Geometrical Constructions Using Compasses Only [4]

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POPULAR LECTURES IN MATHEMATICS SERIES E d itors : I. N . S nfddon

and

M . S tark

Volume 4 GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

TITLES IN THE POPULAR LECTURES IN MATHEMATICS SERIES

Vol. 1 The Method o f Mathematical Induction By I. S. SOMINSKII Vol. 2 Fibonacci Numbers By N . N . V o ro b 'e v

Vol. 3 Some Applications o f Mechanics to Mathematics By V. A. Uspenskii Vol. 4 Geometrical Constructions using Compasses Only By A. N . K o s to v s k ii Vol. 5 The Ruler in Geometrical Constructions By A. S. S m o g o rzh ev sk ii Vol. 6 Inequalities By P. P. K o ro v k in

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY by

A. N. KOSTOVSKII

Translated from the Russian by

HAL1NA MOSS, B.Sc. Translation Editor

IAN N. SNEDDON Sinison Professor of Mathematics in the University of Glasgow

BLAISDELL PUBLISHING N EW Y O R K

COMPANY

LONDON

A DI VI SI ON OF R A N D O M HOUS E

S O LI; D I S T R I B U T O R S I N T H E U N I T E D S T A T E S A N D C A N A D A

Blaisdell Publishing Company 22 East 51st Street, New York 22, N. Y.

Copyright © 1961 P e rg a m o n P ress L td .

A translation of the original volume Geometricheskiye poslroyeniya odnim tsirkidem (Moscow, Fizmatgiz, 1959)

Library of Congress Card Number: 61-11527

Printed in Great Britain by Pergamon Printing and Art Services Limited, London

C O N T E N T S

Page Foreword

vii ix

Introduction

PART

1

CONSTRUCTIONS WITH COMPASSES ALONE

1.

On the possibility of solving geometrical construction problems by means of compasses alonei the basic theorem

2.

Solution of geometrical construction problems by means of compasses alone

14

3.

Inversion and its principal properties

29

4.

The application of the method of inversion to the geometry of compasses

34

P ART

3

2

GEOMETRIC CONSTRUCTIONS BY MEANS OF COMPASSES ALONE BUT WITH RESTRICTIONS

5.

6. 7.

Constructions by means of compasses alone with the opening of the legs restricted from above

47

Constructions by means of compasses alone with the angle restricted from below

65

Constructions using only compasses with constant opening of the legs

69

v

CONTENTS

Page 8.

Constructions with compasses alone on condition that all circles pass through the same point

References

71

79

FOREWORD The a u th o r o f th e p r e s e n t a r t i c l e h a s on many o c c a s io n s g iv e n l e c t u r e s on th e th e o ry o f g e o m e tr ic a l c o n s tr u c tio n s to p a r t i c i p a n t s i n m a th e m a tic a l o ly m p ia d s, w hich h av e been o rg a n iz e d ev e ry y e a r s in c e 1947> f o r th e p u p ils o f seco n d ­ a ry s c h o o ls i n th e c i t y o f L vov. T hese l e c t u r e s s e rv e d a s a fo u n d a tio n f o r th e w r i t i n g o f th e f i r s t p a r t o f t h i s w ork. The second p a r t c o n s i s t s o f i n v e s t i g a t i o n s made by th e a u th o r i n conn ex io n w ith g e o m e tr ic a l c o n s tr u c tio n s c a r r i e d o u t w ith a lim ite d o p en in g o f th e ' l e g s ' . The p r e s e n t a r t i c l e i s w r i t t e n f o r a w ide c i r c l e o f r e a d ­ ers. I t sh o u ld h e lp te a c h e r s and p u p ils o f s e n io r c l a s s e s o f se co n d a ry sc h o o ls to a c q u a in t th e m se lv e s i n g r e a t e r de­ t a i l w ith g e o m e tr ic a l c o n s tr u c tio n s c a r r i e d o u t w ith th e h e lp o f com passes a l o n e . T h is work can s e r v e a s a te a c h in g a id i n th e work o f sc h o o l m a th e m a tic a l c lu b s . I t can a ls o be u sed by s tu d e n ts s tu d y in g e le m e n ta ry m a th em atics i n phy­ s i c s and m ath em atics d e p a rtm e n ts o f t e a c h e r s ' t r a i n i n g c o l­ le g e s and u n i v e r s i t i e s . T ie a u th o r w ish e s to e x p re s s h i s s in c e r e th a n k s to A .S . Kovanko, W.F. Rogachenko and I . F . T eslen k o who r e a d th e m a n u sc rip t c a r e f u l l y and o f f e r e d a g r e a t d e a l o f v a lu a b le a d v ic e .

I N T R O D U C T I O N

G eo m e trica l c o n s tr u c tio n s form a s u b s t a n t i a l p a r t o f a m a th em atical e d u c a tio n . They r e p r e s e n t a p o w e rfu l t o o l o f g e o m e tric a l i n v e s t i g a t i o n s . The t r a d i t i o n o f l i m i t i n g th e t o o l s o f g e o m e tr ic a l con­ s t r u c t i o n to a r u l e r and com passes g oes back to rem o te a n t i ­ q u ity . The famous geom etry o f E u c lid (3 rd c e n tu r y B .C .) was b ased on g e o m e tr ic a l c o n s tr u c tio n s c a r r i e d o u t by com passes and a r u l e r , th e com passes and r u l e r b e in g r e g a rd e d a s e q u iv a le n t in s tru m e n ts ; i t was a m a tte r o f i n d i f f e r e n c e how each se p a ­ r a t e c o n s tr u c tio n was c a r r i e d o u t - by means o f th e com passes and th e r u l e r , o r by means o f th e com passes a lo n e o r by means o f th e r u l e r a l o n e . I t was n o te d a lo n g tim e ago t h a t com passes a r e a more e x a c t, a more p e r f e c t in s tru m e n t th a n a r u l e r . I t was o b serv ed to o t h a t c e r t a i n c o n s tr u c tio n s c o u ld be c a r r i e d o u t by means o f co m passes, w ith o u t th e u s e o f a r u l e r ; f o r exam ple, to d iv id e a c irc u m fe re n c e in t o s i x e q u a l p a r t s , to c o n s tr u c t a p o in t sy m m e tric a l to a g iv e n p o in t w ith r e s ­ p e c t to a g iv e n s t r a i g h t l i n e , and so o n . A tte n tio n was drawn to th e f a c t t h a t i n e n g ra v in g t h i n m e ta l p l a t e s , i n m arking th e d iv id i n g c i r c l e s on a s tro n o m ic a l in s tr u m e n ts as a r u l e o n ly com passes a r e u s e d . The l a t t e r , p r o b a b ly , gave th e im p e tu s to i n v e s t i g a t i o n s on g e o m e tr ic a l c o n s tr u c ­ t io n s c a r r i e d o u t by com passes a l o n e . In th e y e a r 1797, th e I t a l i a n m a th e m a tic ia n L orenzo M asche r o n i, p r o f e s s o r o f th e U n iv e r s it y o f P a v ia , p u b lis h e d a la r g e work 'The Geometry o f Com passes' , w hich was l a t e r t r a n s l a t e d i n t o F ren ch and German. I n t h a t work th e f o llo w ­ in g p r o p o s itio n was proved I A ll c o n s tr u c tio n problem s s o lu b le b.v means o f com passes and a r u l e r can a l s o be s o lv e d e x a c t ly by means o f com­ p a s s e s a lo n e .

ix

INTRODUCTION

T h is s ta te m e n t was p ro v ed i n 1890 by A. A d ler i n an o r i g i ­ n a l w ay, u s in g i n v e r s i o n . He a l s o proposed a g e n e r a l method o f s o lv in g g e o m e tr ic a l c o n s tr u c tio n problem s by means o f com­ p a s s e s a l o n e . I n 1928, th e D anish m a th em aticia n H jelm slev found i n a bookshop i n Copenhagen a book by G. Mohr ( 'The D anish E u c lid ' ) p u b lis h e d i n 1672 i n Amsterdam. In th e f i r s t p a r t o f t h i s book t h e r e i s a f u l l s o lu t io n o f M a sc h e ro n i's p ro b le m . T h u s, i t had been shown a lo n g tim e b e f o r e M aschero n i t h a t a l l g e o m e tr ic a l c o n s tr u c tio n s c a p a b le o f b e in g c a r ­ r i e d o u t w ith com passes and a r u l e r can be c a r r i e d o u t by means o f com passes a l o n e . The d iv i s i o n o f geom etry i n w hich g e o m e tr ic a l c o n s tr u c tio n s by means o f com passes a lo n e a r e s tu d ie d i s c a l l e d 'The Geometry o f th e Com passes' . I n 1833 th e Sw iss g eo m eter Ja co b S t e i n e r p u b lis h e d th e work 'G e o m e tric a l C o n s tru c tio n C a rr ie d o u t w ith th e A id o f a S t r a i g h t L ine and a F ix ed C i r c le ' , i n w hich he i n v e s t i g a t e d m ost f u l l y c o n s tr u c tio n s c a r r i e d o u t w ith th e r u l e r a l o n e . The b a s ic r e s u l t o f t h i s work can be fo rm u la te d a s f o llo w s : E very c o n s tr u c tio n p ro b lem , s o lu b le by means o f com passes and a r u l e r , can be s o lv e d by means o f a r u l e r a lo n e , p ro ­ v id e d t h a t i n th e p la n e o f th e draw ing t h e r e i s a g iv e n c i r c l e w ith f ix e d c e n tr e and r a d i u s . T hus, i n o r d e r to make th e r u l e r e q u iv a le n t to th e com­ p a s s e s , i t i s s u f f i c i e n t to u s e th e com passes o n c e . The g r e a t R u ssia n m a th e m a tic ia n N .I . L o b a c h e v s k ii, i n th e f i r s t h a l f o f th e 19t h c e n tu r y , d is c o v e re d a new g eo m etry , w hich l a t e r became known a s n o n -E u c lid e a n geom etry o r Lobac h e v sk ia n g e o m e try . R e c e n tly , th a n k s to th e e f f o r t s o f a g r e a t number o f s c h o l a r s , e s p e c i a l l y S o v ie t o n e s , th e th e o ry o f g e o m e tr ic a l c o n s tr u c tio n s i n L obachevskian geom etry h as been v ig o r o u s ly d e v e lo p e d . A .S . S m o g o rz h e v sk ii, V .F . Rogachenko, K.K. M okrishchev and o th e r m a th e m a tic ia n s have c a r r i e d o u t i n v e s t i g a t i o n s i n t o c o n s tr u c tio n s i n th e L obachevskian p la n e w ith o u t a r u l e r , show ing th e p o s s i b i l i t y o f e x e c u tin g c o n s tr u c tio n s s i m i l a r to th e c o n s tr u c tio n s o f M ascheroni i n th e E u c lid e a n p la n e . The g r e a t number o f g e n e r a l i n v e s t i g a t i o n s le d to th e f o rm u la tio n i n th e w orks o f o u r s c h o la r s o f q u i t e a f u l l and

x

INTRODUCTION

ex a c t th e o ry o f g e o m e tr ic a l c o n s tr u c tio n s i n th e L obachevsk ia n p la n e , s c a r c e ly i n f e r i o r i n i t s co m p le te n e ss to th e th e o ry o f g e o m e tr ic a l c o n s tr u c tio n s i n th e E u c lid e a n p la n e .

Part 1 C on stru ction s w ith com p asses a lo n e

1.

ON THE POSSIBILITY OF SOLVING GEOMETRICAL CONSTRUCTION PROBLEMS BY MEANS OF COMPASSES ALONE: THE BASIC THEOREM

In t h i s s e c ti o n th e p ro o f o f th e B a sic theo rem o f th e Geome­ t r y o f Compasses w i l l he p r e s c r i b e d . To do s o , i t i s n e c e s ­ s a ry to exam ine th e s o lu t io n s o f c e r t a i n pro b lem s on c o n s tr u c ­ ti o n w ith com passes a l o n e . I t i s c l e a r t h a t we c a n n o t, w ith com passes a l o n e , draw a c o n tin u o u s s t r a i g h t l i n e g iv e n by two p o in ts on i t , a lth o u g h we s h a l l show l a t e r how, u s in g com passes a l o n e , we ca n con­ s t r u c t o n e, tw o , a n d , g e n e r a l l y , any number o f p o i n t s , s i t u ­ a te d as c l o s e l y to g e th e r as d e s ir e d on a g iv e n s t r a i g h t l i n e * . T hus, th e c o n s tr u c tio n o f a s t r a i g h t l i n e i s n o t f u l l y covered by th e M ohr-M ascheroni t h e o r y . In th e geom etry o f co m p asse s, a s t r a i g h t l i n e o r a segm ent i s d e fin e d by two p o in ts and i s n o t g iv e n a s a c o n tin u o u s s t r a i g h t l i n e (drawn w ith a r u l e r ) . The c o n s tr u c tio n o f a s t r a i g h t l i n e i s re g a rd e d a s co m p leted as soon a s any two p o in ts on i t a r e c o n s t r u c t e d . L et u s a g re e to w r i t e th e p h ra s e 'W ith p o in t A a s c e n tr e and r a d iu s BC we d e s c r ib e a c i r c l e ( o r draw an a r c ) ' i n th e s h o r t form : 'We d e s c r ib e th e c i r c l e (A, BC)' o r 'We draw th e c i r c l e (4, BC)' , o r , s h o r t e r s t i l l , 'We d e s c r ib e (A, BC)' . I n s te a d o f th e n o ta tio n ( A , AB) we s h a l l w r i t e (A, B) . F or th e sa k e o f c l a r i t y , we s h a l l - s t i l l mark ( i n d o ts ) s t r a i g h t l i n e s i n th e d ia g ra m s . (T hese s t r a i g h t l i n e s p la y no p a r t i n th e c o n s t r u c t i o n s ) . Problem 1 . To c o n s t r u c t a p o i n t . sy m m e tric a l to a g iv e n p o in t w ith r e s p e c t to th e g iv e n s t r a i g h t . l i n e AH .

*From th e p r a c t i c a l p o in t o f v ie w , th e r e i s no gro u n d to r e g a rd a s t r a i g h t l i n e a s c o n s tr u c te d , i f some o f i t s p o in ts a r e c o n s tr u c te d .

3

4

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

C o n s t r u c t i o n . We d e s c r ib e th e c ir c le s ( A , C)and (fl, C) , i . e . u s in g th e p o i n t s A and B a s c e n t r e s , we draw c i r c l e s p a s s in g th ro u g h th e p o in t C ( F ig . l ) . At th e i n t e r ­ s e c t i o n o f th e s e c i r c l e s we o b ta in p o in t C, . The p o in t C, i s th e r e q u ir e d o n e .

F ig .

1

N o t e . To v e r i f y t h a t t h r e e p o in ts A, B and A’ l i e on one s t r a i g h t l i n e , i t i s n e c e s s a r y to c o n s tr u c t any p o in t C o u ts i d e th e s t r a i g h t l i n e and th e n th e p o in t C, , sym m etri­ c a l to C . O b v io u sly , th e p o in t X l i e s on th e s t r a i g h t l i n e AB i f th e segm ents CX and CtX a r e e q u a l to each o th e r ( F ig .

1) . P roblem 2 . To c o n s t r u c t a seg m en t. 2 . A. . . . an d , i n g e n e r a l , n tim e s g r e a t e r th a n a g iv e n segm ent AA. = r (where n i s any n a t u r a l num ber) . C o n s t r u c t i o n . (Method l ) . K eeping th e o p e n in g o f th e com passes c o n s ta n t and e q u a l to r we d e s c r ib e

POSSIBILITY OP SOLVING PROBLEMSi

BASIC THEOREM

S

th e c i r c l e (-4,, r) and we c o n s tr u c t th e p o in t At , d ia m e tr ic ­ a l l y o p p o s ite to th e p o in t A , f o r w hich p u rp o se we mark o f f th e ch o rd s AB = B C = C A t = r C F ig. 2 ) . The segm ent AAt = 2r . We th e n d e s c r ib e th e c i r c l e (>!,, r) w hich i n t e r s e c t s th e c i r ­ c l e (C, r) a t th e p o in t D . At th e p o in t o f i n t e r s e c t i o n o f th e c i r c l e s {D, r) and (.4,, r) we o b ta in th e p o in t As . The segment AAt = 3r , and so o n . H aving c a r r i e d o u t th e in d ic a t e d c o n s tr u c tio n s n tim e s we c o n s t r u c t th e segm ent AAn — nr . The v a l i d i t y o f th e c o n s tr u c tio n fo llo w s from th e f a c t t h a t compasses w ith an o p en in g e q u a l to th e r a d iu s o f a c i r c l e d iv id e i t s c irc u m fe re n c e i n t o 6 e q u a l p a r t s .

F ig . 2 C o n s t r u c t i o n . (Method 2 ) . We ta k e any p o in t B o u ts id e AAt and we draw th e c i r c l e s (A ,, AB) and (B, r) f w hich m eet a t th e p o in t C ( F i g . 5 ) .

Fig. 5

I f th e c i r c l e s (i4,, r) and (C, BA,) a r e drawn th e y w i l l i n t e r s e c t i n th e p o in t At . The segm ent AAt = 2r . D esB

6

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

c r i b i n g th e c i r c l e s (/4,, r) and (C, BA,) . The segm ent A

8

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

Problem 4 .

To d iv id e th e a r c AB o f a c i r c l e i n h a l f .

C o n s t r u c t i o n . We can assume t h a t th e c e n tr e O o f th e c i r c l e i s known; i t w i l l be shown below (Problem 13) how to c o n s tr u c t th e c e n tr e o f a c i r c l e ( o r a r c ) u s in g com passes a l o n e . P u t t i n g OA — OB = r and AB = a , we d e s c r ib e c i r c l e s (O, a), (A, r) and (/i, r)~ a t th e i n t e r s e c t i o n we o b ta in th e p o in ts C and D ( F i g . 5 ) • We draw th e c i r c l e s (C, B) and (O, A) up to t h e i r i n t e r s e c ­ t i o n a t th e p o in t E . I f , now, th e c i r c l e s (C, OE) and (D, OE) be draw n, t h e n , a t t h e i r i n t e r s e c t i o n , we o b ta in p o in ts X and X, . The p o in t X d iv id e s i n h a l f th e a r c AB , th e p o in t X\ d iv id e s th e a r c t h a t c o m p le te s , w ith th e f i r s t o n e , th e f u l l c i r c l e . I n th e c a s e when th e w hole c i r c l e (O, A) i s draw n, we n ee d draw o n ly one o f th e two c i r c l e s (C, OE) and (D, OE) , w h ic h , a t i t s i n t e r s e c t i o n w ith th e c i r c l e (0, A )', w i l l d e f in e p o in ts X and X, . P r o o f . The f i g u r e s ABOC and ABDO a r e p a r a lle lo g r a m s ; t h e r e f o r e , th e p o in ts C, O and D l i e on th e same s t r a i g h t l i n e (CO\\AB, O D \\A B ). I t f o llo w s from th e i s o s c e l e s t r i a n g l e s CED and CXD t h a t COE = COX = 90° • Thus th e segment OX i s p e r p e n d ic u la r to th e ch o rd AB . C o n s e q u e n tly , i n o r d e r to p ro v e t h a t th e p o in t X d iv id e s th e a r c AB i n h a l f , i t i s s u f f i c i e n t to show t h a t th e segm ent OX = r. I t fo llo w s from th e p a r a lle lo g r a m ABOC t h a t O A '+ B C ' = 2 OB' + 2AB' or r'-\-B C ' = 2r' + 2a', so t h a t BC' = 2a' + rt From th e r i g h t - a n g l e d t r i a n g l e

COE we have

CP - B C = OC* + O P, whence 2a' + r' = a '-\-O P

POSSIBILITY OF SOLVING PROBLEMS«

BASIC THEOREM

9

and o e , = a i 4-r*. F i n a l l y , from th e r i g h t - a n g l e d t r i a n g l e COX we o b ta in o x = Y CX' — OC = V OE‘ — OC* = = y a‘ -f- r* — a‘ = r. As we have a lr e a d y p o in te d o u t , i n th e g eo m etry o f th e com­ p a s se s a s t r a i g h t l i n e i s r e g a rd e d a s c o n s tr u c te d as soon as any two o f i t s p o in ts a r e d e f in e d . In o u r f u r t h e r d is c u s s io n s (Problem s 2 4 , 25 and o th e r s ) we s h a l l have to c o n s t r u c t , w ith com passes a lo n e , o n e , two a n d , i n g e n e r a l , any number o f p o in ts o f th e g iv e n s t r a i g h t l i n e . T h is c o n s tr u c tio n can be c a r r ie d o u t a s f o llo w s ! Problem 5 . On a s t r a i g h t l i n e , d e f in e d by two p o in ts A and f l. c o n s tr u c t one o r s e v e r a l p o i n t s . C o n s t r u c t i o n . We ta k e an a r b i t r a r y p o in t C ( F ig . 6 ) i n th e p la n e , o u ts i d e th e s t r a i g h t l i n e AB . We

P ig . 6 c o n s tr u c t p. p o i n t , C, , sy m m e tric a l to C w ith r e s p e c t to AB (Problem l ) . W ith an a r b i t r a r y r a d iu s r , we d e s c r ib e th e c i r c l e s (C, r) and (C,, r) . At t h e i r i n t e r s e c t i o n we o b ta in th e r e q u ir e d p o in ts X and X t , w hich l i e on th e s t r a i g h t l i n e AB . V ary in g th e s i z e o f th e r a d iu s r , i t i s p o s s ib le to c o n s tr u c t any number o f p o in ts o f th e g iv e n s tra ig h t lin e i X', X[ , etc*. Problem 6 .

C o n s tru c t th e -points o f i n t e r s e c t i o n o f th e

10

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

g iv e n c i r c l e (O, r) and th e s t r a i g h t l i n e /4 and B .

g iv e n by two p o in ts

C o n s t r u c t i o n i n th e c a s e , when th e c e n tr e O does n o t l i e on th e g iv e n s t r a i g h t l i n e AB ( P ig . 7 ) * . We con­ s t r u c t th e p o in t O,, sy m m e tric a l to th e c e n tr e O o f th e g iv e n c i r c l e , w ith r e s p e c t to th e s t r a i g h t l i n e AB (Problem l ) . We d e s c r ib e th e c i r c l e (0,, r) w hich i n t e r s e c t s w ith th e g iv e n c i r c l e a t th e r e q u ir e d p o in ts X and Y . The t r u t h o f th e c o n s tr u c tio n i s obvious from th e symmetry o f th e f i g u r e w ith r e s p e c t to th e g iv e n s t r a i g h t l i n e AB .

Fig. 7 C o n s t r u c t i o n in the case when the centre 0 of the given circle lies on the straight line AB (Fig. 8).

W ith y4.as th e c e n tr e and an a r b i t r a r y r a d iu s d , we d e s c r ib e a c i r c l e , w hich i n t e r s e c t s th e g iv e n c i r c l e a t th e p o in ts C and D . We h a lv e th e a r c s CD o f th e c i r c l e (0, r) (Problem 4 ) • *W ith th e h e lp o f com passes a lo n e i t i s e a sy to check w h eth er th r e e g iv e n p o in ts l i e on one s t r a i g h t l i n e o r n o t ( s e e n o te to Problem 1 ) .

POSSIBILITY OP SOLVING PROBLEMS!

BASIC THEOREM

11

The p o in ts X and Y a r e th e r e q u ir e d o n e s .

F ig . 0 N o t e . th a t

From th e c o n s tr u c tio n d is c u s s e d above i t fo llo w s AX = A O -\-O X

and

A Y = A O — OX.

Problem 7 . To c o n s tr u c t th e p o in t o f i n t e r s e c t i o n o f two s t r a i g h t l i n e s AB and CD . each o f w hich i s g iv e n by two p o in ts . C o n s t r u c t i o n . We c o n s tr u c t p o in ts c, and D, , sy m m etrical to C and D r e s p e c t i v e l y , w ith r e s p e c t to th e g iv e n s t r a i g h t l i n e A3 ( F ig . 9)« We d e s c r ib e c i r c l e s (D,, CC,)

Fig. 9

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

12

and (C, D) and we d e n o te th e p o in t o f t h e i r i n t e r s e c t i o n by E . We c o n s tr u c t th e segm ent x , f o u r th p r o p o r tio n a l to th e segm ents DE, DDt , and CD (P roblem } ) . Now, i f th e c i r c l e s (D, x) and (/)„ x) a r e draw n, we th e n o b ta in th e r e q u ir e d p o in t X a t th e ir in te rs e c tio n . P r o o f . As th e p o in t C, i s sy m m etrical to th e p o in t C , and th e p o in t D, i s sy m m e tric a l to th e p o in t D , th e n , o b v io u s ly , we s h a l l f i n d th e p o in t o f i n t e r s e c t i o n o f th e g iv e n s t r a i g h t l i n e s i f we c o n s t r u c t th e p o in t o f i n t e r ­ s e c t i o n o f th e s t r a i g h t l i n e s CD and C,D, • The f i g u r e C C J )^ i s a p a r a lle lo g r a m , c o n s e q u e n tly , th e p o in ts D. D, and E l i e on th e same s t r a i g h t l i n e (DE\\CC„ OD,|| CC.) . The t r i a n g l e s CDE and XDD, a r e s i m i l a r , t h e r e f o r e DE:DD, = CE:D,X, but CE = CD = ClDI. The segm ent D,.Y = x i s th e f o u r t h p r o p o r ti o n a l o f th e se g ­ m ents DE, DD, and CD . Each c o n s t r u c t i o n a l problem w ith com passes and a r u l e r i n th e E u c lid e a n p la n e i s alw ays r e d u c ib le to th e s o l u t i o n i n a d e f i n i t e o r d e r o f th e f o llo w in g v e ry sim p le b a s ic p ro b lem s: 1.

To draw a s t r a i g h t l i n e th ro u g h two g iv e n p o i n t s .

2.

To d e s c r ib e a c i r c l e o f a g iv e n r a d iu s from a g iv e n c e n tre .

3.

To f i n d th e p o in ts o f i n t e r s e c t i o n o f two g iv e n c i r c l e s

4.

To f i n d th e p o in ts o f i n t e r s e c t i o n o f a g iv e n c i r c l e w ith a s t r a i g h t l i n e g iv e n by two p o i n t s .

5.

To f i n d th e p o in t o f i n t e r s e c t i o n o f two s t r a i g h t l i n e s each o f w hich i s g iv e n by two p o i n t s .

I n o r d e r to p rove t h a t any c o n s tr u c tio n problem w hich can b e so lv e d w ith a r u l e r and com passes can a l s o be so lv e d by means o f com passes a lo n e , i t i s s u f f i c i e n t to show t h a t a l l th e s e b a s ic o p e r a tio n s can be c a r r i e d o u t by means o f com­ p asses a lo n e . The second and t h i r d o p e r a tio n s a r e c a r r i e d o u t d i r e c t l y

POSSIBILITY OF SOLVING PROBLEMS!

BASIC THEOREM

13

By co m passes. The re m a in in g B a sic o p e r a tio n s w ere c a r r i e d o u t i n ProBlems 5 -7 • Suppose t h a t a c e r t a i n c o n s tr u c tio n p ro b lem , s o lu b le By means o f com passes and a r u l e r , h a s to Be so lv e d By means o f com passes a l o n e . L et u s im agine t h i s problem s o lv e d By means o f a r u l e r and co m passes. As a r e s u l t , th e s o lu t io n i s red u c ed to c a r r y in g o u t a c e r t a i n f i n i t e se q u en ce o f th e f iv e B a sic o p e r a t i o n s . H aving c a r r i e d o u t each o f th e s e o p e r a tio n s w ith com passes a lo n e (Problem s 5 -7 ) we a r r i v e a t th e s o lu t io n o f th e o r i g i n a l p ro b lem . Thus a l l c o n s tr u c tio n p ro b le m s, s o lu b le By means o f com­ p a s se s and a r u l e r , can Be so lv e d e x a c t ly By means o f com­ p a s se s a l o n e . The method o f s o lv in g g e o m e tr ic a l c o n s tr u c tio n problem s By means o f com passes a lo n e l e a d s , as a r u l e , to q u i t e comp­ l i c a t e d and le n g th y c o n s t r u c t i o n s , B ut i t h a s g r e a t i n t e r e s t from th e t h e o r e t i c a l p o in t o f v ie w .

2.

SOLUTION OF GEOMETRICAL CONSTRUCTION PROBLEMS BY MEANS OF COMPASSES ALONE

In t h i s s e c t i o n we s h a l l d is c u s s th e s o lu t io n o f c e r t a i n i n t e r e s t i n g problem s i n th e geom etry o f com p asses, a r r iv e d a t m a in ly by th e e f f o r t s o f Mohr, M ascheroni and A d le r . The s o l u t i o n s o f some o f th e s e problem s w i l l be u se d i n th e second p a r t . Problem 8 . To draw a -p e rp e n d ic u la r to th e segm ent AB a t th e p o in t A . C o n s t r u c t i o n . ( F i r s t M e th o d ). K eeping th e o p e n in g o f th e com passes c o n s ta n t and e q u a l to an a r b i t r a r y segm ent r , we draw th e c i r c l e s (.4, r) and (B, r) u n t i l th e y m eet a t th e p o in t O . We d e s c r ib e th e c i r c l e (O, r) and we c o n s tr u c t th e p o in t E on i t , w hich i s d i a m e t r i c a l l y o p p o s ite to th e p o in t B . F o r t h i s we draw th e ch o rd s CD = DE = r ( F i g . 1 0 ) , w here C i s th e p o in t o f i n t e r s e c t i o n o f th e c i r ­ c l e s (B. r) and (O, r ) . The segm ent AE i s p e r p e n d ic u la r to AB . I f we p u t r = A B t th e n AE = V 3 A B , and th e p o in t C c o in c id e s w ith th e p o in t A .

I

F i g . 10

F i g . 11

14

SOLUTION OF PROBLEMS

15

C o n s t r u c t i o n . (Second M e th o d ). We d e s c r ib e th e c irc u m fe re n c e (B, A) ( F i g . l l ) , we ta k e an a r b itr a r y p o in t C on i t and we draw th e c i r c l e (C, A) . L et D b e th e p o in t a t w hich th e s e c i r c l e s i n t e r s e c t . I f now a t h i r d c i r c l e (A,D) i s drawn to i t s i n t e r s e c t i o n w ith th e c i r c l e (C, A) a t th e p o in t E , th e n th e segm ent AE i s p e r p e n d ic u la r to AB . P r o o f . The segm ent AC j o i n s th e c e n t r e s o f th e c i r c l e s {A, D) and (C, A) . DE i s t h e i r common c h o rd ; t h i s means t h a t AC i s p e r p e n d ic u la r to DE and ■$: CAD = •£; CAE ( th e t r i a n g l e ADE i s e q u i l a t e r a l ) . On th e o th e r hand

3; CAD = 3;

=

•

I t fo llo w s from

th e l a s t e q u a tio n s t h a t & C A E = ^£. The s t r a i g h t l i n e AE i s th e ta n g e n t to th e c i r c l e th e p o in t A , so t h a t AE i s p e r p e n d ic u la r to AB . Problem 9 .

To c o n s tr u c t a se g m en t. e q u a l to

(fl, ,4) a t

o f a g iv e n

segment AB ( to d iv id e th e segm ent AB i n t o n e q u a l p a r t s . n = 2, 3 ,...). C o n s t r u c t i o n . ( F i r s t M e th o d ). We c o n s tr u c t th e segment A C = nA B (Problem 2 ) . We d e s c r ib e th e c i r c l e (C, A ) . At th e i n t e r s e c t i o n w ith th e c i r c l e (A, B) we o b ta in th e p o in ts D and D, . The c i r c l e s (D, A) and (O,, A) d e f in e th e p o in t A such t h a t th e segm ent A Y = — n

(F ig . 1 2 ).

I n c r e a s in g th e segm ent AX 2, 3 , and so o n , n tim e s (P ro b ­ lem 2 ) , we c o n s tr u c t th e p o in ts w hich d iv id e th e segm ent AB in to n e q u a l p a r t s . P r o o f . From th e s i m i l a r i t y o f th e i s o s c e l e s t r i a n g l e s ACD and AXD ( th e a n g le A i s a common one) i t fo llo w s t h a t AC: AD = AD: AX or AD' = AB' = A C A X = nAB ■AX.

16

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

Hence A X = -A B . R

The p o in t X l i e s on th e s t r a i g h t l i n e AB . N o t e . F o r l a r g e v a lu e s o f n , th e p o in t X i s poorlyd e f in e d ; th e a r c s o f th e c i r c l e s (D, A) and (£),, A) i n t e r s e c t a t X a t a v e ry sm a ll a n g l e * . I n t h i s c a s e , to d e f in e th e p o in t X , i n s t e a d o f th e c i r c l e (D,, A) , we can draw th e c i r c l e (.4, ED) , w here E i s th e p o in t d i a m e t r i c a l l y o p p o s ite to th e p o in t D, o f th e c i r c l e (A, B) . C o n s t r u c t i o n . (Second M e th o d ). We c o n s tr u c t th e segm ent A C = n A B (P roblem 2 ) . We th e n d e s c r ib e th e c i r c l e s (4, C), (C, A) and (C, AB) w hich i n t e r s e c t a t th e p o in ts D and E . I f we now d e s c r ib e th e c i r c l e s (D, -4) and (C, DE) th e n a t t h e i r i n t e r s e c t i o n we o b ta in th e p o in t X . The se g ­ ment AX = — AD n

(F ig . 1 3 ).

P r o o f . The p o in t X l i e s on th e s t r a i g h t l i n e AC , s in c e AC i s p a r a l l e l to DE and XC i s p a r a l l e l to DE (th e * F o r th e d e f i n i t i o n o f th e a n g le o f i n t e r s e c t i o n o f two c u r v e s , se e S e c tio n 0 , p . 71 b e lo w .

SOLUTION OF PROBLEMS

17

F i g . 13 f ig u r e CEDX i s a p a r a lle l o g r a m ) . From th e s i m i l a r i t y o f th e i s o s c e l e s t r i a n g l e s ACD and AXD we g e t A X = -nA B . We now g iv e th e c o n s tr u c tio n p u t fo rw ard hy S m o g o rz h e v sk ii* . T his c o n s tr u c tio n d i f f e r s from th e p re c e d in g ones i n t h a t th e r e q u ir e d — th p a r t o f th e segm ent AB does n o t l i e on th e g iv e n seg m ent.

*See Bibliography, page

79 •

18

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

C o n s t r u c t i o n . (T h ird M ethod). We c o n s tr u c t AC=znAB (Problem 2 ) . We draw th e c i r c l e s (A C) and (B, AQ u n t i l th e y m eet a t th e p o in t D . The c i r c l e (D, AB) w i l l i n t e r s e c t th e two l a t t e r ones a t th e p o in ts E and H .

The segm ent EH = — AB ( F i g . 1 4 ) .

P r o o f . Prom th e co n g ru en ce o f t r i a n g l e s ABD, ADE and BOH ( th e th r e e s id e s b e in g e q u a l) , i t fo llo w s t h a t Z l = /_EDH . The i s o s c e l e s t r i a n g l e s ADB and EDH a r e s i m i l a r , c o n s e q u e n tly i EH:ED = AB:AD or EH :AB = AB: nAB. F in a lly E H = —AB. n We s h a l l n o te t h a t EK w

Problem 1 0 . segm ent 3 - ).

V «» —1 AB, ii

= ( 2-

f

)^s -

To c o n s tr u c t a segm ent e g u a l to

J_ o f a g iv e n

AB ( d iv id e th e segm ent AB in to 2" e q u a l p a r t s i n = 2.

C o n s t r u c t i o n . ( F i r s t M e th o d ). We c o n s tr u c t th e segm ent AC — 2AB (Problem 2 ) . We draw th e c i r c l e (6\ A) and we d e n o te by D, and D[ th e p o in ts o f i t s i n t e r s e c t i o n w ith th e c i r c l e (A, B) . I f now we draw th e c i r c l e s (D„ A) and (D'„ A) th e n a t t h e i r i n t e r s e c t i o n we o b ta in th e p o in t X ,.

The segm ent BXx =

A X ,= ^ -A B

.

We th e n d e s c r ib e

th e c i r c l e (A,BD,) , and a t th e i n t e r s e c t i o n w ith (C, A) we o b ta in th e p o in t D, and O’, . We draw th e c i r c l e s (D,,A) and (O’,, A) u n t i l th e y m eet a t th e p o in t X , . The segment bx,

= ±

ab

.

19

SOLUTION OF PROBLEMS

F i g . 15 I f we f u r t h e r d e s c r ib e th e c i r c l e s (A, BDt) , (O;, A)

, th e n we g e t th e p o in t

Xt .

(Dt, A) and

The segm ent flA'I = ~ AB

and so o n . P r o o f . From th e s i m i l a r i t y o f th e i s o s c e l e s t r i a n g l e s i4CD,and ADlX l i t fo llo w s t h a t AD,:AC = AX,:A D l or AB:2AB = A X t :AB. Hence AX, = ^ A B . We in tr o d u c e th e n o t a t i o n

AB = a, ADk = mk, k = 1, 2 , 3 , . . .

a

•

The segm ent BDt i s th e m edian o f th e t r i a n g l e ACDlt c o n se ­ q u e n tly

20

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

4 BD\ = 2 AD\ -)- 2 CD\ — AC* o r , i n a n o th e r way, 4 m \ = 2 /IS* - f 2 AC* — AC’ = 2 AS' + AC = 2 Afl* -f- 4AS1.

T h is means t h a t m : = A o : = i + ^ a' = 4 f in d th e in v e r s e p o i n t , C[ , o f C, ( F i r s t ti o n o f th e problem u n d e r d i s c u s s i o n ) . We ment OX = nOC[ . The p o in t X i s in v e r s e o f

P r o o f .

y (Problem 2 ) . We method o f c o n s tru e c o n s t r u c t th e se g ­ th e g iv e n p o in t C

S u b s t i t u t i n g OCt — nOC and OC'l = Q£ i n th e

e q u a tio n OC, •OCJ = r‘ , we g e t

36

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

Problem 1 6 . G iven th e c i r c l e o f in v e r s io n (O, r) and th e s t r a i g h t l i n e AB. w hich does n o t p a ss th ro u g h th e c e n tr e o f i n v e r s i o n , c o n s tr u c t th e c i r c l e w hich i s th e in v e r s e o f th e g iv e n s t r a i g h t l i n e . C o n s t r u c t i o n . We c o n s tr u c t O, , sy m m etrical to th e c e n tr e o f in v e r s i o n O , w ith r e s p e c t to th e s t r a i g h t l i n e AB (Problem l ) . We f i n d th e p o in t 0 [ , in v e r s e o f th e p o in t O, (Problem 15) • The c i r c l e (OJ, O) i s in v e r s e o f th e g iv e n s t r a i g h t l i n e AB ( F i g . 2 8 ) .

i i i

P r o o f . L et C and C be th e p o in ts o f i n t e r s e c t i o n o f th e s t r a i g h t l i n e 0 0 , w ith th e g iv e n s t r a i g h t l i n e AB and th e c i r c l e (OI, O) • I t fo llo w s from th e above c o n s tr u c tio n t h a t 0 0 ,-0 0 ', = r', 0 0 , = 20C, 0 C = 200;, O C ± A B . Hence 0 0 , • 00', = 1 0 C ~ = OC-OC = /•*. A cco rd in g to Theorem 5 f th e c i r c l e (0',, th e s t r a i g h t l i n e AB .

O) i s th e in v e r s e o f

Problem 1 7 . To c o n s t r u c t th e s t r a i g h t l i n e A B . w hich i s th e in v e r s e o f th e g iv e n c i r c l e (O., R) p a s s in g th ro u g h th e

APPLICATION OP METHOD OF INVERSION

centre of inversion

37

O .

C o n s t r u c t i o n . If the given circle intersects the circle of inversion at the points A and B , then the straight line AB is the inverse of this circle. Otherwise, we take the points A, and S, (Fig. 29) on the given circle, and we construct their inverses, A and B (Problem 15)* The straight line AB is the inverse of the given circle (O,, R) • By varying the position of the points At and S, on the given circle, it is possible to construct as many points of this straight line as required. The validity of this construction is self-evident (see Theorem j) •

Fig. 29

Problem 1 8 . A g iv e n c i r c l e (O.. R ) does n o t p a s s th ro u g h th e c e n tr e o f in v e r s i o n O . C o n s tru c t th e c i r c l e w hich i s in v e r s e o f th e g iv e n o n e . C o n s t r u c t i o n . We take the given circle (0,, R) as the circle of inversion and we construct the inverse point O' of the point 0 (Problem 15)• Then we construct the in­ verse point 0, of the point O' with respect to the circle of inversion (O, r) . The point O, is the centre of the circle required (Fig. 50). We take any point A on the given circle (0,, R ) and we find the inverse A ' . The circle (Oa, A ') is in/erse of the given circle (O,, R ) . D

38

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

P r o o f , L e t PP' be th e d i r e c t common ta n g e n t o f th e c i r c l e s (O,, R) and (0„ A‘) and l e t PO’ be p e r p e n d ic u la r to 0 0 , . From th e s i m i l a r i t y o f th e t r i a n g l e s have

UPO'

and OP'O, we

0 0 , : OP' = 0 P :0 0 ' o r, a lte rn a tiv e ly 0 0 , - 0 0 ' = O P-O P'= rt , s in c e th e p o in ts P and P' a r e i n v e r s e . I t fo llo w s from th e l a s t e q u a tio n t h a t th e p o i n t s O a n d O' a r e in v e r s e w ith r e s ­ p e c t to th e c i r c l e o f in v e r s i o n (O, r) « I n th e r i g h t - a n g l e d t r i a n g l e a l titu d e , th e re fo re , 0, 0 - 0, 0' =

0 0 ,P

(

0, /> )*

=

, th e segm ent O'P i s th e

/? « .

T hus, th e p o in t O' i s th e in v e r s e o f th e p o in t O w ith r e s ­ p e c t to th e c i r c l e (O,, R) i f th e l a t t e r i s ta k e n a s th e c i r c le of in v e rs io n . The p o in t O i s g i v e n . In c o n s t r u c t i o n , th e p o in t O'was found f i r s t , th e n th e p o in t O, , th e c e n tr e o f th e r e q u ir e d c i r c l e , was fo u n d .

APPLICATION OP METHOD OF INVERSION

39

In Problem s 15-18 i t was shown how to c o n s t r u c t f i g u r e s to f in d th e in v e r s e o f a p o i n t , a s t r a i g h t l i n e and a c i r c l e , u s in g com passes a l o n e . We now exam ine a g e n e r a l method o f s o lv in g g e o m e tr ic a l c o n s tr u c tio n problem s w ith com passes a l o n e . Each c o n s tr u c tio n c a r r i e d o u t by means o f com passes and a r u l e r g iv e s a f i g u r e i n th e p la n e o f th e d raw in g , c o n s i s t ­ in g o f s e p a r a te p o i n t s , s t r a i g h t l i n e s and c i r c l e s . The i n ­ v erse fig u re ' o f th e f i g u r e w ith r e s p e c t to th e c i r c l e (O, r) , w hich i s ta k e n a s th e c i r c l e o f in v e r s i o n , w ith th e c e n tr e O n o t ly in g on any o f th e s t r a i g h t l i n e s and c i r c l e s o f f i g u r e , c o n s i s t s o n ly o f p o in ts and c i r c l e s . U sing Problem s 1 5 -1 8 , we se e t h a t each o f th e s e p o in ts and s t r a i g h t l i n e s can be c o n s tr u c te d by com passes a l o n e . Now, l e t a c e r t a i n c o n s tr u c tio n p ro b lem , s o lu b le by means o f a r u l e r and co m p asses, be r e q u ir e d to be s o lv e d by means o f com passes a l o n e . L et u s im agine t h a t t h i s problem h a s b een so lv e d by means o f com passes and a r u l e r , a s a r e s u l t o f w hich a c e r t a i n f i g ­ u re h a s been o b ta in e d , c o n s i s t i n g o f p o i n t s , s t r a i g h t l i n e s and c i r c l e s . The c o n s tr u c tio n o f t h i s f i g u r e was r e a l i z e d by c a r r y in g o u t a f i n i t e number o f c o n s tr u c tio n s o f s t r a i g h t l i n e s and c i r c l e s i n a d e f i n i t e o r d e r . L et u s ta k e th e m ost s u i t a b l e c i r c l e o f i n v e r s i o n (0, r) and l e t us c o n s tr u c t th e f i g u r e ' , in v e r s e o f th e f i g u r e (Problem s 1 5 - 1 8 ) . The f i g u r e ' w i l l c o n s i s t o f p o in ts and c i r c l e s o n ly , i f , o f c o u r s e , th e c i r c l e o f in v e r s i o n h a s been chosen su c h , t h a t i t s c e n tr e does n o t l i e on any o f th e s tr a ig h t lin e s o r c ir c le s o f fig u re . I f we now c o n s tr u c t th e in v e r s e o f th e f i g u r e w hich i s ta k e n a s th e r e s u l t i n th e f i g u r e ’ , th e n we a r r i v e a t th e r e q u ir e d re s u lt. We sh o u ld n o te h e r e t h a t we sh o u ld c a r r y o u t th e c o n s tr u c ­ ti o n o f f i g u r e ' i n th e o r d e r i n w hich th e c o n s tr u c tio n o f f ig u r e was c a r r i e d o u t by means o f com passes and a r u l e r . By means o f th e method d e s c r ib e d above i t i s p o s s i b le to s o lv e by com passes a lo n e each c o n s tr u c tio n problem s o lu b le by means o f com passes and a r u l e r . The b a s ic M ohr-M ascheroni r e s u l t h a s been proved once a g a in , t h i s tim e w ith th e h e lp o f th e method o f i n v e r s i o n . The f i v e s im p le s t problem s m entioned a t th e end o f S e c tio n

40

GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

1 can also be solved by the general method. We take the solution of Problem J a.s a n illustration of the general method of solving construction problems. We shall construct the point of intersection of two straight lines A B and C D , each of which is given by two points. We take an arbitrary circle (0, r) in the plane, with centre 0 , not on any of the given straight lines, and we regard it as the circle of inversion. We construct the circles which are the inverses of the given straight ^ines and we mark their point of intersection X ' (Problem 16). We construct the point X inverse of the point X ' (Problem 15)• X is the required point of intersection of the given straight lines A B and C D . Here the figure consists of two given straight lines A B and C D (more exactly, it consists of 4 given points A , B , C and D , through which we mentally draw the given lines); the figure »I>' consists of two circles, inverses of the given straight lines A B and C D . The image taken as the result in figure ' will be the point X ' . The point X inverse of the point X ' is the required result, the point of the inter­ section of the given straight lines. Exactly in the same way, it is possible to solve Problem 6 (the fourth of the simplest problems) - to construct the points of intersection of a given straight line and a given circle. The solution of the problem will be considerably simplified at the same time. The truth of these construc­ tions follows immediately from Theorem 1. Problem 1 9 .

To find the centre of a given circle.

C o n s t r u c t i o n . We take a point O on the given circle, and with an arbitrary radius r we describe a circle (0,r) which intersects the given circle at the points A and B . We take the circle (O , r) as the circle of inversion and we construct the centre of the circle which is the inverse of the straight line A B (Problem 16). In order to carry out the latter construction, we draw the circles (A , 0 ) and (£, O) until they meet at the point O, , we describe the cir­ cle (O,, O) and we mark the points D and D, of its intersect­ ion with the circle of inversion. The circles (D, O) and (£>,, O) define the required centre of the original circle (Fig. 5l).

APPLICATION OP METHOD OF INVERSION

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P r o o f . The p o in ts A and B a r e in v e r s e s o f each o t h e r , s in c e th e y l i e on th e c i r c l e o f i n v e r s i o n . T h u s, th e g iv e n c i r c l e and th e s t r a i g h t l i n e A B a r e m u tu a lly in v e r s e f i g u r e s .

In Problem 16 i t was shown t h a t th e p o in t O' was th e r e ­ q u ire d c e n tr e o f th e g iv e n c i r c l e , w h ich , i n t h i s c a s e , i s th e in v e r s e o f th e s t r a i g h t l i n e A B . The a t t e n t i o n o f th e r e a d e r sh o u ld be drawn to th e s im p li­ c i t y and e le g a n c e o f th e s o l u t i o n o f th e l a s t p ro b le m . In o r d e r to f in d th e c e n tr e o f th e c i r c l e , s i x c i r c l e s have been draw n*. T h is c o n s tr u c tio n i s s im p le r and more e x a c t th a n th e more u s u a l c o n s tr u c tio n w ith a r u l e r and co m p asse s. T his p ro b lem , and a ls o c e r t a i n o th e r p roblem s i n th e g eo ­ m etry o f com p asses, a s , f o r exam ple, P roblem s 3 and 0 (S ec­ ond Method) can be g iv e n to o ld e r p u p ils as e x e r c is e s d u rin g geom etry l e s s o n s . F o r t h i s r e a s o n , we g iv e a p r o o f o f th e c o n s tr u c tio n i n Problem 1 9 , w hich i s n o t b a se d on th e p r i n c i ­ p le o f i n v e r s i o n . P r o o f . The s t r a i g h t l i n e 0 0 , i s p e r p e n d ic u la r to th e chord A B o f th e g iv e n c i r c l e and p a s s e s th ro u g h i t s m id p o in t, *0f c o u r s e , on c o n d itio n t h a t th e r a d iu s r i s made g r e a t e r th a n h a l f o f th e r a d iu s o f th e g iv e n c i r c l e . O th e rw ise , th e r e w i l l be a g r e a t e r number o f c i r c l e s (s e e th e s o l u t i o n o f Problem 13 - c a s e 2 ) .

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GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

t h e r e f o r e th e r e q u ir e d c e n tr e m ust l i e on th e s t r a i g h t l i n e 0 0 , . L et E and F be th e p o in ts o f i n t e r s e c t i o n o f th e s t r a i g h t l i n e 0 0 , w ith th e g iv e n c i r c l e and w ith th e c i r c l e (0,, O ). The segm ent OE i s th e d ia m e te r o f th e g iv e n c i r c l e . E xam ining th e r i g h t - a n g l e d t r i a n g l e s OAE and ODE , whose a l t i t u d e s a r e th e segm ents AH and D K , we f in d OA’ = OE-OH and OO' = OF-OK. T aking in t o a c c o u n t t h a t and

OD--=OA = r .

O f — 2 0 0 ,,

0H = ^ O 0

O A = y O o ; » we o b ta in OE ■OH = OF- OK

or

Hence 001 '= — 2 . Problem 2 0 . ARC .

C lrc n m sc rib e a c i r c l e round a g iv e n t r i a n g l e

C o n s t r u c t i o n . We d e s c r ib e th e c i r c l e [A, 8) and ta k e i t a s th e c i r c l e o f i n v e r s i o n . We c o n s tr u c t th e in v e rs e

APPLICATION OP METHOD OF INVERSION

43

p o in t C ' o f th e p o in t C (Problem 1 5 )• We c o n s tr u c t th e c i r ­ c l e (,Y, A ) , w hich i s th e in v e r s e o f th e s t r a i g h t l i n e BC' (Problem 1 6 ) . (X, A) i s th e r e q u ir e d c i r c l e , c irc u m s c rib e d round th e t r i a n g l e A B C . P r o o f . The p o in t B i s i t s own in v e r s e s in c e i t l i e s on th e c i r c l e o f in v e r s i o n (.4, B ) • The p o in t C ' i s th e i n ­ v e rs e o f th e p o in t C . I t fo llo w s t h a t th e c i r c l e p a s s in g th ro u g h th e g iv e n p o in ts A , B and C i s th e in v e r s e o f th e s t r a i g h t l i n e B C ' . And, a s was shown i n P roblem 1 6 , th e p o in t X i s th e c e n tr e o f th e r e q u ir e d c i r c l e * . N o t e . We can now g iv e th e f o llo w in g method o f s o lv in g Problem 1 8 . We ta k e a r b i t r a r y p o in ts A , B and C on th e g iv e n c i r c l e (0,, R ) and we c o n s tr u c t t h e i r i n v e r s e s , A ' , B ' and C . The c i r c l e , c irc u m s c rib e d round th e t r i a n g l e A ' B ’C , i s th e r e q u ir e d in v e r s e o f th e g iv e n c i r c l e .

* In Problem 16 we c o n s tr u c te d th e c e n tr e o f a c i r c l e , w hich i s in v e r s e o f a g iv e n s t r a i g h t l i n e . T h is c o n s tr u c tio n was used i n s o lv in g Problem s 19 and 2 0 .

Part 2 G eom etric co n stru ctio n s by m eans o f com p asses alon e but w ith restriction s

I n P a r t One o f t h i s book we i n v e s t i g a t e d c o n s tr u c tio n s by means o f com passes a lo n e w hich we can now c a l l th e c l a s s i ­ c a l g eom etry o f c o m p asse s. I n th e th e o r y o f g e o m e tric c o n s tr u c tio n s by means o f com­ p a s s e s a l o n e , th e f r e e u s e o f th e com passes i s alw ays u n d e r­ s to o d ; no r e s t r i c t i o n s a r e p u t on th e s i z e o f th e a n g le made by th e l e g s . W ith such com passes i t i s p o s s i b le to draw c i r c l e s w ith r a d i i a s l a r g e o r a s sm a ll a s we p l e a s e . I t i s w e ll known, how ever, t h a t i n p r a c t i c e , w ith a c tu a l co m p asses, i t i s p o s s i b le to d e s c r ib e c i r c l e s , whose r a d i i a r e no l a r g e r th a n a c e r t a i n le n g th Rmal and no s m a lle r th a n a le n g th flml . The le n g th Rmtl c o rre sp o n d s to th e m axim al, and Rmln to th e m inim al o p e n in g o f th e le g s o f th e g iv e n c o m p asse s. I f we d e n o te by r th e r a d iu s o f a c i r c l e , w hich can be d e s c r ib e d w ith th e s e com p asses, th e f o llo w in g in e q u a l­ i t y alw ays h o ld s : Rmln sS r < We s h a l l sa y t h a t i n t h i s c a s e th e o p en in g o f th e le g s o f th e com passes i s r e s t r i c t e d from below by th e le n g th Rmln , and r e s t r i c t e d from above by th e le n g th RmiI . I n P a r t Two we s h a l l exam ine g e o m e tr ic a l c o n s tr u c tio n s by means o f com passes a ] o n e , when c e r t a i n r e s t r i c t i o n s a r e im­ po sed on th e o p e n in g o f th e l e g s .

5.

CONSTRUCTIONS BY MEANS OP COMPASSES ALONE WITH THE OPENING OP THE LEGS RESTRICTED FROM ABOVE

In t h i s c h a p te r we s h a l l u s e com passes th e o p e n in g o f whose le g s i s r e s t r i c t e d , from above o n ly , by th e g iv e n le n g th Rm t . W ith such com passes i t i s p o s s i b l e to d e s­ c r ib e c i r c l e s whose r a d i i do n o t exceed t h i s l e n g t h . For th e sake o f b r e v i t y , we s h a l l h en c e fo rw a rd w r i t e sim p ly R , in s t e a d o f . I f we d e n o te th e r a d iu s o f a c i r c l e , w hich i t i s p o s s i b le to draw w ith th e s e co m p asses, by r , th e n we alw ay s have 0< r^R . Problem 2 1 . To c o n s tr u c t a se g m en t, w hich i s yi t h p a r t o f a g iv e n segm ent AB ( to d iv id e th e g iv e n segm ent AS in t o 2 . A. 8 ...............2" e q u a l p a r t s ) . I t i s n o t h a rd to v e r i f y t h a t i n th e c a s e when AB

*

i t i s p o s s i b le to u s e th e c o n s tr u c tio n c o n ta in e d i n Problem 10; th e r a d iu s o f th e g r e a t e s t c i r c l e i n t h a t c o n s tr u c tio n i s e q u a l to A C = 2‘ AB C D . To v e r i f y th e i n e q u a l i t y A tfsg i/? o r th e i n e q u a l i t y /?3>2A B , th e c i r c l e (A, R) h a s to be draw n; i f th e p o in t B l i e s on th e c irc u m fe re n c e (A. R) o r o u ts i d e o f i t , th e n r < 2 a b ; i f th e p o in t B l i e s i n s i d e th e c i r c l e , th e n A B < R , a n d , t h e r e f o r e , th e segm ent 2 A B can be c o n s tr u c te d (Problem 2 ) , and compared w ith th e segm ent R by th e means in d ic a t e d a b o v e . * * In th e f i r s t method o f c o n s tr u c tio n o f P roblem 1 0 , i t i s n e c e s s a r y to check t h a t A D „ s i R f o r a l l « = 1 , 2 , 3» • • •

47

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GEOMETRICAL CONSTRUCTIONS USING COMPASSES ONLY

C o n s t r u c t i o n i n th e c a se o f AH /? we c a n n o t d e s c r ib e th e c i r c l e (A. BC) w ith th e g iv e n co m passes. We can o b ta in th e r e q u ir e d p o i n t , h o w ev er, i f we c o n s tr u c t on th e c irc u m fe re n c e o f th e c i r c l e (C, AB) th e p o in t D' , d i a m e t r i c a l l y o p p o s ite to O . The f i g u r e ABD'C i s th e r e q u ir e d p a r a lle lo g r a m . Now, l e t /4C > R and R ( F ig . 3 4 ) . We ta k e an a r b i t r a r y row o f p o in ts 4,, 4 ,........ 4^ .in th e d i r e c t i o n from p o in t A

--A 8 A/t* Aj o —

*