About the Authors Titu Andreescu received his BA, MS, and PhD from the West University of Timisoara, Romania. The topic of his doctoral dissertation was “Research on Diophantine Analysis and Applications.” Professor Andreescu currently teaches at the University of Texas at Dallas. Titu is past chairman of the USA Mathematical Olympiad, served as director of the MAA American Mathematics Competitions (1998–2003), coach of the USA International Mathematical Olympiad Team (IMO) for 10 years (1993–2002), Director of the Mathematical Olympiad Summer Program (1995–2002) and leader of the USA IMO Team (1995–2002). In 2002 Titu was elected member of the IMO Advisory Board, the governing body of the world’s most prestigious mathematics competition. Titu received the Edyth May Sliffe Award for Distinguished High School Mathematics Teaching from the MAA in 1994 and a “Certificate of Appreciation” from the president of the MAA in 1995 for his outstanding service as coach of the Mathematical Olympiad Summer Program in preparing the US team for its perfect performance in Hong Kong at the 1994 IMO. Titu’s contributions to numerous textbooks and problem books are recognized worldwide. Dorin Andrica received his PhD in 1992 from “Babes¸-Bolyai” University in Cluj-Napoca, Romania, with a thesis on critical points and applications to the geometry of differentiable submanifolds. Professor Andrica has been chairman of the Department of Geometry at “Babes¸-Bolyai” since 1995. Dorin has written and contributed to numerous mathematics textbooks, problem books, articles and scientific papers at various levels. Dorin is an invited lecturer at university conferences around the world—Austria, Bulgaria, Czech Republic, Egypt, France, Germany, Greece, the Netherlands, Serbia, Turkey, and USA. He is a member of the Romanian Committee for the Mathematics Olympiad and member of editorial boards of several international journals. Dorin has been a regular faculty member at the Canada–USA Mathcamps since 2001.
Titu Andreescu Dorin Andrica
Complex Numbers from A to. . . Z
Birkh¨auser Boston • Basel • Berlin
Titu Andreescu University of Texas at Dallas School of Natural Sciences and Mathematics Richardson, TX 75083 U.S.A.
Dorin Andrica “Babes¸-Bolyai” University Faculty of Mathematics 3400 Cluj-Napoca Romania
Cover design by Mary Burgess. Mathematics Subject Classification (2000): 00A05, 00A07, 30-99, 30A99, 97U40
Library of Congress Cataloging-in-Publication Data Andreescu, Titu, 1956Complex numbers from A to–Z / Titu Andreescu, Dorin Andrica. p. cm. “Partly based on a Romanian version . . . preserving the title. . . and about 35% of the text”–Pref. Includes bibliographical references and index. ISBN 0-8176-4326-5 (acid-free paper) 1. Numbers, Complex. I. Andrica, D. (Dorin) II. Andrica, D. (Dorin) Numere complexe QA255.A558 2004 512.7’88–dc22 2004051907
ISBN-10 0-8176-4326-5 ISBN-13 978-0-8176-4326-3
eISBN 0-8176-4449-0
Printed on acid-free paper.
c 2006 Birkh¨auser Boston
Complex Numbers from A to. . . Z is a greatly expanded and substantially enhanced version of the Romanian edition, Numere complexe de la A la. . . Z, S.C. Editura Millenium S.R. L., Alba Iulia, Romania, 2001 All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Birkh¨auser Boston, c/o Springer Science+Business Media Inc., 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed in the United States of America. 987654321 www.birkhauser.com
(TXQ/MP)
The shortest path between two truths in the real domain passes through the complex domain. Jacques Hadamard
Contents
Preface
ix
Notation
xiii
1
Complex Numbers in Algebraic Form 1.1 Algebraic Representation of Complex Numbers . . . . . . 1.1.1 Definition of complex numbers . . . . . . . . . . . 1.1.2 Properties concerning addition . . . . . . . . . . . 1.1.3 Properties concerning multiplication . . . . . . . . 1.1.4 Complex numbers in algebraic form . . . . . . . . 1.1.5 Powers of the number i . . . . . . . . . . . . . . . 1.1.6 Conjugate of a complex number . . . . . . . . . . 1.1.7 Modulus of a complex number . . . . . . . . . . . 1.1.8 Solving quadratic equations . . . . . . . . . . . . . 1.1.9 Problems . . . . . . . . . . . . . . . . . . . . . . 1.2 Geometric Interpretation of the Algebraic Operations . . . 1.2.1 Geometric interpretation of a complex number . . . 1.2.2 Geometric interpretation of the modulus . . . . . . 1.2.3 Geometric interpretation of the algebraic operations 1.2.4 Problems . . . . . . . . . . . . . . . . . . . . . .
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1 1 1 2 3 5 7 8 9 15 18 21 21 23 24 27
vi
2
Contents
Complex Numbers in Trigonometric Form
29
2.1
Polar Representation of Complex Numbers . . . . . . . . . . . . . .
29
2.1.1
Polar coordinates in the plane . . . . . . . . . . . . . . . . . .
29
2.1.2
Polar representation of a complex number . . . . . . . . . . .
31
2.1.3
Operations with complex numbers in polar representation . . .
36
2.1.4
Geometric interpretation of multiplication . . . . . . . . . . .
39
2.1.5
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
Roots of Unity . . . . . . . . . . . . . . . . . . . . . . . . .
41
2.2
3
The
nth
2.2.1
Defining the
2.2.2
The
nth
nth
roots of a complex number . . . . . . . . . .
41
roots of unity . . . . . . . . . . . . . . . . . . . . . .
43
2.2.3
Binomial equations . . . . . . . . . . . . . . . . . . . . . . .
51
2.2.4
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
Complex Numbers and Geometry
53
3.1
Some Simple Geometric Notions and Properties . . . . . . . . . . . .
53
3.1.1
The distance between two points . . . . . . . . . . . . . . . .
53
3.1.2
Segments, rays and lines . . . . . . . . . . . . . . . . . . . .
54
3.1.3
Dividing a segment into a given ratio . . . . . . . . . . . . . .
57
3.1.4
Measure of an angle . . . . . . . . . . . . . . . . . . . . . . .
58
3.1.5
Angle between two lines . . . . . . . . . . . . . . . . . . . .
61
3.1.6
Rotation of a point . . . . . . . . . . . . . . . . . . . . . . .
61
3.2
Conditions for Collinearity, Orthogonality and Concyclicity . . . . . .
65
3.3
Similar Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
3.4
Equilateral Triangles . . . . . . . . . . . . . . . . . . . . . . . . . .
70
3.5
Some Analytic Geometry in the Complex Plane . . . . . . . . . . . .
76
3.5.1
Equation of a line . . . . . . . . . . . . . . . . . . . . . . . .
76
3.5.2
Equation of a line determined by two points . . . . . . . . . .
78
3.5.3
The area of a triangle . . . . . . . . . . . . . . . . . . . . . .
79
3.5.4
Equation of a line determined by a point and a direction . . . .
82
3.5.5
The foot of a perpendicular from a point to a line . . . . . . .
83
3.5.6
Distance from a point to a line . . . . . . . . . . . . . . . . .
83
The Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
3.6.1
Equation of a circle . . . . . . . . . . . . . . . . . . . . . . .
84
3.6.2
The power of a point with respect to a circle . . . . . . . . . .
86
3.6.3
Angle between two circles . . . . . . . . . . . . . . . . . . .
86
3.6
Contents
4
5
More on Complex Numbers and Geometry 4.1 The Real Product of Two Complex Numbers . . . . . . . . . . 4.2 The Complex Product of Two Complex Numbers . . . . . . . 4.3 The Area of a Convex Polygon . . . . . . . . . . . . . . . . . 4.4 Intersecting Cevians and Some Important Points in a Triangle . 4.5 The Nine-Point Circle of Euler . . . . . . . . . . . . . . . . . 4.6 Some Important Distances in a Triangle . . . . . . . . . . . . 4.6.1 Fundamental invariants of a triangle . . . . . . . . . . 4.6.2 The distance OI . . . . . . . . . . . . . . . . . . . . . 4.6.3 The distance ON . . . . . . . . . . . . . . . . . . . . 4.6.4 The distance OH . . . . . . . . . . . . . . . . . . . . 4.7 Distance between Two Points in the Plane of a Triangle . . . . 4.7.1 Barycentric coordinates . . . . . . . . . . . . . . . . . 4.7.2 Distance between two points in barycentric coordinates 4.8 The Area of a Triangle in Barycentric Coordinates . . . . . . . 4.9 Orthopolar Triangles . . . . . . . . . . . . . . . . . . . . . . 4.9.1 The Simson–Wallance line and the pedal triangle . . . 4.9.2 Necessary and sufficient conditions for orthopolarity . 4.10 Area of the Antipedal Triangle . . . . . . . . . . . . . . . . . 4.11 Lagrange’s Theorem and Applications . . . . . . . . . . . . . 4.12 Euler’s Center of an Inscribed Polygon . . . . . . . . . . . . . 4.13 Some Geometric Transformations of the Complex Plane . . . 4.13.1 Translation . . . . . . . . . . . . . . . . . . . . . . . 4.13.2 Reflection in the real axis . . . . . . . . . . . . . . . 4.13.3 Reflection in a point . . . . . . . . . . . . . . . . . . 4.13.4 Rotation . . . . . . . . . . . . . . . . . . . . . . . . 4.13.5 Isometric transformation of the complex plane . . . . 4.13.6 Morley’s theorem . . . . . . . . . . . . . . . . . . . 4.13.7 hom*othecy . . . . . . . . . . . . . . . . . . . . . . . 4.13.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . Olympiad-Caliber Problems 5.1 Problems Involving Moduli and Conjugates . . . 5.2 Algebraic Equations and Polynomials . . . . . . 5.3 From Algebraic Identities to Geometric Properties 5.4 Solving Geometric Problems . . . . . . . . . . . 5.5 Solving Trigonometric Problems . . . . . . . . . 5.6 More on the nth Roots of Unity . . . . . . . . . .
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vii
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89 89 96 100 103 106 110 110 112 113 114 115 115 117 119 125 125 132 136 140 148 151 151 152 152 153 153 155 158 160
. . . . . .
161 161 177 181 190 214 220
viii
Contents
5.7 5.8 5.9 6
Problems Involving Polygons . . . . . . . . . . . . . . . . . . . . . . 229 Complex Numbers and Combinatorics . . . . . . . . . . . . . . . . . 237 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . 246
Answers, Hints and Solutions to Proposed Problems 6.1 Answers, Hints and Solutions to Routine Problems . . . . . . . . . . 6.1.1 Complex numbers in algebraic representation (pp. 18–21) . . . 6.1.2 Geometric interpretation of the algebraic operations (p. 27) . . 6.1.3 Polar representation of complex numbers (pp. 39–41) . . . . . 6.1.4 The n th roots of unity (p. 52) . . . . . . . . . . . . . . . . . . 6.1.5 Some geometric transformations of the complex plane (p. 160) 6.2 Solutions to the Olympiad-Caliber Problems . . . . . . . . . . . . . . 6.2.1 Problems involving moduli and conjugates (pp. 175–176) . . . 6.2.2 Algebraic equations and polynomials (p. 181) . . . . . . . . . 6.2.3 From algebraic identities to geometric properties (p. 190) . . . 6.2.4 Solving geometric problems (pp. 211–213) . . . . . . . . . . . 6.2.5 Solving trigonometric problems (p. 220) . . . . . . . . . . . . 6.2.6 More on the n th roots of unity (pp. 228–229) . . . . . . . . . . 6.2.7 Problems involving polygons (p. 237) . . . . . . . . . . . . . 6.2.8 Complex numbers and combinatorics (p. 245) . . . . . . . . . 6.2.9 Miscellaneous problems (p. 252) . . . . . . . . . . . . . . . .
253 253 253 258 258 260 261 262 262 269 272 274 287 289 292 298 302
Glossary
307
References
313
Index of Authors
317
Subject Index
319
Preface
Solving algebraic equations has been historically one of the favorite topics of mathematicians. While linear equations are always solvable in real numbers, not all quadratic equations have this property. The simplest such equation is x 2 + 1 = 0. Until the 18th century, mathematicians avoided quadratic equations that were not solvable over R. √ Leonhard Euler broke the ice introducing the “number” −1 in his famous book Elements of Algebra as “ . . . neither nothing, nor greater than nothing, nor less than nothing . . . ” and observed “ . . . notwithstanding this, these numbers present themselves to the mind; they exist in our imagination and we still have a sufficient idea of them; . . . nothing prevents us from making use of these imaginary numbers, and employing them √ in calculation”. Euler denoted the number −1 by i and called it the imaginary unit. This became one of the most useful symbols in mathematics. Using this symbol one defines complex numbers as z = a + bi, where a and b are real numbers. The study of complex numbers continues and has been enhanced in the last two and a half centuries; in fact, it is impossible to imagine modern mathematics without complex numbers. All mathematical domains make use of them in some way. This is true of other disciplines as well: for example, mechanics, theoretical physics, hydrodynamics, and chemistry. Our main goal is to introduce the reader to this fascinating subject. The book runs smoothly between key concepts and elementary results concerning complex numbers. The reader has the opportunity to learn how complex numbers can be employed in solving algebraic equations, and to understand the geometric interpretation of com-
x
Preface
plex numbers and the operations involving them. The theoretical part of the book is augmented by rich exercises and problems of various levels of difficulty. In Chapters 3 and 4 we cover important applications in Euclidean geometry. Many geometry problems may be solved efficiently and elegantly using complex numbers. The wealth of examples we provide, the presentation of many topics in a personal manner, the presence of numerous original problems, and the attention to detail in the solutions to selected exercises and problems are only some of the key features of this book. Among the techniques presented, for example, are those for the real and the complex product of complex numbers. In complex number language, these are the analogues of the scalar and cross products, respectively. Employing these two products turns out to be efficient in solving numerous problems involving complex numbers. After covering this part, the reader will appreciate the use of these techniques. A special feature of the book is Chapter 5, an outstanding selection of genuine Olympiad and other important mathematical contest problems solved using the methods already presented. This work does not cover all aspects pertaining to complex numbers. It is not a complex analysis book, but rather a stepping stone in its study, which is why we have not used the standard notation eit for z = cos t + i sin t, or the usual power series expansions. The book reflects the unique experience of the authors. It distills a vast mathematical literature, most of which is unknown to the western public, capturing the essence of an abundant problem-solving culture. Our work is partly based on a Romanian version, Numere complexe de la A la . . . Z, authored by D. Andrica and N. Bis¸boac˘a and published by Millennium in 2001 (see our reference [10]). We are preserving the title of the Romanian edition and about 35% of the text. Even this 35% has been significantly improved and enhanced with up-to-date material. The targeted audience includes high school students and their teachers, undergraduates, mathematics contestants such as those training for Olympiads or the W. L. Putnam Mathematical Competition, their coaches, and any person interested in essential mathematics. This book might spawn courses such as Complex Numbers and Euclidean Geometry for prospective high school teachers, giving future educators ideas about things they could do with their brighter students or with a math club. This would be quite a welcome development. Special thanks are given to Daniel V˘ac˘aret¸u, Nicolae Bis¸boac˘a, Gabriel Dospinescu, and Ioan S¸erdean for the careful proofreading of the final version of the manuscript. We
Preface
xi
would also like to thank the referees who provided pertinent suggestions that directly contributed to the improvement of the text. Titu Andreescu Dorin Andrica October 2004
Notation
Z N Q R R∗ R2 C C∗ [a, b] (a, b) z |z| −→ AB (AB) [AB] (AB area[F] Un C(P; n)
the set of integers the set of positive integers the set of rational numbers the set of real numbers the set of nonzero real numbers the set of pairs of real numbers the set of complex numbers the set of nonzero complex numbers the set of real numbers x such that a ≤ x ≤ b the set of real numbers x such that a < x < b the conjugate of the complex number z the modulus or absolute value of complex number z the vector AB the open segment determined by A and B the closed segment determined by A and B the open ray of origin A that contains B the area of figure F the set of n th roots of unity the circle centered at point P with radius n
1 Complex Numbers in Algebraic Form
1.1 1.1.1
Algebraic Representation of Complex Numbers Definition of complex numbers
In what follows we assume that the definition and basic properties of the set of real numbers R are known. Let us consider the set R2 = R × R = {(x, y)| x, y ∈ R}. Two elements (x1 , y1 ) and (x2 , y2 ) of R2 are equal if and only if x1 = x2 and y1 = y2 . The operations of addition and multiplication are defined on the set R2 as follows: z 1 + z 2 = (x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ) ∈ R2 and z 1 · z 2 = (x1 , y1 ) · (x2 , y2 ) = (x1 x2 − y1 y2 , x1 y2 + x2 y1 ) ∈ R2 , for all z 1 = (x1 , y1 ) ∈ R2 and z 2 = (x2 , y2 ) ∈ R2 . The element z 1 + z 2 ∈ R2 is called the sum of z 1 , z 2 and the element z 1 · z 2 ∈ R2 is called the product of z 1 , z 2 . Remarks. 1) If z 1 = (x1 , 0) ∈ R2 and z 2 = (x2 , 0) ∈ R2 , then z 1 · z 2 = (x1 x2 , 0). (2) If z 1 = (0, y1 ) ∈ R2 and z 2 = (0, y2 ) ∈ R2 , then z 1 · z 2 = (−y1 y2 , 0). Examples. 1) Let z 1 = (−5, 6) and z 2 = (1, −2). Then z 1 + z 2 = (−5, 6) + (1, −2) = (−4, 4)
2
1. Complex Numbers in Algebraic Form
and z 1 z 2 = (−5, 6) · (1, −2) = (−5 + 12, 10 + 6) = (7, 16). 1 1 1 (2) Let z 1 = − , 1 and z 2 = − , . Then 2 3 2 1 1 1 5 3 z1 + z2 = − − , 1 + = − , 2 3 2 6 2
and z1 z2 =
1 1 1 1 − ,− − 6 2 4 3
1 7 = − ,− . 3 12
Definition. The set R2 , together with the addition and multiplication operations, is called the set of complex numbers, denoted by C. Any element z = (x, y) ∈ C is called a complex number. The notation C∗ is used to indicate the set C \ {(0, 0)}.
1.1.2
Properties concerning addition
The addition of complex numbers satisfies the following properties: (a) Commutative law z 1 + z 2 = z 2 + z 1 for all z 1 , z 2 ∈ C. (b) Associative law (z 1 + z 2 ) + z 3 = z 1 + (z 2 + z 3 ) for all z 1 , z 2 , z 3 ∈ C. Indeed, if z 1 = (x1 , y1 ) ∈ C, z 2 = (x2 , y2 ) ∈ C, z 3 = (x3 , y3 ) ∈ C, then (z 1 + z 2 ) + z 3 = [(x1 , y1 ) + (x2 , y2 )] + (x3 , y3 ) = (x1 + x2 , y1 + y2 ) + (x3 , y3 ) = ((x1 + x2 ) + x3 , (y1 + y2 ) + y3 ), and z 1 + (z 2 + z 3 ) = (x1 , y1 ) + [(x2 , y2 ) + (x3 , y3 )] = (x1 , y1 ) + (x2 + x3 , y2 + y3 ) = (x1 + (x2 + x3 ), y1 + (y2 + y3 )). The claim holds due to the associativity of the addition of real numbers. (c) Additive identity There is a unique complex number 0 = (0, 0) such that z + 0 = 0 + z = z for all z = (x, y) ∈ C. (d) Additive inverse For any complex number z = (x, y) there is a unique −z = (−x, −y) ∈ C such that z + (−z) = (−z) + z = 0.
1.1. Algebraic Representation of Complex Numbers
3
The reader can easily prove the claims (a), (c) and (d). The number z 1 − z 2 = z 1 + (−z 2 ) is called the difference of the numbers z 1 and z 2 . The operation that assigns to the numbers z 1 and z 2 the number z 1 − z 2 is called subtraction and is defined by z 1 − z 2 = (x1 , y1 ) − (x2 , y2 ) = (x1 − x2 , y1 − y2 ) ∈ C.
1.1.3
Properties concerning multiplication
The multiplication of complex numbers satisfies the following properties: (a) Commutative law z 1 · z 2 = z 2 · z 1 for all z 1 , z 2 ∈ C. (b) Associative law (z 1 · z 2 ) · z 3 = z 1 · (z 2 · z 3 ) for all z 1 , z 2 , z 3 ∈ C. (c) Multiplicative identity There is a unique complex number 1 = (1, 0) ∈ C such that z · 1 = 1 · z = z for all z ∈ C. A simple algebraic manipulation is all that is needed to verify these equalities: z · 1 = (x, y) · (1, 0) = (x · 1 − y · 0, x · 0 + y · 1) = (x, y) = z and 1 · z = (1, 0) · (x, y) = (1 · x − 0 · y, 1 · y + 0 · x) = (x, y) = z. (d) Multiplicative inverse For any complex number z = (x, y) ∈ C∗ there is a unique number z −1 = (x , y ) ∈ C such that z · z −1 = z −1 · z = 1. To find z −1 = (x , y ), observe that (x, y) = (0, 0) implies x = 0 or y = 0 and consequently x 2 + y 2 = 0. The relation z · z −1 = 1 gives (x, y) · (x , y ) = (1, 0), or equivalently x x − yy = 1 yx + x y = 0. Solving this system with respect to x and y , one obtains x =
x2
x y and y = − 2 , 2 +y x + y2
4
1. Complex Numbers in Algebraic Form
hence the multiplicative inverse of the complex number z = (x, y) ∈ C∗ is x 1 y −1 z = = ∈ C∗ . ,− 2 z x 2 + y2 x + y2 By the commutative law we also have z −1 · z = 1. Two complex numbers z 1 = (z 1 , y1 ) ∈ C and z = (x, y) ∈ C∗ uniquely determine z1 a third number called their quotient, denoted by and defined by z x y z1 −1 = z 1 · z = (x1 , y1 ) · ,− 2 z x 2 + y2 x + y2 x1 x + y1 y −x1 y + y1 x ∈ C. , = x 2 + y2 x 2 + y2 Examples. 1) If z = (1, 2), then 1 −2 1 −2 = , , . z −1 = 5 5 12 + 2 2 12 + 2 2 2) If z 1 = (1, 2) and z 2 = (3, 4), then z1 11 2 3 + 8 −4 + 6 , = , . = z2 9 + 16 9 + 16 25 25 An integer power of a complex number z ∈ C∗ is defined by z 0 = 1;
z 1 = z;
z 2 = z · z;
z n = z · z· · · z for all integers n > 0 n times
(z −1 )−n
and = for all integers n < 0. The following properties hold for all complex numbers z, z 1 , z 2 ∈ C∗ and for all integers m, n: 1) z m · z n = z m+n ; zm 2) n = z m−n ; z 3) (z m )n = z mn ; 4) (z 1 · z 2 )n = z 1n · z 2n ; n zn z1 = 1n . 5) z2 z2 When z = 0, we define 0n = 0 for all integers n > 0. e) Distributive law zn
z 1 · (z 2 + z 3 ) = z 1 · z 2 + z 1 · z 3 for all z 1 , z 2 , z 3 ∈ C. The above properties of addition and multiplication show that the set C of all complex numbers, together with these operations, forms a field.
1.1. Algebraic Representation of Complex Numbers
1.1.4
5
Complex numbers in algebraic form
For algebraic manipulation it is not convenient to represent a complex number as an ordered pair. For this reason another form of writing is preferred. To introduce this new algebraic representation, consider the set R × {0}, together with the addition and multiplication operations defined on R2 . The function f : R → R × {0},
f (x) = (x, 0)
is bijective and moreover, (x, 0) + (y, 0) = (x + y, 0) and (x, 0) · (y, 0) = (x y, 0). The reader will not fail to notice that the algebraic operations on R × {0} are similar to the operations on R; therefore we can identify the ordered pair (x, 0) with the number x for all x ∈ R. Hence we can use, by the above bijection f , the notation (x, 0) = x. Setting i = (0, 1) we obtain z = (x, y) = (x, 0) + (0, y) = (x, 0) + (y, 0) · (0, 1) = x + yi = (x, 0) + (0, 1) · (y, 0) = x + i y. In this way we obtain Proposition. Any complex number z = (x, y) can be uniquely represented in the form z = x + yi, where x, y are real numbers. The relation i 2 = −1 holds. The formula i 2 = −1 follows directly from the definition of multiplication: i 2 = i · i = (0, 1) · (0, 1) = (−1, 0) = −1. The expression x + yi is called the algebraic representation (form) of the complex number z = (x, y), so we can write C = {x + yi| x ∈ R, y ∈ R, i 2 = −1}. From now on we will denote the complex number z = (x, y) by x + i y. The real number x = Re(z) is called the real part of the complex number z and similarly, y = Im(z) is called the imaginary part of z. Complex numbers of the form i y, y ∈ R — in other words, complex numbers whose real part is 0 — are called imaginary. On the other hand, complex numbers of the form i y, y ∈ R∗ are called purely imaginary and the complex number i is called the imaginary unit. The following relations are easy to verify:
6
1. Complex Numbers in Algebraic Form
a) z 1 = z 2 if and only if Re(z)1 = Re(z)2 and Im(z)1 = Im(z)2 . b) z ∈ R if and only if Im(z) = 0. c) z ∈ C \ R if and only if Im(z) = 0. Using the algebraic representation, the usual operations with complex numbers can be performed as follows: 1. Addition z 1 + z 2 = (x1 + y1 i) + (x2 + y2 i) = (x1 + x2 ) + (y1 + y2 )i ∈ C. It is easy to observe that the sum of two complex numbers is a complex number whose real (imaginary) part is the sum of the real (imaginary) parts of the given numbers: Re(z 1 + z 2 ) = Re(z)1 + Re(z)2 ; Im(z 1 + z 2 ) = Im(z)1 + Im(z)2 . 2. Multiplication z 1 · z 2 = (x1 + y1 i)(x2 + y2 i) = (x1 x2 − y1 y2 ) + (x1 y2 + x2 y1 )i ∈ C. In other words, Re(z 1 z 2 ) = Re(z)1 · Re(z)2 − Im(z)1 · Im(z)2 and Im(z 1 z 2 ) = Im(z)1 · Re(z)2 + Im(z)2 · Re(z)1 . For a real number λ and a complex number z = x + yi, λ · z = λ(x + yi) = λx + λyi ∈ C is the product of a real number with a complex number. The following properties are obvious: 1) λ(z 1 + z 2 ) = λz 1 + λz 2 ; 2) λ1 (λ2 z) = (λ1 λ2 )z; 3) (λ1 + λ2 )z = λ1 z + λ2 z for all z, z 1 , z 2 ∈ C and λ, λ1 , λ2 ∈ R. Actually, relations 1) and 3) are special cases of the distributive law and relation 2) comes from the associative law of multiplication for complex numbers. 3. Subtraction z 1 − z 2 = (x1 + y1 i) − (x2 + y2 i) = (x1 − x2 ) + (y1 − y2 )i ∈ C.
1.1. Algebraic Representation of Complex Numbers
7
That is, Re(z 1 − z 2 ) = Re(z)1 − Re(z)2 ; Im(z 1 − z 2 ) = Im(z)1 − Im(z)2 .
1.1.5
Powers of the number i
The formulas for the powers of a complex number with integer exponents are preserved for the algebraic form z = x + i y. Setting z = i, we obtain i 0 = 1; i 4 = i 3 · i = 1;
i 1 = i;
i 2 = −1;
i 5 = i 4 · i = i;
i 3 = i 2 · i = −i;
i 6 = i 5 · i = −1;
i 7 = i 6 · i = −i.
One can prove by induction that for any positive integer n, i 4n = 1;
i 4n+1 = i;
i 4n+2 = −1;
i 4n+3 = −i.
Hence i n ∈ {−1, 1, −i, i} for all integers n ≥ 0. If n is a negative integer, we have −n 1 = (−i)−n . i n = (i −1 )−n = i Examples. 1) We have i 105 + i 23 + i 20 − i 34 = i 4·26+1 + i 4·5+3 + i 4·5 − i 4·8+2 = i − i + 1 + 1 = 2. 2) Let us solve the equation z 3 = 18 + 26i, where z = x + yi and x, y are integers. We can write (x + yi)3 = (x + yi)2 (x + yi) = (x 2 − y 2 + 2x yi)(x + yi) = (x 3 − 3x y 2 ) + (3x 2 y − y 3 )i = 18 + 26i. Using the definition of equality of complex numbers, we obtain x 3 − 3x y 2 = 18 3x 2 y − y 3 = 26. Setting y = t x in the equality 18(3x 2 y − y 3 ) = 26(x 3 − 3x y 2 ), let us observe that x = 0 and y = 0 implies 18(3t − t 3 ) = 26(1 − 3t 2 ). The last relation is equivalent to (3t − 1)(3t 2 − 12t − 13) = 0. 1 The only rational solution of this equation is t = ; hence, 3 x = 3, y = 1 and z = 3 + i.
8
1. Complex Numbers in Algebraic Form
1.1.6 Conjugate of a complex number For a complex number z = x + yi the number z = x − yi is called the complex conjugate or the conjugate complex of z. Proposition. 1) The relation z = z holds if and only if z ∈ R. 2) For any complex number z the relation z = z holds. 3) For any complex number z the number z · z ∈ R is a nonnegative real number. 4) z 1 + z 2 = z 1 + z 2 (the conjugate of a sum is the sum of the conjugates). 5) z 1 · z 2 = z 1 · z 2 (the conjugate of a product is the product of the conjugates). 6) For any nonzero complex number z the relation z −1 = (z)−1 holds. z1 z1 7) = , z 2 = 0 (the conjugate of a quotient is the quotient of the conjuz2 z2 gates). 8) The formulas Re(z) =
z+z 2
and
Im(z) =
z−z 2i
are valid for all z ∈ C. Proof. 1) If z = x + yi, then the relation z = z is equivalent to x + yi = x − yi. Hence 2yi = 0, so y = 0 and finally z = x ∈ R. 2) We have z = x − yi and z = x − (−y)i = x + yi = z. 3) Observe that z · z = (x + yi)(x − yi) = x 2 + y 2 ≥ 0. 4) Note that z 1 + z 2 = (x1 + x2 ) + (y1 + y2 )i = (x1 + x2 ) − (y1 + y2 )i = (x1 − y1 i) + (x2 − y2 i) = z 1 + z 2 . 5) We can write z 1 · z 2 = (x1 x2 − y1 y2 ) + i(x1 y2 + x2 y1 ) = (x1 x2 − y1 y2 ) − i(x1 y2 + x2 y1 ) = (x1 − i y1 )(x2 − i y2 ) = z 1 · z 2 . 1 1 1 6) Because z · = 1, we have z · = 1, and consequently z · = 1, yielding z z z (z −1 ) = (z)−1 . z1 1 z1 1 1 7) Observe that = z1 · = z1 · = z1 · = . z2 z2 z2 z2 z2 8) From the relations z + z = (x + yi) + (x − yi) = 2x,
1.1. Algebraic Representation of Complex Numbers
9
z − z = (x + yi) − (x − yi) = 2yi it follows that Re(z) =
z+z 2
and
Im(z) =
z−z 2i
as desired. The properties 4) and 5) can be easily extended to give n n
zk = zk ; 4 ) 5 )
k=1 n
=
zk
k=1
k=1 n
z k for all z k ∈ C, k = 1, 2, . . . , n.
k=1
As a consequence of 5 ) and 6) we have 5 ) (z n ) = (z)n for any integers n and for any z ∈ C. Comments. a) To obtain the multiplication inverse of a complex number z ∈ C∗ one can use the following approach: z x y x − yi 1 = 2 − 2 i. = = 2 2 2 z z·z x +y x +y x + y2 b) The complex conjugate allows us to obtain the quotient of two complex numbers as follows: z1 z1 · z2 (x1 + y1 i)(x2 − y2 i) x1 x2 + y1 y2 −x1 y2 + x2 y1 = = = + i. 2 2 2 2 z2 z2 · z2 x2 + y2 x2 + y2 x22 + y22 Examples. (1) Compute z =
20 5 + 5i + . 3 − 4i 4 + 3i
Solution. We can write z=
(5 + 5i)(3 + 4i) 20(4 − 3i) −5 + 35i 80 − 60i + = + 2 2 25 25 9 − 16i 16 − 9i
75 − 25i = 3 − i. 25 (2) Let z 1 , z 2 ∈ C. Prove that the number E = z 1 · z 2 + z 1 · z 2 is a real number. =
Solution. We have E = z 1 · z 2 + z 1 · z 2 = z 1 · z 2 + z 1 · z 2 = E, so E ∈ R.
1.1.7
Modulus of a complex number
The number |z| = x 2 + y 2 is called the modulus or the absolute value of the complex number z = x + yi. For example, the complex numbers z 1 = 4 + 3i,
z 2 = −3i,
z3 = 2
10
1. Complex Numbers in Algebraic Form
have the moduli
|z 1 | = 42 + 32 = 5,
|z 2 | =
02 + (−3)2 = 3,
|z 3 | =
22 = 2.
Proposition. The following properties are satisfied: (1) −|z| ≤ Re(z) ≤ |z| and −|z| ≤ Im(z) ≤ |z|. (2) |z| ≥ 0 for all z ∈ C. Moreover, we have |z| = 0 if and only if z = 0. (3) |z| = | − z| = |z|. (4) z · z = |z|2 . (5) |z 1 · z 2 | = |z 1 | · |z 2 | (the modulus of a product is the product of the moduli). (6) |z 1 | − |z 2 | ≤ |z 1 + z 2 | ≤ |z 1 | + |z 2 |. (7) |z −1| = |z|−1 , z = 0. z 1 |z 1 | (8) = , z 2 = 0 (the modulus of a quotient is the quotient of the moduli). z2 |z 2 | 9) |z 1 | − |z 2 | ≤ |z 1 − z 2 | ≤ |z 1 | + |z 2 |. Proof. One can easily check that (1)–(4) hold. (5) We have |z 1 · z 2 |2 = (z 1 · z 2 )(z 1 · z 2 ) = (z 1 · z 1 )(z 2 · z 2 ) = |z 1 |2 · |z 2 |2 and consequently |z 1 · z 2 | = |z 1 | · |z 2 |, since |z| ≥ 0 for all z ∈ C. (6) Observe that |z 1 + z 2 |2 = (z 1 + z 2 )(z 1 + z 2 ) = (z 1 + z 2 )(z 1 + z 2 ) = |z 1 |2 + z 1 · z 2 + z 1 · z 2 + |z 2 |2 . Because z 1 · z 2 = z 1 · z 2 = z 1 · z 2 it follows that z 1 z 2 + z 1 · z 2 = 2 Re(z 1 · z 2 ) ≤ 2|z 1 · z 2 | = 2|z 1 | · |z 2 |, hence |z 1 + z 2 |2 ≤ (|z 1 | + |z 2 |)2 , and consequently, |z 1 + z 2 | ≤ |z 1 | + |z 2 |, as desired. In order to obtain inequality on the left-hand side note that |z 1 | = |z 1 + z 2 + (−z 2 )| ≤ |z 1 + z 2 | + | − z 2 | = |z 1 + z 2 | + |z 2 |, hence |z 1 | − |z 2 | ≤ |z 1 + z 2 |. 1 1 1 1 (7) Note that the relation z · = 1 implies |z| · = 1, or = . Hence z z z |z| |z −1 | = |z|−1 . (8) We have z1 = z 1 · 1 = |z 1 · z −1 | = |z 1 | · |z −1 | = |z 1 | · |z 2 |−1 = |z 1 | . 2 2 z z2 |z 2 | 2
1.1. Algebraic Representation of Complex Numbers
11
(9) We can write |z 1 | = |z 1 − z 2 + z 2 | ≤ |z 1 − z 2 | + |z 2 |, so |z 1 − z 2 | ≥ |z 1 | − |z 2 |. On the other hand, |z 1 − z 2 | = |z 1 + (−z 2 )| ≤ |z 1 | + | − z 2 | = |z 1 | + |z 2 |.
Remarks. (1) The inequality |z 1 + z 2 | ≤ |z 1 | + |z 2 | becomes an equality if and only if Re(z 1 z 2 ) = |z 1 ||z 2 |. This is equivalent to z 1 = t z 2 , where t is a nonnegative real number. (2) The 5) and 6) can be easily extended to give properties n n zk = |z k |; (5 ) k=1 k=1 n n
zk ≤ |z k | for all z k ∈ C, k = 1, n. (6 ) k=1 k=1 As a consequence of(5 ) and (7) we have (5 ) |z n | = |z|n for any integer n and any complex number z. Problem 1. Prove the identity |z 1 + z 2 |2 + |z 1 − z 2 |2 = 2(|z 1 |2 + |z 2 |2 ) for all complex numbers z 1 , z 2 . Solution. Using property 4 in the proposition above, we obtain |z 1 + z 2 |2 + |z 1 − z 2 |2 = (z 1 + z 2 )(z 1 + z 2 ) + (z 1 − z 2 )(z 1 − z 2 ) = |z 1 |2 + z 1 · z 2 + z 2 · z 1 + |z 2 |2 + |z 1 |2 − z 1 · z 2 − z 2 · z 1 + |z 2 |2 = 2(|z 1 |2 + |z 2 |2 ). Problem 2. Prove that if |z 1 | = |z 2 | = 1 and z 1 z 2 = −1, then number.
z1 + z2 is a real 1 + z1 z2
Solution. Using again property 4 in the above proposition, we have z 1 · z 1 = |z 1 |2 = 1 and z 1 = Likewise, z 2 =
1 . z1
1 . Hence denoting by A the number in the problem we have z2 z1 + z2 A= = 1 + z1 · z2
so A is a real number.
1 1 + z1 + z2 z1 z2 = = A, 1 1 1 + z1 z2 1+ · z1 z2
12
1. Complex Numbers in Algebraic Form
Problem 3. Let a be a positive real number and let 1 Ma = z ∈ C∗ : z + = a . z Find the minimum and maximum value of |z| when z ∈ Ma . 1 Solution. Squaring both sides of the equality a = z + , we get z a = z +
1 2 1 1 z 2 + (z)2 1 = z+ z+ + 2 = |z|2 + z z z |z|2 |z|
2
=
|z|4 + (z + z)2 − 2|z|2 + 1 . |z|2
Hence |z|4 − |z|2 · (a 2 + 2) + 1 = −(z + z)2 ≤ 0 and consequently |z| ∈ 2
It follows that |z| ∈
a2 + 2 −
−a +
max |z| =
√ √ a 4 + 4a 2 a 2 + 2 + a 4 + 4a 2 , . 2 2
√ √ a2 + 4 a + a2 + 4 , , so 2 2
a+
√ a2 + 4 , 2
min |z| =
−a +
√ a2 + 4 2
and the extreme values are obtained for the complex numbers in M satisfying z = −z. Problem 4. Prove that for any complex number z, 1 |z + 1| ≥ √ or |z 2 + 1| ≥ 1. 2 Solution. Suppose by way of contradiction that 1 |1 + z| < √ and |1 + z 2 | < 1. 2 Setting z = a + bi, with a, b ∈ R yields z 2 = a 2 − b2 + 2abi. We obtain (1 + a 2 − b2 )2 + 4a 2 b2 < 1 and (1 + a)2 + b2 <
1 , 2
and consequently (a 2 + b2 )2 + 2(a 2 − b2 ) < 0 and 2(a 2 + b2 ) + 4a + 1 < 0.
1.1. Algebraic Representation of Complex Numbers
13
Summing these inequalities implies (a 2 + b2 )2 + (2a + 1)2 < 0, which is a contradiction. Problem 5. Prove that
7 7 2 ≤ |1 + z| + |1 − z + z | ≤ 3 2 6
for all complex numbers with |z| = 1. Solution. Let t = |1 + z| ∈ [0, 2]. We have t 2 = (1 + z) · (1 + z) = 2 + 2 Re(z), so Re(z) = Then |1 − z + z 2 | =
We obtain
t2 − 2 . 2
|7 − 2t 2 |. It suffices to find the extreme values of the function f : [0, 2] → R, f (t) = t + |7 − 2t 2 |.
7 7 7 7 2 f = ≤ t + |7 − 2t | ≤ f =3 2 2 6 6
as we can see from the figure below.
Figure 1.1.
Problem 6. Consider the set H = {z ∈ C : z = x − 1 + xi,
x ∈ R}.
Prove that there is a unique number z ∈ H such that |z| ≤ |w| for all w ∈ H.
14
1. Complex Numbers in Algebraic Form
Solution. Let ω = y − 1 + yi, with y ∈ R. It suffices to prove that there is a unique number x ∈ R such that (x − 1)2 + x 2 ≤ (y − 1)2 + y 2 for all y ∈ R. In other words, x is the minimum point of the function
1 f : R → R, f (y) = (y − 1) + y = 2y − 2y + 1 = 2 y − 2 2
2
2
1 1 1 and z = − + i. 2 2 2 Problem 7. Let x, y, z be distinct complex numbers such that hence x =
y = t x + (1 − t)z, Prove that
t ∈ (0, 1).
|z| − |x| |y| − |x| |z| − |y| ≥ ≥ . |z − y| |z − x| |y − x|
Solution. The relation y = t x + (1 − t)z is equivalent to z − y = t (z − x). The inequality
|z| − |y| |z| − |x| ≥ |z − y| |z − x|
becomes |z| − |y| ≥ t (|z| − |x|), and consequently |y| ≤ (1 − t)|z| + t|x|. This is the triangle inequality for y = (1 − t)z + t x. The second inequality can be proved similarly, writing the equality y = t x + (1 − t)z as y − x = (1 − t)(z − x).
2
1 + , 2
1.1. Algebraic Representation of Complex Numbers
1.1.8
15
Solving quadratic equations
We are now able to solve the quadratic equation with real coefficients ax 2 + bx + c = 0,
a = 0
in the case when its discriminant = b2 − 4ac is negative. By completing the square, we easily get the equivalent form b 2 − + 2 = 0. a x+ 2a 4a Therefore
b x+ 2a
√
2 −i
2
− 2a
2 = 0,
√ √ −b + i − −b − i − and so x1 = , x2 = . 2a 2a Observe that the roots are conjugate complex numbers and the factorization formula ax 2 + bx + c = a(x − x1 )(x − x2 ) holds even in the case < 0. Let us consider now the general quadratic equation with complex coefficients az 2 + bz + c = 0,
a = 0.
Using the same algebraic manipulation as in the case of real coefficients, we get b 2 − 2 = 0. a z+ 2a 4a This is equivalent to
z+
b 2a
2 =
4a 2
or (2az + b)2 = , where = b2 − 4ac is also called the discriminant of the quadratic equation. Setting y = 2az + b, the equation is reduced to y 2 = = u + vi, where u and v are real numbers.
16
1. Complex Numbers in Algebraic Form
This equation has the solutions y1,2 = ±
r +u r −u + (sgn v) i , 2 2
where r = || and signv is the sign of the real number v. The roots of the initial equation are 1 (−b + y1,2 ). 2a Observe that the relations between roots and coefficients b c z1 + z2 = − , z1 z2 = , a a as well as the factorization formula z 1,2 =
az 2 + bz + c = a(z − z 1 )(z − z 2 ) are also preserved when the coefficients of the equation are elements of the field of complex numbers C. Problem 1. Solve, in complex numbers, the quadratic equation z 2 − 8(1 − i)z + 63 − 16i = 0. Solution. We have = (4 − 4i)2 − (63 − 16i) = −63 − 16i and 2 b − ac. 2 The equation
r = | | =
632 + 162 = 65,
where =
y 2 = −63 − 16i 65 − 63 65 + 63 +i = ±(1 − 8i). It follows that has the solution y1,2 = ± 2 2 z 1,2 = 4 − 4i ± (1 − 8i). Hence
z 1 = 5 − 12i and z 2 = 3 + 4i. Problem 2. Let p and q be complex numbers with q = 0. Prove that if the roots of the p quadratic equation x 2 + px + q 2 = 0 have the same absolute value, then is a real q number. (1999 Romanian Mathematical Olympiad – Final Round)
1.1. Algebraic Representation of Complex Numbers
17
Solution. Let x1 and x2 be the roots of the equation and let r = |x1 | = |x2 |. Then (x1 + x2 )2 x1 x2 x1 x2 x2 x1 2 p2 = = + + 2 = 2 + 2 + 2 = 2 + 2 Re(x1 x2 ) 2 x1 x2 x2 x1 q r r r is a real number. Moreover, Re(x1 x2 ) ≥ −|x1 x2 | = −r 2 , so
p2 ≥ 0. q2
p is a real number, as claimed. q Problem 3. Let a, b, c be distinct nonzero complex numbers with |a| = |b| = |c|. a) Prove that if a root of the equation az 2 + bz + c = 0 has modulus equal to 1, then b2 = ac. b) If each of the equations Therefore
az 2 + bz + c = 0
and
bz 2 + cz + a = 0
has a root having modulus 1, then |a − b| = |b − c| = |c − a|. c 1 Solution. a) Let z 1 , z 2 be the roots of the equation with |z 1 | = 1. From z 2 = · a z1 c 1 b it follows that |z 2 | = · = 1. Because z 1 + z 2 = − and |a| = |b|, we have a |z 1 | a 2 |z 1 + z 2 | = 1. This is equivalent to 1 1 (z 1 + z 2 )(z 1 + z 2 ) = 1, i.e., (z 1 + z 2 ) = 1. + z1 z2 We find that (z 1 + z 2 )2 = z 1 z 2 , i.e.,
b 2 c − = , a a
which reduces to b2 = ac, as desired. b) As we have already seen, we have b2 = ac and c2 = ab. Multiplying these relations yields b2 c2 = a 2 bc, hence a 2 = bc. Therefore a 2 + b2 + c2 = ab + bc + ca. Relation (1) is equivalent to (a − b)2 + (b − c)2 + (c − a)2 = 0, i.e., (a − b)2 + (b − c)2 + 2(a − b)(b − c) + (c − a)2 = 2(a − b)(b − c).
(1)
18
1. Complex Numbers in Algebraic Form
It follows that (a − c)2 = (a − b)(b − c). Taking absolute values we find β 2 = γ α, where α = |b − c|, β = |c − a|, γ = |a − b|. In an analogous way we obtain α 2 = βγ and γ 2 = αβ. Adding these relations yields α 2 + β 2 + γ 2 = αβ + βγ + γ α, i.e., (α − β)2 + (β − γ )2 + (γ − α)2 = 0. Hence α = β = γ .
1.1.9
Problems
1. Consider the complex numbers z 1 = (1, 2), z 2 = (−2, 3) and z 3 = (1, −1). Compute the following complex numbers: a) z 1 + z 2 + z 3 ; b) z 1 z 2 + z 2 z 3 + z 3 z 1 ; c) z 1 z 2 z 3 ; d) z 12 + z 22 + z 32 ;
e)
z1 z2 z3 + + ; z2 z3 z1
f)
z 12 + z 22 z 22 + z 32
.
2. Solve the equations: a) z + (−5, 7) = (2, −1); c) z · (2, 3) = (4, 5);
b) (2, 3) + z = (−5, −1); z d) = (3, 2). (−1, 3)
3. Solve in C the equations: a) z 2 + z + 1 = 0; b) z 3 + 1 = 0. 4. Let z = (0, 1) ∈ C. Express
n
z k in terms of the positive integer n.
k=0
5. Solve the equations: a) z · (1, 2) = (−1, 3);
b) (1, 1) · z 2 = (−1, 7).
6. Let z = (a, b) ∈ C. Compute z 2 , z 3 and z 4 . 7. Let z 0 = (a, b) ∈ C. Find z ∈ C such that z 2 = z 0 . 8. Let z = (1, −1). Compute z n , where n is a positive integer. 9. Find real numbers x and y in each of the following cases: y−3 x −3 + = i; a) (1 − 2i)x + (1 + 2i)y = 1 + i; b) 3+i 3−i 1 c) (4 − 3i)x 2 + (3 + 2i)x y = 4y 2 − x 2 + (3x y − 2y 2 )i. 2 10. Compute: a) (2 − i)(−3 + 2i)(5 − 4i); b) (2 − 4i)(5 + 2i) + (3 + 4i)(−6 − i); √ 6 √ 6 1−i 8 −1 + i 3 1−i 7 1 + i 16 + ; d) + ; c) 1−i 1+i 2 2 e)
3 + 7i 5 − 8i + . 2 + 3i 2 − 3i
1.1. Algebraic Representation of Complex Numbers
11. Compute: a) i 2000 + i 1999 + i 201 + i 82 + i 47 ; b) E n = 1 + i + i 2 + i 3 + · · · + i n for n ≥ 1; c) i 1 · i 2 · i 3 · · · i 2000 ; d) i −5 + (−i)−7 + (−i)13 + i −100 + (−i)94 . 12. Solve in C the equations: a)
z2
= i;
b)
z2
= −i;
c)
z2
√ 2 1 . = −i 2 2
13. Find all complex numbers z = 0 such that z +
1 ∈ R. z
14. Prove that: √ √ a) E 1 = (2 + i 5)7 + (2 − i 5)7 ∈ R; 19 + 7i n 20 + 5i n + ∈ R. b) E 2 = 9−i 7 + 6i 15. Prove the following identities: a) |z 1 + z 2 |2 + |z 2 + z 3 |2 + |z 3 + z 1 |2 = |z 1 |2 + |z 2 |2 + |z 3 |2 + |z 1 + z 2 + z 3 |2 ; b) |1 + z 1 z 2 |2 + |z 1 − z 2 |2 = (1 + |z 1 |2 )(1 + |z 2 |2 ); c) |1 − z 1 z 2 |2 − |z 1 − z 2 |2 = (1 − |z 1 |2 )(1 − |z 2 |2 ); d) |z 1 + z 2 + z 3 |2 + | − z 1 + z 2 + z 3 |2 + |z 1 − z 2 + z 3 |2 + |z 1 + z 2 − z 3 |2 = 4(|z 1 |2 + |z 2 |2 + |z 3 |2 ). 1 1 16. Let z ∈ C∗ such that z 3 + 3 ≤ 2. Prove that z + ≤ 2. z z 17. Find all complex numbers z such that |z| = 1 and |z 2 + z 2 | = 1. 18. Find all complex numbers z such that 4z 2 + 8|z|2 = 8. 19. Find all complex numbers z such that z 3 = z. 20. Consider z ∈ C with Re(z) > 1. Prove that 1 1 1 − < . z 2 2 √ 1 3 21. Let a, b, c be real numbers and ω = − + i . Compute 2 2 (a + bω + cω2 )(a + bω2 + cω).
19
20
1. Complex Numbers in Algebraic Form
22. Solve the equations: a) |z| − 2z = 3 − 4i; b) |z| + z = 3 + 4i; c) z 3 = 2 + 11i, where z = x + yi and x, y ∈ Z; d) i z 2 + (1 + 2i)z + 1 = 0; e) z 4 + 6(1 + i)z 2 + 5 + 6i = 0; f) (1 + i)z 2 + 2 + 11i = 0. 23. Find all real numbers m for which the equation z 3 + (3 + i)z 2 − 3z − (m + i) = 0 has at least a real root. 24. Find all complex numbers z such that z = (z − 2)(z + i) is a real number.
1 25. Find all complex numbers z such that |z| = . z 26. Let z 1 , z 2 ∈ C be complex numbers such that |z 1 + z 2 | = |z 1 | = |z 2 | = 1. Compute |z 1 − z 2 |.
√ 3 and
27. Find all positive integers n such that √ n √ n −1 − i 3 −1 + i 3 + = 2. 2 2 28. Let n > 2 be an integer. Find the number of solutions to the equation z n−1 = i z. 29. Let z 1 , z 2 , z 3 be complex numbers with |z 1 | = |z 2 | = |z 3 | = R > 0. Prove that |z 1 − z 2 | · |z 2 − z 3 | + |z 3 − z 1 | · |z 1 − z 2 | + |z 2 − z 3 | · |z 3 − z 1 | ≤ 9R 2 . 30. Let u, v, w, z be complex numbers such that |u| < 1, |v| = 1 and v(u − z) w= . Prove that |w| ≤ 1 if and only if |z| ≤ 1. u·z−1
1.2. Geometric Interpretation of the Algebraic Operations
21
31. Let z 1 , z 2 , z 3 be complex numbers such that z1 + z2 + z3 = 0
and
|z 1 | = |z 2 | = |z 3 | = 1.
Prove that z 12 + z 22 + z 32 = 0. 32. Consider the complex numbers z 1 , z 2 , . . . , z n with |z 1 | = |z 2 | = · · · = |z n | = r > 0. Prove that the number E=
(z 1 + z 2 )(z 2 + z 3 ) · · · (z n−1 + z n )(z n + z 1 ) z1 · z2 · · · zn
is real. 33. Let z 1 , z 2 , z 3 be distinct complex numbers such that |z 1 | = |z 2 | = |z 3 | > 0. If z 1 + z 2 z 3 , z 2 + z 1 z 3 and z 3 + z 1 z 2 are real numbers, prove that z 1 z 2 z 3 = 1. 34. Let x1 and x2 be the roots of the equation x 2 − x + 1 = 0. Compute: a) x12000 + x22000 ;
b) x11999 + x21999 ;
c) x1n + x2n , for n ∈ N.
35. Factorize (in linear polynomials) the following polynomials: a) x 4 + 16;
b) x 3 − 27;
c) x 3 + 8;
d) x 4 + x 2 + 1.
36. Find all quadratic equations with real coefficients that have one of the following roots: 5+i a) (2 + i)(3 − i); b) ; c) i 51 + 2i 80 + 3i 45 + 4i 38 . 2−i 37. (Hlawka’s inequality) Prove that the following inequality |z 1 + z 2 | + |z 2 + z 3 | + |z 3 + z 1 | ≤ |z 1 | + |z 2 | + |z 3 | + |z 1 + z 2 + z 3 | holds for all complex numbers z 1 , z 2 , z 3 .
1.2 1.2.1
Geometric Interpretation of the Algebraic Operations Geometric interpretation of a complex number
We have defined a complex number z = (x, y) = x + yi to be an ordered pair of real numbers (x, y) ∈ R × R, so it is natural to let a complex number z = x + yi correspond to a point M(x, y) in the plane R × R.
22
1. Complex Numbers in Algebraic Form
For a formal introduction, let us consider P to be the set of points of a given plane equipped with a coordinate system x O y. Consider the bijective function ϕ : C → P, ϕ(z) = M(x, y). Definition. The point M(x, y) is called the geometric image of the complex number z = x + yi. The complex number z = x + yi is called the complex coordinate of the point M(x, y). We will use the notation M(z) to indicate that the complex coordinate of M is the complex number z.
Figure 1.2.
The geometric image of the complex conjugate z of a complex number z = x + yi is the reflection point M (x, −y) across the x-axis of the point M(x, y) (see Fig. 1.2). The geometric image of the additive inverse −z of a complex number z = x + yi is the reflection M (−x, −y) across the origin of the point M(x, y) (see Fig. 1.2). The bijective function ϕ maps the set R onto the x-axis, which is called the real axis. On the other hand, the imaginary complex numbers correspond to the y-axis, which is called the imaginary axis. The plane , whose points are identified with complex numbers, is called the complex plane. On the other hand, we can also identify a complex number z = x + yi with the −−→ → vector − v = O M, where M(x, y) is the geometric image of the complex number z.
1.2. Geometric Interpretation of the Algebraic Operations
23
Figure 1.3.
Let V0 be the set of vectors whose initial points are the origin O. Then we can define the bijective function ϕ : C → V0 ,
−−→ → − → − → ϕ (z) = O M = − v =x i +y j ,
− → − → where i , j are the vectors of the x-axis and y-axis, respectively.
1.2.2
Geometric interpretation of the modulus
Let us consider a complex number z = x + yi and the geometric image M(x, y) in the complex plane. The Euclidean distance O M is given by the formula OM =
(x M − x O )2 + (y M − y O )2 ,
→ hence O M = x 2 + y 2 = |z| = |− v |. In other words, the absolute value |z| of a complex number z = x + yi is the length of the segment O M or the magnitude of the − → − → → vector − v =x i +y j . Remarks. a) For a positive real number r , the set of complex numbers with moduli r corresponds in the complex plane to C(O; r ), our notation for the circle C with center O and radius r . b) The complex numbers z with |z| < r correspond to the interior points of circle C; on the other hand, the complex numbers z with |z| > r correspond to the points in the exterior of circle C. √ 3 1 i, k = 1, 2, 3, 4, are represented in the Example. The numbers z k = ± ± 2 2 complex plane by four points on the unit circle centered on the origin, since |z 1 | = |z 2 | = |z 3 | = |z 4 | = 1.
24
1.2.3
1. Complex Numbers in Algebraic Form
Geometric interpretation of the algebraic operations
a) Addition and subtraction. Consider the complex numbers z 1 = x1 + y1 i and z 2 = − → − → − → − → → → v 1 = x1 i + y1 j and − x2 + y2 i and the corresponding vectors − v 2 = x2 i + y2 j . Observe that the sum of the complex numbers is z 1 + z 2 = (x1 + x2 ) + (y1 + y2 )i, and the sum of the vectors is − → − → − → → v 1+− v 2 = (x1 + x2 ) i + (y1 + y2 ) j . → → Therefore, the sum z 1 + z 2 corresponds to the sum − v 1+− v 2.
Figure 1.4.
Examples. 1) We have (3 + 5i) + (6 + i) = 9 + 6i; hence the geometric image of the sum is given in Fig. 1.5.
2) Observe that (6 − 2i) + (−2 + 5i) = 4 + 3i. Therefore the geometric image of the sum of these two complex numbers is the point M(4, 3) (see Fig. 1.6).
1.2. Geometric Interpretation of the Algebraic Operations
25
On the other hand, the difference of the complex numbers z 1 and z 2 is z 1 − z 2 = (x1 − x2 ) + (y1 − y2 )i, and the difference of the vectors v1 and v2 is − → − → − → → v 1−− v 2 = (x1 − x2 ) i + (y1 − y2 ) j . → → Hence, the difference z 1 − z 2 corresponds to the difference − v 1−− v 2. 3) We have (−3 + i) − (2 + 3i) = (−3 + i) + (−2 − 3i) = −5 − 2i; hence the geometric image of difference of these two complex numbers is the point M(−5, −2) given in Fig. 1.7.
4) Note that (3 − 2i) − (−2 − 4i) = (3 − 2i) + (2 + 4i) = 5 + 2i, and obtain the point M(−2, −4) as the geometric image of the difference of these two complex numbers (see Fig. 1.8). Remark. The distance M1 (x1 , y1 ) and M2 (x2 , y2 ) is equal to the modulus of the → → complex number z 1 − z 2 or to the length of the vector − v 1−− v 2 . Indeed, → → v 1−− v 2 | = (x2 − x1 )2 + (y2 − y1 )2 . |M1 M2 | = |z 1 − z 2 | = |− b) Real multiples of a complex number. Consider a complex number z = x + i y − → − → → and the corresponding vector − v = x i + y j . If λ is a real number, then the real multiple λz = λx + iλy corresponds to the vector − → − → → λ− v = λx i + λy j . → → Note that if λ > 0 then the vectors λ− v and − v have the same orientation and → → |λ− v | = λ|− v |.
26
1. Complex Numbers in Algebraic Form
→ → → When λ < 0, the vector λ− v changes to the opposite orientation and |λ− v | = −λ|− v |. − → − → Of course, if λ = 0, then λ v = 0 .
Examples. 1) We have 3(1 + 2i) = 3 + 6i; therefore M (3, 6) is the geometric image of the product of 3 and z = 1 + 2i. 2) Observe that −2(−3 + 2i) = 6 − 4i, and obtain the point M (6, −4) as the geometric image of the product of −2 and z = −3 + 2i.
Figure 1.10.
1.2. Geometric Interpretation of the Algebraic Operations
1.2.4
27
Problems
1. Represent the geometric images of the following complex numbers: z 1 = 3 + i; z 2 = −4 + 2i; z 3 = −5 − 4i; z 4 = 5 − i; z 5 = 1; z 6 = −3i; z 7 = 2i; z 8 = −4. 2. Find the geometric interpretation for the following equalities: a) (−5 + 4i) + (2 − 3i) = −3 + i; b) (4 − i) + (−6 + 4i) = −2 + 3i; c) (−3 − 2i) − (−5 + i) = 2 − 3i; d) (8 − i) − (5 + 3i) = 3 − 4i; e) 2(−4 + 2i) = −8 + 4i; f) −3(−1 + 2i) = 3 − 6i. 3. Find the geometric image of the complex number z in each of the following cases: a) |z − 2| = 3; b) |z + i| < 1; c) |z − 1 + 2i| > 3; d) |z − 2| − |z + 2| < 2; e) 0 < Re(i z) < 1; f) −1 < Im(z) < 1; z − 2 1+z = 0; h) ∈ R. g) Re z−1 z 4. Find the set of points P(x, y) in the complex plane such that
√ | x 2 + 4 + i y − 4| = 10. 5. Let z 1 = 1 + i and z 2 = −1 − i. Find z 3 ∈ C such that triangle z 1 , z 2 , z 3 is equilateral. 6. Find the geometric images of the complex numbers z such that the triangle with vertices at z, z 2 and z 3 is right-angled. 7. Find the geometric images of the complex numbers z such that z + 1 = 2. z
2 Complex Numbers in Trigonometric Form
2.1 2.1.1
Polar Representation of Complex Numbers Polar coordinates in the plane
Let us consider a coordinate plane and a point M(x, y) that is not the origin.
The real number r = x 2 + y 2 is called the polar radius of the point M. The direct −−→ angle t ∗ ∈ [0, 2π) between the vector O M and the positive x-axis is called the polar argument of the point M. The pair (r, t ∗ ) is called the polar coordinates of the point M. We will write M(r, t ∗ ). Note that the function h : R×R\{(0, 0)} → (0, ∞)×[0, 2π ), h((x, y)) = (r, t ∗ ) is bijective. The origin O is the unique point such that r = 0; the argument t ∗ of the origin is not defined. For any point M in the plane there is a unique intersection point P of the ray (O M with the unit circle centered at the origin. The point P has the same polar argument t ∗ . Using the definition of the sine and cosine functions we find that x = r cos t ∗ and y = r sin t ∗ . Therefore, it is easy to obtain the cartesian coordinates of a point from its polar coordinates.
Conversely, let us consider a point M(x, y). The polar radius is r = x 2 + y 2 . To determine the polar argument we study the following cases:
30
2. Complex Numbers in Trigonometric Form
Figure 2.1.
a) If x = 0, from tan t ∗ =
y we deduce that x t ∗ = arctan
where
⎧ ⎪ ⎨ 0, for k= 1, for ⎪ ⎩ 2, for
y + kπ, x
x > 0 and y ≥ 0 x < 0 and any y x > 0 and y < 0.
b) If x = 0 and y = 0, then ∗
t =
π/2, for 3π/2, for
y>0 y < 0.
Examples. 1. Let us find the polar coordinates of the points M1 (2, −2), M2 (−1, 0), √ √ M3 (−2 3, −2), M4 ( 3, 1), M5 (3, 0), M6 (−2, 2), M7 (0, 1) and M8 (0, −4).
√ π In this case we have r1 = 22 + (−2)2 = 2 2; t1∗ = arctan(−1) + 2π = − + 4 √ 7π 7π 2π = , so M1 2 2, . 4 4 Observe that r2 = 1, t2∗ = arctan 0 + π = π , so M2 (1, π). √ 7π 3 π 7π ∗ +π = +π = , so M3 4, . We have r3 = 4, t3 = arctan 3 6 6 6 √ π π 3 Note that r4 = 2, t4∗ = arctan = , so M4 2, . 3 6 6 ∗ We have r5 = 3, t5 = arctan 0 + 0 = 0, so M5 (3, 0). √ 3π √ ∗ π 3π We have r6 = 2 2, t6 = arctan(−1) + π = − + π = , so M6 2 2, . 4 4 4 π π Note that r7 = 1, t7∗ = , so M7 1, . 2 2
2.1. Polar Representation of Complex Numbers
Observe that r8 = 4, t8∗ =
31
3π 3π , so M8 1, . 2 2
2π 7π 2. Let us find the cartesian coordinates of the points M1 2, , M2 3, and 3 4 M3 (1, 1). √ √ 1 3 2π 2π We have x1 = 2 cos = 2 − = −1, y1 = 2 sin = 2 = 3, so 3 2 3 2 √ M1 (−1, 3). √ √ 7π 7π 3 2 3 2 Note that x2 = 3 cos = , y2 = 3 sin = − , so 4 2 4 2 √ √ 3 2 3 2 M2 ,− . 2 2 Observe that x3 = cos 1, y2 = sin 1, so M3 (cos 1, sin 1).
2.1.2
Polar representation of a complex number
For a complex number z = x + yi we can write the polar representation z = r (cos t ∗ + i sin t ∗ ), where r ∈ [0, ∞) and t ∗ ∈ [0, 2π ) are the polar coordinates of the geometric image of z. The polar argument t ∗ of the geometric image of z is called the argument of z, denoted by arg z. The polar radius r of the geometric image of z is equal to the modulus of z. For z = 0, the modulus and argument of z are uniquely determined. Consider z = r (cos t ∗ + i sin t ∗ ) and let t = t ∗ + 2kπ for an integer k. Then z = r [cos(t − 2kπ ) + i sin(t − 2kπ )] = r (cos t + i sin t), i.e., any complex number z can be represented as z = r (cos t + i sin t), where r ≥ 0 and t ∈ R. The set Arg z = {t : t ∗ + 2kπ, k ∈ Z} is called the extended argument of the complex number z. Therefore, two complex numbers z 1 , z 2 = 0 represented as z 1 = r1 (cos t1 + i sin t1 ) and z 2 = r2 (cos t2 + i sin t2 ) are equal if and only if r1 = r2 and t1 − t2 = 2kπ , for an integer k. Example 1. Let us find the polar representation of the numbers: a) z 1 = −1 − i, b) z 2 = 2 + 2i, √ c) z 3 = −1 + i 3, √ d) z 4 = 1 − i 3 and determine their extended argument.
32
2. Complex Numbers in Trigonometric Form
a) As in the figure below the geometric image P1 (−1, −1) lies in the third quadrant.
√ Then r1 = (−1)2 + (−1)2 = 2 and t1∗ = arctan
π 5π y + π = arctan 1 + π = + π = . x 4 4
Figure 2.2.
Hence
√ 5π 5π z 1 = 2 cos + i sin 4 4
and Arg z 1 =
5π + 2kπ | k ∈ Z . 4
b) The point P2 (2, 2) lies in the first quadrant, so we can write
√ π r2 = 22 + 22 = 2 2 and t2∗ = arctan 1 = . 4 Hence
√ π π z 2 = 2 2 cos + i sin 4 4
and Arg z =
π
+ 2kπ | k ∈ Z .
4 √ c) The point P3 (−1, 3) lies in the second quadrant, so √ π 2π r3 = 2 and t3∗ = arctan(− 3) + π = − + π = . 3 3 Therefore,
2π 2π + i sin z 3 = 2 cos 3 3
2.1. Polar Representation of Complex Numbers
Figure 2.3.
and Arg z 3 =
2π + 2kπ | k ∈ Z . 3
√ d) The point P4 (1, − 3) lies in the fourth quadrant (Fig. 2.4), so √ 5π π . r4 = 2 and t4∗ = arctan(− 3) + 2π = − + 2π = 3 3
Figure 2.4.
Hence
5π 5π z 4 = 2 cos + i sin , 3 3
and Arg z 4 =
5π + 2kπ | k ∈ Z . 3
33
34
2. Complex Numbers in Trigonometric Form
Example 2. Let us find the polar representation of the numbers a) z 1 = 2i, b) z 2 = −1, c) z 3 = 2, d) z 4 = −3i and determine their extended argument. a) The point P1 (0, 2) lies on the positive y-axis, so π π π r1 = 2, t1∗ = , z 1 = 2 cos + i sin 2 2 2 and Arg z 1 =
π
+ 2kπ | k ∈ Z .
2 b) The point P2 (−1, 0) lies on the negative x-axis, so r2 = 1,
t2∗ = π,
z 2 = cos π + i sin π
and Arg z 2 = {π + 2kπ | k ∈ Z}. c) The point P3 (2, 0) lies on the positive x-axis, so r3 = 2,
t3∗ = 0,
z 3 = 2(cos 0 + i sin 0)
and Arg z 3 = {2kπ | k ∈ Z}. d) The point P4 (0, −3) lies on the negative y-axis, so 3π 3π 3π r4 = 3, t4∗ = , z 3 = 2 cos + i sin 2 2 2
and Arg z 4 =
3π + 2kπ | k ∈ Z . 2
Remark. The following formulas should be memorized: 1 = cos 0 + i sin 0;
i = cos
π π + i sin ; 2 2
3π 3π + i sin . 2 2 Problem 1. Find the polar representation of the complex number − 1 = cos π + i sin π ;
−i = cos
z = 1 + cos a + i sin a,
a ∈ (0, 2π ).
2.1. Polar Representation of Complex Numbers
35
Solution. The modulus is
a a 2 2 |z| = (1 + cos a) + sin a = 2(1 + cos a) = 4 cos2 = 2 cos . 2 2 The argument of z is determined as follows: a π and the point P(1 + cos a, sin a) lies on the first a) If a ∈ (0, π), then ∈ 0, 2 2 quadrant. Hence t ∗ = arctan
sin a a a = arctan tan = , 1 + cos a 2 2
and in this case
a a a z = 2 cos cos + i sin . 2 2 2 π a , π and the point P(1 + cos a, sin a) lies on the b) If a ∈ (π, 2π), then ∈ 2 2 fourth quadrant. Hence a a a t ∗ = arctan tan + 2π = − π + 2π = + π 2 2 2
and z = −2 cos
a a a cos + π + i sin +π . 2 2 2
c) If a = π, then z = 0. Problem 2. Find all complex numbers z such that |z| = 1 and z + z = 1. z z Solution. Let z = cos x + i sin x, x ∈ [0, 2π ). Then z z |z 2 + z 2 | 1= + = z z |z|2 = | cos 2x + i sin 2x + cos 2x − i sin 2x| = 2| cos 2x| hence cos 2x = If cos 2x =
1 1 or cos 2x = − . 2 2
1 , then 2 x1 =
π , 6
x2 =
5π , 6
x3 =
7π , 6
x4 =
11π . 6
36
2. Complex Numbers in Trigonometric Form
1 If cos 2x = − , then 2 π , 3
x5 =
x6 =
2π , 3
x7 =
4π , 3
x8 =
5π . 3
Hence there are eight solutions z k = cos xk + i sin xk ,
2.1.3
k = 1, 2, . . . , 8.
Operations with complex numbers in polar representation
1. Multiplication Proposition. Suppose that z 1 = r1 (cos t1 + i sin t1 ) and z 2 = r2 (cos t2 + i sin t2 ). Then z 1 z 2 = r1r2 (cos(t1 + t2 ) + i sin(t1 + t2 )).
(1)
Proof. Indeed, z 1 z 2 = r1r2 (cos t1 + i sin t1 )(cos t2 + i sin t2 ) = r1r2 ((cos t1 cos t2 − sin t1 sin t2 ) + i(sin t1 cos t2 + sin t2 cos t1 )) = r1r2 (cos(t1 + t2 ) + i sin(t1 + t2 )).
Remarks. a) We find again that |z 1 z 2 | = |z 1 | · |z 2 |. b) We have arg(z 1 z 2 ) = arg z 1 + arg z 2 − 2kπ , where 0, for arg z 1 + arg z 2 < 2π, k= 1, for arg z 1 + arg z 2 ≥ 2π. c) Also we can write Arg (z 1 z 2 ) = {arg z 1 + arg z 2 + 2kπ : k ∈ Z}. d) Formula (1) can be extended to n ≥ 2 complex numbers. If z k = rk (cos tk + i sin tk ), k = 1, . . . , n, then z 1 z 2 · · · z n = r1r2 · · · rn (cos(t1 + t2 + · · · + tn ) + i sin(t1 + t2 + · · · + tn )). The proof by induction is immediate. This formula can be written as n n n n
zk = rk cos tk + i sin tk . k=1
k=1
k=1
k=1
(2)
2.1. Polar Representation of Complex Numbers
37
√ Example. Let z 1 = 1 − i and z 2 = 3 + i. Then √ 7π π 7π π z 1 = 2 cos + i sin , z 2 = 2 cos + i sin 4 4 6 6 and
√ 7π π 7π π z 1 z 2 = 2 2 cos + + i sin + 4 6 4 6 √ 23π 23π = 2 2 cos + i sin . 12 12
2. The power of a complex number Proposition. (De Moivre1 ) For z = r (cos t + i sin t) and n ∈ N, we have z n = r n (cos nt + i sin nt).
(3)
Proof. Apply formula (2) for z = z 1 = z 2 = · · · = z n to obtain z n = r · r· · · r(cos(t + t +· · · + t ) + i sin(t + t +· · · + t )) n times
n times
n times
= r n (cos nt + i sin nt).
Remarks. a) We find again that |z n | = |z|n . b) If r = 1, then (cos t + i sin t)n = cos nt + i sin nt. c) We can write Arg z n = {n arg z + 2kπ : k ∈ Z}. Example. Let us compute (1 + i)1000 . √ π π The polar representation of 1 + i is 2 cos + i sin . Applying de Moivre’s 4 4 formula we obtain √ π π (1 + i)1000 = ( 2)1000 cos 1000 + i sin 1000 4 4 = 2500 (cos 250π + i sin 250π ) = 2500 . Problem. Prove that sin 5t = 16 sin5 t − 20 sin3 t + 5 sin t; cos 5t = 16 cos5 t − 20 cos3 t + 5 cos t. 1 Abraham de Moivre (1667–1754), French mathematician, a pioneer in probability theory and trigonom-
etry.
38
2. Complex Numbers in Trigonometric Form
Solution. Using de Moivre’s theorem to expand (cos t + i sin t)5 , then using the binomial theorem, we have cos 5t + i sin 5t = cos5 t + 5i cos4 t sin t + 10i 2 cos3 t sin2 t + 10i 3 cos2 t sin3 t + 5i 4 cos t sin4 t + i 5 sin5 t. Hence cos 5t + i sin 5t = cos5 t − 10 cos3 t (1 − cos2 t) + 5 cos t (1 − cos2 t)2 + i(sint (1 − sin2 t)2 sin t − 10(1 − sin2 t) sin3 t + sin5 t). Simple algebraic manipulation leads to the desired result. 3. Division Proposition. Suppose that z 1 = r1 (cos t1 + i sin t2 ), Then
z 2 = r2 (cos t2 + i sin t2 ) = 0.
r1 z1 = [cos(t1 − t2 ) + i sin(t1 − t2 )]. z2 r2
Proof. We have z1 r1 (cos t1 + i sin t1 ) = = z2 r2 (cos t2 + i sin t2 ) r1 (cos t1 + i sin t1 )(cos t2 − i sin t2 ) = r2 (cos2 t2 + sin2 t2 ) r1 = [(cos t1 cos t2 + sin t1 sin t2 ) + i(sin t1 cos t2 − sin t2 cos t1 )] r2 r1 = (cos(t1 − t2 ) + i sin(t1 − t2 )). r2 z 1 r1 |z 1 | Remarks. a) We have again = = ; r2 |z 2 | z2 z1 = {arg z 1 − arg z 2 + 2kπ : k ∈ Z}; b) We can write Arg z2 c) For z 1 = 1 and z 2 = z,
1 1 = z −1 = (cos(−t) + i sin(−t)); z r d) De Moivre’s formula also holds for negative integer exponents n, i.e., we have z n = r n (cos nt + i sin nt).
2.1. Polar Representation of Complex Numbers
Problem. Compute
39
√ (1 − i)10 ( 3 + i)5 . z= √ (−1 − i 3)10
Solution. We can write √ 7π π 7π 10 5 π 5 · 2 cos + i sin + i sin ( 2)10 cos 4 4 6 6 z= 10 4π 4π 210 cos + i sin 3 3 35π 5π 5π 35π 10 + i sin cos + i sin cos 2 2 2 6 6 = 40π 40π 210 cos + i sin 3 3 55π 55π cos + i sin 3 3 = cos 5π + i sin 5π = −1. = 40π 40π + i sin cos 3 3
2.1.4
Geometric interpretation of multiplication
Consider the complex numbers z 1 = r1 (cos t1∗ + i sin t1∗ ),
z 2 = r2 (cos t2∗ + i sin t2∗ )
and their geometric images M1 (r1 , t1∗ ), M2 (r2 , t2∗ ). Let P1 , P2 be the intersection points of the circle C(O; 1) with the rays (O M1 and (O M2 . Construct the point P3 ∈ C(O; 1) with the polar argument t1∗ + t2∗ and choose the point M3 ∈ (O P3 such that O M3 = O M1 · O M2 . Let z 3 be the complex coordinate of M3 . The point M3 (r1r2 , t1∗ + t2∗ ) is the geometric image of the product z 1 · z 2 . Let A be the geometric image of the complex number 1. Because O M3 O M2 = , O M1 1
i.e.,
O M2 O M3 = O M2 OA
M1 , it follows that triangles O AM1 and O M2 M3 are similar. and M 2 O M3 = AO z3 In order to construct the geometric image of the quotient, note that the image of z2 is M1 .
2.1.5
Problems
1. Find the polar coordinates for the following points, given their cartesian coordinates: √ a) M1 (−3, 3); b) M2 (−4 3, −4); c) M3 (0, −5); d) M4 (−2, −1); e) M5 (4, −2).
40
2. Complex Numbers in Trigonometric Form
Figure 2.5.
2. Find the cartesian coordinates for the following points, given their polar coordinates: π 3 a) P1 2, ; b) P2 4, 2π − arcsin ; c) P3 (2, π); 3 5 π 3π d) P4 (3, −π); e) P5 1, ; f) P6 4, . 2 2 3. Express arg(z) and arg(−z) in terms of arg(z). 4. Find the geometric images for the complex numbers z in each of the following cases: a) |z| = 2; b) |z + i| ≥ 2; c) |z − i| ≤ 3; 5π 3π π d) π < arg z < ; e) arg z ≥ ; f) arg z < ; 4 2 2 π π π g) arg(−z) ∈ , ; h) |z + 1 + i| < 3 and 0 < arg z < . 6 3 6 5. Find polar representations for the following complex numbers: √ √ √ 3 3 1 1 a) z 1 = 6 + 6i 3; b) z 2 = − + i ; c) z 3 = − − i ; 4 4 2 2 √ d) z 4 = 9 − 9i 3; e) z 5 = 3 − 2i; f) z 6 = −4i. 6. Find polar representations for the following complex numbers: a) z 1 = cos a − i sin a, a ∈ [0, 2π ); b) z 2 = sin a + i(1 + cos a), a ∈ [0, 2π ); c) z 3 = cos a + sin a + i(sin a − cos a), a ∈ [0, 2π ); d) z 4 = 1 − cos a + i sin a, a ∈ [0, 2π ). 7. Compute the following products using the polar representation of a complex number: √ √ 1 3 a) −i (−3 + 3i)(2 3 + 2i); b) (1 + i)(−2 − 2i) · i; 2 2 √ c) −2i · (−4 + 4 3i) · (3 + 3i); d) 3 · (1 − i)(−5 + 5i).
2.2. The n th Roots of Unity
41
Verify your results using the algebraic form. 8. Find |z|, arg z, Arg z, arg z, arg(−z) for √ a) z = (1 − i)(6 + 6i); b) z = (7 − 7 3i)(−1 − i). 9. Find |z| and √ arg z for (1 + i)6 (2 3 + 2i)8 + ; a) z = √ (1 − i)6 (2 3 − 2i)8 1 (−1 + i)4 + √ ; b) z = √ 10 ( 3 −√i) (2 3√ + 2i)4 c) z = (1 + i 3)n + (1 − i 3)n . 10. Prove that de Moivre’s formula holds for negative integer exponents. 11. Compute: a) (1 − cos a + i sin a)n for a ∈ [0, 2π ) and n ∈ N; 1 1 √ b) z n + n , if z + = 3. z z
2.2 2.2.1
The nth Roots of Unity Defining the nth roots of a complex number
Consider a positive integer n ≥ 2 and a complex number z 0 = 0. As in the field of real numbers, the equation Z n − z0 = 0 (1) is used for defining the n th roots of number z 0 . Hence we call any solution Z of the equation (1) an n th root of the complex number z 0 . Theorem. Let z 0 = r (cos t ∗ + i sin t ∗ ) be a complex number with r > 0 and t ∗ ∈ [0, 2π). The number z 0 has n distinct n th roots, given by the formulas √ t ∗ + 2kπ t ∗ + 2kπ Z k = n r cos + i sin , n n k = 0, 1, . . . , n − 1. Proof. We use the polar representation of the complex number Z with the extended argument Z = ρ(cos ϕ + i sin ϕ). By definition, we have Z n = z 0 or equivalently ρ n (cos nϕ + i sin nϕ) = r (cos t ∗ + i sin t ∗ ). We obtain ρ n = r and nϕ = t ∗ +2kπ for k ∈ Z; hence ρ = for k ∈ Z.
√ 2π t∗ n r and ϕk = +k· n n
42
2. Complex Numbers in Trigonometric Form
So far the roots of equation (1) are Zk =
√ n r (cos ϕk + i sin ϕk ) for k ∈ Z.
Now observe that 0 ≤ ϕ0 < ϕ1 < · · · < ϕn−1 < 2π , so the numbers ϕk , k ∈ {0, 1, . . . , n − 1}, are reduced arguments, i.e., ϕk∗ = ϕk . Until now we had n distinct roots of z 0 : Z 0 , Z 1 , . . . , Z n−1 . Consider some integer k and let r ∈ {0, 1, . . . , n − 1} be the residue of k modulo n. Then k = nq + r for q ∈ Z, and ϕk =
t∗ 2π t∗ 2π + (nq + r ) = +r + 2qπ = ϕr + 2qπ. n n n n
It is clear that Z k = Z r . Hence {Z k : k ∈ Z} = {Z 0 , Z 1 , . . . , Z n−1 }. In other words, there are exactly n distinct n th roots of z 0 , as claimed.
roots of a complex number z 0 = 0 are the vertices The geometric images of the √ of a regular n-gon inscribed in a circle with center at the origin and radius n r . To prove this, denote M0 , M1 , . . . , Mn−1 the points with complex coordinates Z 0 , √ Z 1 , . . . , Z n−1 . Because O Mk = |Z k | = n r for k ∈ {0, 1, . . . , n − 1}, it follows that √ the points Mk lie on the circle C(O; n r ). On the other hand, the measure of the arc n th
Mk Mk+1 is equal to arg Z k+1 − arg Z k =
t ∗ + 2(k + 1)π − (t ∗ + 2kπ ) 2π = , n n
for all k ∈ {0, 1, . . . , n − 2} and the remaining arc Mn−1 M0 is 2π 2π = 2π − (n − 1) . n n
Because all of the arcs M0 M1 , M1 M2 , . . . , Mn−1 M0 are equal, the polygon M0 M1 · · · Mn−1 is regular. Example. Let us find the third roots of the number z = 1 + i and represent them in the complex plane. The polar representation of z = 1 + i is √ π π . z = 2 cos + i sin 4 4
2.2. The n th Roots of Unity
43
The cube roots of the number z are √ 2π π 2π π 6 Z k = 2 cos +k + i sin +k , k = 0, 1, 2, 12 3 12 3 or, in explicit form,
√ π π 6 2 cos + i sin , 12 12 √ 3π 3π 6 Z 1 = 2 cos + i sin 4 4 Z0 =
and Z2 =
√ 6
17π 17π 2 cos + i sin . 12 12
Using polar coordinates, the geometric images of the numbers Z 0 , Z 1 , Z 2 are √ π √ √ 3π 17π 6 6 6 , M1 , M2 . 2, 2, 2, M0 12 4 12 The resulting equilateral triangle M0 M1 M2 is shown in the following figure:
Figure 2.6.
2.2.2
The nth roots of unity
The roots of the equation Z n − 1 = 0 are called the n th roots of unity. Since 1 = cos 0 + i sin 0, from the formulas for the n th roots of a complex number we derive that the n th roots of unity are εk = cos
2kπ 2kπ + i sin , n n
k ∈ {0, 1, 2, . . . , n − 1}.
Explicitly, we have ε0 = cos 0 + i sin 0 = 1;
44
2. Complex Numbers in Trigonometric Form
ε1 = cos
2π 2π + i sin = ε; n n
ε2 = cos
4π 4π + i sin = ε2 ; n n ...
εn−1 = cos
2(n − 1)π 2(n − 1)π + i sin = εn−1 . n n
The set {1, ε, ε 2 , . . . , εn−1 } is denoted by Un . Observe that the set Un is generated by the element ε, i.e., the elements of Un are the powers of ε. As stated before, the geometric images of the n th roots of unity are the vertices of a regular polygon with n sides inscribed in the unit circle with one of the vertices at 1. We take a brief look at some particular values of n. i) For n = 2, the equation Z 2 − 1 = 0 has the roots −1 and 1, which are the square roots of unity. ii) For n = 3, the cube roots of unity, i.e., the roots of equation Z 3 − 1 = 0 are given by 2kπ 2kπ εk = cos + i sin for k ∈ {0, 1, 2}. 3 3 Hence √ 3 2π 1 2π + i sin =− +i =ε ε0 = 1, ε1 = cos 3 3 2 2 and √ 4π 1 3 4π + i sin =− −i = ε2 . ε2 = cos 3 3 2 2 They form an equilateral triangle inscribed in the circle C(O; 1) as in the figure below.
Figure 2.7.
2.2. The n th Roots of Unity
45
iii) For n = 4, the fourth roots of unity are εk = cos
2kπ 2kπ + i sin for k = 0, 1, 2, 3. 4 4
In explicit form, we have ε0 = cos 0 + i sin 0 = 1;
ε1 = cos
ε2 = cos π + i sin π = −1 and ε3 = cos
π π + i sin = i; 2 2 3π 3π + i sin = −i. 2 2
Observe that U4 = {1, i, i 2 , i 3 } = {1, i, −1, −i}. The geometric images of the fourth roots of unity are the vertices of a square inscribed in the circle C(O; 1).
Figure 2.8.
The root εk ∈ Un is called primitive if for all positive integer m < n we have = 1.
εkm
Proposition 1. a) If n|q, then any root of Z n − 1 = 0 is a root of Z q − 1 = 0. b) The common roots of Z m − 1 = 0 and Z n − 1 = 0 are the roots of Z d − 1 = 0, where d = gcd(m, n), i.e., Um ∩ Un = Ud . 2kπ 2kπ + i sin , where 0 ≤ k ≤ c) The primitive roots of Z m − 1 = 0 are εk = cos m m m and gcd(k, m) = 1. Proof. a) If q = pn, then Z q − 1 = (Z n ) p − 1 = (Z n − 1)(Z ( p−1)n + · · · + Z n + 1) and the conclusion follows. 2 pπ 2qπ 2 pπ b) Consider ε p = cos + i sin a root of Z m − 1 = 0 and εq = cos + m m n 2qπ a root of Z n − 1 = 0. Since |ε p | = |εq | = 1, we have ε p = εq if and only i sin n
46
2. Complex Numbers in Trigonometric Form
2qπ 2 pπ = + 2r π for some integer r . The last relation is if arg ε p = arg εq , i.e., m n p q equivalent to − = r , that is, pn − qm = r mn. m n On the other hand we have m = m d and n = n d, where gcd(m , n ) = 1. From the relation pn − qm = r mn we find n p − m q = r m n d. Hence m |n p, so m | p. That is, p = p m for some positive integer p and arg ε p =
2 pπ 2 p m π 2 p π = = and ε dp = 1. m md d
Conversely, since d|m and d|n (from property a), any root of Z d − 1 = 0 is a root of Z m − 1 = 0 and Z n − 1 = 0. p c) First we will find the smallest positive integer p such that εk = 1. From the 2kpπ p = 2k π for some positive integer k . That is, relation εk = 1 it follows that m kp = k ∈ Z. Consider d = gcd(k, m) and k = k d, m = m d, where gcd(k , m ) = 1. m k p k pd ∈ Z. Since k and m are relatively primes, we get m | p. We obtain = md m p Therefore, the smallest positive integer p with εk = 1 is p = m . Substituting in the m relation m = m d, it follows that p = , where d = gcd(k, m). d m p If εk is a primitive root of unity, then from relation εk = 1, p = , it gcd(k, m) follows that p = m, i.e., gcd(k, m) = 1. Remark. From Proposition 1.b) one obtains that the equations Z m − 1 = 0 and Z n − 1 = 0 have the unique common root 1 if and only if gcd(m, n) = 1. zn
Proposition 2. If ε ∈ Un is a primitive root of unity, then the roots of the equation − 1 = 0 are εr , εr +1 , . . . , εr +n−1 , where r is an arbitrary positive integer.
Proof. Let r be a positive integer and consider h ∈ {0, 1, . . . , n−1}. Then (εr +h )n = (εn )r +h = 1, i.e., εr +h is a root of Z n − 1 = 0. We need only prove that εr , εr +1 , . . . , εr +n−1 are distinct. Assume by way of contradiction that for r + h 1 = r + h 2 and h 1 > h 2 , we have εr +h 1 = εr +h 2 . Then εr +h 2 (ε h 1 −h 2 − 1) = 0. But εr +h 2 = 0 implies ε h 1 −h 2 = 1. Taking into account that h 1 − h 2 < n and ε is a primitive root of Z n − 1 = 0, we get a contradiction. Proposition 3. Let ε0 , ε1 , . . . , εn−1 be the n th roots of unity. For any positive integer k the following relation holds: n−1
n, if n|k, k εj = 0, otherwise. j=0
2.2. The n th Roots of Unity
47
2π 2π + i sin . Then ε ∈ Un is a primitive root of unity, Proof. Consider ε = cos n n m hence ε = 1 if and only if n|m. Assume that n does not divides k. We have n−1
j=0
ε kj =
n−1
(ε j )k =
j=0
n−1
(εk ) j =
j=0
1 − (ε k )n 1 − (ε n )k = = 0. k 1−ε 1 − εk
If n|k, then k = qn for some positive integer q, and we obtain n−1
ε kj =
j=0
n−1
j=0
qn
εj =
n−1
(ε nj )q =
j=0
n−1
1 = n.
j=0
Proposition 4. Let p be a prime number and let ε = cos a0 , a1 , . . . , a p−1 are nonzero integers, the relation
2π 2π + i sin . If p p
a0 + a1 ε + · · · + a p−1 ε p−1 = 0 holds if and only if a0 = a1 = · · · = a p−1 . Proof. If a0 = a1 = · · · = a p−1 , then the above relation is clearly true. Conversely, define the polynomials f, g ∈ Z[X ] by f = a1 +a1 X +· · ·+a p−1 X p−1 and g = 1 + X + · · · + X p−1 . If the polynomials f, g have common zeros, then gcd( f, g) divides g. But it is well known (for example by Eisenstein’s irreducibility criterion) that g is irreducible over Z. Hence gcd( f, g) = g, so g| f and we obtain g = k f for some nonzero integer k, i.e., a0 = a1 = · · · = an−1 . Problem 1. Find the number of ordered pairs (a, b) of real numbers such that (a + bi)2002 = a − bi. (American Mathematics Contest 12A, 2002, Problem 24)
Solution. Let z = a + bi, z = a − bi, and |z| = becomes z 2002 = z. Note that
√ a 2 + b2 . The given relation
|z|2002 = |z 2002 | = |z| = |z|, from which it follows that |z|(|z|2001 − 1) = 0. Hence |z| = 0, and (a, b) = (0, 0), or |z| = 1. In the case |z| = 1, we have z 2002 = z, which is equivalent to z 2003 = z · z = |z|2 = 1. Since the equation z 2003 = 1 has 2003 distinct solutions, there are altogether 1 + 2003 = 2004 ordered pairs that meet the required conditions. Problem 2. Two regular polygons are inscribed in the same circle. The first polygon has 1982 sides and the second has 2973 sides. If the polygons have any common vertices, how many such vertices will there be?
48
2. Complex Numbers in Trigonometric Form
Solution. The number of common vertices is given by the number of common roots of z 1982 − 1 = 0 and z 2973 − 1 = 0. Applying Proposition 1.b), the desired number is d = gcd(1982, 2973) = 991. Problem 3. Let ε ∈ Un be a primitive root of unity and let z be a complex number such that |z − ε k | ≤ 1 for all k = 0, 1, . . . , n − 1. Prove that z = 0. Solution. From the given condition it follows that (z − εk )(z − ε k ) ≤ 1, yielding |z|2 ≤ z(ε k ) + z · ε k , k = 0, 1, . . . , n − 1. By summing these relations we obtain n|z| ≤ z 2
n−1
k=0
εk
+z·
n−1
ε k = 0.
k=0
Thus z = 0. Problem 4. Let P0 P1 · · · Pn−1 be a regular polygon inscribed in a circle of radius 1. Prove that: a) P0 P1 · P0 P2 · · · P0 Pn−1 = n; π 2π (n − 1)π n b) sin sin · · · sin = n−1 ; n n n 2 π 3π (2n − 1)π 1 c) sin sin · · · sin = n−1 . 2n 2n 2n 2 Solution. a) Without loss of generality we may assume that the vertices of the polygon are the geometric images of the n th roots of unity, and P0 = 1. Consider the 2π 2π + i sin . polynomial f = z n − 1 = (z − 1)(z − ε) · · · (z − εn−1 ), where ε = cos n n Then it is clear that n = f (1) = (1 − ε)(1 − ε 2 ) · · · (1 − ε n−1 ). Taking the modulus of each side, the desired result follows. b) We have kπ 2kπ kπ kπ 2kπ − i sin = 2 sin2 − 2i sin cos n n n n n kπ kπ kπ sin − i cos , = 2 sin n n n
1 − ε k = 1 − cos
kπ hence |1 − ε k | = 2 sin , k = 1, 2, . . . , n − 1, and the desired trigonometric identity n follows from a). c) Consider the regular polygon Q 0 Q 1 · · · Q 2n−1 inscribed in the same circle whose vertices are the geometric images of the (2n)th roots of unity. According to a), Q 0 Q 1 · Q 0 Q 2 · · · Q 0 Q 2n−1 = 2n.
2.2. The n th Roots of Unity
49
Now taking into account that Q 0 Q 2 · · · Q n−2 is also a regular polygon, we deduce from a) that Q 0 Q 2 · Q 0 Q 4 · · · Q 0 Q 2n−2 = n. Combining the last two relations yields Q 0 Q 1 · Q 0 Q 3 · · · Q 0 Q 2n−1 = 2. A similar computation to the one in b) leads to Q 0 Q 2k−1 = 2 sin
(2k − 1)π , 2n
k = 1, 2, . . . , n,
and the desired result follows. 2π 2π Let n be a positive integer and let εn = cos + i sin . The n th -cyclotomic n n polynomial is defined by (x − εnk ). φn (x) = 1≤k≤n−1 gcd(k,n)=1
Clearly the degree of φn is ϕ(n), where ϕ is the Euler “totient” function. φn is a monic polynomial with integer coefficients and is irreducible over Q. The first sixteen cyclotomic polynomials are given below: φ(x) = x − 1 φ2 (x) = x + 1 φ3 (x) = x 2 + x + 1 φ4 (x) = x 2 + 1 φ5 (x) = x 4 + x 3 + x 2 + x + 1 φ6 (x) = x 2 − x + 1 φ7 (x) = x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 φ8 (x) = x 4 + 1 φ9 (x) = x 6 + x 3 + 1 φ10 (x) = x 4 − x 3 + x 2 − x + 1 φ11 (x) = x 10 + x 9 + x 8 + · · · + x + 1 φ12 (x) = x 4 − x 2 + 1 φ13 (x) = x 12 + x 11 + x 10 + · · · + x + 1 φ14 (x) = x 6 − x 5 + x 4 − x 3 + x 2 − x + 1 φ15 (x) = x 8 − x 7 + x 5 − x 4 + x 3 − x + 1 φ16 (x) = x 8 + 1 The following properties of cyclotomic polynomials are well known: 1) If q > 1 is an odd integer, then φ2q (x) = φq (−x).
50
2. Complex Numbers in Trigonometric Form
2) If n > 1, then φn (1) =
p, 1,
when n is a power of a prime p, otherwise.
The next problem extends the trigonometric identity in Problem 4.b). Problem 5. The following identities hold: kπ 1 a) sin = ϕ(n) , whenever n is not a power of a prime; n 2 1≤k≤n−1 gcd(k,n)=1
b)
ϕ(n)
cos
1≤k≤n−1 gcd(k,n)=1
kπ (−1) 2 , for all odd positive integers n. = n 2ϕ(n)
Solution. a) As we have seen in Problem 4.b), kπ kπ kπ 2 kπ kπ kπ k 1 − εn = 2 sin sin − i cos = sin cos + i sin . n n n i n n n We have
1 = φn (1) =
(1 − εnk ) =
1≤k≤n−1 gcd(k,n)=1
1≤k≤n−1 gcd(k,n)=1
⎛ =
2ϕ(n) i ϕ(n)
cos
kπ kπ + i sin n n
⎞
⎜ ⎝ 1≤k≤n−1 gcd(k,n)=1
kπ ⎟ ϕ(n) ϕ(n) sin π + i sin π ⎠ cos n 2 2 ⎛
=
2 kπ sin i n
2ϕ(n) (−1)
ϕ(n) 2
⎞
⎜ ⎝ 1≤k≤n−1 gcd(k,n)=1
sin
ϕ(n) kπ ⎟ ⎠ (−1) 2 , n
where we have used the fact that ϕ(n) is even, and also the well-known relation
1≤k≤n−1 gcd(k,n)=1
k=
1 nϕ(n). 2
The conclusion follows. b) We have 2kπ kπ kπ kπ 2kπ + i sin = 2 cos2 + 2i sin cos n n n n n kπ kπ kπ cos + i sin , k = 0, 1, . . . , n − 1. = 2 cos n n n
1 + εnk = 1 + cos
2.2. The n th Roots of Unity
51
Because n is odd, from the relation φ2n (x) = φn (−1) it follows that φn (−1) = φ2n (1) = 1. Then (1 − εnk ) = (−1)ϕ(n) (1 + εnk ) 1 = φn (−1) = 1≤k≤n−1 gcd(k,n)=1
= (−1)ϕ(n) ⎛
1≤k≤n−1 gcd(k,n)=1
2 cos
1≤k≤n−1 gcd(k,n)=1
⎜ = (−1)ϕ(n) 2ϕ(n) ⎝
cos
kπ kπ + i sin n n
⎞
cos
1≤k≤n−1 gcd(k,n)=1
= (−1)
kπ n
ϕ(n) 2
ϕ(n) ϕ(n) kπ ⎟ cos π + i sin π ⎠ n 2 2
2ϕ(n)
cos
1≤k≤n−1 gcd(k,n)=1
kπ , n
yielding the desired identity.
2.2.3
Binomial equations
A binomial equation is an equation of the form Z n + a = 0, where a ∈ C∗ and n ≥ 2 is an integer. Solving for Z means finding the n th roots of the complex number −a. This is in fact a simple polynomial equation of degree n with complex coefficients. From the wellknown fundamental theorem of algebra it follows that it has exactly n complex roots, and it is obvious that the roots are distinct. Example. 1) Let us find the roots of Z 3 + 8 = 0. We have −8 = 8(cos π + i sin π ), so the roots are π + 2kπ π + 2kπ + i sin , Z k = 2 cos 3 3
k ∈ {0, 1, 2}.
2) Let us solve the equation Z 6 − Z 3 (1 + i) + i = 0. Observe that the equation is equivalent to (Z 3 − 1)(Z 3 − i) = 0. Solving for Z the binomial equations Z 3 − 1 = 0 and Z 3 − i = 0, we obtain the solutions 2kπ 2kπ εk = cos + i sin for k ∈ {0, 1, 2} 3 3 and π π + 2kπ + 2kπ 2 + i sin 2 for k ∈ {0, 1, 2}. Z k = cos 3 3
52
2.2.4
2. Complex Numbers in Trigonometric Form
Problems
1. Find the square roots of the following complex numbers: i 1 a) z = 1 + i; b) z = i; c) z = √ + √ ; 2 2 √ d) z = −2(1 + i 3); e) z = 7 − 24i. 2. Find the cube roots of the following complex numbers: a) z = −i; √ b) z = −27; c) z = 2 + 2i; 3 1 ; e) z = 18 + 26i. d) z = − i 2 2 3. Find the fourth roots of the following complex numbers: √ √ a) z = 2 − i 12; b) z = 3 + i; c) z = i; d) z = −2i; e) z = −7 + 24i. 4. Find the fifth, sixth, seventh, eighth, and twefth roots of the complex numbers given above. 5. Let Un = {ε0 , ε1 , ε2 , . . . , εn−1 }. Prove that: a) ε j · εk ∈ Un , for all j, k ∈ {0, 1, . . . , n − 1}; b) ε−1 j ∈ Un , for all j ∈ {0, 1, . . . , n − 1}. 6. Solve the equations: a) z 3 − 125 = 0; b) z 4 + 16 = 0; c) z 3 + 64i = 0; d) z 3 − 27i = 0. 7. Solve the equations: a) z 7 − 2i z 4 − i z 3 − 2 = 0; c) (2 − 3i)z 6 + 1 + 5i = 0;
b) z 6 + i z 3 + i − 1 = 0; d) z 10 + (−2 + i)z 5 − 2i = 0.
8. Solve the equation z 4 = 5(z − 1)(z 2 − z + 1).
3 Complex Numbers and Geometry
3.1 3.1.1
Some Simple Geometric Notions and Properties The distance between two points
Suppose that the complex numbers z 1 and z 2 have the geometric images M1 and M2 . Then the distance between the points M1 and M2 is given by M1 M2 = |z 1 − z 2 |. The distance function d : C × C → [0, ∞) is defined by d(z 1 , z 2 ) = |z 1 − z 2 |, and it satisfies the following properties: a) (positiveness and nondegeneration): d(z 1 , z 2 ) ≥ 0 for all z 1 , z 2 ∈ C; d(z 1 , z 2 ) = 0 if and only if z 1 = z 2 . b) (symmetry): d(z 1 , z 2 ) = d(z 2 , z 1 ) for all z 1 , z 2 ∈ C. c) (triangle inequality): d(z 1 , z 2 ) ≤ d(z 1 , z 3 ) + d(z 3 , z 2 ) for all z 1 , z 2 , z 3 ∈ C.
54
3. Complex Numbers and Geometry
To justify c) let us observe that |z 1 − z 2 | = |(z 1 − z 3 ) + (z 3 − z 2 )| ≤ |z 1 − z 3 | + |z 3 − z 2 |, from the modulus property. Equality holds if and only if there is a positive real number k such that z 3 − z 1 = k(z 2 − z 3 ).
3.1.2
Segments, rays and lines
Let A and B be two distinct points with complex coordinates a and b. We say that the point M with complex coordinate z is between the points A and B if z = a, z = b and the following relation holds: |a − z| + |z − b| = |a − b|. We use the notation A − M − B. The set (AB) = {M : A − M − B} is called the open segment determined by the points A and B. The set [AB] = (AB) ∪ {A, B} represents the closed segment defined by the points A and B. Theorem 1. Suppose A(a) and B(b) are two distinct points. The following statements are equivalent: 1) M ∈ (AB); 2) there is a positive real number k such that z − a = k(b − z); 3) there is a real number t ∈ (0, 1) such that z = (1 − t)a + tb, where z is the complex coordinate of M. Proof. We first prove that 1) and 2) are equivalent. Indeed, we have M ∈ (AB) if and only if |a − z| + |z − b| = |a − b|. That is, d(a, z) + d(z, b) = d(a, b), or equivalently there is a real k > 0 such that z − a = k(b − z). t k To prove that 2) ⇔ 3), set t = ∈ (0, 1) or k = > 0. Then we have k+1 1−t 1 k z − a = k(b − z) if and only if z = a+ b. That is, z = (1 − t)a + tb and k+1 k+1 we are done. The set (AB = {M| A − M − B or A − B − M} is called the open ray with endpoint A that contains B. Theorem 2. Suppose A(a) and B(b) are two distinct points. The following statements are equivalent: 1) M ∈ (AB; 2) there is a positive real number t such that z = (1 − t)a + tb, where z is the complex coordinate of M;
3.1. Some Simple Geometric Notions and Properties
55
3) arg(z − a) = arg(b − a); z−a 4) ∈ R+ . b−a Proof. It suffices to prove that 1) ⇒ 2) ⇒ 3) ⇒ 4) ⇒ 1). 1) ⇒ 2). Since M ∈ (AB we have A − M − B or A − B − M. There are numbers t, l ∈ (0, 1) such that z = (1 − t)a + tb or b = (1 − l)a + lz. In the first case we are done; for the second case set t =
1 , hence l
z = tb − (t − 1)a = (1 − t)a + tb, as claimed. 2) ⇒ 3). From z = (1 − t)a + tb, t > 0 we obtain z − a = t (b − a), t > 0. Hence arg(z − a) = arg(b − a). 3) ⇒ 4). The relation arg
z−a = arg(z − a) − arg(b − a) + 2kπ for some k ∈ Z b−a
z−a z−a = 2kπ , k ∈ Z. Since arg ∈ [0, 2π ), it follows that k = 0 and implies arg b−a b−a z−a z−a arg = 0. Thus ∈ R+ , as desired. b−a b−a z−a 4) ⇒ 1). Let t = ∈ R∗ . Hence b−a z = a + t (b − a) = (1 − t)a + tb, t > 0. If t ∈ (0, 1), then M ∈ (AB) ⊂ (AB.
1 If t = 1, then z = b and M = B ∈ (AB. Finally, if t > 1 then, setting l = ∈ t (0, 1), we have b = lz + (1 − l)a. It follows that A − B − M and M ∈ (AB. The proof is now complete.
56
3. Complex Numbers and Geometry
Theorem 3. Suppose A(a) and B(b) are two distinct points. The following statements are equivalent: 1) M(z) lies on the line AB. z−a 2) ∈ R. b−a 3) There is a real number t such that z = (1 − t)a + tb. z−a z−a 4) = 0; b−a b−a z z 1 5) a a 1 = 0. b b 1 Proof. To obtain the equivalences 1) ⇔ 2) ⇔ 3) observe that for a point C such that C − A − B the line AB is the union (AB ∪ {A} ∪ (AC. Then apply Theorem 2. Next we prove the equivalences 2) ⇔ 4) ⇔ 5). z−a z−a z−a ∈ R if and only if = . Indeed, we have b−a b − a b − a z−a z−a z−a z−a That is, , or, equivalently, = = 0, so we obtain that b−a b−a b−a b−a 2) is equivalent to 4). Moreover, we have z z 1 z−a z−a 0 a 1 =0 a a 1 = 0 if and only if a b b 1 b−a b−a 0 The last relation is equivalent to z−a b−a
z−a b−a
= 0,
so we obtain that 4) is equivalent to 5), and we are done.
Problem 1. Let z 1 , z 2 , z 3 be complex numbers such that |z 1 | = |z 2 | = |z 3 | = R and z 2 = z 3 . Prove that min |az 2 + (1 − a)z 3 − z 1 | = a∈R
1 |z 1 − z 2 | · |z 1 − z 3 |. 2R
(Romanian Mathematical Olympiad – Final Round, 1984)
Solution. Let z = az 2 + (1 − a)z 3 , a ∈ R and consider the points A1 , A2 , A3 , A of complex coordinates z 1 , z 2 , z 3 , z, respectively. From the hypothesis it follows that the
3.1. Some Simple Geometric Notions and Properties
57
circumcenter of triangle A1 A2 A3 is the origin of the complex plane. Notice that point A lies on the line A2 A3 , so A1 A = |z − z 1 | is greater than or equal to the altitude A1 B of the triangle A1 A2 A3 .
Figure 3.1.
It suffices to prove that A1 B =
1 1 |z 1 − z 2 ||z 1 − z 3 | = A1 A2 · A1 A3 . 2R 2R
Indeed, since R is the circumradius of the triangle A1 A2 A3 , we have 2 2area[A1 A2 A3 ] = A1 B = A2 A3
A1 A2 · A2 A3 · A3 A1 A1 A2 · A3 A1 4R , = A2 A3 2R
as claimed.
3.1.3
Dividing a segment into a given ratio
Consider two distinct points A(a) and B(b). A point M(z) on the line AB divides the segments AB into the ratio k ∈ R \ {1} if the following vectorial relation holds: −→ −→ M A = k · M B. In terms of complex numbers this relation can be written as a − z = k(b − z) or (1 − k)z = a − kb. Hence, we obtain
a − kb . 1−k Observe that for k < 0 the point M lies on the line segment joining the points A and B. If k ∈ (0, 1), then M ∈ (AB \ [AB]. Finally, if k > 1, then M ∈ (B A \ [AB]. z=
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3. Complex Numbers and Geometry
As a consequence, note that for k = −1 we obtain that the coordinate of the mida+b point of segment [AB] is given by z M = . 2 Example. Let A(a), B(b), C(c) be noncollinear points in the complex plane. Then a+b the midpoint M of segment [AB] has the complex coordinate z M = . The cen2 troid G of triangle ABC divides the median [C M] into 2 : 1 internally, hence its complex coordinate is given by k = −2, i.e., zG =
3.1.4
c + 2z M a+b+c = . 1+2 3
Measure of an angle
Recall that a triangle is oriented if an ordering of its vertices is specified. It is positively or directly oriented if the vertices are oriented counterclockwise. Otherwise, we say that the triangle is negatively oriented. Consider two distinct points M1 (z 1 ) and M2 (z 2 ), other than the origin of a complex plane. The angle M 1 O M2 is oriented if the points M1 and M2 are ordered counterclockwise (Fig. 3.2 below). Proposition. The measure of the directly oriented angle z2 M . 1 O M2 equals arg z1 Proof. We consider the following two cases.
Figure 3.2.
a) If the triangle M1 O M2 is negatively oriented (Fig. 3.2), then z2 M O M2 − x O M1 = arg z 2 − arg z 1 = arg . 1 O M2 = x z1 b) If the triangle M1 O M2 is positively oriented (Fig. 3.3), then z2 M , 1 O M2 = 2π − M 2 O M1 = 2π − arg z1
3.1. Some Simple Geometric Notions and Properties
59
since the triangle M2 O M1 is negatively oriented. Thus z2 z1 z2 = arg , = 2π − 2π − arg M 1 O M2 = 2π − arg z2 z1 z1
as claimed.
Figure 3.3.
Remark. The result also holds if the points O, M1 , M2 are collinear. Examples. a) Suppose that z 1 = 1 + i and z 2 = −1 + i. Then (see Fig. 3.4) −1 + i (−1 + i)(1 − i) z2 = = = i, z1 1+i 2 so
π 3π M and M . 1 O M2 = arg i = 2 O M1 = arg(−i) = 2 2
Figure 3.4.
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3. Complex Numbers and Geometry
Figure 3.5.
b) Suppose that z 1 = i and z 2 = 1. Then
z2 1 = = −i, so (see Fig. 3.5) z1 i
3π π M and M . 1 O M2 = arg(−i) = 2 O M1 = arg(i) = 2 2 Theorem. Consider three distinct points M1 (z 1 ), M2 (z 2 ) and M3 (z 3 ). z3 − z1 The measure of the oriented angle M . 2 M1 M3 is arg z2 − z1 Proof. The translation with the vector −z 1 maps the points M1 , M2 , M3 into the points O, M2 , M3 , with complex coordinates O, z 2 − z 1 , z 3 − z 1 . Moreover, we have M 2 M1 M3 = M O M . By the previous result, we obtain 2
3
z3 − z1 M 2 O M3 = arg z − z , 2 1
as claimed. Example. Suppose that z 1 = 4 + 3i, z 2 = 4 + 7i, z 3 = 8 + 7i. Then z2 − z1 4i i(1 − i) 1+i = = = , z3 − z1 4 + 4i 2 2 so M 3 M1 M2 = arg
1+i π = 2 4
and M 2 M1 M3 = arg
2 7π = arg(1 − i) = . 1+i 4
Remark. Using polar representation, from the above result we have z3 − z1 z3 − z1 cos arg z 3 − z 1 + i sin arg z 3 − z 1 = z2 − z1 z2 − z1 z2 − z1 z2 − z1 z3 − z1 (cos M = 2 M1 M3 + i sin M 2 M1 M3 ). z2 − z1
3.1. Some Simple Geometric Notions and Properties
3.1.5
61
Angle between two lines
Consider four distinct points Mi (z i ), i ∈ {1, 2, 3, 4}. The measure of the angle deterz3 − z1 z4 − z2 mined by the lines M1 M3 and M2 M4 equals arg or arg . The proof is z4 − z2 z3 − z1 obtained following the same ideas as in the previous subsection.
3.1.6
Rotation of a point
Consider an angle α and the complex number given by ε = cos α + i sin α. Let z = r (cos t + i sin t) be a complex number and M its geometric image. Form the product zε = r (cos(t + α) + i sin(t + α)) and let us observe that |zε| = r and arg(zε) = arg z + α. It follows that the geometric image M of zε is the rotation of M with respect to the origin by the angle α.
Figure 3.6.
Now we have all the ingredients to establish the following result: Proposition. Suppose that the point C is the rotation of B with respect to A by the angle α. If a, b, c are the coordinates of the points A, B, C, respectively, then c = a + (b − a)ε, where ε = cos α + i sin α. Proof. The translation with vector −a maps the points A, B, C into the points O, B , C , with complex coordinates O, b − a, c − a, respectively (see Fig. 3.7). The point C is the image of B under rotation about the origin through the angle α, so c − a = (b − a)ε, or c = a + (b − a)ε, as desired.
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3. Complex Numbers and Geometry
Figure 3.7.
We will call the formula in the above proposition the rotation formula. Problem 1. Let ABC D and B N M K be two nonoverlapping squares and let E be the midpoint of AN . If point F is the foot of the perpendicular from B to the line C K , prove that points E, F, B are collinear. Solution. Consider the complex plane with origin at F and the axis C K and F B, where F B is the imaginary axis. Let c, k, bi be the complex coordinates of points C, K , B with c, k, b ∈ R. The π maps point C to A, so A has the rotation with center B through the angle θ = 2 complex coordinate a = b(1 − i) + ci. Similarly, point N is obtained by rotating point π K around B through the angle θ = − and its complex coordinate is 2 n = b(1 + i) − ki. The midpoint E of segment AN has the complex coordinate e=
a+n c−k =b+ i, 2 2
so E lies on the line F B, as desired. Problem 2. On the sides AB, BC, C D, D A of quadrilateral ABC D, and exterior to the quadrilateral, we construct squares of centers O1 , O2 , O3 , O4 , respectively. Prove that O1 O3 ⊥ O2 O4 and O1 O3 = O2 O4 . Solution. Let AB M M , BC N N , C D P P and D AQ Q be the constructed squares with centers O1 , O2 , O3 , O4 , respectively.
3.1. Some Simple Geometric Notions and Properties
63
Denote by a lowercase letter the coordinate of each of the points denoted by an uppercase letter, i.e., o1 is the coordinate of O1 , etc. π Point M is obtained from point A by a rotation about B through the angle θ = ; 2 hence m = b + (a − b)i. Likewise, n = c + (b − c)i,
p = d + (c − d)i and q = a + (d − a)i.
It follows that o1 =
a+m a + b + (a − b)i = , 2 2
o3 =
o2 =
b + c + (b − c)i , 2
c + d + (c − d)i d + a + (d − a)i and o4 = . 2 2
Then
o3 − o1 c + d − a − b + i(c − d − a + b) = = −i ∈ iR∗ , o4 − o2 a + d − b − c + i(d − a − b + c) so O1 O3 ⊥ O2 O4 . Moreover, o3 − o 1 o − o = | − i| = 1; 4 2 hence O1 O3 = O2 O4 , as desired. Problem 3. In the exterior of the triangle ABC we construct triangles AB R, BC P, and C AQ such that m( P BC) = m(C AQ) = 45◦ , P) = m( QC A) = 30◦ , m( BC and R) = m( R m( AB AB) = 15◦ .
Figure 3.8.
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3. Complex Numbers and Geometry
Figure 3.9.
Prove that m( Q R P) = 90◦ and R Q = R P.
Solution. Consider the complex plane with origin at point R and let M be the foot of the perpendicular from P to the line BC.
Figure 3.10.
Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter. √ MC From M P = M B and = 3 it follows that MP √ p−m c−m = i and = i 3, b−m p−m
3.2. Conditions for Collinearity, Orthogonality and Concyclicity
65
√ 3b b−c p= √ + √ i. 1+ 3 1+ 3
hence
c+
Likewise,
√ 3a a−c √ + √ i. 1+ 3 1+ 3 Point B is obtained from point A by a rotation about R through an angle θ = 150◦ , q=
c+
√ 3 1 b=a − + i . 2 2
so
p = i ∈ iR∗ , hence Q R ⊥ P R. Moreover, q | p| = |iq| = |q|, so R P = R Q and we are done.
Simple algebraic manipulations show that
3.2
Conditions for Collinearity, Orthogonality and Concyclicity
In this section we consider four distinct points Mi (z i ), i ∈ {1, 2, 3, 4}. Proposition 1. The points M1 , M2 , M3 are collinear if and only if z3 − z1 ∈ R∗ . z2 − z1 Proof. The collinearity of the points M1 , M2 , M3 is equivalent to M 2 M1 M3 ∈ z3 − z1 z3 − z1 {0, π }. It follows that arg ∈ {0, π } or equivalently ∈ R∗ , as z2 − z1 z2 − z1 claimed. Proposition 2. The lines M1 M2 and M3 M4 are orthogonal if and only if z1 − z2 ∈ iR∗ . z3 − z4 Proof. We have M1 M2 ⊥ M3 M4 if and only if (M1 M2 , M3 M4 ) ∈ π 3π z1 − z2 z1 − z2 ∈ ∈ iR∗ . is equivalent to arg , . We obtain z3 − z4 2 2 z3 − z4
π 3π , . This 2 2
Remark. Suppose that M2 = M4 . Then M1 M2 ⊥ M3 M2 if and only if iR∗ .
z1 − z2 ∈ z3 − z2
Examples. 1) Consider the points M1 (2−i), M2 (−1+2i), M3 (−2−i), M4 (1+2i). Simple algebraic manipulation shows that z1 − z2 = i, hence M1 M2 ⊥ M3 M4 . z3 − z4
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3. Complex Numbers and Geometry
2) Consider the points M1 (2 − i), M2 (−1 + 2i), M3 (1 + 2i), M4 (−2 − i). Then we z1 − z2 have = −i hence M1 M2 ⊥ M3 M4 . z3 − z4 Problem 1. Let z 1 , z 2 , z 3 be the coordinates of vertices A, B, C of a triangle. If w1 = % = 90◦ if and only if Re(w1 · w 2 ) = 0. z 1 − z 2 and w2 = z 3 − z 1 , prove that A % = 90◦ if and only if z 2 − z 1 ∈ iR, which is equivalent to Solution. We have A z3 − z1 w1 · w2 w1 w1 = 0. The last relation is equivalent to Re = ∈ iR, i.e., Re −w2 −w2 −|w2 |2 0, i.e., Re(w1 · w2 ) = 0, as desired. Proposition 3. The distinct points M1 (z 1 ), M2 (z 2 ), M3 (z 3 ), M4 (z 4 ) are concyclic or collinear if and only if k=
z3 − z2 z3 − z4 : ∈ R∗ . z1 − z2 z1 − z4
Proof. Assume that the points are collinear. We can arrange four points on a circle in (4 − 1)! = 3! = 6 different ways. Consider the case when M1 , M2 , M3 , M4 are given in this order. Then M1 , M2 , M3 , M4 are concyclic if and only if M 1 M2 M3 + M 1 M4 M3 ∈ {3π, π }. That is, arg
z1 − z4 z3 − z2 + arg ∈ {3π, π }. z1 − z2 z3 − z4
arg
z3 − z4 z3 − z2 − arg ∈ {3π, π }, z1 − z2 z1 − z4
We obtain
i.e., k < 0. For any other arrangements of the four points the proof is similar. Note that k > 0 in three cases and k < 0 in the other three. The number k is called the cross ratio of the four points M1 (z 1 ), M2 (z 2 ), M3 (z 3 ) and M4 (z 4 ). Remarks. 1) The points M1 , M2 , M3 , M4 are collinear if and only if z3 − z2 z3 − z4 ∈ R∗ and ∈ R∗ . z1 − z2 z1 − z4 2) The points M1 , M2 , M3 , M4 are concyclic if and only if k=
z3 − z2 z3 − z4 z3 − z2 z3 − z4 : ∈ R∗ , but ∈ R and ∈ R. z1 − z2 z1 − z4 z1 − z2 z1 − z4
Examples. 1) The geometric images of the complex numbers 1, i, −1, −i are con−1 − i −1 + i : = −1 ∈ R∗ and clearly cyclic. Indeed, we have the cross ratio k = 1−i 1+i −1 − i −1 + i ∈ R and ∈ R. 1−i 1+i
3.3. Similar Triangles
67
2) The points M1 (2 − i), M2 (3 − 2i), M3 (−1 + 2i) and M4 (−2 + 3i) are collinear. −4 + 4i −4 + 4i 1 − i : = 1 ∈ R∗ and = 4 ∈ R∗ . Indeed, k = −1 + i 4 − 4i −1 + i Problem 2. Find all complex numbers z such that the points of complex coordinates z, z 2 , z 3 , z 4 – in this order – are the vertices of a cyclic quadrilateral. Solution. If the points of complex coordinates z, z 2 , z 3 , z 4 – in this order – are the vertices of a cyclic quadrilateral, then z3 − z2 z3 − z4 : ∈ R∗ . z − z2 z − z4 It follows that 1 + z + z2 1 ∗ − ∈ R , i.e., − 1 − z + ∈ R∗ . z z We obtain z + 1z ∈ R, i.e., z + 1z = z + 1z . Hence (z − z)(|z|2 − 1) = 0, hence z ∈ R or |z| = 1. If z ∈ R, then the points of complex coordinates z, z 2 , z 3 , z 4 are collinear, hence it is left to consider the case |z| = 1. Let t = arg z ∈ [0, 2π). We prove that the points of complex coordinates 2π 4π 2 3 4 z, z , z , z lie in this order on the unit circle if and only if t ∈ 0, ∪ , 2π . 3 3 Indeed, π a) If t ∈ 0, , then 0 < t < 2t < 3t < 4t < 2π or 2 0 < arg z < arg z 2 < arg z 3 < arg z 4 < 2π. π 2π , , then 0 ≤ 4t − 2π < t < 2t < 3t < 2π or b) If t ∈ 2 3 0 ≤ arg z 4 < arg z < arg z 2 < arg z 3 < 2π. c) If t ∈
2π , π , then 0 ≤ 3t − 2π < t ≤ 4t − 2π < 2t < 2π or 3 0 ≤ arg z 3 < arg z ≤ arg z 4 < arg z 2 .
In the same manner we can analyze the case t ∈ [π, 2π ). To conclude, the complex numbers satisfying the desired property are 4π 2π ∪ ,π . z = cos t + i sin t, with t ∈ 0, 3 3
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3. Complex Numbers and Geometry
3.3
Similar Triangles
Consider six points A1 (a1 ), A2 (a2 ), A3 (a3 ), B1 (b1 ), B2 (b2 ), B3 (b3 ) in the complex plane. We say that the triangles A1 A2 A3 and B1 B2 B3 are similar if the angle at Ak is equal to the angle at Bk , k ∈ {1, 2, 3}. Proposition 1. The triangles A1 A2 A3 and B1 B2 B3 are similar, having the same orientation, if and only if b2 − b1 a2 − a1 = . (1) a3 − a1 b3 − b1 A1 A2 B1 B2 = and A1 A3 B1 B3 |b2 − b1 | a2 − a1 |a2 − a1 | A = and arg = 3 A1 A2 ≡ B 3 B1 B2 . This is equivalent to |a3 − a1 | |b3 − b1 | a3 − a1 b2 − b1 a2 − a1 b2 − b1 arg . We obtain = . b3 − b 1 a3 − a1 b3 − b 1 Remarks. 1) The condition (1) is equivalent to Proof. We have A1 A2 A3 ∼ B1 B2 B3 if and only if
1 a1 b1
1 a2 b2
1 a3 b3
= 0.
2) The triangles A1 (0), A2 (1), A3 (2i) and B1 (0), B2 (−i), B3 (−2) are similar, but opposite oriented. In this case the condition (1) is not satisfied. Indeed, 1−0 −i − 0 1 b2 − b1 i a2 − a 1 = = = = = . a3 − a1 2i − 0 2i b3 − b1 −2 − 0 2 Proposition 2. The triangles A1 A2 A3 and B1 B2 B3 are similar, having opposite orientation, if and only if a2 − a1 b2 − b1 = . a3 − a1 b3 − b1 Proof. Reflection across the x-axis maps the points B1 , B2 , B3 into the points M1 (b1 ), M2 (b2 ), M3 (b3 ). The triangles B1 B2 B3 and M1 M2 M3 are similar and have opposite orientation, hence triangles A1 A2 A3 and M1 M2 M3 are similar with the same orientation. The conclusion follows from the previous proposition. Problem 1. On sides AB, BC, C A of a triangle ABC we draw similar triangles AD B, B EC, C F A, having the same orientation. Prove that triangles ABC and D E F have the same centroid. Solution. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter.
3.3. Similar Triangles
69
Triangles AD B, B EC, C F A are similar with the same orientation, hence d −a e−b f −c = = = z, b−a c−b a−c and consequently d = a + (b − a)z,
e = b + (c − b)z,
f = c + (a − c)z.
Then
a+b+c d +c+ f = , 3 3 so triangles ABC and D E F have the same centroid. Problem 2. Let M, N , P be the midpoints of sides AB, BC, C A of triangle ABC. On the perpendicular bisectors of segments [AB], [BC], [C A] points C , A , B are chosen inside the triangle such that MC N A P B = = . AB BC CA Prove that ABC and A B C have the same centroid. Solution. Note that from N A P B MC = = AB BC CA AB) = tan( A BC) = tan( B C A). Hence triangles AC B, B A C, it follows that tan(C C B A are similar and we can proceed as in the previous problem.
Problem 3. Let AB O be an equilateral triangle with center S and let A B O be another equilateral triangle with the same orientation and S = A , S = B . Consider M and N the midpoints of the segments A B and AB . Prove that triangles S B M and S A N are similar. (30th IMO – Shortlist)
Solution. Let R be the circumradius of the triangle AB O and let ε = cos
2π 2π + i sin . 3 3
Consider the complex plane with origin at point S such that point O lies on the positive real axis. Then the coordinates of points O, A, B are R, Rε, Rε2 , respectively. Let R + z be the coordinate of point B , so R − zε is the coordinate of point A . It follows that the midpoints M, N have the coordinates zM =
Rε 2 + R − zε R(ε2 + 1) − zε z B + z A = = 2 2 2
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3. Complex Numbers and Geometry
Figure 3.11.
=
−ε(R + z) −Rε − zε = 2 2
and zN =
Now we have
Rε + R + z R(ε + 1) + z −Eε2 + z z A + z B = = = 2 2 2 2 R z− ε = R − zε . = 2 −2ε z A − z S z B − z S = zM − zS zN − zS
if and only if R+z R − zε = . −ε(R + z) R − zε 2 −2ε The last relation is equivalent to ε · ε = 1, i.e., |ε|2 = 1. Hence the triangles S B M and S A N are similar, with opposite orientation.
3.4
Equilateral Triangles
Proposition 1. Suppose z 1 , z 2 , z 3 are the coordinates of the vertices of the triangle A1 A2 A3 . The following statements are equivalent: a) A1 A2 A3 is an equilateral triangle; b) |z 1 − z 2 | = |z 2 − z 3 | = |z 3 − z 1 |; c) z 12 + z 22 + z 32 = z 1 z 2 + z 2 z 3 + z 3 z 1 ; z2 − z1 z3 − z2 d) = ; z3 − z2 z1 − z2
3.4. Equilateral Triangles
71
1 1 1 z1 + z2 + z3 ; + + = 0, where z = z − z1 z − z2 z − z3 3 2π 2π f) (z 1 + εz 2 + ε 2 z 3 )(z 1 + ε 2 z 2 + εz 3 ) = 0, where ε = cos + i sin ; 3 3 1 1 1 g) z 1 z 2 z 3 = 0. z2 z3 z1 Proof. The triangle A1 A2 A3 is equilateral if and only if A1 A2 A3 is similar with same orientation with A2 A3 A1 , or 1 1 1 z 1 z 2 z 3 = 0, z2 z3 z1 e)
thus a) ⇔ g). Computing the determinant we obtain 1 1 1 0 = z1 z2 z3 z2 z3 z1 = z 1 z 2 + z 2 z 3 + z 3 z 1 − (z 12 + z 22 + z 32 ) = −(z 1 + εz 2 + ε 2 z 3 )(z 1 + ε 2 z 2 + εz 3 ), hence g) ⇔ c) ⇔ f). Simple algebraic manipulation shows that d) ⇔ c). Since a) ⇔ b) is obvious, we leave for the reader to prove that a) ⇔ e). The next results bring some refinements to this issue. Proposition 2. Let z 1 , z 2 , z 3 be the coordinates of the vertices A1 , A2 , A3 of a positively oriented triangle. The following statements are equivalent. a) A1 A2 A3 is an equilateral triangle; π π b) z 3 − z 1 = ε(z 2 − z 1 ), where ε = cos + i sin ; 3 3 5π 5π + i sin ; c) z 2 − z 1 = ε(z 3 − z 1 ), where ε = cos 3 3 2π 2π + i sin . d) z 1 + εz 2 + ε 2 z 3 = 0, where ε = cos 3 3 Proof. A1 A2 A3 is equilateral and positively oriented if and only if A3 is obtained π from A2 by rotation about A1 through an angle of . That is, 3 π π (z 2 − z 1 ), z 3 = z 1 + cos + i sin 3 3 hence a) ⇔ b).
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3. Complex Numbers and Geometry
Figure 3.12.
5π The rotation about A1 through an angle of maps A3 into A2 . Similar considera3 tions show that a) ⇔ c). To prove that b) observe that b) is equivalentto ⇔ d), √ √ √ 3 3 3 1 1 1 +i (z 2 − z 1 ) = −i z1 + +i z 2 . Hence b ) z 3 = z 1 + 2 2 2 2 2 2
√ √ 3 3 1 1 z2 + − − i z3 z 1 + εz 2 + ε z 3 = z 1 + − + i 2 2 2 2 2
√ 3 1 = z1 + − + i z2 2 2 √ √ √ 3 3 3 1 1 1 −i z1 + +i z2 + − −i 2 2 2 2 2 2 √ √ 3 3 1 1 z2 − z1 + −i z 2 = 0, = z1 + − + i 2 2 2 2 or b) ⇔ d).
Proposition 3. Let z 1 , z 2 , z 3 be the coordinates of the vertices A1 , A2 , A3 of a negatively oriented triangle. The following statements are equivalent: a) A1 A2 A3 is an equilateral triangle; 5π 5π b) z 3 − z 1 = ε(z 2 − z 1 ), where ε = cos + i sin ; 3 3 π π c) z 2 − z 1 = ε(z 3 − z 1 ), where ε = cos + i sin ; 3 3 2π 2π d) z 1 + ε 2 z 2 + εz 3 = 0, where ε = cos + i sin . 3 3
3.4. Equilateral Triangles
73
Proof. Equilateral triangle A1 A2 A3 is negatively oriented if and only if A1 A3 A2 is a positively oriented equilateral triangle. The rest follows from the previous proposition. Proposition 4. Let z 1 , z 2 , z 3 be the coordinates of the vertices of equilateral triangle A1 A2 A3 . Consider the statements: 1) A1 A2 A3 is an equilateral triangle; 2) z 1 · z 2 = z 2 · z 3 = z 3 · z 1 ; 3) z 12 = z 2 · z 3 and z 22 = z 1 · z 3 . Then 2) ⇒ 1), 3) ⇒ 1) and 2) ⇔ 3). Proof. 2) ⇒ 1). Taking the modulus of the terms in the given relation we obtain |z 1 | · |z 2 | = |z 2 | · |z 3 | = |z 3 | · |z 1 |, or equivalently |z 1 | · |z 2 | = |z 2 | · |z 3 | = |z 3 | · |z 1 |. This implies r = |z 1 | = |z 2 | = |z 3 | and
r2 r2 r2 , z2 = , z3 = . z1 z2 z3 Returning to the given relation we have z1 z2 z3 = = , z2 z3 z1 z1 =
or z 12 = z 2 z 3 ,
z 22 = z 3 z 1 ,
z 32 = z 1 z 2 .
Summing up these relations yields z 12 + z 22 + z 32 = z 1 z 2 + z 2 z 3 + z 3 z 1 , so triangle A1 A2 A3 is equilateral. Observe that we have also proved that 2) ⇒ 3) and that the arguments are reversible; hence 2) ⇔ 3). As a consequence, 3) ⇒ 1) and we are done. Problem 1. Let z 1 , z 2 , z 3 be nonzero complex coordinates of the vertices of the triangle A1 A2 A3 . If z 12 = z 2 z 3 and z 22 = z 1 z 3 , show that triangle A1 A2 A3 is equilateral. Solution. Multiplying the relations z 12 = z 2 z 3 and z 22 = z 1 z 3 yields z 12 z 22 = z 1 z 2 z 32 , and consequently z 1 z 2 = z 32 . Thus z 12 + z 22 + z 32 = z 1 z 2 + z 2 z 3 + z 3 z 1 , so triangle A1 A2 A3 is equilateral, by Proposition 1 in this section.
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3. Complex Numbers and Geometry
Problem 2. Let z 1 , z 2 , z 3 be the coordinates of the vertices of triangle A1 A2 A3 . If |z 1 | = |z 2 | = |z 3 | and z 1 + z 2 + z 3 = 0, prove that triangle A1 A2 A3 is equilateral. Solution. The following identity holds for any complex numbers z 1 and z 2 (see Problem 1 in Subsection 1.1.7): |z 1 − z 2 |2 + |z 1 + z 2 |2 = 2(|z 1 |2 + |z 2 |2 ).
(1)
From z 1 + z 2 + z 3 = 0 it follows that z 1 + z 2 = −z 3 , so |z 1 + z 2 | = |z 3 |. Using the relations |z 1 | = |z 2 | = |z 3 | and (1) we get |z 1 − z 2 |2 = 3|z 1 |2 . Analogously, we find the relations |z 2 − z 3 |2 = 3|z 1 |2 and |z 3 − z 1 |2 = 3|z 1 |2 . Therefore |z 1 − z 2 | = |z 2 − z 3 | = |z 3 − z 1 |, i.e., triangle A1 A2 A3 is equilateral. 1 1 1 Alternative solution 1. If we pass to conjugates, then we obtain + + = 0. z1 z2 z3 Combining this with the hypothesis yields z 12 + z 22 + z 32 = z 1 z 2 + z 2 z 3 + z 3 z 1 = 0, from which the desired conclusion follows by Proposition 1. Alternative solution 2. Taking into account the hypotheses |z 1 | = |z 2 | = |z 3 | it follows that we can consider the complex plane with its origin at the circumcenter of triangle A1 A2 A3 . Then, the coordinate of orthocenter H is z H = z 1 + z 2 + z 3 = 0 = z O . Hence H = O, and triangle A1 A2 A3 is equilateral. Problem 3. In the exterior of triangle ABC three positively oriented equilateral triangles AC B, B A C and C B A are constructed. Prove that the centroids of these triangles are the vertices of an equilateral triangle. (Napoleon’s problem)
Solution.
Figure 3.13.
Let a, b, c be the coordinates of vertices A, B, C, respectively.
3.4. Equilateral Triangles
75
Using Proposition 2, we have a + c ε + bε 2 = 0,
b + a ε + cε 2 = 0,
c + b ε + aε 2 = 0,
(1)
where a , b , c are the coordinates of points A , B , C . The centroids of triangles A BC, AB C, ABC have the coordinates a =
1 (a + b + c), 3
b =
1 (a + b + c), 3
c =
1 (a + b + c ), 3
respectively. We have to check that c + a ε + b ε 2 = 0. Indeed, 3(c + a ε + b ε 2 ) = (a + b + c ) + (a + b + c)ε + (a + b + c)ε 2 = (b + a ε + cε 2 ) + (c + b ε + aε 2 )ε + (a + c ε + bε 2 )ε2 = 0. Problem 4. On the sides of the triangle ABC we draw three regular n-gons, external to the triangle. Find all values of n for which the centers of the n-gons are the vertices of an equilateral triangle. (Balkan Mathematical Olympiad 1990 – Shortlist)
Solution. Let A0 , B0 , C0 be the centers of the regular n-gons constructed externally on the sides BC, C A, AB, respectively.
Figure 3.14.
2π . Let The angles AC A0 C, AB 0 B, B 0 C have the measures of n ε = cos
2π 2π + i sin n n
and denote by a, b, c, a0 , b0 , c0 the coordinates of the points A, B, C, A0 , B0 , C0 , respectively.
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3. Complex Numbers and Geometry
Using the rotation formula, we obtain a = c0 + (b − c0 )ε; b = a0 + (c − a0 )ε; c = b0 + (a − b0 )ε. Thus
b − cε c − aε , b0 = , 1−ε 1−ε Triangle A0 B0 C0 is equilateral if and only if a0 =
c0 =
a − bε . 1−ε
a02 + b02 + c02 = a0 b0 + b0 c0 + c0 a0 . Substituting the above values of a0 , b0 , c0 we obtain (b − cε)2 + (c − aε)2 + (a − bε)2 = (b − cε)(c − aε) + (c − aε)(a − bε) + (a − bε)(c − aε). This is equivalent to (1 + ε + ε 2 )[(a − b)2 + (b − c)2 + (c − a)2 ] = 0. 2π 2π = and we get n = 3. Therefore n = 3 is It follows that 1 + ε + ε 2 = 0, i.e., n 3 the only value with the desired property.
3.5 3.5.1
Some Analytic Geometry in the Complex Plane Equation of a line
Proposition 1. The equation of a line in the complex plane is α · z + αz + β = 0, where α ∈ C∗ , β ∈ R and z = x + i y ∈ C. Proof. The equation of a line in the cartesian plane is Ax + By + C = 0, where A, B, C ∈ R and A2 + B 2 = 0. If we set z = x + i y, then x = z−z y= . Thus, 2i z+z z−z A − Bi + C = 0, 2 2
z+z and 2
3.5. Some Analytic Geometry in the Complex Plane
or equivalently
z
Let α =
A + Bi 2
+z
77
A − Bi + C = 0. 2
A − Bi ∈ C∗ and β = C ∈ R. Then 2 α · z + αz + β = 0,
as claimed.
If α = α, then B = 0 and we have a vertical line. If α = α, then we define the angular coefficient of the line as m=−
A α+α α+α = i. = α−α B α−α i
Proposition 2. Consider the lines d1 and d2 with equations α 1 · z + α1 · z + β 1 = 0 and α 2 · z + α2 · z + β2 = 0, respectively. Then the lines d1 and d2 are: α1 α2 1) parallel if and only if = ; α1 α2 α1 α2 2) perpendicular if and only if + = 0; α2 α2 α1 α2 3) concurrent if and only if = . α1 α2 Proof. 1) We have d1 d2 if and only if m 1 = m 2 . Therefore
α1 + α 1 α2 + α 2 i= i, α1 − α 1 α2 − α 2
α1 α2 = . α1 α2 2) We have d1 ⊥ d2 if and only if m 1 m 2 = −1. That is, α2 α 1 + α2 α 2 = 0, or α1 α2 = 0. + α α2 3) The lines d1 and d2 are concurrent if and only if m 1 = m 2 . This condition yields α1 α2 = . α1 α2 The results for angular coefficient correspond to the properties of slope. α The ratio m d = − is called the complex angular coefficient of the line d of equaα tion α · z + α · z + β = 0.
so α2 α 1 = α1 α 2 and we get
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3. Complex Numbers and Geometry
3.5.2
Equation of a line determined by two points
Proposition. The equation of a line determined by the points P1 (z 1 ) and P2 (z 2 ) is z1 z1 1 z 2 z 2 1 = 0. z z 1 Proof. The equation of a line determined by the points P1 (x1 , y1 ) and P2 (x2 , y2 ) in the cartesian plane is x1 y1 1 x2 y2 1 = 0. x y 1 Using complex numbers we have z1 + z1 2 z2 + z2 2 z +z 1 2 if and only if
That is,
z1 + z1 1 z2 + z2 4i z+z z1 z2 z
z1 − z1 2i z2 − z2 2i z−z 2i z1 − z1 z2 − z2 z−z z1 z2 z
1 1 1
1 1 =0 1
1 1 1
= 0.
= 0,
as desired.
Remarks. 1) The points M1 (z 1 ), M2 (z 2 ), M3 (z 3 ) are collinear if and only if z1 z1 1 z 2 z 2 1 = 0. z3 z3 1 2) The complex angular coefficient of a line determined by the points with coordinates z 1 and z 2 is z2 − z1 m= . z2 − z1
3.5. Some Analytic Geometry in the Complex Plane
79
Indeed, the equation is z1 z2 z3
z1 z2 z3
1 1 1
= 0 ⇔ z 1 z 2 + z 2 z + zz 1 − zz 2 − z 1 z − z 2 z 1 = 0
⇔ z(z 2 − z 1 ) − z(z 2 − z 1 ) + z 1 z 2 − z 2 z 1 = 0. Using the definition of the complex angular coefficient we obtain m=
3.5.3
z2 − z1 . z2 − z1
The area of a triangle
Theorem. The area of triangle A1 A2 A3 whose vertices have coordinates z 1 , z 2 , z 3 is equal to the absolute value of the number z1 i z2 4 z3
z1 z2 z3
1 1 1
.
(1)
Proof. Using cartesian coordinates, the area of a triangle with vertices (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) is equal to the absolute value of the determinant x1 1 = x2 2 x3 Since xk =
zk + zk , 2
yk =
y1 y2 y3
1 1 1
.
zk − zk , 2i
k = 1, 2, 3
we obtain z1 + z1 1 = z2 + z2 8i z3 + z3
z1 − z1 z2 − z2 z3 − z3 z1 i = z2 4 z3
as claimed.
1 1 1 z1 z2 z3
1 =− 4i 1 1 1
z1 z2 z3
z1 z2 z3
1 1 1
,
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3. Complex Numbers and Geometry
It is easy to see that for positively oriented triangle A1 A2 A3 with vertices with coordinates z 1 , z 2 , z 3 the following inequality holds: z1 i z2 4 z3
z1 z2 z3
1 1 1
> 0.
Corollary. The area of a directly oriented triangle A1 A2 A3 whose vertices have coordinates z 1 , z 2 , z 3 is area[A1 A2 A3 ] =
1 Im(z 1 z 2 + z 2 z 3 + z 3 z 1 ). 2
(2)
Proof. The determinant in the above theorem is z1 z2 z3
z1 z2 z3
1 1 1
= (z 1 z 2 + z 2 z 3 + z 3 z 1 − z 2 z 3 − z 1 z 3 − z 2 z 1 )
= [(z 1 z 2 + z 2 z 3 + z 3 z 1 ) − (z 1 z 2 + z 2 z 3 + z 3 z 1 )]
= 2i Im(z 1 z 2 + z 2 z 3 + z 3 z 1 ) = −2i Im(z 1 z 2 + z 2 z 3 + z 3 z 1 ).
Replacing this value in (1), the desired formula follows.
We will see that formula (2) can be extended to a convex directly oriented polygon A1 A2 · · · An (see Section 4.3). Problem 1. Consider the triangle A1 A2 A3 and the points M1 , M2 , M3 situated on lines A2 A3 , A1 A3 , A1 A2 , respectively. Assume that M1 , M2 , M3 divide segments [A2 A3 ], [A3 A1 ], [A1 A2 ] into ratios λ1 , λ2 , λ3 , respectively. Then 1 − λ 1 λ2 λ3 area[M1 M2 M3 ] = . area[A1 A2 A3 ] (1 − λ1 )(1 − λ2 )(1 − λ3 ) Solution. The coordinates of the points M1 , M2 , M3 are m1 =
a 2 − λ1 a 3 , 1 − λ1
m2 =
a 3 − λ2 a 1 , 1 − λ2
m3 =
a 1 − λ3 a 2 . 1 − λ3
(3)
3.5. Some Analytic Geometry in the Complex Plane
81
Applying formula (2) we find that
area[M1 M2 M3 ] =
1 Im(m 1 m 2 + m 2 m 3 + m 3 m 1 ) 2
1 (a2 − λ1 a3 )(a3 − λ2 a1 ) (a3 − λ2 a1 )(a1 − λ3 a2 ) = Im + 2 (1 − λ1 )(1 − λ2 ) (1 − λ2 )(1 − λ3 ) (a1 − λ3 a2 )(a2 − λ1 a3 ) + (1 − λ3 )(1 − λ1 ) 1 1 − λ1 λ2 λ3 = Im (a1 a2 + a2 a3 + a3 a1 ) 2 (1 − λ1 )(1 − λ2 )(1 − λ3 ) =
1 − λ1 λ2 λ3 area[A1 A2 A3 ]. (1 − λ1 )(1 − λ2 )(1 − λ3 )
Remark. From formula (3) we derive the well-known theorem of Menelaus: The points M1 , M2 , M3 are collinear if and only if λ1 λ2 λ3 = 1, i.e., M1 A 2 M2 A 3 M3 A 1 · · =1 M1 A 3 M2 A 1 M3 A 2 Problem 2. Let a, b, c be the coordinates of the vertices A, B, C πofa triangle. It is such that a + known that |a| = |b| = |c| = 1 and that there exists α ∈ 0, 2 b cos α + c sin α = 0. Prove that √ 1+ 2 1 < area[ABC] ≤ . 2 (Romanian Mathematical Olympiad – Final Round, 2003)
Solution. Observe that 1 = |a|2 = |b cos α + c sin α|2 = (b cos α + c sin α)(b cos α + c sin α) = |b|2 cos2 α + |c|2 sin2 α + (bc + bc) sin α cos α =1+
b2 + c2 cos α sin α. bc
It follows that b2 + c2 = 0, hence b = ±ic. Applying formula (2) we obtain
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3. Complex Numbers and Geometry
area[ABC] = =
1 | Im(ab + bc + ca)| 2
1 | Im[(−b cos α − c sin α)b + bc − c(b cos α + c sin α)]| 2 =
= =
1 | Im(− cos α − sin α − bc sin α − bc cos α + bc)| 2
1 1 | Im[bc − (sin α + cos α)bc]| = | Im[(1 + sin α + cos α)bc]| 2 2 1 1 (1 + sin α + cos α)| Im(bc)| = (1 + sin α + cos α)| Im(±icc)| 2 2
1 1 (1 + sin α + cos α)| Im(±i)| = (1 + sin α + cos α) 2 2 √ √ √ √ 1 2 2 π 1 = 1+ 2 sin α + cos α = 1 + 2 sin α + . 2 2 2 2 4 =
√ π 2 π 3π π Taking into account that < α + < we get that < sin α + ≤ 1 and 4 4 4 2 4 the conclusion follows.
3.5.4
Equation of a line determined by a point and a direction
Proposition 1. Let d : αz + α · z + β = 0 be a line and let P0 (z 0 ) be a point. The equation of a line parallel to d and passing through point P0 is α z − z 0 = − (z − z 0 ). α Proof. Using cartesian coordinates, the line parallel to d and passing through point P0 (x0 , y0 ) has the equation y − y0 = i
α+α (x − x0 ). α−α
Using complex numbers the equation takes the form z−z z0 − z0 α+α z+z z0 + z0 − =i − . 2i 2i α−α 2 2 This is equivalent to (α − α)(z − z 0 − z + z 0 ) = (α + α)(z + z − z 0 − z 0 ), or α α(z − z 0 ) = −α(z − z 0 ). We obtain z − z 0 = − (z − z 0 ). α Proposition 2. Let d : αz + α · z + β = 0 be a line and let P0 (z 0 ) be a point. The line α passing through point P0 and perpendicular to d has the equation z − z 0 = (z − z 0 ). α
3.6. The Circle
83
Proof. Using cartesian coordinates, the line passing through point P0 and perpendicular to d has the equation 1 α−α (x − x0 ). y − y0 = − · i α+α Then we obtain z−z z0 − z0 α−α − =i· 2i 2i α+α
z+z z0 + z0 − 2 2
.
That is, (α + α)(z − z 0 − z + z 0 ) = −(α − α)(z − z 0 + z − z 0 ) or (z − z 0 )(α + α + α − α) = (z − z 0 )(−α + α + α + α). We obtain α(z − z 0 ) = α(z − z 0 ) and z − z 0 = αα (z − z 0 ).
3.5.5
The foot of a perpendicular from a point to a line
Proposition. Let P0 (z 0 ) be a point and let d : αz + αz + β = 0 be a line. The foot of the perpendicular from P0 to d has the coordinate z=
αz 0 − α z 0 − β . 2α
Proof. The point z is the solution of the system α · z + α · z + β = 0, α(z − z 0 ) = α(z − z 0 ). The first equation gives
−αz − β . α Substituting in the second equation yields z=
αz − αz 0 = −αz − β − α · z 0 . Hence z=
αz 0 − α z 0 − β , 2α
as claimed.
3.5.6
Distance from a point to a line
Proposition. The distance from a point P0 (z 0 ) to a line d : α · z + α · z + β = 0, α ∈ C∗ is equal to |αz 0 + α · z 0 + β| . D= √ 2 α·α
84
3. Complex Numbers and Geometry
Proof. Using the previous result, we can write −αz 0 − αz 0 − β αz 0 − α · z 0 − β − z0 = D= 2α 2α |α · z 0 + αz 0 + β| |αz 0 + αz 0 + β| = . √ 2|α| 2 αα
=
3.6 3.6.1
The Circle Equation of a circle
Proposition. The equation of a circle in the complex plane is z · z + α · z + α · z + β = 0, where α ∈ C and β ∈ R. Proof. The equation of a circle in the cartesian plane is x 2 + y 2 + mx + ny + p = 0, m2 + n2 m, n, p ∈ R, p < . 4 z+z z−z Setting x = and y = we obtain 2 2i |z|2 + m or z·z+z Take α = proved.
z+z z−z +n +p=0 2 2i
m − ni m + ni +z + p = 0. 2 2
m − ni ∈ C and β = p ∈ R in the above equation and the claim is 2
Note that the radius of the circle is equal to &
n2 m2 + − p = αα − β. r= 4 4 Then the equation is equivalent to (z + α)(z + α) = r 2 . Setting γ = −α = −
n m − i 2 2
3.6. The Circle
85
the equation of the circle with center at γ and radius r is (z − γ )(z − γ ) = r 2 . Problem. Let z 1 , z 2 , z 3 be the coordinates of the vertices of triangle A1 A2 A3 . The coordinate z O of the circumcenter of triangle A1 A2 A3 is 1 1 1 z2 z3 z1 |z 1 |2 |z 2 |2 |z 3 |2 . (1) zO = 1 1 1 z1 z2 z3 z1 z2 z3 Solution. The equation of the line passing through P(z 0 ) which is perpendicular to the line A1 A2 can be written in the form z(z 1 − z 2 ) + z(z 1 − z 2 ) = z 0 (z 1 − z 2 ) + z 0 (z 1 − z 2 ).
(2)
Applying this formula for the midpoints of the sides [A2 A3 ], [A1 A3 ] and for the lines A2 A3 , A1 A3 , we find the equations z(z 2 − z 3 ) + z(z 2 − z 3 ) = |z 2 |2 − |z 3 |2 z(z 3 − z 1 ) + z(z 3 − z 1 ) = |z 3 |2 − |z 1 |2 . By eliminating z from these two equations, it follows that z[(z 2 − z 3 ) + (z 3 − z 1 )(z 2 − z 3 )] = (z 1 − z 3 )(|z 2 |2 − |z 3 |2 ) + (z 2 − z 3 )(|z 3 |2 − |z 1 |2 ), hence
1 z z1 z1
1 z2 z2
1 z3 z2
1 = z1 |z 1 |2
1 z2 |z 2 |2
1 z3 |z 3 |2
and the desired formula follows. Remark. We can write this formula in the following equivalent form: zO =
z 1 z 1 (z 2 − z 3 ) + z 2 z 2 (z 3 − z 1 ) + z 3 z 3 (z 1 − z 2 ) . 1 1 1 z1 z2 z3 z1 z2 z3
(3)
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3. Complex Numbers and Geometry
3.6.2
The power of a point with respect to a circle
Proposition. Consider a point P0 (z 0 ) and a circle with equation z · z + α · z + α · z + β = 0, for α ∈ C and β ∈ R. The power of P0 with respect to the circle is ρ(z 0 ) = z 0 · z 0 + αz 0 + α · z 0 + β. Proof. Let O(−α) be the center of the circle. The power of P0 with respect to the circle of radius r is defined by ρ(z 0 ) = O P02 − r 2 . In this case we obtain ρ(z 0 ) = O P02 − r 2 = |z 0 + α|2 − r 2 = z 0 · z 0 + αz 0 + αz 0 + αα − αα + β = z 0 · z 0 + αz 0 + α · z 0 + β,
as claimed. Given two circles of equations z · z + α 1 · z + α1 · z + β 1 = 0
and
z · z + α2 · z + α2 · z + β2 = 0,
where α1 , α2 ∈ C, β1 , β2 ∈ R, their radical axis is the locus of points having equal powers with respect to the circles. If P(z) is a point of this locus, then z · z + α1 z + α1 · z + β1 = z · z + α2 z + α2 · z + β2 , or equivalently (α1 − α2 )z + (α1 − α2 )z + β1 − β2 = 0, which is the equation of a line.
3.6.3
Angle between two circles
The angle between two circles with equations z · z + α 1 · z + α1 · z + β 1 = 0 and z · z + α2 · z + α2 · z + β2 = 0,
α1 , α2 ∈ C,
β1 , β2 ∈ R,
is the angle θ determined by the tangents to the circles at a common point. Proposition. The following formula β1 + β2 − (α1 α2 + α1 α2 ) cos θ = 2r1r2 holds.
3.6. The Circle
87
Figure 3.15.
Proof. Let T be a common point and let O1 (−α1 ), O2 (−α2 ) be the centers of the circles. The angle θ is equal to O 1 T O2 or π − O 1 T O2 , hence cos θ = | cos O 1 T O2 | =
= =
|r12 + r22 − O1 O22 | 2r1r2
|α1 α1 − β1 + α2 α2 − β2 − |α1 − α2 |2 | 2r1r2
|α1 α1 + α2 α2 − β1 − β2 − α1 α1 − α2 α2 + α1 α2 + α1 α2 | 2r1r2 =
|β1 + β2 − (α1 α2 + α1 α2 )| , 2r1r2
as claimed. Note that the circles are orthogonal if and only if β1 + β2 = α1 α2 + α1 α2 .
88
3. Complex Numbers and Geometry
Problem 1. Let a, b be real numbers such that |b| ≤ 2a 2 . Prove that the set of points with coordinates z such that |z 2 − a 2 | = |2az + b| is the union of two orthogonal circles. Solution. The relation |z 2 − a 2 | = |2az + b| is equivalent to |z 2 − a 2 |2 = |2az + b|2 , i.e., (z 2 − a 2 )(z 2 − a 2 ) = (2az + b)(2az + b). We can rewrite the last relation as |z|4 − a 2 (z 2 + z 2 ) + a 4 = 4a 2 |z|2 + 2ab(z + z) + b2 , i.e., |z|4 − a 2 [(z + z)2 − 2|z|2 ] + a 4 = 4a 2 |z|2 + 2ab(z + z) + b2 . Hence |z|4 − 2a 2 |z|2 + a 4 = a 2 (z + z)2 + 2ab(z + z) + b2 , i.e., (|z|2 − a 2 )2 = (a(z + z) + b)2 . It follows that z · z − a 2 = a(z + z) + b or z · z − a 2 = −a(z + z) − b. This is equivalent to (z − a)(z − a) = 2a 2 + b or (z + a)(z + a) = 2a 2 − b. Finally |z − a|2 = 2a 2 + b or |z + a|2 = 2a 2 − b.
(1)
Since |b| ≤ 2a 2 , it follows that 2a 2 + b ≥ 0 and 2a 2 − b ≥ 0. Hence the relations (1) are equivalent to
|z − a| = 2a 2 + b or |z + a| = 2a 2 − b. Therefore, the points with coordinates z that satisfy |z 2 − a 2 | = |2az + b| lie on two circles of centers C1 and C2 , whose coordinates a and −a, and with radii R1 = √ √ 2a 2 + b and R2 = 2a 2 − b. Furthermore,
C1 C22 = 4a 2 = ( 2a 2 + b)2 + ( 2a 2 − b)2 = R12 + R22 , hence the circles are orthogonal, as claimed.
4 More on Complex Numbers and Geometry
4.1
The Real Product of Two Complex Numbers
The concept of the scalar product of two vectors is well known. In what follows we will introduce this concept for complex numbers. We will see that in many situations use of this product simplifies the solution to the problem considerably. Let a and b be two complex numbers. Definition. We call the real product of complex numbers a and b the number given by a·b =
1 (ab + ab). 2
It is easy to see that a·b =
1 (ab + ab) = a · b; 2
hence a · b is a real number, which justifies the name of this product. The following properties are easy to verify. Proposition 1. For all complex numbers a, b, c, z the following relations hold: 1) a · a = |a|2 . 2) a · b = b · a; (the real product is commutative). 3) a · (b + c) = a · b + a · c; (the real product is distributive with respect to addition). 4) (αa) · b = α(a · b) = a · (αb) for all α ∈ R.
90
4. More on Complex Numbers and Geometry
5) a · b = 0 if and only if O A ⊥ O B, where A has coordinate a and B has coordinate b. 6) (az) · (bz) = |z|2 (a · b). Remark. Suppose that A and B are points with coordinates a and b. Then the real product a · b is equal to the power of the origin with respect to the circle of diameter AB. a+b Indeed, let M be the midpoint of [AB], hence the center of this circle, and 2 1 1 let r = AB = |a − b| be the radius of this circle. The power of the origin with 2 2 respect to the circle is a + b 2 a − b 2 − OM − r = 2 2 2
2
=
(a + b)(a + b) (a − b)(a − b) ab + ba − = = a · b, 4 4 2
as claimed. Proposition 2. Suppose that A(a), B(b), C(c) and D(d) are four distinct points. The following statements are equivalent: 1) AB ⊥ C D; 2) (b − a) · (c − d) = 0; b − a b−a 3) ∈ iR∗ (or, equivalently, Re = 0). d −c d −c Proof. Take points M(b − a) and N (d − c) such that O AB M and OC D N are parallelograms. Then we have AB ⊥ C D if and only if O M ⊥ O N . That is, m · n = (b − a) · (d − c) = 0, using property 5) of the real product. The equivalence 2) ⇔ 3) follows immediately from the definition of the real product. Proposition 3. The circumcenter of triangle ABC is at the origin of the complex plane. If a, b, c are the coordinates of vertices A, B, C, then the orthocenter H has the coordinate h = a + b + c. Proof. Using the real product of the complex numbers, the equations of the altitudes A A , B B , CC of the triangle are A A : (z − a) · (b − c) = 0, B B : (z − b) · (c − a) = 0, CC : (z − c) · (a − b) = 0. We will show that the point with coordinate h = a + b + c lies on all three altitudes. Indeed, we have (h − a) · (b − c) = 0 if and only if (b + c) · (b − c) = 0. The last relation is equivalent to b · b − c · c = 0, or |b|2 = |c|2 . Similarly, H ∈ B B and H ∈ CC , and we are done.
4.1. The Real Product of Two Complex Numbers
91
Remark. If the numbers a, b, c, o, h are the coordinates of the vertices of triangle ABC, the circumcenter O and the orthocenter H of the triangle, then h = a+b+c−2o. Indeed, taking A diametrically opposite A in the circumcircle of triangle ABC, the quadrilateral H B A C is a parallelogram. If {M} = H A ∩ BC, then zM =
b+c z H + z A z H + 2o − a = = , i.e., z H = a + b + c − 2o. 2 2 2
Problem 1. Let ABC D be a convex quadrilateral. Prove that AB 2 + C D 2 = AD 2 + BC 2 if and only if AC ⊥ B D. Solution. Using the properties of the real product of complex numbers, we have AB 2 + C D 2 = BC 2 + D A2 if and only if (b − a) · (b − a) + (d − c) · (d − c) = (c − b) · (c − b) + (a − d) · (a − d). That is, a·b+c·d =b·c+d ·a and finally (c − a) · (d − b) = 0, or, equivalently, AC ⊥ B D, as required. Problem 2. Let M, N , P, Q, R, S be the midpoints of the sides AB, BC, C D, D E, E F, F A of a hexagon. Prove that R N 2 = M Q2 + P S2 if and only if M Q ⊥ P S. (Romanian Mathematical Olympiad – Final Round, 1994)
Solution. Let a, b, c, d, e, f be the coordinates of the vertices of the hexagon. The points M, N , P, Q, R, S have the coordinates
respectively.
m=
a+b , 2
n=
b+c , 2
p=
c+d , 2
q=
d +e , 2
r=
e+ f , 2
s=
f +a , 2
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4. More on Complex Numbers and Geometry
Figure 4.1.
Using the properties of the real product of complex numbers, we have R N 2 = M Q2 + P S2 if and only if (e + f − b − c) · (e + f − b − c) = (d + e − a − b) · (d + e − a − b) + ( f + a − c − d) · ( f + a − c − d). That is, (d + e − a − b) · ( f + a − c − d) = 0; hence M Q ⊥ P S, as claimed. Problem 3. Let A 1 A2 · · · An be a regular polygon inscribed in a circle of center O and radius R. Prove that for all points M in the plane the following relation holds: n
M A2k = n(O M 2 + R 2 ).
k=1
Solution. Consider the complex plane with origin at point O and let Rεk be the coordinate of vertex Ak , where εk are the n th -roots of unity, k = 1, . . . , n. Let m be the coordinate of M. Using the properties of the real product of the complex numbers, we have n
M A2k =
k=1
=
n
k=1 n
(m − Rεk ) · (m − Rεk ) (m · m − 2Rεk · m + R 2 εk · εk )
k=1
= n|m|2 − 2R
n
k=1 2
n
εk · m + R 2 |εk |2 k=1 2
= n · O M + n R = n(O M + R ), 2
2
4.1. The Real Product of Two Complex Numbers
since
n
93
εk = 0.
k=1
Remark. If M lies on the circumcircle of the polygon, then n
M A2k = 2n R 2 .
k=1
Problem 4. Let O be the circumcenter of the triangle ABC, let D be the midpoint of the segment AB, and let E is the centroid of triangle AC D. Prove that lines C D and O E are perpendicular if and only if AB = AC. (Balkan Mathematical Olympiad, 1985)
Solution. Let O be the origin of the complex plane and let a, b, c, d, e be the coordinates of points A, B, C, D, E, respectively. Then d=
a+b a+c+d 3a + b + 2c and e = = . 2 3 6
Using the real product of complex numbers, if R is the circumradius of triangle ABC, then a · a = b · b = c · c = R2. Lines C D and D E are perpendicular if and only if (d − c) · e = 0 That is, (a + b − 2c) · (3a + b + 2c) = 0. The last relation is equivalent to 3a · a + a · b + 2a · c + 3a · b + b · b + 2b · c − 6a · c − 2b · c − 4c · c = 0, that is, a · b = a · c.
(1)
On the other hand, AB = AC is equivalent to |b − a|2 = |c − a|2 . That is, (b − a) · (b − a) = (c − a) · (c − a) or b · b − 2a · b + a · a = c · c − 2a · c + a · a, hence a · b = a · c. The relations (1) and (2) show that C D ⊥ O E if and only if AB = AC.
(2)
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4. More on Complex Numbers and Geometry
Problem 5. Let a, b, c be distinct complex numbers such that |a| = |b| = |c| and |b + c − a| = |a|. Prove that b + c = 0. Solution. Let A, B, C be the geometric images of the complex numbers a, b, c, respectively. Choose the circumcenter of triangle ABC as the origin of the complex plane and denote by R the circumradius of triangle ABC. Then aa = bb = cc = R 2 , and using the real product of the complex numbers, we have |b + c − a| = |a| if and only if |b + c − a|2 = |a|2 . That is, (b + c − a) · (b + c − a) = |a|2 , i.e., |a|2 + |b|2 + |c|2 + 2b · c − 2a · c − 2a · b = |a|2 . We obtain 2(R 2 + b · c − a · c − a · b) = 0, i.e., a · a + b · c − a · c − a · b = 0. It follows that (a − b) · (a − c) = 0, hence AB ⊥ AC, i.e., B AC = 90◦ . Therefore, [BC] is the diameter of the circumcircle of triangle ABC, so b + c = 0. Problem 6. Let E, F, G, H be the midpoints of sides AB, BC, C D, D A of the convex quadrilateral ABC D. Prove that lines AB and C D are perpendicular if and only if BC 2 + AD 2 = 2(E G 2 + F H 2 ).
Solution. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter. Then e=
a+b , 2
f =
b+c , 2
g=
c+d , 2
h=
Using the real product of the complex numbers, the relation BC 2 + AD 2 = 2(E G 2 + F H 2 ) becomes (c − b) · (c − b) + (d − a) · (d − a)
d +a . 2
4.1. The Real Product of Two Complex Numbers
95
1 1 (c + d − a − b) · (c + d − a − b) + (a + d − b − c) · (a + d − b − c). 2 2 This is equivalent to =
c · c + b · b + d · d + a · a − 2b · c − 2a · d = a · a + b · b + c · c + d · d − 2a · c − 2b · d, or a · d + b · c = a · c + b · d. The last relation shows that (a − b) · (d − c) = 0 if and only if AB ⊥ C D, as desired. Problem 7. Let G be the centroid of triangle ABC and let A1 , B1 , C1 be the midpoints of sides BC, C A, AB, respectively. Prove that M A2 + M B 2 + MC 2 + 9M G 2 = 4(M A21 + M B12 + MC12 ) for all points M in the plane. Solution. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter. Then g=
a+b+c , 3
a1 =
b+c , 2
b1 =
c+a , 2
c1 =
a+b . 2
Using the real product of the complex numbers, we have M A2 + M B 2 + MC 2 + 9M G 2 = (m − a) · (m − a) + (m − b) · (m − b) + (m − c) · (m − c) a+b+c a+b+c +9 m− · m− 3 3 = 12|m|2 − 8(a + b + c) · m + 2(|a|2 + |b|2 + |c|2 ) + 2a · b + 2b · c + 2c · a. On the other hand, 4(M A21 + M B12 + MC12 ) b+c b+c c+a =4 m− · m− + m− 2 2 2 c+a a+b a+b · m− + m− · m− 2 2 2 = 12|m|2 − 8(a + b + c) · m + 2(|a|2 + |b|2 + |c|2 ) + 2a · b + 2b · c + 2c · a, so we are done.
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4. More on Complex Numbers and Geometry
Remark. The following generalization can be proved similarly. Let A1 A2 · · · An be a polygon with the centroid G and let Ai j be the midpoint of the segment [Ai A j ], i < j, i, j ∈ {1, 2, . . . , n}. Then n
(n − 2) M A2k + n 2 M G 2 = 4 M Ai2j , i< j
k=1
for all points M in the plane. A nice generalization is given in Theorem 5, Section 4.11.
4.2
The Complex Product of Two Complex Numbers
The cross product of two vectors is a central concept in vector algebra, with numerous applications in various branches of mathematics and science. In what follows we adapt this product to complex numbers. The reader will see that this new interpretation has multiple advantages in solving problems involving area or collinearity. Let a and b be two complex numbers. Definition. The complex number 1 (ab − ab) 2 is called the complex product of the numbers a and b. Note that 1 1 a × b + a × b = (ab − ab) + (ab − ab) = 0, 2 2 so Re(a × b) = 0, which justifies the definition of this product. The following properties are easy to verify: a×b =
Proposition 1. Suppose that a, b, c are complex numbers. Then: 1) a × b = 0 if and only if a = 0 or b = 0 or a = λb, where λ is a real number. 2) a × b = −b × a; (the complex product is anticommutative). 3) a × (b + c) = a × b + a × c (the complex product is distributive with respect to addition). 4) α(a × b) = (αa) × b = a × (αb), for all real numbers α. 5) If A(a) and B(b) are distinct points other than the origin, then a × b = 0 if and only if O, A, B are collinear. Remarks. a) Suppose A(a) and B(b) are distinct points in the complex plane, different from the origin. The complex product of the numbers a and b has the following useful geometric interpretation: 2i · area[AO B], if triangle O AB is positively oriented; a×b = −2i · area[AO B], if triangle O AB is negatively oriented.
4.2. The Complex Product of Two Complex Numbers
97
Figure 4.2.
Indeed, if triangle O AB is positively (directly) oriented, then 2i · area[O AB] = i · O A · O B · sin( AO B) b |a| b = i|a| · |b| · sin arg = i · |a| · |b| · Im · a a |b| 1 b b 1 = |a|2 − = (ab − ab) = a × b. 2 a a 2 In the other case, note that triangle O B A is positively oriented, hence 2i · area[O B A] = b × a = −a × b. b) Suppose A(a), B(b), C(c) are three points in the complex plane. The complex product allows us to obtain the following useful formula for the area of the triangle ABC: ⎧ 1 ⎪ ⎪ (a × b + b × c + c × a), ⎪ ⎪ 2i ⎪ ⎪ ⎪ ⎪ ⎨ if triangle ABC is positively oriented; area[ABC] = ⎪ ⎪ 1 ⎪ ⎪ ⎪ − (a × b + b × c + c × a), ⎪ ⎪ 2i ⎪ ⎩ if triangle ABC is negatively oriented. Moreover, simple algebraic manipulation shows that area[ABC] =
1 Im(ab + bc + ca) 2
if triangle ABC is directly (positively) oriented. To prove the above formula, translate points A, B, C with vector −c. The images of A, B, C are points A , B , O with coordinates a − c, b − c, 0, respectively. Triangles ABC and A B O are congruent with the same orientation. If ABC is positively
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4. More on Complex Numbers and Geometry
oriented, then 1 ((a − c) × (b − c)) 2i 1 1 = ((a − c) × b − (a − c) × c) = (c × (a − c) − b × (a − c)) 2i 2i 1 1 = (c × a − c × c − b × a + b × c) = (a × b + b × c + c × a), 2i 2i
area[ABC] = area[O A B ] =
as claimed. The other situation can be similarly solved. Proposition 2. Suppose A(a), B(b) and C(c) are distinct points. The following statements are equivalent: 1) Points A, B, C are collinear. 2) (b − a) × (c − a) = 0. 3) a × b + b × c + c × a = 0. Proof. Points A, B, C are collinear if and only if area[ABC] = 0, i.e., a × b + b × c + c × a = 0. The last equation can be written in the form (b − a) × (c − a) = 0. Proposition 3. Let A(a), B(b), C(c), D(d) be four points, no three of which are collinear. Then ABC D if and only if (b − a) × (d − c) = 0. Proof. Choose the points M(m) and N (n) such that O AB M and OC D N are parallelograms; then m = b − a and n = d − c. Lines AB and C D are parallel if and only if points O, M, N are collinear. Using property 5, this is equivalent to 0 = m × n = (b − a) × (d − c). Problem 1. Points D and E lie on sides AB and AC of the triangle ABC such that AD AE 3 = = . AB AC 4 Consider points E and D on the rays (B E and (C D such that E E = 3B E and D D = 3C D. Prove that: 1) points D , A, E are collinear; 2) AD = AE . a + 3c a + 3b Solution. The points D, E, D , E have the coordinates: d = ,e= , 4 4 e = 4e − 3b = a + 3c − 3b and d = 4d − 3c = a + 3b − 3c, respectively.
4.2. The Complex Product of Two Complex Numbers
99
Figure 4.3.
1) Since (a − d ) × (e − d ) = (3c − 3b) × (6c − 6b) = 18(c − b) × (c − b) = 0, using Proposition 2 it follows that the points D , A, E are collinear. 2) Note that a − d 1 AD = , = D E e − d 2 so A is the midpoint of segment D E . Problem 2. Let ABC D E be a convex pentagon and let M, N , P, Q, X, Y be the midpoints of the segments BC, C D, D E, E A, M P, N Q, respectively. Prove that X Y AB. Solution. Let a, b, c, d, e be the coordinates of vertices A, B, C, D, E, respectively.
Figure 4.4.
Points M, N , P, Q, X, Y have the coordinates c+d b+c , n= , 2 2 e+a b+c+d +e q= , x= , 2 4 m=
d +e , 2 c+d +e+a y= , 4 p=
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4. More on Complex Numbers and Geometry
respectively. Then a−b 1 y−x = 4 = − ∈ R, b−a b−a 4 hence 1 (y − x) × (b − a) = − (b − a) × (b − a) = 0. 4 From Proposition 3 it follows that X Y AB.
4.3
The Area of a Convex Polygon
We say that the convex polygon A1 A2 · · · An is directly (or positively) oriented if for any point M situated in the interior of the polygon the triangles M Ak Ak+1 , k = 1, 2, . . . , n, are directly oriented, where An+1 = A1 . Theorem. Consider a directly oriented convex polygon A1 A2 · · · An with vertices with coordinates a1 , a2 , . . . , an . Then area[A1 A2 · · · An ] =
1 Im(a1 a2 + a2 a3 + · · · + an−1 an + an a1 ). 2
Proof. We use induction on n. The base case n = 3 was proved above using the complex product. Suppose that the claim holds for n = k and note that area[A1 A2 · · · Ak Ak+1 ] = area[A1 A2 · · · Ak ] + area[Ak Ak+1 A1 ] =
1 1 Im(a1 a2 + a2 a3 + · · · + ak−1 ak + ak a1 ) + Im(ak ak+1 + ak+1 a1 + a1 ak ) 2 2 = +
1 Im(a1 a2 + a2 a3 + · · · + ak−1 ak + ak ak+1 + ak+1 a1 ) 2
1 1 Im(ak a1 + a1 ak ) = Im(a1 a2 + a2 a3 + · · · + ak ak+1 + ak+1 a1 ), 2 2 since
Im(ak a1 + a1 ak ) = 0.
4.3. The Area of a Convex Polygon
101
Alternative proof. Choose a point M in the interior of the polygon. Applying the formula (2) in Subsection 3.5.3 we have area[A1 A2 · · · An ] =
n
area[M Ak Ak+1 ]
k=1
=
n 1
Im(zak + ak ak+1 + ak+1 z) 2 k=1
n n 1
1
Im(ak ak+1 ) + Im(zak + ak+1 z) 2 k=1 2 k=1 n n n n
1 1
1 ak ak+1 + Im z ak + z aj = ak ak+1 , = Im 2 2 2 k=1 k=1 k=1 j=1
=
since for any complex numbers z, w the relation Im(zw + zw) = 0 holds.
Remark. From the above formula it follows that the points A1 (a1 ), A2 (a2 ), . . . , An (an ) are collinear if and only if Im(a 1 a2 + a 2 a3 + · · · + a n−1 an + a n a1 ) = 0. Problem 1. Let P0 P1 · · · Pn−1 be the polygon whose vertices have coordinates 1, ε, . . . , ε n−1 and let Q 0 Q 1 · · · Q n−1 be the polygon whose vertices have coordinates 2π 2π 1, 1 + ε, . . . , 1 + ε + · · · + ε n−1 , where ε = cos + i sin . Find the ratio of the n n areas of these polygons. Solution. Consider ak = 1 + ε + · · · + ε k , k = 0, 1, . . . , n − 1, and observe that n−1 n−1
(ε)k+1 − 1 ε k+2 − 1 1 1 area[Q 0 Q 1 · · · Q n−1 ] = Im ak ak+1 Im · 2 2 ε−1 ε−1 k=0 k=0 n−1
1 = Im (ε − (ε)k+1 − ε k+2 + 1) 2|ε − 1|2 k=0 =
1 2π 1 Im(nε + n) = n sin 2 2 n 2|ε − 1| 2|ε − 1| n π n π π = π 2 sin n cos n = 4 cotan n , 8 sin2 n n−1
ε k+1 = 0 and since k=0
n−1
ε k+2 = 0.
k=0
On the other hand, it is clear that area[P0 P1 · · · Pn−1 ] = n area[P0 O P1 ] =
2π π π n sin = n sin cos . 2 n n n
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4. More on Complex Numbers and Geometry
We obtain π π n sin cos area[P0 P1 · · · Pn−1 ] π = n n π n = 4 sin2 . area[Q 0 Q 1 · · · Q n−1 ] n cotan 4 n
(1)
Remark. We have Q k Q k+1 = |ak+1 − ak | = |ε k+1 | = 1, and Pk Pk+1 = |εk+1 − π εk | = |ε k (ε − 1)| = |ε k ||1 − ε| = |1 − ε| = 2 sin , k = 0, 1, . . . , n − 1. It follows n that Pk Pk+1 π = 2 sin , k = 0, 1, . . . , n − 1. Q k Q k+1 n That is, the polygons P0 P1 · · · Pn−1 and Q 0 Q 1 · · · Q n−1 are similar and the result in (1) follows. Problem 2. Let A1 A2 · · · An (n ≥ 5) be a convex polygon and let Bk be the midpoint of the segment [Ak Ak+1 ], k = 1, 2, . . . , n, where An+1 = A1 . Then the following inequality holds: area[B1 B2 · · · Bn ] ≥
1 area[A1 A2 · · · An ]. 2
Solution. Let ak and bk be the coordinates of points Ak and Bk , k = 1, 2, . . . , n. It is clear that the polygon B1 B2 · · · Bn is convex and if we assume that A1 A2 · · · An is positively oriented, then B1 B2 · · · Bn also has this property. Choose as the origin O of the complex plane a point situated in the interior of polygon A1 A2 · · · An . 1 We have bk = (ak + ak+1 ), k = 1, 2, . . . , n, and 2 n n
1 1 area[B1 B2 · · · Bn ] = Im bk bk+1 = Im (ak + ak+1 )(ak+1 + ak+2 ) 2 8 k=1 k=1 n n n
1 1 1 ak ak+1 + Im ak+1 ak+2 + Im ak ak+2 = Im 8 8 8 k=1 k=1 k=1 n
1 1 ak ak+2 = area[A1 A2 · · · An ] + Im 2 8 k=1 = =
n 1
1 area[A1 A2 · · · An ] + Im(ak ak+2 ) 2 8 k=1
n 1
1 area[A1 A2 · · · An ] + O Ak · O Ak+2 sin Ak O Ak+2 2 8 k=1
≥
1 area[A1 A2 · · · An ]. 2
4.4. Intersecting Cevians and Some Important Points in a Triangle
103
We have used the relations n n
ak ak+1 = Im ak+1 ak+2 = 2 area[A1 A2 · · · An ], Im k=1
k=1
and sin Ak O Ak+2 ≥ 0, k = 1, 2, . . . , n, where An+2 = A2 .
4.4
Intersecting Cevians and Some Important Points in a Triangle
Proposition 1. Consider the points A , B , C on the sides BC, C A, AB of the triangle ABC such that A A , B B , CC intersect at point Q and let p B A = , AC n
C B m = , B A p
AC n = . C B m
If a, b, c are the coordinates of points A, B, C, respectively, then the coordinate of point Q is ma + nb + pc . q= m+n+ p nb + pc ma + pc ,b = and c = Proof. The coordinates of A , B , C are a = n+p m+p ma + nb ma + nb + pc , respectively. Let Q be the point with coordinate q = . We m+n m+n+ p prove that A A , B B , CC meet at Q. The points A, Q, A are collinear if and only if (q − a) × (a − a) = 0. This is equivalent to nb + pc ma + nb + pc −a × −a =0 m+n+ p n+p or (nb + pc − (n + p)a) × (nb + pc − (n + p)a) = 0, which is clear by definition of the complex product. Likewise, Q lies on lines B B and CC , so the proof is complete. Some important points in a triangle. 1) If Q = G, the centroid of the triangle ABC, we have m = n = p = 1. Then we obtain again that the coordinate of G is a+b+c . 3 2) Suppose that the lengths of the sides of triangle ABC are BC = α, C A = β, AB = γ . If Q = I , the incenter of triangle ABC, then, using the known result concerning the angle bisector, it follows that m = α, n = β, p = γ . Therefore the coordinate of I is αa + βb + γ c 1 = [αa + βb + γ c], zI = α+β +γ 2s zG =
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4. More on Complex Numbers and Geometry
1 where s = (α + β + γ ). 2 3) If Q = H , the orthocenter of the triangle ABC, we easily obtain the relations tan C B A = , A C tan B
C B tan A = , B A tan C
AC tan B = . C B tan A
It follows that m = tan A, n = tan B, p = tan C, and the coordinate of H is given by (tan A)a + (tan B)b + (tan C)c zH = . tan A + tan B + tan C Remark. The above formula can also be extended to the limiting case when the π triangle ABC is a right triangle. Indeed, assume that A → . Then tan A → ±∞ 2 (tan B)b + (tan C)c tan B + tan C and → 0, → 0. In this case z H = a, i.e., the tan A tan A orthocenter of triangle ABC is the vertex A. 4) The Gergonne1 point J is the intersection of the cevians A A , B B , CC , where A , B , C are the points of tangency of the incircle to the sides BC, C A, AB, respectively. Then 1 s−γ = , 1 A C s−β
B A
1 C B s − α, = 1 B A s−γ
1 AC s−β = , 1 C B s−α
and the coordinate z J is obtained from the same proposition, where zJ =
rα a + rβ b + rγ c . rα + rβ + rγ
Here rα , rβ , rγ denote the radii of the three excircles of triangle. It is not difficult to show that the following formulas hold: rα =
K , s−α
rβ =
K , s−β
rγ =
K , s−γ
1 (α + β + γ ). 2 5) The Lemoine2 point K is the intersection of the symmedians of the triangle (the symmedian is the reflection of the bisector across the median). Using the notation from
where K = area[ABC] and s =
1 Joseph-Diaz Gergonne (1771–1859), French mathematician, founded the journal Annales de Math´ema-
tiques Pures et Appliqu´ees in 1810. 2 Emile Michel Hyacinthe Lemoine (1840–1912), French mathematician, made important contributions to geometry.
4.4. Intersecting Cevians and Some Important Points in a Triangle
105
the proposition we obtain γ2 B A = , A C β2
α2 C B = , B A γ2
It follows that zK =
β2 AC = . C B α2
α2a + β 2b + γ 2c . α2 + β 2 + γ 2
6) The Nagel3 point N is the intersection of the cevians A A , B B , CC , where B , C are the points of tangency of the excircles with the sides BC, C A, AB, respectively. Then A ,
B A s−γ = , A C s−β
C B s−α = , B A s−γ
AC s−β = , C B s−α
and the proposition mentioned before gives the coordinate z N of the Nagel point N , zN =
1 (s − α)a + (s − β)b + (s − γ )c = [(s − α)a + (s − β)b + (s − γ )c] (s − α) + (s − β) + (s − γ ) s β γ α a+ 1− b+ 1− c. = 1− s s s
Problem. Let α, β, γ be the lengths of sides BC, C A, AB of triangle ABC and suppose α < β < γ . If points O, I, H are the circumcenter, the incenter and the orthocenter of the triangle ABC, respectively. Prove that area[O I H ] =
1 (α − β)(β − γ )(γ − α), 8r
where r is the inradius of ABC. Solution. Consider triangle ABC, directly oriented in the complex plane centered at point O. Using the complex product and the coordinates of I and H , we have ( 1 1 ' αa + βb + γ c area[O I H ] = (I × h) = × (a + b + c) 2i 2i α+β +γ 1 = [(α − β)a × b + (β − γ )b × c + (γ − α)c × a] 4si 1 = [(α − β) · area[O AB] + (β − γ ) · area[O BC] + (γ − α) · area[OC A]] 2s 1' R 2 sin 2C R 2 sin 2A R 2 sin 2B ( = (α − β) + (β − γ ) + (γ − α) 2s 2 2 2 3 Christian Heinrich von Nagel (1803–1882), German mathematician. His contributions to triangle ge-
ometry were included in the book The Development of Modern Triangle Geometry [13].
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4. More on Complex Numbers and Geometry
=
R2 [(α − β) sin 2C + (β − γ ) sin 2A + (γ − α) sin 2B] 4s 1 = (α − β)(β − γ )(γ − α), 8r
as desired.
4.5
The Nine-Point Circle of Euler
Given a triangle ABC, choose its circumcenter O to be the origin of the complex plane and let a, b, c be the coordinates of the vertices A, B, C. We have seen in Section 2.22, Proposition 3, that the coordinate of the orthocenter H is z H = a + b + c. Let us denote by A1 , B1 , C1 the midpoints of sides BC, C A, AB, by A , B , C the feet of the altitudes and by A , B , C the midpoints of segments AH , B H , C H , respectively.
Figure 4.5.
It is clear that for the points A1 , B1 , C1 , A , B , C we have the following coordinates: 1 1 1 z A1 = (b + c), z B1 = (c + a), z C1 = (a + b), 2 2 2 1 1 1 z A = a + (b + c), z B = b + (c + a), z C = c + (a + b). 2 2 2 It is not so easy to find the coordinates of A , B , C . Proposition 1. Consider the point X (x) in the plane of triangle ABC. Let P be the projection of X onto line BC. Then the coordinate of P is given by bc 1 x − 2x +b+c p= 2 R where R is the circumradius of triangle ABC.
4.5 The Nine-Point Circle of Euler
107
Proof. Using the complex product and the real product we can write the equations of lines BC and X P as follows: BC : (z − b) × (c − b) = 0, X P : (z − x) · (c − b) = 0. The coordinate p of P satisfies both equations; hence we have ( p − b) × (c − b) = 0
and
( p − x) · (c − b) = 0.
These equations are equivalent to ( p − b)(c − b) − ( p − b)(c − b) = 0 and ( p − x)(c − b) + ( p − x)(c − b) = 0. Adding the above relations we find (2 p − b − x)(c − b) + (b − x)(c − b) = 0. It follows that ⎡ ⎤ ⎥ 1 c−b c−b 1⎢ p= (x − b) = ⎢ b+x+ b+x+ 2 (x − b)⎥ ⎣ ⎦ 2 2 2 R R c−b − c b bc bc 1 1 b + x − 2 (x − b) = x − 2x +b+c . = 2 2 R R
From the above Proposition 1, the coordinates of A , B , C are 1 bca a+b+c− 2 , z A = 2 R 1 cab z B = a+b+c− 2 , 2 R 1 abc zC = a+b+c− 2 . 2 R Theorem 2. (The nine-point circle.) In any triangle ABC the points A1 , B1 , C1 , A , C , A , B , C are all on the same circle, whose center is at the midpoint of the segment O H , and the radius is one-half of the circumcircle. B,
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4. More on Complex Numbers and Geometry
Proof. Denote by O9 the midpoint of the segment O H . Using our initial assumption, 1 it follows that z O9 = (a + b + c). Also we have |a| = |b| = |c| = R, where R is the 2 circumradius of triangle ABC. 1 1 1 Observe that O9 A1 = |z A1 − z O9 | = |a| = R, and also O9 B1 = O9 C1 = R. 2 2 2 1 1 1 We can write O9 A = |z A − z O0 | = |a| = R, and also O0 B = O9 C = R. 2 2 2 The distance O9 A is also not difficult to compute: 1 1 bca O9 A = |z A − z O9 | = a + b + c − 2 − (a + b + c) 2 2 R =
1 1 R3 1 |bca| = |a||b||c| = = R. 2 2 2 2 2R 2R 2R
1 R. Therefore O9 A1 = O0 B1 = O9 C1 = 2 1 O9 A = O9 B = O9 C = O9 A = O9 B = O9 C = R and the desired property 2 follows.
Similarly, we get O9 B = O9 C =
Theorem 3. 1) (Euler4 line of a triangle.) In any triangle ABC the points O, G, H are collinear. 2) (Nagel line of a triangle.) In any triangle ABC the points I, G, N are collinear. Proof. 1) If the circumcenter O is the origin of the complex plane, we have z O = 0, 1 z G = (a + b + c), z H = a + b + c. Hence these points are collinear by Proposition 3 2 in Section 2.22. α 1 α β γ 2) We have z I = a + b + c, z G = (a + b + c), and z N = 1 − a+ 2s 2s 2s 3 s β γ 1− b+ 1− c and we can write z N = 3z G − 2z I . s s Applying the result mentioned above and properties of the complex product we obtain (z G − z I ) × (z N − z I ) = (z G − z I ) × [3(z G − z I )] = 0; hence the points I, H, N are collinear. Remark. Note that N G = 2G I , hence the triangles OG I and H G N are similar. It follows that the lines O I and N H are parallel and we have the following basic configuration of triangle ABC (in Figure 4.6): 4 Leonhard Euler (1707–1783), one of the most important mathematicians, created a good deal of anal-
ysis, and revised almost all the branches of pure mathematics which were then known, adding proofs, and arranging the whole in a consistent form. Euler wrote an immense number of memoirs on all kinds of mathematical subjects. We recommend William Dunham’s book Euler. The Master of Us All (The Mathematical Association of America, 1999) for more details concerning Euler’s contributions to mathematics.
4.5 The Nine-Point Circle of Euler
109
Figure 4.6.
If G s is the midpoint of segment [I N ], then its coordinate is zGs =
(γ + α) (α + β) 1 (β + γ ) (z I + z N ) = a+ b+ c. 2 4s 4s 4s
The point G s is called the Spiecker point of triangle ABC and it is easy to verify that it is the incenter of the medial triangle A1 B1 C1 . Problem 1. Consider a point M on the circumcircle of the triangle ABC. Prove that the nine-point centers of the triangles M BC, MC A, M AB are the vertices of a triangle similar to triangle ABC. Solution. Let A , B , C be the nine-point centers of the triangles M BC, MC D, M AB, respectively. Take the origin of the complex plane to be at the circumcenter of triangle ABC. Denote by a lowercase letter the coordinate of the point denoted by an uppercase letter. Then a =
m+b+c , 2
b =
m+c+a , 2
c =
m+a+b , 2
since M lies on the circumcircle of triangle ABC. Then b − a a−b b−a = = , c − a a−c c−a and hence triangles A B C and ABC are similar. Problem 2. Show that triangle ABC is a right triangle if and only if its circumcircle and its nine-point circle are tangent. Solution. Take the origin of the complex plane to be at circumcenter O of triangle ABC and denote by a, b, c the coordinates of vertices A, B, C, respectively. Then the
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4. More on Complex Numbers and Geometry
circumcircle of triangle ABC is tangent to the nine-point circle of triangle ABC if and R R2 , that is, |a + b + c|2 = R 2 . only if O O9 = . This is equivalent to O O92 = 2 4 Using properties of the real product, we have |a + b + c|2 = (a + b + c) · (a + b + c) = a 2 + b2 + c2 + 2(a · b + b · c + c · a) = 3R 2 + 2(a · b + b · c + c · a) = 3R 2 + (2R 2 − α 2 + 2R 2 − β 2 + 2R 2 − γ 2 ) = 9R 2 − (α 2 + β 2 + γ 2 ), where α, β, γ are the lengths of the sides of triangle ABC. We have used the formulas γ2 α2 β2 , b · c = R2 − , c · a = R2 − , which can be easily derived a · b = R2 − 2 2 2 from the definition of the real product of complex numbers (see also the lemma in Subsection 4.6.2). Therefore, α 2 + β 2 + γ 2 = 8R 2 , which is the same as sin2 A + sin2 B + sin2 C = 2. We can write the last relation as 1 − cos 2A + 1 − cos 2B + 1 − cos 2C = 4. This is equivalent to 2 cos(A + B) cos(A − B) + 2 cos2 C = 0, i.e., 4 cos A cos B cos C = 0, and the desired conclusion follows. Problem 3. Let ABC D be a cyclic quadrilateral and let E a , E b , E c , E d be the ninepoint centers of triangles BC D, C D A, D AB, ABC, respectively. Prove that the lines AE a , B E b , C E c , D E d are concurrent. Solution. Take the origin of the complex plane to be the center O of the circumcircle of ABC D. Then the coordinates of the nine-point centers are ea =
1 (b + c + d), 2
eb =
1 (c + d + a), 2
ec =
1 (d + a + b), 2
ed =
1 (a + b + c). 2
We have AE a : z = ka + (1 − k)ea , k ∈ R, and the analogous equations for the 1 lines B E b , C E c , D E d . Observe that the point with coordinate (a + b + c + d) lies 3 1 on all of the four lines k = , and we are done. 3
4.6 4.6.1
Some Important Distances in a Triangle Fundamental invariants of a triangle
1 (α + β + γ ), 2 the inradius r and the circumradius R. The numbers s, r, R are called the fundamental invariants of triangle ABC. Consider the triangle ABC with sides α, β, γ , the semiperimeter s =
Theorem 1. The sides α, β, γ are the roots of the cubic equation t 3 − 2st 2 + (s 2 + r 2 + 4Rr )t − 4s Rr = 0.
4.6. Some Important Distances in a Triangle
111
Proof. Let us prove that α satisfies the equation. We have A cos A A A 2, α = 2R sin A = 4R sin cos and s − α = r cotan = r A 2 2 2 sin 2 hence cos2
A A α(s − α) αr = and sin2 = . 2 4Rr 2 4R(s − α)
From the formula cos2
A A + sin2 = 1, it follows that 2 2 αr α(s − α) + = 1. 4Rr 4R(s − α)
That is, α 3 − 2sα 2 + (s 2 + r 2 + 4Rr )α − 4s Rr = 0. We can show analogously that β and γ are roots of the above equation. From the above theorem, by using the relations between the roots and the coefficients, it follows that α + β + γ = 2s, αβ + βγ + γ α = s 2 + r 2 + 4Rr, αβγ = 4s Rr. Corollary 2. In any triangle ABC, the following formulas hold: α 2 + β 2 + γ 2 = 2(s 2 − r 2 − 4Rr ), α 3 + β 3 + γ 3 = 2s(s 2 − 3r 2 − 6Rr ). Proof. We have α 2 + β 2 + γ 2 = (α + β + γ )2 − 2(αβ + βγ + γ α) = 4s 2 − 2(s 2 + r 2 + 4Rr ) = 2s 2 − 2r 2 − 8Rr = 2(s 2 − r 2 − 4Rr ). In order to prove the second identity, we can write α 3 + β 3 + γ 3 = (α + β + γ )(α 2 + β 2 + γ 2 − αβ − βγ − γ α) + 3αβγ = 2s(2s 2 − 2r 2 − 8Rr − s 2 − r 2 − 4Rr ) + 12s Rr = 2s(s 2 − 3r 2 − 6Rr ).
112
4.6.2
4. More on Complex Numbers and Geometry
The distance OI
Assume that the circumcenter O of the triangle ABC is the origin of the complex plane and let a, b, c be the coordinates of the vertices A, B, C, respectively. Lemma. The real products a · b, b · c, c · a are given by a · b = R2 −
γ2 , 2
b · c = R2 −
α2 , 2
c · a = R2 −
β2 . 2
Proof. Using the properties of the real product we have γ 2 = |a − b|2 = (a − b) · (a − b) = a 2 − 2a · b − b2 = 2R 2 − 2a · b,
and the first formula follows. Theorem 4. (Euler) The following formula holds: O I 2 = R 2 − 2Rr. Proof. The coordinate of the incenter is given by zI =
α β γ a+ b+ c 2s 2s 2s
so we can write O I = |z I | = 2
2
=
β γ β γ α α a+ b+ c · a+ b+ c 2s 2s 2s 2s 2s 2s
1 1
2 2 2 2 (α + β + γ )R + 2 (αβ)a · b. 4s 2 4s 2 cyc
Using the lemma above we find that OI2 =
=
1 2 2 γ2 2 2 2 2 (α + β + γ )R + αβ R − 2 4s 2 4s 2 cyc
1 1
1 (α + β + γ )2 R 2 − 2 αβγ 2 = R 2 − 2 αβγ (α + β + γ ) 2 4s 4s cyc 4s
1 αβγ K αβγ = R 2 − 2 · = R 2 − 2Rr, 2s 4K s where the well-known formulas = R2 −
R=
αβγ , 4K
are used. Here K is the area of triangle ABC.
r=
K , s
4.6. Some Important Distances in a Triangle
113
Corollary 5. (Euler’s inequality.) In any triangle ABC the following inequality holds: R ≥ 2r. We have equality if and only if the triangle ABC is equilateral. Proof. From Theorem 4 we have O I 2 = R(R − 2r ) ≥ 0, hence R ≥ 2r . The equality R − 2r = 0 holds if and only if O I 2 = 0, i.e., O = I . Therefore triangle ABC is equilateral.
4.6.3
The distance ON
Theorem 6. If N is the Nagel point of triangle ABC, then O N = R − 2r. Proof. The coordinate of the Nagel point of the triangle is given by β γ α a+ 1− b+ 1− c. zN = 1 − s s s Therefore
α 2 α β 1− 1− +2 1− a·b O N = |z N | = z N · z N = R s s s cyc cyc
γ2 α 2 α β 2 2 +2 1− 1− 1− R − =R s s s 2 cyc cyc 2 α+β +γ α β − 1− 1− γ2 = R2 3 − s s s cyc
α β 2 1− =R − 1− γ 2 = R 2 − E. s s cyc 2
2
2
To calculate E we note that
α+β 1
1
αβ E= 1− γ2 − (α + β)γ 2 + 2 αβγ 2 + 2 γ2 = s s cyc s s cyc cyc cyc =
cyc
γ2 −
1
2αβγ 1 3 αβγ K =− · (2s − γ )γ 2 + α2 + α +8 s cyc s s cyc 4K s cyc =−
cyc
α2 +
1 3 α + 8Rr. s cyc
Applying the formula in Corollary 2, we conclude that E = −2(s 2 − r 2 − 4Rr ) + 2(s 2 − 3r 2 − 6Rr ) + 8Rr = −4r 2 + 4Rr.
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4. More on Complex Numbers and Geometry
Hence O N 2 = R 2 − E = R 2 − 4Rr + 4r 2 = (R − 2r )2 and the desired formula is proved by Euler’s inequality. Theorem 7. (Feuerbach5 ) In any triangle the incircle and the nine-point circle of Euler are tangent. Proof. Using the configuration in Section 4.5 we observe that 1 GI G O9 = = . 2 GN GO
Figure 4.7.
Therefore triangles G I O9 and G N O are similar. It follows that the lines I O9 and 1 1 O N are parallel and I O9 = O N . Applying Theorem 6 we get I O9 = (R − 2r ) = 2 2 R − r = R9 − r , hence the incircle is tangent to the nine-point circle. 2 The point of tangency of these two circles is denoted by ϕ and is called the Feuerbach point of triangle.
4.6.4
The distance OH
Theorem 8. If H is the orthocenter of triangle ABC, then O H 2 = 9R 2 + 2r 2 + 8Rr − 2s 2 . Proof. Assuming that the circumcenter O is the origin of the complex plane, the coordinate of H is z H = a + b + c. 5 Karl Wilhelm Feuerbach (1800–1834), German geometer, published the result of Theorem 7 in 1822.
4.7. Distance between Two Points in the Plane of a Triangle
115
Using the real product we can write O H 2 = |z H |2 = z H · z H = (a + b + c) · (a + b + c)
|a|2 + 2 ab = 3R 2 + 2 a · b. = cyc
cyc
Applying the formulas in the lemma (p. 112) and then the first formula in Corollary 2, we obtain O H 2 = 3R 2 + 2
R2 −
cyc
γ2 = 9R 2 − (α 2 + β 2 + γ 2 ) 2
= 9R 2 − 2(s 2 − r 2 − 4Rr ) = 9R 2 + 2r 2 + 8Rr − 2s 2 .
Corollary 9. The following formulas hold: 2 8 2 1) OG 2 = R 2 + r 2 + Rr − s 2 ; 9 9 9 9 1 1 2) O O92 = R 2 + r 2 + 2Rr − s 2 . 4 2 2 Corollary 10. In any triangle ABC the inequality α 2 + β 2 + γ 2 ≤ 9R 2 is true. Equality holds if and only if the triangle is equilateral.
4.7 4.7.1
Distance between Two Points in the Plane of a Triangle Barycentric coordinates
Consider a triangle ABC and let α, β, γ be the lengths of sides BC, C A, AB, respectively. Proposition 1. Let a, b, c be the coordinates of vertices A, B, C and let P be a point in the plane of triangle. If z P is the coordinate of P, then there exist unique real numbers μa , μb , μc such that z P = μa a + μb b + μc c and μa + μb + μc = 1. Proof. Assume that P is in the interior of triangle ABC and consider the point A PA A B such that A P ∩ BC = {A }. Let k1 = , k2 = and observe that PA AC zP =
a + k 1 z A , 1 + k1
z A =
b + k2 c . 1 + k2
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4. More on Complex Numbers and Geometry
Hence in this case we can write zP =
k1 k2 1 k1 b+ c. a+ 1 + k1 (1 + k1 )(1 + k2 ) (1 + k1 )(1 + k2 )
Moreover, if we consider μa =
1 , 1 + k1
μb =
k1 , (1 + k1 )(1 + k2 )
μc =
k1 k2 (1 + k1 )(1 + k2 )
we have μa + μb + μc =
1 k1 k1 k2 + + 1 + k1 (1 + k1 )(1 + k2 ) (1 + k1 )(1 + k2 ) =
1 + k1 + k2 + k1 k2 = 1. (1 + k1 )(1 + k2 )
We proceed in an analogous way in the case when the point P is situated in the exterior of triangle ABC. If the point P is situated on the support line of a side of triangle ABC (i.e., the line determined by two vertices) zP = where k =
1 k 1 k b+ c =0·a+ b+ c, 1+k 1+k 1+k 1+k
PB . PC
The real numbers μa , μb , μc are called the absolute barycentric coordinates of P with respect to the triangle ABC. The signs of numbers μa , μb , μc depend on the regions of the plane where the point P is situated. Triangle ABC determines seven such regions.
Figure 4.8.
In the next table we give the signs of μa , μb , μc :
4.7. Distance between Two Points in the Plane of a Triangle
I − + +
μa μb μc
4.7.2
II + − +
III + + −
IV + − −
V − + −
VI − − +
117
VII + + +
Distance between two points in barycentric coordinates
In what follows, in order to simplify the formulas, we will use the symbol called “cyclic / f (x1 , x2 , . . . , xn ), the sum of terms considered in the cyclic order. sum.” That is, cyc
The most important example for our purposes is
f (x1 , x2 , x3 ) = f (x1 , x2 , x3 ) + f (x2 , x3 , x1 ) + f (x3 , x1 , x2 ). cyc
Theorem 2. In the plane of triangle ABC consider the points P1 and P2 with coordinates z P1 and z P2 , respectively. If z Pk = αk a + βk b + γk c, where αk , βk , γk are real numbers such that αk + βk + γk = 1, k = 1, 2, then
P1 P22 = − (α2 − α1 )(β2 − β1 )γ 2 . cyc
Proof. Choose the origin of the complex plane at the circumcenter O of the triangle ABC. Using properties of the real product, we have P1 P22 = |z P2 − z P1 |2 = |(α2 − α1 )a + (β2 − β1 )b + (γ2 − γ1 )c|2
= (α2 − α1 )2 a · a + 2 (α2 − α1 )(β2 − β1 )a · b cyc
cyc
γ2 (α2 − α1 )2 R 2 + 2 (α2 − α1 )(β2 − β1 ) R 2 − 2 cyc cyc
= R 2 (α2 + β2 + γ2 − α1 − β1 − γ1 )2 − (α2 − α1 )(β2 − β1 )γ 2 =
=−
cyc
(α2 − α1 )(β2 − β1 )γ 2 ,
cyc
since α1 + β1 + γ1 = α2 + β2 + γ2 = 1.
Theorem 3. The points A1 , A2 , B1 , B2 , C1 , C2 are situated on the sides BC, C A, AB of triangle ABC such that lines A A1 , B B1 , CC1 meet at point P1 and lines A A2 , B B2 , CC2 meet at point P2 . If pk B Ak , = Ak C nk
C Bk mk , = Bk A pk
ACk nk , = Ck B mk
k = 1, 2
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4. More on Complex Numbers and Geometry
where m k , n k , pk are nonzero real numbers, k = 1, 2, and Sk = m k +n k + pk , k = 1, 2, then
1 2 2 2 2 2 2 P1 P2 = 2 2 S1 S2 (n 1 p2 + p1 n 2 )α − S1 n 2 p2 α − S2 n 1 p1 α . S1 S2 cyc cyc cyc Proof. The coordinates of points P1 and P2 are z Pk =
m k a + n k b + pk c , m k + n k + pk
k = 1, 2.
It follows that in this case the absolute barycentric coordinates of points P1 and P2 are given by mk mk nk nk = , βk = = , αk = m k + n k + pk Sk m k + n k + pk Sk pk pk = , k = 1, 2. m k + n k + pk Sk Substituting in the formula in Theorem 2 we find
n2 p2 n1 p1 α2 − − P1 P22 = − S S S S 2 1 2 1 cyc 1
=− 2 2 (S1 n 2 − S2 n 1 )(S1 p2 − S2 p1 )α 2 S1 S2 cyc 1 2 =− 2 2 [S1 n 2 p2 + S22 n 1 p1 − S1 S2 (n 1 p2 + n 2 p1 )]α 2 S1 S2 cyc
1 2 2 2 2 2 = 2 2 S1 S2 (n 1 p2 + p1 n 2 )α − S1 n 2 p2 α − S2 n 1 p1 α S1 S2 cyc cyc cyc γk =
and the desired formula follows.
Corollary 4. For any real numbers αk , βk , γk with αk + βk + γk = 1, k = 1, 2, the following inequality holds:
(α2 − α1 )(β2 − β1 )γ 2 ≤ 0, cyc
with equality if and only if α1 = α2 , β1 = β2 , γ1 = γ2 . Corollary 5. For any nonzero real numbers m k , n k , pk , k = 1, 2, with Sk = m k + n k + pk , k = 1, 2, the lengths of sides α, β, γ of triangle ABC satisfy the inequality
S1
S2
(n 1 p2 + p1 n 2 )2 ≥ n 2 p2 α 2 + n 1 p1 α 2 S2 cyc S1 cyc cyc with equality if and only if
p1 p2 m 1 m 2 n1 n2 = , = , = . n1 n 2 p1 p2 m 1 m2
4.8. The Area of a Triangle in Barycentric Coordinates
119
Applications. 1) Let us use the formula in Theorem 3 to compute the distance G I , where G is the centroid and I is the incenter of the triangle. We have m 1 = n 1 = p1 = 1 and m 2 = α, n 2 = β, p2 = γ ; hence
S1 = m 1 = 3; S2 = m 2 = α + β + γ = 2s; cyc
cyc
(n 1 p2 + n 2 p1 )α 2 = (β + γ )α 2 + (γ + α)β 2 + (α + β)γ 2
cyc
= (α + β + γ )(αβ + βγ + γ α) − 3αβγ = 2s(s 2 + r 2 + 4r R) − 12s Rr = 2s 3 + 2sr 2 − 4s Rr. On the other hand,
n 2 p2 α 2 = α 2 βγ + β 2 γ α + γ 2 αβ = αβγ (α + β + γ ) = 8s 2 Rr cyc
and
n 1 p1 α 2 = α 2 + β 2 + γ 2 = 2s 2 − 2r 2 − 8Rr.
cyc
Then
1 2 (s + 5r 2 − 16Rr ). 9 2) Let us prove that in any triangle ABC with sides α, β, γ , the following inequality holds:
(2α − β − γ )(2β − α − γ )γ 2 ≤ 0. GI2 =
cyc
In the inequality in Corollary 4 we consider the points P1 = G and P2 = I . Then 1 α β γ and α2 = , β2 = , γ2 = , and the above inequality α1 = β1 = γ1 = 3 2s 2s 2s follows. We have equality if and only if P1 = P2 ; that is, G = I , so the triangle is equilateral.
4.8
The Area of a Triangle in Barycentric Coordinates
Consider the triangle ABC with a, b, c the coordinates of its vertices, respectively. Let α, β, γ be the lengths of sides BC, C A and AB. Theorem 1. Let P j (z P j ), j = 1, 2, 3, be three points in the plane of triangle ABC with z P j = α j a + β j b + γ j c, where α j , β j , γ j are the barycentric coordinates of P j . If the triangles ABC and P1 P2 P3 have the same orientation, then α1 β1 γ1 area[P1 P2 P3 ] = α2 β2 γ2 . area[ABC] α3 β3 γ3
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4. More on Complex Numbers and Geometry
Proof. Suppose that the triangles ABC and P1 P2 P3 are positively oriented. If O denotes the origin of the complex plane, then using the complex product we can write 2i area[P1 O P2 ] = z P1 × z P2 = (α1 a + β1 b + γ1 c) × (α2 a + β2 b + γ2 c) = (α1 β2 − α2 β1 )a × b + (β1 γ2 − β2 γ1 )b × c + (γ1 α2 − γ2 α1 )c × a a × b b × c c × a a × b b × c 2i area[ABC] = γ1 α1 β1 = γ1 α1 1 . γ2 γ2 α2 β2 α2 1 Analogously, we find a×b 2i area[P2 O P3 ] = γ2 γ3
b×c α2 α3
2i area[ABC] 1 1
,
a×b 2i area[P3 O P1 ] = γ3 γ1
b×c α3 α1
2i area[ABC] 1 1
.
Assuming that the origin O is situated in the interior of triangle P1 P2 P3 , it follows that area[P1 P2 P3 ] = area[P1 O P2 ] + area[P2 O P3 ] + area[P3 O P1 ] =
1 1 (α1 − α2 + α2 − α3 + α3 − α1 )a × b − (γ1 − γ2 + γ2 − γ3 + γ3 − γ1 )b × c 2i 2i + (γ1 α2 − γ2 α1 + γ2 α3 − γ3 α2 + γ3 α1 − γ1 α3 )area[ABC] = (γ1 α2 − γ2 α1 + γ2 α3 − γ3 α2 + γ3 α1 − γ1 α3 )area[ABC] α1 β1 γ1 1 γ1 α1 = area[ABC] 1 γ2 α2 = area[ABC] α2 β2 γ2 α3 β3 γ3 1 γ3 α 3
and the desired formula is obtained.
Corollary 2. Consider the triangle ABC and the points A1 , B1 , C1 situated on the lines BC, C A, AB, respectively, such that A1 B = k1 , A1 C
B1 C = k2 , B1 A
C1 A = k3 . C1 B
If A A1 ∩ B B1 = {P1 }, B B1 ∩ CC1 = {P2 } and CC1 ∩ A A1 = {P3 }, then area[P1 P2 P3 ] (1 − k1 k2 k3 )2 = . area[ABC] (1 + k1 + k1 k2 )(1 + k2 + k2 k3 )(1 + k3 + k3 k1 )
4.8. The Area of a Triangle in Barycentric Coordinates
121
Figure 4.9.
Proof. Applying Menelaus’s well-known theorem in triangle A A1 B we find that C1 A C B P3 A1 · · = 1. C1 B C A1 P3 A Hence
C1 A C B P3 A = = k3 (1 + k1 ). · P3 A1 C 1 B C A1 The coordinate of P3 is given by
z P3 =
a + k3 (1 + k1 )z A1 = 1 + k3 (1 + k1 )
b + k1 c a + k3 b + k3 k1 c 1 + k1 = . 1 + k3 + k3 k1 1 + k3 + k3 k1
a + k3 (1 + k1 )
In an analogous way we find that z P1 =
k1 k2 a + b + k1 c 1 + k1 + k1 k2
and
z P2 =
k2 a + k2 k3 b + c . 1 + k2 + k2 k3
The triangles ABC and P1 P2 P3 have the same orientation; hence by applying the formula in Theorem 1 we find that area[P1 P2 P3 ] area[ABC]
k1 k2 1 = k2 (1 + k1 + k1 k2 )(1 + k2 + k2 k3 )(1 + k3 + k3 k1 ) 1 =
(1 − k1 k2 k3 )2 . (1 + k1 + k1 k2 )(1 + k2 + k2 k3 )(1 + k3 + k3 k1 )
1 k2 k3 k3
k1 1 k3 k1
Remark. When k1 = k2 = k3 = k, from Corollary 2 we obtain Problem 3 from the Putnam Mathematical Competition.
23r d
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4. More on Complex Numbers and Geometry
Let A j , B j , C j be points on the lines BC, C A, AB, respectively, such that BAj pj = , AjC nj
mj C Bj = , Bj A pj
nj AC j = , Cj B mj
j = 1, 2, 3.
Corollary 3. If P j is the intersection point of lines A A j , B B j , CC j , j = 1, 2, 3, and the triangles ABC, P1 P2 P3 have the same orientation, then m 1 n 1 p1 1 area[P1 P2 P3 ] = m 2 n 2 p2 area[ABC] S1 S2 S3 m 3 n 3 p3 where S j = m j + n j + p j , j = 1, 2, 3. Proof. In terms of the coordinates of the triangle, the coordinates of the points P j are m ja + n jb + pjc 1 z Pj = = (m j a + n j b + p j c), j = 1, 2, 3. m j + n j + pj Sj The formula above follows directly from Theorem 1. Corollary 4. In triangle ABC let us consider the cevians A A , B B and CC such that A B B C C A = m, = n, = p. A C B A C B Then the following formula holds: area[A B C ] 1 + mnp = . area[ABC] (1 + m)(1 + n)(1 + p) Proof. Observe that the coordinates of A , B , C are given by z A =
1 m b+ c, 1+m 1+m
z B =
1 n c+ a, 1+n 1+n
zC =
1 p a+ b. 1+ p 1+ p
Applying the formula in Corollary 3 we obtain
0 1 = n area[ABC] (1 + m)(1 + n)(1 + p) 1
area[A B C ]
=
1 + mnp . (1 + m)(1 + n)(1 + p)
1 0 p
m 1 0
Applications. 1) (Steinhaus6 ) Let A j , B j , C j be points on lines BC, C A, AB, respectively, j = 1, 2, 3. Assume that 2 B A1 = , A1 C 4
C B1 1 = , B1 A 2
AC1 4 = ; C1 B 1
6 Hugo Dyonizy Steinhaus (1887–1972), Polish mathematician, made important contributions in func-
tional analysis and other branches of modern mathematics.
4.8. The Area of a Triangle in Barycentric Coordinates
4 B A2 = , A2 C 1 1 B A3 = , A3 C 2
C B2 2 = , B2 A 4 C B3 4 = , B3 A 1
123
AC2 1 = ; C2 B 2 AC3 2 = . C3 B 4
If P j is the intersection point of lines A A j , B B j , CC j , j = 1, 2, 3, and triangles ABC, P1 P2 P3 are of the same orientation, then from Corollary 3 we obtain 1 4 2 49 1 1 area[P1 P2 P3 ] = 2 1 4 = 3 = . 7 area[ABC] 7·7·7 7 4 2 1 2) If the cevians A A , B B , CC are concurrent at point P, let us denote by K P the area of triangle A B C . We can use the formula in Corollary 4 to compute the areas of some triangles determined by the feet of the cevians of some remarkable points in a triangle. (i) If I is the incenter of triangle ABC we have γ 1+ · β KI = γ 1+ 1+ β =
β α · α γ area[ABC] β α 1+ α γ
2αβγ 2αβγ sr area[ABC] = . (α + β)(β + γ )(γ + α) (α + β)(β + γ )(γ + α)
(ii) For the orthocenter H of the acute triangle ABC we obtain tan C tan B tan A · · tan C tan B tan A area[ABC] = tan C tan B tan A 1+ 1+ 1+ tan B tan A tan C 1+
KH
= (2 cos A cos B cos C)area[ABC] = (2 cos A cos B cos C)sr. (iii) For the Nagel point of triangle ABC we can write s−γ · s−β = s−γ 1+ 1+ s−β 1+
KN
=
s−α s−β · s−γ s−α area[ABC] s−α s−β 1+ s−γ s−α
2(s − α)(s − β)(s − γ ) 4area2 [ABC] area[ABC] = area[ABC] αβγ 2sαβγ =
sr 2 r area[ABC] = . 2R 2R
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4. More on Complex Numbers and Geometry
If we proceed in the same way for the Gergonne point J we find the relation KJ =
r sr 2 area[ABC] = . 2R 2R
Remark. Two cevians A A and A A are isotomic if the points A and A are symmetric with respect to the midpoint of the segment BC. Assuming that B C C A A B = m, = n, = p, A C B A C B then for the corresponding isotomic cevians we have 1 A B = , A C m
B C 1 = , B A n
C A 1 = . C B p
Applying the formula in Corollary 4, it follows that 1 + mnp area[A B C ] = area[ABC] (1 + m)(1 + n)(1 + p) 1 1+ area[A B C ] mnp = = . 1 1 1 area[ABC] 1+ 1+ 1+ m n p Therefore area[A B C ] = area[A B C ]. A special case of this relation is K N = K J , since the points N and J are isotomic (i.e., these points are intersections of isotomic cevians). 3) Consider the excenters Iα , Iβ , Iγ of triangle ABC. It is not difficult to see that the coordinates of these points are z Iα = −
β γ α a+ b+ c, 2(s − α) 2(s − β) 2(s − γ )
α β γ a− b+ c, 2(s − α) 2(s − β) 2(s − γ ) α β γ z Iγ = a+ b− c. 2(s − α) 2(s − β) 2(s − γ ) From the formula in Theorem 1, it follows that β γ α − 2(s − α) 2(s − β) 2(s − γ ) α β γ area[ABC] − area[Iα Iβ Iγ ] = 2(s − α) 2(s − β) 2(s − γ ) α β γ − 2(s − α) 2(s − β) 2(s − γ ) z Iβ =
4.9. Orthopolar Triangles
−1 αβγ = 1 8(s − α)(s − β)(s − γ ) 1 =
1 −1 1
1 1 −1
125
area[ABC]
sαβγ area[ABC] sαβγ area[ABC] 2sαβγ = = 2s R. = 2s(s − α)(s − β)(s − γ ) 4 area[ABC] 2 area2 [ABC]
4) (Nagel line.) Using the formula in Theorem 1, we give a different proof for the socalled Nagel line: the points I, G, N are collinear. We have seen that the coordinates of these points are α β γ z I = a + b + c, 2s 2s 2s 1 1 1 a + b + c, 3 3 3 β γ α a+ 1− b+ 1− c. = 1− s s s zG =
zN Then
α 2s 1 area[I G N ] = 3 α 1− s
β 2s 1 3 1−
1 · area[ABC] = 0, 3 γ 1− s γ 2s
β s
hence the points I, G, N are collinear.
4.9 4.9.1
Orthopolar Triangles The Simson–Wallance line and the pedal triangle
Consider the triangle ABC, and let M be a point situated in the triangle plane. Let P, Q, R be the projections of M onto lines BC, C A, AB, respectively. Theorem 1. (The Simson7 line8 ) The points P, Q, R are collinear if and only if M is on the circumcircle of triangle ABC. 7 Robert Simson (1687–1768), Scottish mathematician. 8 This line was attributed to Simson by Poncelet, but is now frequently known as the Simson–Wallance
line since it does not actually appear in any work of Simson. William Wallance (1768–1843) was also a Scottish mathematician, who possibly published the theorem above concerning the Simson line in 1799.
126
4. More on Complex Numbers and Geometry
Figure 4.10.
Proof. We will give a standard geometric argument. Suppose that M lies on the circumcircle of triangle ABC. Without loss of generality
we may assume that M is on the arc BC. In order to prove the collinearity of R, P, Q, it suffices to show that the angles B P R and C P Q are congruent. The quadrilaterals P R B M and PC Q M are cyclic (since BRM ≡ B P M and M PC + M QC = 180◦ ), hence we have B PR ≡ B M R and C PQ ≡ C M Q. But B M R = 90◦ − AB M = ◦ 90 − MC Q, since the quadrilateral AB MC is cyclic too. Finally, we obtain BMR = 90◦ − MC Q=C M Q, so the angles B P R and C P Q are congruent. To prove the converse, we note that if the points P, Q, R are collinear, then the angles B P R and C P Q are congruent, hence AB M + AC M = 180◦ , i.e., the quadrilateral AB MC is cyclic. Therefore the point M is situated on the circumcircle of triangles ABC. When M lies on the circumcircle of triangle ABC, the line in the above theorem is called the Simson–Wallance line of M with respect to triangle ABC. We continue with a nice generalization of the property contained in Theorem 1. For an arbitrary point X in the plane of triangle ABC consider its projections P, Q and R on the lines BC, C A and AB, respectively. The triangle PQR is called the pedal triangle of point X with respect to the triangle ABC. Let us choose the circumcenter O of triangle ABC as the origin of the complex plane. Theorem 2. The area of the pedal triangle of X with respect to the triangle ABC is given by area[ABC] |x x − R 2 | (1) area[P Q R] = 4R 2
4.9. Orthopolar Triangles
127
Figure 4.11.
where R is the circumradius of triangle ABC. Proof. Applying the formula in Proposition 1, Section 4.5, we obtain the coordinates p, q, r of the points P, Q, R, respectively: 1 bc p= x − 2x +b+c , 2 R 1 x− 2 1 r= x− 2 q=
ca x + c + a , R2 ab x + a + b . R2
Taking into account the formula in Section 2.5.3 we have p p 1 i i q − p area[P Q R] = q q 1 = 4 r−p 4 r r 1
q−p r−p
.
For the coordinates p, q, r we obtain bc 1 p= x − 2x +b+c , 2 R ca 1 x − 2x +c+a , 2 R ab 1 r= x − 2x +a+b . 2 R q=
It follows that
1 cx bx 1 q − p = (a − b) 1 − 2 and r − p = (a − c) 1 − 2 , 2 2 R R
(2)
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4. More on Complex Numbers and Geometry
q−p= Therefore
1 (a − b)(x − c)R 2 2abc
and r − p =
1 (a − c)(x − b)R 2 . 2abc
i q − p area[P Q R] = 4 r−p
=
= = =
q − p r−p cx 1 − 2 (x − c)R 2 R i(a − b)(a − c) 16abc bx 2 1 − 2 (x − b)R R i(a − b)(a − c) R 2 − cx x − c 2 R − bx x − b 16abc i(a − b)(a − c) (b − c)x b − c 2 R − bx x − b 16abc i(a − b)(b − c)(a − c) x 1 2 R − bx x − b 16abc
i(a − b)(b − c)(a − c) (x x − R 2 ). 16abc Proceeding to moduli we find that =
area[P Q R] =
αβγ |a − b||b − c||c − a| |x x − R 2 | |x x − R 2 | = 16|a||b||c| 16R 3
area[ABC] |x x − R 2 |, 4R 2 where α, β, γ are the length of sides of triangle ABC. =
Remarks. 1) The formula in Theorem 2 contains the Simson–Wallance line property. Indeed, points P, Q, R are collinear if and only if area[P Q R] = 0. That is, |x x − R 2 | = 0, i.e., x x = R 2 . It follows that |x| = R, so X lies on the circumcircle of triangle ABC. 2) If X lies on a circle of radius R1 and center O (the circumcenter of triangle ABC), then x x = R12 , and from Theorem 2 we obtain area[ABC] 2 |R1 − R 2 |. 4R 2 It follows that the area of triangle P Q R does not depend on the point X . The converse is also true. The locus of all points X in the plane of triangle ABC such that area[P Q R] = k (constant) is defined by area[P Q R] =
|x x − R 2 | =
4R 2 k . area[ABC]
4.9. Orthopolar Triangles
129
This is equivalent to |x|2 = R 2 ±
4k 4R 2 k = R2 1 ± . area[ABC] area[ABC]
1 If k > area[ABC], then the locus is a circle of center O and radius R1 = 4 4k . R 1+ area[ABC] 1 If k ≤ area[ABC], then the locus consists of two circles of center O and radii 4 1 4k , one of which degenerated to O when k = area[ABC]. R 1± area[ABC] 4 Theorem 3. For any point X in the plane of triangle ABC, we can construct a triangle with sides AX · BC, B X · C A, C X · AB. This triangle is then similar to the pedal triangle of point X with respect to the triangle ABC. Proof. Let P Q R be the pedal triangle of X with respect to triangle ABC. From formula (2) we obtain q−p=
R 2 − cx 1 (a − b)(x − c) 2 . 2 R (x − c)
(3)
Proceeding to moduli in (3), it follows that R 2 − cx 1 |a − b||x − c| |q − p| = . x −c 2R 2
(4)
On the other hand, R 2 − cx 2 R 2 − cx R 2 − cx R 2 − cx R 2 − cx · = · = x −c x −c x −c x −c R2 x− c R 2 − cx R 2 (c − x) = R2, · x −c cx − R 2 hence from (4) we derive the relation =
|q − p| =
1 |a − b||x − c|. 2R
(5)
Therefore
PQ QR RP 1 = = = , C X · AB AX · BC BX · C A 2R and the conclusion follows.
(6)
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4. More on Complex Numbers and Geometry
Corollary 4. In the plane of triangle ABC consider the point X and denote by the triangle with sides AX · BC, B X · C A, C X · AB. Then
A B C
area[A B C ] = area[ABC]|x x − R 2 |.
(7)
Proof. From formula (6) it follows that area[A B C ] = 4R 2 area[P Q R], where P Q R is the pedal triangle of X with respect to triangle ABC. Replacing this result in (1), we find the desired formula. Corollary 5. (Ptolemy’s inequality) For any quadrilateral ABC D the following inequality holds: AC · B D ≤ AB · C D + BC · AD. (8) Corollary 6. (Ptolemy’s theorem) The convex quadrilateral ABC D is cyclic if and only if AC · B D = AB · C D + BC · AD. (9) Proof. If the relation (9) holds, then triangle A B C in Corollary 4 is degenerate; i.e., area[A B C ] = 0. From formula (7) it follows that d · d = R 2 , where d is the coordinate of D and R is the circumradius of triangle ABC. Hence the point D lies on the circumcircle of triangle ABC. If quadrilateral ABC D is cyclic, then the pedal triangle of point D with respect to triangle ABC is degenerate. From (6) we obtain the relation (9). Corollary 7. (Pompeiu’s Theorem9 ) For any point X in the plane of the equilateral triangle ABC, three segments with lengths X A, X B, XC can be taken as the sides of a triangle. Proof. In Theorem 3 we have BC = C A = AB and the desired conclusion follows. The triangle in Corollary 7 is called the Pompeiu triangle of X with respect to the equilateral triangle ABC. This triangle is degenerate if and only if X lies on the circumcircle of ABC. Using the second part of Theorem 3 we find that Pompeiu’s triangle of point X is similar to the pedal triangle of X with respect to triangle ABC and √ AX BX 2R 2 3 CX = = = = . (10) PQ QR RP α 3 Problem 1. Let A, B and C be equidistant points on the circumference of a circle of unit radius centered at O, and let X be any point in the circle’s interior. Let d A , d B , dC be the distances from X to A, B, C, respectively. Show that there is a triangle with 9 Dimitrie Pompeiu (1873–1954), Romanian mathematician, made important contributions in the fields
of mathematical analysis, functions of a complex variable, and rational mechanics.
4.9. Orthopolar Triangles
131
sides d A , d B , dC , and the area of this triangle depends only on the distance from X to O. (2003 Putnam Mathematical Competition)
Solution. The first assertion is just the property contained in Corollary 7. Taking into account the relations (10), it follows that the area of Pompeiu’s triangle of point X is 2 area[P Q R]. From Theorem 2 we get that area[P Q R] depends only on the distance 3 from P to O, as desired. Problem 2. Let X be a point in the plane of the equilateral triangle ABC such that X does not lie on the circumcircle of triangle ABC, and let X A = u, X B = v, XC = w. Express the length side α of triangle ABC in terms of real numbers u, v, w. (1978 GDR Mathematical Olympiad)
Solution. The segments [X A], [X B], [XC] are the sides of Pompeiu’s triangle of point X with respect to equilateral triangle ABC. Denote this triangle by A B C . From relations (10) and from Theorem 2 it follows that √ 2 2 3 1 area[A B C ] = area[P Q R] = area[ABC]|x · x − R 2 | 3 3R 2 √ √ 1 3 α2 3 2 = |x · x − R | = |X O 2 − R 2 |. · (11) 2 4 4 3R On the other hand, using the well-known formula of Hero we obtain, after a few simple computations: area[A B C ] =
1 2 (u + v 2 + w 2 )2 − 2(u 4 + v 4 + w 4 ). 4
Substituting in (11) we find 1 |X O 2 − R 2 | = √ (u 2 + v 2 + w 2 )2 − 2(u 4 + v 4 + w 4 ). 3
(12)
Now we consider the following two cases: Case 1. If X lies in the interior of the circumcircle of triangle ABC, then X O 2 < R 2 . Using the relation (see also formula (4) in Section 4.11) X O2 =
1 2 (u + v 2 + w 2 − 3R 2 ), 3
from (12) we find that 2R 2 =
1 2 1 (u + v 2 + w 2 ) + √ (u 2 + v 2 + w 2 )2 − 2(u 4 + v 4 + w 4 ), 3 3
132
hence
4. More on Complex Numbers and Geometry
√ 3 2 1 2 2 2 α = (u + v + w ) + (u + v 2 + w 2 )2 − 2(u 4 + v 4 + w 4 ). 2 2 2
Case 2. If X lies in the exterior of circumcircle of triangle ABC, then X O 2 > R 2 and after some similar computations we find √ 3 2 1 2 2 2 2 (u + v 2 + w 2 )2 − 2(u 4 + v 4 + w 4 ). α = (u + v + w ) − 2 2
4.9.2
Necessary and sufficient conditions for orthopolarity
Consider a triangle ABC and points X, Y, Z situated on its circumcircle. Triangles ABC and X Y Z are called orthopolar triangles (or S-triangles)10 if the Simson– Wallance line of point X with respect to triangle ABC is perpendicular (orthogonal) to line Y Z . Let us choose the circumcenter O of triangle ABC at the origin of the complex plane. Points A, B, C, X, Y, Z have the coordinates a, b, c, x, y, z with |a| = |b| = |c| = |x| = |y| = |z| = R, where R is the circumradius of the triangle ABC. Theorem 3. Triangles ABC and X Y Z are orthopolar triangles if and only if abc = x yz. Proof. Let P, Q, R be the feet of the orthogonal lines from the point X to the lines BC, C A, AB, respectively. Points P, Q, R are on the same line; that is, the Simson–Wallance line of point X with respect to triangle ABC. The coordinates of P, Q, R are denoted by p, q, r , respectively. Using the formula in Proposition 1, Section 4.5, we have 1 bc p= x − 2x +b+c 2 R ca 1 x − 2x +c+a , q= 2 R 1 ab r= x − 2x +a +b . 2 R We study two cases. 10 This definition was given in 1915 by Romanian mathematician Traian Lalescu (1882–1929). He is
famous for his book La g´eometrie du triangle published by Librairie Vuibert, Paris, 1937.
4.9. Orthopolar Triangles
133
Case 1. Point X is not a vertex of triangle ABC. Then P Q is orthogonal to Y Z if and only if ( p − q) · (y − z) = 0. That is, cx (b − a) 1 − 2 · (y − z) = 0 R or (b − a)(R 2 − cx)(y − z) + (b − a)(R 2 − cx)(y − z) = 0. We obtain R2 R2 R2 R2 R2 R2 2 2 − R − x (y − z) + (b − a) R − c − = 0, b a c x y z hence
1 1 (a − b)(c − x)(y − z) − (a − b)(c − x)(y − z) = 0. abc x yz
The last relation is equivalent to (abc − x yz)(a − b)(c − x)(y − z) = 0 and finally we get abc = x yz, as desired. Case 2. Point X is a vertex of triangle ABC. Without loss of generality, assume that X = B. Then the Simson–Wallance line of point X = B is the orthogonal line from B to AC. It follows that B Q is orthogonal to Y Z if and only if lines AC and Y Z are parallel. This is equivalent to ac = yz. Because b = x, we obtain abc = x yz, as desired. Remark. Due to the symmetry of the relation abc = x yz, we observe that the Simson–Wallance line of any vertex of triangle X Y Z with respect to ABC is orthogonal to the opposite side of the triangle X Y Z . Moreover, the same property holds for the vertices of triangle ABC. Hence ABC and X Y Z are orthopolar triangles if and only if X Y Z and ABC are orthopolar triangles. Therefore the orthopolarity relation is symmetric. Problem 1. The median and the orthic triangles of a triangle ABC are orthopolar in the nine-point circle. Solution. Consider the origin of the complex plane at the circumcenter O of triangle ABC. Let M, N , P be the midpoints of AB, BC, C A and let A , B , C be the feet of the altitudes of triangles ABC from A, B, C, respectively. If m, n, p, a , b , c are coordinates of M, N , P, A , B , C then we have m=
1 (a + b), 2
n=
1 (b + c), 2
p=
1 (c + a) 2
134
and
4. More on Complex Numbers and Geometry
1 1 bc bc a = a + b + c − 2a = a+b+c− , 2 2 a R ca ab 1 1 b = a+b+c− , c = a+b+c− . 2 b 2 2
The nine-point center O9 is the midpoint of the segment O H , where H (a + b + c) 1 is the orthocenter of triangle ABC. The coordinate of O9 is ω = (a + b + c). 2 Now observe that (a − ω)(b − ω)(c − ω) = (m − ω)(n − ω)( p − ω) =
1 abc, 8
and the claim is proved. Problem 2. The altitudes of triangle ABC meet its circumcircle at points A1 , B1 , C1 , respectively. If A1 , B1 , C1 are the antipodal points of A1 , B1 , C1 on the circumcircle ABC, then ABC and A1 B1 C1 are orthopolar triangles. bc ca ab Solution. The coordinates of A1 , B1 , C1 are − , − , − , respectively. Indeed, a b c the equation of line AH in terms of the real product is AH : (z − a) · (b − c) = 0. bc It suffices to show that the point with coordinate − lies both on AH and on the a bc |b||c| R·R circumcircle of triangle ABC. First, let us note that − = = = R, a |a| R hence this point is situated on the circumcircle of triangle ABC. Now, we show that bc the complex number − satisfies the equation of the line AH . This is equivalent to a bc + a · (b − c) = 0. a Using the definition of the real product, this reduces to bc bc + a (b − c) + + a (b − c) = 0 a a or
2 bc R R2 ab c + a (b − c) + +a − = 0. a b c R2
Finally, this comes down to
abc R2 a R2 (b − c) + a − − a bc R2 a relation that is clearly true.
= 0,
4.9. Orthopolar Triangles
135
Figure 4.12.
It follows that A1 , B1 , C1 have coordinates
bc ca ab , , , respectively. Because a b c
bc ca ab · · = abc, a c c we obtain that the triangles ABC and A1 B1 C1 are orthopolar. Problem 3. Let P and P be distinct points on the circumcircle of triangle ABC such that lines A P and A P are symmetric with respect to the bisector of angle B AC. Then triangles ABC and A P P are orthopolar.
Figure 4.13.
Solution. Let us consider p and p the coordinates of points P and P , respectively. It is clear that the lines P P and BC are parallel. Using the complex product, it follows
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4. More on Complex Numbers and Geometry
that ( p − p ) × (b − c) = 0. This relation is equivalent to ( p − p )(b − c) − ( p − p )(b − c) = 0. Considering the origin of the complex plane at the circumcenter O of triangle ABC, we have R2 R2 R2 R2 − − − (b − c) = 0, (p − p ) b c p p so 1 1 2 = 0. − R ( p − p )(b − c) bc pp Therefore bc = pp , i.e., abc = app . From Theorem 3 it follows that ABC and A P P are orthopolar triangles.
4.10
Area of the Antipedal Triangle
Consider a triangle ABC and a point M. The perpendicular lines from A, B, C to M A, M B, MC, respectively, determine a triangle; we call this triangle the antipedal triangle of M with respect to ABC. Recall that M is the isogonal point of M if the pairs of lines AM, AM ; B M, B M ; AB, M M AC ≡ M BC ≡ C M, C M are isogonal, i.e., the following relations hold: M B A, MC A ≡ M C B.
Figure 4.14.
Theorem. Consider M a point in the plane of triangle ABC, M the isogonal point of M and A B C the antipedal triangle of M with respect to ABC. Then area[ABC] |ρ(M )| |R 2 − O M 2 | = , = area[A B C ] 4R 2 4R 2 where ρ(M ) is the power of M with respect to the circumcircle of triangle ABC.
4.10. Area of the Antipedal Triangle
137
Proof. Consider point O the origin of the complex plane and let m, a, b, c be the coordinates of M, A, B, C. Then R 2 = aa = bb = cc and ρ(M) = R 2 − mm.
(1)
Let O1 , O2 , O3 be the circumcenters of triangles B MC, C M A, AM B, respectively. It is easy to verify that O1 , O2 , O3 are the midpoints of segments M A , M B , MC , respectively, and so area[O1 O2 O3 ] 1 = . (2) area[A B C ] 4 The coordinate of the circumcenter of the triangle with vertices with coordinates z 1 , z 2 , z 3 is given by the following formula (see formula (1) in Subsection 3.6.1): zO =
z 1 z 1 (z 2 − z 3 ) + z 2 z 2 (z 3 − z 1 ) + z 3 z 3 (z 1 − z 2 ) . z1 z1 1 z2 z2 1 z3 z3 1
The bisector line of thesegment [z 1 , z 2 ] has the following equation in terms of real 1 product: z − (z 1 + z 2 ) · (z 1 − z 2 ) = 0. It is sufficient to check that z O satisfies this 2 equation as this implies, by symmetry, that z O belongs to the perpendicular bisectors of segments [z 2 , z 3 ] and [z 3 , z 1 ]. The coordinate of O1 is z O1 =
=
mm(b − c) + bb(c − m) + cc(m − b) m m 1 b b 1 c c 1
(R 2 − mm)(c − b) ρ(M)(c − b) = . m m 1 m m 1 b b 1 b b 1 c c 1 c c 1 a = b c
Let
and consider
m 1 α= b c
m b c
1 1 1
,
a b c
1 1 1
m 1 β= c a
m c a
1 1 1
,
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4. More on Complex Numbers and Geometry
m 1 γ = a b
and
m a b
1 1 1
.
With this notation we obtain (αa + βb + γ c) ·
m(ab − ac) − m(ab − ac) + a(bc − bc) = cyc
cyc
= m − m · 0 +
R2
R2
cyc
− a c c
ab ac − = m, = m + R 2 c b cyc a b
cyc
and consequently αa + βb + γ c = m, since it is clear that = 0. We note that α, β, γ are real numbers and α + β + γ = 1, so α, β, γ are the barycentric coordinates of point M. Since z O1 = we have
(c − b) · ρ(M) , α·
z O2 =
(c − a) · ρ(M) , β
z O3 =
(a − b) · ρ(M) , γ ·
z O1 z O1 1 i z O2 z O2 1 4 area[O1 O2 O3 ] z O3 z O3 1 = i area[ABC] · 4 b − c b − c α 1 ρ 2 (M) 1 = · · c − a c − a β αβγ 2 a − b a − b γ ρ 2 (M) c − a c − a 1 = · · 3 αβγ a − b a − b ρ 2 (M) ρ 2 (M) 1 1 = · = · · . 3 2 αβγ αβγ
(3)
4.10. Area of the Antipedal Triangle
139
Relations (2) and (3) imply that |2 αβγ | area[ABC] = . area[A B C ] 4ρ 2 (M)
(4)
Because α, β, γ are the barycentric coordinates of M, it follows that z M = αz A + βz B + γ z C . Using the real product we find that O M 2 = z M · z M = (αz A + βz B + γ z C ) · (αz A + βz B + γ z C )
= (α 2 + β 2 + γ 2 )R 2 + 2 αβz A · z B cyc
AB 2 = (α + β + γ )R + 2 αβ R − 2 cyc
= (α + β + γ )2 R 2 − αβ AB 2 = R 2 − αβ AB 2 . 2
2
2
2
2
cyc
Therefore the power of expressed in the form
M
cyc
with respect to the circumcircle of triangle ABC can be
ρ(M) = R 2 − O M 2 =
αβ AB 2 .
cyc
On the other hand, if α, β, γ are the barycentric coordinates of the point M, then its isogonal point M has the barycentric coordinates given by α =
βγ BC 2 , βγ BC 2 + αγ C A2 + αβ AB 2 γ =
βγ
BC 2
β =
γ αC A2 , βγ BC 2 + αγ C A2 + αβ AB 2
αβ AB 2 . + αγ C A2 + αβ AB 2
Therefore ρ(M ) =
α β AB 2
cyc
αβγ AB 2 · BC 2 · C A2 αβγ AB 2 BC 2 C A2 = . 2 2 2 2 (βγ BC + αγ C A + αβ AB ) ρ 2 (M) On the other hand, we have 2 4 i 2 4 AB 2 · BC 2 · C A2 2 · = · area[ABC] = . = i 4 i R2 =
The desired conclusion follows from the relations (4), (5), and (6).
(5)
(6)
140
4. More on Complex Numbers and Geometry
Applications. 1) If M is the orthocenter H , then M is the circumcenter O and R2 1 area[ABC] = = . 2 area[A B C ] 4 4R 2) If M is the circumcenter O, then M is the orthocenter H and we obtain |R 2 − O H 2 | area[ABC] . = area[A B C ] 4R 2 Using the formula in Theorem 8, Subsection 4.6.4, it follows that |(2R + r )2 − s 2 | area[ABC] . = area[A B C ] 2R 2 3) If M is the Lemoine point K , then M is the centroid G and |R 2 − OG 2 | area[ABC] . = area[A B C ] 4R 2 Applying the formula in Corollary 9, Subsection 4.6.4, then the first formula in Corollary 2, Subsection 4.6.1, it follows that 2(s 2 − r 2 − 4Rr ) α2 + β 2 + γ 2 area[ABC] = = area[A B C ] 36R 2 36R 2 where α, β, γ are the sides of triangle ABC. From the inequality α 2 + β 2 + γ 2 ≤ 9R 2 (Corollary 10, Subsection 4.6.4) we obtain area[ABC] 1 ≤ . area[A B C ] 4 4) If M is the incenter I of triangle ABC, then M = I and using Euler’s formula O I 2 = R 2 − 2Rr (see Theorem 4 in Subsection 4.6.2) we find that 2Rr r |R 2 − O I 2 | area[ABC] = = = . 2 2 area[A B C ] 4R 4R 4R Applying Euler’s inequality R ≥ 2r (Corollary 5 in Subsection 4.6.2) it follows that area[ABC] 1 ≤ . area[A B C ] 4
4.11
Lagrange’s Theorem and Applications
Consider the distinct points A1 (z 1 ), . . . , An (z n ) in the complex plane. Let m 1 , . . . , m n be nonzero real numbers such that m 1 + · · · + m n = 0. Let m = m 1 + · · · + m n .
4.11. Lagrange’s Theorem and Applications
141
The point G with coordinate zG =
1 (m 1 z 1 + · · · + m n z n ) m
is called the barycenter of set {A1 , . . . , An } with respect to the weights m 1 , . . . , m n . In the case m 1 = · · · = m n = 1, the point G is the centroid of the set {A1 , . . . , An }. When n = 3 and the points A1 , A2 , A3 are not collinear, we obtain the absolute barycentric coordinates of G with respect to the triangle A1 A2 A3 (see Subsection 4.7.1): μz 1 =
m1 , m
μz 2 =
m2 , m
μz 3 =
m3 . m
Theorem 1. (Lagrange11 ) Consider the points A1 , . . . , An and the nonzero real numbers m 1 , . . . , m n such that m = m 1 + · · · + m n = 0. If G denotes the barycenter of set {A1 , . . . , An } with respect to the weights m 1 , . . . , m n , then for any point M in the plane the following relation holds: n
m j M A2j = m M G 2 +
j=1
n
m j G A2j
(1)
j=1
Proof. Without loss of generality we can assume that the barycenter G is the origin of the complex plane; that is, z G = 0. Using properties of the real product we obtain for all j = 1, . . . , n, the relations M A2j = |z M − z j |2 = (z M − z j ) · (z M − z j ) = |z M |2 − 2z M · z j + |z j |2 , i.e., M A2j = |z M |2 − 2z M · z j + |z j |2 . Multiplying by m j and adding the relations obtained for j = 1, . . . , n, it follows that n n
m j M A2j = m j (|z M |2 − 2z M · z j + |z j |2 ) j=1
j=1
= m|z M | − 2z M · 2
n
j=1
mjzj
+
n
m j |z j |2
j=1
11 Joseph Louis Lagrange (1736–1813), French mathematician, one of the greatest mathematicians of the
eighteenth century. He made important contributions in all branches of mathematics and his results have greatly influenced modern science.
142
4. More on Complex Numbers and Geometry
= m|z M |2 − 2z M · (mz G ) +
n
m j |z j |2
j=1
= m|z M |2 +
n
m j |z j |2 = m|z M − z G |2 +
j=1
n
m j |z j − z G |2
j=1
= m M G2 +
n
m j G A2j .
j=1
Corollary 2. Consider the distinct points A1 , . . . , An and the nonzero real numbers m 1 , . . . , m n such that m 1 + · · · + m n = 0. For any point M in the plane the following inequality holds: n n
m j M A2j ≥ m j G A2j , (2) j=1
j=1
with equality if and only if M = G, the barycenter of set {A1 , . . . , An } with respect to the weights m 1 , . . . , m n . Proof. The inequality (2) follows directly from Lagrange’s relation (1).
If m 1 = · · · = m n = 1, from Theorem 1 one obtains: Corollary 3. (Leibniz12 ) Consider the distinct points A1 , . . . , An and the centroid G of the set {A1 , . . . , An }. For any point M in the plane the following relation holds: n
j=1
M A2j = n M G 2 +
n
G A2j .
(3)
j=1
Remark. The relation (3) is equivalent to the following identity: For any complex numbers z, z 1 , . . . , z n we have 2 n n z 1 + · · · + z n 2
1
z j − z 1 + · · · + z n . |z − z j |2 = n z − + n j=1 n n j=1 Applications. We will use formula (3) in determining some important distances in a triangle. Let us consider the triangle ABC and let us take n = 3 in the formula (3). We find that for any point M in the plane of triangle ABC the following formula holds: M A2 + M B 2 + MC 2 = 3M G 2 + G A2 + G B 2 + GC 2
(4)
where G is the centroid of triangle ABC. Assume that the circumcenter O of the triangle ABC is the origin of complex plane. 12 Gottfried Wilhelm Leibniz (1646–1716) was a German philosopher, mathematician, and logician who
is probably most well known for having invented the differential and integral calculus independently of Sir Isaac Newton.
4.11. Lagrange’s Theorem and Applications
143
1) In the relation (4) we choose M = 0 and we get 3R 2 = 3OG 2 + G A2 + G B 2 + GC 2 . Applying the well-known median formula it follows that 4 2 (m + m 2β + m 2γ ) 9 α 4 1 1 = [2(β 2 + γ 2 ) − α 2 ] = (α 2 + β 2 + γ 2 ), 9 cyc 4 3
G A2 + G B 2 + GC 2 =
where α, β, γ are the sides of triangle ABC. We find 1 OG 2 = R 2 − (α 2 + β 2 + γ 2 ). 9
(5)
An equivalent form of the distance OG is given in terms of the basic invariants of triangle in Corollary 9, Subsection 4.6.4. 2) Using the collinearity of points O, G, H and the relation O H = 3OG (see Theorem 3.1 in Section 4.5) it follows that O H 2 = 9OG 2 = 9R 2 − (α 2 + β 2 + γ 2 )
(6)
An equivalent form for the distance O H was obtained in terms of the fundamental invariants of the triangle in Theorem 8, Subsection 4.6.4. 3) Consider in (4) M = I , the incenter of triangle ABC. We obtain 1 I A2 + I B 2 + I C 2 = 3I G 2 + (α 2 + β 2 + γ 2 ). 3
Figure 4.15.
On the other hand, we have the following relations: IA=
r A sin 2
,
IB =
r B sin 2
,
IC =
r sin
C 2
,
144
4. More on Complex Numbers and Geometry
where r is the inradius of triangle ABC. It follows that ⎡ ⎛ ⎞ I G2 =
1 ⎢ 2⎜ ⎣r ⎝ 3
1 A sin2 2
+
1 B sin2 2
+
⎤
⎟ 1 2 2 2 ⎥ ⎠ − (α + β + γ )⎦ . C 3 2 sin 2 1
Taking into account the well-known formula sin2
(s − β)(s − γ ) A = 2 βγ
we obtain
1
cyc
A sin2 2 =
=
cyc
βγ βγ (s − α) = (s − β)(s − γ ) cyc (s − α)(s − β)(s − γ )
( s
s '
βγ (s − α) = 2 s βγ − 3αβγ 2 K cyc K
s 1 [s(s 2 + r 2 + 4Rr ) − 12s Rr ] = 2 (s 2 + r 2 − 8Rr ), 2 K r where we have used the formulas in Subsection 4.6.1. Therefore 1 2 1 2 2 2 2 2 IG = s + r − 8Rr − (α + β + γ ) 3 3 1 2 2 1 = s + r 2 − 8Rr − (s 2 − r 2 − 4Rr ) = (s 2 + 5r 2 − 16Rr ), 3 3 9 =
where the first formula in Corollary 2 was used. That is, I G2 =
1 2 (s + 5r 2 − 16Rr ), 9
(7)
hence we obtain again the formula in Application 1), Subsection 4.7.2. Problem 1. Let z 1 , z 2 , z 3 be distinct complex numbers having modulus R. Prove that √ 9R 2 − |z 1 + z 2 + z 3 |2 3 ≥ . |z 1 − z 2 | · |z 2 − z 3 | · |z 3 − z 1 | R Solution. Let A, B, C be the geometric images of the complex numbers z 1 , z 2 , z 3 and let G be the centroid of the triangle ABC. z1 + z2 + z3 The coordinate of G is equal to , and |z 1 − z 2 | = γ , |z 2 − z 3 | = α, 3 |z 3 − z 1 | = β. The inequality becomes √ 3 9R 2 − 9OG 2 ≥ . (1) αβγ R
4.11. Lagrange’s Theorem and Applications
145
Using the formula 1 OG 2 = R 2 − (α 2 + β 2 + γ 2 ), 9 (1) is equivalent to √ √ αβγ 3 4r K √ 3 = 4K 3. = α +β +γ ≥ R R 2
2
2
Here is a proof of this famous inequality, by using Hero’s formula and the AM-GM inequality: & &
(s − α + s − β + s − γ )3 s3 = s K = s(s − α)(s − β)(s − γ ) ≤ s 27 27 s2 (α + β + γ )2 3(α 2 + β 2 + γ 2 ) α2 + β 2 + γ 2 = √ = ≤ = . √ √ √ 3 3 12 3 12 3 4 3 We now extend Leibniz’s relation in Corollary 3. First, we need the following result. Theorem 4. Let n ≥ 2 be a positive integer. Consider the distinct points A1 , . . . , An and let G be the centroid of the set {A1 , . . . , An }. Then for any point in the plane the following formula holds: n2 M G 2 = n
n
M A2j −
j=1
Ai A2k .
(8)
1≤i
Proof. We assume that the barycenter G is the origin of the complex plane. Using properties of the real product we have M A2j = |z M − z j |2 = (z M − z j ) · (z M − z j ) = |z M |2 − 2z M · z j + |z j |2 and Ai A2k = |z i − z k |2 = |z i |2 − 2z i · z k + |z k |2 , where the complex number z j is the coordinate of the point A j , j = 1, 2, . . . , n. The relation (8) is equivalent to n 2 |z M |2 = n
n
(|z M |2 − 2z M · z j + |z j |2 ) −
j=1
|(|z i |2 − 2z i · z k + |z k |2 ).
1≤i
That is, n
n
j=1
|z j |2 = 2n
n
j=1
zM · z j +
1≤i
(|z i |2 − 2z i · z k + |z k |2 ).
146
4. More on Complex Numbers and Geometry
Taking into account the hypothesis that G is the origin of the complex plane, we have n n
zM · z j = zM · z j = n(z M · z G ) = n(z M · 0) = 0. j=1
j=1
Hence, the relation (8) is equivalent to n
|z j |2 = −2
j=1
zi · zk .
1≤i
The last relation can be obtained as follows: n n
1 0 = |z G |2 = z G · z G = 2 zi · zk n i=1 k=1 1 = 2 n
n
|z j | + 2 2
j=1
zi · zk .
1≤i
Therefore the relation (8) is proved.
Remark. The formula (8) is equivalent to the following identity: For any complex numbers z, z 1 , . . . , z n , we have n z 1 + · · · + z n 2 1 1
|z − z j |2 − z − =n n j=1 n
|z i − z k |2 .
1≤i
Applications. 1) If A1 , . . . , An are points on the circle of center O and radius R, then taking in (8) M = O, it follows that
Ai A2k = n 2 (R 2 − OG 2 ).
1≤i
If n = 3 we obtain the formula (5). 2) For any point M in the plane the following inequality holds: n
j=1
M A2j ≥
1 n
Ai A2k ,
1≤i
with equality if and only if M = G, the centroid of the set {A1 , . . . , An }. Let n ≥ 2 be a positive integer, and let k be an integer such that 2 ≤ k ≤ n. Consider the distinct points A1 , . . . , An and let G be the centroid of the set {A1 , . . . , An }. For indices i 1 < · · · < i k let us denote by G i1 ,...,ik the centroid of the set {Ai1 , . . . , Aik }. We have the following result:
4.11. Lagrange’s Theorem and Applications
147
Theorem 5. For any point M in the plane,
n n n 2 2 (n − k) M A j + n (k − 1) M G2 k j=1 k
= kn(n − 1)
1≤i 1 <··· ,i k ≤n
M G i21 ···ik .
(9)
Proof. It is not difficult to see that the barycenter of the set {G i1 ···ik : 1 ≤ i 1 < · · · < i k ≤ n} is G. Applying Leibniz’s relation one obtains n
M A2j = n M G 2 +
j=1
1≤i 1 <···
G A2j ,
(10)
j=1
M G i21 ···ik =
k
n
n M G2 + k 1≤i <···
M Ai2s = k M G i21 ···ik +
s=1
k
k ≤n
GG i21 ···ik
G i1 ···ik Ai2s .
(11) (12)
s=1
Considering in (12) M = G and adding all these relations, it follows that k
G Ai2s = k
1≤i 1 <···
+
1≤i 1 <···
k
1≤i 1 <···
GG i21 ···ik
G i1 ···ik Ai2s .
(13)
Applying formula (8) in Theorem 5 for the sets {A1 , . . . , An } and {Ai1 , . . . , Aik }, respectively, we get n
n2 M G 2 = n
j=1
k 2 M G i21 ···ik = k
k
M A2j −
Ai A2k ,
(14)
1≤i
M Ai2s −
s=1
Ai p Ai2q .
(15)
1≤ p
Taking M = G i1 ···ik in (15), it follows that k
G i1 ···ik Ai2s =
s=1
1
Ai A 2 . k 1≤ p
From (16) and (13) we obtain
k
1≤i 1 <···
G Ai2s = k
1≤i 1 <···
GG i21 ···ik
(16)
148
4. More on Complex Numbers and Geometry
+
1 k 1≤i <···
k ≤n
(17)
Ai p Ai2q
1≤ p
If we rearrange the terms in formula (17), we get k n k n n
1 1 k
2 k G A2j = k GG i21 ···ik + Ai A2j . n n k j=1 1≤i 1 <···
(18)
Remark. The relation (9) is equivalent to the following identity: For any complex numbers z, z 1 , . . . , z n we have
n n n z 1 + · · · + z n 2 2 2 (n − k) |z − z j | + n (k − 1) z− k j=1 k n
= kn(n − 1)
1≤i 1 <···
2 z − z i 1 + · · · + z i k . k ≤n
Applications. 1) In the case k = 2, from (9) we obtain that for any point M in the plane, the following relation holds: (n − 2)
n
M A2j + n 2 M G 2 = 4
j=1
1≤i 1
M G i21 i2 .
In this case G i1 i2 is the midpoint of the segment [Ai1 Ai2 ]. 2) If k = 3, from (9) we get that for any point M in the plane, the relation (n − 3)(n − 2)
n
M A2j + 2n 2 (n − 2)M G 2 = 18
j=1
1≤i 1
M G i21 i2 i3
holds. Here the point G i1 i2 i3 is the centroid of triangle Ai1 Ai2 Ai3 .
4.12
Euler’s Center of an Inscribed Polygon
Consider a polygon A1 A2 · · · An inscribed in a circle centered at the origin of a complex plane and let a1 , a2 , . . . , an be the coordinates of its vertices. By definition, the point E with coordinate zE =
a1 + a2 + · · · + an 2
is called Euler’s center of the polygon A1 A2 · · · An . In the case n = 3 it is clear that E = O9 , the center of Euler’s nine-point circle.
4.12. Euler’s Center of an Inscribed Polygon
149
Remarks. a) Let G(z G ) and H (z H ) be the centroid and orthocenter of the inscribed polygon A1 A2 · · · An . Then zE =
nz G zH n OG OH = and O E = = . 2 2 2 2
Recall that the orthocenter of the polygon A1 A2 · · · An is the point H with coordinate z H = a1 + a2 + · · · + an . b) For n = 4, point E is also called Mathot’s point of the inscribed quadrilateral A1 A2 A3 A4 . Proposition. In the above notation, the following relation holds: n
E Ai2 = n R 2 + (n − 4)E O 2 .
(1)
i=1
Proof. Using the identity (8) in Theorem 4, Section 2.17, n2 · M G 2 = n
n
M Ai2 −
i=1
Ai A2j
1≤i< j≤n
for M = E and M = O, we obtain n2 · E G 2 = n
n
E Ai2 −
i=1
and n 2 · OG 2 = n R 2 −
Ai A2j ,
(2)
1≤i< j≤n
Ai A2j .
(3)
1≤i< j≤n
Setting s =
n
ai , we have
i=1
s s s n − 2 n−2 E G = |z E − z G | = − = · = · O E. 2 n 2 n n
(4)
From the relations (2), (3) and (4) we derive that n
n
E Ai2 = n 2 · E G 2 − n 2 · OG 2 + n 2 R 2
i=1
= (n − 2)2 O E 2 − 4O E 2 + n 2 R 2 = n(n − 4) · E O 2 + n 2 R 2 or, equivalently,
n
E Ai2 = n R 2 + (n − 4)E O 2 ,
i=1
as desired.
150
4. More on Complex Numbers and Geometry
Applications. 1) For n = 3, from relation (1) we obtain O9 A21 + O9 A22 + O9 A23 = 3R 2 − O O92 .
(5)
Using the formula in Corollary 9.2, Subsection 4.6.4, we can express the right-hand side in (5) in terms of the fundamental invariants of triangle A1 A2 A3 : O9 A21 + O9 A22 + O9 A23 =
3 2 1 2 1 R − r − 2Rr + s 2 . 4 2 2
(6)
From formula (5) it follows that for any triangle A1 A2 A3 the following inequality holds: (7) O9 A21 + O9 A22 + O9 A23 ≤ 3R 2 , with equality if and only if the triangle is equilateral. 2) For n = 4 we obtain the interesting relation 4
E Ai2 = 4R 2 .
(8)
i=1
The point E is the unique point in the plane of the quadrilateral A1 A2 A3 A4 satisfying relation (8). 3) For n > 4, from relation (1) the inequality n
E Ai2 ≥ n R 2
(9)
i=1
follows. Equality holds only in the polygon A1 A2 · · · An with the property E = O. 4) The Cauchy–Schwarz inequality and inequality (7) give 3
R · O9 A i
2
≤ (3R 2 )
i=1
3
O9 Ai2 ≤ 9R 2 .
i=1
This is equivalent to O9 A1 + O9 A2 + O9 A3 ≤ 3R.
(10)
5) Using the same inequality and the relation (8) we have 4 4 2
E Ai ≤ 4R 2 · E Ai = 16R 4 R i=1
or, equivalently,
i=1
4
i=1
E Ai ≤ 4R.
(11)
4.13. Some Geometric Transformations of the Complex Plane
151
6) Using the relation
s 2E Ai = 2|e − ai | = 2 − ai = |s − 2ai |, 2 the inequalities (4), (5) become
| − a1 + a2 + a3 | ≤ 6R cyc
and, respectively,
| − a1 + a2 + a3 + a4 | ≤ 8R.
cyc
The above inequalities hold for all complex numbers of the same modulus R.
4.13
Some Geometric Transformations of the Complex Plane
4.13.1
Translation
Let z 0 be a fixed complex number and let tz 0 be the mapping defined by tz 0 : C → C,
tz 0 (z) = z + z 0 .
The mapping tz 0 is called the translation of the complex plane with complex number z 0 .
Figure 4.16.
Taking into account the geometric interpretation of the addition of two complex numbers (see Subsection 1.2.3), we have Fig. 4.16, giving the geometric image of tz 0 (z).
152
4. More on Complex Numbers and Geometry
In Fig. 4.16 O M0 M M is a parallelogram and O M is one of its diagonals. There−→ fore, the mapping tz 0 corresponds in the complex plane C to the translation t− O M0 of −−→ vector O M0 in the case of a Euclidean plane. It is clear that the composition of two translations tz 1 and tz 2 satisfies the relation tz 1 ◦ tz 2 = tz 1 +z 2 . It is also clear that the set T of all translations of a complex plane is a group with respect to the composition of mappings. The group (T , ◦) is Abelian and its unity is t O = 1C , the translation of the complex number 0.
4.13.2
Reflection in the real axis
Consider the mappings s : C → C, s(z) = z. If M is the point with coordinate z, then the point M (s(z)) is obtained by reflecting M across the real axis (see Fig. 4.17). The mapping s is called the reflection in the real axis. It is clear that s ◦ s = 1C .
Figure 4.17.
4.13.3
Reflection in a point
Consider the mapping s0 : C → C, s0 (z) = −z. Since s0 (z) + z = 0, the origin O is the midpoint of the segment [M(z)M (z)]. Hence M is the reflection of point M across O (Fig. 4.18). The mapping s0 is called the reflection in the origin. Consider a fixed complex number z 0 and the mapping sz 0 : C → C,
sz 0 (z) = 2z 0 − z.
If z 0 , z, sz 0 (z) are the coordinates of points M0 , M, M , then M0 is the midpoint of the segment [M M ], hence M is the reflection of M in M0 (Fig. 4.19). The mapping sz 0 is called the reflection in the point M0 (z 0 ). It is clear that the following relation holds: sz 0 ◦ sz 0 = 1C .
4.13. Some Geometric Transformations of the Complex Plane
153
Figure 4.18.
Figure 4.19.
4.13.4
Rotation
Let a = cos t0 + i sin t0 be a complex number having modulus 1 and let ra be the mapping given by ra : C → C, ra (z) = az. If z = ρ(cos t + i sin t), then ra (z) = az = ρ[cos(t + t0 ) + i sin(t + t0 )], hence M (ra (z)) is obtained by rotating point M(z) about the origin through the angle t0 (Fig. 4.20). The mapping ra is called the rotation with center O and angle t0 = arg a.
4.13.5
Isometric transformation of the complex plane
A mapping f : C → C is called an isometry if it preserves distance, i.e., for all z 1 , z 2 ∈ C, | f (z 1 ) − f (z 2 )| = |z 1 − z 2 |. Theorem 1. Translations, reflections (in the real axis or in a point) and rotations about center O are isometries of the complex plane.
154
4. More on Complex Numbers and Geometry
Figure 4.20.
Proof. For the translation tz 0 we have |tz 0 (z 1 ) − tz 0 (z 2 )| = |(z 1 + z 0 ) − (z 2 + z 0 )| = |z 1 − z 2 |. For the reflection s across the real axis we obtain |s(z 1 ) − s(z 2 )| = |z 1 − z 2 | = |z 1 − z 2 | = |z 1 − z 2 |, and the same goes for the reflection in a point. Finally, if ra is a rotation, then |ra (z 1 ) − ra (z 2 )| = |az 1 − az 2 | = |a||z 1 − z 2 | = |z 1 − z 2 |, since |a| = 1.
We can easily check that the composition of two isometries is also an isometry. The set I zo(C) of all isometries of the complex plane is a group with respect to the composition of mappings and (T , ◦) is a subgroup of it. Problem. Let A1 A2 A3 A4 be a cyclic quadrilateral inscribed in a circle of center O and let H1 , H2 , H3 , H4 be the orthocenters of triangles A2 A3 A4 , A1 A3 A4 , A1 A2 A4 , A1 A2 A3 , respectively. Prove that quadrilaterals A1 A2 A3 A4 and H1 H2 H3 H4 are congruent. (Balkan Mathematical Olympiad, 1984)
Solution. Consider the complex plane with origin at the circumcenter, and denote by lowercase letters the coordinates of the points denoted by uppercase letters. If s = a1 + a2 + a3 + a4 , then h 1 = a2 + a3 + a4 = s − a1 , h 2 = s − a2 , h 3 = s − a3 , h 4 = s − a4 . Hence the quadrilateral H1 H2 H3 H4 is the reflection of quadrilateral s A1 A2 A3 A4 across the point with coordinate . 2 The following result describes all isometries of the complex plane. Theorem 2. Any isometry is a mapping f : C → C with f (z) = az + b or f (z) = az + b, where a, b ∈ C and |a| = 1.
4.13. Some Geometric Transformations of the Complex Plane
155
Proof. Let b = f (0), c = f (1) and a = c − b. Then |a| = |c − b| = | f (1) − f (0)| = |1 − 0| = 1. Consider the mapping g : C → C, given by g(z) = az +b. It is not difficult to prove that g is an isometry, with g(0) = b = f (0) and g(1) = a + b = c = f (1). Hence h = g −1 ◦ f is an isometry, with 0 and 1 as fixed points. By definition, it follows that any real number is a fixed point of h, hence h = 1C or h = s, the reflection in the real axis. Hence g = f or g = f ◦ s, and the proof is complete. The above result shows that any isometry of the complex plane is the composition of a rotation and a translation, or the composition of a rotation with the reflection in the origin O and a translation.
4.13.6
Morley’s theorem
In 1899, Frank Morley, then professor of mathematics at Haverford College, came across a result so surprising that it entered mathematical folklore under the name of “Morley’s Miracle.” Morley’s marvelous theorem states that: The three points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle. The theorem was mistakenly attributed to Napoleon Bonaparte, who made some contributions to geometry. There are various proofs of this nice result: J. Conway’s proof, D.J. Newman’s proof, L. Bankoff’s proof, and N. Dergiades’s proof. Here we present the new proof published in 1998, by Alain Connes. His proof is derived from the following result: Theorem 3. (Alain Connes) Consider the transformations of a complex plane f i : C → C, f i (z) = ai z + bi , i = 1, 2, 3, where all coefficients ai are different from zero. Assume that the mappings f 1 ◦ f 2 , f 2 ◦ f 3 , f 3 ◦ f 1 and f 1 ◦ f 2 ◦ f 3 are not translations, i.e., equivalently, a1 a2 , a2 a3 , a3 a1 , a1 a2 a3 ∈ C \ {1}. Then the following statements are equivalent: (1) f 13 ◦ f 23 ◦ f 33 = 1C ; (2) j 3 = 1 and α + jβ + j 2 γ = 0, where j = a1 a2 a3 = 1 and α, β, γ are the unique fixed points of mappings f 1 ◦ f 2 , f 2 ◦ f 3 , f 3 ◦ f 1 , respectively. Proof. Note that ( f 1 ◦ f 2 )(z) = a1 a2 z + a1 b2 + b1 ,
a1 a2 = 1,
( f 2 ◦ f 3 )(z) = a2 a3 z + a2 b3 + b2 ,
a2 a3 = 1,
( f 3 ◦ f 1 )(z) = a3 a1 z + a3 b1 + b3 ,
a3 a1 = 1.
156
4. More on Complex Numbers and Geometry
a1 a3 b2 + a3 b1 =: α , a3 − j a1 a2 b3 + a1 b2 a 2 b3 + b 2 = =: β , Fix( f 2 ◦ f 3 ) = 1 − a2 a3 a1 − j a2 a3 b1 + a2 b3 a3 b1 + b3 = =: γ , Fix( f 3 ◦ f 1 ) = 1 − a3 a1 a2 − j Fix( f 1 ◦ f 2 ) =
a1 b2 + b1 1 − a1 a2
=
where Fix( f ) denotes the set of fixed points of the mapping f . For the cubes of f 1 , f 2 , f 3 we have the formulas f 13 (z) = a13 z + b1 (a12 + a1 + 1), f 23 (z) = a23 z + b2 (a22 + a2 + 1), f 33 (z) = a33 + b3 (a32 + a3 + 1), hence ( f 13 ◦ f 23 ◦ f 33 )(z) = a13 a23 a33 z + a13 a23 b3 (a32 + a3 + 1) + a13 b2 (a22 + a2 + 1) + b1 (a12 + a1 + 1). Therefore f 13 ◦ f 23 ◦ f 33 = idC if and only if a13 a23 a33 = 1 and a13 a23 b3 (a32 + a3 + 1) + a13 b2 (a22 + a2 + 1) + b1 (a12 + a1 + 1) = 0. To prove the equivalence of statements (1) and (2) we have to show that α+ jβ + j 2 γ is different from the free term of f 13 ◦ f 23 ◦ f 33 by a multiplicative constant. Indeed, using the relation j 3 = 1 and implicitly j 2 + j + 1 = 0, we have successively: α + jβ + j 2 γ = α + jβ + (−1 − j)γ = α − γ + j (β − γ ) a a b + a b a 2 a 3 b1 + a 2 b 3 a2 a3 b1 + a2 b3 a1 a3 b2 + a3 b1 1 2 3 1 2 − +j − a3 − j a2 − j a1 − j a2 − j
= =
a1 a2 a3 b2 + a2 a3 b1 − a1 a3 b2 j − a3 b1 j − a2 a32 b1 − a2 a3 b3 + a2 a3 b1 j + a2 b3 j (a2 − j)(a3 − j)
+j
a1 a22 b3 + a1 a2 b2 − a1 a2 b3 j − a1 b2 j − a1 a2 a3 b1 − a1 a2 b3 + a2 a3 b1 j + a2 b3 j (a1 − j)(a2 − j)
b2 j − a2 a3 b1 j 2 − a1 a3 b2 j − a3 b1 j − a2 a32 b1 − a2 a3 b3 + a2 b3 j a3 − j a1 a22 b3 j + a1 a2 b2 j + a1 a2 b3 − a1 b2 j 2 − b1 j 2 + a2 a3 b1 j 2 + a2 b3 j 2 + a1 − j
=
1 a2 − j
4.13. Some Geometric Transformations of the Complex Plane
=
157
1 (a1 b2 j − b1 − a12 a3 b2 j − a1 a3 b1 j − a1 a2 a32 b1 − b3 j (a1 − j)(a2 − j)(a3 − j)
+ a1 a2 b3 j − b2 j 2 + a2 a3 b1 + a1 a3 b2 j 2 + a3 b1 j 2 + a2 a32 b1 j + a2 a3 b3 j − a2 b3 j 2 + a2 b3 j 2 + b2 j 2 + b3 j − a1 a3 b2 j 2 − a3 b1 j 2 + a2 a3 b1 j 2 + a2 a3 b3 j 2 − a1 a22 b3 j 2 − a1 a2 b2 j 2 − a1 a2 b3 j + a1 b2 + b1 − a2 a3 b1 − a2 b3 ) =
1 (−a1 b2 j 2 − a12 a3 b2 j − a1 a3 b1 j − a3 b1 j (a1 − j)(a2 − j)(a3 − j) − a2 a32 b1 − a2 a3 b3 − a1 a22 b3 j 2 − a1 a2 b2 j 2 − a2 b3 ) =−
1 (a 2 a 2 a 2 b2 + a13 a2 a32 b2 (a1 − j)(a2 − j)(a3 − j) 1 2 3
+ a12 a2 a32 b1 + a1 a2 a32 b1 + a2 a32 b1 + a2 a3 b3 + a13 a24 a32 b3 + a13 a23 a32 b2 + a2 b3 ) =−
1 [a2 a32 b1 (1 + a1 + a12 ) + a13 a2 a32 b2 (1 + a2 + a22 ) (a1 − j)(a2 − j)(a3 − j) + a2 b3 (1 + a3 + a13 + a13 a23 a32 )] =−
a2 a32 [a 3 a 3 b3 (1 + a3 + a32 ) (a1 − j)(a2 − j)(a3 − j) 1 2 + a13 b2 (1 + a2 + a22 ) + b1 (1 + a1 + a12 )].
Theorem 4. (Morley) The three points A (α), B (β), C (γ ) of the adjacent trisectors of the angles of any triangle ABC form an equilateral triangle.
Figure 4.21.
Proof. (Alain Connes) Let us consider the rotations f 1 = r A,2x , f 2 = r B,2y , f 3 = 1 1% 1% y= % B, z = C (Figure 4.21). rC,2z of centers A, B, C and of angles x = A, 3 3 3 Note that Fix( f 1 ◦ f 2 ) = {A }, Fix( f 2 ◦ f 3 ) = {B }, Fix( f 3 ◦ f 1 ) = {C } (see Figure 4.22).
158
4. More on Complex Numbers and Geometry
Figure 4.22.
Figure 4.23.
To prove that triangle A B C is equilateral it is sufficient to show, by Proposition 2 in Section 2.4 and above Theorem 3, that f 13 ◦ f 23 ◦ f 33 = 1C . The composition s AC ◦s AB of reflections s AC and s AB across the lines AC and AB is a rotation about center A through angle 6x. Therefore f 13 = s AC ◦ s AB and analogously f 23 = s B A ◦ s BC and f 33 = sC B ◦ sC A . It follows that f 13 ◦ f 23 ◦ f 33 = s AC ◦ s AB ◦ s B A ◦ s BC ◦ sC B ◦ sC A = 1C .
4.13.7
hom*othecy
Given a fixed nonzero real number k, the mapping h k : C → C, h k (z) = kz, is called the hom*othecy of the complex plane with center O and magnitude k. Figures 4.24 and 4.25 show the position of point M (h k (z)) in the cases k > 0 and k < 0. Points M(z) and M (h k (z)) are collinear with the center O, which lies on the line segment M M if and only if k < 0. Moreover, the following relation holds: |O M | = |k||O M|. Point M is called the hom*othetic point of M with center O and magnitude k.
4.13. Some Geometric Transformations of the Complex Plane
159
Figure 4.24.
Figure 4.25.
It is clear that the composition of two hom*othecies h k1 and h k2 is also a hom*othecy, that is, h k1 ◦ h k2 = h k1 k2 . The set H of all hom*othecies of the complex plane is an Abelian group with respect to the composition of mappings. The identity of the group (H, ◦) is h 1 = 1C , the hom*othecy of magnitude 1. Problem. Let M be a point inside an equilateral triangle ABC and let M1 , M2 , M3 be the feet of the perpendiculars from M to the sides BC, C A, AB, respectively. Find the locus of the centroid of the triangle M1 M2 M3 . Solution. Let 1, ε, ε 2 be the coordinates of points A, B, C, where ε = cos 120◦ + i sin 120◦ . Recall that ε 2 + ε + 1 = 0 and ε 3 = 1. If m, m 1 , m 2 , m 3 are the coordinates of points M, M1 , M2 , M3 , we have m1 =
1 (1 + ε + m − εm), 2
160
4. More on Complex Numbers and Geometry
1 (ε + ε 2 + m − m), 2 1 m 3 = (ε 2 + 1 + m − ε 2 m). 2 Let g be the coordinate of the centroid of the triangle M1 M2 M3 . Then m2 =
g=
1 m 1 (m 1 + m 2 + m 3 ) = (2(1 + ε + ε 2 ) + 3m − m(1 + ε + ε 2 )) = , 3 6 2
1 O M. 2 The locus of G is the interior of the triangle obtained from ABC under a hom*oth1 ecy of center O and magnitude . In other words, the vertices of this triangle have 2 1 1 1 2 coordinates , ε, ε . 2 2 2 hence OG =
4.13.8
Problems
1. Prove that the composition of two isometries of the complex plane is an isometry. 2. An isometry of the complex plane has two fixed points A and B. Prove that any point M of line AB is a fixed point of the transformation. 3. Prove that any isometry of the complex plane is a composition of a rotation with a translation and possibly also with the reflection in the real axis. 4. Prove that the mapping f : C → C, f (z) = i · z + 4 − i is an isometry. Analyze f as in problem 3. 5. Prove that the mapping g : C → C, g(z) = −i z + 1 + 2i is an isometry. Analyze g as in problem 4.
5 Olympiad-Caliber Problems
The use of complex numbers is helpful in solving Olympiad problems. In many instances, a rather complicated problem can be solved unexpectedly by employing complex numbers. Even though the methods of Euclidean geometry, coordinate geometry, vector algebra and complex numbers look similar, in many situations the use of the latter has multiple advantages. This chapter will illustrate some classes of Olympiadcaliber problems where the method of complex numbers works efficiently.
5.1
Problems Involving Moduli and Conjugates
Problem 1. Let z 1 , z 2 , z 3 be complex numbers such that |z 1 | = |z 2 | = |z 3 | = r > 0 and z 1 + z 2 + z 3 = 0. Prove that z1 z2 + z2 z3 + z3 z1 = r. z +z +z 1 2 3 Solution. Observe that z1 · z1 = z2 · z2 = z3 · z3 = r 2. Then
z 1 z 2 + z 2 z 3 + z 3 z 1 2 = z1 z2 + z2 z3 + z3 z1 · z1 z2 + z2 z3 + z3 z1 z +z +z z1 + z2 + z3 z1 + z2 + z3 1 2 3
162
5. Olympiad-Caliber Problems
r2 r2 r2 r2 r2 r2 · + · + · z1 z2 + z2 z3 + z3 z1 z1 z2 z2 z3 z3 z1 = · = r 2, z1 + z2 + z3 r2 r2 r2 + + z1 z2 z3 as desired. Problem 2. Let z 1 , z 2 be complex numbers such that |z 1 | = |z 2 | = r > 0. Prove that
z1 + z2 r 2 + z1 z2
2
+
z1 − z2 r 2 − z1 z2
2 ≥
1 . r2
Solution. The desired inequality is equivalent to
r (z 1 + z 2 ) r 2 + z1 z2
2
r (z 1 − z 2 ) + r 2 − z1 z2
2 ≥ 1.
Setting z 1 = r (cos 2x + i sin 2x) and z 2 = r (cos 2y + i sin 2y) yields r (z 1 + z 2 ) r 2 (cos 2x + i sin 2x + cos 2y + i sin 2y) cos(x − y) = . = 2 2 cos(x + y) r + z1 z2 r (1 + cos(2x + 2y) + i sin(2x + 2y)) Similarly,
Thus
r (z 1 − z 2 ) sin(y − x) . = sin(y + x) r 2 − z1 z2
r (z 1 + z 2 r 2 + z1 z2
2
r (z 1 − z 2 ) + r 2 − z1 z2
2 =
cos2 (x − y) sin2 (x − y) + cos2 (x + y) sin2 (x + y)
≥ cos2 (x − y) + sin2 (x − y) = 1, as claimed. Problem 3. Let z 1 , z 2 , z 3 be complex numbers such that |z 1 | = |z 2 | = |z 3 | = 1 and
z2 z 12 z2 + 2 + 3 + 1 = 0. z2 z3 z1 z3 z1 z2
Prove that |z 1 + z 2 + z 3 | ∈ {1, 2}.
5.1. Problems Involving Moduli and Conjugates
163
Solution. The given equality can be written as z 13 + z 23 + z 33 + z 1 z 2 z 3 = 0 or −4z 1 z 2 z 3 = z 13 + z 23 + z 33 − 3z 1 z 2 z 3 = (z 1 + z 2 + z 3 )(z 12 + z 22 + z 32 − z 1 z 2 − z 2 z 3 − z 3 z 1 ). Setting z = z 1 + z 2 + z 3 , yields z 3 − 3z(z 1 z 2 + z 2 z 3 + z 3 z 1 ) = −4z 1 z 2 z 3 . This is equivalent to
z = z 1 z 2 z 3 3z 3
1 1 1 + + z1 z2 z3
−4 .
The last relation can be written as z 3 = z 1 z 2 z 3 [3z(z 1 + z 2 + z 3 ) − 4],
i.e.,
z 3 = z 1 z 2 z 3 (3|z|2 − 4).
2 Taking the absolute values of both sides yields |z|3 = |3|z|2 − 4|. If |z| ≥ √ , then 3 2 3 2 3 2 |z| − 3|z| + 4 = 0, implying |z| = 2. If |z| < √ , then |z| + 3|z| − 4 = 0, giving 3 |z| = 1, as needed. Alternate solution. It is not difficult to see that |z 13 + z 23 + z 33 | = 1. By using the algebraic identity (u + v)(v + w)(w + u) = (u + v + w)(uv + vw + wu) − uvw for u = z 13 , v = z 23 , w = z 33 , it follows that (z 13 + z 23 )(z 23 + z 33 )(z 33 + z 13 ) = (z 13 + z 23 + z 33 )(z 13 z 23 + z 23 z 33 + z 33 z 13 ) − z 13 z 23 z 33 =
z 13 z 23 z 33 (z 13
+
z 23
+
z 33 )
1 z 13
+
1 z 23
+
1 z 33
− z 13 z 23 z 33
= z 13 z 23 z 33 (z 13 + z 23 + z 33 )(z 13 + z 23 + z 33 ) − z 13 z 23 z 33 = z 13 z 23 z 33 − z 13 z 23 z 33 = 0. Suppose that z 13 + z 23 = 0. Then z 1 + z 2 = 0 or z 12 − z 1 z 2 + z 32 = 0 implying z 12 + z 22 = −2z 1 z 2 or z 12 + z 22 = z 1 z 2 .
164
5. Olympiad-Caliber Problems
On the other hand, from the given relation it follows that z 33 = −z 1 z 2 z 3 , yielding z 32 = −z 1 z 2 . We have 1 1 1 |z 1 + z 2 + z 3 |2 = (z 1 + z 2 + z 3 ) + + z1 z2 z3 z1 z2 z2 z3 z3 z1 + + + + + =3+ z2 z1 z3 z2 z3 z1 =3+
z 12
z2 + z1 z2 z2 + z1 z2 z 2 + z 22 z 12 + z 22 + 3 + 3 =3+ 1 . z1 z2 z2 z3 z3 z1 z1 z2
This leads to |z 1 + z 2 + z 3 |2 = 1 if z 12 + z 22 = −2z 1 z 2 and |z 1 + z 2 + z 3 |2 = 4 if + z 22 = z 1 z 2 . The conclusion follows.
Problem 4. Let z 1 , z 2 be complex numbers with |z 1 | = |z 2 | = 1. Prove that |z 1 + 1| + |z 2 + 1| + |z 1 z 2 + 1| ≥ 2. Solution. We have |z 1 + 1| + |z 2 + 1| + |z 1 z 2 + 1| ≥ |z 1 + 1| + |z 1 z 2 + 1 − (z 2 + 1)| = |z 1 + 1| + |z 1 z 2 − z 2 | ≥ |z 1 + 1| + |z 2 ||z 1 − 1| = |z 1 + 1| + |z 1 − 1| ≥ |z 1 + 1 + z 1 − 1| = 2|z 1 | = 2, as claimed. Problem 5. Let n > 0 be an integer and let z be a complex number such that |z| = 1. Prove that n|1 + z| + |1 + z 2 | + |1 + z 3 | + · · · + |1 + z 2n | + |1 + z 2n+1 | ≥ 2n. Solution. We have n|1 + z| + |1 + z 2 | + |1 + z 3 | + · · · + |1 + z 2n | + |1 + z 2n+1 | =
n
(|1 + z| + |1 + z 2k+1 |) +
k=1
≥
n
|z − z 2k+1 | +
k=1
=
n
k=1
as claimed.
n
|1 + z 2k |
k=1 n
|1 + z 2k | =
k=1
(|1 − z 2k | + |z + z 2k |) ≥
n
(|z||1 − z 2k | + |1 + z 2k |)
k=1 n
k=1
|1 − z 2k + 1 + z 2k | = 2n,
5.1. Problems Involving Moduli and Conjugates
165
Alternate solution. We use induction on n. For n = 1, we prove that |1 + z| + |1 + z 2 | + |1 + z 3 | ≥ 2. Indeed, 2 = |1 + z + z 3 + 1 − z(1 + z 2 )| ≤ |1 + z| + |z 3 + 1| + |z||1 + z 2 | = |1 + z| + |1 + z 2 | + |1 + z 3 |. Assume that the inequality is valid for some n, so n|1 + z| + |1 + z 2 | + · · · + |1 + z 2n+1 | ≥ 2n. We prove that (n + 1)|1 + z| + |1 + z 2 | + · · · + |1 + z 2n+1 | + |1 + z 2n+2 | + |1 + z 2n+3 | ≥ 2n + 2. Using the inductive hypothesis yields (n + 1)|1 + z| + |1 + z 2 | + · · · + |1 + z 2n+2 | + |1 + z 2n+3 | ≥ 2n + |1 + z| + |1 + z 2n+2 | + |1 + z 2n+3 | = 2n + |1 + z| + |z||1 + z 2n+2 | + |1 + z 2n+3 | ≥ 2n + |1 + z − z(1 + z 2n+2 ) + 1 + z 2n+3 | = 2n + 2, as needed. Problem 6. Let z 1 , z 2 , z 3 be complex numbers such that 1) |z 1 | = |z 2 | = |z 3 | = 1; 2) z 1 + z 2 + z 3 = 0; 3) z 12 + z 22 + z 32 = 0. Prove that for all integers n ≥ 2, |z 1n + z 2n + z 3n | ∈ {0, 1, 2, 3}. Solution. Let s1 = z 1 + z 2 + z 3 ,
s2 = z 1 z 2 + z 2 z 3 + z 3 z 1 ,
s3 = z 1 z 2 z 3
and consider the cubic equation z 3 − s1 z 2 + s 2 z − s 3 = 0 with roots z 1 , z 2 , z 3 . Because z 12 + z 22 + z 32 = 0, we have s12 = 2s2 .
(1)
166
5. Olympiad-Caliber Problems
On the other hand, s2 = s3
1 1 1 + + z1 z2 z3
= s3 (z 1 + z 2 + z 3 ) = s3 · s1 .
(2)
The relations (1) and (2) imply s12 = 2s3 · s1 and, consequently, |s1 |2 = 2|s3 | · |s1 | = 2|s1 |. Because s1 = 0, we have |s1 | = 2, so s1 = 2λ with |λ| = 1. 1 s2 2λ2 From relations (1) and (2) it follows that s2 = s12 = 2λ2 and s3 = = = λ3 . 2 s1 2λ The equation with roots z 1 , z 2 , z 3 becomes z 3 − 2λz 2 + 2λ2 z − λ3 = 0. This is equivalent to (z − λ)(z 2 − λz + λ2 ) = 0. √ 1 3 2 . The roots are λ, λε, −λε ,where ε = + i 2 2 Without loss of generality we may assume that z 1 = λ, z 2 = λε, z 3 = −λε 2 . Using the relations ε2 − ε + 1 = 0 and ε 3 = −1, it follows that E n = |z 1n + z 2n + z 3n | = |λn + λn ε n + (−1)n λn ε 2n | = |1 + εn + (−1)n ε 2n |. It is not difficult to see that E k+6 = E k for all integers k and that the equalities E 0 = 3,
E 1 = 2,
E 2 = 0,
E 3 = 1,
E 4 = 0,
E 5 = 2,
settle the claim. Alternate solution. It is clear that z 12 , z 22 , z 32 are distinct. Otherwise, if, for example, z 12 = z 22 , then 1 = |z 32 | = | − (z 12 + z 22 )| = 2|z 12 | = 2, a contradiction. From z 12 + z 22 + z 32 = 0 it follows that z 12 , z 22 , z 32 are the coordinates of the vertices of an equilateral triangle. Hence we may assume that z 22 = εz 12 and z 32 = ε2 z 12 , where ε 2 + ε + 1 = 0. Because z 22 = ε 4 z 12 and z 32 = ε 2 z 12 it follows that z 2 = ±ε 2 z 1 and z 3 = ±εz 1 . Then |z 1n + z 2n + z 3n | = |(1 + (±ε)n + (±ε 2 )n )z 1n | = |1 + (±ε)n + (±ε 2 )n | ∈ {0, 1, 2, 3} by the same argument used at the end of the previous solution. Problem 7. Find all complex numbers z such that |z − |z + 1|| = |z + |z − 1||.
5.1. Problems Involving Moduli and Conjugates
167
Solution. We have |z − |z + 1|| = |z + |z − 1|| if and only if |z − |z + 1||2 = |z + |z − 1||2 , i.e., (z − |z + 1|) · (z − |z + 1|) = (z + |z − 1|) · (z + |z − 1|). The last equation is equivalent to z · z − (z + z)|z + 1| + |z + 1|2 = z · z + (z + z) · |z − 1| + |z − 1|2 . This can be written as |z + 1|2 − |z − 1|2 = (z + z) · (|z + 1| + |z − 1|), i.e.,
(z + 1)(z + 1) − (z − 1)(z − 1) = (z + z) · (|z + 1| + |z − 1|).
The last equation is equivalent to 2(z + z) = (z + z) · (|z + 1| + |z − 1|), i.e., z + z = 0, or |z + 1| + |z − 1| = 2. The triangle inequality 2 = |(z + 1) − (z − 1)| ≤ |z + 1| + |z − 1| shows that the solutions to the equation |z + 1| + |z − 1| = 2 satisfy z + 1 = t (1 − z), where t is a real number and t ≥ 0. t −1 , so z is any real number with −1 ≤ z ≤ 1. It follows that z = t +1 The equation z + z = 0 has the solutions z = bi, b ∈ R. Hence, the solutions to the equation are {bi : b ∈ R} ∪ {a ∈ R : a ∈ [−1, 1]}. Problem 8. Let z 1 , z 2 , . . . , z n be complex numbers such that |z 1 | = |z 2 | = · · · = |z n | > 0. Prove that n n
zj Re =0 z j=1 k=1 k if and only if
n
z k = 0.
k=1
(Romanian Mathematical Olympiad – Second Round, 1987)
168
5. Olympiad-Caliber Problems
Solution. Let S= Then S=
n
n n
zj . z j=1 k=1 k
zk
k=1
n
1 · z k=1 k
,
and since z k · z k = r 2 for all k, we have n n
zk zk · S= r2 k=1 k=1 1 = 2 r
n
zk
k=1
n
k=1
zk
1 = 2 r
2 n
zk . k=1
Hence S is a real number, so ReS = S = 0 if and only if
n
z k = 0.
k=1
Problem 9. Let λ be a real number and let n ≥ 2 be an integer. Solve the equation λ(z + z n ) = i(z − z n ). Solution. The equation is equivalent to z n (λ + i) = z(−λ + i). Taking the absolute values of both sides of the equation, we obtain |z|n = |z| = |z|, hence |z| = 0 or |z| = 1. 1 If |z| = 0, then z = 0 which satisfies the equation. If |z| = 1, then z = and the z equation may be rewritten as −λ + i z n+1 = . λ+i −λ + i = 1, there exists t ∈ [0, 2π ) such that Because λ+i −λ + i = cos t + i sin t. λ+i Then z k = cos
t + 2kπ t + 2kπ + i sin n+1 n+1
for k = 0, 1, . . . , n are the other solutions to the equation (besides z = 0).
5.1. Problems Involving Moduli and Conjugates
169
Problem 10. Prove that 6z − i 1 2 + 3i z ≤ 1 if and only if |z| ≤ 3 . Solution. We have 6z − i 2 + 3i z ≤ 1 if and only if |6z − i| ≤ |2 + 3i z|. The last inequality is equivalent to |6z − i|2 ≤ |2 + 3i z|2 , i.e., (6z − i)(6z + i) ≤ (2 + 3i z)(2 − 3i z). We find 36z · z + 6i z − 6i z + 1 ≤ 4 − 6i z + 6i z + 9zz, i.e., 27z · z ≤ 3. Finally, zz ≤
1 9
or, equivalently, |z| ≤ 13 , as desired.
Problem 11. Let z be a complex number such that z ∈ C \ R and 1 + z + z2 ∈ R. 1 − z + z2 Prove that |z| = 1. Solution. We have z z 1 + z + z2 =1+2 ∈ R if and only if ∈ R. 2 2 1−z+z 1−z+z 1 − z + z2 That is, 1 − z + z2 1 1 = − 1 + z ∈ R, i.e., z + ∈ R. z z z The last relation is equivalent to z+
1 1 = z + , i.e., (z − z)(1 − |z|2 ) = 0. z z
We find z = z or |z| = 1. Because z is not a real number, it follows that |z| = 1, as desired. Problem 12. Let z 1 , z 2 , . . . , z n be complex numbers such that |z 1 | = · · · = |z n | = 1 n n
1 . zk · z= z k=1 k=1 k Prove that z is a real number and 0 ≤ z ≤ n 2 .
170
5. Olympiad-Caliber Problems
1 for all k = 1, . . . , n. Because zk n n n n
1 1 = z= zk z k = z, z z k=1 k=1 k k=1 k k=1
Solution. Note that z k =
it follows that z is a real number. Let z k = cos αk + i sin αk , where αk are real numbers. for k = 1, n. Then n n n n
z= cos αk + i sin αk cos αk − i sin αk k=1
k=1
=
n
k=1
2 cos αk
+
k=1
n
k=1
2 sin αk
≥ 0.
k=1
On the other hand, we have z=
n
(cos2 αk + sin2 αk ) + 2
k=1
=n+2
(cos αi cos α j + sin αi sin α j )
1≤i< j≤n
cos(αi − α j ) ≤ n + 2
1≤i< j≤n
n n(n − 1) =n+2 = n2, 2 2
as desired. Remark. An alternative solution to the inequalities 0 ≤ z ≤ n 2 is as follows: n n n n
1 z= zk zk zk = z k=1 k=1 k k=1 k=1 =
n
zk
k=1
n
k=1
zk
2 2 n n
= z ≤ |z k | = n 2 , k=1 k k=1
so 0 ≤ z ≤ n 2 . Problem 13. Let z 1 , z 2 , z 3 be complex numbers such that z 1 + z 2 + z 3 = 0
Prove that Re
1 1 1 + + z1 z2 z3
|z 1 | = |z 2 | = |z 3 |.
and
· Re
1 z1 + z2 + z3
Solution. Let r = |z 1 | = |z 2 | = |z 3 | > 0. Then z1 z1 = z2 z2 = z3 z3 = r 2
≥ 0.
5.1. Problems Involving Moduli and Conjugates
and
171
1 z1 + z2 + z3 z1 + z2 + z3 1 1 + + = = . 2 z1 z2 z3 r r2
On the other hand, we have 1 z1 + z2 + z3 = z1 + z2 + z3 |z 1 + z 2 + z 3 |2 and, consequently, 1 1 1 1 · Re + + Re z1 z2 z3 z1 + z2 + z3 z1 + z2 + z3 z1 + z2 + z3 (Re(z 1 + z 2 + z 3 ))2 = Re · Re = ≥ 0, r2 |z 1 + z 2 + z 3 |2 r 2 |z 1 + z 2 + z 3 |2
as desired. Problem 14. Let x, y, z be complex numbers. a) Prove that |x| + |y| + |z| ≤ |x + y − z| + |x − y + z| + | − x + y + z|. b) If x, y, z are distinct and the numbers x + y − z, x − y + z, −x + y + z have equal absolute values, prove that 2(|x| + |y| + |z|) ≤ |x + y − z| + |x − y + z| + | − x + y + z|. Solution. Let m = −x + y + z,
n = x − y + z,
We have
n+p , 2 a) Adding the inequalities x=
|x| ≤
1 (|n| + | p|), 2
y=
|y| ≤
m+p , 2
p = x + y − z.
z=
1 (|m| + | p|), 2
m+n . 2
|z| ≤
1 (|m| + |n|) 2
yields |x| + |y| + |z| ≤ |m| + |n| + | p|, as desired. b) Let A, B, C be the points with coordinates m, n, p and observe that numbers m, n, p are distinct and that |m| = |n| = | p| = R, the circumradius of triangle ABC. Let the origin of the complex plane be the circumcenter of triangle ABC.
172
5. Olympiad-Caliber Problems
The orthocenter H of triangle ABC has the coordinate h = m + n + p. The desired inequality becomes |h − m| + |h − n| + |h − p| ≤ |m| + |n| + | p| or AH + B H + C H ≤ 3R. This is equivalent to cos A + cos B + cos C ≤
3 . 2
(1)
Inequality (1) can be written as 2 cos or
C A−B 3 A+B cos + 1 − 2 sin2 ≤ 2 2 2 2
C A−B 0 ≤ 2 sin − cos 2 2
2 + sin2
A−B , 2
which is clear. We have equality in (1) if and only if triangle ABC is equilateral, i.e., 2π 2π m = a, n = aε, p = aε2 , where a is a complex parameter and ε = cos + i sin . 3 3 a a a 2 In this case x = − , y = − ε, z = − ε . 2 2 2 Problem 15. Let z 0 , z 1 , z 2 , . . . , z n be complex numbers such that (k + 1)z k+1 − i(n − k)z k = 0 for all k ∈ {0, 1, 2, . . . , n − 1}. 1) Find z 0 such that z 0 + z 1 + · · · + z n = 2n . 2) For the value of z 0 determined above, prove that |z 0 |2 + |z 1 |2 + · · · + |z n |2 <
(3n + 1)n . n!
Solution. a) Use induction to prove that n zk = i k z 0 , for all k ∈ {0, 1, . . . , n}. k Then z 0 + z 1 + · · · + z n = 2n if and only if z 0 (1 + i)n = 2n , i.e., z 0 = (1 − i)n .
5.1. Problems Involving Moduli and Conjugates
173
b) Applying the AM-GM inequality, we have 2 2 n 2 n n 2 2 2 2 |z 0 | + |z 1 | + · · · + |z n | = |z 0 | + + ··· + 0 1 n 2n 2n 2n = 2n · = 2n(2n − 1) · · · (n + 1) = |z 0 |2 · n n n! n 2 (3n + 1)n 2n + (2n − 1) + · · · + (n + 1) n < = , n! n n! as desired. Problem 16. Let z 1 , z 2 , z 3 be complex numbers such that z 1 + z 2 + z 3 = z 1 z 2 + z 2 z 3 + z 3 z 1 = 0. Prove that |z 1 | = |z 2 | = |z 3 |. Solution. Substituting z 1 + z 2 = −z 3 in z 1 z 2 + z 3 (z 1 + z 3 ) = 0 gives z 1 z 2 = z 32 , so |z 1 | · |z 2 | = |z 3 |2 . Likewise, |z 2 | · |z 3 | = |z 1 |2 and |z 3 ||z 1 | = |z 2 |2 . Then |z 1 |2 + |z 2 |2 + |z 3 |2 = |z 1 ||z 2 | + |z 2 ||z 3 | + |z 3 ||z 1 |, i.e., (|z 1 | − |z 2 |)2 + (|z 2 | − |z 3 |)2 + (|z 3 | − |z 1 |)2 = 0, yielding |z 1 | = |z 2 | = |z 3 |. Alternate solution. Using the relations between the roots and the coefficients, it follows that z 1 , z 2 , z 3 are the roots of polynomial z 3 − p, where p = z 1 z 2 z 3 . Hence z 13 − p = z 23 − p = z 33 − p = 0, implying z 13 = z 23 = z 33 , and the conclusion follows. Problem 17. Prove that for all complex numbers z with |z| = 1 the following inequalities hold: √ 2 ≤ |1 − z| + |1 + z 2 | ≤ 4. Solution. Setting z = cos t + i sin t yields √ t 2 2 |1 − z| = (1 − cos t) + sin t = 2 − 2 cos t = 2 sin 2 √ (1 + cos 2t)2 + sin2 2t = 2 + 2 cos 2t 2 t = 2| cos t| = 2 1 − 2 sin . 2 √ 2 t It suffices to prove that ≤ |a| + |1 − 2a 2 | ≤ 2, for a = sin ∈ [−1, 1]. We 2 2 leave this to the reader. and
|1 + z 2 | =
174
5. Olympiad-Caliber Problems
Problem 18. Let z 1 , z 2 , z 3 , z 4 be distinct complex numbers such that Re
z2 − z3 z2 − z1 = Re = 0. z4 − z1 z4 − z3
a) Find all real numbers x such that |z 1 − z 2 |x + |z 1 − z 4 |x ≤ |z 2 − z 4 |x ≤ |z 2 − z 3 |x + |z 4 − z 3 |x . b) Prove that |z 3 − z 1 | ≤ |z 4 − z 2 |. Solution. Consider the points A, B, C, D with coordinates z 1 , z 2 , z 3 , z 4 , respectively. The conditions z2 − z3 z2 − z1 = Re =0 Re z4 − z1 z4 − z3 imply B AD = BC D = 90◦ . Then |z 1 − z 2 | = AB and |z 1 − z 4 | = AD are the lengths of the sides of the right triangle AB D with hypotenuses B D = |z 2 − z 4 |. The inequality AB x + AD x ≤ B D x holds for x ≥ 2. Similarly, |z 2 − z 3 | = BC and |z 4 − z 3 | = C D are the sides of the right triangle BC D, so the inequality B D x ≤ BC x + C D x holds for x ≤ 2. Consequently, x = 2. Finally, AC = |z 3 − z 1 | ≤ B D = |z 4 − z 2 |, since AC is a chord in the circle of diameter B D. Problem 19. Let x and y be distinct complex numbers such that |x| = |y|. Prove that 1 |x + y| < |x|. 2 Solution. Let x = a + ib and y = c + id, with a, b, c, d ∈ R and a 2 + b2 = c2 + d 2 . The inequality is equivalent to (a + c)2 + (b + d)2 < 4(a 2 + b2 ) or (a − c)2 + (b − d)2 > 0, which is clear, since x = y. Alternate solution. Consider points X (x) and Y (y). In triangle X OY we have O X = OY . Hence O M < O X , where M is the midpoint of segment [X Y ]. The x+y coordinate of point M is , and the desired inequality follows. 2 Problem 20. Consider the set A = {z ∈ C : z = a + bi, a > 0, |z| < 1}.
5.1. Problems Involving Moduli and Conjugates
175
Prove that for any z ∈ A there is a number x ∈ A such that z= Solution. Let z ∈ A. The equation z = x=
1−x . 1+x 1−x has the root 1+x
1 − a − ib 1−z = , 1+z 1 + a + ib
where a > 0 and a 2 + b2 < 1. To prove that x ∈ A, it suffices to show that |x| < 1 and Re(x) > 0. Indeed, we have (1 − a)2 + b2 |x|2 = < 1 if and only if (1 − a)2 < (1 + a)2 , (1 + a)2 + b2 i.e., 0 < 4a, as needed. 1 − |z|2 Moreover, Re(x) = > 0, since |z| < 1. |1 + z|2 Here are more problems involving moduli and conjugates of complex numbers. Problem 21. Consider the set A = {z ∈ C : |z| < 1}, a real number a with |a| > 1, and the function f : A → A,
f (z) =
1 + az . z+a
Prove that f is bijective. Problem 22. Let z be a complex number such that |z| = 1 and both Re(z) and Im(z) are rational numbers. Prove that |z 2n − 1| is rational for all integers n ≥ 1. Problem 23. Consider the function f : R → C,
f (t) =
1 + ti . 1 − ti
Prove that f is injective and determine its range. z1 . z2 Problem 25. Prove that for any complex numbers z 1 , z 2 , . . . , z n the following inequality holds: 12 0 |z 1 | + |z 2 | + · · · + |z n | + |z 1 + z 2 + · · · + z n | 1 0 ≥ 2 |z n |2 + · · · + |z n |2 + |z 1 + z 2 + · · · + z n |2 .
Problem 24. Let z 1 , z 2 ∈ C∗ such that |z 1 + z 2 | = |z 1 | = |z 2 |. Compute
176
5. Olympiad-Caliber Problems
Problem 26. Let z 1 , z 2 , . . . , z 2n be complex numbers such that |z 1 | = |z 2 | = · · · = |z 2n | and arg z 1 ≤ arg z 2 ≤ · · · ≤ arg z 2n ≤ π . Prove that |z 1 + z 2n | ≤ |z 2 + z 2n−1 | ≤ · · · ≤ |z n + z n+1 |. Problem 27. Find all positive real numbers x and y satisfying the system of equations √ 1 = 2, 3x 1 + x+y √
1 7y 1 − = 4 2. x+y (1996 Vietnamese Mathematical Olympiad)
Problem 28. Let z 1 , z 2 , z 3 be complex numbers. Prove that z 1 + z 2 + z 3 = 0 if and only if |z 1 | = |z 2 + z 3 |, |z 2 | = |z 3 + z 1 | and |z 3 | = |z 1 + z 2 |. Problem 29. Let z 1 , z 2 , . . . , z n be distinct complex numbers with the same modulus such that z 3 z 4 · · · z n−1 z n + z 1 z 4 · · · z n−1 z n + · · · + z 1 z 2 · · · z n−2 = 0. Prove that z 1 z 2 + z 2 z 3 + · · · + z n−1 z n = 0. Problem 30. Let a and z be complex numbers such that |z + a| = 1. Prove that |z 2 + a 2 | ≥
|1 − 2|a|| . √ 2
Problem 31. Find the geometric images of the complex numbers z for which z n · Re(z) = z n · Im(z), where n is an integer. Problem 32. Let a, b be real numbers with a + b = 1 and let z 1 , z 2 be complex numbers with |z 1 | = |z 2 | = 1. Prove that |z 1 + z 2 | |az 1 + bz 2 | ≥ . 2 Problem 33. Let k, n be positive integers and let z 1 , z 2 , . . . , z n be nonzero complex numbers with the same modulus such that z 1k + z 2k + · · · + z nk = 0. Prove that
1 z 1k
+
1 z 2k
+ ··· +
1 = 0. z nk
5.2. Algebraic Equations and Polynomials
5.2
177
Algebraic Equations and Polynomials
Problem 1. Consider the quadratic equation a 2 z 2 + abz + c2 = 0 b where a, b, c ∈ C∗ and denote by z 1 , z 2 its roots. Prove that if is a real number then c z1 ∈ R. |z 1 | = |z 2 | or z2 b Solution. Let t = ∈ R. Then b = tc and c = (ab)2 − 4a 2 · c2 = a 2 c2 (t 2 − 4). If |t| ≥ 2, the roots of the equation are z 1,2
√
−tac ± ac t 2 − 4 c = = t 2 − 4), (−t ± 2a 2a 2
z1 is a real number. z2 If |t| < 2, the roots of the equation are
and it is obvious that
z 1,2 =
c (−t ± i 4 − t 2 ), 2a
|c| , as claimed. |a| Problem 2. Let a, b, c, z be complex numbers such that |a| = |b| = |c| > 0 and az 2 + bz + c = 0. Prove that √ √ 5−1 5+1 ≤ |z| ≤ . 2 2 hence |z 1 | = |z 2 | =
Solution. Let r = |a| = |b| = |c| > 0. We have |az 2 | = | − bz − c| ≤ |b||z| + |c|, √ 1+ 5 hence . ≤ r |z| + r . It follows that − |z| − 1 ≤ 0, so |z| ≤ 2 2 2 2 On the other √ hand, |c| = | − az − bz| ≤ |a||z| + b|z|, such that |z| + |z| − 1 ≥ 0. 5−1 Thus |z| ≥ , and we are done. 2 Problem 3. Let p, q be complex numbers such that | p|+|q| < 1. Prove that the moduli of the roots of the equation z 2 + pz + q = 0 are less than 1. r |z 2 |
|z|2
178
5. Olympiad-Caliber Problems
Solution. Because z 1 + z 2 = − p and z 1 z 2 = q, the inequality | p| + |q| < 1 implies |z 1 + z 2 | + |z 1 z 2 | < 1. But ||z 1 | − |z 2 || ≤ |z 1 + z 2 |, hence |z 1 | − |z 2 | + |z 1 ||z 2 | − 1 < 0 if and only if (1 + |z 2 |)(|z 1 | − 1) < 0 and |z 2 | − |z 1 | + |z 2 ||z 1 | − 1 < 0 if and only if (1 + |z 1 |)(|z 2 | − 1) < 0. Consequently, |z 1 | < 1 and |z 2 | < 1, as desired. Problem 4. Let f = x 2 + ax + b be a quadratic polynomial with complex coefficients with both roots having modulus 1. Prove that f = x 2 +|a|x +|b| has the same property. Solution. Let x1 and x2 be the complex roots of the polynomial f = x 2 + ax + b and let y1 and y2 be the complex roots of the polynomial g = x 2 + |a|x + |b|. We have to prove that if |x1 | = |x2 | = 1, then |y1 | = |y2 | = 1. Since x1 ·x2 = b and x1 +x2 = −a, then |b| = |x1 ||x2 | = 1 and |a| ≤ |x1 |+|x2 | = 2. The quadratic polynomial g = x 2 + |a|x + 1 has the discriminant = |a|2 − 4 ≤ 0, hence
−|a| ± i 4 − |a|2 y1,2 = . 2 It is easy to see that |y1 | = |y2 | = 1, as desired. Problem 5. Let a, b be nonzero complex numbers. Prove that the equation az 3 + bz 2 + bz + a = 0 has at least one root with absolute value equal to 1. 1 is also a root of the z 1 1 1 equation. Consequently, if z 1 , z 2 , z 3 are the roots of the equation, then , , are z1 z2 z3 the same roots, not necessarily in the same order. 1 1 for some k = 1, 2, 3, then |z k |2 = z k z k = 1 and we are done. If z k = If z k = zk zk for all k = 1, 2, 3, we may consider without loss of generality that Solution. Observe that if z is a root of the equation, then
z1 =
1 , z2
z2 =
1 , z3
z3 =
1 . z1
The first two equalities yield z 1 · z 2 · z 2 · z 3 = 1, hence |z 1 | · |z 2 |2 · |z 3 | = 1. On the a other hand, z 1 z 2 z 3 = − , so |z 1 ||z 2 ||z 3 | = 1. It follows that |z 2 | = 1, as claimed. a Problem 6. Let f = x 4 + ax 3 + bx 2 + cx + d be a polynomial with real coefficients and real roots. Prove that if | f (i)| = 1 then a = b = c = d = 0.
5.2. Algebraic Equations and Polynomials
179
Solution. Let x1 , x2 , x3 , x4 be the real roots of the polynomial f . Then f = (x − x1 )(x − x2 )(x − x3 )(x − x4 ) and | f (i)| =
1 + x12 ·
1 + x22 ·
1 + x32 ·
1 + x42 .
Because | f (i)| = 1, we deduce that x1 = x2 = x3 = x4 = 0 and consequently a = b = c = d = 0, as desired. Problem 7. Prove that if 11z 10 + 10i z 9 + 10i z − 11 = 0, then |z| = 1. (1989 Putnam Mathematical Competition)
Solution. The equation can be rewritten as z 9 =
11 − 10i z . If z = a + bi, then 11z + 10i
2 2 2 2 11 − 10i z = 11 + 220b + 10 (a + b ) . |z| = 11z + 10i 112 (a 2 + b2 ) + 220b + 102 9
Let f (a, b) and g(a, b) denote the numerator and denominator of the right-hand side. If |z| > 1, then a 2 + b2 > 1, so g(a, b) > f (a, b), leading to |z 9 | < 1, a contradiction. If |z| < 1, then a 2 + b2 < 1, so g(a, b) < f (a, b), yielding |z 9 | > 1, again a contradiction. Hence |z| = 1. Problem 8. Let n ≥ 3 be an integer and let a be a nonzero real number. Show that any nonreal root z of the equation x n + ax + 1 = 0 satisfies the inequality 1 n . |z| ≥ n−1 (Romanian Mathematical Olympiad – Final Round, 1995)
Solution. Let z = r (cos α + i sin α) be a nonreal root of the equation, where α ∈ (0, 2π) and α = π. Substituting back into the equation we find r n cos nα + ra cos α + 1 + i(r n sin nα + ra sin α) = 0. Hence r n cos nα + ra cos α + 1 = 0 and r n sin nα + ra sin α = 0. Multiplying the first relation by sin α, the second by cos α, and then subtracting them we find that r n sin(n − 1)α = sin α. It follows that r n | sin(n − 1)α| = | sin α|. The inequality | sin kα| ≤ k| sin α| is valid for any positive integer k. The proof is based on a simple inductive argument on k. | sin α| ≤ Applying this inequality, from r n | sin(n − 1)α| = | sin α|, we obtain 1 1 r n (n − 1)| sin α|. Because sin α = 0, it follows that r n ≥ , i.e., |z| ≥ n . n−1 n−1
180
5. Olympiad-Caliber Problems
Problem 9. Suppose P is a polynomial with complex coefficients and an even degree. If all the roots of P are complex nonreal numbers with modulus 1, prove that P(1) ∈ R if and only if P(−1) ∈ R. P(1) ∈ R. P(−1) be the roots of P. Then
Solution. It suffices to prove that Let x1 , x2 , . . . , x2n
P(x) = λ(x − x1 )(x − x2 ) · · · (x − x2n ) for some λ ∈ C∗ , and 2n 1 − xk λ(1 − x1 )(1 − x2 ) · · · (1 − x2n ) P(1) . = = P(−1) λ(−1 − x1 )(−1 − x2 ) · · · (−1 − x2n ) k=1 1 + xk
From the hypothesis we have |xk | = 1 for all k = 1, 2, . . . , 2n. Then
hence
1 − xk 1 + xk
1 − xk = = 1 + xk
P(1) P(−1)
=
1 xk − 1 1 − xk xk = , =− 1 xk + 1 1 + xk 1+ xk 1−
2n 1 − xk k=1
= (−1)2n
1 + xk
=
2n 1 − xk − 1 + xk k=1
2n 1 − xk P(1) = . 1 + x P(−1) k k=1
P(1) is a real number, as desired. P(−1) Problem 10. Consider the sequence of polynomials defined by P1 (x) = x 2 − 2 and P j (x) = P1 (P j−1 (x)) for j = 2, 3, . . . . Show that for any positive integer n the roots of equation Pn (x) = x are all real and distinct. This proves that
(18th IMO – Shortlist)
Solution. Put x = z + z −1 , where z is a nonzero complex number. Then P1 (x) = − 2 = (z + z −1 )2 − 2 = z 2 + z −2 . A simple inductive argument shows that for all n n positive integers n we have Pn (x) = z 2 + z −2 . n n n The equation Pn (x) = x is equivalent to z 2 + z −2 = z + z −1 . We obtain z 2 − z = n n n n n n z −1 −z −2 , i.e., z(z 2 −1 −1) = z −2 (z 2 −1 −1). It follows that (z 2 −1 −1)(z 2 +1 −1) = n 0. Because gcd(2n −1, 2n +1) = 1 the unique common root of equations z 2 −1 −1 = 0 n and z 2 +1 − 1 = 0 is z = 1 (see Proposition 1 in Section 2.2). Moreover, for any root x2
5.3. From Algebraic Identities to Geometric Properties
181
of equation (z 2 −1 − 1)(z 2 +1 − 1) = 0 we have |z| = 1, i.e., z −1 = z. Also, observe n n that for two roots z and w of (z 2 −1 − 1)(z 2 +1 − 1) = 0 which are different from 1, we have z + z −1 = w + w−1 if and only if (z − w)(1 − (zw)−1 ) = 0. This is equivalent to zw = 1, i.e., w = z −1 = z, a contradiction to the fact that the unique common root n n of z 2 −1 − 1 = 0 and z 2 +1 − 1 = 0 is 1. It is clear that the degree of polynomial Pn is 2n . As we have seen before, all the 2kπ 2kπ +i sin n , roots of Pn (x) = x are given by x = z+z −1 , where z = 1, z = cos n 2 −1 2sπ 2sπ + i sin n , s = 1, . . . , 2n . k = 1, . . . , 2n − 2 and z = cos n 2 +1 2 +1 Taking into account the symmetry of expression z + z −1 , the total number of these 1 1 roots is 1 + (2n − 2) + 2n = 2n and all of them are real and distinct. 2 2 n
n
Here are other problems involving algebraic equations and polynomials. Problem 11. Let a, b, c be complex numbers with a = 0. Prove that if the roots of the equation az 2 + bz + c = 0 have equal moduli then ab|c| = |a|bc. Problem 12. Let z 1 , z 2 be the roots of the equation z 2 + z + 1 = 0 and let z 3 , z 4 be the roots of the equation z 2 − z + 1 = 0. Find all integers n such that z 1n + z 2n = z 3n + z 4n . Problem 13. Consider the equation with real coefficients x 6 + ax 5 + bx 4 + cx 3 + bx 2 + ax + 1 = 0, and denote by x1 , x2 , . . . , x6 the roots of the equation. Prove that 6 (xk2 + 1) = (2a − c)2 . k=1
Problem 14. Let a and b be complex numbers and let P(z) = az 2 + bz + i. Prove that there is a z 0 ∈ C with |z 0 | = 1 such that |P(z 0 )| ≥ 1 + |a|. Problem 15. Find all polynomials f with real coefficients satisfying, for any real number x, the relation f (x) f (2x 2 ) = f (2x 3 + x). (21st IMO – Shortlist)
5.3
From Algebraic Identities to Geometric Properties
Problem 1. Consider equilateral triangles ABC and A B C , both in the same plane and having the same orientation. Show that the segments [A A ], [B B ], [CC ] can be the sides of a triangle.
182
5. Olympiad-Caliber Problems
Solution. Let a, b, c be the coordinates of vertices A, B, C and let a , b , c be the coordinates of vertices A , B , C . Because triangles ABC and A B C are similar, we have the relation (see Remark 1 in Section 3.3): 1 1 1 (1) a b c = 0. a b c That is, a (b − c) + b (c − a) + c (a − b) = 0.
(2)
On the other hand the following relation is clear: a(b − c) + b(c − a) + c(a − b) = 0.
(3)
By subtracting relation (3) from relation (2), we find (a − a)(b − c) + (b − b)(c − a) + (c − c)(a − b) = 0.
(4)
Passing to moduli, it follows that |a − a||b − c| ≤ |b − b||c − a| + |c − c||a − b|.
(5)
Taking into account that |b − c| = |c − a| = |a − b|, we obtain A A ≤ B B + CC . Similarly we prove the inequalities B B ≤ CC + A A and CC ≤ A A + B B , hence the desired conclusion follows. Remarks. 1) If ABC and A B C are two similar triangles situated in the same plane and having the same orientation, then from (5) the inequality A A · BC ≤ B B · C A + CC · AB
(6)
follows. This is the generalized Ptolemy inequality. Ptolemy’s inequality is obtained when the triangle A B C degenerates to a point. 2) Taking into account the inequality (6), we have also B B · C A ≤ CC · AB + A A · BC and CC · AB ≤ A A · BC + B B · C A. It follows that for any two similar triangles ABC and A B C with the same orientation and situated in the same plane, we can construct a triangle of sides lengths A A · BC, B B · C A, CC · AB. 3) In the case when the triangle A B C degenerates to the point M, from the property in our problem it follows that the segments M A, M B, MC are the sides of a triangle, i.e., Pompeiu’s theorem (see also Subsection 4.9.1). Problem 2. Let P be an arbitrary point in the plane of a triangle ABC. Then α · P B · PC + β · PC · P A + γ · P A · P B ≥ αβγ , where α, β, γ are the sides of ABC.
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183
Solution. Let us consider the origin of the complex plane at P and let a, b, c be the coordinates of vertices of triangle ABC. From the algebraic identity bc ca ab + + = 1. (a − b)(a − c) (b − c)(b − a) (c − a)(c − b)
(1)
Passing to the absolute value, it follows that |c||a| |a||b| |b||c| + + ≥ 1. |a − b||a − c| |b − c||b − a| |c − a||c − b|
(2)
Taking into account that |a| = P A, |b| = P B, |c| = PC, and |b −c| = α, |c −a| = β, |a − b| = γ , the inequality (2) is equivalent to PC · P A PA · PB P B · PC + + ≥ 1, βγ γα αβ i.e., the desired inequality. Remarks. 1) If P is the circumcenter O of triangle ABC, we can derive Euler’s inequality R ≥ 2r . Indeed, in this case the inequality is equivalent to R 2 (α + β + γ ) ≥ αβγ . Therefore R2 ≥
αβγ 4R αβγ area[ABC] αβγ = = · = 2R · = 2Rr, α+β +γ 2s 2s 4R s
hence R ≥ 2r . 2) If P is the centroid G of triangle ABC we obtain the following inequality involving the medians m α , m β , m γ : mβ mγ mγ mα mα mβ 9 + + ≥ αβ βγ γα 4 with equality if and only if triangle ABC is equilateral. A good argument for the case of acute-angled triangles is given in the next problem. Problem 3. Let ABC be an acute-angled triangle and let P be a point in its interior. Prove that α · P B · PC + β · PC · P A + γ · P A · P B = αβγ , if and only if P is the orthocenter of triangle ABC.
(1998 Chinese Mathematical Olympiad)
Solution. Let P be the origin of the complex plane and let a, b, c be the coordinates of A, B, C, respectively. The relation in the problem is equivalent to |ab(a − b)| + |bc(b − c)| + |ca(c − a)| = |(a − b)(b − c)(c − a)|.
184
5. Olympiad-Caliber Problems
Let z1 =
ab , (a − c)(b − c)
z2 =
bc , (b − a)(c − a)
z3 =
ca . (c − b)(a − b)
It follows that |z 1 | + |z 2 | + |z 3 | = 1
and
z 1 + z 2 + z 3 = 1,
the latter from identity (1) in the previous problem. We will prove that P is the orthocenter of triangle ABC if and only if z 1 , z 2 , z 3 are positive real numbers. Indeed, if P is the orthocenter, then since the triangle ABC is acute-angled, it follows that P is in the interior of ABC. Hence there are positive real numbers r1 , r2 , r3 such that b = −r2 i, c−a
a = −r1 i, b−c
c = −r3 i, a−b
implying z 1 = r1r2 > 0, z 2 = r2r3 > 0, z 3 = r3r1 > 0, and we are done. Conversely, suppose that z 1 , z 2 , z 3 are all positive real numbers. Because −
z1 z2 = z3
b c−a
2 ,
−
z2 z3 = z1
c a−b
2 ,
−
z3 z1 = z2
a b−c
2
a b c , , are pure imaginary numbers, thus A P ⊥ BC and b−c c−a a−b B P ⊥ C A, showing that P is the orthocenter of triangle ABC.
it follows that
Problem 4. Let G be the centroid of triangle ABC and let R1 , R2 , R3 be the circumradii of triangles G BC, GC A, G AB, respectively. Then R1 + R2 + R3 ≥ 3R, where R is the circumradius of triangle ABC. Solution. In Problem 2, consider P the centroid G of triangle ABC. Then α · G B · GC + β · GC · G A + γ · G A · G B ≥ αβγ , where α, β, γ are the lengths of the sides of triangle ABC. But 1 α · G B · GC = 4R1 · area[G BC] = 4R1 · area[ABC]. 3 Likewise, 1 β · GC · G A = 4R2 · area[ABC], 3
1 γ · G A · G B = 4R3 · area[ABC]. 3
(1)
5.3. From Algebraic Identities to Geometric Properties
185
Hence, the inequality (1) is equivalent to 4 (R1 + R2 + R3 ) · area[ABC] ≥ 4R · area[ABC], 3 i.e., R1 + R2 + R3 ≥ 3R. Problem 5. Let ABC be a triangle and let P be a point in its interior. Let R1 , R2 , R3 be the radii of the circumcircles of triangles P BC, PC A, P AB, respectively. Lines P A, P B, PC intersect sides BC, C A, AB at A1 , B1 , C1 , respectively. Let k1 =
P A1 , A A1
k2 =
P B1 , B B1
k3 =
PC1 . CC1
Prove that k1 R1 +k2 R2 +k3 R3 ≥ R, where R is the circumradius of triangle ABC. (2004 Romanian IMO Team Selection Test)
Solution. Note that k1 =
area[P BC] , area[ABC]
k2 =
area[PC A] , area[ABC]
k3 =
area[P AB] . area[ABC]
α · P B · PC αβγ . Two similar relations for and area[P BC] = 4R 4R1 area[PC A] and area[P AB] hold. The desired inequality is equivalent to But area[ABC] =
R
α · P B · PC β · PC · P A γ · PA · PB +R +R ≥R αβγ αβγ αβγ
which reduces to the inequality in Problem 2. In the case when triangle ABC is acute-angled, from Problem 3 it follows that equality holds if and only if P is the orthocenter of ABC. Problem 6. For any point M in the plane of triangle ABC the following inequality holds: AM 3 sin A + B M 3 sin B + C M 3 sin C ≥ 6 · M G · area[ABC], where G is the centroid of triangle ABC. Solution. The identity x 3 (y − z) + y 3 (z − x) + z 3 (x − y) = (x − y)(y − z)(z − x)(x + y + z)
(1)
holds for any complex numbers x, y, z. Passing to the absolute value we obtain the inequality |x 3 (y − z)| + |y 3 (z − x)| + |z 3 (x − y)| ≥ |x − y||y − z||z − x||x + y + z|.
(2)
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5. Olympiad-Caliber Problems
Let a, b, c, m be the coordinates of points A, B, C, M, respectively. In (2) consider x = m − a, y = m − b, z = m − c and obtain AM 3 · α + B M 3 · β + C M 3 · γ ≥ 3αβγ M G.
(3)
αβγ and the law of sines the desired inequality Using the formula area[ABC] = 4R follows from (3). Problem 7. Let ABC D be a cyclic quadrilateral inscribed in circle C(O; R) having the sides lengths α, β, γ , δ and the diagonals lengths d1 and d2 . Then area[ABC D] ≥
αβγ δd1 d2 . 8R 4
Solution. Take the center O to be the origin of the complex plane and consider a, b, c, d the coordinates of vertices A, B, C, D. From the well-known Euler identity
cyc
a3 =1 (a − b)(a − c)(a − d)
(1)
by passing to the absolute value, it follows that
cyc
|a|3 ≥ 1. |a − b||a − c||a − d|
(2)
The inequality (2) is equivalent to
R3 ≥1 AB · AC · AD
(3)
R 3 · B D · C D · BC ≥ αβγ δd1 d2 .
(4)
cyc
or
cyc
But we have the known relation B D · C D · BC = 4R · area[BC D] and three other such relations. The inequality (4) can be written in the form 4R 4 (area[ABC] + area[BC D] + area[C D A] + area[D AB]) ≥ αβγ δd1 d2 or equivalently 8R 4 area[ABC D] ≥ αβγ δd1 d2 . Problem 8. Let a, b, c be distinct complex numbers such that (a − b)7 + (b − c)7 + (c − a)7 = 0. Prove that a, b, c are the coordinates of the vertices of an equilateral triangle.
5.3. From Algebraic Identities to Geometric Properties
187
Solution. Setting x = a − b, y = b − c, z = c − a implies x + y + z = 0 and x y + y 7 + z 7 = 0. Since z = 0, we may set α = and β = . Hence α + β = −1 z z and α 7 + β 7 = −1. Then x7
α 6 − α 5 β + α 4 β 2 − α 3 β 3 + α 2 β 4 − αβ 5 + β 6 = 1.
(1)
Let s = α + β = −1 and p = ab. The relation (1) becomes (α 6 + β 6 ) − p(α 4 + β 4 ) + p 2 (α 2 + β 2 ) − p 3 = 1.
(2)
Because α 2 + β 2 = s 2 − 2 p = 1 − 2 p, α 4 + β 4 = (α 2 + β 2 )2 − 2α 2 β 2 = (1 − 2 p)2 − 2 p 2 = 1 − 4 p + 2 p 2 , α 6 + β 6 = (α 2 + β 2 )((α 4 + β 4 ) − α 2 β 2 ) = (1 − 2 p)(1 − 4 p + p 2 ), the equality (2) is equivalent to (1 − 2 p)(1 − 4 p + p 2 ) − p(1 − 4 p + 2 p 2 ) + p 2 (1 − 2 p) − p 3 = 1. That is, 1 − 4 p + p2 − 2 p + 8 p 2 − 2 p 3 − p + 4 p 2 − 2 p 3 + p 2 − 2 p 3 − p 3 = 1; i.e., −7 p 3 + 14 p 2 − 7 p + 1 = 1. We obtain −7 p( p − 1)2 = 0, hence p = 0 or p = 1. If p = 0, then α = 0 or β = 0, and consequently x = 0 or y = 0. It follows that a = b or b = c, which is false; hence p = 1. From αβ = 1 and α + β = −1 we deduce that α and β are the roots of the quadratic equation x 2 + x + 1 = 0. Thus α 3 = β 3 = 1 and |α| = |β| = 1. Therefore |x| = |y| = |z| or |a − b| = |b − c| = |c − a|, as claimed. Alternate solution. Let x = a − b, y = b − c, z = c − a. Because x + y + z = 0 and x 7 + y 7 + z 7 = 0, we find that (x + y)7 − x 7 − y 7 = 0. This is equivalent to 7x y(x + y)(x 2 + x y + y 2 )2 = 0. But x yz = 0, so x 2 + x y + y 2 = 0, i.e., x 3 = y 3 . From symmetry, x 3 = y 3 = z 3 , hence |x| = |y| = |z|. Problem 9. Let M be a point in the plane of the square ABC D and let M A = x, M B = y, MC = z, M D = t. Prove that the numbers x y, yz, zt, t x are the sides of a quadrilateral. Solution. Consider the complex plane such that 1, i, −1, −i are the coordinates of vertices A, B, C, D of the square. If z is the coordinate of point M, then we have the identity 1(z − i)(z + 1) + i(z + 1)(z + i) − 1(z + i)(z − 1) − i(z − 1)(z − i) = 0.
(1)
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5. Olympiad-Caliber Problems
Subtracting the first term of the sum from both sides yields i(z + 1)(z + i) − 1(z + i)(z − i) − i(z − 1)(z − i) = −1(z − i)(z + 1), and using the triangle inequality we obtain |z − i||z + i| + |z + 1||z + i| + |z + i||z − 1| ≥ |z − 1||z − i| or yz + zt + t x ≥ x y. In the same manner we prove that x y + zt + t x ≥ yz,
x y + yz + t x ≥ yz
and x y + yz + zt ≥ t x, as needed. Problem 10. Let z 1 , z 2 , z 3 be distinct complex numbers such that |z 1 | = |z 2 | = |z 3 | = R. Prove that 1 1 1 1 + + ≥ 2. |z 1 − z 2 ||z 1 − z 3 | |z 2 − z 1 ||z 2 − z 3 | |z 3 − z 1 ||z 3 − z 2 | R Solution. The following identity is easy to verify z 32 z 12 z 22 + + = 1. (z 1 − z 2 )(z 1 − z 3 ) (z 2 − z 1 )(z 2 − z 3 ) (z 3 − z 1 )(z 3 − z 2 ) Passing to the absolute value we find that
z 12 |z 1 |2 1= ≤ cyc (z 1 − z 2 )(z 1 − z 3 ) cyc |z 1 − z 2 ||z 1 − z 3 | = R2
cyc
1 , |z 1 − z 2 ||z 1 − z 3 |
i.e., the desired inequality. Alternate solution. Let α = |z 2 − z 3 |,
β = |z 3 − z 1 |,
γ = |z 1 − z 2 |.
From Problem 29 in Section 1.1 we have αβ + βγ + γ α ≤ 9R 2 . Using the inequality
(αβ + βγ + γ α)
it follows that
as desired.
1 1 1 + + αβ βγ γα
≥9
1 1 9 1 1 + + ≥ ≥ 2, αβ βγ γα αβ + βγ + γ α R
5.3. From Algebraic Identities to Geometric Properties
189
Remark. Consider the triangle with vertices at z 1 , z 2 , z 3 and whose circumcenter is the origin of the complex plane. Then the circumradius R equals |z 1 | = |z 2 | = |z 3 | and the sides are α = |z 2 − z 3 |,
β = |z 1 − z 3 |,
γ = |z 1 − z 2 |.
The above inequality is equivalent to 1 1 1 1 + + ≥ 2, αβ βγ γα R i.e.,
αβγ 4K 4sr = = . 2 R R R We obtain R ≥ 2r , i.e., Euler’s inequality for a triangle. α+β +γ ≥
Problem 11. Let ABC be a triangle and let P be a point in its plane. Prove that α · P A3 + β · P B 3 + γ · PC 3 ≥ 3αβγ · P G, where G is the centroid of ABC. 2) Prove that R 2 (R 2 − 4r 2 ) ≥ 4r 2 [8R 2 − (α 2 + β 2 + γ 2 )]. Solution. 1) The identity x 3 (y − z) + y 3 (z − x) + z 3 (x − y) = (x − y)(y − z)(z − x)(x + y + z)
(1)
holds for any complex numbers x, y, z. Passing to absolute values we obtain |x|3 |y − z| + |y|3 |z − x| + |z|3 |x − y| ≥ |x − y||y − z||z − x||x + y + z|. Let a, b, c, z P be the coordinates of A, B, C, P, respectively. In (2) take x = z P −a, y = z P − b, z = z P − c and obtain the desired inequality. 2) If P is the circumcenter O of triangle ABC, after some elementary transformations the previous inequality becomes R 2 ≥ 6r · OG. Squaring both sides yields 1 R 4 ≥ 36r 2 · OG 2 . Using the well-known relation OG 2 = R 2 − (α 2 + β 2 + γ 2 ) we 9 obtain R 4 ≥ 36R 2r 2 − 4r 2 (α 2 + β 2 + γ 2 ) and the conclusion follows. Remark. The inequality 2) improves Euler’s inequality for the class of obtuse triangles. This is equivalent to proving that α 2 + β 2 + γ 2 < 8R 2 in any such triangle. The last relation can be written as sin2 A + sin2 B + sin2 C < 2, or cos2 A + cos2 B − sin2 C > 0. That is, 1 + cos 2A 1 + cos 2B + − 1 + cos2 C > 0, 2 2 which reduces to cos(A + B) cos(A − B) + cos2 C > 0. This is equivalent to cos C[cos(A − B) − cos(A + B)] > 0, i.e., cos A cos B cos C < 0.
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5. Olympiad-Caliber Problems
Here are some other problems involving this topic. Problem 12. Let a, b, c, d be distinct complex numbers with |a| = |b| = |c| = |d| and a + b + c + d = 0. Then the geometric images of a, b, c, d are the vertices of a rectangle. Problem 13. The complex numbers z i , i = 1, 2, 3, 4, 5, have the same nonzero modulus and 5 5
zi = z i2 = 0. i=1
i=1
Prove that z 1 , z 2 , . . . , z 5 are the coordinates of the vertices of a regular pentagon. (Romanian Mathematical Olympiad – Final Round, 2003)
Problem 14. Let ABC be a triangle. a) Prove that if M is any point in its plane, then AM sin A ≤ B M sin B + C M sin C. b) Let A1 , B1 , C1 be points on the sides BC, AC and AB, respectively, such that the angles of the triangle A1 B1 C1 are in this order α, β, γ . Prove that
A A1 sin α ≤
cyc
BC sin α.
cyc
(Romanian Mathematical Olympiad – Second Round, 2003)
Problem 15. Let M and N be points inside triangle ABC such that M AB = N AC Prove that
and
MBA = N BC.
AM · AN BM · BN CM · CN + + = 1. AB · AC B A · BC CA · CB (39th IMO – Shortlist)
5.4
Solving Geometric Problems
Problem 1. On each side of a parallelogram a square is drawn external to the figure. Prove that the centers of the squares are the vertices of another square. Solution. Consider the complex plane with origin at the intersection point of the diagonals and let a, b, −a, −b be the coordinates of the vertices A, B, C, D, respectively.
5.4. Solving Geometric Problems
191
Using the rotation formulas, we obtain b = z O1 + (a − z O1 )(−i) or z O1 = Likewise, z O2 =
a − bi , 1+i
z O3 =
−b − ai , 1+i
b + ai . 1+i
z O4 =
−a + bi . 1+i
It follows that π z O2 − z O1 a − bi − b − ai O = arg i = , = arg 4 O1 O2 = arg z O4 − z O1 −a + bi − b − ai 2 so O1 O2 = O1 O4 , and z O4 − z O4 −a + bi + b + ai π = arg = arg i = , O 2 O3 O4 = arg z O2 − z O3 a − bi + b + ai 2 so O3 O4 = O3 O2 . Therefore O1 O2 O3 O4 is a square. Problem 2. Given a point on the circumcircle of a cyclic quadrilateral, prove that the products of the distances from the point to any pair of opposite sides or to the diagonals are equal. (Pappus’s theorem)
Solution. Let a, b, c, d be the coordinates of the vertices A, B, C, D of the quadrilateral and consider the complex plane with origin at the circumcenter of ABC D. Without loss of generality assume that the circumradius equals 1. The equation of line AB is a a 1 b b 1 = 0. z z 1 This is equivalent to z(a − b) − z(a − b) = ab − ab, i.e., z + abz = a + b. Let point M1 be the foot of the perpendicular from a point M on the circumcircle to the line AB. If m is the coordinate point M, then (see Proposition 1 in Section 4.5) z M1 = and
m − abm + a + b 2
m − abm + a + b (m − a)(m − b) d(M, AB) = |m − m 1 | = m − = , 2 2m
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5. Olympiad-Caliber Problems
since mm = 1. Likewise,
(m − c)(m − d) (m − b)(m − c) , d(M, BC) = , d(M, C D) = 2m 2m (m − d)(m − a) , d(M, AC) = (m − a)(m − c) d(M, D A) = 2m 2m (m − b)(m − d) . d(M, B D) = 2m
and
Thus, d(M, AB) · d(M, C D) = d(M, BC) · d(M, D A) = d(M, AC) · d(M, B D), as claimed. Problem 3. Three equal circles C1 (O1 ; r ), C2 (O2 ; r ) and C3 (O3 ; r ) have a common point O. Circles C1 and C2 , C2 and C3 , C3 and C1 , meet again at points A, B, C respectively. Prove that the circumradius of triangle ABC is equal to r . (Tzitzeica’s1 “five-coin problem”)
Solution. Consider the complex plane with origin at point O and let z 1 , z 2 , z 3 be the coordinates of the centers O1 , O2 , O3 , respectively. It follows that points A, B, C have the coordinates z 1 + z 2 , z 2 + z 3 , z 3 + z 1 , hence AB = |(z 1 + z 2 ) − (z 2 + z 3 )| = |z 1 − z 3 | = O1 O3 . Likewise, BC = O1 O2 and AC = O2 O3 , hence triangles ABC and O1 O2 O3 are congruent. Consequently, their circumradii are equal. Since O O1 = O O2 = O O3 = r , the circumradius of triangles O1 O2 O3 and ABC is equal to r , as desired. Problem 4. On the sides AB and BC of triangle ABC draw squares with centers D and E such that points C and D lie on the same side of line AB and points A and E lie opposite sides of line BC. Prove that the angle between lines AC and D E is equal to 45◦ . Solution. The rotation about E through angle 90◦ mappings point C to point B, hence z B − zC i z B = z E + (z C − z E )i and z E = . 1−i z B − z Ai . Similarly, z D = 1−i 1 Gheorghe Tzitzeica (1873–1939), Romanian mathematician, made important contributions in geometry.
5.4. Solving Geometric Problems
193
Figure 5.1.
The angle between the lines AC and D E is equal to arg
1−i π (z C − z A )(1 − i) zC − z A = arg = arg(1 + i) = , = arg zE − zD z B − zC i − z B + z A i −i 4
as desired. Remark. If the squares are replaced in the same conditions by rectangles with centers D and E, then the angle between lines AC and D E is equal to 90◦ − B AD. Problem 5. On the sides AB and BC of triangles ABC equilateral triangles AB N and AC M are drawn external to the figure. If P, Q, R are the midpoints of segments BC, AM, AN , respectively, prove that triangle P Q R is equilateral. Solution. Consider the complex plane with origin at A and denote by a lowercase letter the coordinate of the point denoted by an uppercase letter. The rotation about center A through angle 60◦ maps points N and C to B and M, respectively. Setting ε = cos 60◦ + i sin 60◦ , we have b = n · ε and m = c · ε. Thus p=
b+c , 2
q=
m c·ε = , 2 2
r=
n b bε5 bε2 = = =− . 2 2ε 2 2
To prove that triangle P Q R is equilateral, using Proposition 1 in Section 3.4, it suffices to observe that p 2 + q 2 + r 2 = pq + qr + r p. Problem 6. Let A A B B CC be a hexagon inscribed in the circle C(O; R) such that A A = B B = CC = R.
194
5. Olympiad-Caliber Problems
Figure 5.2.
If M, N , P are midpoints of sides A A , B B , CC respectively, prove that triangle M N P is equilateral. Solution. Consider the complex plane with origin at the circumcenter O and let a, b, c, a , b , c be the coordinates of the vertices A, B, C, A , B , C , respectively. If ε = cos 60◦ + i sin 60◦ , then a = a · ε,
b = b · ε,
c = c · ε.
The points M, N , P have the coordinates m=
aε + b , 2
n=
bε + c , 2
p=
cε + a . 2
It is easy to observe that m 2 + n 2 + p 2 = mn + np + pm; therefore M N P is an equilateral triangle (see Proposition 1 in Section 3.4). Problem 7. On the sides AB and AC of triangle ABC squares AB D E and AC F G are drawn external to the figure. If M is the midpoint of side BC, prove that AM ⊥ E G and E G = 2AM. Solution. Consider the complex plane with origin at A and let b, c, g, e, m be the coordinates of points B, C, G, E, M. b+c Observe that g = ci, e = −bi, m = , hence 2 m−a −(b + c) i = = ∈ iR∗ g−e 2i(b + c) 2 and |m − a| = Thus, AM ⊥ E G and 2AM = E G.
1 |e − g|. 2
5.4. Solving Geometric Problems
195
Figure 5.3.
Problem 8. The sides AB, BC and C A of the triangle ABC are divided into three equals parts by points M, N ; P, Q and R, S, respectively. Equilateral triangles M N D, P Q E, R S F are constructed exterior to triangle ABC. Prove that triangle D E F is equilateral. Solution. Denote by lowercase letters the coordinates of the points denoted by uppercase letters. Then m=
2a + b , 3
n=
a + 2b , 3
p=
2b + c , 3
q=
b + 2c , 3
r=
2c + a , 3
s=
c + 2a . 3
Figure 5.4.
The point D is obtained from point M by a rotation of center N and angle 60◦ . Hence a + 2b + (a − b)ε d = n + (m − n)ε = , 3
196
5. Olympiad-Caliber Problems
where ε = cos 60◦ + i sin 60◦ . Likewise e = q + ( p − q)ε =
b + 2c + (b − c)ε 3
f = s + (r − s)ε =
c + 2a + (c − a)ε . 3
and
Since
c + a − 2b + (b + c − 2a)ε f −d = e−d 2c − a − b + (2b − a − c)ε = =
ε(b + c − 2a + (c + a − 2b)(−ε 2 )) 2c − a − b + (2b − a − c)ε
ε(b + c − 2a) + (c + a − 2b)(ε − 1)) = ε, 2c − a − b + (2b − a − c)ε
we have F D E = 60◦ and F D = F E, so triangle D E F is equilateral. Problem 9. Let ABC D be a square of length side a and consider a point P on the incircle of the square. Find the value of P A2 + P B 2 + PC 2 + P D 2 . Solution. Consider the complex plane such that point A, B, C, D have coordinates zA = Let z P =
√ a 2 , 2
zB =
√ a 2 i, 2
zC = −
√ a 2 , 2
a (cos x + i sin x) be the coordinate of point P. 2
Figure 5.5.
zD = −
√ a 2 i. 2
5.4. Solving Geometric Problems
197
Then P A2 + P B 2 + PC 2 + P D 2 = |z A − z P |2 + |z B − z P |2 + |z C − z P |2 + |z D − z P |2 √
a2 a 2 a π = (z A − z P )(z A − z P ) = 4 + 2 · 2 cos x + 2 cos x + + 2 2 2 2 cyc 3π a2 +2 cos(x + π) + 2 cos x + + 4 = 2a 2 + 0 + a 2 = 3a 2 . 2 4 Problem 10. On the sides AB and AD of the triangle AB D draw externally squares AB E F and ADG H with centers O and Q, respectively. If M is the midpoint of the side B D, prove that O M Q is an isosceles triangle with a right angle at M. Solution. Let a, b, d be the coordinates of the points A, B, D, respectively.
Figure 5.6.
The rotation formula gives a − zO d − zQ = = i, b − zO a − zQ so
b + a + (a − b)i a + d + (d − a)i and z Q = . 2 2 b+d , hence The coordinate of the midpoint M of segment [B D] is z M = 2 a − d + (a − b)i zO − zM = i. = zQ − zM a − b + (d − a)i zO =
Therefore Q M ⊥ O M and O M = Q M, as desired. Problem 11. On the sides of a convex quadrilateral ABC D, equilateral triangles AB M and C D P are drawn external to the figure, and equilateral triangles BC N and AD Q are drawn internal to the figure. Describe the shape of the quadrilateral M N P Q. (23rd IMO – Shortlist)
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5. Olympiad-Caliber Problems
Solution. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter.
Figure 5.7.
Using the rotation formula, we obtain m = a + (b − a)ε,
n = c + (b − c)ε,
p = c + (d − c)ε,
q = a + (d − a)ε,
where ε = cos 60◦ + i sin 60◦ . It is easy to notice that m + p = a + c + (b + d − a − c)ε = n + q, hence M N P Q is a parallelogram or points M, N , P, Q are collinear. Problem 12. On the sides of a triangle ABC draw externally the squares AB M M , AC N N and BC P P . Let A , B , C be the midpoints of the segments M N , P M, P N , respectively. Prove that triangles ABC and A B C have the same centroid. Solution. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter.
5.4. Solving Geometric Problems
199
Figure 5.8.
Using the rotation formula we obtain n = a + (c − a)i and m = a + (b − a)(−i), hence a =
m + n 2a + (c − b)i = . 2 2
Likewise,
2b + (a − c)i 2c + (b − a)i and c = . 2 2 Triangles A B C and ABC have the same centroid if and only if b =
a+b+c a + b + c = . 3 3 Since 2a + 2b + 2c + (c − b + a − c + b − a)i = a + b + c, 2 the conclusion follows. a + b + c =
Problem 13. Let ABC be an acute-angled triangle. On the same side of line AC as point B draw isosceles triangles D AB, BC E, AFC with right angles at A, C, F, respectively.
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5. Olympiad-Caliber Problems
Prove that the points D, E, F are collinear. Solution. Denote by lowercase letters the coordinates of the points denoted by uppercase letters. The rotation formula gives d = a + (b − a)(−i),
e = c + (b − c)i,
a = f + (c − f )i.
Then f =
a + c + (a − c)i d +e a − ci = = , 1−i 2 2
so points F, D, E are collinear. Problem 14. On sides AB and C D of the parallelogram ABC D draw externally equilateral triangles AB E and C D F. On the sides AD and BC draw externally squares of centers G and H . Prove that E H F G is a parallelogram. Solution. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter. Since ABC D is a parallelogram, we have a + c = b + d.
Figure 5.9.
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201
The rotations with 90◦ and centers G and H mapping the points A and C into D d − ai and B, respectively. Then d − g = (a − g)i and b − h = (c − h)i, hence g = 1−i b − ci and h = . 1−i The rotations with 60◦ and centers E and F mapping the point B and D into A and C, respectively. Then a − e = (b − e)ε and c − f = (d − f )ε, where ε = a − bε c − dε cos 60◦ + i sin 60◦ . Hence e = and f = . 1−ε 1−ε Observe that (a + c) − (a + c)i d + b − (a + c)i = =a+c g+h = 1−i 1−i and a + c − (ac )ε a + c − (b + d)ε = = a + c, e+ f = 1−ε 1−ε hence E H F G is a parallelogram. % = 90◦ and let D be the foot Problem 15. Let ABC be a right-angled triangle with C of the altitude from C. If M and N are the midpoints of the segments [DC] and [B D], prove that lines AM and C N are perpendicular. Solution. Consider the complex plane with origin at point C, and let a, b, d, m, n be the coordinates of points A, B, D, M, N , respectively.
Figure 5.10.
Triangles ABC and C D B are similar with the same orientation, hence a−d 0−d ab = or d = . d −0 d −b a+b Then m= Thus
so AM ⊥ C N .
ab b+d 2ab + b2 d = and n = = . 2 2(a + b) 2 2(a + b)
ab −a a π m−a 2(a + b) arg = arg = , = arg − 2 n−0 b 2 2ab + b 2(a + b)
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5. Olympiad-Caliber Problems
Alternate solution. Using the properties of real product in Proposition 1, Section 4.1, and taking into account that C A ⊥ C B, we have ab 2ab + b2 (m − a) · (n − c) = −a · 2(a + b) 2(a + b) 2a + b 2 2a + b 2a + b (a · b) = 0. · b = = a 2(a + b) 2(a + b) 2(a + b) The conclusion follows from Proposition 2 in Section 4.1. Problem 16. Let ABC be an equilateral triangle with the circumradius equal to 1. Prove that for any point P on the circumcircle we have P A2 + P B 2 + PC 2 = 6. Solution. Consider the complex plane such that the coordinates of points A, B, C are the cube roots of unity 1, ε, ε2 , respectively, and let z be the coordinate of point P. Then |z| = 1 and we have P A2 + P B 2 + PC 2 = |z − 1|2 + |z − ε|2 + |z − ε 2 |2 = (z − 1)(z − 1) + (z − ε)(z − ε) + (z − ε 2 )(z − ε 2 ) = 3|z|2 − (1 + ε + ε 2 )z − (1 + ε + ε 2 )z + 1 + |ε|2 + |ε 2 |2 = 3 − 0 · z − 0 · z + 1 + 1 + 1 = 6, as desired. Problem 17. Point B lies inside the segment [AC]. Equilateral triangles AB E and BC F are constructed on the same side of line AC. If M and N are the midpoints of segments AF and C E, prove that triangle B M N is equilateral. Solution. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter. The point E is obtained from point B by a rotation with center A and angle of 60◦ , hence e = a + (b − a)ε, where ε = cos 60◦ + i sin 60◦ . Likewise, f = b + (c − b)ε. The coordinates of points M and N are m=
a + b + (c − b)ε c + a + (b − a)ε and n = . 2 2
It suffices to prove that
m−b = ε. Indeed, we have n−b m − b = (n − b)ε
5.4. Solving Geometric Problems
203
if and only if a − b + (c − b)ε = (c + a − 2b)ε + (b − a)ε 2 . That is, a − b = (a − b)ε + (b − a)(ε − 1), as needed. Problem 18. Let ABC D be a square with center O and let M, N be the midpoints of segments B O, C D respectively. Prove that triangle AM N is isosceles and right-angled. Solution. Consider the complex plane with center at O such that 1, i, −1, −i are the coordinates of points A, B, C, D respectively.
Figure 5.11.
The points M and N have the coordinates m =
i −1 − i and n = , so 2 2
i 1− 2−i a−m 2 = = = i. −1 − i i n−m −1 − 2i − 2 2 Then AM ⊥ M N and AM = N M, as needed. Problem 19. In the plane of the nonequilateral triangle A1 A2 A3 consider points B1 , B2 , B3 such that triangles A1 A2 B3 , A2 A3 B1 and A3 A1 B2 are similar with the same orientation. Prove that triangle B1 B2 B3 is equilateral if and only if triangles A1 A2 B3 , A2 A3 B1 , A3 A1 B2 are isosceles with the bases A1 A2 , A2 A3 , A3 A1 and the base angles equal to 30◦ .
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5. Olympiad-Caliber Problems
Solution. Triangles A1 A2 B3 , A2 A3 B1 , A3 A1 B2 are similar with the same orientab3 − a 2 b1 − a3 b2 − a1 tion, hence = = = z. Then a1 − a2 a2 − a3 a3 − a1 b3 = a2 + z(a1 − a2 ),
b1 = a3 + z(a2 − a3 ),
b2 = a1 + z(a3 − a1 ).
Triangle B1 B2 B3 is equilateral if and only if b1 + εb2 + ε 2 b3 = 0
or
b1 + εb3 + ε 2 b2 = 0.
Assume the first is valid. Then, we have b1 + εb2 + ε 2 b3 = 0 if and only if a3 + z(a2 − a3 ) + εa1 + εz(a3 − a1 ) + ε 2 a2 + ε 2 z(a1 − a2 ) = 0, i.e., a3 + εa1 + ε 2 a2 + z(a2 − a3 + εa3 − εa1 + ε 2 a1 − ε 2 a2 ) = 0. The last relation is equivalent to ⇔ z[a2 (1 − ε)(1 + ε) − a1 ε(1 − ε) − a3 (1 − ε)] = −(a3 + εa1 + ε 2 a2 ), i.e., z=+
a3 + εa1 + ε 2 a2 1 1 = √ (cos 30◦ + i sin 30◦ ), = 2 1−ε (1 − ε)(a3 + εa1 + ε a2 ) 3
which shows that triangles A1 A2 B3 , A2 A3 B1 and A3 A1 B2 are isosceles with angles of 30◦ . Notice that a3 + εa1 + ε 2 a2 = 0, since triangle A1 A2 A3 is not equilateral. Problem 20. The diagonals AC and C E of a regular hexagon ABC D E F are divided by interior points M and N , respectively, such that AM CN = = r. AC CE Determine r knowing that points B, M and N are collinear. (23r d IMO)
Solution. Consider the complex plane with origin at the center of the regular hexagon such that 1, ε, ε 2 , ε 3 , ε 4 , ε 5 are the coordinates of the vertices B, C, D, E, F, A, where √ π π 1+i 3 ε = cos + i sin = . 3 3 2 Since
NE 1−r MC = = , MA NC r
5.4. Solving Geometric Problems
205
Figure 5.12.
the coordinates of points M and N are m = εr + ε5 (1 − r ) and n = ε 2r + ε(1 − r ), respectively. The points B, M, N are collinear if and only if
m−1 ∈ R∗ . We have n−1
m − 1 = εr + ε 5 (1 − r ) − 1 = εr − ε 2 (1 − r ) − 1 √ √ √ −1 + i 3 1 i 3 1+i 3 r− (1 − r ) = − + (2r − 1) = 2 2 2 2 and
√ 1+i 3 n − 1 = ε r + ε(1 − r ) − 1 = −r + (1 − r ) − 1 2 √ 1 3r i 3 =− − + (1 − r ), 2 2 2 3
hence
√ m−1 −1 + i 3(2r − 1) = ∈ R∗ √ n−1 −(1 + 3r ) + i 3(1 − r )
if and only if
√
3(1 − r ) − (1 + 3r ) ·
√
3(2r − 1) = 0.
This is equivalent to 1 − r = 6r 2 − r − 1, i.e., r 2 = 13 . It follows r =
√1 . 3
Problem 21. Let G be the centroid of quadrilateral ABC D. Prove that if lines G A and G D are perpendicular, then AD is congruent to the line segment joining the midpoints of sides AD and BC.
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5. Olympiad-Caliber Problems
Solution. Consider a, b, c, d, g the coordinates of points A, B, C, D, G, respectively. Using properties of the real product of complex numbers we have G A ⊥ G D if and only if (a − g) · (d − g) = 0, i.e., a+b+c+d a+b+c+d a− · d− = 0. 4 4 That is, (3a − b − c − d) · (3d − a − b − c) = 0 and we obtain [a − b − c + d + 2(a − d)] · [a − b − c + d − 2(a − d)] = 0. The last relation is equivalent to (a + d − b − c) · (a + d − b − c) = 4(a − d) · (a − d), i.e., a + d b + c 2 2 2 − 2 = |a − d| .
(1)
Let M and N be the midpoints of the sides AD and BC. The coordinates of points b+c a+d and , hence relation (1) shows that M N = AD and we are M and N are 2 2 done. Problem 22. Consider a convex quadrilateral ABC D with the nonparallel opposite sides AD and BC. Let G 1 , G 2 , G 3 , G 4 be the centroids of the triangles BC D, AC D, AB D, ABC, respectively. Prove that if AG 1 = BG 2 and C G 3 = DG 4 then ABC D is an isosceles trapezoid. Solution. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter. Setting s = a + b + c + d yields g1 =
s−a b+c+d = , 3 3
g2 =
s−b , 3
g3 =
s−c , 3
g4 =
s−d . 3
The relation AG 1 = BG 2 can be written as |a − g1 | = |b − g2 |,
that is,
|4a − s| = |4b − s|.
Using the real product of complex numbers, the last relation is equivalent to (4a − s) · (4a − s) = (4b − s) · (4b − s), i.e., 16|a|2 − 8a · s = 16|b|2 − 8b · s.
5.4. Solving Geometric Problems
207
We find 2(|a|2 − |b|2 ) = (a − b) · s.
(1)
C G 3 = DG 4 if and only if 2(|c|2 − |d|2 ) = (c − d) · s.
(2)
Likewise, we have
Subtracting the relations (1) and (2) gives 2(|a|2 − |b|2 − |c|2 + |d|2 ) = (a − b − c + d) · (a + b + c + d). That is, 2(|a|2 − |b|2 − |c|2 + |d|2 ) = |a + d|2 − |b + c|2 , i.e., 2(aa − bb − cc + dd) = ac + ad + ad + dd − bb − bc − bc − cc. We obtain aa − ad − ad + dd = bb − bc − bc + cc, i.e., |a − d|2 = |b − c|2 . Hence AD = BC.
(3)
Adding relations (1) and (2) gives 2(|a|2 − |b|2 − |d|2 + |c|2 ) = (a − b − d + c) · (a + b + c + d), and similarly we obtain AC = B D.
(4)
From relations (3) and (4) we deduce that ABC D and consequently ABC D is an isosceles trapezoid. Problem 23. Prove that in any quadrilateral ABC D, AC 2 · B D 2 = AB 2 · C D 2 + AD 2 · BC 2 − 2AB · BC · C D · D A · cos(A + C). (Bretschneider relation or a first generalization of Ptolemy’s theorem)
Solution. Let z A , z B , z C , z D be the coordinates of the points A, B, C, D in the complex plane with origin at A and point B on the positive real axis (see Fig. 5.13). Using the identities (z A − z C )(z B − z D ) = −(z A − z B )(z D − z C ) − (z A − z D )(z C − z B )
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5. Olympiad-Caliber Problems
Figure 5.13.
and (z A − z C )(z B − z D ) = −(z A − z B )(z D − z C ) − (z A − z D )(z C − z B ), by multiplication we obtain AC 2 · B D 2 = AB 2 · DC 2 + AD · BC 2 + z + z, where z = (z A − z B )(z D − z C )(z A − z D )(z C − z B ). It suffices to prove that z + z = −2AB · BC · C D · D A · cos(A + C). We have z A − z B = AB(cos π + i sin π ), z D − z C = DC[cos(2π − B − C) + i sin(2π − B − C)], z A − z D = D A[cos(π − A) + i sin(π − A)] and z C − z D = BC[cos(π + B) + i sin(π + B)]. Then z + z = 2Rez = 2AB · BC · C D · D A cos(5π − A − C) = −2AB · BC · C D · D A · cos(A + C) and we are done.
5.4. Solving Geometric Problems
209
Remark. Since cos(A + C) ≥ −1, this relation gives Ptolemy’s inequality AC · B D ≤ AB · DC + AD · BC, with equality only for cyclic quadrilaterals. Problem 24. Let ABC D be a quadrilateral and AB = a, BC = b, C D = c, D A = d, AC = d1 and BC = d2 . Prove that d22 [a 2 d 2 + b2 c2 − 2abcd cos(B − D)] = d12 [a 2 b2 + c2 d 2 − 2abcd cos(A − C)] (A second generalization of Ptolemy’s theorem)
Solution. Let z A , z B , z C , z D be the coordinates of the points A, B, C, D in the complex plane with origin at D and point C on the positive real axis (see the figure in the previous problem but with different notation). Multiplying the identities (z B − z D )[(z A − z B )(z A − z B ) − (z C − z D )(z C − z D )] = (z C − z A ) · [(z B − z A )(z B − z C ) − (z D − z A )(z D − z C )] and (z B − z D )[(z A − z B )(z A − z D ) − (z C − z B )(z C − z D )] = (z C − z A ) · [(z B − z A )(z B − z C ) − (z D − z A )(z D − z C )] yields d22 [a 2 · d 2 + b2 · c2 − (z A − z B )(z A − z D )(z C − z B )(z C − z D ) − (z C − z B )(z C − z D )(z A − z B )(z A − z D )] = d12 [a 2 · b2 + c2 · d 2 − (z B − z A )(z B − z C )(z D − z A )(z D − z C ) − (z D − z A )(z D − z C )(z B − z A )(z B − z C )]. It suffices to prove that 2Re(z A − z B )(z A − z D )(z C − z B )(z C − z D ) = 2abcd cos(B − D) and 2Re(z B − z A )(z B − z C )(z B − z A )(z D − z C ) = 2abcd cos(A − C). We have z B − z A = a[cos(π + A + D) + i sin(π + A + D)],
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5. Olympiad-Caliber Problems
z B − z C = b[cos(π − C) + i sin(π − C)], z D − z A = d[cos(π − D) + i sin(π − D)], z D − z C = c[cos π + i sin π ], z A − z B = a[cos(A + D) + i sin(A + D)], z A − z D = d[cos D + i sin D], z C − z B = b[cos B + i sin B], z C − z D = c[cos 0 + i sin 0]; hence 2Re(z A − z B )(z A − z D )(z C − z B )(z C − z D ) = 2abcd cos(A + D + D + C) = 2abcd cos(2π − B + D) = 2abcd cos(B − D) and 2Re(z B − z A )(z B − z C )(z D − z A )(z D − z C ) = 2abcd cos(π + A + D + π − C + π − D + π ) = 2abcd cos(4π + A − C) = 2abcd cos(A − C), as desired. Remark. If ABC D is a cyclic quadrilateral, then B + D = A + C = π . It follows that cos(B − A) = cos(2B − π ) = − cos 2B and cos(A − C) = cos(2A − π ) = − cos 2A. The relation becomes d22 [(ad + bc)2 − 2abcd(1 − cos 2B)] = d12 [(ab + cd)2 − 2abcd(1 − cos 2A)]. This is equivalent to d22 (ad + bc)2 − 4abcdd22 sin2 B = d12 (ab + cd)2 − 2abcdd12 sin2 A.
(1)
The law of sines applied to the triangles ABC and AB D with circumradii R gives d1 = 2R sin B and d2 = 2R sin A, hence d1 sin A = d2 sin B. The relation (1) is equivalent to d22 (ad + bc)2 = d12 (ab + cd)2 , and consequently
d2 ab + cd . = d1 ad + bc Relation (2) is known as Ptolemy’s second theorem.
(2)
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Problem 25. In a plane three equilateral triangles O AB, OC D and O E F are given. Prove that the midpoints of the segments BC, D E and F A are the vertices of an equilateral triangle. Solution. Consider the complex plane with origin at O and assume that triangles O AB, OC D, O E F are positively orientated. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter. Let ε = cos 60◦ + i sin 60◦ . Then b = aε,
d = cε,
f = eε
and b+c aε + c d +e cε + e = , n= = , 2 2 2 2 Triangle M N P is equilateral if and only if m=
p=
f +a eε + a = . 2 2
m + ωn + ω2 p = 0, where ω = cos 120◦ + i sin 120◦ = ε2 . Because m + ε 2 n + ε 4 p = m + ε 2 n − εp =
1 (aε + c − c + eε 2 − eε 2 − εa) = 0, 2
we are done. We invite the reader to solve the following problems by using complex numbers. Problem 26. Let ABC be a triangle such that AC 2 + AB 2 = 5BC 2 . Prove that the medians from the vertices B and C are perpendicular. Problem 27. On the sides BC, C A, AB of a triangle ABC the points A , B , C are chosen such that B C C A A B = = = k. A C B A C B Consider the points A , B , C on the segments B C , C A , A B such that A C C B B A = = = k. A B C A B C Prove that triangles ABC and A B C are similar. Problem 28. Prove that in any triangle the following inequality is true mα R . ≥ 2r hα Equality holds only for equilateral triangles.
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5. Olympiad-Caliber Problems
Problem 29. Let ABC D be a quadrilateral inscribed in the circle C(O; R). Prove that AB 2 + BC 2 + C D 2 + D A2 = 8R 2 if and only if AC ⊥ B D or one of the diagonals is a diameter of C. Problem 30. On the sides of convex quadrilateral ABC D equilateral triangles AB M, BC N , C D P and D AQ are drawn external to the figure. Prove that quadrilaterals ABC D and M N P Q have the same centroid. Problem 31. Let ABC D be a quadrilateral and consider the rotations R1 , R2 , R3 , R4 with centers A, B, C, D through angle α and of the same orientation. Points M, N , P, Q are the images of points A, B, C, D under the rotations R2 , R3 , R4 , R1 , respectively. Prove that the midpoints of the diagonals of the quadrilaterals ABC D and M N P Q are the vertices of a parallelogram. Problem 32. Prove that in any cyclic quadrilateral ABC D the following holds: a) AD + BC cos(A + B) = AB cos A + C D cos D; b) BC sin(A + B) = AB sin A − C D sin D. Problem 33. Let O9 , I, G be the 9-point center, the incenter and the centroid, respectively, of a triangle ABC. Prove that lines O9 G and AI are perpendicular if and only %= π . if A 3 Problem 34. Two circles ω1 and ω2 are given in the plane, with centers O1 and O2 , respectively. Let M1 and M2 be two points on ω1 and ω2 , respectively, such that the lines O1 M1 and O2 M2 intersect. Let M1 and M2 be points on ω1 and ω2 , respectively, O M and M O M are equal. such that when measured clockwise the angles M 1 1 2 2 1
2
(a) Determine the locus of the midpoint of [M1 M2 ]. (b) Let P be the point of intersection of lines O1 M1 and O2 M2 . The circumcircle of triangle M1 P M2 intersects the circumcircle of triangle O1 P O2 at P and another point Q. Prove that Q is fixed, independent of the locations of M1 and M2 . (2000 Vietnamese Mathematical Olympiad)
Problem 35. Isosceles triangles A3 A1 O2 and A1 A2 O3 are constructed externally along the sides of a triangle A1 A2 A3 with O2 A3 = O2 A1 and O3 A1 = O3 A2 . Let O1 1 be a point on the opposite side of line A2 A3 from A1 , with O 1 A3 A2 = 2 A 1 O3 A2 and 1 O A A A = O A , and let T be the foot of the perpendicular from O 1 2 3 1 to A2 A3 . 2 1 2 3 Prove that A1 O1 ⊥ O2 O3 and that O1 T A 1 O1 =2 . O2 O3 A2 A3 (2000 Iranian Mathematical Olympiad)
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Problem 36. A triangle A1 A2 A3 and a point P0 are given in the plane. We define As = As−3 for all s ≥ 4. We construct a sequence of points P0 , P1 , P2 , . . . such that Pk+1 is the image of Pk under rotation with center Ak+1 through angle 120◦ clockwise (k = 0, 1, 2, . . . ). Prove that if P1986 = P0 then the triangle A1 A2 A3 is equilateral. (27th IMO)
Problem 37. Two circles in a plane intersect. Let A be one of the points of intersection. Starting simultaneously from A two points move with constant speeds, each point travelling along its own circle in the same direction. After one revolution the two points return simultaneously to A. Prove that there exists a fixed point P in the plane such that, at any time, the distances from P to the moving points are equal. (21st IMO)
Problem 38. Inside the square ABC D, the equilateral triangles AB K , BC L, C D M, D AN are inscribed. Prove that the midpoints of the segments K L, L M, M N , N K and the midpoints of the segments AK , B K , B L, C L, C M, D M, D N , AN are the vertices of a regular dodecagon. (19th IMO)
Problem 39. Let ABC be an equilateral triangle and let M be a point in the interior of angle B AC. Points D and E are the images of points B and C under the rotations with center M and angle 120◦ , counterclockwise and clockwise, respectively. Prove that the fourth vertex of the parallelogram with sides M D and M E is the reflection of point A across point M. Problem 40. Prove that for any point M inside parallelogram ABC D the following inequality holds: M A · MC + M B · M D ≥ AB · BC. Problem 41. Let ABC be a triangle, H its orthocenter, O its circumcenter, and R its circumradius. Let D be the reflection of A across BC, let E be that of B across C A, and F that of C across AB. Prove that D, E and F are collinear if and only if O H = 2R. (39th IMO – Shortlist)
Let D be the point Problem 42. Let ABC be a triangle such that AC B = 2 ABC. on the side BC such that C D = 2B D. The segment AD is extended to E so that AD = D E. Prove that B + 180◦ = 2 E EC BC. (39th IMO – Shortlist)
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5. Olympiad-Caliber Problems
5.5
Solving Trigonometric Problems
Problem 1. Prove that π 3π 5π 7π 9π 1 + cos + cos + cos + cos = . 11 11 11 11 11 2 π π Solution. Setting z = cos + i sin implies that 11 11 cos
z + z3 + z5 + z7 + z9 =
z 11 − z −1 − z 1 . = 2 = 2 1−z z −1 z −1
Taking the real parts of both sides of the equality gives the desired result. Problem 2. Compute the product P = cos 20◦ · cos 40◦ · cos 80◦ . Solution. Setting z = cos 20◦ + i sin 20◦ implies z 9 = −1, z = cos 20◦ − i sin 20◦ 8 z2 + 1 z4 + 1 ◦ = z + 1 . Then , cos 40◦ = and cos 20◦ = , cos 80 2z 2z 2 2z 4 P=
(z 2 − 1)(z 2 + 1)(z 4 + 1)(z 8 + 1) (z 2 + 1)(z 4 + 1)(z 8 + 1) = 8z 7 8z 7 (z 2 − 1) =
z 16 − 1 1 −z 7 − 1 = . = 8(−1 − z 7 ) 8 8(z 9 − z 7 )
Solution II. This is a classic problem with a classic solution. Let S cos 20 cos 40 cos 80. Then
=
S sin 20 = sin 20 cos 20 cos 40 cos 80 1 sin 40 cos 40 cos 80 2 1 = sin 80 cos 80 4 1 1 = cos 160 = sin 20. 8 8 =
1 So S = . 8 Note that this classic solution is contrived, with no motivation. The solution using complex numbers, however, is a straightforward computation. Problem 3. Let x, y, z be real numbers such that sin x + sin y + sin z = 0
and
cos x + cos y + cos z = 0.
sin 2x + sin 2y + sin 2z = 0
and
cos 2x + cos 2y + cos 2z = 0.
Prove that
5.5. Solving Trigonometric Problems
215
Solution. Setting z 1 = cos x + i sin x, z 2 = cos y + i sin y, z 3 = cos z + i sin z, we have z 1 + z 2 + z 3 = 0 and |z 1 | = |z 2 | = |z 3 | = 1. We have z 12 + z 22 + z 32 = (z 1 + z 2 + z 3 )2 − 2(z 1 z 2 + z 2 z 3 + z 3 z 1 ) = −2z 1 z 2 z 3
1 1 1 + + z1 z2 z3
= −2z 1 z 2 z 3 (z 1 + z 2 + z 3 )
= −2z 1 z 2 z 3 (z 1 + z 2 + z 3 ) = 0. Thus (cos 2x +cos 2y +cos 2z)+i(sin 2x +sin 2y +sin 2z) = 0, and the conclusion is obvious. Problem 4. Prove that cos2 10◦ + cos2 50◦ + cos2 70◦ =
3 . 2
Solution. Setting z = cos 10◦ + i sin 10◦ , we have z 9 = i and cos 10◦ =
z2 + 1 , 2z
cos 50◦ =
z 10 + 1 , 2z 5
2
cos 70◦ =
z 14 + 1 . 2z 7
The identity is equivalent to
z2 + 1 2z
2 +
z 10 + 1 2z 5
+
z 14 + 1 2z 7
2 =
3 . 2
That is, z 16 + 2z 14 + z 12 + z 24 + 2z 14 + z 4 + z 28 + 2z 14 + 1 = 6z 14 , i.e., z 28 + z 24 + z 16 + z 12 + z 4 + 1 = 0. Using relation z 18 = −1, we obtain z 16 + z 12 − z 10 − z 6 + z 4 + 1 = 0 or equivalently (z 4 + 1)(z 12 − z 6 + 1) = 0. That is, (z 4 + 1)(z 18 + 1) = 0, z6 + 1 which is obvious.
216
5. Olympiad-Caliber Problems
Problem 5. Solve the equation cos x + cos 2x − cos 3x = 1. Solution. Setting z = cos x + i sin x yields cos x =
z2 + 1 , 2z
cos 2x =
z4 + 1 , 2z 2
cos 3x =
z6 + 1 . 2z 3
The equation may be rewritten as z2 + 1 z4 + 1 z6 + 1 − = 1, i.e., z 4 + z 2 + z 5 + z − z 6 − 1 − 2z 3 = 0 + 2z 2z 2 2z 3 This is equivalent to (z 6 − z 5 − z 4 − z 3 ) + (z 3 − z 2 − z + 1) = 0 or (z 3 + 1)(z 3 − z 2 − z + 1) = 0. Finally we obtain (z 3 + 1)(z − 1)2 (z + 1) = 0. x ∈ {2kπ|k ∈ Z} or Thus, z = 1 or z = −1 or z3 = −1 and consequently π + 2kπ |k ∈ Z . Therefore x ∈ {kπ |k ∈ Z} ∪ x ∈ {π + 2kπ|k ∈ Z} or x ∈ 3 2k + 1 π|k ∈ Z . 2 Problem 6. Compute the sums S=
n
q k · cos kx
and
T =
k=1
n
q k · sin kx.
k=1
Solution. We have 1 + S + iT =
n
q k (cos kx + i sin kx) =
k=0
= = =
n
q k (cos x + i sin x)k
k=0
1 − q n+1 (cos x + i sin x)n+1 1 − q cos x − iq sin x
1 − q n+1 [cos(n + 1)x + i sin(n + 1)x] 1 − q cos x − iq sin x
[1 − q n+1 cos(n + 1)x − iq n+1 sin(n + 1)x][1 − q cos x + iq sin x] , q 2 − 2q cos x + 1
5.5. Solving Trigonometric Problems
hence 1+S =
217
q n+2 cos nx − q n+1 cos(n + 1)x − q cos x + 1 q 2 − 2q cos x + 1
and T =
q n+2 sin nx − q n+1 sin(n + 1)x + q sin x . q 2 − 2q cos x + 1
Remark. If q = 1 then we find the well-known formulas n
cos kx =
(n + 1)x nx cos 2 2 x sin 2
sin
k=1
and
n
sin kx =
sin
k=1
(n + 1)x nx sin 2 2 . x sin 2
Indeed, we have n
cos kx =
k=1
=
= and
sin
2 sin
x x (2n + 1)x sin − 2 sin2 2 2 2 2 x 4 sin 2
x (n + 1)x nx (2n + 1)x − sin cos sin 2 2 = 2 2 x x 2 sin sin 2 2
n
sin kx =
k=1
=
cos nx − cos(n + 1)x − (1 − cos x) 2(1 − cos x)
2 sin
sin nx − sin(n + 1)x + sin x 2(1 − cos x)
x x (2n + 1)x x cos − 2 sin cos 2 2 2 2 x 4 sin2 2
(2n + 1)x (n + 1)x nx x − cos sin sin 2 2 2 2 = . = x x 2 sin sin 2 2 Problem 7. The points A1 , A2 , . . . , A10 are equally distributed on a circle of radius R (in that order). Prove that A1 A4 − A1 A2 = R. π π Solution. Let z = cos + i sin . Without loss of generality we may assume that 10 10 3π π R = 1. We need to show that 2 sin − 2 sin = 1. 10 10 cos
218
5. Olympiad-Caliber Problems
In general, if z = cos a + i sin a, then sin a =
z2 − 1 and we have to prove that 2i z
z6 − 1 z2 − 1 − = 1. This reduces to z 6 − z 4 + z 2 − 1 = i z 3 . Because z 5 = i, the 3 iz iz previous relation is equivalent to z 8 − z 6 + z 4 − z 2 + 1 = 0. But this is true because (z 8 − z 6 + z 4 − z 2 + 1)(z 2 + 1) = z 10 + 1 = 0 and z 2 + 1 = 0. Problem 8. Show that cos
π 2π 3π 1 − cos + cos = . 7 7 7 2 (5th IMO)
π π + i sin . Then z 7 + 1 = 0. Because z = −1 and z 7 + 1 = 7 7 (z + 1)(z 6 − z 5 + z 4 − z 3 + z 2 − z + 1) = 0 it follows that the second factor from the 1 . above product is zero. The condition is equivalent to z(z 2 − z + 1) = 1 − z3 The given sum is Solution. Let z = cos
π 2π 3π − cos + cos = Re(z 3 − z 2 + z). 7 7 7 1 1 = . This follows from the wellTherefore, we have to prove that Re 3 2 1−z known: 1 1 Lemma. If z = cos t + i sin t and z = 1, then Re = . 1−z 2 1 1 1 Proof. = = = 1−z 1 − (cos t + i sin t) (1 − cos t) − i sin t cos
=
1 1 = t t t t t t 2 sin2 − 2i sin cos 2 sin sin − i cos 2 2 2 2 2 2
t t t + i cos cos 1 2 2 2 = +i = t t. 2 2 sin 2 sin 2 2 Problem 9. Prove that the average of the numbers k sin k ◦ (k = 2, 4, 6, . . . , 180) is cot 1◦ . sin
(1996 USA Mathematical Olympiad)
Solution. Denote z = cos t + i sin t. From the identity z + 2z 2 + · · · + nz n = (z + · · · + z n ) + (z 2 + · · · + z n ) + · · · + z n
5.5. Solving Trigonometric Problems
=
219
1 [(z n+1 − z) + (z n+1 − z 2 ) + · · · + (z n+1 − z n )] z−1 =
z n+1 − z nz n+1 − z−1 (z − 1)2
we derive the formulas: n
k cos kt =
(2n + 1)t 1 − cos(n + 1)t 2 − , t t 2 sin 4 sin2 2 2
(n + 1) sin
k=1
(2n + 1)t n cos sin(n + 1)t 2 k sin kt = . t − t 2 k=1 4 sin 2 sin 2 2 Using relation (2) one obtains: n
(1)
(2)
2 sin 2◦ + 4 sin 4◦ + · · · + 178 sin 178◦ = 2(sin 2◦ + 2 sin 2 · 2◦ + · · · + 89 sin 89 · 2◦ ) 90 cos 179◦ 90 cos 179◦ sin 90 · 2◦ =− − = 90 cot 1◦ . =2 ◦ 2 ◦ 2 sin 1 sin 1◦ 4 sin 1 Finally, 1 (2 sin 2◦ + 4 sin 4◦ + · · · + 178 sin 178◦ + 180 sin 180◦ ) = cot 1◦ . 90 Problem 10. Let n be a positive integer. Find real numbers a0 and akl , k, l = 1, n, k > l, such that
sin2 nx = a + akl cos 2(k − l)x 0 sin2 x 1≤l
Solution. Using the identities S1 =
n
cos 2 j x =
sin nx cos(n + 1)x sin x
sin 2 j x =
sin nx sin(n + 1)x sin x
j=1
and S2 =
n
j=1
we obtain S12 + S22 =
sin nx 2 sin x
.
220
5. Olympiad-Caliber Problems
On the other hand, S12 + S22 = (cos 2x + cos 4x + · · · + cos 2nx)2 + (sin 2x + sin 4x + · · · + sin 2nx)2
=n+2 (cos 2kx cos 2lx + sin 2kx sin 2lx) 1≤l
=n+2
cos 2(k − l)x,
1≤l
sin nx 2
hence
sin x
=n+2
cos 2(k − l)x.
1≤l
Set a0 = n and akl = 2, 1 ≤ l < k ≤ n, and the problem is solved. Here are some more problems. Problem 11. Sum the following two n-term series for θ = 30◦ : i) 1 +
cos(2θ ) cos(3θ) cos θ cos((n − 1)θ ) + + + ··· + , and cos θ cos2 θ cos3 θ cosn−1 θ
ii) cos θ cos θ + cos2 θ cos(2θ) + cos3 θ cos(3θ) + · · · + cosn θ cos(nθ). (Crux Mathematicorum, 2003)
Problem 12. Prove that 2n π 2n 2π 2n (n − 1)π + cos + · · · + cos 1 + cos n n n 2n = n · 4−n 2 + , n for all integers n ≥ 2. Problem 13. For any integer p ≥ 0 there are real numbers a0 , a1 , . . . , a p with a p = 0 such that cos 2 pα = a0 + a1 sin2 α + · · · + a p · (sin2 α) p , for all α ∈ R.
5.6
More on the nth Roots of Unity
Problem 1. Let n ≥ 3 and k ≥ 2 be positive integers and consider the complex numbers 2π 2π + i sin z = cos n n and θ = 1 − z + z 2 − z 3 + · · · + (−1)k−1 z k−1 .
5.6. More on the n th Roots of Unity
221
n a) If k is even, prove that θ n = 1 if and only if n is even and divides k − 1 or 2 k + 1. b) If k is odd, prove that θ n = 1 if and only if n divide k − 1 or k + 1. Solution. Since z = −1, we have θ=
1 + (−1)k+1 z k . 1+z
a) If k is even, then kπ kπ kπ 2kπ 2kπ sin − i cos sin − i sin 1 − cos 1 − zk n n n n n = = θ= π π π 2π 2π 1+z cos + i sin cos 1 + cos + i sin n n n n n kπ (k − 1)π (k − 1)π n = −i + i sin , π cos n n cos n sin
and
kπ sin n |θ | = cos π n
.
kπ = cos π . |θ | = 1 if and only if sin n n
We have
That is,
kπ π = cos2 n n The last relation is equivalent to sin2
cos
or
cos
2kπ 2π + cos = 0. n n
(k + 1)π (k − 1)π 2(k + 1) cos = 0, i.e., ∈ 2Z + 1 n n n
n 2(k − 1) ∈ 2Z + 1. This is equivalent to the statement that n is even and divides or n 2 k + 1 or k − 1. Hence, it suffices to prove that θ n = 1 is equivalent to |θ | = 1. The direct implication is obvious. Conversely, if |θ | = 1, then n = 2t, t ∈ Z+ and t divides k + 1 or k − 1. Since k is even, numbers k + 1, k − 1 are odd, hence t = 2l + 1 and n = 4l + 2, l ∈ Z. Then (k − 1)π (k − 1)π θ = ±i cos + i sin n n
222
5. Olympiad-Caliber Problems
and θ n = − cos(k − 1)π = 1, as desired. b) If k is odd, then kπ kπ kπ 2kπ 2kπ cos + i sin cos 1 + cos + i sin 1 + zk n n n n n = = θ= π π π 2π 2π 1+z cos cos + i sin 1 + cos + i sin n n n n n kπ n cos k − 1 π + i sin k − 1 π . = π n n cos n π kπ = |θ | = 1 if and only if cos cos . n n cos
We have
That is, cos2
kπ π = cos2 n n
It follows that sin
so
cos
2kπ 2π = cos . n n
(k − 1)π (k + 1)π sin = 0, n n
i.e., n divides k + 1 or k − 1. It suffices to prove that θ n = 1 is equivalent to |θ | = 1. Since the direct implication is obvious, let us prove the converse. If |θ | = 1, then k±1 = nt, t ∈ Z. Then k = nt ±1 and (k − 1)π (k − 1)π t θ = (−1) cos + i sin . n n It follows that θ n = (−1)k±1 (cos(k − 1)π + i sin(k − 1)π ) = (−1)k±1 (−1)k−1 = 1, as desired. Problem 2. Consider the cube root of unity ε = cos
2π 2π + i sin . 3 3
Compute (1 + ε)(1 + ε 2 ) · · · (1 + ε 1987 ). Solution. Notice that ε 3 = 1, ε2 + ε + 1 = 0 and 1987 = 662 · 3 + 1. Then (1 + ε)(1 + ε 2 ) · · · (1 + ε 1987 )
5.6. More on the n th Roots of Unity
=
661
223
[(1 + ε 3k+1 )(1 + ε 3k+2 )(1 + ε 3k+3 )](1 + ε 1987 )
k=0
=
661
[(1 + ε)(1 + ε 2 )(1 + 1)](1 + ε) = (1 + ε)[2(1 + ε + ε 2 + ε 3 )]662
k=0
= (1 + ε)[2(0 + 1)]662 = 2662 (1 + ε) √ √ 662 2 662 1 + i 3 = 2661 (1 + i 3). = 2 (−ε ) = 2 2 Problem 3. Let ε = 1 be a cube root of unity. Compute (1 − ε + ε 2 )(1 − ε 2 + ε 4 ) · · · (1 − ε n + ε 2n ). Solution. Notice that 1 + ε + ε2 = 0 and 1 + ε − ε2 = −2ε2 . Then ⎧ ⎪ ⎨ 1, 1 − ε n + ε 2n = −2ε, ⎪ ⎩ −2ε2 ,
ε3 = 1. Hence 1 − ε + ε2 = −2ε and
if if if
n ≡ 0(mod3), n ≡ 1(mod3), n ≡ 2(mod3),
the product of any three consecutive factors of the given product equals 1 · (−2ε) · (−2ε 2 ) = 22 . Therefore (1 − ε + ε 2 )(1 − ε 2 + ε 4 ) · · · (1 − ε n + ε 2n ) ⎧ 2n ⎪ if n ≡ 0(mod3), ⎨ 23, n 2[ ]+1 = −2 3 ε, if n ≡ 1(mod3), ⎪ ⎩ 2[ n3 ]+2 , if n ≡ 2(mod3). 2 Problem 4. Prove that the complex number z=
2+i 2−i
has modulus equal to 1, but z is not an n th -root of unity for any positive integer n. Solution. Obviously |z| = 1. Assume by contradiction that there is an integer n ≥ 1 such that z n = 1. Then (2 + i)n = (2 − i)n , and writing 2 + i = (2 − i) + 2i it follows that (2 − i)n = (2 + i)n n n n n−1 = (2 − i) + (2 − i) 2i + · · · + (2 − i)(2i)n−1 + (2i)n . 1 n−1
224
5. Olympiad-Caliber Problems
This is equivalent to n n n−2 n−1 (2i − 1) 2i + · · · + (2i) (2i) = (−2 + i) 1 n−1 n
= (−2 + i)(a + bi), with a, b ∈ Z. Taking the modulus of both members of the equality gives 2n = 5(a 2 + b2 ), a contradiction. Problem 5. Let Un be the set of n th -roots of unity. Prove that the following statements are equivalent: a) there is α ∈ Un such that 1 + α ∈ Un ; b) there is β ∈ Un such that 1 − β ∈ Un . (Romanian Mathematical Olympiad – Second Round, 1990)
1 Solution. Assume that there exists α ∈ Un such that 1+α ∈ Un . Setting β = 1 + α n 1 1 n we have β = = = 1, hence β ∈ Un . On the other hand, 1+α (1 + α)n αn α and (1 − β)n = = 1, hence 1 − β ∈ Un , as desired. 1−β = α+1 (1 + α)n 1−β (1 − β)n = 1 and . Since α n = Conversely, if β, 1 − β ∈ Un , set α = β βn 1 (1 + α)n = n = 1, we have α ∈ Un and 1 + α ∈ Un , as desired. β Remark. The statements a) and b) are equivalent with 6|n. Indeed, if α, 1 + α ∈ Un , 2 = (1+α)(1+α) = 1+α+α+|α|2 = then |α| = |1+α| = 1. It follows that 1 = |1+α|√ 3 1 1 1 + α + α + 1 = 2 + α + , i.e., α = − ± i , hence α 2 2 √ 3 2π 2π 1 = cos ± i sin . 1+α = ±i 2 2 6 6 Since (1 + α)n = 1 it follows that 6 divides n.
√ √ 1 3 3 1 Conversely, if n is a multiple of 6, then both α = − + i and 1 + α = + i 2 2 2 2 belong to Un . Problem 6. Let n ≥ 3 be a positive integer and let ε = 1 be an n th root of unity. 2 1) Show that |1 − ε| > . n−1 2) If k is a positive integer such that n does not divides k, then kπ sin > 1 . n n−1
5.6. More on the n th Roots of Unity
225
(Romanian Mathematical Olympiad – Final Round, 1988)
Solution. 1) We have ε n − 1 = (ε − 1)(ε n−1 + · · · + ε + 1) hence, taking into account that ε = 1, we find εn−1 + · · · + ε + 1 = 0. The last relation is equivalent to (ε n−1 −1)+· · ·+(ε−1) = −n, i.e., (ε−1)[ε n−2 +2ε n−3 +· · ·+(n−2)ε+(n−1)] = −n. Passing to the absolute value we find that n = |ε − 1||ε n−2 + 2ε n−3 + · · · + (n − 1)| ≤ |ε − 1|(|ε n−2 | + 2|ε|n−3 + · · · + (n − 1)). Therefore n ≤ |1 − ε|(1 + 2 + · · · + (n − 1)) = |1 − ε|
n(n − 1) , 2
2 . Moreover, equality is not possible since n−1 n−1 the geometric images of 1, ε, . . . , ε are not collinear. 2kπ 2kπ + i sin and obtain 2) Consider ε = cos n n i.e., we find the inequality |1 − ε| ≥
1 − ε = 1 − cos
2kπ 2kπ − i sin . n n
Hence 2kπ kπ 2kπ 2kπ 2 = 2 − 2 cos = 4 sin2 . + sin2 |1 − ε|2 = 1 − cos n n n n Applying the inequality in 1), the desired inequality follows. Problem 7. Let Un be the set of the n th -roots of unity. Prove that ⎧ if n ≡ 0 (mod 4), ⎪ 0, ⎪ ⎪ ⎨ 1 2, if n ≡ 1 (mod 2), ε+ = ⎪ ε −4, if n ≡ 2 (mod 4), ⎪ ε∈Un ⎪ ⎩ 2, if n ≡ 3 (mod 4). Solution. Consider the polynomial f (x) = X n − 1 =
(X − ε).
ε∈Un
Denoting by Pn the product in our problem, we have (ε + i)(ε − i) ε2 + 1 1 ε∈Un Pn = = = ε+ ε ε ε ε∈Un ε∈Un ε∈Un
226
5. Olympiad-Caliber Problems
=
(i + ε)
ε∈Un
(−i + ε)
ε∈Un (−1)n f (0)
=
[(−i)n − 1](i n − 1) f (−1) · f (i) = . n−1 (−1) (−1)n−1
If n ≡ 0 (mod 4), then i n = 1 and Pn = 0. If n ≡ 1 (mod 2), then (−1)n−1 = 1 and Pn = (−i n − 1)(i n − 1) = −(i 2n − 1) = −((−1)n − 1) = −(−1 − 1) = 2. If n ≡ 2 (mod 4), then (−1)n−1 = −1, (−i)n = i n = i 2 = −1, i n = −1, hence (−1 − 1)(−1 − 1) Pn = = −4. −1 If n ≡ 3 (mod 4), then (−1)n−1 = 1 and Pn = (−i n − 1)(i n − 1) = (i 3 − 1)(−i 3 − 1) = −(i 6 − 1) = −((−1)3 − 1) = 2, and we are done. Problem 8. Let ω = cos
2π 2π + i sin , 2n + 1 2n + 1
and let z=
n ≥ 0,
1 + ω + ω2 + · · · + ωn . 2
Prove that: a) Im(z 2k ) = Re(z 2k+1 ) = 0 for all k ∈ N; b) (2z + 1)2n+1 + (2z − 1)2n+1 = 0. Solution. We have ω2n+1 = 1 and 1 + ω + ω2 + · · · + ω2n = 0. Then
or
1 1 + ω + ω2 + · · · + ωn + ωn (ω + ω2 + · · · + ωn ) + = 0 2 2 1 1 + = 0, z + ωn z − 2 2
hence z=
1 ωn − 1 · . 2 ωn + 1
1 1 ωn − 1 = −z. Thus z 2k = z 2k and z 2k+1 = −z 2k+1 . The a) We have z = 2 1 +1 ωn conclusion follows from these two equalities.
5.6. More on the n th Roots of Unity
b) From the relation
227
1 1 z + ωn z − + =0 2 2
we obtain 2z + 1 = −ωn (2z − 1). Taking into account that ω2n+1 = 1, we obtain (2z + 1)2n+1 = −(2z − 1)2n+1 , and we are done. Problem 9. Let n be an odd positive integer and ε0 , ε1 , . . . , εn−1 the complex roots of unity of order n. Prove that n−1
(a + bεk2 ) = a n + bn
k=0
for all complex numbers a and b. (Romanian Mathematical Olympiad – Second Round, 2000)
Solution. If ab = 0, then the claim is obvious, so consider the case when a = 0 and b = 0. We start with a useful lemma. Lemma. If ε0 , ε1 , . . . , εn−1 are the complex roots of unity of order n, where n is an odd integer, then n−1 (A + Bεk ) = An + B n , k=0
for all complex numbers A and B. Proof. Using the identity xn − 1 =
n−1
(x − εk )
k=0
for x = −
A yields B
−
n−1 A An + 1 = − + ε k , Bn B k=0
and the conclusion follows. Because n is odd, the function f : Un → Un is bijective. To prove this, it suffices to show that it is injective. Indeed, assume that f (x) = f (y). It follows that (x − y)(x + y) = 0. If x + y = 0, then x n = (−y)n , i.e., 1 = −1, a contradiction. Hence x = y. From the lemma we have n−1 k=0
(a + bεk2 ) =
n−1 j=0
(a + bε j ) = a n + bn .
228
5. Olympiad-Caliber Problems
n Problem 10. Let n be an even positive integer such that is odd and let 2 ε0 , ε1 , . . . , εn−1 be the complex roots of unity of order n. Prove that n−1
n
n
(a + bεk2 ) = (a 2 + b 2 )2
k=0
for any complex numbers a and b. (Romanian Mathematical Olympiad – Second Round, 2000)
Solution. If b = 0 the claim is obvious. If not, let n = 2(2s +1). Consider a complex a number α such that α 2 = and the polynomial b f = X n − 1 = (X − ε0 )(X − ε1 ) · · · (X − εn−1 ). We have f
α i
a 1 (α − iε0 ) · · · (α − iεn−1 ) = i
α −1 a = (α + iε0 ) · · · (α + iεn−1 ), f − i i
and
hence f
α i
α 2 ). f − = (α 2 + ε02 ) · · · (α 2 + εn−1 i
Therefore n−1
(a + bεk2 ) = bn
k=0
=b f n
n−1 k=0
n−1 a (α 2 + εk2 ) + εk2 = bn b k=0
2 α a 2s+1 n 2 2s+1 2 n + 1] = b +1 f − = b [(α ) i i b 2 2s+1 + b2s+1 n n 2(2s+1) a =b = (a 2 + b 2 )2 . b2s+1
α
The following problems also involve n th roots of unity. Problem 11. For all positive integers k define Uk = {z ∈ C | z k = 1}. Prove that for any integers m and n with 0 < m < n we have U1 ∪ U2 ∪ · · · ∪ Um ⊂ Un−m+1 ∪ Un−m+2 ∪ · · · ∪ Un . (Romanian Mathematical Regional Contest “Grigore Moisil”, 1997)
5.7. Problems Involving Polygons
229
Problem 12. Let a, b, c, d, α be complex numbers such that |a| = |b| = 0 and |c| = |d| = 0. Prove that all roots of the equation c(bx + aα)n − d(ax + bα)n = 0,
n ≥ 1,
are real numbers. Problem 13. Suppose that z = 1 is a complex number such that z n = 1, n ≥ 1. Prove that (n + 1)(2n + 1) |nz − (n + 2)| ≤ |z − 1|2 . 6 (Crux Mathematicorum, 2003)
Problem 14. Let M be a set of complex numbers such that if x, y ∈ M, then
x ∈ M. y
Prove that if the set M has n elements, then M is the set of the n th -roots of 1. Problem 15. A finite set A of complex numbers has the property: z ∈ A implies z n ∈ A for every positive integer n.
a) Prove that z is an integer. z∈A
b) Prove that
for every integer k one can choose a set A which fulfills the above z = k. condition and z∈A
(Romanian Mathematical Olympiad – Final Round, 2003)
5.7
Problems Involving Polygons
Problem 1. Let z 1 , z 2 , . . . , z n be distinct complex numbers such that |z 1 | = |z 2 | = · · · = |z n |. Prove that
z i + z j 2 (n − 1)(n − 2) . z − z ≥ 2 j 1≤i< j≤n i Solution. Consider the points A1 , A2 , . . . , An with coordinates z 1 , z 2 , . . . , z n . The polygon A1 A2 · · · An is inscribed in the circle with center at origin and radius R = |z 1 |. zi + z j The coordinate of the midpoint Ai j of the segment [Ai A j ] is equal to , for 2 1 ≤ i < j ≤ n. Hence |z i + z j |2 = 4O Ai2j Moreover, 4O Ai2j = 4R 2 − Ai A2j . The sum
and
|z i − z j |2 = Ai A2j .
z i + z j 2 z − z i j 1≤i< j≤n
230
5. Olympiad-Caliber Problems
equals
4O Ai2j
1≤i< j≤n
Ai A2j
4R 2 − Ai A2j
1≤i< j≤n
Ai A2j
=
= 4R 2
1≤i< j≤n
1 n − . 2 Ai A2j
The AM − H M inequality gives
1≤i< j≤n
00n 112
2 . Ai A2j
1 ≥ Ai A2j
1≤i< j≤n
Since
Ai A2j ≤ n 2 · R 2 , it follows that
1≤i< j≤n
00n 112 n
2 − 2 2 Ai A j
z i + z j 2 2 z − z ≥ 4R j 1≤i< j≤n i
1≤i< j≤n
≥
4
00n 112 2
n2
1 0 1 0 0n 1 4 2 − n 2 · n2 n (n − 1)(n − 2) − = = , 2 2 n2
as claimed. Problem 2. Let A1 A2 · · · An be a polygon and let a1 , a2 , . . . , an be the coordinates of the vertices A1 , A2 , . . . , An . If |a1 | = |a2 | = · · · = |an | = R, prove that
|ai + a j |2 ≥ n(n − 2)R 2 . 1≤i< j≤n
Solution. We have
|ai + a j |2 =
1≤i< j≤n
=
(ai + a j )(ai + a j )
1≤i< j≤n
(|ai |2 + |a j |2 + ai a j + ai a j )
1≤i< j≤n
n n
n
n + ai a j = n(n − 1)R 2 + ai a j − ai ai = 2R 2 2 i= j i=1 j=1 i=1 n n
2 = n(n − 1)R + ai ai − n R 2 i=1
i=1
2 n
= n(n − 2)R 2 + ai ≥ n(n − 2)R 2 , i=1 as desired.
5.7. Problems Involving Polygons
231
Problem 3. Let z 1 , z 2 , . . . , z n be the coordinates of the vertices of a regular polygon with the circumcenter at the origin of the complex plane. Prove that there are i, j, k ∈ {1, 2, . . . , n} such that z i + z j = z k if and only if 6 divides n. 2π 2π Solution. Let ε = cos + i sin . Then z p = z 1 · ε p−1 , for all p = 1, n. n n We have z i + z j = z k if and only if 1 + ε j−i = εk−i , i.e., ( j − i)π ( j − i)π ( j − i)π 2(k − i)π 2(k − i)π 2cos cos +i sin = cos +i sin . n n n n n The last relation is equivalent to ( j − i)π π 2(k − i)π = = , i.e., n = 6(k − i) = 3( j − i), n 3 n hence 6 divides n. Conversely, if 6 divides n, let i = 1,
j=
n + 1, 3
k=
n +1 6
and we have z i + zl = z k , as desired. Problem 4. Let z 1 , z 2 , . . . , z n be the coordinates of the vertices of a regular polygon. Prove that z 12 + z 22 + · · · + z n2 = z 1 z 2 + z 2 z 3 + · · · + z n z 1 . Solution. Without loss of generality we may assume that the center of the polygon is the origin of the complex plane. Let z k = z 1 ε k−1 , where ε = cos
2π 2π + i sin , n n
k = 1, . . . , n.
The right-hand side is equal to z1 z2 + z2 z3 + · · · + zn z1 =
n
z i z k+1
k=1
=
n
z 12 ε 2k−1 = z 12 · ε ·
k=1
1 − ε 2n = 0. 1 − ε2
On the other hand, z 12 + z 22 + · · · + z n2 =
n
k=1
and we are done.
z i2 =
n
k=1
z 12 ε 2k−2 = z 12
1 − ε 2n =0 1 − ε2
232
5. Olympiad-Caliber Problems
Problem 5. Let n ≥ 4 and let a1 , a2 , . . . , an be the coordinates of the vertices of a regular polygon. Prove that a1 a2 + a2 a3 + · · · + an a1 = a1 a3 + a2 a4 + · · · + an a2 . Solution. Assume that the center of the polygon is the origin of the complex plane and ak = a1 ε k−1 , k = 1, . . . , n, where ε = cos
2π 2π + i sin . n n
The left-hand side of the equality is a1 a2 + a2 a3 + · · · + an a1 = a12
n
ε 2k−1 = a12 ε
k=1
1 − ε 2n = 0. 1 − ε2
The right-hand side of the equality is a12
n
ε 2k = a12 ε 2
k=1
1 − ε 2n = 0, 1 − ε2
and we are done. Problem 6. Let z 1 , z 2 , . . . , z n be distinct complex numbers such that |z 1 | = |z 2 | = · · · = |z n | = 1. Consider the statements: a) z 1 , z 2 , . . . , z n are the coordinates of the vertices of a regular polygon. b) z 1n + z 2n + · · · + z nn = n(−1)n+1 z 1 z 2 · · · z n . Decide with proof if the implications a) ⇒ b) and b) ⇒ a) are true. Solution. We study at first the implication a) ⇒ b). 2π 2π Let ε = + i sin . Since z 1 , z 2 , . . . , z n are coordinates of the vertices of a n n regular polygon, without loss of generality we may assume that z k = z 1 ε k−1
for
k = 1, n.
The relation b) becomes z 1n (1 + ε n + ε 2n + · · · + ε n(n−1) ) = n(−1)n+1 z 1n ε 1+2+···+(n−1) . This is equivalent to
n = n(−1)n+1 ε
1 = (−1)
n+1
n(n−1) 2
, i.e.,
n(n − 1) 2π n(n − 1) 2π cos · + i sin · 2 n 2 n
.
5.7. Problems Involving Polygons
233
We obtain 1 = (−1)n+1 (cos(n − 1)π + i sin(n − 1)π ), i.e., 1 = (−1)n+1 (−1)n−1 , which is valid. Therefore the implication a) ⇒ b) holds. We prove now that the implication b) ⇒ a) is also valid. Observe that |n · (−1)n+1 z 1 z 2 · · · z n | = n|z 1 | · |z 2 | · · · |z n | = n, hence |z 1n + z 2n + · · · + z nn | = n. Using the triangle inequality we obtain n = |z 1n + z 2n + · · · + z nn | ≤ |z 1n | + |z 2n | + · · · + |z nn | = 1 + 1 + · · · + 1 = n, n times
hence the numbers z 1n , z 2n , . . . , z nn have the same argument. Since |z 1n | = |z 2n | = · · · = |z nn | = 1, it follows that z 1n = z 2n = · · · = z nn = a, where a is a complex number with |a| = 1. Numbers z 1 , z 2 , . . . , z n are distinct, therefore there are the n th -roots of a, and consequently the coordinates of the vertices of a regular polygon. Problem 7. Let A, B, C be 3 consecutive vertices of a regular n-gon and consider the point M on the circumcircle such that points B and M lie on opposite sides of line AC. Prove that M A + MC = 2M B cos πn . (A generalization of the Van Schouten theorem; see the first remark below)
Solution. Consider the complex plane with origin at the center of the polygon and let 1 be the coordinate of A1 . 2π 2π If ε = cos + i sin , then ε k−1 is the coordinate of Ak , k = 1, n. n n Without loss of generality, assume that A = A1 , B = A2 and C = A3 . Let z M = cos t + i sin t, t ∈ [0, 2π) be the coordinate of point M. Since point B and M are 4π < t. separated by the line AC, it follows that n Then √ t M A = |z M − 1| = (cos t − 1)2 + sin2 t = 2 − 2 cos t = 2 sin ; 2 t π − M B = |z M − ε| = 2 sin 2 n
and MC = |z M
2π t − − ε | = 2 sin 2 n 2
.
234
5. Olympiad-Caliber Problems
The equality M A + M B = 2MC cos
π n
is equivalent to 2 sin
t 2π t π π t + 2 sin − = 4 sin − cos , 2 2 n 2 n n
which follows using the sum-to-product formula in the left-hand side. Remarks. 1) If n = 3 then we obtain the Van Schouten theorem: For any point M
on the circumcircle of equilateral triangle ABC such that M belongs on the arc AC, the following relation holds: M A + MC = M B. Note that this result also follows from Ptolemy’s theorem. 2) If n = 4, then for any point M on the circumcircle of square ABC D such that B and M lie on opposite sides of line AC, we have the relation √ M A + MC = 2M B. Problem 8. Let P be a point on the circumcircle of square ABC D. Find all integers n > 0 such that the sum Sn (P) = P An + P B n + PC n + P D n is constant with respect to point P. Solution. Consider the complex plane with origin at the center of the square such that A, B, C, D have coordinates 1, i, −1, −i, respectively. Let z = a + bi be the coordinate of point P, where a, b ∈ R with a 2 + b2 = 1. The sum Sn (P) is equal to n
n
n
n
Sn (P) = [(a − 1)2 + b2 ] 2 + [a 2 + (b − 1)2 ] 2 + [(a + 1)2 + b2 ] 2 + [a 2 + (b + 1)2 ] 2 ' ( n n n n n = 2 2 (1 + a) 2 + (1 − a) 2 + (1 + b) 2 + (1 − b) 2 . √ √ 2 2 n+2 n , , we get Set P = A(1, 0). Then Sn (A) = 2 2 + 2 . For P = E 2 2 Sn (E) = 2(2 −
√
n
2) 2 + 2(2 +
√
n
2) 2 .
Since Sn (P) is constant with respect to P, it follows that Sn (A) = Sn (E) or 2 √ n √ n n 2 = 2(2 − 2) 2 + 2(2 + 2) 2 .
n+2 2
+
5.7. Problems Involving Polygons
235
√ n √ n n+2 It is obvious that 2 2 > 2(2 − 2) 2 for all n ≥ 1. We also have 2n > 2(2 + 2) 2 for all n ≥ 9. The last inequality is equivalent to √ n 2+ 2 1 > for n ≥ 9. 4 4 The left-hand side member of the inequality decreases with n, so it suffices to notice that √ 9 2+ 2 1 . > 4 4 Therefore the inequality Sn (A) = Sn (E) can hold only for n ≤ 8. Now it is not difficult to verify that Sn (P) is constant only for n ∈ {2, 4, 6}. Problem 9. A function f : R2 → R is called Olympic if it has the following property: given n ≥ 3 distinct points A1 , A2 , . . . , An ∈ R2 , if f (A1 ) = f (A2 ) = · · · = f (An ) then the points A1 , A2 , . . . , An are the vertices of a convex polygon. Let P ∈ C[X ] be a nonconstant polynomial. Prove that the function f : R2 → R, defined by f (x, y) = |P(x + i y)|, is Olympic if and only if all the roots of P are equal. (Romanian Mathematical Olympiad – Final Round, 2000)
Solution. First suppose that all the roots of P are equal, and write P(x) = a(z−z 0 )n for some a, z 0 ∈ C and n ∈ N. If A1 , A2 , . . . , An are distinct point in R2 such that f (A1 ) = f (A2 ) = · · · = f (An ), then A1 , . . . , An are situated on a circle with center √ (Re(z 0 ), Im(z 0 )) and radius n | f (A1 )/a|, implying that the points are the vertices of a convex polygon. Conversely, suppose that not all the roots of P are equal, and write P(x) = (z−z 1 )(z−z 2 )Q(z) where z 1 and z 2 are distinct roots of P(x) such that |z 1 −z 2 | is minimal. Let l be the line containing Z 1 = (Re(z 1 ), Im(z 1 )) and Z 2 = (Re(z 2 ), Im(z 2 )), 1 and let z 3 = (z 1 + z 2 ) so that Z 3 = (Re(z 3 ), Im(z 3 )) is the midpoint of [Z 1 Z 2 ]. 2 Also, let s1 , s2 denote the rays Z 3 Z 1 and Z 3 Z 2 , and let d = f (Z 3 ) ≥ 0. We must have r > 0, because otherwise z 3 would be a root of P such that |z 1 − z 3 | < |z 1 − z 2 |, which is impossible. Because f (Z 3 ) = 0, lim f (Z ) = +∞,
Z 3 →∞ Z ∈s1
and f is continuous, there exists a point Z 4 ∈ s1 , on the side of Z 1 opposite Z 3 , such that f (Z 4 ) = r . Similarly, there exists Z 5 ∈ s2 , on the side of Z 2 opposite Z 3 , such that f (Z 5 ) = r . Thus, f (Z 3 ) = f (Z 4 ) = f (Z 5 ) and Z 3 , Z 4 , Z 5 are not vertices of a convex polygon. Hence, f is not Olympic.
236
5. Olympiad-Caliber Problems
%+ C %+ E % = 360◦ and Problem 10. In a convex hexagon ABC D E F, A AB · C D · E F = BC · D E · F A. Prove that AB · FC · EC = B F · D E · C A. (1999 Polish Mathematical Olympiad)
Solution. Position the hexagon in the complex plane and let a = B − A, b = C − B, . . . , f = A − F. The product identity implies that |ace| = |bd f |, and the −b −d − f angle equality implies · · is real and positive. Hence, ace = −bd f . Also, a c e a + b + c + d + e + f = 0. Multiplying this by ad and adding ace + bd f = 0 gives a 2 d + abd + acd + ad 2 + ade + ad f + ace + bd f = 0 which factors to a(d + e)(c + d) + d(a + b)( f + a) = 0. Thus |a(d + e)(c + d)| = |d(a + b)( f + a)|, which is what we wanted. Problem 11. Let n > 2 be an integer and f : R2 → R be a function such that for any regular n-gon A1 A2 · · · An , f (A1 ) + f (A2 ) + · · · + f (An ) = 0. Prove that f is identically zero. (Romanian Mathematical Olympiad – Final Round, 1996)
2π 2π Solution. We identify R2 with the complex plane and let ζ = cos + i sin . n n Then the condition is that for any z ∈ C and any positive real t, n
f (z + tζ j ) = 0.
j=1
In particular, for each of k = 1, . . . , n, we have n
f (z − ζ k + ζ j ) = 0.
j=1
Summing over k, we have n n
f (z − (1 − ζ m )ζ k ) = 0.
m=1 k=1
For m = n the inner sum is n f (z); for other m, the inner sum again runs over a regular polygon, hence is 0. Thus f (z) = 0 for all z ∈ C. Here are some proposed problems.
5.8. Complex Numbers and Combinatorics
237
Problem 12. Prove that there exists a convex 1990-gon with the following two properties: a) all angles are equal; b) the lengths of the sides are the numbers 12 , 22 , 32 , . . . , 19892 , 19902 in some order. (31st IMO)
Problem 13. Let A and E be opposite vertices of a regular octagon. Let an be the number of paths of length n of the form (P0 , P1 , . . . , Pn ) where Pi are vertices of the octagon and the paths are constructed using the rule: P0 = A, Pn = E, Pi and Pi+1 are adjacent vertices for i = 0, . . . , n − 1 and Pi = E for i = 0, . . . , n − 1. 1 Prove that a2n−1 = 0 and a2n = √ (x n−1 − y n−1 ), for all n = 1, 2, 3, . . . , where 2 √ √ x = 2 + 2 and y = 2 − 2. (21st IMO)
Problem 14. Let A, B, C be three consecutive vertices of a regular polygon and let us consider a point M on the major arc AC of the circumcircle. Prove that M A · MC = M B 2 − AB 2 . Problem 15. Let A1 A2 · · · An be a regular polygon with the circumradius equal to 1. n Find the maximum value of max P A j when P describes the circumcircle. j=1
(Romanian Mathematical Regional Contest “Grigore Moisil”, 1992)
Problem 16. Let A1 A2 · · · A2n be a regular polygon with circumradius equal to 1 and consider a point P on the circumcircle. Prove that n−1
P A2k+1 · P A2n+k+1 = 2n.
k=0
5.8
Complex Numbers and Combinatorics
Problem 1. Compute the sum 3n−1
k=0
(−1)
k
6n 2k + 1
3k .
Solution. We have 3n−1
k=0
(−1)k
3n−1
6n 6n 3k = (−3)k 2k + 1 2k + 1 k=0
238
5. Olympiad-Caliber Problems 3n−1
6n √ √ 2k 6n 1 3n−1 (i 3) = √ (i 3)2k+1 = 2k + 1 2k + 1 i 3 k=0 k=0 ' √ 6n 1 1 π (6n π = √ Im(1 + i 3) = √ Im 2 cos + i sin 3 3 i 3 i 3 1 = √ Im[26n (cos 2π n + i sin 2π n)] = 0. i 3 n
n cos kα, where α ∈ [0, π ]. Problem 2. Calculate the sum Sn = k k=0 Solution. Consider the complex number z = cos α + i sin α and the sum Tn = n
n sin kα. We have k k=0 Sn + i Tn =
n
n k=0
k
(cos kα + i sin kα) =
=
n
n k=0
k
n
n (cos α + i sin α)k k k=0
z k = (1 + z)n .
(1)
The polar form of complex number 1 + z is α α α + 2i sin cos 2 2 2 α α α = 2 cos cos + i sin 2 2 2 since α ∈ [0, π ]. From (1) it follows that α n nα nα Sn + i Tn = 2 cos cos + i sin , 2 2 2 1 + cos α + i sin α = 2 cos2
i.e.,
α n nα Sn = 2 cos cos 2 2 Problem 3. Prove the identity
and
α n nα sin Tn = 2 cos . 2 2
2 2 n n n n n n − + − ··· + − + − ··· = 2n . 0 2 4 1 3 5 Solution. Denote n n n − + − ··· xn = 0 2 4
and
n n n yn = − + − ··· 1 3 5
and observe that (1 + i)n = xn + yn i.
(1)
5.8. Complex Numbers and Combinatorics
239
Passing to the absolute value it follows that n
|xn + yn i| = |(1 + i)n | = |1 + i|n = 2 2 . This is equivalent to xn2 + yn2 = 2n . Remark. We can write the explicit formulas for xn and yn as follows. Observe that √ n nπ π π n nπ (1 + i)n = 2 cos + i sin = 2 2 cos + i sin . 4 4 4 4 From relation (1) we get n
xn = 2 2 cos
nπ 4
and
n
yn = 2 2 sin
nπ . 4
Problem 4. If m and p are positive integers and m > p, then m m m m + + + + ··· 0 p 2p 3p ' ( ⎛ ⎞ p−1 2 m m
2 ⎜ kπ mkπ ⎟ = cos cos ⎝1 + ⎠. p p p k=1 Solution. We begin with the following simple but useful remark: If f ∈ R[X ] is 2π 2π + i sin is the p th a polynomial, f = a0 + a1 X + · · · + am X m , and ε = cos p p primitive root of unity, then for all real numbers n the following relation holds: a0 + a p x p + a 2 p x 2 p + · · · =
1 ( f (x) + f (εx) + · · · + f (ε p−1 x)). p
(1)
To prove (1) we use the relation 1+ε +ε k
2k
+ ··· + ε
( p−1)k
=
p, if p|k, 0, otherwise,
on the right-hand side. the case when p is odd. Using relation (1) for polynomial f = (1+ X )m = Consider m m m + X + ··· + X m we obtain 0 1 m m m p m 2p 1 + x + x +· · · = ((1+x)m +(1+εx)m +· · ·+(1+ε p−1 x)m ) (2) 0 p 2p p Substituting x = 1 in relation (2) we find m m m 1 Sp = + + + · · · = (2m + (1 + ε)m + · · · + (1 + ε p−1 )m ). (3) p 0 p 2p
240
5. Olympiad-Caliber Problems
2kπ 2kπ + i sin it follows that for all k = 0, 1, . . . , p − 1 p p kπ m mkπ mkπ (1 + ε k )m = 2m cos cos + i sin . p p p
From ε k = cos
Using the relation ε p−k = ε k we find (1 + ε p−k )m = (1 + ε k )m = (1 + ε k )m kπ m mkπ mkπ cos − i sin . = 2m cos p p p Replacing in (3) we obtain ⎤ ⎡ p−1 p−1 p−1 2 2
1 1 Sp = (1 + ε k )m = ⎣ (1 + ε k )m + (1 + ε p−k )m ⎦ p k=0 p k=0 k=1 ⎡ p−1 2
1⎣ m kπ m mkπ mkπ m = cos cos 2 +2 + i sin p p p p k=1 2
p−1
+2m
k=1
kπ cos p
⎤ mkπ ⎦ mkπ − i sin cos p p
m
⎛ ⎞ p−1 2
kπ m mkπ ⎠ 2m ⎝ cos cos 1+2 . = p p p k=1 p
Consider now the case when p is an even positive integer. Because ε 2 = −1 we have ⎤ ⎡ p p−1 p−1 2 −1
1 1 Sp = (1 + ε k )m = ⎣2m + (1 + ε k )m + (1 + ε k )m ⎦ p k=0 p p k=1 k= 2 +1
⎡ p 2 −1 1⎣ m
mkπ kπ m mkπ = 2 + + i sin + cos 2m cos p p p p k=1
p
+
2 −1
2m
k=1
kπ cos p
⎛ =
2m p
⎝1 + 2
⎤ mkπ mkπ ⎦ cos − i sin p p
m
2 −1
p
cos
k=1
kπ p
⎞
m cos
mkπ ⎠ . p
5.8. Complex Numbers and Combinatorics
241
Problem 5. The following identity holds: p−1 kπ n (n − 2m)kπ n n n 2n
cos cos . + + + ··· = p k=0 p p m m+p m + 2p Solution. Let ε0 , ε1 , . . . , ε p−1 be the p th roots of unity. Then p−1
εk−m (1 + εk )n =
k=0
n
n (ε0k−m + · · · + εk−m p−1 ). k k=0
Using the result in Proposition 3, Subsection 2.2.2, it follows that p, if p|(k − m), ε0k−m + · · · + εk−m p−1 = 0, otherwise.
(1)
(2)
Taking into account that εk−m (1 + εk )m 2mkπ kπ n nkπ nkπ 2mkπ − i sin 2 cos + i sin cos = cos p p p p p n kπ (n − 2m)kπ (n − 2m)kπ = 2n cos cos + i sin p p p and using (1) and (2) the desired identity follows. Remark. The following interesting trigonometric relation holds: p−1
k=0
kπ cos p
n sin
(n − 2m)kπ = 0. p
Problem 6. Consider the integers an , bn , cn , where n n n an = + + + ··· , 0 3 6 n n n + + + ··· , bn = 1 4 7 n n n + + + ··· . cn = 2 5 8 Show that: 1) an3 + bn3 + cn3 − 3an bn cn = 2n . 2) an2 + bn2 + cn2 − an bn − bn cn − cn an = 1. 3) Two of integers an , bn , cn are equal and the third differs by one.
(3)
242
5. Olympiad-Caliber Problems
Solution. 1) Let ε be a cube root of unity different from 1. We have (1 + 1)n = an + bn + cn , (1 + ε)n = an + bn ε + cn ε 2 , (1 + ε2 )n = an + bn ε2 + cn ε. Therefore an3 + bn3 + cn3 − 3an bn cn = (an + bn + cn )(an + bn ε + cn ε 2 )(an + bn ε 2 + cn ε) = 2n (1 + ε)n (1 + ε 2 )n = 2n (−ε 2 )n (−ε)n = 2n . 2) Using the identity x 3 + y 3 + z 3 − 3x yz = (x + y + z)(x 2 + y 2 + z 2 − x y − yz − zx) and the above relation it follows that an2 + bn2 + cn2 − an bn − bn cn − cn an = 1. 3) Multiplying the above relation by 2 we find (an − bn )2 + (bn − cn )2 + (cn − an )2 = 2.
(1)
From (1) it follows that two of an , bn , cn are equal and the third differs by one. Remark. From Problem 5 it follows that 1 n nπ 2nπ nπ 1 n 2 + cos + (−1)n cos = 2 + 2 cos , an = 3 3 3 3 3 1 n (n − 2)π (2n − 4)π 2 + cos + (−1)n cos bn = 3 3 3 1 n (n − 2)π = 2 + 2 cos , 3 3 1 n (n − 4)π (2n − 8)π n cn = 2 + cos + (−1) cos 3 3 3 1 n (n − 4)π = 2 + 2 cos . 3 3 It is not difficult to see that an = bn if and only if n ≡ 1
(mod 3),
an = cn if and only if n ≡ 2
(mod 3),
bn = cn if and only if n ≡ 0
(mod 3).
Problem 7. How many positive integers of n digits chosen from the set {2, 3, 7, 9} are divisible by 3? (Romanian Mathematical Regional Contest “Traian Lalescu”, 2003)
5.8. Complex Numbers and Combinatorics
243
Solution. Let xn , yn , z n be the number of all positive integers of n digits 2, 3, 7 or 9 which are congruent to 0, 1 and 2 modulo 3. We have to find xn . 2π 2π + i sin . It is clear that xn + yn + z n = 4n and Consider ε = cos 3 3
xn + εyn + ε 2 z n = ε 2 j1 +3 j2 +7 j3 +9 j4 = (ε2 + ε 3 + ε 7 + ε 9 )n = 1. j1 + j2 + j3 + j4 =n
It follows that xn − 1 + εyn + ε2 z n = 0. Applying Proposition 4 in Subsection 2.2.2 we obtain xn − 1 = yn = z n = k. Then 3k = xn + yn + z n − 1 = 4n − 1 and we find 1 1 k = (4n − 1). Finally xn = k + 1 = (4n + 2). 3 3 Problem 8. Let n be a prime number and let a1 , a2 , . . . , am be positive integers. Consider f (k) the number of all m-tuples (c1 , . . . , cm ) satisfying 1 ≤ ci ≤ ai and m
ci ≡ k (mod n). Show that f (0) = f (1) = · · · = f (n − 1) if and only if n|a j for i=1
some j ∈ {1, . . . , m}. (Rookie Contest, 1999)
2π 2π Solution. Let ε = cos + i sin . Note that the following relations hold: n n m
(X + X 2 + · · · + X ai ) =
X c1 +···+cm
1≤ci ≤ai
i=1
and f (0) + f (1)ε + · · · + f (n − 1)ε n−1 =
ε c1 +···+cm =
1≤ci ≤ai
m
(ε + ε 2 + · · · + εai ).
i=1
Applying the result in Proposition 4, Subsection 2.2.2, we have f (0) = f (1) = · · · = f (n − 1) if and only if f (0) + f (1)ε + · · · + f (n − 1)ε n−1 = 0. This is m equivalent to (ε + ε 2 + · · · + εai ) = 0, i.e., ε + ε2 + · · · + εa j = 0 for some i=1
j ∈ {1, . . . , m}. It follows that εa j − 1 = 0, i.e., n|a j . Problem 9. For a finite set of real numbers A denote by |A| the cardinal number of A and by m(A) the sum of elements of A. Let p be a prime and A = {1, 2, . . . , 2 p}. Find the number of all subsets B ⊂ A such that |B| = p and p|m(B). (36th IMO)
2π 2π + i sin . p p Denote by x j the number of all subsets B ⊂ A with properties |B| = p and m(B) ≡ j (mod p). Solution. The case p = 2 is trivial. Consider p ≥ 3 and ε = cos
244
Then
5. Olympiad-Caliber Problems
p−1
j=0
xjεj =
ε m(B) =
ε c1 +···+c p
1≤c1 <···<≤c p ≤2 p
B⊂A,|B|= p
The last sum is the coefficient of in (X + ε)(X + ε 2 ) · · · (X + ε 2 p ). Taking into account the relation X p − 1 = (X − 1)(X − ε) · · · (X − ε p−1 ) we obtain (X + ε)(X + ε 2 ) · · · (X + ε 2 p ) = (X p + 1)2 , hence the coefficient of X p is 2. Therefore Xp
p−1
x j ε j = 2,
j=0
i.e., x0 −2+x1 ε+· · ·+x p−1 ε p−1 = 0. From Proposition 4, Subsection 2.2.2,it follows 2p −2 that x0 − 2 = x1 = · · · = x p−1 = k. We find pk = x0 + · · · + x p−1 − 2 = p 1 2p hence k = − 2 . Therefore, the desired number is p p 2p 1 x0 = 2 + k = 2 + −2 . p p n
2n + 1 3k 2 is not divisible by 5 for any Problem 10. Prove that the number 2k + 1 k=0 integer n ≥ 0. (16th IMO)
Solution. Since 23 ≡ −2 (mod 5), an equivalent problem is to prove that Sn = n
√ 2n + 1 (−2)k is not divisible by 5. Expanding (1 + i 2)2n+1 and then separat2k + 1 k=0 ing the even and odd terms we get √ √ (1) (1 + i 2)2n+1 = Rn + i 2Sn , n
2n + 1 where Rn = (−2)k . 2k k=0 Passing to the absolute value from (1) it follows that 32n+1 = Rn2 + 2Sn2
(2)
Since 32 ≡ −1 (mod 5), the relation (2) leads to Rn2 + 2Sn2 ≡ ±3
(mod 5).
(3)
Assume by contradiction that Sn ≡ 0 (mod 5) for some positive integer n. Then from (3) we obtain Rn2 ≡ ±3 (mod 5), a contradiction since any square is congruent to 0, 1, or 4 modulo 5.
5.9. Miscellaneous Problems
245
Here are other problems concerning complex numbers and combinatorics. n 2
n Problem 11. Calculate the sum sn = cos kt, where t ∈ [0, π ]. k k=0 Problem following identities: 12. Prove that n n n n 1 n nπ 1) + + + ··· = 2 + 2 2 +1 cos . 0 4 8 4 4 (Romanian Mathematical Olympiad – Second Round, 1981)
n n n 2) + + + ··· = 0 5 10 √ √ nπ 2nπ ( 5 − 1)n 1 n ( 5 + 1)n cos cos 2 + + . = 5 5 5 2n−1 2n−1 Problem 13. Consider the integers An , Bn , Cn defined by An =
n n n − + − ··· , 0 3 6
n n n Bn = − + − + ··· , 1 4 7 n n n − + − ··· . Cn = 2 5 8 The following identities hold: 1) A2n + Bn2 + Cn2 − An Bn − Bn Cn − Cn An = 3n ; 2) A2n + An Bn + Bn2 = 3n−1 . Problem 14. Let p ≥ 3 be a prime and let m, n be positive integers divisible by p such that n is odd. For each m-tuple (c1 , . . . , cm ), ci ∈ {1, 2, . . . , n}, with the property that m
p| ci , let us consider the product c1 · · · cm . Prove that the sum of all these products i=1 m n are divisible by . p Problem 15. Let k be a positive integer and a = 4k − 1. Prove that for any positive integer n, the integer sn =
n n n 2 n 3 − a+ a − a + · · · is divisible by 2n−1 . 0 2 4 6 (Romanian Mathematical Olympiad – Second Round, 1984)
246
5. Olympiad-Caliber Problems
5.9
Miscellaneous Problems
Problem 1. Two unit squares K 1 , K 2 with centers M, N are situated in the plane so that M N = 4. Two sides of K 1 are parallel to the line M N , and one of the diagonals of K 2 lies on M N . Find the locus of the midpoint of X Y as X, Y vary over the interior of K 1 , K 2 , respectively. (1997 Bulgarian Mathematical Olympiad)
Solution. Introduce complex numbers with M = −2, N = 2. Then the locus is the set of points of the form −(w + xi) + (y + zi), where |w|, |x| < 1/2 and |x + y|, |x − √ √ √ y| < 2/2. The result is an octagon with vertices (1+ 2)/2+i/2, 1/2+(1+ 2)i/2, and so on. Problem 2. Curves A, B, C and D are defined in the plane as follows: x A = (x, y) : x 2 − y 2 = 2 , x + y2 y B = (x, y) : 2x y + 2 = 3 , x + y2 C = {(x, y) : x 3 − 3x y 2 + 3y = 1}, D = {(x, y) : 3x 2 y − 3x − y 3 = 0}. Prove that A ∩ B = C ∩ D. (1987 Putnam Mathematical Competition)
Solution. Let z = x + yi. The equations defining A and B are the real and imaginary parts of the equation z 2 = z −1 + 3i, and similarly the equations defining C and D are the real and imaginary parts of z 3 − 3i z = 1. Hence for all real x and y, we have (x, y) ∈ A ∩ B if and only if z 2 = z −1 + 3i. This is equivalent to z 3 − 3i z = 1, i.e., (x, y) ∈ C ∩ D. Thus A ∩ B = C ∩ D. Problem 3. Determine with proof whether or not it is possible to consider 1975 points on the unit circle such that the distances between any two points are rational numbers (the distances being taken along the chord). (17th IMO)
Solution. There are infinitely many points with rational coordinates on the unit circle. This is a well-known result arising from Pythagorean triangles and the corresponding equation: m 2 + n 2 = p2 . Any such point A(x A , y A ) can be represented by a complex number z A = x A + i y A = cos α A + i sin α A
5.9. Miscellaneous Problems
247
where α A is the argument of the complex number z A and cos α A , sin α A are rational numbers. Taking on the unit circle complex numbers of the form z 2A = cos 2α A + i sin 2α A we have for two such points:
|z 2A − z 2B | = (cos 2α A − cos 2α B )2 + (sin 2α A − sin 2α B )2
= 2[1 − cos 2(α B − α A )] = 2 · 2 sin2 (α B − α A ) = 2| sin(α B − α A )| = 2| sin α B cos α A − sin α A cos α B | ∈ Q. Answer: Yes, it is possible. Problem 4. A tourist takes a trip through a city in stages. Each stage consists of three segments of length 100 meters separated by right turns of 60◦ . Between the last segment of one stage and the first segment of the next stage, the tourist makes a left turn of 60◦ . At what distance will the tourist be from his initial position after 1997 stages? (1997 Rio Plata Mathematical Olympiad)
Solution. In one stage, the tourist traverses the complex number √ x = 100 + 100ε + 100ε 2 = 100 − 100 3i, π π where ε = cos + i sin . 3 3 Thus in 1997 stages, the tourist traverses the complex number z = x + xε + xε 2 + · · · + xε 1996 = x
1 − ε 1997 = xε 2 . 1−ε
Hence, the tourist ends up |z| = |xε2 | = |x| = 200 meters away from his initial position. Problem 5. Let A, B, C, be fixed points in the plane. A man starts from a certain point P0 and walks directly to A. At A he turns by 60◦ to the left and walks to P1 such that P0 A = A P1 . After he performs the same action 1986 times successively around points A, B, C, A, B, C, . . . , he returns to the starting point. Prove that ABC is an equilateral triangle, and that the vertices A, B, C, are arranged counterclockwise. (27th IMO)
Solution. For convenience, let A1 , A2 , A3 , A4 , A5 , . . . be A, B, C, A, B, . . . , respectively, and let P0 be the origin. After the k th step, the position Pk will be Pk = 4π 4π + i sin . We easily obtain Ak + (Pk−1 − Ak )ε for k = 1, 2, . . . , where ε = cos 3 3 Pk = (1 − ε)(Ak + ε Ak−1 + ε 2 Ak−2 + · · · + ε k−1 A1 ).
248
5. Olympiad-Caliber Problems
The condition P = P1986 is equivalent to A1986 +ε A1985 +· · ·+ε1984 A2 +ε 1985 A1 = 0, which having in mind that A1 = A4 = A7 = · · · , A2 = A5 = A8 = · · · , A3 = A6 = A9 = · · · , reduces to 662(A3 + ε A2 + ε 2 A1 ) = (1 + ε3 + · · · + ε1983 )(A3 + ε A2 + ε 2 A1 ) = 0, and the assertion follows from Proposition 2 in Section 3.4. Problem 6. Let a, n be integers and let p be prime such that p > |a| + 1. Prove that the polynomial f (x) = x n + ax + p cannot be represented as a product of two nonconstant polynomials with integer coefficients. (1999 Romanian Mathematical Olympiad)
Solution. Let z be a complex root of the polynomial. We shall prove that |z| > 1. Suppose |z| ≤ 1. Then, z n + az = − p, we deduce that p = |z n + az| = |z||z n−1 + a| ≤ |z n−1 | + |a| ≤ 1 + |a|, which contradicts the hypothesis. Now, suppose f = gh is a decomposition of f into nonconstant polynomials with integer coefficients. Then p = f (0) = g(0)h(0), and either |g(0)| = 1 or |h(0)| = 1. Assume without loss generality that |g(0)| = 1. If z 1 , z 2 , . . . , z k are the roots of g, then they are also roots of f . Therefore 1 = |g(0)| = |z 1 z 2 · · · z k | = |z 1 ||z 2 | · · · |z k | > 1, a contradiction. Problem 7. Prove that if a, b, c are complex numbers such that ⎧ ⎪ ⎨ (a + b)(a + c) = b, (b + c)(b + a) = c, ⎪ ⎩ (c + a)(c + b) = a, then a, b, c are real numbers. (2001 Romanian IMO Team Selection Test)
Solution. Let P(x) = x 3 − sx 2 + q x − p be the polynomial with roots a, b, c. We have s = a + b + c, q = ab + bc + ca, p = abc. The given equalities are equivalent to ⎧ ⎪ ⎨ sa + bc = b, (1) sb + ca = c, ⎪ ⎩ sc + ab = a.
5.9. Miscellaneous Problems
249
Adding these equalities, we obtain q = s − s 2 . Multiplying the equalities in (1) by a, b, c, respectively, and adding them we obtain s(a 2 + b2 + c2 ) + 3 p = q or, after a short computation, 3 p = −3s 3 + s 2 + s. (2) If we write the given equations in the form (s − c)(s − b) = b,
(s − a)(s − c) = c,
(s − b)(s − a) = a,
we obtain ((s − a)(s − b)(s − c))2 = abc, and, by performing standard computations and using (2), we finally get s(4s − 3)(s + 1)2 = 0. If s = 0, then P(x) = x 3 , so a = b = c = 0. If s = −1, then P(x) = x 3 + x 2 − 2π 4π 6π 2x − 1, which has the roots 2 cos , 2 cos , 2 cos (this is not obvious, but we 7 7 7 can see that P changes its sign on the intervals (−2, −1), (−1, 0), (1, 2) of the real 1 3 3 line, hence its roots are real). Finally, if s = 3/4, then P(x) = x 3 − x 2 + x − , 4 16 64 which has roots a = b = c = 1/4. Alternate solution. Subtract the second equation from the first. We obtain (a + b)(a−b) = b−c. Analogously, (b+c)(b−c) = c−a and (c+a)(c−a) = a−b. We can see that if two of the numbers are equal, then all three are equal and the conclusion is obvious. Suppose that the numbers are distinct. Then, after multiplying the equalities above, we obtain (a + b)(b + c)(c + a) = 1, and next: b(b + c) = c(c + a) = a(a + b) = 1. Now, if one of the numbers is real, it follows immediately that all three are real. Suppose all numbers are not real. Then arg a, arg b, arg c ∈ (0, 2π ). Two of the numbers arg a, arg b, arg c are contained in either (0, π) or in [π, 2π ). Suppose these are arg a, arg b and that arg a ≤ arg b. Then arg a ≤ arg(a + b) ≤ arg b and arg a ≤ arg a(a + b) ≤ arg(a + b) ≤ arg b. This is a contradiction, since a(a + b) = 1. Problem 8. Find the smallest integer n such that an n × n square can be partitioned into 40 × 40 and 49 × 49 squares, with both types of squares present in the partition. (2000 Russian Mathematical Olympiad)
Solution. We can partition a 2000 × 2000 square into 40 × 40 and 49 × 49 squares: partition one 1960 × 1960 corner of the square into 49 × 49 squares and then partition the remaining portion into 40 × 40 squares. We now show that n must be at least 2000. Suppose that an n × n square has been partitioned into 40 × 40 and 49 × 49 squares, using at least one of each type. Let 2π 2π 2π 2π + i sin and ξ = cos + i sin . Orient the n × n square so that ζ = cos 40 40 49 49
250
5. Olympiad-Caliber Problems
two sides are horizontal, and number the rows and columns of unit squares from the top left: 0, 1, 2, . . . , n − 1. For 0 ≤ j, k ≤ n − 1, and write ζ j ξ k in square ( j, k). If an m × m square has its top-left corner at (x, y), then the sum of the numbers written in it is m m x+m−1
y+m−1
ξ −1 j k x y ζ −1 ζ ξ =ζ ξ . ζ −1 ξ −1 j=x k=y The first fraction in parentheses is 0 if m = 40, and the second fraction is 0 if m = 49. Thus, the sum of the numbers written inside each square in the partition is 0, so the sum of all the numbers must be 0. However, applying the above formula with (m, x, y) = (n, 0, 0), we find that the sum of all the numbers equals 0 only if either ζ n − 1 or ξ n − 1 equals 0. Thus, n must be either a multiple of 40 or a multiple of 49. Let a and b be the number of 40 × 40 and 49 × 49 squares, respectively. The area of the square equals 402 · a + 492 · b = n 2 . If 40|n, then 402 |b and hence b ≥ 402 . Thus, n 2 > 492 · 402 = 19602 ; because n is a multiple of 40, n ≥ 50 · 40 = 2000. If instead 49|n, then 492 |a, a ≥ 492 , and again n 2 > 19602 . Because n is a multiple of 49, n ≥ 41 · 49 = 2009 > 2000. In either case, n ≥ 2000, and 2000 is the minimum possible value of n. Problem 9. The pair (z 1 , z 2 ) of nonzero complex numbers has the following property: there is a real number a ∈ [−2, 2] such that z 12 − az 1 z 2 + z 22 = 0. Prove that all pairs (z 1n , z 2n ), n = 2, 3, . . . , have the same property. (Romanian Mathematical Olympiad – Second Round, 2001)
z1 Solution. Denote t = , t ∈ C∗ . The relation z 12 − az 1 z 2 + z 22 = 0 is equivalent z2 √ a ± i 4 − a2 2 2 and to t − at + 1 = 0. We have = a − 4 ≤ 0, hence t = 2 n 2 2 z a 4−a + = 1. If t = cos α + i sin α, then 1n = t n = cos nα + i sin nα |t| = 4 4 z2 and we can write z 12n − an z 1n z 2n + z 22n = 0, where an = 2 cos nα ∈ [−2, 2]. z 12
Alternate solution. Because a ∈ [−2, 2], we can write a = 2 cos α. The relation − az 1 z 2 + z 22 = 0 is equivalent to z1 z2 + = 2 cos α (1) z2 z1
and, by a simple inductive argument, from (1) it follows that z 1n zn + 2n = 2 cos nα, n z2 z1 Problem 10. Find min
z∈C \R
n = 1, 2, . . . .
Imz 5 Im5 z
5.9. Miscellaneous Problems
251
and the values of z for which the minimum is reached. Solution. Let a, b be real numbers such that z = a + bi, b = 0. Then Im(z)5 = 5a 4 b − 10a 2 b3 + b5 and Imz 5 Im5 z Setting x =
a 2 b
=5
a 4 b
− 10
a 2 b
+ 1.
yields Im(z)5 Im5 z
= 5x 2 − 10x + 1 = 5(x − 1)2 − 4.
The minimum value is −4 and is obtained for x = 1 i.e., for z = a(1 ± i), a = 0. Problem 11. Let z 1 , z 2 , z 3 be complex numbers, not all real, such that |z 1 | = |z 2 | = |z 3 | = 1 and 2(z 1 + z 2 + z 3 ) − 3z 1 z 2 z 3 ∈ R. Prove that π max(arg z 1 , arg z 2 , arg z 3 ) ≥ . 6 Solution. Let z k = cos tk + i sin tk , k ∈ {1, 2, 3}. The condition 2(z 1 + z 2 + z 3 ) − 3z 1 z 2 z 3 ∈ R implies 2(sin t1 + sin t2 + sin t3 ) = 3 sin(t1 + t2 + t3 ).
(1)
π π Assume by way of contradiction that max(t1 , t2 , t3 ) < , hence t1 , t2 , t3 < . Let 6 6 ' π t1 + t2 + t3 π t= ∈ 0, . The sine function is concave on 0, , so 3 6 6 1 t1 + t2 + t3 (sin t1 + sin t2 + sin t3 ) ≤ sin . 3 3 From the relations (1) and (2) we obtain sin(t1 + t2 + t3 ) t1 + t2 + t3 ≤ sin . 2 3 Then sin 3t ≤ 2 sin t. It follows that 4 sin3 t − sin t ≥ 0,
π 1 1 π . Hence sin t ≥ , then t ≥ , which contradicts that t ∈ 0, . 4 2 6 6 π Therefore max(t1 , t2 , t3 ) ≥ , as desired. 6
i.e., sin2 t ≥
(2)
252
5. Olympiad-Caliber Problems
Here are some more problems. Problem 12. Solve in complex numbers the system of equations ⎧ 2 ⎪ ⎨ x|y| + y|x| = 2z , y|z| + z|y| = 2x 2 , ⎪ ⎩ z|x| + x|z| = 2y 2 . Problem 13. Solve in complex numbers the following: ⎧ ⎪ ⎨ x(x − y)(x − z) = 3, y(y − x)(y − z) = 3, ⎪ ⎩ z(z − x)(z − y) = 3. (Romanian Mathematical Olympiad – Second Round, 2002)
Problem 14. Let X, Y, Z , T be four points in the plane. The segments [X Y ] and [Z T ] are said to be connected if there is some point O in the plane such that the triangles O X Y and O Z T are right isosceles triangles in O. Let ABC D E F be a convex hexagon such that the pairs of segments [AB], [C E], and [B D], [E F] are connected. Show that the points A, C, D and F are the vertices of a parallelogram and that the segments [BC] and [AE] are connected. (Romanian Mathematical Olympiad – Final Round, 2002)
Problem 15. Let ABC D E be a cyclic pentagon inscribed in a circle of center O which % = 120◦ , D % = 130◦ , E % = 100◦ . Show that the diagonals B D has angles % B = 120◦ , C and C E meet at a point belonging to the diameter AO. (Romanian IMO, Team Selection Test, 2002)
6 Answers, Hints and Solutions to Proposed Problems
In what follows answers and solutions are presented to problems posed in previous chapters. We have preserved the title of the subsection containing the problem and the number of the proposed problem.
6.1 6.1.1
Answers, Hints and Solutions to Routine Problems Complex numbers in algebraic representation (pp. 18–21)
1. a) z 1 + z 2 + z 3 = (0, 4); b) z 1 z 2 + z 2 z 3 + z 3 z 1 = (−4, 5); c) z 1 z 2 z 3 = (−9, 7); d) z 12 + z 22 + z 32 = (−8, −10); z 2 + z 22 z1 z2 z3 311 65 152 72 e) + + = − = , ; f) 12 , − . z2 z3 z1 130 83 221 221 z 2 + z 32 2. a) z = (7, b) z = (−7, −4); −8); 2 23 ,− ; d) z = (−9, 7). c) z = 13 13
√ √ 3 3 1 1 , z2 = − , − ; 3. a) z 1 = − , 2 2 2 2 √ √ 3 3 1 1 b) z 1 = (−1, 0), z 2 = , , z3 = ,− . 2 2 2 2
254
6. Answers, Hints and Solutions to Proposed Problems
⎧ (1, 0), for ⎪ ⎪ ⎪ n ⎨ (1, 1), for
4. zk = ⎪ (0, 1), for ⎪ k=0 ⎪ ⎩ (0, 0), for 5. a) z = (1, 1);
= 4k; = 4k + 1; = 4k + 2; = 4k + 3.
b) z 1 = (2, 1), z 2 = (−2, −1).
= z 3 = (a 3 − 3ab2 , 3a 2 b − b3 ); z 4 = (a 4 − 6a 2 b2 + b4 , 4a 3 b − 4ab3 ). ⎛& ⎞ & √ √ 2 2 2 2 a+ a +b −a + a + b ⎠ , sgn b , 7. z 1 = ⎝ 2 2 ⎛ & ⎞ & √ √ 2 + b2 2 + b2 a + a −a + a ⎠. , −sgn b z 2 = ⎝− 2 2
6.
z2
(a 2
n n n n
− b2 , 2ab);
8. For all nonnegative integers k we have z 4k = ((−4)k , 0); z 4k+1 = ((−4)k , −(−4)k );
z 4k+2 = (0, −2(−4)k );
z 4k+3 = (−2(−4)k , −2(−4)k ); for k ≥ 0. 1 3 9. a) x = , y = ; b) x = −2, y = 8; c) x = 0, y = 0. 4 4 √ 61 4 11 5 7 − i; e) + i. 10. a) 8 + 51i; b) 4 − 43i; c) 2; d) 4 2 13 13 11. a) −i; b) E 4k = 1, E 4k+1 = 1 + i, E 4k+2 = i, E 4k+3 = 0; c) 1; √ √ √ √ 2 2 2 2 +i , z2 = − −i ; 12. a) z 1 = 2 2 2 2 √ √ √ √ 2 2 2 2 b) z 1 = −i , z2 = − +i ; 2 2 √ 2 2
√ 1+ 3 3−1 − i . c) z 1,2 = ± 2 2 13. z ∈ R or z = x + i y with x 2 + y 2 = 1. 14. a) E 1 = E 1 ; b) E 2 = E 2 . 15. We substitute a formula for the definition of modulus. 16. From the identity z+ we obtain
z +
1 z
3 = z3 +
1 3 ≤ 2 + 3 z + z
1 z
1 1 + 3 z + z z3
or a 3 − 3a − 2 ≤ 0,
d) −3i.
6.1. Answers, Hints and Solutions to Routine Problems
where
a = z +
255
1 , a ≥ 0. z
Since a 3 − 3a − 2 = (a − 2)(a 2 + 2a + 1) = (a − 2)(a + 1)2 , we have a ≤ 2, as desired. 17. The equation |z 2 + z 2 | = 1 is equivalent to |z 2 + z 2 |2 = 1. That is, (z 2 + z 2 )(z 2 + 2 z 2 ) = 1. We find (z 2 + z 2 )2 = 1 or z 2 + z12 = 1. The last equation is equivalent to √ 3 1 4 2 4 4 2 4 2 (z + 1) = z or (z − z + 1)(z + z + 1) = 0. The solutions are ± i ± and 2 2 √ 3 1 ± ± i. 2 2 2 √ 2 18. z ∈ ± , ±i 2 . 3 19. z ∈ {0, 1, −1, i, −i}. 1 1 1 20. Observe that − < is equivalent to |2 − z| < |z|, and consequently (2 − z 2 2 z)(2 − z) < z · z. It follows that 4 < 2(z + z) = 4Re(z), as needed. 21. a 2 + b2 + c2 − ab − bc − ca. √ 7 −6 + 21 + 2i; b) z = − + 4i; c) z = 2 + i; 22. a) z 1,2 = 3√ 6 −2 ± 3 1 13 9 2 d) z 1,2 = + i; e) z = −1, z 2 = −5 − 6i; f) z 2 = − − i. 2 2 2 2 23. m ∈ {1, 5}. 24. z = −2y + 2 + i y, y ∈ R. 25. z = x + i y with x 2 + y 2 = 1. √ 26. From |z 1 + z 2 | = 3 it follows that |z 1 + z 2 |2 = 3, i.e., (z 1 + z 2 )(z 1 + z 2 ) = 3. We obtain |z 1 |2 + (z 1 z 2 + z 1 z 2 ) + |z 2 |2 = 3. That is, z 1 z 2 + z 1 z 2 = 1. On the other hand we have |z 1 − z 2 |2 = |z 1 |2 − (z 1 z 2 + z 1 z 2 ) + |z 2 |2 = 2 − 1 = 1, hence |z 1 − z 2 | = 1. √ 3 1 and noticing that ε3 = 1, we obtain n = 3k, k ∈ Z. 27. Letting ε = − + i 2 2 28. Note that z = 0 is a solution. For z = 0 passing to absolute value we obtain |z|n−1 = |z|, i.e., |z| = 1. The equation is equivalent to z n = i z · z, which reduces to z n = i. The total number of solutions is n + 1. 29. Let α = |z 2 − z 3 |,
β = |z 3 − z 1 |,
γ = |z 1 − z 2 |.
256
6. Answers, Hints and Solutions to Proposed Problems
Since the following inequality, αβ + βγ + γ α ≤ α 2 + β 2 + γ 2 holds, and α 2 + β 2 + γ 2 = 3(|z 1 |2 + |z 2 |2 + |z 2 |2 − |z 1 + z 2 + z 3 |2 ) ≤ 3(|z 1 |2 + |z 2 |2 + |z 2 |2 = 9R 2 , it follows that αβ + βγ + γ α ≤ 9r 2 . 30. Observe that |w| = |v| ·
|u − z| |u − z| = ≤1 uz − 1 |uz − 1|
if and only if |u − z| ≤ |uz − 1|. This is equivalent to |u − z|2 ≤ |uz − 1|2 . We obtain (u − z)(u − z) ≤ (uz − 1)(uz − 1), i.e., |u|2 + |z|2 − |u|2 |z|2 − 1 ≤ 0. Finally (|u 2 | − 1)(|z|2 − 1) ≥ 0. Since |u| ≤ 1, it follows that |w| ≤ 1 if and only if |z| ≤ 1, as desired. 31. z 12 + z 22 + z 32 = (z 1 + z 2 + z 3 )2 − 2(z 1 z 2 + z 2 z 3 + z 3 z 1 ) 1 1 1 = −2z 1 z 2 z 3 = −2z 1 z 2 z 3 (z 1 + z 2 + z 3 ) = 0. + + z1 z2 z3 r2 for k ∈ {1, 2, . . . , n}. Then zk r2 r2 r2 r2 r2 r2 ··· + + + z1 z2 z2 z3 zn z1
32. The relation |z k | = r implies z k =
E=
=
r2 r2 r2 · ··· z1 z2 zn r 2n ·
z1 + z2 z2 + z3 zn + z1 · ··· z1 z2 z2 z3 zn z1 = E, 1 r 2n · z1 z2 · · · zn
6.1. Answers, Hints and Solutions to Routine Problems
257
hence E ∈ R. 33. Notice that z1 · z1 = z2 · z2 = z3 · z3 = r 2 and z 1 z 2 + z 3 ∈ R if and only if z 1 z 2 + z 3 = z 1 · z 2 + z 3 . Then
r2 z1 z2 + z3 z1 z3 + z2 z2 z3 + z1 = = = z1 z2 z3 z1 z2 + r 2 z3 z1 z3 + r 2 z2 z2 z3 + r 2 z1 (z 1 − 1)(z 2 − z 3 ) z2 − 1 z3 − 1 z1 − z2 z1 − 1 = = = = 1. = z1 − z2 (z 2 − z 3 )(z 1 − r 2 ) z1 − r 2 z2 − r 2 z3 − r 2
Hence z 1 z 2 z 2 = r 2 and consequently r 3 = r 2 . Therefore r = 1 and z 1 z 2 z 3 = 1, as desired. 34. Note that x13 = x23 = −1. a) −1; b) 1; c) Consider n ∈ {6k, 6k ± 1, 6k ± 2, 6k ± 3}. 35. a) x 4 + 16 = x 4 + 24 = (x 2 + 4i)(x 2 − 4i) √ √ = [x 2 + ( 2(1 + i))2 ][x 2 − ( 2(1 + i))2 ] √ √ √ √ = (x + 2(−1 + i))(x + 2(1 − i))(x − 2(1 + i))(x + 2(1 + i)). √ 3 1 3 3 3 2 b) x − 27 = x − 3 = (x − 3)(x − 3ε)(x − 3ε ), where ε = − + i. 2 2 √ √ 3 3 3 c) x + 8 = x + 2 = (x + 2)(x + 1 + i 3)(x + 1 − i 3). d) x 4 + x 2 + 1 = (x 2 − ε)(x 2 − ε 2 ) = (x 2 − ε −2 )(x 2 − ε 2 ) √ 3 1 i. = (x − ε)(x + ε)(x − ε)(x + ε), where ε = − + 2 2 36. a) x 2 − 14x + 50 = 0;
b) x 2 −
18 26 x+ = 0; 5 5
c) x 2 + 4x + 8 = 0.
37. We have 2|z 1 + z 2 | · |z 2 + z 3 | = 2|z 2 (z 1 + z 2 + z 3 ) + z 1 z 3 | ≤ 2|z 2 | · |z 1 + z 2 + z 3 | + 2|z 1 ||z 3 |, and likewise, 2|z 2 + z 3 | · |z 3 + z 1 | ≤ 2|z 3 ||z 1 + z 2 + z 3 | + 2|z 2 ||z 1 |, 2|z 3 + z 1 | · |z 1 + z 2 | ≤ 2|z 1 ||z 1 + z 2 + z 3 | + 2|z 2 ||z 3 |. Summing up these inequalities with |z 1 + z 2 |2 + |z 2 + z 3 |2 + |z 3 + z 1 |2 = |z 1 |2 + |z 2 |2 + |z 3 |2 + |z 1 + z 2 + z 3 |2
258
6. Answers, Hints and Solutions to Proposed Problems
yields (|z 1 + z 2 |2 + |z 2 + z 3 |2 + |z 3 + z 1 |2 ) ≤ (|z 1 | + |z 2 | + |z 3 | + |z 1 + z 2 + z 3 |2 ). The conclusion is now obvious.
6.1.2
Geometric interpretation of the algebraic operations (p. 27)
3. a) The circle of center (2, 0) and radius 3. b) The disk of center (0, −1) and radius 1. c) The exterior of the circle of center (1, −2) and radius 3. 1 1 2 d) M = (x, y) ∈ R |x ≥ − ∪ (x, y) ∈ R2 |x < − , 3x 2 − y 2 − 3 < 0 . 2 2 e) M = {(x, y) ∈ R2 | − 1 < y < 0}. f) M = {(x, y) ∈ R2 | − 1 < y < 1}. g) M = {(x, y) ∈ R2 |x 2 + y 2 − 3x + 2 = 0}4. 1 h) The union of the lines with equations x = − and y = 0. 2 4. M = {(x, y) ∈ R2 |y = 10 − x 2 , y ≥ 4}. √ √ 5. z 3 = 3(1 − i) and z 3 = 3(1 + i). 6. M = {(x, y) ∈ R2 |x 2 + y 2 + x = 0, x = 0, x = −1} ∪{(0, y) ∈ R2 |y = 0} ∪ {(−1, y) ∈ R2 |y = 0}. 7. The union of the circles with equations x 2 + y 2 − 2y − 1 = 0
6.1.3
and
x 2 + y 2 + 2y − 1 = 0.
Polar representation of complex numbers (pp. 39–41)
√ 3π 7π ; b) r = 8, t ∗ = ; c) r = 5, t ∗ = π ; 1. a) r = 3 2, t ∗ = 4 6 √ √ 1 1 d) r = 5, t ∗ = arctan + π ; e) r = 2 5, t ∗ = arctan − + 2π . 2 2 √ 16 12 2. a) x = 1, y = 3; b) x = , y = − ; c) x = −2, y = 0; 5 5 d) x = −3, y = 0 e) x = 0, y = 1 f) x = 0, y = −4. 2π − arg z, if arg z = 0, 3. arg(z) = ; 0, if arg z = 0; π + arg z, if arg z ∈ [0, π), arg(−z) = −π + arg z, if arg z ∈ [π, 2π ). 4. a) The circle of radius 2 with center at origin. b) The circle of center (0, −1) and radius 2 and its exterior. c) The disk of center (0, 1) and radius 3.
6.1. Answers, Hints and Solutions to Routine Problems
259
d) The interior of the angle determined by the rays y = 0, x ≤ 0 and y = x, x ≤ 0. e) The fourth quadrant and the ray (OY . f) The first quadrant and the ray (O X . √ √ 3 x, x ≤ 0 and y = 3x, g) The interior of the angle determined by the rays y = 3 x < 0. h) The intersection of the disk of center (−1, −1) and√radius 3 with the interior of 3 x, x > 0. the angle determined by the rays y = 0, x ≥ 0 and y = 3 π 2π 2π π 1 ; b) z 2 = cos + i sin ; 5. a) z 1 = 12 cos + i sin 3 3 3 3 2 4π 5π 4π 5π c) z 3 = cos + i sin ; d) z 4 = 18 cos + i sin ; 3 3 3 3 √ 2 2 e) z 5 = 13 cos 2π − arctan + i sin 2π − arctan ; 3 3 3π 3π f) z 6 = 4 cos + i sin . 2 2 − a) + i sin(2π − a), a ∈[0, 2π );( 6. a) z 1 = cos(2π a π a a ' π b) z 2 = 2 cos · cos − + i sin − if a ∈ [0, π); 2 2 2 2 2 a a a 3π 3π z 2 = 2 cos · cos − + i sin − if a ∈ (π, 2π ); 2 2 2 2 2 ' π( √ 7π 7π c) z 3 = 2 cos a + + i sin a + if a ∈ 0, ; 4 4 4 ( ' √ π π π z 3 = 2 cos a − + i sin a − if a ∈ , 2π ; 4 4 4 a π a ( a ' π d) z 4 = 2 sin cos − + i sin − if a ∈ [0, π); 2 2 2 2 2 a a 5π 5π a z 4 = 2 sin cos − + i sin − if a ∈ [π, 2π ). 2 2 2 2 2 √ 7π 7π + i sin ; b) 4(cos 0 + i sin 0); 7. a) 12 2 cos 4 4 √ 5π 5π π π c) 48 2 cos + i sin ; d) 30 cos + i sin . 12 12 2 2 8. a) |z| = 12, arg z = 0, Arg z = 2kπ , arg z = 0, arg(−z) = π ; √ 11π 13π 11π π , Arg z = + 2kπ , arg z = , arg(−z) = . b) |z| = 14 2, arg z = 12 12 12 12 1 1 5π ; b) |z| = 9 , arg z = π ; 9. a) |z| = 213 + 13 , arg z = 6 2 2 5nπ n+1 c) |z| = 2 cos 3 , arg z ∈ {0, π }. 10. If z = r (cos t + i sin t) and n = −m, where m is a positive integer, then z n = z −m =
1 1 1 cos 0 + i sin 0 = m · = m m z r (cos mt + i sin mt) r cos mt + i sin mt
260
6. Answers, Hints and Solutions to Proposed Problems
=
1 [cos(0 − m)t + i sin(0 − m)t] = r −m (cos(−mt) + i sin(−mt)) rm
= r n (cos nt + i sin nt). n(π − a) n(π − a) n a n 11. a) 2 sin cos + i sin if a ∈ [0, π); 2 2 2 a n(5π − a) n(5π − a) 2n sinn cos + i sin if a ∈ [π, 2π]; 2 2 2 1 nπ b) z n + n = 2 cos . z 6
6.1.4
The n th roots of unity (p. 52) ⎛
⎞ π π + 2kπ + 2kπ √ ⎜ ⎟ 4 1. a) z k = 2 ⎝cos 4 + i sin 4 ⎠, k ∈ {0, 1}; 2 2 π π + 2kπ + 2kπ 2 b) z k = cos + i sin 2 , k ∈ {0, 1}; 2 2 π π + 2kπ + 2kπ 4 c) z k = cos + i sin 4 , k ∈ {0, 1}; 2 2 ⎛ ⎞ 4π 4π + 2kπ + 2kπ ⎜ ⎟ + i sin 3 d) z k = 2 ⎝cos 3 ⎠, k ∈ {0, 1}; 2 2 e) z 0 = 4 − 3i, z 1 = −4 + 3i. 3π 3π + 2kπ + 2kπ + i sin 2 , k ∈ {0, 1, 2}; 2. a) z k = cos 2 3 2 π + 2kπ π + 2kπ b) z k = 3 cos + i sin , k ∈ {0, 1, 2}; 3 3 ⎛ ⎞ π π + 2kπ + 2kπ √ ⎜ ⎟ c) z k = 2 ⎝cos 4 + i sin 4 ⎠, k ∈ {0, 1, 2}; 3 3
d) z k e) z 0 3. a) z k
b) z k
5π 5π + 2kπ + 2kπ = cos 3 + i sin 3 , k ∈ {0, 1, 2}; 3 3 = 3 + i, z 1 = (3 + i)ε, z 2 = (3 + i)ε 2 , where 1, ε, ε 2 are the cube roots of 1. ⎛ ⎞ 5π 5π + 2kπ + 2kπ √ ⎜ ⎟ = 2 ⎝cos 4 + i sin 4 ⎠, k ∈ {0, 1, 2, 3}; 4 4 ⎞ ⎛ π π + 2kπ + 2kπ √ ⎟ ⎜ 4 + i sin 6 = 2 ⎝cos 6 ⎠, k ∈ {0, 1, 2, 3}; 4 4
6.1. Answers, Hints and Solutions to Routine Problems
261
π π + 2kπ + 2kπ 2 + i sin 2 , k ∈ {0, 1, 2, 3}; c) z k = cos 4 ⎛ 4 3π ⎞ 3π + 2kπ + 2kπ √ ⎜ ⎟ 4 d) z k = 2 ⎝cos 2 + i sin 2 ⎠, k ∈ {0, 1, 2, 3}; 4 4 e) z 0 = 2 + i, z 1 = −2 − i, z 2 = −1 + 2i, z 3 = 1 − 2i. 2kπ 2kπ + i sin , k ∈ {0, 1, . . . , n − 1}, n ∈ {5, 6, 7, 8, 12}. 4. z k = cos n n 2π 2π 5. a) Consider ε j = ε j , εk = εk , where ε = cos +i sin . Then ε j ·εk = ε j+k . Let n n r be the remainder modulo n of j + k. We have j + k = p · n + r , r ∈ {0, 1, . . . , n − 1} and ε j · εk = p·n+r = (εn ) p · εr = εr = εr ∈ Un . 1 1 εn b) We can write ε −1 = j = j = εn− j ∈ Un . j = εj ε ε 2kπ 2kπ 6. a) z k = 5 cos + i sin , k ∈ {0, 1, 2}; 3 3 π + 2kπ π + 2kπ b) z k = 2 cos + i sin , k ∈ {0, 1, 2, 3}; 4 4 ⎞ ⎛ 3π 3π + 2kπ + 2kπ ⎟ ⎜ c) z k = 4 ⎝cos 2 + i sin 2 ⎠, k ∈ {0, 1, 2}; 3 3 ⎛ ⎞ π π + 2kπ + 2kπ ⎜ ⎟ d) z k = 3 ⎝cos 2 + i sin 2 ⎠, k ∈ {0, 1, 2}. 3 3 7. a) The equation is equivalent to (z 4 − i)(z 3 − 2i) = 0. b) We can write the equation as (z 3 + 1)(z 3 + i − 1) = 0. c) The equation is equivalent to z 6 = −1 + i. d) We can write the equation equivalently as (z 5 − 2)(z 5 + i) = 0. 8. It is clear that any solution is different from zero. Multiplying by z, the equation is equivalent to z 5 − 5z 4 + 10z 3 − 10z 2 + 5z − 1 = −1, z = 0. We obtain the (2k + 1)π + binomial equation (z − 1)5 = −1, z = 0. The solutions are z k = 1 + cos 5 (2k + 1)π i sin , k = 0, 1, 3, 4. 5
6.1.5
Some geometric transformations of the complex plane (p. 160)
1. Suppose that f, g are isometries. Then for all complex numbers a, b, we have | f (g(a)) − f (g(b))| = |g(a) − g(b)| = |a − b|, so f ◦ g is also an isometry. 2. Suppose that f is an isometry and let C be any point on the line AB. Let f (C) = M. Then M A = f (C) f (A) = AC and, similarly, M B = BC. Thus |M A − M B| = AB.
262
6. Answers, Hints and Solutions to Proposed Problems
Hence A, M, B are collinear. Now, from M A = AC and M B = BC, we conclude that M = C. Hence f (M) = M and the conclusion follows. 3. This follows immediately from the fact that any isometry f is of the form f (z) = az + b or f (z) = az + b, with |a| = 1. 4. The function f is the product of the rotation z → i z, the translation z → z + 4 − i, and the reflection in the real axis. It is clear that f is an isometry. 5. The function f is the product of the rotation z → −i z with the translation z → z + 1 + 2i.
6.2
Solutions to the Olympiad-Caliber Problems
6.2.1
Problems involving moduli and conjugates (pp. 175–176)
Problem 21. At first we prove that function f is well defined, i.e., | f (z)| < 1 for all z with |z| < 1. 2 2 Indeed, we have | f (z)| < 1 if and only if 1+az z+a < 1, i.e., |1 + az| < |z + a| . The last relation is equivalent to (1 + az)(1 + az) < (z + a)(z + a). That is, 1 + |a|2 |z|2 < |a|2 + |z|2 or equivalently (|a|2 − 1)(|z|2 − 1) < 0. The last inequality is obvious since |z| < 1, and |a| > 1. To prove that f is bijective, it suffices to observe that for any y ∈ A there is a unique z ∈ A such that 1 + az f (z) = = y. z+a We obtain ay − 1 = − f (−y), z= a−y hence |z| = | f (−y)| < 1, as desired. Problem 22. Let z = cos ϕ + i sin ϕ with cos ϕ, sin ϕ ∈ Q. Then z 2n − 1 = cos 2nϕ + i sin 2nϕ − 1 = 1 − 2 sin2 nϕ + 2i sin nϕ cos nϕ − 1 = −2 sin nϕ(sin nϕ − i cos nϕ) and |z 2n − 1| = 2| sin nϕ|. It suffices to prove that sin nϕ ∈ Q. We prove by induction on n that both sin nϕ and cos nϕ are rational numbers. The claim is obvious for n = 1. Assume that sin nϕ, cos nϕ ∈ Q. Then sin(n + 1)ϕ = sin nϕ cos ϕ + cos nϕ cos ϕ ∈ Q
6.2. Solutions to the Olympiad-Caliber Problems
263
and cos(n + 1)ϕ = cos nϕ cos ϕ − sin nϕ sin ϕ ∈ Q, as desired. Problem 23. To prove that the function f is injective, let f (a) = f (b). Then
1 + ai = 1 − ai
1 + bi . This is equivalent to 1 + ab + (a − b)i = 1 + ab + (b − a)i, i.e., a = b, as 1 − bi needed. The image of the function f is the set of numbers z ∈ C such that there is t ∈ R with 1 + ti z = f (t) = . 1 − ti
z−1 1 + ti we obtain t = if z = 1. Then t ∈ R if and only if t = t. The 1 − ti i(1 + z) z−1 z−1 last relation is equivalent to = , i.e., −(z−1)(z+1) = (z+1)(z−1). i(1 + z) −i(1 + z It follows that 2zz = 2, i.e., |z| = 1, hence the image of the function f is the set {z ∈ R||z| = 1 and z = −1}, the unit circle without the point with coordinate z = −1. z2 Problem 24. Let = t ∈ C. Then z1
From z =
|z 1 + z 1 t| = |z 1 | = |z 1 t| or |1 + t| = |t| = 1. It follows that tt = 1 and 1 = |1 + t|2 = (1 + t)(1 + t) = 1 + t + t + 1, hence t 2 + t + 1 = 0. Therefore t is a nonreal cube root of unity. Alternate solution. Let A, B, C be the geometric images of the complex numbers z 1 , z 2 , z 1 + z 2 , respectively. In the parallelogram O AC B we have O A = O B = OC, hence AO B = 120◦ . Then z2 z1 = cos 120◦ + i sin 120◦ or = cos 120◦ + i sin 120◦ , z1 z2 therefore z2 2π 2π ± i sin . = cos z1 3 3 Problem 25. We prove first the inequality |z k | ≤ |z 1 | + |z 2 | + · · · + |z k−1 | + |z k+1 | + · · · + |z n | + |z 1 + z 2 + · · · + z n |
264
6. Answers, Hints and Solutions to Proposed Problems
for all k ∈ {1, 2, . . . , n}. Indeed, |z k | = |(z 1 + z 2 + · · · + z k−1 + z k + z k+1 + · · · + z n ) − (z 1 + z 2 + · · · + z k−1 + z k+1 + · · · + z n )| ≤ |z 1 + z 2 + · · · + z n | + |z 1 | + · · · + |z k−1 | + |z k+1 | + · · · + |z n |, as claimed. Denote Sk = |z 1 | + · · · + |z k−1 | + |z k+1 | + · · · + |z n | for all k. Then |z k | ≤ Sk + |z 1 + z 2 + · · · + z n |, for all k.
(1)
|z 1 + z 2 + · · · + z n | ≤ |z 1 | + |z 2 | + · · · + |z n |.
(2)
Moreover, Multiplying by |z k | the inequalities (1) and by |z 1 + z 2 + · · · + z n | the inequalities (2), we obtained by summation: |z 1 |2 + |z 2 |2 + · · · + |z n |2 + |z 1 + z 2 + · · · + z n |2 ≤ |z 1 + z 2 + · · · + z n |
n
|z k | +
k=1
n
|z k |Sk .
k=1
Adding on both sides of the inequality the expression |z 1 |2 + |z 2 |2 + · · · + |z n |2 + |z 1 + z 2 + · · · + z n |2 yields 2(|z 1 |2 + |z 2 |2 + · · · + |z n |2 + |z 1 + z 2 + · · · + z n |2 ) ≤ (|z 1 | + · · · + |z n | + |z 1 + z 2 + · · · + z n |)2 , as desired. Problem 26. Let M1 , M2 , . . . , M2n be the points with the coordinates z 1 , z 2 , . . . , z 2n and let A1 , A2 , . . . , An be the midpoints of segments M1 M2n , M2 M2n−1 , . . . , Mn Mn+1 . The points Mi , i = 1, 2n lie on the upper semicircle centered in the origin and with radius 1. Moreover, the lengths of the chords M1 M2n , M2 M2n−1 , . . . , Mn Mn+1 are in a decreasing order, hence O A1 , O A2 , . . . , O An are increasing. Thus z 1 + z 2n z 2 + z 2n−1 ≤ ≤ · · · ≤ z n + z n+1 2 2 2 and the conclusion follows.
6.2. Solutions to the Olympiad-Caliber Problems
265
Figure 6.1.
Alternate solution. Consider z k = r (cos tk + i sin tk ), k = 1, 2, . . . , 2n and observe that for any j = 1, 2, . . . , n, we have |z j + z 2n− j+1 |2 = |r [(cos t j + cos t2n− j+1 ) + i(sin t j + sin t2n− j+1 )]|2 = r 2 [(cos t j + cos t2n− j+1 )2 + (sin t j + sin t2n− j+1 )2 ] = r 2 [2 + 2(cos t j cos t2n− j+1 + sin t j sin t2n− j+1 )] = 2r 2 [1 + cos(t2n− j+1 − t j )] = 4r 2 cos2 Therefore |z j + z 2n− j+1 | = 2r cos
t2n− j+1 − t j . 2
t2n− j+1 − t j and the inequalities 2
|z 1 + z 2n | ≤ |z 2 + z 2n−1 | ≤ · · · ≤ |z n + z n+1 | are equivalent to t2n − t1 ≥ t2n−1 − t2 ≥ · · · ≥ tn+1 − tn . Because 0 ≤ t1 ≤ t2 ≤ · · · ≤ t2n ≤ π , the last inequalities are obviously satisfied. √ √ Problem 27. It is natural to make the substitution x = u, y = v. The system becomes 1 2 u 1+ 2 =√ , 2 u +v 3 √ 4 2 1 = √ . v 1− 2 u + v2 7 But u 2 + v 2 is the square of the absolute value of the complex number z = u + iv. This suggests that we add the second equation multiplied by i to the first one. We obtain √ u − iv 2 4 2 . u + iv + 2 = √ +i √ u + v2 7 3
266
6. Answers, Hints and Solutions to Proposed Problems
The quotient (u − iv)/(u 2 + v 2 ) is equal to z/|z|2 = z/(zz) = 1/z, so the above equation becomes √ 2 4 2 1 . z+ = √ +i √ z 7 3 Hence z satisfies the quadratic equation √ 2 4 2 2 z − √ +i √ z+1=0 7 3 with solutions
1 2 √ ±√ 3 21
√ 2 2 √ +i √ ± 2 , 7
where the signs + and − correspond. This shows that the initial system has the solutions √ 2 2 2 √ 1 2 2 , y= √ ± 2 , x= √ ±√ 7 3 21 where the signs + and − correspond. Problem 28. The direct implication is obvious. Conversely, let |z 1 | = |z 2 + z 3 |, |z 2 | = |z 1 + z 3 |, |z 3 | = |z 1 + z 2 |. It follows that |z 1 |2 + |z 2 |2 + |z 3 |2 = |z 2 + z 3 |2 + |z 3 + z 1 |2 + |z 1 + z 2 |2 . This is equivalent to z1 z1 + z2 z2 + z3 z3 = z2 z2 + z2 z3 + z2 z3 + z3 z3 + z3 z1 + z1 z3 + z1 z1 + z1 z1 + z1 z2 + z2 z1 + z2 z2,
i.e.,
z 1 z 1 + z 2 z 2 + z 3 z 3 + z 1 z 2 + z 2 z 1 + z 1 z 3 + z 1 z 3 + z 2 z 3 + z 3 z 2 = 0. We write the last relation as (z 1 + z 2 + z 3 )(z 1 + z 2 + z 3 ) = 0, and we obtain |z 1 + z 2 + z 3 |2 = 0, i.e., z 1 + z 2 + z 3 = 0, as desired. Problem 29. Let a = |z 1 | = |z 2 | = · · · = |z n |. Then zk =
a2 , zk
k = 1, n
6.2. Solutions to the Olympiad-Caliber Problems
and z 1 z 2 + z 2 z 3 + · · · + z n−1 z n =
n−1
z k z k+1 =
k=1
=
n−1
k=1
a4 z k z k+1
a4 (z 3 z 4 · · · z n + z 1 z 4 · · · z n + · · · + z 1 z 2 · · · z n−2 ) = 0; z1 z2 · · · zn
hence z 1 z 2 + z 2 z 3 + · · · + z n−1 z n = 0, as desired. Problem 30. Let z = r1 (cos t1 + i sin t1 ) and a = r2 (cos t2 + i sin t2 ). We have 1 = |z + a| =
(r1 cos t1 + r2 cos t2 )2 + (r1 sin t1 + r2 sin t2 )2
=
r12 + r22 + 2r1r2 cos(t1 − t2 ),
so cos(t1 − t2 ) =
1 − r12 − r22 . 2r1r2
Then |z 2 + a 2 | = |r12 (cos 2t1 + i sin 2t1 ) + r22 (cos 2t2 + i sin 2t2 )| = (r12 cos 2t1 + r22 cos 2t2 )2 + (r12 sin 2t1 + r22 sin 2t2 ) = r14 + r24 + 2r12r22 cos 2(t1 − t2 ) = r14 + r24 + 2r1r2 (2 cos2 (t1 − t2 ) − 1) 3 ⎛ ⎞ 4 4 2 − r2 2 1 − r 4 4 1 2 − 1⎠ = 5r1 + r24 + 2r12r22 · ⎝2 2r1r2 =
2r14 + 2r24 + 1 − 2r12 − 2r22 .
The inequality |z 2 + a 2 | ≥
|1 − 2|a|| √ 2
267
268
6. Answers, Hints and Solutions to Proposed Problems
is equivalent to 2r14 + 2r24 + 1 − 2r12 − 2r22 ≥
(1 − 2r12 )2 , i.e., 2
4r14 + 4r24 − 4r12 − 4r22 + 2 ≥ 1 − 4r12 + 4r22 . We obtain (2r22 − 1)2 ≥ 0, and we are done. Problem 31. It is easy to see that z = 0 is a root of the equation. Consider z = a+ib = 0, a, b ∈ R. Observe that if a = 0, then b = 0 and if b = 0, then a = 0. Therefore we may assume that a, b = 0. Taking the modulus of both members of the equation az n = bz n yields |a| = |b| or a = ±b. Case 1. If a = b, the equation (1) becomes (a + ia)n = (a − ia)n . This is equivalent to
1+i n = 1, i.e., i n = 1, 1−i which has solutions only for n = 4k, k ∈ Z. In that case the solutions are z = a(1 + i),
a = 0.
Case 2. If a = −b, the equation (1) may be rewritten as (a − ia)n = −(a + ia)n . That is,
1−i n = −1, i.e., (−i)n = −1, 1+i which has solutions only for n = 4k + 2, k ∈ Z. We obtain
z = a(1 − i),
a = 0.
To conclude, a) if n is odd, then z = 0; b) if n = 4k, k ∈ Z, then z = {a(1 + i)|a ∈ R}, i.e., a line through origin; c) if n = 4k + 2, k ∈ Z, then z = {a(1 − i)|a ∈ R}, i.e., a line through origin.
(1)
6.2. Solutions to the Olympiad-Caliber Problems
Problem 32. Let z 1 = cos t1 + i sin t1 and z 2 = cos t2 + i sin t2 . The inequality |az 1 + bz 2 | ≥ is equivalent to
|z 1 + z 2 | 2
(a cos t1 + b cos t2 )2 + (a sin t1 + b sin t2 )2 1 ≥ (cos t1 + cos t2 )2 + (sin t1 + sin t2 )2 . 2
That is,
2 a 2 + b2 + 2ab cos(t1 − t2 ) ≥ 2 + cos(t1 − t2 ), i.e., 4a 2 + 4(1 − a)2 + 8a(1 − a) cos(t1 − t2 ) ≥ 2 + 2 cos(t1 − t2 ).
We obtain 8a 2 − 8a + 2 ≥ (8a 2 − 8a + 2) cos(t1 − t2 ), i.e., 1 ≥ cos(t1 − t2 ), which is obvious. 1 The equality holds if and only if t1 = t2 , i.e., z 1 = z 2 or a = b = . 2 Problem 33. Let r = |z 1 | = |z 2 | = · · · = |z n | > 0. Then 1 z 1k
+
1 z 2k
+ ··· +
=
1 r 2k
z1k z2k zn k 1 = + + · · · + z nk r 2k r 2k r 2k
(z 1k + z 2k + · · · + z nk ) = 0,
as desired.
6.2.2
Algebraic equations and polynomials (p. 181)
Problem 11. Let r = |z 1 | = |z 2 |. The relation ab|c| = |a|bc is equivalent to |a|bc ab|c| = . aa|a| aa|a| This relation can be written as
b c b c · =− · . a a a a
That is, −(x1 + x2 ) · |x1 x2 | = −(x1 + x2 ) · x1 x2 , i.e., (x1 + x2 )r 2 = |x1 |2 x2 + x1 |x2 |2 .
269
270
6. Answers, Hints and Solutions to Proposed Problems
It follows that (x1 + x2 )r 2 = (x1 + x2 )r 2 , which is certainly true. Problem 12. Observe that z 13 = z 23 = 1 and z 33 = z 43 = −1. If n = 6k + r , with k ∈ Z and r ∈ {0, 1, 2, 3, 4, 5}, then z 1n + z 2n = z r1 + z r2 and z 3n + z 4n = z r3 + z r4 . The equality z 1n + z 2n = z 3n + z 4n is equivalent to z r1 + z r2 = z r3 + z r4 and holds only for r ∈ {0, 2, 4}. Indeed, i) if r = 0, then z 10 + z 20 = 2 = z 30 + z 40 ; ii) if r = 2, then z 12 + z 22 = (z 1 + z 2 )2 − 2z 1 z 2 = (−1)2 − 2 · 1 = −1 and 2 z 3 + z 42 = (z 3 + z 4 )2 − 2z 3 z 4 = 12 − 2 · 1 = −1; iii) if r = 4, then z 14 + z 24 = z 1 + z 2 = 1 and z 33 + z 44 = −(z 3 + z 4 ) = −(−1) = 1. The other cases are: iv) r = 1 then z 1 + z 2 = −1 = z 3 + z 4 = 1; v) r = 3, then z 13 + z 23 = 1 + 1 = 2 = z 33 + z 43 = −1 − 1 = −2; vi) r = 5, then z 15 + z 25 = z 12 + z 22 = −1 = z 35 + z 45 = −(z 32 + z 42 ) = 1. Therefore, the desired numbers are the even numbers. Problem 13. Let f (x) = x 6 + ax 5 + bx 4 + cx 3 + bx 2 + ax + 1 =
6
(x − xk ) =
k=1
6
(xk − x), for all x ∈ C.
k=1
We have 6
(xk2 + 1) =
k=1
6
(xk + i) ·
k=1 4
6
(xk − i) = f (−i) · f (i)
k=1 2
= (i 6 + ai 5 + bi + ci 3 + bi + ai + 1) · (i 6 − ai 5 + bi 4 − ci 3 + bi 2 − ai + 1) = (2ai − ci)(−2ai + ci) = (2a − c)2 , as desired. Problem 14. For a complex number z with |z| = 1, observe that P(z) + P(−z) = az 2 + bz + i + az 2 − bz + i = 2(az 2 + i). It suffices to choose z 0 such that az 02 = |a|i. Let a = |a|(cos t + i sin t),
t ∈ [0, 2π ).
6.2. Solutions to the Olympiad-Caliber Problems
271
The equation az 2 = |a|i is equivalent to π π − t + i sin −t . z 02 = cos 2 2
Set
t π z 0 = cos − 4 2
π t + i sin − , 4 2
and we are done. Therefore, we have P(z 0 ) + P(−z 0 ) = 2(|z|i + i) = 2i(1 + |a|). Passing to absolute values it follows that |P(z 0 )| + |P(−z 0 )| ≥ 2(1 + |a|). That is, |P(z 0 )| ≥ 1 + |a| or |P(−z 0 )| ≥ 1 + |a|. Note that |z 0 | = | − z 0 | = 1, as needed. Problem 15. Let z be a complex root of polynomial f . From the given relation it follows that 2z 3 + z is also a root of f . Observe that if |z| > 1, then |2z 3 + z| = |z||2z 2 + 1| ≥ |z|(2|z|2 − 1) > |z|. Hence, if f has a root z 1 with |z 1 | > 1, then f has a root z 2 = 2z 13 + z 1 with |z 2 | > |z 1 |. We can continue this procedure and obtain an infinite number of roots of f , z 1 , z 2 , . . . with · · · > |z 2 | > |z 1 |, a contradiction. Therefore, all roots of f satisfy |z| ≤ 1. We will show that f is not divisible by x. Assume, by contradiction, the contrary and choose the greatest k ≥ 1 with the property that x k divides f . It follows that f (x) = x k (a + xg(x)) with a = 0, hence f (2x 2 ) = x 2k (a1 + 2k+1 x 2 g(2x 2 )) = x 2k (a1 + xg1 (x)) and f (2x 3 + x) = x k (2x 2 + 1)k (a + (2x 2 + 1)xg(x)) = x k (a + xg2 (x)), where g, g1 , g2 are polynomials and a1 = 0 is a real number. The relation f (x) f (2x 2 ) = f (2x 3 + x) is equivalent to x k (a + xg(x))x 2k (a1 + xg1 (x)) = x k (a + xg2 (x)) which is not possible for a = 0 and k > 0. Let m be the degree of polynomial f . The polynomials f (2x 2 ) and f (2x 3 + x) have degrees 2m and 3m, respectively.
272
6. Answers, Hints and Solutions to Proposed Problems
If f (x) = bm x m + · · · + b0 , then f (2x 2 ) = 2m bm x 2m + · · · and f (2x 3 + x) = 2m bm x 3m + · · · From the given relation we find bm · 2m · bm = 2m bm , hence bm = 1. Again using the given relation it follows that f 2 (0) = f (0), i.e., b02 = b0 , hence b0 = 1. The product of the roots of polynomial f is ±1. Taking into account that for any root z of f we have |z| ≤ 1, it follows that the roots of f have modulus 1. Consider z a root of f . Then |z| = 1 and 1 = |2z 3 + z| = |z||2z 2 + 1| = |2z 2 + 1| ≥ |2z 2 | − 1 = 2|z|2 − 1 = 1. Equality is possible if and only if the complex numbers 2z 2 and −1 have the same argument; that is, z = ±i. Because f has real coefficients and its roots are ±i, it follows that f is of the form (x 2 + 1)n for some positive integer n. Using the identity (x 2 + 1)(4x 4 + 1) = (2x 3 + x)2 + 1 we obtain that the desired polynomials are f (x) = (x 2 + 1)n , where n is an arbitrary positive integer.
6.2.3
From algebraic identities to geometric properties (p. 190)
Problem 12. Let A, B, C, D be the points with coordinates a, b, c, d, respectively. If a + b = 0, then c + d = 0. Hence a + b = c + d, i.e., ABC D is a parallelogram inscribed in the circle of radius R = |a| and we are done. If a + b = 0, then the points M and N with coordinates a + b and c + d, respectively, are symmetric with respect to the origin O of the complex plane. Since AB is a diagonal in the rhombus O AM B, it follows that AB is the perpendicular bisector of the segment O M. Likewise, C D is the perpendicular bisector of the segment O N . Therefore A, B, C, D are the intersection points of the circle of radius R with the perpendicular bisector s of the segments O M and O N , so A, B, C, D are the vertices of a rectangle. Alternate solution. First, let us note that from a + b + c + d = 0 it follows that a + d = −(b + c), i.e., |a + d| = |b + c|. Hence |a + d|2 = |b + c|2 and using properties of the real product we find that (a + d) · (a + d) = (b + c) · (b + c). That is, |a|2 + |d|2 + 2a · d = |b|2 + |c|2 + 2b · c. Taking into account that |a| = |b| = |c| = |d| one obtains a · d = b · c. On the other hand, AD 2 = |d − a|2 = (d − a) · (d − a) = |d|2 + |a|2 − 2a · d = 2(R 2 − a · d). Analogously, we have BC 2 = 2(R 2 − b · c). Since a · d = b · c, it follows that AD = BC, so ABC D is a rectangle. Problem 13. Consider the polynomial P(X ) = X 5 + a X 4 + bX 3 + cX 2 + d X + e
6.2. Solutions to the Olympiad-Caliber Problems
273
with roots z k , k = 1, 5. Then a=−
z1 = 0
b=
and
z1 z2 =
1 2 1 2 z1 − z 1 = 0. 2 2
Denoting by r the common modulus and taking conjugates we also get
0=
z1 =
r2 z1
=
r2 z1 z2 z3 z4, z1 z2 z3 z4 z5
from which d = 0 and 0=
z1 z2 =
r4 r4 z1 z2 z3; = z1 z2 z1 z2 z3 z4 z5
therefore c = 0. It follows that P(X ) = X 5 + e, so z 1 , z 2 , . . . , z 5 are the fifth roots of e and the conclusion is proved. Problem 14. a) Consider a complex plane with origin at M. Denote by a, b, c the coordinates of A, B, C, respectively. As a(b − c) = b(a − c) + c(b − a) we have |a||b − a| = |b(a − c) + c(b − a)| ≤ |b||a − c| + |c||b − a|. Thus AM · BC ≤ B M · AC + C M · AB or 2R · AM · sin A ≤ 2R · B M · sin B + 2R · C M · sin C which gives AM · sin A ≤ B M · sin B + C M · sin C. b) From a) we have A A1 · sin α ≤ AB1 · sin β + AC1 · sin γ , B B1 · sin β ≤ B A1 · sin α + BC1 · sin γ , CC1 · sin γ ≤ C A1 · sin α + C B1 · sin β, which, summed up, give the desired conclusion. Problem 15. Let the coordinates of A, B, C, M and N be a, b, c, m and n, respectively. Since the lines AM, B M and C M are concurrent, as well as the lines AN , B N and C N , it follows from Ceva’s theorem that B M sin AC M sin B AM sin C · · = 1, sin M AC sin M B A sin MC B
(1)
B N sin AC N sin B AN sin C · · = 1. sin N AC sin N B A sin N C B
(2)
By hypotheses, B AM = N AC and M BA = C B N . Hence B AN = M AC and N B A = C B M. Combined with (1) and (2), these equalities imply sin AC M · sin AC N = sin MC B · sin N C B.
274
6. Answers, Hints and Solutions to Proposed Problems
Figure 6.2.
Thus, cos( N C M + 2 AC M) − cos N C M = cos( N C M + 2 N C B) − cos N C M, and hence AC M = N C B. Since B AM = N AC, MBA = C B N and AC N = MC B, the following complex ratios are all positive real numbers: m−a c−a : , b−a n−a
m−b c−b : a−b n−b
and
m−c a−c : . b−c n−c
Hence each of these equals its absolute value, and so BM · BN CM · CN AM · AN + + AB · AC B A · BC CA · CB =
6.2.4
(m − a)(n − a) (m − b)(n − b) (m − c)(n − c) + + = 1. (b − a)(c − a) (a − b)(c − b) (b − c)(a − c)
Solving geometric problems (pp. 211–213)
Problem 26. Let a, b, c be the coordinates of the points A, B, C, respectively. Using the real product of the complex numbers, we have AC 2 + AB 2 = 5BC 2 if and only if |c − a|2 + |b − a|2 = 5|c − b|2 , i.e., (c − a) · (c − a) + (b − a) · (b − a) = 5(c − b) · (c − b). The last relation is equivalent to c2 − 2a · c + a 2 + b2 − 2a · b + a 2 = 5c2 − 10b · c + 5b2 , i.e., 2a 2 − 4b2 − 4c2 − 2a · b − 2a · c + 10b · c = 0.
6.2. Solutions to the Olympiad-Caliber Problems
275
It follows that a 2 − 2b2 − 2c2 − a · b − a · c + 5b · c = 0, i.e., a+b a+c −b · − c = 0. (a + c − 2b) · (a + b − 2c) = 0, so 2 2 The last relation shows that the medians from B and C are perpendicular, as desired. Problem 27. Denoting by a lowercase letter the coordinates of a point with an uppercase letter, we obtain a =
b − kc , 1−k
and
Then
b =
c − ka , 1−k
c =
a − kb 1−k
a =
c − kb (1 + k 2 )a − k(b + c) , = 1−k (1 − k)2
b =
(1 + k 2 )b − k(a + c) a − kc = , 1−k (1 − k)2
c =
b − ka (1 + k 2 )c − k(b + a) . = 1−k (1 − k)2
c − a (1 + k 2 )(c − a) − k(a − c) c−a = , = 2 b −a b−a (1 + k )(b − a) − k(a − b)
which proves that triangles ABC and A B C are similar. Problem 28. Consider the complex plane with origin at the circumcircle of triangle ABC and let z 1 , z 2 , z 3 be the coordinates of points A, B, C. R mα The inequality is equivalent to ≥ 2r hα 2r m α ≤ Rh α , i.e., 2
K 2K mα ≤ R . s α
Hence αm α ≤ Rs. Using complex numbers, we have z 2 + z 3 = |(z 2 − z 3 )(2z 1 − z 2 − z 3 )| 2αm α = 2|z 2 − z 3 | z 1 − 2 = |z 2 (z 1 − z 2 ) + z 1 (z 2 − z 3 ) + z 3 (z 3 − z 1 )| ≤ |z 2 ||z 1 − z 2 | + |z 1 ||z 2 − z 3 | + |z 3 ||z 3 − z 1 | = R(α + β + γ ) = 2Rs. Hence αm α ≤ Rs, as desired.
276
6. Answers, Hints and Solutions to Proposed Problems
Problem 29. Consider the complex plane with origin at the circumcenter O and let a, b, c, d be the coordinates of points A, B, C, D. a+c The midpoints E and F of the diagonals AC and B D have the coordinates 2 b+d and . 2 Using the real product the complex numbers we have AB 2 + BC 2 + C D 2 + D A2 = 8R 2 if and only if (b − a) · (b − a) + (c − b) · (c − b) + (d − c) · (d − c) + (a − d) · (a − d) = 8R 2 , i.e., 2a · b + 2b · c + 2c · d + 2d · a = 0. The last relation is equivalent to b · (a + c) + d · (a + c) = 0, i.e., (b + d) · (a + c) = 0. We find
b+d a+c · = 0, i.e., O E ⊥ O F 2 2
or E = O or F = O. That is, AC ⊥ B D or one of the diagonals AC and B D is a diameter of the circle C. Problem 30. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter and let ε = cos 120◦ + i sin 120◦ . Since triangles AB M, BC N , C O P and D AQ are equilateral we have m + bε + aε 2 = 0,
n + cε + bε 2 = 0,
p + dε + cε 2 = 0,
q + aε + dε 2 = 0.
Summing these equalities yields (m + n + p + q) + (a + b + c + d)(ε + ε 2 ) = 0, and since ε + ε2 = −1 it follows that m + n + p + q = a + b + c + d. Therefore the quadrilaterals ABC D and M N P Q have the same centroid. Problem 31. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter. Using the rotation formula, we obtain m = b + (a − b)ε,
n = c + (b − c)ε,
where ε = cos α + i sin α.
p = d + (c − d)ε,
q = a + (d − a)ε,
6.2. Solutions to the Olympiad-Caliber Problems
277
Let E, F, G, H be the midpoints of the diagonals B D, AC, M P, N Q respectively; then e=
b+d , 2
f =
a+c , 2
g=
b + d + (a + c − b − d)ε 2
a + c + (b + d − a − c)ε . 2 Since e + f = g + h, then E G F H is a parallelogram, as desired. and h =
Problem 32. Consider the points E, F, G, H such that O E ⊥ AB,
O E = C D,
O F ⊥ BC,
O F = AD,
OG ⊥ C D,
OG = AB,
O H ⊥ AD,
O H = BC,
where O is the circumcenter of ABC D. We prove that E F G H is a parallelogram. Since O E = C D, O F = AD and = ADC follows that triangles E O F and ADC are congruent, E O F = 180◦ − ABC hence E F = G H . Likewise F G = E H and the claim is proved. Consider the complex plane with origin at O such that F is on the positive real axis. Denote by a lowercase letter the coordinate of a point denoted by an uppercase letter. We have |e| = C D,
| f | = AD,
|g| = AB,
|h| = BC.
Furthermore, % = A, % F OG = 180◦ − C
G OH = % B,
% H O E = C,
hence f = | f | = AD,
g = |g|(cos A + i sin A) = AD(cos A + i sin A),
h = |h|[cos(A + B) + i sin(A + B)] = BC[cos(A + B) + i sin(A + B)], e = |e|[cos(A + B + C) + i sin(A + B + C)] = C D(cos D − i sin D). Since e + g = f + h, we obtain AD + BC cos(A + B) + i BC sin(A + B) = C D(cos D − i sin D) + AB(cos A + i sin A) and the conclusion follows.
278
6. Answers, Hints and Solutions to Proposed Problems
Problem 33. Consider the complex plane with origin at the circumcenter O of the triangle. Let a, b, c, ω, g, z I be the coordinates of the points A, B, C, O9 , G, I , respectively. Without loss of generality, we may assume that the circumradius of the triangle ABC is equal to 1, hence |a| = |b| = |c| = 1. We have ω=
a+b+c , 2
g=
a+b+c , 3
zI =
a|b − c| + b|a − c| + c|a − b| . |a − b| + |b − c| + |a − c|
Using the properties of the real product of complex numbers, we have O9 G ⊥ AI if and only if (ω − g) · (a − z I ) = 0, i.e., a + b + c (a − b)|a − c| + (a − c)|a − b| · = 0. 6 |a − b| + |b − c| + |a − c| This is equivalent to (a + b + c) · [(a − b)|a − c| + (a − c)|a − b|] = 0, i.e., Re{(a + b + c)[(a − b)|a − c| + (a − c)|a − b|]} = 0. We find that Re{|a − c|(aa + ba + ca − ab − bb − cb) + |a − b|(aa + ba + ca − ac − bc − cc)} = 0. Observe that aa = bb = cc = 1
and
Re(ba − ab) = Re(ca − ac) = 0,
hence the relation (1) is equivalent to Re{|a − c|(ca − cb) + |a − b|(ba − bc)} = 0, i.e., |a − c|(ca + ca − cb − cb) + |a − b|(ab + ab − bc − bc) = 0. It follows that |a − c|[(bb − bc − cb + cc) − (aa − ca − ca + cc)] + |a − b|[(bb − bc − cb + cc) − (aa − ab − ab + bb)] = 0, i.e., |a − c|(|b − c|2 − |a − c|2 ) + |a − b|(|b − c|2 − |a − b|2 ) = 0. This is equivalent to AC · BC 2 − AC 3 + AB · BC 2 − AB 3 = 0.
(1)
6.2. Solutions to the Olympiad-Caliber Problems
279
The last relation can be written as BC 2 (AC + AB) = (AC + AB)(AC 2 − AC · AB + AB 2 ), so AC · AB = AC 2 + AB 2 − BC 2 . We obtain cos A =
1 %= π , , i.e., A 2 3
as desired. Problem 34. (a) Let a lowercase letter denote the complex number associated with the point labeled by the corresponding uppercase letter. Let M , M and O denote the midpoints of segments [M1 M2 ], [M1 M2 ] and [O1 O2 ], respectively. Also let m 1 − o1 m 2 − o2 z = = , so that multiplication by z is a rotation about the origin m 1 − o1 m 2 − o2 m1 + m2 through some angle. Then m = equals 2 1 1 (o1 + z(m 1 − o1 )) + (o2 + z(m 2 − o2 )) = o + z(m − o), 2 2 i.e., the locus of M is the circle centered at O with radius O M . (b) We shall use directed angles modulo π . Observe that Q M1 M2 = Q P M2 = Q P O2 = Q O1 O2 . Similarly, Q M2 M1 = Q O2 O1 , implying that triangles Q M1 M2 and Q O1 O2 are similar with the same orientations. Hence, q − o1 q − m1 = , q − o2 q − m2 or equivalently o1 − m 1 o1 − m 1 (q − m 1 ) − (q − o1 ) q − o1 = = = . q − o2 (q − m 2 ) − (q − o2 ) o2 − m 2 o2 − m 2 Because lines O1 M1 and O2 M2 meet, o1 − m 1 = o2 − m 2 and we can solve this equation to find a unique value for q. Problem 35. Without loss of generality, assume that triangle A1 A2 A3 is oriented counterclockwise (i.e., angle A1 A2 A3 is oriented clockwise). Let P be the reflection of O1 across T . We use the complex numbers with origin O1 , where each point denoted by an uppercase letter is represented by the complex number with the corresponding lowercase
280
6. Answers, Hints and Solutions to Proposed Problems
letter. Let ζk = ak / p for k = 1, 2, so that z → ζk (z − z 0 ) is a similarity through angle P O1 Ak with ratio O1 A3 /O1 P about the point corresponding to z 0 . Because O1 and A1 lie on opposite sides of line A2 A3 , angles A2 A3 O1 and A2 A3 A1 have opposite orientations, i.e., the former is oriented counterclockwise. Thus, angles P A3 O1 and A2 O3 A1 are both oriented counterclockwise. Because P A 3 O1 = 2 A2 A3 O2 = A2 O3 A1 , it follows that isosceles triangles P A3 O1 and A2 O3 A1 are similar and have the same orientation. Hence, o3 = a1 + ζ3 (a2 − a1 ). Similarly, o2 = a1 + ζ2 (a3 − a1 ). Hence, o3 − o2 = (ζ2 − ζ3 )a1 + ζ3 a2 − ζ2 a3 = ζ2 (a2 − a3 ) + ζ3 (ζ2 p) − ζ2 (ζ3 p) = ζ2 (a2 − a3 ), or (recalling that o1 = 0 and t = 2 p) a2 − a3 1 a2 − a3 o3 − o2 = ζ2 = = . a 1 − o1 p − o1 2 t − o1 Thus, the angle between [O1 A1 ] and [O2 O3 ] equals the angle between [O1 T ] and 1 [A3 A2 ], which is π/2. Furthermore, O2 O3 /O1 A1 = A3 A2 /O1 T , or O1 A1 /O2 O3 = 2 2O1 T /A2 A3 . This completes the proof. Problem 36. Assume that the origin O of the coordinate system in the complex plane is the center of the circ*mscribed circle. Then, the vertices A1 , A2 , A3 are represented by complex numbers w1 , w2 , w3 such that |w1 | = |w2 | = |w3 | = R. 2π 2π + i sin . Then ε2 + ε + 1 = 0 and ε3 = 1. Suppose that P0 Let ε = cos 3 3 is represented by the complex number z 0 . The point P1 is represented by the complex number (1) z 1 = z 0 ε + (1 − ε)w1 . The point P2 is represented by z 2 = z 0 ε2 + (1 − ε)w1 ε + (1 − ε)w2 , and P3 by z 3 = z 0 ε 3 + (1 − ε)w1 ε 2 + (1 − ε)w2 ε + (1 − ε)w3 = z 0 + (1 − ε)(w1 ε 2 + w2 ε + w3 ). An easy induction on n shows that after n cycles of three such rotations, we obtain that P3n is represented by z 3n = z 0 + n(1 − ε)(w1 ε 2 + w2 ε + w3 ).
6.2. Solutions to the Olympiad-Caliber Problems
281
In our case, for n = 662 we obtain z 1996 = z 0 + 662(1 − ε)(w1 ε 2 + w2 ε + w3 ) = z 0 . Thus, we have the equality w1 ε 2 + w2 ε + w3 = 0.
(2)
This can be written under the equivalent form w3 = w1 (1 + ε) + (−ε)w2 .
(3)
π π Taking into account that 1 + ε = cos + i sin , the equality (3) can be translated, 3 3 using the lemma on p. 218, into the following: the point A3 is obtained under the π rotation of point A1 about center A2 through the angle . This proved that A1 A2 A3 is 3 an equilateral triangle. Problem 37. Let B(b, 0), C(c, 0) be the centers of the given circles and let A(0, a), X (0, −a) be their intersection points. The complex numbers associated to these point are z B = b, z C = c, z A = ia and z X = −ia, respectively. After rotating A through angle t about B we obtain a point M and after rotating A about C we obtain the point N . Their corresponding complex numbers are given by formulas: z M = (ia − b)ω + b = iaω + (1 − ω)b and z N = iaω + (1 − ω)c.
Figure 6.3.
The required result is equivalent to the following: the bisector lines l M N of the segments M N pass through a fixed point P(x0 , y0 ). Let R be the midpoint of the segment
282
6. Answers, Hints and Solutions to Proposed Problems
1 M N . Then z R = (z M + z N ). A point Z of the plane is a point of l M N if and only if 2 the lines R Z and M N are orthogonal. By using the real product of complex numbers we obtain zM + zN z− · (z N − z M ) = 0. 2 This is equivalent to z · (z N − z M ) =
1 (|z N |2 − |z M |2 ). 2
By noting that z = x + i y we obtain x(c − b)(1 − cos t) − y(c − b) sin t =
1 (|z N |2 − |z M |2 ). 2
After an easy computation we obtain |z M |2 = 2b2 + a 2 − 2b2 cos t − 2ab sin t and |z N |2 = 2c2 + a 2 − 2c2 cos t − 2ac sin t. Thus, the orthogonality condition yields x(1 − cos t) − y sin t = (b + c) − (b + c) cos t − a sin t. This can be written in the form (x − b − c)(1 − cos t) = (y − a) sin t. This equation shows that the point P(x0 , y0 ) where x0 = b + c, y0 = a is a fixed point of the family of lines l M N . The point P belong to the line through A parallel to BC and it is the symmetrical point of X with respect to the midpoint of the segment BC. This follows from the equality b+c zP + zX = . 2 Problem 38. Let A(1+i), B(−1+i), C(−1−i), D(1−i) be the vertices of the square. Using the symmetry of the configuration of points, with respect to the axes and center O of the square, we will do computations for the points lying in the first quadrant. √ √ 3 − 1), M(( 3 − 1)i). The Then L , M are represented by the complex numbers L( √ √3 − 1 3−1 +i . Since K is represented by midpoint of the segment L M is P 2 2 √ 1 √ 2 − 3 K (−i( 3 − 1)), the midpoint of AK is Q +i . In the same way, the 2 2
6.2. Solutions to the Olympiad-Caliber Problems
283
Figure 6.4.
−2 + √3 i i + and the midpoint of B L is S + . It midpoint of AN is R 2 2 2 2 5π . For any point is sufficient to prove that S R = R P = P Q and S R P = R P Q = 6 X we denote by Z X the corresponding complex number. We have 2 − √3
√ √ R S 2 = |Z S − Z R |2 = (−2 + 3)2 = 7 − 4 3, √ 2 √ √ 3−1 i 3−1 2− 3 2 2 R P = |Z P − Z R | = +i − − 2 2 2 2 √ √ √ √ 2 2 3 − 3 3 − 2 (2 3 − 3)2 + (2 3 − 2)2 = = +i 2 2 4 √ √ 28 − 16 3 = = 7 − 4 3. 4 √ Using reflection in O A, we also have P Q 2 = R P 2 = 7 − 4 3. For angles we have √ √ √ 3−2 3 2−2 3 (2 − 3) + ·0 2 2 cos S RP = √ 7−4 3 √ √ √ (12 − 7 3)(7 + 4 3) 3 = . √ √ =− 2 2(7 − 4 3)(7 + 4 3) √ 5π 3 5π This proves that S RP = . In the same way, cos RPQ = − and RPQ = . 6 2 6
284
6. Answers, Hints and Solutions to Proposed Problems
Problem 39. Let 1, ε, ε 2 , be the coordinates of points A, B, C, M, respectively, where ε = cos 120◦ + i sin 120◦ .
Figure 6.5.
Consider point V such that M E V D is a parallelogram. If d, e, v are the coordinates of points D, E, V , respectively, then v = e + d − m. Using the rotation formula, we obtain d = m + (ε − m)ε
and
e = m + (ε2 − m)ε 2 ,
hence v = m + ε2 − mε + m + ε 4 − mε 2 − m = m + ε 2 + ε − m(ε 2 + ε) = m − 1 + m = 2m − 1. This relation shows that M is the midpoint of the segment [AV ] and the conclusion follows. Problem 40. Consider the complex plane with origin at the center of the parallelogram ABC D. Let a, b, c, d, m be the coordinates of points A, B, C, D, M, respectively. It follows that c = −a and d = −b. It suffices to prove that |m − a| · |m + a| + |m − b||m + b| ≥ |a − b||a + b|,
6.2. Solutions to the Olympiad-Caliber Problems
285
or |m 2 − a 2 | + |m 2 − b2 | ≥ |a 2 − b2 |. This follows immediately from the triangle inequality. Problem 41. Let the coordinates of A, B, C, H and O be a, b, c, h and o, respectively. Consequently, aa = bb = cc = R 2 and h = a + b + c. Since D is symmetric to A with respect to line BC, the coordinates d and a satisfy d −b a−b (1) = , or (b − c)d − (b − c)a + (bc − bc) = 0. c−b c−b Since
R 2 (b − c) R 2 (b2 − c2 ) and bc − bc = , bc bc by inserting these expressions in (1), we obtain that b−c =−
d=
−bc + ca + ab k − 2bc = , a a
R 2 (−a + b + c) R 2 (h − 2a) = , bc bc where k = bc + c + ab. Similarly, we have d=
e= Since
k − 2ca , b = =
e=
R 2 (h − 2b) , ca
d e f
d e f
1 1 1
e−d = f −d
(b − a)(k − 2ab) ab
(c − a)(k − 2ca) ca R 2 (c − a)(a − b) = × a 2 b2 c2
=
f =
k − 2ab c
e−d f −d
and
f =
R 2 (h − 2c) . ab
2 R (a − c)(h − 2b) abc −(ck − 2abc) (h − 2c) (bk − 2abc) −(h − 2b) R 2 (a − b)(h − 2c) abc
−R 2 (b − c)(c − a)(a − b)(hk − 4abc) a 2 b2 c2
and h = R 2 k/abc, it follows that D, E and F are collinear if and only if = 0. This is equivalent to hk − 4abc = 0, i.e., hh = 4R 2 . From the last relation we obtain O H = 2R.
286
6. Answers, Hints and Solutions to Proposed Problems
Problem 42. Let the coordinates of A, B, C, D and E be a, b, c, d and e, respectively. the ratio Then d = (2b + c)/3 and e = 2d − a. Since AC B = 2 ABC,
a−b c−b
2 :
b−c a−c
is real and positive. It is equal to (AB 2 · AC)/BC 3 . On the other hand, a direct computation shows that the ratio c−b 2 e−c : b−c e−b is equal to 1 × (b − c)3 =
(b − a) + 2(c − a) 3
2
4(b − a) − (c − a) 3
4 4 (b − a)2 (c − a) AB 2 · AC = , + − 27 27 (b − c)3 BC 3
which is a real number. Hence the arguments of (e − c)/(b − c) and (c − b)2 /(e − b)2 , B and 2 E namely, EC BC, differ by an integer multiple of 180◦ . We easily infer that B = 2 E B = 2 E either EC BC or EC BC − 180◦ , according to whether the ratio is positive or negative. To prove that the latter holds, we have to show that AB 2 · AC/BC 3 is greater than 4/27. Choose a point F on the ray AC such that C F = C B.
Figure 6.6.
we have C Thus Since C B F is isosceles and AC B = 2 ABC, F B = ABC. AB F and AC B are similar and AB : AF = AC : AB. Since AF = AC + BC,
6.2. Solutions to the Olympiad-Caliber Problems
287
AB 2 = AC(AC + BC). Let AC = u 2 and AC + BC = v 2 . Then AB = uv and BC = v 2 − u 2 . From AB + AC > BC, we obtain u/v > 1/2. Thus AB 2 · AC u 4v2 (u/v)4 (1/2)4 4 = 2 = > = , 3 2 3 2 2 3 3 27 BC (v − u ) (1 − u /v ) (1 − 1/4) and the conclusion follows.
6.2.5
Solving trigonometric problems (p. 220)
Problem 11. (i) Consider the complex number z=
1 (cos θ + i sin θ). cos θ
From the identity n−1
zk =
k=0
1 − zn 1−z
(1)
we derive 1 (cos nθ + i sin nθ) 1− nθ 1 cos (cos kθ + i sin kθ) = 1 cosk θ k=0 1− (cos θ + i sin θ) cos θ 1 cos θ − (cos nθ + i sin nθ) n−1 θ sin nθ cosn θ − cos nθ cos = = + i . −i sin θ sin θ cosn−1 θ sin θ cosn−1 θ
n−1
It follows that n−1
cos kθ k=0
cosk θ
=
sin nθ sin θ cosn−1 θ
and we just have to substitute θ = 30◦ . (ii) We proceed in an analogous way by considering the complex number z = cos θ (cos θ + i sin θ ). Using identity (1) we obtain n
k=1
zk =
z − z n+1 . 1−z
288
6. Answers, Hints and Solutions to Proposed Problems
Hence n
cosk θ (cos kθ + i sin kθ )
k=1
=
cos θ (cos θ + i sin θ) − cosn+1 θ(cos(n + 1)θ + i sin(n + 1)θ )
sin2 θ − i cos θ sin θ cos θ (cos θ + i sin θ) − cosn+1 θ(cos(n + 1)θ + i sin(n + 1)θ ) =i sin θ(cos θ + i sin θ) ' cosn+1 θ(cos nθ + i sin nθ) ( = i cotanθ − sin θ sin nθ cosn+1 θ cosn+1 θ cos nθ = + i cotanθ − sin θ sin θ It follows that
n
sin nθ cosn+1 θ sin θ
cosk θ cos kθ =
k=1
Finally, we let θ =
30◦
in the above sum.
Problem 12. Let ω = cos
2π 2π + i sin n n
for some integer n. Consider the sum Sn = 4n + (1 + ω)2n + (1 + ω2 )2n + · · · + (1 + ωn−1 )2n . For all k = 1, . . . , n − 1, we have 2kπ 2kπ kπ + i sin = 2 cos 1 + ω = 1 + cos n n n k
and (1 + ωk )2n = 22n cos2n
Sn = 4n + = 4n 1 + cos2n
π n
kπ kπ cos + i sin n n
n−1
+ cos2n
k=1
2π n
(1 + ωk )2n
+ · · · + cos2n
(n − 1)π n
On the other hand, using the binomial expansion, we have Sn =
n−1
k=0
kπ kπ (cos 2kπ + i sin 2kπ ) = 4n cos2n . n n
Hence
(1 + ωk )2n =
n−1
2n k=0
+
2n k ω + 1
.
(1)
6.2. Solutions to the Olympiad-Caliber Problems
289
+
2n 2k 2n nk 2n 2n ω + ··· + ω + ω(2n−1)k + 2 n 2n − 1 2n =n
2n−1 n−1
2n
2n 2n 2n +n +n + ω jk · 0 n 2n j j=1 k=0 i=n
2n−1
2n 1 − ω jn 2n 2n · = 2n + n + = 2n + n . j j n n 1 − ω j=1
(2)
i=n
The relations (1) and (2) give the desired identity. Problem 13. For p = 0, take a0 = 1. If p ≥ 1, let z = cos α + i sin α and observe that z 2 p = cos 2 pα + i sin 2 pα, z −2 p = cos 2 pα − i sin 2 pα and cos 2 pα =
z 2 p + z −2 p 1 = [(cos α + i sin α)2 p + (cos α − i sin α)2 p ]. 2 2
Using the binomial expansion we obtain 2p 2p 2p 2 p−2 2 p 2p sin2 p α. cos α sin α + · · · + (−1) cos 2 pα = cos α − 2p 2 0 Hence cos 2 pα is a polynomial of degree p in sin2 α, so there are a0 , a1 , . . . , a p ∈ R such that cos 2 pα = a0 + a1 sin2 α + · · · + a p sin2 p α for all α ∈ R, with ap =
2p 2p 2p 2p − (−1) p−1 + (−1) p−2 + · · · + (−1) p 0 2 4 2p = (−1) p
6.2.6
2p 2p 2p + + ··· + = 0. 0 2 2p
More on the n th roots of unity (pp. 228–229)
Problem 11. Let p = 1, 2, . . . , m and let z ∈ U p . Then z p = 1. Note that n − m + 1, n − m + 2, . . . , n are m consecutive integers, and, since p ≤ m, there is an integer k ∈ {n − m + 1, n − m + 2, . . . , n} such that p divides k.
290
6. Answers, Hints and Solutions to Proposed Problems
Let k = k p. It follows that z k = (z p )k = 1, so z ∈ Uk ⊂ Un−m+1 ∪ Un−m+2 ∪ · · · ∪ Un , as claimed. Remark. An alternative solution can be obtained by using the fact that (a n − 1)(a n−1 − 1) · · · (a n−k+1 − 1) (a k − 1)(a k−1 − 1) · · · (a − 1) is an integer for all positive integers a > 1 and n > k. Problem 12. Rewrite the equation as bx + aα n d = . ax + bα c d Since |c| = |d|, we have = 1 and consider c d = cos t + i sin t, c It follows that
t ∈ [0, 2π ).
bxk + aα = uk , axk + bα
(1)
where
t + 2kπ t + 2kπ + i sin , n n The relation (1) implies that u k = cos
xk =
bαu k − aα , b − au k
k = 0, n − 1.
k = 0, n − 1.
To prove that the roots xk , k = 0, n − 1 are real numbers, it suffices to show that xk = xk for all k = 0, n − 1. Denote |a| = |b| = r . Then 1 r2 r2 ·α· ·α − bαu k − aα b uk a xk = = r2 r2 1 b − au k − · b a uk =
αa − bαu k = xk , au k − b
k = 0, n − 1,
as desired. Problem 13. Differentiating the familiar identity n
k=0
zk =
x n+1 − 1 x −1
6.2. Solutions to the Olympiad-Caliber Problems
291
with respect to x, we get n
nx n+1 − (n + 1)x n + 1 . (x − 1)2
kx k−1 =
k=1
Multiplying both sides by x and differentiating again, we arrive at n
k 2 x k−1 = g(x),
k=1
where
n 2 x n+2 − (2n 2 + 2n − 1)x n+1 + (n + 1)2 x n − x − 1 . (x − 1)3 Taking x = z and using |z| = 1 (which we were given), we obtain g(x) =
|g(z)| ≤
n
k 2 |z|k−1 =
k=1
n(n + 1)(2n + 1) . 6
(1)
On the other side, taking into account that z n = 1, z = 1, we get g(z) =
n(nz − (n + 2)) n(nz 2 − 2(n + 1)z + n + 2) = . (z − 1)3 (z − 1)2
(2)
From (1) and (2) we therefore conclude that (n + 1)(2n + 1) |z − 1|2 . 6 x Problem 14. Setting x = y ∈ M yields 1 = ∈ M. For x = 1 and y ∈ M we obtain y 1 −1 = y ∈ M. y If x and y are arbitrary elements of M, then x, y −1 ∈ M and consequently |nz − (n + 2)| ≤
x y −1
= x y ∈ M.
Let x1 , x2 , . . . , xn be the elements of set M and take at random an element xk ∈ M, k = 1, n. Since xk = 0 for all k = 1, n, the numbers xk x1 , xk x2 , . . . , xk xn are distinct and belong to the set M, hence {xk x1 , xk x2 , . . . , xk xn } = {x1 , x2 , . . . , xn }. Therefore xk x1 · xk x2 · · · xk xn = x1 x2 · · · xn , hence xkn = 1, that is, xk is an n th root of 1. The number xk was chosen arbitrary, hence M is the set of the n th -roots of 1, as claimed.
292
6. Answers, Hints and Solutions to Proposed Problems
Problem 15. a) We will denote by S(X ) the sum of the elements of a finite set X . Suppose 0 = z ∈ A. Since A is finite, there exists positive integers m < n such that z m = z n , whence z n−m = 1. Let d be the smallest positive integer k such that z k ∈ 1. Then 1, z, z 2 , . . . , z d−1 are different, and the dth power of each is equal to 1; therefore m 6 these numbers are the dth roots of unity. This shows that A \ {0} = Un k , where k=1
U p = {z ∈ C| z p = 1}. Since S(U p ) = 0 for p ≥ 2, S(U1 ) = 1 and U p ∩ Uq = U( p,q) we get
S(A) = S(Un k ) − S(Un k ∩ Unl ) k
+
k
S(Un k ∩ Unl ∩ Un s ) + · · · = an integer.
k
b) Suppose that for some integer k there exists A =
m 6
Un k such that S(A) = k.
k=1
Let p1 , p2 , . . . , p6 be the distinct primes which are not divisors of any n k . Then S(A ∪ U p1 ) = S(A) + S(U p1 ) − S(A ∩ U p1 ) = k − S(U1 ) = k − 1. Also S(A ∪ U p1 p2 p3 ∪ U p1 p4 p5 ∪ U p2 p4 p6 ∪ U p3 p5 p6 ) = S(A) + S(U p1 p2 p3 ) + S(U p1 p4 p5 ) + S(U p2 p4 p6 ) + S(U p3 p5 p6 ) − S(A ∩ U p1 p2 p3 ) − · · · + S(A ∩ U p1 p2 p3 ∩ U p1 p4 p5 ) + · · · − S(A ∩ U p1 p2 p3 ∩ U p1 p4 p5 ∩ U p2 p4 p6 ∩ U p3 p5 p6 ) = k + 4 · 0 − 4S(U1 ) −
6
S(U pk ) + 10S(U1 ) − 5S(U1 ) + S(U1 )
k=1
= k − 4 + 10 − 5 + 1 = k + 2. Hence, if there exists A such that S(A) = k, then there exist B and C such that S(B) = k − 1 and S(C) = k + 2. The conclusion now follows easily.
6.2.7
Problems involving polygons (p. 237)
Problem 12. Suppose that such a 1990-gon exists and let A0 A1 · · · A1989 be its ver−−−−→ tices. The sides Ak Ak+1 , k = 0, 1, . . . , 1989 define the vectors Ak Ak+1 which can be represented in the complex plane by the numbers z k = n k wk ,
k = 0, 1, . . . , 1989
6.2. Solutions to the Olympiad-Caliber Problems
293
2π 2π + i sin . Here A1990 = A0 and n 0 , n 1 , . . . , n 1989 represents 1990 1990 2 a permutation of the numbers 1 , 22 , . . . , 19902 . 1989
−−−−→ Because Ak Ak+1 = 0, the problem can be restated as follows: find a permuta-
where w = cos
k=0
tion (n 0 , n 1 , . . . , n 1989 ) of the numbers 12 , 22 , . . . , 19902 such that 1989
n k w k = 0.
k=0
Observe that 1990 = 2 · 5 · 199. The strategy is to add vectors after suitable grouping of 2, 5, 199 vectors such that these partial sums can be directed toward the suitable result. To begin, let consider the pairing of numbers (12 , 22 ), (32 , 42 ), . . . , (19882 , 19892 ) and assign these lengths to pairs of opposite vectors respectively: (wk , wk+995 ),
k = 0, . . . , 994.
By adding the obtained vectors, we obtain 995 vectors of lengths 22 − 12 = 3; 42 − 32 = 7; 62 − 52 = 11; . . . ; 19892 − 19882 = 3979 which divide the unit circle of the coordinate plane into 995 equal arcs. Let B0 = 1, B1 , . . . , B994 be the vertices of the regular 995-gon inscribed in the unit circle. We intend to assign the lengths 3,7,11, . . . ,3979 to the unit vectors −→ −→ −→ O B 0 , O B 1 , . . . , O B 994 such that the sum of the obtained vectors is zero, We divide 995 lengths into 199 groups of size 5: (3, 7, 11, 15, 19), (23, 27, 31, 35, 39), . . . , (3963, 3967, 3971, 3975, 3979). 2π 2π 2π 2π + i sin , ω = cos + i sin be the primitive roots of unity 5 5 199 199 of order 5 and 199, respectively. Let P1 be the pentagon with vertices 1, ζ, ζ 2 , ζ 3 , ζ 4 . 2kπ ,k = Then we rotate P1 about the origin O with coordinates through angles θk = 199 1, . . . , 198, to obtain new pentagons P2 , . . . , P198 , respectively. The vertices of Pk+1 are ωk , ωk ζ, ωk ζ 2 , ωk ζ 3 , ωk ζ 4 , k = 0, . . . , 198. We assign to unit vectors defined by the vertices Pk of the respective lengths: Let ζ = cos
294
6. Answers, Hints and Solutions to Proposed Problems
Figure 6.7.
2k + 3, 2k + 7, 2k + 11, 2k + 15, 2k + 19
(k = 0, . . . , 198).
Thus, we have to evaluate the sum: 198
[(2k + 3)ωk + (2k + 7)ωk ζ + (2k + 1)ωk ζ 2 + (2k + 15)ωk ζ 3 + (2k + 19)ωk ζ 4 ]
k=0 198
2kωk (1 + ζ + ζ 2 + ζ 3 + ζ 4 ) + (3 + 7ζ + 11ζ 2 + 15ζ 3 + 19ζ 4 )
k=0
198
ωk .
k=0
Since 1 + ζ + ζ 2 + ζ 3 + ζ 4 = 0 and 1 + ω + ω2 + · · · + ω198 = 0, it follows that the sum equals zero. Problem 13. It is convenient to take a regular octagon inscribed in a circle and note its vertices as follows: A = A0 , A1 , A2 , A3 , A4 = E, A−3 , A−2 , A−1 . π 2π = about the center O We imagine a step in the path like a rotation of angle 8 4 of the circ*mscribed circle of the octagon. In this way, a path is a sequence of such rotations, submitted to some conditions. If the rotation is counterclockwise we add the π π angle ; if the rotation is clockwise we add the angle − . The starting point is A0 , 4 4 which is represented by the complex number z 0 = cos 0 + i sin 0. Any vertex Ak of 2kπ 2kπ + i sin . It is convenient to work only the octagon is represented by z k = cos 8 8 2kπ with the angles , −4 ≤ k ≤ 4. But these k’s are integers considered mod 8, such 8 that z 4 = z −4 and A4 = A−4 .
6.2. Solutions to the Olympiad-Caliber Problems
295
Figure 6.8.
We may associate to a path of length n, say (P0 P1 · · · Pn ), an ordered sequence (u , u 2 , . . . , u n ) of integers which satisfy the following conditions: a) u k = ±1 for any k = 1, 2, . . . , n; more precisely u i = ±1 if the arc(Pk−1 Pk ) is π π and u k = −1 if the arc(Pk−1 Pk ) is − ; 4 4 b) u 1 + u 2 + · · · + u k ∈ {−3, −2, −1, 0, 1, 2, 3} for all k = 1, 2, . . . , n − 1; c) u 1 + u 2 + · · · + u n = ±4. For example, the sequence associated with the path (A0 , A−1 , A0 , A1 , A2 , A3 , A4 ) is (−1, 1, 1, 1, 1, 1). From now on we consider only sequences that satisfy a), b), c). It is obvious that conditions a), b), c) define a bijective function between the set of paths and the set of sequences. For any sequence u 1 , u 2 , . . . , u n and any k, 1 ≤ k ≤ n, we call the sum sk = u 1 + u 2 + · · · + u k a partial sum of the sequence. It is easy to see that for any k, sk is an even number if and only if k is even. Thus, a2n−1 = 0. Thus we have to prove the formula for even numbers. For small n we have a2 = 0, a4 = 2; for example, only sequences (1, 1, 1, 1) and (−1, −1, −1, −1) of length 4 satisfy conditions a)–c). In the following we will prove a recurrence relation between the numbers an , n even. The first step is to observe that if sn = ±4, then sn−2 = ±2. Moreover, if (u 1 , u 2 , . . . , u n−2 ) is a sequence that satisfies a), b) and sn−2 = ±2 there are only two ways to extend it to a sequence that satisfy c) as well: either the sequence (u 1 , u 2 , . . . , u n−2 , +1, +1) or the sequence (u 1 , u 2 , . . . , u n−2 , −1, −1). So if we denote by xn the number of sequences that satisfy a), b) and sn = ±2, then n is even and an = xn−2 .
296
6. Answers, Hints and Solutions to Proposed Problems
Let yn denote the number of sequences which satisfy a), b) and sn = 0. Then n is even and we have the equality yn = xn−2 + 2yn−2 .
(1)
This equality comes from the following constructions. A sequence (u 1 , . . . , u n−2 ) for which sn−2 = ±2 gives rise to a unique sequence of length n with sn = 0 by extending it either to (u 1 , . . . , u n−2 , 1, 1) or (u 1 , u 2 , . . . , u n−1 , −1, −1). Also, a sequence (u 1 , . . . , u n−2 ) with sn−2 = 0 gives rise either to sequence (u 1 , . . . , u n−2 , 1, −1) or (u 1 , . . . , u n−2 , −1, 1). Finally, every sequence of length n with sn = 0 ends in one of the following “terminations”: (−1, −1), (1, 1), (1, −1), (−1, 1). The following equality is also verified: xn = 2xn−2 + 2yn−2 .
(2)
This corresponds to the property that any sequence of length n for which sn = ±2 can be obtained either from a similar sequence of length n−2 by adding the termination (1, −1) or the termination (−1, 1), or from a sequence of length n − 2 for which sn−2 = 0 by adding the termination (1,1) or the termination (−1, −1). Now, the problem is to derive an = xn−2 , from relations (1) and (2). By subtracting (1) from (2) we obtain xn−2 = xn −yn , for all n ≥ 4, n even. Thus, yn−2 = xn−2 −xn−4 . Substituting the last equality in (2) we obtain the recurrent relation: xn = 4xn−2 − 2xn−4 , for all n ≥ 4, n even. Taking into account that x n = an+2 , we obtain the linear recurrent relation an+2 = 4an − 2an−2 , n ≥ 4, (3) with the initial values a2 = 0, a4 = 2. The sequence (an ), n ≥ 2, n even is uniquely defined by a2 = 0, a4 = 2 and relation (3). Therefore, to answer the question, it is sufficient to prove that the sequence √ √ 1 (c2n )n≥1 , c2n = √ ((2 + 2)n−1 − (2 − 2)n−1 ) obeys the same conditions. This is 2 a straightforward computation. Problem 14. Consider the complex plane with origin at the center of the polygon. Without loss of generality we may assume that the coordinates of A, B, C are 1, ε, ε2 , 2π 2π + i sin . respectively, where ε = cos n n Let z M = cos t + i sin t, t ∈ [0, 2π ) be the coordinate of point M. From the hypoth4π . Then esis we derive that t > n √ t M A = |z M − 1| = (cos t − 1)2 + sin2 t = 2 − 2 cos t = 2 sin ; 2
6.2. Solutions to the Olympiad-Caliber Problems
& M B = |z M − ε| = &
297
t π 2π = 2 sin − ; 2 − 2 cos t − n 2 n
4π t 2π 2 − 2 cos t − = 2 sin − ; n 2 n 2π π AB = |ε − 1| = 2 − 2 cos = 2 sin . n n
MC = |z M − ε | = 2
We have
π t π − − 4 sin2 2 n n 2π 2π − cos t − 2 cos n n 2π 2π 2π 2π − t− + t− n n n n sin = −2 · 2 sin 2 2 2π t t − = M A · MC, = 2 sin · 2 sin 2 2 n M B 2 − AB 2 = 4 sin2
as desired. Problem 15. Rotate the polygon A1 A2 · · · An so that the coordinates of its vertices are the complex roots of unity of order n, ε1 , ε2 , . . . , εn . Let z be the coordinate of point P located on the circumcircle of the polygon and note that |z| = 1. The equality n zn − 1 = (z − ε j ) j=1
yields |z n − 1| =
n j=1
|z − ε j | =
n
P Aj.
j=1
Since |z n − 1| ≤ |z|n + 1 = 2, it follows that the maximal value of
n
P A2j is 2 and
j=1
is attained for z n = −1, i.e., for the midpoints of arcs A j A j+1 , j = 1, . . . , n, where An+1 = A1 . Problem 16. Without loss of generality, assume that points Ak have coordinates ε k−1 for k = 1, . . . , 2n, where π π ε = cos + i sin . n n Let α be the coordinate of the point P, |α| = 1. We have P Ak+1 = |α − ε k |
298
6. Answers, Hints and Solutions to Proposed Problems
and P An+k+1 = |α − ε n+k | = |α + ε k |, for k = 0, . . . , n − 1. Then n−1
P A2k+1 · P A2n+k+1 =
k=0
n−1
|α − ε k |2 · |α + ε k |2
k=0 n−1
=
[(α − ε k )(α − ε k )][(α + ε k )(α + ε k )]
k=0
=
n−1
(2 − αε k − αε k )(2 + αε k + αε k )
k=0
=
n−1
(2 − α 2 ε 2k − α 2 ε 2k ) = 2n − α 2
k=0
n−1
ε 2k − α 2 ·
k=0
= 2n − α 2 ·
ε 2n
−1
ε2
−1
− α2 ·
n−1
ε2k
k=0
−1 = 2n, ε2 − 1
ε 2n
as desired.
6.2.8
Complex numbers and combinatorics (p. 245)
Problem 11. Let us consider the complex number z = cos t + i sin t and the sum n 2
n tn = sin kt. Observe that k k=0 sn + itn =
n 2
n k=0
k
(cos kt + i sin kt) =
n 2
n k=0
k
(cos t + i sin t)k .
In the product (1 + X )n (1 + z X )n = (1 + (z + 1)X + z X 2 )n we set the coefficient of X n equal to obtain
n n
n! zs = (1) (z + 1)s z r . k s k!s!r ! 0≤k,s≤n 0≤k,s,r ≤n k+s+r =n s+2r =n
k+s=n
The above relation is equivalent to n 2
n k=0
k
7n8
2
n 2k (z + 1)n−2k z k . z = 2k k k=0 k
The trigonometric form of the complex number 1 + z is given by t t t t t 2 t 1 + cos t + i sin t = 2 cos + 2i sin cos = 2 cos cos + i sin , 2 2 2 2 2 2
(2)
6.2. Solutions to the Olympiad-Caliber Problems
299
since t ∈ [0, π ]. From (2) it follows that 7n8
sn + itn =
2
n 2k 2k
k=0
hence
k
t 2 cos 2
n−2k
nt nt cos + i sin 2 2
7n8
sn =
2
n 2k k=0 7n8
tn =
2k
k
t 2 cos 2
2
n 2k k=0
2k
k
2 cos
t 2
n−2k cos
nt , 2
sin
nt . 2
n−2k
,
Remark. Here we have a few particular cases of (2). 1) If z = 1, then n 2
n k=0
k
7n8
2
n 2k n−2k 2n 2 = = . 2k k n k=0
2) If z = −1, then
⎧ 0, ⎪ ⎪ n 2 ⎨
n (−1)k = ⎪ k n n ⎪ k=0 ⎩ (−1) 2 , n/2
1 3) If z = − , then 2
if n is odd, if n is even.
7 8
n 2 2
n 2k k k n n−k 2 . (−1) 2 = (−1)k k 2k k k=0 k=0
n
Problem 12. 1) In Problem 4 consider p = 4 to obtain n n n 2n nπ π n + + + ··· = cos 1 + 2 cos 0 4 8 4 4 4 n nπ 1 n 2 + 2 2 +1 cos . = 4 4 2) Let us consider p = 5 in Problem 4. We find that n n n 2n π n 2π n nπ 2nπ + + +··· = 1 + 2 cos + 2 cos . cos cos 0 4 8 5 5 5 5 5 Using the well-known relations √ 5+1 π cos = 5 4 the desired identity follows.
and
2π cos = 5
√ 5−1 4
300
6. Answers, Hints and Solutions to Proposed Problems
Problem 13. 1) Let ε be a cube root of unity different from 1. We have (1 − ε)n = An + Bn ε + Cn ε 2 ,
(1 − ε 2 )n = An + Bn ε2 + Cn ε
hence A2n + Bn2 − Cn2 − An Bn − Bn Cn − Cn An = (An + Bn ε + Cn ε 2 )(An + Bn ε 2 + Cn ε) = (1 − ε)n (1 − ε 2 )n = (1 − ε − ε 2 + 1)n = 3n . 2) It is obvious that An + Bn + Cn = 0. Replacing Cn = −(An + Bn ) in the previous identity we get A2n + An Bn + Cn2 = 3n−1 .
Problem 14. For any k ∈ {0, 1, . . . , p − 1}, consider xk = c1 · · · cm , the sum of m
ci ≡ k (mod p). all products c1 · · · cm such that ci ∈ {1, 2, . . . , n} and i=1
2π 2π + i sin , then If ε = cos p p
(ε + 2ε 2 + · · · + nε n )m =
c1 · · · cm ε c1 +···+cm =
c1 ,...,cm ∈{1,2,...,n}
p−1
xk εk .
k=0
Taking into account the relation ε + 2ε 2 + · · · + nεn =
nεn+2 − (n + 1)ε n+1 + ε nε = 2 ε −1 (ε − 1)
(see Problem 9 in Section 5.4 or Problem 13 in Section 5.5) it follows that p−1
nm = xk εk . (ε − 1)m k=0
(1)
On the other hand, from ε p−1 + · · · + ε + 1 = 0 we obtain that 1 1 = − (ε p−2 + 2ε p−3 + · · · + ( p − 2)ε + p − 1), ε−1 p hence nm n m p−2 = − (ε + 2ε p−3 + · · · + ( p − 2)ε + p − 1)m . (ε − 1)m p Put (X p−2 + 2X p−3 + · · · + ( p − 2)X + p − 1)m = b0 + b1 X + · · · + bm( p−2) X m( p−2) ,
6.2. Solutions to the Olympiad-Caliber Problems
n m nm = − (y0 + y1 ε + · · · + y p−1 ε p−1 ), (ε − 1)m p
bk .
and find
where y j =
301
(2)
k≡ j (mod p)
From (1) and (2) we get x0 − r y0 + (x1 − r y1 )ε + · · · + (x p−1 − r y p−1 )ε p−1 = 0, n m where r = − . From Proposition 4 in Subsection 2.2.2, it follows that x0 −r y0 = p x1 − r y1 = · · · = x p−1 − r y p−1 = k. Now it is sufficient to show that r |k. But pk = x0 + · · · + x p−1 − r (y0 + · · · + y p−1 ) = (1 + 2 + · · · + n)m − r (b0 + · · · + bm( p−2) ) = (1 + 2 + · · · + n)m − r (1 + 2 + · · · + ( p − 1))m ,
and we obtain pk =
n(n + 1) 2
m −r
p( p − 1) 2
m .
Since the right-hand side is divisible by pr , it follows that r |k. √ Problem 15. Expanding (1 + i a)n by binomial theorem and then separating the even and odd terms we find √ √ (1 + i a)n = sn + i atn . (1) Passing to conjugates in (1) we get √ √ (1 − i a)n = sn − i atn .
(2)
From (1) and (2) it follows that √ √ 1 (3) [(1 + i a)n + (1 − i a)n ]. 2 √ √ The quadratic equation with roots z 1 = 1+i a and z 2 = 1−i a is z 2 −2z+(a+1) = 0. It is easy to see that for any positive integer n the following relation holds: sn =
sn+2 = 2sn+1 − (1 + a)sn .
(4)
Now, we proceed by induction by step 2. We have s1 = 1 and s2 = 1 − a = 2 − 4k = 2(1 − 2k), hence the desired property holds. Assume that 2n−1 |sn and 2n |sn+1 . From (4) it follows that 2n+1 |sn+2 , since 1 + a = 4k and 2n+1 |(1 + a)sn .
302
6.2.9
6. Answers, Hints and Solutions to Proposed Problems
Miscellaneous problems (p. 252)
Problem 12. Using the triangle inequality, we have 2|z|2 = |x|y| + y|x|| ≤ |x||y| + |y||x|, so |z|2 ≤ |x| · |y|. Likewise, |y|2 ≤ |x| · |z|
and
|z|2 ≤ |y||x|.
Summing these inequality yields |x|2 + |y|2 + |z|2 ≤ |x||y| + |y||z| + |z||x|. This implies that |x| = |y| = |z| = a. If a = 0, then x = y = z = 0 is a solution of the system. Consider a > 0. The system may be written as ⎧ 2 ⎪ ⎪ x + y = z2, ⎪ ⎪ a ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 2 y + z = x 2, ⎪ a ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ z + x = 2 y2. a Subtracting the last two equations gives 2 2 x − y = (y 2 − x 2 ), i.e., (y − x) y + x + = 0. a a Case 1. If x = y, then x = y =
z2 . The last equation implies a z+
This is equivalent to 2 hence
z2 z4 = 2 3. a a
z 3 a
=
z + 1, a
z −1 ± i z = 1 or = . a a 2 If z = a, then x = y = z = a is a solution of the system.
6.2. Solutions to the Olympiad-Caliber Problems
If
−1 ± i z = , then a 2
303
z −1 ± i √2 = 1 = = , a 2 2
which is a contradiction. 2 2 2 Case 2. If x + y = − , then − = z 2 . We obtain z = ±i and a = |z| = 1. a a a Consider z = i; then x = (x + y) − (y + z) + z = 2z 2 − 2x 2 + z = −2 + i − 2x 2 or equivalently, 2x 2 + x + 2 − i = 0. 1 Then x = i or x = − − i. Since |x| = a = 1, we have x = i. Then y = 2x 2 − z = √ 2 −2 − i and |y| = 5 = a = 1, so the system has no solution. The case z = −i had the same conclusion. Therefore, the solutions are x = y = z = a, where a ≥ 0 is a real number. Problem 13. In any solution (x, y, z) we have x = 0, y = 0, z = 0 and x = y, y = z, z = x. We can divide each equation by others and obtain new equations: x 2 + y 2 = yz + zx, y 2 + z 2 = x y + zx, z 2 + x 2 = x y + yz.
(1)
By adding them one obtains the equality x 2 + y 2 + z 2 = x y + yz + zx.
(2)
After subtracting equations (1), the second from the first, one obtains x + y + z = 0. By squaring this identity one obtains an improvement of (2): x 2 + y 2 + z 2 = x y + yz + zx = 0.
(3)
Using (3) in (1) one obtains x 2 = zy,
y 2 = zx,
z2 = x y
(4)
and also x 3 = y 3 = z 3 = x yz. x3
It follows that x, y, z are distinct roots of the same complex number a = x yz. From = y 3 = z 3 = x yz = a we obtain √ √ √ x = 3 a, t = ε 3 a, z = ε2 3 a, (5)
304
6. Answers, Hints and Solutions to Proposed Problems
where ε 2 + ε + 1 = 0, ε3 = 1. When introduce relations (5) in the first equation of the original system, one obtains a 3 (1 − ε)(1 − ε 2 ) = 3. Taking into account the computation (1 − ε)(1 − ε 2 ) = 1 − ε − ε 2 + 1 = 3, we have a 3 = 1. Hence, we obtain using (5) that (x, y, z) is a permutation of the set {1, ε, ε2 }. Problem 14. Suppose that the triangles O X Y and O Z T are counterclockwise oriented, and let x, y, z, t be the coordinates of the points X, Y, Z , T and let m be the coordinate of O. As these are right isosceles triangles we have x − m = i(y − m), z−m = i(t −m). It follows that m(1−i) = x −i y = z−it. We deduce x −z = i(y−t). x − iy , and the triangles Reciprocally, if x −i y = z −it, the coordinate of O is m = 1−i O X Y and O Z T are right and isosceles. Let a, b, c, d, e, f be the coordinates of the given hexagon in that order. We can write a − ib = c − ie, b − id = e − i f . It follows that a + d = c + f , i.e., AC D F is a parallelogram. Multiplying the first equality by i, we obtain b − ic = e − ia, i.e., BC and AE are connected. Problem 15. By standard computations, we find that on the circ*mscribed circle the
sides of the pentagon subtend the following arcs: AB= 80◦ , BC= 40◦ , C D= 80◦ ,
D E= 20◦ and E A= 140◦ . It is then natural to consider all these measures as multiples 2π 2π of 20◦ that correspond to the primitive 18th roots of unity, say ω = cos + i sin . 18 18 We thus assign, to each vertex, starting from A(1), the corresponding root of unity: B(ω4 ), C(ω6 ), D(ω10 ), E(ω11 ). We shall use the following properties of ω: ω18 = 1,
ω9 = −1,
ωk = ω18−k ,
ω6 − ω3 + 1 = 0.
(A)
We need to prove that the coordinate coordinate of the common point of the lines B D and C E is a real number. The equation of the line B D is z z 1 4 (1) ω4 1 = 0, ω 10 10 ω ω 1 and the equation of the line C E is z 6 ω 11 ω
z ω6 ω11
1 1 1
= 0.
(2)
6.2. Solutions to the Olympiad-Caliber Problems
305
Figure 6.9.
The equation (1) can be written as follows: z(ω14 − ω8 ) − z(ω4 − ω10 ) + (ω12 − ω6 ) = 0 or zω8 (ω6 − 1) + zω4 (ω6 − 1) + ω6 (ω6 − 1) = 0. Using the properties of ω we derive a simplified version of (1): zω4 + z + ω2 = 0.
(1 )
In the same way, equation (2) becomes zω + z − ω3 (ω4 − 1) = 0.
(2 )
From (1 ) and (2 ) we obtain the following expression for z: z=
ω−1 −ω7 + ω3 − ω2 −ω6 + ω2 − ω = −1 + . = 4 6 ω −ω ω ω5
To prove that z is real, it will suffice to prove that it coincides with its conjugate. It is easy to see that ω−1 ω−1 = 5 ω ω5 is equivalent to ω4 − ω5 = ω4 − ω5 , i.e., ω14 − ω13 = ω4 − ω5 , which is true by the properties of ω given in (A).
Glossary
Antipedal triangle of point M: The triangle determined by perpendicular lines from vertices A, B, C of triangle ABC to M A, M B, MC, respectively. Area of a triangle: The area of triangle with vertices with coordinates z 1 , z 2 , z 3 is the absolute value of the determinant z1 z1 1 i = z2 z2 1 . 4 z3 z3 1 Area of pedal triangle of point X with respect to the triangle ABC: area[P Q R] =
area[ABC] |x x − R 2 |. 4R 2
Argument of a complex number: If the polar representation of complex number z is z = r (cos t ∗ + i sin t ∗ ), then arg(z) = t ∗ . Barycenter of set { A1 , . . . , An } with respect to weights m1 , . . . , m n : The point G 1 with coordinate z G = (m 1 z 1 + · · · + m n z n ), where m = m 1 + · · · + m n . m Barycentric coordinates: Consider triangle ABC. The unique real number μa , μb , μc such that z P = μa a + μb b + μc c, where μa + μb + μc = 1. Basic invariants of triangle: s, r, R
308
Glossary
Binomial equation: An algebraic equation of the form Z n + a = 0, where a ∈ C∗ . Ceva’s theorem: Let AD, B E, C F be three cevians of triangle ABC. Then, lines AD, B E, C F are concurrent if and only if AF B D C E · · = 1. F B DC E A Cevian of a triangle: any segment joining a vertex to a point on the opposite side. Concyclicity condition: If points Mk (z k ), k = 1, 2, 3, 4, are not collinear, then they are concyclic if and only if z3 − z2 z3 − z4 : ∈ R∗ . z1 − z2 z1 − z4 Collinearity condition: M1 (z 1 ), M2 (z 2 ), M3 (z 3 ) are collinear if and only if R∗ .
z3 − z1 ∈ z2 − z1
Complex coordinate of point A of cartesian coordinates (x, y): The complex number z = x + yi. We use the notation A(z). a+b Complex coordinate of the midpoint of segment [ AB]: z M = , where A(a) 2 and B(b). Complex coordinates of important centers of a triangle: Consider the triangle ABC with vertices with coordinates a, b, c. If the origin of complex plane is in the circumcenter of triangle ABC, then: • the centroid G has coordinate z G =
1 (a + b + c); 3
• the incenter I has coordinate z I =
αa + βb + γ c , where α, β, γ are the sides α+β +γ
length of triangle ABC;
• the orthocenter H has coordinate z H = a + b + c; rα a + rβ b + rγ c , where rα , rβ , rγ rα + rβ + rγ are the radii of the three excircles of triangle;
• the Gergonne point J has coordinate z J =
α2a + β 2b + γ 2c ; α2 + β 2 + γ 2 α β γ = 1− a+ 1− b+ 1− c; s s s
• the Lemoine point K has coordinate z K = • the Nagel point N has coordinate z N
• the center O9 of point circle has coordinate z O9 =
1 (a + b + c). 2
Glossary
309
Complex number: A number z of the form z = a + bi, where a, b are real numbers √ and i = −1. 1 Complex product of complex numbers a and b: a × b = (ab − ab). 2 Conjugate of a complex number: The complex number z = a −bi, where z = a +bi. Cyclic sum: Let n be a positive integer. Given a function f of n variables, define the cyclic sum of variables (x1 , x2 , . . . , xn ) as
f (x1 , x2 , . . . , xn ) = f (x1 , x2 , . . . , xn ) + f (x2 , x3 , . . . , xn , x1 ) cyc
+ · · · + f (xn , x1 , x2 , . . . , xn−1 ) De Moivre’s formula: For any angle α and for any integer n, (cos α + i sin α)n = cos nα + i sin nα. Distance between points M1 (z1 ) and M2 (z2 ): M1 M2 = |z 2 − z 1 |. Equation of a circle: z · z + α · z + α · z + β = 0, where α ∈ C and β ∈ R. Equation of a line: α · z + αz + β = 0, where α ∈ C∗ , β ∈ R and z = x + i y ∈ C. Equation of a line determined by two points: If P1 (z 1 ) and P2 (z 2 ) are distinct points, then the equation of line P1 P2 is z1 z1 1 z 2 z 2 1 = 0. z z 1 Euler’s formula: Let O and I be the circumcenter and incenter, respectively, of a triangle with circumradius R and inradius r . Then O I 2 = R 2 − 2Rr. Euler line of triangle: The line determined by the circumcenter O, the centroid G, and the orthocenter H . Extend law of sines: In a triangle ABC with circumradius R and sides α, β, γ the following relations hold: β γ α = = = 2R. sin A sin B sin C Heron’s formula: The area of triangle ABC with sides α, β, γ is equal to
area[ABC] = s(s − α)(s − β)(s − γ ), where s =
1 (α + β + γ ) is the semiperimeter of the triangle. 2
310
Glossary
Isometric transformation: A mapping f : C → C preserving the distance. Lagrange’s theorem: Consider the points A1 , . . . , An and the nonzero real numbers m 1 , . . . , m n such that m = m 1 + · · · + m n = 0. For any point M in the plane the following relation holds: n
m j M A2j = m M G 2 +
j=1
n
m j G A2j ,
j=1
where G is the barycenter of set {A1 , . . . , An } with respect to weights m 1 , . . . , m n . √ Modulus of a complex number: The real number |z| = a 2 + b2 , where z = a + bi. Morley’s theorem: The three points of adjacent trisectors of angles form an equilateral triangle. Nagel line of triangle: The line I, G, N . nth roots of complex number z0 : Any solution Z of the equation Z n − z 0 = 0. nth roots of unity: The complex numbers εk = cos
2kπ 2kπ + i sin , k ∈ {0, 1, . . . , n − 1}. n n
The set of all these complex numbers is denoted by Un . Orthogonality condition: If Mk (z k ), k = 1, 2, 3, 4, then lines M1 M2 and M3 M4 are orthogonal if and only if z1 − z2 ∈ iR∗ . z3 − z4 Orthopolar triangles: Consider triangle ABC and points X, Y, Z situated on its circumcircle. Triangles ABC and X Y Z are orthopolar (or S-triangles) if the Simson– Wallance line of point X with respect to triangle ABC is orthogonal to line Y Z . Pedal triangle of point X: The triangle determined by projections of X on sides of triangle ABC. Polar representation of complex number z = x + yi: The representation z = r (cos t ∗ + i sin t ∗ ), where r ∈ [0, ∞) and t ∗ ∈ [0, 2π ). Primitive nth root of unity: An n th root ε ∈ Un such that ε m = 1 for all positive integers m < n. Quadratic equation: The algebraic equation ax 2 + bx + c = 0, a, b, c ∈ C, a = 0. 1 Real product of complex numbers a and b: a · b = (ab + ab). 2 Reflection across a point: The mapping sz 0 : C → C, sz 0 (z) = 2z 0 − z. Reflection across the real axis: The mapping s : C → C, s(z) = z.
Glossary
311
Rotation: The mapping ra : C → C, ra (z) = az, where a is a given complex number. Rotation formula: Suppose that A(a), B(b), C(c) and C is the rotation of B with respect to A by the angle α. Then c = a + (b − a)ε, where ε = cos α + i sin α. Similar triangles: Triangles A1 A2 A3 and B1 B2 B3 of the same orientation are similar if and only if b2 − b1 a2 − a1 = . a3 − a1 b3 − b1 Simson Line: For any point M on the circumcircle of triangle ABC, the projections of M on lines BC, C A, AB are collinear. Translation: The mapping tz 0 : C → C, tz 0 (z) = z + z 0 . Trigonometric identities sin2 x + cos2 x = 1, 1 + cot2 x = csc2 x, tan2 x + 1 = sec2 x; addition and subtraction formulas: sin(a ± b) = sin a cos b ± cos a sin b, cos(a ± b) = cos a cos b ∓ sin a sin b, tan a ± tan b , 1 ∓ tan a tan b cot a cot b ∓ 1 cot(a ± b) = ; cot a ± cot b double-angle formulas: tan(a ± b) =
sin 2a = 2 sin a cos a =
2 tan a , 1 + tan2 a
cos 2a = 2 cos2 a − 1 = 1 − 2 sin2 a = tan 2a =
2 tan a ; 1 − tan2 a
triple-angle formulas: sin 3a = 3 sin a − 4 sin3 a, cos 3a = 4 cos3 a − 3 cos a, tan 3a =
3 tan a − tan3 a a; 1 − 3 tan2
1 − tan2 a , 1 + tan2 a
312
Glossary
half-angle formulas:
1 − cos a a = , 2 2 1 + cos a a , cos2 = 2 2 1 − cos a sin a a = ; tan = 2 sin a 1 + cos a sum-to-product formulas: sin2
a+b a−b cos , 2 2 a+b a−b cos a + cos b = 2 cos cos , 2 2 sin(a + b) ; tan a + tan b = cos a cos b difference-to-product formulas: sin a + sin b = 2 sin
a+b a−b cos , 2 2 a−b a+b cos a − cos b = −2 sin sin , 2 2 sin(a − b) ; tan a − tan b = cos a cos b product-to-sum formulas: sin a − sin b = 2 sin
2 sin a cos b = sin(a + b) + sin(a − b), 2 cos a cos b = cos(a + b) + cos(a − b), 2 sin a sin b = − cos(a + b) + cos(a − b). Vieta’s theorem: Let x1 , x2 , . . . , xn be the roots of polynomial P(x) = an x n + an−1 x n−1 + · · · + a1 x + a0 , where an = 0 and a0 , a1 , . . . , an ∈ C. Let sk be the sum of the products of the xi taken k at a time. Then an−k sk = (−1)k , an that is, an−1 x1 + x2 + · · · + xn = , an an−2 x1 x2 + · · · + xi x j + xn−1 xn = , an ... a0 x1 x2 · · · xn = (−1)n . an
References
[1] Adler, I., A New Look at Geometry, John Day, New York, 1966. [2] Andreescu, T., Andrica, D., 360 Problems for Mathematical Contests, GIL Publishing House, Zal˘au, 2003. [3] Andreescu, T., Enescu, B., Mathematical Treasures, Birkh¨auser, Boston, 2003. [4] Andreescu, T., Feng, Z., Mathematical Olympiads 1998–1999, Problems and Solutions from Around the World, The Mathematical Association of America, 2000. [5] Andreescu, T., Feng, Z., Mathematical Olympiads 1999–2000, Problems and Solutions from Around the World, The Mathematical Association of America, 2002. [6] Andreescu, T., Feng, Z., Lee, G. Jr., Mathematical Olympiads 2000–2001, Problems and Solutions from Around the World, The Mathematical Association of America, 2003. [7] Andreescu, T., Gelca, R., Mathematical Olympiad Challenges, Birkh¨auser, Boston, 2000. [8] Andreescu, T., Kedlaya, K., Mathematical Contests 1996–1997, Olympiads Problems and Solutions from Around the World, American Mathematics Competitions, 1998.
314
References
[9] Andreescu, T., Kedlaya, K., Mathematical Contests 1997–1998, Olympiads Problems and Solutions from Around the World, American Mathematics Competitions, 1999. [10] Andrica, D., Bis¸boac˘a, N., Complex Numbers from A to Z (Romanian), Millenium, Alba Iulia, 2001. [11] Andrica, D., Bogdan, I., A Formula for Areas in Terms of Complex Numbers (Romanian), Revista de Matematic˘a Transylvania, 3(1999), 3–14. [12] Andrica, D., Varga, C., V˘ac˘aret¸u, D., Selected Topics and Problems in Geometry (Romanian), PLUS, Bucharest, 2002. [13] Baptist, Peter, Die Entwicklung der Neueren Dreiecksgeometrie, Wissenschaftsverlag, Mannheim, 1992. [14] Baker, H. F., Principles of Geometry, Vol. 1–3, University Press, Cambridge, 1943. [15] B˘alun˘a, M., Becheanu, M., Romanian Mathematical Competitions, Romanian Mathematical Society, Bucharest, 1997. [16] Becheanu, M., International Mathematical Olympiads 1959–2000. Problems. Solutions. Results, Academic Distribution Center, Freeland, USA, 2001. [17] Berger, M., G´eom´etrie, CEDUC Nathan Paris, 1977–1978. [18] Berger, M. et al., Probl`emes de g´eom´etrie comment´es et redig´es, Paris, 1982. [19] Brˆanzei, D., Notes on Geometry, Paralela 45, Pites¸ti, 1999. [20] Brumfiel, C. E. et al., Geometry, Addison-Wesley, Reading, MA, 1975. [21] Coxeter, H. S. M., Introduction to Geometry, John Willey & Sons, New York, 1969. [22] Coxeter, H. S. M., Greitzer, S. L., Geometry Revisited, Random House, New York, 1967. [23] Deaux, R., Introduction to the Geometry of Complex Numbers, Ungar, New York, 1956. (Deaux, R., Introduction a` la g´eom´etrie des nombres complexes, Brussels, 1947.) [24] Dinc˘a, M., Chirit¸a˘ , M., Complex Numbers in High School Mathematics (Romanian), All Educational, Bucharest, 1996. [25] Engel, A., Problem-Solving Strategies, Springer-Verlag, New York, 1998. [26] Fano, G., Complementi di geometria, Felice Gilli, Turin, 1935.
References
315
[27] Fenn, R., Geometry, Springer-Verlag, New York, 2001. [28] Gleason, A. M., Greenwood, R. E., Kelly, L. M., The William Lowell Putnam Mathematical Competition. Problems and Solutions: 1938–1964. The Mathematical Association of America, 1980. [29] Hahn, L., Complex Numbers & Geometry, The Mathematical Association of America, 1994. [30] Johnson, R. A., Advanced Euclidean Geometry, New York, 1960. [31] Kedlaya, K. S., Poonen, B., Vakil, R., The William Lowell Putnam Mathematical Competition 1985–2000. The Mathematical Association of America, 2002. [32] Kutepov, A., Rubanov, A., Problems in Geometry, MIR, Moscow, 1975. [33] Lalescu, T., La g´eom´etrie du triangle, Librairie Vuibert, Paris, 1937. [34] Lozansky, E., Rousseau, C., Winning Solutions, Springer-Verlag, New York, 1996. [35] Mihalca, D. et al., Quadrilateral Geometry (Romanian), Teora, Bucharest, 1998. [36] Mihalescu, C., The Geometry of Remarkable Elements (Romanian), Editura Tehnic˘a, Bucharest, 1957. [37] Mih˘aileanu, N. N., Using Complex Numbers in Geometry (Romanian), Editura Tehnic˘a, Bucharest, 1968. [38] Modenov, P. S., Problems in Geometry, MIR, Moscow, 1981. [39] Modenov, P. S., Parkhomenko, A. S., Geometric Transformations, Academic Press, New York, 1965. [40] Moisotte, L., 1850 exercices de math´ematique, Bordas, Paris, 1978. √ [41] Nahin, P. J., An Imaginary Tale. The Story of −1 (Romanian), Theta, Bucharest, 2000. [42] Nicula, V., Complex Numbers (Romanian), Scorpion 7, Bucharest, 1999. [43] Pedoe, D., A Course of Geometry for Colleges and Universities, Cambridge University Press, Cambridge, 1970. [44] Pompeiu, D., The Mathematical Works (Romanian), Academiei, Bucharest, 1959. [45] Prasolov, V. V., Problems of Plane Geometry, 2 volumes, Nauka, Moscow, 1986.
316
References
[46] Retali, V., Biggiogero, G., La geometria del triangolo (cap. XXIV din Enciclopedia delle matematiche elementari, vol. II, parte I, Milan, 1937). [47] S˘al˘agean, Gr. S., The Geometry of Complex Plane (Romanian), Promedia-Plus, Cluj-Napoca, 1997. [48] Schwerdtfeger, H., Geometry of Complex Numbers, University of Toronto Press, Toronto, 1962. [49] Sergyeyev, I. N., Foreign Mathematical Olympiads, Nauka, Moscow, 1987. [50] Stanilov, G., Kuchnov, Y., Gjorgjev, V., Vectors and Plane Geometrical Transformations, Narodna Prosveta, Sofia, 1979. [51] Tomescu, I. et al., Problems from High School Mathematical Olympiads (1950– 1990) (Romanian), Editura S¸tiint¸ific˘a, Bucharest, 1992. [52] Tomescu, I. et al., Balkan Mathematical Olympiads 1984–1994 (Romanian), Gil, Zal˘au, 1996. [53] Tonov, I. K., Complex Numbers (Bulgarian), Narodna Prosveta, Sofia, 1979. [54] Yaglom, I. M., Complex Numbers in Geometry, Academic Press, New York, 1968.
Index of Authors
Acu, D.: 5.7.9 Anca, D.: 5.2.9 Andreescu, T.: 1.1.9.16, 2.2.4.8, 5.1.8, 5.1.25, 5.1.28, 5.3.10, 5.5.9, 5.9.10, 5.9.11 Andrei, Gh.: 5.1.12, 5.1.15, 5.1.25 Andrica, D.: 5.1.26, 5.3.9, 5.4.4, 5.5.10, 5.6.11, 5.7.14, 5.7.15, 5.9.9 Becheanu, M.: 5.6.6 Bencze, M.: 5.1.2, 5.1.5, 5.2.2 Berceanu, B.: 5.7.9 Bis¸boac˘a, N.: 5.3.11, 5.7.2, 5.7.4, 5.7.8 Bivolaru, V.: 5.5.12 Burtea, M.: 5.4.30, 5.9.12 Caragea, C.: 5.4.27 Ceteras¸, M.: 5.7.7 Chirit¸a˘ , M.: 5.1.32, 5.3.8
Cipu, M.: 5.9.7 Cocea, C.: 5.3.3, 5.3.6, 5.4.16, 5.7.16 Cucurezeanu, I.: 5.2.5 Culea, D.: 1.1.9.29 Dadarlat, M.: 5.1.4 Diaconu, I.: 5.2.12 Dinc˘a, M.: 5.3.8, 5.4.31 Dospinescu, G.: 5.8.14 Dumitrescu, C.: 5.1.33 Enescu, B.: 5.2.3, 5.9.14 Feher, D.: 5.7.14 Gologan, R.: 5.2.8 Iancu, M.: 5.4.33 Ioan, C.: 5.1.31 Ionescu, P.: 5.6.15
318
Index of Authors
Jinga, D.: 5.1.3, 5.1.6, 5.2.12 Manole, Gh.: 5.1.11 Marinescu, D.: 5.2.13 Miculescu, R.: 4.11.1 Mihet¸, D.: 1.1.8.2, 5.4.7, 5.4.8, 5.4.28, 5.8.15 Mortici, C.: 5.2.9, 5.4.40 M¨uler, G. E.: 5.4.32
Pop, M. L.: 5.1.29 Pop, M. S.: 5.6.8 Pop, V.: 5.4.29 R˘adulescu, S.: 5.1.30 S˘aileanu, I.: 5.4.39 Sz¨or¨os, M.: 5.2.13
Nicula, V.: 3.6.3.1
S¸erb˘anescu, D.: 5.3.5, 5.9.15
Panaitopol, L.: 1.1.7.5, 4.1.2, 5.4.7, 5.4.8, 5.4.28 Pˆars¸e, I.: 5.8.1 Petre, L.: 5.4.25, 5.4.26 Piticari, M.: 5.9.13
T¸ena, M.: 5.6.1, 5.7.3, 5.6.7 Vlaicu, L.: 5.2.11 Zidaru, V.: 5.6.14
Subject Index
absolute barycentric coordinates, 116 additive identity, 2 algebraic representation (form), 5 angle between two circles, 86 antipedal triangle, 136 area, 79 associative law, 2
complex product, 96 Connes, Alain, 155 cross ratio, 66 cyclotomic polynomial, 49
barycenter of set {A1 , . . . , An }, 141 binomial equation, 51 Bretschneider relation, 207
De Moivre, 37 difference, 3 directly oriented triangle, 80 distance, 53 distance function, 53 distributive law, 4
Cauchy–Schwarz inequality, 150 centroid, 58 centroid of the set {A1 , . . . , An }, 141 circumcenter, 91 circumradius, 94 closed segment, 54 commutative law, 2 complex conjugate, 8 complex coordinate, 22 complex plane, 22
Eisenstein’s irreducibility criterion, 47 equation of a circle, 84 equation of a line, 76 equilateral triangles, 70 Euclidean distance, 23 Euler line of a triangle, 108 Euler’s center, 148 Euler’s inequality, 113 Euler, Leonhard, 108 excircles of triangle, 104
320
Subject Index
extended argument, 31 Feuerbach point, 114 Feuerbach, Karl Wilhelm, 114 fundamental invariants, 110 geometric image, 22 Gergonne point, 104 Gergonne, Joseph-Diaz, 104 Hlawka’s inequality, 21 hom*othecy, 158 imaginary axis, 22 imaginary part, 5 imaginary unit, 5 isogonal point, 136 isometry, 153 isotomic, 124 Lagrange’s theorem, 140 Lagrange, Joseph Louis, 141 Lalescu, Traian, 132 Leibniz’s relation, 145 Leibniz, Gottfried Wilhelm, 142 Lemoine point, 104 Lemoine, Emile Michel Hyacinthe, 104 Mathot’s point, 149 median, 58 Menelaus’s theorem, 81 midpoint of segment, 58 modulus, 9 Morley’s theorem, 155 Morley, Frank, 155 multiplicative identity, 3 multiplicative inverse, 3 Nagel line of a triangle, 108
Nagel point, 105 Napoleon, 74 nine-point circle of Euler, 106 n th roots of a complex number, 41 n th roots of unity, 43 n th -cyclotomic polynomial, 49 open ray, 54 open segment, 54 orthocenter, 90 orthopolar triangles, 132 Pappus’s theorem, 191 pedal triangle, 126 polar argument, 29 polar coordinates, 29 polar radius, 29 Pompeiu triangle, 130 Pompeiu’s theorem, 130 Pompeiu, Dimitrie, 130 power of a point, 86 primitive (root), 45 product of z 1 , z 2 , 1 Ptolemy’s inequality, 130 Ptolemy’s theorem, 130 purely imaginary, 5 quadratic equation, 15 quotient, 4 real axis, 22 real multiple of a complex number, 25 real part, 5 real product, 89 reflection in the real axis, 152 reflection in the origin, 152 reflection in the point M0 (z 0 ), 152 regular n-gon, 42 rotation formula, 62
Subject Index
rotation with center, 153
sum of z 1 , z 2 , 1
S-triangles, 132 second Ptolemy’s theorem, 210 set of complex numbers, 2 similar triangles, 68 Simson, Robert, 125 Simson–Wallance line, 125 Spiecker point, 109 Steinhaus, Hugo Dyonizy, 122 subtraction, 3
translation, 151 Tzitzeica’s five-coin problem, 192 Tzitzeica, Gheorghe, 192 unit circle, 29 Van Schouten theorem, 233 von Nagel, Christian Heinrich, 105 Wallance, William, 125
321
