Advanced book on Mathematics Olympiad

236 4 Geometry and Trigonometry

Solution.Let 1 , 2 ,...,kbe thekth roots of unity, that is,j=cos^2 jπk +isin^2 jπk ,
j= 1 , 2 ,...,k. The sum

s 1 + 2 s+···+ks

is equal tokifkdividess, and to 0 ifkdoes not divides. We have

∑k

j= 1

( 1 +j)n=

∑n

s= 0

(

n

s

)⎛

⎝

∑k

j= 1

js

⎞

⎠=k

∑nk

j= 0

(

n

jk

)

.

Since

1 +j=2 cos

jπ

k

(

cos

jπ

k

+isin

jπ

k

)

,

it follows from the de Moivre formula that

∑k

j= 1

( 1 +j)n=

∑k

j= 1

2 ncosn

jπ

k

(

cos

nj π

k

+isin

nj π

k

)

.

Therefore,

(

n

0

)

+

(

n

k

)

+

(

n

2 k

)

+··· =

2 n

k

∑k

j= 1

cosn

jπ

k

(

cos

nj π

k

+isin

nj π

k

)

.

The left-hand side is real, so we can ignore the imaginary part and obtain the identity
from the statement. 

And now a problem given at an Indian Team Selection Test for the International
Mathematical Olympiad in 2005, proposed by the first author of the book.

Example.For real numbersa, b, c, dnot all equal to zero, letf:R→R,

f(x)=a+bcos 2x+csin 5x+dcos 8x.

Suppose thatf(t)= 4 afor some real numbert. Prove that there exists a real numbers
such thatf(s) <0.

Solution.Letg(x)=be^2 ix−ice^5 ix+de^8 ix. Thenf(x)=a+Reg(x). Note that

g(x)+g

(

x+

2 π

3

)

+g

(

x+

4 π

3

)

=g(x)

(

1 +e^2 πi/^3 +e^4 πi/^3

)

= 0.

Therefore,