Advanced book on Mathematics Olympiad

Geometry and Trigonometry 655

cosx,cosy,cos(x+ky)can be obtained by adding polynomials of lower degrees, and

eventually multiplying them by one of the three functions.

Hence it suffices to show that the property is invariant under multiplication by

cosx,cosy, and cos(x+ky). It can be verified that this follows from

2 cos(ax+by)cosx=cos((a+ 1 )x+by)+cos((a− 1 )x+by),

2 cos(ax+by)cosy=cos(ax+(b+ 1 )y)+cos(ax+(b− 1 )y),

2 cos(ax+by)cos(x+ky)=cos((a+ 1 )x+(b+k)y)+cos(a− 1 )x+(b−k)y).

So for cos(ax+by)to be a polynomial in cosx,cosy, and cos(x +ky), it must be such

a sum with a single term. This can happen only ifbis divisible byk.

The answer to the problem is thereforek=± 1 ,± 3 ,± 9 ,± 11 ,± 33 ,±99.

(proposed by R. Gelca for the USA Mathematical Olympiad, 1999)

663.Clearly, this problem is about the addition formula for the cosine. For it to show

up we need products of sines and cosines, and to obtain them it is natural to square the

relations. Of course, we first separateaanddfrombandc. We have

(2 cosa+9 cosd)^2 =(6 cosb+7 cosc)^2 ,

(2 sina−9 sind)^2 =(6 sinb−7 sinc)^2.

This further gives

4 cos^2 a+36 cosacosd+81 cos^2 d=36 cos^2 b+84 cosbcosc+49 cosc^2 ,

4 sin^2 a−36 sinasind+81 sin^2 d=36 sin^2 b−84 sinbsinc+49 sinc^2.

After adding up and using sin^2 x+cos^2 x=1, we obtain

85 + 36 (cosacosd−sinasind)= 85 + 84 (cosbcosc−sinbsinc).

Hence 3 cos(a+d)=7 cos(b+c), as desired.

(Korean Mathematics Competition, 2002, proposed by T. Andreescu)

664.The first equality can be written as

sin^3 a+cos^3 a+

(

−

1

5

) 3

− 3 (sina)(cosa)

(

−

1

5

)

= 0.

We have seen before that the expressionx^3 +y^3 +z^3 − 3 xyzfactors as

1

2

(x+y+z)[(x−y)^2 +(y−z)^2 +(z−x)^2 ].