Advanced book on Mathematics Olympiad

Geometry and Trigonometry 659

which means that

cottjsint=

1

r

−cost, forj= 1 , 2 , 3.

If sint =0, then cott 1 =cott 2 =cott 3. There are only two possible values that
t 1 ,t 2 ,t 3 can take between 0 and 2π, and so two of thetjare equal, which is ruled out
by the hypothesis. It follows that sint=0. Then on the one hand,rcost− 1 =0,
and on the other, cost=±1. This can happen only if cost=1 andr=1. Therefore,
z 1 z 2 z 3 =r^3 cost=1, as desired.

671.Consider the complex numberω=cosθ+isinθ. The roots of the equation

(

1 +ix

1 −ix

)n

=ω^2 n

are preciselyak=tan(θ+kπn),k= 1 , 2 ,...,n. Rewriting this as a polynomial equation
of degreen, we obtain

0 =( 1 +ix)^2 −ω^2 n( 1 −ix)n

=( 1 −ω^2 n)+ni( 1 +ω^2 n)x+···+nin−^1 ( 1 −ω^2 n)xn−^1 +in( 1 +ω^2 n)xn.

The sum of the zeros of the latter polynomial is

−nin−^1 ( 1 −ω^2 n)

in( 1 +ω^2 n)

,

and their product

−( 1 −ω^2 n)

in( 1 +ω^2 n)

.

Therefore,

a 1 +a 2 +···+an

a 1 a 2 ···an

=nin−^1 =n(− 1 )(n−^1 )/^2.

(67th W.L. Putnam Competition, 2006, proposed by T. Andreescu)

672.More generally, for an odd integern, let us compute

S=(cosα)(cos 2α)···(cosnα)

withα= 22 nπ+ 1. We can letζ=eiαand thenS= 2 −n

∏n

k= 1 (ζ

k+ζ−k). Sinceζk+ζ−k=

ζ^2 n+^1 −k+ζ−(^2 n+^1 −k),k= 1 , 2 ,...,n, we obtain