. COROLLARY. Two distinct circles cannot have more than two points in common.

7.3 FORMULA FOR MID-LINE OF AN ANGLE-SUPPORT 7.3.1 COMMENT. We now start to prepare the ground for our treatment of angles. Earlier on we found that mid-points have a considerable role. Now we shall find that midlines of angle-supports, dealt with in 3.6, have a prominent role as well. Given any angle-support |BAC, if we take any number k > 0 there are unique points P1 and P2 on [A, B and [A,C respectively, such that |A, P1 | = k, |A, P2 | = k. Thus P1 and P2 are the points of [A, B and [A,C on the circle C (A; k). Then |BAC = |P1 AP2 and it is far more convenient to work with the latter form. We first prove a result which will enable us to deal with the mid-lines of angle-supports by means of Cartesian coordinates. With a frame of reference F = ([O, I , [O, J ), let P1 , P2 ∈ C (O; 1) be such that P1 ≡F (a1 , b1 ), P2 ≡F (a2 , b2 ). Then the mid-line l of |P1 OP2 has equation (b1 + b2 )x − (a1 + a2 )y = 0 when P1 and P2 are not diametrically opposite, and equation a1 x + b1 y = 0 when they are.

Proof . When P1 and P2 are not diametrically opposite, their mid-point M is not O and we have l = OM. As M has  1 1 coordinates 2 (a1 + a2 ), 2 (b1 + b2 ) , the line OM has equation (b1 + b2 )x − (a1 + a2 )y = 0. When P1 is diametrically opposite to P2 , l is the line through O which is perpendicular to OP and this has the given equation.

P2 P3

M O

Figure 7.3.

P1 Q

108

Circles; their basic properties

Ch. 7

With the notation of the last result, let Q ≡F (1, 0) and sl (Q) = P3 where P3 ≡F (a3 , b3 ). Then a3 =

(a1 + a2 )2 − (b1 + b2 )2 2(a1 + a2 )(b1 + b2 ) , b3 = , 2 2 (a1 + a2 ) + (b1 + b2 ) (a1 + a2 )2 + (b1 + b2 )2

when P1 and P2 are not diametrically opposite, and a3 = b21 − a21 , b3 = −2a1 b1 , when they are. Proof . For l ≡ ax + by + c = 0, we recall from 6.6.1 that sl (Z0 ) ≡ x0 −

2a 2b (ax + by + c), y − (ax + by + c) . 0 0 0 0 0 a2 + b2 a2 + b2

When P1 and P2 are not diametrically opposite, l ≡ (b1 + b2 )x − (a1 + a2 )y = 0. Thus for it x0 = 1, y0 = 0, a = b1 + b2 , b = −(a1 + a2 ), c = 0 and so a3 =

(a1 + a2 )2 − (b1 + b2 )2 2(a1 + a2 )(b1 + b2 ) , b3 = . 2 2 (a1 + a2 ) + (b1 + b2 ) (a1 + a2 )2 + (b1 + b2 )2

When P1 and P2 are diametrically opposite, l ≡ a1 x + b1 y = 0. Thus for it x0 = 1, y0 = 0, a = a1 , b = b1 , c = 0, so we have a3 = b21 − a21 , b3 = −2a1 b1 , as a21 + b21 = 1.

7.4 POLAR PROPERTIES OF A CIRCLE 7.4.1 Tangents from an exterior point Let P be a point exterior to a circle C . Then two tangents to the circle pass through P. Their points of contact are equidistant from P. Proof . Let the circle have cenT1 tre O and length of radius a. Let |O, P| = b, so that b > a. Choose the point U ∈ [O, P so that x = O P |O, U| = a2 /b. As b > a, then U V x < a < b so U ∈ [O, P]. Erect a perpendicular to OP at U and T2 mark off on it a distance  Figure 7.4.  a 2 . y = |U, T1 | = a 1 − b

Sec. 7.4

Polar properties of a circle

109

By Pythagoras’ theorem, |O, T1 |2 = |O, U|2 + |U, T1 |2 = x2 + y2 =

a4 a4 2 + a − = a2 , b2 b2

so that T1 ∈ C . Let V be the mid-point of O and P, so that V ∈ [O, P] and |O, V| = b2 . Then |U, V| = ±(|O, V| − |O, U|) = ±( 12 b − x). Again by Pythagoras’ theorem, |V, T1 |2 = |U, V|2 + |U, T1 |2 = ( 21 b − x)2 + y2

2

a2 a2 2 1 = 2b− + a 1 − 2 = 14 b2 . b b Thus T1 is on the circle C1 with centre V and radius length 12 b. Note that C1 also passes through O and P. Then ∠OT1 P is an angle in a semi-circle of C1 , so that by 7.2.1 it is a right-angle. Thus by 7.1.1 PT1 is a tangent to C at T1 . By a similar argument, if we take T2 so that U is the mid-point of T1 and T2 , then PT2 is also a tangent to C at T2 . We note that T1 and T2 are both on the line which is perpendicular to OP at the point U . By Pythagoras’ theorem |P, T1 |2 = |O, P|2 − |O, T1 |2 = |O, P|2 − |O, T2 |2 = |P, T2 |2 , and so |P, T1 | = |P, T2 |. There cannot be a third tangent PT3 as then T3 would be on C and C1 , whereas by 7.2.3 these circles have only two points in common.

7.4.2 The power property of a circle For a fixed circle C (O; k) and fixed point P ∈ C (O; k), let a variable line l through P meet C (O; k) at R and S. Then the product of distances |P, R||P, S| is constant. When P is exterior to the circle, |P, R||P, S| = |P, T1 |2 , where T1 is the point of contact of a tangent from P to the circle. Proof . By the distance formula Z ≡ (x, y) is on C (O; k) if and only if x2 + y2 − 2 k = 0. If P ≡ (x0 , y0 ) and l has Cartesian equation ax + by + c = 0, by 6.4.1 points Z on l have parametric equations of the form x = x0 + bt, y = y0 − at (t ∈ R). Now l also has Cartesian equation √

b c a x+ √ y+ √ = 0. a2 + b2 a2 + b2 a2 + b2

110

Circles; their basic properties

Ch. 7

√ √ Thus as we we may replace a and b by a/ a2 + b2 and b/ a2 + b2 , without loss of generality we may assume that a2 + b2 = 1. Then the point Z on the line lies on the circle if (x0 + bt)2 + (y0 − at)2 − k2 = 0, that is if t 2 + 2(bx0 − ay0 )t + x20 + y20 − k2 = 0. If t1 ,t2 are the roots of this equation, then t1t2 = x20 + y20 − k2 . As for R and S we have x1 = x0 + bt1 , y1 = y0 − at1 , x2 = x0 + bt2 , y2 = y0 − at2 , so |P, R| = |t1 |, |P, S| = |t2 |. Thus |P, R||P, S| = |t1 t2 | = |x20 + y20 − k2 |, which is constant. When P is exterior to the circle, the roots of the quadratic equation are equal if (bx0 − ay0 )2 = x20 + y20 − k2 , and the repeated root is given by t = −(bx0 − ay0 ). Then for a point of intersection T1 of the line and circle, we have for the coordinates of T1 x = x0 − (bx0 − ay0 )b,

y = y0 + (bx0 − ay0 )a.

Hence |P, T1 |2 = (x − x0 )2 + (y − y0 )2 = (bx0 − ay0 )2 = x20 + y20 − k2 = |O, P|2 − k2 . It is also easy to give a synthetic proof as follows. We first take P interior to the circle. Let M be the mid-point of R and S so that M is the foot of the perpendicular from O to RS. Then P is in either [R, M] or [M, S]; we suppose that P ∈ [R, M]. S

T1 M

R P

R

M

O

O

P

S Figure 7.5.

Then |P, R||P, S| = (|M, R| − |P, M|) (|M, S| + |P, M|) = |M, R|2 − |P, M|2     = |M, R|2 − |P, O|2 − |O, M|2 = |M, R|2 + |O, M|2 − |P, O|2 = |O, R|2 − |P, O|2 = k2 − |P, O|2 ,

Sec. 7.4

Polar properties of a circle

111

and this is fixed. We continue with the case where P is exterior to the circle, and may suppose that |P, R| < |P, S|, as otherwise we can just interchange the points R and S. As P is outside the circle, by 7.1.1 it is outside the segment [R, S] on the line RS. Then we have |P, R||P, S| = (|P, M| − |M, R|) (|P, M| + |M, S|) = |P, M|2 − |M, R|2     = |P, O|2 − |O, M|2 − |M, R|2 = |P, O|2 − |O, M|2 + |M, R|2 = |P, O|2 − |O, R|2 = |P, O|2 − |O, T1 |2 = |P, T1 |2 .

7.4.3 A harmonic range Let T1 and T2 be the points of contact of the tangents from an exterior point P to a circle C with centre O. If a line l through P cuts C in the points R and S, and cuts T1 T2 in Q, then P and Q divide {R, S} internally and externally in the same ratio. Proof . We use the notation of S T1 7.4.1 and first recall that T1 T2 M cuts OP at right-angles at a point Q U . Then, by 7.2.1(ii), the circle R P C2 on [O, Q] as a diameter passes U O through U . We let M be the midpoint of R and S; then by 4.1.1 T2 OM ⊥ MQ, and so M also lies Figure 7.6. on the circle C2 . We have |P, R||P, S| = |P, T1 |2 by 7.4.2, |P, T1 |2 = |P, U||P, O| by the proof of Pythagoras’ theorem in 5.4.1, and |P, U||P, O| = |P, Q||P, M|, by the 7.4.2 applied to the circle C2 . On combining these we have |P, R||P, S| = |P, Q||P, M|. We cannot have l ⊥ OP as that would make l T1 T2 whereas l meets T1 T2 . Then, with the notation of 7.4.1, l is not a tangent to C1 at P so, by 7.1.1 l must meet C1 at a point H. We are supposing that l is not either of PT1 , PT2 and so H is not T1 or T2 . We let K be the foot of the perpendicular from H to OP. Then by 4.3.3 K ∈ [P, O] and by the proof of Pythagoras’ theorem in 5.4.1 |P, H|2 = |P, K||P, O|. If we had K ∈ [P,U ] we would have |P, K| < |P, U| and so |P, H|2 = |P, O||P, K| < |P, O||P, U| = |P, T1 |2 . From this it would follow that |O, H|2 = |O, P|2 − |P, H|2 > |O, P|2 − |P, T1 |2 = a2 , and make H exterior to the circle. But H is the foot of the perpendicular from O to l, and by 7.1.1 this would cause l to have no point in common with the circle. This cannot occur and so we must have K ∈ [O,U ]. By a similar argument it then follows that H is interior to the circle C and so l meets C in two points R and S. By 7.2.1(iv) every point of the circle C is in the closed half-plane H1 with edge PT1 and which contains O. By 2.2.3 H1 contains U ∈ [P, O and then it also contains

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T2 ∈ [T1 ,U . Similarly every point of C is also in the closed half-plane with edge PT2 and which contains T1 . It follows that every point of C lies in the interior region I R(|T1 PT2 ). Now every point of [P, R is in this interior region and so Q is. It follows that Q ∈ [T1 , T2 ] and so by 7.1.1 is interior to the circle; we thus must have Q ∈ [R, S] by 7.1.1 again. We let x = |P, R|, y = |P, S|, z = |P, Q|, and without loss of generality assume |P, R| < |P, S| so that x < y. As P is outside the circle it is outside the segment [R, S]; as Q is on the segment [R, S], it follows that 0 < x < z < y. Then in turn 2 1 1 1 1 1 1 1 = + , − = − , xy = (x + y)z, 2 z x y x z z y z−x y−z x y |P, R| |P, S| = , = , = . xz zy z−x y−z |R, Q| |S, Q| In the above we have assumed that l is not the line OP. When it is we have a simple case; l cuts the circle in points R1 , S1 such that [R1 , S1 ] is a diameter. Then taking |P, R1 | < |P, S1 |, with the notation of 7.4.1 we have that |S1 , P| b + a |S1 , U| a + a2 /b = , = , |P, R1 | b − a |U, R1 | a − a2 /b and these are equal.

7.5 ANGLES STANDING ON ARCS OF CIRCLES 7.5.1 Let P, Q, R, S be points of a circle C (O; k) such that R and S are on one side of the line PQ. Then |∠PRQ|◦ = |∠PSQ|◦ . (i)When O ∈ PQ, |∠PRQ|◦ = 90. (ii)When O ∈ PQ and R is on the same side of PQ as O is, then |∠PRQ|◦ = 1 ◦ 2 |∠POQ| . (iii) When O ∈ PQ and R is on the opposite side of PQ from O, then |∠PRQ|◦ is equal to half of the degree-measure of the reflex-angle with support |POQ. Proof . We are given that R ∈ PQ, S ∈ PQ. Now S ∈ QR as by 7.1.1 a line cannot meet the circle in more than two points; for this reason also S ∈ RP. Thus S cannot be on a side-line of the triangle [P, Q, R]. Moreover, neither can S be in [P, Q, R] but not on a side, as then by the cross-bar theorem we would have S ∈ [P,V ] for some point V in [Q, R] but not at an end-point. Then V would be interior to the circle and P is on the circle, so by 7.1.1 every point of [P,V ], other than P, is interior to the circle; this would make S interior to the circle whereas it is on it. Thus as S ∈ [P, Q, R] we must have at least one of the following :(a) S is on the opposite side of QR from P, (b) S is on the opposite side of RP from Q, (c) S is on the opposite side of PQ from R,

Sec. 7.5

Angles standing on arcs of circles

113

and of course (c) is ruled out by assumption. R R

S O

U

P

Q

O

P

V

Q

R P

V

Q

Q

O

P

R

W S Figure 7.7.

We suppose that (a) holds as in the first figure; the case of (b) is treated similarly. Then there is a point U ∈ [P, S] ∩ QR. As U ∈ [P, S], U must be an interior point for the circle and hence we must have U ∈ [Q, R]. By 7.4.2 |U, P| |U, R| = , |U, Q| |U, S| and we also have |∠PUR|◦ = |∠QUS|◦ as these are opposite angles. By 5.3.2, the triangles [U, P, R], [U, Q, S] are similar. In particular |∠PRU|◦ = |∠QSU|◦ . The first diagram in Fig. 7.7 deals with this general case. In (i) when O ∈ PQ, that we have a right-angle comes from 7.2.1 and there is a diagram for this in Fig. 7.2. When O ∈ PQ we let V be the mid-point of {P, Q} and for the second case (ii), as in the second diagram in Fig. 7.7, we take R to be point in which [V, O meets the circle. Then by 5.2.2, Corollary, and 4.1.1(i) |∠VOP|◦ = 2|∠VRP|◦ , |∠VOQ|◦ = 2|∠VRQ|◦ . But [R,V ⊂ I R(|PRQ) so that |∠VRP|◦ + |∠VRQ|◦ = |∠PRQ|◦ . Moreover [O,V ⊂ I R(|POQ) so that |∠VOP|◦ + |∠VOQ|◦ = |∠POQ|◦ . By addition we then have that |∠POQ|◦ = 2|∠PRQ|◦ . For the third case (iii), as in the third diagram in Fig. 7.7, we take R to be point in which [O,V meets the circle and W = O a point such that O ∈ [V,W ]. Then by 5.2.2, Corollary, and 4.1.1(i) |∠WOP|◦ = 2|∠WRP|◦ , |∠WOQ|◦ = 2|∠WRQ|◦ . But [R,W ⊂ I R(|PRQ) so that |∠WRP|◦ + |∠WRQ|◦ = |∠PRQ|◦ . Moreover [O,V ⊂ I R(|POQ) so that by 3.7.1 |∠WOP|◦ + |∠WOQ|◦ is equal to the degree-measure

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of the reflex- angle with support |∠POQ. By addition we then have that the degreemeasure of this reflex-angle is equal to 2|∠PRQ|◦ .

Definition . If the vertices of a convex quadrilateral all lie on some circle, then the quadrilateral is said to be cyclic. COROLLARY. Let [P, Q, R, S] be a convex cyclic quadrilateral. Then the sum of the degree-measures of a pair of opposite angles is 180. Proof . Using the fourth diagram in Fig. 7.7, we first we note that |∠RPQ|◦ + |∠PQR|◦ + |∠QRP|◦ = 180. Next as [S, Q ⊂ I R(|PSR), we have |∠PSR|◦ = |∠PSQ|◦ + |∠QSR|◦ . But |∠PSQ|◦ = |∠QRP|◦ , |∠QSR|◦ = |∠RPQ|◦ . Hence |∠PSR|◦ + |∠PQR|◦ = |∠RPQ|◦ + |∠PQR|◦ + |∠QRP|◦ = 180.

7.5.2 Minor and major arcs of a circle Definition . Let P1 , P2 ∈ C (O; k) H6 be distinct points such that P P 1 P2 O ∈ P1 P2 . Let H5 , H6 be the H5 W closed half-planes with edge P1 P2 , with O ∈ H5 . Then O C (O; k) ∩ H5 , C (O; k) ∩ H6 , are called, respectively, the maFigure 7.8. jor and minor arcs of C (O; k) with end-points P1 and P2 . The point P ∈ C (O; k) is in the minor arc with end-points P1 , P2 if and only if [O, P] ∩ [P1, P2 ] = 0. / Proof . Let P be in the minor arc. Then O ∈ H5 , P ∈ H6 so [O, P] meets P1 P2 in some point W . As W ∈ [O, P] we have |O, W| ≤ k so by 7.1.1 W ∈ [P1 , P2 ]. Conversely suppose that W ∈ [P1 , P2 ] so that |O, W| ≤ k. Choose P ∈ [O,W so that |O, P| = k. Then as |O, W| ≤ |O, P| we have W ∈ [O, P] so that as O ∈ H5 we have P ∈ H6 .

7.6 SENSED DISTANCES 7.6.1 Sensed distance Definition . If l is a line, ≤l is a natural order on l and Z1 , Z2 ∈ l, then we define Z1 Z2≤l by  |Z1 , Z2 |, if Z1 ≤l Z2 , Z1 Z2≤l = −|Z1 , Z2 |, if Z2 ≤l Z1 , and call this the sensed distance from Z1 to Z2 . In knowing this rather than just the distance from Z1 to Z2 we have extra information which can be turned to good

Sec. 7.6

Sensed distances

115

account. It can have negative as well as positive and zero values and it is related to the distance as Z1 Z2≤l = ±|Z1 , Z2 | or equivalently |Z1 Z2≤l | = |Z1 , Z2 |. We note immediately the properties: Z1 Z1≤l = 0,

(7.6.1)

Z2 Z1≤l = −Z1 Z2≤l ,

(7.6.2)

in all cases. We can add sensed distances on a line and have the striking property that Z1 Z2≤l + Z2 Z3≤l = Z1 Z3≤l ,

(7.6.3)

for all Z1 , Z2 , Z3 ∈ l. This is easily seen to hold by (7.6.1) when any two of the three points coincide, as e.g. when Z1 = Z2 it amounts to 0 + Z1 Z3≤l = Z1 Z3≤l . Suppose then that Z1 , Z2 , Z3 are all distinct and suppose first that Z1 ≤l Z2 . Then by 2.1.3 we have one of the cases (a) Z3 ≤l Z1 ≤l Z2 ,

(b) Z1 ≤l Z3 ≤l Z2 ,

(c) Z1 ≤l Z2 ≤l Z3 .

In (a) we have Z1 Z2≤l = |Z1 , Z2 |, Z2 Z3≤l = −|Z2 , Z3 |, Z1 Z3≤l = −|Z1 , Z3 |, and as Z1 ∈ [Z3 , Z2 ], |Z3 , Z1 | + |Z1 , Z2 | = |Z3 , Z2 |, which is −Z1 Z3≤l + Z1 Z2≤l = −Z2 Z3≤l . In (b) we have Z1 Z2≤l = |Z1 , Z2 |, Z2 Z3≤l = −|Z2 , Z3 |, Z1 Z3≤l = |Z1 , Z3 |, and as Z3 ∈ [Z1 , Z2 ], |Z1 , Z3 | + |Z3 , Z2 | = |Z1 , Z2 |, which is Z1 Z3≤l − Z2 Z3≤l = Z1 Z2≤l . In (c) we have Z1 Z2≤l = |Z1 , Z2 |, Z2 Z3≤l = |Z2 , Z3 |, Z1 Z3≤l = |Z1 , Z3 |, and as Z2 ∈ [Z1 , Z3 ], |Z1 , Z2 | + |Z2 , Z3 | = |Z1 , Z3 |, which is Z1 Z2≤l + Z2 Z3≤l = Z1 Z3≤l . Next suppose that Z2 ≤l Z1 . Then on interchanging Z1 and Z2 in the cases just proved we have Z2 Z1≤l + Z1 Z3≤l = Z2 Z3≤l , for all Z3 ∈ l and by (7.6.2) this gives −Z1 Z2≤l + Z1 Z3≤l = Z2 Z3≤l . This completes the proof of (7.6.3) which shows that addition of sensed distances on a line is much simpler than addition of distances. We next relate sensed distances to the parametric equations of l noted in 6.4.1,Corollary. Suppose that W0 ≡ (u0 , v0 ), W1 ≡ (u1 , v1 ) are distinct points on l and that W0 ≤l W1 . Then for points Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ) on l we have x1 = u0 + s1 (u1 − u0 ), y1 = v0 + s1 (v1 − v0 ), x2 = u0 + s2 (u1 − u0 ), y2 = v0 + s2 (v1 − v0 ), and we recall that Z1 ≤l Z2 if and only if s1 ≤ s2 . Moreover, by the distance formula |Z1 , Z2 | = |s2 − s1 ||W0 , W1 |.

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From these we conclude that Z1 Z2≤l = (s2 − s1 )|W0 , W1 |.

(7.6.4)

In particular the simplest case of parametric representation in relation to sensed distances is when we additionally take |W0 , W1 | = 1 as we then have Z1 Z2≤l = s2 − s1 . When we consider the reciprocal natural order on l we note that Z1 Z2≥l = −Z1 Z2≤l , so that changing to the reciprocal natural order multiplies the value by −1. As well as adding sensed distances on one line we can multiply or divide them. Now for Z1 , Z2 , Z3 , Z4 in l, Z3 Z4≥l Z1 Z2≥l = −Z3 Z4≤l (−1)Z1 Z2≤l = Z3 Z4≤l Z1 Z2≤l , so this sensed product is independent of which natural order is taken. Similarly, when Z1 = Z2 , we can take a ratio of sensed distances −Z3 Z4≥l Z3 Z4≤l Z3 Z4≥l s4 − s3 = = = , s2 − s1 Z1 Z2≤l −Z1 Z2≥l Z1 Z2≥l and see that this sensed ratio is independent of whichever of ≤l , ≥l is used. When the line l is understood, we can relax our notation to Z3 Z4 Z1 Z2 and ZZ3 ZZ4 for these 1 2 products and ratios. If for Z1 , Z2 , Z ∈ l we take the parametric equations x = x1 + t(x2 − x1 ), y = y1 + t(y2 − y1 ), (t ∈ R), then we have that x = u0 + [s1 + t(s2 − s1 )](u1 − u0 ), y = v0 + [s1 + t(s2 − s1 )](v1 − v0 ), and by (7.6.4) we have that Z1 Z ≤l = t(s2 − s1 )|W0 , W1 |, ZZ2≤l = (1 − t)(s2 − s1 )|W0 , W1 |, and so t Z1 Z = . ZZ2 1 − t

(7.6.5)

Our main utilisation of these concepts is through sensed ratios; for example (Z1 , Z2 , Z3 , Z4 ) is a harmonic range when Z1 Z3 /Z3 Z2 = −Z1 Z4 /Z4 Z2 . It is convenient to defer the details until Chapter 11. However we make one use of sensed products in the next subsection.

Sec. 7.6

Sensed distances

117

7.6.2 Sensed products and a circle The conclusion of 7.4.2 can be strengthened to replace |P, R||P, S| by PR PS. In fact the initial analytic proof gives this but it also easily follows from the stated result as PR PS = −|P, R||P, S| when P is interior to the circle while PR PS = |P, R||P, S| when P is exterior to the circle. We now look to a converse type of result. Suppose that Z1 , Z2 and Z3 are fixed non-collinear points. For a variable point W let Z1W meet Z2 Z3 at W and W W W Z1 = W Z2 W Z3 . Then W lies on the circle which passes through Z1 , Z2 and Z3 . Z1 Proof. Without loss of generality we may take our frame of reference so that Z1 ≡ (0, y1 ), Z2 ≡ W (x2 , 0), Z3 ≡ (x3 , 0), and we Z2 Z3 take W ≡ (u, v), W ≡ (u , 0). W Figure 7.9.

u

= y1 u/(y1 − v), and so, first of all,



y1 u y1 u W Z2 W Z3 = x2 − x3 − . y1 − v y1 − v

Then it is easily found that

The line W W has parametric equations x = u + s(u− u ), y = 0+ s(v− 0), with s = 0 giving W and s = 1 giving W . Thus W W = |W , W|. The point Z1 has parameter given by y1 = sv and so s = y1 /v; then W Z1 = yv1 |W , W|. It follows that  

2 y u y y 2 1 1 1 u− W W W Z1 = |W , W| = + v2 v v y1 − v  

2 u +1 . = y1 v y1 − v On equating the two expressions we have  



2 y1 u y1 u u x3 − = y1 v +1 , x2 − y1 − v y1 − v y1 − v which we re-write as y1 (x2 + x3 )u y1 u2 (y1 − v) = y1 v − x2 x3 + . 2 (y1 − v) y1 − v On multiplying across by y1 − v we obtain y1 (u2 + v2 ) − y1 (x2 + x3 )u − (y21 + x2 x3 )v + y1 x2 x3 = 0, and this is the equation of a circle.

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7.6.3 Radical axis and coaxal circles In 7.4.2 our proof showed that if a line through the point Z meets the circle C (Z1 , k1 ) at the points R and S then ZR ZS = |Z1 , Z|2 − k12 depends only on the circle and the point Z. We call this expression the power of the point Z with respect to this circle. To cater for degenerate cases, when k1 = 0 we also call |Z1 , Z|2 the power of Z with respect to the point Z1 . We let C1 denote either C (Z1 , k1 ) or Z1 and similarly consider C2 which is either C (Z2 , k2 ) or Z2 . We ask for what points Z its powers with respect to C1 and C2 are equal. This occurs when |Z1 , Z|2 − k12 = |Z2 , Z|2 − k22 which simplifies to 2(x2 − x1 )x + 2(y2 − y1 )y + x21 + y21 − k12 + x22 + y22 − k22 = 0. If Z1 = Z2 this is the equation of a line which is called the radical axis of C1 and C2 . It is always perpendicular to the line Z1 Z2 and it passes through any points which C1 and C2 have in common. More generally we also ask for what points Z its powers with respect to C1 and C2 have a constant ratio. For a real number λ which is not equal to 1 we consider when   (7.6.6) |Z1 , Z|2 − k12 = λ |Z2 , Z|2 − k22 .   When λ = 0 this yields C1 and by considering μ |Z1 , Z|2 − k12 = |Z2 , Z|2 − k22 as well, we also include C2 . Now (7.6.6) expands to x1 − λ x2 y1 − λ y2 x2 + y21 − k12 − λ (x22 + y22 − k22 ) x−2 y+ 1 = 0, 1−λ 1−λ 1−λ and on completing the squares in both x and y it becomes     y1 − λ y2 2 x1 − λ x2 2 + y− x− 1−λ 1−λ  2    1 = k1 + (x1 − x2 )2 + (y1 − y2 )2 − k12 − k22 λ + k22 λ 2 . 2 (1 − λ ) x2 + y2 − 2

This quadratic expression in λ is postive when |λ | is large, so it has either a positive minimum, or its minimum is 0 attained at λ1 , say, or it has a negative minimum and so has the value 0 at λ2 and λ3 , say, where λ2 < λ3 . In the first of these cases (7.6.6) always represents a circle and in the second case it represents a circle for all λ = λ1 and a point for λ = λ1 . In the third case it represents a circle when either λ < λ2 or λ > λ3 , it represents a point when either λ = λ2 or λ = λ3 , and it represents an empty locus when λ2 < λ < λ3 . Thus it is the equation of a circle, a point or an empty locus. Suppose that we consider two of these loci, corresponding to the values λ4 and λ5 of λ . They will then have equations x1 − λ4 x2 y1 − λ4 y2 x2 + y21 − k12 − λ4 (x22 + y22 − k22 ) x−2 y+ 1 1 − λ4 1 − λ4 1 − λ4 2 2 2 x1 − λ5 x2 y1 − λ5 y2 x + y1 − k1 − λ5 (x22 + y22 − k22 ) x−2 y+ 1 x2 + y2 − 2 1 − λ5 1 − λ5 1 − λ5 x2 + y2 − 2

= 0, = 0.

Sec. 7.6

Sensed distances

119

On subtracting the second of these from the first, and simplifying, we find that their radical axis is the line with equation   λ4 − λ5 2(x2 − x1 )x + 2(y2 − y1 )y + x21 + y21 − k12 + x22 + y22 − k22 (1 − λ4 )(1 − λ5 ) =0. As we can cancel the initial fraction we see that these loci have the same radical axis as did the original pair C1 and C2 . For this reason all the loci considered are said to be coaxal.

Exercises 7.1 Prove that a circle cannot have more than one centre. [Hint. If O and O1 are both centres, consider the intersection of OO1 with the circle.] 7.2 Give an alternative proof of 7.2.1(iv) by showing that if (x − k)2 + y2 = k2 , where k > 0, then x ≥ 0. 7.3 Prove that if the point X is interior to the circle (O; k), l is a line containing X , and M = πl (X ), then M is also an interior point of this circle. Deduce that l is a secant line. Show too that if Y is also interior to this circle, then every point of the segment [X ,Y ] is also interior. 7.4 Show that if A, B,C are non-collinear points, there is a unique circle to which the side-lines BC,CA, AB are all tangents. 7.5 Let Z1 ≡ (x1 , 0), Z2 ≡ (x2 , 0) and Z3 ≡ (x3 , 0) be distinct fixed collinear points and Z3 not the mid-point of Z1 and Z2 . For W ∈ Z1 Z2 let l be the mid-line of |Z2 WZ1 . Find the locus of W such that either l, or the line through W perpendicular to l, passes through Z3 . 7.6 Show that the locus of mid-points of chords of a circle on parallel lines is a diameter. 7.7 Show that if two tangents to a circle are parallel, then their points of contact are at the end-points of a diameter. 7.8 Show that if each of the side-lines of a rectangle is a tangent to a given circle, then it must be a square. 7.9 Consider the circle C (O; a) and point Z1 ≡ (x1 , 0) where x1 > a > 0, so that Z1 is an exterior point which lies on the diametral line AB, where A ≡ (a, 0), B ≡ (−a, 0). Show that for all points Z ≡ (x, y) on the circle, |Z1 , A| ≤ |Z1 , Z| ≤ |Z1 , B|.

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7.10 For 0 < a < b, suppose that A ≡ (0, a), B ≡ (0, b). Show that the circles C (A; a) and C (B; b) both have the axis OI as a tangent at the point O, and that A ∈ [O, B]. Verify that every point of C (A; a), other than O, is an interior point for C (B; b). [Hint. Consider the equations of C (A; a) and C (B; b).] 7.11 Use Ex.6.3 to establish the equation of the mid-line l in 7.3.1 when P1 and P2 are not diametrically opposite.

8

Translations; axial symmetries; isometries COMMENT. In this chapter we introduce translations and develop them and axial symmetries. These will be useful in later chapters. It is more convenient to frame our proofs for isometries generally.

8.1 TRANSLATIONS AND AXIAL SYMMETRIES 8.1.1 Definition . Given points Z1 , Z2 ∈ Π, we define a translation tZ1 ,Z2 to be a function tZ1 ,Z2 : Π → Π such that, for all Z ∈ Π, tZ1 ,Z2 (Z) = W where mp(Z1 ,W ) = mp(Z2 , Z).

Z

Z1 T W

Z2

Figure 8.1.

Translations have the following properties:(i) If Z1 ≡ (x1 , y1 ) Z2 ≡ (x2 , y2 ) Z ≡ (x, y), W ≡ (u, v), then tZ1 ,Z2 (Z) = W if and only if u = x + x2 − x1 , v = y + y2 − y1 . (ii) In all cases |tZ1 ,Z2 (Z3 ), tZ1 ,Z2 (Z4 )| = |Z3 , Z4 |, so that each translation preserves all distances. (iii) For each W ∈ Π the equation tZ1 ,Z2 (Z) = W has a solution in Z, so that each translation is an onto function. (iv) Each translation tZ1 ,Z2 has an inverse function tZ−1 = tZ2 ,Z1 . 1 ,Z2 (v) The translation tZ1 ,Z1 is the identity function on Π. Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50008-2 © 2016 Elsevier Ltd. All rights reserved.

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(vi) If Z1 = Z2 , Z ∈ Z1 Z2 and W = tZ1 ,Z2 (Z), then [Z1 , Z2 ,W, Z] is a parallelogram.

Proof . (i) For mp(Z1 ,W ) ≡





1 1 1 1 (x1 + u), (y1 + v) , mp(Z2 , Z) ≡ (x2 + x), (y2 + y) 2 2 2 2

and these are equal if and only if u = x + x2 − x1 , v = y + y2 − y1 . (ii) For if W3 = tZ1 ,Z2 (Z3 ), W4 = tZ1 ,Z2 (Z4 ), then u4 − u3 = x4 + x2 − x1 − (x3 + x2 − x1 ) = x4 − x3 , v4 − v3 = y4 + y2 − y1 − (y3 + y2 − y1 ) = y4 − y3 . It follows that |Z3 , Z4 | = |W3 , W4 |. (iii) By (i) the equation tZ1 ,Z2 (Z) = W has the solution given by x = u+x1 −x2 , y = v + y1 − y2 . (iv) By (ii) and (iii) the equation tZ1 ,Z2 (Z) = W has a unique solution and this is denoted by Z = tZ−1 (W ). The correspondence from Π to Π given by W → Z is the 1 ,Z2 inverse of tZ1 ,Z2 and is a function. As by the proof of (iii) x = u + x1 − x2 , y = v + y1 − y2 , by (i) this inverse function is tZ2 ,Z1 . (v) For if Z1 = Z2 , in (i) we have u = x + x1 − x1 = x, v = y + y1 − y1 = y. (vi) We denote by T the common mid-point mp(Z1 ,W ) = mp(Z2 , Z). First we note that W = Z, as mp(Z1 , Z) =mp(Z2 , Z) would imply Z1 = Z2 . As Z ∈ Z1 Z2 , we have T ∈ Z1 Z2 and hence W ∈ Z1 Z2 . It follows that T ∈ ZW as otherwise we would have Z1 ∈ ZW, Z2 ∈ ZW and so Z ∈ Z1 Z2 . The triangles [Z1 , T, Z2 ], [W, T, Z] are congruent in the correspondence (Z1 , T, Z2 ) → (W, T, Z) by the SAS-principle. Hence the alternate angles ∠W Z1 Z2 , ∠Z1W Z have equal degree-measures and so Z1 Z2 W Z. Similarly Z1 Z Z2W . Axial symmetries have the following properties:(i) In all cases |sl (Z3 ), sl (Z4 )| = |Z3 , Z4 |, so that each axial symmetry preserves all distances. (ii) Each axial symmetry sl has an inverse function s−1 l = sl .

Proof . (i) We note that by 6.6.1, sl (Z3 )

1 1 2 2 2 2 [(b − a )x3 − 2aby3 − 2ac], 2 [−2abx3 − (b − a )y3 − 2bc] , ≡ 2 a + b2 a + b2 sl (Z4 )

1 1 2 2 2 2 [(b − a )x4 − 2aby4 − 2ac], 2 [−2abx4 − (b − a )y4 − 2bc] , ≡ 2 a + b2 a + b2

Sec. 8.2

Isometries

123

and thus |sl (Z3 ), sl (Z4 )|2 =

1



[(b2 − a2 )(x3 − x4 ) − 2ab(y3 − y4 )]2  +[−2ab(x3 − x4 ) − (b2 − a2 )(y3 − y4 )]2  2 1 [(b − a2 )2 + 4a2 b2 ][(x3 − x4 )2 + (y3 − y4 )2 ] = 2 2 2 (a + b )  +[−4ab + 4ab](b2 − a2 )(x3 − x4 )(y3 − y4 ) (a2 + b2 )2

= (x3 − x4 )2 + (y3 − y4 )2 . (ii) For if m is the line through Z which is perpendicular to l and W = sl (Z), then W ∈ m and so πl (W ) = πl (Z). Then πl (W ) = mp(W, Z) so Z = sl (W ). This shows that the function sl is its own inverse.

8.2 ISOMETRIES 8.2.1 Definition A function f : Π → Π which satisfies |Z1 , Z2 | = |f (Z1 ), f (Z2 )| for all points Z1 , Z2 ∈ Π, is called an isometry of Π. Each translation and each axial symmetry is an isometry. Proof . This follows from 8.1.1. Each isometry f has the following properties:(i) The function f : Π → Π is one-one. (ii) For all Z1 , Z2 ∈ Π, f ([Z1 , Z2 ]) = [ f (Z1 ), f (Z2 )], so that each segment is mapped onto a segment, with the end-points corresponding. (iii) For all distinct points Z1 , Z2 ∈ Π, f ([Z1 , Z2 ) = [ f (Z1 ), f (Z2 ) , so that each half-line is mapped onto a half-line, with the initial points corresponding. (iv) For all distinct points Z1 , Z2 ∈ Π, f (Z1 Z2 ) = f (Z1 ) f (Z2 ), so that each line is mapped onto a line. If f (Z) ∈ f (Z1 ) f (Z2 ) then Z ∈ Z1 Z2 . (v) If Z1 , Z2 , Z3 are noncollinear points, then f ([Z1 , Z2 , Z3 ]) ≡ [ f (Z1 ), f (Z2 ), f (Z3 )]. (vi) Let Z3 ∈ l and H1 , H2 be the closed half-planes with common edge l, with Z3 ∈ H1 . Let H3 , H4 be the closed half-planes with common edge f (l), with f (Z3 ) ∈ H3 . Then f (H1 ) ⊂ H3 , f (H2 ) ⊂ H4 . (vii) The function f : Π → Π is onto. (viii) In (vi), f (H1 ) = H3 , f (H2 ) = H4 .

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(ix) If l and m are intersecting lines, then f (l) and f (m) are intersecting lines. If l and m are parallel lines, then f (l) and f (m) are parallel lines. (x) If the points Z1 , Z2 and Z3 are non-collinear, then |∠Z2 Z1 Z3 |◦ = |∠f (Z2 )f (Z1 )f (Z3 )|◦ . (xi) If l and m are perpendicular lines, then f (l) ⊥ f (m). (xii) If a point Z has Cartesian coordinates (x, y) relative to the frame of reference F = ([O, I , [O, J ), then f (Z) has Cartesian coordinates (x, y) relative to the frame of reference ([ f (O), f (I) , [ f (O), f (J) ).

Proof . (i) If Z1 = Z2 then |Z1 , Z2 | > 0 so that |f (Z1 ), f (Z2 )| > 0, and so f (Z1 ) = f (Z2 ). (ii) If Z1 = Z2 the result is trivial, so suppose that Z1 = Z2 . Then for all Z ∈ [Z1 , Z2 ], we have |Z1 , Z|+|Z, Z2 | = |Z1 , Z2 | and so |f (Z1 ), f (Z)| +|f (Z), f (Z2 )| = |f (Z1 ), f (Z2 )|. It follows by 3.1.2 and 4.3.1 that f (Z) ∈ [ f (Z1 ), f (Z2 )] and so f ([Z1 , Z2 ]) ⊂ [ f (Z1 ), f (Z2 )]. f (Z2 ) W Z Z1

Z2 f (Z1 ) Figure 8.2.

Next let W ∈ [ f (Z1 ), f (Z2 )]. Then |f (Z1 ), W| ≤ |f (Z1 ), f (Z2 )| = |Z1 , Z2 |. Choose the point Z ∈ [Z1 , Z2 so that |Z1 , Z| = |f (Z1 ), W|; as |Z1 , Z| ≤ |Z1 , Z2 | then Z ∈ [Z1 , Z2 ]. Moreover |f (Z1 ), f (Z)| = |Z1 , Z| = |f (Z1 ), W|. Then f (Z) and W are both in [ f (Z1 ), f (Z2 ) and at the same distance from f (Z1 ) so f (Z) = W . Thus W is a value of f at some point of [Z1 , Z2 ]. Hence f ([Z1 , Z2 ]) = [ f (Z1 ), f (Z2 )]. (iii) By (i) f (Z1 ) = f (Z2 ). Suppose that Z ∈ [Z1 , Z2 . Then either Z ∈ [Z1 , Z2 ] or Z2 ∈ [Z1 , Z]. It follows from part (ii) of the present theorem, that then either f (Z) ∈ [ f (Z1 ), f (Z2 )] or f (Z2 ) ∈ [ f (Z1 ), f (Z)]. Thus f (Z) ∈ [ f (Z1 ), f (Z2 ) and so f ([Z1 , Z2 ) ⊂ [ f (Z1 ), f (Z2 ) . If W ∈ [ f (Z1 ), f (Z2 ) choose Z ∈ [Z1 , Z2 so that |Z1 , Z| = |f (Z1 ), W|. Then by the last paragraph f (Z) ∈ [ f (Z1 ), f (Z2 ) and as |f (Z1 ), f (Z)| = |f (Z1 ), W|, we have f (Z) = W . Thus W is a value of f at some point of [Z1 , Z2 . Hence f ([Z1 , Z2 ) = [ f (Z1 ), f (Z2 ) .

Sec. 8.2

Isometries

125

(iv) Take Z3 = Z1 so that Z1 ∈ [Z2 , Z3 ]. Then Z1 Z2 = [Z1 , Z2 ∪ [Z1 , Z3 . Hence f (Z1 Z2 ) = f ([Z1 , Z2 ) ∪ f ([Z1 , Z3 ) = [ f (Z1 ), f (Z2 ) ∪ [ f (Z1 ), f (Z3 ) = f (Z1 ) f (Z2 ), as f (Z1 ) ∈ [ f (Z2 ), f (Z3 )]. If f (Z) ∈ f (Z1 ) f (Z2 ), then by the foregoing there is a point Z4 ∈ Z1 Z2 such that f (Z4 ) = f (Z) and then as f is one-one Z = Z4 ∈ Z1 Z2 . (v) For |Z2 , Z3 | = |f (Z2 ), f (Z3 )|, |Z3 , Z1 | = |f (Z3 ), f (Z1 )|, |Z1 , Z2 | = |f (Z1 ), f (Z2 )|, so by the SSS-principle, these triangles are congruent in the correspondence (Z1 , Z2 , Z3 ) → ( f (Z1 ), f (Z2 ), f (Z3 )) . (vi) Suppose that f (H1 ) is not a subset of H3 . Then there is some Z4 ∈ H1 such that f (Z4 ) ∈ H4 , f (Z4 ) ∈ f (l). Then f (Z3 ) and f (Z4 ) are on opposite sides of f (l), so there is a point W on both f (l) and [ f (Z3 ), f (Z4 )]. By (ii) there is a point Z ∈ [Z3 , Z4 ] such that f (Z) = W , and then by (i) and (iv) Z ∈ l. But this implies that Z4 ∈ H1 and so gives a contradiction. Hence f (H1 ) ⊂ H3 and by a similar argument f (H2 ) ⊂ H4 . (vii) Take distinct points Z1 , Z2 in Π. If W ∈ f (Z1 ) f (Z2 ), then by (iv) f (Z) = W for some Z ∈ Z1 Z2 and so W is a value of f . f (Z2 ) Z

H1 Z2

H3

W f (Z1 )

Z1 Figure 8.3.

Suppose then that W ∈ f (Z1 ) f (Z2 ) and let H3 be the closed half-plane with edge f (Z1 ) f (Z2 ) which contains W . Let H1 be the closed half-plane with edge Z1 Z2 such that by (vi) f (H1 ) ⊂ H3 . Take a point Z ∈ H1 such that |∠Z2 Z1 Z|◦ = |∠f (Z2 )f (Z1 )W|◦ and |Z1 , Z| = |f (Z1 ), W|. Then by the SAS-principle [Z1 , Z2 , Z] ≡ [ f (Z1 ), f (Z2 ),W ], and so by (v) [ f (Z1 ), f (Z2 ), f (Z)] ≡ [ f (Z1 ), f (Z2 ),W ]. In particular |∠f (Z2 )f (Z1 )f (Z)|◦ = |∠f (Z2 )f (Z1 )W|◦ . As f (Z) ∈ H3 , W ∈ H3 we then have f (Z) ∈ [ f (Z1 ),W . But by the congruence we also have |f (Z1 ), f (Z)| = |f (Z1 ), W|. It follows that f (Z) = W and so W is a value of f .

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(viii) Let W ∈ H3 . Then by (vii) W = f (Z) for some Z ∈ Π. If W ∈ f (l) then by (iv) Z ∈ l ⊂ H1 . If W ∈ f (l) then W ∈ H4 and by (vi) we cannot have Z ∈ H2 as that would imply W ∈ H4 . Thus again Z ∈ H1 . In both cases W is a value f (Z) for some Z ∈ H1 . (ix) By part (iv) f (l), f (m) are lines. If Z belongs to both l and m, then f (Z) belongs to both f (l) and f (m) so these have a point in common. On the other hand, if l m suppose first that l = m. Then f (l) = f (m) and so f (l)

f (m). Next suppose that l = m; then l ∩ m = 0. / We now must have f (l) ∩ f (m) = 0, / as if W were on both f (l) and f (m), by (iv) we would have W = f (Z) for some Z ∈ l, W = f (Z0 ) for some Z0 ∈ m. But by (i) Z = Z0 so we would have Z on both l and m. (x) By (v) the triangles [Z1 , Z2 , Z3 ], [ f (Z1 ), f (Z2 ), f (Z3 )] are congruent, and so corresponding angles have equal degree-measures. (xi) If l and m are perpendicular, let Z1 be their point of intersection, and let Z2 , Z3 be other points on l and m respectively. Then as in part (x), ∠Z2 Z1 Z3 is a right-angle and so its image is also a right-angle. (xii) For any line l and any point Z we recall that πl (Z) denotes the foot of the perpendicular from Z to l. For any point Z ∈ Π, let U = πOI (Z) and V = πOJ (Z). Let O = f (O), I = f (I), J = f (J). Then O = I , O = J and O I ⊥ O J so that we can take ([O , I , [O , J ) as a frame of reference. Let H1 , H2 be the half-planes with edge OI, with J ∈ H1 , and H3 , H4 the half-planes with edge OJ, with I ∈ H3 . Similarly let H1 , H2 be the half-planes with edge O I , with J ∈ H1 , and H3 , H4 the half-planes with edge O J , with I ∈ H3 . H4 H3

Z Z

V

V J H1

J O

I

U

O

I

U

H2 Figure 8.4.

If (x, y) are the Cartesian coordinates of Z relative to ([O, I , [O, J ), then  x=

|O, U|, if Z ∈ H3 , −|O, U|, if Z ∈ H4 .

But if Z = f (Z), U = f (U ) we have U ∈ O I , and if Z ∈ OI we have ZU ⊥ OI and hence Z U ⊥ O I . It follows that U = πO I (Z ). Moreover f (H3 ) = H3 , f (H4 ) =

Sec. 8.3

Translation of frame of reference

127

H4 . Then if (x , y ) are the Cartesian coordinates of Z relative to ([O , I , [O , J ), when Z ∈ H3 we have Z ∈ H3 and so x = |O , πO I (Z )| = |O , U | = |O, U| = x. Similarly when Z ∈ H4 we have Z ∈ H4 and so x = −|O , πO I (Z )| = −|O , U | = −|O, U| = x. Thus x = x in all cases, and by a similar argument y = y.

8.2.2 If l = ml(|BAC), then sl ([A, B ) = [A,C and sl ([A,C ) = [A, B . Proof . We prove sl ([A, B ) = [A,C as the other then follows. As A ∈ l we have sl (A) = A and so by 8.2.1(iii) sl ([A, B ) = [A, D for some point D. Suppose first that A, B,C are collinear. When C ∈ [A, B we have l = AB, and so sl (P) = P for all P ∈ [A, B . As [A, B = [A,C the conclusion is then immediate. On the other hand when A ∈ [B,C] so that |BAC is straight, l is the perpendicular to AB at A. Then if A = mp(B, D) we have sl ([A, B ) = [A, D , and [A, D = [A,C as D ∈ [A,C . Finally suppose that A, B,C are non-collinear. Now take D ∈ [A,C so that |A, D| = |A, B|. If M = mp(B, D) by 4.1.1(iv) we have that l = AM and as sl (B) = D then sl ([A, B ) = [A, D = [A,C .

8.3 TRANSLATION OF FRAME OF REFERENCE NOTATION. By using 8.2.1(iii), (vi) and (xi), we showed in 8.2.1(xii) that for any frame of reference F = ([O, I , [O, J ) and any isometry f , F = ([ f (O), f (I) , [ f (O), f (J) ) is also a frame of reference, and that Cartesian coordinates of Z relative to F are also Cartesian coordinates of f (Z) relative to F . We denote F by f (F ). For any frame of reference F = ([O, I , [O, J ), let Z0 ≡F (x0 , y0 ) and F = tO,Z0 (F ). Then if Z ≡F (x, y) we have Z ≡F (x − x0 , y − y0 ). Proof . By 8.2.1(xii), tO,Z0 (Z) has coordinates (x, y) relative to F , and by 8.1.1(i) it also has H4 H3 V Z coordinate (x + x0 , y + y0 ) relative to F . Thus for all (x, y) ∈ Z R × R the point with coordiV J nates (x + x0 , y + y0 ) relative to O F has coordinates (x, y) relaU I tive to F . On replacing (x, y) by H1 J (x − x0 , y − y0 ), we conclude that O I U H2 the point with coordinates (x, y) Figure 8.5. relative to F has coordinates (x − x0 , y − y0 ) relative to F .

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Exercises 8.1 If T is the set of all translations of Π, show that (T , ◦) is a commutative group. 8.2 If I is the set of all isometries of Π, show that (I , ◦) is a group. 8.3 Given any half-lines [A, B , [C, D show that there is an isometry f which maps [A, B onto [C, D . 8.4 Show that each of the following concepts is an isometric invariant:- interior region of an angle-support, triangle, dividing a pair of points in a given ratio, mid-point, centroid, circumcentre, orthocentre, mid-line, incentre, parallelogram, rectangle, square, area of a triangle, circle, tangent to a circle. 8.5 For any line l, sl [C (O; k)] = C (sl (O); k) so that, in particular, if O ∈ l then sl [C (O; k)] = C (O; k).

9

Trigonometry; cosine and sine; addition formulae COMMENT. In this chapter we go on to deal fully with reflex-angles as well as with wedge and straight ones, we define the cosine and sine of an angle and we deal with addition of angles. As a vitally convenient aid to identifying the two angles with a given support |BAC, we start by introducing the notion of the indicator of an angle.

9.1 INDICATOR OF AN ANGLE 9.1.1 Definition . If α is an angle with support |BAC, we call the other angle with support |BAC the co-supported angle for α , and denote it by co − sp α . α

co − sp α Figure 9.1. Co-supported angle.

i(α ) C

C B

A

B A i(α )

Figure 9.2. Angle indicators.

Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50009-4 © 2016 Elsevier Ltd. All rights reserved.

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Definition . Referring to 2.3.3, for each angle support |BAC let l = ml(|BAC) as in 3.6 and 4.1.1. When A ∈ BC, we call l ∩ I R(|BAC) and l ∩ E R(|BAC) the indicators of the wedge-angle (|BAC, I R(|BAC)) and of the reflex-angle (|BAC , E R(|BAC)), respectively. When A ∈ [B,C] we call l ∩ H1 , l ∩ H2 the indicators of the straight-angles (|BAC, H1 ), (|BAC, H2 ), respectively. The null-angle with support |BAB has indicator [A, B . The full-angle with support |BAB has indicator [A, D where D = A and A ∈ [B, D]. In each case an indicator is a half-line of l with initial point the vertex A. We denote the indicator of an angle α by i(α ). COMMENT. The first use we make of the concept of indicator is in defining the cosine and sine of any angle.

9.2 COSINE AND SINE OF AN ANGLE 9.2.1 Definition . Consider an angle α with support |BAC such that the indicator i(α ) lies in a closed half-plane H1 with edge AB. Then we define cos α and sin α as follows. Take any point P = A on [A,C , let Q ∈ [A, B be such that |A, Q| = |A, P| and R ∈ H1 be such that |A, R| = |A, P| and AR ⊥ AB.

R V A

C P

i(α )

Q H1 U B

T Figure 9.3. Cosine and Sine.

Let U,V be the feet of the perpendiculars from P to AB = AQ and AR respectively. Then we define cos α =

|A, P| − |Q, U| |A, P| − |R, V| , sin α = . |A, P| |A, P|

It follows from this definition that if H2 is the other half-plane with edge AB and if we take T ∈ H2 so that |A, T| = |A, P| and AT ⊥ AB, then cos(co − sp α ) =

|A, P| − |Q, U| |A, P| − |T, V| , sin(co − sp α ) = . |A, P| |A, P|

COMMENT. Two comments on this definition are in order. First we note that when A, B,C are collinear, H1 and H2 are not uniquely determined above but are interchangeable with each other, so that the angles α and co-spα are not uniquely determined. Our second comment is that to show that cos α , sin α are well-defined it is first necessary to use the ratio results for triangles to show that the values of cos α and sin α do not depend on the particular point P ∈ [A,C taken, and then to show that if the arms [A, B and [A,C are interchanged the outcome is unchanged.

Sec. 9.2

Cosine and sine of an angle

131

To help us in our study of angles, it is convenient to fit a frame of reference to the situation in the definition. We take O = A, I = B and J = O a point in H1 so that OI ⊥ OJ. We let H3 , H4 be the closed half-planes with edge OJ, with I ∈ H3 . H3 H4 C R With k = |A, P| = |O, P|, let Q V P i(α ) be the point on [O, I = [A, B J A = O such that |O, Q| = k, and let R H1 S be the point on [O, J such that B = I U Q H2 |O, R| = k. Choose S, T so that O = mp(Q, S), O = mp(R, T ). T Figure 9.4.

The cosine and sine of an angle are well-defined. Proof . (i) When A, B and C are collinear there are two cases to be considered. One case is when A ∈ [B,C] so that |BAC is straight. Then each of α , co − sp α is a straight-angle and as P = S, we have U = S, V = A and so cos α = cos(co − sp α ) = −1, sin α = sin(co − sp α ) = 0. A second case is when C ∈ [A, B so that one of α , co − sp α is a null-angle with indicator [A, B and the other is a full-angle with indicator [A, B1 where A is between B and B1 . Both of the indicators are in H1 and H2 , but as P = Q we have U = Q, V = A and so cos α = cos(co − sp α ) = 1, sin α = sin(co − sp α ) = 0. Thus in neither case does the ambiguity affect the outcome. (ii) We now use the ratio results for triangles to show that the values of cos α and sin α do not depend on the particular point P ∈ [A,C chosen. Take k1 > 0 and let P1 , Q1 , R1 be the points in [O, P , [O, Q , [O, R , respectively, each at a distance k1 from O. Let U1 = πOI (P1 ), V1 = πOJ (P1 ). Suppose first that P ∈ OI, P ∈ OJ. As PU P1U1 , by 5.3.1 |O, P| |O, U| = , |O, U1 | |O, P1 | and so

|O, U| |O, U1 | = . k k1

Now if P ∈ H3 so that U ∈ [Q, O] and so |O, U| = k − |Q, U|, by 2.2.3(iv) P1 ∈ H3 and similarly |O, U1 | = k1 − |Q1 , U1 |. On inserting this we get that k − |Q, U| k1 − |Q1 , U1 | = . k k1

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Ch. 9

When instead P ∈ H4 , we have O ∈ [Q,U ] so |O, U| = |Q, U| − k and similarly |O, U1 | = |Q1 , U1 | − k1 . On inserting these we obtain k − |Q, U| k1 − |Q1 , U1 | = k k1 again. When P ∈ OI we have either P = Q or P = S. When P = Q, we have P1 = Q1 and the formula checks out. It checks out similarly in the cases when P is R, S or T . By a similar proof we find that k − |R, V| k1 − |R1 , V1 | = . k k1 Thus it makes no difference to the values of cos α and sin α if P is replaced by P1 . H4

H3

H4 R

R V V1 S

O

H3

P

J P1 U1 I

H1 U Q H2

P

V J

i(α ) S

O

X I

sl (H1 ) i(α )

H1 U Q H2

Y T

T Figure 9.5.

W Figure 9.6.

(iii) It remains to show that if the arms [A, B and [A,C are interchanged the outcome is unchanged. Let l = midlQOP so that sl (OQ) = OP and sl (H1 ) is a closed half-plane with edge OP. As i(α ) ⊂ H1 we have sl (i(α )) ⊂ sl (H1 ); but as i(α ) ⊂ l, sl (i(α )) = i(α ) and thus i(α ) ⊂ sl (H1 ). If W = sl (R) then W ∈ sl (H1 ) and as OQ ⊥ OR we have OP ⊥ OW . Moreover X = sl (U ) = πOP (Q) and Y = sl (V ) = πOW (Q) satisfy |P, X| = |Q, U|, |W, Y| = |R, V|. Hence k − |P, X| k − |Q, U| k − |W, Y| k − |R, V| = , = . k k k k This completes the proof.

9.2.2 Polar coordinates For Z = O, let k = |O, Z| and the angle α have support |IOZ and indicator i(α ) in H1 . Then if Z ≡F (x, y), x = k cos α , y = k sin α .

Sec. 9.3

Angles in standard position

133

Proof . Let Q, R be the points where C (O; k) meets [O, I and [O, J , respectively; then Q and R have Cartesian coordinates (k, 0) and (0, k), respectively. Let U, V be the feet of the perpendiculars from Z to the lines OI and OJ, so that these have Cartesian coordinates (x, 0) and (0, y) respectively. Now O ≡ (0, 0) and Z ≡ (x, y) so by the distance formula (x − 0)2 + (y − 0)2 = k2 . Thus x2 + y2 = k2 , so that x2 ≤ k2 and as k > 0 we have x ≤ k. Then by the distance formula  |Q, U| = (k − x)2 + (0 − 0)2 = k − x, as k − x ≥ 0, and similarly |R, V| = k − y. Hence k − |Q, U| k − (k − x) x = = , k k k k − |R, V| k − (k − y) y sin α = = = . k k k

cos α =

Thus x = k cos α , y = k sin α . We refer to k and α as polar coordinates of the point Z with respect to F .

9.2.3 With the notation of 9.2.1, let α be an angle with support |IOP = |QOP and indicator i(α ) in H1 . Then we have the following properties:(i) For all α , cos2 α + sin2 α = 1. (ii) For P ∈ Q1 , cos α ≥ 0, sin α ≥ 0; for P ∈ Q2 , cos α ≤ 0, sin α ≥ 0; for P ∈ Q3 , cos α ≤ 0, sin α ≤ 0; for P ∈ Q4 , cos α ≥ 0, sin α ≤ 0.

Proof . (i) As in the proof in 9.2.1, cos α = ±

|O, U| |O, V| , sin α = ± . |O, P| |O, P|

Now when O, U, P, V are not collinear they are the vertices of a rectangle and so |O, V| = |U, P|. Then by Pythagoras’ theorem |O, U|2 + |U, P|2 = |O, P|2 , and the result follows. It can be verified directly when P is any of Q, R, S, T . (ii) This follows directly from details in the proof in 9.2.1.

9.3 ANGLES IN STANDARD POSITION 9.3.1 Angles in standard position COMMENT. The second use that we make of the concept of indicator of an angle is to identify angles with respect to a frame of reference.

134

Trigonometry; cosine and sine; addition formulae H4 R

P

J S

O

I

J

i(α ) H1 Q H2

S

O

H4

α S

H1 Q H2

I

T

T

i(α )

H3 i(α )

H4 R P

H3

Ch. 9

H4 R

H3 R J OI

i(α ) H1 Q H2

H3

J

I

H1 Q H2

O

S

P

T

T

P

Figure 9.7.

Definition . We recall from 3.7 our extension of degree-measure to reflex-angles. Let |QOP be a non-straight support, and let H1 , H2 be the closed-half planes with edge OQ, with P ∈ H2 . Let α be the reflex angle with support |QOP, so that i(α ) ⊂ H1 . Let S be the point such that O = mp(Q, S). Let β be the wedge or straight angle with support |SOP. i(α )

H4

R

α

S

H3

β

J I O

Q H1 H2

P T Figure 9.8. Measure of a reflex angle.

Then we defined the degree measure of α by |α |◦ = 180 + |β |◦ . In particular if P = Q, then β is a straight angle, α is the full angle with support |POP = |QOQ and indicator [O, S , and |α |◦ = 360. Definition . Given a frame of reference F = ([O, I , [O, J ), we denote by A ∗ (F ) the set of angles α with arm [O, I and with indicator i(α ) ⊂ H1 . These angles are said to be in standard position with respect to F .

Sec. 9.3

Angles in standard position

135

If α and γ are different angles in A ∗ (F ), then |α |◦ = |γ |◦ . Proof . This is evident if α and γ are both wedge or straight angles and hence, by addition of 180, if they are both reflex or straight. If α is wedge or straight and γ is reflex, then |α |◦ ≤ 180, |γ |◦ > 180. NOTATION. Given any real number x such that 0 ≤ x ≤ 360, we denote the angle α ∈ A ∗ (F ) with |α |◦ = x by xF . Thus the null, straight and full angles in A ∗ (F ) are denoted by 0F , 180F and 360F , respectively.

9.3.2 Addition of angles COMMENT. Given angles α , β ∈ A ∗ (F ) we wish to define two closely related forms of addition, the first suited to angle measure as to be dealt with in Chapter 12 and the second suited to more general situations. As we make more use of the latter we employ for it the common symbol +, and ⊕ for the former. As α ⊕ β is to be an angle we need to specify its support and its indicator; similarly for α + β . Definition . Let α , β be angles in A ∗ (F ) with supports |QOP1 , |QOP2 , respectively. Let l be the midline of |P1 OP2 and let P3 = sl (Q). Then α ⊕ β is an angle with support |QOP3 for which i(α ⊕ β ) ⊂ H1 , so that α ⊕ β ∈ A ∗ (F ). This identifies α ⊕ β uniquely except when P3 = Q; in this case both the null angle 0F and the full angle 360F have support |QOQ and we define α ⊕ β to be this full angle 360F in every case except when α and β are both null; in the latter case we define the sum to be this null angle 0F . We call α ⊕ β the sum of the angles α and β . H4 H3 P2 P3

R

β

J

P1

α O

S

I

H1 Q H2

T Figure 9.9. Addition of angles.

For all angles α , β ∈ A ∗ (F ), (i) cos(α ⊕ β ) = cos α cos β − sin α sin β , (ii) sin(α ⊕ β ) = sin α cos β + cos α sin β .

Proof . On using the notation of 7.3.1 and above, we have a1 = cos α , b1 = sin α , a2 = cos β , b2 = sin β , a3 = cos(α ⊕ β ), b3 = sin(α ⊕ β ).

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We note that in 7.3.1 (a1 + a2 )2 + (b1 + b2 )2 = 2(1 + a1 a2 + b1 b2 ), as a21 + b21 = a22 + b22 = 1. Then, by 7.3.2, when P1 and P2 are not diametrically opposite, cos(α ⊕ β ) − cos α cos β + sin α sin β =

(a1 + a2 )2 − (b1 + b2 )2 + 2(−a1 a2 + b1 b2 )(1 + a1 a2 + b1 b2 ) , 2(1 + a1 a2 + b1 b2 )

and the numerator here is equal to a21 + a22 − b21 − b22 − 2a21 a22 + 2b21 b22 = 2a21 + 2a22 − 2 − 2a21 a22 + 2(1 − a21 )(1 − a22 ) = 0. Similarly sin(α ⊕ β ) − sin α cos β − cos α sin β 2(a1 + a2 )(b1 + b2 ) − 2(a1 b2 + a2 b1 )(1 + a1 a2 + b1 b2 ) = , 2(1 + a1 a2 + b1 b2 ) and the numerator here is equal to twice a1 b1 + a2 b2 − a1 b1 (a22 + b22 ) − a2 b2 (a21 + b21 ) = 0. When P1 and P2 are diametrically opposite, cos(α ⊕ β ) − cos α cos β + sin α sin β = b21 − a21 − a1 (−a1 ) + b1 (−b1 ) = 0, sin(α ⊕ β ) − sin α cos β − cos α sin β = −2a1 b1 − b1 (−a1 ) − a1 (−b1 ) = 0.

9.3.3 Modified addition of angles COMMENT. In 9.3.2 we clearly exercised a choice in specifying what α ⊕ β should be when P3 = Q. The choice made there is what suits length of a circle and area of a disk which will be treated in Chapter 12, and that was the reason for the choice made. We now define modified addition α + β of angles, which is easier to use. Definition . Let A (F ) = A ∗ (F ) \ {360F }, so that A (F ) is the set of all nonfull angles in A ∗ (F ). We denote by ∠F QOP = ∠F IOP the unique angle in A (F ) with support |QOP = |IOP. Definition . Let α , β be angles in A (F ) with supports |QOP1 , |QOP2 . Let l be the midline of |P1 OP2 and let P3 = sl (Q). Then α + β is the angle in A (F ) with support |QOP3 . Note that when P3 = Q we have α + β = 0F . We call α + β the modified sum of the angles α and β . For all α , β ∈ A (F ), cos(α + β ) = cos α cos β − sin α sin β , sin(α + β ) = sin α cos β + cos α sin β .

Proof . This follows immediately from 9.3.2 as cos 360F = cos 0F , sin 360F = sin 0F .

Sec. 9.3

Angles in standard position

137

Modified addition + of angles has the following properties:(i) For all α , β ∈ A (F ), α + β is uniquely defined and lies in A (F ). (ii) For all α , β ∈ A (F ), α + β = β + α . (iii) For all α , β , γ ∈ A (F ), (α + β ) + γ = α + (β + γ ). (iv) For all α ∈ A (F ), α + 0F = α . (v) Corresponding to each α ∈ A (F ), there is a β ∈ A (F ) such that α + β = 0F .

Proof . (i) This is evident from the definition. (ii) This is evident as the roles of P1 and P2 are interchangeable in the definition. (iii) We note that by the last result cos[(α + β ) + γ ] = cos(α + β ) cos γ − sin(α + β ) sin γ and then cos[(α + β ) + γ ] = [cos α cos β − sin α sin β ] cos γ − [sin α cos β + cos α sin β ] sin γ , while cos[α + (β + γ )] = cos α cos(β + γ ) − sin α sin(β + γ ) = cos α [cos β cos γ − sin β sin γ ] − sin α [sin β cos γ + cos β sin γ ], and these are equal. Similarly sin[(α + β ) + γ ] = sin(α + β ) cos γ + cos(α + β ) sin γ = [sin α cos β + cos α sin β ] cos γ + [cos α cos β − sin α sin β ] sin γ , while sin[α + (β + γ )] = sin α cos(β + γ ) + cos α sin(β + γ ) = sin α [cos β cos γ − sin α sin β ] + cos α [sin β cos γ + cos β sin γ ], and these are equal. Thus (α + β ) + γ and α + (β + γ ) are angles in A (F ) with the same cosine and the same sine and so by 9.2.2 they are equal. (iv) When β = 0F , in the definition we have P2 = Q and then l is the midline of |QOP1 and so P3 = P1 . Thus α and α + 0F are both in A (F ) and they have the same support, so they must be equal. (v) Given any angle α ∈ A (F ) with support |QOP1 , let P2 = sOI (P1 ) and β be the angle in A (F ) with support |QOP2 . Then l = OI is the midline of |P1 OP2 and so in the definition P3 = sl (Q) = Q. Thus α + β has support |QOQ and so it is 0F . COMMENT. The properties just listed show that (A (F ), +) is a commutative group. Because of this the familiar properties of addition, subtraction and additive cancellation apply to it.

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9.3.4 Subtraction of angles

Definition . For all α ∈ A (F ), we denote the angle β in 9.3.3(v) by −α . The difference γ − α in A (F ) is defined by specifying that γ − α = γ + (−α ). In this way we deal with subtraction.

H3

H4 P1

R

−α

J

S

i(α )

α O I

P2

H1 Q H2

T Figure 9.10.

For all α ∈ A (F ), cos(−α ) = cos(co − sp α ) = cos α ,

sin(−α ) = sin(co − sp α ) = − sin α .

Proof . With P2 as in the proof of 9.3.3(v), we have cos(−α ) =

k − |Q, U| k − |R, V1 | , sin(−α ) = , k k

and |R, V1 | = |T, V| = 2k − |R, V|. We use this in conjunction with 9.2.1.

9.3.5 Integer multiples of an angle Definition . For all n ∈ N and all α ∈ A (F ), nα is defined inductively by 1α = α , (n + 1)α = nα + α ,

for all n ∈ N.

We refer to nα as integer multiples of the angle α . For all α ∈ A (F ), (i) cos(2α ) = cos2 α − sin2 α = 2 cos2 α − 1 = 1 − 2 sin2 α , (ii) sin(2α ) = 2 cos α sin α .

Proof . These are immediate by 9.3.3 and 9.2.3.

9.3.6 Standard multiples of a right-angle The angles 90F , 180F , 270F have the following properties:-

Sec. 9.4

Half angles

139

(i) cos 90F = 0, sin 90F = 1, cos 180F = −1, sin 180F = 0, cos 270F = 0, sin 270F = −1. (ii) 2(90F ) = 180F , 2(180F ) = 0F so that −180F = 180F , and 90F + 270F = 0F so that 270F = −90F . (iii) For all α ∈ A (F ), cos(α + 90F ) = − sin α , sin(α + 90F ) = cos α , cos(α + 180F ) = − cos α , sin(α + 180F ) = − sin α , cos(α + 270F ) = sin α , sin(α + 270F ) = − cos α .

Proof . (i) These follow immediately from 9.2.1. (ii) These follow from 9.2.1 and 9.3.4. (iii) These follow immediately from 9.3.3 and (i) of the present theorem.

9.4 HALF ANGLES 9.4.1 Definition . Given any angle α ∈ A (F ) with support |QOP, its indicator i(α ) meets C (O; k) in a unique point P which is in H1 . Then the wedge or straight angle in A (F ) with support |QOP is denoted by 12 α and is called a half-angle. Given any angle α ∈ A (F ), the equation 2γ = α has exactly two solutions in A (F ), these being 12 α and 12 α + 180F . Proof . In the definition we have 1 H3 H4 i(α ) 2 α = ∠F QOP and take P so that O = mp(P , P ). Let β = P R ∠F QOP so that β = 12 α + 1 2α 180F . Then sOP (Q) = P so that J α H1 by 9.3.3, 2( 12 α ) = α , 2β = α . H2 S O Q I Now suppose that γ , δ ∈ A (F ) and 2γ = 2δ = α . Then cos 2δ = P cos 2γ so that 2 cos2 δ − 1 = T 2 cos2 γ − 1, and hence cos δ = Figure 9.11. ± cos γ . Then also sin2 δ = sin2 γ so sin δ = ± sin γ . Moreover sin 2δ = sin 2γ so 2 sin δ cos δ = 2 sin γ cos γ . We first suppose that α = 180F so that cos α = −1 and so cos γ = 0. Then if cos δ = cos γ we must have sin δ = sin γ , and so δ = γ . Alternatively we must have cos δ = − cos γ , sin δ = − sin γ and so δ = γ + 180F .

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Trigonometry; cosine and sine; addition formulae

Ch. 9

If α = 180F then cos δ = 0, so that sin δ = ±1 and so δ is either 90F or 270F . COMMENT. Our definition of a half-angle is the standard one for the angles we deal with, but it would not suit angles which we do not consider, for example ones with degree-magnitude greater than 360. The latter are difficult to give an account of geometrically. For us 12 α + 12 β and 12 (α + β ) need not be equal; we shall deal with such matters in 12.1.1. Because of this, there is a danger of error if half-angles are used incautiously. For any angles α , β ∈ A (F ), if γ = 12 α + 12 β and δ = 12 α − 12 β , then γ + δ = α and γ − δ = β . Proof . As we are dealing with a commutative group, we have      γ + δ = 12 α + 12 β + 12 α + − 12 β      = 12 α + 12 α + 12 β + − 12 β = α + 0F = α . Similarly

γ −δ = =

1









α + 12 β − 12 α + − 12 β   1  1   21 1 2α + 2β + −2α + 2β

= β.

9.5 THE COSINE AND SINE RULES 9.5.1 The cosine rule NOTATION. Let A, B,C be non-collinear points. Then for the triangle [A, B,C] we denote by a the length of the side which is opposite the vertex A, by b the length of the side opposite B, and by c the length of the side opposite C, so that a = |B, C|, b = |C, A|, c = |A, B|. We also use the notation

α = ∠BAC, β = ∠CBA, γ = ∠ACB. Let A, B,C be non-collinear points, let D = πBC (A) and write x = |B, D|. Then with the notation above, 2ax = a2 + c2 − b2 when D ∈ [B,C] or C ∈ [B, D], while 2ax = b2 − a2 − c2 when B ∈ [D,C]. Proof . When D ∈ [B,C] we have |D, C| = a − x, and when C ∈ [B, D], |D, C| = x − a. In both of these cases, by Pythagoras’ theorem used twice we have |A, D|2 = |A, B|2 − |B, D|2 = c2 − x2 , |A, D|2 = |A, C|2 − |D, C|2 = b2 − (a − x)2 . On equating these we have c2 − x2 = b2 − a2 + 2ax − x2 , giving 2ax = c2 + a2 − b2 . When B ∈ [D,C] we have |D, C| = a + x, so by the formulae for |A, D|2 above we have c2 − x = b2 − (a + x)2 . This simplifies to 2ax = b2 − a2 − c2 .

Sec. 9.5

The cosine and sine rules A

141

A

A B B

B C

C

D

D

D

C

Figure 9.12.

The cosine rule. In each triangle [A, B,C], cos α =

b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 , cos β = , cos γ = . 2bc 2ca 2ab

Proof On returning to the last proof, we note that when D ∈ [B,C we have |B, D| x = , |B, A| c

cos β = while x=

c2 + a2 − b2 , 2a

and so cos β =

c2 + a2 − b2 . 2ca

Similarly, when B ∈ [D,C] we have cos β = − while x=−

x |B, D| =− , |B, A| c

c2 + a2 − b2 , 2a

and this gives the same conclusion.

9.5.2 The sine rule The sine rule In each triangle [A, B,C], sin β sin γ sin α = = . a b c

Proof . By the cosine rule cos2 α 1 − sin2 α (b2 + c2 − a2 )2 = = , a2 a2 4a2 b2 c2

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Trigonometry; cosine and sine; addition formulae

Ch. 9

so that sin2 α 4b2 c2 − (b2 + c2 − a2 )2 = a2 4a2 b2 c2 2 2 2(b c + c2 a2 + a2 b2 ) − (a4 + b4 + c4 ) = . 4a2 b2 c2 As the right-hand side here is symmetrical in a, b and c we must have sin2 β sin2 γ sin2 α = = . a2 b2 c2 As the sines of wedge-angles are all positive, we may take square roots here and the result follows.

9.5.3 In a triangle [A, B,C], let the mid-line of |BAC meet [B,C] at D and let d1 = |A, D|. Then 2bc 1 cos α . d1 = b+c 2 Proof . By 5.5 c |B, D| = , |D, C| b so that

c a. b+c On applying the sine rule to the triangle [A, B, D] we have that |B, D| =

ca 1 d1 = , sin β b + c sin 12 α and so d1 =

1 ca sin β ca sin β b ca sin α b 2bc cos α . = = = 1 1 1 b + c sin 2 α b + c b sin 2 α b + c a sin 2 α b+c 2

9.5.4 The Steiner-Lehmus theorem, 1842 A Suppose that we are given a triangle [A, B,C], that the mid-line of |CBA meets CA at E, that the mid-line of |ACB meets AB at F, and that |B, E| = |C, F|. We then wish to show that the triangle is isosceles. This is known as the Steiner-Lehmus theorem.

F

E

B

C

Figure 9.13. Steiner-Lehmus theorem.

Sec. 9.6

Cosine and sine of angles equal in magnitude

Proof . By the last result we have d2 =

2ca c+a

cos 12 β , d3 =

2ab 1 a+b cos 2 γ .

143 Then



 b2 c2 21 21 β − γ cos cos (c + a)2 2 (a + b)2 2   2 2 b c 2 (1 + cos β ) − (1 + cos γ ) = 2a (c + a)2 (a + b)2 

 c2 b2 c2 + a2 − b2 a2 + b2 − c2 2 = 2a − 1+ 1+ (c + a)2 2ca (a + b)2 2ab   b2 c bc2 − = a c−b+ (a + b)2 (c + a)2   bc 2 2 [c(c + a) − b(a + b) ] = a c−b+ (a + b)2 (c + a)2   bc 2 2 2 = a(c − b) 1 + [a + b + c + 2ab + bc + 2ca] (a + b)2 (c + a)2

d22 − d32 = 4a2

Then b < c implies that d2 > d3 .

9.6 COSINE AND SINE OF ANGLES EQUAL IN MAGNITUDE 9.6.1 If angles α , β are such that |α |◦ = |β |◦ , then cos α = cos β and sin α = sin β . Conversely if cos α = cos β and sin α = sin β , then |α |◦ = |β |◦ unless one of them is null and the other is full. Proof . Let F = ([O, I , [O, J ) and α have support |QOP and indicator in H1 , where |O, P| = |O, Q| = k. Let F = ([O , I , [O , J ) and β have support |Q O P and indicator in H1 , where |O , P | = |O , Q | = k. Q H4 H3 P

Q

J H1

J O

I

P H2 Figure 9.14.

O

I

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Trigonometry; cosine and sine; addition formulae

Ch. 9

We suppose first that |α |◦ = |β |◦ ≤ 90 so that P ∈ Q1 , P ∈ Q1 . Then U ∈ [O, Q], V ∈ [O, R], U ∈ [O , Q ], V ∈ [O , R ]. The triangles [O,U, P], [O ,U , P ] are congruent by the ASA-principle, so |O, U| = |O , U |, |O, V| = |O , V |. Then |Q, U| = |Q , U |, |R, V| = |R , V |. Hence cos α = cos β , sin α = sin β . Similar arguments work in the case of the other three quadrants of F . Conversely, let cos α = cos β , sin α = sin β . Suppose first that cos α ≥ 0, sin α ≥ 0. Then P ∈ Q1 , P ∈ Q1 . But |Q, U| = |Q , U |, |R, V| = |R , V | and so |O, U| = |O , U |, |U, P| = |U , P |. By the SSS-principle, the triangles [O,U, P], [O ,U , P ] are congruent so |α |◦ = |β |◦ , unless we have a degeneration from a triangle and one angle is null and the other is full. A similar argument works for the other three quadrants of F .

Exercises 9.1 Prove that for all angles α ∈ A ∗ (F ), −1 ≤ cos α ≤ 1, −1 ≤ sin α ≤ 1. 9.2 Let C1 be the circle with centre O and radius of length k. Let Z1 ≡ (k cos θ , k sin θ ), Z2 ≡ (−k, 0), Z3 ≡ (k, 0), so that Z1 is a point on this circle, and [Z2 , Z3 ] is a diameter. Let C2 be the circle with [Z1 , Z3 ] as diameter. Find the coordinates of the second point in which C2 meets the line Z2 Z3 . How does this relate to 4.3.3? 9.3 If D is the mid-point of the side [B,C] of the triangle [A, B,C] and d1 = |A, D|, prove that 4d12 = b2 + c2 + 2bc cos α . Deduce that 2d1 > a if and only if α is an acute angle. 9.4 Prove the identities cos α + cos β = 2 cos( 12 α + 12 β ) cos( 12 α − 12 β ), cos α − cos β = −2 sin( 12 α + 12 β ) sin( 12 α − 12 β ), and find similar results for sin α + sin β and sin α − sin β . 9.5 Show that

3 sin 270F + sin 210F = − , 2

and yet 3 2 sin[ 12 (270F + 210F )] cos[ 12 (270F − 210F )] = . 2

10

Complex coordinates; sensed angles; angles between lines COMMENT. In this chapter we utilise complex coordinates, develop sensed angles and rotations, complete our formulae for axial symmetries and identify isometries in terms of translations, rotations and axial symmetries. We go on to establish more results on circles and consider a variant on the angles we have been dealing with.

10.1 COMPLEX COORDINATES 10.1.1 We now introduce the field of complex numbers (C, +, .) as an aid. This has an added convenience when doing coordinate geometry. We recall that any z ∈ C can be written 2 uniquely  in the form z = x + ıy, where x, y ∈ R and ı = −1. We use the notations |z| = x2 + y2 , z¯ = x − ıy for the modulus or absolute value, and complex conjugate, respectively, of z. As well as having the familiar properties for addition, subtraction, multiplication and division (except division by 0), these have the further properties: z = z, z1 z2 = z1 z2 , ∀ z, z1 , z2 ∈ C; z = z iff z ∈ R; |z1 z2 | = |z1 ||z2 |, |z| = |z|, ∀ z, z1 , z2 ∈ C; |z| = z iff z ∈ R and z ≥ 0; z 1 zz = |z|2 , ∀ z ∈ C; = 2 ∀ z = 0. z |z|

Definition . Let F = ([O, I , [O, J ) be a frame of reference for Π and for any point Z ∈ Π we recall the Cartesian coordinates (x, y) of Z relative to F , Z ≡F (x, y). If z = x + ıy, we also write Z ∼F z, and call z a Cartesian complex coordinate of the point Z relative to F . When F can be understood, we can relax our notation and denote this by Z ∼ z. Complex coordinates have the following properties:(i) |z2 − z1 | = |Z1 , Z2 | for all Z1 , Z2 . Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50010-0 © 2016 Elsevier Ltd. All rights reserved.

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Ch. 10

(ii) If Z1 = Z2 , then Z ∈ Z1 Z2 if and only if z − z1 = t(z2 − z1 ) for some t ∈ R. (iii) If Z1 = Z2 , then Z ∈ [Z1 , Z2 if and only if z − z1 = t(z2 − z1 ) for some t ≥ 0. (iv) If Z1 = Z2 , then Z ∈ [Z1 , Z2 ] if and only if z − z1 = t(z2 − z1 ) for some t such that 0 ≤ t ≤ 1. (v) For Z1 = Z2 and Z3 = Z4 , Z1 Z2 Z3 Z4 if and only if z4 − z3 = t(z2 − z1 ) for some t ∈ R \ {0}. (vi) For Z1 = Z2 and Z3 = Z4 , Z1 Z2 ⊥ Z3 Z4 if and only if z4 − z3 = tı(z2 − z1 ) for some t ∈ R \ {0}.

Proof . (i) For |z2 − z1 |2 = |x2 − x1 + ı(y2 − y1 )|2 = (x2 − x1 )2 + (y2 − y1 )2 = |Z1 , Z2 |2 . (ii) For z − z1 = t(z2 − z1 ) if and only if x − x1 + ı(y − y1 ) = t[x2 − x1 + ı(y2 − y1 )]. If this happens for some t ∈ R, then x − x1 = t(x2 − x1 ), y − y1 = t(y2 − y1 ). By 6.4.1, Corollary (i), this implies that Z ∈ Z1 Z2 . Conversely if Z ∈ Z1 Z2 , by the same reference there is such a t ∈ R and it follows that z − z1 = t(z2 − z1 ). (iii) and (iv). In (ii) we have Z ∈ [Z1 , Z2 when t ≥ 0 by 6.4.1, Corollary, and similarly Z ∈ [Z1 , Z2 ] when 0 ≤ t ≤ 1. (v) By 6.5.1, Corollary (ii), Z1 Z2 and Z3 Z4 are parallel only if −(y2 − y1 )(x4 − x3 ) + (y4 − y3 )(x2 − x1 ) = 0.

(10.1.1)

We note that as Z1 = Z2 we must have either x1 = x2 or y1 = y2 . Suppose first that z4 − z3 = t(z2 − z1 ) for some t ∈ R. Then x4 − x3 + ı(y4 − y3 ) = t(x2 − x1 ) + ıt(y2 − y1 ), and so as t is real, x4 − x3 = t(x2 − x1 ), y4 − y3 = t(y2 − y1 ). Then − (y2 − y1 )(x4 − x3 ) + (y4 − y3 )(x2 − x1 ) = −(y2 − y1 )t(x2 − x1 ) + t(y2 − y1 )(x2 − x1 ) = 0, so that (10.1.1) holds and hence the lines are parallel. Conversely suppose that the lines are parallel so that (10.1.1) holds. When x1 = x2 , we let x4 − x3 t= , x2 − x1 so that x4 − x3 = t(x2 − x1 ). On inserting this in (10.1.1), we have −t(y2 − y1 )(x2 − x1 ) + (y4 − y3 )(x2 − x1 ) = 0,

Sec. 10.1

Complex coordinates

147

from which we have y4 − y3 = t(y2 − y1 ). When x2 − x1 = 0, by (10.1.1) we must have x4 − x3 = 0. We now let t=

y4 − y3 , y2 − y1

so that y4 − y3 = t(y2 − y1 ). For this t we also have, trivially, x4 − x3 = t(x2 − x1 ). Thus in both cases x4 − x3 = t(x2 − x1 ), y4 − y3 = t(y2 − y1 ), and so on combining these x4 − x3 + ı(y4 − y3 ) = t(x2 − x1 ) + ıt(y2 − y1 ). Thus z4 − z3 = t(z2 − z1 ). (vi) By 6.5.1, Corollary (i), these lines are perpendicular if and only if (y2 − y1 )(y4 − y3 ) + (x2 − x1 )(x4 − x3 ) = 0.

(10.1.2)

Suppose first that z4 − z3 = tı(z2 − z1 ) for some t ∈ R. Then x4 − x3 + ı(y4 − y3 ) = ıt(x2 − x1 ) − t(y2 − y1 ), and so as t is real, x4 − x3 = −t(y2 − y1 ), y4 − y3 = t(x2 − x1 ). Then (y2 − y1 )(y4 − y3 ) + (x2 − x1 )(x4 − x3 ) = (y2 − y1 )t(x2 − x1 ) − t(x2 − x1 )(y2 − y1 ) = 0, so that (10.1.2) holds, and hence the lines are perpendicular. Conversely suppose that the lines are perpendicular so that (10.1.2) holds. When x1 = x2 , we let y4 − y3 t= x2 − x1 so that y4 − y3 = t(x2 − x1 ). On inserting this in (10.1.2), we have (y2 − y1 )t(x2 − x1 ) + (x2 − x1 )(x2 − x1 ) = 0, from which we have x4 − x3 = −t(y2 − y1 ). When x2 − x1 = 0, by (10.1.2) we must have y4 − y3 = 0. We now let t=−

x4 − x3 , y2 − y1

so that x4 − x3 = −t(y2 − y1 ). For this t we also have, trivially, y4 − y3 = t(x2 − x1 ). Thus in both cases x4 −x3 = −t(y2 −y1 ), y4 −y3 = t(x2 −x1 ), and so on combining these x4 − x3 + ı(y4 − y3 ) = −t(y2 − y1 ) + ıt(x2 − x1 ) = ıt[x2 − x1 + ı(y2 − y1 )]. Thus z4 − z3 = tı(z2 − z1 ).

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10.2 COMPLEX-VALUED DISTANCE 10.2.1 Complex-valued distance The material in 7.6 is long established; we can generalise those concepts of sensed distances and sensed ratios as follows.

Definition . Let F be a frame of reference for Π. If Z1 ∼F z1 , Z2 ∼F z2 we define Z1 Z2F = z2 − z1 , and call this a complex-valued distance from Z1 to Z2 . We then consider also ZZ3 ZZ4F , a ratio of complex-valued distances or complex ratio when 1 2F Z1 = Z2 . We show that this latter reduces to the sensed ratio Z3 Z4≤l Z1 Z2≤l when Z1 , Z2 , Z3 , Z4 are points of a line l. As in 7.6.1 we suppose that l is the line W0W1 where W0 ≡ (u0 , v0 ) and W1 ≡ (u1 , v1 ), and has parametric equations x = u0 + s(u1 − u0 ), y = v0 + s(v1 − v0 ). By 10.1.1(ii) l then has complex parametric equation z = w0 + s(w1 − w0 ). If Z1 , Z2 , Z3 , Z4 have parameters s1 , s2 , s3 , s4 , respectively, then z2 − z1 = (z2 − w0 ) − (z1 − w0 ) = (s2 − s1 )(w1 − w0 ), z4 − z3 = (s4 − s3 )(w1 − w0 ), and so Z3 Z4F s4 − s3 = . s2 − s1 Z1 Z2F By 7.6.1 this is equal to the sensed ratio. This shows that for four collinear points a ratio of complex-valued distances reduces to the corresponding ratio of sensed distances. COMMENT. We could make considerable use of this concept in our notation for the remainder of this chapter but in fact we use it sparingly.

10.2.2 A complex-valued trigonometric function For Z0 ∼F z0 and F = tO,Z0 (F ), let I0 = tO,Z0 (I); we recall from 8.3 that Z ∼F z − z0 . Then if Z = Z0 , Z ∼F z and θ = ∠F I0 Z0 Z, by 9.2.2 we have x − x0 = r cos θ , y − y0 = r sin θ , where r = |Z0 , Z| = |z − z0 |. It follows that z − z0 = r(cos θ + ı sin θ ).

Sec. 10.2

Complex-valued distance

H4 H3 J0

Z

149

H4 H3

J0

θ

Z0 I0 J O

Z

Z0 I0

H1

J O I

H2

I

Z α1

H1 H2

Figure 10.1.

If Z1 = Z0 , Z1 ∼F z1 and α = ∠F I0 Z0 Z1 , then by this x1 − x0 = k cos α , y1 − y0 = k sin α , where k = |Z0 , Z1 |. On inserting this in 6.3.1 Corollary, we see that Z0 Z1 = {Z ≡ (x, y) : (x − x0 ) sin α − (y − y0 ) cos α = 0} . When Z0 Z1 is not parallel to OJ we have that cos α = 0 and this equation of the line Z0 Z1 can be re-written as y − y0 = tan α (x − x0 ) where tan α = sin α / cos α . We call tan α the slope of this line. Notation. For any angle θ we write cis θ = cos θ + ı sin θ . The complex-valued function cis has the properties:(i) For all θ , φ ∈ A (F ), cis (θ + φ ) = cis θ .cis φ . (ii) cis 0F = 1. (iii) For all θ ∈ A (F ), cis1 θ = cis (−θ ). (iv) For all θ ∈ A (F ), cis θ = cis (−θ ), where z¯ denotes the complex conjugate of z. (v) For all θ , |cis θ | = 1.

Proof . (i) For cis θ .cis φ = (cos θ + ı sin θ )(cos φ + ı sin φ ) = cos θ cos φ − sin θ sin φ + ı[sin θ cos φ + cos θ sin φ ] = cos(θ + φ ) + ı sin(θ + φ ) = cis (θ + φ ). (ii) For cis 0F = cos 0F + ı sin 0F = 1 + ı0 = 1.

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Ch. 10

(iii) For by (i) and (ii) of the present theorem, cis θ .cis (−θ ) = cis (θ − θ ) = cis 0F = 1. (iv) For the complex conjugate of cos θ + ı sin θ is cos θ − ı sin θ = cos(−θ ) + ı sin(−θ ). (v) For |cis θ |2 = cos2 θ + sin2 θ = 1.

10.3 ROTATIONS AND AXIAL SYMMETRIES 10.3.1 Rotations Definition . Let Z0 ∼F z0 , t be the translation tO,Z0 , F = t(F ) and I0 = t(I). Let α ∈ A (F ). The function rα ;Z0 : Π → Π defined by Z ∼F z, Z ∼F z , rα ;Z0 (Z) = Z if z − z0 = (z − z0 )cis α , is called rotation about the point Z0 through the angle α .

Z

θ

Z H4 H3

θ J0 Z0

α I0

H1

J O

Z1

H2

I Figure 10.2.

If rα ;Z 0 (Z) = Z we have the following properties:(i) In all cases |Z0 , Z | = |Z0 , Z|, and hence in particular rα ;Z0 (Z0 ) = Z0 . (ii) If Z = Z0 , θ = ∠F I0 Z0 Z and θ = ∠F I0 Z0 Z , then θ = θ + α . (iii) If Z0 ∼F z0 , Z ∼F z, Z ∼F z , then rα ;Z0 has the real coordinates form x − x0 = cos α .(x − x0 ) − sin α .(y − y0 ), y − y0 = sin α .(x − x0 ) + cos α .(y − y0 ),

Sec. 10.3

Rotations and axial symmetries

151

which has the matrix form





cos α − sin α x − x0 x − x0 = . y − y0 sin α cos α y − y0

Proof . (i) For |z − z0 | = |(z − z0 )cis α | = |z − z0 ||cis α | = |z − z0 |. (ii) For by 10.2.2 z − z0 = |z − z0 |cis θ , z − z0 = |z − z0 |cis θ , and so |z − z0 |cis θ = |z − z0 |cis θ cis α . Hence cis θ = cis θ cis α = cis (θ + α ) and so θ = θ + α by 9.2.2. (iii) Now x − x0 + ı(y − y0 ) = (cos α + ı sin α )[x − x0 + ı(y − y0 )] and so x − x0 = cos α .(x − x0 ) − sin α .(y − y0 ), y − y0 = sin α .(x − x0 ) + cos α .(y − y0 ). COMMENT. The rotation rα ;Z0 is characterised by (i) and (ii), as the steps can be traced backwards. Why a frame of reference F is prominent in this characterisation stems from the need to identify the angles α , θ , θ .

10.3.2 Formula for an axial symmetry

Z l H4 H3 J0

α

Z1

sl (Z)

Z0 I0 J O

H1 I

H2 Figure 10.3.

The form of equation of a line noted in 10.2.2 can be used in the formula in 6.6.1(iii) for an axial symmetry. However, for practice with complex-valued coordinates we deduce the result independently.

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Ch. 10

Let l be the line Z0 Z1 , Z0 ∼F z, F = tO,Z0 (F ), I0 = tO,Z0 (I) and α = ∠F I0 Z0 Z1 . Then sl (Z) = Z where Z ∼F z, Z ∼F z , z − z0 = (¯z − z¯0 )cis 2α , so that sl has the real coordinates form x − x0 = cos 2α .(x − x0 ) + sin 2α .(y − y0 ), y − y0 = sin 2α .(x − x0 ) − cos 2α .(y − y0 ), and so has the matrix form

cos 2α x − x0 = y − y0 sin 2α

sin 2α − cos 2α



x − x0 y − y0

.

Proof . To find a formula for Z = sl (Z) we first show that if W = πl (Z) and W ∼F w then     z − z0 z − z0 (z1 − z0 ), z − w = ıℑ w − z0 = ℜ (z1 − z0 ) . z1 − z0 z1 − z0 To start on this we note that z − z0 =

    z − z0 z − z0 z − z0 (z1 − z0 ) = ℜ (z1 − z0 ) + ı ℑ (z1 − z0 ). z1 − z0 z1 − z0 z1 − z0

If we now define w by   z − z0 (z1 − z0 ) w − z0 = ℜ z1 − z0 then W ∈ Z0 Z1 as w − z0 is a real multiple of z1 − z0 . But then   z − z0 (z1 − z0 ), z−w = ı ℑ z1 − z0 so W is on a line through Z which is perpendicular to Z0 Z1 . Thus W is the foot of the perpendicular from Z to Z0 Z1 . From this, as z + z = 2w, we have z − z = z − w − (z − w) = −2(z − w) so   z − z0 z − z = −2ı ℑ (z1 − z0 ). z1 − z0 As z1 − z0 = kcis α for some k > 0, we then have   z − z0 z − z = −2ı ℑ kcis α = −2ı{ℑ[(z − z0 )cis (−α )]}cis α kcis α = −[(z − z0 )cis (−α ) − (¯z − z¯0 )cis α ]cis α = −(z − z0 ) + (¯z − z¯0 )cis 2α

Sec. 10.4

Sensed angles

153

and so z − z0 = (¯z − z¯0 )cis 2α . Hence x − x0 + ı(y − y0 ) = [x − x0 − ı(y − y0 )](cos 2α + ı sin 2α ), so that x − x0 = cos 2α .(x − x0 ) + sin 2α .(y − y0 ), y − y0 = sin 2α .(x − x0 ) − cos 2α .(y − y0 ). We can express this in matrix form as stated. We denote sl by sα ;Z0 as well.

10.4 SENSED ANGLES 10.4.1 Definition . For F = tO,Z0 (F ), let I0 = tO,Z0 (I). Then if Z1 = Z0 , Z2 = Z0 , we let θ1 = ∠F I0 Z0 Z1 and θ2 = ∠F I0 Z0 Z2 . We define the sensed-angle F Z1 Z0 Z2 to be θ2 − θ1 .

Z2

θ2

H4 H3

Z1 θ − θ 2 1

J0 Z0

I0

H1

J O

θ1

I

H2 Figure 10.4.

Sensed angles have the following properties. Throughout Z0 ∼F z0 , Z1 ∼F z1 , Z2 ∼F z2 . (i) If the points Z1 and Z2 are both distinct from Z0 , and φ is the sensed-angle F Z1 Z0 Z2 , then Z0 Z2F z2 − z0 |Z0 , Z2 | cis φ . = = z1 − z0 |Z0 , Z1 | Z0 Z1F 0 (ii) The sensed-angle F Z1 Z0 Z2 is wedge or reflex according as ℑ zz21 −z −z0 is positive 1 or negative, and this occurs according as 2 [(y2 − y0 )(x1 − x0 ) − (x2 − x0 )(y1 − y0 )] is positive or negative.

154

Complex coordinates; sensed angles; angles between lines

(iii) If the points Z1 and Z2 are both distinct from Z0 , then F Z1 Z0 Z2 = −F Z2 Z0 Z1 . (iv) If Z1 , Z2 , Z3 are all distinct from Z0 , then F Z1 Z0 Z2 + F Z2 Z0 Z3 = F Z1 Z0 Z3 . (v) If φ = F Z1 Z0 Z2 , then rφ ;Z0 ([Z0 , Z1 ) = [Z0 , Z2 . (vi) In 10.3.1(ii), F ZZ0 Z = α . Z2 H4 H3 J0 Z0 J O

θ2

Z1

θ1

φ

I0

H1 H2

I

Z2 Z3

θ3 − θ2

θ3 H4 H3

J O

θ2 J0 Z0 H1

I

θ1 I0

Z1

θ2 − θ1

H2

Figure 10.5.

Proof .(i) For z1 − z0 = |z1 − z0 |cis θ1 , z2 − z0 = |z2 − z0 |cis θ2 and so z2 − z0 |z2 − z0 |cis θ2 |z2 − z0 | cis (θ2 − θ1 ). = = z1 − z0 |z1 − z0 |cis θ1 |z1 − z0 |

Ch. 10

Sec. 10.4

Sensed angles

155

(ii) From (i)

z2 − z0 |z2 − z0 | = sin(θ2 − θ1 ), z1 − z0 |z1 − z0 | and this is positive or negative according as θ2 − θ1 is wedge or reflex. Moreover ℑ

|z1 − z0 |2 ℑ

z2 − z0 = ℑ[(z2 − z0 )(z¯1 − z¯0 )] = (y2 − y0 )(x1 − x0 ) − (x2 − x0 )(y1 − y0 ). z1 − z0

(iii) For the first is θ2 − θ1 and the second is θ1 − θ2 . (iv) For if θ1 = ∠F I0 Z0 Z1 , θ2 = ∠F I0 Z0 Z2 , θ3 = ∠F I0 Z0 Z3 , then

θ2 − θ1 + (θ3 − θ2 ) = θ3 − θ1 . (v) For if Z ∈ [Z0 , Z1 , then z = z0 +t(z1 − z0 ) = z0 +t|z1 − z0 |cis θ1 for some t ≥ 0. Hence rθ2 −θ1 ;Z0 (Z) = Z where z − z0 = (z − z0 )cis (θ2 − θ1 ) = t|z1 − z0 |cis θ1 cis (θ2 − θ1 ) = t|z1 − z0 |cis θ2 . Thus Z ∈ [Z0 , Z2 . (vi) For θ − θ = α . If the points Z1 and Z2 are both distinct from Z0 and φ = F Z1 Z0 Z2 , then |Z1 , Z2 |2 = |Z0 , Z1 |2 + |Z0 , Z2 |2 − 2|Z0 , Z1 ||Z0 , Z2 | cos φ .

Proof . For by (i) in the last result, z2 − z0 = so that

|Z0 , Z2 | (cos φ + ı sin φ )(z1 − z0 ), |Z0 , Z1 |



 |Z0 , Z2 | (cos φ + ı sin φ ) − 1 (z1 − z0 ). z2 − z1 = |Z0 , Z1 |

Then

 |Z1 , Z2 |2 =

2  2  |Z0 , Z2 | |Z0 , Z2 | |Z0 , Z1 |2 , cos φ − 1 + sin φ |Z0 , Z1 | |Z0 , Z1 |

and the result follows on expanding the right-hand side here. For a non-collinear triple (Z0 , Z1 , Z2 ), let α be the wedge-angle ∠Z1 Z0 Z2 and φ be the sensed angle F Z1 Z0 Z2 . Then cos φ = cos α so that |φ |◦ = |α |◦ when φ is wedge, and |φ |◦ = 360 − |α |◦ when φ is reflex. Proof . By the last result, |Z1 , Z2 |2 = |Z0 , Z1 |2 + |Z0 , Z2 |2 − 2|Z0 , Z1 ||Z0 , Z2 | cos φ , while by the cosine rule for a triangle in 9.5.1 |Z1 , Z2 |2 = |Z0 , Z1 |2 + |Z0 , Z2 |2 − 2|Z0 , Z1 ||Z0 , Z2 | cos α . Hence cos φ = cos α so that sin2 φ = sin2 α and hence sin φ = ± sin α . The result follows from 9.3.4 and 9.6.

156

Complex coordinates; sensed angles; angles between lines

Ch. 10

10.5 SENSED-AREA 10.5.1 For points Z0 ≡F (x0 , y0 ), Z1 ≡F (x1 , y1 ), Z2 ≡F (x2 , y2 ) such that Z1 = Z0 , Z2 = Z0 , and θ = F Z1 Z0 Z2 we have (i) 1 2 |Z0 , Z1 ||Z0 , Z2 | sin θ

= 12 [(x1 − x0 )(y2 − y0 ) − (x2 − x0 )(y1 − y0 )],

1 2 |Z0 , Z1 ||Z0 , Z2 | cos θ

= 12 [(x1 − x0 )(x2 − x0 ) + (y1 − y0 )(y2 − y0 )].

(ii)

Proof . By 8.3 and 9.2.2, if k1 = |Z0 , Z1 |, k2 = |Z0 , Z2 |, then x1 − x0 = k1 cos θ1 , y1 − y0 = k1 sin θ1 , x2 − x0 = k2 cos θ2 , y2 − y0 = k2 sin θ2 . Then by 9.3.3 and 9.3.4, k1 k2 sin(θ2 − θ1 ) = k2 sin θ2 k1 cos θ1 − k2 cos θ2 k1 sin θ1 = (y2 − y0 )(x1 − x0 ) − (x2 − x0 )(y1 − y0 ). Similarly k1 k2 cos(θ2 − θ1 ) = k2 cos θ2 k1 cos θ1 + k2 sin θ2 k1 sin θ1 = (x2 − x0 )(x1 − x0 ) + (y2 − y0 )(y1 − y0 ).

10.5.2 Sensed-area of a triangle For an ordered triple of points (Z1 , Z2 , Z3 ) of points and a frame of reference F , if Z1 ≡F (x1 , y1 ), Z2 ≡F (x2 , y2 ) and Z3 ≡F (x3 , y3 ), we recall from 6.6.2 and 10.5.1(i) δF (Z1 , Z2 , Z3 ) defined by the formula

δF (Z1 , Z2 , Z3 ) = 12 [x1 (y2 − y3 ) − y1 (x2 − x3 ) + x2 y3 − x3 y2 ] = 12 [(x2 − x1 )(y3 − y1 ) − (x3 − x1 )(y2 − y1 )] ⎛ ⎞ x1 y1 1 1 = det ⎝ x2 y2 1 ⎠ . 2 x3 y3 1 By 6.6.2, when Z1 , Z2 , Z3 are non-collinear |δF (Z1 , Z2 , Z3 )| is equal to the area of the triangle [Z1 , Z2 , Z3 ]. In this case we refer to δF (Z1 , Z2 , Z3 ) as the sensed-area of the triangle [Z1 , Z2 , Z3 ], with the order of vertices (Z1 , Z2 , Z3 ). This was first introduced by Möbius in 1827.

Sec. 10.5

Sensed-area

157

Note that

δF (Z1 , Z2 , Z3 ) = δF (Z2 , Z3 , Z1 ) = δF (Z3 , Z1 , Z2 ) = −δF (Z1 , Z3 , Z2 ) = −δF (Z2 , Z1 , Z3 ) = −δF (Z3 , Z2 , Z1 ), so that its value is unchanged if Z1 , Z2 , Z3 are permuted cyclically, and its value is multiplied by −1 if the order of these is changed. We note also that 10.4.1(ii) can be restated as that the sensed-angle F Z1 Z0 Z2 is wedge or reflex according as ℑ

Z0 Z2F z2 − z0 =ℑ z1 − z0 Z0 Z1F

is positive or negative, and this occurs according as δF (Z0 , Z1 , Z2 ) is positive or negative.

10.5.3 A basic feature of sensed-area A basic feature of sensed-area is given by the following. Let the points Z3 ≡ (x3 , y3 ), Z4 ≡ (x4 , y4 ), Z5 ≡ (x5 , y5 ) be such that x3 = (1 − s)x4 + sx5 , y3 = (1 − s)y4 + sy5 , for some s ∈ R. Then for all Z1 , Z2 ,

δF (Z1 , Z2 , Z3 ) = (1 − s)δF (Z1 , Z2 , Z4 ) + sδF (Z1 , Z2 , Z5 ). For ⎛ x1 y1 1 y2 x2 δF (Z1 , Z2 , Z3 ) = det ⎝ 2 (1 − s)x4 + sx5 (1 − s)y4 + sy5 ⎞ ⎛ ⎛ y1 1 x1 x1 1 1 ⎠ ⎝ ⎝ x x y 1 + det = det 2 2 2 2 2 (1 − s)x4 (1 − s)y4 1 − s sx5 ⎞ ⎞ ⎛ ⎛ x1 y1 1 x1 y1 1 1 1 = (1 − s) det ⎝ x2 y2 1 ⎠ + s det ⎝ x2 y2 1 ⎠ 2 2 x4 y4 1 x5 y5 1

⎞ 1 ⎠ 1 (1 − s) + s ⎞ y1 1 y2 1 ⎠ sy5 s

= (1 − s)δF (Z1 , Z2 , Z4 ) + sδF (Z1 , Z2 , Z5 ).

10.5.4 An identity for sensed-area An identity that we have for sensed-area is that for any points Z1 , Z2 , Z3 , Z4 ,

δF (Z4 , Z2 , Z3 ) + δF (Z4 , Z3 , Z1 ) + δF (Z4 , Z1 , Z2 ) = δF (Z1 , Z2 , Z3 ).

158

Complex coordinates; sensed angles; angles between lines

Ch. 10

For the left-hand side here is equal to ⎞ ⎞ ⎞ ⎛ ⎛ ⎛ x4 y4 1 x4 y4 1 x4 y4 1 1 1 1 det ⎝ x2 y2 1 ⎠ + det ⎝ x3 y3 1 ⎠ + det ⎝ x1 y1 1 ⎠ 2 2 2 x3 y3 1 x1 y1 1 x2 y2 1 ⎛ ⎛ ⎛ ⎞ ⎞ ⎞ x4 y4 1 x4 y4 1 x4 y4 1 1 1 1 = det ⎝ x2 y2 1 ⎠ − det ⎝ x1 y1 1 ⎠ + det ⎝ x1 y1 1 ⎠ 2 2 2 x3 y3 1 x3 y3 1 x2 y2 1 ⎛ ⎛ ⎞ ⎞ x4 x4 y4 1 y4 1 1 1 = det ⎝ x2 − x1 y2 − y1 0 ⎠ + det ⎝ x1 y1 1 ⎠ 2 2 y3 1 x3 x2 y2 1 ⎛ ⎛ ⎞ ⎞ x4 x4 y4 1 y4 1 1 1 = det ⎝ x2 − x1 y2 − y1 0 ⎠ + det ⎝ x1 y1 1 ⎠ 2 2 y3 1 x3 x2 − x1 y2 − y1 0 ⎞ ⎞ ⎛ ⎛ x4 x4 y4 1 y4 1 1 1 = det ⎝ x2 − x1 y2 − y1 0 ⎠ − det ⎝ x2 − x1 y2 − y1 0 ⎠ 2 2 y3 1 y1 1 x3 x1 ⎛ ⎞ x4 y4 1 1 = det ⎝ x2 − x1 y2 − y1 0 ⎠ 2 x3 − x1 y3 − y1 0 ⎛ ⎛ ⎞ ⎞ x1 x1 y1 1 y1 1 1 1 = det ⎝ x2 − x1 y2 − y1 0 ⎠ = det ⎝ x2 y2 1 ⎠ , 2 2 x3 − x1 y3 − y1 0 x3 y3 1 and this is equal to the right-hand side. This was first proved by Möbius.

10.6 ISOMETRIES AS COMPOSITIONS 10.6.1

J0

H4 H3

Z2

Z1 Z0

J

I0

J0

H4 H3

Z1 Z0

J H1

H1 O

I

O

H2 Figure 10.6.

I

H2

I0 Z2

Sec. 10.6

Isometries as compositions

159

Let F = ([O, I , [O, J ) and F1 = ([Z0 , Z1 , [Z0 , Z2 ) be frames of reference. Let tO,Z0 (I) = I0 , tO,Z0 (J) = J0 , F = ([Z0 , I0 , [Z0 , J0 ) and α = ∠F I0 Z0 Z1 . Then there is a unique isometry g such that g([O, I ) = [Z0 , Z1 ,

g([O, J ) = [Z0 , Z2 .

When F Z1 Z0 Z2 is a wedge-angle and so is a right-angle 90F , g = rα ;Z0 ◦ tO,Z0 . When F Z1 Z0 Z2 is a reflex-angle 270F and so its co-supported angle is a rightangle, g = s 1 α ;Z0 ◦ tO,Z0 . 2

Proof . Without loss of generality we take |O, I| = |O, J| = |Z0 , Z1 | = |Z0 , Z2 | = 1. Let Z0 ∼F z0 , Z1 ∼F z1 , Z2 ∼F z2 and note that I ∼F 1, J ∼F ı, z1 − z0 = cis α . As Z0 Z1 ⊥ Z0 Z2 , by 10.1.1(vi) we have z2 − z0 = ı(z1 − z0 ) when F Z1 Z0 Z2 is a wedge-angle,

(10.6.1)

z2 − z0 = −ı(z1 − z0 ) when F Z1 Z0 Z2 is a reflex-angle.

(10.6.2)

and In case (10.6.1) we take the transformation Z = g(Z) where z = z0 + zcis α = z0 + (z + z0 − z0 )cis α . Then for z = t ≥ 0, z = z0 + t(z1 − z0 ) so g([O, I ) = [Z0 , Z1 . Similarly for z = ıt (t ≥ 0), z = z0 + t(z2 − z0 ) so g([O, J ) = [Z0 , Z2 . In case (10.6.2) we take the transformation Z = g(Z) where z = z0 + z¯cis α = z0 + (z + z0 − z0 )cis α . Then for z = t ≥ 0, z = z0 + t(z1 − z0 ) so g([O, I ) = [Z0 , Z1 . Similarly for z = ıt (t ≥ 0), z = z0 + t(z2 − z0 ) so g([O, J ) = [Z0 , Z2 . This establishes the existence of g. As to uniqueness, suppose that f is also an isometry such that f (F ) = F1 . Then by 8.2.1(xii), if Z ∼F z we have f (Z) ∼F1 z, g(Z) ∼F1 z, and so f (Z) = g(Z) for all Z ∈ Π. COROLLARY. Let f be any isometry. Then f can be expressed in one or other of the forms (a) f = rα ;Z0 ◦ tO,Z0 , (b) f = s 1 α ;Z0 ◦ tO,Z0 . 2

Proof . In the theorem, take Z0 = f (O), Z1 = f (I), Z2 = f (J) and consequently f is equal to the function g as defined in the proof.

160

Complex coordinates; sensed angles; angles between lines

Ch. 10

10.7 ORIENTATION OF A TRIPLE OF NONCOLLINEAR POINTS 10.7.1 Definition We say that an ordered triple (Z0 , Z1 , Z2 ) of non-collinear points is positively or negatively oriented with respect to F according as the sensed angle F Z1 Z0 Z2 is wedge or reflex. By 10.4.1(ii) this occurs according as δF (Z0 , Z1 , Z2 ) is positive or negative. Definition . Let F = ([O, I , [O, J ) and F1 = ([Z0 , Z1 , [Z0 , Z2 ) be frames of reference. We say that F1 is positively or negatively oriented with respect to F according as (Z0 , Z1 , Z2 ) is positively or negatively oriented with respect to F . The special isometries have the following effects on orientation:(i) Each translation preserves the orientations with respect to F of all non-collinear triples. (ii) Each rotation preserves the orientations with respect to F of all non-collinear triples. (iii) Each axial symmetry reverses the orientations with respect to F of all noncollinear triples.

Proof . (i) Let f = tZ0 ,Z1 and Z2 ∼F z2 , Z3 ∼F z3 , Z4 ∼F z4 . Then z 3 = z3 + (z1 − z0 ),

z 2 = z2 + (z1 − z0 ),

so that z 3 − z 2 = z3 − z2 , and similarly z 4 − z 2 = z4 − z2 . Hence z 4 − z 2 z4 − z2 = , z 3 − z 2 z3 − z2 and so by 10.4.1(ii) the result follows. (ii) Let f = rα ;Z0 . Then by 10.3.1 z 2 − z0 = (z2 − z0 )cis α , z 3 − z0 = (z3 − z0 )cis α , and so Hence

z 3 − z 2 = (z3 − z2 )cis α , z 4 − z 2 = (z4 − z2 )cis α . z 4 − z 2 z4 − z2 = , z 3 − z 2 z3 − z2

and so by 10.4.1(ii) the result follows. (iii) Let f = sα ;Z0 . Then by 10.3.2 z 2 − z0 = (z¯2 − z¯0 )cis 2α , z 3 − z0 = (z¯3 − z¯0 )cis 2α ,

Sec. 10.7

Orientation of a triple of non-collinear points

and so

161

z 3 − z 2 = (z¯3 − z¯2 )cis 2α , z 4 − z 2 = (z¯4 − z¯2 )cis 2α .

Hence

z 4 − z 2 z4 − z2 = , z 3 − z 2 z3 − z2

so that ℑ

z4 − z2 z 4 − z 2 = −ℑ , z3 − z2 z3 − z2

and so by 10.4.1(ii) the result follows. Let F , F1 be frames of reference and Z3 , Z4 , Z5 non-collinear points. Let θ = F Z4 Z3 Z5 and φ = F1 Z4 Z3 Z5 . Then |φ |◦ is equal to |θ |◦ or 360 − |θ |◦ , according as F1 is positively or negatively oriented with respect to F . Proof . We use the notation of 10.6.1. When F1 is positively oriented with respect to F , we recall that for f (Z) = Z with z = z0 + zcis α , we have f (F ) = F1 . On solving this for z and then interchanging z and z , we see that f −1 (Z) ∼F (z − z0 )cis (−α ). Then by 8.2.1(xii), Z = f ( f −1 (Z)) ∼F1 (z − z0 )cis (−α ). Letting Z j ∼F z j , Z j ∼F1 z j we then have z j = (z j − z0 )cis (−α ). Thus z 5 − z 3 (z5 − z0 )cis (−α ) − (z3 − z0 )cis (−α ) z5 − z3 = = . z 4 − z 3 (z4 − z0 )cis (−α ) − (z3 − z0 )cis (−α ) z4 − z3 But by 10.4.1(i), z − z 3 |Z3 , Z5 | z5 − z3 |Z3 , Z5 | cis θ , 5 cis φ . = = z4 − z3 |Z3 , Z4 | z4 − z 3 |Z3 , Z4 | Thus cis φ = cis θ and so |φ |◦ = |θ |◦ . When F1 is negatively oriented with respect to F , we take instead f (Z) = Z with z = z0 + z¯cis α . Now f −1 (Z) ∼F (¯z − z¯0 )cis α and so z 5 − z 3 (z¯5 − z¯0 )cis (α ) − (z¯3 − z¯0 )cis (α ) z5 − z3 = = . z 4 − z 3 (z¯4 − z¯0 )cis (α ) − (z¯3 − z¯0 )cis (α ) z4 − z3 Thus cis φ = cis θ = cis (−θ ) and so |φ |◦ = |(−θ )|◦ = 360 − |θ |◦ . Let F and F1 be frames of reference. Then the ratios of complex-valued distances

ρ=

Z3 Z4F1 Z3 Z4F , σ= , Z1 Z2F Z1 Z2F1

defined in 10.2.1, satisfy σ = ρ when F1 is positively oriented with respect to F , and σ = ρ¯ when F1 is negatively oriented with respect to F .

162

Complex coordinates; sensed angles; angles between lines

Ch. 10

Proof We use the notation of 10.6.1. In the case (10.6.1) z = z0 + zcis α so that σ=

z 4 − z 3 (z4 − z3 )cis α = = ρ. z 2 − z 1 (z2 − z1 )cis α

In the case (10.6.2) z = z0 + z¯cis α so that

σ=

z 4 − z 3 (z¯4 − z¯3 )cis α = = ρ¯ . z 2 − z 1 (z¯2 − z¯1 )cis α

10.8 SENSED ANGLES OF TRIANGLES, THE SINE RULE 10.8.1 Definition . For any non-full angle θ , we denote by θF the angle in A (F ) such that |θF |◦ = |θ |◦ . |α |◦ Z1

c H4 H3

Z1

|α |◦ |γ |◦ |β |◦

a

b Z3

c H4 H3

H1 I

H2

Z2 a

Z3 |γ |◦

Z2 J O

|β |◦

b

J O

H1 I

H2

Figure 10.7.

NOTATION. For non-collinear points Z1 , Z2 , Z3 , we use as standard notation |Z2 , Z3 | = a, |Z3 , Z1 | = b, |Z1 , Z2 | = c, b c a u= , v= , w= , c a b

α = F Z2 Z1 Z3 , β = F Z3 Z2 Z1 , γ = F Z1 Z3 Z2 . Note that by comparison with 9.5.1 we are now using sensed-angles instead of wedgeangles.

Sec. 10.8

Sensed angles of triangles, the sine rule

163

For non-collinear points Z1 , Z2 , Z3 if

α = F Z2 Z1 Z3 , β = F Z3 Z2 Z1 , γ = F Z1 Z3 Z2 , and θ = αF , φ = βF , ψ = γF , then θ + φ + ψ = 180F . Proof . For z1 − z2 c z2 − z3 a z3 − z1 b = cis θ , = cis φ , = cis ψ . z2 − z1 c z3 − z2 a z1 − z3 b On multiplying these together, we find that −1 = cis θ .cis φ .cis ψ = cis (θ + φ + ψ ). As cis 180F = −1 it follows that θ + φ + ψ = 180F . With the above notation, the lengths of the sides and the sensed-angles of a triangle [Z1 , Z2 , Z3 ] have the properties:(i) In each case 1 , 1 − ucis α and two pairs of similar identities obtained from these on advancing cyclically through (u, v, w) and (α , β , γ ). vcis β =

(ii) In each case

sin α sin β = , a b and two pairs of similar identities obtained from these on advancing cyclically through (a, b, c) and (α , β , γ ). c = b cos α + a cos β ,

Proof . (i) For z3 − z1 = ucis α .(z2 − z1 ) so that z3 − z2 = (1 − ucis α )(z1 − z2 ), while z1 − z2 = vcis β .(z3 − z2 ), which give (1 − ucis α )vcis β = 1. (ii) From (i) 1 1 − u[cos α + ı sin α ] = [cos β − ı sin β ], v so equating real parts gives c = b cos α + a cos β , while equating imaginary parts gives sin α /a = sin β /b. This result re-derives the sine rule for a triangle. If Z1 , Z2 , Z3 are distinct points, then Z3 Z1F Z2 Z1F + = 1. Z3 Z2F Z2 Z3F

Proof . For

z1 − z3 z1 − z2 + = 1. z2 − z3 z3 − z2

164

Complex coordinates; sensed angles; angles between lines

Ch. 10

10.9 SOME RESULTS ON CIRCLES 10.9.1 A necessary condition to lie on a circle In this section we provide some results on circles which are conveniently proved using complex coordinates. Let Z1 , Z2 be fixed distinct points, and Z a variable point, all on the circle C (Z0 ; k). Z2 Let F = tO,Z0 (F ) and β α = ∠F I0 Z0 Z1 , β = ∠F I0 Z0 Z2 Z1 H4 H3 1 and γ = 2 (β − α ). As Z varies Z J0 on the circle, in one of the open α half-planes with edge Z1 Z2 Z0 I0 the sensed angle F Z1 ZZ2 is H1 equal in measure to γ , while J in the other open half-plane O I H2 with edge Z1 Z2 it is equal in measure to γ + 180F . Note that Figure 10.8. 2γ = F Z1 Z0 Z2 . Proof . Now z1 − z0 = kcis α , z2 − z0 = kcis β and if θ = F I0 Z0 Z, then z − z0 = kcis θ . We write φ = F Z1 ZZ2 so that z2 − z |Z, Z2 | = lcis φ , where l = . z1 − z |Z, Z1 | Then lcis φ =

cis β − cis θ , cis α − cis θ

while on taking complex conjugates here lcis (−φ ) =

cis (−β ) − cis (−θ ) cis α cis θ − cis β . = cis (−α ) − cis (−θ ) cis β cis θ − cis α

By division cis 2φ =

cis β = cis (β − α ). cis α

Thus 2(cis φ )2 = (cis γ )2 so that cis φ = ±cis γ . Thus either cis φ = cis γ or cis φ = cis (γ + 180F ), and accordingly ℑ

z2 − z z2 − z = l sin γ or ℑ = l sin(γ + 180F ). z1 − z z1 − z

As sin γ > 0, the first of these occurs when Z is in the half-plane with edge Z1 Z2 in which ℑ zz21 −z −z > 0, and the second when Z is in the half-plane with edge Z1 Z2 in which −z ℑ zz12 −z < 0.

Sec. 10.9

Some results on circles

165

10.9.2 A sufficient condition to lie on a circle Let Z1 , Z2 be fixed distinct points and Z a variable point. As Z varies in one of the half-planes with edge Z1 Z2 , for the sensed angle θ = F Z1 ZZ2 let |θ |◦ = |γ |◦ where γ is a fixed nonnull and non-straight angle in A (F ), while as Z varies in the other half-plane with edge Z1 Z2 , let |θ |◦ = |γ + 180F |◦ . Then Z lies on a circle which passes through Z1 and Z2 . Proof . We have

Z2 H4 H3

Z1 Z

J0 Z0 I0

J O

H1 H2

I

Figure 10.9.

z2 − z = tcis γ z1 − z

for some t ∈ R \ {0}. Then z=

z2 − tz1 cis γ 1 − tcis γ

so that with cot γ = cos γ / sin γ , 1 1 z − (z1 + z2 ) − ı cot γ .(z2 − z1 ) 2 2 1 z2 − tz1 cis γ 1 − (z1 + z2 ) − i cot γ .(z2 − z1 ) = 1 − tcis γ 2 2 1 (z − z )[1 + tcis γ − ı cot γ (1 − tcis γ )] 2 1 = 2 1 − tcis γ 1 (z2 − z1 )[sin γ (1 + tcis γ ) − ı cos γ (1 − tcis γ )] = 2 sin γ (1 − tcis γ ) 1 (z2 − z1 )[sin γ + ı(t − cos γ )] = 2 sin γ (1 − tcis γ ) and this has absolute value

|z2 − z1 | . 2|sin γ |

This shows that Z lies on a circle, the centre and length of radius of which are evident.

10.9.3 Complex cross-ratio Let Z2 , Z3 , Z4 be non-collinear points and C the circle that contains them. Then Z ∈ Z3 Z4 lies in C if and only if ℑ

(z − z3 )(z2 − z4 ) = 0. (z − z4 )(z2 − z3 )

166

Complex coordinates; sensed angles; angles between lines

Ch. 10

When this holds and Z and Z2 are on the same side of Z3 Z4 , then (z − z3 )(z2 − z4 ) > 0. (z − z4 )(z2 − z3 )

Proof . The given condition is equivalent to z − z4 z2 − z4 =t , z − z3 z2 − z3

(10.9.1)

for some t = 0 in R. Let G1 , G2 be the open half-planes with common edge Z3 Z4 , with Z2 ∈ G1 . Let θ = F Z3 Z2 Z4 and φ = F Z3 ZZ4 . Suppose first that (10.9.1) holds. For Z ∈ G1 , ℑ

z − z4 z2 − z4 and ℑ z − z3 z2 − z3

must have the same sign and so t > 0; it follows that φ = θ . For Z ∈ G2 , ℑ

z − z4 z2 − z4 and ℑ z − z3 z2 − z3

must have opposite signs and so t < 0; it follows that φ = θ + 180F . By 10.9.2 Z ∈ C in both cases. Conversely let Z ∈ C . Then by (10.9.1) for Z ∈ G1 we have φ = θ , while for Z ∈ G2 we have φ = θ + 180F and the result now follows. 3 )(z2 −z4 ) The expression (z−z (z−z4 )(z2 −z3 ) is called the cross-ratio of the ordered set of points (Z, Z2 , Z3 , Z4 ).

10.9.4 Ptolemy’s theorem, c. 200A.D.

Z Let Z2 , Z3 , Z4 be non-collinear points and C the circle that contains them. Let Z ∈ C be such that Z and Z3 are on opposite sides of Z2 Z4 . Then |Z, Z4 ||Z2 , Z3 | + |Z, Z2 ||Z3 , Z4 | = |Z, Z3 ||Z2 , Z4 |.

Z2 W Z3 Z4 Figure 10.10.

Proof . By multiplying out, it can be checked that (z − z4 )(z2 − z3 ) + (z − z2 )(z3 − z4 ) = (z − z3 )(z2 − z4 ).

Sec. 10.9

Some results on circles

167

This is an identity due to Euler and from it (z − z4 )(z2 − z3 ) (z − z2 )(z4 − z3 ) + = 1. (z − z3 )(z2 − z4 ) (z − z3 )(z4 − z2 )

(10.9.2)

By 10.9.3 both fractions on the left are real-valued. As Z and Z3 are on opposite sides of Z2 Z4 , there is a point W of [Z, Z3 ] on Z2 Z4 . Then W is an interior point of the circle, and so W ∈ [Z2 , Z4 ] as the only points of the line Z2 Z4 which are interior to the circle are in this segment. It follows that Z2 and Z4 are on opposite sides of ZZ3 . Then [Z, Z2 ], [Z3 , Z4 ] are in different closed half-planes with common edge the line ZZ3 , so they have no points in common. It follows that Z and Z2 are on the one side of Z3 Z4 so the first of the fractions in (10.9.2) is positive, and so equal to its own absolute value. But [Z, Z4 ] and [Z3 , Z2 ] are in different closed half-planes with common edge ZZ3 so they have no point in common. It follows that Z and Z4 are on the one side of Z2 Z3 , so the second fraction in (10.9.2) is positive and so equal to its own absolute value. Hence |(z − z4 )(z2 − z3 )| |(z − z2 )(z4 − z3 )| + = 1. |(z − z3 )(z2 − z4 )| |(z − z3 )(z4 − z2 )| This is known as Ptolemy’s theorem From the original identity (10.9.2) with Z1 replacing Z we can deduce that for four distinct points Z1 , Z2 , Z3 , Z4 Z1 Z4F Z2 Z3F Z1 Z2F Z4 Z3F + = 1. Z1 Z2F Z2 Z4F Z1 Z3F Z4 Z2F This can be expanded as |Z2 , Z3 | |Z4 , Z3 | |Z1 , Z2 | |Z1 , Z4 | cis α cis β + cis γ cis δ = 1, |Z1 , Z3 | |Z2 , Z4 | |Z1 , Z3 | |Z2 , Z4 | where

α = F Z3 Z1 Z4 , β = F Z4 Z2 Z3 , γ = F Z3 Z1 Z2 , δ = F Z4 Z2 Z3 . From this we have that |Z1 , Z2 | |Z4 , Z3 | |Z1 , Z4 | |Z2 , Z3 | cis (αF + βF ) + cis (γF + δF ) = 1. |Z1 , Z3 | |Z2 , Z4 | |Z1 , Z3 | |Z2 , Z4 | We get two relationships on equating the real parts in this and also equating the imaginary parts. NOTE. For other applications of complex numbers to geometry, see Chapter 11 and Hahn [8].

168

Complex coordinates; sensed angles; angles between lines

Ch. 10

10.10 ANGLES BETWEEN LINES 10.10.1

Motivation

Since cos(180F + θ ) = − cos θ , sin(180F + θ ) = − sin θ , we have that tan(180F + θ ) = tan θ . Thus results that tan θ is constant do not imply that θ is an angle of constant magnitude. To extract more information from such situations, we develop new material. This also deals with the rather abrupt transitions in results such as those in 10.9.1 and 10.9.2.

10.10.2

Duo-sectors

Let l1 , l2 be lines intersecting at a point Z1 . When l1 = l2 , let Z2 , Z3 ∈ l1 with Z1 between Z2 and Z3 , and let Z4 , Z5 ∈ l2 with Z1 between Z4 and Z5 . Then the union Z3 Z1 Z5 ) I R(|Z Z2 Z1 Z4 ) ∪ I R(|Z we shall call a duo-sector with side-lines l1 and l2 ; we shall denote it by D1 . Similarly Z3 Z1 Z4 ) I R(|Z Z2 Z1 Z5 ) ∪ I R(|Z is also a duo-sector with side-lines l1 and l2 , and we shall denote it by D2 . l2 D2 D1 D1 Z4

Z4

l1 Z2

Z3

Z2 Z3

Z1

D1

D1 Z5

l3 Z5

Figure 10.11.

Z1 l4 D2

Figure 10.12.

The mid-line l3 of |Z2 Z1 Z4 is also the mid-line of |Z3 Z1 Z5 and it lies entirely in D1 . The mid-line l4 of |Z2 Z1 Z5 is also the mid-line of |Z3 Z1 Z4 and it lies entirely in D2 . We call {l3 , l4 } the bisectors of the line pair {l1 , l2 } and use l3 to identify D1 , l4 to identify D2 . In Figure 10.12 suppose that |∠Z2 Z1 Z4 |◦ = |∠Z3 Z1 Z5 |◦ = x and |∠Z4 Z1 Z3 |◦ = y then we have 2x + 2y = 360 and so x + y = 180. It follows that 12 x + 12 y = 90 and so l3 ⊥ l4 . WhenδF (Z1 , Z2 , Z4 ) > 0 we note that D1 = {Z ∈ Π : δF (Z1 , Z2 , Z)δF (Z1 , Z4 , Z) ≤ 0}, D2 = {Z ∈ Π : δF (Z1 , Z2 , Z)δF (Z1 , Z4 , Z) ≥ 0}, and get a similar characterisation when δF (Z1 , Z2 , Z4 ) < 0. When l1 = l2 , we take D1 = l1 ; we could also take D2 = Π but do not make any use of this. EBSCO Publishing : eBook Collection (EBSCOhost) - printed on 1/30/2016 1:40 PM via NORTH CAROLINA STATE UNIV LIBRARIES AN: 1134257 ; Barry, Patrick D..; Geometry with Trigonometry Account: s5822915

Sec. 10.10

10.10.3

Angles between lines

169

Duo-angles

When l1 , l2 are distinct lines, intersecting at Z1 , we call the pairs ({l1 , l2 }, D1 ), ({l1 , l2 }, D2 ), duo-angles, with arms l1 , l2 ; in this D1 , D2 are the duo-sectors of 10.10.2. We denote these duo-angles by αd , βd , respectively. We call the bisector l3 the indicator of αd , and the bisector l4 the indicator of βd . We define the degree-magnitudes of these duo-angles by |αd |◦ = |∠Z2 Z1 Z4 |◦ = |∠Z3 Z1 Z5 |◦ , |βd |◦ = |∠Z2 Z1 Z5 |◦ = |∠Z3 Z1 Z4 |◦ . If l1 ⊥ l2 we have that |αd |◦ = |βd |◦ = 90, and we call these right duo-angles. When l1 = l2 we take αd = ({l1 , l2 }, l1 ) to be a duo-angle with arms l1 , l1 , and call it a null duo-angle. Its indicator is l1 , and we define its degree-measure to be 0. We do not define a straight duo-angle. Thus the measure of a duo-angle γd always satisfies 0 ≤ |γd |◦ < 180. When l1 = l2 we define sin αd = sin(∠Z2 Z1 Z4 ) = sin(∠Z3 Z1 Z5 ), cos αd = cos(∠Z2 Z1 Z4 ) = cos(∠Z3 Z1 Z5 ), sin βd = sin(∠Z2 Z1 Z5 ) = sin(∠Z3 Z1 Z4 ), cos βd = cos(∠Z2 Z1 Z5 ) = cos(∠Z3 Z1 Z4 ). For a right duo-angle these have the values 1 and 0, respectively. When l1 and l2 are not perpendicular, we can define as well tan αd =

sin αd cos αd ,

sin βd tan βd = cos βd . If αd is a null duo-angle we define sin αd = 0, cos αd = 1, tan αd = 0.

10.10.4

Duo-angles in standard position

D2

H4

H3 l

D2

m

J O

H1 I

D1

D1 l

H4 J

I

H2

H1 H2

O

D2 D1 Figure 10.13.

m H3 D1

D2

170

Complex coordinates; sensed angles; angles between lines

Ch. 10

We extend our frame of reference F by taking in connection with the line pair {OI, OJ} a canonical pair of duo-sectors D1 and D2 , with D1 the union of the first and third quadrants Q1 and Q3 , and D2 the union of the second and fourth quadrants. For any line l through the origin O, we consider the duo-angle αd with side-lines OI and l, such that the indicator m of αd lies in the duo-sector D1 , that is the bisector of the line-pair {OI, l} which lies in the duo-sector of αd also lies in D1 . We denote by DA (F ) the set of such duo-angles, and we say that they are in standard position respect to F . If l = OI and Z4 ≡ (x4 , y4 ) is a point other than O on l, then so is the point with coordinates (−x4 , −y4 ); thus, without loss of generality, we may assume that y4 > 0 in identifying l as OZ4 . Then Z4 ∈ H1 and |αd |◦ = |∠IOZ4 |◦ , cos αd = cos(∠IOZ4 ) = 

x4

, x24 + y24 y4 sin αd = sin(∠IOZ4 ) =  . x24 + y24

When αd is not a right duo-angle, we have tan αd =

y4 . x4

We identify l = OI as OZ4 where Z4 ≡ (x4 , 0) and x4 > 0. Thus for the null duoangle in standard position we have cos αd = 1, sin αd = 0, tan αd = 0. We denote this null duo-angle by 0dF and the right duo-angle in standard position by 90dF . We now note that if αd , βd ∈ DA (F ) and tan αd = tan βd , then αd = βd . Proof . For this we let αd , βd have pairs of side-lines (OI, OZ4 ), (OI, OZ5 ), respectively, where |O, Z4 | = |O, Z5 | = k, and either y4 > 0 or x4 > 0, y4 = 0, and similarly either y5 > 0 or x5 > 0, y5 = 0. Then neither αd nor βd is 90dF and y4 y5 2 = , x + y24 = x25 + y25 = k2 . x4 x5 4 If y4 = 0 then y5 = 0 and both duo-angles are null. Suppose then that y4 = 0 so that y4 > 0; it follows that y5 > 0. Then k2 = x25 + y25 =

y25 2 y2 y2 x4 + y25 = 52 (x24 + y24 ) = 52 k2 . 2 y4 y4 y4

Hence y25 = y24 , and so y5 = y4 . It follows that x5 = x4 .

Sec. 10.10

10.10.5

Angles between lines

171

Addition of duo-angles in standard position

To deal with addition of duoangles in standard position, let Q ≡ (k, 0), R ≡ (0, k) for some k > 0, and αd have side-lines OQ and OZ4 , βd have side-lines OQ and OZ5 , where |O, Z4 | = |O, Z5 | = k and both have their indicators in D1 . Without loss of generality, we may suppose that either y4 > 0 or y4 = 0, x4 > 0, and similarly with respect to (x5 , y5 ).

D2 Z

5

R

D1

Z6

βd

J S

αd I

O

D1

T

Z4 Q

D2

Figure 10.14. Addition of duo-angles.

Then the line through Q which is parallel to Z4 Z5 will meet the circle C (O; k) in a second point, which we denote by Z6 ≡ (x6 , y6 ). The line through Q parallel to Z4 Z5 has parametric equations x = k + t(x5 − x4 ), y = t(y5 − y4 ), and so meets the circle again when t = 0 satisfies [k + t(x5 − x4 )]2 + [t(y5 − y4 )]2 = k2 . This yields t =−

2k(x5 − x4 ) , (x5 − x4 )2 + (y5 − y4 )2

and so we find for (x6 , y6 ) that x6 = k

(y5 − y4 )2 − (x5 − x4 )2 (x5 − x4 )(y5 − y4 ) , y6 = −2k . (x5 − x4 )2 + (y5 − y4 )2 (x5 − x4 )2 + (y5 − y4 )2

(10.10.1)

We define the sum αd + βd = γd , where γd has side-lines OQ and OZ6 and has its indicator in D1 . When Z4 = Z5 we take QZ6 as the line through Q which is parallel to the tangent to the circle at Z4 . This is analogous to the modified sum of angles. It can be checked that y6 x5 y4 + x4 y5 x6 x4 x5 − y4 y5 − − = 0, = 0. k k2 k k2

(10.10.2)

To see this we first note that (y5 − y4 )2 + (x5 − x4 )2 = 2[k2 − (x4 x5 + y4 y5 )]. Then the numerator in (x5 − x4 )(y5 − y4 ) x5 y4 + x4 y5 − x4 x5 + y4 y5 − k2 k2

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is equal to k2 [(x5 − x4 )(y5 − y4 ) + x5 y4 + x4 y5 ] − (x5 y4 + x4 y5 )(x4 x5 + y4 y5 ) = k2 (x5 y5 + x4 y4 ) − [x4 y4 (x25 + y25 ) + x5 y5 (x24 + y24 )] = (k2 − k2 )(x5 y5 + x4 y4 ) = 0. Similarly the numerator in 1 (y5 − y4 )2 − (x5 − x4 )2 x4 x5 − y4 y5 − 2 k2 − (x4 x5 + y4 y5 ) k2 equals k2 [(y5 − y4 )2 − (x5 − x4 )2 − 2(x4 x5 − y4 y5 )] + 2(x4 x5 + y4 y5 )(x4 x5 − y4 y5 ) = k2 [y25 + y24 − x25 − x24 ] + 2(x24 x25 − y24 y25 ) = k2 [y25 + y24 − x25 − x24 ] + 2[x24 (k2 − y25 ) − y24 y25 ] = k2 [y25 + y24 − x25 − x24 ] + 2[x24 k2 − y25 (x24 + y24 )] = k2 [y25 + y24 − x25 − x24 + 2x24 − 2y25 ] = 0. To apply these we note that by 10.10.4 sin αd =

y4 x4 y5 x5 , cos αd = , sin βd = , cos βd = . k k k k

The sum γd = αd + βd has side-lines OQ and OZ6 , and we sub-divide into two major cases. First we suppose that x5 y4 + x4 y5 > 0 or equivalently |αd |◦ + |βd |◦ < 180. Then y6 > 0 and we have y6 x6 sin γd = , cos γd = . k k It follows from (10.10.2) that sin(αd + βd ) = sin αd cos βd + cos αd sin βd , cos(αd + βd ) = cos αd cos βd − sin αd sin βd . Secondly we suppose that x5 y4 + x4 y5 < 0 or equivalently |αd |◦ + |βd |◦ > 180. Then y6 < 0 so we have sin γd = −

y6 x6 , cos γd = − . k k

It follows from (10.10.2) that − sin(αd + βd ) = sin αd cos βd + cos αd sin βd , − cos(αd + βd ) = cos αd cos βd − sin αd sin βd . There is a further case when x5 y4 + x4 y5 = 0 and we obtain these formulae according as x4 x5 − y4 y5 is positive or negative, respectively. Thus the addition formulae for sine and cosine of duo-angles are more complicated than those of angles.

Sec. 10.10

10.10.6

Angles between lines

173

Addition formulae for tangents of duo-angles

(i) We first note that if αd , βd ∈ DA (F ) and αd + βd = 90dF , neither duo-angle being null or right, then tan αd tan βd = 1. For we have that x6 = 0, so that by (10.10.2) x4 x5 − y4 y5 = 0 and thus y4 y5 = 1. x4 x5 (ii) Next we note that, as tan αd = sin αd / cos αd , it follows from the above addition formulae for cosine and sine that tan(αd + βd ) =

tan αd + tan βd , 1 − tan αd tan βd

provided that 0 does not occur in a denominator, that is provided none of αd , βd , αd + βd is a right duo-angle; this can be done separately for the cases considered in 10.10.5. In fact this addition formula for the tangent function can be verified without subdivision into cases, as y6 (y5 − y4 )(x5 − x4 ) = −2 , x6 (y5 − y4 )2 − (x5 − x4 )2 and we wish to show that this is equal to y5 /x5 + y4 /x4 x4 y5 + x5 y4 = . 1 − y4 y5 /x4 x5 x4 x5 − y4 y5

(10.10.3)

On subtracting the first of these expressions from the second, we obtain a quotient the numerator of which is equal to (x4 y5 + x5 y4 )[(y5 − y4 )2 − (x5 − x4 )2 ] + 2(x4 x5 − y4 y5 )(y5 − y4 )(x5 − x4 ) = (x4 y5 + x5 y4 )[y25 + y24 − x25 − x24 + 2(x4 x5 − y4 y5 )] + 2(x4 x5 − y4 y5 )[x4 y4 + x5 y5 − (x4 y5 + x5 y4 )] = (x4 y5 + x5 y4 )[y25 + y24 − x25 − x24 ] + 2(x4 x5 − y4 y5 )(x4 y4 + x5 y5 ) = (x4 y5 + x5 y4 )[y25 + y24 − x25 − x24 ] + 2(x24 x5 y4 − y24 x4 y5 + x25 x4 y5 − y25 x5 y4 ) = (x25 + y25 − x24 − y24 )(x4 y5 − x5 y4 ) = 0, as x24 + y24 = x25 + y25 = k2 . This identity then implies the standard addition formula for the tangents of duo-angles. (iii) We also wish to show that tan(αd + 90dF ) =

−1 , tan αd

when αd is neither null nor right. For with x5 = 0, y5 = k (10.10.1) gives x6 = k

(k − y4 )2 − x24 −x4 (k − y4 ) , y6 = 2k 2 , 2 2 x4 + (k − y4 ) x4 + (k − y4 )2

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Complex coordinates; sensed angles; angles between lines

so that

Then

10.10.7

Ch. 10

2x4 (k − y4 ) y6 = . x6 (k − y4 )2 − x24 y4 (k − y4 ) y4 (k − y4 ) y6 y4 =2 2 =2 4 = −1. 2 2 x6 x4 k − 2ky4 + y4 − x4 2y4 − 2ky4

Associativity of addition of duo-angles

With the notation of 10.10.5, suppose that αd , βd and γd are duo-angles in DA (F ), with pairs side- lines (OQ, OZ4 ), (OQ, OZ5 ), (OQ, OZ6 ), respectively. We wish to consider the sums (αd + βd ) + γd and αd + (βd + γd ). We suppose that αd + βd has side-lines (OQ, OZ7 ) and that (αd + βd ) + γd has side-lines (OQ, OZ9 ). Similarly we suppose that βd + γd has side-lines (OQ, OZ8 ) and αd + (betad + γd ) has side-lines (OQ, OZ10 ). Then by (10.10.2) applied several times we have that y7 = y9 = = x9 = =

x5 y4 + x4 y5 x4 x5 − y4 y5 , x7 = , k k x x −y y x y +x y x6 y7 + x7 y6 x6 5 4 k 4 5 + 4 5 k 4 5 y6 = k k x6 (x5 y4 + x4 y5 ) + (x4 x5 − y4 y5 )y6 , k2 x4 x5 −y4 y5 4 y5 x6 − x5 y4 +x y6 x7 x6 − y7 y6 k k = k k (x4 x5 − y4 y5 )x6 − (x5 y4 + x4 y5 )y6 . k2

Similarly y8 = y10 = = x10 = =

x6 y5 + x5 y6 x5 x6 − y5 y6 , x8 = , k k x5 x6 −y5 y6 5 y6 y4 + x4 x6 y5 +x x8 y4 + x4 y8 k k = k k (x5 x6 − y5 y6 )y4 + x4 (x6 y5 + x5 y6 ) , k2 x x −y y x y +x y x4 x8 − y4 y8 x4 5 6 k 5 6 − y4 6 5 k 5 6 = k k x4 (x5 x6 − y5 y6 ) − y4 (x6 y5 + x5 y6 ) . k2

From these we can see that Z9 = Z10 and so we have that (αd + βd ) + γd = αd + (βd + γd ). Thus addition of duo-angles is associative on DA (F ).

Sec. 10.10

10.10.8

Angles between lines

175

Group properties of duo-angles; sensed duo-angles

We note the following properties of addition of duo-angles:(i)Given any duo-angles αd , βd in DA (F ), the sum αd + βd is a unique object γd and it lies in DA (F ). (ii)Addition of duo-angles is commutative, that is

αd + βd = βd + αd , for all αd , βd ∈ DA (F ). (iii) Addition of duo-angles is associative on DA (F ). (iv)The null angle 0dF is a neutral element for + on DA (F ). (v)Each αd ∈ DA (F ) has an additive inverse in DA (F ). Proof . (i) This is evident from the definition. (ii) This is evident as the definition is symmetrical in the roles of the two duoangles. (iii) This was established in 10.10.7. (iv) For αd + 0dF = αd , for all αd ∈ DA (F ). (v) With the notation of 10.10.5 let Z5 = sOJ (Z4 ) so that Z5 ≡ (−x4 , y4 ), and let δd be the duo-angle in DA (F ) with arms OI, OZ5 . Then, straightforwardly, αd + δd = 0dF . Thus this duo-angle δd is an additive inverse for αd in DA (F ). We denote it by −αd . These properties show that we have a commutative group. We note that y4 = sin αd , k y4 tan(−αd ) = − = − tan αd . x4 sin(−αd ) =

cos(−αd ) = −

x4 = − cos αd , k

If α = ∠F QOZ4 is a wedge-angle in A (F ), with Z4 ≡ (x4 , y4 ) and y4 > 0, we recall that −α = ∠F QOZ6 where Z6 = sOI (Z4 ) ≡ (x4 , −y4 ). If αd is the duo-angle in DA (F ) with side-lines (OQ, OZ4 ) then −αd is the duo-angle in DA (F ) with side-lines (OQ, OZ5 ) where Z5 = sOJ (Z4 ) ≡ (−x4 , y4 ). This inverse angle and inverse duo-angle are linked in that OZ5 = OZ6 and so |−α |◦ = |−αd |◦ + 180. We define βd − αd = βd + (−αd ), and this is the duo-angle in standard position with side-lines OQ and OZ7 , where Z7 ≡ (x7 , y7 ) is the point where the line through Q and parallel to Z5 sOJ (Z4 ) meets the circle C (O; k) again. We call βd − αd the difference of two duo-angles and also a sensed duo-angle with side-lines OZ4 , OZ5 and denote it by F (OZ4 , OZ5 ). If F is any frame of reference obtained from F by translation, we also define F (OZ4 , OZ5 ) = F (OZ4 , OZ5 ). Earlier names for this were a ‘complete angle’ and a ‘cross’; see Forder [7] for applications and exercises, and Forder [6, pages 120–121, 151–154] for applications,

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Ch. 10

the terminology used being ‘cross’. Sensed duo-angles were also used by Johnson [9, pages 11 –15] under the name of ‘directed angles’. We have tan βd − tan αd , tan(βd − αd ) = 1 + tan αd tan βd provided none of αd , βd , βd − αd is a right duo-angle. For a coordinate formula to utilise this we replace x4 by −x4 in (10.10.3) and translate to parallel axes through Z1 . Thus for γd = F (Z1 Z4 , Z1 Z5 ) we have tan γd = when γd is not right, and 1+

y5 −y1 y4 −y1 x5 −x1 − x4 −x1 1 y4 −y1 1 + yx55 −y −x1 x4 −x1

y5 − y1 y4 − y1 =0 x5 − x1 x4 − x1

when it is.

10.10.9

An application

For fixed points Z4 and Z5 , consider the locus of points Z such that F (ZZ4 , ZZ5 ) has constant magnitude. If it is a right duo-angle we will have 1+

y5 − y y4 − y = 0, x5 − x x4 − x

and so the points Z ∈ Z4 Z5 lie on the circle on [Z4 , Z5 ] as diameter. Otherwise, we have that y5 −y y4 −y x5 −x − x4 −x y4 −y = 1 − λ , 1 + yx55 −y −x x4 −x for some λ = 1, and then the points Z ∈ Z4 Z5 lie on a circle which passes through Z4 and Z5 . In fact we obtain a set of coaxal circles through Z4 and Z5 . This should be compared with 7.5.1 and 10.9.1.

10.11 A CASE OF PASCAL’S THEOREM, 1640 10.11.1 Let Z1 ,W1 , Z2 ,W2 be distinct points on the circle C (O; k). Then Z1W2 W1 Z2 if and only if F Z2 OW2 = F Z1 OW1 . Proof . We let z1 ∼ kcis θ1 , z2 ∼ kcis θ2 , w1 ∼ kcis φ1 , w2 ∼ kcis φ2 . Then Z1W2 and W1 Z2 are parallel if and only if kcis φ2 − kcis θ1 =t kcis θ2 − kcis φ1

Sec. 10.11

A case of Pascal’s theorem, 1640

177

for some t = 0 in R. By 9.4.1 the left-hand side is equal to cos φ2 − cos θ1 + ı(sin φ2 − sin θ1 ) cos θ2 − cos φ1 + ı(sin θ2 − sin φ1 ) −2 sin( 12 φ2 + 12 θ1 ) sin( 12 φ2 − 12 θ1 ) + 2ı cos( 12 φ2 + 12 θ1 ) cos( 12 φ2 − 12 θ1 ) = −2 sin( 12 θ2 + 12 φ1 ) sin( 12 θ2 − 12 φ1 ) + 2ı cos( 12 θ2 + 12 φ1 ) cos( 12 θ2 − 12 φ1 ) = =

sin( 12 φ2 − 12 θ1 ) cis ( 12 φ2 + 12 θ1 ) sin( 12 θ2 − 12 φ1 ) cis ( 12 θ2 + 12 φ1 ) sin( 12 φ2 − 12 θ1 ) sin( 12 θ2 − 12 φ1 )

cis ( 12 φ2 + 12 θ1 − 12 θ2 − 12 φ1 ).

Thus cis ( 12 φ2 + 12 θ1 − 12 θ2 − 12 φ1 ) =

sin( 12 θ2 − 12 φ1 ) sin( 12 φ2 − 12 θ1 )

t.

But the absolute value here is 1, so the right-hand side is ±1. Thus we have either 1 1 1 1 2 φ2 + 2 θ1 − 2 θ2 − 2 φ1

= 0F ,

or 1 1 1 1 2 φ2 + 2 θ1 − 2 θ2 − 2 φ1

= 180F .

In each case, we have that φ2 − φ1 = θ2 − θ1 and so F Z2 OW2 = F Z1 OW1 . Z3 If (Z1 ,W1 ), (Z2 ,W2 ), (Z3 ,W3 ) are distinct pairs of points all on a circle and such that Z1W2

W1 Z2 and Z2W3 W2 Z3 then Z1W3 W1 Z3 . Proof . This follows immediately from the last subsection. It is a case of what is known as Pascal’s theorem.

W3

Z1

W2 W1

Z2

Figure 10.15. A case of Pascal’s theorem

COROLLARY. If (Z1 ,W1 ), (Z2 ,W2 ), (Z3 ,W3 ), (Z4 ,W4 ) are four distinct pairs of points all on a circle and such that Z1W2 W1 Z2 , then Z1W4 W1 Z4 .

Z2W3 W2 Z3 ,

Z3W4 W3 Z4 ,

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Ch. 10

Figure 10.16. Very symmetrical cases.

Proof . For from the first two we deduce that Z1W3 W1 Z3 and on combining this with the third relation we obtain the conclusion. NOTE. Clearly this last result can be extended to any number of pairs of points on a circle.

10.11.2 Starting more generally than in the last subsection, for pairs of distinct points let (Z1 ,W1 ) ∼ (Z2 ,W2 ) if and only if Z1W2 W1 Z2 . Then clearly the relation ∼ is reflexive and symmetric. We ask when it is also transitive and thus an equivalence relation. Now if Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ), W1 ≡ (u1 , v1 ), W2 ≡ (u2 , v2 ), we have (Z1 ,W1 ) ∼ (Z2 ,W2 ) if and only if (v2 − y1 )(x2 − u1 ) = (u2 − x1 )(y2 − v1 ).

(10.11.1)

Similarly we have (Z2 ,W2 ) ∼ (Z,W ) if and only if (v − y2 )(x − u2 ) = (u − x2 )(y − v2 ).

(10.11.2)

We wish (10.11.1) and (10.11.2) to imply that (v − y1 )(x − u1 ) = (u − x1 )(y − v1 ).

(10.11.3)

From (10.11.1) we have that v2 x2 − u2 y2 = u1 v2 − u2 v1 + x2 y1 − x1 y2 + x1 v1 − y1 u1 , and from (10.11.2) v2 x2 − u2 y2 = uv2 − u2 v + x2 y − xy2 + vx − uy, so together these give vx − uy = u1 v2 − u2 v1 + x2 y1 − x1 y2 + x1 v1 − y1 u1 − uv2 + u2 v − x2 y + xy2 . We need for (10.11.3) that vx − uy = vu1 − uv1 + y1 x − x1 y − y1 u1 + x1 v1

Sec. 10.11

A case of Pascal’s theorem, 1640

179

and so our condition for transitivity is got by equating the two right- hand sides here. This turns out to be δF (Z1 , Z2 , Z) = δF (W1 ,W2 ,W ). Now (10.11.2) and (10.11.3) simultaneously give a transformation under which Z → W as we see by writing them as y − v2 v − y2 = , u − x2 x − u2

v − y1 y − v1 = . u − x1 x − u1

(10.11.4)

On solving for u and v in this we obtain u = v =

y−v1 y−v2 y2 − y1 + x1 x−u − x2 x−u 1 2 y−v1 x−u1

y−v2 − x−u 2

x−u2 1 x2 − x1 + y1 x−u y−v1 − y2 y−v2 x−u1 y−v1

2 − x−u y−v2

, .

To utilize this transformation we consider loci with equations of the form 2h

y − v2 y − v1 y − v2 y − v1 + 2g +2f + c = 0. x − u2 x − u1 x − u2 x − u1

(10.11.5)

Under the transformation this maps into the locus with equation 2h

v − y2 v − y1 v − y2 v − y1 + 2g +2f + c = 0. u − x2 u − x1 u − x2 u − x1

(10.11.6)

On clearing the equation (10.11.5) of fractions we obtain 2h(y − v2 )(y − v1 ) + 2g(y − v2 )(x − u1 ) + 2 f (y − v1 )(x − u2 ) + c(x − u1 )(x − u2 ) = 0, from which we see that W1 and W2 are both on this locus. This equation can be rearranged as cx2 + 2(g + f )xy + 2hy2 − (2gv2 + 2 f v1 + cu1 + cu2 )x− (2gu1 + 2 f u2 + 2hv1 + 2hv2 )y + 2hv1 v2 + 2gu1 v2 + 2 f u2 v1 + cu1 u2 = 0. (10.11.7) Similarly we see that Z1 and Z2 are on the locus given by (10.11.6), and the equation for it becomes cu2 + 2(g + f )uv + 2hv2 − (2gy2 + 2 f y1 + cx1 + cx2 )u− (2gx1 + 2 f x2 + 2hy1 + 2hy2 )v + 2hy1 y2 + 2gx1 y2 + 2 f x2 y1 + cx1 x2 = 0. (10.11.8) We note that W1 maps to Z1 and W2 maps to Z2 under the transformation in which Z maps to W To identify all the loci that can occur in (10.11.7) and (10.11.8) would take us beyond the concepts of the present course, so we concentrate on when they represent circles.

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Complex coordinates; sensed angles; angles between lines

Ch. 10

Now (10.11.7) is a circle when c = 2h = 0 and g = − f . The equation then becomes g  g  x2 + y2 + (v1 − v2 ) − u1 − u2 x + (u2 − u1 ) − v1 − v2 y h h g + (u1 v2 − u2 v1 ) + u1 u2 + v1 v2 = 0. (10.11.9) h This is the set of circles which pass through the points W1 and W2 , a set of coaxal circles. The corresponding equation for the second locus is  g  g u2 + v2 + (y1 − y2 ) − x1 − x2 u + (x2 − x1 ) − y1 − y2 v h h g + (x1 y2 − x2 y1 ) + x1 x2 + y1 y2 = 0, (10.11.10) h and this gives the set of coaxal circles passing through Z1 and Z2 . We can take an arbitrary circle from the first coaxal set and then there is a unique one from the second set corresponding to it. If we take Z1 , Z2 ,W1 ,W2 to be concyclic we get just one circle and that is the classical case; it occurs when the remaining coefficients in the two equations are pairwise equal.

Z

W2

Z1 W

Z2 W1

Figure 10.17. Pascal result for two circles.

10.11.3 Instead of using parallelism of lines as the basis of the relation in 10.11.2, we could take instead a fixed line o with equation lx + my + n = 0, and let (Z1 ,W1 ) ∼ (Z2 ,W2 ) if the lines Z1W2 and W1 Z2 meet on o. The results are like those in 10.11.2 and the transformation corresponding to (10.11.4) is n(v2 − y) − l(u2 y − v2 x) n(y2 − v) − l(x2 v − y2 u) = , n(x2 − u) + m(x2 v − y2 u) n(u2 − x) + m(u2 y − v2 x) n(y1 − v) − l(x1 v − y1 u) n(v1 − y) − l(u1 y − v1 x) = . n(x1 − u) + m(x1 v − y1 u) n(u1 − x) + m(u1 y − v1 x)

Exercises 10.1 Prove the result of Varignon (1731) that if A, B,C, D are the vertices of a convex quadrilateral and P = mp(A, B), Q = mp(B,C), R = mp(C, D), S = mp(D, A),

Sec. 10.11

A case of Pascal’s theorem, 1640

181

then P, Q, R, S are the vertices of a parallelogram. 10.2 If Z1 ∼ z1 , Z2 ∼ z2 and Z3 ∼ z3 are non-collinear points show that z1 (¯z3 − z¯2 ) + z2 (¯z1 − z¯3 ) + z3 (¯z2 − z¯1 ) = 4iδF (Z1 , Z2 , Z3 ) = 0. 10.3 Let A ∼ a, B ∼ b, C ∼ c be non-collinear points and P ∼ p a point such that AP, BP, CP meet BC, CA, AB at D ∼ d, E ∼ e, F ∼ f , respectively. Show for sensed ratio that p(b¯ − a) ¯ + p(a ¯ − b) + ba¯ − ab¯ BD = , p(a¯ − c) ¯ + p(c ¯ − a) + ac¯ − ca¯ DC and hence prove Ceva’s theorem that BD CE AF = 1. DC EA FB 10.4 Let A ∼ a, B ∼ b, C ∼ c be non-collinear points. Given any point P ∼ p, show that as (c − a)/(b − a) is non-real there exist unique real numbers y and z such that p − a = y(b − a) + z(c − a), and so p = xa + yb + zc where x + y + z = 1. Show that if AP meets BC it is in a point D ∼ d such that d=

r 1 b+ c, 1+r 1+r

where r = z/y. Hence prove Ceva’s theorem that if D ∈ BC, E ∈ CA, F ∈ AB are such that AD, BE, CF are concurrent, then BD CE AF = 1. DC EA FB 10.5 Let A ∼ a, B ∼ b, C ∼ c be non-collinear points. If D ∼ d, E ∼ e where d=

λ μ 1 1 b+ c, e = c+ a, 1+λ 1+λ 1+μ 1+μ

and DE meets AB it is in a point F ∼ f where f=

ν 1 a+ b, 1+ν 1+ν

and λ μν = −1. Prove Menelaus’ theorem that if D ∈ BC, E ∈ CA, F ∈ AB are collinear, then BD CE AF = −1. DC EA FB

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Ch. 10

10.6 Let A, B,C be non-collinear points and take D ∈ BC, E ∈ CA, F ∈ AB such that BD CE AF = r, = s, = t. BC CA AB Let l, m, n be, respectively, the lines through D, E, F which are perpendicular to the side-lines BC, CA, AB. Show that l, m, n are concurrent if and only if (1 − 2r)|B, C|2 + (1 − 2s)|C, A|2 + (1 − 2t)|A, B|2 = 0. 10.7 If R(Z0 ) is the set of all rotations about the point Z0 , show that (R(Z0 ), ◦) is a commutative group. 10.8 Show that the composition of axial symmetries in two parallel lines is equal to a translation, and conversely that each translation can be expressed in this form. 10.9 Prove that sφ ;Z0 ◦ sθ ;Z0 = r2(φ −θ );Z0 . 10.10 Prove that sφ ;Z0 ◦ rθ ;Z0 = sφ − 1 θ ;Z0 . Deduce that any rotation about the point Z0 2 can be expressed as the composition of two axial symmetries in lines which pass through Z0 . 10.11 Let F1 ∼ F2 if the frame of reference F2 is positively oriented with respect to F1 . Show that ∼ is an equivalence relation. 10.12 Prove the Stewart identity (z4 − z1 )2 (z3 − z2 ) + (z4 − z2 )2 (z1 − z3 ) + (z4 − z3 )2 (z2 − z1 ) = − (z3 − z2 )(z1 − z3 )(z2 − z1 ). Interpret this trigonometrically. 10.13 Prove DeMoivre’s theorem that (cos α + ı sin α )n = cos(nα ) + ı sin(nα ), for all positive integers n and all angles α ∈ A ∗ (F ), where ı is the complex number satisfying ı2 = −1. 10.14 Suppose that l, m, n are distinct parallel lines. Let Z1 , Z2 , Z3 , Z4 ∈ l with Z1 = Z2 , Z3 = Z4 . Suppose that Z5 , Z6 ∈ n, Z1 Z5 , Z2 Z5 meet m at Z7 , Z8 , respectively, and Z3 Z6 , Z4 Z6 meet m at Z9 , Z10 , respectively. Prove that then Z9 Z10 Z3 Z4 = . Z7 Z8 Z1 Z2

Sec. 10.11

A case of Pascal’s theorem, 1640

183

10.15 If [Z1 , Z2 , Z3 , Z4 ] is a parallelogram, W is a point on the diagonal line Z1 Z3 , a line through W parallel to Z1 Z2 meets Z1 Z4 and Z2 Z3 at W1 and W2 respectively, and a line through W parallel to Z1 Z4 meets Z1 Z2 and Z3 Z4 at W3 and W4 , respectively, prove that

δF (W,W4 ,W1 ) = δF (W,W3 ,W2 ). 10.16 If Z1 = Z2 and δF (Z1 , Z2 , Z3 ) = −δF (Z1 , Z2 , Z4 ), prove that the mid-point of Z3 and Z4 is on Z1 Z2 . 10.17 Suppose that Z1 , Z2 , Z3 , Z4 are points no three of which are collinear. Show that [Z1 , Z3 ] ∩ [Z2 , Z4 ] = 0/ if and only if

δF (Z1 , Z2 , Z4 ) δF (Z2 , Z1 , Z3 ) < 0 and < 0. δF (Z3 , Z2 , Z4 ) δF (Z4 , Z1 , Z3 )

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11

Vector and complex-number methods 11.1 EQUIPOLLENCE 11.1.1 Definition . An ordered pair (Z1 , Z2 ) of points in Π is said to be equipollent to the pair (Z3 , Z4 ), written symbolically (Z1 , Z2 ) ↑ (Z3 , Z4 ), if mp(Z1 , Z4 ) = mp(Z2 , Z3 ). Thus ↑ is a binary relation in Π × Π. Equipollence has the properties:(i) If Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ), Z3 ≡ (x3 , y4 ), Z4 ≡ (x4 , y4 ), then (Z1 , Z2 ) ↑ (Z3 , Z4 ) if and only if x1 + x4 = x2 + x3 , y1 + y4 = y2 + y3 , or equivalently x2 − x1 = x4 − x3 , y2 − y1 = y4 − y3 . (ii) Given any points Z1 , Z2 , Z3 ∈ Π, there is a unique point Z4 such that (Z1 , Z2 ) ↑ (Z3 , Z4 ). (iii) For all Z1 , Z2 ∈ Π, (Z1 , Z2 ) ↑ (Z1 , Z2 ). (iv) If (Z1 , Z2 ) ↑ (Z3 , Z4 ) then (Z3 , Z4 ) ↑ (Z1 , Z2 ). (v) If (Z1 , Z2 ) ↑ (Z3 , Z4 ) and (Z3 , Z4 ) ↑ (Z5 , Z6 ), then (Z1 , Z2 ) ↑ (Z5 , Z6 ). (vi) If (Z1 , Z2 ) ↑ (Z3 , Z4 ) then (Z1 , Z3 ) ↑ (Z2 , Z4 ). (vii) If (Z1 , Z2 ) ↑ (Z3 , Z4 ), then |Z1 , Z2 | = |Z3 , Z4 |. (viii) For all Z1 ∈ Π, (Z1 , Z1 ) ↑ (Z3 , Z4 ) if and only if Z3 = Z4 . (ix) If Z1 = Z2 and Z3 ∈ l = Z1 Z2 , then (Z1 , Z2 ) ↑ (Z3 , Z4 ) if and only Z4 ∈ l, |Z1 , Z2 | = |Z3 , Z4 | and if ≤l is the natural order for which Z1 ≤l Z2 , then Z3 ≤l Z4 . Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50011-2 © 2016 Elsevier Ltd. All rights reserved.

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(x) If Z1 = Z2 and Z3 ∈ Z1 Z2 , then (Z1 , Z2 ) ↑ (Z3 , Z4 ) if and only if [Z1 , Z2 , Z4 , Z3 ] is a parallelogram.

Proof . (i) By the mid-point formula,



x1 + x4 y1 + y4 x2 + x3 y2 + y3 , , mp(Z2 , Z3 ) ≡ , , mp(Z1 , Z4 ) ≡ 2 2 2 2 and the result follows immediately from this. (ii) By part (i) it is necessary and sufficient that we choose Z4 so that x4 = x2 + x3 − x1 , y4 = y2 + y3 − y1 . (iii) This is immediate as x1 + x2 = x1 + x2 , y1 + y2 = y1 + y2 . (iv) This is immediate as x2 + x3 = x1 + x4 , y2 + y3 = y1 + y4 . (v) We are given that x1 + x4 = x2 + x3 , y1 + y4 = y2 + y3 , x3 + x6 = x4 + x5 , y3 + y6 = y4 + y5 . By addition (x1 + x6 ) + (x3 + x4 ) = (x2 + x5 ) + (x3 + x4 ), so by cancellation of x3 + x4 we have x1 + x6 = x2 + x5 . Similarly y1 + y6 = y2 + y5 and so the result follows. (vi) For by (i) above we have x1 + x4 = x3 + x2 , y1 + y4 = y3 + y2 . (vii) For by (i) above (x2 − x1 )2 + (y2 − y1 )2 = (x4 − x3 )2 + (y4 − y3 )2 , and now we apply the distance formula. (viii) For if x1 = x2 , y1 = y2 , then (i) above is satisfied if and only if x3 = x4 , y3 = y4 . Z3

Z1

Z2

Z4

Z2

Z1

T

O I Z3

Z4

Figure 11.1.

(ix) For suppose first that (Z1 , Z2 ) ↑ (Z3 , Z4 ). Then Z3 ∈ l and mp(Z3 , Z4 ) ∈ l, so Z4 ∈ l. By (vii) above we have |Z1 , Z2 | = |Z3 , Z4 |. Suppose first that l is not perpendicular to OI and that, as in 6.4.1(ii), the correspondence between ≤l and the natural order ≤OI , under which O ≤OI I, is direct. Then πOI (Z1 ) ≤OI πOI (Z2 ), so x1 ≤ x2 . By (i) above x3 ≤ x4 , and so by this argument traced in reverse we have Z3 ≤l Z4 . If the correspondence is indirect, we have x2 ≤ x1 , x4 ≤ x3 instead. When l is perpendicular to OI, we project to OJ instead and make use of the y-coordinates. Conversely suppose that Z4 ∈ l, |Z1 , Z2 | = |Z3 , Z4 | and Z3 ≤l Z4 . Now l has parametric equations x = x1 + t(x2 − x1 ), y = y1 + t(y2 − y1 ) (t ∈ R). Suppose that Z3 and Z4 have parameters t3 ,t4 , respectively, so that x3 = x1 + t3 (x2 − x1 ),

y3 = y1 + t3 (y2 − y1 ),

x4 = x1 + t4 (x2 − x1 ),

y4 = y1 + t4 (y2 − y1 ).

Sec. 11.2

Sum of couples, multiplication of a couple by a scalar

187

Recall that Z1 , Z2 have parameters 0 and 1 and 0 < 1. As in the last paragraph above, if l is not perpendicular to OI and the correspondence between ≤l and ≤OI is direct, then x1 < x2 , x3 < x4 ; hence t3 < t4 and we obtain this same conclusion when the correspondence is inverse. When l is perpendicular to OI we project to OJ instead, and use the y-coordinates. Moreover |Z3 , Z4 |2 = [(t4 − t3 )(x2 − x1 )]2 + [(t4 − t3 )(y2 − y1 )]2 = (t4 − t3 )2 |Z1 , Z2 |2 . Hence |t4 − t3 | = 1, and so as t3 < t4 we have t4 = 1 + t3 . Then x1 + x4 = 2x1 + (1 + t3 )(x2 − x1 ), x2 + x3 = x2 + x1 + t3 (x2 − x1 ) and these are equal. Similarly y1 + y4 = 2y1 + (1 + t3 )(y2 − y1 ), y2 + y3 = y2 + y1 + t3 (y2 − y1 ) and these are equal. By (i) above we now have (Z1 , Z2 ) ↑ (Z3 , Z4 ). (x) If [Z1 , Z2 , Z4 , Z3 ] is a parallelogram, then mp(Z1 , Z4 ) = mp(Z2 , Z3 ). Conversely suppose that Z1 = Z2 , Z3 ∈ Z1 Z2 and mp(Z1 , Z4 ) = mp(Z2 , Z3 ). Then Z1 Z2 , Z3 Z4 have equations −(y2 − y1 )(x − x1 ) + (x2 − x1 )(y − y1 ) = 0, −(y4 − y3 )(x − x3 ) + (x4 − x3 )(y − y3 ) = 0. By (i) above, x2 − x1 = x4 − x3 , y2 − y1 = y4 − y3 , so these lines are parallel. Similarly Z1 Z3 , Z2 Z4 have equations −(y3 − y1 )(x − x1 ) + (x3 − x1 )(y − y1 ) = 0, −(y4 − y2 )(x − x2 ) + (x4 − x2 )(y − y2 ) = 0, and by (i) above x3 − x1 = x4 − x2 , y3 − y1 = y4 − y2 , so that these lines are parallel. Thus [Z1 , Z2 , Z4 , Z3 ] is a parallelogram.

11.2 SUM OF COUPLES, MULTIPLICATION OF A COUPLE BY A SCALAR 11.2.1 Z3

Z2

Z1 O I

T O

Z4

Z1 Figure 11.2.

Z1

(t > 0) Z4

O I

(t < 0)

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Definition . For O ∈ Π, let V (Π; O) be the set of all couples (O, Z) for Z ∈ Π. We define the sum (O, Z1 ) + (O, Z2 ) of two couples to be (O, Z3 ) where mp(O, Z3 ) = mp(Z1 , Z2 ), so that (O, Z1 ) ↑ (Z2 , Z3 ). Thus + is a binary operation in V (Π; O). We define the product by a number or scalar t.(O, Z1 ), of a number t ∈ R and a couple, to be a couple (O, Z4 ) as follows.When Z1 = O we take Z4 = O for all t ∈ R. When Z1 = O we take Z4 to be in the line l = OZ1 and with |O, Z4 | = |t||O, Z1 |; furthermore if ≤l is the natural order for which O ≤l Z1 , we take O ≤l Z4 when t ≥ 0, and Z4 ≤l O when t < 0. Thus product by a number is a function on R × V (Π; O) into V (Π; O). COMMENT. To prove by synthetic means the basic properties of couples listed in 11.2.2 and 11.3.1, would be very laborious in covering all the cases. We establish instead initial algebraic characterizations which allow an effective algebraic approach. If O is the origin and Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ), then (i) (O, Z1 ) + (O, Z2 ) = (O, Z3 ) where Z3 ≡ (x1 + x2 , y1 + y2 ). (ii) t.(O, Z1 ) = (O, Z4 ) where Z4 ≡ (tx1 ,ty1 ).

Proof . (i) For this we have 0 + x3 = x1 + x2 , 0 + y3 = y1 + y2 . (ii) We verify this as follows. Let (x4 , y4 ) = (tx1 ,ty1 ). When (x1 , y1 ) = (0, 0) clearly we have (x4 , y4 ) = (x1 , y1 ). When (x1 , y1 ) = (0, 0), clearly Z4 ∈ OZ1 while |O, Z4 |2 = (tx1 )2 + (ty1 )2 = t 2 |O, Z1 |2 . Now if l is not perpendicular to OI and the correspondence between the natural order ≤l and the natural order ≤OI on OI, under which O ≤OI I, is direct then x1 < x2 . Thus when t > 0, we have tx1 > 0 and so O ≤l Z4 ; when t < 0, we have tx1 < 0 and so Z4 ≤l O. When the correspondence between the natural orders is inverse, we reach the same conclusion. When l is perpendicular to OI we project to OJ instead.

11.2.2 Vector space over R Definition . A triple (V , +, .) is said to be a vector space over R if the following hold:(i) First, + is a binary operation in V . (ii) For all a, b, c ∈ V , (a + b) + c = a + (b + c). (iii) There is an o ∈ V such that for all a ∈ V , a + o = a, o + a = a. (iv) Corresponding to each a ∈ V , there is some −a ∈ V such that (−a) + a = o, a + (−a) = o.

Sec. 11.2

Sum of couples, multiplication of a couple by a scalar

189

(v) For all a, b ∈ V , a + b = b + a. (vi) Next, R × V → V is a function. (vii) For all a ∈ V and all t1 ,t2 ∈ R, t2 .(t1 .a) = (t2 t1 ).a. (viii) For all a, b ∈ V and all t ∈ R, t.(a + b) = t.a + t.b. (ix) For all a ∈ V and all t1 , ,t2 ∈ R, (t1 + t2 ).a = t1 .a + t2 .a. (x) For all a ∈ V , 1.a = a. We then have the following result. (V (Π; O), +, .) is a vector space over R. Proof . (i) This has been covered already in 11.2.1. (ii) Now (O, Z1 ) + (O, Z2 ) = (O, Z4 ) where (x4 , y4 ) = (x1 + x2 , y1 + y2 ). Then [(O, Z1 ) + (O, Z2 )] + (O, Z3 ) = (O, Z4 ) + (O, Z3 ) = (O, Z5 ), where (x5 , y5 ) = (x4 + x3 , y4 + y3 ) = ((x1 + x2 ) + x3 , (y1 + y2 ) + y3 ) . Similarly (O, Z2 ) + (O, Z3 ) = (O, Z6 ) where (x6 , y6 ) = (x2 + x3 , y2 + y3 ), and so (O, Z1 ) + [(O, Z2 ) + (O, Z3 )] = (O, Z1 ) + (O, Z6 ) = (O, Z7 ) where (x7 , y7 ) = (x1 + x6 , y1 + y6 ) = (x1 + (x2 + x3 ), y1 + (y2 + y3 )). Clearly Z5 = Z7 . (iii) For any Z1 ∈ Π, (O, Z1 ) + (O, O) = (O, Z2 ) where (x2 , y2 ) = (x1 + 0, y1 + 0) = (x1 , y1 ), so that Z2 = Z1 . Similarly (O, O) + (O, Z1 ) = (O, Z3 ) where (x3 , y3 ) = (0 + x1 , 0 + y1 ) = (x1 , y1 ), so that Z3 = Z1 . (iv) Now (O, Z1 ) + (O, Z2 ) = (O, Z3 ), (O, Z2 ) + (O, Z1 ) = (O, Z4 ) where (x3 , y3 ) = (x1 + x2 , y1 + y2 ) and (x4 , y4 ) = (x2 + x1 , y2 + y1 ). Clearly Z3 = Z4 . (v) If (x2 , y2 ) = (−x1 , −y1 ), then (O, Z1 ) + (O, Z2 ) = (O, Z3 ) where (x3 , y3 ) = (x1 − x1 , y1 − y1 ) = (0, 0); hence Z3 = O. Similarly (O, Z2 ) + (O, Z1 ) = (O, Z4 ) where (x4 , y4 ) = (−x1 + x1 , −y1 + y1 ) = (0, 0); hence Z4 = O. (vi) This was covered in 11.2.1. (vii) For t1 .(O, Z1 ) = (O, Z2 ) where (x2 , y2 ) = (t1 x1 ,t1 y1 ). Then t2 . (t1 .(O, Z1 )) = t2 .(O, Z2 ) = (O, Z3 ) where (x3 , y3 ) = (t2 (t1 x1 ),t2 (t1 y1 )). Also (t2 t1 ).(O, Z1 ) = (O, Z4 ) where (x4 , y4 ) = ((t2 t1 )x1 , (t2 t1 )y1 ). Thus Z3 = Z4 . (viii) For (O, Z1 ) + (O, Z2 ) = (O, Z3 ) and t.[(O, Z1 ) + (O, Z2 )] = t.(O, Z3 ) = (O, Z4 ) where (x3 , y3 ) = (x1 + x2 , y1 + y2 ), (x4 , y4 ) = (t(x1 + x2 ),t(y1 + y2 )). Also t.(O, Z1 ) = (O, Z5 ), t.(O, Z2 ) = (O, Z6 ) where (x5 , y5 ) = (tx1 ,ty1 ), (x6 , y6 ) = (tx2 ,ty2 ). Moreover (O, Z5 ) + (O, Z6 ) = (O, Z7 ) where (x7 , y7 ) = (x5 + x6 , y5 + y6 ) = (tx1 + tx2 ,ty1 + ty2 ). Hence Z4 = Z7 . (ix) For t1 .(O, Z1 ) = (O, Z2 ), t2 .(O, Z1 ) = (O, Z3 ), (t1 + t2 ).(O, Z1 ) = (O, Z4 ) and (O, Z2 ) + (O, Z3 ) = (O, Z5 ) where (x2 , y2 ) = (t1 x1 ,t1 y1 ), (x3 , y3 ) = (t2 x1 ,t2 y1 ) and (x4 , y4 ) = ((t1 + t2 )x1 , (t1 + t2 )y1 ). Moreover (x5 , y5 ) = (x2 + x3 , y2 + y3 ) = (t1 x1 + t2 x1 ,t1 y1 + t2 y1 ). Clearly Z4 = Z5 . (x) For 1.(O, Z1 ) = (O, Z2 ) where (x2 , y2 ) = (1.x1 , 1.y1 ) = (x1 , y1 ). Thus Z2 = Z1 .

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11.3 SCALAR OR DOT PRODUCTS 11.3.1 Z2

Z1

Z2 Z1

O

πOZ1 (Z2 )

πOZ1 (Z2 ) O

Figure 11.3.

Definitions . We define a scalar product, or dot product, (O, Z1 ).(O, Z2 ) as follows. If Z1 = O then (O, Z1 ).(O, Z2 ) = 0; otherwise Z1 = O and we set  |O, Z1 ||O, πOZ1 (Z2 )|, if πOZ1 (Z2 ) ∈ [O, Z1 , (O, Z1 ).(O, Z2 ) = −|O, Z1 ||O, πOZ1 (Z2 )|, if πOZ1 (Z2 ) ∈ OZ1 \ [O, Z1 . Clearly the scalar product is a function on V (Π; O) × V (Π; O) into R. The norm a of a vector a = (O, Z) is defined to be the distance |O, Z|. The scalar product has the following properties:(i) If Z j ≡ (x j , y j ) for j = 1, 2 then a.b = (O, Z1 ).(O, Z2 ) = x1 x2 + y1 y2 . (ii) For all a, b ∈ V (Π; O), a.b = b.a. (iii) For all a, b, c ∈ V (Π; O), a.(b + c) = a.b + a.c. (iv) For all a, b ∈ V (Π; O) and all t ∈ R, t.(a.b) = (t.a).b. (v) For all a = o, a.a > 0, while o.o = 0. √ (vi) For all a, a = a.a.

Proof . (i) If Z1 = O, then x1 = y1 = 0 so that x1 x2 + y1 y2 = 0 as required. Suppose then that Z1 = O. Write l = OZ1 and let m be the line through the point O which is perpendicular to l. Define the closed half-plane H5 = {X : πl (X ) ∈ [O, Z1 } and let H6 be the other closed half-plane with edge m. Now l ≡ −y1 x+x1 y = 0 and m ≡ x1 x + y1 y = 0.

m H6

H5

l Z1

J O

I Figure 11.4.

Then as Z1 ∈ H5 , H5 = {Z ≡ (x, y) : x1 x + y1 y ≥ 0},

H6 = {Z ≡ (x, y) : x1 x + y1 y ≤ 0}.

Sec. 11.3

Scalar or dot products

191

But by 6.6.1(ii),

y1 x1 πl (Z2 ) ≡ x2 + 2 (−y1 x2 + x1 y2 ), y2 − 2 (−y1 x2 + x1 y2 ) y1 + x21 y1 + x21

x1 x2 + y1 y2 x1 x2 + y1 y2 . = x1 2 , y1 2 2 x1 + y1 x1 + y21



Thus 

2

2

|O, Z1 | |O, πl (Z2 )| = =

(x1 x2 + y1 y2 )2 x21 (x21 + y21 )2 (x1 x2 + y1 y2 )2 , (x21 + y21 )

2 2 (x1 x2 + y1 y2 ) + y1 (x21 + y21 )2

so that |O, Z1 ||O, πl (Z2 )| = |x1 x2 + y1 y2 |. If Z2 ∈ H5 so that (O, Z1 ).(O, Z2 ) = |O, Z1 ||O, πl (Z2 )|, and x1 x2 + y1 y2 ≥ 0 so that |x1 x2 + y1 y2 | = x1 x2 + y1 y2 , clearly (O, Z1 ).(O, Z2 ) = x1 x2 + y1 y2 . If Z2 ∈ H6 \ m we have πl (Z2 ) ∈ l \ [O, Z1 . Then (O, Z1 ).(O, Z2 ) = −|O, Z1 ||O, πl (Z2 )| and x1 x2 + y1 y2 ≤ 0, so that |x1 x2 + y1 y2 | = −(x1 x2 + y1 y2 ). Clearly again (O, Z1 ).(O, Z2 ) = x1 x2 + y1 y2 . (ii) Let a = (O, Z1 ), b = (O, Z2 ). Then by (i) of the present theorem, a.b = x1 x2 + y1 y2 , b.a = x2 x1 + y2 y1 , and clearly these are equal. (iii) Let a = (O, Z1 ), b = (O, Z2 ), c = (O, Z3 ). Then by 11.2.1(i) b + c = (O, Z4 ) where Z4 ≡ (x2 + x3 , y2 + y3 ). Then by (i) above a.(b + c) = x1 (x2 + x3 ) + y1 (y2 + y3 ), while a.b + a.c = (x1 x2 + y1 y2 ) + (x1 x3 + y1 y3 ), and these are equal. (iv) Let a = (O, Z1 ), b = (O, Z2 ). Then t.(a.b) = t(x1 x2 + y1 y2 ). But by 11.2.1(ii), t.a = (O, Z4 ) where Z4 ≡ (tx1 ,ty1 ) and so (t.a).b = (tx1 )x2 + (ty1 )y2 , which is equal to the earlier expression. (v) If a = (O, Z) then a.a = x2 + y2 . This is positive when (x, y) = (0, 0), and equal to 0 for x = y = 0. (vi) This follows immediately. NOTE. Note that 11.2.2(i) to (v) make (V , +) a commutative group. In textbooks on algebra it is proved that there is not a second element which has the property (iii); we shall refer to o as the null vector. It is also a standard result that for each a ∈ V there is not a second element with the property (iv); we call −a the inverse of a. Subtraction − is defined by specifying the difference b − a = b + (−a); then − is a binary operation on V . If a = (O, Z1 ), b = (O, Z2 ), then −a = (O, Z3 ) where Z3 ≡ (−x1 , −y1 ), and consequently b − a = (O, Z4 ) where Z4 ≡ (x2 − x1 , y2 − y1 ). Thus (O, Z2 ) − (O, Z1 ) = (O, Z4 ) if and only if (Z1 , Z2 ) ↑ (O, Z4 ).

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COMMENT. Now that we have set up our couples we call (O, Z) a position vector −→ with respect to the point O, and we adopt the standard notation OZ for (O, Z). Position vectors can be used for many geometrical purposes instead of Cartesian coordinates, or complex coordinates and complex-valued distances. We would note −−→ −−→ −−→ that by 6.1.1(iv) and 11.2 Z0 = mp(Z1 , Z2 ) if and only if OZ0 = 12 (OZ1 + OZ2 ); by −−→ −−→ −−→ −−→ 10.1.1(v) that Z1 Z2 Z3 Z4 if and only if OZ4 − OZ3 = t(OZ2 − OZ1 ) for some t = 0 −−→ in R, and by 9.7.1(ii) and 6.5.1 Corollary (ii) that Z1 Z2 ⊥ Z3 Z4 if and only if (OZ1 − −−→ −−→ −−→ OZ2 ).(OZ3 − OZ4 ) = 0. Most importantly, from parametric equations of a line x = x1 + t(x2 − x1 ), y = y1 + t(y2 − y1 ) (t ∈ R), we have that Z ∈ Z1 Z2 if and only if −−→ −−→ −−→ −−→ −→ −−→ OZ = OZ1 + t(OZ2 − OZ1 ) = (1 − t)OZ1 + t OZ2

(11.3.1)

for some t ∈ R. COMMENT. It is usual, in modern treatments, to define vectors to be the equivalence classes for equipollence. This defines free vectors. Position vectors are then defined by taking a specific point O in Π so that we have a pointed plane, and then concentrating on the representatives of the form (O, Z) for the vectors. But if our objective is to introduce position vectors, it is wasteful of effort to set up the free vectors, and in fact the use of free vectors and subsequent specialisation to position vectors can be a confusing route to position vectors.

11.4 COMPONENTS OF A VECTOR 11.4.1 Components Given non-collinear points Z1 , Z2 , Z3 , we wish to obtain an expression −−→ −−→ −−→ Z1 Z = pZ1 Z2 + qZ1 Z3 . For this we need (x2 − x1 )p + (x3 − x1 )q = x − x1 , (y2 − y1 )p + (y3 − y1 )q = y − y1 . We obtain the solutions p= and so have

δF (Z1 , Z, Z3 ) δF (Z1 , Z2 , Z) , q= , δF (Z1 , Z2 , Z3 ) δF (Z1 , Z2 , Z3 )

δF (Z1 , Z, Z3 ) −−→ δF (Z1 , Z2 , Z) −−→ −−→ Z1 Z2 + Z1 Z3 . Z1 Z = δF (Z1 , Z2 , Z3 ) δF (Z1 , Z2 , Z3 )

Sec. 11.4

Components of a vector

193

11.4.2 Areal coordinates Given non-collinear points Z1 , Z2 , Z3 , the position vector of any point Z of the plane −→ −−→ −−→ −−→ can be expressed in the form OZ = pOZ1 + qOZ2 + rOZ3 , with p + q + r = 1. This is equivalent to having q, r such that q(x2 − x1 ) + r(x3 − x1 ) = x − x1 , q(y2 − y1 ) + r(y3 − y1 ) = y − y1 . These equations have the unique solution q=

δF (Z, Z3 , Z1 ) δF (Z, Z1 , Z2 ) , r= , δF (Z1 , Z2 , Z3 ) δF (Z1 , Z2 , Z3 )

and now we take p = 1 − q − r so that by 10.5.4 p=

δF (Z, Z2 , Z3 ) . δF (Z1 , Z2 , Z3 )

For non-collinear points Z1 , Z2 , Z3 , for any Z we write

α = δF (Z, Z2 , Z3 ), β = δF (Z, Z3 , Z1 ), γ = δF (Z, Z1 , Z2 ), and call (α , β , γ ) areal point coordinates of Z with respect to (Z1 , Z2 , Z3 ). Note that we have p=

α β γ , q= , r= , δF (Z1 , Z2 , Z3 ) δF (Z1 , Z2 , Z3 ) δF (Z1 , Z2 , Z3 )

and α + β + γ = δF (Z1 , Z2 , Z3 ). These were first used by Möbius in 1827.

11.4.3 Cartesian coordinates from areal coordinates With the notation in 11.4.2, we have (y2 − y3 )x − (x2 − x3 )y = 2α − x2 y3 + x3 y2 , (y3 − y1 )x − (x3 − x1 )y = 2β − x3 y1 + x1 y3 , and if we solve these we obtain x=

y1 α + y2 β + y3 γ x1 α + x2 β + x3 γ , y= . δF (Z1 , Z2 , Z3 ) δF (Z1 , Z2 , Z3 )

11.4.4 The representation in 11.4.2 is in fact independent of the origin O. For we have x = px1 + qx2 + rx3 , y = py1 + qy2 + ry3 ,

194

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Ch. 11

and so for any point Z0 ≡F (x0 , y0 ), x − x0 = p(x1 − x0 ) + q(x2 − x0 ) + r(x3 − x0 ), y − y0 = p(y1 − y0 ) + q(y2 − y0 ) + r(y3 − y0 ). −−→ −−→ −−→ But Z ≡F (x − x0 , y − y0 ), where F = tO,Z0 (F ). Hence Z0 Z = pZ0 Z1 + qZ0 Z2 + −−→ rZ0 Z3 , with p + q + r = 1. −→ NOTATION. Where a vector equation is independent of the origin, as in OZ = −−→ −−→ −−→ pOZ1 + qOZ2 + rOZ3 , with p + q + r = 1, it is convenient to write this as Z = pZ1 + qZ2 + rZ3 with p + q + r = 1. In particular, in (11.3.1) we write Z = (1 − t)Z1 + tZ2 . Z1

Z2

O

Z Z3

Figure 11.5.

Figure 11.5 caters for when O and Z1 are taken as origins, a similar diagram would cater for when Z0 and Z1 are origins, and then a combination of the two would give the stated result.

11.4.5 −−→ We also use the notation δF (Z1 , Z2 , pZ4 + qZ5 + rZ6 ) for δF (Z1 , Z2 , Z3 ) where OZ3 = −−→ −−→ −−→ pOZ4 + qOZ5 + rOZ6 and p + q + r = 1. We can then write the conclusion of 10.5.3 as

δF (Z1 , Z2 , (1 − s)Z4 + sZ5 ) = (1 − s)δF (Z1 , Z2 , Z4 ) + sδF (Z1 , Z2 , Z5 ). The more general result

δF (Z1 , Z2 , pZ4 + qZ5 + rZ6 ) = pδF (Z1 , Z2 , Z4 ) + qδF (Z1 , Z2 , Z5 ) + rδF (Z1 , Z2 , Z6 ),

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195

where p + q + r = 1, can be deduced from this. For

δF (Z1 , Z2 , pZ4 + qZ5 + rZ6 )

q r Z5 + Z6 = δF Z1 , Z2 , pZ4 + (1 − p) 1− p 1− p

q r Z5 + Z6 = pδF (Z1 , Z2 , Z4 ) + (1 − p)δF Z1 , Z2 , 1− p 1− p   q r δF (Z1 , Z2 , Z5 ) + δF (Z1 , Z2 , Z6 ) . = pδF (Z1 , Z2 , Z4 ) + (1 − p) 1− p 1− p In this we have used the fact that q r q+r 1− p + = = = 1. 1− p 1− p 1− p 1− p

11.5 VECTOR METHODS IN GEOMETRY There is an informative account of many of the results of this chapter contained in Coxeter and Greitzer [5], dealt with by the methods of pure geometry. Some results are very basic, involving just collinearities or concurrencies, or ratio results. We start by showing how vector notation can be used to prove such results in a very straightforward fashion.

11.5.1 Menelaus’ theorem, c. 100A.D. On using the notation of sensed distances and sensed ratios in 7.6.1, we note that for noncollinear points Z1 , Z2 and Z3 , let Z4 ∈ Z2 Z3 , Z5 ∈ Z3 Z1 and Z6 ∈ Z1 Z2 . Then Z4 , Z5 and Z6 are collinear if and only if

Z6 Z1 Z5

Z2 Z4 Z3 Z5 Z1 Z6 = −1. Z4 Z3 Z5 Z1 Z6 Z2

Z2 Z4 Z3 Proof . Let Z4 = (1 − r)Z2 + rZ3 , Figure 11.6. Z5 = (1 − s)Z3 + sZ1 , Z6 = (1 − t)Z1 + tZ2 . Since Z4 , Z5 and Z6 are collinear, we have that Z6 = (1 − u)Z4 + uZ5 , for some real number u. Then (1 − t)Z1 + tZ2 = (1 − u)[(1 − r)Z2 + rZ3 ] + u[(1 − s)Z3 + sZ1 ].

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As the coefficients on each side add to 1, by the uniqueness in 11.4.2 we can equate coefficients and thus obtain 1 − t = su, t = (1 − u)(1 − r), r(1 − u) = −u(1 − s). On eliminating u we obtain s t u 1 r =− (1 − u) = −1, 1−s 1−t 1−r 1−u u and so

s t r = −1. 1−r 1−s 1−t This yields the stated result. This is known as Menelaus’ theorem.

11.5.2 Ceva’s theorem and converse, 1678 A.D. For non-collinear points Z1 , Z2 and Z3 , let Z4 ∈ Z2 Z3 , Z5 ∈ Z3 Z1 and Z6 ∈ Z1 Z2 . If Z1 Z4 , Z2 Z5 and Z3 Z6 are concurrent, then Z2 Z4 Z3 Z5 Z1 Z6 = 1. Z4 Z3 Z5 Z1 Z6 Z2

(11.5.1)

Proof . Denoting the point of concurrency by Z0 , we have Z4 = (1 − u)Z0 + uZ1 = (1 − r)Z2 + rZ3 , Z5 = (1 − v)Z0 + vZ2 = (1 − s)Z3 + sZ1 , Z6 = (1 − w)Z0 + wZ3 = (1 − t)Z1 + tZ2 , for some u, v, w, r, s,t ∈ R. Then u 1−r r Z1 + Z2 + Z3 , 1−u 1−u 1−u s v 1−s Z1 − Z2 + Z3 , Z0 = 1−v 1−v 1−v 1−t t w Z1 + Z2 − Z3 . Z0 = 1−w 1−w 1−w Z0 = −

On equating the coefficients of Z1 , Z2 and Z3 , in turn, we obtain u s 1−t − = = , 1−u 1−v 1−w v t 1−r =− = , 1−u 1−v 1−w 1−s w r = =− . 1−u 1−v 1−w

Z1 Z6

Z2

Z5

Z4 Figure 11.7.

Z3

Sec. 11.5

Vector methods in geometry

197

From this

1−v t 1−w r 1−u s = , = , = , 1−t 1−w 1−r 1−u 1−s 1−v and so by multiplication t r s = 1. 1−t 1−r 1−s Thus we obtain our conclusion. This is known as Ceva’s theorem. In fact we also have that u 1−u v 1−v w 1−w =− , =− , =− , s 1−v t 1−w r 1−u which gives uvw = −rst. This is Z0 Z4 Z0 Z5 Z0 Z6 Z2 Z4 Z3 Z5 Z1 Z6 =− . Z0 Z1 Z0 Z2 Z0 Z3 Z2 Z3 Z3 Z1 Z1 Z2

CONVERSE of Ceva’s theorem. Conversely, for non-collinear points Z1 , Z2 and Z3 , let Z4 ∈ Z2 Z3 , Z5 ∈ Z3 Z1 and Z6 ∈ Z1 Z2 . If (11.5.1) holds and Z2 Z5 and Z3 Z6 meet at a point Z0 , then Z1 Z4 also passes through Z0 . To start our proof we note that we have Z5 = (1 − v)Z0 + vZ2 = (1 − s)Z3 + sZ1 , Z6 = (1 − w)Z0 + wZ3 = (1 − t)Z1 + tZ2 . Hence Z0 =

s v 1−s 1−t t w Z1 − Z2 + Z3 , Z0 = Z1 + Z2 − Z3 . 1−v 1−v 1−v 1−w 1−w 1−w

It follows that 1−t v t 1−s w s = ,− = , =− , 1−v 1−w 1−v 1−w 1−v 1−w from which

1−v s = , (1 − s)t = vw. 1−t 1−w

On eliminating s between these, we obtain (1 − v)t 2 + (v − w)t − vw(1 − w) = 0. We then obtain two pairs of solutions, t = w, s = 1 − v, and t = −v

1 − vw 1−w , s= . 1−v 1−w

The first pair of solutions leads to v = w = 0 and so Z5 = Z6 = Z0 = Z1 , which we regard as a degenerate case. With Z4 = (1 − r)Z2 + rZ3 , we are given that st 1−r = , r (1 − s)(1 − t)

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and so have Z4 =

Ch. 11

st (1 − s)(1 − t) Z2 + Z3 . st + (1 − s)(1 − t) st + (1 − s)(1 − t)

With the second pair of solutions above, we obtain that Z0 =

w(1 − v) 1 − vw v(1 − w) Z1 − Z2 Z3 , (1 − v)(1 − w) (1 − v)(1 − w) (1 − v)(1 − w)

and also that Z4 = so that Z0 =

v(1 − w) w(1 − v) Z2 + Z3 , v + w − 2vw v + w − 2vw

1 − vw v + w − 2vw Z1 − Z4 . (1 − v)(1 − w) (1 − v)(1 − w)

As the sum of the coefficients of Z1 and Z4 is equal to 1, Z1 Z4 passes through Z0 . This proves the result. To obtain a formula for Z0 we note that on solving the second pair of solutions above for v and w, we obtain the pair of solutions v = 1, w = 1;

v=−

(1 − s)(1 − t) st , w=− . 1−t s

To see this, note that 1−w = − so that

  1 − v 1 + 1−v v t 1−v v t

1−v 1−v t, w = 1 + t, v v

= s,

i.e.

1 − v − (1 − v)t = s. − 1−v v t

Thus either v = 1 and consequently w = 1, or st 1−t , t = s, i.e. v = −v 1−t and hence

(1 − s)(1 − t) . s The first pair lead to Z5 = Z2 , Z6 = Z3 , another degenerate case, while the second pair lead to w=−

Z0 =

s(1 − t) st (1 − s)(1 − t) Z1 + Z2 + Z3 . 1 − t + st 1 − t + st 1 − t + st

(11.5.2)

Because of the condition (11.5.1) the coefficients in (11.5.2) could be given in several different forms.

Sec. 11.5

Vector methods in geometry

199

11.5.3 Desargues’ perspective theorem, 1648 A.D. Let (Z1 , Z2 , Z3 ) and (Z4 , Z5 , Z6 ) be two pairs of non-collinear points. Let Z2 Z3 and Z5 Z6 meet at W1 , Z3 Z1 and Z6 Z4 meet at W2 , and Z1 Z2 and Z4 Z5 meet at W3 . Then W1 ,W2 ,W3 are collinear if and only if Z1 Z4 , Z2 Z5 , Z3 Z6 are concurrent. Proof . Suppose that Z1 Z4 , Z2 Z5 , Z3 Z6 meet at a point Z0 . Then Z4 =

(1 − u)Z0 + uZ1 ,

Z5 =

(1 − v)Z0 + vZ2 ,

Z6 =

(1 − w)Z0 + wZ3 ,

W2

Z2 Z1 W3 Z3 Z4 Z6

W1

Z5 Figure 11.8.

for some u, v, w, ∈ R. On eliminating Z0 between the second and third of these, we obtain that (1 − w)Z5 − (1 − v)Z6 = v(1 − w)Z2 − w(1 − v)Z3 , from which we obtain that 1−w 1−v v(1 − w) w(1 − v) Z5 − Z6 = Z2 − Z3 . v−w v−w v−w v−w Now the sum of the coefficients of Z5 and Z6 is equal to 1, so the left-hand side represents a point on the line Z5 Z6 . Similarly, the sum of the coefficients of Z2 and Z3 is equal to 1, so the right-hand side represents a point on the line Z2 Z3 . Thus this must be the point W1 . By a similar argument based on the third and first lines, we find that 1−u 1−w w(1 − u) u(1 − w) Z6 − Z4 = Z3 − Z1 w−u w−u w−u w−u must be the point W2 , and by a similar argument based on the first and second lines, we find that 1−u u(1 − v) v(1 − u) 1−v Z4 − Z5 = Z1 − Z2 u−v u−v u−v u−v must be the point W3 .

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Vector and complex-number methods

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Then by repeated use of 10.5.3 and 11.4.5

δF (W1 ,W2 ,W3 ) v(1 − w) w(1 − v) u(1 − w) w(1 − u) Z2 − Z3 , − Z1 + Z3 , = δF v−w v−w w−u w−u

v(1 − u) u(1 − v) Z1 − Z2 u−v u−v   v(1 − w) w(1 − u) u(1 − v) w(1 − v) −u(1 − w) −v(1 − u) − δF (Z1 , Z2 , Z3 ) = v−w w−u u−v v−w w−u u−v = 0. This shows that W1 ,W2 and W3 are collinear. This is known as Desargues’ perspective or two-triangle theorem. Conversely, let W1 = (1 − l)Z2 + lZ3 = (1 − m)Z5 + mZ6 , W2 = (1 − p)Z3 + pZ1 = (1 − q)Z6 + qZ4 , W3 = (1 − r)Z1 + rZ2 = (1 − s)Z4 + sZ5 . From the third of these we deduce that (1 − r)Z1 − (1 − s)Z4 = sZ5 − rZ2 , and from this 1−s s r 1−r Z1 − Z4 = Z5 − Z2 , s−r s−r s−r s−r so that this must be the point of intersection of Z1 Z4 and Z2 Z5 . By a similar argument, we deduce from the second equation that 1−m m l 1−l Z2 − Z5 = Z6 − Z3 , m−l m−l m−l m−l and so this must be the point of intersection of Z2 Z5 and Z3 Z6 . By a similar argument, we deduce from the first equation that 1− p 1− p q p Z3 − Z6 = Z4 − Z1 , q− p q− p q− p q− p and so this must be the point of intersection of Z3 Z6 and Z1 Z4 . We are given now that W1 ,W2 and W3 are collinear, so that W3 = (1 − t)W1 + tW2 , for some t ∈ R. Then (1 − t)[(1 − l)Z2 + lZ3 ] + t[(1 − p)Z3 + pZ1 ] = (1 − r)Z1 + rZ2 , (1 − t)[(1 − m)Z5 + mZ6 ] + t[(1 − q)Z6 + qZ4 ] = (1 − s)Z4 + sZ5 . Since the points Z1 , Z2 , Z3 are not collinear we can equate the coefficients in the first line here, and obtain that pt = 1 − r, (1 − t)(1 − l) = r, (1 − t)l + t(1 − p) = 0,

Sec. 11.5

Vector methods in geometry

201

and since the points Z4 , Z5 , Z6 are not collinear we can equate the coefficients in the second line, and obtain that qt = 1 − s, (1 − t)(1 − m) = s, (1 − t)m + t(1 − q) = 0. Now for Z2 Z5 and Z3 Z6 to meet Z1 Z4 in the same point, we need to have p 1−r =− , s−r q− p and from this

s−r 1−r =− . p q− p

But we have from above

1−r 1−s = , p q

as a common value of t, and so need s−r 1−s =− q q− p or equivalently q(1 − r) = p(1 − s), and we have already noted that this is so. It follows that Z1 Z4 , Z2 Z5 and Z3 Z6 are concurrent.

11.5.4 Pappus’ theorem, c. 300A.D. Let the points Z1 , Z2 , Z3 lie on one line, and the points Z4 , Z5 , Z6 lie on a second line, these two lines intersecting at some point Z0 . Suppose that Z2 Z6 and Z5 Z3 meet at W1 , Z3 Z4 and Z6 Z1 meet at W2 , and Z1 Z5 and Z4 Z2 meet at W3 . Then the points W1 ,W2 ,W3 are collinear.

Z3 Z2 Z1 Z4

Z5

Z6

Figure 11.9.

Proof . We have that Z2 = (1 − p)Z0 + pZ1 , Z3 = (1 − q)Z0 + qZ1 , Z5 = (1 − u)Z0 + uZ4 , Z6 = (1 − v)Z0 + vZ4 , for some p, q, u, v ∈ R. On eliminating Z0 from the equations for Z2 and Z5 , we find that (1 − u)Z2 − (1 − p)Z5 = p(1 − u)Z1 − u(1 − p)Z4 , and so

u(1 − p) p(1 − u) 1− p 1−u Z2 + Z4 = Z1 + Z5 . 1 − pu 1 − pu 1 − pu 1 − pu

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This must be the point W3 then. Similarly, on eliminating Z0 from the equations for Z3 and Z6 we have that (1 − v)Z3 − (1 − q)Z6 = q(1 − v)Z1 − (1 − q)vZ4 , and so (1 − q)v q(1 − v) 1−q 1−v Z3 + Z4 = Z1 + Z6 . 1 − qv 1 − qv 1 − qv 1 − qv This must be the point W2 then. Now from the equations for Z2 and Z3 we have that pZ3 − qZ2 = (p − q)Z0 , and from the equations for Z5 and Z6 we have that uZ6 − vZ5 = (u − v)Z0 . On combining these, we have that (v − u)(pZ3 − qZ2 ) = (q − p)(uZ6 − vZ5 ). From this we have that u(q − p) q(v − u) p(v − u) (q − p)v Z6 + Z2 = Z3 + Z5 . qv − pu qv − pu qv − pu qv − pu This must then be the point W1 . However, the left-hand sides of the representations for W1 ,W2 and W3 contain four points Z1 , Z2 , Z5 , Z6 and we wish to reduce this to three non-collinear points. For this purpose we eliminate Z5 . From the equations for Z5 and Z6 we have that uZ6 − vZ5 = (u − v)Z0 , while from the equation for Z2 we have Z2 − pZ1 = (1 − p)Z0 . Combining these gives u (u − v) (Z2 − pZ1 ). Z5 = Z6 − v v(1 − p) On substitution, this gives that W3 =

pu(1 − v) v−u u(1 − p) Z1 + Z2 + Z6 . v(1 − pu) v(1 − pu) v(1 − pu)

We note that the sum of the coefficients for each of W1 ,W2 ,W3 in terms of Z1 , Z2 and Z6 is equal to 1, and so by repeated use of 10.5.3 and 11.4.5 we have that

Sec. 11.5

Vector methods in geometry

203

δF (W1 ,W2 ,W3 ) is equal to u(q − p) q(1 − v) 1−q q(v − u) Z2 + Z6 , Z1 + Z6 , δF qv − pu qv − pu 1 − qv 1 − qv

pu(1 − v) v−u u(1 − p) Z1 + Z2 + Z6 v(1 − pu) v(1 − pu) v(1 − pu) q(v − u) q(1 − v) u(1 − p) = δF (Z2 , Z1 , Z6 ) qv − pu 1 − qv v(1 − pu) q(v − u) 1 − q pu(1 − v) δF (Z2 , Z6 , Z1 ) + qv − pu 1 − qv v(1 − pu) u(q − p) q(1 − v) v − u + δF (Z6 , Z1 , Z2 ) qv − pu 1 − qv v(1 − pu) qu(1 − v)(v − u) [−q(1 − p) + p(1 − q) + q − p]δF (Z1 , Z2 , Z6 ) = v(qv − pu)(1 − qv)((1 − pu) =0. This shows that W1 ,W2 and W3 are collinear. This is known as Pappus’ theorem.

11.5.5 Centroid of a triangle If Z4 , Z5 , Z6 are the mid-points of {Z2 , Z3 }, {Z3 , Z1 }, {Z1 , Z2 }, respectively, then with the notation of 11.5.2 we have that r = s = t = 12 , and the condition (11.5.1) in the converse of Ceva’s theorem holds. Note that [Z3 , Z6 ] is a cross-bar for the interior region I R(|Z3 Z2 Z1 ) and so [Z2 , Z5 meets [Z3 , Z6 ] in a point Z7 , which is thus on both Z2 Z5 and Z3 Z6 . It follows that it is also on Z1 Z4 . Thus the lines joining the vertices of a triangle to the mid-points of the opposite sides are concurrent. The point of concurrence Z7 is called the centroid of the triangle, and for it by 11.5.2 we have Z7 = 13 Z1 + 13 Z2 + 13 Z3 . (11.5.3)

11.5.6 Orthocentre of a triangle Let Z8 , Z9 , Z10 be the feet of the perpendiculars from Z1 to Z2 Z3 , Z2 to Z3 Z1 , Z3 to Z1 Z2 , respectively. Then with the notation of 11.5.2 we have that r= Hence 1−r =

c a b cos β , s = cos γ , t = cos α . a b c

c cos β + b cos γ − c cos β b a − c cos β = = cos γ . a a a

Similarly

c a cos α , 1 − t = cos β . b c It follows that the condition (11.5.1) in the converse of Ceva’s theorem is true. 1−s =

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Ch. 11

Repeating an argument that we used in 7.2.3, suppose now that Z9 Z11 m and n are any lines which Z3 are perpendicular to Z3 Z1 and Z1 Z10 Z1 Z2 , respectively. If we had m

n, then we would have m ⊥ Z8 Z3 Z1 , n m so that n ⊥ Z3 Z1 ; but already n ⊥ Z1 Z2 so Z3 Z1 Z1 Z2 ; Z2 as Z1 , Z2 , Z3 are not collinear, Figure 11.10. Orthocentre of triangle. this gives a contradiction. Thus m is not parallel to n and so these lines meet in a unique point Z11 . In particular the lines Z2 Z9 , Z3 Z10 must meet in a unique point Z11 and then by the converse of Ceva’s theorem, Z1 Z4 will pass through Z11 . Thus the lines through the vertices of a triangle which are perpendicular to the opposite side-lines are concurrent. The point of concurrence Z11 is called the orthocentre of the triangle. By (11.5.2) we thus have Z11 = a a a a b cos γ c cos β c cos γ cos α b cos α cos β Z + Z + Z3 1 2 a a a a a a c cos β + c cos γ cos α c cos β + c cos γ cos α c cos β + c cos γ cos α

=

a cos γ cos α c cos β cos γ cos α cos β Z1 + Z2 + Z3 . b cos β + cos γ cos α cos β + cos γ cos α b cos β + cos γ cos α (11.5.4)

We could also proceed in this special case as follows. The argument is laid out for the case in the diagram, with β and γ acute angles and (Z1 , Z2 , Z3 ) positive in orientation. The other cases can be treated similarly. Now by 5.2.2 applied to [Z10 , Z2 , Z3 ], |∠Z11 Z2 Z4 |◦ = |∠Z10 Z2 Z3 |◦ = 90 − |γ |◦ so that |Z11 , Z8 | = tan ∠Z11 Z2 Z8 = cot γ . |Z2 , Z8 | But |Z2 , Z8 | = c cos β and so |Z11 , Z8 | = c cos β cot γ . Thus δF (Z11 , Z2 , Z3 ) = cot γ and since δF (Z1 , Z2 , Z3 ) = 12 ac sin β , we have that

1 2 ac cos β

δF (Z11 , Z2 , Z3 ) = cot β cot γ . δF (Z1 , Z2 , Z3 ) As similar results hold in the other two cases, we have by 11.4.2 that Z11 = cot β cot γ Z1 + cot γ cot α Z2 + cot α cot β Z3 . That the sum of the coefficients is equal to 1 follows from the identity tan α + tan β + tan γ = tan α tan β tan γ

Sec. 11.5

Vector methods in geometry

205

for the angles of a triangle. For, using the notation of 10.8.1, |α |◦ + |β |◦ + |γ |◦ = 180, and so αF + βF + γF = 180F so that − tan γF = tan(180F − γF ) = tan(αF + βF ) tan αF + tan βF = , 1 − tan αF tan βF whence the result follows by multiplying across and rearranging. This formula fails in the case of a right-angled triangle. From our two methods we have two formulae for Z11 , but we further note that cos(αF + γF ) = cos(180F − βF ), cos αF cos γF − sin αF sin γF = − cos βF , cos α cos γ + cos β = sin α sin γ . On using this with the sine rule, the two formulae for the orthocentre are reconciled.

11.5.7 Incentre of a triangle Let Z1 , Z2 , Z3 be non-collinear points. By 5.5.1 the mid-line of |Z2 Z1 Z3 will meet [Z2 , Z3 ] in a point Z12 where Z12 = (1 − r)Z2 + rZ3 , and c r = . 1−r b By similar arguments the mid-line of |Z3 Z2 Z1 will meet [Z3 , Z1 ] in a point Z13 where Z13 = (1 − s)Z3 + sZ1 , and a s = , 1−s c and the mid-line of |Z1 Z3 Z2 will meet [Z1 , Z2 ] in a point Z14 where Z14 = (1 − t)Z1 + tZ2 , and b t = . 1−t a The product of these three ratios is clearly equal to 1 so (11.5.1) is satisfied. By the cross-bar theorem, [Z2 , Z13 will meet [Z3 , Z14 ] in a point Z15 and so Z2 Z13 , Z3 Z14 meet in Z15 . It follows that Z1 Z12 also passes through the point Z15 . Thus the mid-lines of the angle-supports |Z2 Z1 Z3 , Z3 |Z3 Z2 Z1 , |Z1 Z3 Z2 for a triangle Z1 [Z1 , Z2 , Z3 ] are concurrent. The Z15 perpendicular distances from this point Z15 to the side-lines of the triangle are equal by Ex.4.4, so the circle with Z15 as centre and length of radius Z2 these common perpendicular Figure 11.11. Incentre of triangle. distances will pass through the feet of these perpendiculars.

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This circle is called the incircle for the triangle; its centre Z15 is called the incentre of the triangle. The three side-lines are tangents to the circle with the points of contact being the feet of the perpendiculars. For the incentre, by (11.5.2) we have the formula Z15 =

a b c Z1 + Z2 + Z3 . a+b+c a+b+c a+b+c

(11.5.5)

11.6 MOBILE COORDINATES NOTE. In standard vector notation, the vector product or cross product takes us out of the plane Π and into solid geometry. Sensed-area gives us half of the magnitude of the vector product and we use that instead. Without the vector product, however, we have not got orientation of the plane Π by vector means. We go on to supply this lack. However the standard vector operations can be awkward in dealing with perpendicularity and distance, and can involve quite a bit of trigonometry, so we also set out a method of reducing unwieldy calculations.

11.6.1 Grassmann’s supplement of a vector Given any Z = O, we show that there is a unique W such that |O, W| = |O, Z|, OW ⊥ OZ, δF (O, Z,W ) > 0.

Proof . With Z ≡ (x, y), W ≡ (u, v) these require x2 + y2 = u2 + v2 , ux + vy = 0, xv − yu > 0. By the middle one of these

   x y     −v u  = 0,

so the rows of this are linearly dependent. Thus we have r(x, y) + s(−v, u) = (0, 0), for some (r, s) = (0, 0). We cannot have r = 0 as that would imply W = O and so Z = O. Then s s x = v, y = − u, r r so by the first property above x2 + y2 =

s2 2 (x + y2 ). r2

Thus we have either s/r = 1, so that u = −y, v = x, for which 2δF (O, Z,W ) = x2 + y2 > 0, or we have s/r = −1, so that u = y, v = −x, for which 2δF (O, Z,W ) = −(x2 + y2 ) < 0. Thus the unique solution is u = −y, v = x.

Sec. 11.6

Mobile coordinates

207

−→⊥ −−→ For any Z ∈ Π, Z = O, we define OZ = OW where Z ≡ (x, y), W ≡ (−y, x), and −→ call this the Grassmann supplement of OZ. This clearly has the properties −−→⊥ −−→⊥ −−→ −−→ (OZ1 + OZ2 )⊥ = OZ1 + OZ2 , −→ −→ (kOZ)⊥ = k(OZ)⊥ , −→⊥ −→ (OZ )⊥ = −OZ.

11.6.2 In F we take |O, I| = |O, J| = 1. If |O, Z| = 1 and θ is the angle in AF with support |IOZ, then we recall from 9.2.2 that Z ≡ (x, y) where x = cos θ , y = sin θ . As I ≡ (1, 0), J ≡ (0, 1), we note that − → − → −→⊥ − → − → − →⊥ − → −→ OI = OJ, OZ = cos θ OI + sin θ OJ, OZ = − sin θ OI + cos θ OJ. −−→ − → − → Suppose that we also have OW = cos φ OI + sin φ OJ. Then by 11.4.1 we have −−→ −→ −→⊥ OW = rOZ + sOZ where r = cos φ cos θ + sin φ sin θ , s = sin φ cos θ − cos φ sin θ . By the addition formulae we recognise that r = cos F ZOW, s = sin F ZOW.

11.6.3 Handling a triangle Although for a triangle [Z1 , Z2 , Z3 ], we have the vector form for the centroid as (11.5.3) and for the incentre as (11.5.5) where as usual a = |Z2 , Z3 |, b = |Z3 , Z1 |, c = |Z1 , Z2 |, neither this formula for the incentre, nor the more awkward formula for the orthocentre, are convenient for applications and generalization. In 7.2.3 we noted that a unique circle passes through the vertices of a triangle [Z1 , Z2 , Z3 ]. It is called the circumcircle of this triangle and its centre is called the circumcentre. It is possible to find an expression for the circumcentre in terms of the vertices as in 11.4.4 but it is tedious to cover all the cases. For these reasons we consider the following way of representing any triangle. Z3 Z1 W 90◦

Z

O Figure 11.12. Grassmann supplement.

W1 Z2 w1 − z2 = p(z3 − z2 ); z1 − w1 = qı(z3 − z2 )

208

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Given non-collinear points Z1 , Z2 , Z3 , by 11.4.1 we can express −−→ −−→ −−→ −−→ −−→ −−→ OZ1 = OZ2 + p1 (OZ3 − OZ2 ) + q1 (OZ3 − OZ2)⊥ , for unique real numbers p1 and q1 , not both equal to 0. This is the geometrical background but the manipulations are simpler if we use complex coordinates instead. In (11.6.1) for any Z = O we take Z ∼ z where z = x + ıy and W ∼ w where w = u + iv. Then we note that ız = −y + ıx and so W = Z ⊥ if and only if w = ız. Then for Z1 ∼ z1 , Z2 ∼ z2 , Z3 ∼ z3 we write z1 − z2 = p1 (z3 − z2 ) + q1 ı(z3 − z2 ) = (p1 + q1 ı)(z3 − z2 ),

(11.6.1)

that is (x1 − x2 , y1 − y2 ) = (p1 (x3 − x2 ) − q1 (y3 − y2 ), p1 (y3 − y2 ) + q1 (x3 − x2 )) . We coin the name mobile coordinates of the point Z1 with respect to (Z2 , Z3 ) and F , for the pair (p1 , q1 ).  It follows immediately that |Z1 , Z2 | =

p21 + q21 |Z2 , Z3 |, and as from (11.6.1) z1 −  z3 = (p1 − 1 + q1 ı)(z3 − z2 ) we also have |Z3 , Z1 | = (p1 − 1)2 + q21 |Z2 , Z3 |. From z3 − z1 p1 − 1 + q1 ı z1 − z2 z2 − z3 1 , , = = p1 + q1 ı, = z2 − z1 p1 + q1 ı z3 − z2 z1 − z3 1 − p1 + q1 ı with α = F Z2 Z1 Z3 , β = F Z3 Z2 Z1 , γ = F Z1 Z3 Z2 , we have that (p1 − 1 + q1 ı)(p1 − q1 ı) p1 + q1 ı 1 − p1 + q1 ı  , cis β =  , cis γ =  . cis α =  p21 + q21 (p1 − 1)2 + q21 p21 + q21 (1 − p1 )2 + q21 We also have that 

c p21 + q21 cis β = cis β , a  b 1 − p1 − ıq1 = (p1 − 1)2 + q21 cis γ = cis γ , a   p1 + ıq1 =

p1 (p1 − 1) + q21 + ıq1 =

=

p21 + q21

(p1 − 1)2 + q21 cis α

bc cis α . a2

(11.6.2)

Thus we have in terms of p1 and q1 , the ratios of the lengths of the sides and cosines and sines of the angles. Moreover, it is easily calculated that

δF (Z1 , Z2 , Z3 ) =

q1 |Z2 , Z3 |2 , 2

so the orientation of this triple is determined by the sign of q1 .

Sec. 11.6

Mobile coordinates

209

Note too that if z − z2 = (p + qı)(z3 − z2 ), z − z2 = (p + q ı)(z3 − z2 ), then |z − z| = |p − p + (q − q)ı||z3 − z2 |, and so |Z, Z | =



 (p − p)2 + (q − q)2 |Z2 , Z3 | = a (p − p)2 + (q − q)2 .

(11.6.3)

11.6.4 Circumcentre of a triangle To make our notation consistent we now use the notation Z3 Z5 [Z1 , Z2 , Z3 ] for the triangle Z1 [A, B,C] in 7.2.3 and Z16 Z4 for its circumcentre D. As Z6 usual we denote by Z4 , Z5 , Z6 Z16 the mid-points of the sides Z2 [Z2 , Z3 ], [Z3 , Z1 ], [[Z1 , Z2 ], respectively. We note that points Z on the perpendicular bisector of [Z2 , Z3 ] have complex coordinates of the form Figure 11.13. Circumcentre of triangle. z = z2 + ( 12 + qı)(z3 − z2 ), where q ∈ R. We also have z6 = 21 (z1 + z2 ). As we are to have ZZ6 ⊥ Z1 Z2 we must have z2 + ( 12 + qı)(z3 − z2 ) − 12 (z1 + z2 ) = ır(z1 − z2 ), for some real number r. Then z2 + ( 12 + qı)(z3 − z2 ) − 12 [2z2 + (p1 + q1 ı(z3 − z2 )] = ır(z1 − z2 ), [ 12 + qı − 12 (p1 + q1 ı) − ır(p1 + q1 ı)](z3 − z2 ) = 0, 1 1 2 p1 − rq1 − 2

+ ( 12 q1 + rp1 )ı = qı

Thus we must have q = 12 q1 + rp1 , rq1 = 12 p1 − 12 , 1 −1 and then q = from which we obtain the solutions r = p2q 1 Thus the circumcentre has complex coordinate

z16 = z2 +

1 2 2 2q1 [p1 + q1 − p1 ].

  p2 + q21 − p1 1 1+ 1 ı (z3 − z2 ). 2 q1

(11.6.4)

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By Pythagoras’ theorem |z16 − z2 |2 = |z4 − z2 |2 + |z4 − z16 |2 = { 12 |z3 − z2 |2 }2 + {|q|2 |z3 − z2 |}2 = { 14 + [ =

q21 + p21 − p1 2 ] }|z3 − z2 |2 2q1

a2 {(q21 + p21 − p1 )2 + q21 }. 4q21

Thus the radius length of the circumcircle is  a q21 + [p21 + q21 − p1 ]2 . |z16 − z2 | = 2|q1 |

(11.6.5)

In fact we can also derive the formula (11.6.5) slightly differently as follows. Directly from (11.6.4) we have that 2

p1 + q21 p1 1 ı|2 a2 |z16 − z2 |2 = |1 + − 4 q1 q1 

2  2 p1 + q21 p1 1 a2 = 4 1+ − q1 q1 =

 a2  2 q1 + [p21 + q21 − p1 ]2 . 2 4q1

We can deduce from this material a formula for the circumcentre in terms of areal coordinates. For by (11.6.2) p21 − p1 + q21 =

bc bc cos α , q1 = 2 sin α , a2 a

so that z = z2 + 12 (1 + cot α ı)(z3 − z2 ). Then by (11.6.1) we have x − x2 = 12 (x3 − x2 ) − 12 cot α (y3 − y2 ), y − y2 = 12 cot α (x3 − x2 ) + 12 (y3 − y2 ). From this we have that

δF (Z, Z2 , Z3 ) = δF (Z − Z2 , O, Z3 − Z2 )   = 14 cot α (x3 − x2 )2 + (y3 − y2 )2 = 14 a2 cot α . As δF (Z1 , Z2 , Z3 ) = 12 q1 a2 = 12 bc sin α we have 1 a2 cot α δF (Z, Z2 , Z3 ) = , δF (Z1 , Z2 , Z3 ) 2 bc sin α and by use of the sine rule this is seen to be equal to 1 cos α . 2 sin β sin γ

Sec. 11.6

Mobile coordinates

211

By cyclic rotation we can write down the other two coefficients and so have Z16 =

1 cos α 1 cos β 1 cos γ Z1 + Z2 + Z3 . 2 sin β sin γ 2 sin γ sin α 2 sin α sin β

That the sum of the coefficients here is equal to 1 follows from the identity sin 2α + sin 2β + sin 2γ = 4 sin α sin β sin γ , for the angles of a triangle. For sin 2αF + sin 2βF + sin 2γF = sin(αF + βF ) cos(αF − βF ) + 2 sin γF cos γF = 2 sin γF [cos(αF − βF ) + cos γF ] = 2 sin γF [cos(αF − βF ) − cos(αF + βF )] = 2 sin γF .2 sin αF sin βF .

11.6.5 Other distinguished points for a triangle For the centroid Z7 of [Z1 , Z2 , Z3 ] we have by 11.5.3 that z7 = 13 (z1 + z2 + z3 ) and so z7 − z2 = 13 (z1 + z2 + z3 ) − z2 = 13 (z1 − z2 ) + 13 (z3 − z2 ) = 13 (p1 + q1 ı)(z3 − z2 ) + 13 (z3 − z2 ) = 13 (p1 + 1 + q1 ı)(z3 − z2 ).

(11.6.6)

This gives the complex coordinate of the centroid. We next turn to the orthocentre of this triangle. Points Z on the line through Z1 perpendicular to Z2 Z3 have complex coordinates of the form z =z2 + (p1 + qı)(z3 − z2 ) = z3 + (p1 − 1 + qı)(z3 − z2 ) p1 − 1 + qı (z1 − z2 ). =z3 + p1 + q1 ı For Z3 Z to be also perpendicular to Z1 Z2 we also need the coefficient of z1 − z2 here to be purely imaginary. Thus we need (p1 − 1)p1 + qq1 = 0 i.e.

and so obtain z11 = z2 + p1

q=−

(p1 − 1)p1 , q1

p1 − 1 ı (z3 − z2 ), 1− q1

(11.6.7)

as the complex coordinate of the orthocentre. It takes more of an effort to deal with the incentre. The point with complex coordinate z2 + |Z21,Z3 | (z3 − z2 ) lies on the half-line [Z2 , Z3 at a unit distance from Z2 .

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Similarly the point with complex coordinate z2 + |Z21,Z1 | (z1 − z2 ), lies on the half-line [Z2 , Z1 at a unit distance from Z2 . We wish to consider the mid-point of these points with complex coordinates z2 +

1 1 (z3 − z2 ), z2 + (z1 − z2 ). |Z2 , Z3 | |Z2 , Z1 |

Now z1 − z2 = (p1 + ıq1 )(z3 − z2 ) so that |z1 − z2 | = |p1 + ıq1 ||z3 − z2 | = Then



p21 + q21 |z3 − z2 |.

p1 + ıq1 z3 − z2 z1 − z2 = . |Z2 , Z1 | p21 + q21 |Z2 , Z3 |

The mid-point of these two points has complex coordinate ⎤ ⎡ 1 ⎣z3 − z2 +  1 (z1 − z2 )⎦ z2 + 2|Z2 , Z3 | p21 + q21 ⎡ ⎤ + q ı 1 p 1 ⎣z3 − z2 + 1 = z2 + (z3 − z2 )⎦ 2|Z2 , Z3 | 2 p1 + q21 ⎡ ⎤ p + ıq 1 1 1 ⎣1 +  ⎦ (z3 − z2 ). = z2 + 2|Z2 , Z3 | 2 2 p +q 1

1

Then points on the midline of |Z1 Z2 Z3 then have complex coordinates of the form ⎡ ⎤ p + q ı r 1 1 ⎣1 +  ⎦ (z3 − z2 ), z2 + 2|Z2 , Z3 | 2 2 (p + q 1

1

for real numbers r. By a similar argument the point with complex coordinate z3 + |Z21,Z3 | (z2 − z3 ) lies on the half-line [Z3 , Z2 at a unit distance from Z3 . Also the point with complex coordinate z3 + |Z31,Z1 | (z1 − z3 ), lies on the half-line [Z3 , Z1 at a unit distance from Z3 . We wish to consider the mid-point of these points with complex coordinates z3 +

1 (z2 − z3 ), |Z2 , Z3 |

z3 +

1 (z1 − z3 ). |Z3 , Z1 |

As preparation we note that z1 = z2 + (p1 + ıq1 )(z3 − z2 ) = z3 + (p1 − 1 + ıq1 )(z3 − z2 )  z1 − z3 = (1 − p1 − ıq1 )(z2 − z3 ), |Z1 , Z3 | = (1 − p1 )2 + q21 |Z2 , Z3 |, z1 − z3 1 − p1 − ıq1 z2 − z3 =  . |Z3 , Z1 | [ (1 − p )2 + q2 |Z2 , Z3 | 1 1

Sec. 11.6

Mobile coordinates

213

The mid-point sought is

⎤ ⎡ − ıq 1 1 − p 1 1 ⎣z2 − z3 +  (z2 − z3 )⎦ z3 + 2|Z2 , Z3 | 2 (1 − p1 ) + q21 ⎡ ⎤ − ıq 1 1 − p 1 1 ⎣1 +  ⎦ = z3 + 2|Z2 , Z3 | 2 2 (1 − p ) + q 1

1

Then points on the midline of |Z2 Z3 Z1 have complex coordinates of the form ⎛ ⎞ s ⎝ 1 − p1 + q1 ı ⎠ (z2 − z3 ) z3 + 2|Z2 , Z3 | (p − 1)2 + q2 1

1

for real numbers s. For a point of intersection of the two mid-lines we need ⎛ ⎞ r ⎝1 + p1 + q1 ı ⎠ (z3 − z2 ) z2 + 2|Z2 , Z3 | (p21 + q21 ⎛ ⎞ 1 − p − q ı s 1 1 ⎝1 +  ⎠ (z2 − z3 ) = z3 + 2|Z2 , Z3 | 2 2 (p1 − 1) + q1 ⎛ ⎞ − q ı s 1 − p 1 1 ⎝1 +  ⎠ (z3 − z2 ). = z3 − 2|Z2 , Z3 | 2 2 (p − 1) + q 1

1

Then z3 − z2 + ⎡ ⎤ −r p 1 − p + ıq − ıq s 1 1 1 1 +⎣ (1 +  (1 +  − )⎦ (z3 − z2 ) = 0, 2|Z2 , Z3 | 2|Z , Z | 2 2 2 2 3 2 p +q (1 − p ) + q 1

1

1

1

s p1 + ıq1 1 − p1 − ıq1 −r )− (1 +  (1 +  1+ ) = 0. 2|Z2 , Z3 | 2|Z , Z | 2 3 p21 + q21 (1 − p1 )2 + q21 Equating to 0 the imaginary part we obtain sq1 rq1 + = 0, − p21 + q21 (1 − p1 )2 + q21 and the real part

⎛ ⎛ ⎞ ⎞ r p 1 − p s 1 ⎝1 +  1 ⎝1 +  ⎠− ⎠ = 0. 1− 2|Z2 , Z3 | 2|Z , Z | 2 2 2 3 p +q (1 − p )2 + q2 1

1

1

1

214

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Ch. 11

Assuming q1 = 0 we have the pair of equations 

⎛

1

p21 + q21 ⎞ ⎛

r− 

1 (1 − p1 )2 + q21 ⎞ 1 − p1 ⎠

⎠ r + ⎝1 +  ⎝1 +  p1 2 2 p1 + q1 (1 − p1 )2 + q21

s = 0,

s = 2|Z2 , Z3 |.

(11.6.8)

We denote by D1 the value of the determinant of coefficients on the left-hand side   √1  p21 +q21   1 + √ p1  p2 +q2 1

and have

 D1 =

1

     

−√ 1+

1 (1−p1 )2 +q21 √ 1−p12 2 (1−p1 ) +q1

 (1 − p1 )2 + q21 + p21 + q21 + 1   . p21 + q21 (1 − p1 )2 + q21

We denote by D2 the value of the determinant   0 −√ 1 2 2  (1−p1 ) +q1   2|Z2 , Z3 | 1 + √ 1−p1  (1−p1 )2 +q2 1

     

which has the value D2 = 

2|Z2 , Z3 | (1 − p1 )2 + q21

.

Then we have the solution  2|Z , Z | p21 + q21 2 3 D2  . r= = D1 (1 − p1 )2 + q21 p21 + q21 + 1 Thus the point of intersection from this bisector is 

⎡ ⎤ p21 + q21 p1 + ıq1 ⎦ ⎣1 +   (z3 − z2 ), z2 +  2 2 2 (1 − p1 )2 + q1 + p1 + q1 + 1 p21 + q21 which is thus the incentre Z15 .

(11.6.9)

Sec. 11.6

Mobile coordinates

215

11.6.6 Euler line of a triangle With the notation Z7 , Z16 , Z11 for the centroid, circumcentre and orthocentre, respectively, of a triangle [Z1 , Z2 , Z3 ] we have the formulae z7 − z2 = 13 (p1 + 1 + q1 ı)(z3 − z2 ),

1 p2 − p1 + q21 ı (z3 − z2 ), 1+ 1 z16 − z2 = 2 q1

p1 − 1 z11 − z2 = p1 1 − ı (z3 − z2 ). q1 It is straightforward to check that 2 1 3 z16 + 3 z11

= z2 + 23 (z16 − z2 ) + 13 (z11 − z2 ) = z7 ,

and so Z7 ∈ Z16 Z11 . Thus we have shown that the centroid, circumcentre and orthocentre of any triangle are collinear. This is a result due to Euler, after whom this line of collinearity is named the Euler line of the triangle.

11.6.7 Similar triangles For any two triangles [Z1 , Z2 , Z3 ] and [Z1 , Z2 , Z3 ] suppose that we have z1 = z2 + (p1 + q1 ı)(z3 − z2 ), z 1 = z 2 + (p 1 + q 1 ı)(z 3 − z 2 ), where (p1 , q1 ), (p 1 , q 1 ) are couples of real numbers, neither being (0, 0). Then these triangles are similar in the correspondence (Z1 , Z2 , Z3 ) → (Z1 , Z2 , Z3 ) if and only if p 1 = p1 , q 1 = ±q1 . Proof . First suppose that p 1 = p1 , q 1 = q1 so that z1 = z2 + (p1 + q1 ı)(z3 − z2 ), z 1 = z 2 + (p 1 + q 1 ı)(z 4 − z 2 ). Then we have that |F Z3 Z2 Z1 |◦ = |F Z3 Z2 Z1 |◦ , |F Z1 Z3 Z2 |◦ = |F Z1 Z3 Z2 |◦ . It follows by 5.3.2 that the measures of the corresponding angles of these triangles are equal, and so the triangles are similar, with the lengths of corresponding sides proportional. Moreover, the triples (Z1 , Z2 , Z3 ) and (Z1 , Z2 , Z3 ) are similarly oriented. Z3 Z3

Z1 W1

Z1

W1

Z2

Z2 Figure 11.14. Similar triangles.

216

Vector and complex-number methods

Ch. 11

Next suppose that z1 = z2 + (p1 + q1 ı)(z3 − z2 ), z 1 = z 2 + (p1 − q1 ı)(z 3 − z 2 ). By an analogous argument the triangles are still similar, and now the triples (Z1 , Z2 , Z3 ) and (Z1 , Z2 , Z3 ) are oppositely oriented. Conversely, suppose that [Z1 , Z2 , Z3 ] and [Z1 , Z2 , Z3 ] are similar triangles in the correspondence (Z1 , Z2 , Z3 ) → (Z1 , Z2 , Z3 ). Let W1 be the foot of the perpendicular from Z1 to Z2 Z3 , and from parametric equations of Z2 Z3 choose p1 ∈ R so that w1 = z2 + p1 (z3 − z2 ). Then |Z2 , W1 | = |p1 |, |Z2 , Z3 | and p1 is positive or negative according or not as W1 is on the same or opposite side of Z2 as Z3 is on the line Z2 Z3 , that is according as the wedge-angle ∠Z3 Z2 Z1 is acute or obtuse. As W1 Z1 ⊥ Z2 Z3 we can find q1 ∈ R so that z1 − w1 = q1 ı(z3 − z2 ). Then |Z1 , W1 | = |q1 |, |Z2 , Z3 | and q1 is positive or negative according as (Z1 , Z2 , Z3 ) is positively or negatively oriented. As the lengths of the sides of the two triangles are proportional, we have |Z2 , Z3 | = k|Z2 , Z3 |, |Z3 , Z1 | = k|Z3 , Z1 |, |Z1 , Z2 | = k|Z1 , Z2 |, for some k > 0. Let W1 be the foot of the perpendicular from Z4 to the line Z5 Z6 . Then the triangles [Z1 , Z2 ,W1 ] and [Z1 , Z2 ,W1 ] are similar, so we have that |Z2 , W1 | |Z1 , Z2 | = = k. |Z2 , W1 | |Z1 , Z2 | It follows that |Z2 , W1 | = k|Z2 , W1 | = k|p1 ||Z2 , Z3 | = |p1 ||Z2 , Z3 |. But W1 is on the same side of the point Z2 on the line Z2 Z3 as Z3 is if the wedge-angle ∠Z3 Z2 Z1 is acute, and on the opposite side if this angle is obtuse. Hence we have that w 1 − z 2 = p1 (z 3 − z2 ). As Z1 W1 ⊥ Z2 Z3 , we have z 1 −w 1 = jı(z 3 −z 2 ) for some j ∈ R, and then |Z1 , W1 | = |j||Z1 , Z3 |. But |Z1 , W1 | |Z1 , Z2 | |Z2 , Z3 | = = , |Z1 , W1 | |Z1 , Z2 | |Z2 , Z3 | so |Z1 , W1 | |Z1 , W1 | = = |q1 |. |Z2 5, Z3 | |Z2 , Z3 | Hence j = ±q1 and we are to take the plus if (Z1 , Z2 , Z3 ) and (Z1 , Z2 , Z3 ) have the same orientation, the minus if the opposite orientation. Thus our mobile coordinates (p1 , q1 ) are intimately connected with similarity of triangles.

Sec. 11.6

Mobile coordinates

217

11.6.8 Centroids of similar triangles erected on the sides of a triangle Given an arbitrary triangle [Z1 , Z2 , Z3 ], if we consider points Z4 , Z5 and Z6 defined by z4 =z2 + (p1 + q1 ı)(z3 − z2 ), z5 = z3 + (p1 + q1 ı)(z1 − z3 ), z6 =z1 + (p1 + q1 ı)(z2 − z1 ), for some non-zero real numbers p1 and q1 , then we have triangles [Z2 , Z3 , Z4 ], [Z3 , Z1 , Z5 ], [Z1 , Z2 , Z6 ] erected on the sides [Z2 , Z3 ], [Z3 , Z1 ] and [Z1 , Z2 ], respectively, which are similar to each other and have the same orientation as each other. We denote the centroid of [Z1 , Z2 , Z3 ] by Z7 so that z7 = 13 (z1 + z2 + z3 ) and the centroid of [Z4 , Z5 , Z6 ] by Z7 so that z 7 = 13 [z2 + (p1 + q1 ı)(z3 − z2 ) + z3 + (p1 + q1 ı)(z1 − z3 ) + z1 + (p1 + q1 ı)(z2 − z1 ), = 13 [z2 + z3 + z1 ] + 13 (p1 + q1 ı)[(z3 − z2 ) + (z1 − z3 ) + (z2 − z1 )] = 13 (z1 + z2 + z3 ). Hence Z7 is the same point as Z7 .

(i)

First we denote the centroid of [Z2 , Z3 , Z4 ] by Z7 and then (i)

z7 = 13 [z2 + z3 + z4 ] = 13 [z2 + z3 + z2 + (p1 + q1 ı)(z3 − z2 )]. (ii)

Secondly we denote the centroid of Z3 , Z1 , Z5 ] by Z7 and then (ii)

z7 = 13 [z3 + z1 + z3 + (p1 + q1 ı)(z1 − z3 )]. (iii)

Finally we denote the centroid of [Z1 , Z2 , Z6 ] by Z7 (iii)

z7

and then

= 13 [z1 + z2 + z1 + (p1 + q1 ı)(z2 − z1 )].

Then we have that (i)

(ii)

(iii)

z7 + z7 + z7

= 13 [(2z2 + z3 + 2z3 + z1 + 2z1 + z2 ) + (p1 + q1 ı)(z3 − z2 + z1 − z3 + z2 − z1 )] = z1 + z2 + z3 . (i)

(ii)

(iii)

It follows that the centroid of the triangle [Z7 , Z7 , Z7 ] is the same point as the centroid Z7 of the original triangle [Z1 , Z2 , Z3 ].

218

Vector and complex-number methods

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Z5 Z5 Z3

Z3

Z1

Z1

Z6 Z6 Z2

Z4 Z2

Z4

Figure 11.15. Similar triangles on sides of triangle.

11.6.9 Circumcentres of similar triangles on sides of triangle In a more complicated fashion than in the last subsection, for an arbitrary triangle [Z1 , Z2 , Z3 ] suppose that we take points Z4 , Z5 and Z6 so that z4 =z2 + (p1 + q1 ı)(z3 − z2 ), z3 = z1 + (p1 + q1 ı)(z5 − z1 ), z2 =z6 + (p1 + q1 ı)(z1 − z6 ),

(11.6.10)

so that we have similar triangles once again on the sides of the original triangle but now in the correspondences (Z2 , Z3 , Z4 ) → (Z1 , Z5 , Z3 ) → (Z6 , Z1 , Z2 ). We (i) (ii) (iii) let Z16 , Z16 , Z16 be the circumcentres of these three similar triangles, so that by (1.6.4)we have

p21 − p1 + q21 1 (i) 1+ z16 − z2 = ı (z3 − z2 ), (11.6.11) 2 q1

p2 − p1 + q21 1 (ii) 1+ 1 ı (z5 − z1 ), (11.6.12) z16 − z1 = 2 q1

1 p2 − p1 + q21 (iii) ı (z1 − z6 ). (11.6.13) z16 − z6 = 1+ 1 2 q1 1 But by (11.6.9) we have z5 − z1 = p1 +q (z3 − z1 ) and z2 − z1 = (1− p1 − q1 ı)(z6 − z1 ), 1ı 1 so that z6 − z1 = 1−p1 −q1 ı (z2 − z1 ).

Sec. 11.6

Mobile coordinates

219

Also, by (11.6.12) (iii)

(ii)

(iii)

z16 + (p1 + q1 ı)(z16 − z16 ) (iii)

(ii)

= (1 − p1 − q1 ı)z16 + (p1 + q1 ı)z16 (iii)

(ii)

(iii)

(ii)

= (1 − p1 − q1 ı)(z16 − z1 ) + (p1 + q1 ı)(z16 − z1 ) + (1 − q1 ı + z1 ı)z1 = (1 − p1 − q1 ı)(z16 − z1 ) + (p1 + q1 ı)(z16 − z1 ) + z1     p2 − p1 + q21 p2 − p1 + q21 = 21 1 − 1 ı (z2 − z1 ) + 12 1 + 1 ı (z3 − z1 ) + z1 q1 q1   2 2 2 p1 − p1 + q21 1 1 p1 − p1 + q1 = 2 (z2 + z3 ) + 2 ı(z3 − z1 ) − ı(z2 − z1 ) q1 q1 p2 − p1 + q21 = 12 (z2 + z3 ) + 12 1 ı(z3 − z1 − (z2 − z1 )) q1 p2 − p1 + q21 ı(z3 − z2 ) = 12 (z2 + z3 ) + 12 1 q1 (i)

=z16 . (i)

(ii)

(iii)

It follows that the triangle [Z16 , Z16 , Z16 ] is also similar to the similar triangles (iii) (ii) (i) above, in the correspondence (Z2 , Z3 , Z4 ) → (Z16 √, Z16 , Z16 ). In the particular case when p1 = 1/2, q1 = 3/2, the similar triangles are all equilateral triangles, that is all three sides have equal lengths. In this case the last result is known as Napoleon’s theorem. It is easier to prove than the more general case, as for it we can work just with centroids.

11.6.10

Triangle with vertices the mid-points of sides of given triangle

For a triangle [Z1 , Z2 , Z3 ] let Z4 , Z5 , Z6 be the mid-points of the the sides [Z2 , Z3 ], [Z3 , Z1 ], [Z1 , Z2 ], respectively. Suppose that z1 = z2 + (p1 + q1 ı)(z3 − z2 ). Now z4 = 12 (z2 + z3 ) = z2 + 12 (z3 − z2 ), z5 = 12 (z3 + z1 ) = z2 + 12 (z3 − z2 ) + 12 [z2 + (p1 + q1 ı(z3 − z2 )] =z2 + 12 (1 + p1 + q1 ı)(z3 − z2 ), z6 = 12 (z1 + z2 ) = 12 [z2 + (p1 + q1 ı(z3 − z2 ) + z2 ] = z2 + 12 (p1 + q1 ı)(z3 − z2 ). Then z6 − z5 = 12 (p1 + q1 ı − 1 − p1 − q1 ı)(z3 − z2 ) = − 12 (z3 − z2 ), z4 − z5 = z2 + 12 (z3 − z2 ) − z2 − 12 (1 + p1 + q1 ı)(z3 − z2 ) = − 12 (p1 + q1 ı)(z3 − z2 ) = (p1 + q1 ı)(z6 − z5 ). Thus the triangle [Z4 , Z5 , Z6 ] is similar to the original triangle [Z1 , Z2 , Z3 ].

(11.6.14)

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Vector and complex-number methods

Ch. 11

of the triangle [Z , Z , Z ] is given by By (11.6.4) the circumcentre Z16 4 5 6

z 16

 = z5 +

1 2

 p21 + q21 p1 1+ ı − ı (z6 − z5 ), q1 q1

(11.6.15)

and if we relate this to the original triangle it becomes z 16

  p1 p21 + q21 = z2 + ı (− 12 )(z3 − z2 ) 1+− + q1 q1   p21 + q21 1 1 1 p1 = z2 + 2 1 + p1 + q1 ı − 2 + 2 − ı (z3 − z2 ) q1 2q1   p1 p2 + q21 ı (z3 − z2 ). (11.6.16) = z2 + 12 12 + p1 + q1 ı + 12 ı − 1 q1 2q1 1 1 2 (1 + p1 + q1 )ı(z3 − z2 ) + 2

We now find another point on this circle with centre (11.6.16). The foot Z8 of the perpendicular from Z1 onto Z2 Z3 has complex coordinate z8 = z2 + p1 (z3 − z2 ). We seek the circumcentre of the triangle [Z4 , Z8 , Z6 ]. For the moment let us take z4 = z8 + (p2 + q2 ı)(z6 − z8 )

(11.6.17)

from which z4 = (1 − p2 − q2 ı)z8 + (p2 + q2 ı)z6 = (1 − p2 − q2 ı)[z2 + p1 (z3 − z2 )] + (p2 + q2 ı)[z2 + 12 (p1 + q1 ı)(z3 − z2 )] = (1 − p2 − q2 ı + p2 + q2 ı)z2 + [p1 (1 − p2 − q2 ı) + (p2 + q2 ı) 12 (p1 + q1 ı)](z3 − z2 )   = z2 + p1 (1 − p2 ) + 12 (p2 p1 − q2 q1 ) + [−p1 q2 + 12 (p2 q1 + p1 q2 )]ı] (z3 − z2 ) = z2 + 12 (z3 − z2 ). Thus p1 (1 − p2 ) + 12 (p2 p1 − q2 q1 ) + [−p1 q2 + 12 (p2 q1 + p1 q2 )]ı = 12 , from which 1 2 (p2 q1 − p1 q2 ) =

0,

p1 (1 − p2 ) + 12 (p2 p1 − q2 q1 ) = 12 .

As p1 = 0 we have q2 = pp21 q1 . As 2p1 − 2p1 p2 + p2 p1 − q2 q1 = 1 we have 2p1 − p1 p2 = 1 + q2 q1 . Hence 2p1 − p1 p2 = 1 + pp21 q21 and so 2p1 − 1 = p2 ( q2 =

q21 + p1 ), p1

p2 =

(2p1 − 1)p1 , p21 + q21

q1 (2p1 − 1)p1 (2p1 − 1)q1 = . p1 p21 + q21 p21 + q21

Sec. 11.6

Mobile coordinates

Then p22 + q22 =

221

(2p1 − 1)2 (p21 + q21 ) (2p1 − 1)2 = . (p21 + q21 )2 p21 + q21

By (11.6.4) the circumcentre Z16 of the triangle [Z1 , Z2 , Z3 ] has complex coordinate   p1 p21 + q21 1 z16 = z2 + 2 1 − ı + ı (z3 − z2 ), q1 q1 (ii)

and so the circumcentre Z16 of of the triangle [Z4 , Z8 , Z6 ] has complex coordinate   p2 p22 + q22 (ii) 1 z16 = z8 + 2 1 − ı + ı (z6 − z8 ) q2 q2    p2 p2 + q22  1 ı 2 (p1 + q1 ı)(z3 − z2 ) − p1 (z3 − z2 ) = z2 + p1 (z3 − z2 ) + 12 1 − ı + 2 q2 q2   2 2 p2 p + q2 1 ı) 2 (−p1 + q1 ı) (z3 − z2 ) = z2 + p1 + 12 (1 − ı + 2 q2 q2 = z2 + p1 (z3 − z2 )+   p22 + q22 p2 p22 + q22 p2 1 + 4 −p1 + q1 − q1 + [p1 − p1 + q1 ]ı (z3 − z2 ). (11.6.18) q2 q2 q2 q2 From the values of p22 + q22 and q2 , and also of p2 /q2 we have −q1 p2 +q2

1 −1) 1 1 + q1 qp11 = −(2p1 − 1) + p1 = 1 − p1 . −q1 (2p p21 +q21 q1 (2p1 −1) Similarly 2

p1

Hence (ii) z16

p22 +q22 q2

+ q1 qp22 =

p2 p2 + q22 (2p1 − 1)2 p21 + q21 p1 − p1 2 = −p1 2 + p1 q2 q2 q1 p1 + q21 (2p1 − 1)q1 p1 p1 p1 p1 p1 = − (2p1 − 1) + p1 = −2p1 + + p1 q1 q1 q1 q1 q1 p1 = (1 − p1 ) . q1 

 p1 = z2 + p1 (z3 − z2 ) + −p1 + 1 − p1 + [(1 − p1 ) + q1 ]ı (z3 − z2 ) q1   2 2 p1 p − q1 ]ı (z3 − z2 ) = z2 + 14 4p1 + 1 − 2p1 + [ − 1 q1 q1   p1 p2 + q21 + 2q1 ]ı (z3 − z2 ). (11.6.19) = z2 + 12 12 + p1 + 12 [ − 1 q1 q1 1 4

(ii)

This is the same as (11.6.16) and so z16 = z 16 . The reason we took the particular foot of a perpendicular Z8 is it simplify the calculations a little, but the result holds just as well for the feet of the perpendiculars Z9 and Z10 as well. For we can give the simplifying to both of these as well as

222

Vector and complex-number methods

Ch. 11

follows. Starting with our usual notation z1 = z2 + (p1 + ıq1 )(z3 − z2 ) let us seek a corresponding relationship z3 = z1 + (p3 + ıq3 )(z2 − z1 ) based on the same triangle. Then we have (p3 + q3 ı − 1)z1 = −z3 + (p3 + ıq3 )z2 = (p3 + ıq3 − 1)z2 − (z3 − z2 ), −1 (z3 − z2 ) = z2 + (p1 + q1 ı)(z3 − z2 ), z1 = z2 + p3 + q3 ı − 1 −1 , p1 + q1 ı = p3 + q3 ı − 1 −1 p1 − q1 ı −p1 + q1 ı p3 + q3 ı − 1 = . = 2 p1 + q1 ı p1 − q1 ı p1 + q21 −p1 p3 − 1 = 2 , p1 + q21 p1 , q3 = 2 p1 + q21 p3 = 1 −

11.6.11

p1 p21 + q21 − p1 = . p21 + q21 p21 + q21

(11.6.20)

The nine-point circle Z11

Z9 Z17

Z10 = Z18

Z1

Z19 Z3

Z16

Z5 Z4

Z6 Z8 Z2

Figure 11.16. Nine-point circle.

We continue with the situation in 11.6.10 where we identified three further points which lie on the circumcircle of the triangle [Z4 , Z5 , Z6 ] with vertices the midpoints Z4 , Z5 and Z6 of the sides [Z2 , Z3 ], [Z3 , Z1 ], [Z1 , Z2 ], respectively, of the original triangle Z1 , Z2 , Z3 . In this subsection we identify three other points which lie on this circumcircle. Let Z17 be the mid-point of the orthocentre and the vertex Z1 in the original triangle, Z18 be the mid-point of the orthocentre and the vertex Z2 , and Z19 be the mid-point of the orthocentre and the vertex Z3 . We seek the circumcentre of the triangle [Z4 , Z8 , Z18 ].

Sec. 11.6

Mobile coordinates

223

Now z4 = z2 + 12 (z3 − z2 ), z8 = z2 + p1 (z3 − z2 ), and by the formula in 11.6.5 for

ı (z3 − z2 ). From these we have that the orthocentre, z11 = z2 + p21 1 − p1q−1 1

p1 − 1 1 p1 − 1 1 z18 = 2 [z2 + z2 + p1 1 − ı (z3 − z2 ) = z2 + + p1 (1 − ı)(z3 − z2 ). q1 2 q1 (iii)

We let Z16 be the circumcentre of the triangle [Z4 , Z8 , Z18 ] and for the present write z4 = z8 + (p4 + ıq4 )(z18 − z8 ) = (1 − p4 − ıq4 )z8 + (p4 + ıq4 )z18 1 p1 − 1 = (1 − p4 − ıq4 )[z2 + p1 (z3 − z2 )] + (p4 + ıq4 )[z2 + p1 (1 − ı)](z3 − z2 ) 2 q1   1 p1 − 1 = (1 − ıq4 + ıq4 )z2 + (1 − p4 − ıq4 )p1 + (p4 + ıq4 ) p1 (1 − )ı (z3 − z2 ) 2 q1   p1 − 1 = z2 + p1 1 − p4 − ıq4 + 12 (p4 + ıq4 )[1 − ı] (z3 − z2 ) q1   p1 − 1 1 1 = z2 + p1 1 − p4 − q4 ı + 2 (p4 + q4 ı) − 2 (p4 + q4 ı) ı (z3 − z2 ) q1   p1 − 1 p1 − 1 ı + 12 q4 (z3 − z2 ) = z2 + p1 1 − 12 p4 − 12 q4 ı − 12 p4 q1 q1   p1 − 1 1 p1 − 1 − 2 (q4 + p4 )ı (z3 − z2 ) = z2 + p1 1 − 12 p4 + 12 q4 q1 q1 = z2 + 12 (z3 − z2 ). Hence   p1 − 1 1 p1 − 1 − 2 (q4 + p4 )ı = p1 1 − 12 p4 + 12 q4 q1 q1 so that

1 2

  p1 − 1 p1 − 1 1 1 1 q4 + p4 = 0, p1 1 − 2 p4 + 2 q4 = . q1 q1 2

Then q4 = − p1q−1 p4 and 1 p1 − 1 q4 = q1 p1 − 1 p1 − 1 p4 = −p4 − q1 q1 − 12 p1 p4 + 12

p4 =

1 p1 − 1 2 − p1 , −p4 + q4 = ( 12 − p1 ) , 2 q1 p1   1 − 2p1 1 (p1 − 1)2 = − 2, −p4 1 + , p1 p1 q21 (2p1 − 1)q21 , p1 (q21 + (p1 − 1)2 )

p1 − 1 2p1 − 1 q21 2 q1 p1 [q1 + (p1 − 1)2 ]   p1 − 1 2 (2p1 − 1)2 q41 2 2 ) p4 + q4 = 1 + ( . p1 p21 [q21 + (p1 − 1)2 ] q4 = −

224

Vector and complex-number methods

Then (p1 − 1) 2 2 ) p4 , p24 + q24 = p24 + ( q1

p1 − 1 2 p24 + q24 p2 = 4 1+( ) q4 q4 q1

−q1 p1 − 1 2 ) = p4 . 1+( p1 − 1 q1

1 −q1 2p1 − 1 p1 − 1 2 ) = 1 + ( p1 1 + ( p1q−1 )2 p1 − 1 q1 1

2p1 − 1 q1 . =− p1 p1 − 1 (iii)

By (11.6.4) the circumcentre Z16 of [Z4 , Z8 , Z18 ] where z4 = z8 + (p4 + q4 ı)(z18 − z8 ) is given by   p4 p2 + q24 (iii) ı − ı (z18 − z8 ) z16 = z8 + 12 1 + 4 q4 q4 ⎡ ⎤ 2 (p −1) p24 (1 + 1q2 ) q1 ⎦ 1 ı (z18 − z8 ) = z8 + 12 ⎣1 + ı+ q4 p1 − 1  

(p1 − 1)2 −q1 q1 1+ ı (z18 − z8 ) ı + = z8 + 12 1 + p4 p1 − 1 p1 − 1 q21   2p1 − 1 q1 q1 q1 1 = z8 + 2 1 − ı+ ı+ ı (z18 − z8 ) p1 p1 − 1 p1 − 1 p1 − 1

  2p1 − 1 q1 = z8 + 21 1 + + 1 ı (z24 − z8 ) − p1 − 1 p1 q1 = z8 + 21 [1 − ı](z18 − z8 ). p1 Continuing from this we have

Ch. 11

Sec. 11.6

(iii)

Mobile coordinates

225

q1 p1 − 1 ı)[z2 + 12 p1 (1 − ı) − z2 − p1 ](z3 − z2 ) p1 q1 q1 p1 − 1 = z2 + p1 (z3 − z2 ) + 12 (1 − ı)[ 12 p1 (1 − ı) − p1 ](z3 − z2 ) p1 q1   p1 (p1 − 1) q1 1 = z2 + p1 + 12 (1 − ı)[ p1 − 12 ı − p1 ] (z3 − z2 ) p1 2 q1   q1 p1 − 1 ı] (z3 − z2 ) = z2 + p1 1 + 12 (1 − ı)[− 12 − 12 p1 q1   q1 p1 − 1 = z2 + p1 1 − 14 (1 − ı)(1 + ı) (z3 − z2 p1 q1   q1 p1 − 1 p1 − 1 q1 = z2 + p1 1 − 14 [1 + +( − )ı] (z3 − z2 ) p1 q1 q1 p1   2 2 p1 − 1 p1 − p1 − q1 1 + ı] (z3 − z2 ) = z2 + p1 1 − 4 [1 + p1 p1 q1   2 2 q21 1 2p1 − 1 1 p1 + q1 − p1 − 4[ ı−2 ı] (z3 − z2 ) = z2 + p1 1 − 4 p1 p1 q1 p1 q1   p2 + q21 p1 = z2 + 14 + 12 p1 − 14 [ 1 − − 2q1 ]ı (z3 − z2 ) q1 q1   2 2 p1 + q1 p1 1 ı + ı + 2q1 ı (z3 − z2 ). = z2 + 4 1 + 2p1 − q1 q1

z16 = z2 + p1 (z3 − z2 ) + 12 (1 −

. The same argument applies to Z By (11.6.16) this is the same as that for Z16 17 and Z19 . Thus we have identified six extra points on the circle which passes through the mid-points of the sides of the original triangle, which is named from this property. There is an interesting article on this circle in Wikipedia. By a coincidence we have Z10 = Z18 in our diagram.

11.6.12

Parametric equations of lines.

For general distinct points Z4 and Z5 and general Z ∈ Z4 Z5 we have that z4 = z2 + (p4 + q4 ı)(z3 − z2 ), z5 = z2 + (p5 + q5 ı)(z3 − z2 ), z = z2 + (p + qı)(z3 − z2 ), and then z = z4 + s(z5 − z4 ) = (1 − s)z4 + sz5 ,

(s ∈ R)

= (1 − s)[z2 + (p4 + q4 ı)(z3 − z2 )] + s[z2 + (p5 + q5 ı)(z3 − z2 )] = (1 − s + s)z2 + [(1 − s)(p4 + q4 ı) + s(p5 + q5 ı)](z3 − z2 ) = z2 + {(1 − s)p4 + sp5 + [(1 − s)q4 + sq5 ]ı}(z3 − z2 ).

226

Vector and complex-number methods

Ch. 11

11.7 SOME WELL-KNOWN THEOREMS NOTE. The advantage of mobile coordinates is that they located a point with respect to a triangle using just two instead of three numbers and they also behave like rescaled Cartesian coordinates. None the less they can lead to unwieldy expressions as in this section and it is a good idea when possible to check the algebraic manipulations using a computer software programme.

11.7.1 Feuerbach’s theorem, 1822 We recall that the nine-point circle has radius-length equal to the distance from Z16 to Z4 . From the formula   ı 1 (p1 − 1 − q1 ı) (z3 − z2 ) z16 − z4 = 2 (p1 + q1 ı) 1 − 2q1 1 = (p1 + q1 ı)[q1 + (1 − p1 )ı](z3 − z2 ) 4q1

we note that |z 16 − z4 | =

a a |p1 + q1 ı||q1 + (1 − p1 )ı| = 4|q1 | 4|q1 |



p21 + q21



(p1 − 1)2 + q21 ,

and so this is the radius-length of the nine-point circle. We denote this radius length by r1 . Our formula in 11.6.5 for the incentre of a triangle is very awkward to apply because of the complicated term in the denominator. However by eliminating the surds in the denominator in two steps, by multiplying above and below by a conjugate surd of the denominator, we obtain the more convenient formulation that    p1 + p21 + q21 1 1 1   = + p21 + q21 − (p1 − 1)2 + q21 . (11.7.1) 2 2 2 2 2 2 2 1 + p + q + (p − 1) + q 1

1

1

1

In fact once we know the form of this we can establish it more directly and easily by noting that        1 2 2 2 2 2 2 2 2 p1 + q1 + (p1 − 1) + q1 1 + p1 + q1 − (p1 − 1) + q1 2 1+  

2    1 + p21 + q21 − (p1 − 1)2 + q21 = 12 = p1 +



p21 + q21 .

We note that the right-hand side in (11.7.1) must be positive.

Sec. 11.7

Some well-known theorems

227

Recalling from 11.6.5 that the incentre Z15 has complex coordinate  p1 + p21 + q21 + q1 ı   (z3 − z2 ), z15 = z2 + 1 + p21 + q21 + (p1 − 1)2 + q21 we re-write this as z15 − z2 ⎡ ⎤     q ı 1 ⎦ (z3 − z2 ) 1 1 + p21 + q21 − (p1 − 1)2 + q21 ⎣1 + = 2 2 2 p1 + p1 + q1   

   1 2 2 2 2 2 2 = p1 + q1 − p1 ı (z3 − z2 ). 1 + p1 + q1 − (p1 − 1) + q1 q1 + 2q1 From this, the foot Z20 of the perpendicular from the incentre Z15 to the line Z2 Z3 has complex coordinate     1 1 + p21 + q21 − (p1 − 1)2 + q21 q1 (z3 − z2 ) z20 = z2 + (11.7.2) 2q1 and so the length of radius of the incircle is equal to       a p21 + q21 − p1 . 1 + p21 + q21 − (p1 − 1)2 + q21 2 We denote this radius length by r2 .

Z11

Z9 Z17

With these preparatory results, we can now show that the nine-point circle and the incircle meet at just one point and they have a common tangent there.

Z1

(11.7.3)

Z19 Z3

Z16

Z1 0 = Z18

Z5 Z15 Z4

Z6

Z8

Z2 Figure 11.17. Feuerbach’s theorem.

It is clear from diagrams that r1 > r2 and known from earlier proofs. We will first give the proof in this case and then a proof that r4 ≥ r1 cannot occur. ,Z The half-line [Z16 15 has points Z = Z16 + s(Z15 − Z16 ) where s ≥ 0 and will meet the nine-point circle at a point z21 = z 16 +

r1 (z15 − z 16 ). |z15 − z 16 |

228

Vector and complex-number methods

Ch. 11

It will also meet the nine-point circle at the point z22 = z 16 −

r1 (z15 − z 16 ). |z15 − z 16 |

of the nine-point circle and so The mid-point of these two points is the centre Z16 [Z21 , Z22 ] is a diameter of this nine-point circle. has points Z = Z − t(Z − Z ) and meets the Similarly the half-line [Z15 , Z16 15 15 16 incircle at the point r2 (z − z15 ). z = z15 − |z16 − z15 | 16

Then clearly this is the point Z21 again and the corresponding point z24 = z15 +

r2 (z − z15 ) |z 16 − z15 | 16

will be a diametrically opposite point on the incircle. Then the nine-point circle and the incircle have a common tangent at the point Z21 . It is clear from elementary geometry that, if two circles touch at just one point, the centres of the two triangle and the point of tangency must be collinear. The following is a skeleton picture of a possible situation.

Figure 11.18. Now we turn to the supposition that r2 ≥ r1 . We first suppose that r2 > r1 . We treat the incircle as above and take the point z24 = z15 +

r2 (z − z15 ). |z16 − z15 | 16

Then Z24 is a point on the incircle and the point z25 = z15 −

r2 (z − z15 ). |z16 − z15 | 16

is such that [z24 , z25 ] is a diameter of the incircle.

Sec. 11.7

Some well-known theorems

229

For the nine-point circle we take the point Z26 such that z26 = z 16 +

r1 (z − z15 ), |z15 − z 16 | 16

Z Then Z26 is a point of the line Z16 15 which lies on the nine-points circle and the corresponding point r1 (z − z15 ), z27 = z 16 − |z15 − z 16 | 16

also lies on the nine-point circle for which [z26 , z27 ] is a diameter. But in fact the points Z24 and Z26 coincide and we then have the situation that Z27 ∈ [Z24 , Z25 ]. Thus we have that the incircle has diameter [Z24 , Z25 ] and the nine-point circle has diameter [z24 , z27 ] and [z24 , z27 ] ⊂ [z24 , z25 ]. It follows that the nine-point circle lies inside or on the the incircle. But the incircle lies inside or on the triangle [Z1 Z2 Z3 ] and it follows that the nine-points circle lies inside or on the triangle. But this yields a contradiction as the nine-points circle passes through the mid-points Z4 , Z5 , Z6 of the sides of this triangle. =Z There remains the case when r2 = r1 and now we must have Z16 15 and the nine-points circle must coincide with the incircle. In this case the two circles do not meet in a unique point. This result is known as the basic form of Feuerbach’s theorem.

11.7.2 The Wallace-Simson line, 1797 We take a triangle [Z1 , Z2 , Z3 ] and for a point Z let W1 ,W2 ,W3 be the feet of the perpendiculars from Z to the side-lines Z2 Z3 , Z3 Z1 , Z1 Z2 , respectively.

W3 W3

Z

Z1

Z1

Z Z3 W2

W1

Z3 Z2

W2 W1

Z2

Figure 11.19(a). A right sensed duo-angle.

(b). A Simson-Wallace line.

Using notation like that in 11.6.3?, we suppose that z1 − z2 = (p1 + q1 ı)(z3 − z2 ), z − z2 = (p + qı)(z3 − z2 ). Then z = z2 + p(z3 − z2 ) + qı(z3 − z2 ), and so w1 = z2 + p(z3 − z2 ). Next w2 = z3 + s(z1 − z3 ), for some s ∈ R. Hence z − w2 = (p − 1 + qı)(z3 − z2 ) −

230

Vector and complex-number methods

Ch. 11

1 s(z1 − z3 ). But z1 − z3 = (p1 − 1 + q1 ı)(z3 − z2 ), so z3 − z2 = p1 −1+q (z1 − z3 ). On 1ı   − s (z1 − z3 ). We wish the coefficient inserting this, we have that z − w2 = pp−1+qı 1 −1+q1 ı

of z1 − z3 to be purely imaginary, and so take s = (p−1)(p1 −1)+qq1 (z1 − z3 ). (p1 −1)2 +q21

(p−1)(p1 −1)+qq1 . (p1 −1)2 +q21

Hence w2 = z3 +

Thirdly, w3 = z2 + t(z1 − z2 ),for some t ∈ R. Hence z −w3 = (p +qı)(z3 − z2 ) − 1 (z1 − z2 ) and so z − w3 = pp+qı − t (z1 − z2 ). We t(z1 − z2 ). But z3 − z2 = p1 +q 1ı 1 +q1 ı choose t so that the coefficient of z1 − z2 is purely imaginary. Thus t = pp1 +qq1 (z1 − z2 ). p21 +q21

yields w3 = z2 + From these expressions for w1 , w2 , w3 we note that

pp1 +qq1 p21 +q21

which

(p−1)(p1 −1)+qq1 w2 − w1 z3 − z2 + (p1 −1)2 +q21 (z1 − z3 ) − p(z3 − z2 ) = pp1 +qq1 w3 − w1 2 2 (z1 − z2 ) − p(z3 − z2 ) p1 +q1

=

1−

1 −1)+qq1 (p1 − 1 + q1 ı) p + (p−1)(p (p1 −1)2 +q21 . pp1 +qq1 (p1 + q1 ı) − p p2 +q2 1

1

real part of this has numerator (p21 + q21 − p1 )(p2 + q2 ) − (p21 + q21 − p1 )p + q1 q, the imaginary part has numerator q1 (p2 + q2 ) − q1 p − (p21 + q21 − p1 )q. If θ =

The and F W3W1W2 then for θ to have a constant magnitude it is necessary and sufficient that q1 (p2 + q2 ) − q1 p − (p21 + q21 − p1 )q = k[(p21 + q21 − p1 )(p2 + q2 ) − (p21 + q21 − p1 )p + q1 q]. This can be re-written as 



2   (p2 + q21 − p1 )2 1 2 a2 p21 + q21 − p1 2 1+ 1 p− − + q− a 2 2q1 4 q21  

2 

k(p21 + q21 − p1 ) 1 2 q1 2 = p− a + q+ q1 2 2(p21 + q21 − p1 ) "  q21 a2 . 1+ 2 − 4 (p1 + q21 − p1 )2

On using 11.6.1 we infer that as k varies this gives the family of coaxal circles which pass through Z2 and Z3 . For W1 ,W2 ,W3 to be collinear, it is necessary and sufficient that the expression be real. On equating its imaginary part to 0 we obtain q1 (p2 + q2 ) − q1 p − (p21 + p2 +q2 −p

q21 − p1 )q = 0. On writing this as p +qq −p = 1 q11 1 , we note from the formula for a circumcentre in 11.6.4? that it holds when Z lies on the circumcircle of the triangle[Z1 , Z2 , Z3 ]. This latter result is due to Wallace, but Simson’s name has for a long time been associated with it. 2

2

Sec. 11.7

Some well-known theorems

231

11.7.3 The incentre on the Euler line of a triangle We suppose that we have the mobile coordinates z1 − z2 = (p1 + q1 ı)(z3 − z2 ), where p1 and q1 are real numbers and q1 = 0. Then z1 − z3 = (p1 − 1 + q1 ı)(z3 − z2 ), and as in 11.6.3 we have z1 − z2 c z2 − z3 1 − p1 + q1 ı a = p1 + q1 ı = cis β , = = cis γ , 2 2 z3 − z2 a z1 − z3 (1 − p1 ) + q1 b z3 − z1 p1 − 1 + q1 ı b = cis α , = z2 − z1 p1 + q1 ı c 2

2

where we are using our standard notation. Then p21 + q21 = ac2 , (1 − p1 )2 + q21 = ba2 . We recall that the orthocentre, √ centroid and incentre have mobile coordinates p1 + p1 (1−p1 ) p1 +1 ı, 3 q1

+ q31 ı,

Now 1 − 2p1 = Moreover

√p1 +

p21 +q21 +q1 ı

√

respectively.

1+ p21 +q21 (1−p1 )2 +q21 2 2 −b2 b2 −c2 , p1 = c +a ,1 − a2 2a2

p1 =

a2 +b2 −c2 , p1 2a2

+1 =

c2 +a2 −b2 +2a2 . 2a2

2

2 c2 c2 c + a2 − b2 (c2 + a2 − b2 )2 − 4c2 a2 2 − p = − =− 1 2 2 2 a a 2a 4a4 [(c + a)2 − b2 ][(c − a)2 − b2 ] =− , 4a4

q21 =

while  c2 + a2 − b2 c (c + a)2 − b2 + = , p1 + p21 + q21 = 2a2 a 2a2   c b a+b+c 1 + p21 + q21 + (1 − p1 )2 + q21 = 1 + + = , a a a so that

1+



1+

p1 +



p21 + q21 (c + a)2 − b2 c+a−b  = = , 2a p21 + q21 + (1 − p1 )2 + q21 2a(a + b + c) 

q21 1 [(c + a)2 − b2 ][(c − a)2 − b2 ]  =− 3 4a a+b+c p21 + q21 + (1 − p1 )2 + q21 =−

1 (c + a − b)[(c − a)2 − b2 ]. 4a3

The determinant for collinearity, on multiplying the middle column by q1 , is       (c2 +a2 −b2 )(a2 +b2 −c2 )   c2 +a2 −b2 1 2 4   4a 2 −b2 ][(c−a)2 −b2 ] ,  c2 +a22a 2 2 [(c+a) −b +2a  − 1  6a2 12a4   2 2  c+a−b − (c+a−b)[(c−a) −b ] 1  2a

4a3

232

Vector and complex-number methods

Ch. 11

and this is a non-zero multiple of       2 + a2 − b2 2 + a2 − b2 )(a2 + b2 − c2 )   c (c 1    c2 + a2 − b2 + 2a2 −[(c + a)2 − b2 ][(c − a)2 − b2 ] 3  ,    a(c + a − b) −a(c + a − b)[(c − a)2 − b2 ] 1  the value of which is 4(ca5 − c3 a3 − ba5 + a3 b3 + c3 ba2 − b3 ca2 ). This factorizes as 4a2 (b − c)(c − a)(a − b)(a + b + c) and so the incentre lies on the Euler line if and only if the triangle is isosceles.

11.7.4 Miquel’s theorem, 1838

Z1

Z5

Z3 Z5 Z3

Z1 Z4

Z6

Z6 Z2

Z4 Z2

Figure 11.20(a). Miquel’s theorem.

Figure 11.20(b). Miquel’s theorem.

Let Z1 , Z2 , Z3 be non-collinear points, and Z4 ∈ Z2 Z3 , Z5 ∈ Z3 Z1 , Z6 ∈ Z1 Z2 be distinct from Z1 , Z2 and Z3 . Let C1 , C2 , C3 be the circumcircles of [Z1 , Z5 , Z6 ], [Z2 , Z6 , Z4 ], [Z3 , Z4 , Z5 ], respectively. Then C1 , C2 , C3 have a point in common. Proof . Suppose that these circles have centres the points W1 ,W2 ,W3 , respectively. We first assume that C2 and C3 meet at a second point Z7 = Z4 . If Z7 = Z6 the result is trivially true, so we may exclude that case. As Z1 , Z2 , Z6 are collinear, we have z1 − z6 = −ν (z2 − z6 ),for some non-zero ν in R. As Z2 , Z4 , Z6 , Z7 are concyclic, by 10.9.3 z2 −z4 6 we have zz27 −z −z6 = ρ z7 −z4 , for some non-zero ρ in R. As Z2 , Z3 , Z4 are collinear, we have z2 − z4 = −λ (z3 − z4 ), for some non-zero λ in R. As Z3 , Z4 , Z5 , Z7 are concyclic, z3 −z5 4 we have zz37 −z −z4 = σ z7 −z5 , for some non-zero σ in R. On combining these we have z1 −z6 z3 −z5 z7 −z6 = νρλ σ z7 −z5 . It follows by 10.9.3 that Z1 , Z5 , Z6 , Z7 are concyclic. We suppose secondly that C2 and C3 have a common tangent at Z2 . It is convenient to suppose that z1 = z2 + (p1 + ıq1 )(z3 − z2 ) and w2 = z2 + (p + ıq)(z3 − z2 ). Then the foot Z7 of the perpendicular from W2 to Z2 Z3 has complex coordinate z7 = z2 + p(z3 − z2 ), and hence z4 = z2 + 2p(z3 − z2 ). Then the mid-point Z8 of Z3 and Z4 has complex coordinate z8 = z2 + (p + 12 )(z3 − z2 ). It follows that for the centre W3 of C3

Sec. 11.7

Some well-known theorems

233

we have w3 = z2 + (p + 12 + ıq )(z3 − z2 ), for some real number q . But Z4 ,W2 ,W3 are collinear, so that    2p 0 1    p q 1  = 0,   p + 1 q 1  2   p− 12 p− 1 and from this q = p qı(z3 − z2 ). Hence w3 = z2 + p + 12 + p 2 qı (z3 − z2 ). From p− 21 1 p (p + qı)(z3 − z2 ), and since z3 − z2 = p1 −1+ıq1 (z1 − z3 ), we have p− 1 1 −1−ıq1 ) (z1 − z3 ). Thus if Z9 is the foot of the perpendicular w3 = z3 + p 2 (p+ıq)(p (p1 −1)2 +q21 p− 1 1 −1)+qq1 from W3 to Z3 Z1 , we have z9 = z3 + p 2 p(p (z1 − z3 ), and so (p1 −1)2 +q21

this w3 − z3 =

z5 = z3 +

2p − 1 p(p1 − 1) + qq1 (z1 − z3 ). p (p1 − 1)2 + q21

Similarly w2 = z2 + (p + ıq)(z3 − z2 ), z3 − z2 = (p+ıq)(p1 −ıq1 ) (z1 − z2 ). p21 +q21

1 p1 +ıq1 (z1 − z2 ),

and so w2 = z2 +

It follows that for the foot Z10 of the perpendicular from W2 to

+qq1 pp1 +qq1 Z1 Z2 we have z10 = z2 + ppp21+q 2 (z1 − z2 ), and so z6 = z2 + 2 p2 +q2 (z1 − z2 ). 1 1 1 1 We have   2p − 1 p(p1 − 1) + qq1 (z1 − z3 ), z1 − z5 = 1 − p (p1 − 1)2 + q21   pp1 + qq1 (z1 − z2 ), z1 − z6 = 1 − 2 2 p1 + q21 pp1 + qq1 z4 − z6 = 2p(z3 − z2 ) − 2 2 (z1 − z2 ) p1 + q21   pp1 + qq1 (p1 + ıq1 ) (z3 − z2 ), = 2 p− 2 p1 + q21   2p − 1 p(p1 − 1) + qq1 z4 − z5 = (2p − 1)(z3 − z2 ) − (z1 − z3 ) p (p1 − 1)2 + q21   p(p1 − 1) + qq1 2p − 1 p− (p − 1 + q ) = 1 1 (z3 − z2 ). p (p1 − 1)2 + q21 1 −z5 )(z4 −z6 ) As z1 − z2 = (p1 + ıq1 )(z3 − z2 ), from these combined we have that (z (z1 −z6 )(z4 −z5 ) is a real multiple of     +qq1 +qq1 p − ppp21+q p − ppp11−ıq (p1 − 1 + ıq1 ) 2 (p1 + ıq1 ) (p1 − 1 + ıq1 ) 1 1 1    = −1)+qq1 1 −1)+qq1 p − p(p p − p(pp11−1−ıq (p1 + ıq1 ) (p1 − 1 + ıq1 ) (p1 + ıq1 ) (p −1)2 +q2 1 1

1

−q1 (ıp + q)[(p1 − 1)2 + q21 ] (p1 − 1)2 + q21 = , = −q1 (ıp + q)[p21 + q21 ] p21 + q21 and this is real. It follows that Z1 , Z4 , Z5 , Z6 are concyclic. This is known as Miquel’s theorem.

234

Vector and complex-number methods

Ch. 11

11.8 ISOGONAL CONJUGATES 11.8.1 Isogonal conjugates Definition. Given non-collinear points Z1 , Z2 , Z3 , we say that half-lines [Z1 , Z4 [Z1 , Z5 are isogonal conjugates with respect to the angle-support |Z2 Z1 Z3 if the sensed angles F Z2 Z1 Z4 , F Z5 Z1 Z3 , have equal magnitudes. Z3 Z1 Z5 Z4 Z2 Figure 11.21. Isogonal conjugates.

To see how this operates, we first suppose that Z4 and Z5 are both on the line Z2 Z3 and that λ1 λ2 1 1 Z2 + Z3 , Z5 = Z2 + Z3 , Z4 = 1 + λ1 1 + λ1 1 + λ2 1 + λ2 for real numbers λ1 and λ2 . We recall that then Z2 Z4 Z2 Z5 = λ1 , = λ2 . Z4 Z3 Z5 Z3 Then z4 − z1 z3 − z1 / = z2 − z1 z5 − z1

λ1 λ2 1 1 1+λ1 z2 + 1+λ1 z3 − z1 1+λ2 z2 + 1+λ2 z3 − z1

z2 − z1

z3 − z1

is positive and so on multiplying across by (1 + λ1 )(1 + λ2 ) z2 − z1 + λ1 (z3 − z1 ) z2 − z1 + λ2 (z3 − z1 ) z2 − z1 z3 − z1 z3 − z1 z2 − z1 = + λ1 λ2 + λ1 + λ2 z3 − z1 z2 − z1 is real. On subtracting λ1 + λ2 it follows that z2 − z1 1 + λ1 λ2 z3 − z1 (z2 − z1 )/(z3 − z1 ) is real. We write

z2 − z1 = u + vı, z3 − z1

(u, v ∈ R),

Sec. 11.8

Isogonal conjugates

235

where v = 0 as Z1 ∈ Z2 Z3 . With this notation we have that u + vı + λ1 λ2 uu−vı 2 +v2 is real v 2 2 so that v − λ1 λ2 u2 +v 2 = 0, and hence λ1 λ2 = u + v . We deduce that λ1 λ2 =

|Z1 ,Z2 |2 . |Z3 ,Z1 |2

As λ1 λ2 > 0, the points Z4 and Z5 must be both in [Z2 , Z3 ] or both in Z2 Z3 \ [Z2 , Z3 ].

11.8.2 Concurrency Given non-collinear points Z1 , Z2 , Z3 , suppose now that for Z4 , Z5 ∈ Z2 Z3 the halflines [Z1 , Z4 , [Z1 , Z5 are isogonal conjugates with respect to the angle-support |Z2 Z1 Z3 , for Z6 , Z7 ∈ Z3 Z1 the half-lines [Z2 , Z6 , [Z2 , Z7 are isogonal conjugates with respect to the angle-support |Z3 Z2 Z1 , and for Z8 , Z9 ∈ Z1 Z2 the half-lines [Z3 , Z8 , [Z3 , Z9 are isogonal conjugates with respect to the angle-support |Z1 Z3 Z2 . Then by 11.8.1, with an obvious notation we have λ4 λ5 μ6 μ7 ν8 ν9 =

|Z1 ,Z2 |2 |Z2 ,Z3 |2 |Z3 ,Z1 |2 |Z3 ,Z1 |2 |Z1 ,Z2 |2 |Z2 ,Z3 |2

= 1.

Now if Z1 Z4 , Z2 Z6 , Z3 Z8 are concurrent, by Ceva’s theorem we have λ4 μ6 ν8 = 1. It follows that λ5 μ7 ν9 = 1, which is the condition (11.5.1) for Z1 Z5 , Z2 Z7 , Z3 Z9 . Then if two of these are concurrent, the third must pass through their point of intersection.

11.8.3 Symmedians If in 11.8.2 [Z1 , Z4 , [Z2 , Z6 , [Z3 , Z8 are the median half-lines of [Z1 , Z2 , Z3 ], then their isogonal conjugates [Z1 , Z5 , [Z2 , Z7 , [Z3 , Z9 are called the symmedians of the triangle. Now the mid-points Z4 , Z6 , Z8 lie in the segments [Z2 , Z3 ], [Z3 , Z1 ], [Z1 , Z2 ], respectively. Hence the points Z5 , Z7 , Z9 lie in the segments [Z2 , Z3 ], [Z3 , Z1 ], [Z1 , Z2 ], respectively. By the cross-bar theorem the symmedian lines Z1 Z5 , Z2 Z7 must meet and then Z3 Z9 must pass through their point of intersection. This point of concurrency is called the symmedian point of the triangle. To identify the symmedians further, suppose that O be the centre of the circumcircle of the Z3 triangle [Z1 , Z2 , Z3 ], and W1 the Z1 mid-point of {Z2 , Z3 }. Let the line OW1 meet the circumcirW1 cle, on the opposite side of Z2 Z3 from Z1 , at the point U1 , [Z1 ,W1 O meet the circle again at Z4 , and Z2 Z4 the line through Z4 parallel to Z2 Z3 meet the circle again at Z5 . Then |Z2 Z1 Z3 and |Z4 Z1 Z5 have U1 the same mid-line Z1U1 as each Z5 other. Thus [Z1 , Z5 is isogonal Figure 11.22. Symmedians. conjugate of the median halfline [Z1 , Z4 and so is the corresponding symmedian.

236

Vector and complex-number methods

Ch. 11

Exercises 11.1 Show that if Z1 = Z2 , the points Z on the perpendicular bisector of [Z1 , Z2 ] are −−→ −−→ −−→ −−→ −→ those for which OZ = 12 (OZ1 + OZ2 ) + t(OZ2 − OZ1)⊥ , for some t ∈ R. 11.2 If l = Z1 Z2 , then Z4 = πl (Z3 ) if and only if Z4 = Z1 + s(Z2 − Z1 ) where s=

−−→ −−→ −−→ −−→ (OZ3 − OZ1 ).(OZ2 − OZ1 ) , −−→ −−→

OZ2 − OZ1 2

−−→ −−→ −−→ −−→ and also if and only if OZ4 = OZ3 + t(OZ2 − OZ1 )⊥ , where t =−

−−→ −−→ −−→ −−→ (OZ3 − OZ1 ).(OZ2 − OZ1 )⊥ . −−→ −−→

OZ2 − OZ1 2

−−→ −−→ −−→ −−→ 11.3 If l = Z1 Z2 , then Z4 = sl (Z3 ) if and only if OZ4 = OZ3 +t(OZ2 − OZ1 )⊥ , where −−→ −−→ −−→ −−→ (OZ3 − OZ1 ).(OZ2 − OZ1 )⊥ . t = −2 −−→ −−→

OZ2 − OZ1 2 11.4 If Z2 ∈ [Z0 , Z1 and l is the mid-line of the angle support |Z1 Z0 Z2 , then Z ∈ l if and only if −→ −−→ OZ =OZ0 

⊥ −−→ −−→ −−→ −−→ 1 +t −−→ −−→ (OZ2 − OZ0 ) − −−→ −−→ (OZ1 − OZ0 ) ,

OZ2 − OZ0

OZ1 − OZ0

1

for some t ∈ R. −→ −−→ −→ −−→ −→ −−→ 11.5 Prove that OZ − OZ1 = cos α (OZ − OZ1 ) + sin α (OZ − OZ1 )⊥ , represents a rotation about the point Z1 , through the angle α . −→ 11.6 For any vector OZ, with Z ≡F (x, y), define −→ −→ OZ = OU,

−→ −→ OZ = OV ,

where U ≡F (x, −y), V ≡F (y, x), so that U = sOI (Z), V = sl (Z), l being the mid-line of |IOJ. Prove that (i)

(ii)

−−→ −−→ −−→ −−→ (OZ1 + OZ2 ) = OZ1 + OZ2 , −→ −→ (kOZ) = k(OZ) ,

Sec. 11.8 (iii)

(iv)

(v) (vi)

(vii)

(viii)

Isogonal conjugates

237

−→ −→ (OZ ) = OZ, −−→ −−→ −−→ −−→ (OZ1 + OZ2 ) = OZ1 + OZ2 , −→ −→ (kOZ) = k(OZ) , −→ −→ (OZ ) = OZ, −→⊥ −→ (OZ ) = OZ , −→⊥ −→ (OZ ) = −OZ .

Prove that axial symmmetry in the line through the origin, with angle of incli−−→ −→ −→ nation α , is given by OZ = cos 2α OZ + sin 2α OZ . 11.7 For non-collinear points Z1 , Z2 , Z3 , suppose that −−→ −−→ −−→ −−→ ⊥ −−→ −−→ −−→ −−→ ⊥ OZ4 = OZ2 + (OZ1 − OZ2 ) , OZ5 = OZ2 − (OZ3 − OZ2) . Show that then Z1 Z5 ⊥ Z3 Z4 , and in fact −−→ −−→ −−→ −−→⊥ OZ5 − OZ1 = OZ4 − OZ3 . 11.8 Prove that δF (Z1 , Z2 , Z3 ) + δF (Z1 , Z4 , Z3 ) = δF (Z1 , Z5 , Z3 ), where (Z1 , Z2 ) ↑ (Z4 , Z5 ). 11.9 If a point Z be taken on the circumcircle of an equilateral triangle [Z1 , Z2 , Z3 ], prove that when Z is on the opposite side of Z2 Z3 from Z1 , |Z, Z1 | = |Z, Z2 | + |Z, Z3 |. 11.10 For a triangle [Z1 , Z2 , Z3 ], if Z4 is the foot of the perpendicular from Z1 to Z2 Z3 , show that Z4 is the mid-point of the orthocentre and the point where the line Z1 Z4 meets the circumcircle again. 11.11 Deduce the Pasch property of a triangle from Menelaus’ theorem.

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12

Trigonometric functions in calculus COMMENT. In 3.7.1 and 9.3.1 we extended degree-measure to reflex angles, but the only use we have made of this hitherto has been for the purposes of notation in Chapter 9. In this chapter we relate it to the length of an arc of a circle, defining π in the process. In doing this we make a start on calculus for the cosine and sine functions, introduce radian measure of angles and can go on to derive Maclaurin series. We also derive the area of a disk. In doing this we assume from the calculus definition of the length of a rectifiable curve, the Riemann integral and a formula for area in terms of a line integral; this is done so as to save labour, although it would be possible to handle arc-length of a circle and the area of a disk more elementarily.

12.1 REPEATED BISECTION OF AN ANGLE 12.1.1 Addition ⊕ is associative on A ∗ (F ). Proof . On using 9.3.2 instead of 9.3.3, as in the proof of 9.3.3(iii) we see that cos[(α ⊕ β ) ⊕ γ ] = cos[α ⊕ (β ⊕ γ )], sin[(α ⊕ β ) ⊕ γ ] = sin[α ⊕ (β ⊕ γ )]. Thus (α ⊕ β ) ⊕ γ and α ⊕ (β ⊕ γ ) are the angles in A ∗ (F ) with the same cosine and the same sine, so they are the same angle except perhaps when P3 = Q and either angle could be 0F or 360F . Thus (α ⊕ β ) ⊕ γ = α ⊕ (β ⊕ γ ), except perhaps when either side is a null or full angle. But when each of α , β , γ is 0F , then each of (α ⊕ β ) ⊕ γ , α ⊕ (β ⊕ γ ) is 0F and so they are equal. If any of α , β , γ is non-null, and either of (α ⊕ β ) ⊕ γ , α ⊕ (β ⊕ γ ) is 0F or 360F then both of them must be 360F , and again they are equal. Geometry with Trigonometry http://dx.doi.org/10.1016/B978-0-12-805066-8.50012-4 © 2016 Elsevier Ltd. All rights reserved.

240

Trigonometric functions in calculus

Ch. 12

Definition . Given any angle α ∈ A ∗ (F ), in 9.4.1 we defined 12 α . Now for any integer n ≥ 1, 21n α is defined inductively by 1 1 α = α, 1 2 2

1

n ≥ 1,

and for all

2n+1

1 1 α = [ n α ]. 2 2

For all α ∈ A ∗ (F ), | 12 α |◦ = 12 |α |◦ . Proof . We use the notation of 9.4.1. From the definition of midlQOP this is immediate when P ∈ H1 , so we suppose that P ∈ H2 , P = S, P = Q. Then |α |◦ = 180 + |∠SOP|◦ = |∠QOP |◦ + |∠P OS|◦ + |∠SOP|◦ = |∠QOP |◦ + |∠P OP|◦ = 2|∠QOP |◦ by 3.7.1.

i(α )

For this, note that [P, P ] meets OQ in a point G. If H = mp(Q, P), then P ∈ OH and if H5 , H6 are the closed halfplanes with common edge OH, with Q ∈ H5 , then as H ∈ [Q, P] we have P ∈ H6 . Then [P , P] ⊂ H6 so G ∈ H6 . But [O, Q ⊂ H5 , [O, S ⊂ H6 so G ∈ [O, S . Thus [O, S ⊂ I R(|P OP).

P

S

α

G

Q H1 H2

I

P T Figure 12.1.

P2

Proof . By the commutativity P = N 3 and associativity of ⊕,

= ( 12 α ⊕ 12 α ) ⊕ ( 12 β ⊕ 12 β )

J O H

⊕ β ) = 12 α ⊕ 12 β .

( 21 α ⊕ 12 β ) ⊕ ( 12 α ⊕ 12 β )

H5 R

H6

If α , β ∈ A ∗ (F ) are such that |α |◦ + |β |◦ ≤ 360, then 1 2 (α

H4 H3

S

H4 H3 R

W P1 P1

J O I

H1 Q H2

= α ⊕β, T so we show that under the conP3 P2 1 ◦ ◦ dition |α | + |β | ≤ 360, if 2 α ⊕ Figure 12.2. 1 2 β has support |QOP3 then P3 ∈ H1 . Let [O, P1 = i(α ), [O, P2 = i(β ) with |O, P 1 | = |O, P 2 | = |O, Q| = k. Then 12 α and 1 ∗ 2 β are the angles in AF with supports |QOP1 , |QOP2 , respectively.

Sec. 12.2

Circular functions

241

Without loss of generality we suppose that |α |◦ ≤ |β |◦ , so that by the above, | 12 α |◦ ≤ | 12 β |◦ . As | 12 α |◦ + | 12 β |◦ ≤ 180, then | 12 α |◦ ≤ 90, | 12 β |◦ ≤ 180, so that P1 ∈ H1 , P2 ∈ H1 and [O, P1 ⊂ I R(|QOP 2 ). Let the angle in A ∗ (F ) with degree measure 12 (|α |◦ + |β |◦ ) have support |QON where |O, N| = k. Then as 12 (|α |◦ + |β |◦ ) ≤ 180 we have N ∈ H1 . Let the angle in A ∗ (F ) with degree measure 14 (|α |◦ + |β |◦ ) have support |QOW, where |O, W| = k. Then W ∈ H1 and [O, P1 ⊂ I R(|QOW) as 12 |α |◦ ≤ 14 (|α |◦ +|β |◦ ). Thus |∠P 1 OW|◦ = |∠QOW|◦ − |∠QOP 1 |◦ = 14 (|α |◦ + |β |◦ ) − 12 |α |◦ = 14 (|β |◦ − |α |◦ ). As 14 (|α |◦ + |β |◦ ) ≤ 12 |β |◦ we have [O,W ⊂ I R(|QOP 2 ). Then |∠WOP 2|◦ = |∠QOP 2 |◦ − |∠QOW|◦ = 12 |β |◦ − 14 (|α |◦ + |β |◦ ) = 14 (|β |◦ − |α |◦ ). As |∠P 1 OW|◦ = |∠WOP 2 |◦ we have OW = midlP1 OP2 . By a similar argument |∠QOW|◦ = 14 (|α |◦ + |β |◦ ) = |∠WON|◦ , so OW = midlQON. Then N = sOW (Q) and so 12 α ⊕ 12 β has support |QON. When 12 (|α |◦ + |β |◦ ) < 180 this gives that P3 = N and so 12 α ⊕ 12 β = 12 (α ⊕ β ). The remaining case is when |α |◦ + |β |◦ = 360. Then 12 (|α |◦ + |β |◦ ) = 180 so 12 α ⊕ 1 1 2 β = 180F , while α ⊕ β = 360F so 2 (α ⊕ β ) = 180F .

COROLLARY 1 . If α , β ∈ A ∗ (F ) are such that |α |◦ + |β |◦ ≤ 360, then for all integers n ≥ 1, 1 1 1 (α ⊕ β ) = n α ⊕ n β . 2n 2 2 COROLLARY 2 . If α , β ∈ A ∗ (F ) are such that |α |◦ + |β |◦ ≤ 360, then |α ⊕ β |◦ = |α |◦ + |β |◦ .

Proof . In the course of the proof we saw that, under the given conditions, | 12 α ⊕ 12 β |◦ = 12 (|α |◦ + |β |◦ ), 12 α ⊕ 12 β = 12 (α + β ). From these | 12 (α ⊕ β )|◦ = 12 (|α |◦ + |β |◦ ), so that |α ⊕ β |◦ = |α |◦ + |β |◦ .

12.2 CIRCULAR FUNCTIONS 12.2.1 Definition . Let x ∈ R satisfy 0 ≤ x ≤ 360. Then if for α ∈ A ∗ (F ) we have |α |◦ = x, we write c(x) = cos α , s(x) = sin α . By 9.6, c and s do not depend on F .

242

Trigonometric functions in calculus

Ch. 12

The functions c and s have the following properties:(i) For 0 ≤ x ≤ 360, 0 ≤ y ≤ 360, x + y ≤ 360, c(x + y) = c(x)c(y) − s(x)s(y), s(x + y) = s(x)c(y) + c(x)s(y). (ii) For 0 ≤ x ≤ 360, c(x)2 + s(x)2 = 1. (iii) For 0 ≤ x ≤ 360, c(x) = 1 − 2s( 12 x)2 , s(x) = 2c( 12 x)s( 12 x).

Proof . These follow immediately from the definition of c and s, and the formulae for cos(α ⊕ β ) and sin(α ⊕ β ). Definition . For 0 ≤ x ≤ 360 and integers n ≥ 3, let  x   x  n s  2n+1  . un (x) = 2 s n+1 , vn (x) = 2 x 2 c 2n+1 n

COMMENT. When n ≥ 1, un is the sum of the areas of 2n+1 non-overlapping congruent triangles, each having its vertex at the centre of the circle/arc with radius length 1, and each having the end-points of its base on the circle/arc.

Figure 12.3.

When n ≥ 1, vn is the sum of the areas of 2n non-overlapping congruent triangles, each having its vertex at the centre of the circle/arc, and each having its base-line a tangent to the circle/arc. Our reason for taking n ≥ 3 is that then x 360 = 22.5 < 45. ≤ 2n+1 16

Figure 12.4

Sec. 12.2

Circular functions

243

The sequences un , vn have the following properties:(i) (un ) is non-decreasing. (ii) (vn ) is non-increasing. (iii) For 0 < x ≤ 360 and n ≥ 2, we have 0 < un (x) ≤ vn (x). (iv) As n → ∞, un (x) and vn (x) tend to a common limit μ (x) and un (x) ≤ μ (x) ≤ vn (x).

Proof . (i) For un (x) = 2n+1 s

 x   x  c n+2 ≤ un+1 (x). 2n+2 2

(ii) For vn (x) =



 

x x c 2n+2 2n+2  x 2 2c 2n+2 − 1

2n+1 s

 = vn+1 (x)

2c

c 





x 2 2n+2  x 2

2n+2

−1

> vn+1 (x)

x 2 x 2 x 2 x as c( 2n+2 ) < 1 and so 2c( 2n+2 ) − 1 < c( 2n+2 ) , while c( 2n+1 ) > 0 and thus x 2 2c( 2n+2 ) − 1 > 0. (iii) For  x  un (x) = vn (x)c n+1 ≤ vn (x). 2 (iv) By (i), (ii) and (iii), un (x) ≤ v3 (x) so we have a non-decreasing sequence which is bounded above. Hence there is some μ (x) such that limn→∞ un (x) = μ (x). It x follows that s( 2n+1 ) → 0 (n → ∞) and hence  x 2  x  c n+1 = 1 − 2s n+2 → 1 (n → ∞). 2 2

But then

 x −1 vn (x) → 1 (n → ∞), = c n+1 un (x) 2

and so as well limn→∞ vn (x) = μ (x). As un (x) is non-decreasing and vn (x) is nonincreasing we must have un (x) ≤ μ (x) ≤ vn (x). For x ≥ 0, y ≥ 0, x + y ≤ 360,

μ (x + y) = μ (x) + μ (y). Proof . For



 x x+y y  n s + = 2 2n+1 2n+1 2n+1  x   y   y   x  = 2n s n+1 c n+1 + c n+1 2n s n+1 2 2 2 2 → μ (x).1 + 1.μ (y) (n → ∞).

un (x + y) = 2n s

244

Trigonometric functions in calculus

Ch. 12

COMMENT. This equation for μ is referred to as Darboux’s functional equation. There is a number p > 0 such that for 0 ≤ x ≤ 360, μ (x) = px. Proof . We start by writing out a well known proof that any function μ which satisfies Darboux’s functional equation on [0, 360] satisfies μ (r) = r μ (1) for all rational numbers r in [0, 360]. On first taking y = x we have μ (2x) = 2μ (x); then taking y = 2x, we get μ (3x) = 3μ (x); next taking y = 3x, we get μ (4x) = 4μ (x); and proceeding this way we evidently get

μ (nx) = nμ (x) for all n ∈ N and 0 ≤ x ≤

360 . n

This can be confirmed by induction on n. On applying this with n = m, x = 1/m for any m ∈ N we deduce that

1 μ (1) = mμ m so that

μ



1 1 = μ (1). m m

Now on taking n and x = 1/m we see that



n n 1 1 1 μ = μ n. = nμ = n. μ (1) = μ (1). m m m m m Thus μ (r) = r μ (1) for every rational number r with 0 < r ≤ 360, and this extends to the case r = 0 also as putting x = y = 0 in Darboux’s functional equation gives μ (0) = 0. It remains to show that μ (y) = yμ (1) for each irrational number y with 0 < y < 360. Choose a decreasing sequence of rational numbers tn (y) which converge to y. By the case for rational numbers, we have μ (tn ) = tn μ (1). Now tn = (tn − y) + y so by Darboux’s functional equation

μ (tn ) = μ (tn − y) + μ (y), and so

μ (y) = tn μ (1) − μ (tn − y). We wish to make n → ∞ in this and take limits. Now as defined μ (z) > 0 for 0 < z < 360. Then if 0 < x < y < 360 as μ (x)+ μ (y− x) = μ (y) and μ (y − x) > 0, we must have μ (x) < μ (y). Thus μ is an increasing function. Hence as z decreases to 0, μ (z) decreases and is positive and so the onesided limit l = lim μ (z) = lim μ (z) z→0+

z→0,z>0

Sec. 12.3

Derivatives of cosine and sine functions

245

1 must exist. But the formula μ ( 1n ) = μ (1) n shows that μ ( n ) → 0 (n → ∞) and so we must have l = 0. As tn − y → 0 it follows that μ (tn − y) → 0 (n → ∞) and then

lim μ (y) = lim [tn μ (1)] + lim μ (tn − y)

n→∞

n→∞

n→∞

yields μ (y) = yμ (1) + 0. To conclude we take p = μ (1) and note that this is positive as shown above.

12.2.2 Definition of π Definition . We denote μ (360) by π . Then 360p = π so that p = μ (x) =

π 360

and so

π x. 360

12.3 DERIVATIVES OF COSINE AND SINE FUNCTIONS 12.3.1 With the notation of 12.1,

s(t) π = . t→0+ t 180 lim

Proof . We have seen above that for 0 < x ≤ 360, taking n = 2 in 12.2.1(iv), 0 0 is so small that x+ h < 360, x− h > 0. Then c(h) − 1 s(h) c(x + h) − c(x) c(x)c(h) − s(x)s(h) − c(x) = = c(x) − s(x) . h h h h But

c(h)−1 h

=−

2s( h2 )2 h

→ 0 (h → 0+) and so

π c(x + h) − c(x) →− s(x) (h → 0+). h 180 Also c(x) = c(x − h + h) = c(x − h)c(h) + s(x − h)s(h), s(x) = s(x − h + h) = c(x − h)s(h) + s(x − h)c(h). As c(h)2 + s(h)2 = 1 we can solve these equations by Cramer’s rule to obtain c(x − h) = c(x)c(h) + s(x)s(h), s(x − h) = s(x)c(h) − c(x)s(h). From the first of these, by a similar argument to that above, it follows that

π c(x − h) − c(x) →− s(x) (h → 0+). −h 180 π These results combined show that c (x) = − 180 s(x). The other result is proved similarly.

12.4 PARAMETRIC EQUATIONS FOR A CIRCLE 12.4.1 Area of a disk The disk D(O; k)= {Z : |O, Z| ≤ k} has area π k2 . Proof . We use a well-known expression for area as the value of a line integral; see e.g. Apostol [1, Volume II, page 383]. If γ is the path with equation x = kc(t), y = ks(t) (0 ≤ t ≤ 360), then γ traverses the circle C (O, k) once. Moreover if O ≡ (0, 0), P ≡ (kc(t), ks(t)) , T ≡ (kc (t), ks (t)) =   π ks(t) π kc(t) − 180 , 180 , we have that

π 2 2 π 2 k [c (t) + s2 (t)] = k > 0. 360 360 Now C (O; k) is the boundary of D(O; k) and so the area of D(O; k) is given by δF (O, P, T ) = # 1 2

γ

x dy − y dx =

1 2

=

1 2

# 360 0

k2 [c(t)s (t) − s(t)c (t)] dt

# 360 π 0

180

k2 dt = π k2 .

Sec. 12.4

Parametric equations for a circle

247

12.4.2 Length of an arc of a circle For P ∈ C (O; k) let α be an angle in A ∗ (F ) with support |QOP. Let |α |◦ = θ and γ be the curve with parametric equations x = kc(t), y = ks(t) (0 ≤ t ≤ θ ). Then γ has π θ k. the end-points Q and P and has length 180 Proof . The length of γ is given by # θ

[kc (t)]2 + [ks (t)]2 dt =

0

π θ k. 180

NOTE. In the foregoing result, when θ = 360 , γ is the circle C (O; k) and has length 2π k. When θ = 180, γ is a semi-circle and has length π k. For other cases, we note that the line QP has equation s(θ )(x − k) − [c(θ ) − 1]y = 0. The left-hand side in this has the value −ks(θ ) when x = y = 0, and when x = kc(t), y = ks(t) the value k{s(θ )[c(t) − 1] − [c(θ ) − 1]s(t)} = k[−2s(θ )s2 ( 12 t) + 2s2 ( 12 θ )s(t)]   = 4ks( 12 θ )s( 12 t) s( 12 θ )c( 12 t) − s( 12 t)c( 12 θ ) = 4ks( 12 θ )s( 12 t)s( 12 θ − 12 t) > 0, for 0 < t < θ . When θ < 180, we have −s(θ ) < 0 so that Z ≡ (kc(t), ks(t)) is in the closed halfplane H6 with edge QP which does not contain O. Thus, recalling 7.5, γ is the minor arc with end-points Q and P. When 180 < θ < 360, −s(θ ) > 0 so Z is in the closed half-plane H5 with edge QP which contains O. Thus γ is the major arc with end-points Q and P.

12.4.3 Radian measure π Definition . If, for any angle α , x = |α |◦ then 180 x is called the radian measure of r α , and denoted by |α | . We also define, for 0 ≤ x ≤ 2π ,





180 180 C(x) = c x , S(x) = s x , π π and then by the chain rule have C (x) = −S (x), S (x) = C(x), for 0 < x < 2π .

248

Trigonometric functions in calculus

Ch. 12

12.5 EXTENSION OF DOMAINS OF COSINE AND SINE 12.5.1 NOTE. If we define E(x) by E(x) = C(x) + ıS(x) (0 ≤ x ≤ 2π ), where ı2 = −1, then by 12.2.1(i) (12.5.1) E(x + y) = E(x)E(y) (0 ≤ x ≤ 2π ), and by 12.4.3

E (x) = ıE(x) (0 < x < 2π ).

It is natural to enquire if the definition of E can be extended from [0, 2π ] in such a way as to preserve its basic properties. From (12.5.1) we note that in particular E(x + π ) = −E(x) (0 ≤ x ≤ π ), as E(π ) = −1, and if we specify that this is to hold for all x ∈ R then the domain of definition of E becomes extended to R. It follows that E(x + 2π ) = E(x) for all x ∈ R and so for all n ∈ Z, E(x + 2nπ ) = E(x). By the chain rule, for 0 < x < 2π , E (x + 2nπ ) = E (x) = ıE(x) = ıE(x + 2nπ ). Thus for each n ∈ Z, E (x) exists and satisfies E (x) = ıE(x) for 2nπ < x < (2n+ 2)π . Moreover for 0 < h < 2π , S(h) E(2nπ + h) − E(2nπ ) E(h) − 1 C(h) − 1 = = +ı → ı (h → 0), h h h h while for −2π < h < 0, E(2nπ + h) − E(2nπ ) E(2π + h) − 1 −E(π + h) − 1 = = h h h E(π + h) − E(π ) =− → −E (π ) (h → 0) h = ı. Thus E (2nπ ) exists for each n ∈ Z and as its value there is ı = ıE(2nπ ), we see that E (x) = ıE(x) for all x ∈  R. We note that |E(x)| = C(x)2 + S(x)2 = 1 for 0 ≤ x ≤ 2π and then as E has period 2π , |E(x)| = 1 for all x. Hence E(x) = 0 for all x ∈ R. Then for any fixed y ∈ R, d E(x + y) E(x)ıE(x + y) − E(x + y)ıE(x) = = 0, dx E(x) E(x)2

Sec. 12.5

Extension of domains of cosine and sine

249

so for some complex constant c, E(x + y) = cE(x), for all x ∈ R. On putting x = 0 we see that c = E(y) and so E(x + y) = E(x)E(y) for all x, y ∈ R. From E (x) = ıE(x) ∀ x ∈ R, we deduce that C (x) = −S(x), S (x) = C(x), ∀ x ∈ R. From these the usual Maclaurin series for C(x) and S(x) (or E(x)) can be derived and tables of approximations to them prepared; see e.g. Apostol [1, Volume I, pages 435– 436]. By introducing arctan x, π could be expressed as the sum of infinite series and these used to prepare approximations efficiently; see e.g. Apostol [1, Volume I, pages 253–255, 284–285].

12.5.2 Given a frame of reference F , let Q ≡ (1, 0) and consider the points Z of the unit circle C (O; 1) and the angles θ = ∠F QOZ in A (F ). If s = |θ |r then Z ∼ cis θ = C(s) + ıS(s). Any number t ∈ R such that C(t) + ıS(t) = C(s) + ıS(s) is called a radian measure of θ ; collectively we shall call them the multiple radian measures of θ . They have the form t = s + 2nπ ı where n ∈ Z. We write t ∼ ∠F QOZ. Suppose now that t1 ∼ ∠F QOZ1 = θ1 and t2 ∼ ∠F QOZ2 = θ2 . Then E(t1 ) = C(t1 ) + ıS(t1 ) = cos θ1 + ı sin θ1 , E(t2 ) = C(t2 ) + ıS(t2 ) = cos θ2 + ı sin θ2 , and from this by multiplication we have that E(t1 + t2 ) = E(t1 )E(t2 ) = cos(θ1 + θ2 ) + ı sin(θ1 + θ2 ). Thus t1 + t2 ∼ ∠F QOZ1 + ∠F QOZ2 and so addition of multiple measures corresponds to addition of angles. Given any t ∈ R, then γ (u) = E(ut) (0 ≤ u ≤ 1) gives a curve on the unit circle, with initial point Q and terminal point Z. If t ∈ [−2π , 2π ], γ winds around the origin and the curves generated in this way give us some pictorial feel for these multiple radian measures of angles.

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List of axioms AXIOM A1 . Each line is a proper non-empty subset of Π. For each set {A, B} of two distinct points in Π, there is a unique line in Λ to which A and B both belong. AXIOM A2 . Each natural order ≤l has the properties:(i) A ≤l A for all points A ∈ l; (ii) if A ≤l B and B ≤l C then A ≤l C; (iii) if A ≤l B and B ≤l A, then A = B; (iv) for any points A, B ∈ l, either A ≤l B or B ≤l A. AXIOM A3 . Open half-planes G1 , G2 with common edge l have the properties:(i) Π \ l = G1 ∪ G2 ; (ii) G1 and G2 are both convex sets; (iii) if P ∈ G1 and Q ∈ G2 , then [P, Q] ∩ l = 0. / AXIOM A4 . Distance has the following properties:(i) |A, B| ≥ 0 for all A, B ∈ Π; (ii) |A, B| = |B, A| for all A, B ∈ Π; (iii) if Q ∈ [P, R], then |P, Q| + |Q, R| = |P, R|; (iv) given any k ≥ 0 in R, any line l ∈ Λ, any point A ∈ l and either natural order ≤l on l, there is a unique point B ∈ l such that A ≤l B and |A, B| = k, and a unique point C ∈ l such that C ≤l A and |A, C| = k. AXIOM A5 . Degree-measure | |◦ of angles has the following properties:(i) In all cases |α |◦ ≥ 0; (ii) if α is a straight-angle, then |α |◦ = 180;

252

List of axioms

(iii) if ∠BAC is a wedge-angle and the point D = A lies in the interior region I R(|BAC), then |∠BAD|◦ + |∠DAC|◦ = |∠BAC|◦ , while if |BAC is a straight angle-support and D ∈ AB, then |∠BAD|◦ + |∠DAC|◦ = 180; (iv) if B = A, if H1 is a closed half-plane with edge AB, and if the half-lines [A,C and [A, D in H1 are such that |∠BAC|◦ = |∠BAD|◦, then [A, D = [A,C ; (v) if B = A, if H1 is a closed half-plane with edge AB, and if 0 < k < 180, then there is a half-line [A,C in H1 such that |∠BAC|◦ = k. AXIOM A6 . If triangles T and T , with vertices {A, B,C} and {A , B ,C }, respectively, are such that |C, A| = |C , A |, |A, B| = |A , B |, |∠BAC|◦ = |∠B A C |◦ , then T

≡

(A,B,C)→(A ,B C )

T .

AXIOM A7 . Given any line l ∈ Λ and any point P ∈ l, there is at most one line m such that P ∈ m and l m.

Bibliography [1] T.M. Apostol. Calculus, volume I and II. Blaisdell, second edition, 1967. [2] J.K. Baumgart et al., editors. Historical Topics for the Mathematics Classroom. National Council of Teachers of Mathematics, second edition, 1989. [3] F. Cajori. A History of Elementary Mathematics. Macmillan, 1896. [4] H.S.M. Coxeter. Introduction to Geometry. Wiley, third edition, 1969. [5] H.S.M. Coxeter and S.L. Greitzer. Geometry Revisited. Mathematical Association of America, 1967. [6] G.H. Forder. The Foundations of Euclidean Geometry. Cambridge, Dover, 1931, 1958. [7] G.H. Forder. Higher Course Geometry. Cambridge, second edition, 1949. [8] L. Hahn. Complex Numbers and Geometry. Mathematical Association of America, 1994. [9] R.A. Johnson. Advanced Euclidean Geometry (Modern Geometry). Houghton Mifflin, Dover, 1929, 1960. [10] W. Ledermann and S. Vajda. Handbook of Applicable Mathematics: Combinatorics and Geometry, volume V, part A. Wiley, 1985. [11] C.M. McGregor, J.J.C. Nimmo, and W.W. Stothers. Fundamentals of University Mathematics. Horwood Publishing (begun as Albion Publishing, renamed), Chichester, West Sussex, 1994. [12] G.C. Smith. Introductory Mathematics: Algebra and Analysis. Springer, 1998. [13] R.E. Wheeler and E.R. Wheeler. Modern Mathematics for Elementary School Teachers. Brooks/Cole, ninth edition, 1995.

Index Z3 Z4 , Z1 Z2

116 (O, Z1 ) + (O, Z2 ), 188 (O, Z1 ).(O, Z2 ), 190 (Z1 , Z2 ) ↑ (Z3 , Z4 ), 185 −αd , 175 0F , 135 0dF , 170 180F , 135 360F , 135 90dF , 170 AB, 21 C(x), 247 S(x), 247  ≡ T (A,B,C)→(A  ,B ,C  ) T , 49 Z = pZ1 + qZ2 + rZ3 , p + q + r = 1, 194 Z ≡F (x, y), 82 Z ∼F z, 145 [A, B , 25 [A, B,C, D], 32 [A, B,C], 31 [A, B], 23 Δ[A, B,C, D], 75 Δ[A, B,C], 73 Λ, 21 Π, 21 α + β , 136 α ⊕ β , 135 αd + βd , 171 αd , 169 ∠BAC, 30 ∠F IOP, 136 |α |◦ d , 169 |α |◦ , 40 |α |r , 247 |BAC, 29

βd − αd , 175 cis θ , 149 cos α , 130 cos αd , 169 co − sp α , 129 cot γ , 165 δF (Z1 , Z2 , Z3 ), 94 |A, B|, 35 1 2n α , 240 Z3 Z4F Z1 Z2F

, 148 γ − α , 138 ≤l , 22 A (F ), 136 A ∗ (F ), 134 DA (F ), 170 D1 , 168 I R(|BAC), 29 V (Π; O), 188 F Z1 Z0 Z2 , 153 ml(|BAC), 46 μ (x), 243 Z1 Z2≤l , 114 Z1 Z2F , 148 Z3 Z4 Z1 Z2 , 116 π , 245 πl , 57 ρ (l, A, ≤l ), 25 sin α , 130 sin αd , 169 tan α , 149 tan αd , 169 1 2 α , 139 θF , 162 b − a, 191 −→ OZ, 192

Index −→⊥ OZ , 207 c(x), 241 f (F ), 127 i(α ), 130 l ≡F ax + by + c = 0, 87 l m, 54 l ⊥ m, 44 nα , 138 rα ;Z0 , 150 s(x), 241 sα ;Z0 , 153 sl , 58 t ∼ ∠F QOZ, 249 t.(O, Z1 ), 188 tZ1 ,Z2 , 121 xF , 135 C (O; k), 103 E R(|BAC), 30 F (OZ4 , OZ5 ), 175 adjacent sides convex quadrilateral, 32 adjacent vertices convex quadrilateral, 32 alternate angles, 54 angle arms, 30 co-supported, 129 degree-measure, 40 duo, 169 full, 30 half, 139 indicator, 130 multiple radian measures, 249 multiples, 138 null, 30 radian measure, 247 reflex , 30 right, 44 sensed, 153 straight, 30 vertex, 30 verticall opposite, 41

wedge, 30 angle- support arms, 29 angle-support, 29 mid-line, 46 straight, 29 vertex, 29 angles alternate, 54 corresponding, 62 difference, 138 modified sum, 136 subtraction, 138 sum, 135 supplementary, 41 area convex quadrilateral, 75 polygonal region, 78 triangle, 73 areal coordinates points, 193 arms angle, 30 angle-support, 29 ASA principle, 52 axes reference, 81 axial symmetry, 58 bisectors, 168 Cartesian complex coordinate, 145 Cartesian coordinates, 82 Cartesian equation line, 87 centre circle, 103 centroid, 203 Ceva’s theorem, 196 chord circle, 105 circle, 103 centre, 103 chord, 105

255

256 diameter, 105 exterior point, 103 interior point, 103 major arc, 114 minor arc, 114 nine-point, 225 non-secant line, 105 radius, 103 secant, 105 tangent, 105 circumcentre, 207 circumcircle, 207 closed half-planes, 27 co-supported angle, 129 coaxal, 119 collinear, 22 common edge, 26 complex ratio, 148 complex-valued distance, 148 concurrent, 22 congruent triangles, 49 contact point tangent, 105 convex polygon, 76 convex polygon interior, 76 convex quadrilateral, 32 adjacent sides, 32 adjacent vertices, 32 area, 75 diagonal, 32 diagonal line, 32 opposite sides, 32 opposite vertices, 32 perimeter, 32 side, 32 side-line, 32 vertex, 32 convex set, 26 coordinate Cartesian complex, 145 coordinates

Index Cartesian, 82 mobile, 208 polar, 133 corresponding angles, 62 cosine rule, 141 cross, 175 Cross-bar theorem, 40 cross-ratio, 166 cyclic quadrilateral, 114 degree-measure angle, 40 reflex angle, 46 DeMoivre’s theorem, 182 Desargues’ perspective theorem, 199 diagonal convex quadrilateral, 32 diagonal line convex quadrilateral, 32 diameter circle, 105 difference vectors, 191 difference of angles, 138 direct correspondence natural orders, 90 disk, 246 distance, 35 complex-valued, 148 sensed, 114 dot product vectors, 190 drop perpendicular, 57 duo-angle, 169 indicator, 169 null, 169 right, 169 sensed, 175 duo-angles;sum, 171 duo-sector, 168 edge common, 26 equilateral triangle, 219

Index equipollent point pairs, 185 erected perpendicular, 44 Euclidean geometry, 61 Euler line, 215 exterior point circle, 103 exterior region, 30 external bisector, 71 Feuerbach’s theorem, 229 frame of reference, 81 frame reference orientation, 160 full-angle, 30 geometry Euclidean, 61 Grassmann supplement, 207 half-angle, 139 half-line, 25 half-lines opposite, 25 half-plane closed, 27 open, 26 harmonic mean, 99 harmonic range, 71 hypotenuse, 70 incentre, 206 incircle, 206 indicator angle, 130 duo-angle, 169 indirect correspondence natural orders, 90 inequality triangle, 55 initial point, 25 interior convex polygon, 76 interior point

circle, 103 interior region, 29 inverse vector, 191 isogonal conjugates, 234 isometry, 123 isosceles triangle, 51 length, 35 length of radius, 103 line, 21 Cartesian equation, 87 different sides of, 27 one side of, 27 parametric equations, 92 lines perpendicular, 44 major arc circle, 114 mean harmonic, 99 Menelaus’theorem, 195 mid-line angle-support, 46 mid-point, 38 minor arc circle, 114 Miquel’s theorem, 232 mobile coordinates, 208 modified sum angles, 136 mp(A, B), 38 multiple radian measures, 249 multiples angle, 138 Napoleon’s theorem, 219 natural order, 22 natural orders direct correspondence, 90 naturalorders indirect correspondence, 90 nine-point circle, 225 non-secant line

257

258 circle, 105 norm vector, 190 null duo-angle, 169 null vector, 191 null-angle, 30 open half-plane, 26 opposite vertex and side in triangle, 31 half-lines, 25 opposite sides convex quadrilateral, 32 opposite vertices convex quadrilateral, 32 order natural, 22 reciprocal, 23 orientation frame reference, 160 point triple, 160 origin, 81 orthocentre, 204 Pappus’ theorem, 201 parallel, 54 parallelogram, 63 parameter point on line, 92 parametric equations line, 92 Pascal’s theorem, 177 perimeter convex quadrilateral, 32 triangle, 31 perpendicular drop, 57 perpendicular erected, 44 perpendicular lines, 44 plane, 21 point, 21 initial, 25 point pairs equipollent, 185

Index point triple orientation, 160 points areal coordinates, 193 polar coordinates, 133 polygon convex, 76 polygonal region area, 78 position vector, 192 power, 118 principle ASA, 52 SAS, 50 SSS, 53 product sensed, 116 product by a scalar vector, 188 projection, 57 Ptolemy’s theorem, 167 Pythagoras’ theorem, 69 quadrants, 81 quadrilateral convex, 32 cyclic, 114 radian measure angle, 247 radical axis, 118 radius circle, 103 range harmonic, 71 ratio complex, 148 sensed, 116 reciprocal order, 23 rectangle, 63 reference axes, 81 frame, 81 reflex angle

Index degree-measure, 46 reflex-angle, 30 region exterior, 30 interior, 29 right duo-angle, 169 right-angle, 44 rotation, 150 rule cosine, 141 sine, 141 SAS principle, 50 scalar product vectors, 190 secant circle, 105 segment, 23 sensed distance, 114 sensed duo-angle, 175 sensed product, 116 sensed ratio, 116 sensed-angle, 153 sensed-area triangle, 156 set convex, 26 side convex quadrilateral, 32 triangle, 31 side- line convex quadrilateral, 32 side-line triangle, 31 similar triangles, 68 sine rule, 141 slope, 149 space vector, 188 square, 78 SSS principle, 53 standard position;angles, 134 standard position;duo-angles, 170 Steiner-Lehmus theorem, 142

straight angle- support, 29 straight-angle, 30 subtraction vectors, 191 subtraction of angles, 138 sum angles, 135 vectors, 188 sum of duo-angles, 171 supplementary angles, 41 support angle, 30 symmedian point, 235 symmedians, 235 tangent circle, 105 contact point, 105 theorem Ceva’s, 196 Cross-bar, 40 DeMoivre’s, 182 Desargues’ perspective, 199 Feuerbach’s, 229 Menelaus’, 195 Miquel’s, 232 Napoleon’s, 219 Pappus’, 201 Pascal’s, 177 Ptolemy’s, 167 Pythagoras’, 69 Steiner-Lehmus, 142 translation, 121 triangle, 31 area, 73 equilateral, 219 isosceles, 51 sensed-area, 156 side, 31 side-line, 31 triangle inequality, 55 triangles congruent, 49 similar, 68

259

260 vector inverse, 191 norm, 190 null, 191 position, 192 product by a scalar, 188 vector space, 188 vectors difference, 191 dot product, 190 scalar product, 190 subtraction, 191 sum, 188 vertex angle, 30 angle- support, 29 convex quadrilateral, 32 vertex and side in triangle opposite, 31 vertex of triangle, 31 verticall opposite angle, 41 Wallace-Simson line, 230 wedge-angle, 30

Index