Modenov-Prob...Mir-1981 by Ybalja

rr. c . MOflEi-iOB

3ARAHW nO TEOMETPHH

MOCKBA Â«H A Y K A *

PS. MODENOV

PROBLEMS IN GEOMETRY Translated from the Russian by George Yankovsky

MIR PUBLISHERS â&#x20AC;¢ MOSCOW

First published 1981 Revised from the 1979 Russian edition

H a cmeAUucKOM nabuee

CONTENTS PREFACE

CHAPTER I. VECTOR ALGEBRA Sec. 1. Vectors in the plane (solved p r o b le m s )................................................ Sec. 2. Vectors in space (solved p r o b lem s)........................................................ Sec. 3. Vectors in the plane and in space (problems with hints and a n sw ers)...........................................................................................................

11 14 30

CHAPTER II. ANALYTIC GEOMETRY Sec. 1. Application o f analytic geometry(solved p ro b lem s).......................... Sec. 2. Application of analytic geometry(problems with hints and answers) 1. Plane g e o m e t r y ...................................................................................... 2. Solid g e o m e t r y .......................................................................................

44 66 66 78

CHAPTER III. THE USE OF COMPLEX NUMBERS IN PLANE GEOMETRY Sec. 1. Solved p r o b le m s.......................................................................................... Sec. 2. Problems with hints and a n s w e r s .......................................................

82 253

CHAPTER IV. INVERSION Sec. Sec. Sec. Sec. Sec. Sec.

1. 2. 3. 4. 5. 6.

Iversion defined. Properties of i n v e r s i o n .......................................... Problems involving inversion ................................................................ Mapping of regions under in v e r s io n ................................................... Mechanical inversors: the Peaucelliercell and the Hart cell . . The geometry o f M a s c h e r o n i............................................................... Inversion of s p a c e ....................................................................................

281 285 297 308 309 313

CHAPTER V. BASIC DEFINITIONS, THEOREMS AND FORMULAS Sec. 1. Determinants o f order three Sec. 2. Vector algebra . .

334 373

Contents Sec. 3. Analytic g e o m e t r y ..................................................................................... Sec. 4. Complex numbers .....................................................................................

LIST OF SYMBOLS

347 377

..............................................................................................

387

APPENDIX. LIST OF BASIC FORMULAS FOR REFERENCES

390

B IB LIO G R A PH Y ...............................................................................

394

NAME IN D E X ........................................................................................ 395 SUBJECT IN D E X ....................................................................................396

PREFACE

This text offers certain general methods of solving problems in elemen tary geometry and is designed for teachers of mathematics in secondary schools and also for senior students. The present text includes material that goes beyond the scope of mathe matics curricula for secondary schools (the use of complex numbers in plane geometry, inversion, pencils of circles and others). The book consists of five chapters. The first four chapters deal with the application of vector algebra, analytic geometry, complex numbers and the inversion transformation to geometric problems. Chapter V contains a list of the basic definitions and formulas used in the first four chapters. Before starting a new chapter, the reader is advised to refresh his memory with the appropriate material of Chapter V. Some of the derivations of formulas given in Chapter V are familiar to senior students of secondary school. More detailed theoretical material can be found in the bibliography at the end of the book. I wish here to remark on a supplement to vector algebra that was brought to my attention in 1930 by Professor Ya. S. Dubnov, my teacher at Moscow State University. It is that vector algebra in the plane has not been deve loped to the point that vector algebra in space has, and in order to remedy this situation in an oriented plane it is necessary to introduce the rotation of a vector through an angle of +n/2 (designated [a]) and also a pseudo scalar (or cross) product a x b [or (a, b)] of a vector a by a vector b. Note that the linear vector function Ax of a vector argument x possessing the property that ^ x j_ x for any vector x has the form Ax — X[x](X is an arbitrary number) in the plane, and Ax -- [a, x] (a is an arbitrary vector) in space. The cross product of vectors in the plane and in space may be defined as a polylinear scalar function (of two vectors in the plane and of three vectors in space) which is antisymmetric with respect to any pair of vectors — in the plane we have A(x, y) = —A(y, x);

Preface

in space we have A(x, y,z) = —A (y, x, z), A(x, y, z) = - A (z , y, x), A(x, y, z) = —^(x, z, y) — and is normed (that is, it becomes +1 for some base). This product may be defined as the result of two operations (in the plane and in space) (a, b) = [a] . b, (a, b, c) = [a, b] . c. Although a free vector in geometry constitutes the class of all equiva lent directed line segments, I will permit myself, in this book (in accord ance with a very solid tradition) to identify a vector and a directed line segment as equal (as, for example, in arithmetic, where one regards as equal the fractions pjq and npjnq9 where /?, q, n are natural numbers). For this reason, in this text, two directed line segments that are collinear, have the same length, and are in the same direction will be termed equi valent or equal. The idea of using complex numbers in plane geometry came to me in connection with some very interesting lectures on the theory of analytic functions delivered at Moscow University by Professor A. I. Markushevich, and also with a book on that subject by Markushevich. Also, since the 1940 s, papers have appeared regularly in mathematical journals in many countries illustrating how the use of complex numbers in plane geometry makes for rather simple solutions to complicated problems by relating the solutions to basic geometric transformations that are nor mally studied in secondary school (motion, the similarity transformation, circular transformations, including inversion). A book by R. Deaux [2] appeared in France devoted specially to the problems taken up in Chapter III of this book. Since this methodology is not all represented in Soviet textbooks, I have given detailed explana tions and calculations of the procedures. In this text I have made use of the work of R. Deaux, R. Blanchard, Gourmagschieg, V. Jebeau and others. I believe that the contents of Chapter III is added proof of how much elementary mathematics loses if complex numbers are not brought into the picture. A consideration of the most elementary functions of a complex variable, z

a i± b _ ^ a d _ bc # az + d

zf = az + b(a # 0),

z' = QZ ° {ad — be ^ 0), cz + d z' = az + b(a ^ 0),

embraces the isometric transformations of the first and second kind (zf = az + b, z' = az + b9 where \a\ = 1), similarity transformations of

Preface

the first and second kind (z' = az + b, z' = az + 6, a ^ 0) and circular transformations (the case of a linear fractional function; in particular, the inversion z' = a/z). Chapter IV gives a survery of the properties of inversion of a plane and space and various applications (inversors, the geometry of Mascheroni, and the mapping of regions under inversion). In particular, detailed con sideration is given to various stereographic projections of a sphere onto a plane and the construction of conformal maps of a spectrum of meri dians and parallels of the sphere. The final chapter, Chapter V, contains a list of basic definitions, for mulas and the bibliography. The bibliography contains books in which the reader will find proofs of the formulas used in this text; they include textbooks on vector algebra, analytic geometry, the theory of geometric transformation^, and the theory of functions of a complex variable. The general methods for solving geometric problems described in this text are closely interrelated: it will be recalled that vector algebra is close ly related with analytic geometry. The basic formulas used in Chapter III are derived on the basis of facts taken from analytic geometry; the linear fractional function of a complex variable contains within it the inversion transformation; the inversion transformation can be reliably studied by the methods of analytic geometry, and so forth. I would like to point out that the drawing on the cover of the book (it is the same as that in Fig. 114) is a copy of a photograph of a model that I constructed to illustrate the stereographic projection of a sphere onto a plane under which the parallels and meridians pass into a hyper bolic pencil of circles and an associated elliptical pencil of circles. Figures 107 and 108 were done in the same manner. During the writing of this text I received valuable advice from Professor V. A. Ilin and Corresponding Member of the USSR Academy of Sciences S. V. Yablonsky, to whom I express my deep gratitude. Very profound and valuable advice was obtained from the reviewer of the Nauka Publish ing House; practically all his suggestions were incorporated in the final version of the manuscript. It goes without saying that the general methods of solving elementarygeometry problems given in this text do not exhaust the range of such methods. For instance, mention may be made of a very powerful analytic method for applying trilinear coordinates in the plane, and tetrahedral coordinates in space (the trilinear coordinates of a point on a projective-Euclidean plane are the projective coordinates of proper points of such a plane, provided that all four fundamental points of the projective system

Preface

of coordinates are also proper points; the same goes for space as well). The limited scope of this book did not allow for the inclusion of that method. And there are of course other general methods, which, unforÂ tunately, have not been discussed in our textbooks or teaching literature (for example, synthetic methods of solving problems with the use of isoÂ metric, similarity, affine and projective transformations). However, I am sure that this situation will be remedied in time. P. S. Modenov The present edition was prepared after the author died. The material of the book has been re-examined and brought into accord with generally accepted terminology and notation. A small number of inaccuracies in the Russian edition have been corrected and the bibliography has been expanded. â&#x20AC;˘

Chapter I

VECTOR ALGEBRA Sec. 1. Vectors in the plane (solved problems) Problem 1. Given the angles where M is the midpoint of BC. Solution.

B, C of A ABC. Find /_<p = /^BAM ,

AM \\(AB + AC) and so cos <p

->— > AB(AB + AC) \AB\ \AB + ~AC\

AB2 + A B -A C c yc2 + b2 + 26c cos A c + 6 cos A ]fb2 + c2 + 2be cos A

and since 6 : c = sin B : sin C, it follows that sin C + sin B sin cos ysin2i? + sin2 C + 2 sin 2? sin C cos A Problem 2. Given the angles A, B, C of A ABC. Let M be the midpoint of segment AB, and let Z> be the foot of the bisector of /_C. Find the ratio (CDM): (ABC) and also (p = /_ DCM. Solution. ah + 6a a + b CD = -------- 9 a+ b 2 where a = CB, b == CA. Consequently (<tb + bz, a + b) (C D M )= — (CD, CM) 4(a + b) (b - a) (a, b) 4(a + b) whence (CDN) _ a - b (ABC) ~ 2(a + b)

(a - b) (ABC) 2(a + b)

Problems in Geometry

which can also be written as (CDM) {ABC)

sin A — sin B 2(sin A + sini?)

Furthermore, since CD\\{ab + fca), CA/|j(a + b), it follows that {ab + ba) (a + b) ► s tp = -------------------------| ab + fta | | a + b| _ ~

ab2+ ba2 + ab(a -f- b) cos C Y2a2b2 + 2a2b2 cos C f a 2 + b2 + lab cos7c

(a + b) cos (C/2) _ 1fa2 + b2 + lab cos C

(sin A + sin B) cos (C/2) f sin2 A + sin2 B + 2 sin A sin B cos C

_ {ab + fea, a + b) _ {b — a) (a, b) |ab + 6aj |a + b| ab /2(1 + cosC) fa 2 + fc2 + 2

cos C

(sin B — sin A) sin (C/2) l/sin2 A + sin2 B + 2 sin A sin B cos C Problem 3. Given the interior angles A, B, C of A ABC; M is the mid point of segment BC, N is the foot of the altitude dropped from point C to side AB, and 0 is the point of intersection of the straight lines AM and CN. Find cos cp, where = /_AOC. Solution. Orientate the plane with the base a, b, where a = CB, b = CA. Then C/V | j [a — b]. Indeed, the vector [a — b] = [BA] is perpendicular to the straight line AB and forms acute angles with the vectors a and b since [a — b] a = —[b] a = (a, b) > 0, [a — b] b - [a] b - (a, b) > 0. Furthermore, AM }} — — b

a — 2b.

The desired /_(p is the angle between the vectors AM and CN; consequently, cos c p - f r - bH a ~ 2 b ) _ —2[a]b — [b]a I[a — b]| |a — 2b| a - b| | a - 2b| _ - 2 (a, b) + (a, b) _ (a, b) |a — b| |a — 2b| | a - b| |a — 2b|

Vector Algebra

ab sin C Ya2 + b2 — lab cos C Y&2 + 462 —4ab cos C sin A sin B sin C ]/sin2/4 4-sin2# —2sin A sinB cosC /sin2/t 4-4sin2# —4sim4 sin# cos C Problem 4. Given vectors a = CB, b ™ CA. Find the vector x = CO, where O is the centre of a circle circumscribed about A ABC. Solution. From the relations x2 = (a — x)2 = (b — x)2 we find xa - a2/2, xb = b2/2 and, hence, by the Gibbs formula b2 x = xa •a* + xb •b* = — a* + — b* 2 2 where a*, b* is the reciprocal (or dual) basis of the basis a, b: [b] [a] b* (b, a) (a, b) Thus, b2[a] - a2[b] (», b) Remark. If we take x in the form x = 2a + /*b, then from the relations xa = a2j 2, xb = b2/2 we obtain Xa2 + ju•ab = a2/2, 2-ab +Hb2 = b2l2, whence ^ _ a2b2 — b2• ab _ a2b2 — a2•ab ~ 2(aW - (ab)2)’ ** ~~ 2(a262 - (ab)2) and so x _ 1 a2b2 - b 2- ab & 1 a262 - a 2-ab ~~ 2" a2Z>2 - (ab)2 a T a262 - (ab)2 Problem 5. Given the cross products (a, x) = /?, (b, x) = q of vector x into the noncollinear vectors a and b. Express the vector x in terms of the vectors a, b and the numbers p , q.

Problems in Geometry

Solution. Let . [b] a* = - ......., b* (b, a) be the reciprocal basis of a, b. Then

[a] (a, b)

(xa*)a + (xb*)b = ( x ) a + ( - i- J - , x ) V(b, a) ) V(a, b) ) (b,x) a , x) b (b, a) (a, b) (b, a) (a, b)

p b — qa

~ (aT b T Problem 6. Two forces = {2, 3} and F2 — {4, 1} are specified rela tive to a general Cartesian system of coordinates. Their points of appli cation are, respectively, A = (1,1) and B = (2, 4). Find the coordinates of the resultant and the equation of the straight line / containing it. Solution. The coordinates of the resultant F are 6 and 4. Now let M(x, y) be an arbitrary point of /. Then the moment of the resultant about point M is equal to zero. This moment is equal to the sum of the moments (AL4, Fj) and (MB, F2) of component forces (the cross product of vectors is distributive). Since MA = {1

x, 4 — >’}, it follows that X, 1 — y}, MB - { 2 -\l - x 2 2 —x 4 ( m a , f,) = yg \ (MB, F2) = fg 4 -y 1 |i -y 3 and, hence, the equation of the straight line / is 1- x 2 2 -x 4 0 + 1 ■y 3 1 4 —y 1 or 4x — 6y + 13 0. Sec. 2. Vectors in space (solved problems) Problem 1. The plane angles of a trihedral angle OABC are a = /_BOC, b — /_ COA, c — / AOB. The interior dihedral angles of the given tri hedral angle are: A = B(OA)C, B = C(OB)A, C = A(OC)B* A trihedral angle OA*B*C* that is the reciprocal of the trihedral angle OABC is a trihedral angle constructed in the following manner: ray OA* * The symbol B(OA)C is used to denote a dihedral angle with edge OA, in the half planes o f which are points B and C.

Vector Algebra

is perpendicular to the rays OB and OC and forms an acute angle with ray OA. The rays OB* and OC* are constructed in similar fashion. Let a*, b*, c* be the plane angles of the trihedral angle OA*B*C* and let A*, B*, C*, be its interior dihedral angles. 1°. Knowing, a, b, c, find cos A, cos B, cos C. 2°. Prove that a* — n — A, b* — n — B, c* = n — C. 3°. Prove that A* — n — a, B* = n — b, C* =- 7C— c. 4°. Knowing A, B, C, find cos a, cos b, cos c. 5°. Prove that sin A sin B. sin C A sin a sin b sin c sin a sin b sin c where 1 cos b cos c \ ,/2 cos b 1 cos a I cos c cos a 1 /

A =

— 1 + 2 cos a cos b cos c — cos2a

cos 2b — cos2c

(this relation is called the theorem o f sines for a trihedral angle OABC)l\ 6°. Prove that the sine theorem for the trihedral angle OABC (see item 5°) may be written in the form sin A sin B sin C A* sin a sin b sin c A where A* = ]f\ + 2 cos a* cos b* cos c* — cos2 a* — cos2 b* — cos2 c* = V I — 2 cos A cos B cos C — cos2 A — cos2 B — cos2 C. Solution. 1°. Let eA, ea, ea be the direction vectors of the rays CM, OB, OC (ex t t OA, e2 ft OB, e3tfOC). Then the vectors e1, e2, e3 of the reci procal basis of the basis eA, e2, e3 are the direction vectors of the rays OA*, OB*, OC*. We assume the vectors eA, e2, e3 to be unit vectors and lay them off from point O; then their endpoints Ex, E2, will lie on the ray OA, — »— > —> OB, OC respectively. Through O draw a plane %perpendicular to the ray OC. Let E\ and E2 be orthogonal projections of the points Ex and E2 on the plane n. Then C =- A(OC)B = /_E° OE$. Consider the vectors e? = OEl,

= OE\.

Or for a spherical triangle cut out o f a sphere with centre O by a trihedral angle.

Problems in Geometry

We have e? = e2 + 2e3,

eg = e2 + /*e3.

Forming the scalar product of both sides of each of these relations by the vector e3, we obtain 0 = cos b + X9 0 — cos a + /z. so that eg = ex — e3 cos b, and consequently eV-eS _ cos C = |Cl| ie 2|

eg = e2

e3 cos a

(ex — e3 cos b) (e2 — e3 cos a) V (e i “ e3 COS ^ )2 ]f(e 2 ““ e3 C0S fl) 2

cos c — cos b cos a — cos a cos b + cos b cos a ]f I — 2 cos26 + cos2b ][ 1 — 2 cos2a + cos2 a cos c — cos a cos b sin a sin b In similar fashion we calculate cos A and cos B. Thus, , cos a — cos b cos c COS A = ---------------------- ;----------------- 9 sin b sin c „ cos b — cos c cos a cos B = --------------------------sin c sin a „ cos c — cos a cos b cos C = -------------------------sin a sin b 2°. The formulas obtained in item 1° can be rewritten thus: cos A =

|[eie2]|![e1e3]|

cos B =

|[e2e1]i|[e2e3]| [e3eiHe3e2] cos C = --------------- • l[e3ei]| |[e3e2]j Note that in this notation the vectors el5 e2, e3 need not necessarily be regarded as unit vectors because when e1? e2, e3 are replaced respectively by Xel9 ^e2, ve3, where X > 0, it > 0, v > 0, the right-hand members of these relations remain unchanged. Thus, el5 e2, e3 may be regarded as —

> —

any direction vectors of the rays OA, OB, OC.

Vector Algebra

These last relations can be rewritten as q2• ^3■^1 gl •e2 cos = -------------5 cos B = — —■:— j cos C = -------------|e2| |e3| |e3|(e1! le1)je2|

(a)

Indeed, [e3ei] e2-e3

[eie2] e,e„e. teie2] e!e2e3

®1®2®3 |Jf? eiL i c1e2e3

|e2| |e3|

[eiealteiej!] = cos A ]t©ie3] | |[e1e2]|

and similarly for the other two formulas. Note also the formulas for cos a*, cos b*9 cos c*: cos a* =

(e3, e1) (e2, e3) cos b* = *21 |e3| |e*| |e,>| : |e*| |eA|

cos c’

(e1, e2) — •

(b>

From formulas (a) and (b) we conclude that cos A = — cos a*9 cos B = — cos b*9 cos C = — cos c* and since all the angles A , B, C, a*, fc*, c* lie in the interval (0, n)9it follows that a* = n — A, b* = n — B9 c* = 7i — C. These relations can also be derived geometrically. 3°. Writing down the formulas obtained above, a2 • a

. p i

p i ■p 2

cos A = — -------, cos B= --------------, cos C= -----------|e2| |e3j !e3j (e1! le1!|e2| we have for the trihedral angle OA*B*C* Qjb

A ik

63*6]

1 * ^ 2

cos A* = ------------ , cos B* = ------ — cos C* = ------, |e2| |e3| |e3| le^ |ex| ;e2| and since 'ej = |e2| = je3; = 1, it follows that cos A* = — cos a9 cos B* = — cos b9 cos C* = — cos c9 whence A* = n — a,

B* = n — b9

C* = n — c.

4°. Applying the formulas of item 1° to the trihedral angle OA*B*C*> we obtain cos a* — cos b* cos c* cos A* = sin b* sin c* 2 -8 1 0

Problems in Geometry

or cos (n — a) =

cos(7r — A) — cos (n — B) cos (n — C) sin (7r — B) sin (k — C)

or cos ^4 + cos 5 cos C sin 1? sin C The formulas for cos b and cos c are derived in similar fashion. Thus cos A + cos B cos C cos a sin B sin C cos 5 + cos C cos cos b = sin C sin A cos C + cos A cos 2? cos c = sin A sin 5°. From the formula cos a — cos b cos c cos A = sin b sin c we get cos a =

sin A = VI — cos2^4 __ ^1 + 2 cos a cos &cos c — cos2a —cos2b—cos2c sin b sin c whence sin A A sin a sin b sin c sin a sin B . sin C have the same value. Thus The ratios — and sin c sin b sin C A sin B sin A sin b sin c sin a sin b sin c sin a

sin b sin c

6°. Let us write down the sine theorem for the trihedral angle OA*B*C*, which is the reciprocal of the trihedral angle OABC: sin A* sin B* sin C* A* sin c* sin a*sin b* sin c* sin a* sin b* or (see items 2° and 3°) sin b sin a sin A sin B

sin c sin C

sin A sin B sin C

Vector Algebra

where A*

1 cos b* cos c* cos b* cos a* i cos c* cos a* 1

1/2

y 1 + 2 cos a* cos b* cos c* — cos2 a* — cos2 b* - cos2 c* = ][ 1 — 2 cos A cos B cos C — cos2 >4 — cos2 2? — cos2 C. From the equations sin A -------

A ------------------9

sin a sin a sin b sin c sin a __ A* sin A sin A sin B sin C we obtain the following by thermwise division: sin2 A __ A sin A sin B sin C A sin3/4 sin2fl A* sin a sin b sin c A* sin3 a whence sin A _ A* sin a A Similarly, sin B __ sin C __ A* sin b sin c A To summarize, sin A sin B sin C __ A* sin a sin b sin c A Problem 2. Given, in a parallelepiped, the lengths a, b, c of its edges OA, OB, OC and the plane angles between them: /_BOC = a, Z COA = 0, Z AOB = y. 1°. Find the length d of the diagonal OD of the parallelepiped. 2°. Find cosines of the angles <pl5 q>2, cp3 formed by the diagonal OD and the edges OA, OB, OC. 3°. If the coordinates u, v, w of the projections of OD on the axes OA, —> OB, OC are also given, then prove that d = y au + bv + cw.

Problems in Geometry

4°. Express the length d of the diagonal OD in terms of the angles a, p, y and the coordinates u, v, w of the projections of the directed line —> — >— >— > segment OD on the axes OA, OB, OC. 5°. Express the volume V of the given parallelepiped in terms of a, b, c, a, P, y. 6°. Two rays OP and OQ emanate from point O; the first ray forms — >— >— > with the axes OA, OB, OC the angles a,, Pl9 yx\ the second ray forms the angles a2, p2, y2. Express the cosine of the angle 6 between the rays OP and OQ in terms of c l , p, y, otl9 pl9 yl9 cc2, p2, y2. 7°. Find the shortest distance between the straight lines OD and AB if a, b, c, c l, P, y are given. 8°. Find the distance 3 between point D and straight line AB. Solution. 1°. Consider the vectors a = OA, b = OB, c — OC, d = OD. Then d= a + b + c whence d == /d* = /(a + b + c)2 — ][a2 + b2 + c2 | 2be cos a | 2ca cos ji + lab cos y. 2°. ad a(a + b + c) cos <px = la! |d| Ia| |d| a + b cos y + c cos p a2 + ab cos y + ac cos P ad d The formulas for cos <p2 and cos cpz are derived in similar fashion. Thus a + b cos y + c cos p cos <p1 = -------------73 a cos y + b + c cos a cos <p2 = ----------------------------- r, d

a cos P + b cos a + c cos cp3 = ------- ---------------------d 3°. From the formulas of item 2° we have u = d cos cp1 = a + b cos y + c cos p, v = d cos q>2 = a cos y + b + c cos c l, w = d cos cpz = a cos P + b cos a + c.

Vector Algebra

Multiplying both sides of each of these formulas by a, b, c respectively and adding the resulting formulas termwise, we get au + bv + cw = a(a + b cos y + c cos ft) + b(a cos y + b + c cos a) + c(a cos P + b cos a + c) = d2. 4°. From the formulas of item 2° it follows that a + b cos y + c cos P = w, a cos y + ft + c cos a — v, a cos P + 6 cos a + c = w. Solving this system for a, 6, c, we obtain 0

u cos y cos P v 1 cos a » \v cos a 1 ! 1 u cos P 1 i ~ cos y v cos a * o \ 0 t I cos p w 1 1 cos y w cos y 1 v 9 cos P cos a w

where 5=

1 cos y cos P cos y 1 cos a cos P cos a 1 = ^ 1 + 2 cos a cos P cos y — cos2 a — cos2 p

cos* y,

and, consequently, on the basis of the formula d2 =r=au + bv + cw we have u cos y cos P |j 1 u cos p 1 cos y u ll v 1 cos a + v j cos y v cos a + w cos y 1 v w cos a 1 cos P cos a w ! COS P w 1 1 cos y cos p u cos y 1 cos a v ~S cos P cos a 1 w u v w 0

Problems in Geometry

so that

a2 5°. F2= (a ,b , c)2 = ba ca a2 ab cos y ac cos p

ab ac b2 be cb c2 06 cos y ac cos p | b2 cos a 1= be cos a c2 |

1/2

cos cos y 1 cos a cos P cos a 1 1

0 0Crt

cos y cos /? u \ 1 cos y 1 cos a t? cos P cos a 1 w v w 0 / u

and so V = abc ]f 6 = abc f 1 + 2 cos a cos P cos y — cos2 a — cos2 p —cos2 y. 6°. Let us consider the vectors p = OP, q = OQ, assuming that Ipl = Iql = L

Expand the vector p in terms of the basis a, b, c:

P = Pi* +PJ> +Pz c. Taking the scalar product of both sides of this equation into a, b, c in succession, we obtain cos <*1 = Pi + p2 cos y — p3 cos P, cos & = Pi cos y + P 2 + P 2 cos a, cos yi = pi cos P + p2 cos a + p2. whence cos a2 cos y cos P ! cos Pi 1 cos a » cos cos a 1 1 cos oti cos P cos y cos Pi cos a 5 cos P cos yx 1 1 cos y cos <Xi cos y 1 cos Pi • cos P cos a cos y2

Vector Algebra

Furthermore, let a*, b*, c* be the reciprocal basis of a, b, c. Expand the vector q in terms of the basis a*, b*, c*: q = qxa* + q2b* + q2c*. Taking the scalar product of both sides of this relation by a, b, c in succes sion, we obtain qx = cos a 2, q2 = cos /J2, qz = cos y2. We get q = cos a2a* + cos P>2b* -|- cos y2c*. We now find cos(p, q) - cos 0 = pq - (pxa + p 2b + pzc) foa* + q2b* + q3c*)

—PlVl + + Pdfi!§ I cos clx cos y cos ft cos ax cos P cos a21cos px 1 cos a + cos p 2 cos y cos px cos a cos P cos yt 1 | cos yx cos a 1

1 cos y cos ax + cos y2 cos y 1 cos px cos P cos a cos yx

1 cos y cos P cos ax cos y 1 cos a cos px cos P cos a 1 cos yr cbs a cos p2 cos y2 0

The formula for cos 0 can be written more compactly: 1 cos y cos P cos olx cos y 1 cos a cos px _ q cos p cos a 1 cos yi cos a2 cos p2cos y2 cos 0 7°. The shortest distance d between two noncollinear straight lines in space is equal to the length of the projection of any line segment MXM29 whose ends lie on the given lines, onto the common perpendicular to the given straight lines. Thus, in item 7° M1 = 0, A/o = By M1M2 =

= b.

Now, the direction of the common perpendicular to the straight lines OD and AB is given by the vector product of the vector a + b + c = OD into the vector b — a = AB: [a + b + c, b — a] = 2[a, b] + [c, b] + [a, c].

Problems in Geometry

Thus, d=

|b(2[a, b] + [c, b] + [a, c])l _ JaLb, c| |2[a,b] + [c,b] + [a,c]|

where (see item 5°) |a, b, c| = abc /<5, T = 4[a, b]2 + [c, b]2 + [a, c]2 + 4([a, b]-[c, b]) + 2([c, b]-[a, c]) + 4([a, b] •[a, c]) = 4a2b2 sin2 y + c2b2 sin2 a + a2c2 sin2 P + 4b2ac(cos P — cos a cos 7) + 4a2bc(cos a — cos P cos y) —2c2ab(cos y — cos a cos /?). VS And so d = abc —— • \t 8°. Suppose e is the direction vector of the straight line / and B is an arbitrary point in space. Take some point A on I and let AB = a. Then the distance d from point B to / is found from the formula d = - |[a, e]|/|e|. Indeed, |[a, e]| - |a| |e| sin cp |e| d. In particular, if e is the unit vector, then d = |[a, e]|. In determining the distance from point D to the straight line AB, note that the direction vector of that line is equal to a — b, and since AD = = b + c, it follows, using the formula

that j

l[(» ~ b), (b + c)]| _ | [a, b] + [a, c] + [c, b]j _ Kj j la - b| la - b| ]fT2

where 7\ = a2b2 sin2y + b2c2 sin2a + c2a2 sin2P + 2a2bc(cos a — cos p cos y) + 2b2ca (cos p — cos y cos a) — 2c2ab (cos y — cos a cos /?), T2 = a2 + b2 — 2ab cos y. Problem 3. Given the plane angles /_ BOC = a, /_ COA = b, /_ AOB= — c of the trihedral angle OABC. The ray / emanates from point O and — >— >— > forms with the edges OA, OB, OC of the given trihedral angle the equal angles cp. Find tan cp.

Vector Algebra

Solution. On the edges of the given trihedral angle, choose points A, B, C so that OA = OB = OC = 1 and then consider the unit vectors a = OA, b = OB, c = OC. Let OP be a directed line segment of length 1 such that /_ POA — = /_ POB = Then

POC = <p. We now consider the unit vector x = OP.

|xa = xb — xc| — cos <p and, by the Gibbs formula, x = (xa) •a* + (xb) •b* + (xc) •c* = (a* + b* + c*) cos <p where a*, b*, c* is the reciprocal (or dual) basis of a, b, c; that is,

(*)

tb>C l , b* K* = lC’ ®] , c* [“’ bl • -----(a, b, c) (a, b, c) (a, b, c) Raising both sides of (*) to a scalar square, we obtain 1 = cos2 <p(a* + b* + c*)2 whence „* = a’

1 + tanV = a*2 + b*2 + c*2 + 2b*c* + 2c*a* + 2a*b* =

where Tx = sin2# + sin2b + sin2c + 2(cos b cos c — cos a) + 2(cos c cos a — cos b) + 2(cos a cos b — cos c), T2 = (a, b, c)2 = 1 + 2 cos a cos b cos c — cos2a — cos2b — cos2c. Thus Ti - T2 T, Vdn2<p = 1 t2 t2 Tx — T2 —- sin2a + sin2fe + sin2c + 2(cos b cos c — cos a) + 2(cos c cos a — cos b) + 2(cos a cos b — cos c) — 1 — 2 cos a cos b cos c + cos2# + cos2b + cos2c = 2[1 —(cos a + cos b + cos c) + (cos b cos c + cos c cos a + cos acosb) — cos a cos b cos c] = 2(1 — cos a) (1 — cos b) (1 — cos c) = 16 sin2 — sin2 — sin2 — >

tan2 q> =

Tx - T 2

16 sin2 — sin2 — sin2 — 1 cos b cos c cos b 1 cos a cos c cos a 1

Problems in Geometry

The angle q> may be regarded as an acute angle (because if q> is obtuse the direction of the line segment OP may be reversed). Thus A. a . b . c 4 sm — sin — sm — 2 2 2 tan cp = ------------- --------------where 1 cos b cos c A = cos b 1 cos a cos c cos a 1 = 1 + 2 cos a cos b cos c — cos2a — cos2& — cos2c. Problem 4. Given in a tetrahedron OABC the lengths of the edges OA = a, OB = b, OC = c and the plane angles / BOC = a, /_ COA = fi9 /_AOB = y. Let PQ be the common perpendicular to the straight lines OA and BC (point P lies online OA, point Q on line BC). Find the ratios J )P _ = x OA

Jsp

Solution. Consider the vectors a = OA, b = OB, c = OC, p

OP,

q — OQ, t = 5 0 , s = P 0 . Then p = 2a, t = fi(c — b). Furthermore, p + s —t —b = 0 whence s = — 2a + p(c — b) + b. Since the vector s is perpendicular to the vectors a and b — c, we have as = 0, (b — c) s = 0 a(— 2a + p(c — b) + b) = 0, (b - c) ( - 2a + /i(c - b) + b) = 0 or ka2 — /ia(c — b) = ab, — 2a(c — b) + ju(c — b)2 = — b(c — b).

Vector Algebra

From the resulting system of equations that are linear in A and /i, we obtain these unknowns: ft cos y(c2 + ft2 — 2be cos a)—(c cos P—b cos y)(c cos a —ft) a c2 sin2/? + ft2 sin2y — 2ftc(cos a — cos P cos y)

— c cos a + ft + cos y(c cos P — b cos y) c2 sin2/? + ft2 sin2y — 2ftc(cos a — cos /? cos y)

Problem 5. We give here a vectorial derivation of the basic formulas used in the theory of axonometry (axonometric projections). Suppose Oxy Oy9Oz are three pairwise perpendicular axes. Let us consider the plane % that intersects these axes in the points A, B9C respectively. We assume the space to be oriented by the ordered triple of axes Ox, Oy9Oz. Denote by O' the orthogonal projection of point O on the plane n — ABC9 by al9 a2, a3 the angles of the axes Ox, Oy9Oz with the plane n9 and by Pl9 P2 >Pa the respective angles BO'C, CO'A, AO'B. Prove that 1°. cos /?i = — tan a2 tan a3, cos /?2 = — tan a3 tan ctl9 cos /?3 = — tan olx tan a2. T

sin p9 _________ 1_______ sin Pt __= ___sin p2 _ sin axcos ax sin a2cos a2 sin a3 cos a3 cos ax 00s a2 cos a3

sin 2Pi _ sin 2/?2 sin 2/?3 _ ^ s^n ai s*n a2 s*n a3 cos2 ax cos2 a2 cos2a3 cos2 ax cos2 a2 cos2 a3 4°. There exists a triangle, the lengths of whose sides are proportional to cos2ax, cos2a2, cos2a3; one of these triangles is a triangle formed by the feet of the altitudes of the triangle ABC (SchlomilcKs theorem). Solution. 1°. OB OC = {OO' + O'B) (OO' + O'C) = OO'2 + O'B-O'C - 0

<*>

since 00'_L O'C, OO' _L O'B, OB J_ OC; consequently, from the relation (*) we obtain O'B tan a2- O'C tan a3 -f O'B• O'C cos /?x = 0 whence cos Pi = — tan a2 tan a3. The formulas cos /?2 = — tan a3 tan ax, cos /?3 = — tan ax tan a2

Porblems in Geometry

are derived in similar fashion. 2°. [ M , OC] = [ 0 0 + OB, OO + ~OC] = [O'O, OC] + [OB, O'O) + [OB, OC] whence [O'B, O C] OOf = [OB, 0 C \-0 0 '.

(**)

Furthermore, to 7!, trc j-O O ' - | [ ^ , (TC]\ 0 0 ' = 0 'i? •0 'C sin

•0 0 ' = 0 Z? cos a2•OC cos a3•0 0 ' sin Pl9

[0B,^C] 0 0 ' = Oi!'OC’0 0 ' c o s | - -

= OB OC-OO' sin «!

and, hence, by virtue of (**) we have sin & cos a2 cos a3 = sin ax. Consequently, sin Pi _ sin a2 cos ax

1 cos ax cos a2 cos a3

In similar fashion, proof is given that sin p2 __ sin j?3 1 sin a2cos a2 sin a3 cos a3 cos oq cos a2 cos a3 3°. The solution follows from items 1° and 2°. 4°. Rewriting the relations of item 3° in the form cos2 oq cos2 a2 sin 2 ^ ----~ -j

sin 2

--j

cos2 a3

sin 2 |j8 3 —

and noting that 2 ( f t - f ) + 2 ( f c - f ) + 2 ( f c - | ) = «, we conclude that there exists a triangle A with angles 2 | Indeed>

0i > Jt/2, 02 > jr/2,

> n/2

Pi ~ n ~2

)’

Vector Algebra

since all plane angles of the trihedral angle OABC are equal to n/2 and therefore ft, ft, f t are obtuse angles: a sphere with diameter AB passes through the point O {/_ AOB = 7i/2), the plane ABC intersects the sphere along a great circle since AB is the diameter of the sphere, the projection O' of point O on the plane ABC will be inside that circle and, hence, the angle AO'B = f t is obtuse (similarly n/2 < f t < n, n/2 < f t < n). From these relations it follows that cos 2ax, cos2a2, cos 2a3are proportional to the sines of the angles of the triangle A, and so also to the lengths of its sides. Finally, we will now prove that one of these triangles is formed by the feet of the altitudes of AABC. Indeed, ABC is an acute-angled triangle because the lengths of its sides AB = ]jOA2 + OB2, BC = ]f(?B2 + O C \ CA = ]fOC2 + OA2 and, hence, for example, AB2 + BC2 > AC2 (angle B is acute) and so forth. Furthermore, since OA _L OBC, it follows that OA J_ BC and, hence, on the basis of the theorem of three perpendiculars AO' ± BC and, simi larly, BO' _L CA, CO' 1 AB, that is O' is the point of intersection of the altitudes of A A B C . Let Al9 Bx, Cx be the feet of the altitudes of A ABC. What we then have is that since / ACxO' — /_ ABxO' = n/2, it follows that points A, Cx, O', Bx lie on one circle (with diameter AO'). From this it follows that /_ CxftO' — /_ CxAO' (both angles are intercepted by the arc CjO'). But / C ^ O ' = 71 - B and so Z C1B1Or = % - B. 2 Similarly, the points Bx, O', Ax, C lie on one circle,, / AiftO ' = /_ AxCO'== n = ------ B. Thus, BxO' is the bisector of the interior angle Bx of A A ^ C ^ 2 Now, from A 0 'B 1C1 we have

and since Ax + Bx + Ci — n it follows that

Similarly,

Problem 6. On the base of the polyhedron OACBPQ lies a rectangle OACB. The edge PQ is parallel to the plane of the rectangle, and the

Problems in Geometry

orthogonal projections Px and Qx of points P and Q on the plane OABC are such that PxO = PXA = QXB = QXC (this kind of polyhedron is called a wedge). Given the lengths of the sides of the rectangle OACB: OA — ay OB — b, the length c of segment PQ and the altitude h = PPX= QQX. Compute the cosine of the dihedral angle A(CQ) B = <p. Solution. Introduce a rectangular Cartesian system of coordinates Oxyz> choosing the directions of the a- and j ’-axes along OA and OB respectively. The coordinates of the vertices of the wedge in the chosen system are: O = (0,0,0), A = (a, 0, 0), C = (a, b, 0), B = (0, by0),

(X T -" )From this we find the vectors C A ^ { 0 , - b , 0}ft{0, —1,0} = x, CQ = | - -

• C~ h , AJ t t {

a, c — b, 2 h} = y,

{— a , 0 , 0 } | | { — 1, 0 , 0 } = *.

We also find \c - b 2 h\ j2A - a i - 1 0 | 10 0 fi c — b 2 h {Hi - a [y, z] = i 0 o ! ’ jo - 1 and, consequently, we have [y, x]-[y,z] cos (p = Y4h2 l[y, *]| l[y. z]| [y, x] =

"", H “'

1 a c — b\ 1 . 1 0 J = *0, ~ 2h’

a(c — b) + a2 )[4fi£ + (c — b)2

Sec. 3. Vectors in the plane and in space (problems with hints and answers) 1. Given two vectors CA = b, CB — a ^ 0. Let P be the orthogonal projection of point A on the straight line BC. Find the vector CP. Answer. < ^ - a .

Vector Algebra

2. Four points A, B ,C ,D are located (in the plane or in space) so that AD ± BC, BD ± CA. Prove that CD _L AB. Hint. Assuming DA — rl5 DB = r2, DC = r3, we find — r3) = 0,. r2(r3 — r2) = 0, whence r f a — r2) = 0. 3. An arbitrary point O is joined to the centroid G of A ABC and a point P is constructed such that OP = 3OG. Let A', B', C be points symmetric to the point P about the points A, B, C. Prove that O is the centre of gravity of the four points A', Bf, C', P. 4. Let ABC be an arbitrary triangle. We consider 12 vectors whose initial points are the centre of an inscribed circle and the centres of escribed circles, and the end points are the points of contact of circles and the sides of the triangle. Prove that the sum of these vectors is equal to the vector OH, where O is the centre of the circle (ABC), and H is the orthocentre of the triangle ABC. 5. ABC is an arbitrary nondegenerate triangle lying on an oriented plane; P is an arbitrary point lying in the plane of the triangle. The vectors; '—>— >— > PA, PB, PC are coplanar and, hence, are linearly dependent: otPA + fiPB + yPC = 0 (at least one of the numbers a, /?, y is different from zero). Prove that a: fi: y = (PBC):(PCA): (PAB). Hint. Project the zero vector ctPA + fiPB + yPC onto the straight line BC in the direction of the straight line PA. Let Px be the point into which are projected the points P and A. Then PP^B + yP^C - 0 . 6 . Let / be the centre of a circle (/) inscribed in a triangle ABC; let D, E, F be the points of contact of the circle (I) with the sides BC, CA, AB; let P be an arbitrary point lying in the plane of A ABC; let Pa, Ph, Pc be orthogonal projections of point P on the sides BC, CA, AB. Prove that the circle passing through the centroids of the triangles EPaFy

FPbD, DPCE has a diameter equal to

IP.

Hint. Let G be the centroid of the triangle DEF, and let Ga, Gb, Gc be the centroids of the triangles EPaF, FPhD, DPCE; then /? = j

(7 i+ 7 ? + 7 ? ) =

(IE + JF + ID + DPJ == IG

DP.

Problems in Geometry

Thus, the point Ga is constructed in the following manner: from point G lay off a directed line segment equal to one third of the projection IP on the side BC. From this it immediately follows that the points Ga, Gh, Gc lie on the circle whose diameter is obtained by laying off from point G a vector determined by the directed segment * IP. 7. ABC is an arbitrary triangle; A l 9 B l 9 C1 are the midpoints of its sides BC, CA, AB; A2, B2, C2 are the feet of its altitudes; O is the centre of the circle (ABC); G is the centroid of A ABC (the point of intersection of its medians); H is the orthocentre (the point of intersection of the altitudes of A ABC); A 3 ,B 3, C3 are the midpoints of the segments AH,BH, CH; A a,B a,C 4 are the second points of intersection of the circle (ABC) with the altitudes/!//, BH, CH (Fig. 1). Prove that: 1°. The points O, G, H are collinear and OG —- — GH. 2 2°. The points A l 9 B l 9 Cl 9 A 29 B2, C2, A 3 , B3 , C3 lie on one circle called the Euler circle of A ABC or the nine-point circle (0 9) of A ABC. 3°. The points Aa, B4, C4 are symmetric to the orthocentre H with respect to the straight lines BC, CA, AB. Hint. This problem may be solved vectorially. First prove that if O is the centre of the circle (O) circumscribed about A ABC, then OA 4+ OB 4 OC — OH. (The vectorial solution is left to the reader to carry out; below we give another proof of all these propositions by resorting to complex numbers.) For the present, note the following simple synthetic solution. 1°. In the case of a homothetic transformation ( G, —

, the points

A, B, C go into the points A l9 Bl 9 C1 and, hence, the altitudes of the triangle ABC (which are now regarded as straight lines) go into the midperpendic ulars of the sides BC, CA, AB (the midperpendicular of a line segment is a straight line perpendicular to the line segment at its midpoint). From this it follows that the point H of intersection of the altitudes of A ABC goes into the point O of intersection of the midperpendiculars of its sides. Thus, in the case of a homothetic transformation

, point H goes

into point O, which means the points G, H, and O are collinear and OG = GH/2. 2°. In the case of a homothetic transformation goes into the circle (A ^ C x ) and, hence, the centre of the circle ( A ^ C j) is the image of the centre O of the circle (ABC) that is, the midpoint 0 9

Vector Algebra

C«

Fig. 1

of segment OH. Since 0 9 is the midpoint of segment OH, and the points O and H are projected respectively onto the side BC into the points Ax and A2, it follows that the point 0 9 is projected into the midpoint of the line segment AXA 2 and, hence, 0 '9AX= 0'9 A2, where 0 9 is the projection of 0 9 on the straight line BC. But if the projections of inclined lines are equal so also are equal the inclined lines themselves, hence, 0 9 AX= 0 9 A2. Similarly, proof is given that 0 9 BX= 0 9 B2, 0 9 CX= 0 9 C2 and since 0 9 AX— 0 9 BX= 0 9 CX [0 9 is the centre of the circle (AXBXCX)], it follows that the circle (AXBXCx) also passes through the points A2, B2, C2. Note that the radius of the circle (AXBXCX) is half the radius R of the circle (ABC). Now let us consider the homothetic transformation

Under

this transformation, the point O goes into the point 0 9, and the circle (ABC) goes into a circle with centre 0 9 and radius Rj2, that is, into the circle (AxBxCj). But under the homothetic transformation

—- j , the points

A, B, C go into the points A3, B3, C3 and since the points A, B, C lie on the circle (ABC), it follows that the points A3, B3, C3 lie on the Euler circle (AXBXCX) = (0 9). 3°. Finally, let us consider the homothetic transformation (H, 2). Under this transformation, the circle (AXBXCX) goes into a circle with centre O and radius R, that is, into the circle (ABC). On the other hand, under the transformation (H, 2), the points A2, B2, C2go into the points Ax, Biy C4, which are symmetric to the point H with respect to the straight lines BC, CA, AB. But the points A2, B2, C2 lie on the Euler circle and, hence, the points Ait B4, C4 lie on the circle (ABC). ii hl()

Problems in Geometry

8. Suppose H, G, O, and 0 9 are, respectively, the orthocentre of A ABC, its centroid the centre of the circumscribed circle (ABC), and the centre of the Euler circle; let A l 9 Bu Cx be the midpoints of the sides BC, CA, AB; let AfI, Bn, Cn be the feet of the altitudes; let A', B', C be points symmetric to the vertices A, B, C of A ABC with respect to its sides BC, CA, AB; let A ", B", C" be points in which the altitudes AH, BH, CH intersect the circumscribed circle (ABC); let O' be the centre of the circle (A'B'C') = = (O'); let a, /?, y be the projections of the point 0 9 on the sides BC, CA, AB of A ABC. Then: 1°. The triangle A'B'C' is an image of the triangle afiy under the homothetic transformation (G, 4).

2°. 4 0 9a = AA", 40 9fi = BB", 40 9y = CC ". 3°. For the circumscribed circle (ABC) to be tangent with the altitude dropped from A onto BC, it is necessary and sufficient that the centre 0 9 of the Euler circle lie on the side BC. 4°. Let 0 9 be a point symmetric to the point 0 9 with respect to the centre co of the circle (oc/fy); then the centres O and O' of (ABC) and (A'B'C') are symmetric with respect to the point 0 9. 5°. If the centre O' of the circle (A'B'C') lies inside A ABC, then it is a point that has the property that the shortest distances from O' to each vertex A, B, C with a preliminary shift along the straight line to the opposite side are equal. 6 °. If the point O' lies on the side BC, then the point lies on the.4 circle (A'B'C'). Hint. 1°. 4G0 9 = GH, 4 0 9x = 2(HAU + OAx) = HA" + AH = = HA" + A" A' = HA'. 2°. 4 0 9a = HA' = AA". 3°. This is a consequence of item 2°. 4°. On the basis of item 1° we have Geo = — GO' and besides, 4

but we also have

5°. If P is an arbitrary point lying within A ABC, then, denoting by A2, B2, C2 the points of intersection of the straight lines PA', PB', PC' with the sides BC, CA, AB, we conclude that the lengths of the polygonal lines PA2A = PA', PB2B = PB', PC2C = PC' will be the shortest routes from point P to the vertices A, B, C with preliminary displacements to

Vector Algebra

the opposite sides. For these minimal distances to be equal, it is necessary and sufficient that PA' = PB' = PC', that is, that the point P coincide with the point O'. 6° This is a consequence of item 5°. 9. Given three noncoplanar vectors OA = a, OB = b, OC = c. Let S be the centre of a sphere passing through the points 0 9 A, By C. Find the vector OS = x. a2 [b, c] + b2[c, a] + c2[a, b] Answer, x = 2(a, b, c) 10. Given the vectors OA = a, OB = b, OC = c. The vectors b and c are noncollinear. Let H be the orthogonal projection of point A on the plane OBC. Find the vector OH = h. Answer, h = a —

-?)_ [b, c]. [b, c? 11. Given four vectors d.

OA = a,

It is also given that the vectors a, b, c are noncoplanar and that the straight line OD intersects the plane ABC at some point M. Find the vector OM = m. (a, b, c) , Answer. ------------------------------------- d. (d, b, c) + (d, c, a) + (d, a, b) 12. Given three vectors OA

The vectors a and b are noncollinear. Let H be the orthogonal projection of the point C on the plane OAB. Find the vector CH = x. Answer, x = ----— [a, bl. [a, b f 13. Find the vector x if three noncoplanar vectors a, b, c and their scalar products into x are known: xa = p, Answer.

xb = q,

p[b, c] + q[c, a] + r[a, b] (a, b, c)

xc = r.

Problems in Geometry

14. Find the vectors x and y if we know their sum x + y = a, the scalar product (x, y) = p and the vector product [x, y] = b. Answer. If a 4 < 4(b2 + pa2), then there are no solutions. If a 4 = = 4(b2 + pa2), then there is one solution: x

a , [a,b] — t, y a2

[a, b] a2

If a4 > 4(b2 + pa2), then there are two solutions: _ X l~

a2 + Ya* — 4(b2 +pa2) , [a, b] 2a 2 *+

_ yi “

a2 — ]/a4— 4(b2 +pa2) _ 2a2 “

[a, b] c f~

a2 — j/a4 — 4(b2 +pa2) [a, b] 2a2 8 ~ ~ a2 a2 + 1[a* — 4(b2 -\-pa2) n , [a, b] a * 15. ABCD is an arbitrary tetrahedron located in oriented space; P is —

> —

an arbitrary point. The vectors PA, PB, PC, PD are linearly dependent: ocPA+PPB + yPC + 8 PD = 0 (at least one of the numbers a, /?, y, 8 is different from zero). Prove that a: P: y: 5 = (PBCD): (APCD): (ABPD): (ABCP). Hint. Project the zero vector olPA + fiPB + yPC + 8 PD on the straight line AB by planes parallel to the plane PCD. The points P, C, D are pro jected into the single point Pj and we obtain aP±A + PP±B = 0, and so on. 16. Given a trihedral angle OABC such that among its plane angles BOC, COA, AOB there is not more than one angle equal to n/2. Prove vectorially that the three straight lines that pass through the vertex O, lie in the planes of the faces BOC, COA, AOB, and are perpendicular to the edges OA, OB, OC, respectively, lie in one plane. Hint. If a, b, c are the direction vectors of the edges, then the direction vectors of the indicated straight lines are [a, [b, c]], [b, [c, a]], [c, [a, b]]. Their sum is zero. 17. Prove vectorially that if all edges of the trihedral angle OABC, all plane angles of which are right angles, are cut by a plane that does not pass through its vertex O, then the point of intersection of the altitudes

Vector Algebra

of the triangle obtained in the section coincides with the projection of the point O on the plane of that triangle. 18. ABCD is an arbitrary tetrahedron: A', B', C', D' are the orthogonal projections of its vertices on the planes of the opposite faces. Given the radius vectors DA = r„ DB = r2, DC = r3. 1°. Find the radius vectors DA' = t[, DB' = r2, DC' = r^, DD' = and prove that if the straight lines AA' and BB' lie in one plane, then AB J_ CD, and conversely. 2°. If it is also given that AC — AD = BC = BD, then prove that the straight lines AA' and BB' lie in one plane. Let H be the point of intersection of these straight lines, and let K be the point of intersection of the straight lines CC' and DD'. Prove that the plane AHB intersects the segment CD at its midpoint / and the plane CKD intersects the segment AB at its mid point J. Prove that the point H is located on the straight line IJ. 19. Given, in a pyramid OABC, the length of the edge OA = a and the plane angles /_ BOC = a, /_ COA = P, /_ AOB = y. A sphere (5) is tangent to the face BOC at the point O and passes through the point A. Find its radius x. Solution. Consider the vector x = OS. Since the vector x is perpendic ular to the plane BOC, it follows that x = 2[b, c], where b = OB, c = OC. From the equality OS = SA, we find x2 = (a — x)2, where a = OA, whence xa = a2 /2. Substituting 2[b, c] for x in this equation, we obtain 2(a, b, c) = a2 j 2 , whence X =

2(a, b, c)

and, consequently, OS = x =

= a2 [b, c] . The length x of the vector x is equal to the radius R (of the 2(a, b, c) sphere S): a2 \[bc]\ a2bc sin a R \x\ = x = 2|(a,b,c)| 2|(a, b, c)| a2bc sin a 2abc 1^1 + 2 cos a cos p cos y — cos2a —cos2/? —cos2y 'finis, R —-

a sin a . .v.— ' = 2 ]/1 + 2 cos a cos /? cos y — cos2a — cos2/? — cos2 y ■

■ - ----- --- —:— ■

•

20. Given, in a tetrahedron OABC, the lengths of the edges OA = a, OB — b, OC — c and the plane angles at the vertex O: /_ BOC = a,

Problems in Geometry

/_ COA = P, /_ AOB = y. Find the radius x of the sphere (5) circum scribed about the given pyramid. Solution. Let x = OS, a = OA, b = OB, c = OC; then x2 = (x — a)2 = (x — b)2 = (x — c)2 whence xa = a2 j 2 , xb = b2 j 2 , xc = c2/ 2 . By the Gibbs formula, x = xa-a* + xb-b* + xc-c*, where a*, b*, c* is the reciprocal triple of the triple a, b, c, we find x = ^ a* + ^ b* + ^ c * = * » . . P t e j L + c2ta>■»] 2 2 ‘ 2 2(a, b, c) 2(a, b, c) 2(a, b, c) and, hence, „ _ ife _ V V

c] + Z>2[c, a] + c2[a, b]}2 _ 21(a, b, c)|

ff 2 j(a, b, c)| ’

where T = a \ b, c]2 + b*[c, a]2 + c4[a, b]2 + 2a 262[b, c]*[c, a] + 262c2[c, a]*[a, b] + 2cW[a, b]-[b, c] = axb2 c2 sin2a + bxc2 c? sin2/? + c*a2 b2 sin2y + 2a2 b2(bc cos a- ac cos ft — c2ab cos y) + 2b2 c2(ca cos /?• ba cos y — a2bc cos a) + 2c2 a2{ab cos y • cb cos a — b2ca cos /?) = <fib2 c2 sin2a -f b*c2 a2 sin2/? + c*a2 b2 sin2y + 2 a2 b2 c2(cos a cos /? — cos y) + 2 b2 c2 a2(cos p cos y — cos cl\ + 2c2 a2 b2(cos y cos a — cos P) = a2 b2 c2 {a2 sin2a + b2 sin2/? + c2 sin2y + 2a&(cos a cos P — cos y) + 26c(cos P cos y — cos a) + 2ca(cos y cos a — cos /?)} = a2 b2 c2 M, |0 »>b. c)| = abc ][ 1 + 2 cos a cos P cos y — cos2a — cos2/? — cos2y = abc • d. To summarize: 1fM X~ 2A 21. Given the plane angles a — Z_ BOC, b = Z. COA, c = /_ AOB of a trihedral angle OABC and the interior dihedral angles A, B ,C that

Vector Algebra

are respectively opposite the plane angles. In item 6 ° of example 1, Sec. 2, proof was given that sin a sin b sin c A -------- - = --------- = ---------= ------9 sin A sin B sin C A* where A is the volume of the tetrahedron constructed on the unit vectors OA, OB, OC, that is, A = (1 + 2 cos a cos b cos c — cos2a — cos26 — cos2 c) 1/2 and A* = (1 — 2 cos A cos B cos C — cos2A — cos2B — cos2C)1/2 Consider the following special cases: 1°. a = A, b = B, c = C. Prove that in this case at least one of the angles A, B, C is equal to n/2 {Hint: A 2 — A * 2 = 0); furthermore, two edges are perpendicular to a third; two dihedral angles of the trihedral angle OABC are equal to n/2, and the third one is arbitrary. 2°. a — A, b = n — B, c = n — C. The reasoning is the same as in item 1°. 3°. a = A, b = B, c = n — C. Prove that in this case cos a cos b cos c and cos A cos B cos C have opposite signs. The plane angle c is found from the relation cos A cos B cos c ----------------------1 + sin A sin B 4°. a = n — A, b = n — B, c — n — C. Piove that the edges of the trihedral angle OABC are pairwise perpendicular. 22. ABCD is an arbitrary tetrahedron. Let

1 °.

x - [DB, DC], y = [DC, DA], z = [DA, DB], t = [AC, AB]. Prove that x + y + z + t = 0.

Derive from this fact that with any tetrahedron ABCD it is possible to associate three spatial quadrangles whose sides are perpendicular to the faces of the tetrahedron and the lengths of the sides of each of the spatial quadrangles are proportional to the areas of the faces of the tetrahedron. Let lia, hb, hc, hd be the altitudes of the tetrahedron ABCD, let (AB), (AC), (AD), (BC), (BD), (CD) be the interior dihedral angles of the tetra hedron, which angles are adjacent to the edges AB, AC, AD, BC, BD, CD. Prove that 0O 1 cos (CD) cos (DB) cos (BC) y

K ~ K ~ ” /T ' 1 J ____ 2 cos{CD) _ hi h% hahb hi

2 cos(AB)

hchd

Problems in Geometry

4°.

2 V«

cos(/L5)

«A? ’ hi J cos(AD)

cos (BC)

cos(2?Z>)

cos (CD) hahh

hA h jic hA KK hji* + - 1 cos (CD) cos (BD) cos (BC) 5°. cos (CD) - 1 cos (AD) cos (AC) = 0 . cos (AB) cos (BD) cos (AD) -1 -1 cos (BC) cos (AC) cos (AB) Hint. Form the scalar product of the equation x + y + z + t = 0 b y x ; write down the equation x + y + z + t = 0 as x + y = — z — t and square both sides; square the equation x + y + z + t = 0 ; eliminate h in the equations obtained in item 1°. 23. Prove that the six planes passing through the midpoints of the edges of the tetrahedron ABCD and perpendicular to the opposite edges of the tetrahedron pass through the single point M (Mongers point). Prove that the Monge point is symmetric to the centre O of the sphere (O) = (ABCD) with respect to the centroid G of the tetrahedron ABCD. Hint. Let OA = rl 9 OB = r2, OC = r3, OD = r4. Then the equations of the planes indicated in the statement of the problem may be written thus: + r, + r , + r 4Lj(r< l ,_ ~ _r*), =_ A / ^ y (/J = 1; 2, 3,4). 24. Given a tetrahedron ABCD and a point M. Let the straight line passing through M parallel to the straight line AB intersect the faces CD A and CDB at the points P and Q. Prove that the sum of the scalar products M P • MQy the terms of which are made up as indicated for all six edges of the tetrahedron ABCD, is equal to the power of the point M with respect to the sphere (ABCD). 25. Given a tetrahedron ABCD and a point M. Prove that the sum of the powers of the vertex A with respect to the spheres with diameters MC, MB 9 MD is equal to the sum of the squares of the lengths of all edges of the tetrahedron ABCD (and similarly for the vertices B, C, D). 26. Find the tetrahedron if we know the areas of its faces and also that the altitudes intersect in a single point (such a tetrahedron is said to be orthocentric). Solution. Lagrange investigated this problem analytically (J. L. Lagrange Mem. Acad. Berlin, 1773, p. 160; OEuvres, t. Ill, p. 662) and formulated it slightly differently: “Find a tetrahedron of greatest volume for which the areas of all its faces are given”. From the formulas he obtained it follows that the desired tetrahedron is an ortho centric tetrahedron, as was also pointed out by

Vector Algebra

Serret (P. Serret.—J. de Liouville, 1862, p. 377), who proved this fact geometrically (without, however, going into the definition of such a tetra hedron). In connection with this problem, Lagrange obtained a fourthdegree equation and proved that it has at least one positive root; he com puted the lengths of the edges of the tetrahedron as functions of that root, but did not consider the conditions under which a tetrahedron with such lengths of edges exists. Finally, an English mathematician Iyenger (Iyenger, The Mathematics Student, 1947, p. 104) solved the same problem by reduc ing it to a fourth-degree equation; he carried out a full investigation and proved that for such a tetrahedron to exist it is necessary and sufficient that the sum of the areas of any three faces be greater than the area of the fourth face. We now give the solution proposed by Marmion (A. Marmion, Mathesis, 1953, p. 69; problem No. 3253 proposed by V. Th6bault). As we know, the sum of four vectors x, y, z, t perpendicular to the edges of the tetrahedron ABCD and directed outwards (the lengths of the vectors are equal to the areas of the faces of the tetrahedron) is equal to zero (see example 22). From this it follows that there exists a closed spatial quadrangle A 1 B1 C1 D1 whose sides are such that AXBX = x, B1 C1 = y, C1 D1 = z, D1 A 1 = t and, consequently, the sum of the areas of any three faces of the tetrahedron ABCD exceeds the area of the fourth face. Putting DA = a, DB = b, DC = c, we conclude that for the vectors x, y, z we can take the following vectors: [b, c], y = [c, a], z = ^ -[a , b]. 2 2 2 From the vectors x, y, z, t we can generate 6 spatial quadrangles of the type A 1 B1 C1 D1; all tetrahedrons A 1 B 1 C 1 D1 will have equal volume V1: x=

y , = 4 “ (x >y> z) = — ([b, C], [c. a], [a, b]) 6 48 = ^ 48 ( a>b>c)2 = i48( 6 F)2 = i4 - v s •

From this it follows that the tetrahedrons ABCD and A 1 B 1 C1 D1 assume the maximum value simultaneously. On the other hand, the volume of any tetrahedron is equal to 2/3 of the product of the areas of two of its faces into the sine of the dihedral angle <p between them divided by the length of the edge of that dihedral angle, and so Tr 2 area A ^A C V area A A ^ Q . V1 = --------------- -- - ----------------—-—- sin cp, 3 AXCX where cp is the size of the dihedral angle with edge AiCi. Since the factor in front of sin cp does not depend on cp, it follows that Vx is a maximum when cp = n/2. Similar reasoning carried out with respect to the other

Problems in Geometry

faces A1B1D1 and C1B1D1 leads to the fact that the dihedral angle formed by these faces must also be a right angle. But since A±BXLDBC, BXCX1.ACD, C A 1 BDA, DXAX 1 CAB it follows that AB 1 A&Dto BC _L B& Ato CD 1 C ^ A , DA 1 and, consequently, AB _L CD, BC 1 AD, that is, the tetrahedron ABCD of maximum volume for which the areas of its faces are given is ortho centric. The coefficient of sin <p in the expression for Vx depends solely on one variable AXCX= A. Applying the Heron formula for the areas of A AxByCx and AAxDxCx and noting that AXBX= x, B1C1 = y, CXDX= z, DXAX = t, we find that the square of the expression area A ^ A C ra re a A A1C1Dl A& multiplied by 28 is equal to m 2 ) = [V - ( x + y f] [A2 - (x - y f] [A2 - (t + z f ][A2 - (t - zf] {A) A2 Putting A2 = n, we find the following expression for the logarithmic deri vative of the function F(p): fM . = 1 F(n) V -(x + y f

II — (x — y f 1

n — (t + z f

n - (t - z f

Now let (x — y)2 < (t — z)2 < (x + j ')2 ^ (I + z f since it must be true that (t — z f < AXC\ = H, (x + y f > AXC\ = n (only under this condition do triangles A1B1C1and A±DXCXexist with lengths of sides y, x, A and t, z, A). The derivative F \n) has only one root in the interval ((t — z)2, (x + y f ) and therefore there is only one value of A for which the function F (A 2) attains an extremum, a maximum. Knowing A, we can construct A Ai^ C j and A A ^ D ^ Putting one against the other so that the dihedral angle B1{A1C-j)D1 is equal to n/2, we construct the tetra hedron (and there is only one such tetrahedron to within isometry) that cor responds to the tetrahedron ABCD with greatest volume. Furthermore, by virtue of the relations AB 1 AXCXDX, BC JL B & A ^

CD 1

DA 1 D1B1Cl

Vector Algebra

and AB + BC + CD + DA = 0, we conclude that AB, BC, CD, DA must be proportional to the areas of the faces A ^ D ^ A & D ^ B j C ^ , C1D1B1 so that, denoting proporÂ tionality factor by k , we have AB = k iA & D J , BC = k iA & D J , CD = k iB & A J , DA = k iC & B J , where, for instance, (A ^ C ^) is the nonoriented area of face A ^ C ^ and on the basis of the foregoing, we have 3 27 V = â&#x20AC;&#x201D; k*Vf = -----k? V4, 4 64 whence we find the proportionality factor k = _ i_ = _ 4 _ 3V )[3Fi The lengths of the edges AB, BC, CD, DA have been determined and they must be perpendicular to the planes A ^ D ^ A ^ D ^ B ^ C ^ , C x D ^ respectively. To summarize: up to isometric transformations (motions), there is only one tetrahedron that satisfies the statement of the problem.

CHAPTER II

ANALYTIC GEOMETRY Sec. 1. Application of analytic geometry (solved problems) Problem 1. The sides BC, CA, AB of a triangle ABC are divided by the points P, Q, R in the ratios PP __ ^

= v* RB

Find (PQR) (ABC) Solution. We introduce a general Cartesian coordinate system in the plane, putting C = (0, 0), A = (1, 0), B = (0, 1). Then P = ( 0, - L - V I l+ X ) sequently, =

(PQR) {ABC)

= (-J L -;

U+Ai

(A ,

)

— — 1+ X

n 1+ n 1

— v1+ V 1+ V

- - - - - )

* =

ll+v

and,

1 + vj

con-

1 + Xp v (1 + A) (1 + //) (1 + v)

Corollary. For the points P, Q, R, lying on the sides BC, CA, AB of A ABC, to be collinear, it is necessary and sufficient that the following equa tion hold: BP _C Q _AR__

BP CQ AR

PC QA

CP AQ

(Menelaus’ theorem). Problem 2. A triangle ABC is inscribed in a circle. Prove that the points P, Q, R of intersection of tangents to the circle at the points A, B, C are respectively collinear with the sides BC, CA, AB (this is special case o f the Brianchon theorem).

Analytic Geometry

Proof‘ Note that all the ratios CQ P= — * PC are negative. Furthermore, BP _ sin C sin B consequently, PP PC

AR PP

PC P^4

sin P sin C

sin2 C sin2 P

6*'

Z>2 Similarly, /* = — —, v = ----- and so An v = — 1 (see problem 1). c2 fl2 Problem 3. Three circles with centres A, B, C and radii Rl9 R2, P 3 are located so that each of them lies outside the other two. Let P, Q, R be the centres of exterior similitude of these circles taken in pairs. Prove that the points P, Q, R are collinear. D D

_ Ri *2 Proof A = —- = __ ^ _ ^3 v - AR h V Tb PC QA whence Afi v = —1, Problem 4. The straight lines AAl9 BBl9 CCX belong to one pencil (in particular, they pass through one point 0). Prove that the points P, Q, R of intersection of the corresponding sides PC and P 1C1, CA and CxAl9 AB and AXBX of the triangles ABC and AXBXCX are collinear ( Desargues’ theorem). Solution. Consider A OBC and the transversal BXPCX (the transversal o f a triangle is any straight line lying in the plane of the triangle). On the basis of Menelaus’ theorem we have OBx BP CCX — >— > — > BXB PC CxO Similarly, OCx CQ AAX — > — >— > = - l , CXCQA AxO — >— > — > OAx AR BBX AXA RB BxO

Problems in Geometry

Multiplying these equalities term by term, we get ■>— >— > BP CQ AR — >— > — > PC QA RB It is now left to the reader to prove Desargues’ theorem for the case where AAX, BBl9 CCx belong to one and the same improper pencil, that is to say, are parallel. — > — >— > Problem 5.1°. The sides BC, CA, AB of A ABC are divided in the ratios

PC QA RB Suppose the straight lines BQ and CR intersect in the point Al9 the straigh lines CR and AP in the point Bl9 and the straight lines AP and BQ in the point Ci. Find the ratio

*— (ABC) 2°. Consider the special case A = p = v = 2. 3°. Under what necessary and sufficient condition do the straight lines AP, BQ, CR belong to the same proper pencil? 4°. Under what necessary and sufficient condition do the straight lines AP, BQ, CR belong to the same improper pencil? Solution. 1°. Introduce a general Cartesian coordinate system and set C = (0, 0), A = (\, 0), B = (0, 1). Then

The equations of the straight lines AP, BQ, CR are: x y 1 = 0 or jc + (1 + X)y — 1 = 0 (straight line B1C1 or AP) 0 ------1 +A

= 0 or (1 + /i) x + py — n — 0 (straight line CLA±or BQ), 1 + li

—- — — V— 1 = 0 or vx — y = 0 (straight line AXBX or CR) 1+ v 1+ v 0 0 1

Analytic Geometry

whence 1 + A —1 2

1 1 +A*

—J“ 0 V -1 (A 1 B1 C1) _ (.ABC) (1 -J- [i -f- f i \ ) (1 -f- v -f- A v) (1

A -f- /iA)

= _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ H v - I ) 2_ _ _ _ _ _ _ _ _ _ _ _ _ (1 + n + fiv) (1 + v + Av) (1 + A + /«A)

Remark. By solving the systems of the straight lines BxCl9 CxAl9 AXBU we could have found the vertices of the triangle AXB^CX\ vti 1 + n + nv 1 + JJ +/iV 1 1 + v + vA 1 + A/z 1 -f- A -f- AjU 1 +

V v + vA, 1 A + A/i

and then \i

(A1 B1 C1) {ABC)

fiv 1 + /I + p v

1 V 1 + V+ Av A^i 1 1 -j- A -f- A /i (1 + A + A /i) (1 + n + /tv) (1 + v + vA)

By subtracting from the last column the sum of the first two and simpli fying, we obtain the very same result. 2°. For A = n = v = 2, we find — - - 1— = — • (ABC) 1 3°. The straight lines AP, BQ, CR belong to a single proper pencil if and only if A/z v = 1,1 + A+ A/z ^ 0 (by multiplying both sides of the in equality 1 + A + X\x # 0 by v ^ 0, we obtain 1 + v + vA # 0; then, multiplying by n ^ 0 , we obtain 1 + ^ + / i v ^ 0). 4°. The straight lines AP, jB<2, CR belong to the same improper pencil, if and only if A// v = 1, 1 + A + A^ = 0 (in this case, the lines CR are pairwise parallel, and there are no coincident lines). Combining items 3° and 4°, we obtain a theorem: if the sides BC, CA, AB of A arc divided in the ratios

Problems in Geometry

then the straight lines AP, BQ, CR belong to a single pencil (proper or im proper) if and only if X p v = 1 ( Ceva's theorem). Problem 6 . Prove that if a circle inscribed in A ABC is tangent to the sides BC, CA, AB at the points, respectively, P, Q, jR, then the straight lines AP, BQ, CR intersect in one point. Solution. X

p -b

P -a p -b

(p = (a + b + c)/2, a, b, c are the lengths of the sides of A ABC), whence X p v = \, 1 + X + X p ^ O . Problem 7. Two distinct points A and B are fixed in a plane. Also fixed is a real number k. Find the locus of points M for each of which M A2 + MB2 = k. Solution. We introduce a rectangular Cartesian system of coordinates. Let A = (xl5 y j , B = (x2, y 2), M = (x, y). Using the formula for the distance between two points in a plane, we get (x - xxy + ( y - J ,)2 + (x - x 2)2 + ( y - y 2)2 = k or

or MO 2 = — _ — A B \ 2 4 where O is the midpoint of segment AB. From the last relation, it follows that if k < AB2/2, then the specified locus is empty. If k = AB2j2, then the specified locus consists of the single point O, the midpoint of AB. If k > AB2! 2, then the given locus of points M is a circle with centre O and radius ]f2k — AB2/2. Problem 8 . Three points A, B, C are fixed in a plane. Also fixed is a real number k. Find the locus of points M for each of which M A2 + MB2 + MC2 = k. Solution. Introduce in the plane of A ABC a rectangular Cartesian coor dinate system. In this system, let A = (xl5 Ji), B = (x2, y 2), C = (x3, j 3), M = (x, y).

Analytic Geometry

Applying the formula for the distance between two points, we obtain (x - Xj)2 + o - yyf + (x - x 2f + ( y - y 2 ) 2 + (x - x3)2 + ( y - y 3f = k or yi + y 2 + j 3 j 2j

[(x3 - x 2)2 + (y3 — jo )2 + (xa - x3)2 + (Ji - Js )2 + (*2 - xi)2 + (J 2 - J 1)2] = k.

Since Xl

^ 2 are the coordinates of G, the 3 3 point of intersection of the medians of the triangle ABC, and X2

0 3 — Xo)2 + ( j 3 — j 2)2 = a2, (xi - x3)2 + (yx — j 3)2 = b \

(x 2 - Xi)2 + 0 2 - >’l)2 = c2, where 0 , ft, c are the lengths of the sides BC, C>4, >45 of the triangle, it follows that the last relation may be rewritten thus: MG 2 = — [3k - (a2 + b2 + c2)]. From this it follows that if k < (a2 + ft2 + c2 )j3, then the desired locus of points M is empty. If k = (<a2 + ft2 + c2)/3 then the locus of points M consists of the single point G, which is the point of intersection of the me dians of the triangle ABC. If k > (a2 + ft2 + c2)/3, then the locus is a circle with centre G and radius j/3k — {a2 + ft2 + c2)/3. Problem 9. Fixed in the plane are two distinct points Cx and C2 and also a real number k . Find the locus of points Af for which MC\ - MCI = k. Solution. Let us introduce a rectangular Cartesian coordinate system and take C±C2 for the x-axis and the midpoint O of segment CXC2 for the co ordinate origin. In that coordinate system, let Ci = (—a, 0), C2 = (a, 0), M = (x, y). Then the given relation takes the form k (x + a) 2 + y 2 — (x — a) 2 — y 2 = k or x = — 9 4 a which is the equation of a straight line perpendicular to CXC2 (the x-axis). 4-810

Problems in Geometry

Problem 10. The radical axis of two nonconcentric circles (Cl9 r±) and (C2, r2) is the locus of points M for each of which its powers with respect to the circles are equal. Set up the equation of the radical axis of the cir cles (Cj, r2) and (C2? r2) assuming CXC2 to be the ;c-axis and taking for the coordinate origin O the midpoint of the line segment CXC2. Solution. In our coordinate system, C2= ( —a, 0), C2=(a, 0), M = (x, y). The relation or MCI - r\ = MCI - r\ or MCI - MCI = r\ - r\ takes the form (x + a)2 + y2 — (x — a f — y2 = r\ — r\ or o

x = n ^ l. 4 a This is a straight line perpendicular to the straight line C1C2 (also see the preceding problem). Remark. Consider the construction of a radical axis of two circles depend ing on their different mutual positions. (1) The circles (Cx) and (C2) intersect. Their radical axis is [the straight line / that passes through the points P and Q (their points of intersection). Indeed, the powers of both points P and Q with respect to (C2) and (C2) are zero and hence the points P and Q belong to the radical axis of those circles; therefore the radical axis itself is the straight line PQ. (2) The circles (C2) and (C2) are located so that each lies outside the other. Draw one of four straight lines tangent to the given circles; Pk, Qk (k = 1, 2, 3, 4) are points of contact; Pk Qk are segments of the tangents. Let the point M k bisect the segment Pk Qk. Then the point M k lies on the radical axis of the given circles. Indeed, M k Pi = M k Q% = k 9 where k is the power of the point M k with respect to both circles. Thus, the mid points of the four segments Pk Qk of the common tangents to the circles (Cx) and (C2) bounded by the points of tangency Pk and Qk lie on the ra dical axis of the circles. To construct the radical axis, it suffices to construct two of the points Ml9 M 2, M3, A/ 4 or even one (and then drop from it a perpendicular to the line of centres CXC2). Remark. If the nonconcentric circles (Cj) and (C2) do not intersect, their radical axis does not have any points in common with either of them. Indeed, suppose that the radical axis / of the circles (Cx) and (C2) has a

Analytic Geometry

common point A with circle (Cj). Then the power of point A with respect to the circle (Cx) is equal to zero, and since point A lies on the radical axis of (Cj) and (C2), it follows that the power of A with respect to the circle (C2) is also zero and therefore A also lies on the circle (C2). This is a contradic tion: the circles (Cx) and (C2) have a common point A , which runs counter to the assumption. It is left to the reader to prove a stronger statement: if each of the cir cles (Cj), (C2) lies outside the other, then (Cj) and (C2) lie on different sides of their radical axis, and if one of the circles (Q), (C2) lies inside the other (see below), then (Cx) and (C2) lie on one side of their radical axis [in all instances we are dealing with two nonconcentric circles (Cx) and (C2)]. (3) The circle (Cx) is a zero circle (point Cx) and lies outside the circle (C2). Construct the tangent lines CxTx and C2T2 drawn from the point Cx to the circle (C2) (7\ and T2 are the points of tangency). Suppose Mx and M 2 are the midpoints of segments CXTX and C2T2; the straight line MXM2 is the radical axis of the circles (C2) and (CV). One could construct a single tangent line CxTly the radical axis is a straight line passing through the midpoint M x of segment CXTXperpendicularly to the straight line CXC2. (4) The circles (Q) and (C2) are zero circles. Their radical axis is the per pendicular bisector of the line segment CXC2. (5) The circles (Cx) and (C2) are nonzero and are tangent to one another externally or internally. Their radical axis is the tangent line / at their common point T. Indeed, the powers of any point M lying on the tangent line / with respect to (CA) and (C2) are equal to M T2, which means they are equal. (6) The zero circle (Cj) “lies” on the nonzero circle (C2). The radical axis of these circles is the tangent line to the circle (C2) at the point Cv Indeed, the point Cx belongs to the radical axis since its powers with respect to (C2) and (C2) are zero, and, besides, the radical axis is perpen dicular to the straight line CXC2. (7) The circle (Cj) is a zero circle, the circle (C2) is nonzero and the point Cj lies inside (C2); also the circles are nonconcentric (Cx ^ C2). Let us construct some circle (C) tangent to the straight line CXC2 at the point Cx and such that it intersects (C2) at the points A and B. Then the straight line AB intersects the straight line CXC2 at the point P lying on the radical axis of the circles (C2) and (C2). Indeed, the power of the point P with re spect to the circle (C2) is equal to PA-PB. But this product is equal to PC\> that is, to the power of the point P with respect to the zero circle (Cx). The radical axis of the circles (Cx) and (C2) is a straight line passing through P and perpendicular to the straight line CXC2. (8) The nonconcentric circles (Cx) and (C2) are nonzero and the circle (C,) lies inside the circle (C2). Let us construct some circle (C) which inter sects (C2) at the points A and B, and the circle (C2) at the points A ' and B'. Let P be the point of intersection of the straight lines AB and A'Bf. Then P belongs to the radical axis of the circles (Q) and (C2), since its powers

Problems in Geometry

c = (PA) • (PB) and o' = (PA') • (PB') with respect to the circles (Ci) and (C2) are equal: (PA) • (PB) = (PA') • (PB'); this is the power of the point P with respect to the circle (C). The radical axis / of the circles (Q) and (C2) is a straight line passing through the point P perpendicularly to the straight line CXC2. Problem 11. Fixed in a plane are two distinct points A and B and a positive number k not equal to unity. 1°. Prove that the locus of points M for which the following equation holds, MA , -----= k , MB is a circle (Cfc)*, the centre Ck of which lies on the straight line AB. Intro duce a rectangular Cartesian coordinate system taking for the origin the midpoint O of AB, and for the x-axis the straight line AB. On the x-axis find the coordinate x k of the centre Ck of the circle (Ck) if we know the coordinates —a and a of the points A and B (we assume a > 0, that is, the positive direction of the coordinate axis is chosen from A to B). Also find the radius R k of the circle (Ck). 2°. Prove that the centre Ck of the circle (Ck) is an external point of the line segment AB. 3°. Prove that for any k (0 < k ^ 1) the circle (Ck) does not have com mon points with the perpendicular bisector (or midperpendicular) s of A B : for 0 < k < 1, the circle (Ck) and the point A lie to one side of the midperpendicular ^ of AB [and, consequently, B and the circle (Ck) lie on different sides of s], and for k > 1, the point A and the circle (Ck) lie on different sides of s [and, hence, B and the circle (C*) lie on the same side of s]. 4°. Prove that if 0 < k < 1, then point A lies inside the circle (Ck) and point B outside it. If k > 1, then, conversely, point A lies outside (Ck) and B lies inside (Ck). 5°. Prove that the circles (Ckl) and (Ck2) (0 < # 1, 0 < fc2 # 1) are symmetric with respect to the midperpendicular s of segment AB if and only if kxk 2 = 1. 6 °. Let Pk and Qk be points of intersection of the circle (Ck) with the straight line AB. Prove that the ordered quadruplet of points A, B, Pk, Qk is harmonic, that is, APk. AQk K

o i

* A circle thus defined is called a circle o f Apollonius. The points A and B are termed limit points , or Poncelet points o f the circles (Ck) (if k varies from 0 to + o o , with the exception o f k = 1, then we obtain a family o f circles (Ck); the case of k = 1 is clearly associated with the straight line: the midperpendicular o f points A and B).

Analytic Geometry

7°. Prove that if S is an arbitrary point of the midperpendicular s of AB9 and T is the point of contact of any tangent (any one of two) drawn from point S to the circle (Ck), then S T = SA = SB. In other words, the midperpendicular s of AB is the radical axis of any one of the circles (Ck) (0 < k ^ 1), including the zero circles A and B. 8°. Prove that any one of the circles (S) passing through points A and B intersects all circles (Ck) (0 < k ^ 1) orthogonally, that is, the tan gents to the circles (S) and (Ck) at any one of two points of their intersec tion are mutually perpendicular. The set (Cfc) of all circles of Apollonius given by MA where k takes on all positive values (this set includes the midperpendicular of AB as well) is termed a hyperbolic pencil o f circles. Points A and B are called the limit points of the pencil, or Poncelet points. The set of all circles passing through points A and B (this set includes the straight line AB as well) is called an elliptic pencil o f circles which is conjugate to the hyperbolic pencil. The points A and B are termed base points of the elliptic pencil. Remark. A hyperbolic pencil of circles may be defined in a variety of ways (all the definitions given below are equivalent). (1) A hyperbolic pencil r of circles is the image of a family of all con centric circles (0) under inversion. Here, the pencil of straight lines with centre O is transformed into an elliptic pencil E of circles which is conjugate to the hyperbolic pencil r . (2) A hyperbolic pencil r of circles is a set of all circles with a common radical axis (that is, a set such that the radical axes of any two circles coin side); in this case the radical axis does not intersect the circles. (3) A hyperbolic pencil r of circles is a set of all circles (PQ) with dia meter PQ, where the points P and Q are harmonic conjugates of the limit points A and B (any two distinct points may be taken for A and B). The set of all circles passing through A and B then form an elliptic pencil E of circles conjugate to the hyperbolic pencil r. (4) A hyperbolic pencil r of circles is a stereographic projection (see Chapter IV) from an arbitrary points of the sphere of the set of all circles of the sphere, the planes of which are perpendicular to any diameter N S of the sphere. In this case, the projections A and B of points A and S are limit points of the pencil r , and the stereographic projections of the set of great circles of the sphere that pass through the points N and S form an elliptic pencil E of circles conjugate to the pencil r . (5) A hyperbolic pencil r of circles with limit points A (—a90) and B (a9 0) is a family of circles given by the equation A / + pep = 0 ,

Problems in Geometry

where / = x2 + y2 — 2px + a2 = 0, <p = x2 + y 2 — 2qx + a2 = 0,

p ^ q

are any two circles of the family (at least one of the numbers A and p is nonzero). (6) A hyperbolic pencil r of circles is a set of circles specified by the equation x2 + y 2 — 2px + 1 = 0 , where p is a parameter that takes on all values that do not exceed 1 in abso lute value. Then the conjugate elliptic pencil is given by the equation x 2 + ^2 _ 2qx — 1 = 0 , where the parameter q takes on all real values; the fundamental points of the pencils r and E are (± 1,0). There are other definitions as well. 9°. Let 0 < k x < k 2 < 1. Prove that the circle (Ckl) is inside the circle (Cki). But if 1 < kx < k2, then the circle (Cki) is inside the circle (Ckl). 10°. Prove the validity of the following method of constructing a circle of Apollonius: given Poncelet points A and B and one of the points Md of the circle of Apollonius not lying either on the straight line AB or on the midperpendicular of the segment AB; at the point M0 draw a tangent to the circle (ABM0). If C is the point of intersection of that tangent with AB, then the circle of Apollonius passing through M0 is a circle with centre C and radius CM0. Consider the case where the point M0lies on the straight line AB but is distinct from the points A, B and the midpoint O of AB. 11 °. Find the locus of points M for each of which „, 0 MA , ^ lli. ------> where 0 < k # 1, MB 112- ~ ~ < k, where 0 < k ^ 1. MB 12°. Prove that two nonconcentric circles (C!) and (C") without com mon points define uniquely a hyperbolic pencil of circles to which they belong. How are the limit points of the pencil constructed? Consider the case where one or both circles are zero circles. 13°. Fixed in the plane are two distinct points A and B. Given a straight line /. On / find points M such that the ratio MA MB assumes maximum and minimum values. Investigate the question depend ing on the position of the straight line / relative to the points A and B. 14°. Given: a plane n and a straight line / intersecting the plane in the point A. Given an angle cp that the straight line / forms with the plane n

Analytic Geometry

(if / is not perpendicular to the plane n, then <p is an acute angle between / and its projection /' on the plane n; if / is perpendicular to the plane n, ip = 90°). A point B different from A is fixed on /. Let PQ be an arbitrary line segment lying in space. Draw through points P and Q straight lines collinear with / and let P' and Q' be the points of intersection of these lines with the 7r-plane. The line segment P fQ' is called the parallel projection of PQ on the rc-plane in the direction of /. The ratio

is called the deformation ratio of segment PQ under its parallel projection on the 7r-plane in the direction of the straight line / (we will simply write deformation ratio because in this problem we consider the parallel pro jection only on the ;r-plane in the direction of the straight line /). Let there be a fixed positive number k. In the 7r-plane, find the locus of points M each of which has the following property: the deformation ratio of any nonzero segment lying on the straight line BM is equal to k . Solution. 1°. A = (—a, 0), B = (<a, 0), M = (x, y). Employing the for mula for the distance between two points, we obtain K(* + a f + / —a f + y 2 or )t2 + l \ 2 -f y2 , = -----------4a2 k* , --------A:2 -

(k 2 — l ) 2

this is the equation of a circle with centre at the point Ck(xk, 0), where xk= a

k2 + 1 k2 -

and with radius Rk

2ak I*2 - i f

2°. From the equation xk

k2 + 1 k2 -

it follows that |xk| > a. 3°. If the circle (Ck) had at least one point M0 in common with the mid perpendicular of AB, then for that point we would have M0A = k M 0B

Problems in Geometry

£2 i i 1. If 0<A:< 1, then x k= a ------- <0; k2- 1 hence, the points Ck and A lie on one side of the midperpendicular of AB and thus the point A and (Ck) lie on one side of the midperpendicular. If k > 1, then x k > 0; hence, the points B and Ck lie on one side of the midperpendicular of segment AB, and since the circle (Ck) does not have any points in common with that midperpendicular, it follows that the point B and (Ck) lie on one side of the midperpendicular of AB. 4°. The power of the point A with respect to the circle (Ck) is equal to

but this contradicts the condition

k2 + n 2 k2- i )

4a2 k 2 _ 4a2 k 2 (k2 - l)2 ~ k 2 - 1

If 0 < k < 1, then ga < 0, that is, point A lies inside (Ck) and, hence, point B lies outside (Ck) since the points A and B lie on different sides of the midperpendicular s of segment AB, and the point A and (Ck) lie (for 0 < k < 1) on one side of the midperpendicular of AB; if k > 1, then g a > 0 and A lies outside (Cfc). The power g b of point B with respect to (Ck) is k2+ 1 \ 2 k2 - 1

4a2 k2 _ 4a2 (k2 — l)2 ~ 1 — k2

If 0 < k < 1, then g b > 0, and if k > 1, then g b < 0. 5°. Prove that if 0 < k x < k 2 < 1, then R kl < Rk2, and if 1 < k x < k 2, then Rkl > Rki. Indeed, if 0 < k x < k 2 < 1, then -ftjfci ” Rk2

2ak1_____ 2ak2 ^ {k1 k 2) (1 -|- k1k 2) ^ ^ Y ^ k 2" i - k \ ~ (i - k\) (i - k2)

and if 1 < k x < k 2, then R -R

2akl k2 “ k \

2ak'2 = 2a 1

* i - i

~ kl^ 1 + k l k ^ > Q 1) ( * | - 1)

(*i -

Thus, the circles (Ckl) and (Ck2) may be symmetric with respect to the mid perpendicular s of segment AB in one of the following cases: 0 < k x < 1 < k 2,

0 < k2 < 1 < kv

Both cases are investigated in the same way. For example, let 0 < k ± < 1, k 2 > 1; then 2ak1 2ak2 Rkt — k \-\' 1~k\

Analytic Geometry

A necessary condition for the circles (Cftl) and (Ckt) (in the case of 0 < < k x < 1 < k 2) to be symmetric with respect to the midperpendicular s of segment AB is that their radii be equal: Rki — 2akx _ 2ak2 1 - k \ ~ ~ k f^ I 9 k x k\ — k x — k 2 + k\ k 2 = 0, (ki + k 2) (kk k 2 — 1) = 0, and since k x + k 2 > 0, it follows that k xk 2 — 1 = 0 , whence kjc2 = 1. This condition is also sufficient for the circles (Ckl) and (Cki) to be sym metric with respect to the midperpendicular s of segment AB. Indeed, assuming that k xk 2 = 1, we have x kl = 2a

1 +kl

k l + 1 = 2a w + i kt-l

«-

l-k\

~ x kt.

Consequently the points Ch and Ckt are symmetric with respect to the point O or, what is the same thing, with respect to the midperpendicular s of segment AB. But from the condition k xk 2 = 1 it follows that Rkl = R kt, which means the circles (Cfcl) and (Ckt) themselves are symmetric with respect to the straight line s. Remark. Everything stated in this item also follows from the fact that the given equation is equivalent to the following equation: MB _ 1 M A ~ lc 6°. One of the points A , B lies inside the circle (Ck) and other outside the circle; to put it differently, one of the points A, B is an interior point of the diameter PkQk of the circle (PkQk) and the other is an exterior point. From this it follows that one of the points P k9 Qk is an interior point of segment AB and the other is an exterior point. Therefore the numbers

PkB QkB have different signs. But the absolute values of these ratios are equal to k> since the points Pk and Qk lie on the circle (Ck) and, hence, belong to the given locus. Thus, APu.AQt PkB QkB

Problems in Geometry

To summarize: the points Pk and Qk are harmonic conjugates of the points A and B. Conversely, if the points Pk and Qk are harmonic conjugates of the pair A, By then AP- =

( = k ),

PkB QkB that is, the points Pk and Qk belong to the circle (Cfc) of the hyperbolic pencil of circles with limit points A and B (for the indicated value of k). From this it follows that the circle constructed on the segment PkQk as diameter is the circle (Ck). The proposition just proven that A, B, Pk, Qk is a harmonic set of four points permits the following elegant construction of circles of Apollonius: through points A and B draw two straight lines sx and s2 perpendicular to the straight line AB (Fig. 2) and on them fix points 5Xand S2 that are distinct from points A and B respectively. On the straight line ^2 lay off, on different sides of the straight line AB, congruent line segments BPk and BQ'k. The straight lines 5X/* and 51£>* intersect AB at the points Pk and Qk. Then the straight lines S,A, SXB, SLPk, SxQk form a harmonic set of four points. The circle with diameter PkQk is the circle of a hyperbolic pencil with limit points A and B. Figure 3 gives yet another method for constructing a circle of Apollo nius: let / be a tangent to the circle Q at the point S; N is a point on the circle Q diametrically opposite to point 5. Construct a family of chords PkQk of the circle Q parallel to its diameter SN. Let A'B' be the diameter of the circle Q that is perpendicular to the diameter SN, and let A and B be the projections of the points A ’ and B' on the straight line / from the point N. Denote by Pk and Qk the projec tions of points Pk and Qk on the straight line I from point N . The set of four points B, Pk, Qk is a harmonic set since the quadruplet of straight lines NA', NB', NPk, NQ'k is harmonic (/_A'NB'=nl2 and NBf and NA' are bisectors of Z.Pk^Q'k)The circle with diameter PkQk is a circle of Apollonius with limit points A and B. This method of constructing circles of Apollonius is connected with the so-called stereographic projection of a sphere on a plane (this projection is used in the preparation of geographical maps). 7°. Let 5(0, b) be an arbitrary point of the midperpendicular s of seg ment AB. Since all points of the midperpendicular s of AB are external points of the circle (Ck), it follows that from the point 5 it is possible to

Analytic Geometry

Problems in Geometry

Analytic Geometry

draw two tangents to the circle (Ck). Let S T be one of them and let T be the point of tangency. The power of point S with respect to the circle (Ck) is equal to ST 2: a = ST 2 =

4a2k 2 k2 + 1 \ 2 + b2 = a2 + b2 k2 — I j (k2 - 1 ) *

and, consequently, ST = ]fa2 + b2 = )[0 A2 + OS2 = SA = SB. We see that the length of segment S T is independent of k, and, hence, all circles (Ck) and the zero circles A and B have a common radical axis, the midperpendicular s of AB. 8°. From what was proved in item 7° it follows that the circle whose centre is any point S of the midperpendicular s of AB and whose radius is equal to SA = SB orthogonally intersects the circle (Ck), since SA = = SB = ST, where T is the point of contact of the tangent (either one of the two) to (Ck) drawn from point S. Any circle passing through the points A and B has its centre on the midperpendicular of segment AB and its radius is equal to SA = SB. 9°. Let 0 < k x < k 2 < 1. The circles (Ckl) and (Ck%) do not have any points in common for k x ^ k 2 because if they did have a common point M 0, then the following relations would hold true: . _ m 0a _ M0B 1

—

but this is not true (k± ^ k 2). In the case of 0 < k x < k 2 < 1, point A lies inside both (Ckl) and (Ck2), but since these circles do not have any points in common, one of them lies inside the other. But if 0 < k x < k 2 < 1 we have R ^ < R ki (see item 5°); this means the circle (Ckl) lies inside (Ck%). But if 1 < k x < k 2, then the point B lies inside (Cfcl) and (Ck2) (see item 4°), the circles (Cfcl) and (Cki) do not have any points in common and R kl > R k2 (see item 5°); which means (Cfc2) lies inside (Ckl). 10°. (1) Construct a circle (ABM0). By what was proved in item 8°, the circle (Ck) of Apollonius must intersect (at point M 0) (ABM0) ortho gonally. Therefore the radius of (Ck) will lie on the tangent to (ABM0) at the point M0. On the other hand, the centre Ck of the circle (Ck) lies on line AB and therefore it is the point of intersection of the straight line AB with the tangent to (ABM0) at the point M0. The value of k that cor responds to the circle (Ck) is: M 0A m 0b because the point M 0 lies on (Ck).

Problems in Geometry

(2) If point M0 lies on the straight line AB but differs from the points A, B, O, then it is one of the points Pk, Qk of intersection of (Ck) with that straight line. The other point Ar0 of intersection of (Ck) with AB is found from the relation (ABM qN0) = â&#x20AC;&#x201D;1, whence (OM0)-(ON0) = a \ where O is the midpoint of AB,, the point N0 is the image of point M 0 under inversion with respect to the circle constructed on AB as a diameter. The construction of point N0 is given in Figs. 4 and 5; the point N0 belongs to the circle of Apollonius passing through point M0 because from the relation (OM0y(O N 0) = a2 it follows that (ABM0 N0) = - 1 , AM 0 _ A N 0 BM0 ~ B N 0(~ The construction of point A^0, which is the harmonic conjugate of point M 0 with respect to A, B may be carried out in a great variety of ways (without recourse to inversion). If a straight line parallel to AB is given in a plane, then that construction may be performed even with a straightedge alone. 11Â°. The inequality MA , ------> k MB or MA2 - k2 MB2 > 0 is equivalent to the following: (1 - k 2) x2 + (1 - k 2) y 2 + 2a{\ + k 2) x + a2 (1 - k2) > 0.

Analytic Geometry

If 0 < k < 1, then by dividing the left member of this inequality by 1 — k 2 and simplifying we obtain

This inequality is satisfied by the coordinates of all points (x, y) lying outside the circle (Ck). If k > 1, we get

This inequality is satisfied by the coordinates of all points lying within (Ck). In similar fashion, proof is given that for 0 < k < 1 the inequality AIA

MB is satisfied by all points M lying inside (Ck), and for k > 1, by all points lying outside (Ck) 12°. Let s be the radical axis of the circles (C') and (C") and let O be the point at which the radical axis intersects the straight line C'C" of cen tres of the given circles. Denote by Pk, Qk, Pk\ Qk respectively the points of intersection of the circles (C') and (C") with the straight line C'C". Since the point O lies on the radical axis of (C') and (C"), it follows that its powers with respect to these circles are equal : a = {OP'k)(OQ'k) = (OP;;)-(OQ'k') (these powers are positive since the radical axis does not intersect the given circles and, hence, point O lies outside the circles). Let us now construct on the straight line C'C" the points A and B dis tant ][a from O. Then OA‘»= OB2 = (OPk) ■(OQk) = (OP’k )-(OQk’) = a. From these relations it follows that {ABP'k Q'k) = - 1, (ASP? Q'k' ) = - 1, that is, (C') and (C") are circles of Apollonius for values k equal to P U _ QjA k „ = PL'A _ Qk A PLB Q’kB Pk B Qk B 13°. Construct the midperpendicular s of segment AB. Suppose that a straight line / intersects this midperpendicular but does not pass through the points A and B [case (1) see Fig. 6]. Let S be the point of intersection of straight lines / and s. Now construct a circle with centre S and radius S A ; it intersects the straight line / in two points M± and M2. Draw a tan-

Problems in Geometry

gent line to the circle at point Mv Let P1 be the point of intersection of this tangent and the straight line AB. Construct a circle with centre at the point Px and with radius PxM^ This is a circle of Apollonius with the limit points A and B. Thus M XA â&#x20AC;&#x201D; k x > 1. M XB All the points of the straight line I lie outside the circle (Ckl) and so MA for all points of / different from Ml9 the ratio -----is less than kl9 that MB MA is, at the point Mx the ra tio -----is a maximum. Similarly it can be proved MB MA that this ra tio ----- is minimal at the point M2. MB Figure 6 gives distinct positions of the straight line / with respect to the points A and B: (1) The straight line / does not pass through the points A and B and is not perpendicular to AB: at the point M x we have the maximum of the ratio

(

\ and at M 2 the minimum (this case was examined above

m b)

in detail). (2) The straight line / is perpendicular to AB, does not pass through the points A and B and is located on the side of the midperpendicular of

Analytic Geometry

AB on which point B lies. At Mx we have the maximum, and there is no minimum. (3) The straight line / is perpendicular to AB, does not pass through the points A, B and is located on that side of the midperpendicular of AB on which the point A lies. At point M 2 we have the minimum, and there is no maximum. (4) The straight line / passes through the point B but does not come to coincidence with AB, the minimum is at Af2, and there is no maximum. (5) The straight line / passes through point A but does not coincide with AB, the minimum is at point A ( the ratio in this case is zero^; the V MB ) maximum is at point Mv (6) The straight line / passes through point B and is perpendicular to AB. There is neither maximum nor minimum. (7) The straight line / passes through point A and is perpendicular to MA AA \ ---- = â&#x20AC;&#x201D; =01, and there is no maximum.

(

) (8) The straight line / coincidesMB with AB the straight line A B. The minimum is at point A

= 0 V and there is no maximum. \M B ) 14Â°. The locus of points M in the rc-plane is found from the condition MA MB This same locus of points M in space is a sphere S of Apollonius with limit points A and B. If 0 < k < 1, then the sphere S contains within itself the point A, and since A lies in the rc-plane, the sphere S and thera-plane intersect along the circle a, which is the desired locus of points M. But if k > 1, then the sphere S contains within it the point B and here three cases are possible: either the sphere intersects the 7r-plane or is tangent to the 7r-plane, or does not have a single point in common with the 7r-plane. Let us investigate these cases. On the straight line AB we assume a positive direction from point A to point B. Then AP =

AB,

- ~ k2 -

where P is the centre of the sphere S. The distance d from the centre P of sphere S to the rc-plane is k2

d â&#x20AC;&#x201D; --------- AB sin <p. k2 -

The sphere S intersects the 7r-plane if and only if its radius k r= AB k2 -

Problems in Geometry

is greater than d : k 'A B kVAB . ---------- > ----- sin w k2 - 1 k2 — 1 or sin q> < \/k. When this condition is complied with, the sphere S intersects the rc-plane along the circle <x, which is the locus of points M. If sin (p = 1jk, then the sphere S is tangent to the 7i-plane at the point M, which is the desired locus of points (this point M coincides with the point A if and only if \jk, then the sphere S and the rc-plane do not inter sect and the desired locus of points M is empty. Sec. 2. Application of analytic geometry (problems with hints and answers) 1. Plane geometry 1. Fixed in a plane are two distinct points A and B. Given: a real number k. Find the locus of points M for each of which 2 MA2 + 3MB2 = k. Answer. If k < — AB2, then the desired locus of points M is empty; 6 if k = - - AB2, the locus consists of a single point, P, which divides the directed line segment AB in the ratio 3/2:

if k > —AB2, the desired locus of points M is a circle with centre at the point P and with radius R =

— 6AB2.

2. Fixed in a plane are two distinct points A and B. Given: a real number k. Find the locus of points M for each of which 2MA2 - 3MB2 = k.

Analytic Geometry

Answer. If k > 6AB2, the desired locus of points M is empty; if k = 6 AB2, the locus of points M consists of the single point P, which divides the directed line segment AB in the ratio —3/2: AP _ r—

3, _y

if k < 6AB2, the desired locus of points M is a circle with centre P and radius R = J/6AB2 — k. 3. Fixed in a plane are three points A , P, C. Find the locus of points M for each of which MA2 + MB2 = MC2. Hint. Introduce a rectangular Cartesian system of coordinates. Answer. If the angle C of the triangle ABC is obtuse, the desired locus of points is empty; if C = n\ 2, the locus of points M consists of the single point D symmetric to the point C with respect to the midpoint of AB; if C is an acute angle, the locus is a circle with centre D and radius ]fa2 + b2 — c2 (<a, b, c are the lengths of the sides of &ABC). 4. The bisector of the interior angle A of S^ABC divides the side PC into line segments BD = 4, DC = 2. Find the lengths of the sides AB and AC if we know that they are given by integers and that the altitude dropped from vertex A to side BC exceeds JflO. Solution. It is given that the bisector AD divides the side BC into seg ments of length 4 and 2; consequently, BD/DC = 2, that is AB/AC 2. The locus of points A for which this relation holds is a circle of Apollo nius of radius 2ak \'k2 ^ T \ where 2a = CB = 6, k = 2, that is, R

6 2

= 4 with centre P lying

on the straight line BC and having the coordinate k * + l 3i•—5 = 5,< OP = a ---------= A:2 - 1 3 where O is the midpoint of segment BC. If a rectangular Cartesian system of coordinates is introduced and BC is taken for the x-axis, and the mid point O of BC as the origin, then the equation of the circle (P, 4) takes the form ( * - 5 ) 2 + /-= 1 6 .

Problems in Geometry

Let us consider the straight line y = 3. This line intersects the circle (P, 4) at the points A^5 — |/7, 3), A2(5 + )[l, 3). From this it follows that A,C = 1^20 - 4 /7 < 3.1, A2C = K20 + 4 /7 > 5.5. Since AH> /lO >3, it follows that point A must lie on the lesser arc AXA2of the circle (P, 4); AXC < AC < A2C, hence 3.1 < AC < 5.5, and since it is given that the length of AC must be a whole number, it follows that AC = 4, then AB = 8, or AC = 5 and AB = 10. Supplementary questions. Suppose the bisector AD of the interior angle A of A ABC divides the side BC into segments BD = 4, and DC = 2. (a) What is the maximum value of the altitude AH7 (Answer. 4.) (b) Within what range do the lengths of the sides AB and AC vary? (Answer. 2 < < 6, 4 < AB < 12.) (c) Prove that AC and AB increase if the point A describes a circle of AB Apollonius---- = 2 from the point D to its diametrically opposite AC point. (d) Prove that the length AD of the bisector of angle A varies from 0 to 8. (e) Given the length m of a median emanating from A. What is a neces sary and sufficient condition for the existence of a triangle with given BD = 4, DC = 2, /»? How is such a triangle constructed? (Answer. 1 < m < 9.) 5. Find the images of the circles of Apollonius (see problem 11) / fc2 + l \ 2 , , 4a2 k 2 l x — a ---------- } + y2= -------V k2 - I J (k2 — i)4 under the inversion I = (A ,A B 2): x + a = 4a2

X a (X + a)2 + Y2

4a2

(X + a)2 + T2

Under this inversion, into what do the circles of an elliptic pencil con jugate to the hyperbolic pencil under consideration go? Answer. (r'):(X — a)2 + Y 2 = (2ajk)2\ this is a family of concentric circles (C^) with common centre B(a, 0); the radius of the circle (C*) is equal to lajk. The circle E of the elliptic pencil conjugate to the pencil r with centre at the point 5(0, s) goes into the radical axis of the circle E and the circle of inversion, that is, into the straight line a(X - a) + sY - 0. 6. Prove that the equations x2 + y2 — 2px + a2 = 0, x 2 + y 2 — 2qy — a2 = 0,

Analytic Geometry

where a is a fixed positive number, and p and q are real parameters, are, respectively, the equations of the hyperbolic pencil of circles and the conjugate elliptic pencil of circles. What are the fundamental points of these pencils? Into what lines do the circles r and E go under the inversion / = (A, 4a2), where A = (—a, 0). Answer. The fundamental points are A(—a, 0) and B(a, 0). Under the inversion / = (A, 4a2), the circle f goes into the circle (X - a2) + Y2 = 4a2 - ~ ° P+ a with centre at the point B(a, 0) and radius R' = 2a

P__ a . Note that

only under the values of the parameter p, p < —a and p > a, the equation of r is the equation of a real circle. Under the same inversion, the circle E goes into the straight line a(X — a) + p Y = 0 passing through point B(a, 0); this is the radical axis of the circle E and of the circle of inversion. If we consider the circle f as a circle of Apollonius, then k and p ,, .. A:2 + 1 , p —a 1 are connected by the relation p = a --------- whence-------- = — and, con____ k2 - 1 p + a k2

sequently, R' = 2a 1 - ---- - = 1 p a 7. Prove that if

/ = x2 + y2 — 2p1 x + a2 = 0,

(p = x2 + y2 — 2p2 x + a2 = 0

are two distinct circles of a hyperbolic pencil (px ^ p2) with limit points (± a, 0), then any circle of this pencil may be given by the equation ¥

+h<P = 0,

where A and p are numbers. Conversely, for any A and p of which at least one is nonzero, the equation A/ + pep — 0 is the equation of a circle of a hyperbolic pencil (for A = —p ^ 0 , the equation of the midperpen dicular of a line segment bounded by limit points). Consider the special case of / = x2 + y2, cp = * — a. 8. State and prove a similar location for two distinct circles / = x2 + y 2 — 2qx y — a2 = 0, <p == x2 + y 2 — 2q2y — a2 = 0 of the elliptic pencil E. 9. Let M be an arbitrary point in the plane. Prove that the difference between the powers of point M with respect to the circles (O) and (O') with centres O and O' is equal to 2(00') • (KM), where K is the orthogonal projection of point M on the radical axis of (O) and (O').

Problems in Geometry

10. AAXand BB1are medians of A ABC; CC1is the altitude. The straight lines AA1,B B 1,C C 1 form A A 2 B2 C2. Find the ratio (A2b 2c 2) (ABC) (cot B — cot A)2 Answer. 3(2 cot B + cot A) (cot 5 + 2 cot A) 11. A', B', C’ are the feet of bisectors of the interior angles of A ABC. Knowing the lengths a ,b ,c of the sides BC, CA, AB of A ABC, find the ratio (A 'B 'C ) (ABC) . 2abc Answer.-----------------------------(b + c) (c + a) (a + b) 12. A’, B ’, C are points of tangency of a circle inscribed in A ABC with the sides BC, CA, AB. Prove that (A'B'C ') r (ABC) ~ 2R 13. A', B', C' are the feet of the altitudes of A ABC. Given: the angles A, B, C of the triangle. Find the ratio (A'B'C ') ABC For what necessary and sufficient condition do we have A'B'C '\[A BC f Answer. 2 cos A cos B cos C; A'B'C'\\ABC if and only if A ABC is obtuse-angled. 14.. Given in the plane three circles (A), (B), (C) each of which lies outside the other two. Let P, Q, R be the centres of internal similarity of the pairs (B), (C); (C),(A); (A),(B). Find (PQRl (ABC) knowing that the radii of (A), (B), (C) are respectively equal to Rlf R2, R2. 2Ri R2 R2 Answer. (Rs + R») (Rj + Bi) (Ri + R >) 15. Given A ABC. Through point D lying on the straight line BC are drawn straight lines DP and DQ, which are parallel, respectively, to AB and AC. Prove that (DQC) + (PBD) = 0. 16. Through an arbitrary point lying inside A ABC are drawn straight lines parallel to its sides. These lines divide A ABC into six parts, three of which are triangles with areas Slt S2, S3. Find the area of A ABC.

Analytic Geometry

Answer. (][ + V S2 + ]f S3)2. 17. Let A ", B", C" be points symmetric to the vertices A , B, C of triangle yli?C with respect to the feet of the bisectors of its interior angles. Given: the lengths a ,b ,c of the sides BC, CA, AB of triangle ABC. Compute the ratio (A" B” C") {ABC) 8abc Answer. 3 + (b + c) (c + a) (a + b) 18. The straight line (and also its portion inside the triangle) that is symmetric to the median with respect to the interior-angle bisector emanat ing from that vertex is termed the cimedian of the triangle. Given: the lengths c and b of the sides AB and AC of A ABC. In what ratio does the cimedian emanating from vertex A divide the directed line segment BC1 Prove that the three cimedians of the triangle intersect in one point (the Lemoine point). c2 Answer. — • b2 19. In a right triangle, construct the cimedian emanating from the vertex of the right angle. Answer. The perpendicular dropped from the vertex of the right angle to the hypotenuse. 20. Express, in terms of the lengths a, b, c of the sides BC, CA, AB of A ABC, the area of A PQR, where P, Q, R are the projections of the centroid G of the triangle on its sides. 4 (a2 + b2 + c2) S3 Answer. where S is the area of A ABC. 9a2b2c2 21. The bisectors of the interior angles of A ABC intersect the opposite sides BC, CA, .42?respectively at the points A', Bf, C . The points A ", B", C” are symmetric to the points A \ B', C' with respect to the respective vertices A ,B ,C of A ABC. Given: the lengths a, b, c of the sides of A ABC. Find the ratio (A"B" C,r) (ABC) 2 abc Answer. 6 (b + c) (c + a) (a + b) 22. Let M and N be the midpoints of the medians BD and CE of A ABC. The area of A ABC is equal to S. Compute the area of the quadrangle HMNC. Answer. — S. 16

Problems in Geometry

23. The straight lines AO, BO and CO intersect the sides BC, CA, AB of triangle ABC in the points P, Q, R respectively. Prove that m . + °Q + 0A ~AP

~BQ

= L

C7*

24. Given two straight lines / and m lying on an oriented plane and intersecting in the point A. Through an arbitrary point O not lying either on a line /, or on a line m a straight line n is drawn that intersects / and m in the points B and C respectively. Prove that the sum 1 (iOAB)

1 (OCA)

does not depend on the choice of n. 25. Through the vertex A of a parallelogram ABCD is drawn an arbi trary straight line that intersects the diagonal j3D in a point E, and the straight lines BC and CD intersect in points F and G respectively. Prove that the line segment AE is the mean proportional between the lines EF and EG. 26. Let a, p, y be points symmetric to some single point O with respect to the midpoints of the sides BC, CA, AB of a triangle ABC. Prove that the straight lines Act, Bp and Cy pass through the same point P. Also prove that if point O describes some line r , then point P describes a line r f that is homothetic to line F. Where does the homothetic centre lie? What is the homothetic ratio? Answer. The homothetic centre lies in the centroid M of A ABC; MP

27. A straight line / intersects the sides BC, CA, AB of a triangle ABC in the points , j8, y respectively. Let a', P', y' be points symmetric to the points a, p, y respectively about the midpoints of the sides BC, CA, AB. Prove that the points a', P', y' lie on one straight line. 28. Let A 'B 'C be a triangle that is obtained if through each vertex of a triangle ABC we draw a straight line parallel to the opposite side; ct, p, y are points taken respectively on the sides BC, CA, AB. Prove that if the straight lines Act, Bp, Cy pass through a single point, then the straight lines A' , B'P, C y also pass through one point. 29. Drawn through the midpoint of each diagonal of a convex quad rangle is a straight line parallel to the other diagonal. The point of inter section of the lines thus drawn is joined to the midpoints of the sides o^ the given quadrangle. Prove that the quadrangle is thus partitioned into parts of equal size. cl

Analytic Geometry

30. The points P and Q divide the directed sides BC and CA of A ABC in the given ratios Aand fi. Suppose the straight lines AP and BQ intersect in a point O. Find the ratios: (!)

(2) (ABC)

(ABC)

(3).< 4 2 ® , (ABC)

Answer. (1) ------ - ------ ; (2)------- ——----------- ; (3) ---------- 1 + A + A/r (1+A) (1 +A+A/i) (1 4~a0 (1 +A+A/z) 31. A certain point O lies in the plane of a parallelogram ABCD (AC and BD are the diagonals). Prove that: 1°. If the point O lies inside the parallelogram ABCD, then the sum of the areas of A OAB and A OCD is equal to the sum of the areas of the triangles OBC and ODA. 2°. No matter where the point O lies in the plane of parallelogram ABCD> AO AC is equal in size either to the sum or the difference of A OAB and A OAD. 32. Points A', Br, C' are taken on the sides BC, CA and AB of A ABC. Let Al9 Bl9 Ci and A2, B2, C2be the images of the points A, B, C respectively under homothetic transformations with the same homothetic ratio and with homothetic centres in the points C , A ’, B' and B', C', A Prove that A A ^ C x and A A2B2C2 have one and the same centroid. 33. Let M1 and M 2 be two arbitrary points lying on side BC of A ABC, and let N be an arbitrary point lying on side AB. Denote the points of intersection of the straight lines NMX and NM 2 with side AC by Px and P2 and the points of intersection of the straight lines PMX and PM2 (where P is an arbitrary point of the straight line AC) with the side AB by Q±and Q2. Prove that BC, PxQ2, and Qi^ 2 belong to the same pencil. 34. Suppose P and Q are two points lying in the plane of A ABC; AiBxCx and A2B2C2 are triangles symmetric to A ABC atjout the points P and Q respectively. Let oe, ft, y be points of intersection of the straight lines AxA29 BxB2, CxC2 respectively with the lines BC, CA, AB. Prove that the points c l, p , y are collinear (Fig. 7). Proof. Let a', fi', y' be points of intersection of the straight lines A a. Bp, Cy with PQ. Then a', P’, y' are the midpoints of the line segments Act, BP, Cy. These points lie on the sides of A A'B'C', where A', B', Cr are the midpoints of the sides BC, CA, AB, respectively. Under the homo thetic transformation (G, —2), where G is the point of intersection of the medians of A ABC, the points a', /?', y' go into the points a", p", y", which also lie on one straight line and on the lines BC, CA, AB, respec tively. Here, Coe = 2B'ot', Bet" = 2B'ol' and, hence, Ca = Bctn, whence

Problems in Geometry

and similarly for the other two sides. Since the points a", /?", y" are collinear, it follows that

and, hence, — » — >— > Bcc CP Ay _ — » — >— > ~ otC PA yB Consequently, the points a, /?, y are also collinear. 35, Let the straight lines A and A' intersect the sides of A ABC in the points Al9 Bl9 Cx (line A) and A'x, B[, C[ (line A'), respectively. Prove that the lines A ^ , B ^ i, CXA[ intersect, respectively, the straight lines AB, BC, CA in the points p s,p » p * lying on one line and the lines A[Bl9 B[Cl9 C[Aj intersect the lines AB, BC, CA respectively in the points Qz, Qx, C?2> which also lie on one straight line. The straight lines PiP2Pz and QiQzQz are called the Brocardians XRrocard lines) of the straight lines A and A' with respect to A ABC. 36. Given: the lengths a, b, c of the sides BC,CA, AB of A ABC. Through point C is drawn a bisector CC0 of the interior angle C. Drawn

Analytic Geometry

through point A is the median AA0 to the side BC and through point B the altitude BB0 to side CA. These lines form A AxBxCx. Find the ratio (ABC) J [b(a2 + b2 - c2) - a(b2 + c 2 - a2) ] 2 Answer. -------------------------------------------------------------—— • (a2b + bz — be2 + lab2) (— a 2 + 3b2 + c2) (a + lb) 37. Let A ^ C x and A2B2C2 be triangles that are the images of A ABC under the homothetic transformations (P, k), (Q, k). Denote by a, /?, y the points of intersection of the straight lines AXA2, BXB2, CXC2 with the straight lines BC, CA, ^42? respectively. Find the ratio

(aft?) (v45C) Answer. —&(& + 1). 38. Prove that if three diagonals of a hexagon (not necessarily convex) have a common midpoint, then any two of the opposite sides are parallel. 39. From an arbitrary point Ax lying on side BC of a triangle ABC draw a straight line A-fix parallel to BA up to intersection with CA at point Bx\ then draw a straight line BxCx parallel to BC up to intersection with line AB at point Cx and, finally, draw line CXA2 parallel to AC up to intersection with line BC at point A2. Prove that if Ax is the midpoint of segment BC, then points Ax and A2 coincide; otherwise, continue the process, that is, draw line A2B2 parallel to BA, line B2C2 parallel to CB, and line C2A3 parallel to AC. Prove that the path closes on itself, that is, point A3 coincides with point Av 40. Join vertices A, B, C, D of a parallelogram ABCD with the midpoints of the sides BC, CD, DA, AB so that a parallelogram is formed that lies inside the parallelogram ABCD. Prove that the area of the thus formed parallelogram is equal to 1/5 the area of the parallelogram ABCD. Another such parallelogram is formed if we join the points A, B, C, D with the midpoints of the sides CD, DA, AB, BC. Prove that the common part of these two small parallelograms is a centrally symmetric octagon and has an area equal to 1/6 that of the parallelogram ABCD. 41. ABCD is an arbitrary convex homogeneous lamina. Each of the sides of the quadrangle ABCD is divided into three equal parts: AAX = = A2A2 = A2B, BBx = BxB2 = B2C, CCx = CXC2 = C2D, DDX = D±D2 = D2A. The straight lines A2Bly B2CX, C2Dl9 D2AX form a parallelogram. Prove that the centroid of the lamina ABCD coincides with the point of intersection of the diagonals of the parallelogram. Hint. Take the diagonals of the parallelogram for the coordinate axes. 42. Given, with respect to a general Cartesian system of coordinates, two points Mx(xl9y^, M2(x2,y 2) and the straight line ax + by + c = 0 . It is given that the points M x and M2 do not lie on the given straight line

Problems in Geometry

and that the third line M XM2 intersects the line ax + by + c = 0 at some point M. Find the ratio X in which the point M (x,y) divides the line ----- ► segment MXM2. A , axx + byx + c _r. *1 + Xx2 y x + Xy2 Answer. X = ------ - - - ■- — . Hint. jc - ---------“ , y = ax 2 by2 + c 1 -f-A 1 -f- ^ these coordinates must satisfy the equation ax + bx + c = 0 . 43. Taking advantage of the result of the preceding problem, prove that if the straight line / does not pass through any one of the vertices of triangle ABC and intersects its sides BC, CA, AB respectively at the points P, Q, R , then the product of the ratios

at which the points P, Q, R divide the directed line segments BC, CA, AB, is equal to —1 : Xpv = — 1 . Hint. Introduce into the plane of the triangle ABC a general Cartesian system of coordinates; in this system, let A = (xx, y x), B = (x2,y 2), C = (x2, ^3) and let ax + by + c = 0 be the equation of the straight line /. Then , ax 2 + £y2 + c . + 6*3 + c axx + byx + c A— 1 > fi * ^ '* ax 3+ + c axx + b y x + c ax2 + by2 + c Remark. The converse is true as well: if Xfiv ~ —1, then the points P, Q, R which divide directed segments BC, CA, AB in the ratios X, \i, v are collinear. Indeed, suppose the line PQ intersects AB in the point R'. Denote by v' the ratio in which the point R' divides the directed line segment AB. Then X\iV = —1. From this and from the equation Xfiv = = —1 it follows that v = v', that is,

R'B

Hence, the points R and R' coincide. 44. A second-order curve given by the equation F(x, y) — ax2 + 2bxy + cy2 + 2dx + ley + f = 0

Analytic Geometry

does not pass through a single vertex of the triangle ABC and intersects the sides BC, CA, AB at the points A l9A2; Bl9B2; Cl9 C2 respectively. Prove that the product of the ratios in which the points Al9 A2; Bx, B2; Cl9 C2 respectively divide the directed line segments BCf CA, AB: AC2

*i =

C2B is equal to 1: X1Xi n1n2 v2 v2 = 1. Conversely: if on the sides of triangle ABC are chosen points Alf A2 (on BC), Bl9 B2 (on CA), Cl9 C2 (on AB) and if the product of the ratios in which these points divide the directed line segments BC, CA, AB is equal to 1, then the points Ax, A2, Bl9 B29 Cx, C2 lie on one and the same second-order curve. Hint. The numbers Xx and k2 are found from the equation J *1 + Xx2 ' y x + Xy2 \ = Q I 1+ A I +X j where B(xlt >>,), C(x2,y 2) or ( * i + Xx2 \ 2 , + Xx2 y t + ).y2 , „ + Xy2 \ 2 + 2d X2 1 +X

iTTFj

2e * ± ± * * 1.

+ f ,

1 +X or F{x2, yi) P + • •. + F(xx, yx) = 0. By Yieta’s theorem. ^>1^>2 —

F (x x, y i) F(X2, y 2)

and so on. The proof of the converse is similar to the proof given in the remark concerning problem 43. 45. Suppose an algebraic curve / of order n does not pass through a single vertex of A ABC and intersects each of the sides of the triangle, BC, CA, AB in n points A if Bi9 Ct (/ = 1,2, . . . , n). Then the product of the ratios BA§ CBt

AC{

A f i BiA C(B is equal to (—1)" ( Carnot's theorem). Does the converse hold true?

Problems in Geometry

Hint. The proof is similar to that given in the hint referring to prob lem 44. Generally speaking, the converse for n > 2 is not true. 46. Given in the plane are m points p. (i = 1 , 2 , . . . , m). An algebraic curve / of order n does not pass through a single one of the points Pt and intersects the straight lines PXP^ iV V - • •, Pm-1 PmPi ln n points respectively A ? \ A f \ A j ”~\ AW (/ = 1, 2, . . . , m). Find the product of the ratios

4 n- 1} Pm

^P l

Answer. (—l)mn (a generalization of Carnot’s theorem; see problem 45). 47. Al 9 A 2 9 A3; Bl 9 B2 9 B3; Cl 9 C2, C3 are arbitrary points which lie, respectively, on the sides BC, CA, AB of A ABC and are the interior points of the sides. Introduce a general Cartesian coordinate system into the plane of A ABC. Since the general equation of a third-order curve contains 10 coefficients, there exists a third-order curve that passes through 9 points A i9 Bi9Ci (/ = 1,2,3). By the Carnot theorem, the product of the ratios in which the points A i9 B i9 Ct divide BC 9 CA9 AB, respectively, is equal to (—l )3 = —1, and yet all these ratios are positive end their product cannot be a negative number. Wherein lies the error? 2. Solid geometry 1. AXA 2 A 3 A 4 is an arbitrary tetrahedron; Bl 9 B 29 B39 B4 are the centroids of its faces A 2 A3 A4, A^A3 A4, AiA 2 A49 A^A2 A 3 9 Ci29 Ci3, C*4, C239 C249 C3 4 are midpoints of its edges AXA 29 AyA 39 AXA 49 A 2 A3, A 2 A 49 A 3 A4. The straight lines AxBl 9 A 2 B2, A 3 B39 A 4 B4 (and also the line segments themselves) are called medians o f the tetrahedron A ^ A q A ^ The straight lines Q 2C34, C1 3 C24, C2 3 C1 4 (and also the line segments themselves) are termed bimedians o f the tetrahedron A 1 A 2 A 3 A4. Prove that four medians and three bimedians of the tetrahedron AXA2A3A4 pass through one and the same point G, called the centroid o f the tetra hedron A1A2A3A4; here, A f i : GBt = 3 :1 and the bimedians are bisected by the point G. Assuming the radius vectors of the points A x are equal to OAt = ti (i = 1,2, 3, 4), find the radius vectors r l7 of the midpoints of the edges AxAj, the radius vectors rijk of the centroids of the faces AxAjAk, and the radius vector r of the centroid G. _ r, + iy + rk *,• + *j T = ri + r2 + r3 4" r4 Answer, r.. rijk

Analytic Geometry

2. Prove that the six planes passing through the edges A xAj of the tetra hedron Ax A 2 A 3 A 4 and the midpoints of the opposite edges pass through the centroid G of the tetrahedron. Hint. A iA jC ki and A kA xCij intersect along the bimedian CklCxj. 3. Prove that the six planes that pass through the edges of the tetra hedron and divide its volume into two intersect in a single point. 4. OA, OB, OC are the edges of a parallelepiped; A', B', C , O’ are the vertices symmetric to the vertices A, B ,C ,0 about the centre of the paralle lepiped. Prove the following statements: 1°. The diagonal OO' of the parallelepiped is divided into three equal parts by the planes ABC and A'B'C'. 2°. The diagonal OOf intersects the planes of the triangles ABC and A'B'C' in their centroids. Hint. Set O = (0, 0,0), A = (1,0, 0), B = (0, 1, 0), C = (0,0, 1). 5. Let Ga, Gb, Gc, Gd be the centroids of the faces BCD, CDA, ABD, ABC of the tetrahedron ABCD. Prove the following statements: 1°. If a straight line k intersects the faces BCD, CDA, ADB, ABC in the points coa, cob, (oc, o d, respectively, then the midpoints of the segments A oa, Bcob, Co)c, D od are coplanar. 2°. The tetrahedrons A 1 B1 C1 D1 and A 2 B2 C2 D2 are images of the tetra hedron ABCD under the homothetic transformations (Pl 9 —1/2), (P2, — 1/2), where P 1 and P2 are arbitrary points. Prove that the straight lines AXA2, B±B2, C±C2, DxD2 intersect the faces BCD, CDA, DAB, ABC in the points a, p, y, 3 lying in one plane. Proof. Let G be the centroid of the tetrahedron ABCD; let a', /?', y', 3' be the points of intersection of the straight line PXP2 with the straight lines Aa, Bp, Cy,D3; let wa, (ob, wc, cod be points at which the following lines intersect: Got and Gaa'; Gp and Gbp'; Gy and Gcy'; G3 and Gd3'. — »— »— >— > The points a', jS', / , 3' divide the directed line segments A a, BP, Cy, D3 in the ratio of 2 : 1 and therefore lie in the planes of the faces of the tetra hedron Ga Gb Gc Gd. The point G divides segment AGa in the ratio of 3 : 1, and therefore coa is the midpoint of segment Gaot' and Got = 3Go>a. On the basis of item 1°, the points coa, cob, coc, a)d lie in one plane, and hence so also do the points a, p, y, 3 obtained from the points coa, cob, coc, a>d under the homothetic transformation (G, 3). 3°. The images co', a)b, w'c, co'd of the points coa, cob, coc, cod under the homothetic transformations (Ga, — 1/2), (Gb, — 1/2), (Gc, —1/2), (Gd, — 1/2) are coplanar. AG ota' 3 G jx Hint. See item 2°, —— = 2 A'Ga GGa a'A " \ GaA' 6 . Given: four arbitrary points A ',B ', C',D ' lying, respectively, in the faces BCD, CDA, DAB, ABC of a tetrahedron ABCD. Let (A l 9 Bl 9 Cl 9 DJ,

Problems in Geometry

(A2, B2, C2, D2), (A3, B3, C3, D3) be images of the points A, B, C, D under a homothetic transformation with ratio k and, respectively, with centres (D', A', B', C'), (C\ D', A', B'), (B \ C', D', A') so that D'A1 _

~~ D'A

A'B1 __ ^ A'B

KCi ^ B'C

C'D1 CD

7 C'A

= k and so forth.

Prove that the tetrahedrons Ax B1 Cx Dl9 A2B2 C2D2, A3Bz C3 Dz have a common centroid. 7. Let Pbe an arbitrary point lying inside a tetrahedron ABCD. Denote by A ',B ',C ',D ' the points of intersection of the straight lines PA, PB, PC, PD with the opposite faces. Prove that PA'

+ " + ÂŁ ? > 12, PB' PC' PD' AP BP CP DP nt -----------------------------> 81. PA' PB' P C PD'

Under what condition do we have equality? 8. Let a and a', b and b', c and c' be the respective points in which the edges BC and AD, CA and DB, AB and DC of a tetrahedron ABCD intersect an arbitrary plane. Denote by Ga, Gb, Cc, Gd the centroids of the triangles a'be, b'ca, c'ab, a'b'c'. Construct the points A', B', C', D' such that A A' = 3AGn, BB' = 3BGb, C C = 3CGr, DD' - 3DGd Prove that the points A', B', C , D' lie in a single plane. 9. T = ABCD is an arbitrary tetrahedron; a/ty<5 is spatial quadrangle whose sides are equal and parallel to the medians of the given tetrahedron. Through the midpoints of the sides of the quadrangle a/fy<5 draw planes parallel respectively to the faces of the tetrahedron (so that if a/? # A A', where AA' is a median of the tetrahedron ABCD, then the plane passing through the midpoint of segment a/? is parallel to the plane BCD and so on). The four planes thus drawn form a tetrahedron Tv Prove that the volume of Tx is eight times that of T and that the tetrahedrons TÂą and cuflyd have a common centroid. 10. Draw through the vertices A ',B ', C ',D ' of a tetrahedron A'B'C'D' parallel lines dx, d2, dz, dA. Let ABCD be a tetrahedron homothetic to the tetrahedron A'B'B'D' under a homothetic transformation with ratio k. Denote by Al9 Bl9 Cl9 D1 the points of intersection of the lines dl9 d2, d3, d4 respectively with the planes BCD, CDA, ABD, ABC. Prove that Ai Bx Ci Dx = - k \ 2 k + 1) ABCD .

11. Through the vertices A, B,C, D of a tetrahedron ABCD draw planes a, b, c, d parallel to some plane m. Through the vertices A', B', C 9 D 9

Analytic Geometry

of the tetrahedron A'B'C'D' draw planes a', b ' 9 c\ d' parallel to some plane m'. Here the planes m and m' are chosen so that the straight lines (a9 a'), (b, b'), (c9 c' ) 9 (d, d') lie in one plane (the P-plane). Prove that when the planes m and m' are changed (however, the change must be such that coplanarity of the indicated four lines is preserved at all times), the P-plane will rotate about a fixed point. 12. Straight lines joining the vertices A, B, C9D of a tetrahedron ABCD with point P intersect its opposite faces in the points A \ B \ C ' 9 D'. On segments A A \ B B \ CC", DDf consider points A \ 9 &19 Cl9 D1 such that AAâ&#x20AC;&#x2122;

= k

A ,A 1 and the points A2, A3, A 4 in which the straight lines AxBl 9 AxCl 9 AXDX intersect the plane BCD. Prove that the areas of the triangles A 2 CD, A 3 DB, A4BC are equal. Consider the case k â&#x20AC;&#x201D; 2. 13. Given a tetrahedron T = ABCD and four straight lines a ,b 9 c,d that pass through one and the same point P and are parallel to the four given lines. Let line a intersect the planes BCD, CDA 9 DAB, ABC in the points ll 9 m4 9 n3 9 p2; line b in the points l29 mx, h4,/?3, line c in the points /3, m29 nl 9 /?4, and line d in the points /4, w3, n2 9 pv Prove that, generally, there exists only one position of point P under which the quadruplets of points (ll 9 ml 9 nl 9 px) (l2 9 m2 9 n2 9 p 2 ) 9 (/3, m3, w3, /?3), (/4, w4, w4, p4) are coplanar. Consider the case where the straight lines a9 b9 c9 d are parallel to the medians of the tetrahedron ABCD.

CHAPTER III

THE USE OF COMPLEX NUMBERS IN PLANE GEOMETRY Sec. 1. Solved problems Before examining the problems, make a brief study of the theoretical material given in Chapter V, Sec. 4. Since the method applied here is not readily available and is effective and simple in the solution of many problems of plane geometry, we have selected a large number of problems and have provided solutions; there are also exercises in the unaided solving of problems (with hints and answers) Problem 1. ABC is an arbitrary triangle, G is the point of intersection of its medians (centroid); H is the point of intersection of the altitudes (orthocentre); O is the centre of a circumscribed circle (O) = (ABC)', A l9 Bx, Cx are the midpoints of the sides BC, CA, AB respectively; A2, B2, C2 are the feet of the altitudes; Az, B3, C3 are the corresponding midpoints of the line segments AH, BH, CH (Euler’s points); A4, B4, C4 are the corres ponding points symmetric to the orthocentre H about the straight lines BC, CA, AB. Prove that: 1°. The points O, G, H are collinear and OH = 30G. 2°. The points Al9 Bl9 Cl9 A2, B2, C2, A3, B3, C8 lie on one and the same circle (0 9) (Euler's circle or the nine-point circle o f A ABC). The centre of the circle 0 9 is the midpoint of the line OH, and the radius of the circle (0 9) is equal to R/2, where R is the radius of the circle circumscribed about the triangle ABC. 3°. The points A4, B4, C4 lie on the circumscribed circle. Solution 1°. Take (ABC) for the unit circle. Let a, b, c be the corres ponding affixes of the points A, B, C. Prove that a + b + c is the affix of the orthocentre H of triangle ABC. The slope of the straight line BC is c —b c —b x —-------- — ------------- = — be. E -b J ____ 1 c b The 6lope of the line joining vertex A(a) and point M and having the affix a + b + c is b+ c b+ c — be. x b c

Complex Numbers in Plane Geometry

whence x -j- x 9 = 0 . Hence, AM ± BC. In similar fashion it is proved that BM _L CA and CM ± AB. Hence, M —the point with affix a + b + c—coincides with point H (the point of intersection of the altitudes of A ABC). Furthermore, the affix of point G is a+ b+ c --------------- 5 3 whence and also from the fact that the affix of point //is equal to a + b + c it follows that OH = 30G. That is, the points O, G, H are collinear. 2°. The midpoints Q of the sides BC9 CA 9 AB have affixes b+ c c+a a+ b ax —-------- 9 b1 = --------* Ci —--------• Let E be the midpoint of line OH. Its affix is a+ b+ c e ------------------2 Since s — ax = aj2 9 e — bx = bj2, e — c1 = c/ 2 and \a\ = |6 | = \c\ = 1, it follows that Ifi - flil = Ifi - bx\ = |6 - cx\ = 1/2 = Rj2 (R = 1) that is. EAX= EBX= ECt = R/2. Thus, the midpoints A l 9 B l 9 Cx of lines BC, CA9 AB are at equal dis tances—R/2—from the midpoint of line OH and, hence, lie on circle (A ^ C x ) with centre E((a + b + c)/2) and radius R/2. Furthermore, the equations of the straight lines AH and BC are of the form z — a = bc(z — a), z — b = — bc(z — ~b) or z — bcz = a — abc, z + bcz = b + c.

Problems in Geometry

Forming the half-sum of these equations, we find the affix z= a 2of point ^ 2: be a+ b+ c be -------p -- G__ a2 = 2 la la ’ whence be 2a

EA2 = |a2 — e| =

1 - JL 2

”

(|a| = \b\ — \c\ = 1). Similarly, proof is given that EB2 = R/29 EC2 = R/2. Thus, the points Al9 Blf Cl9 A2, B2, C2 lie on a circle whose centre is the midpoint of line OH and whose radius is equal to R/2. Furthermore, the affixes of the midpoints A3, B3, C3 of lines AH, BH, CH are, respectively, ci Q a3 = --------

Q = e H---- >

whence EA3 = \a3 - e| = \a/2\ = 1/2 = R/2 and, similarly, EB3 = ECZ = R/2. Thus, all nine points Ak, Bk, Ck (k = 1,2, 3) lie on the circle (E, R/2)9 whose centre E is the midpoint of line OH and whose radius is R/2. 3°. Let us now go back to the equation of the straight line AH, z — a = bc('z — a), that passes through point A perpendicular to BC. Now find the affixes of the points of intersection of this line with the unit circle zz = 1. Provided \z\ = 1, we have . \/ ---------1 1 z — a = be \ z a or z — a = — be

z —a az

Complex Numbers in Plane Geometry

One of the roots of this equation is z = a (the affix of point A); the other is be the fact that point N with this affix belongs to the unit circle follows he ‘ 1 - H We will prove that the midpoint of line HN coincides with point A2. Indeed, the affix of this midpoint is , , , be a -r b -V c— — , a be ---------- ~------ — = 8 - — = *22 2a Thus, point N coincides with point A4. In similar fashion, proof is given that the altitudes emanating from vertices B and C intersect the circle {ABC) in the points 1?4 and C4,which are symmetric to the orthocentre H with respect to the sides CA and AB. Problem 2. (Boutain points). Suppose A ABC is inscribed in the unit circle (0). Let zl9 z2, z3 be the affixes of the points A ,B ,C . A Boutain point for A ABC is a point such that when it is chosen as the unit point (that is, a point whose affix is z 1), the equation cr3 = z1z2z3 = 1 holds true. Prove that for a given circle (0) and A ABC inscribed in it, there are three Boutain points on the circle (0) that form an equilateral triangle. Proof. Take the point a (|a| = 1) for the new unit point. Then the new from the equality

affixes of the points A, B, C will be — , — 9 — • The point a will be a a a the Boutain point if the product of these affixes is equal to 1: Z1 Z2 Z30 — 11 or a31 = (7, -----This equation has three roots: 3,

£21^3.

3 __

where /cr3 is any one of three values of the cubic root of <r3, and e is 3 any one of two imaginary values ^1, that is, either

- I + 1/3 £ = ------ -—-—

-1 -//3 e = --------— -—

For example, let e = — f_ + 1 /3 _ cos

]2 0 o

j sjn |2o°,

Problems in Geometry

then c2 = cos 240° + i sin 240°, and to construct the Boutain points it is necessary first to construct the point for the 3 __

chosen value )[<t3, then multiply it by £ (a rotation through 120°) and then once again by e (another rotation through 120°). We obtain the vertices of an equilateral triangle with affixes 3 ___

3 __

3 ____

f<73, 6jfo3, e2]faa. Problem 3. Given an arbitrary triangle ABC. Take on the circle (ABC) an arbitrary point M (Fig. 8). Let Al9 Bx and Cx be the orthogonal pro jections of M on the straight lines BC, CA, and AB. Prove that the points Al9 Bl9 Cx are collinear. Taking the circle (ABC) for the unit circle and assuming that the affixes of the points A9B9C, M are equal, respectively, to zl9 z29 z39z09 set up the equation of the straight line A ^ C x (the Simson line for the point M with respect to the triangle ABC). Solution. The equation of the line BC and the straight line passing through point M perpendicularly to BC is of the form z — z2 — z2z3(z — Z2), Z Zq z2z3(z %o) or z + z2z3z = z2 + z39 Z

Z2Z3Z

—

Z2Z3Zq.

Forming the half-sum of these equations, we find the affix z = ax of the orthogonal projection of point M on the straight line BC: = — (z0 + Z2 + z3 - Z2Z3Z0). 2 Similarly we can find the affix bt of point Bt :

(*)

bx = -

(z0 + z3 + Zj — z3z1"z0). 2 Let us find the slope of the straight line A1B1; we have — b\ — — [z2 \ ,

-X

z3z 0(z2

1 /

Zj}]

z3 \

1 (z2 - Zj)(z0 - Z3) _ •

= -2- ( I , - * , X I - V i - -2 f e — ) ( 1 - - ) = 7 ----------

a, - J , - - ( 1-----M ( - --------- 1 ) - — 2 Vz2 z j [ z 0 z3 / 2

- Zi)(z. - z j — zxz2z3

Complex Numbers in Plane Geometry

and so *_ A = *l 03ZO* ai bi zo The equation of the straight line AXBX may be written as Z

Cl i

—

G3Z

q{ z

&i)

or, taking into account equation (*), z — “ (z0 -\- z2

z2z3z 0)

„ . A _ r * _ i ( ± + j . + ± _ i ) i . L 2 \ Z0 z2 z3 z 2z 3 / J This equation can be simplified: z — • <73 Z 0Z =

1 / / Z2Z3\ -(z0 + Z2 + Zs ---------------

z0)

2 V

/ — + — + — 2z0 I Z 0 Z2 Z3

Z iZ 2Z3

or z

tr3z0z

1( | i z2z3 I Z 0 + z2 + z3 — 2 V zo

<t3

ZXZ3

ZiZ2

or z — asz 0z = — (z0 + ox — <r2z0 — 0**0). 2

( 1)

where <Tl —

®2 — ^1^2 “I"

”1“ Z3Z1,

O* —

The symmetry of this equation with respect to zx, z2, zs permits asserting that line also passes through point C,. Incidentally, this becomes evident by substituting into the left and right members of equation (1) the affix Cl = -- (z0 + zx + Z2 — ZjZjjZo) 2 of point Cj (the results will be the same; check this!) or by seeing that the following equation holds: aj ax 1 bx bi 1 = 0 cx c\ 1 (this is left to the reader to verify).

Problems in Geometry

Ifz0is the unit point, then the equation of the Simson line (1) assumes the form z -< 7 3z = — ( l + a 1 — a2 - ( 7 3),

(2)

and if we take the Boutain point for the unit point (that is, <x3 = 1), then z

((ti - <r2).

(3)

Remark. It will be proved below that if point M does not lie on the circle (ABC), then its projections Al9 Bl9 Cx on the sides BC, CA, AB do not lie on one straight line. Problem 4. Prove that if for the unit point we take the Boutain point M with respect to &ABC inscribed in the unit circle (O), then the Simson line corresponding to point M will be collinear with the diameter of the unit circle passing through M (see problems 2 and 3). Solution. The equation of the Simson line will have the form z-z=j((7

(3)

which means it will be collinear with the x-axis or the diameter of the unit circle passing through point M because M is the unit point of the A>axis. Note the converse: if the Simson line for point M of (ABC) with respect to A ABC is collinear with the diameter of the circle (the diameter passing through M), then M is the Boutain point. True enough, for if M is the unit point, then the slope of the Simson line is equal to <r3 and if the Simson line is parallel to the diameter passing through M , that is, parallel to the x-axis, then o3 = 1, since the slope of the x-axis is equal to 1. Problem 5. Let us consider A ABC inscribed in the unit circle (0). Prove that: 1°. The point P with affix <x2 is symmetric to the orthocentre H of A ABC with respect to the diameter 8 of the unit circle, which diameter is parallel to the Simson line constructed for the unit point with respect to A ABC. 2°. The point Q with affix a3 is symmetric to the unit point with respect to the diameter 8. Solution. 1°. The equation of the diameter 8 is of the form Z

— <73Z — 0

[see equation (2) of problem 3]. The affix of the projection of the ortho centre H on this diameter is found from the system of equations z — o^z = 0, z — o1 = — cr2(z — d±)

Complex Numbers in Plane Geometry

or z

—

0, +

3g ±

or, since dx = — , it follows that Z

—

Z +

<73Z = 0, ff3 Z =

C7X +

G2.

Adding these equations, we find the affix A of the projection of the orthocentre H on the diameter S: X—

G2 2

The affix co of the point symmetric to point H with respect to the di ameter S is found from the equation (O + Gx Gx + <72 2 2 whence ”

co =

’

2°. The equation of the perpendicular dropped from the unit point to the diameter 8 is of the form z — 1 = — g z ( z — 1). Solving this equation together with the equation of the unit circle zz = 1, we obtain

z The roots of this equation are: z = 1, z = g 3 . Problem 6. 1°. Let A l9 Bl9 CL be the orthogonal projections of point P on the sides BCy CAf AB of A ABC. We take the centre O of {ABC) for the origin. Let Rzl9 Rz2, Rz3 be the affixes of the points A , B9C, where R is the radius of the circle (ABC) = (|zx| = \z2\ = [z3| = 1). Let p be the affix of point P. Find the affixes al9 bl9 cx of points Al9 Bl9 CL and then express the area ( A ^ C J of the oriented A A lBlC1 in terms of the area ------> {ABC) of the oriented A ABC9 in terms of the lengths a9b9c of the sides BC9CAy AB and in terms of the affix p of point P.

Problems in Geometry

Consider the following special cases: 2°. The point P coincides with the centroid G of A ABC. 3°. The point P coincides with the centre I of the circle inscribed in A ABC. 4°. The point P coincides with the centre O of circle (ABC). 5°. The point P coincides with the orthocentre H of A ABC. Solution 1°. The equation of line BC is z — z2R = — z2z3(z — z 2R) (4) z

+ z2z3z = R(z 2 + z3)

or z + z2z3z = R(ol — zx). (5) The equation of the perpendicular dropped from point P to line BC is of the form z - p = z2z3(z - p)2 or 2 Z^Z^Z = p Z2Z3/7 or z — z2z„z = p — ZyOzp. (6) Forming the half-sum of equations (5) and (6), we can find the affix z = ax of point Ax: ox = — (R<rx + P — Rzx — z xoip). Similarly, bx =

(Ro 1 + P ~ Rz2 — z 2o3p),

Ci =

- -

(Rox + p — R z3 — ZaPsP)-

From this we get — (Rcr1 + p —Rzx—~zx<j3p) i (Rdx+ p —R z x—zxd3p) 1 2 2

Complex Numbers in Plane Geometry

Rzt -f z xa3p R z x + zfixp 1 / Rz 2 + z 2a3p R z 2 + z2d% p 1 16 Rz 3 + z 3<T3p R J3 + z3d3p 1 z"l Zj 1

Zj Z i 1 l /p z 2 z 2 1 + 16 Z3 Z 3 1

z 2 z2 1 z3 1

W - PP

4R2

R* _ Q p 2 (ABC) = -------------(ABC). 4R2

Note that OP2 — R2is the power o> of point P with respect to the circle (ABC). Therefore we finally have (A & C J = ---- (ABC). 4R2 2°. If point P coincides with point G (the point of intersection of the medians of A ABC), then OP2 = OGz = pp = — (Rzy + Rz., + R z3) (R~z\ + R z 2 + R z 3) R2 = — (3 + zt z 2 + z 2z x + z 2z 3 + z3z 2 + z3z x + ZxZ3). But a 2 = (Rz 2 —

Rz3) (R z 2 — RY3) — R2[2 — (z2z 3 + z 3z 2)],

whence Z2z 3

z3z 2 — 2

and, similarly. z3Zj -)- z1z 3 — 2 zxz 2 +

Z2Z l

R2 -

Problems in Geometry

that

/j2 bp rp\ a2 + b2 + c2 3 + 2 - — + 2 -------- 1-2--------- 1 = R2 R2 R2 * 2/ Consequently, a2 + b2 + c2 R2 - OG2 =

~ -t (

and therefore sp

-1 -

hP -4- (p

But R

°bC 4\(ABC)\

hence

3°. If point P coincides with the centre / of the circle (I) inscribed in A ABC, then pp = OP2 — Ol2 = R2 — 2/?r (Euler's formula, see problem 33 below) and, hence. _ (jR2 „ 2/?r) (^ 5 C )= — (ABC) (^ A C ,) 4/P 2R <, 1 ( ^ 0 1 a° c 2M2fC)|

abc(a + b + c)

Thus, (M C i)

4Q4,8C)3 abc(a + 6 + c)

Remark. Note the formula (^ A A ) _ r (,4£C) 2/? ’ where AX, B X, Cx are the projections of point / on the sides 2?C, C4, ,41?. 4°. If the point P coincides with the point O, then pp = OP2 = 0 and, consequently, ( A ^ C ,) = — (^BC) = A (^BC) 4/?2 4

Complex Numbers in Plane Geometry

(this is also immediately clear on the basis of elementary-geometry reasoning). 5°. If the point P coincides with the orthocentre H of triangle ABC, we have R2 — OH2 R2 — (3OG)2 (AJB1C1) = (ABC) = (ABC) 4R2 4R2 R2 - 9 I R 2 -

a2 + b 2 + c 2)

4R2 f a2 + b 2+ c2 V 4R2

(ABC) =

a2 + b2 + c2 - 8R2 (ABC) 4R2

a2 + b2 + c2 - 2 \ (ABC) a2b2c2 4(ABC)2

Q2 b2 Since —° — sin2A, = sin* B, -----= sin2 C, it follows that 4R2 4R2 4R2 (A1B1C1) (ABC)

a2 -4- W - L ------------ — 2 — sin2A + sin2B + sin2 C — 2 4TP 1 — cos 2i4 1 — cos IB t 1 — cos 2C H-------- 7 I-------- 1 2 2 2

= ---- (1 + cos 2C + cos 2A + cos 2B) 2 = ---- [2 cos2 C + 2 cos (v4 + 5) cos 04 — 5)] 2 = — cos2C—cos 04 + .6) cos (A — B) = — cos2C + cos Ceos 04 — 2?) = cos C[cos(y4 — B) — cos C] = cos C[cos 04 — B) + cos (A + B)] = 2 cos A cos B cos C. Thus, if A1B1C1 is a triangle that is orthocentric with respect to A ABC (that is Alf Bl9 Cx are the feet of the altitudes of A ABC), then (W i) (ABC)

2 cos A cos B cos C,

Problems in Geometry

whence, incidentally, it follows that if triangle ABC is an acute-angled triangle, then A ABC and A AXBXCXhave the same orientation (Fig. 9), and if it is an obtuse-angled triangle, then opposite orientation (Fig. 10).

Problem 7. Given A ABC and a straight line /. Let Ax, Bx, Cx be the orthogonal projections of points A, B, C on /. Prove that the lines that pass through the points Ax, Bx, Cx and are respectively perpendicular to the straight lines BC,CA,AB intersect in a single point Q (called the orthopole of the straight line / with respect to the triangle ABC) (Fig. 11). Taking {ABC) for the unit circle and assuming / is given by Z — z0 = x(~z -

z0),

\x\ = 1,

(7)

find the affix o of the orthopole Q of / with respect to A ABC. The affixes of the vertices of A ABC are equal to z l9 z 2, z3 respectively. Solution. The equation of the straight line passing through point A perpen dicularly to / is of the form z — z x = — x( "z -

Z j).

(8 )

Rewriting equations (7) and (8) as Z — K Z — Z0 — X Z 0,

z + xz = zx + x z x and forming the half-sum, we find the affix z = ax of point Ax: ax = — (z0 + zx + x z x — x z 0). 2 The equation of the straight line passing through point Ax perpendicularly to BC is of the form z - fli = Z j Z * ( z - a x)

Complex Numbers in Plane Geometry

or z — J (z0 + z, + x z \ - XZ0) = Z^ 3 ( z - — ( z 0 + J - + — - — j j or ...

Z —

z *,z3z

If 2 V

, X Zi

= — z„ + Zj H---------- x z 0 Z2Z3

„ ^ -

°3

Z 2Z3Z0 ^

-------------- z2z3z 0 ----------1------------- •

/Q \

(9)

Zx X X ) Similarly, the equation of the perpendicular dropped from point B1 on AC is of the form 2 —Z3Zx¥ = - U + z2 + -

- XZ0- Z*1 - z3z1z„ - ^ + . (10) z 2 x x ) Subtracting equation ( 10) from (9) termwise, we find the number co that is conjugate to the affix co of the point of intersection of lines (9) and (10): 2 v

z a(z i ~ z i )

z 2

= [Zl“22+X(~— “j l-

+ zs ( — — — ) — - z*)z<> (^1 - z2)] \ Z2 zi J x J or, cancelling z, — z2, Z3Z i -J- Z3Z2 _ Z;3Z0 \ 1 (. x + Z3 Z 0 — z&> = — 1 -------2 V ZjZ; ZiZo ■ rj' whence co

\ Z3

Z jZ 2Z3

Z jZ 2

^ /

or CO 2 U 3

* /

whence CO =

1 / (T2

* , - \ — - — + z0 - XZo] \ Oz oz }

1 But a 2 = — * ct3 = — and so O3 03 CO

= — (o'1 — — + zo — * z0j

( 11)

The symmetry of this equation with respect to zl5 z2, za permits stating that the perpendicular dropped from point Cx on line AB will also pass

Problems in Geometry

through the point with affix co defined by ( 11), that is, ( 11) defines the affix of the orthopole of the straight line / with respect to A ABC. Inci dentally, this is evident from the following: write down the equation of the perpendicular dropped from Cx on the line AB and convince yourself that the number co given by ( 11) satisfies the equation of that perpendicular. In particular, if / passes through the centre O of the unit circle (O) = = {ABC), then the affix co of its orthopole with respect to A ABC is

and if the Boutain point is taken for the unit point, then o> = ~ ( a 1 — x).

(13)

Figure 11 depicts the construction of the orthopole Q of line /with respect to A ABC. Remark. There is an elegant synthetic proof (which belongs to the French geometer R. Deaux) that the straight lines lx, l2, /3 that pass through the points Al9 Bl9 Cxand are perpendicular to the lines BC, CA9AB respectively intersect in one point Q. Let A2, B29 C2 be the midpoints of sides BC, CA, AB, then the radical axes of the circles (A2, A2CX), (B2, B2AX), (C2, C2Bt) taken pairwise are the straight lines ll912, /3, and therefore they intersect in the single point Q (Fig. 12). Problem 8. Given a triangle ABC and a straight line /. The circle (O) = {ABC) is taken to be the unit circle; zl9 z2, z3 are the affixes of the points A, B,C, respec tively. The line / is given by the self conjugate equation Hz 4 - az = b (14) where a ^ 0 and b is a real number. Find the affix co of the orthopole Q of line / with respect to the triangle ABC (see the preceding problem). Solution. The slope of the given straight line is equal to x — — ~ a and the slope of the line perpendicular to / is equal to x' =

• a The equation of the straight line that passes through point A perpendi cularly to / is of the form z - Zl = — (z — z x)

Complex Numbers in Plane Geometry

or (15)

az — az = azx — a zx.

Adding the equations (14) and (15) termwise, we find the affix ax of the projection A1 of point A on I: ai =

az1 — a z1 + b Id

The equation of the straight line passing through point A1 perpendicularly to BC is of the form z — a1 = z2z3(z — a j or azx — a z x + b /_ a zx — az1 + b ' z ------------------- — Z2z 3 I z 2a 2a 0 or a z x — azx + b azt — a zt + b (16) z z2z3z — ^2^3 25 2a In similar fashion we can write down the equation of the perpendicula to AC that passes through the projection Bx of point B onto /: z — ZjjZiZ

az2 — a z 2 + b — la

az2 Z3Zi

— az2 + b 2a

(17)

Subtracting equation (17) from (16) term by term, we get the number dj, which is the conjugate of the affix co of the orthopole Q of line / with respect to triangle ABC:

(zi -

z 2)

zzco

(zx - z2) a + ( - ---------] a ___________ V*2 ) 2d z2 ( — - — ) a + bzz(zx - z2) V *2 *1 /______________ + 2a

and, cancelling zx — z2, we obtain a zzco = -

a *1^2 2d

z& i + Z2) a„ +, bz3 -------------Z1Z2__________ 2a

Problems in Geometry

2zz

CO —

a 2aaz

1 1

2z x

i N > N 1I to l

whence _

ac* la

2 ^

b 2a

and, hence, " =

2 V

- + -)• a a)

(18)

If / passes through the centre of the unit circle, then 6 = 0 and for mula (18) becomes CO =

(19)

and if the unit point is a Boutain point (a3 = 1), then co

1 ( t r H — 2 \ a

)■

(20)

The equations (19) and (20) coincide, respectively, with (12) and (13) of the preceding problem, since — = — x, where x = — ~ is the slope _ a_ a of / given by the equation az + az = b. Problem 9. Let Alt Bu Cx be the feet of the altitudes of A ABC inscribed in the circle {ABC) = (O), which we take to be the unit circle. Points P, Q, R are chosen on the straight lines AAU BBU CCy so that AAX AP Find the ratio (PQR) {ABC)

Express this ratio in terms of the interior angles A, B ,C of the given A ABC and in terms of X. Consider the special cases: (1) X — 1, (2) X = - 1, (3) X = 1/ 2, (4) X = -1 /2 , (5) X = 2, (6) X = - 2 .

Complex Numbers in Plane Geometry

Solution. Take the circle (O) as the unit circle. Let zl9 z2, z3 be the affixes of the points A, B,C, respectively. The equations of the straight lines BC and AAX are z + z2z3z = z2 4“ Za, Z2Z3 Z

ZoZoZ — Zi

Combining these equations term by term, we find the affix z = ax of point Ax: ai = and then from the relation AAX

= A

AP we find the affix p of point P : cii - zx — A, p - Zi Ap = <7i -

Zj — Xp

Azj, + Azr

+ Azx = y

whence 1 Z'

<r3 \ , A — 1

P ~ 2x Similarly,

z f ) + ~~A f

O’3 '\ . A — 1 I+ A 4 ,1

(T3 )I , A - 1 0-1 - “7 + I A

wlicre q and r are the affixes of points Q and R. We now find ___g3 , 2Azf ^ (/'(>«)=

^3 2;.za2 <r3 2Az§

- z 4- — F " 1 2A — A

r, + ^ 2A

. i n i r, + £ i A 2A

72 A- 1 1 <^2 1 + —+ 2kaa A 2Act3 A - 1 1 I 0-2 -r A *2 2Ac73 A- 1 1 <^2 Z| 1 + 2A<r3 1 A *3 2Ac73

4 ~1 2A<t3 '

Problems in Geometry

100

^3 i_ A - 1 2Azf 1 A

2X<j 3 +

LA - 1 2Az|

4A2

z\ 1

2A<r3

zi A — 1 _1 — 1 A z2

4 2A«r»

A- 1 1 A z3

,2 Z2

A- 1 z2 2Azl 1 A

z-2 'z zf 1 -Z2— “ 2‘ zl 1

A— 1 1

(73(A -1 ) zf2 z f1 1 2A2 z3“ 2 Z3-1 1

Zl zf 1 A- 1 z2 zf 1 ' 2AV, Z3 z§ 1

Zl Z f*

(A - l )2 z2 z f 1 A2

(21)

Z3 Z3 - 1

Furthermore, z r 2 z2 1 z f 2 zf 1

1 z2 zf

Z3-2 z3 1

1 z§ z£

z?)(z§ - z2)(zl - zf)

1 zl z|

---- -- (z2 — Zi)(z3 — z2)(z3 — Zi)(z2 + Zi)(z3 + z2)(z3 + 2l) <j\ 1 2i 2 l l 2 2 2 l ( 2 2 + z i ) (Z3 + Z2) (Z3 + Zl) ol 1 2 3 2§ Z1 1 Z2 1

<*3

z3 1 =

2l 22

(<?1 -

2 i)(ffi -

Z2) ( f f i -

Zz)

1 4 (ABC) 03 I

(<7i

a2 — aq) :=

—

and the first term in round brackets in (21) is /(l — ad1) i ■(ABC). A2

(22)

Complex Numbers in Plane Geometry

101

Subtracting the second term, we get Z f2 Z l1

z2~ 2 Z2 1

*1 1 *1

4/ = ------ {ABC), <*3

*2 1 *2

“ a3

z f 2 Z3 1 1

*3 1 *3

so that the second term in round brackets in (21) is equal to 2/(2 - 1)

(ABC).

(23)

Furthermore, Zi z \ 1

1 Zy Zy

1 r2 z 2

z \ 1 = ^3

1 Z3

z%z| 1

= —4ioz(ABC),

and so the third term in round brackets in (21) is 2/(2 - 1)

(ABC).

(24)

— ~(ABC).

(25)

22 Finally, the last term is equal to -4 i—

From the formulas (21)-(25) it follows that (PQR)

i f

/(I — aydi) [ 2/(2 — 1) 22 22 +

2/(2

22 —W y 422

- 4 /; a - I )2 ] (ABC) 22

(2 - l)2 '

whence ( /15C)

422

( - D ( - t)

(ABC)

102

Problems in Geometry

But adx — 1 is the power of the orthocentre H of A ABC with respect to the circle (ABC). We have od1 — 1 = (zx + z2 + z3) (Zi + z 2 + z3) — 1 = 2 + (zxz 2 + z2z\) + ( z 2z 3 + z 3 z 2) + ( z 3 z i Furthermore, AB2 = c2 = (z2 — z2) (z 2 — Zi) = 2 - (zxz 2 + z2zi), whence ZjZ2 + z2Z! = 2 — c2 = 2 — 4 sin2 C, since

Z iZ g ).

—— = 2R = 2. sin C In similar fashion we find z2~z3 + z3z 2 = 2 — 4 sin2 A, ZgZ^ + ^1^3 = 2 — 4 sin2 P and therefore (see problem 6) Gj&x — 1 = 4(2 — (sinM + sin2 B + sin2 C)) = — 8 cos A cos B cos C, and formula (26) becomes = (PQK) _ 2 cos A cos B cos C A _ J \A _ A V 07) ** (ABC) ~ A2 aJ\ A/ ’ 1 ' (1) If A = 1, then the points P, Q, P coincide, respectively, with the feet of the altitudes of triangle ABC. Answer, p = 2 cos >4 cos P cos C. (2) If A = —1, then the points P, £?, P are symmetric to the feet Alf Bl9 Cj of the altitudes of A ABC with respect to its vertices A, B, C. Answer, p = 6 + 2 cos ^4 cos P cos C. (3) If A = 1/2, then the points P, Q, P are symmetric to the vertices ^4, P, C of A ABC with respect to its sides. Answer, p = 3 + 8 cos A cos P cos C. (4) If A = —1/2, then the points P, g , P are obtained from the points A 1, P l5 Cx via the homothetic transformations (A, —2), (P, —2), (C, —2). Answer, p = 1 5 + 8 cos ^4 cos P cos C. (5) If A = 2, then P, Q, P are the midpoints of the altitudes AAl9 PPX, CCX of A ABC. Answer, p = — cos A cos P cos C. 2 (6) If A = —2, then the points P, (?, P are symmetric to the midpoints A2, P 2> C2 of the altitudes y4/tl5 BBl9 CC± of A ABC with respect to its vertices. Answer, p = 3 -----cos A cos P cos C.

Complex Numbers in Plane Geometry

103

Problem 10. Let A', B', C be points symmetric to the vertices A, B ,C of A ABC with respect to its sides BC, CA, AB. What relationship be tween the angles A, B ,C of A ABC is necessary and sufficient for the straight lines AB', BC', CA' to pass through one point (Fig. 13)?

Solution. Take (ABC) for the unit circle. We find the affix b' of point B'. The equation of the straight line AC is z + z3zxz = zx + z3. The equation of the straight line passing through point B perpendicularly to AC is Z

Z 2 — ZgZ^ (Z

Z 2)

or Z3Z1 • *2 From the equation of AC and from this equation we find the affix of the projection of point B on line AC: Z

—

--- ZgZAZ

z H 2i+ 2,+ z’ ~ ~ t y

The affix b' of point B', which is symmetric to point B with respect to AC, is found from the relation b' + z2 2

whence b' = zx + z3

*3^1 ^2

Problems in Geometry

104

From this we have z2 Z3Z1

b’ = z \ + z 3 —

za z2 zxzz

We now set up an equation of the straight line AB': z z

z l

b’ bx 1 or z ( z x — b') + (b' — z x) z + z ^ ' — 6 'Z i = 0 .

We have — -, 1 Z3 - z1 z2 z x — b = ----------------------Zi

^ 3 Z1

z 3z x

_1____ 1_ u, Z1 —

z 2 =

z 3( z 2 —

Z3(Z i — Z2)

------------------------- =

-------------------------------------- f

—

Z 1=

Z j)

---------------------------------------9

Z3Z1

_ Z3Zt \ 1 [ — Z3 + *1 Zx V *2 /

Z3 + Zx -

z3 + Zl — *2 /— ZX= Zi Z3Z1

- 1 + * =__ *1 z2 Zl *3

- ( Z x - ■Z2) + z3 ( — Z3 V *2

_^3 _J_ ^3 Zl z2

_^2_ z3

■(z1 - z 2) + z 3^ “ ^ ZiZ2 *3

*i /

= (zx - z2) ( — +

V Z3

and the equation of AB' takes the form 1

Z3Z1

*+ ^3

J3 _ = 0 *1*2

or z2z + z3zxz — zi — zxz2 = 0 .

ZxZt

Complex Numbers in Plane Geometry

105

In similar fashion we can write down the equations of the straight lines BC’ and CA'. Thus, Z .J I

+ z\zxz — zl

zxz2 = 0, (AB')

—

z3z 4 - zfz2z — z\ — z2z3 = 0, (BC) zxz + z|z3z — z\ — z3zx = 0.

(CA')

Denote by K, L, M the points of intersection of the lines B C and CA', CA' and AB', AB' and BC. Then z2

z 3z x

z§ +

z xz 2

727 z3 zx z2 .2 Zf + Z2Z3 —2 —

z x Z2Z3

(KLM) =

z-.22 + Z3ZX - 4) (zi4 - 4 )

o-3(z2z§ -

Z?z£ + Z]Z2Z3 + z\z\

Furthermore, z 2 z |z x

A = z3

Z1Z2

z\ Zl

z xz 2

z2z3

zx zfz3 zl + Z3ZX -7^7 73^ Z1Z2 ZiZ2Z - 4 4 ( 4 - 4 ) - 4 4 ^ 4 2 - *i)

4 (4 -

z xz 3z 2 +

z\z\

z xz 2z 3

z\z\— zxz\z% — z\z%

z ? z lz 3 — +

z\z3zx(z\ -

z f) +

z\(z\

z f) -

z\yt

- zlzxz2(z2 — zx) = (z2 — zx)(zfz2 + z\z\ — zfzfz3 + z*3z2 + z\zx + z\z\zx -I- z\z2z\ — z\z\ — z§z|zx — z§z2zf — z\z\ — zlzxz2) = (z2 — zx)[zfz2(zx — z3) + zf(zf

—

Z2Z§(zf - 4 ) — Zxz|(zf — 4 )

: - (z2 - zx)(zx - z3)(zfz2 + zfzx + zfz3 — z2zxz3 =

(z 2 —

0*2 -

z x) ( z x -

z x) ( z x —

z 3) [ z ! ( z | -

z 3) ( z 2 —

z i)

z 3) ( z f z 2 +

z fz 3 +

z 2z 3

zl)

z x( z f -

ZjZ2Zx(zx — z3)]

zfz§

z xz 3

z 2z 3 ( z i —

-|-Z3ZxZ2)

z\ ) ]

z xz l + z xz 2z 3 + z xz l + z l z 3 + z 2z § )

(Z2 - Zx)(z3 - zx)(z3 — z2)(zfz2 + z|zx + z\z%+ zfz2+zfzx+z?z3+ z xz2z3). Consider the product (/..

z 3) ( z 3 +

z x) ( z x + =

z 2) = 2 Z i Z 2z 3 +

—

<t 3

z?z2 +

z |Z i

from this it follows that the last factor so (hat A = (z2 - Z j ) ( z 3 - z x) ( z 3 - z 2)

z |z 3 +

z 3z 2 +

— o3 — c3 = (<rx<r2 -

2 ( t 3) .

z |z x +

z\z3.

— 2<r8

106

Problems in Geometry

Thus, {KLM) =

(z2

Z1) 2(ZS —

—

Z ^ { z z

—

Z 2) 2 (< T1<T2

— 2(73) 2

<r3(z2z§ — zi)(z3zi — z^Cz^l — z;j)

Note that (z3 - z2)(z3 - z1)(z2 - zx) 1 zx z? = 1 z2 zf = 1 z\

1 zt

Zl *1

z 2 1 z2 = *3 z2 z 2 1

— 4icr3(ABC),

Z3 Z3 1

*3 1 *3

so that (KLM) = -

4ic3(o‘1(T2 — 2<t3)2(ABC)2 (z2zi - z?)(z3z2 - z i ^ z 2 - zl)

The lines A B \ BC \ CA' pass through one point if and only if axa2 — 2<t3 = 0 or Gx

— — 2 = 0, <^3

G l (Tl

— 2 — 0,

that is, c r ^ = 0 / / 2 = 2 = 2R* (R = 1) or O H = R y2. Incidentally, it follows from this that the triangle ABC is an obtuse-angled triangle since OH > R, that is, the orthocentre H lies outside the circle (ABC). A triangle for which OH = R )[2 may be constructed as follows: con struct two concentric circles (O) and (O') with radii 1 and ^2 (see Fig. 13). On (O') take an arbitrary point H and on (O) an arbitrary point A. Divide the line segment OH in the ratio 1 :2 : OG : GH= 1: 2. Then G is the point of intersection of the medians of the triangle ABC. Join A and G and on the extension of segment AG beyond point G lay off segment GAX= AG/2. The point Ax is the midpoint of side BC and therefore, by drawing a line through Ax perpendicular to the straight line OAx we obtain points B and C as points of intersection with the circle (0).

Complex Numbers in Plane Geometry

107

The condition that makes the construction possible is that point Ax lie inside circle (0). Since point is the image of point A under the homothetic transformation

-----j > and under that homothetic transforma

tion the circle (O) goes into the circle (O"), the radius of which is equal i Vi Vi Vi to -y, and the centre O" lies on segment OH, 0 0 " = — + — = ~ , it follows that the circles (O) and (0") intersect in points P and Q. The point A can therefore describe an open arc PRQ of the circle (0). From the triangle OPO" we have PO"2 = OP2 + 0 0 "2 - 20P-00"cos(a/2), where a = /_ POQ. And since PO" = 1/2, 0 0 " = ^2/2, OP = 1, it follows that =

a i i1/2 _ cos —

whence a cos — 2

_5_

4]fi

and, consequently, 25 9 cos a = 2 - ---------1 = — t a = / POQ = arc cos (9/16). 32 16 The condition OH 2 = 2R2 may be rewritten in a different form. Since OH2 = 9R2 — {a2 + b2 + c2) (see problem 6), it follows that 9* 2 - (a2 + b2 + c2) = 2R2 or a2 + b2 + c 2 = 1R2. And, in yet another form (since a = 2R sin A and so forth), sin2 A + sin2 B + sin2C = 7/4 or

cos A cos B cos C = —1/8 (from this it also follows that ABC is an obtuse-angled triangle). Problem 11. Inscribed in a circle (0) is a regular 14-gon:

108

Problems in Geometry

Let us consider the triangle T = ABC with vertices A = A 7, B = Al9 C — A 3. The angles of this triangle intercept arcs AXA39 A3A7, A7A± and, hence, A = n p y B = 2n/79 C = 4n/l (Fig. 14) (that is to say, the angles of tri

angle T form a geometric progression with ratio 2). Denote by (/fl) the circle escribed in the angle A of A ABC, and by H the orthocentre of A ABC. Prove that: 1°. OH = 01 a = R ]/2 (R is the radius of (0)). 2°. R = 2ra (ra is the radius of circle (Ia)). 3°. IaH = R. 4°. a2 + b 2 + c2 = 1R2 (here, a9 b9 c are the lengths of sides BCy CA9 AB of the triangle; do not confuse with the affixes of its vertices!). 5°. OIaHAQ is a parallelogram whose centre coincides with the centre of the Euler circle of triangle ABC.

Complex Numbers in Plane Geometry

109

6°. The midpoint P of segment HA6 coincides with one of the points of intersection of (O) and (Od) ((09) is the Euler circle of A ABC).

T . The triangles AIaH, HBIa, IaHC are similar. Determine their orientation (which pairs have the same orientation and which are oppositely oriented). 8°. The straight lines BC, CA and AB intersect line HIa in points sym metric to the points A, B, C with respect to the bisectors of the angles C, A, B of triangle ABC (for angles C and B take the bisectors of the ex terior angles). 9°. The squares of the lengths of the sides of A OIaA&and the squares of the lengths of the sides of A A3IaH form a geometric progression with ratio 2 . Solution. 1°. Take (O) as the unit circle. Give point A± the affix 1 (that s, A± is the unit point). Then the affixes ak of points A k are (k — 1) 7T (k — 1) n ak = cos + i sin 7 1 9

* = 1 , 2 , 3, 4, 5, 6 , 7, 8, 9, 10, 11, 12, 13, 14. The affixes of the vertices A = A7>B = Al9 C = A3 are 6n , . . 6n a = a7 = cos------b i sin — > 7 7 b = a1 = 1, 2n , . . 2n c — a3 — cos------ b i sin----7 7

Their sum is equal to the affix h of orthocentre H of A ABC: h= whence

27T 1

6n 1

,. . 2n . 6n sin ------ b sin — 7 7

C O S ----------- b C O S ----------- b I

(

)•

Problems in Geometry

Let -

x = 2

, _ 4ft , „ 6n cos------ b 2 cos------ b 2 cos — 1 1 1

then ft * sin — 7 —

= 2

. . ft 2ft , . . ft 4ft , ^ . ft 6ft sin — cos------ b 2 sin — cos — + 2 sm — cos — 7 7 7 7 7 7

. ft , . 3ft . 3 ft , . 5ft . 5ft , .. f t — sin ------- b sin -----------sm ----------- b sin ---------- sm ----------- b sm ft = — sm — 7 7 7 7 7 7

and, consequently, jc = —1 ; therefore OH 2 = 3 — 1 = 2 , whence =

( /? = ! ) .

Now let us determine the affix ra of the centre Ia of the circle (/,). The bisector of the interior angle B is BA5 since point A 5 bisects arc CA. From this it follows that the bisector of the exterior angle B is the straight line A1Z B since the points A5 and A12 are diametrically opposite points of the circle ( O) , and therefore BA5 _L BA12. The bisector of the interior angle A is AA2 = A7A2. T o summarize: Ia is the point of intersection of the lines A 2A7 and AXA12. Now set up the equations of these lines. The slope of A2A7 is 1, since A2A7 1| A±A8, and A ^ g is the real axis. Hence, the equation of the straight line A2A7 is of the form ft . . . ft \ c o s ------- b i sm — I

(

/ n . . 7r \ — I c o s ----------i sm —

7 /

z — z = 2i sin

(28)

The slope of the straight line AXA12 is #12

#12

1 —

#12

and, consequently, the equation of AXA12 is

7 /

#12

Complex Numbers in Plane Geometry

111

or i ( 4;r . . An \ z — 1 = I cos— - + i sin — J (z — 1).

(29)

Solving the system (28)-(29), we find the affix ra of point Ia From equation (28), — ~ • 71 z = z — 2i sin — 7 and equation (29) takes the form « ( An . . . An \ ( . 7u \ z — 1 = 1 cos-------b i sin---- I | z — 2/ sin ------- 1 1 , V 7 7 M 7 ; . . . 7r ( An , . . An \ 2/ sin— I cos-------b / sin---- I j ________ 7 V 7__________ 7 j . . 9 2n . 27t 27t 2 sm2 -------- 2/ sm ----- co s----7 7 7 . .

(

An \

/sin — • cos------- b *sin----- 1 = 1 a. 7 V 7 7 j 1 . .

2n (

2n \

i sin---- cos--------- b i sin----- 1 7 V 7 i J 7T / 2n t . . 2n \ sm — I cos-------b i sin----- I 7 V 7 1 ) = 1+ . 27T sm---7 2 ;r cos 1+

n 7

2 cos —

+ i sin — 7

From this / .

Oil = ItJ 2 =

2tT \

I 2 cos — + c o s---- I I 7 7 j

+ sin2

4 cos2 a cos22 ^ i a ^ 2»n 9 2n . 9 2n A ----- b 4 cos — co s-------b cos2 ---- + sm2 ----7 7 7 7 7 n 4 cos2

Problems in Geometry

112

2^1

COS

+ 2 ^CO S

COS - -

4 cos2— 7 2n An 6n\ 3 + 2 cos--------cos-------- COS----- I

(

7 4 cos2— 7 271 2 ft 3 + 2 ^2 cos ------- cos----7 7

7; 671 4rc\ ------- cos----- I 2 _________ 7^/

4 cos2-717 ^ 271 , 47T t 6tt 1 But c o s ---- + cos-------- b cos---- = -------, therefore 7 7 7 2 2 tt 3 + 4 cos + 1 1 + cos __________ 7 = 2, on= cos* 4 cos2— 7 whence OIa = ][2 = R f 2 = OH. . ^4 n f . n . 2n . 3n \* 2 . ra = p tan - - = p ta n ---- I sjn------b sin------ b sin---- ) • 2 14 V 7 7 7 j Let 7T

. 2 7 1

3 tT

v = sin----- b sin-------b sin---1 1 1 Multiplying both sides of this equation by 2 sin

we obtain

^ . n ^ . n n , . . 7r . 2n , - . n . 3n 2y sin---- = 2 s in ------sin------ b 2 sin ---- sin --------b 2 s in ---- sin —14 14 7 14 7 14 7 n 3n 3n 5n 5n In n = COS-------- COS------- b COS-------- COS------- b COS------ cos — = cos — 14 14 14 14 14 14 14 * If the radius of the circle is equal to 1, the chord subtending arc a is equal to 2 sin (a/2). Angle A is an inscribed angle intercepting the arc 2tt/7, hence AXA3 = BC — = 2 sin (7T/7) and similarly for the other two sides. The lengths o f the sides may also be found by the sine theorem: a = 2R sin A = 2 sin (n jl) and so forth.

Complex Numbers in Plane Geometry

113

Hence, y

n c o t-----

— 2

and so n 1 A n 1 ra = ta n ----------c o t------= — 14

whence R = 2ra. 3°. Furthermore, IaM = h — xa {h and xa were computed in item 1°), whence 2k

cos---. 2k , 6k 7 . . 2n h — za = c o s ------ b cos-----------------------b / sin---7 7 _ 7T 7 7 7 2 cos — 2*7 7 -

2 tt

2 cqs — c o s -------- b 2 c o s ----- c o s -------- c o s -----7 7 7 7 7 , . . 2tt = --------------------------------------------------------------- h zsin----71 1 2^ cos — 7 7T 3ft , 5tt 2ft cos----- b COS-------b cos--------1 — cos-----7 7 7 7 .2 n = -------------------------------------------------------- + isin----ft 7 2 cos — 7 7 6ft 4ft 2ft . 2ft — cos--------cos---------cos--------- 1 — cos-----7 7 7 7 . . . 2tt ---------------------- b isin-----7T / 2 cos — 7 7

271 1 271 ------ 1 — cos-----cos----2 7 . 2;r 2 7 . . 2tt --------------------------- b i sin — =>------------------------- b i sin -----7 * 7 2~ cos — 2 cos — 1

7 4tt

cos---- ■+ cos---2 cos---- cos — 7 7 . 2 k 1 7 . 2 k -------------------------- b * sin — = ------------------------- b *s m -----~ ft 7 7T 7 2 cos — ■ 2 cos— 7

7 5k

. .

= cos-------b i s m ---- = cos----- 4- i s m ------ = 7

8 810

Problems in Geometry

114

Thus, h — Tn

cos

. . 5n i sin = a,6 1

and so HIa = |h T .I = 1 = R. 2n n 3tt 4°. a2 + b2 + c2 = 4 sin2— + 4 sin2 + 4 sin2 ~7 7 7 2n 4;t , , 66n ir \ h i — COS C O S ------------ h i — COS 7 7 7 / =

2,1 + cos — + c o s - ~ - j J = 2 [3 - ( - 1/ 2)] = 1—1R2. 2 r 3, — I/ cos----

5°. The directed line segments 7fl/7 and CM6 are equivalent, that is to say, they are parallel, equal in magnitude, and in the same direction; since h — xa = a6 (see item 3°); hence IaHA%0 is a parallelogram. The centre of this parallelogram is the midpoint of the line segment OH, that is, the centre of the Euler circle of triangle ABC. 6°. The affix of the point P is 2n 6tc 5tt h + a Q _ 1 I"jt , 1 + COS------- b COS------- b COS-----■ 1 1 1 2 "2 L . 6k . 57T 2n + sin---- + sin---+ / 7 7 /J 7 27T . . 7T 71 .( 1 : --- 1 — cos ----- b / 1 2 s in ---- -b sin — 7 7 2 L 7 I ■whence I h + a* I2 )2 (2 )1 ^ „n . . 0 2n , A . 2n . n , . _ n \ 2 cos----- b cos2----- b4 sin2 — + 4 sin — sin -----bsin2— | 7 7 7 1 1 1 ) —

=T [(1"cosf + sinV +““f

- i( -

= i r [ 2 “ 2 cosy + 2 ( 1 ~ cos 4 “ ) + 2 ( C0Sf ' “ C0S' 7 L) ] 1 n , , 4tt , 7C 3tt \ 1 f = — I 1 —c o s b 1 — cos-------b cos cos 1= — • 2 = 1 2 V 7 7 7 1) 2 h + ^6

= 1,

Complex Numbers in Plane Geometry

115

That is, the midpoint P of segment HAe lies on (O). The centre 0 9 of the Euler circle of A ABC is the midpoint of segment OH. Thus, 0 9P is 1 P the midline of triangle OHAb and, hence, 0 9P = — OA6 = — . From this 2 2 it follows that point P also lies on the Euler circle (0 9). 7°. We will prove that A AIaH and A HBIa are similar and have the same orientation. We have |ai h 1 Tfl

h Ta lj

—

h h - Tfl 0 = (fl7 - A ) ( l - T j + ( A - T j 2.

—h 1 h

Ta 1

Furthermore, 6k

cos----- \~ i sin------ 1

. . 2k

. .

cos------cos------- i sin-------- ism— 7 7 7 7 2k

= —1 — cos---- i sm — 7 7 2k 2 ;r co s---cos 7 . . K 1 1 -------------------i 7 . sm . - 71- = 11 — t* = 11 — ------- i sin — > ^ K 7 K 7 2 cos — 2 cos so I ®71 3K 3k (Ji — Ta)2 = ai2 = cos-------- .b ^• s•m ------= — cos--------i s. m. -----7 7 7 7 /7

cos — /i) (1 — rfl) = ^1 + cos ---- + i sin — - j

f 1 + cos — y 7

2k .

------- b 1sm— 2 cos. .

co s-------b i sm ----+ i Sin — ) -----77 2 cos — 2k

. .

cos----- b i sin 7 7 3k t . . 3k = I 2 cos2----- b 2/ sm — cos— 1 ---------------------- = cos------b i s m ---v 7 1 1 ) 2co a JL 7 7 ( -

9 K

116

Problems in Geometry

Consequently, A = (It - Taf + (a7 - h) (1 - T0) = 0. ------ > ------> ------ > ------ > To summarize: AIaH[[HBIa and the triangles AIaH and HBIa are similar and have the same orientation (precisely for the order in which the vertices are specified). We can similarly prove that A AIaH and A are similar and have the same orientation. 8 °. We will prove, for example, that point B*9 which is symmetric to point B = Ax with respect to the bisector of the interior angle A , lies on the straight line HIa. The equation of the bisector of the interior angle A is . n z — z = 2i sin---7 The equation of the perpendicular dropped from point AL = B on that bisector is of the form z + z = 2, whence we find the affix bx of the projection Bx of point B on the bisector of angle A: , i , . • 71 b± == 1 i sin ---7 The affix b* of point B* can be found from the relation i* + i whence b* = 1 + 2/ sin

71 1 ’

Furthermore, , 5n . . 5n h — ra = c o s -------h i s i n ---- > 7 7 2n c o s ---Tfl — 6 * = 1 H---------------b i sin n---- 1 — 2i sin 7C 7 7 2 cos — 7 1 2k 2k . . 2n 5k . . 5k c o s ----c o s --------- i s in -----cos------1 sin — 7 . . n 1 1 1 1 =z---------------- / sin — —------------------------- = r -------------------------^ 71 7 Tt ~ 71 2 cos — 2 cos — 2 cos — 7 7 7

117

Complex Numbers in Plane Geometry

and, hence, the ratio h — ra n ------- - = — 2 cos — ta-b * 1 is a real number, and so the points H , /fl, B* lie on one straight line. It is left to the reader to prove the other two propositions of this item. 9°. Let us consider the triangle OIaAQy We have

Oil

T„ 6 =

OA%= 1, 2n \ 2 cosn +sm 2— 1+ 2 cos 7 cos-

cos2

7 . 2 ------- b sin2— n 7 4 cos2 — 7

= 1 +• cos

. Q 7t . 7t 2n A 2k . n 7T #w n 7T 4 cos2 ----- h 4 cos — co s------- \- cos2 ------ b 4 cos2 — sin2 7_________7 7_________ 7__________7_____ 1

4 cos2 --7 1 + 2 ^ 1 + c o s ^ -j + 2 ^cos - ~ + cos— j

4 cos2 ~ ~

~ f

6 k\

3 + 2 1 co s------- cos---------cos------| V 7 7 7; 4 cos2— 7 f 2^ 271 1\ 3 + 2 I c o s-------b c o s ---- 4------I V 7 1 2 ) 4 cos2 - 1

2rc 4 + 4 cos = 2.

4cos2-— 7

And, finally, from the parallelogram OIaHA6 we have (the sum of the squares of the diagonals of a parallelogram is equal to the sum of the -squares of the four sides):

118

Problems in Geometry

Oil + HA§ + 0 A \ + IaH* = OH 2 + IaA l But Oil = HA% = 2 (see items 1° and 5°), 0 A 6 = IaH = 1, OH = f2 ; consequently, 2 + 2 + 1 + 1 —2 + /^ i, whence IaA% = 4. The triangles and A6IaH are equal, and so the squares of the their sides IaH2, also form a geometric progression with ratio 2. Problem 12. Given the linear fractional function az + ft (30) u = --------- J cz + d where a, ft, c, d are fixed complex numbers and ad — be ^ 0, c ^ 0 ; z is the argument and u the value of the function. Prove that if |c| ^ \d\ then the image of the unit circle under this trans formation is a circle; find its radius and the affix of the centre. Solution. Transform the function u as follows: be — ad a , be — ad ^ _ a ^ az + b a c c (cz + d) c cz + d c (1) The transformation z

z H----- takes the unit circle (QJ = (0, 1)*

into the circle (£22) (2) The transformation z +

(Fig. 15). 1

consists in symmetry with

z + respect to the real axis Ox, under which symmetry the circle (Q2)= goes into the circle ((23) =

and subsequent inversion (see chapter

IV) of (&3) with respect to (&+ under which inversion the circle (Q3) goes into the circle (&4); to construct (&4) it suffices to draw the straight line OQ3 and to construct the images P f and Q of the endpoints P and Q of the diameter that cuts out that line on (&3). The circle constructed on segment P'Q' as a diameter is precisely the circle (04). Let us compute * The symbol (z0i R) will be used to denote a circle o f radius R , the affix o f whose centre is z 0.

Complex Numbers in Plane Geometry

119

Fig. 15

the radius of (fi4) and the affix of its centre. To do this, note that (&3) passes into (G4) under the inversion with respect to the circle (fix) and under the homothetic transformation (0 , \ / g ), where g is the power of the point O with respect to the circle ($23). But under the homothetic transformation (O, l/<r), the centre of (&3) goes into the centre of (0 4); hence the affix co4 of the centre Q4 of (0 4) is equal to G

and the radius is equal to the product of the radius R ^ — 1 of the circle (Q3) by the modulus of the homothetic ratio, that is, by 1/\ g \: 1 _1_ i £4 — Rs (*3 = I)kl Furthermore, dd — cc * a = OQl - R j= —- — - 1 c c cc * In the general case, under inversion with respect to the point Oy we have (O M )-(O M ') = k , where k ^ 0 is the power o f the inversion, and, hence, (OM) _

( ON) "

cr ’

where k is the second point o f intersection o f the straight line O M with circle being inverted; # 0, since \d\ ^ \c\.

Problems in Geometry

120

SO that

cd (Q*) = ( s r i f cc (3) The transformation

cc \ |ddj —cc\)

be — ad c2

z + —

d z H-----

consists in turning the plane through an angle of arg - - ---- — and in a c2 homothetic transformation with centre O and ratio

. After Id2 these two transformations, the circle (0 4) goes into the circle (£5), the affix co5 of whose centre is co5 = q 4

be — ad

cd dd — cc

be — ad <

d c

be — ad dd — cc

and the radius is *5 =

cc dd — cc

be — ad dd — cc

be — ad c2

Thus

d be — ad c dd — cc

be — ad dJ — c7\)

(4) Finally, the transformation of translation be — ad c2

--------------------- ---------------

a c

>—

z +

be — ad c2 , d z H----c

carries the circle (Q5) into (fi6) of the same radius: R6 = R5, and the affix of the centre Q6 of (0 6) is equal to 0)q — cc |

! |c|2

Id2 |C2|

a d be — ad bd — ac 1-------I= = H' c c dd — ac dd — cc Id 2

= 1.

Complex Numbers in Plane Geometry

121

To summarize, then: as a result of the transformation

the circle

u = a z + b , a d - b e * 0 , c * 0 , Ic| # |d\, cz + d = (0 , 1) goes into the circle

bd — ac I be — ad dd — cc Idd — cci ------- >■ ■■> ~ Problem 13. Suppose BCAXA2, C A B ^ , ABC1C2 are squares with the same orientation and constructed on the sides BC, CA, AB of A ABC, the orthocentre of which is H. Denote by (O) = {ABC) the circle passing through the points A, B, C. Let P, 0 , be the respective centres of the ---------------------- >.

--------------------- k.

squares A ^ C 'C " , B±C2A' A ", C ^ B 'B " having the same orientation as the first three squares. Denote by Pl9 P2, P 3 the orthogonal projections of point P on the sides BC, CA, AB of A ABC, by Ql9 Q2, Q3 the orthogonal projections of point Q on the same sides and by R v R2, R3 the orthogonal projections of point R on the same sides BC, CA, AB of A ABC. Prove that the sum of the forces PPi + PP*

PPz + QQi

QQl

QQs

RRi

RR*

(3 1 )

is equal to the vector OH', where H' is the orthocentre of the triangle whose vertices are the feet of the altitudes of the given triangle ABC. Prove that the straight line on which the resultant of these forces * lies passes through the endpoint of the directed line segment 4OH. Now if the squares of the second triplet have an orientation opposite that of the first three squares, then the sum of the forces (31) belongs to the straight line OH'. Set up the equations of these two straight lines and take the circle (O) = — {ABC) as the unit circle (Fig. 16). Solution. 1°. Let a, b, c be the affixes of the points A, B, C. Find the affixes p, q, r of the points P, Q, R. Denoting by au bl9 cl9 a2, b2, c2 the affixes of the points Al9 Bl9 Cl9 A2, B2, C2, we have b2 = c + CB2 = c + iCA = c + i{a — c) = ia + (1 — /) c,

(32)

ax = c + CAX = c — iCB = c — i{b — c) = {1 + /) c — ib. Also, p - bo + - - B.iA1 — —- B2A1 = b2 + - - (ax — b2) 2 2 2

(ax — b2)

_ bu2 +, •— 1 a1 „ --— 1 b2 t ------------------------tai , lb* 1 + 1 b2 — 2 2 2 2 2 Force is a sliding vector.

1- i

122

Problems in Geometry

Fig. 16

and, taking into account (32), we have p = —— [ia + (1 - /) c] + 1----- [(1 + i)c-ib)]= 2c+ — ' - - a - —--- b. 2 2 2 2 (33) Similarly (we can carry out a circular permutation of the letters a, b, c)> we can find 1+ q = 2a + ----- - b 2 2 r = 2b + -— - c 2

1+/ 2

Note here that it follows from (33) and (34) that

Complex Numbers in Plane Geometry

123

and, hence, A ABC and A PQP have a common centroid (the point of intersection of the medians). Now, from the equations of the straight lines BC and PPl9 z — b — — be (z — ~b)> z — p = be (z — p) or z + bc~z = b + c, z — bc'z = p — bcp9 we find the affix px of point Pt : Pi = —

( b + c + p — bep);

and, hence, PP\ = P i — P = — (b + c + p — bep) — p = — ( b + c — p — bep). 2 2 Similarly, PPz = p » - P = - - { c + a - p - cap), PP$ = Pz — P = - - (a + b — p — abp), 2 whence PP1 -\-PP2t+PP^ = 01 ---- — P — ~~ &2 P‘ Similarly, QQi + QQz + QQ3 =

- — q - - - o-2q 9

RR± + RR2 + RR3 = o1 ----— r -----—<r2r. 2 2 Adding these relations term by term, we can find the principal vector of the resultant: bf ~ PPx + PP2 + PP$ + QQi + 6 6 2 + QQz + RRi + RR 2 + PP$ 3Gi — ^ - { p + q + r ) — -l- a2 (p + q + T) = 3crx - — P + q + r 2 3 z z

Problems in Geometry

124

3 p + q + r ------<x2------ -------2 3

3(7!

9 a+ b+ c 3

3 ^ a + b + c 02 -------------2 3

= 30, - { S - {

- 3 (« , - { * - - - » .* ) ■

where g = (a + b + c)/3 is the affix of the point of intersection of the medians of A ABC (or, what is the same, the point of intersection of the medians of /\PQR). But since a + b + c = a±, the sum h' may be transformed as follows:

Let us now prove that h' is the affix of the orthocentre of A AhBhCh formed by the feet of the altitudes of the given triangle. The equations of BC and of the altitude from A to BC are of the form z + bcz be z — bcz — a — a Adding, we find the affix ah of point Ah:

and, similarly, If ca \ bh = ~ 2 I "1 ” T ) *

If Ck ~ ~2

ab \ ” ~ )

The slope of BhCh is ab bh ~~ ch bh — c H

__C_ ab

a(b2 — c2) be c2 - b 2 abc

-a\

The equation of the altitude dropped from vertex Ah to side BhCh is

or — a2z = — a1 2

be 2a

a 2c*i +

2be

Complex Numbers in Plane Geometry

125

or 1 o , ------a*G, 1 2- -----------------, a* ~ b*c2 z — a2z—= — 2 2 2o3

(35)

In similar fashion we can write down the equation of the altitude from 5k to ChAh: iB1 1 lb- , b* — c2a2 z — b2z — — o , ------ o2o, i -----------------2 2 2os

(36)

Subtracting (36) from (35) term by term, we obtain /t2 ax 1 /La 2\ b* — a* + c \b 2 — a2) (6s - a2) z = - - (A2 - a2) ox --------------- ---------------2

2 ct3

or, cancelling 62 — a2, -

-j, h

1 _ C7 2

a 2 + b2 + c2 2o3 1Gi 2

1 _ O' i 2

o\ — 2o-2 2o3

trf , o2 1_ i ---- 1-------= — CTi 2<t3 (73 2

‘

crgOj , _ b 0i 2

3 _ ^ 2^1 1 /0 — = — 0i ------r — = - - (30i 2 2 2

— v 02)

and, hence, h' =

(3 0 ! -

OjOa).

We now set up the equation of the straight line to which the resultant thus found belongs. Since the nine indicated forces emanate three at a time from a single point, their resultant is equal to the sum of three forces: PP' = PPX + PP2 + PP3y QQf = QQi + C02 + <203,

PR' = RRi -j- RR 2 + RRz> these forces are laid off, respectively, from the points PyQ, R. From the relation 1 _ P’ - P = ° i -

Problems in Geometry

126

we find the affix p ' of point P' and, similarly, the affixes of the points Q' and R': P' = < 7 i - — / 7 - y G2P, 1 - — q 2

1 _ — 0*2q, 2

1 1 _ r' = *i “ — r ------ <r2r. 2 2

Remark. If on a plane (which we consider oriented by the introduction of a rectangular coordinate system) there is given a set of forces A kBk {k = 1, 2, . . w), then the principal vector of their resultant can be found as the sum of the free vectors: K= *= 1 The straight line to which the sum of forces belongs is a locus of points M (x,y) for which n £ mom,, AkBk = 0. A:= 1

Since, in the plane, it is natural to regard as the moment of force F the cross product momMF = (MT, F), where T is the point of application of the force F, it follows that the equation of the straight line to which the resultant of the system of forces A kBk belongs is of the form n

ij ak a'k 1; = °» bk b'k l!

x y

(37)

where (ak, a'k) = Ak, (bk, b'k) = Bk. Equation (37) is, generally, an equation of the first degree; hence, (37) is, generally speaking, an equation of a straight line [the left-hand member of (37) vanishes if and only if the principal vector of the system of forces is zero].

127

Complex Numbers in Plane Geometry

The equation of the straight line containing the vector of the sum of forces A kBk may be written differently by introducting the affixes ak and bk of points A k and Bk: 2

Z 1

ak a_k 1 = 0 . i =l bk bk 1

(38)

Applying this equation to the given problem, we have z 'z 1

1 1 _ z p <7,------ p ------ a2p 2 2

p p 1 =

______ 1_ 1 _ z P oi - - - p ~ — <r2P 2 2

_____ 1 _ Z P gi - — g2P 2

p V 1

1 1

z f(-{

i — ---- G o P

3 2

«•») + * ( - { P + 1 1 _ + PGi — P<Ti + - - o2p 2 ---- GiP2~ 2 ~2

Adding together three similar expressions that result from this one by substituting q for p and then r for q, and equating this sum to zero we obtain the equation of the desired line in the form (also note th a t d + q + r = a + b + c = <r1) z ( T f 1_

52<Tl)

o2(p 2 +

q 2 + r2) - — a2(p2 + q2 + r2) = 0

(

“

0l + 2

ffl° 2 )

Z +

(

ffl “

CTl<T2) ;

+ y o2(P2 + q 2 + r 2) — — o2(p2 + q2 + r2) = 0 or (3dj— (TjO^z — (3(7!— djffa)? + a2(p2 + q2 + r2) — a2(p2 + q2+ r 2) = 0.

Problems in Geometry

128

We then find

/>» = 4c2 - — + — + 2ac{i - 1) - 2cb(i + 1) + ab, 2 2 •t 2 •2 q2 = 4a2 — 1- —— (- 2ba{i — 1) — 2ac(i + 1) + be, 2 2 •2 •2 r 2 = 4fr2 - — + — + 2cb(i - 1) - 26a(/ + 1) + ca, 2 2 and, hence, p2 + q 2 + r2 = 4(af - 2a2) - 3<t2 = 4a? - 1la 2. From this it follows that P2 + q2 +

= 4 a| — l l a 2

and the equation of the straight line assumes the form ' (3ai

—

c^Gg) z

—

(3(7!

—

c^a2)z + o 2(4a? — l l a 2) — a2(46?

—

11g2) = 0

or (3 gj — a1G2) z — (3a! — a ^ j z + 4(a?o2 — a 2o?) = 0.

(39)

If instead of z we put into the left-hand member the affix of the endpoint of the directed line segment 4OH, that is, 4a1? we obtain 4(3 gx — aiG2) ax — 4(3ax — a ^ )

g1 +

4a?o2 — 4a2o?

= 120^ ! — 4a?G2 — l2c1o1 -f 4aL6? + 4o[<j 2 ~~ 4a26? = 0

That is, the carrier of the resultant passes through the endpoint of the directed line segment 4OH.

-------------- ----------------- *.

2°. Now suppose the orientations of the squares AXB2C C ", B ^ A 'A " , C^AzB’B" are the same but are opposite to the orientations of each of the squares BCA1A2, CABXB2, ABCXC2. In order to keep Fig. 16 as simple as possible, only the centres P*, Q*, R* ------------- >. -------------- >■ --------------->■

of the squares A ^ C 'C " , B ^ A 'A " , C1A2B,Bn are constructed; the points P*9 Q*, R* are symmetric to the corresponding points P, Q and R with respect to AXB2, BXC2, CXA2. The figure shows only three of the nine forces; ------> ------> ------ > namely the following forces are constructed: P*Pf, P*P$, P*P3*, where P*>P*>P* are the orthogonal projections of the point P* on the sides PC, CA, AB.

Complex Numbers in Plane Geometry

129

The affix p* of point P* is P* = b2 + — B2Ai + — B2AX 2 2 = h + " (fli - h ) + V 2 2 =

[(1 + 0 C - ib] +

— b2) = ~ ~ °i + 2 2

[ia + (1 - 0 c] = — ~ b + —^— 0 . 2 2

Similarly 1+

q* = r*

b, c,

where <7* and r* are the affixes of points Q* and R*. Note that in this case as well, p* + q* + r* = a + b + c = < tv Now, from the equations of the lines BC and P*P?9 z + fccz = b + c, z — 6cz = p* — bcp*, we find the affix p* of point P * : />* = ~ (b + c + p* — bcp*) 2 and, hence, p*p* — p* _ p* = J_ q, _(_ c _ p* _ bcp*). 2 Similarly, p * p * = — (c _|_ a _ p* _ ca/>*),

2 P V 3* = — (a + b - p* - abp*) 2 and, consequently, P*P* 9 -810

P*P* + P*P* = a 1 -

— p * - - L o-2p *

130

Problems in Geometry

and Q*Qt + 0 * 0 ! + Q*Q! = ° 1 - 4 q* -

\ <r2q*> 2

R*R* + R*R? + R*R* = crx - — r* - — < t27*, 2 2 ----- >' -----and so the principal vector of the sum of nine forces P*P*> P*P2* ,... is equal to C = 3cTi — -~-(p* + q* +/•*) —— <r2 (p* + q* + 7*) 2 2 ,

— 3(7x

3 (JI 2

1 _ 3 ^2a l — ~ &1 2 2

1 _ 1 °’2(71 — ~ (3c! 2 2

0 2Ox) — ll i

where ti is the affix of the orthocentre of A AhBhCh. As in the case of item 1°, let us now consider the sums of nine forces taken three at a time: P*P'* = P*P f + P*P* + P*P*9 ¥ & * = ~ Q * Q t + 0 * 0 ! +~Q*Q!,

R*R'* = R*R? + R*R? + R*R*. Let us find the affix p'* of point P'*: P * ? * = p ’* - p * = <T1 - ^ - p * 2

a2p*,

whence we find p'* and, similarly, q'* and r'*:

The equation of the straight line carrying the resultant is of the form z p * p f* z q* q ' * z r* r’* z p* p'* -1- z q* q'* + z 7* 7 ’* 1 1 1 1 1 1 1 1 1

131

Complex Numbers in Plane Geometry

We have z p* p' *

p *

C l -

z p* a 1 — ■— OiP*

c 2p *

1 z P* Ol - — ViP* 2

1 1_ * Z 1* F * = z P * i—— P ~ ~ G*P 2 2

1 1

= ^Ap** - ffl + A a2p*jz +

= ( i

_3_ 2

A <r2p* - —/>*) z

+ P*o1 — 1*0 1 + A <j2p *2 — A f f 2p*». 2 2 If we write down two other similar expressions and then add them and equate the result to zero, we obtain the equation of that line in the case of item 2°;

^A Ox -

A o=a j Z + ^3<rx- A

+ <7!?! — (Tjffi +

A ^ (p * 2 +

q* 2 + r*2) —

- A <Tj j Z A ct2(p*2 +

?*2 + r*2) = 0

or (3aj — 0±02) z — (3 ^ — 0X02) z + ^(P*2 + tf*2 + r*2) — ^ 2(P*2 + q *2 + r *2) = 0 Furthermore, \2 / 1_ ; 1 _i_ / \ 2 - I 1- ' b 1 1+ 2 I 2 4+ i ____j Cfl + 2 2 J + _L (fl2 + 62 + C 2) = ff2, 2 o-2(P*2 + tf*2 + r*2) = oza2. And therefore <*2(P*2 + q*2 +

"r*2) = <72^2,

which means our equation takes the form (3d1! — 0 ^ 2) z — (3oi — 0X02) z = 0.

132

Problems in Geometry

Problem 14. Let P, Q, R be the orthogonal projections of point M on the sides BC, CA, AB of triangle ABC (Fig. 17). Denote by A', B', C* the points obtained by inversion of the midpoints A0, B0, C0 of the segments MA, MB, MC with the circle of inversion (PQR) and by A", B", C", the triangle formed by the polar lines of the points A0, B0, C0 with respect to the same circle {PQR). Prove that (A'B'C')2 = 1 (A"B"C")-(PQR). What is the necessary and sufficient condition for A A'B'C' and A PQR to have opposite orientations? Solution. We take (PQR) for the unit circle. Let zl9 z2, z3 be the affixes of the points P, Q, R and let fi be the affix of point M. The slope of the straight line MP is p — z1 and hence the slope of line BC \ i - z1 and the equation of line BC is of the form z—

J* — *i (z - Zj). f l — zi

(40)

Complex Numbers in Plane Geometry

133

Similarly, the equation of line CA is * -* 1 = - - ^ —

(* -* ■ )• (41) P - *2 Subtracting equation (41) from (40) term by term, we find the complex number z = c; it is the conjugate of the affix c of point C: z2 - * = ( a - * . _ \H -z 2 n - z j

c + z, H -Z i

- z2 U— zi

or Zi + ZA ZlZ2 )

- 1 zxz2 V ZlZ2 Zo z, — (/« - * 1) - Z2) 0 < - z'l) 0 * - z 2) Cancelling z2 — zx and multiplying both sides by (p— z x) (ji — z 2), we get z l)

(z2 - zxy0

z . + z 8 \ . - , P/Z - 1 (p - ij) (p - z 2) = [ - p ----- -— b —1- ^ L ) C H--------------» V ZlZ2 zxz2 / Z iZ 2 whence

(

h ± lC ] c ZiZ2 )

\ i -

Z1Z2

( - /i - p z xz 2 +

(/Z - Zi) (n - z2) - Z!Z2(/l/t - 1)

+ z2) c = /I2 - p (zx + z 2) + (2 - p/i) ZjZ,.

This means that c=

nz — n (zx + z 2) + (2 — nn) z xz 2

and, hence. p2 - p(zx + Z2) + (2 - nn) z1z2 — H ~ /iz1z2 + zx + z2 or — n2 + n(zi + z2) + (nn - 2) zxz2 H + fi zxz2 — zx — z2 The affix c0 of the midpoint of segment MC is 1 i —^ + K zi + z2) + ( 0 ~ 2 ) z1z2\ <(i I P~r ) 2 V fi + V zxz2 - (zx + z2) } _

(pp - 1) zxz2 /i + /zzxz2 — Zx — z2

134

Problems in Geometry

and the affix c' of point C' obtained from C0 by inversion with the circle of inversion (PQR) is H + f i z ^ z — zx — z 2 ft + pztz3 — z1 — z3 1 Co

(mP - 1) z t z t

fiji - 1

We find the affixes a' and b' of points A' and B' in similar fashion: B + PZ2 Z» ~ Z 2 - Z3 * a —-------------------------HH 1

H + fizi z 1

b' =

ftp. -

We now find (A'B’C ) . a’ a ' \ b’ b' 1 c' c' 1

ft -f- ftZ3Z3 Z2 Zg ft + pz3zx — z3 — zt 4(fift—l )2 |ft + pZiZ2 — Zj — z2 ftz3z3 — Z2 — Z3 flZzZ3 pz3zt — z3 — zx /JZ3ZX 4 { fi ft- \ f f . I Z^Z 2 — Z i Z2 f l Z - ^ Z ^ jU(73Zi 1*02?! 4 (^ -1 f

|K 7 g Z 2

fi(^2z z

ft-(- flZ3Z3 Z2 Zg 1| p + ftz3z\ — z 3 —z x 1! p + ftz{zz — z x — z 2 1 z2

—z 3 — Zx 2"2

^1 "t"

^ 1 't“ ^2 0 1 + ^ 3

/ i f f 3 Zx +

z3 1

^ 1 ■(■^ 2

IIG&Z

f ia ^ i

z 3

/Zff3z 2 + Z2 jUff3Z2 + z 2 1 4(ftn—l )2 pa3z 3 + z3 /iff3z3 + z 3 1 ■

f t p a 3<f3

4(ftp ~ l ) s _

z 2 z2 1 + 1 Z3 z3 1 /(I - fi/i) 4(1 - w ?

■

z2 z 2 1 Z3 Z S 1 . z2 I *3 *3 1

(W ) 1—

Furthermore, the slope of the straight line OA0 is (ftp — 1) z2z3 (ftp — 1) z 2z 3 «O^0 = ft + pz3z3 — z2 z3 p-\- ftz3z 3 z 2 z 3 (ftp — 1) z2z3 ftp 1 ft + ftz3z3 — Z2 — Z3 /i + /Zz2z3 — z2 — z3

- *

Z.>Z. 243

Complex Numbers in Plane Geometry

135

From this it is easy to find the equations of the polars of points A0 and B0 with respect to the circle (PQR); these are straight lines that pass through points A' and Bf respectively and are perpendiculars to the lines OA' and OB' ; /I PZ2 Z2 Z2 & + Hz2z 3 z 2 z 3 - z2z3 (I -z -----2— Hi* - 1 Hi* - 1 and H+ 1— z 3— z 1 H + HZ3 Z1 — z 3 — z 1 Hi* - 1 Hi* — 1 The affix z = c" of point C", the point of intersection of these lines, can be found from the equation _____1 _ ( 2 _ H + i*z &3 ~ Z2 ~ z a\ H + H ^ 2*3 ~ z~a — z 3 z2z3 V Hi* — 1 ) Hi*— 1 = _ 1 / _ H + i*zszi — z3 — \ + H + HZ3 Z1 — z 3 —z x % V Hi* — 1 / Hi* — 1 or Z ^ 2 /j + /^ 2z3 — z2 - z 3 = ___ z _ 2 n + /iz3zx — z3 — zx Z2Z3

Z2Z3(HH (z2 — zx) z

Z3Z1

0’s

Z3Z1(HH — 1) zi ]■

whence c" = — ^ — - (H - z3). HH ~ 1 Similarly, a” —

z~~~r (h — Zi) /i/t - 1

b" =

3—— (/i — z2) HH ~ 1

and

m> that

( l"/i"C")

H ~ zi H - z i 1 / 4 4 ( p ji- i f H ~ *2 H — Z2 1 H - z 3 ji-~ z 3 1 Zi z x 1 4 i Z2 z 2 1 (HH- i f 4 Zj Z3 1

- (Pfi*). (HH— I)2

136

Problems in Geometry

Thus (A'B'C') = ~^PQR)~, (.A"B"C") = ------ (PQR), I - fiH (1 - n n f whence (A 'B 'C 'f=

(p QRf (1 - w ?

= _L 4

The triangles A'B'C' and PQR have the same orientation if and only if 1 — pp > 0, that is, the power of point M with respect to the circle (PQR) is negative; in other words, if and only if the point M lies inside (PQR). In Fig. 17, M lies inside (PQR) and, indeed, A'B'C'i\PQR. Problem 15. Given A ABC. Through an arbitrary point P of circle (ABC) circumscribed about A ABC are drawn lines parallel to the sides BC, CA, AB. Let A', B', C' be the respective second points of intersection of these lines with the circle (ABC). Denote by A", B", C" the points symmetric to points A ,B ,C about the straight lines B'C', C'A', A'B'. 1°. Prove that A ABC and A A"B"C" are congruent but have opposite orientations. 2°. Prove that 0 0 " = PH, where O and O" are centres of (ABC) and (A"B"C") and H is the orthocentre of A ABC (Fig. 18). Solution. Version One. 1°. Take (ABC) for the unit circle and assign to point P an affix of 1 (that is, let P be the unit point). Then the affixes a ', b', c' of points A', B', C' are a — z2 z3, b — z2Zi, c — z±z2, where zl9 z2, z3 are the affixes of the vertices A, B, C of the triangle. The slope of line B'C' is Zj z2 - zx z3 ^ zt(z2 — z3) ~Zi Z2 Z1

.1 Zi \ =

— z f Z2 z 3 =

— Zi (T3

and so the equation of line B'C' is of the form Z — Z x z3 = — Zx <72 ( — Z x Z 3) z

or Fig. 18

Z +

Zjl C73

= zx z2 + zx z3. (42)

Complex Numbers in Plane Geometry

137

The equation of a perpendicular dropped from point A to B'C' is z — zx = zx a3(z — Zj) or z — z1cr3 z = z1 — <j3.

(43)

Combining equations (42) and (43) term by term, we find the affix z = ax of the orthogonal projection A1 of point A on the line B'C': ax = — (zx + zx z2 + zx z3 — cr3) = --- (zx — z2 z3 + <j2 — c3). 2 2 The affix a" of point A ", symmetric to A with respect to B'C', is found from the relation *1 + a"

whence Cl — 2flj

Zx — (72

(73

Z2 Z3.

In similar fashion we find the affixes b" and c" of points B" and C": b — (7% (73 Z3ZX, C = (72 Ct3 zx Z3. From these relations it follows that the centre O" of circle (A"B"C") has the affix O = (72 (73, and the radius is equal to 1 since | a " - o " | - \b" - o " \ = \c” — o"\ = 1 = !)• ------- * Let us consider 1\A qB'qC3 the affixes of the vertices of which are equal respectively to —z2z3, —z3zl5 —zxz2. The triangle A'qBqCq is symmetric --------> ------- > to A A'B'C' with respect to the point 0 , and, hence, the triangle A"B"C" is obtained by a translation of /\A'0B'0C'0 via the directed line segment OTt where the affix t of point T is equal to <t2 — (73. Furthermore, (73 ... — ^2 ?3 — — &3 ^ 1? ^3 ^1 — ^3 Z2) Zj Z3 — C3 Z3. *1 Therefore A^o^oCo is obtained from A A*B*C* (the affixes of the vertices of which are z l5 z 2, z 3) by a rotation about 0 through the angle arg (— a3).

Problems in Geometry

138

--------> ------> The triangle A*B*C* is symmetric to AABC about line OP (the real axis ------► ------- >■ Ox), and therefore A ABC and A A*B*C* are equal and have opposite --------> --------> orientations. But AA*B*C* is equal to A A qBqCq and has the same orien tation; A A qBqCo is equal to A A 'B 'C and has the same orientation, while A A "B ”C" is equal to A A qBqCq and has the same orientation. Conse quently, A B C l\A "B "C " and A ABC = A A”B"C" (the symbol = here signifies congruence). 2°. The affix of point O" is equal to o" = <r2 — <r3, whence OO” = |<72 — 0*31 = |<73|

= ! — - 1 \ = \s l - I H ^ I - l\= P H . \ff3 I Version Two. 1°. Take {ABC) for the unit circle. Let zly z2, z3, p be the affixes of points A, B, C, P respectively. Take the Boutain point of A ABC for the unit point so that a3 = 1. The affixes a', b', c’ of points A ’, B \ C' are Z2Z3> »b/ = Z3 Zl ct= ----Zl Z2 a , = --------P P P or (c3 = 1) a’ = p z x, b' = p z 2, c' = p z 3. The slope of line B’C is b’ — c = p z 3 — p z 3 b' — c' pz3 —pz3

- J — = - ZL (a3 = 1) z2 z3p 2 p2

and the equation of B’C is of the form z

—

z 2 =

—

zi p \ z

—

p z 2)

or z+

~p2 z = p z 2 + p z 3.

(44)

The equation of the perpendicular dropped from point A on line B C is or

z - zx =

p \ z - Zl)

(45) ~p2Combining the equations (44) and (45) term by term, we find the affix z = ax of projection AL of point A on the straight line B'C Z — Zj

p 2 z

z x

Complex Numbers in Plane Geometry

139

The affix a" of point A", symmetric to point A about B 'C \ is found from the relation + a" = *i 2 whence a" = lay — z, = p z 3 + p z 3 — ~p2. Similarly, b" = p z 3 + p z x — p 2. We now find a" — b" = p ( z 2 — z , ) , a" — b" —P(z2 - H) and, hence, A"B"2 = (a" - b") (d" - b") = PP(?2 - H) (z 2 - zi) = AB2. That is. A"B" = AB. In similar manner it can be proved that B"C" = BC, C"A" = CA, that is A ABC and A A"B"C" are congruent. ------> ------- > In order to prove that A ABC and AA"B"C" have opposite orienta tions, it suffices to prove that d" 1 A = z2 b" 1 = 0 . ^3 c" 1 We have Zy p(Z2 + Z3) - P2 1 Zi — #>Zi 1 Zl pip 1 - Zl) 1 == A = Z2 P(Z3 + Zl) — P2 1 Z2 pia 1— Z2) 1 = z2 — pz2 1 Zg p(Zy + Z3) - P2 1 z3 P in - z3) 1 H — PH 1 and so A ABC = A A"B"C" 2°. From the formulas a" = p z 2 + p z 3 - p 2 = b" = c" =

and ABC

A"B"C".

>(<?! - z ^ ) - p 2 = - p z y + p a y - p 2, — p z i + p t i - p 2, — p z 3 + ~pdy — p 2

it follows that the affix o" of the centre O" of the circle (A"B"€") is o" = pdy - p 2 and the radius is equal to 1 (|— p z x\ = |— p z 2| = |— p z 3\ = 1), whence ()()" — \o"\ = |pay — P 21 - \p\ Wy — p\ = \&i — p\ = I<Sy — p\ — PH.

Problems in Geometry

140

Remark, From the relation o" = p{dx — p) it follows that the directed line segment 0 0 " is equivalent to the directed line segment obtained by symmetry in the x-axis of the directed line segment PH and a subsequent rotation through the angle argp (the x-axis is a straight line passing through point O and the Boutain point). ------► Problem 16. On the sides BC, CA, AB are constructed triangles A'BC, ------> ------> B'CA, C A B , which are similar and have the same orientation. Let P be an arbitrary point lying on the circle (O) = {ABC), The directed line segments OA', OB', O C rotate about point O through angles that are respectively equal to {OP, OA), {OP, OB), {OP, OC) (these angles are oriented). Let OA", OB", OC" be the respectively rotated segments. Prove that the centroid G" of A A"B"C" is symmetric to the centroid G of A ABC about the diameter of the circle {O), which diameter is parallel to the Simson line constructed for point P with respect to A ABC, Solution, We take the circle (O) = {ABC) for the unit circle, and the point P for the unit point. Let zl9 z2, z3 be the affixes of the points A, B, C. Denote by a', b', c' the affixes of the points A', B', C , Since the triangles ------> ------> ------> A'BC, B'CA, C'AB are similar and have the same orientation, it follows, assuming cp = {CB, CA'),

CA' CB ’

we will have a = z3 + p{cos <p + / sin (p) (z2 — z3) = az2 + (1 — a) z3, where a = p(cos <p + i sin cp), and similar expressions for b' and c' with the same value of a: b' = otz3 + (1 — a) zu c' = OLZi + (1 — a) z2. The affixes a", b", c" of the points A", B", C" are a" = zx a',

b" = z2 b',

c" = z3 c'.

That is, a" = zxz2a + (I — a) zLz3, b" = z2z3 a + (1 — a) z2 zl9 c" = z3 zxa + (1 — a) z3 z2.

Complex Numbers in Plane Geometry

141

From this we find the affix g" of the centroid of A A"B"C": ,, _ a" + b" + c" C2 where <r2 = zx z2 + z2z3 + z3 zx. The equation of the diameter of (0), which diameter is parallel to the Simson line constructed for the unit point P with respect to A ABC, is of the form (see problem 3) z — <73 z = 0 . The ends of that diameter have affixes l/og (tfo3 has two values; each of them satisfies the equation z — a2z' = 0). In order to prove that the points G and G" are symmetric about the diameter z — cr3 z = 0, it suffices to prove that AODG and AODG" are similar but have opposite orientations (D is one of the ends of the diameter z — <j3 z = 0), that is, that 0 0 1 A = g g" 1 = 0 ]fc3 l/ff3 1 (for/< 73, any one of the two values may be taken; ][g3 is the conjugate of that value). We then have A = ][oz g - 1/V3g" = 1faz (g - c3g")

Problem 17. The altitudes of an arbitrary triangle ABC intersect the circle (O) = {ABC) in the points Al9 Bl9 Cx; A \ B', C9are points symmetric to point P9 which lies on (0), with respect to the straight lines OA, OB, OC; A", jB'\ C" are points symmetric to the point P with respect to the lines OA', OB', OC'; a, /?, y are points symmetric to the points A "9B", C" about the line OP. Prove that the points A2, B2, C2, which are symmetric to the points a, y with respect to the tangents to (0) at the points Al9 Bx, Ci, form a triangle A2B2C2 that is homothetic to A AxBiCi with ratio equal to 2; the centre Q of this homothetic transformation belongs to (he circle (0 ). Solution. Take (0) = {ABC) for the unit circle and P for the unit point of the complex-variable plane. Let z1,[z2,z 3 be the affixes of the points A, B, C. The equation of the straight line BC is of the form z + z2z3~z = z2 + z3.

142

Problems in Geometry

The equation of an altitude dropped from the vertex A on the line BC is 2 — Zx =

Z2 Z3( z

—

Z j.

Solving this equation together with the equation zz — 1 of the unit circle, we get 1 1 Z - Z i = Z2Z3 / ----------

or z — Zi =

Z* 2a

Zt Z

One of the roots of this equation is z = zx (the affix of point A); the other is z = ax

Z2 Z3 Z1

(the affix of point AJ. Similarly, bi = -

zl z 2

*3 *1 *2

The equation of the line OA is z — z\ z = 0 and the equation of a perpendicular dropped from point P on the same line is z - 1 - - zf (z - 1). Solve this equation together with the equation zz = 1 of the unit circle: z — 1 = — z\

----- 1j ,

that is, z — 1 == zf -----

One of the roots of this equation is z = 1 (the affix of point P), the other is the affix a' of point A': z = a' = zf. Similarly, b' = zh

cf = z l

are the affixes of points B' and C". The equation of the straight line OA' is z — z\ ~z = 0 and the equation of the straight line passing through point P perpen dicularly to line OA' is z - 1 = - z\ (z - 1).

Complex Numbers in Plane Geometry

143

From this and from the equation zz = 1 of the unit circle we find the affix a" of point A": z —1 z— that is, z — 1 = z? z One of the roots of this equation is z = 1 (the affix of point P), the other is z = a" = z\. Similarly, b" = zi, c" = z l where b” and c" are the affixes of the points B,r and C". The slope of the straight line OP is equal to 1. Hence, the equation of the straight line passing through point A ” perpendicularly to line OP is of the form z - a" = - (z - a"). Solving this equation together with the equation zz = 1 of the unit circle, we find the affix A of point a:

One of the roots of this equation is z = a" (the affix of point A"); the other is

which is the affix of point a. In similar fashion we find the affixes /z and v of the points p and y: 1 1 The slope of line OAx is <h_ = q2 = 4 4 <*i 1 4 and so the equation of the tangent line to the unit circle (O) at the point A1 is of the form

Problems in Geometry

144

or -2 -2

z2 z3 — -------- Z 2

(46)

The equation of the perpendicular dropped from the point a on this tangent line is of the form z ---- i = 4- ^4 ( z - z t ) 4 or -2 -2

Z2 Z3 — __

(47)

— 4-

Combining equations (46) and (47) term by term, we find the affix a* of the projection of point a on the tangent line: z = a? =

Z? Zo

24 The affix a2 of point A2, which is symmetric to point a with respect to the tangent to the unit circle at point Al9 is found from the relation ^1 + a2

= a* 9

whence 'l ♦ a2 — 2a2 —

^ 2 Z3 2 — — 2O — ------ ~\—

^ ^ Z 2 Z3 ---------= — 2 -------a3.2

In similar fashion we find the affixes b2 and c2 of points B2 and C2: L b 2 —

9 Z3 “ 1 2

2 03,

^ c2 —

Z1 Z2 2 — 2 -------- — 0^3.

From these relations it follows that the straight line A XA2 passes through point Q with affix q = o\ since the midpoint of segment A2Q has the affix a2 + q z3 z2 — = --------= 2 zx which is to say the midpoint of A2Q coincides with point Av Point Q lies on the unit circle since \q\ = 1.

Complex Numbers in Plane Geometry

145

In similar fashion, we can prove that the points Bx and Cx are, respec tively, the midpoints of line segments B2Q and C2 Q. Thus, QA 2 _QB 2__QC2 QA±

QB±

QC1

That is, A A 2 B2 C2 is an image of A A 1 B1 C1 under a homothetic transfor mation with centre Q lying on {ABC) and with homothetic ratio 2. Problem 18. 1°. Through the vertices A l 9 A2, A 3 of A A YA 2 A 3 lying on an oriented plane, draw parallel lines intersecting the given straight line A at the points Pl 9 P2, P3; note that the angle from the line A to the lines A±Pl 9 A 2 P2, A 3 P3 is equal to a (Fig. 19). Through points Bi, P 29 P3 draw straight lines ll 9 /2, 13 that intersect the respective sides A 2 A3, A 3 A l 9 A±A2; note that angles reckoned from the straight lines A 2 A 39 A 3 Al 9 A 1 A 2 to the straight lines ll 9 /2, l3 are all equal to /?. Prove that the lines ll 9 /2, l3 form a triangle QiQ 2 Qz, which is similar to the triangle AXA 2 A3; the factor of proportionality is sin(a + P) j sin a | Consider the following special cases; 2°. P = 7i — a. 3°. p = 0. Solution. 1°. Take the circle (O) = (A 1 A 2 A3) for the unit circle. Let r j5 z2, z3 be the affixes of the points A l 9 A 29 A 3 and let az + a ~z b be the equation of the straight line A (a ^ 0 and b is a real number). Put X = cos 2a + i sin 2a, ju = cos 2p + i sin 2p. The equation of line AXPX may be written in the form a(z — Zj) + 2 a (z — z x) = 0 . Indeed, the slope of the straight line A is a x = -----a iuul the slope of the line (48) is

I lenee X In

Ml()

(48)

146

Problems in Geometry

and so = cos a + / sin a. That is, the angle from the straight line A to the straight line (48) is equal to a. If we took the other value of ][X9 ^X = — (cos a + i sin a) = cos(a + n) + / sin(a + n)9 then the angle from the straight line A to the straight line (48) would turn out equal to n + a, that is, it would be congruent to a modulo it. From the system of equations of the straight lines A and AxPl9 that is, from the system of equations az + a z = b9 az + Xa z = azx + Aa z l9 we find the affix px of point Px: z =Pi =

— a

whence

A / 1 A — 1 \ zx

)•

b a zi\_ _ A f Xa X a ) 1 — A\

b Xa

azx\ a j

Complex Numbers in Plane Geometry

147

The affixes p2 and pz of points P2 and P 3 have similar expressions: lb , , a~z2\ I f P2 ------ Z2 - - + A — 2 , 1 —x V ci a ) P3 Furthermore, the equation of line A2A3 is of the form z — z2 = — z2z3(~z — z 2) and so the equation of line 7Xmay be written in the form z - Pi = — HZ2 zz(z - p 2) or Zb + Zd = _ _» 3 - _ A- ( _ ± + z _ __L 1 —1 v a azx ) [ 1 —1 \ zx la or

, ZiZ h

<1 - A)(z + „ w ) = z, - ^ + a

+ az1

la ) J (49)

The equations of the straight lines /2 and /3 are written down in similar fashion: n (1

X) (z +t p....z 3z 2 z ) — z2 a

^ , , -2-) = *, -

tyzaZi , Paa3 |---------------------j----- —----------::—* (50) az2 z2

+ di _ #z3

_ « ? .. z3

(so a

Multiplying both sides of equation (49) by —z l9 both sides of (50) by z2, and adding termwise, we obtain (1 - 1) (z2 - Zj) z = 4 — Z1 — ~ (Z2 — Zl) + Z3(Z2 — Zl) lp — P -G- (z2 — Zl)

whence we get the affix q3 of point Q3, the point of intersection of the lines lx and /2: lb a Z1 + z2 -------- \- Ip Z3 — p c3 a a

O - l---------------- / / — - <X3 +

a1 1 —1

(>1 /2 — l ) z 3

148

Problems in Geometry

The affixes

and q2 of the points Q± and Q2 have similar expressions: Ab a a i ----------------- <r3 + (Aft — 1) zx a a <h r r i » Xb a o-i-----------p — a3 + (X^i - 1) z2 a a <?2 = 1 —X

From the last three relations it follows that the points Ql 9 Q2, Q3 are obtained from the points Al 9 A 2 9 A 3 by a linear transformation of the first kind: q = mz + n, where m

Xfi — 1 1-X Xb a —------ p — <r3 a a

and so A A tA 2 A 3 and A Q 1 Q2 Q3 are similar and have the same orientation (see Fig. 19): n is the affix of point 0 \ into which the point O passes under the transformation; in other words, (O') = (QiQ^Qz)- The factor of proportionality is _ [ ^P — 1 _ lcos | 1 —X

+ 2/0 + / sin(2a + 2/?) — 11 11 — cos 2a — i sin 2a|

12 sin2(a +

— 2/ sin(a + /?) cos(a + /?)| |2 sin2 a — 2 i sin a cos a| p)

| sin(a + P)| | sin(a + /?) — 1cos (a + /?) __ sin(a + /?) |sina| |sina — /co sa| sin a 2°. If p = 7T— a (or /? = — a), then A 6i0203 degenerates into a point. We have the following theorem: if through the vertices Al 9 A 2 9 A 3 of A A xA 2 A 3 lying on an oriented plane we draw parallel lines intersecting the given straight line A at points P l 9 P 29 P3, the angle from the straight line A to the straight lines A±Pl 9 A 2 P2, A 3 P3 being equal to a, and then through the points P l 9 P2, P 3 we draw straight lines ll 9 /2, 4 cutting the sides A 2 A g9 A 3 A l 9 AxA 29 the angles from the straight lines A 2 A3, A 3 A l 9 AXA 2 to the straight lines ll 9 4 , 4 respectively being equal to a, then the lines 4, 4> 4 pass through one point (Fig. 20).

Complex Numbers in Plane Geometry

149

In particular, if a = n/2 9 P = n/2 , we obtain the theorem on the orthopole of a straight line with respect to a triangle: if P l 9 P2, P3 are the orthogonal projections of the vertices A l 9 A2, A 3 of A A XA 2 A 3 on a given straight line A, l h e n the lines that pass through the points Pl 9 P2, P3 and that are perpenÂ dicular to the straight lines A 2 A3, A 3 A l 9 AXA 2 respectively, intersect in one point Q (the so-called orthopole o f the straight line A with respect to /\A xA 2 A3\ Fig. 21). Vâ&#x20AC;&#x2122;. Suppose P = 0. Then we obtain the following theorem: if through lhe vertices A l 9 A2, A 3 of A A XA 2 A 3 we draw parallel straight lines interÂ s e c t i n g the given line A in the points Pl 9 P 2 9 P3, and then draw through the points Pl 9 P2) P3 lines ll 9 /2, l3 respectively parallel to lines A 2 A3, A 3 Al 9 ------- > then the lines ll 9 /2, 13 form a triangle Q1 Q2 Q3 that is equal to the ------- >triiingle AYA 2 A 3 and has the same orientation (Fig. 22).

Problems in Geometry

150

Fig. 23

Problem 19. Let A 1 A 2 A 3 A4 be a convex quadrangle inscribed in a circle (0). Denote by A12, A23, ^ 34, ^ 4i the midpoints of its sides ^i/42, ^ 2^ 3? ^ 3^ 45A 4 Av Let <5 be some diameter of (0). Denote by a12, a23, a34, a41 the orthogonal projections of the points A12, A<&,A 349 A 41 on the diameter <5. Draw through the points a12, a23, a34, a41 straight lines p349 p l 4 9 p 129 P23 that are respectively perpendicular to the straight lines A 3 A4, A 4 A l 9 A±A2, A 2 A3. Prove that the lines p 349 P149 P12, p 23 form a quadrangle B 1 B 2 B3 B4 similar to the given one but with orientation opposite that of the quadrangle AXA 2 A 3 A 4 (Bx is the point of intersection of the lines p 23 and p34, B2 is the point of intersection of the lines P1 4 and P34, B3 is the point of intersection of the lines p i 2 and P149 B 4 is the point of intersection of the lines p 12 and p23; Fig. 23). Solution. Regard (0 ) as the unit circle, and take the .x-axis so that the diameter S lies on it. Let z l 9 z2, z3, z4 be the respective affixes of the points A l 9 A 29 A3, A4. The midpoint A 1 2 of segment A 4 A 2 has the affix

and its projection aI2 on the line S (the x-axis) has the affix zi + z2 H------ 1----( 1|+ Zj z2) (z j+ z2) ________ Z1 Z2 4 4z4 z2 The equation of the perpendicular P3 4 dropped from the point a12 to the line A 3 A 4 (the slope of which is equal to —z3z4) is of the form z t 12 = z3 z4(z f 12) a\2 ~f~ ^12 T“ = — 2— _

Complex Numbers in Plane Geometry

151

or (1 + Zi z2) (zl + z2) 4Zj z2

(1 + *i z2) (zi + z 2) z ------------------------------ = z3z4 4zx z, L

or Z

Z aZ

(zx + z2) (1 + ^ z2) (1 - z3 z4) 4zj z2

(52)

Similarly, the equation of the perpendicular /?23 dropped from the point a 14 to the line A2A3 is of the form _ _ (Zl + Z4) (1 + Zj z4) (1 — z2 z3) z — z2 z3 z = --------(53) 4z4 z4 Multiplying both sides of equation (52) by z2, both sides of (53) by —z4, and adding termwise, we obtain the affix z — bx of point Bx, the point of intersection of the lines fi3i and /?23: (Zj + z2) (1 + Zx z2) (1 — z3 z4) - ( z 4+ z4) (1 + z 4 z4) (1 - z 2 z3) (z2 - z4) = 4zx 1 = —- (Zl + z2 + zj z2 + zl z4 — zx z3 z4 — z2z3 z4 — zf z2z3z4 4zj -z2Zi Z3 z4 z4 z4 Zx Z4 Z4 ZJ -(- Zj Z2 Z3 + Z 2 Z3 Z4 -fZ | z4 z2 z3+zf z2 z3 z4) =

4zt

[z2 - z4 + zj(z2 - z4) +

Zx z3(z2 -

z4)

+ Zx(z| - z\) - zx z2 z3 z4(z2 - z4)], whence bi =

1 4~

-j-

zx z3 -f- zt z2 -j- z4 z 4 — Zx z2z3 z4 _ 4z4

— ff4 4zx

where o-x = z4

Z3 + Z4,

^4 — ^2 ^3 ^4* Similarly we find the affixes b2, b3, bt of points B,, B3, Bx: b3 — 1 - <?4 1 ffl4z2 h 4 b3 =

1 - <r4 4z3 1 4

K =

1 - <t4 4z4

4 ’

152

Problems in Geometry

To summarize: bk = T + - — 4 4

( * = 1, 2,3,4).

From this relation it follows that the points Bl9 B2, B39 B4 lie on the circle (BXB2B3B4)9the affix of whose centre is equal to aJA and the radius is 11 — — <t4!/4. Since |<r4| = 1, it follows that the radius can vary from 0 to 1/2 (cr4 can assume the values ± 1). Thus, if o4 — - 1, that is the Boutain point of the quadrangle is the unit point, then the quadrangle BXB2B3B4 contracts to a point: all the lines fi34, P23 pass through the centroid G (trJA) of the system of points Al9 A2, A3, A4, to which are assigned equal masses. If a1 = 0, that is, if the centroid of four points Al9 A2, A39 A 4 coincides with the centre O of the circle (A1A2A3A4) = (0), then (AXA2A3A4) and (BXB2B3B4) are concentric circles. Finally, if the unit point does not coincide with any one of the four Boutain points of the quadrangle A1A2A3A4 (these Boutain points form the vertices of a square inscribed in the circle (0 )), then the quadrangle --------- > B1B2B3B4 does not degenerate. It is an image of the quadrangle AXA2A3A4 under a similarity transformation of the second kind:

We first take the points A[, A29 A 3, A4 that are symmetric respectively to the points Al9 A2, A3, A4 with respect to the jt-axis, and then the quad rangle A[A»A3A4 is rotated about point 0 through the angle arg ----4 ----------------> and the rotated quadrangle A " A2A3 A4 is subjected to ahomothetic trans formation with centre 0 and ratio |1 — a 4!/4; we get a quadrangle ---------------- > d x 'A2 'A3 'A4 '; finally, the last quadrangle is subjected to a parallel transla tion determined by the directed line segment OG, where the affix g of point G is equal to <7j /4 (G is the centroid of the system of four points Al9 A2, A39 A4 --------- > to which are assigned equal masses). We obtain a quadrangle B4B2B3B4. Of all these transformations, only the first one (symmetry about the ----------- ---------------> x-axis) changes the orientation and, hence, and the quadrangles A1A2A3A4 and BlB2B3B4 are similar. Problem 20. A triangle ABC is inscribed in a circle with centre 0 ; A 0, B0, C0 are the centres of the circles (OBC), (OCA), (OAB); Al9 Bl9 C4 are points symmetric to the points A0, B0, C0 about BC, CA, AB respec tively. Prove that the orthocentre H of A ABC is the centre of one of the circles (S) tangent to the straight lines BxCl9 CxAl9 AXBX and that the

Complex Numbers in Plane Geometry

153

circle (5) passes through the centre of the Euler circle of AABC. Prove that the radius of (5) is equal to ~ OH. Proof. Assume {ABC) to be the unit circle. Draw tangent lines to {ABC) at the points A , B, C. They form a triangle A2B2C2. The circle {ABC) = {O) is inscribed in A A 2B2C2 if ABC is an acute-angled triangle, and escribed if ABC is an obtuse-angled triangle; if, for example, the angle C is obtuse, then the circle {O) is escribed in the angle C2 of triangle A2B2C2. In the quadrangle OBA2C the angles B and C are equal to 7i/2, and so a circle can be circumscribed about it; the segment OA2 becomes a diameter of that circle and, hence, its centre is the midpoint of segment OA2. But the circle {OBC) coincides of course with the circle {OBA2C); hence, the centre of {OBC) is the midpoint A0 of segment OA2. Let zl9 z2, z3 be the affixes of the points A, B,C. Then the affix a3 of the midpoint A3 of segment BC is a y ??J r J s 5 2 and since point A2 is obtained from A3 by inversion with the inversion circle {ABC), it follows that the affix a2 of point A2 is _ 1_ 2___ #3

~Z 2 +

The affix o0 of point A0 will therefore be 1 a0 — —----- ——• Z2 +

The affix a± of point Al9 symmetric to point A0 about the line BC, is found from the relation ao + al — #3> whence 1 —

Z2 + Z3~" Z T — — = 2 + Z3

Z2 +

Zo ZQ "

Z 3 -------------

Z2 +

- — 01~

Z2 ZZ —Zi=ax— z2 + z3

<*2 Z2 +

The affixes bx and cL of points BL and Cx are similar in form. Thus, 1 _ ^2 Cl = C i-----------(7l y^2 a2t Z2 +

t>i — G1 Z3

Cl Zi

u<*2 — C1 + Zi < t2 Cl + z2

1 — 2 1 _ — c 2 c2. 2

y O’2 b 2,

Problems in Geometry

154

From this it follows that A ^ A C i is obtained from A A 2B2C2 via the similarity transformation 1 U = t T i ----------- (To Z . 2 This is a similarity transformation of the second kind: first symmetry is executed about the jc-axis (z -* z), then a rotation about the point O through the angle arg(—g 2) 9 and then a homothetic transformation with centre O and ratio |tr2|/2 = laJ/ 2 ; finally, a translation defined by the directed line segment OH (since o1 is the affix of point H). As a result of these transformations, A A 2B2C2 goes into A ^i^iC l5 ------- N which is similar to A A 2B2C2 but has opposite orientation. Furthermore, since under the transformation 1 U = 17, ----------- Go Z 2 “ the centre O of the circle {ABC) that is tangent to the sides of A A 2B2C2 goes into the orthocentre H of A ABC, it follows that the circle (5), into which the circle (O) passes, will touch the sides of A A ^ C x (into which A A2B2C2 passes), and point H will be the centre of (S). Under the succession of transformations resulting in the transformation 1 U = (J1 ----------- Go

the radius of {ABC) changes only under the homothetic transformation {09 /ffil/2); consequently, the radius of (5) is equal to 2 2 since the radius R of {ABC) is equal to 1. Problem 21. Let AXA2AZA4 be an arbitrary quadrangle inscribed in a circle {O) and let P be an arbitrary point. Denote by P12, P1Z, Pu , P23, P 24, P 34 the points symmetric to point P about the straight lines AXA2, A xAZ9 A xAa, A2A39 A2A49 A zA4. Let R 9 S9T be the respective midpoints of segments P12, P ^ 9P139 P24, P14, P2Z9 and let O' be a point symmetric to the centre O of (O) about the centroid of the system of four points Al9 A29 AZ9 A4 (to which are assigned equal masses; Fig. 24). Prove that the points R 9 S 9T, O' lie on one straight line. Solution. Assume {O) to be the unit circle. Let zl9 z2, z3, z4 be the respective affixes of the points Al9 A29 Az, A4. Denote by g19 g 29 g Z9 g 4 the basic symmetric polynomials of the (complex) numbers z i9 i = 1,2, 3,4: — Z1 +

z2 +

Z3 +

C 2 = Z X Z 2 ~f- Z j Z z +

Z4>

^2 Z Z +

Z 2 Z4 +

Z3 Z4>

Complex Numbers in Plane Geometry

— Z2 ZZ Z4 +

<74 = Zi Z2

Z \ Z3 Z4 +

155

Z 1 Z 2 Z 4 4 ~ Z 1 Z 2 Z 3>

Z4.

The affix g of the centroid G of the system of points Al9 A2, A3, A4 is equal to a j 4, hence, the affix o' of point O', which is symmetric to point O about point G, is equal to o’ = ffJ2. The equation of the straight line ALA2 is z + zl z2 z = z1 + z2.

(54)

The equation of the perpendicular dropped from point P to the line AXA2 is of the form z - p = z1 z2(z - p) or z - z1z2z = p - zxz2p , (55) where p is the affix of point P. Adding termwise equations (54) and (55), we find the affix z = of the projection P& of point P on the line AXA2: P lZ =

(^ 1 + ^ 2

+ P ~

Z l ^2 P)-

Problems in Geometry

156

The affix p12 of point P12, which is symmetric to point P about line AXA29 is found from the relation P + Pi 2

= Pl2,

whence Pi 2 = 2P% — p = z1 + z 2 — z1z2p .

(56)

Of similar form is the affix p34 of point ^345 which is symmetric to point P about line A3A4: P m = *3 + z4 - z3 z4p . (57) From (56) and (57) we find the affix r of the midpoint R of segment ^12^34: (58)

r = — k i - (z 4z2 + z3 z4)p]. The slope of the line O'R is —r

x =

-----^ f a - (z1 z2 + z3 z4) p ]

2 2 2 L and therefore the equation of O'R is z7

P rr — •—— <J4 2 p

*3*4/

p = — <r4> P

(*-a

This equation may be rewritten thus: 2pz — 2p o4~z + <r3p — pa1 = 0 .

(59)

Since this equation involves only symmetric polynomials of the affixes zl9 z2, z3, z4 of points Al9 A2, A3, A4, the equations of the straight lines O'S and O'T will be the same as equation (59), that is, the points O', S, T, R lie on one straight line. Incidentally, we can see directly that the affixes s and t of points S and T, s = ~

1= —

[ ° i — (* 1 *3 + *2 *4) P ]

“ (zi z4 + Z2za) P]

satisfy the equation (59). Remark. Since equation (59) passes into an equivalent equation if p is replaced by Xp, where X is an arbitrary real number, it follows that

Complex Numbers in Plane Geometry

157

the straight line (59) does not change if point P describes a straight line passing through the centre O of circle (O) (provided, of course, the point O itself is excluded from the line). Problem 22. Let A1,B 1,C 1 be projections of the vertices A ,B ,C of A ABC on the diameter <5 of the circle {ABC), Denote by A2,B 2, C2 the points symmetric to points Al9 Bx, C1 respectively with respect to the mid perpendiculars of the sides BC ,C A ,A B of A ABC, Let A39 B3, C3 be the midpoints of segments BC, CA, AB, and let A 4, I k C\ be the midpoints of segments A2A3, B2B3, C2C3, Prove that A ABC and A ^ A C 4 are similar and have opposite orientations. Prove that if the diameter <5of (ABC) = (O) rotates about the point 0 , then the centre S of the circle (S) = (AaB^Ca} describes a circle (0) that is concentric with (ABC), and the rotations are in opposite directions. The radius of (S) is equal to — OH, where H 4 is the orthocentre of A ABC. The radius of (0) is equal to R/4, where R is the radius of (ABC) (Fig. 25). Solution. Take (O) = (ABC) to be the unit circle, and let the diameter & be the x-axis. Let zl9 z2, z3 be the affixes of the points A, B, C. The affix aL of the projection Ax of point A on the straight line <5 is

Since line BC has a slope of —z2z3, it follows that the equation of the straight line OA3 is of the form z z2z3 z and the equation of the perpendicular dropped from point Ax to line OA3 is.

From the system of equations z

z2z3 z — 0 , (60)

we find the affix z = a* of the projection A f of point Ax on the midperpen dicular OA3 of segment BC:

The affix a2of point A2is found from cii + a2

Problems in Geometry

158

whence a2 — 2a* — ax 2

_ 5 L ± Ii

The affix aAof midpoint A 4 of segment is #2 + #3 # 4 = ----------------- 9 2 where a3is the affix of the midpoint A3 of segment BC, that is,

Fig. 25

± i *. 2

a3 To summarize:

“I ('•+

z‘)= H"* +S)'

£z7 + Z! + O f similar form are the affixes bt and c4 of points Bt and C4. We have a\

l + 4 i ,’

/ a3 , (72 04 = ---- 1----- 9 4

(61)

4za

*3 . ^2 c4 — 0---r • 4

4 z3

From this it follows that the points At, Bt, C4 are obtained from the points A, B, C via the similarity transformation (7o

(To —

U= — + — Z 4

of the second kind, and, hence, A ABC and A A A Q have opposite orien tations (and are similar). From the relations (61) it also follows that the points A4, B4, C4 lie on the circle (5) with centre S whose affix s is s= < tJ4

159

Complex Numbers in Plane Geometry

and the radius of the circle (S) is equal to 1*2 1_ l*ll _ 1

The distance d from the centre S of (/f4#4C4) to the centre O of (O) = (ABC) is I - * d = ksl (R = 1). 4

4 “

This means that when the diameter 8 is rotated about the point O, the centre S of (/f4Z?4C4) describes a circle (f2) whose radius is equal to — = Rj4, where R is the radius of (ABC). If the diameter S is turned 4

through an angle a, then the affix of its end, which had an affix equal to 1, will have an affix P = cos a + / sin a; and if this point with affix p is taken as the new unit point, then the new affixes of the points A, B ,C 2

will be — » — » — j and the new affix of the new centre S* of the new triangle A fB fC f will be <r3//?3. This means the affix of point S* in the initial system will be <r3//J2. From this it follows that the radius OS of the circle (Q) will rotate in a direction opposite that of the diameter 8, and the angular velocity of rotation of OS will be twice the angular velocity of rotation of 8. Remark. Let us also find the ratio (a 4b 4c 4) (ABC) We have *3 *3 , *2 °± 7l i •° —2 z-, i1 4 z4 4 4~+ 4z1 7 (a 4b 4c 4) = 4

4 *~4z2 4 + T4 Za 1 *3

I 7

4 z2

z 2

4 T 4z3 4 J i *1 1 i O2 O2 Z2 22 1 4 16 z 3 Z3 1 whence

k 2|2 (ABC) = 16

kil - (ABC) = 16

OH2 (ABC) ~r<r (6 2 )

(A4B4C4) (ABC) ~

OH2 16

Problems in Geometry

160

------------—

From this it also follows that A^B^C* if ABC. Problem 23. Given a triangle T such that there exists a straight line t intersecting its sides at angles equal to the angles of the triangle, that is *, (AB, t) - (AC, AB) = A, (BC, t) = (BA, BC) = B,

(63)

(CA, t) = (CB, CA) = C. 1°. Find the angles of the triangle T. 2°. Through the points A, B ,C draw straight lines parallel to the line r. Denote by Ax, Cx, Bx, respectively, the points of intersection of these lines with the straight lines BC, CA, AB. Prove that A ACB and A B xCiAl9 are similar where A <--> Bx, C <-> Ci, B <— ►Ai, and that they have opposite orientations. 3°. A transversal xx is constructed for A B XC1A1 in the same way as the transversal t was constructed for A ABC. Let the straight lines passing through the points Al9 Bx, Cx respectively and parallel to the straight line xx intersect the lines BXCX, CXAX, AXBX respectively, in the points A2, B2, C2. Prove that A ABC and A C2B2A2 are homothetic and have the same orientation, that is, prove that the straight lines AC2, BB2, CA2 pass through the same point S and AB\\ C2B2, BC\ \ A2B2, CA 11 A2C2. 4°. Prove that the diameter of the circle (ABC) passing through the point S is perpendicular to the straight line t (Fig. 26). Solution. 1°. Suppose that C < B < A. By the Chasles theorem we find B = (BA, BC) = (BA, r) + (t, BC) = A - B, whence A = 2B. Furthermore, C = (CB, CA) = (CB, t) + (t, CA) = B - C * The symbol (p , q) is used to denote the oriented angle from straight line p to straight line q {p and q lie on an oriented plane). If a is one o f the values of the angle (p , q), then all values of this angle are given by the formula (A q) = « + kn, where

k assumes all integral values; stated differently, (p} q) = a (mod n).

The Chasles theorem for three straight lines p ,q ,r lying on an oriented plane holds: (A q) + (A r) = ( a r) (mod n). The relation (63) is to be understood as follows: one of the values (CA, r) is equal to C, one of the values (AB, BC) is equal to B, and so on.

Complex Numbers in Plane Geometry

161

whence B= 2C . Thus A = 2B = 4C and so n C = —> 7

4tt 7

2°. Assume {ABC) to be the unit circle, and A the unit point. The ver tices A, By C of A ABC will be the vertices of a regular heptagon AC'CB'A'QB inscribed in the circle (ABC). Setting a = cos (2n/l) + + i sin (27t/7) we find the affixes a, c \ c, b'> a', co, b of the vertices of this heptagon: a — 1, c = a (= a1), c = a2, 6' = a3, a' = a4, cu == a5, b = a®. The quadrangle is a rhombus because we can take the straight line for the straight line t (then all of the relations (63) of the statement of the problem will be fulfilled) and, hence, by construction we have t - AAX||CC', CB\\AC', C'C = C'A. The quadrangle CA,BB1 is also a rhombus since its opposite sides are parallel: A'B | | t || CB1,A B B 1 | | A'C and A'C = A'B. Similarly, we see that the quadrangle B'BCXA is a rhombus as well. From this we can find the affixes of the points A ^ C c , since A^C + A ^ A = A^C', it follows that c — ax + a — ax = c' — al9 1 1 -8 1 0

162

Problems in Geometry

whence al — c + a —- c' = a2 + I Furthermore, from the relation

or.

A'C \ A B =- A'BX we find c — a’ + b — a’ = bx — a and, consequently. bx — c + b — a! = a2 + a6

or.

Finally, from the relation ¥

+]F b = B T ,

we find a —b' + b — bf — c, ~ b \ c1 —- a + b — // — 1 + ar>— a3. To summarize: = 1 — a + a2, bx = a2 — a4 + ot6,

The triangles ACB and if and only if the determinant

— 1 — a3 -f- a6. -> are similar and have opposite orientations a bx 1 c cx 1

is equal to zero. We i a2 —a4 -|- a6 A = a2 1 —a3 + a6 a8 1 — a + a2

b 1 have 1 a2 —a4 + a8 1 1 — 1 a5 1 —a3 + a* 1 1 a 1- a + a 2 1

---• 1 — a3 + a* + a6 — a® + a7 + a3 — a* + a7 — a -f a4 + a

a7 - 1

a7 + ot9 - a11 = 0.

— nr® — nr7

Now we find the centre of similarity* of A ACB and A^iCV^i. Let us consider the similarity transformation of the second kind under which * The centre o f similarity of two mirror-similar triangles (that is similar triangles with opposite orientations) is understood to be the fixed point o f a similarity transformation of the second kind that carries one o f these triangles into the other.

Complex Numbers in Plane Geometry

163

the points Ax and B5 go into points B and A (then point C, goes into C). Let z be the affi x of an arbitrary point M of the plane, and let M \u) be its image under the indicated similarity transformation. Then z u 1 bx 1 1 = 0 ax 1 or u(ax — bx) + z( I — a) + abx — ax = 0 or n(l — a -f- a4 — aB) + z(l — a) + a3 — a5 + a7 — 1 + a — a2 = 0, cl

whence, cancelling 1 — a, we have u( a5 + a4 + I) + z + a(a3 + a2 + 1) = 0. The fixed point of the similarity transformation satisfies the condition z(a5 + a4 + I) + z + a4 -f- a3 + a = 0, (64) whence (passing over to conjugate numbers) we have z(a5 + a4 + 1) + z + a4 -1- a3 + a = 0 or, multiplying the left-hand side by a7 = 1, z(a2 + a3 + 1) + z + a3 + a4 + a6 --- 0. From the last relation we find z = —z(a3 f a3 + 1) — a3 — a4 — a6 and equation (64) takes the form —z(a2 r a3 + 1) (a5 + a4 -[- 1) — (a5 + a4 + 1) (a3 + a4 + a®) + z + a3 + a4 + a = 0 or z( I — a7 — a6 — a2 — a8 — a7 — a3 — a5 — a4 — 1) = a8 + a9 + a11 -f a7 + a8 -f- a10 + a3 + a4 + a6 — a3 — a4 — a or z(—a® -- a5 — a4 — a3 — a2 — a — 2) = a6 + a4 + a3 + a2 +■ a + 1. But a7 == 1, and since a — 1 ^ 0, it follows that a6 + a5 -f- a4 -f a3 + a2 + a + 1 = 0 and the last relation takes the form

Problems in Geometry

164

whence z = a5. That is, the fixed point of the~similarity transformation of the second kind --------> ------ > that carries A BXCXA^ into A ABC is the point Q. Setting a5 + a4 + 1 =- A, a3 + a2 + 1 r~ p, we find the relation Xu + z +

olh =

(65)

that relates the affix z of the arbitrary point M of the plane with the affix u of its image M ' under the similarity transformation at hand. The points Bx and Ax go into points A and B under this similarity transformation, so that +afi, bx + <xfx a= ,----- 9 b X whence ...

a ~ b = <*i

------------------

bx X

and, thus, A\B\ â&#x2013; = WAB However, a5 + a4 + l = a 2 + a 3 + l. a5 + a4 + 1 -- a2 + a3 + 1; consequently, |A|2 = (a5 + a4 + 1) (a2 + a3 + 1) ~ a7 + a8 + a5 + a6 + a7 + a4 + a2 + a3 + 1 = 2 + a6 + a6 + a4 + a3 + a2 + a + 1 = 2, whence A\BX_Y2 9 __ AB ~ â&#x20AC;&#x2122; A1B1 ~~ ](2 ---------------------------- * is the proportionality factor that carries A B ^ A i into A ACB; it is, equal to 1//2.

Complex Numbers in Plane Geometry

165

3°. Let Ox be the centre of the circle (AXBXCX). The point 0 1 is the image of point O under the indicated similarity transformation (item 2°). There fore, the affix ox of point Ox is found from equation (65) by setting u = 0 in the equation: ox ~ — a(a3 -f a2 + 1). Let us consider a regular heptagon inscribed in a circle ( A ^ C j) with the vertices arranged in the following order: A qCqQqA qBqBqCo, where A0 = Bx, B0 Cx, C0 = Ax. Such a regular heptagon exists since the angles of A BXAXCX are

and, hence, AXBXis one of the sides of the heptagon and Cx is its third ver tex. The affixes of the vertices of the heptagon are Oi + (a0 — °i) 1 (k = 1, 2, 3, 4, 5, 6, 7). Expanded, they are: a0 = bx = a2 — a4 + a8, c0 = = a2 — a + L ct)0 = — a4 — a3 — a + (a2 + a6 + a3 + a) a2 — —a4 — a3 — a + a4 + a8 + a6 + a* = oc\ a'0 = — a4 — a3 — a + (a2 + a8 + a3 + a) a3 = — a4 — a3 — a + a5 + a9 + a* + <*4 — a8+ a5 —a3 + a*— a, bo = — a4 — a3 — a + (a2 -f a6 + a3 + a) a4

= — a4 — a3 — a + a6 + a10 -f a7 + a6= a® + a5 — a4 — a -f- 1,

*o =

= a6 — a3 + 1,

ci =-- — a4 — a3 — a + (a2 + a6 + a3 + a) a6 = — a4 — a3 —■a + a8 + a12 + a9 + a7 = a6 — a4 — a3 + a2 + 1. From this it follows that Q0— Q and that this point is the centre of the simi larity transformation under which A C 2A2B2 goes into That these triangles are similar and have opposite orientations follows from the -------------------> .

------------------- > .

fact that the construction of A C2A2B2 on the basis of ABiCxAx is exactly the same as the construction of A B ^ A x on the basis of AACB; the

166

Problems in Geometry

figures under consideration with two circles circumscribed about A ABC and l\A 1B1Cl and all the corresponding triangles are samilar. Furthermore, since CqA zA qBqis a rhombus, it follows that Cq “ Qq b0, whence — Qq 4~ Co feo - oc2 - a 4 + a6 + a 2 — a + 1 - a* - a5 + a4 + a - 1 = 2<x* - a5. Again, BzA0CqB0 is a rhombus and so a0 bz ~ Cq whence b2 = a$ -(- b0

= a 2 — a4 + a6 + a8 — a3 + 1 — a5 + a4 + a* — a 2 — 1 = 2a6 — a5. Finally, since CzC0AqB0 is a rhombus, it follows that c0 cz = Gq b09 when ce CZ — Cq -f* 6 q

= a2 — a + 1 + a8 — a3 + 1 — a6 — a5 + a* — a2 + a = 2 — a5.

Thus, a2 = 2a2 — a5,

b2 = 2a6 - a5, c2 = 2 — a5 and so A C2B2A2 is obtained from the triangle ABC (the affixes of whose vertices are 1, a6, a2) via a homothetic transformation with centre Q and ratio 2 since c2 — 0) 2 - a5 - a5 2 z,, a — 0) 1 - a5 b2 - - 0) 2a® — a5 - a5 b - 0) ~ a®— a5 a2 ~- (0 c — 0)

2a2 - 2a5 - 2 -a2 — a5

4°. The slope of the straight line OQ is equal to (X5 — = a'° = a*. a5

Complex Numbers in Plane Geometry

167

The straight line t passes through the points A and B' and so its slope is 1 - a3

1 - a3

1 - a3 ~ i __ 1 a3

a3.

The sum of the slopes of the straight lines OQ and x is 0 and so OQ _L x. Problem 24. 1°. On the sides of the hexagon AyA2A3A4AbA6are construct ed equilateral triangles ---------->.---- :----- > ----------> —--------> ----------> ----------► AyA2Al9 A2A3A29 A3A4A39 A4AbA49 AbAeAb, A^AyAB (66) which have the same orientation. Prove that the ends P[9 P3, Pb of the directed line segments OP[9 OP3, OPb (O is an arbitrary point), which are ------ -------- >.-----respectively equipollent to the directed line segments A[A49 A3Ab, AbA2y form an equilateral triangle T = P[P3Pb9 which has an orientation oppo site that of any one of the triangles (66). In particular, A T can degenerate into a point. 2°. Prove the converse: namely, that if an arbitrary equilateral triangle T — P 1P3P 5 and a point O are chosen in a plane and directed line segments AyA4f A3A3, AbA2 are constructed thatjare respectively equipollent to the directed line segments OP[9 OP3', OPb (the position of the directed line ------- > .------- > -------->

segments A[A49 A3A£, AbA2 are otherwise arbitrary), then, by choosing another arbitrary point Al9 it is possible to construct the points A2, A3, A49 Ab, A6 such that the hexagon A[A2A3A4AbAb is obtained from the hexagon A1A2A3A4AbA6 by the construction indicated in item 1°. Prove that if the point Ay is changed in the plane, then the principal diagonals AyA4, A3A69 AbA 2 of the hexagon A lA2A3A4A bA6 (which will change to gether with any change in point Ay) will rotate about three points Ol9 0 2, 0 3. How are these points constructed? 3°. The equilateral triangles A ;A 'A ;',A 'A ;A '', A'3A ’4A'3'9 A 'A 'A 4'9 A'bA'A'b\ A ,bA[A,bf

(67)

are constructed so that they all have the same orientation and the orien tation of any one of them is opposite that of any one of the triangles (66). Prove that ------ > ------ > ------ > ------> ------> ------ > A\Ay = A2A39 A2A2 = A3A4, A3A3 = A4Ab9 ----- > ------> ------> ------ > ------> ------> A4A4 = A bA69 AbAb = A6Al9 A€A6' = AyA2 where == is the sign of equipollency of (directed) line segments.

Problems in Geometry

168

4°. Prove that A['A'a' = A W

= A W

= 3OG,

where G is the centroid of the triangle T* = M14A/36M52 and M liy Af36, A/52 are the ends of the directed line segments OMUy OM36, OM52 laid off from an arbitrary point O and respectively equipollent to the directed line ------> ------> ----- > segments AÂA9 AÂ^ AÂ^. 5°. Prove that the midpoint of the principal diagonal AXA4 of the hexagon A ^ A ^ A ^ A q coincides with the point/C14, in which segments A['A'2' and AZ'A$' intersect and are bisected; the midpoint of A3A6 coincides with the point K3S, in which segments A z’ 'A a and A's A[' intersect and are bisected; finally, the midpoint of A 5A2 coincides with the point A^52, in which seg ments A Â ^ and A2A'^ intersect and are bisected (Fig. 27). Solution. We introduce in the plane a rectangular Cartesian system of coordinates Oxy. Denote by a and a the imaginary roots of the equation jc3 + 1 = 0 : a

l+ijfa 2

’

“

1-/K3 2

Let alt a2, a3)a4, a5, a6, p[, p'3, p3, a[, a2, a3, a[, a3, a«, be the respective affixes of the points A u As, A3, At, A5, A„, P[, P3, P3, A[, A3, A3, A4, A3, A3. Then «2 — fli = <x(a[ - at), o3

a3 = oc(n2

@z)y

o3 — oi(a3

a3).

as - a 4 = ct(a4 — at), a6 - a 3 = a(as — a5), — a6 = <x(a3 - a6). From this we find (note that since a2 — a + 1 = 0 it follows that a — 1 + + a = 0, whence 1 — a = a) a. = aa[ + aa4, a3 — aa'3 + a a2, a4 = aa3 + a a3, o5 = aa4 + aa4, = afls + a^s. a4 = cta'i -f aa8.

(68)

Complex Numbers in Plane Geometry

169

Fig. 27

Multiplying these relations by a5, a4, a3, a2, a, 1 respectively and adding, we obtain (note that aa = 1, a3 = a3 = —1) a'2 — ab = a (a'6 — aJ) + a (a'A— flOSince a’4 — a[ = p[, ai — a3 = Pz9 of2 ~~ afb = p b, it follows that pi = 05P* + or

pi = <*/>3 + (1 - a) = />J

Problems in Geometry

170

or />5 ~ P'l = a (pZ - p[).

(69)

From this it follows that AP[PZPZ (provided it does not degenerate into a point) is equilateral because from relation (69) it follows that the directed line segment P[P$ is obtained by a rotation of the directed line segment P[PZ through the angle tt/3 (a = (1 + i J/3)/2 cos (tt/3 + / sin (it/3)). ------> The orientation of A P1P3P5 is opposite that of any one of the triangles of (66): this follows from relation (69) and, for example, from the relation a-i — at = a (a[ — ax); since aa = 1, it follows that a [ - a x = a (a2 — a t),

that is, the directed line segment AXA[ is obtained by a rotation of the directed line segment AXA2 through the angle —tt/3 (a = cos (—n/3) + + 1 sin (-tt/3)). The case p[ -- pZ is also possible; then also p[ = pZ, that is, A P[PZPZ contracts to a point. This occurs if and only if the directed principal di agonals A[Ai AZAZ, AZAZ of the hexagon A[AZAZAZAZAZ are equipollent: a [ a '4 - azaz = azaz.

2°. Let P[PZP* be an arbitrary equilateral triangle and O an arbitrary --------^

> ------- >

point in the plane; let A[A4 AZAZ, AZAZ be arbitrary directed line segments ,

respectively equipollent to the line segments OP[, OPZ, OPZ: OP[ - A[Ai OPZ = A fzA l OPZ = AZAZ

(70)

Choose an arbitrary point Ax in the plane. Construct the following equi lateral triangles with the same orientation: ------y ► » y > > AXA2AX, A2A3A2, A3A4A3, A4Ar>A4f A^A^A^, AqAxA$ but with orientation opposite that of A P{PZPZ- On the basis of item 1°, I with the aid of the construction given in item 1°, the hexagon AxA2A3A4AhA^ leads to the chosen points A[, AZ, AZ, A\, AZ, AZ and then to the chosen triangle P[P’zP i 3°. The affixes a[, aZ,aZ,aZ, aZ,aZof the points A[, AZ, AZ, AZ, AZ, AZ are connected with the affixes a", a2 , a2 , a4", aZf, aZ' of the points A x , A2 ,

Complex Numbers in Plane Geometry

171

A 3', A'a A'b\ Ab by relations of the form (68) in which the positions of only a and a need be interchanged: a2 — aa[f + cta[9 a'z = aa2 + aa2, a\ = aaz + ctaZ9 ab == 0W4' + aaA, = otab -f-

010r5,

a[ = a^e' -|- a^g* From these relations and from the relations (68) we find ata" = a2 — owj = a3a — 02& + <*i&= afl3 — ota2 + &Q\ and so o-y

-|-

Similarly a2 ~ a2 ~

# 3

* =

£ 4

# 3

“F ^ 4 ?

# 4

“ t“ ^

(71)

^ 6 ’

tffi* “ a% “t~al> a* = a< — + a2. From these relations it follows that Oy Q\ ~

a2,

Cl2

Q2 =

0 A,

aA= a6

^5,

£6*

#5 = ^1

^65

= @2

^1»

and therefore A-yA" = A2A39 ^ 2^2 = ^3^49 AZA3 = A4A5, A4A4

AbAe, AbAb

AgAj, A qA q

—

AyA2.

4°. From the relations (71) it also follows that ai' — a" = ai' — 03' = ^2' — ^ 5' = ^4 —

— <*1+

— #3- (72)

Problems in Geometry

172

Consequently, A ^A 4 = A ^A i' = a7

a ;\

If we construct the directed line segments ------- >

y ------- >.

OMu = AjAf, OAfgg = AgAg, OAf$.> = A$A2, then the affixes m14, m36, m52 of the points M u , Mw, Afsa will be and so the affix g of the centroid of A

will be

g = 0*4 — Oi + </6 — + a, - «s)/3. From this and from the relations (72) it follows that A['Al'

A'3'a \' = A'^A” = 3OcT

5°. The midpoints of segments

AtA4 have the same affix Qi + a« 2

[see formulas (71)] and, hence, these midpoints coincide with the point Kl4. The remaining propositions in item 5° can be proved in similar fashion. Problem 25. Let Av Blt Ct be the orthogonal projections of the point P ------► on the sides BC, CA, AB of triangle ABC. Let us construct A CXBXQX ------> similar to A BCP but with opposite orientation. Let A', B \ C be the second points of intersection of the straight lines PA, PB, PC with the circle (ABC) = (O). 1°. Prove that the triangles AXCXQX and CAP are similar and have op posite orientations. 2°. Prove that the triangles BXAXQX and ABP are similar and have op posite orientations. 3°. Prove that the triangles AXBXCX and A'B'Cf are similar and have the same orientation. Find the proportionality factor. 4°. Prove that the point Qx is a point obtained by inversion with the circle of inversion (.AXBXCx) (or by symmetry with respect to the straight line AXBXCXfor the case when the points Ax, Bx, Cx lie on one straight line) of the midpoint T, of line segment PP*9 where P* is the image of point P under inversion with the circle of inversion (ABC). 5°. Find the affix of the fixed point of the similarity transformation ------- > ------- > that carries one of the triangles AXBXCXand A'B'C' into the other. 6°. For what position of point P does it coincide with point Qx (Fig. 28)?

Complex Numbers in Plane Geometry

173

Fig. 28

Solution. 1°. Take (O) — (ABC) for the unit circle. Let z„ zt, z3 be the respective affixes of the points A, B, C. The equation of the straight line BC is of the form Z

— z3 = —z., z3 (z — zs)

or z + z2z3z = z2 + z3.

(73)

The equation of the perpendicular dropped from point P to line BC is z - p = z.,z3(z - p ) or z-

Z2Z3 Z

= p — z2z3p,

(74)

where p is the affix of point P. Adding termwise the equations (73) and (74), we find the affix z = ax of point A1: ax = —- (z2 + z3 + p — z2z3p).

Problems in Geometry

174

Similarly, we can find the affixes t 1 , y

b i

Cl =

and c, of points Bx and C ,:

(Z3 +

- J - (Z i

+ P

Z3Z ]/7),

z2 + p - z

, z 2p ) .

Since it is given that A BCP and A C1B1Q1 are similar but have opposite orientations, it follows that the affix of point Q1 can be found from the condition b cx 1 c b2 1 = 0 P <h 1 or ------ - - (z1 + z 2 + p - zxz2p) 1 2 ~ (Z3 + zl + P

—

Z3Zxp) 1

or 1

—

^2 1 *3

Z1 + Z2

P — z& p 1

Z3 + Z1 + P ~~ z 3zlP 1

Subtract the second row from the first row of the determinant to get - — — z2 — z3 + (z3 — z2) pzl 0 Z2Z3

— Z3 p

z3 +

Cancelling z3 — z2 # 0, we get

Z i + p — z 3Z j p

Complex Numbers in Plane Geometry

175

Multiplying the left-hand side by z2z3, we obtain -s + Zl + p — z3z,p - 2q1 — (z, - ztzzp) (z,p - I) - 0 or Z3 + Z 1 + P — z%Z\P — 2<7i — z2ztp -f r, — a3p %— z.tz3p -= 0 whence 9i

* i)-

2°. The symmetry of the right-hand side with respect to z l 9 z3, z3 permits asserting that (item 1°) A

A y C 1Q l

and A

CAP

are similar and have oppo

site orientations and that (item 2°) A P y Ay Qy and A A B P are similar and have opposite orientations. At the same time we found the affix of point Qy. 3°. Associated with the directed line segment Aybx is the complex number

Let us find the affixes a \ b \ c' of the points A \ B \ C . The equation of the straight line PA is of the form P — z\ Zi — —----- ~ ( z - Zj). P - *i Solving this equation together with the equation zz = 1 of the unit circle (ABC), we obtain z

P - Z 1

Z — Zy

P - Zi

ZZy

One of the roots of this equation is naturally z --- Zy (the affix of point A); the other root (the affix a' of point A') and be found from the equation P — Zy

P — Zy

ZZy

whence z --= a

1 P - Zy Zi P ~~ Zi

176

Problems in Geometry

and b' and c' have similar expressions. Thus, l p - z 1 a = -------- : ---Zi ~ p - Zl

b, = _ _1. P - z2 ^ Z2 p — ~Z2

C' = -

Z3 P — Z3 To the directed line segment ^f'5' there corresponds the complex number

b, _

, = J _ P - Z i _ _ L P - Z2 zi P — z i z2p z2 =

z*(pP

z iP

z i* a ) ~

zi(PP ~ ZiP ~ Z*P +

z i z 2)

ZjZ2(p - z ,) (p — z 2) _ z2pp — ZiZ2p — p + Zi — Z|PP + p + ztz2p — Zjj ZiZ2(p - Zj) (p - z 2) _ pp(z2 — zt) — (z2 — zt) __ ZiZ2(p — Zl) (P — Z2)

(z2 - Zt)(pp - 1) ZlZ2(P - Zl) (P -

z 2)

From this it follows that b’ - a' = __________ 2 ( p p - 1)__________ *1 - flj

ZjZ2(p -

Zj) (p -

z 2) (z3p -

2(pp — 1)

<r3(P - Zi) (p — z 2) (p J

O’2 — t

z 3)

2(pp-l)

a3(p 3 — ^ p 2 + d2p - d3) ’

Gl a2 = — 5 it follows that eo

and since a,

b’ - a ' — fli

2 (pp -

<r3P3 - <r2P2 +

(76) —1

Denoting the right-hand side by A, we have t 2 (p p -l) <r3P3 — <r2p 2 + atp — 1 and then obtain b’ - a' = A(b! - a!>

177

Complex Numbers in Plane Geometry

and, similarly (since A is a symmetric function of zl9 z2, z3), c' - b' - A(cx - bx)9 a' — c' = Afa — cx). --------> -------->. Consequently, A A'B'C' and A A1B1C1 are similar and have the same orientation. On the basis of (75), the proportionality factor may be written as 2\OP2 - R 2\ 2\OP2 — /?2| ~ Ip - z x\ |p - z t \ |P - z t | “ PA-PB-PC 4°. Let us consider the similarity transformation that carries A A,B,C, into A A’B'C’. Under this transformation, the centre Ox of the circle (A& C ,) goes into the centre O’ = O of the circle (A'B'C'), and the point H„ the point of intersection of the altitudes of A A,B,C„ goes into H', the point of intersection of the altitudes of A A'B'C'. Since the indicated similarity transformation is of the form z' = Xz, + p and since under this transformation the directed line segment OxA, goes into the directed line segment O'A' = OA', it follows that a’ = X(a, — ox). (77) where a', au o1 are the respective affixes of the points A', A„ Ox. Simi larly, b' = Mb, (78) C' = MCl — O,),} where b', c', b„ c, are the respective affixes of the points B', C', B„ C, From (77) and (78) we find o' + b' + c' — Mai + b, + cx — SOi). Now, using the earlier obtained expressions for a', b', c', we have o’ + b ' + c ' =

JL P ~ J ± _ J _ P ~

Z\ P

where = z 2z3(p

Z2 P

Z2 ___1_ P - Z3 Z3 P Z3

_L x a3 Y

(79)

z,) (p - z 2) (p - z 3)

+ Z-MP - z2) (P ~ z 3) (P + Zi) + zxz2(p - z3) (p - z,) (p - z 2), Y — (p — z x) (p — z 2) (p - z 3). 12-810

Problems in Geometry

'178

Furthermore, z 2z 3( p

z 2) ( p

Z 3) ( p

rO =

--== Z j Z 3p - p

(z2 +

z , z 3[ p a —

( z 2 + z ;.) > +

\ z 3z 3 p l -

(z , +

z 3) p p

z 3) p

1] (/> -

Z , z 3] ( p

z ,)

Z i)

ffj/;2 +

(z tz2 +

Z iZ j)

zt.

The other two terms in the numerator of (79) are of the form Z3 Z1 ~P% P - (z3 + Z2) pp + p - o3pa + ( z 2z 3 + ZjZj) p - z 2, zLz2p ap — (Zj + z2)pp + p — o3p a + (ZjZj + ZjZ2) p - Zj. Adding the last three expressions, we obtain X = pp2<r9 — 2(7!/?/? + 3/? — 3(7.,/?2 — 2(T2p — To summarize: , . ,, , , -pp-O i + 2ffipp - 3p -f 3<r3pa - 2<s2p -f Ox a + b + c —------------------- -3--------------------- ------------o*(P - * i) (/> - - 2) iP - ^ 3) Using the expressions for au bu cx obtained above, we get 1 «i + ^ + c2 = ' (2(71 + 3/? - cr2/?); 2 {p p

and since A — °*(P

it follows that the relation

z'i) (P - z%) (P -

a' + bf + c' — A (^! +

+ cx — 36?x)

— 3*! — --- (a! + bf + c ) — (^i + bx + Ci) A may be rewritten thus: - to , =

+ 2°,PP ~ » + 3» ^ * ~

+ ». —

PP - 1 3/>V + 3o3p3 — 3<r2p 4- 3*7, pp - 1

hence, O, —

P*P — o3 p a - a ? p 2(pp - 1) 4

— Oy

From this we obtain 9i - Ox

o 3p 2

—

<r2P

+ P

< ri

p %p

—

o 3p 2

o ,p

— <Tj

2(PP - 1)

O a P P * — O i P P %Jr P i P + Q i P P — Q ^ + Q i P — p — O

2(PP -

x - I ^ P + o ^ - o -iP + O x

Complex Numbers in Plane Geometry

179

a3pp3 — a2pp2 -rOxPP —p _ o3p(p3 — <bP2 + aiP — a3) 2(pp ~ 1) 2(PP - I) <r3p(p — Zj) (p - z 2) (p - z 3) 2(pp —1) The affix t of the midpoint T of segment PP* is

iK ) We have pp + 1 t - o 1= 2p

PP + 1 2P

P2P — < t3p 2 + <r2p — o', 2(pp - 1)

p2p 2 — 1 —p2p 2 + a3p3 — <r2p 2 + ffjp _ a3p 3 — a2p 2 + o,p — 1 Tpipp - J) 2P (PP - 1 ) <r3(P ~ Z j) (P ~ z't) <P ~ z 3)

2p (p p - 1) whence : <y3(P - Zj) (P ~ z2) (p - z3) t — 01 —-----------------------------------2P(PP - 1) and therefore (?i - <h) (tj - o,) =

Ip — *iI8 \p — ^>12 !p — z3\2 _ 4(pp - I)2

But this means that the points T and Qx are obtained one from the other by inversion with the circle of inversion with centre Ox and radius |p — Z ,| Ip — Z 2 1Ip — z 3| 2 \pp — 11

PA-PB-PC (R = 1). 2 | OP2 - R2 1

The radius of the circle (AXBXCX) — (Ox) can be found from the simi larity relation z' = Az1 = p that carries the circle (AXBXC}) into the circle (A' B' C'). Since the radius of (A'B'C') is R — 1, it follows that the radius Rx of (AXBXCX) is R 1

R |A|

! o3(P — Zi) (p — z >) (P — z3) _ PA-PB-PC _ | 2(Pp - 1) ~ 2 \OP2 - R2\ P'

5°. The affix co of the fixed point Q of the similarity transformation that carries one of the triangles, , into the other, A' B' C', is found from

180

Problems in Geometry

the reasoning that under this transformation A OxAx Q goes A OA'Q so that 0 1 a f ax 1 = 0, CO C0

into

whence o,a

co —

Oi + a - ax

Furthermore, a

Zl ~ P

Z2 — Z ^ + p — Z2ZzP

- ax = ------- --------------------------- -----ZiP -

2zj — 2p — zvztp — zxz%p — zxpp + a3p 2 + z3 + z3 + p — ztz3p 2{z{p - 1) Z\

<>i+a'— ax = :

p*p - g8p a + a2p - ffi

I- ffi

zt — p +

2(pp — 1)

---------—

2(p p -

—P

—

------------------ —

1) (zlP - 1)

— (Tjp — ZiPP 2(zlP - 1)

<r»p*

- <T2p - ztpp + <raP2 2fop - 1)

(P2P % - a3p % + a & p * - axzxp

pip

+ <r3p s - atp -f a l + p p z l — p 2p + PP<J\ — p 2p<r2 — Zippp* + <r»PP* — z1 + p — a1 + pa 2 + Zxpp — a3p2) 2pp(zx - p) - (zL - p) + gaP2(^i ~ P) ~ OiP(zx - p) -<x3p3(zt - p ) 2(PP - 1) (ZiP - 1) (zi ~ p)(2pp - 1 + g2P2 - ffiP - JaP8) 2(PP - l)(ziP - 1 ) and, consequently, P*P — o'sP2 + <r2p — ffi Zi

— P

_________2(pp - 1)______ z,p — 1___ (zi —p) (2pp — 1 + ff2P2 — ffiP —ff3P3) 2(PP - 1)(zjP - 1) _______P2P — g3P2 + ff2p ~ gi______ 2(pp — 1) - 0 3P3 - <t2P2 + crtp - 1)

Complex Numbers in Plane Geometry

181

But P2P — G-iP1 + <*lP — <*1 = 2ox(pp — 1), <r3P* — <*iP'

4 - <?iP

— 1 = 2p(pp - !)(/ — o,)

and so 20i(/>P - 1)

0) — ----- =------------- ———---------------- “ ------------------ . 2(pp - 1) — 2p(pp — 1)(/ - o,) 1 - p(t - 0t) 6°. The points P and Qt coincide if and only if p = qlt that is, P =

- (0-jP* — OtP + P + ffi) 2

(80)

0 , p 2 — ff2Jp —P + 0i — 0.

From this. a*p%—

— p — 0! •= 0

— P*03 0g

=0.

that is, p* — 0tp + 08 0, and (80) takes the form — (p* + alp* + 0 ! — 20, p* + 2/>*02 — 20j02p) ^3 — 1 (0 8p2 — 0 10 a/> -h 0 !) —p + 0 i = 0 <*3 or p* —

2 0

jp* + ( 0 ? -I-

0 2

)p 2 — ( 0 X0

3) p +

0 1 0 3

(81)

It is clear that p = is a root of this equation. Dividing the polynomial in the left-hand side of (81) by p — <rl9 we obtain p 2 — Oi/?2 + <x2p — <73 = 0 or (p — z j (/? — z2) (/> — z3) = 0. Thus, the remaining roots of equaton (81) are P

—

1, p = z2, p = z3.

Assuming that P does not coincide with aoy of the vertices of the given triangle, we conclude that the only possibility is p = ov Thus, the points P

J 82

Problems in Geometry

and Q 1 coincide if and only if P is the orthocentre of the given triangle ABC. Problem 26. Take an arbitrary point P on a circle (ABC) = (O). Let A* be the point symmetric to point A with respect to the straight line OP, and let A' be the point symmetric to P with respect to the straight line OA*. Construct points B' and C' in similar fashion. Denote by A", B",C,r points respectively symmetric to the points A', B', C’ with respect to lines jBC, CA, AB. 1°. Prove that the triangles ABC and A"B"C" are similar and have the same orientation. 2°. Prove that A ABC is inscribed in A A " B " C namely, that the triplets of points A , B”, C"; B, C", A"; C, A", B" lie on one straight line. 3°. Prove that the orthocentre H of A ABC is the centre of the circle (A"B"C"). 4°. Find the radius R" of (j4"2P'C") and indicate a method for construct ing it if we know the position of the unit point on the unit circle (ABC) and the position of point P on that circle. 5°. What is the maximum value of R"1 What position of point P is associated with this maximum value of P"? 6°. For what positions of the point P do all the points A", B ", C" coincide with the point //? 7°. For how many and what positions of point Pare the triangles A ”B"C” and ABC congruent? 8°. Find the affix of the centre of the similarity transformation of A ABC and A A nB " C \ assuming (ABC) to be the unit circle and know ing the affixes a, b, c, p of the points A, B ,C , P. 9°. Prove that if the point P describes the unit circle (ABC), then the centre Q' of the similarity transformation of A ABC and A A ,,B ,,Cn de scribes the orthocentroidal circle of A ABC, that is, the circle (GH) con structed on the line segment GH as a diameter (H is the orthocentre of A ABC, G is its centroid, hence the term orthocentroidal). 10°. For how many and what positions of the point P do the points G and Q' coincide? 11°. For how many and what positions of the point P do the points H and Q' coincide? 12°. What line does the centroid G" of A A"B"C'f describe if point P describes the circle (ABC)1. 13°. Prove that the affix of the centre Q' of the similarity transformation of A ABC and A A"B"Cff is connected with the affix of the centroid G" of A A"B"G” by a linear fractional relation, which, consequently, deter mines a certain circular transformation of the plane. Prove that this trans formation leaves in place the orthocentroidal circle (GH) of the triangle ABC. Find the fixed points of this transformation. 14°. Prove that the centroid G" of A A nB nCn, which centroid cor responds to the centre Q[ of similarity of A ABC and A A"B/fC", is the

Complex Numbers in Plane Geometry

183

second point of intersection with the circle (GH) of the straight line 0 }Q*, where Ol is a point symmetric to the point O about the midpoint K of segment GH, and Q* is a point symmetric to the point Q' about the mid perpendicular of segment GH. Solution. 1°. We take {ABC) for the unit circle, and Boutain point of A ABC for the unit point. Let a, b, c, p, a', b', c , a", b", c" be the re spective affixes of the points A ,B ,C ,P , A f,B f, C , A ",B ",C ". Then ~ abc — 1. The equation of the straight line passing through point A and perpendicular to the straight line OP is of the form z — a — —pl (z — a). Solving this equation together with the equation of the unit circle zz we get 2 z —a z — a = p * -------az

One of the roots of this equation, z = a, is the affix of point A, the other root is z = a* ----- p2ja. which is the affix of point A*. The slope of the straight line perpendicular to the straight line OA* is z - p = x(z - p ),

P‘ a

The equation of the straight line passing through point P perpendicularly to OA* is of the form P4 * ~ P = ------ (* - P l a2 Solving this equation together with the equation of the unit circle zz = 1, we find the affix a' of point A':

One of the roots of this equation is naturally z = p (the affix of point P), the other is the^ affix of the point A \ that is, a! — p2/a2. In similar fashion we find the affixes of the points B' and C . We thus have PL c2 '

a’ The equation of BC is z + bcz = b + c.

184

Problems in Geometry

The equation of the perpendicular dropped from point A ' to the straight line BC is

Adding these last equations and the equation of BC termwise, we find the affix z = a'l of the projection of point A' on line BC:

We find the affix a" of point A " from the relation

That is,

whence (since a3 = 1) a" = b + c — a

= a1 — a( 1 + p 3). n3 P6 In similar fashion we find b" and c". Thus, a" = ffj — (1 +p~*)a, (82) b" = <rx - (1 +p~3)b, c" = 0 i — (1 +P~3)cAnd so A A"B"C" is obtained from A ABC by a similarity transformation of the first kind: (83) 2 " = (Tx - (1 + p - * ) Z . Hence the oriented triangles ABC and A"B"C" are similar and have the same orientation. 2°. We now prove that A ABC is inscribed in A A"B"C ". We will prove that the straight line B"C " passes through point A. To do this it suffices, for example, to prove that a — b‘ a — c"

Complex Numbers in Plane Geometry

185

is a real number. Taking advantage of formulas (82), we find a — b" _ a — + b{\ + p~9) _ bp 9 — c _ a — c" a — + c(l + p 9) cp~3 — b We have 1 3 1 p3 -----b c cp9 — b bp 3 — c u = —-------- — = ------------ — — u. 1 3 __ 1 bp9 — c cp 3 — b c b

In a similar manner, proof is given that line C 'A " passes through point B and line A"B" passes through point C. 3°. The image of point O under the similarity transformation (83) is a point with affix aly that is, the orthocentre H of A ABC. Under the simi larity transformation (83), A ABC goes into A A"B''C” and so the centre 0 of {ABC) goes into the centre of (A''B''C"), that is, the point H. 4°. The radius R" of (A"B"C,f) is equal to | 1 + p~3| = '| 1 + p3| and, hence, may be con structed as follows (Fig.29): draw a tangent line to the circle (O) at the point P and through the unit point Q (a Boutain point); draw the straight line 1 parallel to that tangent line; the second point of intersection, point Q, of the line / with the unit circle will have an affix p2. Through point Q draw a straight linew parallel to line PQ, the second point of intersection, point T, of line m and the Fig. 29 unit circle will have the affix p3. Construct a rhom bus with sides OQ and OT. Its diagonal OS — |1 + p3| = R ", where R" is the radius of (A"B"C"). 5°. Let p = cos (p + / sin <p, then 1 + p3 = 1 + cos 3and, hence, R" = | 1 + p 31= /( I + cos 3(p)2 + sin2<p = 2 | cos (3<p/2)!. From this it follows that R" is a maximum [twice the radius R of {ABC)] if and only if jcos (3<p/2)| = 1, whence 3<p/2 == ± kn and, consequently, : <Pi = 0, <Ps = 2tt/3, <^5 = 4 n/3. For <?! = 0 we obtain the unit point, that is, the Boutain point 7\ of A ABC; Ts and T5 are the other two Boutain points [A TxT^Tb is an equilateral triangle inscribed in the circle {ABC)]. Thus, R" assumes the maximum value, equal to R" = 2 = 2R if and only if the point P coincides with one of the three Boutain points of A ABC.

Problems in Geometry

186

6°. The points A ", B '\ C" coincide with the point H if and only if /?3 + 3

I 1 ~ 0, that is, p — 1. The affixes of these points T2, T4, have the arguments (p2 = */3, </>4 == n, <p6 = 5n/3. These points T2, r 4, Tb together with the Boutain points of A ABC form a regular hexagon TxT2T3TATbT%inscribed in the circle {ABC). Thus, al! the points A ”, B", C" coincide if and only if the point P coincides with one of the three vertices T2, TA, T6 of a regular hexagon inscribed in the circle (ABC), whose three other vertices are the Boutain points Tl9 T3, Tb of A ABC. 7°. The triangles ABC and A"B"C” are congruent under six positions of the point P. Indeed, A ABC and A A"B"C'f are congruent if and only if | p3 f 11= 1, that is, 2 | cos (3<p/2) | ~ 1, whence cos (3<p/2) — ±1/2, 3<p/2 == 2kn ± tt/3, 3<p/3 — 2kn ±2n/3; consequently, 4 4 , - kn ± — n. cp - kn ± ... <P 3 9 9 3 On the circle (ABC) there are altogether six points with such arguments and they are obtained by rotations of the radii OTl9 OT3, OTb (Tl9 T3, Th are the Boutain points of A ABC) through the angles ±40° (rotations of the radii OTu OT29 OT3 through the angles ±80° lead to the same six points). 8°. From the relation z" = <7j — (1 + p " 3)r, (84) which define a similarity transformation that carries A ABC into A A "B nC,r9it follows that the affix to' of the fixed point Q' of the similari ty transformation (84) is co = - -------2 + /T 3 [in relation (84), put z” = z \ 9°. From the relation obtained in item 8° it follows that if point P describes a unit circle, then the point Q' describes the ciicle (£') obtained from the unit circle via the linear fractional transformation z = — —. 2 + z We will now prove that (Q') is an orthocentroidal circle of the triangle ABC, that is, a circle with diameter GH. The affix k of the midpoint K of segment GH is

187

Complex Numbers in Plane Geometry

Furthermore, \z ' — k\ - | - " 1 - —(Ti | - | --------i------Lj 2 Jr z 3| i 3(2 -(- z) =3

i 1j .... .w . izi i2 + z \ ; 2 -I z 3 -I2-M

= ]?_i[ 3’

hence, KQ' r= 0///3 = GHI2. Consequently, the point describes a circle with diameter GH. 10°. The points Q' and G coincide if and only if 0\ 2+P3 ~ 3 ’ whence p3 = 1, that is, the point P coincides with one of the points Tu Th. IT. The points Q' and H coincide if an only if

whence p3 = —1, that is, the point P coincides with one of the points t 4, r 6. 12°. The centroid G" of A A"B/fC'f corresponds to the centroid G of A ABC under the similarity transformation * " - * 1 - 0 +/>■*)*• That is, the affix g" of point G" is *" = * 1 - 0 +/> *) *3—

V3 (2-/»-*)•

From this we have -3 \ i* " - * i ■= i— (2-/»-•)

2cr, 3

That is, if the point P describes the unit circle, then the point G" describes the orthocentroidal circle (GH) of A ABC. 13°. Eliminating p from the relations (O

2 + />-*”’

8" = y (2 - P 3)>

188

Problems in Geometry

we obtain 2+ p 3

—y , co

whence 3g" co' — 4co'<71 + a? = 0.

In this relation, ax is fixed and so it expresses g" as a linear fractional func tion of co' (both points G" and Q with affixes g” and co' lie on a circle with diameter GH) and therefore this function transforms the orthocentroidal circle (GH) of A ABC into itself. We find the fixed points of this transfor mation from the equation 3z2 — 4axz + o\ — 0, whence 2 — CTj, Z — Oi/3 are the affixes of the points H and G. 14°. The affix ox of point GAis The slope of line GH is equal to

4 3Hence, the equation of the straight

line passing through point Q' parallel to GH is z — to' — -a_l- (z — co') or 2

<*1

2 =

CO —

(85)

The equation of the midperpendicular of line segment GH is *1 <*1

2— or

z + - l- z = — ffj. trA

(86)

Adding termwise the equations (85) and (86), we find the affix z = 7r of the projection of point G' on the midperpendicular of segment GH:

Complex Numbers in Plane Geometry

189

The affix co* of the point Q* can be found from the relation <h

co' +

CO* ----------—n— 2

whence CO

S ' 01

but a/ —

01 2 + p 3’

Hence, 01

to*

+ P

■3

We have 0i to* g" 0i S* g" 1 1

4 4 0i aU 2 - p 3) <7i o1 — 3 3 2 + p~ 3 3 4 _ 4 _ <r, «. 0i L (2 -/7~3) (7l o1 — 2+/T3 3 3 3 1

1 4 3 = 0]01 4 T 1

1 2+ p 3 -0.

1_ 2-}V* 0

3<2 + ' ‘) 0

Problem 27. Let P, 0 , R be the orthogonal projections of point M on the sides BC, CA, of AABC; let A0,B 0, C0 be the midpoints of segments M A, AfP, MC, and let A', B', Cf be the points obtained by inver sion of the points A0, B0, C0 with the circle of inversion (PQR). Prove that the ratio of the area of A A'B'C' to the area of A PQR is equal to the ratio of the square of the radius of (PQR) to the power of the point M with respect to that circle with sign reversed (Fig. 30). Solution. Take (PQR) as the unit circle. Let zl9 z2, z3i z0 be the respective affixes of the points P, Q, R9M . Since the radius of (PQR) is regarded as equal to 1, it follows that the power of the point M with respect to the

190

Problems in Geometry

circle (PQR) is equal to a — z0 z 0 — 1. We thus have to prove that (A'B'C) _ 1 (PQR) ~ a The equation of the straight line MQ is of the form z — z 0 ==

*0 ^2 / - - (z Z« -

—

-v

Z0)

or ( z n — z-i) z — ( z 0

z — z 0z2 — z 0 Z j . Z0 " The slope of the straight line AQ is equal to — —— — - and therefore the z0 —z% equation of this line is of the form — z 2)

(z 0 — z 2) z + (z0 — z2) z — z2 z 0 + z 2z0 — 2. Similarly, we can obtain the equation of the straight line AR: (z 0 — z 3) z -f- (z0 — z3) z = z3 z 0 + z 3 z0 — 2 . Solving the system of equations (87) and (8 8 ), point A : j z 2 z 0 + 7 2z0 — 2 a A i z3 z 0 z :iz0 2 A I Zq — z2 z0 1^0

^3 -o

(87) (8 8 )

we find the affix z = a of z0 — z, | z0 z3 1 z2 j I

191

Complex Numbers in Plane Geometry

We have A' = (z2z 0 + z 2z0 — 2) (z0 — z3) — (z3"z0 + z3z0 — 2) (z0 — z2) ”

z 2Z qZ q

-j~ z 2Z q

z3ZoZ0

2z0

z 2z 3Z q

z 3z

2Z q

-f- 2z3

z 3Zq -f- 2z0 -f- z2z3z 0 ~h z3z2Zq

2z2

=-- (z2 - z3)z0z 0 + 4 ( ' --------1 ) + z« ( Z‘“------------ — 2(z2 — r3) \ z* z3 / V Z3 z2 /

(■

Z? . --------------h

= (z2 — z3) Zo'z'o

Z2 + z 3 - ... —

Z„

Z2Z3

= (z2 - z3)

- 2)

Z2Z3 zqZqZ2z3 — Zq 4" (z2 ~h ^3 )

2z2z3

^2^3

A = ( z 0 — z 2) (z0 - z3) - (z0 - z2) (z 0 - z 3) = ZqZq

ZqZ3

z 2z0

z3z 0 -|- z0z 3 -)- z2z 3

+ ZjiZa

= z 0(z2 - z3) + z„ (

------- M

V z3

z2^ 3

-z.3 ____ Z 2

_|_

z2 /

= (z2 - z3)

Z0 + Z2Z3Z0

Z2ZS Thus — z | + (z2z3z 0 + z2 + z3) z0 — 2z2z3 Zq +

Z 2Z 3 Z 0

z 2

The affix a0 of the midpoint A0 of segment AM is Z0

__ Z0 ~\~Z 2Z 3Z 0 Z 0

Z 2Z 0

Z 3Z 0

~ f'Z 2Z3Z0 Z 0 ~ h ^ 0 ^ 2 ~f~^0^3

2(z0 + z2z3z 0 -

2 z 2Z3

- 2s) _ ___ Z2Z3( z 0Zq * 0 "t"

The affix a' of the point A' is equal to l/a 0, that is, 1_ z 0 H-----— zo ---- -Z q - f - Z 2Z3Z q Z2 Z2Z3 Z2 Z3 a = 1 Z0 Z — 1 (ZflZo - 1) Z9Z0

Z2Z?,Zt)

--- (z0 + Z0Zj<X3 + Zx — <Ti).

Problems in Geometry

192

We now have o' = - 1- (z 0 + Z^Zydz + z \ — IT,). a Similarly, b' = — (z0 + z0z2ff3 + z2 - ^l), (7

- (z0 + (7

b' = —

Z o Z 30>3

ffl)»

( z 0 + ZoZg^g + z 2 —

<7 C' — ---- (Z0 + Z0Z3(73 + Z3 — 0^) G

and, hence, (A 'B C ) = .

a' I b’ b' 1 c' c' 1

z o “f“ z o z i ° ,3 ' i ~ z i — ° i

4a2

Zo+ZqZ^j + z x &1 1

Zo+ ZoZ2<734-z2—ax z 0+ z 0z8ff3+ z'a 1 Z0~KZ0Z3°’3"l'Z3—^1 z 0+ZoZ3ff3+ z 3—a t 1

zo*>3+zi ZqZ ^ + Zj I ;

^ 2 ^0Z2°3 i~Z2 Z0Z2^3 "hz 2 ^ ; ! Z0Z3^3H~Z3 Z0Z3^3~KZ3 ^ I

/ 4(J2

IZ|

z l

Zi z \

1 ■

Z0 Z» j Z 2 Z2 1,

z 2 Z2 1

! z 3 Z3 lj

z3 z 3 1

= -

1 (PQR) G

whence (A'B'C') _ 1 (PQR) “ (7 ’ Problem 28. AXA2A3 is an arbitrary triangle; P is an arbitrary point. Through P draw two mutually perpendicular straight lines S and 8'. Let these lines intersect side A2A3 at the points B2 and B3, respectively, and the altitude (dropped from point Ar on side A2A3) at points B2 and B3. In similar fashion construct the points C3 and Cl9 C3 and C[ for side A3AX and the altitude dropped from vertex A2 on that side; also construct the points Dx and D2y D[ and D2 for side AXA2 and the altitude dropped from point A3 on side AXA2. Prove that the centroids of the three groups of points B2y B39 B2, B3\ C3, Cl9 C3', C[; Dl9 D2y D[yD2 lie on a single straight line (to all indicated twelve points are assigned identical masses, see Fig. 31).

Complex Numbers in Plane Geometry

193

Solution. Take (AXA 2 A3) for the unit circle. Let zl5 z2, z3,/? be the re spective affixes of the points A l 9 A2, A3, P. Let us write down the equations of the straight lines 6 and S' in the form z —p = t(z — p) or z — tz = p — rp, (89) z — p = — t(z — p) or z + t ’z = p + xp, (90) where | t | = 1. The equation of the straight line A 2 A3 is of the form z + z 2z3z = z2 + z 3. (91) Subtracting (89) from (91) term by term, we obtain the number z = ft2, which is the conjugate of the affix b2 of point B2: b2

Z2 + Z 3 — P+'c'P Z 2 Z3 +

(92)

whence 1

+ 1 - p

+ 1 p

t (~ -2 + Zg

Z 2Z 3 +

— +

Z2 Z3

z2z3p ) -{- z, z3p

(93)

Substituting —t for t in the formulas just obtained, we find the affix b3 of point B 3 and J>3: _ 1 — 't(z2 + z3 z2z3p) + z 2 z 3 p b3 — 9 Z2ZZ T _ _ z2 + z 3 — p — zp bz —

1 3 -8 1 0

Z2ZZ

194

Problems in Geometry

From these formulas we find the number b23, which is the conjugate of the affix b23 of the midpoint B2Z of segment B2B3: ^ = b 2 + bz = J _ / z 2 + z 3 — p + T p + z 2 + z3 - p - x p \ 2

2 V

z 2 z3 +

z2z3 — t

)

= z 3 z3(z2+ z3) — p z 2 z3 — t2 p

z \z \

whence 1

1 \

Z3 )

Z2Z3

-----I ------- 1---- I — p --------- — P

_ Z2z3 \ Z2 "

2 | Z§

_ z\ z l p +

p z 2 z3—(z2+ z 3) T2

z2 z§

Furthermore, the equation of the straight line passing through point Ax perpendicularly to line A2A3 is of the form * - *1 = Z2Z3(Z - Z\)

or z 2 Z3

(94) Z1 Subtracting (94) from (89) term by term, we find the affix z = bz-> which is the conjugate of affix b2 of point B2: z — Z2 Z3 Z

p -

Z2 Z3

zp - z 1

b* = z 2 Z3

and, similarly, p + x p — Zt + —

b3 =

______________ £i_ Z2 Z3 + T

The number £ 23*which is the conjugate of the affix b23 of the midpoint £23 of segment B2B3y is Z2 Z3

p - z p - z±

p + tp — Zi + *2*3

Z1 Z2 Z3

*2 *3 +

72 72 * 2 *3 o — z2 z3p — a3 .----------x2p Z% Z% — T2

Complex Numbers in Plane Geometry

195

whence l _ — p z 2 Z3

_1_

pz\ Z%— Zx T2 -f

J i— i , z|z§ T2 1

°3

Z2Z , 2 ^3

T2 —

-- «

Z 2Z 3 p t 3

z| z| — T2

-2 -2 Z2 Z3

The affix A of the centroid of the set of points B2, 2?3, 2?3 is equal to the affix of the midpoint of segment B2ZB23, the ends B23 and B23 of which are the midpoints of segments B2B3 and B2B3:

■ (-¥ )

2z| z\ p ^ _ ^23 + ^23

2(z| Z§ — T2)

(95)

Let us now prove that the point with affix A lies on the straight line A given by the equation* 2z^2p + ^ - dx j + 2z^2p +

— <rx j

" ( 2' , + ^ ) ( 2 ? + ^ ) _ ',‘ ” ‘

<96)

z 2 z3/ _ (Ti z \z \ — a3 — 2p t 2 2(z| z| — T2)

(97)

We have

A =

z \z \

t2 V

if-1 -- 1 ) Ul 4 t2y

* The equation of this straight line could have been set up by similarly determining the affix /i of the centroid o f the second set o f points C3, Clf C3, C\ : _ 2 2 2z3 z ip

2(zi zf -

t2)

also, by determining the affix v o f the set o f points D l t D2tD i9 Z>2*itis possible to verify the validity of the necessary and sufficient condition o f collinearity o f the three points A A 1

P P 1 V v 1

= 0.

Prooiems in Geometry

196

Let us compute the left-hand side of (96), setting z = A, z = A (the de nominator z\ z\ — x2is dropped for the time being): ^ 2p z\ z| — t 2ax + x2 *3

)

+ ^ 4 z l ^ - o s - 2 p x2j ^2p + z— - <7! j = 4p p z \ z\ + 2p z\ z\ — — 2p — zf zf — 2p x2o’! — Oi <r3 + t2 cr3 + 2p x%-2— + ^ Zi

—

- °j. x2^ -^1

ff3

2t 2

+ 2/> z2 z§ ** + t 2 - I z2z§ Z2

(J3

<r3

2| zi — 2p a3 — t 2 + (T3 o’! — 4p px2 — 2p - — f- 2p x2 ax 0*3

®*3

= 4/> p(z! Z§ — T2) + zf z\

(zi Z§ — T2) ^3

^ (z| zi - T2) + t2 Thus, the left-hand side is, for z = A, equal to <,p- + l - ^ +

2 / |+

<73

(Z2 Z2 -

T2)*

^ . <73

The right-hand side of equation (96) is x2

cr3

which is the same as the left-hand side. The symmetry of equation (96) proves the theorem. At the same time, (96) is the equation of straight line on which lie three centroids of three sets of points: B2, B3, B2, B3\ C3, Ci, C3, Cii Di, D2, Di, D>. From this theorem there follows a corollary (the Droz-Farny theorem) (see Fig. 32): if through the orthocentre H of triangle AXA2A3 we draw two mutually perpendicular straight lines, which cut off on sides A2A3, A3Ax, A xA2 segments B2B3, C3CX, DXD2, then the midpoints of segments B2B3, C3Ci , DxD2 lie on one straight line. Of course, this theorem can be proved in a straightforward fashion, and its proof is technically not so elegant as that given in the problem generalizing the Droz-Farny theorem. In a direct proof of the Droz-Farny theorem, one should bear in mind that Ci is the affix of the orthocentre H of the triangle A i A2A3.

Complex Numbers in Plane Geometry

197

It is obvious that the Droz-Farny line is obtained from the straight line A given in the generalization (for the same case P = H) under the horn othe tic tran sfor mat ion

('■ i) :

the points B2, B3 merge with the

point 77, and the centroid of the set of points B2, Bz, B2, B3 is the midpoint of segment HB23, where B23 is the midpoint of segment B2B3. And similarly for the points C3', C[ and D[, D2. Problem 29. Let 7 be the centre of a circle (/) inscribed in A ABC ; let 7), F, F be points of tangency of (7) with the sides BC, C/4, AB, respec tively; let H be the orthocentre of ADEF; let <5 be some diameter of (7); let D \ E \ F ' be points symmetric to the corresponding points 7), E, F with respect to the straight line <5. Prove that if Pa9Pb9Pc are points sym metric to an arbitrary point P with respect to the sides F 'F ', F'D \ D'E' of A 7)'F'F' and a, /?, y are points symmetric to the points Pa9 Pb9 Pc respec tively with respect to the sides BC, CA, AB of A ABC, then the centre Q of the circle (a/fy) has the property that the Simson line of point Q with respect to AD EF is parallel to the straight line <5. Prove that if we construct a directed line segment QN equivalent to a directed line segment HP*j2, where P* is a point symmetric to point P with respect to the straight line 5, then the radius of (a/ty) will be equal to 2IN. Solution. Take (7) = (DEF) for the unit circle and take the diameter 5 for the real axis. Let zv z2, z3, p be the affixes of the points Dy E, F, P. Then the affixes b and c of points B and C are 2 z 3 Zj

> c—

Z3 4 ~ Z1

2 z1 Z1

z2 Z2

and their conjugates are ~b

2 Z3 +

_ > c Z1

2 Z1 +

. Z -1

198

Problems in Geometry

Let us find the affix pa of point Ptt. The affixes of the points D', E’, F' are ~zlt z 2, z 3. The slope of the straight line E'F' is

_L_ J. Z2 Z2

Z 3 __

z2 z2

z3 _ z3

The equation of the straight line Z -

z2 z3

Z2 =

— ”2

(Z -

Z2)

or Z +

Z 2 Z 3 Z — Z j - |- Z 3.

(98)

The equation of the perpendicular dropped from point P to this line is z-p

—

%2^3

< J ~ P )

or z

z 2z3 z — p

z2 z 3p.

(99)

From equations (98) and (99) we find the affix p* of the projection of point P on the straight line E 'F ' : P* = ~r (*a + z 3 + P - *2 *3 P)2 The affix pa of point Pa, which is symmetric to point P about the straight line E'F', can be found from the relation P +„Po- = P* = — (z2 + z 3 - z 2 z3p + p), z z whence Pa —

Z2 Z3 p.

The slope of the straight line ID is z1f z 1 = zf, and the slope of BC is equal to —zf. The equation of BC is Z — ZX = - zf(z — z 2) or z + zf z = 2zx. (100 The equation of the perpendicular dropped from point Pa on the side BC is Z - P a

= zl&

-P a )

Complex Numbers in Plane Geometry

199

or 2 — z\ Z = pa — z\ p a.

(101)

From equations (100) and (101) we find the affix p , of the projection of point Pa on line BC: P- = \ (p‘ ~ z' p ‘ + 2z^ and from the relation X \ Pa ~ Pa ~ “ (Pa 2 2

A p a + 2ZX)

we find the affix X of point a : X = 2z1 — z?(z2 + z3 — z2z3p) = 2zj — zx(zx z2 + z1zs — a3p) = 2zj — Zy{a2 — z2z3 — <r3p) = <r3 + zx(2 — <r2 + p<r3). The affixes p and v of points P and y have similar expressions: p = <xs + z2(2 — cr2 + pff3), V = ff3 + z3(2 — <r2 + p<r3). Thus we see that the points a, /?, y lie on a circle, the affix of whose centre is equal to a3, and since |er3| = 1, it follows that this centre lies on the circle (DEF). The radius of the circle (a/ty) is P — |2 — o3

po31— |2 — c2 + p<r3| |2ct3 —

2 —— + — ^3 O’3

+ p|

= |2<r3 - ox + p | + -d -O i) k 3l The fact that the Simson line corresponding to the point with affix a3 is parallel to the real axis follows from problem 3. The affix p* of point P*, which is symmetric to point P about the straight line <5, is equal to p* = p. To the directed line segment HP* there corre sponds a complex number p* — ax — p — av * The directed line segment HP* 12 = QN is associated with the complex number (p — <x1)/2 = n — a3y where n is the affix of point N. From this we have n — < j 3 + (p — Ci)/2. Since n is the affix of point N and (/) is taken as the unit circle, it follows that IN = |n|

+ — (P - Oi) 2

* Let the points A and B have affixes a and b, respectively. We will say that the directed line segment AB is associated with the complex number b — a.

200

Problems in Geometry

Fig. 33

Problem 30. On the circle (O) take four arbitrary points Al9 A2, A39 A4. Take any one of six segments joining these points in pairs, for example, segment AXA2. Construct an isosceles triangle A1A2A12 with vertex A 12 and base AXA2 so that its centroid coincides with the point O. Let A[2 be a point symmetric to point A12 about the straight line A3A4. Prove that the six points thus obtained, A[29 A[39 A[49 A23) A24, A349 lie on one circle, (S), that is concentric with (O); note that the radius of (S) is four times the distance from point O to the centroid G of the system of four points Al9 A2, A3, A4 (Fig. 33). Solution. Take (O) for the unit circle and let zl9 z2, z3, z4 be the respective affixes of points Al9 A29 A39 A4. The affix a12 of the midpoint of segment AXA2 is a12 = (z4 + z2)/2; hence, the affix a12 of vertex A12 of the isosceles triangle AXA2A12 (AxAl2 = A2A12)9the centroid of which coincides with O, is ai2 —

(zi

^2)-

The equation of the straight line A3A4 is z + z3z4~z = z3 -[- z4.

(102)

The equation of the perpendicular dropped from the point A12 to the straight line A3A4 is z + z x + z 2 = z3 z4(z + z x + Zo) or z — z3z4 z = — z1 — z2 + z 3z4 z 1 + z 3 z4 z 2. (103)

Complex Numbers in Plane Geometry

201

Adding equations (102) and (103) termwise, we find the affix z = a*2 of the projection of point A12 on A3A4: a*2 = — (z3 -j- z4

z2 + z3 z4 z x -f- z3z4 Zg).

The affix a[2 of point A[2 is found from the relation 012 + 012

or*012 ~ Z1 ~~ Z2 + 012 2

Y (Z3 + Z4 “ Zl ~ Z2 + Z3 Z4 Z4 + Z3 Z4 Z2),

whence 012 ~

Z3 +

Z4 +

Z 3 Z4 Z 1 +

Z 3 Z4 2 2 == Z 3 Z a { 2 1 +

Z 2 +

Z3 +

2 4) =

Z 3 Z A &1*)

where ax = zx + z2 + z3 + z4. From this relation it follows that

OA'12 = \a’12\ = Io-jI (|z3| = |z4| = 1, |ctx| = |<tx|). Similarly, OA[3 = OA[4 = 0 ^ 3 = Oa ; 4 = OA34 = laj. Thus, the points ^ 2, d[3, A[4, A23, A24, A34 lie on one circle, whose radius is equal to p = |o-J = 4|o’1/4|. But gJ4 is the affix of the centroid of the system of four points Al9 A2, A3, A4; hence, \gJ4\ = 0(7. Thus, p = 40G. Problem 31. Let D, E, F be points of tangency of the sides FC, CA, ^42? of AABC with an inscribed ciicle. Consider an arbitrary diameter S of the circle (I) = (DEE). Draw through points 2)*, F*, F*, in which the altitudes of A DEF intersect the circle (DEE), straight lines parallel to the straight line <5, and let A \ B \ C be points at which these lines intersect the circle (/) a second time. Denote by A *, 2?*, C* points symmetric to the points A \ B \ C", respectively, with respect to the sides BC, CA, AB of A ABC. Prove that the centroids G and G* of A D F F and AA*B*C* are symmetric about the orthopole a> of the straight line 8 with respect to A DEF (Fig. 34). Solution. Take {DEF) for the unit circle, and let the diameter 8 be the real axis Ox (point O coincides with point I). Denote by zl9 z2, z3 the respective affixes of the points D, F, F. The equation of the straight line passing through point D perpendicularly to line F F is Z — Zi =

z2z3(z

—

zx).

Problems in Geometry

202

Fig. 34

Solving this equation together with the equation of the unit circle zz = 1, we find the affix z — d* of the point D*. Indeed, = Z2Z3 ^ ~ ~

Z,.

Zi — ----------- ( Z - Z i ) .

or Z* Z*

zx z

Naturally, one of the roots of this equation is z == zx (the affix of point Z)), the other root is Zi which is the affix of point D*. Similarly we find the affixes e* _

_ El E1 } f * = — Z2

of points E* and F*. The equation of the straight line passing, for example, through the point D* parallel to line <5 is of the form Zj_

Z2 Zg

Complex Numbers in Plane Geometry

203

Solving this equation jointly with the equation of the unit circle zz = 1, we find the affix of point A': Z 2 Z3 Z1 z_ H,-----= — 1+ , ----, Zl 2 Z2 Z3

z±z + z 2z3 = zxz + z 2z3 Z\

Z2 Z3 Z

One of the roots of this equation, z = — z2z j z l9 is the affix d* of point Z>*, the other root is equal to the affix a' of point A': a' = z 2 Z3

Similarly, Z3

Z1 Z2

where b’ and c' are the affixes of points B' and C" respectively. The slope of ID is equal to z1j z 1 = z\; hence, the equation of BC (the slope of which is — z\) is 2 — Zi = - z\{~Z — Zj) or z + zI z = 2z2. (104) The equation of the perpendicular dropped from point A ' to line BC is _

2 ( ;

Z 2 Z3

Z --------- = Zl \ Z --------z2 z3 V z1 or z — z\ z = — ---- c3.

(105)'

Z 2 Z3

From (104) and (105) we find the affix a of the projection of point A' on line BC: “ = - ( 2Z1 + - - ^ 3 2 V z2z3

)• /

The affix a* of point v4* is found from the relation a' + a* = a 2

Problems in Geometry

204

whence 0 * = 2z1 — cr3

and, similarly, b* = 2z2 — <r3,

c* = 2z3— <r3,

where fe* and c* are the affixes of the points B* and C* respectively. From the relations obtained it follows that d* —cr3 __ 2zx — 2<t3 Z1

and, similarly, that b* — <t3 = 2 c* — <r3 = 2 Z2

<X3

<J3

That is, the point P with affix = cr3 is the centre of the homothetic trans formation under which AD EF goes into A A*B*C* (see Fig. 34). Furthermore, the affixes g and g* of points G and G* respectively are

The midpoint Q of segment GG* has the affix co = (a1 — cr3)/2 and, hence, coincides with the orthopole of the straight line 6 (the real axis) [formula (11), problem 7, z0 = 0, x = 1]. Remark. Let us find the affix of the point symmetric to point A' about/)/: the equation of the perpendicular dropped from point A' to D l is of the form

or z

Z1 Z

— — -— (- <t3.

The equation of D l is z — z \ z = 0.

Complex Numbers in Plane Geometry

205

From this we find the affix z = X of the projection of point A' on line DI:

The affix p of the point symmetric to point A' about DI is found from the relation ^ r ~ S- = a 2 or

whence V= =PThus, all points symmetric to points A', B \ C about the straight lines DI, El, FI coincide with the same point P, which is the centre of the homothetic transformation r = (P, 2) that carries A DEF into AA*B*C*. Problem 32. The circle {DEF) = (I) inscribed in A ABC is taken as the unit circle; D9E, F are the respective points of tangency of the straight lines BC, CA, AB with the circle; zl5 z2, z3 are the respective affixes of the points D, F, F. It is required to find: 1°. The affix o of centre O of the circle (O) = (ABC). 2°. The radius R of the circle {ABC). 3°. The radius p of the Euler circle of A ABC. 4°. The affix h of the orthocentre 77 of A ABC. 5°. The affix e of the centre 0 9 of the Euler circle (0 9) of A ABC. 6°. Prove that the circle inscribed in A ABC is tangent to the Euler circle constructed for that triangle at some point <P0 (called the Feuerbach point). Find the affix cp0 of the Feuerbach point <P0. 7°. Prove that the three circles {Ia), {Ib), (7C) — they are escribed circles of A ABC at the angles A, B ,C — also touch the Euler circles of A ABC at the points <Pl9 <P2, # 3 (also called Feuerbach points). Find the affixes xa, t ^ Tc °f the centres 7fl, h ,Ic of the circles (7fl), (7ft), (7C). Find the affixes tl912, t3 of the points Tl9 T2, T3 of tangency of the circles (7a), {Ib)9(7C) to BC, CA, AB. Find the radii of these circles (Fig. 35). Find the affixes <Pi><P2 ><?3 of the Feuerbach points &l9 <P2, # 3. Solution. 1°. The affix of the midpoint of segment EF is equal to (z2 + z3)/2 and since point A is obtained from this midpoint by inversion with respect to the unit circle (7), it follows that the affix a of point A is 2 a = -3----- — . *2 + ^3

206

Problems in Geometry

Similarly, we can find the affixes b and c of points B and C : t 2 2 b = ----------- , c = —----- 3—. z3 + z\ ZL + z 2 The affix o of the centre O of the circle (O) = {ABC) is found from the system of equations (o — a) (o — a) = (o — b)(b — b), (0 — b) (d — b) = (o — c) (b — c)

or — o(a — b) — o(a — b) = bb — aa,

(106)

— o(b — ~c)— o(b — c) = cc — bb.

(107)

We have _ v 2 a — b = --------[z2 -\r z3 a —b —

2 2(z1 - z2) = -----------—-------z3 + zx (z2 + z3) (z3 -f- zx) — 2zaZl =

Zj 4" Z3

2z« (za ~ zi) (Zj + Z3) (z3 + Zj)

207

Complex Numbers in Plane Geometry

bb — aa z3

4z.z. (Z2 +

(“3 + ^ l ) 2

= 4za

zx(z| + 2 z2 z3 + z|) — z2(zf + 2 z3 zx + zf)

= 4z«

Z 3) 2

(Z 3 +

Zi zf + Zxzf — Z2zf — Z2 zf

^ l)2 ( z 2 +

4 z 3 [Zj z2(z2 —

(Z3 + Zj)'“ ( Z 2 + Z3 ) 2

(Z 3

Z3) 2

Zj) — zf (z2 — Zj)] (z2 + z3)2

Z j)2

= 4z3(z2 — z2) (zt z2 — zf) . (z3 + Z j ) 2 (z2 + z3)2 The equation (106) takes the form 2(z2 — zt) —o (Z2 + Z3) (z3 + Zl)

2 (Z2

4z3(z2 - z i ) ( z 1z2-z § )

z 3 (z 2 — Zj)

+ Z3>(Z3 + Zi)

(z2 + Z3)2 (z3 + Zj)2

or 2 z3(zj z2 — zf)

O — Z§ 0 =

(108)

(z2 + z3) (z3 + Zj)

Similarly, equation (107) is transformed as follows: 0 — zf 0 =

2zi(z8 z3 - zf) (z3 +

Z l)

(109)

(zx + z2)

Subtracting (108) from (109) term by term, we find the complex number 0, which is the conjugate of the affix 0 of point O: (zl - zf) 0 = 3

2Zl(z2 Za ~ zt)

(Zl + z3) (Zi + z2)

2 z3(z! z2 - zf) (z2 + z3) (z3 + zx)

= (z2 + z3) (2<r3 — 2zf) — (zt + z2) (2<r3 — 2zf) (Z2 + Z3) (Z8 + Zi) (Zj + z2) _ 2<t3(z3 — Zj) + 2z2(zf — zf) + 2 z3 zx(zf I— zf) 5

0102

whence / , x — ( z 3 + Z j) o =

2(T3 + 2z2(z3 + z 3 z 1 + zf) + 2 z3 z 1( z 3 + zx) -----------------------------------------------------------------------------------

01 0*2

(T3

4<t3 + 2z2z\ + 2z2 zf + 2zt zf + 2z3 zf

01 (T2 =

<r3

z2 Z3 + Zi z2 z3 + z2 zf + z2zf + z x zf + z3 zf

01 0*2 = 2

Z2 Z3(z3 + Zj) + z3 zt(z3 + zx) +

01 02

Z y Z 2( Z y

+ z3)

Problems in Geometry

208

Consequently, 2<7o

o= 1

(73

and from this we have 2 5. 2d2 O = _________ __ o-! (73 (72

2(Ti (73 (7i (72 (73

r2 °3

U’a

2°. jR2 = \o — a\2 = (o — a) ( 0 — a) = do + ad — do — oa 4(7! (72 (73

i*2 4

01 02

( z 2 + Z 3) ( Z 2 + Z 3)

+ •

(cra (J2-

4(7! (73

(CTj (72— (73) 2

0 .3

Z 3)2

_j_ Z 2 Z 3( f f 1 0 2

<72 -

(0 1 <72 — <73) 2

ff3)

Z2 Z 3( z 2 - f - Z3) <72

z 2z 3( z 2 + z 3) ( z 3 + z 1) ( z 1 + z 2) — ( Z a + Z a ) ^ ; , — z 2z 3( z 2 +

z 3)(T2

(z2 + z3)2 (<7! <t2 — 03) z2 z3(zt +

01 02 03

(Z is + Z s X ffiffg —

(z2 + 23)2 Ol ^2 - 03)

(01 <r2 - 03)2

Z2 Zo (7o

( z 2 "j- Z3) <7j

0 3)

4(7o

(7j (72— (73 Z 2 + Z 3

(<7X <72— ff 3) ( z 2 + z 3)

(ff! <72 - 03)2 CTi<72 ff3

(73 Z2 + Z 3

(7i (7o

Zo2 Z < ^3

(z 2

<73) 2

(7i (72

Z3)

(zx -(- Z2)

<73

0?2 + Zs) (O’! 0-2 “ ff3)

<73) 2

Z1 (73

(7! (73

0 1 C72 <73

(Z2 + Z 3) (<7!<72— (73)

<7:j ( z 2

( z 2 + Z 3) (<7j <72

<7i (72 (73

(73

(75

((7X<72— 03)2

<7i <72 — <73

(<7X(72 —C73)2

4ff3 (t7x <72 -

Z3)

(<7j 0 2 — 0 3) 2

And so Ri =

z2 Z3 02

= / 0 3) 2

We will now prove that the number

2<73

V <73 — <7X <72 /

0 3)

20£

Complex Numbers in Plane Geometry

is real and positive. We have 1 ^

03___________ ^3

0 2 &3

<*3

^ 01 02

Hence, A is a real number. Done differently, we have A=

_____ ?____ = _____ L _ = — i _ . l - a i— 1 - °1°1 1 - kil2

c3 - <Ti G2

0 3

We now prove that \c±\ < 1. Indeed, since all angles of AD EF are always acute, it follows that the orthocentre of ADEF, the affix of which is equal to cl9 lies inside ADEF and, hence, also inside the circle (DEF). But we assumed the radius of (DEF) to be 1, and so \cx\ < 1, and, hence, A > 0; now since R2 = 4A2, it follows that R = 2A, that is, _ 2o-3 2 R = - - - - - - - -— = — . C3

Ci C2

CiCi

3°. The radius p of the Euler circle of A ABC is _R 0*3 = ___ 1_____ 2 <r3 — c2 1 — \ci\2

_ 1 c’

where c is the power of the orthocentre H' of AD EF with respect to the circle (DEF). 4°. Since the sum of the directed line segments OA + OB + O C = OH, where O is the centre of the circle (O) = (ABC), and H is the orthocentre of A ABC, it follows that a — o + b — o + c — o = h — o. whence h = a + b + c — 2o, where h is the affix of H. We have h

z 2 Z3

Z 2 Z 3 (Z 3 +

_|_

zz zi Z 3 + Z1

Zl) (Z1

Z 2)

zi z l

(Z2

1 4 -8 1 0

z 2

02 —

Z3 Zl(Zl +

Z») (Z3

c1 c2

Z 2) ( Z 2 + Z 3 ) + Z l Z 2 (Z 2 + Z 3> ( Z 3 + Zl ) ( Zl +

Z 2)

Zl )

Problems in Geometry

210

2<Ti 03 01 02

z2z3(4 + <r2) + z3zx(zl + <r2) + z x z2(z| +ff2)

(z2 + Za) (z3 + *i) (Zi + z2) ai ff3 ~r «t| 0! 02 03

2ffj ff3 01 02 03

2<rx <r3 01 02 03 02 0i 03 01 02 02

Hence, h = 2

02 01 0*3 01 02 03

5°. The affix e of centre 0 9 of the Euler circle of A ABC is equal to £ = ( ^ 7 -°)^ sjnce

q 9 js

the midpoint of segment OH. Thus, 0i 0*3 4~ °*3 0102 03

01 02

6°. The equation of the radical axis of the circle (DEF) (zz — 1 = 0) and of the Euler circle, the equation of which is 0I = 0, (^3 — CTi GT2)2

(z - e) (z - e) is of the form

zz — 1 — (z — e) (z — 1) +

^3 = 0 (<r3 -o-jff,)2

or — 1 + ez + ez — e~e

(<73 - ffiffjj)2

Solving this equation together with the equation zz = 1 of the circle (DEF), we find - 1 + — + «z - es + ------^ ------ = 0, Z ((T3 -ffi(T2)2 and since *5 <r2 ___1_

*1 01 02

*3 it follows that ee~ = a\ 0 |/( 0 3 — ax 0 2)2 and, hence,

—ee +

03 (ff3 — (Tj <r2) 2

_2 —2

0102

(<r3 — <7X<t2)2

03 + 01 02 <r3 — cr1 <j2

( 110)

Complex Numbers in Plane Geometry

211

Equation (110) takes the form +

-1 = 0 G3

Gi G2

or ■i. + l2 + -..??■ »■. = o z a3 0i <t2

az2 H------”— z + e = 0. 0’s 0"l 0"2

( I ll)

The discriminant of this equation is equal to zero: ^

o \ a \

A = -------------------------s s = —

<72) 2

o \o \

---------------------------------------------- = 0 . (<73 —

G x G 2) 2

(g 3 —

G xG 2) 2

Consequently, equation (111) has equal roots. This means that the radical axis of the circle (I) and of the Euler circle (0 9) of A ABC has a single point <*>o in common both with the circle (/) and the circle (0 9), that is, the circles (0 9) and (/) are tangent at the point <P0- The affix <p0 of point <P0 (the point of tangency) is found from equation (111): ^

Gi G2 (g 3

G 1 G 2)

G £

Gi G2

(Ji (^2

0*1

And so we have G2

<Po = — •

<*1

Remark. If for the unit point on the circle (DEF) we take a Boutain point of A DEF, then g 3 = 1 and the affix cp0 of the Feuerbach point <P0 may be written thus 0i 7°. We find the affixes xa, xb, zc of the centres of the circles (Ia), (Ib), (Ic) escribed in the angles A, B, C of A ABC. The slope of the straight line IB is equal to b/b and, hence, the slope of the straight line perpendicular to IB is equal to —b/b, that is, _ _b X~ b

Zl +*3 z 1 Jr~z3

z3.

Problems in Geometry

212

and the equation of the bisector of the exterior angle B is of the form _2_ ^1 +

Z1 ++ Z3 *3 ./

or 4zj z3 z -f Zj z2 z = Z1 + z3

( 112)

In similar fashion we can write down the equation of the bisector of the exterior angle C: , 4zx z2 z + zx z2 z = Zi +

(113)

From the system of equations (112) and (113) we find the affix z = of the point /a: (Z3 -

Z2) Ta = 4(T3 ( —

---------- — 1—

V ^+ Z g (z3 — z2)

= 4u3

z1 + z 3J

^3 ^2 (zi + z2) + z3)

whence =

4<73 Oi + z2) (zx + z3)

Similarly, n =

4<t3

:> Tc (Z2 + Zi) (z2 + z3)

4<r3

(z3 + Zj) (z3 + z2)

Since (z 2 + z 3) 0 3 + Zj) ( z t

+ z2) =

al a3

— a 3,

these formulas may be rewritten thus: 4<73 t, +I Z _3)%— • 4a3 (z2 (o-i - *l), <7l <r2 <r3 °2 °3 ** =

4g3 4cr, (z3 + Zi) = <Ti er2 <r3 0*1 <72 (73 4<r3 &1 &2

(Zl + ^2)

(ffl -

Z2),

4o-3 (tfl - z3). &i cr2 (73

Complex Numbers in Plane Geometry

213

or thus ^divide the numerator and the denominator of the fraction by <rs and substitute dx for

)■

r„ =

- 1

• ( Z2 + Z3)

- ks + z l) = Zr =

k il2- i 4 k il2

■k i + z2) = •

k i I2 — i 4 k il2- 1 4 k il2- i

kx - *i)> k i - z2), k i - -3).

or, finally, thus: ra = — 2fl(oi — zj) =— 2/? o’!+ 2R zu zb = — 2/?(o'1 — z2) =— 2/? o'!+ 2i?z2, t c = — 2-R(oi — z3) =— 2/? 01+ 2R z3. From this it follows that the centre of the circle (IJblc) has the affix 4oj o3 4ff! — 2R o 1 = Gi <J2 C3 (Ti (Ti 1 and the radius of the circle (IaIbIc) is equal to 2R. We now find the affixes tl912, f3 of the points 7\, T2, ^3 of contact of the circles (7fl), (Ib), (7C) with the sides BC, C4, ,42? respectively. Since the line segments TXD and BC have a common midpoint, it follows that t x -{- zx = b + c, whence _ 3z2z3+ z 1z3+ z 1z2—z? * a _L_ _ 2zj z3 2z2 zx z,1 = z. /i — 0 + c — z1 — — r Z i + ^3 *2+ ^ 1 " Oi + Z 3 ) (zx + z2) 2 z 2 z3 + (T2 — zf 2(73 + 1 < 2 — \ = *1 k i + z2) (zx + z3) (zx + z3)(zx + z2) Similarly, 2 o -3 + z2 o-2 — zi , _ 2 o-3 + z3 o2 — z3 t3 = ^2 — (z2 + Zi) (z2 + z3) (z3 + Zi) (z3 + z2) We now find the radii ra, rb>rc of the circles (70), (Ib), (Ic). We first find 2o 3 + zx o 2 4(7, za — t, = k i + z2) (zx + z3) k i + Z 2 ) ( z 1 + z3) = 2zxz2 z, — zx(z2Z3 + Z3 zx + Z i z2) + zx z

k i + ^2) k i + z3) ZaZa-ZjZg-ZjZa+zf _ z ^ -z ^ -z ^ -z ,) k j - z 2) k i-^ s ) zi — —— :— —---------------- —zi = z i (zi+ z2) k i + z3> k i + z ^ k i + z3) k i + z2) k i + z3)

Problems in Geometry

214

From this it follows that - h Zl

ra ID

(Z l (Z l

Z 2) ( Z 1

Zo)

Z 2) ( z x

Z q)

and, similarly, = (z2 - *i) (z2 - Za) r = (23 - zx) (z3 - z2) (*2 + Zl) (z2 + *3)’ C (z3 + Zl) (z3 + z2) If for the unit circle we take the circle (Ia) = (DaEaFa) escribed in the angle A of A ABC [Da9 Ea9 Fa are the points of tangency of this circle (7a) with the straight lines BC, CA, AB], then the proof of the fact that this circle touches the Euler circle constructed for A ABC will be precisely the same as the proof that the circles (/) and (0 9) are tangent; however, now it is necessary to assign the affixes zl9 z2, z3 to the points Z>fl, Ea>Fa. Then the affixes a, b, c of the points A, B ,C will remain _ z z

_ ? u

Z 2

_? c Z 1

__ Z x

_ + Z2

and even the affix <p± of the Feuerbach point &l9 in which the circles (7a) and (0 9) are tangent, will be <Pi = — , where a1 and <r2 are expressed n terms of the affixes zl9 z2, z3 of the points Da, Ea, Fa. We have to express (pl9 <p2i <p3 in terms of the affixes zl9 z2, z3 of the points 7), E9F. The fore going simplifies this problem because the fact (established above) that the circles (7a), (7^), (7C) are tangent to the circle (0 9) will simplify the computations (see below). The equations of the circle (7a) and the circle (0 9) are (z - ra) (z - T a) - r l = 0 (z — c) (z — e) — p2 = 0 By what has been proved, these circles are tangent at the Feuerbach point <Pl9 the affix of which is, consequently, found from the equation 'a

Z -* a r2 'a Z — Ta

+ Ta ~

— — + «, z —e + Ta — e = 0,

rl(z — t) — p \z — Ta) + [z2 — (Ta + e) z + xa e] (za — e) = 0, (ia — e) z 2 + [rl — p2 — (xfl + e) (fa — b)] z — £r2+xa p2+ t a £(fa—e)= 0.

Complex Numbers in Plane Geometry

215

Since the circles (Ia) and (0 9) have only one point in common, it follows that this quadratic equation has equal roots, which are the affixes (px of the point 4>x of tangency of the circles (/fl) and (0 9): r? — n2

(ta +

<Pl =

)

(fa -

rl-p* 2 (6 - t a)

2(e - ta)

Ta + £

zfz?z| (zi - z2)2 (zx - z3)2 ____________________________ ( Z i + Z2) 2 (zx + Z3) 2 (zx + Z2) 2 (z8 + z3)2 (z3 4- Z i ) 2 (Z i +

f e

Z2 +

Z t ) (z% +

4zx(z2 + z3)

z 3) 2

Z3)

(z3 + zx)

4zx z2 z3 , i r 2 L(zx + Z 2) (zx +

(zx + z2) (z2 + z3) (z3 +

Z i) _

(z2 z3 + z3 zx + zxz2)2 1 , ,_______________________ Z j) (zx + Z2) (z2 + z3) (z3 + zx) J

_ _1_ (zt — z2)2 (zx — z3)2 (z2 + z3)2 — zf zf zf 2 (zx + z2) (z2 + z3) (z3 + zx) (z2 + z3 — zx)2 |

1 4zxz2z3(z2 + z3) + (z2Z 3 + Z3 zx + zxz2)2 2 ( Z X + Z2) (Z g + Z3 ) (z3 + zx)

Furthermore, Ol — z2) (zx — z3) (z2 + z3) — zx z2z3 = (zx — z2 — z3) (zx z2+ z x z3—z2 z3)^ '

(zx — z3) (z2+ z 3) + z x z2 z3= (z x—z2—z3) (zx z2+ z xz3—z2 z3) + 2<x3.

Z2)

Hence, (zx - z2)2 (zx - z3)2 (z2 + z3)2 - z? zf zf = (z2 + z3 - zx)2 X (zxz2 4- zxz3 — z3 z3)2 — 2zxz2 z3(z2 4- z3 —zx) (zx z, + zx z3 — z2 z3), and so 1 (z24-z3—zx)2 (zxz24 - z xz 3 —z2z3)2—2zx z2z3(z24-z3- z x) (zxz24 z xz3 - zz3 <Pi= 2

(z2 4- z3) (z3 4- zx) (zx 4- z2) (z2 4- z3 — zx)2

4 z xz 2z 3( z 2 4- z3) +

(zxz24- z2z34- z3zx)2 = J_

(z2 4- z3) (z3 4- zx) (zx 4- z2)

(zxz2 4- zxz3 — z2z3)2

(z2 4- z3) (z3 + zx) (zx + z2) i

2 Z j Z 2 Z’3 ( z 2 +

(z2 + Z3)

0 3

+ *l)

^ 3) (* 1

2(<7j<72— (73)

1 2

a\ (Tjj — ^3

^1 ^2

+ ^2 )

^ 3 (^ 1

+ *3> (*3 +

<t|

z 2> O 2

Z j Z2

Zx Z3

0*1 ^2

+ *3“

* l)

Z2 Z3

^2 "1" ^3^1

Zi z 2 + Zj z3

g \

^ 3 ) ________

^1 ^3

Gi O’2

(Tz

&1 & 2

4“

* l) O l

<TZ

<7j 02

z 2 z3

Problems in Geometry

216

In similar fashion we find

~~ O3

<P3

The formula as follows:

+ ZS Z2 Z1 Z 2 Z 1 4“ Z 2 — Z S

Z3 Z1

Oi G2

o\ 0\ o2

~~ ° 3

®2

<Pz

°3

ZS Z1

<73

for the affixes of the Feuerbach points are better

&i a 2

<p2

°2

0\ o2 —

Z2 Z3

0 \ g2

<NCl

(Pi

+ Z2 Z1 Z S 4“ Z 1

<p2

+ °3

°3 03

Gi O’2

+ *2*3 Z 3 4~ Z 1

Z2 Z1

Gi O’2

+ Z1 Z2 Z2 Z3 Z 2 + Z 3 —Z 1

Z1 Z3

Gi G2

+ *3*2 Z1 + Z2 —

Z1 Z3 Z2

Zi Z2 9

or as: <Pi = £ + P

l Z1 Z2

Z 2 4~~ Z 3

<p2 = £ + p

Z2 Z1

(p3 = £ + P

Z2 Z3

Z3 Z1 +

Z2 Z3

Z3 Z1 Z2

Z3 Z2

Z1 +

Z1 Z2

Z2 — Z3

or as: <Pl = « + PUU

<P2 = « + PW2,

<P3 = £ + pW3,

where _

111 —

zxz2 +

Zj z 3 — z 2 z 3

Z2 +

U2 =

Z 2 Z 1 4 “ Z 2Z 3 Z3

Mo =

Z3 — Z1 Z3 Z1

i Z1 — Z2

Z3 Z1 4 “ Z3 Z2

Z1 Z2

Z1 + Z2 — z 3

217

Complex Numbers in Plane Geometry

[Note that the coefficients of p are complex numbers equal to unity in absolute value, for example, _ z\ Z2 Wi —

Z1Z3 ;

z 2 ~r z 3

—

Z1 Z2 J _ Z2

Z 1 Z3 +

Z2 Z3

J _______ L Z3

Z2Z3

Z 2Z 3

--------------------- —-Z l Z2 +

Z1 Z3 —

Z2 H

and, hence, u1u1 = 1. This of course immediately follows from the fact that \<pk — e\ = p {k = 1, 2, 3).] Problem 33. ABC is an arbitrary triangle; {ABC) is the circle circum scribed about it with centre O; the radius of circle (O) is equal to R; (/) is the circle inscribed in the triangle ABC; I is its centre and r the radius. Let d be the distance between the centres O and I of the circles circumscribed about and inscribed in A ABC. Prove that d 2 = = R 2 _ 2 R r.

Solution. Let D, E, F be the points of contact of the sides BC, CA, AB and the circle (/). For the unit circle, take the circle with centre /. Then the affixes of the points D, E, F, will be rzl9 rz2, rz3. In the preceding problem we obtained the following expression for the affix o: . _ o

&3 > <J1 <T2 03

where al9 o2, (t3 are the basic symmetric polynomials of the affixes of the vertices of the triangle. However, since these affixes are now taken in the form rzl9 rz2, rz3, we obtain the following expressions for the affix o of point O and for the radius R of the circle {ABC): 2(7, 20~i 0~3 r, R = (7i C2 0"3 &3 0"! (T2 Thus, o = — Rol9 o = — Rdx and so d 2 - OP = o-o = R?c1d1 = R?

= R%( 1 ~ °2 \ = r 4 l _ = R2—2Rr. V ^3 / V -R / Problem 34. Construct a triangle ABC if we know the points A0, 2?0, C0 of intersection of the bisectors of the interior angles A, B ,C with the circle {O) = {ABC) (Fig. 36). °3

Problems in Geometry

218

Solution. Take the circle (ABC) for the unit circle (O) = (ABC) = = (A0 B0 C0) circumscribed about the given triangle ABC. Let zu z2, z3 be the respective affixes of the points A, B, C. The bisector of the interior angle A of A ABC inscribed in the circle {ABC) bisects by a point A0 the arc BC subtended by the chord BC; note that the points A 0 and A are located on different sides of line BC. The situation of the points B0 and C0 is similar (see the statement of the problem). Therefore, for the affixes a0 = z3, *o = K* 3 Zi, c0 = i *2 014) of the points A0, B0, C0 values must be chosen (each of the radicals has two values) so that the points A and A0 lie on different Fig. 36 sides of line BC, the points B and B0 lie on different sides of line CA, and the points C and C0 lie on different sides of line AB. It is under these conditions that we have to solve the system (114), which we rewrite as follows: z3z3 ==

=L bo,

Zi z2

Cq.

(115)

From the system (115) it follows that z\ z\z\ = alblc% so that Z\

Z3 —

Qq bo Cq.

In the case Z\ Z2 Zq = Qq

(116)

we have ___________ Cqa0 Z\ — » z2 Qq bQ

a0 b0 z3 . Cq

(117)

In the case Zi Zo Z3 =

QqbQCq

(118)

we have Zi

bQCQ Q0

CqQq — , z3 ^0

^0^0 , ^0

(119)

Now we have to establish which of the two solutions (117) or (119) (or neither or both) constitutes a solution to the problem. Let us investigate the values of zu z2, z3 given by formulas (117). Since the foot of the perpen-

Complex Numbers in Plane Geometry

219 Zo

j £

dicular dropped from point O on line BC yields the affix — ---- —, it 2 follows that the equation of line BC may be written in the form —L_ +

= 2

Z 2 ~f~ Z3

Z 2 “f” Z Z

or z

+ — ■-z — - 1 - 0 .

Z2 + Z3

z 2 +

( 120)

z 3

We have c0a0 a0b0 *2 + z3 = ■— i ------O0 Co

(i+

_boCo z 2 T~ Z 3 — --------

~ ( * i + cS), D0 C0

6p + eg 0q b0Cq

c ')

*0 and the equation (120) of line BC takes the form

f(z) = -----bo£o----z + a_oboC_o j _ l = o. a0(bl + Co) b% + c§ We find /(a 0) and /(z,) = / ( ^ £2j : /(a.) =

^ 0 Cp

bo +

/(*i) =

b p Cq

(b p

b o ~ { - Cq

fljeg alibi + eg)

bo +

Cq)2

0q _ j ^ (0q - fro) (0q ~ c%) b\ + eg 0g(Ag + eg)

From the last two relations we have f ( zi) == (al—bl) (al—eg) _ _ (fl0—60) (a0—e0) (bp—Cp) (a0+fr0) (0o+eQ) /(0b) “ 02o(6o - Co)2 al(bo - e0)3 but ( A qB qC q) — ^

0o 0o 1 60 *o 1 Cq

i 4 Qq bp Cq

al 1 a0 bp 1 bp eg 1 e0

- — :----------(c0 “ 0o) (c0 - *o) (bo ~ 0o) 4a0 b0 Cp

Problems in Geometry

220

and, consequently, 4 a0boCo(AoBoCo) = i

(cq — a0) (c<) — bo) (b0 — a0) =

4/Vz0ioô(ôôô)-

Thus, (a0 -}~ b0) (dp 4~ Co) (b0 c0) f(z l) 4icio bQCq(AqB qCq) a&b0 - c0)4 f(ao) We consider the point A * with affix a* = — a0, which point is diametri cally opposite the point A0 on the circle {ABC) = (A0 B0 C0). We have (*o + b0) (a0 + c0) (b0 - c0) = (b0 - fl?) (c0 - a%) (b0 - c0) 4/tf0b0c0(A* B0 Co). To summarize: f&i) = - 16(^0 B0 Co) (A* B0 Co) (b0~ c 0)4 /K ) = - 16(^o B0 Co) {At B0 C0) [ - &o Co --12. (121) L{bo ~ Co)2] The number u = b0Co/(bo — c0)2 is a real number. Indeed, 1 —

___

0^*0

/_L _ '^ 0

b o Cq

___

M 2 ~~ (*• - co)2 Co)

Therefore, u2 > 0 (u # 0). Now note that A A qB0C0, whose vertices are the points of intersection of the bisectors of the interior angles A, B, C of AABC with the circle (ABC), is always an acute-angled triangle. Indeed, the interior angles A0, B0, C0of A A 0B0C0are respectively equal to 71 B+ C A 2 2' ~ 2 C+A ~~2 A+B 2

~2 n

~~ 2

B 2’ C 2

irrespective of whether A ABC is acute, obtuse or right-angled. From this it follows that the diameter A0A$ cuts the chord B0C0 and, ------> ------> hence, A A 0B0Co and A A ^B 0C0 have opposite orientations. Thus, the num-

Complex Numbers in Plane Geometry

221

bers (AqBqCq) and (A*BqC^) are of opposite sign, and from formula (121) it follows that f(zi)lf(a0) > 0, that is, the points A 0 and A lie on one side of the chord BC. Thus, the values (117) do not afford any solution to the problem. The required solution of system (115) is given by formulas (119), provided that A AqBqCq is acute-angled. If it is obtuse-angled or right-angled, the problem does not have any solution. The points A, B9C are points of intersection of the altitudes of A A 0 B0 C0 and the circle (ABC). Indeed, the slope of line B 0 C0 is equal to —b0 c0 and so the equation of the perpendicular dropped from point A 0 to the line B0 Cq is of the form z a0 = b0 Cq(z ab solving this equation together with the equation zz = 1 of the unit circle, we obtain z

a0 — b0 Cq j

V z

I, <*o)

_ b0 Co(z a0) z — a0 = ------------------ . a0z One of the roots of this equation is naturally z = a0 (the affix of point A0)y the other is z = — bQColaQ, which is the affix of point A : Z\ —

bo Cq/(2 q.

Similarly, proof can be given that [z2 — coa o/bo> ~3== a0 bo/co, which are the affixes of points B and C respectively. Remark. From what has been proved it follows that formulas (114) should be written as follows: a0 = — ]fz2 ][z39

b0 = — ]fz3 J(z l 9

c0 = — Vzx ^ z2,

(122)

where we take the same value for ][zx, / z 2, ]fz3 (in that case, for example, Vzi ]fzi = Kz2 ]fZ 2 — z2» Yzs Yzs — Z 3 9 but if for ]f zx we take different values in the last two formulas of (122), then ^zx ][z1 = — zx). Problem 35. Construct A ABC if we are given the points A 1 ,B 1 ,C 1; these are the points of intersection of its altitudes with the circle (O) = = (ABC) circumscribed about A ABC [(O) = (ABC) = (A^C ^)]. Solution. Take (A ^ C ^ ) = (ABC) as the unit circle. Let al 9 bl 9 cx be the affixes of the points A l 9 B l 9 Cl 9 and let z l9 z2, z3 be the affixes of the vertices A , B, C of the desired triangle ABC. Since the slope of the line AB is equal to —z 1 z 29 it follows that the equation of the altitude dropped from vertex C on the side AB is of the form

222

Problems in Geometry

Solving this equation together with the equation zz = 1 of the unit circle, we obtain /1 z3 — zx z2[ Vz

1\ I’ z3)

(Z -

*3=

z 3) .

Z3 Z

One of the roots of this equation is naturally equal to z = z3 (the affix of point C). The other is z = ct = — zx z2/z3,

(123)

which is the affix of point Cx. In similar fashion we find the affixes 0i = — z2 z3/zx,

bt = — z3 zx/z2

(124)

o f the points 2?x and Ax. Let us solve the resulting system for zx, z2, z3. Multiplying these equations together pairwise, we obtain 4 = bi ci> 4 = ci ai> 4 = 0i bl9 whence Zi = y^i

z2 = ]/cx |/"tfx, Z3 — V"01 K^l>

(125)

Z2 =

Yc1 f a 1,

Zi = -

- / c x fa x.

Z\ =

V bi Vci> *2 = Y h V clt *2 = - V bi Y c i, *2 = Y b i Y ci> *2 = - Y b i Y c 1. *2 = Y b i Y c i> *2 = Y'bi Y c i» *2 =

zt =

zi ~ Zl = Zi = Zi =

?3 =

,Fi

f ^ ]fct.

S 1

Zl =

and since each of the products of two radicals in the right-hand member of each equation has two values, the system (125) has 8 solutions. Assuming that in each of the equations (125), any value is taken for each radical K , J/^i, Yci (but one and the same value in each equation), we can write all the solutions as follows:

f« l f* l.

Y a i> z3 = - Y<*i Y ^ i, fc i /fli. z3 = - Y a x f £ ; .

- f c x ffli.

z3 =

fa f^ i.

fc i ffli,

z3 =

f ax f />i,

- f<a f a ,

z3 =

fa i f*i,

fc i fai.

2 3

= — f« i f*l- .

(127)

Complex Numbers in Plane Geometry

223

From the system (123), (124) it follows that dxbxCx = — zxz2zz and so all number triples (126) fail to serve as solutions to the system (124). Now any row of relations (127) is a solution of the system (123), (124). Indeed* *1 Z 2 _

][b

- ] f a i

Z2 Z 2 Z 3 __ __

)fc i

Va!

I fo

_ U

)[<*! Y a t ] f b i -

Z3 Z1 =

1 /c x f c x

] fa i

V bi

] fc .

y b i

- yCl

y a

y Cl _ b i

and also for the other three relations (127). To summarize, then, there exist four triangles that satisfy the condition of the problem. Let us construct, for example, a triangle corresponding to the first row of solutions (127). To do this, draw to the circle (A ^ C x ) any one of two tangent lines parallel to the straight line QAX(& is the unit point); the point Pi, which is the point of contact of the tangent drawn to the circle (^j^C j), has as its affix one of the values ] (Fig. 37). In similar fashion, construct the points Qx and Rl9 whose affixes are the values 1fbx and ]fc±. To construct a point with the affix Yb± ]/ci, draw through the point Q a straight line parallel to the line QxBxl the second point P9 the point of intersection of the drawn line and the circle (AxBxCx), has the affix Y^i Yci• Finally, point A , which is symmetric to point P with respect to the centre O of (AxBxC^) has the affix ~~Ybi Yci-The points B and C with affixes —Yci an(* —Yai Y&i are constructed in similar fashion. The other three triangles that satisfy the conditions of the problem are: AQR, PBR, PQC. Problem 36. Inscribed in the unit circle is A ABC, the affixes of whose vertices are zl9 z2, z3. Find the affixes t 0, rl9 t 2, t 3 of the centre of the circle (/) inscribed in that triangle (affix t 0) and of the centres of the circles (Ia)9 (/*), (/c) escribed in that triangle (in the angles A9B9C9 respectively). Solution. The centre / of the circle (/) inscribed in A ABC is the point of intersection of the bisectors of the interior angles. These bisectors inter sect the circle (ABC) in the points A0, B0, C0; note that /\A QBQC0 is always an obtuse-angled triangle (see problem 34). The affixes a09 b09 c09 of the points A0, B0, C0 are expressed by the formulas (122) of problem 34: Qq =

YZ2 YZ3’ *0 =

YZ3 YZ1* Co

YZ1 Y^2>

(128)

where the same values are taken in all formulas for the square roots J/z^ Yz 2 >YzZ’ note that these values for Yzi> / z2, Yzz must always be taken so that formulas (128) define just the points of intersection of the bisectors

224

Problems in Geometry

of the interior angles A, B, C of the given A ABC with (ABC). Namely, for |f z l9 ][z2, J/z3 we have to take the values so that the inequality \][z1 + + V Z 2 + Yzz\ < 1 holds, that is, so that the triangle whose vertices have the affixes |f zl9 Y z2, ]f z3 is acute-angled. Indeed, then l^o +

b 0 + c0)| = | - Y z 2 f z 3 — V z3 Y z i — Y z i V Z 2 \ = |V*2 Y Z3 + V z3 Y Z1 + Y Z1 Y Z2\ = \ Y z i Y zz + Y zz Y z i + Y zi Y z 2 iKzi + Y z 2 + Y z I lfzl YZ2 VzZ1 3

jY z 2

Y zs + Y

z 3

+ Kzil/*a

— i / zi + y"z2 + v £3! < i* That is, A A 0BqC0 is acute-angled' [there are only two such choices for the square roots of zl9 z2, z3; if one of them is denoted by Yzi> Yz 2> Yzs> then the other will be —|/zl5 —J/z2, —]/z3, and any one of these choices yields the formulas (128)].

Complex Numbers in Plane Geometry

225

To summarize, then, AA0, BB0, CC0 are the bisectors of the interior angles A, B, C of A The slope of the bisector AA0 is equal to Y" = zi YzzY zs,

a a — Z,

YzzVz3

and, hence, the equation of the straight line AA0 is z — Zi =

Z jI/z ss | / z

(z — Z!>

or z — Z! Vz2 /z ,z = z1 — / z 2 / z 3.

(129)

The equation of BB0 is z — z2 ]/z3 J/zx z = z2 — ]/z3 / z x.

(130)

From the system (129), (130) we find the affix t 0of the centre I of the circle (/) inscribed in A ABC: t 0(z2 V*i ~ zi If**) = -

Y zz - / zi) ~ Yzi f z2 Vza (z» ~ zi)>

or To K z i /

( / z2 -

|/ z i) =

Z iZ 2 ( K z g —

/z j)

- fz j Jfz2 |/z3 [(l^z2)2 - ( /z ,) 2] That is, To = - YZ1 VZi - f zS( / z2 + /A ). and, finally. To =

—

f z 2 / z 3 — ][z3 ^Zj

—

/ z x l/z2 =

+ c0.

( 131)

If we now take the point B$, which is symmetric to the point B0 with res pect to the centre O of the circle {ABC), then the affix of the point B* will be equal to 6* == —b02LndBB* will be the bisector of the exterior angle of the triangle at the vertex B. The slope of BB* is ?2 + ^0 ^ 2 ~T bo

- YzaYz

= — Z* / Zs / Zl.

Yzs f z and the equation of line BB* is z - z2 =

z2 / z 3 f z x (z - z2)

or z + z2 YzzYzi z = 1 5 -8 1 0

Z2 +

YzaYzv

'(132)

226

Problems in Geometry

From the equations (129) and (132) we find the affix the circle (Ia) escribed in angle A of triangle ABC: Ti

(zx } f z 2

Z2 Yzx) = ZZZX (fz 2 +

Yzl)

~ Y zl

of the centre Ia of

f Z 2 Yz 3

( z 2

Zl)

or Tj j/zj / z a (/z x + / z 2) =

Z jZ g

(]/zx + ]/z2) -

f z , 1fz3 [(/z2)2 - fl/z^2],

that is, Z1 =

/ zi

KZ2 — YZ3 ( / Z2 —

/ z l) =

f Zl f Z2 — Y Zs Y z z

fz s/z x ,

and, finally, Tj = — / z 2 f z 3 + f z 3 Yzi +

za = a0 — b 0 — c0.

f zi f

Similarly we find t 2 and r3. Thus, T0 =

—

Zx —

Y Z1 Y Z3 — Y Z3 Y Z1 — Y Z1 Y Z2 = | / z 2 V z 3 "I" K z 3 V

"i" K Z1 K

z2 —

«0 +

b0 +

C o,

x2 =

fz-t / z 3 —Yz3 Yzx + Yzx f z 2 = —a0 + ho — c0,

Ta =

Y z 3 ] / z 3 + / z 3 Y Z1 — Y z 1 f z 2 — —

— h0 +

C0 .

From formulas (133) it follows that *0 + Zl _ „ T« + T2 _ A — Oo, ---- ~------— 2

Zo + z3

c0.

That is, the midpoints of the segments IIa, IJb, 1IC coincide respectively with the points A0, B0, C0 in which the bisectors of the interior angles of A ABC intersect the circle (ABC) --- (O) circumscribed about Is ABC (Fig. 38). From formulas (133) it also follows that *2 +

------------ —

a0,

—

oo,

Zx + Z 2

—

Cq,

that is, the midpoints of the segments I J a, IaIb coincide respectively with the points A$,B$,C* in which the bisectors of the exterior angles A, B, C of A ABC cuts the circle {ABC) = (O) circumscribed about A ABC. The points A*,B$9C$ are symmetric respectively to the points A0, B0, C0 about the centre O of (0)=(ABC) so that A0A$9 B0B£, C0C£ are diameters of that circle. Incidentally, from that it follows that the circle (0)=(ABC)= = (A0B0C0) = (A*B*C*) is the Euler circle of A IJbh and therefore the radius of the circle (/fl4 /c) is twice the radius of the circle {ABC) [this has already been proved analytically in problem 32, item 7° (see solution)].

Complex Numbers in Plane Geometry

227

Finally, since IA _L IbIc, IB _L IC _L IaIb, it follows that I is the orthocentre of A IJbh> furthermore, since (tx + t 2 + t 3)/3 = — (a0 + (- b0 -f c0)/3, it follows that the centroid G* of A IJbh is symmetric to the centroid G of A A qBqC0 about the centre O of the circle (ABC), or the point G* is the centroid of A ^*B$C*, which is symmetric to A A 0B0C0 with respect to the point O. Problem 37. ABC is an arbitrary triangle and P is an arbitrary point. Prove that the straight lines a!, b', c', which are symmetric to the straight lines a = AP, b = BP, c = CP about the bisectors AI, BIy C l of the inte rior angles A yByC of A ABC also pass through one and the same point Q. The points P and Q are said to be isogonally conjugate with respect to the triangle ABC (Fig. 39). Prove that if the circle (ABC) is taken as the unit circle and if the affixes of the points A, B, C, P, Q are equal respectively to zl9 z2, z3, p , q, then they are connected by the relation p + q + o»P q = <?!

(134)

(this relation was obtained by the English mathematician Morley). Express q in terms of zl9 z2, z3, p. Solution. A0, B0, C0, the points of intersection of the bisectors of the interior angles A , B, C of A ABC with the circle (ABC), always form an acute-angled triangle A0B0C0. If for ]fz l9 ]/z2, ]/z3 we choose values such that !Yz± + J/z2 + |/z31< 1, then the affixes a0, b0, c0 of points A0, B0, C0 arc ai>=— Vz2 )[z99 b0 = — Yz9 Yzl9 c0 = — Yzi Y z 2

228

Problems in Geometry

[problem 36, formulas (128)]. The slope of the bisector of the interior angle A of A ABC is .*■ + Zl + ]f^2

= *, V*

yf,

and the equation of this bisector is z — zx I!zn ]fz3 z = z1— ][z2 ]fz3

(135)

[see problem 36, equation (129)]. The equation of the perpendicular dropped from point P on this line is of the form Z -P

= — Zi ]fz 2 ]fz 3 ( z -

or z +

Zj,

]fz2 ]fz3 z =~-p + zx f z 2 1fz3 p.

(136)

Adding the equations (135) and (136) term by term, we find the affix z —p' of projection P' of point P on the bisector of the interior angle A of A ABC: p' = ~~

(Z i

+ P - ]fza ][z3 + zx ][z3 ]/z3p).

The affix p f of the point P *, which is symmetric to point P about the bi sector A l of the interior angle A of A ABC is found from the relation p + p* 2

= p'

( Z i + p —Vzz ]fz3 +

][z2 ][z3p).

Complex Numbers in Plane Geometry

229

whence P t = Z1 V*3 + 1^2 1^3 PThe equation of the line >42^* may be written thus: 2 Z />f Z

ZX JP f

1 1 1 or Z

*1 — f * 2

V *3

Zx If z 2 ][ z 2 p

* Zr z\ -][z2 ][z2 + z 1 ]fz2 ][zs p = 0 1 1 1 or ()fz2 ]fz3 - z 1]fz2 ]fz2p )z + ( - ][z2]fz2 + zx y z2]fz2p ) z — zx y z 2 y z2 + y z 2 yz2p + z x Vz2yz2 — ]fz2yz2p = o or, multiplying both sides of the last equation by y z2 Yz2, (1 — z\p)z + (—z2z3 + zxz2z2p)~z — zx + p + z xz2z2 — z2z2p = 0.

(137)

In similar fashion we can write down the equation of the straight line sym metric to line BP about the bisector BI of the interior angle 2? of A ABC: (1 — z 2p) z + (—z2zx + zxz2z2p) z — z2 + p + z 2z2zx — z2zxp

0. (138)

The affix of point Q, the point of intersection of the lines (137) and (138) that are symmetric to the lines AP and BP about the bisectors A I and BI of the interior angles A and B of A ABC, may be found by solving the system of equations (137) and (138). Since A =

1 -

* l/>

z 2p

— *2*3 +

*1*2*3~P

Z3Zi + zxz2z2p

— — Z3 Z1 -f" (Top +

Z2p — Z 2Z2p p

z2(z2 — Z j)

* 2*3

z2pp (z2 -

G*P Z j) =

Z*P +

ZxZ2p p

z2 (z2 — Z i )

(1 -

pp).

it follows that A ^ 0 if and only if 1 — PP ^ 0, that is \p\ ^ 1; in other words, when the point P does not lie on the circle (ABC). (1) Assuming that \p\ ^ 1, that is that the point P does not lie on the circle (ABC), we conclude that the system (137), (138) has a unique solu tion for z and z : z =*•- q, z —q. This solution could be found, for example, via the Cramer formulas, but we will take a somewhat different route; namely, by substituting the value z — q, z = q into the equations (137),

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230

(138) and subtracting the second equation from the first term by term, we get ( z 2 — z ±)pq + z3 (zt — z2)q + z* - zt + z3 ( — — — ) + ^ ( ^ 1- ^ 2) = 0

V *1

*2 /

or Z1 — Z2 pq + - z J - (zi - z2) - —— ( z l-z t)+ z 2p(zi—z2) = 0. Z1Z2 Z1Z2 Cancelling zt — z2 and multiplying both sides by zxz29 we obtain pq + <r9(p + q ) ^ a* = 0.

(139)

Now this is the Morley relation. Indeed, passing to conjugate numbers, we have P q + 0z(P + q) “ 0 2 = 0. (140) But <78 = l/a3, <r2 =

J 0 2 and so

P + q + 03Pq = 0 i (141) From the relations (139) and (141) it is easy to express q in terms of p. From (139) we find 0 2 - pq _ q = 02

and (141) takes the form p + q + p(0 2 -p q -

P) = 0 i

or q (i - pp) =

P2 -

0 2

p —p +

whence =

p - p +01 1 — pp From the symmetry of this expression with respect to zl9 z2, z3 it follows that the straight line C as well passes through the point Q. (2) If point P lies on the circle {ABC) but does not coincide with any of the points A ,B 9C9 then the straight lines (137) and (138) are collinear; as will be evident from what follows, there is a third line collinear with them: it is the line symmetric to line CP with respect to the straight line C l (Fig. 40). The slopes of these lines are: 09

p2 -

0 2

1 ^2^3 02P 1 - z\p

Zi - P 1- Z i P

= 03 Z

P - Zi p(zi- p )

03 P

—09Pt

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231

and if we take a Boutain point for the unit point, then x = —p. Setting p = cos a + i sin a, we find x = cos (7i — a) + i sin (n — a). The affixes of the ends of a diameter of the unit circle (ABC), which has such a slope, are ]/x = ± [ c°s

+ ' S'” ( t - t ) ] '

From this it follows that if the point P describes the circle (ABC), then this diameter will turn through an angle n in a direction opposite that of the radius OP (Fig. 41). If the Euclidean plane is completed to a projectiveEuclidean plane by points at infinity, or ideal points, we can then assume that if the point P / lies on (ABC) but does not coincide with any one of the vertices A, B, C of A ABC, then it will be associated with the ideal point of the projective-Euclidean plane. The Morley relation p + q + OzPq= <7i does not, of course, hold in this case since no complex number can be as sociated with ideal points. Finally, from geometric reasoning it is clear that if point P lies on line BC but does not coincide with B or C, then point A will be isogonally conjugate to it (the same goes for the lines CA and AB). Finally, if point P coincides with one of the vertices of A ABC, for example

Problems in Geometry

232

P = A, then any point of BC will be isogonally conjugate to it (and the same goes for the cases P — B and P = C). This can also be verified ana lytically: let p = az2 + /?z8, where a and p are real and a + P = 1 (that oc B is, the point P lies on the straight line BC). Then p = ----- 1----- and, Z2

assuming g = zl9 q = — , we will have zi p + q + o-iP q = «z2 + Pz2 + zx + ztz,z3 (■“ •- + - - ) ™ V z2 zs / zt — <*Z2 +

PZZ +

Z1 +

aZ 3 +

^ Z2 — Z1 +

(a +

P)Z2 + =

(a +

Z1 +

P ) Z3

Z2 +

Z3 ~

Note that the centre / of the circle (/) inscribed in A /4#C, and the cen tres Ia, /*, l c of the circles (/fl), (/fc), (Ic) escribed in the angles A, B, C of that triangle are isogonal conjugates of themselves with respect to A ABC. It will be proved below, in problem 41, that besides the points /, /a, Ib, Ic there are no points that are isogonal conjugates of themselves with respect to A ABC. We also note that if we eliminate from the Euclidean plane the circle (ABC) circumscribed about A ABC and eliminate the lines BC, CA> ABy then the isogonal correspondence between the points P(p) and Q(q) will be a one-to-one mapping (P *-* Q) or a one-to-one and involutory trans formation (involution) described by the relation p + q h p qoz — ox. If the Euclidean plane is completed to a projective-Euclidean plane, and the straight lines BC, CA, AB are deleted, then the correspondence between the points P(p) and Q(q) will again be one-to-one, but then the relation p + q + pqoz — ox will refer only to the ideal points P(p) and Q(q). Finally, throughout the projective-Euclidean plane, the mapping of P(p) on Q(q) will no longer be a one-to-one mapping and the relation p + q + pq<Tz= gx will again be valid only for the proper points P(p) and Q(q) (ideal points do not have affixes). Problem 38. The circle (O) = {ABC) circumscribed about A ABC is taken as the unit circle; zl9 z2, z3 are the affixes of the vertices A, By C. Find the affixes of the points that are isogonal conjugates of the following points with respect to A ABC: 1°. The point G, the point of intersection of the medians of A ABC. 2°. The orthocentre H of A ABC. 3°. The centre 0 9 of the Euler circle of A ABC. Solution. Let us take advantage of the formula of problem 37; P + P<*2 ~ — <r»P* PP - 1

Complex Numbers in Plane Geometry

233

which defines the affix q of point Q ; this is the image of point P with affix p under an isogonal transformation with respect to A ABC. We have the following. 1°. Since the affix g of point G is equal to <t3/3, it follows that the affix / of the image L of point G under an isogonal transformation with respect to A ABC is aL

? .\_

(7o

(7i

(7a

(7? + -------------3 (To

<*1

— 3

9<73

9 2

-1- 2 ' 3 1 + ~q 9<--7o ^ 3(7j ?1<*1 <^1 - 9 9 The point L, which is the isogonal conjugate of point G with respect to A ABC, is termed the Lemoine point. Thus, the affix / of the Lemoine point I of A ABC is I _2 3 o~i - 9 2°. The affix of the point that is isogonally conjugate to the orthocentre H with respect to the triangle ABC is _ of _ ^3_______ q (TiO1! “ 1 <T101 — 1 that is, the orthocentre H and the centre O of the circle (ABC) are isogo nally conjugate with respect to A ABC (the straight lines AO and AH are symmetric about the line AI and similarly BO and BH are symmetric about B l, CO and CH are symmetric about Cl). 3°. The affix of the point isogonally conjugate to the centre 0 9 of the Euler circle (0 9) with respect to A ABC is 2

o^i _ 4

5i_ + a\ ____ ?i_ 2 2<t3 4(t3*2 o1o1 4 ffi a\ 2 4o3 _ j

a2cr1 — 2xx CTiffi -

Problems in Geometry

234

Problem 39. Prove that the midpoint M of a segment whose ends are points Px and Ql9 which are isogonal conjugates (with respect to A ABC) of the ends P and Q of any diameter PQ of the circle (Q), which is con centric with the circle (ABC), describes (as the diameter PQ rotates) a circle that touches the tangents drawn from the orthocentre H of A ABC to the circle (ABC), Triangle ABC is assumed to be obtuse, and the radius p of the circle (£) is assumed to be less than OH [only in this case is it pos sible to draw tangents from point H to (ABC)], Solution, Take (ABC) for the unit circle; let zx, z2, zz be the respective affixes of the points A, B, C, Let PQ be an arbitrary diameter of (Q), and let p and —p be the respective affixes of the points P and Q. Denoting by px and qx the affixes of points Px and Ql9 which are the res pective isogonal conjugates of the points P and Q with respect to A ABC, we will have (see problem 37) 2 _ 02P ~ P + <*1 Pi 1 ~ pp <*sP2 + 02P + p + <T1 <h 1 - pp From these we find the affix m of the midpoint M of segment PXQX: °dP* + ffi m = Pi + <7i 2 1 -P P From this relation it follows that if point P describes a circle (&) which is concentric with the circle (ABC), then 1 —/>/>=constant [1 — pp — = —< 7 , where a is the power of the point P with respect to (ABC)] and, hence, point M describes a circle (T), the affix of whose centre is t=

1 - pp

and the radius is OP2 °3P* \ \ - pp\ 11 — o p 2 i Knowing the affixes of the points H and T, we can find the complex number Ri =

corresponding to the directed line segment HT: t — <7,

—

<*i ----- 01 = 1 - pp

OP2 0-1P P ___ ~ =T — 0V 1 - OP2 1 - pp

Associated with the directed line segment HO is the complex number Hence HT HO

OP2 OP2 - 1

Complex Numbers in Plane Geometry

235

From this relation it follows that point T is the image of the centre O of the circle (ABC) under a homothetic transformation with centre 77 and OP2 r a tio ----------- . Therefore the circle (T7) described by point M touches OP2 - 1 the tangents drawn to (ABC) from the orthocentre 77 of A ABC (only here is the condition used that A ABC is an obtuse-angled triangle). Remarks. (1) The radius Rx of (T) is equal to the radius 1 of (ABC) if and only if OP2 1 - OP2 whence, obviously. OP2 1 - OP2 (2) The radius Rx of (T) is equal to the radius of the Euler circle if and only if

(3) The radius RÂą of (T) is equal to the radius of ((2) if and only if

Problem 40. Given in a plane an arbitrary triangle ABC. The circle (ABC) = (O) is taken as the unit circle. Let t 0 be the affix of the centre 7 of the circle inscribed in A ABC, and let xi9 t 2, t 3 be the affixes of the centres Ia, Ih, Ic of the circles (7fl), (Ib), (Ic) escribed in the angles A, B9 C of the given A ABC. Prove that the orthopoles of the straight lines 01, OIa9 OIb9 01 c with respect to A ABC are, respectively, the Feuerbach points <P0, <Pl9 &2, # 3 of A ABC: the points 4>0, 4>2, are, respectively, the points in which the Euler circle of A ABC is tangent to (7), (7fl), (lb)9 (7C). By proceeding from the foregoing, find the affixes q>09 <Pi, <P2 ><Ps of the Feuerbach points # e, 4>l9 0 2, <J>3. The affixes of the vertices A , B9 C of A ABC are equal to zl9 z2, z3 respectively (see Fig. 35). Solution. Let t 0, t 15 t 2, t 3 be the respective affixes of the points 7, Ia9 Ihy Ic. On the basis of problem 7, the orthopoles of the straight lines OI9 OIa9 OIb9 01c with respect to A ABC have the affixes (142) (143)

Problems in Geometry

236

< p* =

044)

<Ps =

(*! - ^ y )

(145)

since the slopes of the lines 07, 0 Ia, OIb, OIc are respectively, Xq —

Tj_

^1 — “ > ^2 — _ i ^3 ~

We now have to prove that <po, <p[, </>2, </>£ are the affixes of the respective Feuerbach points. Let us first consider the formula (142):

Let og be the affix of the centre of the circle (0 9) of A ABC. Since o9 = = — , it follows that 2 /

^3To

(p0

“

~ — 9 o °9

whence , , I ^0

°9 I —

1 2

" “

and so the point lies on the Euler circle. By the Euler formula for the distance between the centres of a circumscribed circle and an inscribed circle of the triangle, Ol2 = R2 — 2Rr = 1 — 2r (because we have R = 1), where r is the radius of (/) inscribed in A A B C . Since 012 = t 0t 0, it fol lows from the last formula that

The equation of the circle (7) is 4 We now prove that the point <P'0 with affix

(146)

Complex Numbers in Plane Geometry

237

lies on the circle (/). We have <Po -

= -

T0 ,

2t0

and since r0 is the fixed point of an isogonal transformation with respect to A ABC [see problem 37, formula (141)], it follows that 2 t0 +

a 3 fg — a x =

whence *i ~

t0 =

T?> 2

and therefore ,

</>0 “

y:)To 2t 0

_ *i

T0 —

0*1 , 0*3Tp 2 2 <r3f0

ffsio 2t0

( t 0t 0 -

2t0

1).

From this we get _«TSTo 2f0

<Po - To :

( t 0t 0

and, hence, (<Po -

T o) ( 9 i -

T o) =

(T ° T(>

—

r 2.

That is, the point <P'0 with affix </>' lies on the circle (/), the equation of which is of the form (146). Furthermore, the slope of the straight line I<Pq is <P0 -

T o ____ 2 f §

<Po ~~

9 T0

The slope of the straight line 0 9<P'Q is *i 2 2

<Po - ~<Po

and this means the points 0 9, #o, /lie on one straight line. To summarize, then: the point <P'0 with affix </>o lies both on the circle (/) and on the Euler circle (0 9), and, besides, the points #o, h 0 9 lie on one straight line. Now if two circles have one common point (in this case, $o) and their centres

Problems in Geometry

238

(in the given case, 7 and 0 9) lie on the same line as that common point, then they are tangent to one another at precisely that point. But the point of tangency of the circles (7) and (0 9) is precisely the Feuerbach point <P0, and so the points 4>9 and <P0 coincide and, hence, q>'0 = (p0. To derive the other three formulas <Pl — ^P19 ^2 = <p2, <Ps = <Ps, where q>[9 <p2, (p's are given by (143), (144), (145), and <p1? <p2, <p3, are the affixes of the Feuerbach points <Pl9 <P2, <P3, we first derive the formulas Oil = R2 + 2raR9 (147) Oil = R2 + 2rbR,

(148)

01* = R2 + 2 rcR

(149)

for the distances between the centres of the circumscribed circle and escrib ed circle (it is of course sufficient to prove only the first one of them). Let us consider the inversion [7fl, r%|, whose centre is the centre Ia of the circle (Ia) escribed in the angle A of triangle ABC, and the power is equal to ra. Let the circle (Ia) be tangent to the straight lines BC, CA, AB respec tively at the points Pl9 P2, P3 (see Fig. 35). Under the inversion [7fl, r%]9the points A, B, C go respectively into the points'^*, B*, C*9 in which the fol lowing straight lines intersect: IaA and P2P3; IaB and 7>37>1; IaC and 7>17>2. The points A*, B*9 C* are the respective midpoints of the segments R 2 R& P3Pi,PiP2. Thus, the circles (ABC) and (A fB fC f) go into each other under this inversion with the circle of inversion (7>17>27>3)=(7a).The circle (AfBfC *) is the Euler circle of A 7>17>2P3 and therefore the radius of (AfBfC *) is equal to R'a=rJ2. On the other hand, the ratio R'JR of the radii of (A?BfC*) and (ABC) is equal to \kja\9 where a is the power of the centre Ia with respect to the circle (ABC), and k is the power of the inversion at hand, that is k = r*. Indeed, suppose M and M ' are corresponding points under the inversion at hand [point M on the circle (ABC) and point M ' on the circle (A*B*C*)]. Then (IaM )(IaM') = r*a ( = k ) . (150) Let M y be the second point of intersection of the line IaM with the circle (ABC). Then (IaM )(IaM1) = c. (151) From the relations (150) and (151) we find (hM ‘) ChM 1)

Incidentally, this means that the circle (0*)=(A*B*C*) may be obtained from the circle (ABC) by a homothetic transformation with the centre of similitude (homothetic centre) I a and ratio k j a . (Of course, the correspon-

Complex Numbers in Plane Geometry

239

dence of points under this homothetic transformation and under the inver sion that we considered is different.) Since a homothetic transformation is a similarity transformation, it follows that g

K R

o il- *T

Since the point Ia lies outside the circle (ABC) [the midpoint of segment IIa lies on the circle (ABC) — see problem 36], it follows that OIa > R and, hence. 2R

O il ~ R2 ’

whence Oil = R *+ 2Rra. We assumed /J = l. Besides, Oil = t 1t 1 since hence, ra =

is the affix of point Ia;

(txt l — 1). We now prove that the point

with affix

<Pi lies both on the Euler circle and on the circle (/a). We have r

<P1

2t,

and since Ia is a fixed point of an isogonal transformation with respect to A ABC, it follows that 2xx + <r3 ff — <rx = 0, whence *1 — *3*1 Tl = 2 and therefore ,

<*i

*1 — *3*1

2 tx

2 *

3*1

*3*1

2t 2

From this we have ? ;-? ! = — ( v i- i) 2ti

*3*1

2t x

■ ( V i - l ) . (152)

Problems in Geometry

240

and so (<P'i - Tj) (ipl - Tj) = That is, the point

(*1*1 - l)2 4

with affix <p[ lies on the circle (/fl) whose equation is (z - Ti) (z —

= r\.

Furthermore, 0 9 = aJ2 is the affix of the centre of (0 9). Therefore, / 1 / fi \ 1 a3t1 2 - -2 « . = whence | ^ - o 9| = 1/2 and therefore the point <P[ lies also on the Euler circle, the equation of which may be written down in the form |z — o9\ = 1/2. The slope of the straight line Ia <P[ is <Pl - T, =

2ti__________ =

V' ~ ^

^ Tl (t.T, - 1) 2fx

Tf 3

The slope of the straight line 0 9<P[ is

Consequently, the points 0 9, Ja and lie on one straight line. Thus, the point also lies on the circle (0 9) and on the circle (Ia) and the centres 0 9 and Ia of these circles lie on the same straight line as their common point <P’i\ hence, (Ia) and (0 9) are tangent at the point <P[ with affix q>[. But the point of tangency of (/a) and (0 9) is the Feuerbach point &l9 and so the points and <P± coincide and (p[ — (px. In the same way we can prove that cp'2 and cp3, that is, the affixes of the orthopoles of the straight lines OIb and OIc with respect to A ABC are the affixes of the Feuerbach points ^>2 and tf>3. That is, <P'2 and are points of tangency of the circles (Ib) and (Ic) with the Euler circle (0 9) of A ABC. Problem 41. Let zl9 z2, z3 be the affixes of the points A, B, C in a system of coordinates in which (ABC) = (O) is taken as the unit circle. It is re quired to prove that:

Complex Numbers in Plane Geometry

241

1°. The affixes t (), zl9 r 2, t 3 of the centres of the circles inscribed and escribed in A ABC are given by the equation T4 — 2 (7 2I 2 +

8(73 T +

<?2 — 4(72(73 =

2°. The affixes ij/0, i//l9 ^ 2, <A3 of the points !P2, symmetric to the orthocentre H of A ABC with respect to the Feuerbach points are found from the equation (4cf2 — a\)+ 4^1j/3 + 2a1a1ij/2 + 4al\// + 4<72 — a\ = 0. These points *F09 *Fl9 *F29 X V3 lie on the circle (ABC) (see Fig. 42). Solution. 1°. Let T be the centre of the circle inscribed in A ABC (or the centre of one of the escribed circles). Any one of these points has the characteristic property that it coincides with the point that is isogonally conjugate to it with respect to A ABC. It is easy to verify the sufficiency of this condition geometrically; namely, any one of the indicated four points T is fixed under an isogonal transformation. We will now prove that there are four such points. Indeed, for point Twith affix t( |t| # 1) to coincide with its conjugate point, it is necessary and sufficient that the Morley relation P + q + <r3 P ~q' = hold for p = q = t, that is, 2t + (73 T2 = at. For example, we will prove that it holds for the affix T# =

1f z 2 1 ^ 3

If z 3 1! z t

][z2

of the centre of the circle (/) inscribed in A ABC (see problem 36). We have - = _ _ _ J________ }_________1 _ = _ 0 K*2 1^3 ]/zz fzj ]fz1 ]fz2

+ KZ2 + fcg 1lzx Vz2 )[zz

or To = — [*1 + Z2 + Z3 + 2 (^z2 ]fz3 + ]/z3 ]/zx + /z j / z 2)]. <r3 From this we have 2t0 + o3xg = 2(— f z 2 ]/z3 — 1lz3 ]fz2 — ]!z2 j/z2) + zx + z2 + z3 + H]fz2 1fz 3 + 1fz3 j/zj + |/zj ]/z2) = Zj + z2 + z3 = alt and so also for the other affixes zl9 t 2, t 3 [formulas (133) of problem 36]. Adjoining to the equation 2t + a3 t 2 — ax = 0 (153) the equation obtained by equating to zero the conjugate number of the left-hand side, we obtain 2t -f a3 z2 — ax = 0, (154) 16-810

Problems in Geometry

242

Fig. 42

whence X

T <7

and the equation (153) takes the form — 2 f f 1 ff 3 Ta +

t 4 <r|

Lx + <x3 ---------------------------------- —

= 0

or 8r +

0*3

t* +

—

<T3

t 4 -

4ox = 0

or T 4 — 2(7 2 T 2 +

8 0 -3

T +

— 4(7! (73 =

Since this is a fourth-degree equation, it follows that there cannot be more than four fixed points under an isogonal transformation. But the centre of the inscribed circle in A ABC and the centres of the circles escribed in this triangle are fixed under an isogonal transformation with respect to A ABC. Hence, the fixed nature of a point under the isogonal transfor-

Complex Numbers in Plane Geometry

243

mation is the characteristic property of the centres of the circles (I), (Ia), (Ib), (Ic) (there are no other fixed points under an isogonal transformation with respect to A ABC). 2°. The Feuerbach point # is an orthopole, with respect to A ABC, of the straight line OIk that passes through the centre O of the circle (ABC) and the centre Ik of the corresponding circle tangent to the lines BC, CA> AB. If the straight line is given by the equation z — z0 = A (z — z 0), then its orthopole with respect to A ABC has the affix /‘ = - | f f i - y + Zo - ^ 0j (see problem 7). In particular, if the line at hand passes through the coor dinate origin, then H = —■Oi — o^)The affixes q>k of the Feuerbach points <Pk corresponding to the centres Tk of the four circles (Tk), each of which is tangent to the lines BC, CA, AB, will be (see problem 40) ?* = ~

(155>

where zk is the affix of the point Tk (incidentally, it follows from this that the slope of line OTk is equal to 2k = — and so — = Xk). We will ** ** denote the affixes xk and <pk of the points Tk and 4>k that correspond to each other by t and cp. Then the relation (155) can be rewritten as <P = —

—

(156)

Let us find the affix \j/ of point lF, which is symmetric to point H with respect to point 0: + * _ 1 (n „ f \ 2~ ~ " t r “ * 7 j' whence ik = - a 3 - . T

(157)

Since |^| = 1, the point W lies on the circle (ABC). Incidentally, this also follows immediately from the fact that under the homothetic transforma-

244

Problems in Geometry

tion (H, 2) the Euler circle of A ABC goes into the circle (O) = (ABC), and since all Feuerbach points lie on the Euler circle, their images TV *Pl9 ^ 2, under the homothetic transformation (H, 2) lie on the circle {ABC). Let us eliminate r and f from the equations (157), (153), (154). We obtain an equation for finding if/. From the relation (157) we have f = — ^ T dz and, hence, the relations (153) and (154) take the form 2r + — (T1 — 0 —2 ^ t <t3 + <t3 t 2 — dx — 0

(158) (159)

<r3 + 2 t — ax = 0, ovr2 — 2 \j/ x <x3 — = 0.

(160) (161)

It remains to write the resultant of these equations as equal to zero. But this can be done differently: multiply Eq. (161) by —if/2 and add it to Eq. (160) to get (2^3 az + 2) t = (T1 — whence = Substituting this value of

~ 1 2ip* <r3 + 2

for example, into equation (161), we obtain

_ G\ — 2<71(X1^ 2 + C73^4 **

4(^«r3 + l)^

2^<t3 gi 2

— (Tj — 0 + 1)

or ( ^ 1 — 4(T1ff3) \p4 — 4ff16:3^ 3 — 2cr1a 1^3^2 — 4<T1a 3i/' +

— 4^ - 0

or, cancelling (j3, (<rf — 4(71d:3)^4 — 4<71i/f3 — 2(71d:1^ 2 — 4(7^ + a2 — 4<t1(73 =- 0 or (<t2 — 4<t2) 1/f4 — 4<r1i/f3 — 2

— 4a1ij/ + cr2 — 4<72 = 0.

Note that the coefficients of this equation that are equidistant from the ends are conjugate in pairs: the first and the last, the second and the fourth (an antireciprocal equation). Problem 42. Let M be a variable point of a circle (O) of radius R circum scribed about ABC\ P is a point symmetric to the orthocentre H of A ABC with respect to the diameter of (O) parallel to the Simson line

Complex Numbers in Plane Geometry

245

for the point M with respect to A ABC; A', B \ C' are points symmetric to M about the straight lines OA, OB, OC. Prove that the point Q, which is symmetric with the orthocentre //' of A A'B'C' with respect to point P describes a circle concentric with (O) and having a radius OQ equal to O Q = ^ O I O I a-Orh O!c, where / is the centre of the circle inscribed in A ABC, and h , h , lc are the centres of the circles escribed in that triangle (Fig. 43). Solution. Take (O) as the unit circle and let zly z2, z3, z0 be the respective affixes of the points A, B, C, M. The equation of the straight line AB is z + Z]Z2z = zx + z2. (162) The equation of the perpendicular dropped from point M to line AB is z — z0 = ZiZ2 (z — z 0)

or z — ZjZ2z = z0 — ZjZ2z0.

(163)

Adding the equations (162) and (163) termwise, we find the affix mz of the projection Af3 of point M on line AB:

246

Problems in Geometry

Similarly, the affix m 2 of the projection M 2 of point M on line AC: m 2 = — (z0 + Z 1 + Z 3 — ZC3 Z0 ) 2 The slope of the straight line M 2 M 3 is m3 — m2 _ z2 — z 3 — ZjZpjzz — z3) m3 — m2 z 2 — z 3 — z Yz3( z 2 — z 3) (z2 - z3)

(z2 — z3)( 1 - Z3Z o) (z2 -

z 3) ( l -

ZZ Z2 Z2 Z0

Z 1Z0)

Zq — Z1 Z\

and so the equation of the diameter of the circle parallel to the Simson line of point M with respect to A ABC is of the form z = - 3- z z0

(164)

ZqZ — gzz = 0.

(165)

or Incidentally, this equation could have been written at once by proceeding from the equation of the Simson line and dropping the absolute term of the equation [problem 3, equation (1)]. The equation of the perpendicular dropped from point H to the straight line (165) is of the form <r3 -x z ~ *i = -------(z — ffi). zo Adding the equations (164) and (166) term by term, we obtain 2

z — a1 =

(166)

<7l<73 9

Zo whence z=

<r2 zo }

This is the affix of the projection of point I f on the straight line (164). The affix p of point P is found from the relation 2

= X

Complex Numbers in Plane Geometry

247

whence p — o-2z0. Furthermore, the equation of line OA is z = z\z or z — z\z — 0. The equation of the straight line passing through point M perpendicular to line OA is z — Z0 = —z\ (z — z„). Solving this equation together with the equation zz = 1 of the unit circle (ABC), we obtain

One of the roots of this equation is naturally z = z0 (the affix of point M) the other is z — a' = zfz 0, which is the affix of point A'. In similar fashion we find the affixes b' and c' of points & and C': b ' — zpo, c' = ziz0, and, hence, the affix b! of the orthocentre H ' of A A'B'C': h ' = (zf + z l + z|)z0. The affix q of point Q, which is symmetric to point H ’ with respect to point P, is found from the relation

q + h' whence q = 2p — h' — 2 <r2z 0 — (z\ + z\ + z§)z0 = z 0 [2<r2 — (zx + z 2 + z3)2 + 2(Z2Z3 + z3zt + zxz2)] = z 0(2<r2—a\+ 2a^ = z 0 (4<r2 - «rf). Thus, q = (4<t2 — a l)z0. From this it follows that if point M with affix z0 describes a circle (ABC), then point Q describes a circle (12), in the opposite direction, that is con-

248

Problems in Geometry

centric with {ABC). The radius p of (£2) is P =

4<t2 —

a l\.

The affixes of the points I, /„, Ib, Jc are: T0 =

— K Z 2 Y Z2 — 1 /Z3 f z l

— ] f Zl ] ^ 2

*0 +

Tl = — ]fz2 Kz3 + Zz3 Kzi + Kzi Vz2 = a0 — b0 — c0, T2

= /z2 KZ3 —Kz3

T3 =

f z 2 f z 3 +

+ )fz1 KZ2 = —Oo + * 0 —Co,

^ Z 3 K Z1 —

f z l Kz 2

— Oo — ^ 0 +

C0 .

From this we have T0TiT2T3 = [eg - (a0 + Z»o)2] [eg - (a0 - 60)2] = (eg — ag — bl — 2a0b0) (eg — al — bl + 2a0b0) = (eg - flg - b l f - Aalbl = (a% + b%- eg)2 - 4agig = (flo + ^o + eg)2 — 4 (fegcg + eg ag + eg&g) = (zjz2 + z2z3 + z3Zi)2 — 4 (zgz^ +

z 2z 2z 3

+ 'z|z3Zj) = af — 4<r1a3

(see also problem 41). And so we get ToTiT2t 3 = al — 4a,a3 — - - - 4y = a\ a\

4° 2

and, thus, | V i T 2 t 3| =

I to T ^ jT a j =

4<t2| =

or P=

| t (,1 | t ,| ;t 2| [ t j |

or OQ = O IO Ia OIb OIc. If the affixes of the points A 9 B9 C are taken in the form Rzl9 Rz29 Rz29 where \z±\ = \z2\ = |z3| = 1, then we obtain OQ =

0 1 'O h '< > h 'O h

Problem 43. The points A l9 A29 AZ9 A4 lie on a single circle (O) = = (A1A2A3A4) which is taken as the unit circle. Denote by ipl9 (p29<p3, ty* the affixes of the Feuerbach points <Pl9 tf>2, ^>3, ^4 that lie on the circles in scribed in the triangles A2A3A49 AXA3A4, A XA2A49 AXA2A3. Prove that the midpoints of the segments Ax<Pl9 A2<P29 A3<P3. A4<P4 lie on one circle. Find the affix of the centre of this circle and its radius.

Complex Numbers in Plane Geometry

249

Solution. Let zl9 z2, z3, z4 be the respective affixes of the points Alf A2, A3, A4. Then the affix <p1 of the Feuerbach point <P4 of the circle inscribed in A A2A3A4 is computed from the formula (see problem 32, item 6°) ZoZ3 | z2z4 ~b z*>z4 (Pi = ----------------- "— Z2 + Z3

and so the affix pL of the midpoint of segment Al<Pl is __ <P i ~ h Z l

Z2Z4 ~t~ Z3~4

z l z 2 4 ~ Z 1Z 3 ~ h Z l Z 4 ~t~ Z 2Z 9

2(z2 + z3 + z4)

2(o1 —zty

where er2 — z i z 2 +

Z 1Z 3 +

Zl Z\ +

Z 2Z Z +

Z ZZ \ +

Z 3Z t*

(Tl = Z4 ~j~ z2 ~h z3 ~h z4We obtain similar expressions for the affixes jx2, /t3, j.i4 of the midpoints of the segments A2<P2>A3<P3, A4<P4: g

Pi ~ -------------- 5 Pz -2 (g 4 —

z 2)

2 {g 1 —

» Pt -z 3)

2 (

—

• z 4)

From this it follows that the points pk (k= 1, 2, 3, 4) lie on the circle ((2), the equation of which is u — ------------------- , 2

— 2z

where z describes the unit circle. Performing an inversion of this circle with the circle of inversion (0), we obtain the circle 1 2 — 2z &x 2 v= — -------= 2 z. u 2 2 2 g

This is the circle (0'), the affix of whose centre is 2 g J g 2 and the radius is equal to 2/|<t2|. The circle (G) may be obtained from (0') by the same inver sion and also by a similarity transformation. From this we can find the affix of the centre of (G) and its radius. But we can also take advantage of the formulas of problem 12 that yield the affix of the centre, bd — ac (o = — — dd — cc and the radius be — ad

Problems in Geometry

250

of (Q) into which the unit circle (O) goes under the linear-fraction trans formation u = — Z tL t cz + d Let us consider the transformation

a d - b e * o.

(167)

(168)

20*! — 2z

Here [compare with (167)], a = 0, b — er2, c = — 2, d = 2a1 and so the affix co of the centre of (£) on which the midpoints of the segments Ai&i, A 2 & 2 9 A3*P39 A4<P4 lie is 20-201 &2&1 co — 2(* 1 *1 1) 4(01*1 ~ 1) and the radius is i ___*2 — 2at 4axCi — 4 ~ ~2 *1*1 ~ 1 Problem 44. Let Bx and Cx be the points of intersection of the bisectors of the interior angles B and C of A ABC with the circle (ABC) = (O). Consider the sum of the directed line segments OBx + OCx — ON. Prove that IN JL BC, where / is the centre of the circle inscribed in A ABC. Also prove that the segment IN is equal to the radius of (ABC) = (O). Solution. Take (O) as the unit circle. Let zl9 z2, z3 be the respective affixes of the points A9B, C. The affixes al9 bl9 cx of the points A l9 Bl9 Cx of intersection of the circle (ABC) and the bisectors of the interior angles of A ABC are given by the relations 01 =

— ] f Z2 V Z3,

*1 =

— V z 3 ) f ZU

C =

— fr ,

where we take for the roots ]fzlt ]Jz2, J[z3 values such that ^ < 1. The affix n of point N is

]fz2 ,

3\<.

n = - ]fz3 ]jzl - /z j )Jz,, and since the affix of point I is equal to T0 = - jfz2 Vz3 - ]}z3 1(7X - fz j )[z2,

it follows that the directed line segment IN is associated with the complex number ][z%]fz3, whence IN = \fz~3 ][z3\ = 1 = R. The slope of the straight line IN is

The slope of line BC is equal to x ’ — — z2z3 and so % + x ’ 0, and therefore the straight lines BC and IN are mutually perpendicular .

251

Complex Numbers in Plane Geometry

Problem 45. The angles of A ABC form a geometric progression with ratio 2. Prove that the midpoints of its sides and of the feet of the altitudes are six vertices of a regular heptagon (Fig. 44). Solution. Take (ABC) for the unit circle and assign to point A (the vertex of the smallest angle of the triangle) an affix of 1. Then, assuming a = cos(27t/7) + i sin(2;r/7), we conclude that 1, a, a2, a3, a4, a5, a6 are the affixes of the vertices of the regular heptagon inscribed in the circle (ABC). Under the conditions of the problem, the

Fig. 44 2k

angles of the triangle are equal to A, 2A and 4A, whence A = ---- ; therefore 7 the affixes of the points B and C are equal, respectively, to a4 and a5. Let Al9 Bly Cx be the respective midpoints of the sides BC, CA, AB, and let al9 bl9 cL be the affixes of these points. Then a4 + a5 a5 + 1 1 +a4 h = The slope of line BC is equal to * bc —

— a4a5 = — a9

— r/2

(since a7 = 1) and, hence, the equation of the straight line passing through point A perpendicularly to BC is z — 1 = ct2(z — 1). The equation of BC is - a1 nr4 = - a2(z - a4). z— Solving the system of equations z — 1 = a2(z — 1) z — a4 — — a2(z — a4), we find 2z — 1 — a4 = — a2 + a2 = — a2 + a5, whence 1 — a2 + a4 + a5 z = a2 = -----------------------,

252

Problems in Geometry

which is the affix of the io o tA 2 of the altitude dropped from vertex A to side BC. Then 1 —a5 xAC ^ = — a5. A 1 - a5 The equation of AC is z — 1 “ — a5(z — 1). The equation of the altitude dropped from vertex B to side AC is z — ola = <x5(z — a4). From the last two equations we find 2z — 1 — a4 = a5 — a, whence , 1 — a + a4 + a5 2 And 1 - a4 AB 1 - a4 The equation of AB is z — 1 — a4(z - 1). The equation of the altitude dropped from vertex C to side AB is z —a5 — a4(z — a5). From the last two equations we find the affix c2 of the projection of point C on line AB: 2 c2 — 1 — a5 -- a4 — a = a4 — a6, whence 1 + a4 + a5 — a6 c2 = • Consider the affix o9 of point Og obtained from point O under the homothetic transformation ^<7,----—j, that is, the affix of the centre 0 9 of the Euler circle of A ABC. Since the affix h of the orthocentre of A ABC is equal to 1 + a4 + a5 and 0 9 is the midpoint of OH, it follows that 1 + a4 + a6 2

Complex Numbers in Plane Geometry

253

We now find

0\ c2

1 — a2 + a4 + a5 2 1 — a + a4 + a5 On --2 a4 + a5 1 + a4 + o9 — 2 2 1 + a4 + a5 — a6 O9 2 1 + a4 + 1 + a4 On --2 2 1 + a5 1 4- a4 + o9 — 2 2 oti

—

1 + a4 + a5 2 1 -f a4 + a5 2 1 a5 T ’ 1 + a4 2 a5 a5 _ T ’ a4 a5 2' + ftC1

a 2’

a° 2

These numbers form a geometric progression with ratio 1/a; and the moduli of all these differences are equal to — . This means that each of the directed 2

line segments C9A2, 0 9B2, C?9A i , O9 C2 , 0 9Ci, 0 9B i ,

beginning with the second, is obtained from the preceding one by a rotation in one and the same direction through an angle of 2n/7, that is, A2, B2r A 1, C2, Clf 2?x are the six successive vertices of a regular heptagon inscribed in the Euler circle of A ABC. Sec. 2. Problems with hints and answers 1. The angles A, B, C of a triangle are connected by the relation C = = 3B = 9A ; AA \ BB \ C C are the altitudes of the triangle, H is its orthocentre, / is the centre of an inscribed circle; r is its radius, (O) = {ABC) is a circle circumscribed about A ABC ; R is its radius, and a>b, c are the lengths of the sides BC, AC, AB. Prove that: 1°. H A' + HB' - H C = R/2. 2°. OP + OH2 - 5R2. 3°. be + ca + ab — R2^ 13. 4°. cos A + cos B f cos C = (1 + V13)/4. 5°. 0/* = J l* (5 -/l3 )/2 . 6°. r - if(]/f3 - 3)/4. 7°. OH2 - * 2(1 - 8 cos ^ cos 5 cos C) - /*2(5 + / 13)/2.

Problems in Geometry

2 54

8°. cos A cos 2? cos C = — (3 + /l3)/16. 9°. sin A sin flsinC = K(13 - 3 ^l3)/128. 10°. sin A + sin B + sin C = /(13 + 3^ 13)/8. Solution. The points A, B, C may be regarded as the respective vertices of a regular 26-gon inscribed in a circle of radius R y since A = nj13, B = 37r/13, C = 9tt/13 (Fig. 45). First note that for any natural number, cos(tt/2/13) = cos((26 — n) tt/13) = — cos((13 — n) tt/13) and, furthermore. IOtt 12tt i . 2tt , 4tt , 6tc , 8tt , 1 + COS-------bCOS--------bCOS----- + cos — - + COS + COS 13 13 13 13 IT ~13~ , 14tc , 16k 18tc , 20n , 227r , 24rc + cos--------b cos--------b cos,------- b cos--------b cos--------b cos---13 13 13 13 13 13 2tt 4k 6k , 87T IOtt ,127c cos---- + C O S -------- b C O S -------b C O S ------ b cos--------bC O S -----13 13 13 13 13 13 =

147T

COS

167T

1Sk

20k

--------bC O S ---------bC O S -------- b COS 13 13 13 13

22k

bC O S -----13 24tc + cos = ~ 1/2. 13

Set 6k 2k 8K + COS X = c o s ----- + cos 13 13 13 —

107T

(169)

1271

y == c o s ----- + cos + COS -----. 13 13 13 —

Then cos A + cos B + cos C = —y. Proceeding from the identities 2 cos2a = 1 + cos 2a, 2 cos a cos p = cos(a + fi) + cos(a *we obtain the equation 4x2 + 2x — 3 = 0, whence - 1 + /1 3 4

Complex Numbers in Plane Geometry

255

and so

This can also be obtained from (169). 1°. ABC is an obtuse-angled triangle (C = 9tt/1 3), hence, HA' = |2J? cos B cos C\ = — 2R cos B cos C — — R[cos(B — C) + cos(fi + C)], HB' = — 2?[cos(C — A) + cos(C + A)], HC' = /?[cos(^ - B) + cos(A + £)]. From this we obtain HA' + HB' - HC' = R ( - x - y ) = Rj2. 2°. OP + OH2 = R2 - 2Rr + 9R2 - a2 - b2 - c2, __ . A . B . C 2Rr — 8R2sin — sin — sin — 2 2 2 = 2R2( - 1 + cos A + cos B + cos C) = 2R2( - 1 - y)T

Problems in Geometry

256

a2 + b2 + c2 = 4 R2(sin2A + sirrfi + sin2C) = 4/?2(3 - cos*A - cos3B - cos2C) = 2R\3 - x). and, thus. OP + OH2 = /{2(6 + 2jc + 2y) = 5/?2. 3°. fee + ca + ab = 4/?2(sin B sin C + sin C sin A + sin A sin B) = 2rt2[cos(fi - C) - cos(fl + C) + cos(C - 4) - cos(C + A) + cos(/4 - B) - cos(/J + B)\ = 2/?2(x - >>) = IP /H . The derivation of the other relations is left to the reader. In deriving them, take advantage of the basic formulas relating to trigonometry and the geometry of any triangle; also take advantage of the specific nature of the given triangle (A = 7t/13, B = 37r/13, C ^ 97r/13). It is also possible to introduce a complex system of coordinates by 26 __ taking {ABC) as the unit circle. Then all the 26 values of ^I will constitute the affixes of the vertices of the regular 26-gon mentioned earlier. These affixes are ak ^ cos-------b i sin----13 13

(k ~ 0, 1, 2, 3 ,..., 23, 24, 25).

We can take point A as the unit point a = a0 — 1. The points B and C will have the affixes f 871 . . 8tt 6n , . . 6n b — ag = cos-------b i sin , c = a 6^cos - ---- b i sm ---13

(compare this with problem 45 above). 2. A quadrangle A BCD is inscribed in a circle (O). Perpendiculars AA2 and BB2 are dropped on CD from points A and B; perpendiculars BBX and CCl are dropped on DA from points B and C; perpendiculars CC2 and DD2 are dropped on AB from points C and D; and, finally, from points D and A we drop perpendiculars DDt and AAX on BC. Prove that (see Fig. 46): 1°. The segments ALA2, B±B2, CXC2, DXD2 are equal and lie on straight lines passing through point F, which is symmetric to point O with respect to the centroid of the system of points A, B, C, D. 2°. The quadrangles AlB1C1D1and A2B2C2D2are similar to the quadrangle ABCD and are inscribed in circles with centre F (I. Langer). Hint. 1°. Take (O) = {ABCD) for the unit circle. Let a ,b ,c , d be the respective affixes of the points A, B,C, D. The affix of point F is f — Oi/2 {ox = a + b + c + d).

Complex Numbers in Plane Geometry

257

F ig . 4 6

The affixes of the projections A1 and A2 of point A on the lines BC and CD are +b + c —

a ,-\{ a + c + d -

whence fli — a2 =

(a — c)(b — d), f — a, = ——{ad + be). 2a

Furthermore, — a2 _

/ —

_ bed

<*l — a* f — <*i a so that the line ALA2 passes through point F. Besides,

and similarly for 2°. We have

AxA2 =--- I ax — a* I = --- ^C* BD 2 CXC2,

—/ ------1— 2a

(ad + be), a2 = f — 2a

- (ab + cd).

From this (and from similar relations for bly b2y cly e2y dly d2) it follows that the quadrangles >11J61C1D1 and A2B2C2D> are inscribed in a circle ----L> with centre F(f). Suppose the quadrangle ABCD is convex. The transfor17 - 81

258

mation

Problems in Geometry

' = / — —- (ad+bc)

carries quadrangle ABCD into the quadr-

1 angle A ^ C ^ . The transformation 2 " = / ------ (ab + cd) 2 carries the 2 ----- > V quadrangle ABCD into the quadrangle A2B2C2D2. And so the quadrangles AXBXC1D1, A2B2C2D2 are convex, have an orientation opposite that of quadrangle ABCD, and all quadrangles ABCD, A^C -^D ^ A 2B2C2D2 are similar. The radii of the circles in which the quadrangles A1B1C1Dl and A2B2C2D2 are inscribed are equal, respectively, to — | ad + be | and 2

— \ab + cd\ 2

or R\coz(AD, BC) | and

R\ cos (AB, CD) | (R = 1).

Indeed, from the relations _ 1 I/ a _L b C1,\ + + cc ----— —— + b0A _L

2 V

a j

we find a i-d ! =

2ad

(a - d) (ad + be)

AXDX= — AD | ad + be |, 2 whence - | ad + bc\ = 2

— y4Z)

(cos 04a £C)|.

Remark. The affix a3 of the projection A3 of point A on BD is 1 {ac + bd). a3 — f~ 2a Hence, — (ax + a2 + a3) 3

- f ----- -,

Complex Numbers in Plane Geometry

259

where a2 = ab + ac + ad + be + b d + cd. From this it follows that if the quadrangle Q (that is, ABCD) is inscribed in the circle (O), then the centroids of the projections of its vertices on the sides of the triangle, formed by the other three vertices of Q, form a quadrangle Q' similar to Q but with opposite orientation. Here, the quadrangle Q' is inscribed in a circle whose centre is symmetric to the point O with respect to the centroid of the quadrangle Q. The radius of this circle is equal to the distance from the centroid of the quadrangle Q to the centroid of the six points at which the circle (O) intersects the straight lines passing through some point of (O) parallel to the sides and diagonals of the quadrangle Q (the solution of Gourmagschieg). 3. Straight lines AP, BP, CP drawn from an arbitrary point P cut the sides BC, C A , AB of A ABC in points A,, B„ C, and divide these sides in the ratios: v. AjC

B,A

C,B

Find the ratios z=

x =

PC, Hint. Apply the Menelaus theorem to A AA,B and to the transver sal CC,. Answer, x = (1 + /I) v, y — (1 + /0 A, z = (1 + v) \i. Complex num bers can also be used. 4. A triangle A,B,CXis inscribed in A ABC (that is, point A, lies on line BC, point B! on line CA and point C, on line AB). 1°. Prove that A ABC and l ^ A ^ C , have a common centroid if and only if the ratios ,

X =

Ta x

---------II =

A1C

Cfit ------------------- V =

AC, ------------------- ----

BXA

C,B

are equal: A = ji = v. 2°. Assuming that A ABC and A A ,B ±C, have a common centroid (A = n — v), prove that A,B,C,

A2 - A + 1 (A + l ) 2

ABC 3°. Suppose A A 2B2C2 is circumscribed about A ABC and has a common centroid with A ABC. Prove that the ratio A', in which the points A, B, C

260

Problems in Geometry

divide the directed line segments B2C2, C2A2, A2B2, is then equal to 1//,, that is, the inverse (in magnitude) ratio in which the points Al9 Bl9 C1 divide BC, CA, AB. Find A1BiC1 A2B2C2 and prove that (ABC)2 = (A & C J (A2B2C2) (R. Deaux). 5. The triangles ABC and A1B1C1 are homothetic under the homothetic transformation with centre S and ratio

We know that A A'B'C' is inscribed in A ABC and circumscribed about A îîQ . Prove that we then have (A'B'C')2 = (ABC) ( A ^ C j) (PilattPs theorem). 6. Given a triangle ABC. Its sides are oriented as follows: BC, CA, AB. Prove that on the axes BC,CA, AB there is only one triplet of points AUBX, Cx such that (BA2) ^ (CBX) = (AC2) and the straight lines AAl9 BBl9 CCX intersect in one point. Also prove that the points Al9 Bx, Ct are the respective interior points of the segments BC, CA, AB. Hint. Use the Ceva theorem. Investigate the resulting cubic equation [(BA,) = (CBj) - (ACj) = x] for x (S. Vatricant). 7. Let I9Ia9Ih9Ic be the centre of the circle (/) inscribed in A ABC, and the centres of the circles (Ia), (Ib), (7C) escribed in the triangle ABC. Let R be the radius of the circle (O) = (ABC), and let A be a point located on segment HHXat a distance of (2/3) HHt from H\ //is the ortho centre of l\ABC , and H1 is the orthocentre of A îîC i, whose vertices are the feet of the altitudes of A ABC. Denote by L the centroid of a system of four points obtained from the points /, Ia, Jb9 Ic by inversion with respect to the circle (O). Prove that the points O, N , L lie on one straight line and that OP OIa•OIb•OIc - 3R2•ON, O N O L = ~ - R 2. 3 8. A triangle ABC is inscribed in a circle (O). Construct right-angled ------ ^ > ------ > triangles AO A', BOB', COC' (in all three triangles, the angle O is equal

Complex Numbers in Plane Geometry

261

Fig. 47

to n/2) with the same orientation. Let the circles (A', A'A), (B',B'B), (C', C'C) cut the sides AB and AC, BC and BA, CA and CB in the points Ba, Ca; Cb,A b; AC,B C. Prove that the triangles ABaCa, BCbAb, CAcBc have the same Euler circle (Fig. 47) (R. Blanchard). Hint. Take (O) as the unit circle. Let a, b, c, a', b \ c' be the affixes of the points A, B,C, A', B', C . Then a! = ia, b '^ib , c'=ic,

AA' = ]f2.

The equation of the circle (A', A'A) is (z — ia) (z + ia) — 2. The equation of AB is z + abz = a + b, whence z + abz = a + b, z = a + b — abz. and equation (170) takes the form (z — ia) (a + b — abz + ia) — 2 or (z — ia) (abz — a — b — ia) + 2 = 0.

(170)

Problems in Geometry

262

One of the roots of this equation is z = a, the other is ba = b + i(a + b). Similarly, the affix ca of point Ca is ca = c + i(a + c). The affix ha of the orthocentre Ha of A ABaCa is found from the relation OHa = OA' + A'Bm+ A'Ca + A 1A. That is, ha = ia + b + i(a + b) — ia + c + i(a + c) — ia -f- a —ia=al( 1+ i)—ai. The affix a9 of the midpoint of HaA', that is, of the centre of the Euler circle of AABaCa, is (7i(l + i) — ai + ai 1+ i (— bg — C9). 9. Given the affixes z1$z2,z 3 of the vertices AyB ,C of is.ABC. Findthe affix z0 of the centre of the circle (ABC). Answer * z0 = Zl J l ^ 3

+ Zg

+ 73

~ *i)

*1 z i 1

z2 z 2 1 Z3 Z 3 1

Hint. (z0 — z{) (z0 — z x) = (z0 — z2) (z 0 — z 2) = (z 0 — z3) (z 0 — z 3). 10. Let Al9 Bl9 Cx be points symmetric to the centre O of (O) = (ABC) circumscribed about IsABC with respect to the sides j5C, CAf AB respec tively. Prove that ISABC and lsA 1B1C1are symmetric about the midpoint 0 9 of segment OH, where H is the orthocentre of ISABC (Hamilton's theorem). Hint. Take (ABC) for the unit circle. Let zx, z2, z3 be the affixes of point A, B, C. Then the affixes of the points Al9 Bl9 Cx will be z2 + z3, z3 + zl9 Zjl + z2. The affixes of the midpoints of segments AAlf BBX and CCX are equal to (zx + z2 + z3)/2 = a j l , and this is the affix of the midpoint of segment OH. 11. Given in the plane a similarity transformation consisting of a rotation about a point O through angle a ^ nn (n an integer) and of homothetic transformations with centre O and ratio k. Under this similarity trans formation, let IsABC go into IsA'B'C'. Denote by Ax, Bl9 Cx the respective points of intersection of the straight lines BC and B C , CA and C A ', AB and A'B'. What must the position of O be with respect to A ABC ------- >

-------->-

so that A ABC and A A 1B1C1 are similar and have the same orientation?

Complex Numbers in Plane Geometry

263

Hint. Let a ,b ,c be the affixes of the points A ,B ,C in an arbitrary system of coordinates with origin at point O. Setting p = &(cos a + + i sin a), we find the affixes a' = pa, b' = /?&, c' = pc of points A \ B', C'. From the equations of the lines BC and B'C', (b — c) z — (b — c)~z + b~c — be = 0, p(b — c) z —p(b — c) ~z + pp(b>c —

— 0,

we find the affix z = ax of point Ax be be p(p — 1) a1 = b —c p —p and similar expressions for b1 and cx: ^

ca — ca p(p — 1) ~c — a p ~ P

ab — ab p(p —1) a —b p —P

The oriented triangles ABC and AXBiQ are similar and have the same orientation if and only if a at 1 b bx 1 = 0 . c 1 Substituting into this equation the expressions for al9 bl9 cx in terms of a, b9c, p and simplifying, we obtain aa(b — c) + bb(c — a) + cc(a — b) = 0, and this means (see the preceding problem) that O is the centre of the circle (ABC). Remark. If we now take (O) = (ABC) for the unit circle, then p (p -

ax — bx

p - p

(b + c),

P(P - - (c + a). P- P (a + b). P- P

and, hence,

is the image of A A2B2C2 (which “ supplements”

A ABC) under the similarity transformation with centre O, angle of rotation a rg fc ^

Problems in Geometry

264

and proportionality factor ][k2 — 2k cos a + 1 2 P(P ~ 1) | sin a | P- P where k = |p (R. Blanchard). 12. Prove that if in a triangle ABC the median ma issuing from angle A is equal to the length la of the bisector of the same angle A, ma = la, then A ABC is an isosceles triangle and 6 = c. Solve the problem analytically (by computation) and geometrically. 13. A1A2A9A4ASA6A7 is a regular convex heptagon. Prove that the circle circumscribed about the triangle formed by the lines AXA2, A3A5, A4A7 passes through the vertex A6 (V. Thebault). 14. Prove that if the angles of A ABC are 4n n A= C= 7 79 and if a, b, c denote the lengths of the sides BC, CA, AB, then the following relations hold true: 1°. a2 + 62 + c 2 = 7R2. 2°. OH = R)[2 (see problem 11). 3°. The angles of A A'B'C', whose vertices are the feet A \ B \ C’ of the altitudes of the given triangle, are A

= A.

4°. cos A = — , cos B = —, cos C = ----—= ----- —. 2a 2b 2b 2c 5°. be = ^(6 + c)y be = c2 — a2, ac = b2 — a2 and, conversely, if the first two relations hold, then /I = n il, B = 2njl, C = 4tt/7. 6°. cos ^4 cos 5 cos C = —1/8, sin /f sin 5 sin C = 1/7/8. 36 a . 3c + 6 • 2 ^^ = 3-------0 —c sm2 sin2 5 = ------, sin2 C — ----4a 46 4c 8°. sin2 A sin2 B sin2 C + cos2 A cos2 B cos2 C — 1/8. 9°. cos2 A + cos2 B + cos2 C = 5/4. 10.° sin2 A + sin2 B + sin2 C = 7/4. 11°. cos A + cos B + cos C = - — ----- ~ ~----a 2 2(c —a) 12°. cos2 B cos2 C + cos2 C cos2 A + cos2 A cos2 B = 3/8. 13°. tan A tan B tan C = — Yl. 14°. ha = hb + h c. 15°. hi + h\ + hi = (a2 + 62 + c2)/2. 16°. BA’ A’C =

CB’- B 'A = ab-, AB' C'B 4

Complex Numbers in Plane Geometry

265

17°. If Qa, Qb, Qc are the feet of the bisectors of the interior angles of A ABC, and Q'a, Qb, Q'c are the feet of the bisectors of the exterior angles of the triangle, then AQa 1 QO

/~%2

b2 a2 be I i

. ^L \/a T

D /^ 2

BQb i

r * r \ 2 ___

be a+ c

— a, CQc =

ac = b — a. a+ b

•

19°. BQb CQc = (c — a) (b — a) — a \ 20°. AQ'a = b + c. 21°. BQ'b2 = a2 - a b + b2. 22°. CQ'2 = b2 + be + c2. 23°. AQ'2 + BQ'b2 + CQ'2 = 4b2 + 4c2 - 2a2 = 8m2a. 24°. The radii of circles of Apollonius are equal to the sides of A ABC (the circles of Apollonius of the triangle are (QaQ’a)> (QbQb)> (QcQ'c) cons tructed on the segments QaQ'a, QbQ'b>QCQC as diameters). Let A tBtCt — Tt be the triangle formed by tangents at points A, B ,C to the circle (ABC) circumscribed about A ABC = T. Prove that: ------> ------> 25°. The triangles ABC and A tBtCt are similar and have opposite orien tations; the proportionality factor is equal to 1/2. 26°. The altitudes A A \ BB', C C of A ABC bisect the sides B tCt, CtA t, A tBt of AT,. 27°. The Euler circle of AT, passes through the orthocentre of AT. ------> *-----^ 28°. The triangles IIbIc and ABC are similar, have the same orientation, and the proportionality factor is equal to 1/2. 15. Given a triangle ABC, G is the point of intersection of the medians. Let Au Bl9 Cx be the images of the points A, B ,C under the homothetic transformation (G, — 2). Let BCA', CAR, A B C

(171)

be equilateral triangles having the same orientation; let B C T ',

CAB'', A B C '

(172)

also be equilateral triangles having the same orientation; note that the orientation of any one of the triangles (171) is opposite that of any one of the triangles (172). Prove that the triangles B 'C A l9

C A 'B 19 A'B'Cx

(173)

are equilateral and have the same orientation as that of the triangles (172), and the triangles B " C " \, C"A"Bl9 A"B"C x (174) are equilateral and have the same orientation as that of the triangles (171) (Eig. 48).

266

Problems in Geometry

3 _____

Hint. If a is any one of the imaginary values of Y —I, then the affixes of the points A \ B \ C', A ", B '\ C" are expressed in terms of the affixes of the points A, B ,C by the relations af = etc + a by b' = cca + ac, c' = CLb + a a; a" = a c + a6, b" — aia + ac, c" = a6 + acr. The affixes of the third vertices of the equilateral triangles constructed ------>on the segments B’C' and i?"C" and with orientation the same as ABC A" ------> and ABC A', respectively, are ac' + cub' = b + c — a ac" + ab" = b + c — a and, hence, these third vertices coincide with the point A1 (R. Deaux). 16. Prove that if the angles A, B ,C in A ABC are connected by the relation sin A = cos I? tan C,

Complex Numbers in Plane Geometry

267

then the altitude A A \ the median BB', and the bisector CC' intersect in one point. Hint. The angles B and C are acute. Hence, the foot A' of the altitude AA' is an interior point of segment BC and BA' A'C

c cos B b cos C

From there on use Ceva’s theorem. Supplementary question. What theorem results from what has been proved if we apply an isogonal transformation to point M in which the altitude A A \ the median BB', and the bisector CC' intersect? (after Morley). Hint. Since, under an isogonal transformation, the centre O of {ABC) goes into the orthocentre H of Is ABC (and conversely), we get the following theorem: if angles A, B, C in Is ABC are connected by the relation sin A — = cosf?tanC, then the diameter AO of the circle (O) = {ABC), the bisector CC', and the cimedian BB* (the straight line symmetric to the median BB' with respect to the bisector of the interior angle B) intersect in one point. 17. Inscribed in the unit circle is a triangle ABC, the affixes of whose vertices are respectively equal to zl9 z2, z3. Given a point P whose affix is azj + pz2 + yz2 p = --------------------------------- , a + P+ y where a, p, y are real numbers. Let Al9 Bl9 Cx be points in which the straight lines AP, BP, CP intersect the sides BC, CA, AB respectively. Find (M C i) {ABC) Hint. The affixes of the points Al9 Bl9 Cx are fiz2 + yz3 t yzs + a z j a z x + 0 z2 a, — ---------------, ol , cx = -------------- . P + y y + « <* + P Answer.-----------= -------------------------------. {ABC) {P + y){y + * ) { x + P ) 18. Prove that a necessary and sufficient condition for the orthogonal projections Bx and Cx of the vertices B and C of A ABC on the sides AC and AB respectively to lie on the same straight line as the centroid G of A ABC is the equality 3a2 = b2 + c2 or a2 = b2 + c2, where a, b9c arc the lengths of the sides BC9CA9AB of A ABC. _ Hint. Take {ABC) as the unit circle. Prove the equations z2z3 + z3z2= 2 — a2 and so forth, where zl9 z2, z3 are the affixes of the vertices of the triangle, and a9b9c are the lengths of its sides.

268

Problems in Geometry

19. Let zl9 z 2, z 3 be the affixes of the vertices of A ABC inscribed in the unit circle. Prove that the equation z\ = z 2z 3 is a necessary and sufficient condition for A ABC to be an isosceles triangle: AC — AB. 20. ABC is a right-angled triangle (C = n/2); (/) is a circle inscribed in the triangle. Let Al9 Bl9 Cx be the orthocentres of the triangles IBC, ICA, IAB. Prove that the length of the projection A$B$ of segment A1Bl on the hypotenuse AB is equal to the diameter of (/). The length of the projection B*C* of segment BXCX on leg BC is equal to the diameter of the circle (Ih) escribed in angle B, and the length A£C?, the projection of segment AXCXon leg AC, is equal to the diameter of the circle (Ia) escribed in angle A. Hint. Take (/) for the unit circle. Let P9Q, R be the points of tangency of (/) with the sides BC, CA, AB, and let their affixes be 1, /, a respectively. Then the affixes of the points A, B ,C are 2 2a/ , 2 2a a = z------= -------- , b - ------== -------- , a + i a + / 1+ a 1+ a 2 c= - 1+ / 1+ 7 since the points A 9B9C are images, under inversion with respect to circle (/), of the points A \ B \ C (the points of intersection of the straight lines IA y IB, IC with lines RQ, PR, PQ, respectively) (Fig. 49). Now, the affix ax of the orthocentre Ax of A IBC is ax = —----------and so forth the 1+a calculations are rather unwieldy). 21. ABC is a right-angled triangle (C = n/2). The circle {ABC) ={0) is taken as the unit circle. The point M(z0) describes this circle. Let Q

Complex Numbers in Plane Geometry

269

be the midpoint of segment OM and let / be the Simson line for point M, which line is constructed with respect to A ABC. Find the locus of points P symmetric to Q with respect to /. Answer. The segment A'B' obtained by a parallel translation of diameter ABy the translation being defined by the directed line segment

(Fig. 50). 22. Given in the plane two distinct points A(zj) and B(z2). Find the affix p' P of point P' which is symmetric to point P{p) about AB. Hint. The affix /?' is found from the 5 Pt equation (Fig. 51) ! zi z\ 1 = 0, where pL = (p + />')/2. Z2 Z 2 P' \ P i P i 1| Answer. pr — (z, ~z2 — z2 z\ + p(z2 — Fig. 51 - z i))l(z 2 - z i)23. Two distinct points ^(zj) and B(z2) are given in a plane. Prove that if the point M(X) describes the unit circle (with the exception of a point with affix X = — 1), then the point M 'l

| describes the mid-

V 1 + A )

perpendicular of segment AB. 24. Let Dy Ey F be the points of tangency of the circle (7) inscribed in A ABC with sides BCy CA, AB. Denote by Al9 Bl9 Cx the midpoints of the medians DA0, EB09 FC0 of the triangles ADI, BEI, CFI emanating from the vertices Z>, E, F (A0, B0, C0 are the respective midpoints of the segments AI, BI, Cl). Let H be the orthocentre of A DEF. Prove that (Fig. 52) IA i =

III IB ,= 4 sin(/4/2) f

IH 4 sin(^/2)

1C, - -

IH 4sin(C/2) *

Hint. Take (DEF) as the unit circle; denote by z,9z29z%the affixes of the points D, E9F. Then IA ,= and so on.

k il 2\z2 + Zz\

IA =

___ 2 ^ _ Iz2 + z s \

k? I = k i I

270

Problems in Geometry

25. ABC is an arbitrary triangle; (ABC) â&#x20AC;&#x201D; (O) is the circle circumÂ scribed about it; Q is an arbitrary point lying on that circle. Through point A draw a straight line parallel to line OQ (Fig. 53)4 Let A' be the

Fig. 53

Complex Numbers in Plane Geometry

271

second point of intersection of this line with the circle (ABC), and let P be the point symmetric to point A' about the diameter of (O) parallel to BC. Prove that: 1°. A similar construction carried out for the points B and C leads to the same point P. 2°. The Simson line of point P with respect to A ABC is parallel to the line OP. 26. Let A1,A 2,A 3,A 4 be four straight lines lying in the same plane; none are parallel and no three pass through one point. Any three of them, say Ax, A2, A2, form a triangle. Suppose Ci2 3 is a circle circumscribed about this triangle. Construct similar circles C124i, C134, ^ 234- Prove that: 1°. The centres of the circles C234, C134, C124, C123 iie on one and the same circle C1234. 2 . The circles C234, C134, Ci24, C123, CA34 pass through the same point y4 (Fig. 54). To the lines Ax, A2, Az, A4 add a fifth straight line A5, but do this so that no three lines of the five belong to a single pencil. Eliminating one of these lines, say A5, we construct circle C1234 and, simi larly, eliminating lines Ax, A2, A3, A4 in turn, we construct the circles C2345, ^1345? Cl245> Cl235 (Fig. 55). Prove that: 3°. The centres of these five circles lie on one and the same circle C12345. 4°. The circles C2345, C1345, C1245, C1235, C1234 pass through the same point yB, which, however, does not generally lie on the circle C12345. Prove similar statements for any number of straight lines Ax, A2, . . ., An. Hint. Here, in brief form, is the solution to this problem given by a 9th form student of a Moscow secondary school. Let xl9 x2, . . . , xn be the affixes of points symmetric to the point O (an arbitrary point taken as the origin) with respect to the straight lines Ax, A2, . .., An. Set tj = — Xj/xj. Consider the expression a m==___________ xi A_____________ |___________ X2J2__________

(A h) (h

A)* •• (A

(A h) (h A)- • • (A A)

From this relation it follows that and

t^n.i “f"

i+1

A+l @n+l,i

Qn,i= ( 1)” &n art.n-i-l> where an = txt2. .. tn. The equation of the straight line A} is X = fA X ~ Xj).

(175) (176)

Problems in Geometry

272

Z2 —

— # 2 .1 — # 3 .2

^ 3 # 3 |1

h ~h h~ h and, similarly, the affixes of the points of intersection of the lines Al9 A3 and A29 J 3are a3t2 — f2#3.i and #3,2 — fi# 3,i- From this it follows that the points of intersection of the lines Al9 A2, A3 taken in pairs lie on the circle x = a3t2 — ^#3.i (|f| = 1), the affix of whose centre is a3 2 and the radius is \a3 i|. This circle is said to be a circle associated with three lines Al9 A29 A3, and its centre is a point associated with these three straight lines. Also consider the straight line A4. From the relation a3 2 = a4 3 — t4a4t2 it follows that the affix a3t2 of the centre of the circle associated with the lines Al9 A2, A3 may be taken in the form a4t3 — t4a4y2 and, similarly, the affixes

Complex Numbers in Plane Geometry

273

of the centres of the three circles associated with triplets of straight lines (Al9 A2, d4), (Al9 A3, ^ 4,) (A2, 2I3, 2I4) will be tf4f3 t3a ^2, 04,3 ^*04,2> ^ 4,3 ~ ^ 4,2* From this it follows that the centres of these four circles (circumscribed about the triangles formed by the straight lines (Al9 A2, A3), (Al9A29A4)9 (Al9A39A4)9 (A29 A39 Aa)) lie on the circle x =~- a4%3 — ta4t2, the affix of whose centre is a4 3 and the radius is equal to |a4f2|. We will call this circle the circle associated with four straight lines Al9 A29 A39 A4 and its centre the point associated with these lines. Continuing this reasoning, we arrive at the following theorem: if n straight lines Al9 A29. .., A„ are given in a plane, then, by eliminating one line at a time from this set An9. . . , A2, Al9 we obtain n groups of lines with n — 1 lines in each group; the centres of the n circles associated with these n groups belong to one circle r „: X=

„-i - ta„' „_2

(|f| = 1),

(177)

the affix of whose centre is an, n_i and the radius is |a„t n_2\. We call this circle r n the circle associated with n straight lines Al9 A2, . . ., An9 and its centre is called the point associated with these lines. The equation of the circle r„_x associated with the n — 1 lines Al9 A29. A„_x is of the form X = an_1, „_2 - ta„ „ _ 3 (|r| = 1). (178)

..,

The complex number gn,n-3\ an, n- 2/ has a modulus equal to 1 [this follows from (176)]. If this number is substi tuted into (178) instead of /, we obtain tn

u n ,2 n -1

n -

a n t1

which is an expression symmetric in the indices 1 , 2 (this is because the first position is always occupied by n). From this it follows that the n circles associated with n systems of lines having n — 1 lines in each system (these systems are obtained by a successive elimination of one of the n lines from the system Al9 A29. .., An) pass through a single point with affix yn. Remark. For n = 4, the modulus of the complex number a n, 2 __ fl4 ,2 0/1.1

0 4 ,1

is equal to 1, hence the circle T4, X =

<74j3

f 0 4>25

associated with the four lines Al9 A2, d3, d 4 [that is, containing the centres of the four circles circumscribed about the triangles (Al9 A29 A3), (Al9 A2t d 4), 1H

810

274

Problems in Geometry

(Al9 A39 A4)9 (A29 A39 A4)] also passes through the point with affix y4 â&#x20AC;&#x201D;

tf4 ,3

**4,2

**4,2â&#x20AC;&#x2DC;

**4,1

A similar situation does not hold when n > 4, because, for example, the modulus of the complex number C^ ~ is no longer (generally speaking), **5,1

equal to 1. 27. AXA2A3A4 is an arbitrary quadrangle (not necessarily convex). --------> --------> --------- > -------- > Squares AXA2PXP29 A2A3P3P49 A3A4P$P39 A4AxPnP3 are constructed on its sides; the squares have the same orientation; also constructed on its sides are the four squares AXA2P\P29 A2A3P3P49 A3A4P3P39 A4AxP^Pg9 which also have the same orientation, but the orientation of any one of the squares of the first group is opposite that of any one of the squares of the second group. Let Bl9 B29B39 B4 be the centres of the squares of the first group, and let B[9B29B3f B4 be the centres of the squares of the second group. Prove that:

Complex Numbers in Plane Geometry

275

1°. Segments BXB3 and B2B4 are equal and mutually perpendicular (the quadrangle B1B2B3B4 that satisfies this condition is termed a pseudo square). 2°. B[B2B3B4 is also a pseudosquare. 3d. The segments B4B3 and B2B\ have the same midpoint C\; the seg ments B[B3 and B2B4 have the same midpoint C2. Let C3 and C4 be the respective midpoints of the segments AXA3 and A2A4. Prove that CXC2C3C4 is a square (Fig. 56). [Solve this problem by using complex numbers; also solve it by the methods of analytic geometry and the methods of vector algebra.] 28. A1A2A3A4A5A(iA7A8 is an arbitrary octagon (not necessarily convex). On its sides, construct the squares A,A2PXP29 A2A3P3P4, A3A4P5P6, A4A5P7Ps, ^ 5^ 6^VPio> d6/47Pi1PJ2, A7AsP13P14, A g A ^^P ^ so that all of them have the same orientation, and also construct another eight squares A1A2PlP2, A2A3P3P4, A3A4PbP69 A4A$P7P8, A5AeP9P 105 A6A7Pn P 12, A7A8P13Pl4, AgA^lsP^, also having the same orientation and such that the orientation of any one of the squares of the first group is opposite that of any one of the squares of the second group. Let Bl9 B2, B39 B49 B59B69 B7, Bg be the respective centres of the squares of the first group, and B[9B29 B3, B4, B'h, B B 7, Bg the respective centres of the squares of the second group. Prove that: 1°. The midpoints Cl9 C2, C3, C4 of the principal diagonals * of the octagon BxB2B3B4BbB8B7Bg form a pseudosquare. 2°. The midpoints C[9C2, C3, C4 of the principal diagonals of the octagon B[B2B3B4B$BQBiBg also form a pseudosquare. 3°. The sets of points A 1, A2, A3, A 4, A5, A6, A7, A8, Bi, B2, B3, B49 i?5, B69 B79 Bg, Ci» Q? c 3, C4, c;, a , g , c; have a common centroid. 4°. The midpoints of the segments CXC3 and C2C4 coincide respectively with the midpoints of the segments C2C4 and C[C3. * The principal diagonals o f the octagon B ^ B g B ^ B ^ B jB 8 are the segments BxBh9 B2B6, B3B~, B4B6 that join opposite vertices (and also the straight lines on which the segments lie).

Problems in Geometry

276

5°. A necessary and sufficient condition that C XC 2C 3C 4 and, hence, also C [ C 2C 3C 4 be squares (with the indicated order of their vertices) is that the two sets of points Ay, A 3, A-0, A 7 and A 2, A 4, A 0, A s have a common centroid. 6°. The squares C 4C 2C 3C 4 and C [ C 2C 3C 4 coincide ( C x — C [ , C 2 — C 2, C 3 — C3, C4 = C4) if and only if the two sequences of the principal diagonals of the octagon A 1A 2A 3A 4A 5A 6A 7A a are bisected by their points of intersection. 7°. If the points Cl9 C3, C 2, C 4 and the points C [ 9 C3, C2, C 4 coincide, then the quadrangles A 1A 3A 5A 7 and A 2A 4A 6A S are parallelograms, and conversely. Solve this problem through the use of (1) complex numbers, (2) vector algebra, and (3) analytic geometry. 29. Inscribed in a circle (O) is a triangle ABC. Prove that the points P, Q, R of intersection of the tangent lines, drawn at the points A, B,C to (0), with the straight lines BC, CA, AB, respectively, lie on one straight line (Fig. 57). 30. Through points A = (zx), B = (z2), C — (z3) lying on the unit circle draw straight lines parallel to a given straight line A , which has slope X. Let A y , By, Cy be the second points of intersection of these lines with the circle ( A B C ) ; A 2, B 2, C 2 are points symmetric to the points A lHB y , C y with respect to the straight lines B C , C A , A B respectively. Draw through points A 2, B 2, C 2 straight lines parallel to the line A , and denote the points of intersection of these lines with the lines B C , C A , A B respectively by P, Q, R. Prove that the points P, Q, R lie on one straight line, and set up the equation of that line (Fig. 58). Answer. X(2X3 + ax )? — 1) z + X2(2 f- cr1X — ) 3) z =

26 +

0 2 X5 — Gy X 1 F (1

— Oy (J>) X3 —

G 2 )}

F GyX +

31. Four points A y , A 2, A 3, A 4 are taken on a circle (O). Let P be the point of intersection of the lines A 4A 3 and A 2A 4 ; let Q be the point of inter section of the lines A 4A 2 and ^ 3^ 4; let R be the point of intersection of the tangents to (O ) at the points A y and A 4 ; let S be the point of intersection of the tangents to the circle (O) at the points A 2 and A 3. Prove that the points P, Q, R, S lie on one straight line (Fig. 59). Hint. Take (O) as the unit circle. 32. A triangle ABC is inscribed in a circle (O). Tangent lines are drawn to (O) at the points A, B,C; these lines form (called the tangential triangle of tsABC). Prove that: 1°. The points P, Q, R of intersection of the straight lines: P == (BC, BqCq) Q = (CA, C0A0), R = (AB, A0B0) lie on one straight line m. 2°. The lines AA0, BB0, CCQpass through the single point M.

Complex Numbers in Plane Geometry

277

3°. The point M is the pole of the straight line m with regard to the circle (O) (or, what is the same, m is the polar of point M with respect to this circle) (Fig. 60). Hint. Take (O) for the unit circle. 33. AXA2A3 is an arbitrary triangle lying in an oriented plane; P is an arbitrary point lying in the plane of that triangle but not lying on any one of its sides, nor on the circle {A4A2A3). Let S l9 B2, B3 be the orthogonal projections of point P on the lines A2As, A3Al9 A1A2; Cl9 C2, C3 are the orthogonal projections of point P on the straight lines B2B39 B3Bl9 BXB29 and Dl9 D2, D3 are the orthogonal projections of point P on the straight lines C2C3, C3Cl9 CLC2 respectively. Prove that A A tA2A3 and A D 1D2D3 are similar and have the same orientation (Fig. 61). Hint. Take point P for the coordinate origin. Let ak9 bk, ck9 dk (k = == 1,2, 3) be the respective affixes of the points Akf Bk, Ck9 Dk. Express bk in terms of ak; ck in terms of bk; dk in terms of ck. 34. On the circle (O) take six arbitrary points Al9 A2, A39 A4, A59 A6. Prove that the three points P, Qy R , P — (A1A2f A4A5), Q — (A2A3, A5A6), R = (A3A4j A6A4),

Problems in Geometry

278

the p oints of intersection of the straight lines, lie on one straight line (Pascal's theorem) (Fig. 62). Hint. Take (O) for the unit circle. 35. Prove that the points A(z1) and B(z2), where zx and z 2 are the roots of the equation z2 + 2pz + q = 0 (pand q are complex numbers and p2— — q 7^ 0), lie on the straight line passing through the origin of coordinates if and only if one of the following conditions is valid: (1) P = 0; (2) p t^O, qjp2 is a real number and (qjp2) < 1; here, if p # 0 , 0 ^ qjp2 < 1, then the points A and B lie on one ray emanating from the coordinate origin 0 \ and if p ¥= 0, qjp2 < 0, then the points A and B lie on opposite rays emanating from the origin. 36. Given a cubic equation: a0z 3 + 3ax z2 + 3a2z + a3 = 0 . Let z1? z2, z3 be its roots. Prove that a necessary and sufficient condition that the points A(zx), B(z2), C(z3) be collinear is as follows: either ( 1) 2a\ — 3a0 axa2 + eft az = 0 or (2) -------(Pojh---- <h)-------- which is a real number less th a n —1/4. (2a\ — 3a0ax a2 + dba3)2 Hint. Under the translation transformation z= C -^ ao (Tschirnlraus transformation), the given equation takes the form C3 + 3/?C + q = 0 , where _

a0 a2 — a\

_ 2a\ — 3a0 ax a2 + a%a3

The sum of the roots t l9 C2, C3 of the last equation is zero, and since (Ci + C2 + C3)/3 = 0 is the affix of the centroid of the system of points A'(Ci), B'(Q , C'(C3), it follows that this centroid coincides with the coordi nate origin. If the points A, B9C are collinear, then the points A \ B'y C'

Complex Numbers in Plane Geometry

279

are also collinear (and conversely). If the points A', B \ C' are collinear, then, by the foregoing, they lie on a straight line that passes through the origin. 37. Let a general Cartesian system of coordinates Oxy be introduced in the plane. The affine transformation of the plane is a correspondence under which the coordinates x', y' of the image M'(x', y') of a point M(x, y) are expressed in terms of the coordinates *, y of the preimage M (x9y) of the point M '(x', y') by the linear relations x' = axx + hxy + cl9 y' = a2x + b2y + c2, where al9 bly cl9 a29 b2, c2 are arbitrary real numbers and axb2 — a2b1 ^0. Prove that: 1°. Every affine transformation may be written as z' = otz + pz + y, where z and z' are the respective affixes of the points M and M', and |a| ^ \fi\. Conversely, if |a| ^ \P\9 then the relation z9 = ccz + p~z + y defines a certain affine transformation. 2°. There exists an affine transformation (and only one) which carries any three noncollinear points A(zx), B(z2), C(z3) into the three noncollinear points A'(zi), B '(z 2), C'(z'3), respectively. 3°. A triangle ABC is said to be metaparallel to A A'B'& if the straight lines that pass through the points A9B, C and are parallel respectively to the lines 2?'C', C 'A \ A'B' intersect in one point. Let z9 = <xz + + P'z + y be an affine transformation under which A ABC goes into AA'B'C'. Prove that A ABC is metaparallel to A A'B’C if and only if: (1) a is a pure imaginary number or, what is the same, *1

is a real number. 4°. Prove that the concept of metaparallelism is symmetric, but not reflexive and not transitive. -----> 5°. Prove that if A ABC is metaparallel to two of the three triangles ------> ------> ------^ A'B’C , B'C 'A\ C'A'B\ then it is metaparallel to the third one as well -----> ------> (in this case we say that A ABC and AA'B'C' are three-times metaparallel). 6°. Given a triangle ABC in a plane: A = (zj), B = (z2), C = (z3).

280

Problems in Geometry

Find the affix z3 of point C' if we know that A ABC is three-times meta------> parallel to A A'B'C' with vertices A ’ = (0), = C' — (zg). 7°. A triangle ABC is said to be orthologic to A A'B'C' if the straight lines that pass through the points A, B, C and are respectively perpendicular to the straight lines B'C', C'A', A'B' intersect in one point. Prove that -----> ------> AABC is orthologic to A A'B'C' if and only if (1) a is a real number, or, what is the same, Z1 Z1 Z2 Z 2 *3 z*

( 2)

1 1 1

is a pure imaginary number. 8°. Prove that the concept of orthologicality is reflexive and symmetric but is not transitive. 9°. Prove that if A ABC is orthologic to two of the triangles A'B’C', -----> ------> B'C'A', C'A'B', then it is orthologic also to the third triangle (in this case we say that AABC and A A’B'C' are three-times orthologic). Zl - Z i z2 1 ^2 — z2 z3 1 Answer. 6°.

— Z3 Z i

4 =

Zi ZX 1 z2 z 2 1 z3 z3 1

Chapter IV

INVERSION Sec. 1. Inversion defined. Properties of inversion Let us adjoin to the set of all points of the Euclidean plane a single ele ment which we will call the ideal point or the point at infinity of the plane n. Let us agree that any straight line of the rc-plane passes through the point at infinity and that this point does not belong to any finite figure. The Euclidean plane supplemented by a single point at infinity (with the indi cated agreements) is termed a Euclidean circular plane or, simply, a circular plane. Straight lines lying in the circular plane will sometimes be called circles of infinite radius. We will also regard points as circles; such “circles” will be called circles of zero radius or zero circles. Two inter secting straight lines have two common points, one proper point and the other the point at infinity. Two parallel lines have only one point in com mon: the point at infinity; we will say that two parallel straight lines or two circles of infinite radius meet at the point at infinity. Suppose 0 is a fixed proper point of a circular plane, and k is a fixed real number not zero. An inversion [0, k] with pole 0 and power k of the rc-plane is a one-to-one transformation of that plane under which each proper point M of the rc-plane distinct from point 0 is associated with a proper point M f lying on the straight line OM and such that O M O M ' = (OM) (<OM') = k. With the pole O of inversion we associate the point at infinity 0 ', and with the point O' we associate the point O. Every inversion I is an involutory transformation, that is, / 2 ^ E, where E is the identical transformation. This follows from the fact that if under inversion / a point M ' is the image of a point M, then M is the image of M'. If the power k of the inversion [0, k] is positive, then the circle K with centre O and radius ]fk is termed the circle o f inversion, and the inversion itself is also called a symmetry with respect to the circle K. Under the inversion [0, &], where k > 0, each point M of the circle of inversion is invariant, that is, its image M' coincides with the point itself. To construct the image M' of point M lying outside the circle K of inversion [0, &], where k > 0, we draw from point M a tangent M T to the circle K (T is the point of tangency); M' is the projection of T on the straight line OM (two tangents may be drawn, M T and M T , the

282

Problems in Geometry

point M' is the point of intersection of straight lines OM and TT') (Fig. 63). To construct the image M ' of point M lying inside the circle K of inversion [O, k\9 k > 0, draw through M a straight line perpendicular to the straight line OM. Let T be any one of the points of intersection of this perpendicular with the circle K. Then the tangent to the circle K at the point T will intersect the straight line OM in the point M ' (Fig. 64). Thus, under the positive inversion [Oyk\9 k > 0, points lying outside the circle of inversion K pass into interior points of that circle, and points lying inside the circle K pass into points lying outside the circle. For negative inversion [O, fc], k < 0, the circle K with centre O and radius ^\k\ is invariant; however, each one of its points is noninvarian and passes into a point diametrically opposite it. To construct the image M of a point M under negative inversion [O, k]9 k < 0, we construct the image M* of point M under the negative inversion [0, |&j], M ' is symmeÂ tric to point A/* with respect to point O. From this it follows that under negative inversion as well the set of all points lying outside the circle K passes into the set of interior points of circle K and the set of all interior points of circle K goes into the set of all exterior points of K (Fig. 65 and Fig. 66). Let [0, k\ be an inversion with pole O and power k. If C is an arbitrary circle that does not pass through the pole of inversion, then its image C' under the inversion [0, k\ is again a circle that does not pass through the pole of inversion. Proof. Let M be an arbitrary point of the circle C, and let M ' be its image under the inversion [0, k]9 that is, OM- OM' = k; here it is well to point out that there is a frequently used notation OM -OM ' = &, where we speak of the product of the lengths (OM) and (OM') of directed line segments. Since the straight line OM has a common point M with the circle C, this line has a second common point N (possibly M = N) with the circle C. The product OM- ON â&#x20AC;&#x201D; OM- ON = a is the power of the pole O of inversion with respect to the circle C (Fig. 67). From

Inversion

283

the last two equations it follows that

Thus, the point M ' is the image of point N of the circle C under the homothetic transformation (0, kja). But under the homothetic transforÂ mation (0, k ja \ circle C goes into circle C'. If point M describes a circle C, then the point N too describes the same circle C, and, hence, its image M ' under the homothetic transformation (0, kja) describes the circle C . Thus, the circle C is the image of C also under the homoÂ thetic transformation (0, kja), where a is the power of the pole 0 of inversion with respect to the circle C* and under the inversion [0, k]. * Note that the inversion [0 , ÂŁ ] and the homothetic transformation (0 , fc/rr) carry circle C into one and the same circle C'; however, the points lying on circle C are transformed by the inversion [0 , k ] and the homothetic transformation (0 , kja) into different points of circle C u n d e r the inversion [0, k], point M goes into point M ' of circle C', and under the homothetic transformation (0 , kfo), point M goes into point N 'y where N' is the second point o f intersection of straight line O M ' and circle C'.

284

Problems in Geometry

If circle C passes through the pole of inversion 0 , then its image under the inversion [0, A] is a straight line that does not pass through the pole of inversion and is perpendicular to the straight line joining the pole of inversion and the centre of circle C. If the straight line does not pass through the pole of inversion 0 , then its image under the inversion [0, A] is a circle that passes through the pole of inversion. If the straight line passes through the pole of inversion 0 , then under the inversion [0, A] it passes into itself. Under inversion, the tangency of the circles is retained. This follows from the fact that under inversion, a circle passes into a circle and the inversion is a one-to-one transformation (to explain this more fully: under inversion, two tangent circles can go into two tangent circles or into a circle and a tangent line to the circle, or into two parallel straight lines). If the pole of inversion [0, A] lies outside circle C, and C is the inverse, then the set of points lying inside circle C goes into the set of points lying inside circle C' (and conversely). The set of points lying outside (C) goes into the set of points lying outside (C') (and conversely) (see Fig. 67). If the pole 0 of inversion [0, A] lies inside circle C, and C' is the inverse of (C), then the set of points lying inside (C) goes into the set of points lying outside circle C', and the set of points lying outside (C) goes into the set of points lying inside circle C'. Remark. When investigating the mapping of regions under an inversion, it is useful to bear in mind the following. Let us consider the inversion [0, A] where A > 0. Let M be an arbitrary proper point of the circular rc-plane, and let M' be its image under the inversion [0, A], that is, {O^-OMT) = OM OM’ = A. From this relation it follows that if the point M moves along the ray OM receding from the pole, then the point M f will move in the opposite direc tion since the product OM- OMr must remain constant and equal to A (the points M and M ' will meet on the circle of inversion). The same occurs in the case of the opposite ray to the one we considered. In concrete cases (see below), this reasoning may be utilized profitably when investi gating the mapping of regions under an inversion and when finding in variant regions, that is, regions that go into themselves under the inver sion in question. The angle between two intersecting circles is preserved under an inver sion, but the orientation of the angle is reversed (inversion is a conformal transformation of the second kind). The proof of this proposition is given below when we consider the inversion of space.

Inversion

285

Remark. A linear fractional transformation of the plane of a complex, variable z' =

cz + d

a d - b e * 0, c = 0

may be rewritten in the form , a A t A be — ad z — -----\---------- —, where A = -----------. c . d c2 c z H----c It may be interpreted geometrically as follows: d 1°. zx = z H-----is a translation. c 2C. z2 = — is an inversion with respect to the unit circle with subsequent symmetry with respect to the x-axis *. 3°. z3 = A z 2 is a similarity transformation with centre O : a rotation about the origin of coordinates through an angle arg A and a homothetic transformation (O, jA ). 4°. z' = z3 +

is a translation again. c From this it follows that a linear fractional transformation is a conformal transformation of the first kind (that preserves orientation of angles), si nee the transformations 1°, 3°, 4° preserve orientation of angles, and the inver sion z2 — — with subsequent symmetry with respect to the x-axis also preserves orientation of angles. Sec. 2. Problem, involving inversion Problem 1. Let A' and Bf be images of the points A and B under the inversion [0, k\. Express the length of segment A'B' in terms of the lengths of the segments ABy OA, OB and in terms of k. It is assumed that the points A and B are distinct from point O. Solution. Suppose that the points O, A and B do not fie on one straight fine. Let k > 0. Then the points A' and B' fie, respectively, on the rays OA and OB, and we have OA •OA' = OB OB’ = k. I*al lzil = L arg -2 = —arg z x (mod 2n).

Problems in Geometry

286

From this it follows that OA’ _ OB ~

OB’ OA '

Hence, A OAB and A OB’A’ are similar (but with opposite orientations) (Fig. 68). From the similarity of these triangles it follows that A ’B’ AB

OA’ OA’-O ^ OA O W _ k _ \k\ ~ OB~ OB-OA ~ OA-OB ~ OA-OB OA-OB*

and, hence, OAOB’ This formula is also true if the points O, A and B lie on one straight line and in the case of k < 0. Problem 2. Prove that it is possible to circumscribe a circle about a convex quadrangle ABCD if and only if the product of the diagonals of the quadrangle is equal to the sum of the products of its opposite sides ( Ptolemy's theorem): AC-BD = 'AB CD + BC-AD. I. Suppose we can circumscribe a circle K about a quadrangle ABCD; we will then prove that the relation AC-BD = AB-CD + BC-AD holds true. Consider the inversion [A , 1], Under this inversion, the circle K goes into the straight line K ', and the points B, C, D go into the points B \ C", D’ lying on that straight line. The point C will lie between points B’ and D \ and therefore (Fig. 69) B C + C D ’ = B ’D’

Inversion

287

or, on the basis of the preceding problem, _ * c __. + CD _ BD A B A C ^ AC AD ~ A B -A D ’ whence follows the relation AC-BD — AB-CD + BC-AD. II. Suppose the relation AC-BD — AB-CD -\- BC-AD is valid. Let us then prove that the quadrangle ABCD is convex and a circle can be cir cumscribed about it. We consider the inversion [A, 1]. Let B', C \ D' be the inverses of the points B, C, D. From the relation AC-BD = AB-CD + B C A D we have BC CD _ BD AB AC AC-AD ~~ A B -A D ’ and from this relation and the result of problem 1 it follows that B'C' + f C'D' = B’D’. Hence, the points B \ C', D' lie on one straight line and the point C lies between the points B' and D'. From this it follows that the points B, C, D lie on one circle (that passes through point A), which is the image of the straight line K' under the inversion [A, 1]. Since the ray AC lies inside A formed by the rays AD and AB, it follows that AC is a diagonal of the quadrangle ABCD, and this means that the quadrangle is convex (under the indicated order of its vertices). Problem 3. Inscribed in a circle K is an equilateral triangle ABC. Let O be a point not lying on the circle K. Prove that there is a triangle with sides OA, OB and OC. Prove that if the point O lies on the circle K, then the sum of two of its segments OA, OB, OC is equal to the third. Proof. Suppose O does not lie on the circle K (Fig. 70). Consider the inversion [ O, 1]. Circle K goes into circle K \ and the points A, B, C go

288

Problems in Geometry

into the points A', B', C' that lie on K' and, hence, do not lie on one straight line. On the basis of problem 1, we have OA BC BC OA- OB-OC OB-OC CA OC OA _

O BC A OA- OB-OC

(i)

AB _ OCAB OA OB ~~ OA- OB-OC and, hence, B'C' : C fA' : A'B' = OA : OB : OC since BC = CA = AB. Thus, the segments OA, OB, OC are proportional to the sides B'C', C'A', A'B' of A A'B'C' and, hence, there is a triangle with sides OA, OB, OC (this triangle is similar to A A'B'C'). Suppose point O lies on the circle K, for example, on the arc AC, that does not contain B (Fig. 71). Under the inversion [O, 1], circle K goes into line A", the points A, B, C go into the points A', B', C' that lie on K', and the point B' will lie between points A' and C'; thus, A'B' + B'C' = O'C' and since relation (1) is valid in this case as well, it follows that OB = OA + OC. Remark. The theorem holds if point O is chosen arbitrarily in space. The proof is analogous. Problem 4. Two circles Cx and C2 with centres Ox and 0 2 are externally tangent to each other. The straight line / touches both circles at distinct

Inversion

289

F ig . 71

points A and B. Construct a circle tangent to the two given circles and the straight line /. Solution. Let us consider the inversion [A, AB*]. Under this inversion, circle C2 goes into itself because if an arbitrary straight line is drawn through point A , the line intersecting circle C2 in points M and M \ then AM -A M 1= AB2. Circle Cx goes into line C[ parallel to line / and tangent to circle C2 (Fig. 72). Thus, the problem reduces to constructing a circle that is tangent to circle C2 and to two tangent lines I and C[ parallel to it. There are two such circles. Let K[ be one of these circles, let A[ be the point of tangency of circles K{ and C2, and let B[ be the point of tangency of K[ to line C[. Let Ax be the second point of intersection of line AA[ with circle C2; point Ax is the image of point A[ under the inversion [A, AB*]. Let Bx be the point of intersection of line AB[ with circle Cx (point Bx is disÂ tinct from point A); point Bx is the inverse of point B[. The points Ax and Bx are points of contact of the desired circle with the circles C2 and Cx respectively. The centre Px of one of the desired circles is the point of intersection of the straight lines 0 2AX and OxBl9 and the radius is equal to PXAX= PXBX. The second circle is constructed in similar fashion.

F ig . 72

19â&#x20AC;&#x201D;810

290

Problems in Geometry

F ig . 7 3

Problem 5. Two circles Cx and C2 intersect in points A and B. A point C, different from A and B and lying outside circles Cx and C2, is taken on a straight line AB. Construct a circle that passes through point C and is tangent to circles Cx and C2 (Fig. 73). Solution. Consider the inversion [C, &], where k = CA-CB. Under this inversion, each of the circles C1 and C2 goes into itself, and the desired circle goes into a straight line tangent to the circles C1 and C2. Since the two intersecting circles Cx and C2 have two common tangents K[ and K!ly it follows the problem has two solutions. Let A[ and B[ be the points of tangency of line K[ with circles Cx and C2. Denote by Ax the second point of intersection of line CA[ with circle C2, and by Bx the second point of intersection of line CB[ with circle C2. One of the desired circles passes through points C, Ax and Bx. The second circle is constructed in similar fashion (Fig. 73). Problem 6. Given in a A ABC the radius r of an inscribed circle and the radius R of a circumscribed circle. Find the distance d between their centres. Solution. If, under the inversion [O, &], circle C goes into circle C \ then circle C goes into circle C' also under the homothetic transformation (O, k/a), where a is the power of the point O with respect to the circle C. Consider the inversion [P, r2], where P is the centre of the circle inscribed in the given triangle, and r is the radius. Under this inversion, the points of the inscribed circle are fixed (since the inscribed circle is the circle of inversion). The vertices of the given triangle under the inversion [P, r2J go into the midpoints of the sides of A AxBxCXy whose vertices are the points of contact of the sides of A ABC with the circle inscribed in A ABC (Fig. 74). The radius of the circle passing through the midpoints of the sides of the indicated triangle is equal to r/2. Thus, the circle {ABC)

Inversion

291

circumscribed about triangle ABC, whose radius is R, goes into a circle of radius rj2 under the inversion [P, r2]. Since the power of the inversion under consideration is k = r2, and the power of the point P with respect to the circle {ABC) is equal to a — d2 — R2, it follows that r/2 _ r2_ r2 1r ~ |d2 - * 2| ~ R2 - d2’ whence d2 = R2 - 2Rr (Euler’s formula). Problem 7. N and S are two diametrically opposite points of a circle C; I is a straight line tangent to C at point S. From an arbitrary point O lying outside circle C but not lying on the tangent to C at N, we draw to circle C tangent lines OA and OB (A and B are the points of tangency). Let O', A f and Bf be the projections from point N on the straight line / of points O, A and B. Prove that O' is the midpoint of segment A'B' (Fig. 75). Proof. Under the inversion [N, N S 2], circle C goes into line / and circle K with centre O and radius OA = OB, which is orthogonal to circle C, goes into circle K which is orthogonal to line I; hence, the centre of K ' lies on /; on the other hand, the centre of K f also lies on line NO, and for this reason the projection O' of point O from point N on line / is the centre of K'. Hence, A'B' is a diameter of K and O' is the centre of K ', and so A'O' = O'B'.

292

Problems in Geometry

F ig . 7 5

Problem 8. Circles CÂą and C2 do not have any points in common. Use inversion to transform them into two concentric circles and indicate how, under this inversion, the regions Du Z>2, L>3 into which the circles Cl, c 2 divide the plane, are transformed (Fig. 76). Solution. Construct some circle K that orthogonally cuts both circles Cx and C2. To do this, draw some circle M that cuts both circles Cx and C2 at the points P, Q and R, S respectively. Let O be the point of intersection of the straight lines PQ and R S (Fig. 77). Then the segments m of tangents drawn from point O to the circles Cx and C2 will be equal and, hence.

F ig . 7 6

Inversion

293

F ig . 7 7

the circle K with centre O and radius m will intersect both circles Cx and C2 orthogonally. Denote by A and B the points of intersection of K with the line of centres of Cx and C2. Then under the inversion [A, AB2], the circles Q and C2 go into two concentric circles with centre B. Indeed, the circle K goes into the straight line K ' that passes through point B and since K is orthogonal to the circles C2 and C2, it follows that line K' will be orthogonal to C2 and C[ into which circles Cx and C2 go; that is, the centres of C[ and C2 must lie on the straight line K \ But they also must lie on the straight line AB, hence, the centres of C[ and C2 coincide with point B. Furthermore (see Fig. 76), since the centre A of inversion lies inside circle Cl9 the region Dx of points lying inside Cx goes into the region D[ of points lying outside C[. Since the centre A of inversion lies outside C2, the region D2 of points lying inside C2 will go into the region D'2 of points lying inside C2. Hence the region Z)3 of points lying outside Cx and C2 will go into a plane annulus bounded by the circles C[ and C2. Consider the case where the circle C2 is put inside circle C2 (see Fig. 77). In this case, the circle K that cuts C2 and C2 orthogonally goes into the line K' = BC, where C is the second point of intersection of K and the circle of inversion. If T and L are the points of intersection of K with Cx and C2 and if the straight lines TA and LA intersect K' in the points T' and L \ then the images of and C2 are the concentric circles C'2 with centre B and radii B T and B L Furthermore, since point A lies inside C2, it follows that the region D2 of points lying inside C2 goes, under the inverÂ sion [Ay AB2], into the region D2 of points lying outside C2. The region D3 of points lying outside Cx goes into the region D'3 of points lying inside C[, and this means that the eccentric annulus Dx bounded by C2 and C2 goes

294

Problems in Geometry

into annulus D[ that is bounded by the concentric circles C[ and C2 (Fig. 77). Problem 9. Construct a circle tangent to three given circles Cl9 C2, C3 (Apollonian problem). Solution. Consider only the case where each of the circles Cl9 C2, C3 lies outside the other two. Let us perform an inversion under which Cx and C2 go into two concentric circles C[ and C'2 (problem 8). Suppose that the pole A of this inversion [A, AB2] lies inside Cl9 then R[>R2 where Ri and R2 are the respective radii of the circles C[ and C2. Since C3 lies outside Cx and C2, it follows that its image C3 under the inversion [A, AB2] will lie inside the annulus formed by C[ and C'2 since points lying outside Cx go into points lying inside C*l9 and points lying outside C2 go into points lying outside C2 (Fig. 78). We split into two sets all the circles tangent to the concentric circles C[ and C2: circles of radius (/?( — R2)/2, each of which is externally tangent to C2 and internally tangent to C[, and the circles of radius (i?( + R£!2, each of which is internally tangent to C[ and C2. Only four circles K{, K'2, K3, Kl belong to the first set, the first two of which are externally tangent to circle C3, and K3 and Kl are internally tangent. To construct the circles K[ and K f2 it suffices to construct their centres: these are the points of inter section of the circles ( 0 \ (R[ — R2)/2) and ( 0 '\ Rl + (/?( — R 2)I2), where O' and O" are the respective centres of C[ (or C2) and C3. The centres of the circles K3 and Kl are the points of intersection of the circles (O', (R[~ -tf')/2 ) and (O", (R' - RQ/2 - R£. Similarly, from among the circles of the second set that are internally tangent to the circles C[ and C2 there are only four circles SJ, S'2, S3, S'4 that are externally tangent to circle C3 (5( and S%) and internally tangent

F i g . 78

Inversion

295

to C3 (S 3 and Si). The centres of the circles S[ and S'2 are points of inter section of the circles (O', + Ri)/2) and (O", (Ri + R£)/2 + Ri), and the centres of the circles Si and Si are the points of intersection of the circles (O', (R[ + J?£)/2) and (O", (R[ + R'2) /2 - Ri). All these eight circles K{, Ki, Ki, K4', S[, Si, Si, Si are constructed in Fig. 78. Their images

296

Problems in Geometry

Kl9 K2, Ks, K±, Sl9 S2, S2, S4 under the inversion [A, ,42?2] will be tangent to the three given circles Cl9 C2, C3. Thus, in the case at hand, the problem has eight solutions. Figure 79 gives the positions of the circles Kt and S t with respect to the circles Cl9 C2, C3 on eight separate figures. Problem 10. Prove that the Euler circle of A ABC is tangent to the circle (I) inscribed in that triangle and is tangent to the three circles (4 ), (4 ), (/c) escribed in that triangle (the points of tangency are <P0, <Pl9 <P2, # 3 and are called Feuerbach points). Solution. Let (/) and (Ia) be the respective circles (one inscribed in the given triangle and the other escribed in Z_A). Denote by R and S the points of tangency of (/) and (4) with side BC (Fig. 80). Then BS = CR (= p —c9 where p is the semiperimeter of A ABC). Let A', B', C' be the respective midpoints of the sides BC9 CA9 AB. Denote by A", the orthogonal projec tion of point A on the side BC, and by Q the point of intersection of side BC with bisector IIa of /^BAC. The set of points A, Q, I, Ia is an harmonic set of four points, that is, the double ratio of these four points is equal to —1. Hence the points A ", Q, R, S, which are the orthogonal projections of the points A, Q, I, Ia on line BC, will also be an harmonic set of four points. Point A ' is the midpoint of segment R S (since A' is the midpoint of BC and BS = CR); hence, A'Q -A'A" = A'R2. Let us consider the inversion [A', A'R?\. Under this inversion, the circles (I) and (Ia) are invariant (since the circle of inversion is a circle with diameter R S and is orthogonal to both circles). From the relation A'Q-A' A" — A'R2 it follows that under this inversion the point A" goes into point Q. On the other hand, point A" (the foot of the altitude from A onto BC) lies on the Euler circle, the point A' (the midpoint of side BC) also lies on the Euler circle and, hence, under the inversion [A', A'R2] the Euler

F ig . 8 0

Inversion

297

circle goes into a straight line that passes through point Q and is antiparallel to the straight line B'C' with respect to /_ C'A'B'* or into a straight line antiparallel to BC with respect to /_ CAB (because BC || B'C', CA\\ ]|C'A', AB || A'B'). But the straight line that passes through point Q and is antiparallel to line BC (with respect to BAC), is a straight line symÂ metric to line BC with respect to the bisector of /_BAC, that is, the second common tangent co' (interior) to the circles (/) and (7fl). Let and be points of tangency of co' with (7) and (7fl). Under the inversion [A', A'R2], the straight line co' is transformed into the Euler circle (co), and the points #o and <P[ into the points # 0 and #1 that lie on (7) and (7fl). In these points <2>0 and <Pl9 the Euler circle is tangent to the inscribed circle (7) and to the escribed circle (7fl) because the straight line co' is tangent to (7) and (7a) in the points d>o and <P[; # 0 and are the points of intersection of the straight lines A'<P'0 and A'<P[ with the Euler circle or the second points of intersection of these lines A'<P'0 and A'&[ with (7) and (7fl). Figure 80 depicts the construction of thirteen points A', B', C , A", B", C", A" B" ', C" <P0, # 2, # 3 (all of them lie on the Euler circle). Sec. 3. Mapping of regions under inversion Problem 1. Suppose a circle C lies outside the circle K of the inversion [O, r2], where r is the radius of K. Let C' be the image of circle C under the inversion [O, r2]. Then the set DÂą of all points lying inside C is mapped one to one onto the region D[ of all points lying inside C'; the set 7)2 of all points lying outside circles C and K is mapped one to one onto the set D'2 of all points lying inside K but outside C'. The connected region D2 U D'2 is invariant under the inversion [O, r2] (Fig. 81).

* Straight lines that are symmetric with respect to the bisector o f / _ CAB are termed antiparallel with respect to that angle.

298

Problems in Geometry

F ig . 8 2

Problem 2. A circle C is externally tangent to a circle K of the inverÂ sion [0, r2]. Let C' be the inverse of C. The correspondence of the regions under the inversion [0 , r 2] is similar to that considered in problem 1 (Fig. 82). Problem 3. A circle C cuts the circle K of the inversion [0, r2] but the centre 0 of the inversion lies outside C. The correspondence of the regions under the inversion is indicated in Fig. 83. The regions D2 U D2, Ds u D'Zy Z>4 U Di, Dq u Dq are simply connected and invariant under the inversion [0, r2]. They are divided by the circle of inversion K into the regions D29 D2\ DS9 D%; Z)4, D^; 0 6, D'Qthat pass into one another under the inversion at hand (Fig. 83).

Problem 4. A circle C passes through the centre of the inversion [0, r2] and cuts the circle of inversion K. Under the inversion [0, r2], the image C' of circle C is a straight line passing through the points of intersection of C and K. The correspondence of the regions under the inversion is shown in

Inversion

299

Fig. 84

Fig. 84. The regions Dx U D[ and D5 u D'b are simply connected and are invariant under the inversion [09r2]. The circle of inversion K divides them into the regions Dx and D[, D5 and D 's, which go into one another under the inversion at hand (Fig. 84). Problem 5. The circle of inversion K lies inside circle C. The image of the connected region Dl9 which consists of points lying outside circle C is a simply connected region D[, which consists of points lying inside circle C', where C is the inverse of C with respect to circle K. The image of region D2, which consists of points lying outside K, but inside C, is the region D'29 which consists of points lying inside K but outside C'. The connected region D2 U D2 is invariant under the inversion under consideration (Fig. 85).

Fig. 85

300

Problems in Geometry

Problem 6. A circle C intersects the circle of inversion K orthogonally. In this case the image C of circle C (under the inversion with respect to circle K) coincides with the circle itself: C = C'. The region which con sists of all points lying inside C but outside K goes into the region D[, which consists of all points lying inside C and inside K. The region Dx UD'l9 which consists of all points lying inside C is invariant under the inversion under consideration. Let P and Q be points in which the circles C and K

intersect. Draw rays OP and OQ. The region D 2, which is bounded by arc PQ of circle C and the radii OP and OQ of circle K goes into the region Z)2, which is also bounded by arc PQ of circle K (the complement of the first arc of this circle) and by the prolongations of the radii OP and OQ beyond points P and Q . Finally, the region Dz consisting of points lying outside K and outside the angle POQ goes into region Dz, which consists of all points lying inside K but outside /_ POQ . The region Dz u Dz is simply connected and invariant under the inversion with respect to circle K (Fig. 86). Problem 7. A circle C passes through the centre O of the circle of inver sion K and lies inside K. The image C of circle C is a straight line that does not intersect K. The region Dl9 which consists of all points of the half-plane on the side of straight line C that does not contain circle K , goes into a set D[ of all points lying inside C. The region Z>2, which consists of all points lying out side K and located in the half-plane (reckoned from line C ) that contains AT, goes into the set D'2 of all points lying inside circle K but outside circle C. The connected region D2 u D ’2 is invariant under the inversion at hand (Fig. 87). Problem 8. A circle C goes through the centre O of the circle of inver sion K and is tangent to that circle. The image C" of circle C under the inver sion with respect to circle K is the common tangent to C and K. The cor respondence of regions Dx and D[, D2 and D2 is similar to the preceding problem. The region D2 U D2 is simply connected and is invariant under the inversion with respect to the circle K (Fig. 88).

Inversion

301

Problem 9. A circle C intersects the circle of inversion K and the centre O of inversion (that is, the centre of circle K) lies inside C. The corresponÂ dence of regions Dx and D[, D2 and D'2, Dz and Z>3, Z>4 and Z>4 is shown in Fig. 89. The regions Dx U D[, and Z)4 U D\ are simply connected and are invariant under the inversion with respect to circle K. Problem 10. Given three equal circles Cl9 C2, C3 that pass through one point O and intersect at angles of n/3. Take point O as the pole of inversion and take the circle of inversion K so that it intersects Cl9 C2, C3 and so that the points of intersection lie inside K. The images C[9C2, C3 of the given cirÂ cles Cl9 C2, C3 are straight lines passing through the points of intersection of circle K with each of the circles Cl9 C2, C3. The given circles Cl9 C2, C3, their images C[9 C2, C3 (under the inversion with respect to the circle K), and the circle K itself divide the plane into 24 regions. The correspondence of regions under the inversion of the plane with respect to circle K is shown in Fig. 90. The regions D7 U D?, D8 u D89 D9 u D9y D10 u A'o* A i U U D[i, Dx2 U D'12 are simply connected and invariant under the inver-

Fig. 89

302

Problems in Geometry

sion at hand. The regions Dk and D'k (k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) go into one another under this inversion. Problem 11. Consider the cardioid p = 1 + cos <p, the equation of which is given in polar coordinates. Under the inversion [ 0 , 1], where O is the pole of the polar system of coordinates, the image of the cardioid is a parabola whose equation in that same polar coordinate system is of the form 1 P --------------------------1 + cos q> If a rectangular Cartesian system of coordinates is introduced and the polar axis is taken as the x-axis, then the last equation becomes y2 = 1 — 2x . Figure 91 depicts a parabola constructed on the basis of this equation. The cardioid p = 1 + cos <p, the parabola y2 = 1 — 2x, and the circle of inver sion K divide the plane into six regions. The correspondence of these regions under the inversion with respect to circle K is depicted in Fig. 91. The regions Dx U D[ and D3 U D3 are simply connected and invariant under the inversion with respect to circle K Region D2, which is bounded by a part of the cardioid and an arc of the parabola, goes into region D'2, which is made up of points lying outside the cardioid and the parabola. Conversely, under the inversion at hand, region D2 goes into region D2.

Inversion

303

Fig. 9

Problem 12. We consider Pascal’s limagon, the equation of which in polar coordinates is of the form p = m +72 cos cp.

We assume that 0 < n < m. Then we have p > 0 for all values of cp, and, hence, the limagon considered here is a closed curve without self-inter sections, inside which curve is the pole O of the polar system of coordinates. Under the inversion [O, 1], where O is the pole of the polar system of coor dinates, the limagon goes into a curve, whose equation in that same polar system is of the form 1 P= m -------- i-------------------+ n cos cp or P P= 1 + e cos (p where p = \jm , e = n/m. Since from the conditions 0 < n < m it follows that 0 < e < 1, the last equation is the equation of an ellipse for which O is one of the foci, p is a parameter, and e is the eccentricity. For example, take m = 3/2, n = 1. Then the equation of Pascal’s limagon is 3 , p = --------- b COS cp. 2 and the equation of the ellipse, which is the image of this limagon under the inversion [O, 1], is 1 --------- b COS

304

Problems in Geometry

If a rectangular Cartesian system of coordinates is introduced, and the polar axis is taken as the x-axis, then the last equation becomes

which is an ellipse with centre at the point (—4/5, 0) and with semi-axes equal to a = 6/5, b= 2/^5. Figure 92 depicts the limagon under consid eration and the ellipse, which intersects the limagon p = 1.5 + cos cp for values of q> that satisfy the equation 1 3_ + COS (p = 2 3 , C O S (p ’ -------\2 whose cos (p = ------- ; and, thus, assuming that 0^<p<2n, we have 2 cp = 2n/3, <p = 47t/3 (two points of intersection). For these values of (p we have p = 1 and, hence, the circle of inversion K passes through the points of intersection of the limagon and the ellipse. The limagon, the ellipse, and the circle of inversion K divide the plane into six regions. The correspondence of these regions is shown in Fig. 92. The regions Dx U D[ and D2 U D'2 are simply connected and invariant under inversion, and under the inversion at hand they pass into one anoth-

Inversion

305

er: D i ^ D[, Z>2 ^ The region Z)3, which is bounded by an arc of the lima^on and an arc of the ellipse, contains the centre O of inversion (the pole of the polar coordinate system), goes into the region D3, which consists of all points lying outside the ellipse and the limagon. Conversely, under the inversion [O, 1], the region Z>3 goes into the region Z>3. Problem 13. Let us consider Pascal’s limagon once again in polar coor dinates; it is given by the equation p = m + n cos cp9 but it is now assumed that 0 < m < n. The equation p = 0 now has the solution cos cp = — mjn and if we assume 0 ^ cp < 2k, then the last equation yields two values for<p: cp = arccos (—m/ri), cp = 2 n — arccos ( —mjn). Therefore, when 0 < m < n the limagon passes through the pole twice. In the case at hand, this is a curve with self-intersection; it forms a loop inside the rest of the curve. Under the inversion [O, 1], where O is the pole of the polar system of coordinates, the limagon goes into a curve whose equation in the same polar system of coordinates is of the form 1 P = — ----------m + n cos cp or H

1 + e cos cp

where p= 1/m, e = njm. Since we now have e > 1, the curve specified by this equation is a hyperbola with eccentricity e and parameter p (half the focal chord). Let us now take, for example, m = 1, n = 2 (Fig. 93). The equations of the limagon and its image (hyperbola) under the inversion [O, 1] are p = 1 + 2 cos cp,

p = ---------1 + 2 cos cp

Transform the equation of the hyperbola by introducing a rectangular Cartesian system of coordinates in which the x-axis is the polar axis. We then have

9 2 0 -8 1 0

306

Problems in Geometry

Fig. 93

This is a hyperbola with centre (2/3, 0) and semi-axes a = 1/3, b= l/]/3. Since bja = |A3, it follows that the asymptotes of the hyperbola are inclined to the #-axis at angles of ±7r/3. When cp = ±2n/3 in the equation of the limagon p = 1 + 2 cos <p, the radius vector p vanishes. Therefore the asym ptotes of the hyperbola are parallel to tangents to the loop of the limagon at the coordinate origin. The limagon, the hyperbola, and the circle of inver sion K divide the plane into 14 regions. The correspondence Dk^±D£ of these regions is depicted in Fig. 93. The regions Dz UDz, Z>4 U D\, D7 U D'7 are simply connected and are invariant under our inversion; also the circle of inversion divides them into the regions Dz and Dz, Z>4 and Z>4, D7 and D7i which go into one another under the inversion. The limagon and the hyperbola intersect in seven points; three of them lie also on the circle of inversion; the other four P, P', Q, Q' do not lie on the circle of inversion and correspond to one another {P ^l P \ Q Q') under inversion (symmetry with respect to the circle K). Using the straight lines OPP' and OQQ\ we can partition the regions Dx and D[ into three * ** *** * ** *** regions: Dl9 Dl9 and D[, D[, which will correspond to one another under the inversion at hand. These same lines divide the regions Z>3, Dz and Z>4, Z)4 into two parts each, and they correspond to each other under the inversion [0, 1]. Problem 14. The equation of the cissoid of Diodes in polar coordinates may be written as _ 2 sin2 <p ( 2) cos <p

Inversion

307

Fig. 94

The straight line x = 2 is a vertical asymptote of the cissoid because lim

p = + oo;

<p -» ^— o 2

and from the equation (2) it follows that lim 71

<p -» — —0 2

x =

lim

p cos q>=

lim

(2 sin2 q>) = 2.

<p-» ------- 0

<p - » ------o 2

Under the inversion [0,1] (0 is the pole of the polar system of coordinates), the cissoid is transformed into a curve whose equation in the same polar sys tem is of the form cos cp 2 sin2 <p or 2'y2 = x. This is a parabola for which the polar axis serves as the axis. The circle of inversion K passes through two points in which the cissoid and parabola intersect. (Note that the circle of inversion does not intersect the asymptote x = 2 of the cissoid.) The cissoid, the parabola, and the circle of inversion divide the plane into eight regions Dk9 D ’k (k = 1,2, 3, 4), which pass into one another under the inversion at hand: Dk Dk. The regions Dx U D[ and Z>4 U D\ are simply connected and are in variant under the inversion [<9, 1] (Fig. 94).

308

Problems in Geometry

Sec. 4. Mechanical inversors: the Peaucellier cell and the Hart cell Suppose MPM'Q is a rhombus. The point O is a point equidistant from points P and Q. Also suppose the rhombus is hinged, point O being fixed and joined by hinges with P and Q (the Peaucellier cell). Then if M describes some curve L, the point M ' describes a curve L', //' which is the image of L under the inversion [O, OP2 - PM 2] (Fig. 95). Proof, The points M 9M ' and O are at equal dis tances from the points P and Q, and for that reason lie on the midperpendicular of segment PQ, Let us construct the circle (P, PM), The product OM-OM' 0 is equal to OP2 — PM 2. Indeed, let S be the mid point of segment PQ ; then OM-OM' = (OS + SM) (OS + SM') = (OS + SM) (OS - SM) = OS2 - SM 2 = OP2 - SP2 - (MP2 - SP2) = OP2 - PM 2. Note, in particular, that if point M describes a circle passing through point O, then point M ' de scribes a straight line. Thus, the Peaucellier cell makes it possible, mechanically, to transform circular mo tion into rectilinear motion. Let ABCD be an antiparallelogram (that is, ABDC is an isosceles trape zoid, BD and AC are its parallel sides, AD and BC the diagonals). Fix a point O on segment AB and a point M on segment AD, and on segment 2?C fix a p o in t M 'so that the points O, M, and M f belong to a single straight line parallel to BD ||AC. The points A, B, C, D, O have hinges; the point O is fixed (this is Hart's cell). If under these conditions we deform the anti parallelogram ABCD, then the product OM-OM' will remain constant, that is, if point M describes a curve L, then point M ' will describe a curve obtained from L by an inversion with the pole O (Fig. 96)# Proof. If the antiparallelogram ABCD is hinged, then, as will readily be seen, when it is deformed it remains an antiparallelogram (this stems from the fact that an antiparallelogram is characterized by the equalities AB = CD, and AD = Fig. 96

Inversion

309

= BC). On the other hand, the straight lines OM and O M \ which are parallel to the bases BD and AC of the trapezoid in the original position of the figure, will all the time remain parallel to them. Indeed, AM _ AO BM' __ BO ~AD ~~ AB BC ~ A B ' When the antiparallelogram ABCD is deformed, these proportions are preserved. In particular, from what has been said, the points O, M and M ' remain on one straight line. Furthermore, from similar triangles we find OM _ AO_ OM' __ AC BD ~ AB 9 OB ~~ AB â&#x20AC;&#x2122; whence O M 'O M ' = -C^ - ? J L .B D A C . AB* But a circle can be circumscribed about the trapezoid ABCD, and so AB2 + B D -AC = AD2 And, hence, BD AC = AD2 - AB2. Finally, we get OM O W = 0A 0 B (AD* - AB*). AB* Therefore point M' is obtained from point M under the inversion [O, k], where _ OA OB(AD2 - AB2) k = --------------------------------------------. AB2 Sec. 5. The geometry of Mascheroni We will assume that a straight line is specified if two distinct points of the line are fixed in the plane. A circle is regarded as specified if the folÂ lowing are given: the centre and also a point lying on the circumference of the circle, or three points on the circumference. We will denote a circle with centre O and radius OM as (O, OM). A circle specified by three points A, B, C lying on the circumference will be symbolized as (ABC). We now consider the solution of a number of basic problems involving construction with a compass alone. Most of the solutions involve the use of inversion. * Starting mainly with these constructions, which were carried out without the aid of inversion, the Danish mathematician G. Mohr (17th century)

Problems in Geometry

310

succeeded in proving that with a compass alone it is possible to perform all constructions that can be performed with compass and straight edge. This proposition was independently proved at the end of the 18th cen tury by the Italian mathematician L. Mascheroni, and it was precisely Mascheroni’s work that became known in Europe (Mohr’s book had gone un noticed). For this reason, constructions peformed with only a compass are frequently associated with the name of Mascheroni (the geometry of Mas cheroni, Mascheroni constructions and so forth). We now consider several construction problems involving only a com pass. Problem 1. A line segment AB is specified by its endpoints. Construct, on the extension of segment AB, beyond point B, a point C such that AC = = n-AB, where n is a natural number. Solution. We construct the circles (A, AB) and (B, BA). Let P be one of the points of their intersection. Construct a circle (P, PA)\ let Q be the second point of intersection of this circle with (B, BA). We construct the circle (Q, QB)\ let R be the second point of intersection of this circle with (P, BA). The point R lies on the extension of segment AB beyond B, and AB = BR (Fig. 97). True enough, the indicated construction yields the vertex R of a regular hexagon; R is opposite vertex A and the hexagon is inscribed in (B, BA). Continuing similar constructions, we can construct point C that lies on the extension of segment AB beyond point B and such that AC = n-AB, where n is any natural number.

Fig. 97

Fig. 98

Problem 2. Construct with compass alone the image M of a point M ' under the inversion [O, r2]. Solution. 1°. OM > r/2. If point M lies on the circle of inversion, then its image M ' coincides with itself. Therefore, let us assume that point M lies either inside the circle of inversion (Fig. 98) or outside it (Fig. 99). In both cases we construct the circle (M, MO). Let X and Y be the points of intersection of this circle with the circle of inversion. We construct the circles (X , XO) and (F, YO); the second point of intersection of these circles is the point M f. Indeed, since the points X and Y are symmetric about the

Inversion

311

Fig. 99

straight line MO, and the circles (X, XO) and ( Y, YO) are congruent, it follows that the second point M ' of their intersection lies on the radial line OM. Furthermore, if H is a point diametrically opposite point O on the circle (M, MO), and E is the point of intersection of X Y and OM (E is the midpoint of segment OM' ; neither point H nor point E need be con structed, they are introduced merely to aid the proof), then r2 = OX2 = O H O E = OMOM' . 2°. OM < r/2. In this case (M, MO) does not intersect the circle of inversion. We construct on the radial line OM a point N such that rrO M = ON and such that ON > rjl. Construct point N' obtained by the inversion [O, r2] from point N (case 1°) and then construct on the radial line ON' a point M ' such that OM' = rrON'; then ___ ___ ON ___ O M •OM' = ------ •nON' = ON•ON' = r2. n From the solution of this problem there follows the possibility of con structing with compass alone a point C lying on segment AB and such that n*AC = AB. Indeed, construct on the extension of segment AB, beyond point B, a point C' such that AC' — rrAB (problem 1). Let C be the image of point C' under the inversion [A, AB2] (problem 2). Point C is the desired one. Indeed, AC AC' = AB2, AC nAB = AB2, n-AC = AB. Problem 3. Using only a compass, construct a circle C as the image, under the inversion [O, r2], of the straight line C that does not pass through the pole O of the inversion. The straight line is assumed to be given by two distinct points X and Y lying on it.

Problems in Geometry

312

'i---=----t.t--h~--~-iP

Fig. 100

Solution. The circle C' passes through point 0. Let P be the second point of intersection of the circles (X, XO) and (Y, YO). Then P is symmetric to point 0 with respect to the line XY. Construct (with compass alone) the image P' of point P under the inversion [0, r 2]. The desired circle C' is the circle (P', P'O) (Fig. 100). Indeed, if E is the point of intersection of the straight lines XY and OP [E is the midpoint of OP and £' is a point diammetrically opposite to point 0 on circle (P', P'O)] then

r 2 = OP·OP' = OE·OE'.

Problem 4. Using only a compass, construct the point of intersection of two straight lines, one of which is given by the points A and B, the other by the points C and D. Solution. Construct an arbitrary circle S in the plane. Construct (with compass alone) the circles K 1 and K2 , which are inverses of the given lines with respect to the circle S. Let M' be the point of intersection of the circles K 1 and K 2 • Construct (with compass alone) the inverse M of point M' with respect to the circle S. The point M is just the point of intersection of the given straight lines. Problem 5. Construct (with compass alone) the centre of a circle specified by three points A, B, C lying on that circle (the circle itself is not drawn). Solution. Let P and Q be the points of intersection of the circles (A, AB) and (B, BA). The straight line PQ is the midperpendicular of segment AB. Let L and M be the points of intersection of (B, BC) and (C, CB). The line LM is the midperpendicular of segment BC. The centre 0 of the circle (ABC) is the point of intersection of the straight lines PQ and LM (problem 4). Problem 6. On a straight line given by two points A and B, lay otr from point B segments BC1 and BC2 equal to the given segment PQ. Solution. Construct a circle (B, PQ). The problem reduces to seeking the points of intersection of that circle with line AB. Construct (with compass alone) the images K 1 and K2 of line AB and the circle (B, PQ) under inversion with an arbitrary circle of inversion. Let c; and C~ be the points of intersection of the circles K 1 and K2 • The inverses C1 and C~ of points c~ and c~ are the desired points. Problem 7. Take three given line segments PQ, MN, RS and construd a fourth, proportional to them, that is, such that PQ· MN= RS· XY.

Inversion

313

Solution. 1°. PQ ,p MN. Construct circles (0, PQ) and (0, MN), where 0 is an arbitrary point in the plane. On (0, PQ) take an arbitrary point A and construct (with compass alone) a point Bin which the straight line OA intersects (0, iVtN) (take the point B which lies on the radial ~

line OA). Construct a circle (0, RS) and on it take an arbitrary point C. Construct a circle (ABC) (problem 5). Let D be the second point of intersection of this circle with line OC (it is constructed with a compass alone). Then

OA· OB = OC· OD or

PQ· MN

RS· OD;

OD is the desired line segment. 2°. PQ = MN. In this case we consider segments PQ and 2MN (segment 2MN is constructed with compass alone). Construct (with compass alone) the segment X1 Y1 such that PQ· 2MN= RS· X1 Y1 (case 1°). The desired segment XY

X1 Y1 is constructed by compass alone (see 2 remark at the end of the solution of problem 2). Problem 8. Given in the plane two points A and B. Construct (with compass alone) a point C such that the angle ABC is equal to 90°. Solution. On the straight line AB we construct (with compass alone) the line segment BD = AB. Let C be any one of the points of intersection or the circles (A, AD) and (D, DA). The point C is the desired point. Problem 9. Given a segment AB with points A and B. Let n be an arbitrary natural number not the square of any natural number. Construct tlw segment ABVn. Solution. Construct (with compass alone) a point C such that L ABC= 90°. Find the point P of intersection of BC with the circle (B, BA). Then AP = ABV2. Then construct (with compass alone) a point D such that L APD = 90° and construct the point Q of intersection of the straight line PD with the circle (P, AB). Then PQ = ABV3 and so on. From the results of this problem it follows that it is possible with compass alone to construct on a circle (specified by three points) the vertices of a regular triangle, a square, a regular pentagon and hexagon. Sec. 6. Inversion of space The inversion [0, k] of space (more precisely, of spherical space with a single ideal point, or point at infinity, adjoined) is defined in the same w;~y as the inversion of a plane. The inversions [0, k] of space, like those

314

Problems in Geometry

of a plane, separate into positive inversions (k > 0) and negative inver sions (k < 0) depending on the sign of the power k of inversion. Under a positive inversion [<0, k], a sphere S0 with centre 0 and radius ]fk (termed the sphere o f inversion) consists of invariant points. The set of points lying outside the sphere S0 is mapped into the set of interior points of 50, and the set of all points lying inside S0 is mapped onto the set of all points lying outside S0. If k < 0, then a sphere with centre 0 and radius ^\k\ is invariant, and each point of it is mapped into a point diametrically opposite on the sphere S0. If the inversion [0, k] is positive, then all spheres intersecting the sphere S0 of inversion orthogonally, and only such spheres, are invariant. The inverse of a sphere not passing through the centre of inversion is a sphere. The tangency of spheres and the tangency of a sphere and a plane are preserved under inversion. If a sphere S with centre Q passes through the centre 0 of inversion, then its image, under the inversion [0, &], is a plane orthogonal to the straight line OQ, which plane does not pass through the pole of inversion, and, conversely, a plane that does not pass through 0 goes into a sphere that passes through 0. The inversion of space is a conformal transformation, that is, under the inversion, angles between tangent planes in the points of intersection of two spheres are preserved. A circle that does not pass through the pole of inversion goes into a circle. Indeed, every such circle may be regarded as the line of intersection of two spheres not passing through the pole of inversion. Under the inversion of space, angles are preserved between intersecting circles (which, generally speaking, lie in distinct planes). Indeed, let a and b be tangents to the circles Cx and C2 in the point of their inter section M. It suffices to prove that under the inversion [0, k] the angle is preserved between the straight lines a and b. We consider only that case where the straight lines a and b do not pass through the pole of inversion 0. Let a* and b* be straight lines that are respectively parallel to straight lines a and b and that pass through the pole of inversion 0 . Under the inversion [0, £], straight lines a and b go into the circles a’ and b' that are respectively tangent to the straight lines a* and b* at point 0 . Therefore, the angle between the circles a’ and b' is equal to the angle between the straight lines a* and b*, that is, it is equal to the angle between a and b because a* || a, 6*|| b. * * We now introduce a number of geometric constructions and analytic derivations associated with stereographic projection in connection with the use of this mapping of a sphere on a plane when making maps

Inversion

315

of the earth. We dwell solely on the geometric construction of the depiction of the network of meridians and parallels and on the analytic investigation of the nature of their distribution (under stereographic projections). Let us recall the basic definitions and properties of a stereographic projection in connection with the inversion transformation. Let A and B be the ends of some diameter AB of a sp h ere^ ; ax andZ?! are planes tangent to Qx at the points A and B. Let M be an arbitrary point of Qx distinct from point A . Denote by M ' the point in which the straight line AM intersects the plane bx. The correspondence under which point M is associated with point M f is termed the stereographic projection of the sphere Qx on the plane bx (Fig. 101). Let us consider a section Q of the sphere Qx cut by the plane ABM (Fig. 102). From the similarity of A ABM and A ABM' we find AM- AM ' = AB2. From this relation it follows that the image M f of point M under the stereographic projection of the sphere Qx on the plane bx is also the image of point M under the inversion / = [A, AB2] with pole A and the power of inversion equal to AB2; in other words, point M ' is the image of point M under an inversion with the sphere of inversion cox the centre of which is point A and radius AB. To construct the image M ' of point M under the inversion I it is first convenient to construct sections Q and co of the spheres Qx and a>1 by the plane ABM. Then join point M with point A and draw a straight line passing through point M perpendicular to straight line AM. Let P be the point in which this line intersects the circle co; then the tangent to co at the point P will cut line AM at point M'. This construction is true of any point M lying inside the circle co. If point M lies on co, then its image is the point M itself. If M lies outside co, then to construct the image M ' under the inversion with respect to circles co, we construct the tangent MQ drawn from point M to co (Q is the point of tangency); the orthogonal projection M ' of point Q on line AM is precisely the image of point M under the inversion I. All these construcÂ tions can be carried out in any section Q of the sphere Qx by a plane

316

Problems in Geometry

passing through the straight line AB. Incidentally, they can also be carried out in space by drawing the appropriate tangents not to the circles Q and a), but to the spheres QÂą and a>x. From the foregoing we arrive at the following important conclusion: on the set of all points of (with the exception of point A), a stereographic projection coincides with an inversion, the pole of which is the centre of projection A, and the power of the inversion is equal to the square of the diameter of in other words, a stereographic projection on the set of all points of (with the exception of point A) coincides with an inversion with the sphere of inversion col9 the centre of which is point A, and the radius is equal to the diameter of Qv From this it follows that the geometric properties of a stereographic projection may be obtained from the familiar properties of inversion, namely: under an inversion (and, hence, under a stereographic projection) the circles that lie on the sphere and do not pass through point A go into circles (lying in the plane b2). Since inversion is a conformal transÂ formation (that is, angles are preserved), it follows that a stereographic projection has the same property, so that using a stereographic projection makes it possible to construct a conformal map of the globe (it is not, however, possible to construct a map of the earth on which distances are preserved). Let us now see how one constructs the centre of circle C", into which circle C, which lies on the sphere and does not pass through p oints, is mapped. Let us first suppose that C is not a great circle of We construct a cone K tangent to along the circle C. Let P be the vertex of the cone. We construct a sphere T with centre P that passes through circle C (Fig. 103). This sphere intersects Qx orthogonally. Under the inversion I = [A, AB2], the sphere goes into the plane bl9 and the sphere Tgoes into the sphere T \ which will intersect the plane bx orthoÂ gonally. From this it follows that the centre M ' of T' must lie on the plane But, on the other hand, the centre of sphere T must also lie on line AP

Inversion

317

since sphere T is the image of A sphere T under the inversion [A, AB2]. Thus, the centre M ' of Q sphere T' is a point in which the straight line AP intersects the plane bv Furthermore, the circle C is the line of intersection of the spheres Qx and T, hence 6 the image C" of circle C (both Tf M T{ B under the inversion / and under the stereographic projection Fig. 104 under consideration) is the intersection of the sphere T and the plane bx. But since the centre M ' of the sphere T lies in the plane bl9 it follows that C' is a great circle of sphere T', and therefore the centre of circle C coincides with the centre M ' of sphere T , that is, it is the projection M ' on plane bÂą of the vertex P of the cone K that is tangent to the sphere Q1 along the circle C. Note that the projection M ' of point P on the plane bx is not the image of point P under the inversion / = [A, AB2]. To construct the centre and the radius of the circle C", let us consider the section Q of sphere Q1 by the plane ABP (Fig. 104). Let T[9M and T% be the centre projections of points Tl9 P, T2 from point A on the straight line b (Tx and T2 are the points of tangency of tangents drawn from point P to the circle Q). Then M is the centre of the circle C' and its radius is equal to MT[ = M T2. Let us now examine the case where the circle C is a great circle of Qx and does not pass through point A. In this case, on the basis if the general properties of inversion, the stereographic projection C of the circle C on the plane bx will again be a circle. Geometrically, its centre and radius are constructed as follows: the plane n in which circle C lies goes, under the inversion I = [A9AB2], into the sphere n'9 the centre of which lies on a straight line passing through point A perpendicular to plane n. On the other hand, since the plane n is orthogonal to the sphere Ql9 it follows that the sphere n must be orthogonal to the plane bx and, hence, the centre M of the sphere n' must lie in the plane bv Thus, the centre M of the sphere n and, hence, of the circle C' as well (which is a great circle of the sphere n') is the point of intersection with the plane bt of a straight line passing through point A perpendicularly to the plane n in which the circle C lies. Figure 105 shows a section Q of the sphere cut by a plane passing through straight line AB and through the perpendicular dropped from point A to the plane n in which circle C lies; DE is a diameter of circle C along which the drawn plane intersects C. Joining point A with points D and E, we obtain, in the intersection with line b, the points D' and E \ which are maps of the endpoints of diameter DE of circle C; the straight line^L, which is perpendicular to the diameter DE, intersects (as indicated

318

Problems in Geometry

above) the straight line b in the point M \ which is the centre of circle C". Since M'D' = M 'E ' = M 'A, it follows that to construct the centre and the radius of circle C' we need not draw the straight lines AD and AE: by dropping the perpendicular AL from point A on the diameter DE of circle C we obtain both the centie M ' of circle C" and the radius AM ' of that circle. Also note that the orthogonal projection L of point i on the straight line DE lies on the circle with diameter OA (O is the midpoint of line segment AB). This reasoning is used below in the geometric consÂ truction of a spectrum of meridians when constructing maps of the western and eastern hemispheres of the earth. It remains to consider the case where the circle C passes through the centre A of projection. In this case, the stereographic projection of circle C is the straight line C' along which the plane b1 intersects the plane n (in which circle C lies) (Fig. 106). Let us now examine the construction of maps. The geographic coordinates of a point lying on the earthâ&#x20AC;&#x2122;s surface are called the latitude (p and the longitude 0. The lines on which the latitude (p is the same, (p = constant, are termed parallels. These are sections of the earth by planes perpendicular to the axis N S of the poles. The lines on which the longitude 8 has the same value, 0 = constant, are termed the meridians of the earth. They are semicircumferences of the great circles of the earth, the boundary points of which are the poles N and S . The network of meridians and parallels of latitude on the earthâ&#x20AC;&#x2122;s surface is an orthogonal network. Figure 107 depicts, in axonometric projection, the parallels of latitude and the meridians of a sphere. Ordinarily, two methods are employed in the construction of a map o f the earth with the use of stereographic projection: in one method the map of the northern and southern hemispheres is done as follows. Let N and S be the north pole and south pole, respectively, and let n and s be the planes tangent to the earth at these poles. To construct a stereographic projection of the northern hemisphere, one projects this hemisphere from the south pole S on the plane tangent to the earth at the north pole, and to construct a map of the southern hemisphere, one projects that

Inversion

Fig. 107

319

Fig. 108

hemisphere from point N on the plane s. The result is a map made up of two circles: one is a map of the northern hemisphere, the other is a map of the southern hemisphere. Of course, this construction is subjected to a similarity transformation. Figure 108 gives, in axonometric projection, just such a construction of the map of the southern hemisphere. In such a mapping of the earth, the parallels of latitude of the southern hemisphere, for example, are projected into concentric circles lying in the plane s and having the common centre S ; the equator o1 is projected into the circle a inside which are the projections of all parallels. The semi meridians of the southern hemisphere (or, similarly, of the northern hemi sphere) are projected into radii of circles (on maps, this construction is of course supplemented by a similarity transformation). If we change the longitude 9 at equal intervals (from 0° to 180°), then the corresponding radii of the circles a will turn in succession through one and the same angle. Figure 109 depicts the construction of meridians under a change

Fig. 109

Fig. 110

320

Problems in Geometry

780' 165'

15 105° go - ^

105° 30° 75' F ig . I l l

in longitude 0 of 15°. When constructing parallels corresponding to identical intervals of latitude (from 0° to 90°) one can do as follows: construct the section of the sphere (Fig. 110) by some plane passing through the straight line NS. Consider, for example, the southern hemisphere. Let PSQ be the semicircle of the southern hemisphere along which the drawn plane intersects the hemisphere, and let s be the straight line of the intersection of this plane with the plane sx. Divide the semicircle PSQ into several equal parts (in Fig. 110 they are indicated by points with latitudes 0°, 15°, 30°, 45°, 60°, 75°, 90°) ; then the projections of the points of division from point N on straight line s with the same latitudes (0°0°, 15°15°, 30°30°, 45°45°, 60°60°, 75*75°, 90°90°) yield the diameters of the parallels. Construction of the parallels can be carried out together with that of the meridians. Just such a construction has been carried out in Fig. 110. Since the circle a (the representation of the equator) is homothetic to circle (NS) with diameter N S (the homothetic ratio is equal to 2), it follows that when constructing meridians and parallels, one need not construct the circle (NS). That is the construction made in Fig. I ll for both hemispheres. This geometric method of constructing meridians and parallels brings us to the conclusion that the distribution of meridians or, as we shall say, the spectrum of meridians, is uniform and the spectrum of parallels expands as we approach the representation a of the equator (for the time being ^ve note this property visually for parallels). We will give an analytic proof that the spectrum of parallels expands near the equator. Let M be an arbitrary point of the sphere Q, (p the latitude of point M , and M ' the stereographic projection of point M from point N on the plane sv Let us now construct a section of the sphere Q1 by the plane N SM (Fig. 112). Since

321

Inversion

Fig. 112

Fig. 113

and, consequently, (we assume that N S = 1)

Now let us prove that if we take three points with latitudes <p1 < q>2 < <P3 such that (p2 — <Px = <p3 — <p2, then the corresponding values of the func tion x , that is,

will be connected by the relation x 1 — x 2 > x 2 — x3. This then is proof that the spectrum of representation of parallels is expanded near the repre sentation of the equator (Fig. 113). The inequality x± — x 2 > x 2 — x3t which we want to prove, is equivalent to 2jc2 < x± + x3 or

or 1 — sin cp2 cos <p2

2 cos <p2 cos <Ps - <P1 2

That is, 1 — sin <p2 ^ _______ cosq)2_______ cos 2 1 -8 1 0

C0S<P?JZh-+ sin <p2 2

Problems in Geometry

322

— <Pi

Since cos <p2 > 0, cos -

> 0, sin (p2 > 0, it follows that the inequa

lity is equivalent to the following: (1 — sin q>2) ^cos <Pa~<Pi1 + sin q>2jc c o s 2 (p2 or (D., (toy cos —-----— + sin q>2 < 1 + sin <p2. That is. __ (P s-(pi ^ , 1. cos----------< 2 This inequality is valid since 0° < <Pz - <Pl < 90°; the inequality x x — — x 2 > x 2 — x3 is equivalent to it and, hence, it too is valid. Remark. The very same result can be obtained by using derivatives. We have

— (i+ i)- ( t +t )

—

< 0.

Consequently, on the interval [0, n/2], the function x = cot ( - - + — | V4 2) is a decreasing function. Furthermore,

* =

)

>0,

(pe [0, n/2].

and so on the interval [0, n/2] the graph of the function x = is convex down and this means that on that interval the Jensen inequality holds true: x2 <

X! + X2

Inversion

323

That is, xx — x 2 > x 2 — x 3 > 0 (x± — x 2 > 0, x 2 — x 3 > 0 since 0° ^ < (Px < q>2 < <Ps < 90°, and on the interval [0, ic/2] the function x = = cot ( — + — ) is a decreasing function). V4 2) The established fact of the spectrum of parallels becoming sparser as one approaches the equator can also be substantiated approximately: the accompanying table contains the values of the function x = cotl — + +

for the following values of q>: 0°, 15°, 30°, 45°, 60°, 75°, 90° and

the values of Ax = x t — x t +1 (/ = 1,2, 3,4, 5, 6). We see that the values of Ax diminish as the latitude increases, that is, as we recede from the equator <p

0°

1,000

15°

0.7673

30°

0.5774

45°

0.4142

60°

0.2679

75°

0.1317

90°

0.0000

0.2327 0.1899 0.1632 0.1463 0.1362 0.1317

Let us now examine the stereographic projections of the western and eastern hemispheres. Let a be the meridian from which we reckon the longitudes 0, and W V the diameter of the earth perpendicular to the plane in which the meridian a lies. The stereographic projection of the western hemisphere (Fig. 114 shows the lower half of the sphere) is obtained by projecting the western hemisphere from point W on the plane j8, which is tangent to the western hemisphere at the point V. Since such projection coincides with the inversion 1 = [W ,W V2\ for points of the sphere Ql9 it follows that the meridians and the parallels, as forming an orthogonal network on the earth, will project into an orthogonal network of circles on the plane Here, the meridians of each of the hemispheres (in Fig. 114 we have the western hemisphere) will project into arcs of circles passing through the points N and S for which these points serve as boundary points. Note that the spectrum of representations of the meridians (of any one of the hemispheres) is exactly the same as the spectrum of parallels

324

Problems in Geometry

F ig . 1 1 4

under the foregoing stereographic projection of parallels of the northern and southern hemispheres. However, the centres of the arcs representing the meridians of each of the hemispheres must be constructed in accord with Fig. 115 which shows that construction (a map of the western and eastern hemispheres). As for parallels, their construction should be based on the following reasoning: since meridians and parallels form an ortho gonal network, this network must also remain orthogonal in the repre sentation of the sphere under a stereographic projection. In particular, the parallels must be orthogonal to the representation of their principal meridian a. Besides, the centres of parallels lie on the axis N S of the N and S poles. From this we obtain the following method of a geometric construction of a parallel passing through a given point M of the principal meridian a : at point M draw a tangent to the circle a (to the representation of the principal meridian). Suppose P is the point in which this tangent line intersects the straight line NS; then the circle with centre at point P and radius PM will be the representation of the parallel passing through point M (Fig. 116). In Fig. 117 are constructed the representations of the parallels (and meridians, in accord with Fig. 115, that correspond to the values of latitude 0°, 15°, 30°, 45°, 60°, 75° for both hemispheres — western and eastern). From Fig. 117 it is evident that the spectrum of parallels expands near the poles. We will now prove this analytically. Let q> be the latitude of point M (Fig. 116). Setting N S = 2, we have PM = cot <p, OP = cosec q> and, hence, v = OK = OP — PM = cosec cp — cot cp = ----- CQS ^ = tan — • sin cp 2 This function y = tan((p/2) is an increasing function on the interval [0, tz/2] and as <p varies from 0 to n/2, the function y — tan(<p/2) increases from 0 to 1. We will prove that the values of the latitudes <pl9<p2 9<p3 , which have equal increments (on the interval 0 ^ cpx < (p2 < cp3 ^ n/2; cp2 —

F ig . 11 7

326

Problems in Geometry

— q>! = q>3 — (p2 > 0), that is, <p2 — (<px + <?3)/2, are associated with the values of the function Ti = tan (<pJ2), y2 = tan(<?>2/2), y 3 = tan((/>3/2) such that y 2 — yt < y 3 — >’2 or that 2y2 < y\ + y3. Such is the proof that the spectrum of parallels becomes less dense near the poles. Thus, we have to prove that 2 tan(<p2/2) < tan(<pj/2) + tan(<p2/2), that is, 2 sin(<jg2/2) < s in ^ ! + <p3)/2) cos(<p2/2) cos(('/)1/2) cos(<p3/2) or 2 sin(^>2/2) cos(q>2/2)

sin <p2 cosOx/2)cos (< p 3 1 2 ) ’

which is equivalent to cos(<px/2) cos(<p3/2) < cos2(<p2/2) or < 1 + cos </>2,

COS<j»2 +COS

that is. 2

This inequality is valid since 0 < (<p3 — <px)/2 < n/2; the inequality y2 < (yx + y3)/2 is equivalent to it and, hence, is true as well. The same values can be obtained if one uses derivatives. We have y = tan(<jo/2). ---------------> 0, 2 cos2((p/2)

<Pe [0, n/2].

and so the function y = tan(<p/2) increases on the interval [0, nj2]. Fur thermore, y

sin(W2) > o 2 cos3(<p/2)

e [0, tt/2].

consequently, the graph of the function y = tan(<p/2) on the interval [0, 7c/2] is convex down, and therefore on that interval the following Jensen inequality holds true: y 2 < (ji + y 3)/2, where yl9y2,yz have the aboveindicated values.

327

Inversion

Finally, the fact that the spectrum of parallels expands near the poles is confirmed by the following table too, in which are given the values of the function y = tan(<p/2) at 15° intervals of latitude and the values of Ay = J i +1 — y% O' = 152, 3,4, 5,6). y

<p 0°

0.0000

15°

0.1317

30°

0.2679

45°

0.4142

60°

0.5774

75°

0.7673 \

90°

1.0000

0.1317 0.1362 0.1463 0.1632 0.1899 0.2327

Incidentally, all of this immediately follows from the fact that the spec trum of the function y = tan(<p/2) repeats “in inverse order” , as q> varies from 0° to 90°, the spectrum of the function x = cot

<P_ 2 )

^compare the tables of the values of the functions y = tan —and x =

If we complete the representations of the meridians and parallels to full circles when constructing maps of the western and eastern hemispheres (in stereographic projection), then we obtain a set of all circles passing through the points N and S (meridians) and a set of all circles orthogonal to the representations of the meridians (parallels) (Fig. 118). The set of all circles passing through two fixed points N and S is called an elliptic pencil o f circles; the points N and S are then termed base points. The set of all circles, each of which is orthogonal to all circles of the elliptic pencil with base points N and S is termed a hyperbolic pencil o f circles, which is the conjugate of the elliptic pencil; the points N and S are termed limit points of the hyperbolic pencil, or Poncelet points of such a pencil. When representing a network of meridians and parallels under a stereographic projection of the western and eastern hemispheres, we have to

328

Problems in Geometry

Fig. 118

isolate from the elliptic pencil of circles and the conjugate hyperbolic pen cil that part contained within the circle of the elliptic pencil, whose dia meter is the line segment NS, where TVand S are base points of the elliptic pencil. A stereographic projection of a network of meridians and parallels may be carried out by chosing, on a sphere, any two diametrically opposed points A and B and projecting the sphere from point A on a plane tangent to the sphere Qx at the point B. Since a stereographic projection is a con formal mapping of the sphere on a plane, it follows that the orthogonal network of meridians and parallels will be mapped into the same kind of orthogonal network of two pencils of circles: the meridians are mapped into an elliptic pencil with base points N and S, and the parallels into the conjugate hyperbolic pencil with limit points N and S. The projection of the principal meridian is some circle K that passes through the points N and S, and N S is a chord of that circle. A map of one of the hemispheres (western or eastern) will lie inside circle K. The construction of the map of the other hemisphere is similar. If we isolate any one of the circles (say circle a) from the hyperbolic pencil of circles and regard it as a representation of the equator (and regard the Poncelet point lying inside the chosen circle a as a represen tation of the pole), then we obtain a map of the northern or southern hemisphere. Representations of the parallels are all the ciicles of the hyper bolic pencil that lie inside the circle a, and the semimeridians are arcs of

Inversion

329

the conjugate elliptic pencil that lie inside the circle a and emanate from the pole (located inside a). In this situation, these semimeridians will join up into an arc of one circle if the sum of the longitudes (east and west) is equal to 180° (or, to put it differently, if the difference between the posi tive and negative longitude is equal to 180°). Such is the qualitative picture of representation of meridians and parallels under an arbitrary stereographic projection of the western, eastern, northern and southern hemispheres. We will now show how an exact geometric construction is performed of the spectrum of meridians and parallels under an arbitrary stereographic projection. Let us begin with the construction of the spectrum of meridians and parallels when representing the western (or eastern) hemisphere; we assume the latitude and longitude to vary in 15° intervals. Consider the section Q = ANBS of the sphere by a plane passing through the dia meters AB and SN (Fig. 119). Let N and S be projections of points# and S from point A on a tangent to the section Q at the point B. Since the meridians pass one into another (and the parallels are invariant) under rotation of the sphere about the axis SN, it follows that any circle with chord N S may be taken for the representation a of the principal meridian with a supplement to complete the meridian to a full circle. Since a stereo graphic projection is a conformal mapping of the sphere on a plane, in order to construct the representation of the meridians in equal intervals of longitude, one has to construct arcs of circles whose centres lie on the midperpendicular of line segment NS, the centres lying inside the repre sentation a of the principal meridian and forming a sequence of equal angles. For example, if the longitude changes in 15° intervals, then we obtain the representations of thirteen meridians (the circle a is a repre sentation of two meridians with longitudes of 0° and 180°), which intersect in succession at 15° angles. In Fig. 120, the constructions have been carried out in superposed planes: the plane of the circle Q is brought to coincidence with the plane tangent to the sphere Qx at the point B (the coincidence is attained by rotating about the straight line along which these two planes intersect). For a geometric construction of the representation of these meridians, note that the angle between two intersecting circles is equal to the angle between their radii drawn to the point of intersection of the circles. For this reason, to construct the radii and centres of the representations of the meridians, turn the radius NO of the circle a about point # in a succes sion of, say, 15° angles. To do this, construct the semicircle CDE with centre N, whose diameter CE is tangent to the circle a at the point N (see Fig. 119). Divide the arc CDE into 12 equal parts and project the points of division from point N onto the midperpendicular of line segment NS. These projections T are the centres of the representations of the meridians (which correspond to 15° intervals of longitude), and TN are their radii.

Problems in Geometry

Fig. 119

330

Inversion

331

Note that the idea of conformity of a stereographic projection that was made use of in this construction could have been utilized earlier in the construction of the spectrum of meridians in a simpler case. To construct the representations of parallels at, say, 15Â° intervals of latitude, one should take advantage of the method of constructing the

332

Problems in Geometry

centre of a stereographic projection of a circle that was given above for an arbitrary diameter AB of the sphere Qx with respect to the circle lying on that sphere. Let OT be a radius of the circle Q perpendicular to S ^ . Divide the arc S j^TNj^ into 12 equal parts. Let M be one of the points of division (say, the closest one to Nx). The tangent to the circle Q at point M intersects the straight line Sx in the vertex P of a cone tangent to the sphere Qx along a parallel passing through point M. The projection P' of point P from point A onto the straight line N S will be the centre of the representation of that parallel. If PL is the tangent, for example, to the representation a of the principal meridian (L is the point of tangency), then PL is a radius of the representation of the parallel at hand (and P is its centre). Thus, Fig. 120 depicts the representations of half the parallels and half the equator (the centre of the representation of the equator is the point of intersection of the straight line N S with the straight line passing through point A parallel to NXS^). Let us now consider the construction of the spectrum of meridians and parallels under a stereographic projection, say, of the northern hemisphere for an arbitrary position of the diameter AB of the sphere Qx with respect to the equator. Again construct the section Q of sphere by the plane AN1BS1 and let DE be a diameter of the circle Q along which the plane of the section cuts the plane of the equator. Let N and S be the projections of the points N-l and Sx from point A on the tangent line to the circle Q at the point B. To make a representation of the equator, drop from point A a perpendicular to the diameter DE. Let M be the point of intersection of this perpendicular with the tangent to the circle Q at the point B. Then M is the centre of the representation of the equator, and MA is the radius of that representation. We will perform all subsequent constructions in the superposed planes: the plane of the circle Q with the plane tangent to the sphere Qx at point B (as above, coincidence is attained by turning about the straight line along which these two planes intersect). To construct the representations of parallels at, say, 15° intervals of latitude, divide the arc EN1 of the circle Q into 6 equal parts. Let L be one of the points of division (for example, the one closest to E). If Q is the point of intersection of the tangent to the circle Q at point L with the straight line SxNl9 then Q is the vertex of a cone that is tangent to the sphere ^ along the parallel passing through point L. The projection R of point Q on the straight line BN from point A is the centre of the repre sentation of the parallel that passes through point L. Now the radius of the representation of this parallel is equal to the line segment RE of the tangent drawn to the representation of any meridian (that is, to any circle passing through the points N and S). The foregoing method is used to construct the parallels of the northern hemisphere for 15-degree intervals of latitudes from 90° to 0°.

Inversion

333

To construct the representations of meridians, note (we have already mentioned this) that when the sphere is rotated about the axis N ^ , the meridians pass into one another (while the parallels are invariant). There fore, in order to construct semimeridians we can take, for the two principal semimeridians that correspond to longitudes of 0° and 180° and form a single arc of the circle, any arc of the circle that passes through the points N and S and lies inside the representation of the equator. Construct a semicircle with centre N, divide it, say, into 12 equal parts and project the points of division from point N onto the midperpen dicular of the line segment SN (exactly the same approach was used above to construct the representations of the meridians under a stereographic projection of the western and eastern hemispheres). The projections will be the centres of semimeridians emanating from point N and forming a succession of 15-degree angles. The semimeridians, the difference of the longitudes of which is equal to 180° (180° — 0° = 180°, 165° — (—15°)= = 180°, 150° — (— 30°) = 180° and so on) are joined in the representation into one single arc of the circle, which arc lies inside the representation of the equator. This occurs, for instance, in the most elementary case as well (it is shown in the figure), where the representations of the meridians (that is, the radii of a circle) for the values of the longitudes (the difference of which is equal to 180°) are joined into a diameter of the circle, which is the representation of the equator. Of course, the same also occurs on the sphere Q1 on any one of the two hemispheres (northern or southern).

CHAPTER V

BASIC DEFINITIONS, THEOREMS AND FORMULAS Sec. 1. Determinants of order three

A third-order determinant is introduced by the following relation: <*i bx cx A=

02 ^2 ^2 a3 ^3 ^3

—

01^ 2^3

2 ^ 3 ^ 1 " 1“

3 ^ 1 ^ 2 “ 1“

Q J * 1 CZ

1^ 3 ^ 2

ah b i9 cv (i = 1,2, 3) are termed the elements of the determinant. Basic properties: 1°. A determinant remains unchanged under a transposition, that is, under an interchange of rows and columns: a 2 03

bl b 2 b3 Cl c2 C3

Gl h

bz C2 b a* 3 C3

2°. A determinant preserves its absolute value but changes sign when any two columns are interchanged; for example:

C0O «$» «?

01 b± Cl b i ax c1 b% a<± C2 = — 02 b<2 C3 #3 ^3 C3

A determinant remains unchanged under a cyclic permutation of its columns: Ci 0i b± bi cx 0i 02 b% C2 = b% C2 02 = c3 02 b% 63 C3 03 CI3 b$ c3 C3 03 63 0 i bx Ci

and changes sign if the cyclic order of its columns is disrupted [see (3)]. 3°. If two columns of a determinant are the same, the determinant is equal to zero. 4°. If all the elements of some column of a determinant are equal to zero, the determinant is zero.

Basic Definitions, Theorems and Formulas

335

5°. A determinant is a linear function of its columns. This means that the following relation holds true: k x bx cx Xax + jukx bx c1 01 *1 01 Xa2 + fxk2 b2 c2 = X ci2 b2 c2 + ^ k 2 b2 c2 kz *3 cz Xa3 + H^a h c3 03 * 3 03 and analogous relations for the second and third columns. 6°. A determinant remains unchanged if a linear combination of any two columns is added to another column, that is. ax + Xbx + [icx bx cx 01 *1 01 a2 + Xb2 + nc2 b2 c2 = <x2 b2 c2 a2 -(- Xb3 -t- nc3 b3 c3 03 * 3 03 and analogously for the second and third columns. In particular, a deter minant remains unchanged if any column is added to or subtracted from another column. 7°. For a determinant to be equal to zero, it is necessary and sufficient that its columns be linearly dependent, that is, that there exist numbers A, n, v among which at least one is nonzero and such that Xax + \ibx + \cx = 0, Xa2 + i*b2 + vc2 — 0,

(7)

Xaz + /x*3 + vcz = 0. In other words, for a determinant to be equal to zero, it is necessary and sufficient that one of the columns be a linear combination of the other two, for example: cx = Xax + \ibi, c2 = ?m2 + \ib2,

(8)

cz— kaz + fibz. 8°. Multiplication of determinants is carried out with the aid of the following formula: *1

ct2 b2 c2

y 2 Z2

y& z 3

01*1 + * 1 * 2 02*1 +

^1 * 3

*2*2 +

02*3

03*1 + * 3 * 2

03*3

yi + * 1 ^ 2 + 0 1 J 3 axzx + b 1z 2 + c1zz 02^1 + b2y 2 + c2yz 0 2 Z 1 + * 2 Z2 + C2 Zz • (9) 0s7i + bzy 2 + czyz azzx + bzz2 + czzz

Problems in Geometry

336

Since a determinant remains unchanged under the transpose operation, we can obtain another three modes of multiplication of determinants if one of the factors of the left-hand side of (9) is transposed. The cofactor of an element of a determinant is the coefficient of tha element in formula (1). For example, the cofactor Cx of element cx in determinant (1) is Ci = a2b3 a3b2. The cofactor A2 of element a2 of the determinant (1) is A 2 = b3Ci b±c3 and so forth [in all, there are 9 cofactors A i9 Bh Ct (/ = 1,2, 3)]. A determinant is equal to the sum of the products of the elements of any column into the corresponding cofactors: A — a^A-! + a2A2 + a3A3, and similarly for the second and third columns. A determinant is equal to the sum of the products of the elements of any row into the corresponding cofactors: A = a1A1 + b1B1 + cxCl9 and similarly for the second and third rows. The sum of the products of the elements of any column into the corres ponding cofactors of the elements of another column is equal to zerot, for example: d\Bx + a2B2 + a3Bs = 0 (and there are five other analogous relations). The sum of the products of the elements of any row into the corresponding cofactors of the elements of another row is equal to zero, for example: axA 2 + b1B2 + 0 ^ 2 = 0 (and there are five other analogous relations). The minor of an element of a determinant is the determinant obtained by deleting the row and column at the intersection of which that element lies. For example, the minor of element cx in determinant (1) is the determinant a2 b2 — tt2b3 #3^2 #3 b3 and the minor of element a2 of determinant (1) is the determinant ^ Cl = bxcz — b3cl9 and so on. ^3

The cofactor A of any element of a determinant is equal to the minor M of that element multiplied by (— l)i+J, where i and j are, respectively.

Basic Definitions, Theorems and Formulas

337

the number of the row and the number of the column, at the intersection of which that element lies: A = ( - 1)i+J M. Sec. 2. Vector algebra A vector is a directed line segment. Vectors are symbolized as follows: a, b, x, r ,. .. or AB, CD, FQ,. .. A vector AB is said to be nonzero if the points A and B are distinct; a vector AB is said to be a zero vector if the points A and B coincide. The zero vector is designated by the symbol 0. Two nonzero vectors AB and CD are said to be collinear if the straight lines AB and CD are collinear, that is, either parallel or coincident. The zero vector is assumed to be collinear with any other vector. If the vectors a and b are collinear, then we can write a|| b. If the vectors a and b are collinear and in the same direction, we can write a jf b, and if they have opposite directions we write afj,b. The magnitude (or absolute value) of a vector AB is the length of the line segment AB. This can be symbolized as: \AB\, AB, |a|, or a. The nonzero vectors a and b are said to be equal if affb and |a| = |b|. The zero vector is assumed to be different from any nonzero vector. Zero vectors are all assumed to be equal. The nonzero vectors AB, CD, EF are said to be coplanar if the straight lines AB, CD and EF are coplanar to one and the same plane (a straight line and a plane are said to be coplanar if the line is either parallel to, or lies in, the plane). If there is at least one zero vector among the vectors — > — >—> AB, CD, EF, then they are regarded as being coplanar. To lay off a vector a from a given point A means to construct a directed line segment AB equal to the vector a. The vector —a = BA is said to be the opposite of vector a = AB. The sum a + b of the vectors a and b is a vector which is constructed as follows: from an arbitrary point A lay off the vector a: ~AB = a and from the point B lay off the vector b: B C = b. 2 2 -8 1 0

338

Problems in Geometry

Then a + b = AC. The properties of a sum of vectors are: a + (b + c) = (a + b) + c, a + 0 = a, a + ( - a) = 0, a + b = b + a. The difference a — b between two vectors is a vector x such that b + x = a. To construct the difference a — b, lay off vectors a and b from some point O: OA = a, OB = b. Then a — b = BA. The product Aa o f a number A ^ 0 by a vector a ^ 0 is a vector defined as follows: |Aa| — \X\ |a|, and we have Aa ffa if A> 0, and Aa j j a if A< 0. If either A = 0 or a = 0, then, by definition, Aa = 0. The properties of a product of a number (scalar) by a vector are: 1 a = a, A(^a) = (Aju) a, A(a + b) = Aa + Ab, (A

= Aa -f~ fia. j|

If a || b ^ 0, then the ratio — is a number A such that Ab = a. If b a |al 0 # a||b # 0, then — = + — , the sign being if a ttb and the sign b |b| being

” if a | j, b. If a = 0, b # 0, then — = 0. b A linear combination of the vectors a, b, c, . . . is the sum Aa + pb + + vc + . . . The vectors a, b, c, . . . are said to be linearly dependent if there are num bers A, p, v such that at least one of them is different from zero and such that Aa + pb + vc + . . . = 0. If this equation is possible only for A = p = = v = . . . = 0, then the vectors a, b, c are said to be linearly independent. For the vectors a and b to be collinear, it is necessary and sufficient that they be linearly dependent. For the vectors a, b, c to be coplanar, it is necessary and sufficient that they be linearly dependent.

Basic Definitions, Theorems and Formulas

339

A basis el9 e2 in a plane is an ordered pair of noncollinear vectors el9 e2 coplanar to that plane. If the vectors of the basis are unit vectors and are mutually perpendicular, the basis is said to be orthonormal. Ordinarily, the vectors of an orthonormal basis are designated as i, j : i ± i , |ii = ui = i. Suppose a basis el5 e2 has been introduced in a plane; a is an arbitrary vector coplanar to that plane. There exists a pair of numbers x, y (and only one pair) such that a - xeA+ ye2. The coefficients x and y of ej, and e2 in this expansion of the vector a with respect to the basis el5 e2 are called the components (or coordinates) of vector a in (or with respect to) the basis el9 e2. If the vector a in the basis el9 e2has components x, y , then we write a — {x, y) or a {x, y}. In space, the basis el5 e2, e3 is formed by an ordered triplet of noncoplanar vectors el9 e2, e3. If the vectors of a basis are unit vectors and orthogonal in pairs, then the basis is said to be orthonormal The vectors of an orthonormal basis are ordinarily designated as i, j, k: i l j , j l k , k_Li, |i| = |j| = |k| = l. If we introduce a basis el5 e2, e3 in space and let a be an arbitrary vector, then there is only one triplet of numbers x 9 y , z such that a = xex + ye2 + ze3. The numbers x, y , z are called the components (or coordinates) of the vector a in the basis e2, e3. If the vector a has components x, y 9 z, then we write a = {x, y 9 z} or a (x, y , z}. Two vectors a and b are equal if and only if their corresponding compo nents are equal. If, in the basis el5 e2, we have a = {x,j}, b = {x’,y'}, then a - f b = {x + x', y + / } , a - b = {x - x \ y - y'}. An = {Ax, Ay}. If, in the basis el5 e2, e3, we have a = {x, y, z}, b = {x\ / , z'}, then a + b = {x + x', y + y', z + z'}, a — b = {x — x', y — y \ z — z'}, /la = {Ax, Xy9Az}.

340

Problems in Geometry

A necessary and sufficient condition for two vectors a = {x , y } and b = {x \ y'} to be collinear in a plane is the equation x9 / A necessary and sufficient condition for two vectors a = (x, y, z} and b = {x' y \ z'} to be collinear in space is the condition y z | |z x x y x’ y / z’ \ iz' x ’ A necessary and sufficient condition for the coplanarity of three vectors a = {x ,y , z}, b = {x ’, y z'}, c = {x", y", z"} is the equation x y x' y

z z 9 = o.

A scalar product a-b (or ab) of two nonzero vectors a and b is the product of their magnitudes by the cosine of the angle between them: ab = |a| |b| cos <p. If a = 0 or b = 0, then, by definition, ab = 0. The properties of a scalar product of two vectors are: a2 = aa > 0 if a # 0, a2 = 0 if a = 0, ab = ba, (Aa) b - A(ab),

a(b + c) = ab + ac. The scalar product aa (ordinarily we write a2) is called the scalar square of the vector a. If, in an orthonormal basis i, j in the plane, a = {x,y},

b = { * ',/} .

then ab = xx' + y y \ and if * = {x>y}, b = { * ',/} in an arbitrary basis e2, e2, then ab = guxx' + gn (xy' + x'y) + g22yy'.

Basic Definitions, Theorems and Formulas

341

where £ij ei e7* The collection of scalar products = e, ey- of the basis vectors is termed the fundamental tensor of that basis. If, in the orthonormal basis i, j, k, a = {x, y, z}, b = {x', / , z'}, then ab = x*' + yy' + zz\ and if a = {x, y, z), b = {jt\ / , z'}

in the arbitrary basis el9 e2, e3, then ab = gn xx' + g12 yy' + g33 zz' + g12(xyf + x'y) + gniyz' + y ’z) + g*i(zx' + z’x \ where gxi = e; ey (the fundamental tensor o f the basis el5 e2, e3). A plane is said to be oriented if a basis el5 e2(with a fixed order of vectors) is introduced in the plane and the whole set of pairs a, bof noncollinear vectors is partitioned into two subsets in the following manner. If an ordered pair a, b of noncollinear vectors a and b have the same orientation as that of the basis ex, e2; a>MT eJL>e2, that is, OAB 1 1 OEi#,, where OA = a, OB = b, OEx= elf OE2 = e2. then this ordered pair a, b is called the right-hand pair, or the pair with posi tive orientation. But if a, b and el5 e2 have opposite orientations, „ a, b U e 1? e2, ------ ^ ------ > that is, (with notation the same as above) OAB f [ OExE2, then the ordered pair a, b is called the left-hand pair, or the pair with negative orientation. Suppose the noncollinear vectors a and b lie in an oriented plane. The cross (or pseudoscalar) product (a, b) of vector a and vector b is a number whose absolute value is equal to the area of a parallelogram with sides OA = a, OB = b and which is positive if the ordered pair a, b is a righthand pair, and negative if that pair is a left-hand pair. If a j| b, then, by definition, ab = 0. The properties of a cross (or pseudoscalar) product are: (a, b) = — (b, a), (a, b + c) = (a, b) + (a, c), (A a, b) = A(a, b).

342

Problems in Geometry

Suppose, in an oriented plane, there are given two noncollinear vectors a and b, the angle between them being equal to a. The angle cpformed by vector a and vector b is the angle cp = a if a, b is a right-hand pair, and the angle cp = —a if a, b is a left-hand pair. If the vectors a and b are nonzero vectors and a ] f b, then, by definition. <p = 0, and if a b, then, by definition, cp = n. The angle in the orthonormal basis i, j, then xx' + yy9 COS cp = . ]fx2 + y 2 ]fx'2 + y'2

b = { * ',/} . xy9 — x'y sm cp = ■■■■■■■ ]fx2 + y2 J/V2 + y '2

If the nonzero vectors a and b are given by their components in an arbitrary basis el9 e2, then gn x x 9 + g12(xy' + x'y) + g22yy' COS (p = ‘ r......................................... ,7......... ■— ■ = ; , Vgn X2 + 2g12 xy + g22y2 Vgn x ’2 + 2g12 x 'y ’ + g22y '2 sm cp — .-----------------------—

Ygr-(xy'- - x'y)._ .—..

. Zy Yg n x 2 + 2g12 xy + g22y 2 Yg n x'2 + 2g12 x' / + g22y '2

where g is the Gram determinant: £ll ^12 = («i e2)2. g 21 <§22 Our formulas specify a set of values of the angle cp of the form {cp + + 2k 7i}, where k assumes all integral values and cp is some value of the angle formed by vector a and vector b. Suppose vector a lies in a plane oriented by the basis el9 e2. The vector Up is a vector obtained by rotation of vector a through the angle cp (if 0 < cp < n, then the pair a, a^ is a right-hand pair; and if —n < cp < 0, then it is a left-hand pair; if cp = 0, then a^ = a, if cp = n9then a^ = —a). The vector obtained by rotation of vector a through the angle n/2 is symbolized thus: [a]. We have the following formula: (a, b) = [a] b. g=

If, in the orthonormal basis !, j, a = {x,y},

b = {x9, / } ,

343

Basic Definitions, Theorems and Formulas

then a— y sin q>, x sin <p + y cos <p}, [a] = { - y, *}, (a,b) = * y x' y' If the vectors a = {x, y } and b = { x \ / } are specified in an arbitrary basis el9 e2, then (a, b) =

\x y

Two bases el9 e2 and e1, e2 in a plane are said to be dual bases if when i = j, [0 when / / j. The vectors of dual bases are connected by the relations W [ej ^ [e2] _ (ea.ej)' (el5 e2) ' 1 (e2, e1) ’ 2

[e1] (e1, e2)

In other notation: the vectors of dual bases a, b and a*, b* are connected by the relations (a,>*) = (a, b*) = 1, (a, b*) = (a*, b) = 0, [b] [b, a]

b* _

[a] (a, b)

[b*] , b (b* a*)'

[a*] (a*, b*)

An expansion of an arbitrary vector a in terms of the bases e1; e2.and e1, e2 is of the form a = (ae^-ej + (ae2)-e2, a = (aei)^1 + (ae2)-e2, where in parentheses we have scalar products of the vectors ( Gibbs' formulas). The numbers a1 = ae1, a2 = ae2, that is, the coefficients of ej and e2 in the expansion of the vector a in terms of the vectors ej, e2 are called the contravariant components of the vector a with respect to the basis ex, e2, and the numbers a± = aex, a> = ae2, that is, the coefficients of e1, e2 in the exapansion of vector a in terms of the basis e1, e2, are called the cova riant components of the vector a with respect to the basis ex, e2. If one of the vectors is specified in the basis el5 e2 by its contravariant components. a = x e f+ ye2 = {x, y).

344

Problems in Geometry

and the other by its covariant components, b = jc'e1 + / e * = [x\ / ] , then ab = xx' + yy'. [a] = f g [ - y , x ] . = U ? n cos + Yg sin (p) + yg.,2 cos <p] (when <p = 7i/2 we get the preceding formula). A space is said to be oriented if a basis e1? e2, e3 has been introduced and the set of triples of noncoplanar vectors has been partitioned into two subsets. If an ordered triple a, b, c of noncoplanar vectors has the same orientation as basis el9 e2, e3, then it is termed a right-hand triple, or a triple with positive orientation. If the ordered triples a, b, c and el5 e2, e3 have opposite orientations, then the triple a, b, c is termed a left-hand triple, or a triple with negative orientation. The triple scalar product (a, b, c) of an ordered triple a, b, c of noncoÂ planar vectors lying in an oriented space is a number whose absolute value is equal to the volume of a parallelepiped with edges OA = a, OB = b, OC = c (O is an arbitrary point) and which is positive if the triple a, b, c is right-hand, and negative if the triple a, b, c is left-hand. If the vectors a, b, c are coplanar, then, by definition, (a, b, c) = 0. The properties of a triple scalar product of three vectors are: (a, b, c) - (b, c, a) - (c, a, b) - - (b, a, c), (Aa + pb, c, d) = A(a, c, d) + ju(b, c, d) (and analogously for the second and third factors). If the space is oriented by means of the orthonormal basis i, j, k and, in this basis, a = {x, y, z}, b = {*', / , z'}, c = z"), then x y z (a, b, c) = x ' y* z â&#x20AC;&#x2122; x" y" z" and, in an arbitrary basis, (a, b, c) = Yg x

y //

Basic Definitions, Theorems and Formulas

345

where g is the Gram determinant £=

£ n

£12

£13

£21

£22

£23

£31

£32

£33

(ei, e2, e3)2.

Suppose two noncollinear vectors a and b are specified in an oriented space. The vector product [a, b] of vector a by vector b is a vector defined by the following conditions: (1) |[a, b]| = |a || b| sin <p, where cp is the angle formed by vector a and vector b; (2) [a, b] _L a, [a, b] l b ; (3) the ordered triple a, b [a, b] is a right-hand triple. If a || b, then, by definition, [a, b] = 0. A vector product has the following properties: [a, b] = - [b, a], [2a, b] = 2[a, b], [a, b + c] = [a, b] + [a, c], [a, b] c = a[b, c] = (a, b, c). If, in an orthonormal basis i, j, k, a = {x, y, z}9 b = {*', / , z'}. then z x ■ ' I}. z' x' liy *

[•>»]=IKz,

' / If

Two bases el9 e2, e3and id e1, e2, e3are ; said to be dual bases if when i = j 9 when i ^ j. The vectors of dual bases are connected by the relations _ [e2e3] _ [e3e J _ _ (ex e2 es) (ei e2 e3) [ei e2 e3) _ 1

[e2 e3] (e1 e2 e3) ’

_ 2

[e^e1] (e1e2 e3) ’

_ 3

[e1 e2] '(e1 e2 e3)

Other notation: the dual bases a, b, c and a*, b*, c* are defined by the relations (a, a*) = (b, b*) = (c, c*) = 1, (a, b*) = (a*, b) = (b, c*) = (b*, c) = (c, a*) = (a, c*) = 0

Problems in Geometry

346

and, furthermore, c* = (abc)’ (abc) [b* c*] [c* a*] ^ [a* b*] (a* b* c*) 5 (a* b* c*) ’ (a* b* c*) Any vector a is expanded in terms of the bases el9 e2, e3 and e1, e2, e3 as follows: a = (ae1) •ex + (ae2)*e2 + (ae3)-e3, a = (aej)•e1 + (ae2)-e2 + (ac3)-e3, where the numbers in parentheses are scalar products of vectors. These are the Gibbs formulas. The numbers a1 = ae* are called the contravariant components (coordi nates) o f the vector a in the basis el9 e2, e3, and the numbers at = aef are called the covariant components (coordinates) o f the vector a in the same basis. If the vectors a and b are specified by contravariant components in an arbitrary basis el5 e2, e3, a = xej + ye2 + ze3 = {x, y, z), b = x'Ci + y'e2 + z'e3 = {x \ y ’, z'}, then the covariant components of the vector product a *

b *

(abc)’

x y\] (z' x' ’ X' y 'W We note here a number of other identities: ac ad ac ad I (a, b) (c, d) = ,[«, b][c, d] = be bd be bd| [a[b, c]] = b(ac) - c (ab), [a, b]2 + (a b)2 = a2b2 (the parentheses in the last two formulas contain scalar products); [[a, b] [c, d]] = c(a, b, d) - d(a, b, c) = b (a, c, d) - a(b, c, d), a(b, c, d) + b(c, a, d) + c(b, d, a) + d(a, c, b) = 0, [a [b, e ]]+ [b [c , a]] + [c [a, b]] = 0, ax ay az (a, b, c) (x, y, z) = bx by bz ex cy cz (a, a) (a, b) (a, c) (a, b, c)2 = (b, a) (b, b) (b, c) (c, a) (c, b) (c, c) ([a, b], [b, c], [c, a]) = (a, b, c)2.

[ • , b ] = y * f K z,

Basic Definitions, Theorems and Formulas

347

There are problems in this text that also involve what are known as slid ing vectors. A sliding vector is also a directed line segment; however, the equality of such vectors is defined as follows: two nonzero sliding vectors AB an CD are said to be equal if the lengths of segments AB and CD are equal, if AB and CD are in the same direction, and if they lie on the same straight line. A sliding vector co is given in a plane by its components x , y and by the moment z = (r, co) with respect to some point O (r = OM, where M is an arbitrary point of the straight line); (r, to) is the cross (or pseudo scalar) product. Sec. 3. Analytic geometry Let us fix a point O in the plane and a basis el9 e2. Lay off vectors el5 e2 from the point O : OE1 = el9

OE2 = e2.

The collection of straight lines Ox = OEl9 Oy = OE2 on which vectors e1? e2 of the basis are laid off from O is termed a general Cartesian system o f coordinates in the plane. The line Ox is the axis o f abscissas, or x-axis, and the line Oy is the axis o f ordinates, or the y-axis. The point O is the origin o f coordinates'. Ox and Oy are the axes o f the coordinate system. A general Cartesian system of coordinates Oxyz in space is defined in similar fashion. If an orthonormal basis i, j (respectively i, j, k) is introduced in a plane (respectively in space) and a point O is fixed, then the corresponding system of coordinates is said to be rectangular Cartesian. The radius vector r of point M is a directed line segment: r - OM. The general Cartesian coordinates x ,y o f a point M in the general Carte sian system of coordinates (<0, el9 e2) in the plane are the coordinates of its radius vector r

= OM in the basis el9 e2: r = xeL + ye2.

The coordinates o f point M in the rectangular Cartesian system of coordinates (O, i, j) in the plane are the coordinates x, y of its radius vector r = OM in the basis i, j: r = xi + y\.

Problems in Geometry

348

Similarly defined are the general Cartesian and rectangular Cartesian coordinates x 9y 9z of point M in space: r = OM = xe± + je 2 + ze3. r = OM = xi + y] + zk. If the point M has coordinates x 9y , then we write M(x, y) or M = (x, y) [in space: M(x, y , z) or M — (x, y, z)]. If rx and r2 are the radius vectors of points A and B, then AB = r2 — r2. In coordinates: = { x 2 — x l9 y 2 — yi}

(in the plane),

AB = {x2 — xl9 y 2 — yi, z2 — Zi}

(in space),

where A = (x l9 yx)9 B = (x 2, y 2) (in the plane); A = (xl9 yl9 zx), B = (x2, y 2, z2) (in space). The length d o f a line segment AB whose endpoints are specified by radius vectors rx and r2 are computed from the formula d = |r2 - Til. In a' rectangular Cartesian system of coordinates: d = ] f( x 2 — xx)2 (y2 — y xf

(in the plane),

d = ][(x 2 - x2)2 (y2 - }\)2 + (z2 - Zj)2

(in space)

In a general Cartesian system of coordinates: d =]fgn (x2 - Xx)2 + 2g12(x2 - xx) (y2 - yx) + g22 (y2 - y x)2 (in the plane) d2 = gn (x2 - xx)2 + g22 (y2 - y j 2 + £33 (^2 - ZiY + 2g12 (x2 - xx) (y2 — y x) + 2g23 (y2 — y\) (z2 - z^ + 2^31 (z2 — zx) (x2 — xx) (in space), where gl7 are the components of the metric tensor introduced in Sec. 2. In particular, the distance r from the point to the origin of coordinates in a rectangular Cartesian system of coordinates: r = ]/x2 + y 2 r = 1/x2 + y 2 + z2

(in the plane) (in space).

Basic Definitions, Theorems and Formulas

349

and in a general Cartesian system of coordinates: r = ]fgnx2 + 2g12 xy + g22y2

(in the plane)

r = U n x 2 + g22yz + gi3 z2 + 2g12xy + 2g2syz + g31zx (in space). Suppose a rectangular Cartesian system of coordinates has been intro duced in the plane. Let us consider an arbitrary point M different from the coordinate origin. The first polar coordinate of point M is the length of segment OM. The second polar coordinate of the point M is the angle q> formed by vector i and the radius vector OM of the point M. If x, y are the coordinates of point M in a rectangular Cartesian system of coordinates, then x = p cos cp, y = p sin q>; x x . y y p = ][x 2 + J 2, cos line segment MXM2 is the number .

m Jm

mm2 No matter what the number X ^ — l and no matter what the nonzero direct------> ed line segment MXM29 there is one and only one point M which divides the line segment M XM2 in the ratio X. If ii and r2 are the radius vectors of the points Mx and M 2 then the radius vector r of point M is defined by the relation

r = ri + ^ . 1+ X In particular, the radius vector of the midpoint of the line segment is equal to the half-sum of the radius vectors of its endpoints: 2

In a general Cartesian system of coordinates, the coordinates of the point M are expressed in terms of the coordinates of the points M x and Af2 by the relations *i + Xx2^ y_i+ 4 v2(in the plane). i + ;/’ i + x~ + Xx2 yx + Xy2 zx + Xz2 --------------------------- 9 V— ------------------------------ 9 Z= ------------------(in space), 1+ X 1+2 1+2

Problems in Geometry

350

and the coordinates of the midpoint of the line segment are: x=

y =

(in the plane),

*1 + X2 + z 2 (in space). 2 2 Points A, B,C, . . . are called collinear if there is a straight line on which they all lie. Points A, B ,C ,D , . . . are said to be coplanar if there is a plane in which all the points lie. If rl9 r2, r3 are the radius vectors of the points A, B,C , then a necessary and sufficient condition for their collinearity is of the form (*i - r3, r2 - r3) = 0 and, in a general Cartesian system of coordinates, “ *3 7i — 73 0 X2 *3 y 2 73 or *i yi 1 *2 72 1 = 0 , *3 ys i where A = (xl9 7i), B = (x29y2), C = (x3, y9). If rl5 r2, r3, r4 are the radius vectors of the points A, B,C , D, then a necessary and sufficient condition for their coplanarity is of the form (fi — r4, r2 - r4, r3 — r4) = 0, and, in a general Cartesian system of coordinates, ------------------------9

*i “ *4

7i - >4zi - *4

*2 “

J 2 — 7 4 *2 — *4

-o .

*3“ * 4 73 “ 74 *3 “ * 4 where A = Oi> 7i> Zi)9 B = ( x 2 , y 29 z2), C = ( x 3 , 73>Z * ) 9 D = ( x A9 yA9 z4). A triangle ABC is said to be a collection of three points A, B,C . An oriented triangle ABC is an ordered set of three points A ,B ,C . If the points A ,B ,C are noncollinear, then A ABC is said to be nondegenerate, and if they are collinear, then the triangle is degenerate. The area (ABC) of a nondegenerate oriented A ABC lying in an oriented plane is a number whose absolute value is equal to the area of A ABC, it is positive if A ABC has a right-hand orientation (that is, the ordered pair CA,CB has a right-hand orientation), and is negative if A ABC has

Basic Definitions Theorems and Formulas

351

a left-hand orientation. If the points A, B, C are collinear, then, by defi nition, we agree that {ABC) = 0. The area (ABC) of the oriented A ABC is computed from the formulas 1 1 (ABC) = — (CA, CB), (ABC) = — (rx - r3, r2 - r3) (r1? r2, r3 are the radius vectors of the points A , B, C). In a general Cartesian system of coordinates.

h 2

* i- * 3

x*— * 3

yi - ys _V9 2 y* -

7i 1

*2 72 1 *3 73 1

In a rectangular Cartesian system of coordinates j y% i 1 (ABC) = — * 1 ~ * 3 J ' l - J ’j ~ X.2 y 2 i * 2 — *3 72— 7S

*3 7 3 1

To compute the area S of A ABC, take the right-hand sides in absolute value: Vg Xi yi 1 *1 *3 J l J3 = T -m o d S = -~- mod *2 7s 1 * 2 - *3 yz — y* *3 7s 1 7i 1 S = — mod = —- mod 2 2 x2 j>2 1 ^ 2 -^ 3 ^ 2 -^ 3 *3 7a 1

In these formulas, A = (xlf jj), 5 ; The ratio

(* 2, 72) ,

C = (*3, J 3).

of the nondegenerate A PQR and A ABC lying in a plane is a number whose absolute value is equal to the ratio of the areas A PQR to the area of A ABC and which is positive if A PQR and A ABC have the same orientation (that is, RP, RQ | t CA, CB) and negative if A PQR and A ABC have opposite orientations. If PQR is a degenerate triangle and ABC is a nondegenerate triangle, then, by definition, PQR ABC

Problems in Geometry

352

We have the formula PQR >

ABC

(PQR)

(ABC)

{if A PQR and A ABC lie in an oriented plane). The barycentric coordinates , p, y of a point M with respect to a non degenerate oriented triangle are the numbers cl

MBC CL =

n ,

ABM

AMC =

--------- -

ABC ABC ABC A tetrahedron ABCD is a set of four points A , B, C, D in space. An orient ed tetrahedron ABCD is an ordered set of four points A, B, C, D in space. ------> If the points A, B,C, D are noncoplanar, then the tetrahedron ABCD is said to be nondegenerate and if they are coplanar, it is said to be dege nerate. ------> If a nondegenerate tetrahedron ABCD lies in an oriented space, then it has a right-hand, or positive, orientation if ordered triples of vectors DA, DB, DC and e1? e2, e3 have the same orientation. But if these ordered ------> triples of vectors have opposite orientations, then the tetrahedron ABCD has a left-hand, or negative, orientation. ------> The volume (ABCD) of a nondegenerate oriented tetrahedron ABCD lying in an oriented space is a number whose absolute value is equal to the volume of the tetrahedron ABCD and which is positive if the te trahedron ABCD has a right-hand orientation and is negative if the ------re orientation is left-hand. If ABCD is a degenerate tetrahedron, then, by definition, we assume that (ABCD) — 0. ------y The volume (ABCD) of the oriented tetrahedron ABCD lying in an ori ented space is computed from the formula {ABCD) = — (DA, DB, DC). 6 If rl5 r2, r3, r4 are the radius vectors of the points A, B,C, D, then (ABCD) = — (rL - r4, r2 - r4, r3 - r4). 6

Basic Definitions, Theorems and Formulas

353

In a general Cartesian system of coordinates, xx - x t y1 - yx zx - z4 . (ABCD) = — ^ - ^ y 2 - y 4 z8 - z4 = ^ 6 *s — *4 y» — y* zz — z*

71 *i *2 J'a *2 *3 *4 >-4 Z4

i 1 1 ’ 1

and in a rectangular Cartesian coordinate system, *i

(ABCD) = — 6

> 4 Z1

*2 IX3

In these formulas, A = (xu y lt Zi), 5 =

*4 J2 -

— Xx ^ 3

*2 — *4 y 4 Z-3 ^4 y4

(*2, y 2, z8),

*i zx 1 _1_ *2 y2 z2 1 6 *3 y» z3 1 *4 y* z* 1

C = (x3, y3, z3),

Z) = fo , y t, z4)

The ra/io P gi?5 ABCD of a nondegenerate oriented tetrahedron PQRS to a nondegenerate oriented ------> tetrahedron ABCD is a number whose absolute value is equal to the ratio of the volume of the tetrahedron PQRS to the volume of the tetrahedron ------> ------> ABCD and which is positive if PQRS and ABCD have the same orientation (that is, the ordered triples SP, SQ, SR and DA, DB, DC have the same ------ > ------ > orientation), and negative if PQRS and ABCD have opposite orientations. ------> ------> If PQRS is a degenerate tetrahedron and ABCD is a nondegenerate one, then, by definition, we assume that PQRS

ABCD If the tetrahedrons PQ RS and ABCD lie in an oriented space, then PQRS ABCD 2 3 -8 1 0

(PQRS) (ABCD) '

Problems in Geometry

3 54

The barycentric coordinates a, /?, y, <5 of a point M with respect to a non degenerate oriented tetrahedron ABCD are the numbers MBCD «

---------------- ,

A M CD =

ABMD y =

ABCD

—

c <5 =

ABCD

ABCM -

ABCD

The area 5, in space, of A ABC whose vertices are given by the radius vectors r2, r3 is computed from the formula S =

Kt(ri - *») (** - r3)]2.

In a rectangular Cartesian coordinate system: S

= ---]f A\ + A\ + A2,

where }’i - y t a - z3 5 A% — Z1 — ^ 3 *1 — * 3 j z2 Z Z X2 -*3 y s — y 3 - z3 and in a general Cartesian coordinate system:

Ax =

——

Yg11 ^1 + g 22^ i +

—

*i — X a y , , — *2 — xs y 2 — y3

2g 12 ^1^2 + 2g23A2A3 + 2g31A3Av

In these formulas, A = (*i, yl9 zO,

£ = (*2, .y2, ^2),

C = (*3>

z3)> £,v = cV .

If we introduce a general Cartesian coordinate system in the plane, then the equation o f any straight line lying in the plane is the first-degree equation Ax -f By C — 0 and, conversely, any first-degree equation Ax + By + C = 0 (where A2 + B2 # 0) in any general Cartesian coordinate system is an equation of that straight line. The direction vector a = PQ of a straight line p is any nonzero vector, collinear with that straight line (vector PQ ^ 0 and the straight line p are said to be collinear if the straight lines PQ and p are collinear, that is either parallel or coincident; the zero vector is assumed to be collinear with any straight line). For a straight line specified in a general Cartesian coordinate system by the equation Ax + By -j- C — 0,

Basic Definitions, Theorems and Formulas

355

the vector a = {—B, A} is a direction vector. A necessary and sufficient condition of the collinearity of a vector a = = {l,m} and a straight line A x+ B y+ C = 0 in a general Cartesian coordinate system is of the form Al -j- Bm = 0. For the coordinates of all points (x, y) lying to one side of the straight line given by the equation Ax -j- By -f C — 0 with respect to a general Cartesian coordinate system, the following in equality is valid: Ax By -\- C > 0 (positive half-plane); and for the coordinates x , y of all points (x, lying to the other side of that straight line, the followng inequality is valid: Ax -J-* By -f C <! 0 (negative half-plane). The vector n = {A, B} is termed the principal vector o f the straight line specified with respect to a general Cartesian coordinate system by the equa tion Ax + By + C = 0. If it is laid off from any point M 0 of the straight line in question, M qP = n, then the point P will lie in the positive half plane. If A and B are regarded as covariant components of the vector, then the vector [A, B\ is said to be normal to the straight line specified with respect to a general Cartesian coordinate system by the equation Ax + By + C = 0. In a rectangular Cartesian coordinate system, the principal vector n = {A, B} of a straight line Ax + By + C = 0 is the normal vector to that line. The equation of a straight line passing through point determined by the radius vector r± and having the direction vector a is of the form (r - rl9 a) = 0. If in a general Cartesian coordinate system, a = {/, m j, M 1 = (x^ jx), then the last equation becomes x - x 1y - y ± = 0 I m or

x - *i _ y —y± l

(if / = 0, then this notation is to be understood as x — x2 = 0, and if m = 0, then as y — y x = 0).

Problems in Geometry

356

If / 7* 0, the last equation may be rewritten in the form y — yi = k(x — Xx), where k — m/l is termed the slope o f the straight line (in a general Carte sian coordinate system). The slope of the straight line that passes through two points (xl9 yx) and (x2, y 2) and is noncollinear with thej-axis is com puted from the formula * = -^ = 2 L *2

(in a general Cartesian coordinate system). In a rectangular Cartesian coordinate system, the slope is k = tan a, where a is the angle from the x-axis to the straight line under consideration. The equation of the straight line cutting the jy-axis in the point (O, b) and having slope k is, in a general Cartesian coordinate system, of the form y = kx + b. The equation of a straight line that does not pass through the coordinate origin and cuts the axes in points (a9 0) and (0 , b) is of the form

( intercept form o f the equation o f a straight line). If a straight line passes through a point M x given by the radius vector rx and has the direction vector a, then the radius vector r of any point of it can be represented as r = rx + fa, where t takes on all real values. This equation is called the parametric equa tion o f the straight line. The number t is the coordinate of point M on the line in question provided that Mx is the coordinate origin and a is the basis vector: t

r — r, a

M XM a

Suppose in a general Cartesian coordinate system we have r = {x, y}9 ri = {*i> yi}> a = {/, w}; then we obtain the parametric equations of the straight line in the form x = x x + It,

y = yi + mt.

Basic Definitions, Theorems and Formulas

357

Let rx and r2 be the radius vectors of two distinct points Mx and Mt. Then the equation of the straight line M ±M 2 is of the form (r - rl9 r2 - rx) = 0. In a general Cartesian coordinate system. x - *1 y - y i = o *2

y* yi -

or x y 1 *i yi i = 0 *2 ^2 1 or X

— x1= y — y1

x* ~ x 1

y2 — yx

(if any one of the denominators is equal to zero, then this notation is to be understood in the sense that the numerator is also zero). The equation of a straight line passing through point M x given by radius vector rx and having the normal vector n is n(r — rj) = 0. If in a rectangular Cartesian coordinate system rx = {xl9 y x}, n = = {A yB}y r{x, y }, then the equation n(r — = 0 becomes A(x - Xj) + B(y - yf) = 0. Such also is the form of the equation n(r — = 0 in general Cartesian coordinates if t x = {xl9 y x}9 r = {x9 y] but n = [A, B] (covariant coor dinates). A necessary and sufficient condition that two straight lines specified by the equations Ax -f By + C = 0, A 'x + B'y + C = 0 with respect to a general Cartesian coordinate system intersect, are parallel, or are coincident is that the system of equations have only one solution, no solution, or an infinity of solutions, respectively. If these straight lines intersect, then to find the coordinates of their point of intersection we have to solve the following system of equations: Ax “f- By -|- C — 0, A'x + B'y + C' = 0.

358

Problems in Geometry

A pencil o f straight lines is the set of all straight lines passing through the same point S (proper pencil; the point S is called the centre o f the pencil) or the set of all parallel straight lines (improper pencil). If general Cartesian coordinates are used to specify the equations of two straight lines Ax + By + C = 0 A 'x + B 'y + C ' = 0 that intersect in a point S, then the equation of the pencil with centre S is of the form oc(Ax + By + C) + P(A'x + B'y -f- C') — 0, where a and ft take on all values with at least one nonzero value. If the straight lines Ax -f By -f~ C — 0, A 'x + B'y + C' = 0 are parallel, then the preceding equation is the equation or an improper pencil to which the given lines belong; note that all possible values are taken for a and P but such that at least one of the coefficients of x and y is nonzero: ocA -f PA' y ccB -f- BB . A necessary and sufficient condition that three straight lines given in a general Cartesian coordinate system by the equations Ax 4“ By

+ C — 0,

A'x + B'y + C ' - 0 , A "x + B"y + C" = 0 belong to a single pencil is the equality A B C A' B' C' = 0. A" B" C" Here, the straight lines in question belong to a single proper pencil if at least one of the determinants A' B’ 5 A" B"

A" B" 9 A B

A B A' B'

is different from zero, and they belong to one improper pencil if all the determinants are equal to zero.

Basic Definitions, Theorems and Formulas

359

If in a rectangular Cartesian coordinate system two straight lines are given by the equations Ax -f- By -J- C = 0, A 'x + B'y + C' = 0, then the cosines of the angles (plt2 between them are determined from the formula , A A '+ B B ' COS <pl f 2 = ± --^===r-• YA2 + B2 Y ^ '2 + B'2 In general Cartesian coordinates, 8 u AA' + g12(AB' + A'B) + g22BB' COS (p 1 2 = ± —7— ----y • I^ 2 + 2 g12AB + g22B2 YgnA'2 + 2 g12A'B' + g22B'2 The angle <p formed by the straight line Ax + By + C = 0 and the straight line A'x + B'y + C' = 0 in rectangular Cartesian coordinates is found from the relations AB' - A'B AA' + BB' cos (p » sm <p — Y a 2 + B2 ][A'2 + B'2 Y a 2 + b 2 y A 't+ B '* and if the straight lines are not perpendicular, then it suffices to know only the value of tan<p: AB' - A'B tan cp = ----------------AA' + BB' In general Cartesian coordinates we have g n AA' + g12(AB' + A'B) + g22BB' cos q> YgnA2 + 2 g 12AB + g22B2 ]fguA '2 + 2g12A'B' + g22B'2 = gnAA' - 2g12(AB' + A'B) + gnBB'_______________ Yg2A2 - 2 g 12A B + g nB2 ] f ^ A ^ ~ 2g12A'B' + g llB’2 AB' - A'B sm cp = f f. ■ -— , ]fg y gllA2 + 2g12AB + g22B2 ]fguA’2 + 2g12A'B' + g 22B'2 = _________________Vg(AB' - A'B)_________________ y g22A2 - 2 g 12A B + gllB2 ][g22A'2 - 2g12A'B’ + gn B'2 ’ and if the straight lines are not mutually perpendicular, then AB' - A'B tan <p = —7=--------------- ------------------------------Yg(gllAA' + g12(AB' + A'B) + g22BB' = _________ fg {A B ' - A'B)_________ . g22A A ’ - 2g12(AB' + A'B) + gu BB'

Problems in Geometry

360

Ordinarily, for the angle formed by the straight line Ax + By + C = 0 and the straight line A'x + B'y + C' = 0 one takes the set of values {(p + kn} (k assumes all integral values), where cp is one of the values of the angle formed by the straight lines. If, in a rectangular Cartesian coordinate system, the straight lines p and p' have slopes k and k ' and if they are not mutually perpendicular, then the tangents of the angles q>12 between p and p' are computed from the formula k' ~~k tan <plf2 = ± 1 + kk' and in a general Cartesian system of coordinates V g fr - k)

tan q>li2 = ±

' g u — g u ( k + k ’) + g z jc k '

The tangent of the angle <p formed by straight line p and straight line p' is k' — k 1 + kk' and in a general Cartesian coordinate system tan <p =

1fg(k’- k ) gn - gi2(k + k') + g22kkf If the straight lines p and p ’ have, in general Cartesian coordinates, the slopes k and k \ then a necessary and sufficient condition for their collinearity is that k = k\ A necessary and sufficient condition of perpendicularity is that £n — g^{k + k f) + g22k k f = 0, tan <p =

and, in rectangular Cartesian coordinates, 1 + kk’ = 0 or k k ' = - 1. If a straight line is given by the equation Ax + By + C = 0 in rectangu lar Cartesian coordinates, then the distance d from point M0(x0, y0) to this straight line is computed from the formula \Ax0 + By0 + C| = ----------................. ][A2 + B2 a

r -

and if the vector n = {A, B}, which is normal to the straight line under consideration, is a unit vector: A2 + B2 = 1 (in this case the equation Ax + By + C = 0 is said to be normal), then d = \Axq + By0 + C|.

Basic Definitions, Theorems and Formulas

361

In general Cartesian coordinates, j ______\Ax0 + By0 + C\ _y — \Ax0 + Byp + Cj\ ~ g™AB+g**B* ~ 18 M - 2 g12A B + g n B* ’ and if the vector [A, B] is a unit vector, then d = \Ax0 + Byo + C |. If the straight line is given by the equation (a, r - r 1) = 0, then the distance d from the point M 0 with radius vector r0 to this straight line is d = )(a, rt - r0)| ^ |a| and if a is a unit vector, then </= |(a, rx — r0)|. If in rectangular Cartesian coordinates, a = {/, m}, ro = {*o>Jo}> fi = {*i, Ji}, then mod *i - *o Ji - Jo I m d= Vi* + mr and if the vector a is a unit vector (12 + m2 = 1), then - *o y ± - yQI / m I where the symbol mod x denotes the absolute value of the number x. In general Cartesian coordinates, fgm od * i — * o y± — y n m I d= ]fgul* + 2gn lm + g2Zm2 and if a = {/, m} is a unit vector, then mod

d = ] fg mod

J i-J . / m If the straight line is given by the equation n(r - rO = 0, then the distance d from the point M 0 with radius vector r0 to this straight line is computed from the formula j !n(r, - r0)| a = --------------- f

Problems in Geometry

362

and if the vector n is a unit vector, then d = Info - r0)|. If in rectangular Cartesian coordinates r0 = {x0, j 0}, rx = {xx, j x), ft = {A, B}, then \A(xl — x0) + B(yx — jo) | d ~ )[A*+~B* and if the vector {A, B} is a unit vector, then d = |A(x0 - xx) + B(y0 - j x)j. In general Cartesian coordinates, d=

Ign^fa — *o) + gizU(y! - Jo) + B(x\ — x0)] + g22B(yL — j 0)j VgiiA2 + 2g12AB + g22B2

and if \{A,B}\ = 1, then d=

~ x0) + g12[A(y1 - j 0) + B(x1 - x0)] + g22B(yx - j 0)|.

If the vector n is given by covariant components, n = [A, B], then d=

M(xx - x0) + B(y1 — j 0)| _ ][g^A* + 2g12AB + g22B2

\A(X! — x0) + B(yl — j 0)| ^ 8 ]fg22A2 - 2gn AB + g nB* *

and if |n| = 1, then d = \A(xt - x0) + B(y\ - jo)|. The area {ABC) of an oriented A ABC whose sides are specified with respect to a general Cartesian coordinate system by the equations (BC): Axx + Bxy + Cx = 0, (CA):

A 2x

-f- B2y -j- C2 = 0,

(AB):

A 3x +

B3y

C3 = 0,

is computed from the formula Ax Bl Cx 2 A2 B2 C2 (ABC)

VF 2

A3 B3 c 3 1 A2 Bg A Bt A3 B3 U i Bx A2 B2 a 2 b2

Basic Definitions, Theorems and Formulas

363

and, in rectangular Cartesian coordinates, from the formula Ai Bi Ci 2 A2 B2 C2 1 A3 B3 C3 A2 B2 ^3 &3 A x ;;2 A3 B3 A ! B, A2 B2 The area S of A ABC is Ai BÂą Ci A 2 &2 C2 A 3 B 3 C3 S = Vg B2 A 3 Bs Ai B " m o d (r2 A 2 Bi \ \ A Z B3 Ai Bi and, in rectangular Cartesian coordinates, Ax Bx Cx A 2 B 2 C2 A 3 Bs C3 1 A 3 B3 IA 2 B 2 Ai B} S = T mod I A 3 B2 Ai Bx A 2 Bi If a general Cartesian coordinate system Oxyz is introduced in space, then the equation of any plane in the first-degree equation A x -f- By -(- Cz -{-.0 = 0,

II)

il)

and, conversely, any first-degree equation Ax -}- By -f- Cz -{- O = 0 (where A 2 + B 2 + C2 ^ 0) in any general Cartesian system of coordinates is the equation of the plane. For the coordinates of all points (*, y, z) lying to one side of a plane that is specified with respect to a general Cartesian system of coordinates by the equation A x 4" By -f- Cz -f- 0 = 0, the following inequality is valid: Ax -f- By -J- Cz -j- O > 0 (positive half-space), and for the coordinates of all points (x ,y ,z) lying to the other side of that plane, the following inequality is valid: Ax -{- By -f Cz -f- O <: 0 (negative half-space).

Problems in Geometry

364

The vector n = {A, B,C } is called the principal vector o f the plane speci fied by the equation Ax "j- By -j- Cz -f- 2) — 0. ---- ** If it is laid off from any point M0 of the plane, M0P = n, then point P lies in the positive half-space. If A, B ,C are regarded as covariant components, then the vector [A, B, C] is normal to the plane Ax + By + Cz + D = 0. In a rectangular Cartesian system of coordinates, the principal vector n = {A, B, C} of the plane specified by the equation Ax + By + Cz + D = = 0 is a vector normal to that plane. The equation of a plane passing through point M 1 defined by the radius vector rx and normal to the vector n is of the form n(r - rj) - 0. If in general Cartesian coordinates rx = {xl9 y l9 zx}9n[A, B, C], r = {x9y 9z}9 then the equation n(r — r2) = 0 becomes A(x — A'i) + B(y —y j + C(z — zL) = 0. In rectangular Cartesian coordinates, d

= {A9B9C} = [A9B9C].

The vector PQ =£ 0 and the plane n are said to be coplanar if the straight line PQ is either parallel to, or lies, in the plane n. The zero vector is assumed to be coplanar with any plane. For a vector a = {/, m9n} and a plane Ax + By + Cz + D = 0 to be coplanar, it is necessary and sufficient that the following equality be valid: Al -f- Bin -f- Cn — 0 (the coordinates are general Cartesian). The equation of a plane passing through a point with two noncollinear vectors a and b is of the form (r - r 1? a, b) = 0.

and coplanar

In general Cartesian coordinates,

H 1 H

y - yi z —> = 0, mi m2 n2

k k where a = {ll9 ml9 n ^ 9 b = {/2, m2, «2},

tx =

{xl9 yl9 z j , r = {x9y 9z).

Basic Definitions, Theorems and Formulas

365

The equation of a plane passing through two points Afifo) and Af2(r2) ------> and coplanar with the vector a Ik M XM2 is of the form (r - rl9 r2 - rl9 a) = 0. In general Cartesian coordinates. x - x x y — y\ z - Zi *2

— I

>’2—y± m

z2— = 0, n

where M x = (xl9 y l9 Zj), M 2 = (x29y2, z2), a = {/, m9n}. The equation of a plane passing through three noncollinear points M2(t2), M3(r3) has the form (r — r3, rx — r3, r2 — r3) = 0. and, in general Cartesian coordinates, x — x 3 y — ys z — z3 *i — *3 T1 - T 3 z1 — z3 = 0, * 2 — * 3 >J2 —y* z2— z3 or x y z 1 ^1 J i 1 _ Q *2 T2 ^2 1 x 3 x 3 z3 1

where M x = (*l5 z*), M2 = (x2, y 2, z2), M3 = (x3, >>3, z3). The equation of a plane passing through a point Mjfo) (and coplanar with two noncollinear vectors a and bean be written in parametric form: r = r± + wa + fb. In this equation, u and v are the coordinates of the point M in a general Cartesian coordinate system in the plane at hand, in which the origin is the point Af1(r1) and the basis is a, b. The parametric equations of a plane in a general Cartesian coordinate system are: X = Xt +

+ vl2.

y = yi + umX + vm2,

z = zx + uitx + 1m2y where rx = {xx, y u z j , a = {/x, mu wx}, b = {/2, m2, «2}.

366

Problems in Geometry

The parametric equation of a plane passing through points M^rri, M2(i2) and coplanar with the vector a Ik MJM2is of the form r = rx + «a + v(r2 — r^. In general Cartesian coordinates, X = Xi + Ul + v(x 2 — Xj), y = y i + um + v(y 2 — ji)-

2 = zx + un + v(z 2 — Zj), where Mx = (x1; y lt zx), M 2 = (x2, y 2, z2), a = {/, m ,«}. The parametric equation of a plane passing through three noncollinear points M fa ), M2(r2), M 3(r3) is of the form r = r3 + — r3) + v(r2 — r3)

and, in general Cartesian coordinates, x = x3 + u(xj, — x3) + v(x2 — x3), y = y* + u(yi - y 3) + v(y2 - y a), z = z3 + u(z1 — z3) + v(z 2 — z3),

where M x = (x3, y u zx) M 2 = (x2, y 2, z2) M = (x3, y 3, z3). If the plane does not pass through the coordinate origin of a general Cartesian coordinate system and cuts the coordinate axes in the points (a, 0,0), (0, b, 0), (0,0, c), then its equation can be written as -^ + ^ + ^ = 1 a b c (the intercept form o f the equation o f the plane). For two planes given by the equations Ax By -J-' Cz -f- D — 0, A x - \~ B y - \- C z - { - D — 0 with respect to a general Cartesian coordinate system to intersect, it is necessary and sufficient that the vector (15 C C A ||5 ' C ' B' I] C A' be nonzero: a ^ 0. In this case (a # 0), the vector a is the direction vector of the straight line along which the planes in question intersect. For the two planes Ax -f- By -f Cz -f- D — 0, A'x + B'y + C'z + D' = 0

■5T

Basic Definitions, Theorems and Form ulas

367

to be parallel, it is necessary and sufficient that the vector a be equal to zero but that at least one of the determinants A D B D C D A' D' ’ B' D' C D' be different from zero. For the two planes Ax -I- By -J- Cz -f- £) — 0, A 'x + B'y + C'z + D' = 0 to be coincident, it is necessary and sufficient that the corresponding coefficients of the equations of the planes be proportional: A ' = kA, B' = kB, C = kC, D9 = kD. Three planes specified with respect to a general Cartesian coordinate system by the equations Ax -f" By -f- Cz -f- D — 0, A'x + B'y + C z + D ' = 0 , A"x + B"y + C"z + D" = 0 have only one point in common if and only if the following inequality is valid: A B C A' B' C' * 0 . A" B" C" To find the coordinates of this point it is necessary to solve the system of equations of the three given planes. A pencil o f planes is a set of all the planes passing through one straight line / (proper pencil). The straight line / is called the axis o f the pencil. An improper pencil of planes is the set of all parallel planes. If two planes given by equations with respect to a general Cartesian system of coordinates. Ax -f- By -j- Cz -f- D — 0, A'x + B'y + C 'z + D ' = 0 intersect along a straight line /, then the equation of the pencil of planes with the axis / is of the form oc(Ax + By + Cz + D) + fi(A'x + B'y + C'z + D') = 0, where a and /? take on all possible values with at least one of them being different from zero.

368

Problems in Geometry

If the planes are parallel, then the last equation is the equation of the improper pencil of planes to which they belong, and for a and P one takes all possible values with the exception of those for which all the coefficients of x , y , z, that is, ocA + p A \

<xB + PB',

aC + p C

are equal to zero. A bundle (or sheaf) o f planes is the set of all planes that pass through the same point S (proper bundle) or the set of all planes coplanar with one straight line (improper bundle). The point S is termed the centre o f the bundle. If three planes specified by equations in general Cartesian coordinates, Ax + By + Cz + D = 0 , A 'x

B'y -f- C'z -}- D; = 0,

A"x + B"y + C"z + D" = 0, have the same common point 5, then the equation of the bundle of planes with centre S is of the form <x(Ax + By + Cz + D) + P(A'x + B'y + C z + D) + y(A"x + B"y + C z + D") = 0, where a, p , y take on all values, at least one of which is nonzero. If the three indicated planes are coplanar with one and the same straight line, but do not pass through one straight line, then the last equation is the equation of the improper bundle to which the three given planes belong; and for a, p9y one takes all possible values, with the exception of those for which all the coefficients of x, y, z, that is, ocA + PA' + yA",

ctB + p V + yB",

<xC + fiC' + PC"

are equal to zero. The distance d from the point M 0(x09y09 z0) to the plane given by the equation Ax -f- By -J- Cz -f Z) â&#x20AC;&#x201D; 0 in rectangular Cartesian coordinates is computed from the formula j _ 1Ax0 + By0 -j- Cz0 + D I â&#x20AC;&#x153;

y A 2 + B 2 + C

and if the vector n = {A, B, C} is the unit vector (the equation Ax + By + + Cz + D = 0 is then said to be normal)9 then d = |A xq + By0 + Cz0 + D\.

Basic Definitions Theorems and Formulas

369

In general Cartesian coordinates, ____________ 1Ax0 + By0 -f- Cz0 + D 1 _____ " + g2*B2 + g33C2 + 2g12AB + 2g23BC + 2g31CA _

Vg\Ax$ + £ll _ g%l gz\

By0 -f~ Cz0 + ^1 gl2 gl9 A \ g22 #23 B #32 #33 C A B C 0 /

and if the vector [A, B, C] is the unit vector, then d = |Ax0 + By0 + Cz0 -f- D\. The distance d from the point M0(r0) to the plane given by the equation 0 is computed from the formula d _ [nfa-Tp)!

(n, r— rj) =

M and if

n is

’

the unit vector, then d = \n(rx - r0)|.

If the vector nis given by covariant components in a general Cartesian system of coordinates: n= [A, B, C], and the vectors r0 and are given by contravariant coordinates (components): *0 =

{ * o > J o . Z0} ,

*1 =

{ * 1, J j , Z l } .

then d =

____________ \A (x i -

*o) +

B (yi

Jo) +

C (Z i -

Z0) ____________

IfgnA2 + g225 2 + g33C2 + 2g12AB + 2g™BC + 2g31CA _ fg \A(xi

*o) +

/ _

B(yt -

Jo)

+ C(z1 -

gll gl2 gl3 d \ &21 H>22 g23 & g$l &32 g23 C A B C 0 /

and if the vector n = [A, B, C] is the unit vector, then d =

\A(xx — X q)

B(yl — j o )

C(zx — z0)|.

In rectangular Cartesian coordinates, d=

—810

~ *o) + -S(ji - Jo) + C(z1 — z0)| VA2 + £ 2 + C2

2 0) |

Problems in Geometry

370

and if the vector n = {A, B , C] = [A, B, C] is the unit vector, then d = \A(x± - x 0) + B fa - yo) + C(zi - z0) |. The cosines of the angles <plt2 between two planes specified by the equations Di(r — ri) = 0, n2 (r — r2) = 0 are computed from the formula iDiDol cos <plf2 = ± I«il |n2|

If the planes are given by the equations A x -f- By -j- Cz -f- D =■ 0 , A 'x + B'y + C z + D' = 0 with respect to a rectangular Cartesian coordinate system, then AA' + BB' + C C COS <p12 = + -7-........................... -. ][a 2 + B2 + C2 Y a '2 + B'2 + C'2 In general Cartesian coordinates we have _ , ([A,B,C], [ A \B \C '] ) . cos <^i 2 — ± ------------------------------; I[A, B, C] I |[A', B', C ) \ ([A, B, C], [A\ B', C']) = g*AA' + g*BB’ + g™CC’ + g12(AB’ + A'B) + g™(BC' + B'C) + ga,(CA' + C'A) gll Sit Sit A

_ _.A.

Sti Stt Sts B E £*i Stt Stt O A' B' C' 0

IN, B, C]\ = YguA2 + gl2B2 + g™C2 + 2g '2AB + 2g*BC + 2g31CA / _

Ye IN', B', C']|

8ll 8 l 2 8 l 3 A 821 822 823 B # 3 1 &32 833 C A B C 0

1/2

= YgnA'2 + g*F* + gMC'2 ^ 2 ^ 'B '~ + 2 g ^ B ’Cr :\-2g2lC'A' / 811 S12 813 A' \ 1/2 _ 821 822 822 B' £31 £32 8%3 C' V A’ Bf C 0

Basic Definitions Theorems and Formulas

371

The parametric equation of a straight line passing through point and having the direction vector a is of the form r = fi + fa. where f is the coordinate of the point A/(r) on that line if we take point for the origin and vector a for the basis vector: t =

M XM 1— . a

In a general Cartesian coordinate system, the parametric equations of a straight line are written as follows: * =

tl ,

y = yi + tm, z = zx + tn ,

where Mx = (xl9 yl9 zx), a - {/, m, n}. The parametric equation of the straight line passing through two points A f ^ ) and M2(r2) is of the form r = Ti + f(r2 -

rx),

and, in general Cartesian coordinates, X = X \ 4 - t ( X o — A'x),

y = yi + t(y i- yi), z = zx + t(z2 — Zx).

Two straight lines are said to be coplanar if they lie in one plane. For two straight lines r = rx + fa, r2 = r2 + fb to be coplanar, it is necessary and sufficient that the following equality hold: (r2 — rlf a, b) = 0 or, in general Cartesian coordinates, *2 — *1 y* — yi z2 zx | = 0, m± /i *i m2 no k where M1 = (xly yu zt), M2 = (x2, y2, z2), a = {/„ mu n j , b = {/2, nu, n2). Remark. The equations of a straight line are often written in the form x — x x _ y — y x _ z — zx I n m

Problems in Geometry

372

(canonical equations o f a straight line). If one of the denominators is zero, then this notation is to be understood as follows: the numerator is also equal to zero. For example, the system of equations x —2 _ y —5 _ z —1 3 0 ~ 4~~ is to be understood as follows: y - 5 = o,

x — 2 __ z — 1 3 _ 4 The distance d from the point M0(r0) to the straight line given by the equation r = rx + /a is computed from the formula d = l(ri - r0, «)l |a| In rectangular Cartesian coordinates, ]fA2 + A\ + A\ |/72 + m2 + n2 where A± =

yi-yo zi—zo m

Ao —

* i-x o y i-y o i m

-Xo

and, in general Cartesian coordinates. d=

+ g22A\ + g™A% + 2g12A1A2 + 2g™A2A3 + 2g^A3A1 ]fgiJ2 + 8 2 2 m2 + g 3 3 n2 + 2 g12lm + 2 g23mn + 2 g91nl /

8 ll 8l2 8 13 At \

_ 8 21 8 2 2

822

831 832 #33

A\ A 2 A 3 0

1/2

\fgiJ2 + gi3m2 + g33n2 + 2g12lm +2g23nm+2g3lnl The shortest distance d between the noncollinear straight lines r = rj + fa,

r = r2 + fb

is d = Kr2 ~ ri>a. b)l l[ab]|

Basic Definitions, Theorems and Formulas

373

If in rectangular Cartesian coordinates we have *x = {*1, yi, Zi}, r2 = {x2, y 2, z3}, a = {/1( mlf nx}, b = {l2, m2, n2}, then * 2 - * i

mod d=

J 2 - J 1

^ 2 - ^ 1

h mx 12 M2 ][d\ + S*+ SI

nx ^2

where mx nx m2 n2

»1 h > <^3 = h "ii n2 l2 h m2 and, in general Cartesian coordinates. X2 — y 2 - yi z2 — zx mod mx h m2 «2 h d= U ]lgn Sl + g*b\ + g™b\ + l g liS1S2 + 2 g™S2S,3 + 2g31<53(51 *1 =

s* =

y z - y i z2 m1 h m2 n2 h _ 1/2 < 5 i gll gl2 giz g2l g22 #23 &31 gs2 £33 ^3 < <53 0 A straight line in space can be specified by the equations o i two inter secting planes: *2 — * 1

mod

Ax + By + Cz + D = 0, A 'x + B’y + C'z + Z>' = 0 (general Cartesian coordinates). Its direction vector is

C A C' A'

To reduce the equations of the straight line Ax + By + Cz + D = 0, A 'x + B'y + C z + D' = 0

Problems in Geometry

374

to parametric form, we have to find some solution x l9 y l9 zx of the system; and then the parametric equations of the given straight line are B C X = Xt + t B' Cf ’ y

)’i + t

— zt + I

C A C Af ’ A B A9 B'

and the canonical equations are X — Xj

Z - Zi

y -y i C A C' A'

B C B' C'

A B A' B'

A necessary and sufficient condition for the plane Ax + By + Cz -f- D = = 0 and the straight line x = x x + It, y = y x + mt, z = zx + nt, given by equations with respect to general Cartesian coordinates, to intersect, be parallel, or for the line to lie in the plane is as follows: | intersection

condition A l + Bm + Cn # 0

parallelism

A l + Bm + Cn = 0 Ax, + By, + Cz, + D * 0

straight line lies in plane

Al + Bm 4- Cn = 0 Axx + Byx + C zx 4- D = 0

The angle between the straight line r = rA+ fa and the plane n(r — r0) = 0 is given by the formula sin <p =

Iani a In!

If, in rectangular Cartesian coordinates, a straight line is given by the equations x = Xi + It, y = y1 + mt, z = zx + nt

Basic Definitions, Theorems and Formulas

375

and a plane by the equation Ax + By + Cz + D — 0, then Al + Bm + Cn sm q> — + B2 + C2 V/2 + ro2 + /I2 In general Cartesian coordinates we have Al + Bm + Cn sin <p = -— ------- , Vk Vt s where 7i = gJM2 + g22£ 2 + g^C 2 + 2g'2AB + 2g23BC + 2g31CA

gll gl2 glZ d g21 g22 #23 & £31 £32 #33 c A B C 0

T2 = gul2 + g22m2 + gZ3n2 + 2g12lm + 2g2Zmn + 2g31nl. A necessary and sufficient condition for a line and a plane to be parallel is of the form [A,B, C) || {Urn, n} or {Agu + Bg12 + Cg13, Ag21 + Bg22 + Cg“ Ag31 + Bg32 + C^33}|| {/, m, n} or Agn + Bg'2 + Cg13 - £/, v4g21 + #£22 + Cg23 = km, k ^ 0, >te31 + *£32 + Cg33 = *n, or £ 11/ + = kA, g2il + £ 22™ + £ 23" = kB, k ^ 0, gzil + £32™ + g^n = kC9 and, in rectangular Cartesian coordinates, A — A7, 5 = km9 C = kn. The equation o f a circle (C, r) with centre C(a, ft) and radius r in rec tangular Cartesian coordinates is of the form (x — a)2 + (y - £>)* — r2 = 0,

376

Problems in Geometry

and if the centre of the circle is the origin of coordinates, then x 2 + y2 + z2 = 0. Sometimes the point C(a, b) is regarded as a circle of zero radius (the zero circle). The equation of a zero circle C(a, b) in rectangular Cartesian coordinates is of the form (x - a)2 + ( y - b)2 = 0. For the coordinates x, y of any point (x, y) lying outside the circle (x - a)2 + ( y - b ) 2 - r 2 = 0 we have the inequality o = (x — a)2 + (y — b)2 — r2 > 0, and for all points M(x, y) lying inside that circle we have a = (x — a)2 + (y — b)2 — r2 < 0. The number a is called the power o f the point M(x, y) with respect to the circle (C, r) and is equal to G = d2- r \ where d is the distance from point M to the centre C of the circle (C, r). If an arbitrary straight line intersecting the circle (C, r) in two points A and B is drawn through M, then a = (MA, MB) = MA-MB. The equation o f a sphere (5, r) with centre S(a, b, c) and radius r in rectangular Cartesian coordinates is of the form (x — a)2 + ( y — b)2 + (z — c)2 — r2 = 0, and if the centre of the sphere lies in the coordinate origin, then x 2 + y 2 + z2 — r2 = 0. The equations of zero spheres are: (x — a)2 + (y — b)2 + (z — c)2 = 0, x 2 + y2 + z2 = 0 For the coordinates x 9y , z of any point M lying outside the sphere (x - a)2 + ( y - b)2 + (z - c)2 - r2 = 0 we have the inequality g = (x — a)2 + (y — b)2 + (z — c)2 — r2 > 0, and for all points M (x, y , z) lying inside the sphere we have g = (x — a)2 + (y — b)2 + (z — c)2 — r2 < 0.

Basic Definitions, Theorems and Formulas

377

The number a is called the power o f the point M with respect to the sphere (S, r) and is equal to cr = d2 — r2 = (MA, MB) = M A-M B, where d is the distance between the points S and M, and A and B are points in which an arbitrary straight line passing through point M inter sects the sphere (S, r). Sec. 4. Complex numbers In the set of complex numbers x + yi (x and y assume all real values), the sum and product are defined as follows: (x + yi) + (*' + y'i) = (x + s') + (y + ? ) i, (x + yi) (pc' + y'i) = (xx' — yy') + (xy' + x'y) i; here, the real number x may be written as x + 0-j>; in particular, 1 = 1 h + 0/, 0 = 0 + 0 - /. It is obvious that 0 + z = z,

1 • z = 1 and 0 • z = 0

for all complex numbers. The set of complex numbers contains all the real numbers (x + 0 - / = = x + 0 = x) and also the number i (0 + 1• i = 1 • / = /), the square of which, by virtue of the definition of a product of complex numbers, is equal to —1: i2 = - 1 . The operations of subtraction and division are defined as the inverses of addition and multiplication; if z = x + yi, then —z = (—x) + (—y) i. The following properties are obvious: 2 +

(Z' +

2") =

(2 +

2 ') + 2 " ,

2 + ( - 2) = 0, z + z' = z f + z. z(z'z") = (zz') z",

22' = Z’Z Z (z '

Z " )

ZZ' +

2Z " ,

where z, z', z" are arbitrary complex numbers. The modulus (absolute value) \z\ = p of the complex number z — x + yi is the principal root (positive real root) Y x2 + y 2: W = P = V*1 + 7*-

Problems in Geometry

378

The argument (or amplitude) q>= arg z of the complex number z = x + + yi ^ 0 is the number (p defined by the relations JC x y_ ____y _ _ sin (p cos cp = — \[x* + y2’ j/*2 + y* P P The argument of the number z ^ 0 has an infinity of values. If cp is one of the values of the argument of z ^ 0, then all the values arg z are con tained in the formula arg z = q> + 2kny where k takes on all integer values. This relation is frequently written as follows: arg z = cp(mod 2 n ) (read: “the argument z is congruent to cp modulo 2rc”). If z = 0, then p = 0, and cp is any number. For two complex numbers z ^ 0 and z' # 0 to be the same, it is necessary and sufficient for their moduli to be equal, 1*1 = l*'l, and for their arguments to be congruent modulo 2n: arg z = arg z'(mod 2n). From the preceding formulas it follows that z = x + yi = p (cos '), then zz' = pp' [cos((p + <p') + i sin(cp + <p% — = — [cos(')], z' / 0, z p 1*1 I* I, arg(zz') = arg z + arg z'(mod 2n), J*l izi arg

— arg z — arg z'(mod 2n), zn

= pw(cos ncp + / sin n(p)

z' ^ 0,

Basic Definitions, Theorems and Formulas

379

(n an integer); in particular, if |z| = p = 1, then (cos ) (the de Moivre formula). If z ^ 0 and n is a natural number, then 2k + l .s l.„ <p _ +_ 2A:7u _ j \. )jz = y p(cos <p + / sin (p)- ■ f c ( c o s ^ ± _ fc = 0, 1,2,. .. , n — 1. Two complex numbers z = x + yi and z = x — yi are said to be conj ugate. They have the following properties of conjugacy: z = z, z + z' = ~z + z', z — z' = z — z', zz' = zz'. If m= — (z' 7^ 0), then u = — z' z' M = 1*1, arg z — — arg z (mod 27r). Suppose a rectangular Cartesian coordinate system is introduced in a plane. With every complex number z = x + y i we associate a point Af(jc, y). This correspondence is one-to-one. The number z is called the affix of the point M. A point with affix z is here symbolized thus: M{z) or M = (z). If z1 = x ± + yfi, z2 = x 2 + jv , z 3 = x 3 + are the affixes of the points A, B ,C in rectangular Cartesian coordinates, then the area {ABC) of an oriented £±ABC is computed from the formula (ABC) = — 4

Z l

1 1 1

In particular, a necessary and sufficient condition for the collinearity of three points A{z1), B(z2), C(z3) is of the form Zi z\ 1 z2 z 2 1 - 0 . z3 z 3 1

380

Problems in Geometry

The transformation under which a point M(z) is associated with a point M \z '), where z' = az + b. is a similarity transformation o f the first kind (that is, a transformation that does not change the orientation of the plane). Indeed,

and transition from point M(z) to point M f(z') is performed as a translation — z -j-----and a transformation z' = azl9 which consists in a rotation a about the origin through an angle arg a and a homo the tic transformation with centre O and ratio \a\. All these instances are similarity transformations of the first kind. A transformation under which a point M(z) is associated with a point M'(z'), where z' = az + b, is a similarity transformation of the second kind (that is, a transformation that reverses the orientation). Indeed, this trans formation consists in a symmetry zx = z with respect to the x-axis and a transformation z' = az± + b that reduces to a translation, a rotation, and a homothetic transformation. Of all these transformations, only symmetry with respect to the x-axis changes the orientation of the plane. From the foregoing it follows that if two triangles ABC and PQR are given via the affixes of their vertices, A = fe),

£ = (zg), C = (z3),

P = («i),

Q = (w2), R = (w3)>

then a necessary and sufficient condition that A ABC and i\PQ R be similar and have the same orientation is zx

z2 m2 1 = 0, z3 *^3 i and a necessary and sufficient condition that A ABC and A PQR be similar but have opposite orientations is «i 1 Zg Ug

z 3

«3

Basic Definitions, Theorems and Formulas

381

Let us consider two distinct points M x(z^) and A/2(z2). On the basis of the foregoing, point M(z) lies on the straight line MXM2 if and only if z

Zi *i 1 = 0. z2 z2 1 This equation can therefore be called the equation of the straight line MXMV It can be transformed to z

Z1 /—

zi ~ -------- ZT (z

—

z l)*

We will call the ratio x=

the complex slope o f the straight line M XM2. Note that Z2 ~ Zl = 1. *2 - *1 Thus, the equation of the straight line M XM2can be written down as Z — Zx =

X (Z

—

Z7! ) ,

where |x| = 1. Conversely, any equation of the form Z — Zx = x(z — Zj), where |x| = 1, is an equation of a straight line. Indeed, since \x\ = 1, it follows that x = cos <p + i sin <p. Setting up the equation of the straight line passing through two points with affixes zx and zx + cos - - + / sin — , we obtain the equation 2 2 Z ~ Zjl = x (z — z’j).

In particular, note the equation of the straight line that passes through the coordinate origin: z = xz, where x = cos (p + / sin (p = — zi

382

Problems in Geometry

and zx ^ 0 is the affix of any point (zx) of the straight line under consi deration. Note that the straight line z = xz passes through points of the unit circle (the unit circle is a circle with centre at the coordinate origin and with radius equal to 1) with affixes ][x (J/x always has two values: they are the affixes of the endpoints of the diameter of the unit circle). Indeed, if ]/x is either of the values of this radical, then for z = ]fx the equation z = xz becomes an equality (the left-hand side is ]/x; the right-hand side is, x]fx = J/x). The two straight lines Z — zx = x(z — Zj), \x\ = 1, z — z2 = x'(z — Z2),

\x'\ = 1,

are collinear if and only if x = x'. True enough, these lines are collinear if and only if the system of equations z — xz = z1 — x z l9 z — x'z = z2 — x z 2 in z, z either has no solution or has an infinitude of solutions, and this occurs if and only if 1 —x = 0 or x = x \ 1-x ' The two straight lines p and q given by the equations z — Zx = x(z — Zj), |x | = 1, z — *2 = x'(z — Z2), |x'| = 1, are perpendicular if and only if x + x' = 0. Indeed, let us consider the straight lines p ' and q \ which are collinear respectively with p and q, but which pass through the coordinate origin : p': z = xz, q': z = x ’z The straight lines p and q are perpendicular if and only if the lines p ' and q' are perpendicular. Suppose p ‘ and q' are perpendicular. On p ’ take a point with the affix z0 ^ 0. Then by virtue of the relation arg(izG) = arg i + arg z0 = ----- f- arg z0(mod In) 2 ,

Basic Definitions, Theorems and Formulas

383

the point with affix zz0 lies on the straight line q\ We have z0 — x z o,

iz0 = x iz0

or z0 = xz'o,

iz0 - — x f i z 0

or Zq

— xz o,

Zo —

Z q.

Hence, x z 0 — — x'z0 and since z 0 ^ 0, it follows that x = — x \ whence x + x' = 0. Conversely, let x + x f = 0. Let us prove that the straight lines p' and qf are mutually perpendicular. Draw through the origin a straight line p* perpendicular to q', and let x* be the complex slope of p*. Then x * + + x' = 0, and since x + x' = 0, it follows that x = x* and, hence, the straight lines p* and p' are coincident, that is p' J_ q', whence p _L q. Remark. In analytic geometry, the slope k of a straight line that is noncollinear with the y -axis is the tangent of the angle of inclination of that line to the jc-axis. If the complex slope of a straight line that is noncollinear with the y-axis is equal to x = cos cp + / sin <p, then the angle a of inclination of that line to the *-axis is equal to (p!2 (since a straight line passing through the coordinate origin and having a complex slope x passes through points with affixes Yx). Now we find 1 —x . <p ----- = — i tan — = — ik ; 1+ x 2 consequently, , A —x k = z ------ . 1+ x Conversely, i —k x = ------ . i ~f~ k The angle formed by the straight line Z —

x(z

— Z j)

and the straight line z — z2 = x \ z — z?)

Problems in Geometry

384

is equal to y x* arg — (mod n). yx Indeed, the straight lines z = xz, z = x'z, which are collinear with the given lines, pass through points with affixes |/x and |/V respectively and so the angle formed by the two straight lines is equal to arg 1/V — arg jfx = arg

- (mod n).

The equation of any straight line can be written as Az + Bz + C = 0, where C is a real number and B — A ^ 0. Conversely, any such equation is an equation of a straight line provided C is a real number and B = A # 0 Proof. Let Px + Qy + R = 0 be the equation of a straight line in rec tangular Cartesian coordinates. Since X =

(z + z ) 9

y = — (* - 2) = — (Z - Z), 21 2 it can be rewritten thus: “ P(z + Z) + — Q(z - 2) + R = 0 or P - Qi z + 2

2 '.'

z +R = 0

or A z+ A z + C = 0

(C = 2R).

Conversely, setting A — P — Qi, A = B = P + Qi, we can rewrite the equation Az + B z + C = 0 as (P — Qi) (X + yi) + (P + Qi) (X — yi) + C = 0

Basic Definitions. Theorems and Formulas

385

or 2Px + 2 Q y + C = 0, which is a first-degree equation. The equation Az + A z + C = 0 is called a self-conjugate equation o f a straight line since the left-hand side of the equation is a real function of x and y: u = Az -f- A z -j- C, u = A z + Az + C = u. If a straight line is given by the self-conjugate equation Az -f- Bz C — 0, where B = A ^ 0 and C is a real number, then the distance d from point (z0) to this line is , \Az0 + B z0 + C\ a = ----------------------. 2\A\ Indeed, the equation of the straight line passing through point (z0) per pendicularly to the given straight line is of the form Z

— z0 =

—

(

z — z0)

or Az — Bz — ZqA + z 0B = 0. From the system Az + B z + C = 0, Az — Bz — Az0 + B z o = 0 we find the affix of the projection of point (ze) on the given straight line: Az0 — B~z0 — C z = 2A whence Az0 -j- B zq -j- C z9 — z = 2A and so \Az%+ B z %+ C| d = \ z Q- 2 ' I = 2\A\ 21—SIS

386

Problems in Geometry

If a straight line is given by the self-conjugate equation Az -f- Bz -f- C — 0, where B = A ^ 0 and C is a real number, then for all points (z) lying to one side of that straight line we have Az -f Bz -j- C > 0 (positive half-plane), and for all points (z) lying to the other side of that line we have Az -f- Bz -j- C 0 (negative half-plane). If (z0) is any point lying on the straight line Az -f- Bz -{- C = 0, where B = A ^ 0 and C is a real number, then the point (z0 + B) lies in the positive half-plane since A(z0 -f- B) + B (z0 + B) + C — Az0 + B zq + C + AB + BB = AB + BB = BB + BB = 2BB = 2\B\* > 0. If the sides of A ABC are given by the self-conjugate equations (BC): Axz -j- B±z -j- C\ = 0, ((7/4) : A2z BoZ -f- C2 = 0, (AB): A3z + B3z -f- C3 = 0, where Ck are real numbers and Bk = A k ^ 0, then the area (ABC) of the

oriented A ABC is computed from the formula Ai B L Cx 2 A 2 B2 C2 A3 ^ 3 C 3 i (ABC) 4 | a 2 b 2 A3 B3 j Ax Bt \ a 3 b 3 A l Bx \A , B2 The result remains the same if the left-hand sides of the equations (BC), (CA), (AB) are multiplied by any complex numbers different from zero. The area^S of A ABC is computed from the formula A i Bx Cx 2 mod A2 B2 C2 1 A 3 Bz j A\ j\ 4 mod I \ A2 B% ' \ A a B3 A-l B- j a 2 b 2 J

LIST OF SYMBOLS vector AB a vector with origin at point A and terminus at point B: a ray (radial line) with origin at A and passing through B 0 zero vector 1*1 = a magnitude (absolute value) of vector a \AB\ =- AB a ij b â&#x2013; TM aU b/ â&#x20AC;˘Tib *9 [*] (a, b)l a x bf (a, b, c) ab = a-b a2 [a, b] e1, e2 e1, e2, e3

i, J i, j, k a*, b* a*, b*, c* (O, e1? e2) (O, e2, e3) ( O , i, j)

magnitude of vector AB vectors a and b are collinear vectors a and b are collinear and in the same direction vectors a and b are collinear and in opposite directions a vector obtained from vector a by a rotation through an angle <p in an oriented plane a vector obtained from vector a by a rotation through the angle n/2 in an oriented plane a pseudo scalar (or cross) product of vector a by vector b in an oriented plane triple scalar product (or cross product or mixed product) of three vectors in oriented space scalar product of vectors a and b scalar square of vector a vector product of vector a by vector b in oriented space general basis of vectors in a plane dual basis of the basis el5 e2 general basis of vectors in space dual basis of the basis e1? e2, e3 orthonormal basis of vectors in a plane orthonormal basis of vectors in space dual basis of the basis a, b in a plane dual basis of the basis a, b, c in space general Cartesian system of coordinates in a plane general Cartesian system of coordinates in space rectangular Cartesian system of coordinates in a plane

Problems in Geometry

388

(O, i, j, k) rectangular Cartesian system of coordinates in space {* ,y} vector in a plane specified by contravariant com ponents (or coordinates) x, y {*> y, z) vector in space specified by contravariant com ponents (or coordinates) x , y, z [*, jO vector in a plane specified by covariant compo nents (or coordinates) x, y [x, y, z] vector in space specified by covariant components (or coordinates) x, y , z (x, y) point in a plane specified by coordinates x , y (x, y, z) point in space specified by coordinates x, y , z AB (1) a line segment, (2) a straight line passing through points A and B, (3) the magnitude of the vector AB, (4) the length of the line segment AB (AB) oriented length of a line segment (length of seg ment AB on an oriented straight line with ap pended sign) ABC oriented triangle (ABC) area of oriented A ABC (area with appended sign) ABCD oriented tetrahedron

------ ^ (ABCD) volume of oriented tetrahedron ABCD (volume with appended sign)

ABC \ \ A'B'C'

triangles with the same orientation

A B C \i A'B'C' ABC i t A'B'C' triangles with opposite orientations (a, b) (1) the straight line of intersection of planes a and b, (2) the oriented angle from straight line a to straight line b C(AB)D; sometimes dihedral angle with edge AB in the half-planes simply (AB) of which lie points C and D (ABCD) =

anharmonic (or cross) ratio of the points A, B, C, D

% Kronecker delta: S{ = ^ ^°r * ^ 1 for i = j Sij fundamental tensor specified by covariant com ponents

List o f Symbols g iJ g

(O, r) (ABC) O,

(O)

I, (I)

4 ’ 4> 4> (^a)> (4), (4) Ot, (o 9)

R r ra, rb, rc (ABCD) o = a (M, (O, r))

(O, k) [O, k) \A arg z z GO (*o> R)

± II # tt, U IT & 19

°*

&Z9

mod *

(H‘>

fundamental tensor specified by contravammi components Gram determinant circle with centre at point O and radius r circle passing through points A, B, C* the centre of a circle circumscribed about a In angle, and the circle itself the centre of a circle inscribed in a triangle with centre at point /, and the circle itself the centres of circles inscribed in A ABC, and the circles themselves the centre of the nine-point circle (Euler’s circle), and the circle itself the radius of a circle (ABC) the radius of a circle (I) inscribed in a triangle the radii of circles (/a), (/*), (Ic) escribed in a triangle sphere passing through points A, B, C, D the power of a point M with respect to a circle (O, r); a — MO2 — r2 a homothetic transformation with centre at point O and ratio k inversion with centre at point O and power of inversion k modulus of a complex number z argument of a complex number z the conjugate complex of z point having affix z circle of radius R , the affix of the centre of which is z0 sign of equivalence sign of a one-to-one mapping sign of a one-to-one correspondence sign of perpendicularity sign of parallelism (collinearity) sign of equality and parallelism signs of collinearity and identical direction sign of collinearity and opposite direction symmetric polynomials of affixes zl9 z2, z3 symmetric polynomials of affixes zl9 z2, z8, z4 absolute value of x

♦Thus, the notation (ABC) has the following distinct meanings: (1) the (oriented) area o f A ABC: (2) a circle passing through A , B, C (no confusion can result because the first meaning o f the symbol is a number, the second is a figure).

Appendix

LIST OF BASIC FORMULAS FOR REFERENCES Quadratic equations -f p x + q = 0; x 12 -

v c = (1• *1,2

ax2 + bx

—b±

Vb2 ~ 2a

—4ac

(a * 0).

n -k ± Y k r-a c ax2 + 2kx + c *= (0; x 1>2 = ----------------------

(a ^ 0).

Progressions (a) Arithmetic progression. 1. General term of an arithmetic progression: <*n = <*i + (n - 1) d. 2.

The sum of n terms of an arithmetic progression: c

(

<*n\

+ (/! - 1) d '

--------- 2-------where d is the difference. (b) Geometric progression. 1. General term of geometric progression:

“n = 2. The sum of n terms of geometric progression:

Sn =

ux - unq

1 - qn

t------- ' ui -------=

where # is the common ratio of the progression {q /

qn — 1

1).

List of Basic Formulas 3.

391

The sum of an infinitely decreasing geometric progression:

"i q'

Logarithms 1. The notation loga N = N > 0) so that a 2. loga 1 =* 0.

log N

is equivalent to the notation ax = N (a > 0, a /

= N.

3. loga a = 1* 4. loga(AT- M ) = logfl AT + loga M.

N 5. loga ---- = M

= loga N

A/- 6- loga Af" = « loga W (V > 0). 7. logaK^Vr= — loga AT. 8. log* TV = n

__ loga AT

loga * Relationships between trigonometric functions of an angle sin a cosa 1. sin2a -f cos2a = 1 . 2 . ------- = tan a. 3. —— = cot a. 4. sin a • cosec a*= 1. 5 cos a sina cos a • sec a = 1. 6. tan a • cot a = 1. 7. 1 + tan2a = sec2 a. 8. 1 + cot2 a = co sec2a.

Table of Signs and Selected Values of Trigonometric Functions

sin a

cos a

tan a

—

cot a

III

60°

90°

270°

360°

-1

360° k

11 Ill IV 0°

O © 00

Quadrants

Function

45° |

}f2

V£ 2

Vi 2

V 1 "2

vv 3 1'T

lr3‘

K 3

Table of Reduction Formulas Angle —a

90° =F a

180° =f a

270°

Function^'"'"sin cos tan cot

— sin a + cos a — tan a — cot a

+ cos a ± sin a — cos a ± sin a =f tan a db cot a ± tan a | dF cot a

— cos a =F sin a ± cot a ± tan a

sin a + cos a =F tan a d1 cot a

Problems in Geometry

392

Transformations of trigonometric expressions 1. sin(a ± p) = sin a cos P ± cosa sin p; cos(a ± P) = cos a cos

q= sin a sin P;

tan a ± tan P tan (a ± /J) = ——------ — 1=Ftan a tan P 2. sin 2a = 2 sin a cos a; cos 2a = cos2a — sin2a = 1 — 2 sin2a = 2 cos2a — 1; tan 2a =

3. sin

2 tan a 1 — tan2 a

i . ± y i r p iM1- . ± y i

+ COS a

COS a

1 — cos a

sin a

1 + cos a

tan

+ cos a x 2 tan —

1 - tan2

-; cos * =

4. sin * :

2 tan tan jc =

1 + tan2

1 + tan2 —

1 — tan2

1 -f cos 2a 1 — cos 2a 5. cos2 a = ---------------; sin2 a = -------------- .

a ± P a ip P 6. sin a ± sin P = 2 s m ------- c o s ---------;

a -f- P a —p cos a + cos p *= 2 c o s --------c o s ---------;

a ^ . a + 0 . a —, cos a — cos p *= — 2 s m --------s i n --------

„ . a + 0 . P —a ‘ 2 s m --------s m ---------;

sin(ft ± a) „ sin(a ± 0) tan a ^ tan 0 *= -------------- ; cot a ± cot p 1 sin a sinp cos a cos /5 7. cos m* -cos nx — — [eos(m — n) x + cos(m + n) x ];

sin mjc sin nx = — [cos(/n — n) x — cos(#i + n) x ];

List o f Basic Formulas

Inverse trigonometric functions n

1 . ------- < arc sin

sin(arc sin x ) = x.

2. 0 ^ arccos x ^ ny cos(arccos x) = x. 3 .

n 2

------------- arctan x < —, tan(arctan x) = x.

4. 0 < arccot x < n, cot(arccot x) = x.

Elementary trigonometric equations 1. sin x =» a, x = ( — I)* arc sin a 4- nn. 2. cos x = af x = ± arc cos a 4- 2tt/i. 3. tan x — a, x = arc tan a -f rc/i. 4.

cot * = at x = arc cot

4 - w i.

Symmetric Polynomials For three numbers z l9 zZt z3 <Tl — •Z’l 4“ ^2 4" ^3» 0*2 =

Z 1Z 2 +

°3

Z 1 Z 2 Z 3*

^ 2 Z 3 4 " ^ 3 Z 1»

For four numbers z x> z2, z3, z4 <Ti =

Z i +

Z2 +

^3 +

Z 4>

az = z^a + riZ3 4- z xzk + r2z3 -f zaz4 -f z3z4, <*3 =

Z 1Z 2Z 3 +

z i z 2z 4 +

z i z 3z 4 +

°4 = Zl Z2Z3Z4*

Kronecker delta

Zzz zz ^

393

BIBLIOGRAPHY 1 Coxetcr, H. S. M., Greilzer, S. L. Geometry Revisited, Random House, New York, 1967 (New Mathematical Library, Vol. 19). 2. Deaux, R. Introduction to the Geometry o j Complex Numbers, Ungar, New York, 1956. 3. Dubnov, Ya. S. Basic Vector Calculus, Parts 1 and 2, (Osnovy vektornogo ischisleniya), Gostekhizdat, 1950, 1952 Moscow (in Russian). 4. Golovina, L. I., Yaglom, I. M. Induction in Geometry, Mir Publishers, Moscow, 1979. 5. Johnson, R. A. Advanced Euclidean Geometry (Modern Geometry), Dover, New York, 1960. 6. Modenov, P. S., Parkhomenko, A. S., Geometric Transformations, Academic Press, New York, 1965. 7. Pedoe, D. A Course o f Geometry for Colleges and Universities, Cambridge University Press, Cambridge, 1970. 8. Schwerdtfeger, H. Geometry o f Complex Numbers, University o f Toronto Press, Toronto, 1962. 9. Wooton, W., Beckenbach, E. R., Fleming, F. J. Modern Analytic Geometry, Houghton Mifflin, Boston, 1975. 10. Yaglom, 1. M. Complex Numbers in Geometry, Academic Press, New York, 1968. 11. Yaglom, I. M. Geometric Transformations I-IV, Random House, New York, 1962, 1968, 1973 (New Mathematical Library, Volumes 8,21, 24). 12. Yaglom, I. M. Geometric Transformations (Geometricheskie preobrazovaniya), Vol. 2, Chap. II (in Russian), Gostekhizdat, Moscow, 1956. 13. Yaglom, I. M. A Simple Non-Euclidean Geometry and Its Physical Basis, Chapter II and Supplement, S. Springer-Verlag, New York, 1979.

NAME INDEX Apollonius of Perga 52

Kroneckcr 393

Beckenbach, E.R . 394 Brianchon, C. J. 44 Blanchard, R. 8, 261, 264 Boutain 85 Brocard 74

Lagrange, J. L. 41 Langer, I. 256 Lemoine 71, 233

Carnot 77 Ceva 48 Chasles, Michel 160 Coxeter, H. S . M. 394 Deaux, R. 8, 260, 266, 394 Desargue, Girard 45 D iodes 306 Dubnov, Ya. S. 7, 394 Euclid o f Alexandria 281 Euler, Leonhard 291 Feuerbach, K. W. 205 Fleming, F. J. 394 Gibbs 343, 346 Golovina, L. I. 394 Gourmagschieg 8, 259 Gram 342, 345 Greitzer, S. L. 394 Hamilton 262 Hart 308

Markushevich, A. I. 8 Marmion, A. 41 Mascheroni, Lorenzo 310 Menelaus 44 Modenov, P. S. 394 Mohr, Georg 310 Monge 40 Morley 227, 267 Parkhomenko, A. S. 394 Pascal, Blaise 278 Peaucellier 308 Pedoe, D . 394 Pilatti 260 Poncelet, J. V. 327 Ptolemy 286 Schlomilch 27 Schwerdtfeger, Hans 394 Serret, J. A. 41 Simson 86

Thebault, V. 41, 264 Tschirnhaus 278 Vatricant, S. 260 Vieta 390

Ilin, V. A. 9 Iyenger 41

Wooton, W. 394

Jebeau, V. 8 Jensen 322 Johnson, R. A. 394

Yablonsky, S. V. 9 Yaglom, 1. M. 394

SUBJECT INDEX affine transformation 279 affix of a point 379 algebra, vector 1 Iff* analytic geometry 44ff, 347ff problems with hints and answers 66ff antiparallel lines 297 antiparallelogram 308 antireciprocal equation 244 Apollonian problem 294 Apollonius, circle of 52 argument is congruent t o . . . modulo . . . 378 axis, radical 50 barycentric coordinates 352, 354 base points 53, 327 bases, dual 343, 345 basis 339 orthogonal 339 orthonormal 339 Brianchon theorem 44 bimedians of tetrahedron 78 Boutain points 85 Brocard lines 74 Brocardians 74 bundle of planes 368 centre o f 368 improper 368 proper 368 canonical equations o f a line 372 Carnot's theorem 77 centroid o f tetrahedron 78 centroid o f a triangle 32 Ceva's theorem 48 cimedian 71 Chasles theorem 160 circle(s) o f Apollonius 52 o f infinite radius 281 o f inversion 281 Euler 32 nine-point 32

orthocentroidal 182 unit 382 zero 281, 376 o f zero radius 281 circular plane 281 cissoid o f D iodes 306 cofactor 336 collinear points 350 complex numbers 377ff absolute value o f 377 amplitude of 378 argument o f 378 conjugate 379 modulus o f 377 trigonometric form of 378 use o f in plane geometry 82ff contravariant components 343, 346 coordinates barycentric 352, 354 general Cartesian system o f 347 origin of 347 rectangular Cartesian system of 347 coplanar points 350 covariant components 343, 346 cross product 341 definitions 334flf deformation ratio 55 de Moivre formula 379 Desargueâ&#x20AC;&#x2122;s theorem 45 determinant(s) o f order three 334ff elements o f 334 Gram 342, 345 diagonal, principal 275 D iod es, cissoid o f 306 directed line segments, equal (or equiÂ valent) 8 direction vector 354 Droz-Famy theorem 196 dual bases 343, 345 elliptic pencil, o f circles 53 equation (s)

Subject Index antireciprocal 244 canonical (of a line) 372 elementary trigonometric 393 intercept form o f (of a line) 356 intercept form o f (o f a plane) 366 parametric (of a line) 356 quadratic 390 self-conjugate 386 o f a sphere 376 equipollency, sign o f 167 Euclidean circular plane 281 Euler circle 32 Euler’s formula 291

Feuerbach point(s) 205, 296 formula(s) 334ff de Moivre 379 Euler’s 291 Gibbs’ 343, 346 formula(s) (cond.) list of basic 390ff Vieta’s 390 function(s) inverse trigonometric 393 trigonometric (relationships between) 391 fundamental tensor 341

geometry analytic 44ff, 347ff problems with hints and answers 66ff of Mascheroni 309ff plane 66ff solid 78ff Gibb’s formulas 343, 346 Gram determinant 342, 345

Hamilton’s theorem 262 harmonic quadruplet (set) 52 Hart cell 308 hyperbolic pencil o f circles 53

ideal point 281 identical transformation 281 inequality, Jensen 322 inversion 28Iff, 284 mapping of regions under 297ff problems involving 285ff o f space 313ff inversors, mechanical 308 involuntary transformation 281 isogonally conjugate points 227

397

Jensen inequality 322 Kronecker delta 393 latitude 318 Lemoine point 71, 233 lima^on, Pascal’s 305 limit points 52, 53, 327 line(s) Brocard 74 Simson 86 logarithms 391 longitude 318 mapping o f regions under inversion 297 Mascheroni, geometry o f 3091T mechanical inversors 308 medians o f a tetrahedron 78 Menelaus* theorem 44 meridian 318 metaparallel triangle 279 metaparallelism 279 midperpendicular 32 minor 336 mod a: 361 Monge's point 40 Morley relation 230

nine-point circle 32 number(s) complex (see complex number) 82„ 377fF oriented space 344 orthocentre 32 orthocentric tetrahedron 40 orthocentroidal circle 182 orthogonal basis 339 orthologic triangle 280 orthologicality 280 orthonormal basis 339 orthopole of line 94, 149 pair left-hand 341 right-hand 341 with negative orientation 341 with positive orientation 341 parallels 318 parallel projection 55 Pascal’s limagon 305

398

Subject Index

Pascal’s theorem 278 Peaucellier cell 308 pencil o f circles, elliptic and hyperbolic 53, 327 pencil o f lines 358 pencil o f planes 367 Pilatti’s theorem 260 plane circular 281 Euclidean circular 281 oriented 341 plane geometry 66 point(s) affix o f 379 base 53, 327 Boutain 85 collinear 350 coplanar 350 Feuerbach 205, 296 ideal 281 at infinity 281 isogonally conjugate 227 Lemoine 71, 233 limit 52, 53, 327 M onge’s 40 Poncelet 52, 53, 327 power of 376 unit 85 Poncelet points 52, 53, 327 power o f a point 376 problem, Apollonian 294 product cross 341 pseudoscalar 341 scalar 340 triple scalar 344 progression(s) 390 arithmetic 390 geometric 390 projection parallel 55 stereographic 315 pseudoscalar product 341 pseudosquare 275 Ptolemy’s theorem 286 radical axis 50 radius vector 347 ratio, deformation 55

scalar product 340 Schlomilch’s theorem 27 self-conjugate equation 385 sheaf of planes (see bundle of planes) 368 Simson line 86

sliding vector 347 slope complex (of a line) 381 o f a line 356 solid geometry 78fF sphere equation o f 376 o f inversion 314 stereographic projection 58, 315 symbols, list o f 387 symmetric polynomials 393 symmetry 281

tangential triangle 276 tensor, fundamental 341 tetrahedron, degenerate and nondegenerate 352 theorem(s) 334ff Brianchon 44 Carnot’s 17 Ceva's 48 Chasles 160 Desargue’s 45 Droz-Farny 196 Hamilton’s 262 Menelaus’ 44 Pascal’s 278 Pilatti's 260 Ptolemy’s 286 Schlomilch’s 27 o f sine 15 transformation(s) affine 279 identical 281 involuntary 281 similarity 380 Tschirnhaus 278 transversal 45 triangle(s) degenerate 350 metaparallel 279 mirror-similar 162 nondegenerate 350 oriented 350 orthologic 280 tangential 276 trigonometric expressions, transformations of 392 triple left-hand 344 right-hand 344 with negative orientation 344 with positive orientation 344 triple scalar product 344 Tschirnhatis transformation 278

Subject Index

unit circle 382 unit point 85

vector(s) 337 collinear 337 components of 339 coordinates of 339 coplanar 337 direction 354 equal 337 linearly dependent 338 linearly independent 338 magnitude of 337 nonzero 337 normal 355

399

in the plane (solved problems) 1Iff principal 355, 364 sliding 347 in space (solved problems) 14ff zero 337 vector algebra llff, 337 Vietaâ&#x20AC;&#x2122;s formulas 390 wedge 30

zero circles 281, 376 zero vector 337

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