Advanced book on Mathematics Olympiad

4.2 Trigonometry 239

Thenf(cost)=csintfor some constantcandtin( 0 ,π), i.e.,f(x)=c

√

1 −x^2 for
allx∈(− 1 , 1 ). It follows thatfis an even function. But then in the equation from the
statementf( 2 x^2 − 1 )= 2 xf (x)the left-hand side is an even function while the right-
hand side is an odd function. This can happen only if both sides are identically zero.
Therefore,f(x)=0 forx∈[− 1 , 1 ]is the only solution to the functional equation. 

We continue with a problem that was proposed by Belgium for the 26th International
Mathematical Olympiad in 1985.

Example.Letx, y, zbe real numbers such thatx+y+z=xyz. Prove that

x( 1 −y^2 )( 1 −z^2 )+y( 1 −z^2 )( 1 −x^2 )+z( 1 −x^2 )( 1 −y^2 )= 4 xyz.

Solution.The conclusion is immediate ifxyz=0, so we may assume thatx, y, z =0.
Dividing through by 4xyzwe transform the desired equality into

1 −y^2

2 y

·

1 −z^2

2 z

+

1 −z^2

2 x

·

1 −x^2

2 x

+

1 −x^2

2 x

·

1 −y^2

2 y

= 1.

This, along with the condition from the statement, makes us think about the substitutions
x=tanA,y=tanB,z=tanC, whereA, B, Care the angles of a triangle. Using the
double-angle formula

1 −tan^2 u

2 tanu

=

1

tan 2u

=cot 2u

we further transform the equality into

cot 2Bcot 2C+cot 2Ccot 2A+cot 2Acot 2B= 1.

But this is equivalent to

tan 2A+tan 2B+tan 2C=tan 2Atan 2Btan 2C,

which follows from tan( 2 A+ 2 B+ 2 C)=tan 2π=0. 

And now the problems.

675.Leta, b, c∈[ 0 , 1 ]. Prove that

√

abc+

√

( 1 −a)( 1 −b)( 1 −c)≤ 1.

676.Solve the equationx^3 − 3 x=

√

x+2 in real numbers.