### IMO 1963

**Problem B2**
Prove that cos π/7 - cos 2π/7 + cos 3π/7 = 1/2.

**Solution**

Consider the roots of x^{7} + 1 = 0. They are e^{iπ/7}, e^{i3π/7}, ... , e^{i13π/7} and must have sum zero since there is no x^{6} term. Hence, in particular, their real parts sum to zero. But cos7π/7 = - 1 and the others are equal in pairs, because cos(2π - x) = cos x. So we get cos π/7 + cos 3π/7 + cos 5π/7 = 1/2. Finally since cos(π - x) = - cos x, cos 5π/7 = - cos 2π/7.

Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

5th IMO 1963

© John Scholes

jscholes@kalva.demon.co.uk

21 Sep 1998

Last corrected/updated 24 Sep 2003