geometry and complex numbers

35RESONANCE January 2008

GENERAL ARTICLE

Keywords

Complex numbers, plane geom-

etry, Ptolemy’s theorem, Euler-

line, nine-point circle.

Anant R Shastri is a

Professor at IIT, Bombay.

His research inerests are in

algebraic topology and

algebraic geometry. He is

also keen in math

education and music.

Complex Numbers and Plane Geometry*

Anant R Shastri

* This article is based on a talk

given to an audience consisting

mainly students of class IX and

X, at Nehru Science Centre un-

der the aegis of Bombay Asso-

ciation for Science Education

and Bombay Math. Colloq. on

25th Jan. 2003. An earlier ver-

sion of this article was published

in Bona Mathematica, Vol.14

Nos.1–2, 2003.

T h e r e p r e s e n t a t io n o f c o m p le x n u m b e r s a s p o in t s

o f t h e E u c lid e a n p la n e n a t u r a lly le a d s t o a tw o -

w a y in t e r a c t io n b e tw e e n g e o m e t r y a n d n u m b e r s .

T h e g e o m e t r y o f t h e p la n e h a s a v e r y d e e p in -

° u e n c e in t h e s t u d y c o m p le x a n a ly t ic fu n c t io n s .

I n t h is a r t ic le , w e illu s t r a t e t h e o t h e r w a y a s p e c t

b y a fe w s im p le -m in d e d a p p lic a t io n o f c o m p le x

n u m b e r s t o g iv e e le g a n t s o lu t io n s o f p r o b le m s

in p la n e g e o m e t r y , s u c h a s P t o le m y 's T h e o r e m ,

E u le r -lin e a n d N in e -p o in t C ir c le T h e o r e m .

1 . I n t r o d u c t io n

W e b e g in w ith a sto ry ta k e n fro m th e fa m o u s b o o kO n e T w o T h ree...In ¯ n in ity b y G G a m o w : T h e re w a s ay o u n g a n d a d v en tu ro u s m a n w h o fo u n d a m o n g h is g rea t-

g ra n d fa th e r's p a p e rs a p ie ce o f p a rch m e n t th a t rev ea le dth e lo c a tio n o f a h id d e n tre a su re . T h e in stru c tio n s re a d :

\ S a il to ... N o rth la titu d e a n d ... W est lo n gitu d e w h ereth o u w ilt ¯ n d a d eserted isla n d . T h ere lieth a la rgem ea d o w , n o t pen t, o n th e n o rth sh o re o f th e isla n d w h eresta n d eth a lo n ely oa k a n d a lo n ely p in e. T h ere th o u w iltsee a lso a n o ld ga llo w s o n w h ich w e o n ce w ere w o n t toh a n g tra ito rs. S ta rt th o u fro m th e ga llo w s a n d w a lk toth e oa k co u n tin g th y step s. A t th e oa k th o u m u st tu rnrig h t by a righ t a n gle a n d ta ke th e sa m e n u m ber o f step s.P u t h ere a sp ike in th e gro u n d . N o w m u st th o u retu rn toth e ga llo w s a n d w a lk to th e p in e co u n tin g th y step s. A tth e p in e th o u m u st tu rn le ft by a righ t a n gle a n d see th a tth o u ta kest th e sa m e n u m ber o f step s, a n d p u t a n o th ersp ike in to th e gro u n d . D ig h a lf-w a y betw een th e sp ikes;th e trea su re is th ere."

T h e sto ry is th a t w h e n th e y o u n g m a n ¯ n a lly la n d e d o n

36 RESONANCE January 2008

GENERAL ARTICLE

th e isla n d , e v e n th o u g h h e c o u ld ¯ n d th e m e a d o w a n d

th e tw o tre e s a s d escrib e d , th e o ld g a llo w s h a d to ta llyd isa p p ea re d ! T h e sto ry en d s w ith th e a d v e n tu ro u s m a nre tu rn in g e m p ty -h a n d e d a fter so m e d e sp era te d ig g in g a tra n d o m .

N o w , th e p o in t G a m o w w a n ts to m a k e is th a t if th ey o u n g m a n k n e w a b it o f m a th em a tic s, p a rtic u la rly th eu se o f c o m p lex n u m b e rs, h e co u ld h a v e fo u n d th e trea -su re . W e o n ly a d d th a t, ev en if th e m a n h a d n o t d e -

sp a ire d a n d ju st m a d e a sin g le a tte m p t to g u ess th elo c a tio n o f th e g a llo w s a n d fro m th e re o n w a rd , ca rrie do u t th e in stru c tio n s g iv e n in th e p a rch m en t, h e w o u ldh a v e g o t th e trea su re a s w ell a s th e sa tisfa ctio n (p erh a p sfa lse ly ) o f g u essin g th e p o sitio n o f th e g a llo w s c o rre ctly .

T ry to ¯ g u re it o u t y o u rse lf b efo re re a d in g th e so lu tio nso th a t y o u w ill h a v e th e sa tisfa c tio n o f ¯ n d in g th e trea -su re .

I w o u ld lik e to em p h a sis th e fa c t th a t, it is n o t v e ry d if-

¯ c u lt to so lv e th is p ro b le m th ro u g h e lem e n ta ry sch o o lg e o m e try, e ith e r. H o w e v e r, in th is a rtic le , w e sh a ll see av e ry n e a t so lu tio n to th is p ro b le m , u sin g c o m p le x n u m -b e rs. T h e id ea b e h in d th is c a n a lso b e u se d to so lv em a n y p la n e g e o m etry p ro b lem s a s w e ll. A s a n illu stra -

tio n , w e sh a ll p ro v e P to lem y 's T h e o re m a n d N in e-P o in tC ircle T h e o re m .

A t th is p o in t, h e re is a q u estio n w h ich d o es n o t n e e d a n y

k n o w led g e o f c o m p le x n u m b e rs n o r a n y m a th e m a tics.T ry to a n sw e r it to y o u rself:

W h ich sid e o f th e lin e fro m p in e tree to oa k tree lies th etrea su re?

W e a ssu m e th a t th e rea d er h a s so m e fa m ilia rity w ithb a sic o p e ra tio n s o f a d d itio n a n d m u ltip lic a tio n o f c o m -p lex n u m b e rs. H o w ev er w e sh a ll b rie ° y re ca ll th em in

S e ctio n s 2 a n d 3 a n d th e n p resen t a so lu tio n o f G a m o w 'sp ro b le m in S e ctio n 4 . In S e c tio n 5 , w e sh a ll re c a ll so m e


GENERAL ARTICLE

m o re p ro p erties o f c o m p lex n u m b ers th o u g h n o t e v e ry -

th in g re c a lle d is p u t to u se im m ed ia te ly . T h e la st th reese c tio n s a re d e v o ted to d isc u ssio n o f p la n e g e o m etryp ro b le m s u sin g c o m p le x n u m b ers. T h e ch o ic e o f th e sep ro b le m s is p u re ly a m a tte r o f ta ste .

2 . B a s ic s o f C o m p le x N u m b e r s

A c o m p le x n u m b e r is a n e x p re ssio n o f th e fo rm x + {y

w h e re x a n d y a re re a l n u m b e rs. It is im p o rta n t tou n d e rsta n d th a t tw o co m p le x n u m b e rs

z 1 = x 1 + {y 1 ; z 2 = x 2 + {y 2

a re eq u a l if a n d o n ly if x 1 = x 2 a n d y 1 = y 2 : F o r a c o m -

p lex n u m b er z = x + {y ; x is c a lled th e rea l pa rt o f z

a n d y is c a lled th e im a gin a ry pa rt o f z : A re a l n u m b e r r

is id e n ti e d w ith a c o m p lex n u m b e r w h o se re a l p a rt is r

a n d th e im a g in a ry p a rt is 0 : T h e a d d itio n a n d m u ltip li-c a tio n o f a n y tw o c o m p le x n u m b ers is d e ¯ n e d b y :

z 1 + z 2 := (x 1 + x 2 ) + {(y 1 + y 2 );

z 1 z 2 := (x 1 x 2 ¡ y 1 y 2 ) + {(x 1 y 2 ¡ y 1 x 2 ):

A ll th e sta n d a rd la w s o f a rith m etic w h ich h o ld fo r re a l

n u m b ers h o ld fo r c o m p lex n u m b ers a s w e ll. N o te th a tth e re a re tw o v e ry sp e cia l co m p le x n u m b ers v iz., § { su chth a t

(§ {)2 = ¡ 1 :

T h e sy m b o l { is th e G re ek le tter io ta ' a n d is p ro n o u n ce dey e'.

S in ce th e re a l a n d im a g in a ry p a rts o f a c o m p lex n u m b e rd e term in e th e c o m p lex n u m b e r u n iq u e ly , it fo llo w s th a tth e set o f a ll c o m p lex n u m b e rs z = x + {y is in a o n e-to -o n e c o rre sp o n d e n ce w ith th e o rd e red p a irs (x ; y ) o f re a ln u m b ers x ; y : T o e a ch c o m p le x n u m b e r, z = x + {y ; w e

c a n a ssig n a p o in t v iz ., (x ; y ) in th e 2 -d im e n sio n a l co -o rd in a te p la n e . O f co u rse , w e sh o u ld ch o o se th e o rig in


GENERAL ARTICLE

Figure2. Polarcoordinates.

Figure 1. Argand diagram.

a n d th e tw o p e rp e n d icu la r a x e s b efo reh a n d fo r th is to

m a k e sen se . It sh o u ld a lso b e n o ted th a t it is a c o m -m o n ly a c ce p te d p ra ctic e to ch o o se th e tw o a x e s in su cha w a y th a t th e d ire ctio n o f m o v in g fro m th e p o sitiv e x -a x is to th e p o sitiv e y -a x is is c o u n te rc lo ck w ise. O n ce th ea x e s h a v e b e e n ch o se n , ea ch p o in t o f th e p la n e d e ¯ n e s a

u n iq u e c o m p lex n u m b e r. A t th is sta g e w e re m a rk th a tth e fre ed o m in ch o o sin g th e o rig in a n d th e a x e s c o m e sa s a b ig b o n u s, a s illu stra te d b y v a rio u s e x a m p le s in th isa rtic le .

T h e g ra p h ic a l rep re se n ta tio n o f c o m p le x n u m b e rs is co m -m o n ly k n o w n a s th e A rg a n d d ia g ra m , (F igu re 1 ), a fte rth e p u b lish e d w o rk (1 8 0 6 ) o f J e a n R o b ert A rg a n d o fG e n e v a , e v e n th o u g h G a u ss h a d u se d th is id e a in h is

th e sis a b o u t eig h t y ea rs b e fo re .

T h e p o la r c o o rd in a te re p re se n ta tio n o f p o in ts o n th ep la n e c a n a lso b e e x p lo ite d to rep rese n t c o m p lex n u m -b e rs. A c c o rd in g ly, e a ch n o n -z ero c o m p lex n u m b e r z w ill

h a v e a u n iq u e e x p ressio n (r c o s µ ; r sin µ ), w h ere r is ap o sitiv e re a l n u m b e r eq u a l to th e d ista n c e o f z fro m 0 ,a n d 0 · µ < 2 ¼ is th e a n g le su b ten d e d b y th e lin e seg -m en t [0 ; z ] w ith th e p o sitiv e re a l a x is m e a su red c o u n te r-c lo ck w ise a n d in ra d ia n s. H e re r re p resen ts th e d ista n ce

o f z fro m th e o rig in . T h e a n g le µ is ca lled th e a rg u m en to f th e n u m b e r z a n d d e n o te d b y a rg z (F igu re 2 ). T h e


GENERAL ARTICLE

Figure 3. Addition of com-

plex numbers.

n u m b er co s µ + { sin µ is a lso d e n o te d b y e {µ : In th is n o ta -

tio n , w e h a v e z = r e {µ , w h e re jz j = r a n d a r g z = µ : F o rth e co m p le x n u m b er 0 itse lf, w e h a v e r = 0 b u t a rg 0 isn o t d e ¯ n e d . A lso th e set o f p o in ts z w ith jz j = 1 is th eu n it circle (F igu re 2 ).

3 . G e o m e t r ic W a y o f A d d in g a n d M u lt ip ly in g

T h e in tera ctio n b etw ee n th e g e o m e try o f th e p la n e a n d

th e a rith m e tic o f c o m p lex n u m b e rs b e g in s a s so o n a sw e h a v e re p resen te d th e co m p le x n u m b ers b y p o in ts inth e p la n e . L e t u s see h o w to in te rp ret a d d itio n a n dm u ltip lica tio n g e o m e tric a lly .

S ta rt w ith tw o c o m p lex n u m b ers z 1 ; z 2 w h ich w e w a n tto a d d . If o n e o f th e m is 0 th e re is n o t m u ch to d o . S im -ila rly , if o n e is a re a l m u ltip le o f th e o th e r, sa y z 2 = r z 1

th e n z 1 + z 2 = (1 + r )z 1 a n d th e re fo re , it esse n tia llya m o u n ts to a d d in g tw o re a l n u m b ers. S o , let u s co n -

sid e r th e m o st im p o rta n t ca se w h e n z 1 ; z 2 a re n o t a re a lm u ltip le o f e a ch o th er. If P 1 , P 2 a re p o in ts re p re se n te db y z 1 ; z 2 th e n th is is th e sa m e a s sa y in g th a t O , P 1 , P 2

a re n o n -c o llin e a r re sp ec tiv ely .

W e th en co n stru c t th e p a ra lle lo g ra m w ith O , P 1 , P 2 a sth re e o f its v ertice s a n d O P 1 , O P 2 a s a d ja c e n t sid es.T h e fo u rth v e rtex Q w ill th e n re p re se n t th e su m z 1 + z 2 :

T o p ro v e th is, y o u m a y d ro p p e rp e n d icu la rs o n to th ex a n d y a x e s fro m th e p o in ts P 1 , P 2 a n d Q a n d a rg u ew ith so m e a p p ro p ria te sim ila r tria n g le s (F igu re 3 ).

T h u s w e se e th a t in a d d in g th e n u m b e r z 1 to th e n u m b e rz 2 ; w e m o v e th e lin e se g m e n t [0 ; z 1 ] p a ra llel to itse lf, till

th e p o in t c o rre sp o n d in g to 0 c o in c id e s w ith th e p o in tz 2 a n d th en ju st lo o k a t w h ere th e p o in t z 1 itse lf h a sla n d e d . T h u s w e c a n c a ll th e o p e ra tio n o f a d d in g z 1 a spa ra llel tra n sla tio n b y z 1 :

In p a rticu la r, if a lin e se g m en t [p ; q ] is p a ra llelly tra n s-la te d so th a t th e p o in t p fa lls o n th e o rig in , th e n th e


GENERAL ARTICLE

Figure 4. Multiplying com-

plex numbers.

p o in t q w ill fa ll o n th e p o in t rep re se n tin g th e n u m b e rq ¡ p : T h is o b serv a tio n is th e b e g in n in g o f th e so -c a lle d

v e cto r-m e th o d ' in p la n e g e o m e try. T h is is n o th in g v erysp ec ia l a b o u t c o m p lex n u m b e rs. T h e tru e stre n g th , a sw e sh a ll se e, c o m es fro m th e g eo m e tric in te rp re ta tio n o fm u ltip lica tio n .

S o , le t u s w o rk o u t h o w to c o n stru ct th e p o in t rep rese n t-in g th e p ro d u c t z w o f tw o co m p le x n u m b e rs z ; w : O n cea g a in , w e c o n sid e r o n ly th e im p o rta n t ca se w h en 0 ; z ; w

a re n o n -c o llin e a r, a n d z ; w a re b o th n o n -rea l. C o n sid e rth e tria n g le w ith v e rtic e s 0 ; 1 ; z : W e c o n stru c t a n o th e rtria n g le ¢ (0 ; w ; q ) w h ich is sim ila r to ¢ (0 ; 1 ; z ) so th a tth e sid e [0 ; 1 ] co rresp o n d s to th e sid e [0 ; w ] a n d th e tw otria n g le s a re b o th la b e led in th e c o u n te rclo ck w ise se n se .

U sin g sim ila r tria n g le s, it is n o t d i± c u lt to v e rify th a tth e p o in t q n o w re p re sen ts z w ; in a sim ila r w a y a s inth e c a se o f a d d itio n . W e le a v e th is a s a n e x e rc ise to th ere a d e r. H o w ev er, if w e u se p o la r c o o rd in a te s a n d a littleb it o f trig o n o m e try, th e n th e v eri c a tio n th a t q = z w

c a n b e ca rrie d o u t a s fo llo w s (F igu re 4 ): L e t

z = (r 1 co s µ 1 ; r 1 sin µ 1 );

w = (r 2 c o s µ 2 ; r 2 sin µ 2 ):

W e ro ta te th e tria n g le (0 ; 1 ; z ) th ro u g h a n a n g le µ 2 too b ta in th e tria n g le ¢ (0 ; w 0; z w 0), sa y , w h e re w 0= e {µ 2 : It

fo llo w s th a t w 0is a p o in t o n th e lin e jo in in g 0 a n d w a n dsin c e th is tria n g le is sim ila r to ¢ (0 ; w ; q ); it a lso fo llo w sth a t z w 0 w ill lie o n th e lin e jo in in g 0 a n d q : F u rth er, b ysim ila rity o f tria n g les, it fo llo w s th a t th e le n g th o f [0 ; q ]is e q u a l to r 2 jz w j = r 1 r 2 : T h e refo re

q = r 1 r 2 (co s(µ 1 + µ 2 ); r 1 r 2 sin (µ 1 + µ 2 )):

N o w u se th e fo rm u la e

c o s(µ 1 + µ 2 ) = c o s µ 1 co s µ 2 ¡ sin µ 1 sin µ 2 ;

sin (µ 1 + µ 2 ) = sin µ 1 c o s µ 2 + c o s µ 1 sin µ 2 ;


GENERAL ARTICLE

Figure 5. Multiplication by i

and –i.

to co n clu d e th a t q = z w :

T h e a d v a n ta g e o f th is p ro o f is th a t it g iv e s u s a c o m -p lete g eo m e tric m e a n in g o f m u ltip lica tio n b y a c o m p lexn u m b er w = (r 2 c o s µ 2 ; r 2 sin µ 2 ); v iz ., w e ro ta te th e lin e

[0 ; z 1 ] th ro u g h a n a n g le µ 2 a n d th e n ex p a n d / c o n tra c t itb y a fa c to r r 2 : In p a rtic u la r, if w is a p o sitiv e re a l n u m -b e r th e n th e re is n o ro ta tio n to b e p erfo rm e d . If it isa n e g a tiv e rea l n u m b e r, th e n w e h a v e to ro ta te b y a na n g le ¼ w h ich c o rre sp o n d s to ch a n g in g th e sig n . If th e

n u m b er w is a t a u n it d ista n c e fro m 0 th en th e re is n osc a lin g fa c to r.

In p a rtic u la r, w e se e th a t m u ltip lic a tio n b y { h a s th ee ® e ct o f ro ta tin g th e v e c to r th ro u g h 9 0 ± in th e c o u n te r-

c lo ck w ise se n se (F igu re 5 ). C lea rly m u ltip lic a tio n b y ¡ {

a m o u n ts to ro ta tio n o f th e v e c to r in th e o th er d irec tio nth ro u g h 9 0 ±:

4 . S o lu t io n t o G a m o w 's T r e a s u r e

L e t u s rep re se n t th e m a p o f th e isla n d b y c o m p lex n u m -b e rs. O f co u rse , w e a re fre e to ch o o se o u r a x e s a n d w h a t

is b e tter th a n to ch o o se th e lin e jo in in g th e tw o tre es a sth e rea l a x is. N o w , c le a rly, h a lf-w a y b etw ee n th e tw otre es sh o u ld b e a g o o d ch o ic e fo r th e o rig in . T h e n it re -a lly sh o u ld n o t m a tte r w h e th er th e p o sitio n o f th e o a ko r th a t o f p in e is ch o se n a s th e n u m b e r 1 sa y , th e p in e .

T h e n n a tu ra lly th e p o sitio n o f th e o a k w ill refe r to ¡ 1 :

N o w let ¡ d e n o te th e p o sitio n o f th e g a llo w s, w h ich isn o t k n o w n . T h e p o in t is th a t it d oes n o t m a tter: ca rryo u t th e rest o f th e in stru ctio n s a n d y o u a rrive a t a n a n -sw e r in d epen d en t o f th is u n kn o w n qu a n tity. W e fee l th a ty o u sh o u ld still try th is p ro b lem o n y o u r o w n . A t th is

sta g e w e sh a ll g iv e y o u a h in t { u se th e fa ct th a t m u l-tip lica tio n b y { c o rre sp o n d s to tu rn in g a v e cto r th ro u g ha rig h t a n g le in th e a n ti-c lo ck w ise d ire c tio n . R ea d B o x1 fo r th e co m p le te so lu tio n , o n ly a fter y o u h a v e trie de n o u g h to g e t th e a n sw e r o n y o u r o w n .


GENERAL ARTICLE

B o x 1 .

The position S 1 of the ¯rst spike is found as follows: The vector representing the distanceand the direction from the gallows to the oak is ¡ 1 ¡ ¡: Therefore, the vector representingthe direction and the distance from the oak to the ¯rst spike is got by multiplying by ¡ {;

viz. , {(1 + ¡) : Since this vector has t o originate at the oak, we see t hat S 1 = {(1 + ¡) ¡ 1 :

Likewise, t he p osition of the second spike is given by S 2 = {(1 ¡ ¡) + 1 : The midpoint ofthe line segment [S 1 , S 2 ] is then { where the treasure is!

S c h o o l G e o m e t r y S o lu t io n t o G a m o w 's P r o b le m : Draw perpendiculars S 1 Q 1 andS 2 Q 2 to t he line j oining O(ak) and P(ine) . Using similar triangles show that the lengthof OP is equal to the sum of the lengths of these two p erpendiculars. If TM is theperp endicular to OP from t he midpoint T of S 1 S 2 then M bisects OP and we have TM= OM = MP. (In particular, the position of T(reasure) is determined independently ofthe position of t he gallows.)

5 . M o r e A b o u t C o m p le x N u m b e r s

N o w th a t w e h a v e e n o u g h m o tiv a tio n to stu d y c o m p lexn u m b ers, le t u s q u ick ly g o th ro u g h a fe w m o re b a sicp ro p e rtie s o f th e m .

F o r z = x + {y ; w e sh a ll u se th e n o ta tio n :

< (z ) = x ; = (z ) = y ;

to d en o te th e rea l a n d im a g in a ry p a rt o f z re sp e c tiv ely .T h e co m p lex co n ju ga te a n d th e m od u lu s o f z a re re sp e c -tiv e ly d e ¯ n e d b y

¹z = x ¡ {y ; jz j := (x 2 + y 2 )1 = 2 :

S o m e o f th e im p o rta n t p ro p e rties o f th ese o p e ra tio n s a reliste d b elo w :

1 . j¹z j = jz j; z ¹z = jz j2 ; ( ¹z ) = z :

2 . jz 1 z 2 j = jz 1 jjz 2 j:

3 . j< (z ) · jz j; j= (z )j · jz j:

4 . C o sin e R u le: jz 1 + z 2 j2 = jz 21 j + jz 2 j2 + 2 < (z 1 ¹z 2 ):


GENERAL ARTICLE

5 . P a ra lle lo g ra m L a w : jz 1 + z 2 j2 + jz 1 ¡ z 2 j2 = 2 (jz 1 j2 +jz 2 j2 ):

6 . T ria n g le In e q u a lity : jz 1 + z 2 j · jz 1 j + jz 2 j

a n d eq u a lity h o ld s if a n d o n ly if o n e o f th e z j is a n o n -n e g a tiv e re a l m u ltip le o f th e o th e r.

7 . C a u ch y 's In e q u a lity :¯¯¯¯¯

nX

j = 1

z j w j j

¯¯¯¯¯

2

·

ÃnX

j = 1

jz j j2

! ÃnX

j = 1

jw j j2

!

:

P ro p e rtie s 1 ,2 ,3 , a re e a sily v e ri e d . P ro p e rty 4 is v eri-

¯ e d d irec tly b y e x p a n d in g

jz 1 + z 2 j2 = (z 1 + z 2 )(z 1 + z 2 ):

P ro p e rtie s 5 a n d 6 a re th e n ea sy c o n seq u e n c es o f 4 . T h ela st o n e is th e o n ly n o n -triv ia l o n e w h ich c a n a lso b ed e d u c ed fro m 4 in d u c tiv e ly .

6 . A F e w S im p le G e o m e t r ic P r o b le m s

N o w w e sh a ll c o n sid e r a fe w g eo m etric p ro b lem s a n dse e k a n sw e rs to th e m u sin g c o m p lex n u m b e rs, in th efo rm o f q u estio n a n d a n sw e r. T h e y a re a rra n g e d in su ch

a w a y th a t th e a n sw er to o n e q u e stio n w ill h elp to a n sw e rsu b seq u e n t o n e s. Y o u a re e n c o u ra g e d to try e a ch o fth e m o n y o u r o w n b efo re re a d in g th e a n sw e rs.

Q u e s t io n 1 : W h e n a re tw o lin e se g m en ts in th e p la n ep a ra lle l to e a ch o th e r?

Q u e s t io n 2 : W h e n a re th re e p o in ts A , B , C in th e

p la n e c o llin e a r?

Q u e s t io n 3 : If tw o c o m p le x n u m b e rs z 1 a n d z 2 re p -re se n t tw o p o in ts P 1 a n d P 2 re sp e c tiv ely in th e p la n e ,h o w d o y o u rep re se n t o th e r p o in ts o n th e lin e p a ssin g

th ro u g h P 1 a n d P 2 ? W h a t is th e re p resen ta tio n fo r th em id p o in t o f th e se g m e n t [P 1 , P 2 ]?


GENERAL ARTICLE

Q u e s t io n 4 : S h o w th a t th e lin e se g m e n t jo in in g th e

m id p o in ts o f tw o sid es o f a tria n g le is p a ra llel to th eth ird sid a n d is h a lf its siz e.

Q u e s t io n 5 : S h o w th a t th e m id p o in ts o f th e sid e s o f

a tria n g le fo rm a n o th er tria n g le w h ich is sim ila r to th eg iv e n o n e.

Q u e s t io n 6 : S h o w th a t th e m id p o in ts o f sid es o f a n y

q u a d rila tera l fo rm a p a ra lle lo g ra m .

Q u e s t io n 7 : G iv e n fo u r p o in ts A ,B ,C ,D in th e p la n esu ch th a t A B ? B C a n d A B ? A D , sh o w th a t th e m id -p o in t o f [C ,D ] is e q u id ista n t fro m A a n d B .

Q u e s t io n 8 : S h o w th a t in a rh o m b u s th e d ia g o n a ls a rep e rp en d ic u la r to e a ch o th er.

Q u e s t io n 9 : G iv en a n y q u a d rila te ra l, e rec t sq u a re s e x -te rn a lly o n e a ch o f th e sid e s. S h o w th a t th e c e n te rs o fth e se sq u a res fo rm a q u a d rila te ra l w h o se d ia g o n a ls a reo f eq u a l len g th a n d p erp en d ic u la r to ea ch o th er.

Q u e s t io n 1 0 : G iv e n fo u r d istin c t p o in t in C ; w e d e¯ n eth e ir cro ss-ra tio to b e th e n u m b e r g iv e n b y

[z 1 : z 2 : z 3 : z 4 ] :=

µz 1 ¡ z 3

z 1 ¡ z 4

¶

/

µz 2 ¡ z 3

z 2 ¡ z 4

¶

:

S h o w th a t th e fo u r p o in ts lie o n a circ le o r a stra ig h tlin e if a n d o n ly if th e ir c ro ss ra tio is a re a l n u m b e r.

A s a n a p p lic a tio n o f so m e o f th ese sim p le o b se rv a tio n s,w e sh a ll g iv e a refre sh in g p ro o f o f P to lem y 's th e o re m .

P t o le m y 's T h e o r e m : If A , B , C , D a re v e rtic es o f ac y c lic q u a d rila te ra l th en

jA C jjB D j = jA B jjC D j + jA D jjB C j.

P r o o f: If z 1 ; z 2 ; z 3 ; z 4 a re th e co m p le x n u m b e rs re p -re se n tin g A , B , C , D , listed c y c lic a lly , th e n it fo llo w s


GENERAL ARTICLE

th a t [z 1 : z 3 : z 2 : z 4 ] =

µz 1 ¡ z 2

z 1 ¡ z 4

¶

/

µz 3 ¡ z 2

z 3 ¡ z 4

¶

is

a n e g a tiv e rea l n u m b er. T h a t is (z 1 ¡ z 2 )(z 3 ¡ z 4 ) =r (z 1 ¡ z 4 )(z 3 ¡ z 2 ); w h ere r < 0 : T h ere fo re

j(z 1 ¡ z 2 )(z 3 ¡ z 4 )j + j(z 1 ¡ z 4 )(z 3 ¡ z 2 )j= j(z 1 ¡ z 2 )(z 3 ¡ z 4 ) ¡ (z 1 ¡ z 4 )(z 3 ¡ z 2 )j= j(z 1 ¡ z 3 )(z 2 ¡ z 4 )j

w h ich y ield s P to lem y 's th eo rem . Ä

N o w c o n sid e r th e fo llo w in g p ro b le m w h ich c a n b e so lv e din se v e ra l w a y s. W e ¯ n d th e so lu tio n u sin g P to lem y 'sth e o re m m o st e le g a n t.

E x a m p le (F igu re 6 )

S u p p o se fo r n ¸ 4 ; A 1 ,A 2 ; : : : ; A n b e th e v ertic es o f are g u la r n -g o n in th a t o rd er su ch th a t

1

jA 1 A 2 j=

1

jA 1 A 3 j+

1

jA 1 A 4 j:

W h a t is th e v a lu e o f n ?

S o lu t io n : P u t tj = jA 1 A j j; j = 2 ; 3 ; 4 ; 5 : B ec a u se A 1 A 2 ¢ ¢ ¢A n is a re g u la r p o ly g o n .

W e h a v e

t2 = jA 2 A 3 j = jA 3 A 4 j; t3 = jA 2 A 4 j = jA 3 A 5 j; t4 = jA 2 A 5 j;

e tc.. A p p ly P to lem y 's th eo rem to th e cy c lic q u a d rila t-e ra l A 1 A 2 A 3 A 5 to g e t

t2 t3 + t2 t5 = t3 t4 :

A lso th e g iv e n c o n d itio n is e q u iv a le n t to

t3 t4 = t2 t4 + t2 t3 :

T h e re fo re t2 t5 = t2 t4 w h ich m e a n s t4 = t5 : T h is is p o s-sib le if a n d o n ly if n = 7 :

Figure 6. Using Ptolemy’s

theorem.


GENERAL ARTICLE

F in a lly h e re a re th e a n sw ers to th e q u e stio n s th a t w e

ra ise d e a rlie r.

A n s w e r 1 . T h e se g m e n ts A B a n d C D a re p a ra lle l if a n do n ly if th e tw o n u m b ers A ¡ B a n d C ¡ D a re re a l m u lti-

p les o f ea ch o th er. T h is is th e sa m e a s sa y in g th a t th e

se g m e n ts A B a n d C D a re p a ra lle l if a n d o n ly ifA ¡ B

C ¡ Dis a rea l n u m b e r.

A n s w e r 2 . T a k e A = D in th e a b o v e q u estio n . W e see

th a t th e p o in ts a re c o llin ea r if a n d o n ly ifA ¡ B

A ¡ Cis a

re a l n u m b e r.

A n s w e r 3 . A n y p o in t o n th e lin e is re p re se n te d b y(1 ¡ t)z 1 + tz 2 , w h e re t ra n g e s o v er rea l n u m b e rs. F o r0 · t · 1 ; w e g e t a ll p o in ts in sid e th e lin e se g m en t

[P 1 ; P 2 ]:

T h e m id p o in t c o rre sp o n d s to th e v a lu e t = 1 = 2 ; i.e .,z 1 + z 2

2:

A n s w e r 4 . R e p resen t th e v ertic es o f th e tria n g le b y

th re e c o m p lex n u m b ers z 1 ; z 2 ; z 3 : S a y th e se g m e n t [z 2 ; z 3 ]c o rre sp o n d s to th e b a se . T h e n th e m id p o in ts o f th e tw o

leg s a re g iv e n b yz 1 + z 2

2;

z 1 + z 3

2: T h e refo re, th e le n g th

o f th e lin e se g m en t jo in in g th em is e q u a l to¯¯¯¯

z 1 + z 2

2¡

z 1 + z 3

2

¯¯¯=

jz 2 ¡ z 3 j

2:

T h a t th is seg m e n t is p a ra lle l to th e b a se fo llo w s fro mth e p re v io u s e x e rc ise.

A n s w e r 5 . F o llo w s fro m 4 .

A n s w e r 6 . E a sy.

A n s w e r 7 . H e re w e fo llo w a m e th o d sim ila r to th e

o n e fo llo w e d in tre a su re h u n t. C h o o se th e m id p o in t o fA B a s th e o rig in , A = ¡ 1 a n d B = 1 : W ith re sp e c t to


GENERAL ARTICLE

Figure 7. Diagonals of a

rhombus are perpendicu-

lar.

Figure 8. External squares

on a quadilateral.

th is c o o rd in a te ch o ic e , it fo llo w s th a t C = 1 + {y 1 ; D =¡ 1 + {y 2 : T h e refo re th e m id p o in t o f [C ,D ] is g iv e n b y{(y 1 + y 2 )= 2 w h ich is cle a rly e q u id ista n t fro m A = ¡ 1a n d B = 1 :

A n s w e r 8 . (F igu re 7 ) H e re jz j = jw j: T h ere fo re

(z + w ){(z ¡ w ) = ¡ {z ¹z + {w ¹w ¡ {(¡ z ¹w + w ¹z ) = 2 = (w ¹z ),w h ich is a re a l n u m b er.

T h e re fo re {(z ¡ w ) is p a ra lle l to z + w w h ich is th e sa m ea s sa y in g th a t z ¡ w is p erp en d ic u la r to z + w :

A n s w e r 9 . (F igu re 8 ) L et w j ; j = 1 ; 2 ; 3 ; 4 ; b e th e re -sp ec tiv e c en tre s o f th e sq u a re s. It su ± c e s to sh o w th a tw 1 ¡ w 4 = § {(w 2 ¡ w 3 ): T h e m id p o in t o f th e se g m en t[z 1 ; z 2 ] is g iv e n b y (z 1 + z 2 )= 2 : T h e lin e p e rp e n d ic u la r toth is a n d o f h a lf its le n g th is [0 ; {(z 2 ¡ z 1 )= 2 ]: T h e re fo re , it

fo llo w s th a t 2 w 1 = (z 1 + z 2 )+ (z 2 ¡ z 1 ){: L ik e w ise w e h a v e ,2 w 2 = (z 2 + z 3 ) + (z 3 ¡ z 2 ){; 2 w 3 = (z 3 + z 4 ) + (z 4 ¡ z 3 ){;a n d 2 w 4 = (z 4 + z 1 ) + (z 1 ¡ z 4 ){:

O b se rv e th a t (1 + {){ = { ¡ 1 a n d (1 ¡ {){ = 1 + {:

T h e re fo re , 2 (w 3 ¡ w 1 ) = (z 3 ¡ z 1 )(1 ¡ {) + (z 4 ¡ z 2 )(1 + {);a n d 2 (w 4 ¡ w 2 ) = (z 4 ¡ z 2 )(1 ¡ {) + (z 1 ¡ z 3 )(1 + {): T h u s,w e se e th a t (w 3 ¡ w 1 ) = (w 4 ¡ w 2 ){ a s req u ired .

A n s w e r 1 0 : O b se rv e th a t th e c ro ss-ra tio o f fo u r p o in tsd e p e n d s u p o n th e o rd e r in w h ich w e ta k e th e m . H o w -e v e r, ch e ck th a t if o n e o f th e m is rea l th e n a ll a re re a l

to o . If a n y th ree o f th em a re c o llin ea r, sa y z 2 ; z 3 ; z 4 th e n

w e k n o w th a tz 2 ¡ z 3

z 2 ¡ z 4

is a re a l n u m b e r. B u t th en z 1 lie s

o n th e sa m e lin e if a n d o n ly ifz 1 ¡ z 3

z 1 ¡ z 4is rea l, i.e., if a n d

o n ly if [z 1 : z 2 : z 3 : z 4 ] is re a l.

C o n sid er th e ca se w h en n o th re e a re c o llin e a r. L e t C b e

th e c ircle th ro u g h th re e o f th e m sa y z 2 ; z 3 ; z 4 : T h e n w ek n o w th a t a p o in t z in th e p la n e lie s o n C if a n d o n lyif th e tw o a n g le s \ z 3 z z 4 a n d \ z 3 z 2 z 4 a re e q u a l o r a d d


GENERAL ARTICLE

Figure 9. The nine-point

circle and the Euler line.

u p to ¼ : T h is is th e sa m e a s sa y in g th a t th e cro ss ra tio

[z : z 2 : z 3 : z 4 ] is re a l. It w ill b e p o sitiv e o r n e g a tiv ea c c o rd in g a s z ; z 2 a re o n th e sa m e o r o p p o site sid e o f th ech o rd [z 3 ; z 4 ]:

7 . T h e N in e -P o in t C ir c le

W e sh a ll n o w p rese n t a p ro o f o f th e c e le b ra te d n in e -p o in tc irc le th e o re m :

T h e o r e m A : T h e circ le p a ssin g th ro u g h th e m id p o in tso f th e sid es o f a tria n g le p a sse s th ro u g h th e fe et o f th ea ltitu d es a n d th e m id p o in ts o f th e lin e se g m en ts jo in in gth e o rth o c en tre to th e v e rtic e s.

T h u s g iv e n a n y tria n g le, n in e o f th e g eo m e trica lly m ea n -in g fu l p o in ts a sso c ia ted to th e tria n g le a re a ll fo u n d o n asin g le c irc le. T h a t circle d ese rv e s to b e c a lle d th e n in e -

p o in t c irc le a sso cia te d to th e tria n g le . O f c o u rse fo rsp ec ia l tria n g les so m e o f th e se p o in ts th e m se lv es m a yo v e rla p , th e w o rst ca se b e in g th a t o f a n e q u ila te ra l tri-a n g le .

W e w ill u se m o st o f th e re su lts th a t w e h a v e d isc u sse dso fa r, in p ro v in g T h e o re m A . In fa ct, a lo n g th e w a y ,w e sh a ll a lso p ro v e a n o th e r im p o rta n t th e o re m d u e toE u ler.

T h e o r e m B ( E u le r L in e ) : (F igu re 9 ) T h e c irc u m -c en tre O , th e c en tre o f g ra v ity G , th e ce n tre N o f th en in e-p o in t circ le a n d th e o rth o c e n tre H o f a tria n g le a rea ll co llin ea r. M o re o v e r, w e h a v e

O G :O N :O H = 2 :3 :6 .

P ro o f o f T h e o re m B : W e ch o o se th e o rth o ce n tre O o f

th e g iv e n tria n g le to b e th e o rig in o f th e p la n e . L et th eth re e v e rtic e s z 1 ; z 2 ; z 3 o f th e tria n g le b e re p re se n te d b yth e c o m p le x n u m b ers z 1 ; z 2 ; z 3 : T h en it fo llo w s th a t

jz 1 j = jz 2 j = jz 3 j = : r:


GENERAL ARTICLE

U sin g th e fa c t th a t th e d ia g o n a ls o f a rh o m b u s a re p e r-

p e n d ic u la r to e a ch o th e r, (see Q .7 in th e p re v io u s se c -tio n ), w e co n clu d e th a t

(z i ¡ z j ) ? (z i + z j ):

B y d e¯ n itio n , th e cen tro id is g iv e n b y

G :=z 1 + z 2 + z 3

3:

W e c la im th a t th e o rth o c en tre H is g iv e n b y

H = z 1 + z 2 + z 3 :

F o r, th e o rth o c en tre is th e u n iq u e p o in t w h ich sa tis¯ e sth e p ro p e rty

H ¡ z 1 ? (z 2 ¡ z 3 ); H ¡ z 2 ? (z 3 ¡ z 1 ); H ¡ z 3 ? (z 1 ¡ z 2 ):

A n d it is e a sily v eri e d th a t H = z 1 + z 2 + z 3 sa tis¯ e sth e se c o n d itio n s.

T h u s w e h a v e p ro v e d th a t O , G a n d H a re c o llin ea r a n da lso th a t

O G :O H = 1 :3 .

In th e litera tu re, O G H is c a lle d th e E u ler lin e. L e t n o w

P 1 =z 2 + z 3

2; P 2 =

z 3 + z 1

2; P 3 =

z 1 + z 2

2

b e th e m id p o in ts o f th e th ree sid es o f th e g iv e n tria n -

g le . L e t N b e th e c en tre o f th e circ le C p a ssin g th ro u g hP 1 ; P 2 ; P 3 : T h u s N is th e u n iq u e p o in t w h ich is e q u id is-

ta n t fro m P 1 ; P 2 ; P 3 : W e v e rify th a tH

2=

z 1 + z 2 + z 3

2h a s th is p ro p e rty , i.e.,

jH

2¡ P ij =

jz ij

2=

r

2:

T h u s, N = H = 2 : T h is co m p le tes th e p ro o f o f T h eo remB . Ä


GENERAL ARTICLE

W e n o w tu rn o u r a tten tio n to T h e o re m A . L e t u s ¯ rst

p ro v e th a t th e p o in t Q j w h ich is m id p o in t o f H Z j is o nth e c irc le C fo r e a ch j = 1 ; 2 ; 3 :

B u t th e n

Q j =H + z j

2:

T h e re fo re ,

jN ¡ Q j j =¯¯

¡z j

2

¯¯

=r

2:

(In fa c t, N ¡ Q j = ¡ (N ¡ P j ) sh o w s th a t Q j a re a n tip o d a lto P j w ith re sp ec t to N ).

W e sh a ll n o w sh o w th a t W j a re a lso o n C .

F o r th is, w e o b se rv e th a t H W 1 a n d O P 1 a re b o th p e r-p e n d ic u la r to th e lin e z 2 z 3 : A n d N is th e m id p o in t o fth e lin e O H . T h e re fo re, a s se e n in Q u estio n 6 , N is eq u i-

d ista n t fro m P 1 a n d W 1 : T h e refo re W 1 is o n C . L ik ew iseW 2 a n d W 3 a re a lso o n C . T h is c o m p lete s th e p ro o f o fT h e o re m A . Ä

8 . S o m e M o r e G e o m e t r ic P r o b le m s

T h e re a d e r is w elco m e to try e a ch o n e o f th e fo llo w in gq u estio n s. W e a d v ise th a t o n ly a fte r try in g e n o u g h o ra fte r o b ta in in g a so lu tio n o n o n e 's o w n , o n e m a y re a d

th e so lu tio n s g iv e n b e lo w w h ich a re b a se d o n th e u se o fc o m p le x n u m b e rs. In d ee d , m a n y o f th ese p ro b le m s c a nb e so lv ed b y so m e w h a t d i® e ren t m eth o d s. H o w ev er, th eo n ly so lu tio n I k n o w o f P ro b le m 7 is v ia P ro b le m 6 .

[Q .1 ] S h o w th a t th ree p o in ts rep re se n ted b y z 1 ; z 2 ; z 3

fo rm a n e q u ila te ra l tria n g le if a n d o n ly if z 2 ¡ z 1 =! (z 2 ¡ z 3 ), w h e re ! = e § ¼ {= 3 = ¡ 1 § i

p3

2:

[Q .2 ] If z 1 + z 2 + z 3 = 0 ; jz j j = 1 ; th e n sh o w th a tf z 1 ; z 2 ; z 3 g fo rm s th e v erte x set o f a n eq u ila tera l tria n g le .

[Q .3 ] S h o w th a t th e ce n ters o f th e e q u ila te ra l tria n g le s

e rec te d ex te rn a lly o n th e th re e sid e s o f a g iv en tria n g lefo rm a n e q u ila te ra l tria n g le.


GENERAL ARTICLE

[Q .4 ] S u p p o se w 1 ; w 2 ; w 3 2 C re p resen t th ree n o n -c o lli-

n e a r p o in ts a n d w 1 + w 2 + w 3 = 0 : If a 1 ; a 2 ; a 3 a re re a ln u m b ers su ch th a t a 1 w 1 + a 2 w 2 + a 3 w 3 = 0 , th e n sh o wth a t a 1 = a 2 = a 3 : (T h is is e q u iv a le n t to sa y th a t ifth re e p la n a r fo rc e s a c tin g o n a p o in t a re k e e p in g it ine q u ilib riu m th en b y sc a lin g a ll th e th re e fo rc es b y th e

sa m e fa cto r a n d o n ly b y th e sa m e fa c to r, th e p o in t w illstill b e in e q u ilib riu m .)

[Q .5 ] L e t w 1 ; w 2 ; w 3 2 C n f 0 g b e su ch th a t w 1 + w 2 + w 3 =

0 : P ro v e th a t th e fo llo w in g sta te m e n ts a re e q u iv a le n t.

(i) w 21 + w 2

2 + w 23 = 0 :

(ii) w 1 w 2 + w 2 w 3 + w 3 w 1 = 0 :

(iii)1

w 1

+1

w 2

+1

w 3

= 0 :

(iv ) jw 1 j = jw 2 j = jw 3 j:

(v ) 0 ; w 1 ; w 1 + w 2 fo rm th e v ertice s o f a n eq u ila tera l tri-

a n g le .

(v i) w 1 ; w 2 ; w 3 fo rm th e v ertic es o f a n e q u ila te ra l tria n -g le .

[Q .6 ] P ro v e th a t th ree d istin c t p o in ts z 1 ; z 2 ; z 3 in th ep la n e fo rm th e v e rtic e s o f a n eq u ila tera l tria n g le if a n do n ly if z 2

1 + z 22 + z 2

3 = z 1 z 2 + z 2 z 3 + z 3 z 1 :

[Q .7 ] S h o w th a t if w 1 ; w 2 ; w 3 a re p o in ts d iv id in g th eth re e sid e s o f th e tria n g le ¢ (z 1 ; z 2 ; z 3 ) in th e sa m e ra tio ,th e n th e tria n g le ¢ (w 1 ; w 2 ; w 3 ) is e q u ila te ra l if a n d o n lyif th e tria n g le ¢ (z 1 ; z 2 ; z 3 ) is e q u ila te ra l.

A n s w e r s :

[A .1 ] T h e tria n g le is eq u ila te ra l if a n d o n ly if jz 2 ¡ z 1 j =jz 2 ¡ z 3 j a n d th e a n g le b e tw e e n th e tw o se g m e n ts is eq u a lto § ¼ = 3 :

[A .2 ] C o n sid e r th e iso sc ele s tria n g le z 1 ; z 1 + z 2 ; 0 : T h e


GENERAL ARTICLE

g iv e n c o n d itio n im p lie s th a t it is e q u ila te ra l. T h ere fo re

th e a n g le b etw e en z j a n d z k m u st b e 2 ¼ = 3 : T h is m ea n sz 2 = z 1 ! ; z 3 = z 1 ! 2 w h e re ! 6= 1 is a c u b e ro o t o fu n ity . O b se rv e th a t 1 ; ! ; ! 2 fo rm th e v erte x se t o f a ne q u ila te ra l tria n g le. T h ere fo re , z 1 ; z 2 ; z 3 a lso fo rm a ne q u ila te ra l tria n g le.

[A .3 ] L e t z 1 ; z 2 ; z 3 d en o te th e v e rtic e s o f th e g iv e n tri-a n g le la b eled clo ck w ise. T h e c e n te r w 1 o f th e e q u ila t-e ra l tria n g le ra ise d o n th e sid e [z 1 ; z 2 ] is su ch th a t if

M 1 = z 1 + z 2

2 is th e m id p o in t o f [z 1 ; z 2 ] th en [M 1 ; w ] is

p e rp en d ic u la r to th e sid e [z 1 ; z 2 ] a n d is o f len g th1

2p

3tim e s th e le n g th o f th e sid e [z 1 ; z 2 ]: T h ere fo re

w 1 =z 1 + z 2

2+

{(z 2 ¡ z 1 )

2p

3:

L ik e w ise ,

w 2 =z 2 + z 3

2+

{(z 3 ¡ z 2 )

2p

3; w 3 =

z 3 + z 1

2+

{(z 1 ¡ z 3 )

2p

3:

V e rify th a t w 2 ¡ w 1 = ! (w 3 ¡ w 1 );

w h e re ! = e ¼ {= 3 =1 + {

p3

2:

[A .4 ] W e h a v e

w 1 + w 2 + w 3 = 0 = a 1 w 1 + a 2 w 2 + a 3 w 3 :

M u ltip ly th e ¯ rst o n e b y a j ; j = 1 ; 2 a n d su b tra c t fro mth e se c o n d to g e t

(a 2 ¡ a 1 )w 2 + (a 3 ¡ a 1 )w 3 = 0 = (a 1 ¡ a 2 )w 1 + (a 3 ¡ a 2 )w 3 :

O b se rv e th a t n o n e o f th e p o in ts w j is 0 : T h e re fo re , ifb o th th e se e q u a tio n s w e re n o n triv ia l, th en to g e th e r th eyim p ly th a t w 1 ; w 2 ; w 3 a re co llin ea r. W e c o n c lu d e th a t

o n e o f th e e q u a tio n s is triv ia l, sa y th e ¯ rst o n e , i.e .,(a 2 ¡ a 1 ) = (a 3 ¡ a 1 ) = 0 :


GENERAL ARTICLE

Address for Correspondence

Anant R Shastri

Indian Institute of Technology

Bombay

Mumbai 400 085, India.

Email: ars@ math.iitb.ac.in

Suggested Reading

[1] G Gamow, One, Two Three ... Infininity, A Mentor Book, The New

American Library, 1947. Or Facts and Speculations of Science, Dover

Publication, 1963.

[2] R Anant Shastri, An Introduction to Complex Analysis, Macmillan India

Ltd., 1999.

[A .5 ] (i) , (ii)

0 = (w 1 + w 2 + w 3 )2 = w 21 + w 2

2 + w 23 ¡ 2 (w 1 w 2 + w 2 w 3 + w 3 w 1 ):

(ii) , (iii) M u ltip ly a n d d iv id e b y w 1 w 2 w 3 :

(iii) , (iv ) w ¡ 1j = ¹w j = jw j j2 : W e h a v e w 1 + w 2 + w 3 = 0 ,

i.e ., ¹w 1 + ¹w 2 + ¹w 3 = 0 : T h e refo re, w e h a v e

jw 1 j2

w 1

+jw 2 j2

w 2

+jw 3 j2

w 3

= 0 :

N o w a p p e a l to th e p rev io u s e x e rcise .

T h e eq u iv a len c e o f (iv ) , (v ) a n d (v i) is e a sy . (R em e m b e rw 1 + w 2 + w 3 = 0 ).

[A .6 ] P u t w 1 = z 2 ¡ z 1 ; w 2 = z 3 ¡ z 2 ; w 3 = z 1 ¡ z 3 : T h e nz 1 ; z 2 ; z 3 fo rm th e v erte x se t o f a n e q u ila tera l tria n g le ifa n d o n ly if so d o w 1 ; w 2 ; w 3 : N o w th e g iv en c o n d itio n fo rz j is e q u iv a le n t to (ii) o f th e p re v io u s e x erc ise .

[A .7 ] P u t w 1 = tz 1 + (1 ¡ t)z 2 ; w 2 = tz 2 + (1 ¡ t)z 3 ; w 3 =tz 3 + (1 ¡ t)z 1 : T h e n v erify th a tX

w 21 ¡

Xw 1 w 2 = [t2 ¡ (1 ¡ t)2 ¡ t(1 ¡ t)](

Xz 2

1 ¡X

z 1 z 2 ):

S in ce t2 + (1 ¡ t)2 ¡ t(1 ¡ t) > 0 fo r a ll t 2 R ; w e a red o n e.

A c k n o w le d g e m e n t

T h e a u th o r w o u ld lik e to th a n k h is c o lle a g u e P ro f. KD J o sh i (D e p a rtm en t o f M a th em a tics, IIT , B o m b a y ) fo rm a n y u se fu l d iscu ssio n s a n d e sp ec ia lly fo r b rin g in g p ro b -

lem s 6 a n d 7 to h is n o tic e .

geometry and complex numbers

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