Advanced book on Mathematics Olympiad

Algebra 395

holds true forα= 2 nπ+ 1 , 2 n^2 π+ 1 ,..., 2 nnπ+ 1. Hence the equation

(

2 n+ 1

1

)

xn−

(

2 n+ 1

3

)

xn−^1 +···+(− 1 )n

(

2 n+ 1

2 n+ 1

)

= 0

has the roots

xk=cot^2

kπ

2 n+ 1

,k= 1 , 2 ,...,n.

The product of the roots is

x 1 x 2 ···xn=

( 2 n+ 1

2 n+ 1

)

( 2 n+ 1

1

)=

1

2 n+ 1

.

So

cot^2

π

2 n+ 1

cot^2

2 π

2 n+ 1

···cot^2

nπ

2 n+ 1

=

1

2 n+ 1

.

Because 0< 2 kπn+ 1 <π 2 ,k= 1 , 2 ,...,n, it follows that all these cotangents are posi-
tive. Taking the square root and inverting the fractions, we obtain the identity from the
statement.
(Romanian Team Selection Test for the International Mathematical Olympiad, 1970)

163.A good guess is thatP(x)=(x− 1 )n, and we want to show that this is the case. To
this end, letx 1 ,x 2 ,...,xnbe the zeros ofP(x). Using Viète’s relations, we can write

∑

i

(xi− 1 )^2 =

(

∑

i

xi

) 2

− 2

∑

i<j

xixj− 2

∑

i

xi+n

=n^2 − 2

n(n− 1 )

2

− 2 n+n= 0.

This implies that all squares on the left are zero. Sox 1 =x 2 = ··· =xn=1, and
P(x)=(x− 1 )n, as expected.
(Gazeta Matematica ̆(Mathematics Gazette, Bucharest))

164.Letα, β, γbe the zeros ofP(x). Without loss of generality, we may assume that
0 ≤α≤β≤γ. Then

x−a=x+α+β+γ≥0 and P(x)=(x−α)(x−β)(x−γ).

If 0≤x≤α, using the AM–GM inequality, we obtain

−P(x)=(α−x)(β−x)(γ−x)≤

1

27

(α+β+γ− 3 x)^3