Advanced book on Mathematics Olympiad

618 Geometry and Trigonometry

On the other hand,Pis obtained by rotatingBaroundAby−α, so its coordinate is
p =bω ̄. Similarly, the coordinate ofQisq =cω. It is now straightforward to
check that

q−p

z− 0

=ω−

1

ω

,

a purely imaginary number. Hence the linesPQandAOform a 90◦angle, which is the
desired result.
(USA Team Selection Test for the International Mathematical Olympiad, 2006, solu-
tion by T. Leung)

602.∏ In the language of complex numbers we are required to find the maximum of
n
k= 1 |z−
k|aszranges over the unit disk, where=cos 2 π
n +isin

2 π

n. We have

∏n

k= 1

|z−k|=

∣

∣∣

∣∣

∏n

k= 1

(z−k)

∣

∣∣

∣∣=|zn−^1 |≤|zn|+^1 =^2.

The maximum is 2, attained whenzis annth root of−1.
(Romanian Mathematics Competition “Grigore Moisil,’’ 1992, proposed by D. An-
drica)

603.First solution: In a system of complex coordinates, place each vertexAk,k =
0 , 1 ,...,n−1, atk, where=e^2 iπ/n. Then

A 0 A 1 ·A 0 A 2 ···A 0 An− 1 =|( 1 −)( 1 −^2 )···( 1 −n−^1 )|.

Observe that, in general,

(z−)(z−^2 )···(z−n−^1 )=

1

z− 1

(z− 1 )(z−)···(z−n−^1 )

=

1

z− 1

(zn− 1 )=zn−^1 +zn−^2 +···+ 1.

By continuity, this equality also holds forz=1. Hence

A 0 A 1 ·A 0 A 2 ···A 0 An− 1 = 1 n−^1 + 1 n−^2 +···+ 1 =n,

and the identity is proved.

Second solution: Choose a pointPon the ray|OA 0 , whereOis center of the circumcircle
of the polygon, such thatA 0 is betweenOandP.IfOP=x, then the last problem in
the introduction showed thatPA 0 ·PA 1 ···PAn− 1 =xn−1. Hence

A 0 A 1 ·A 0 A 2 ···A 0 An− 1 =lim

x→ 1

xn− 1

x− 1

=n.