Let f(x) = cos(a1 + x) + 1/2 cos(a2 + x) + 1/4 cos(a3 + x) + ... + 1/2n-1 cos(an + x), where ai are real constants and x is a real variable. If f(x1) = f(x2) = 0, prove that x1 - x2 is a multiple of π.
Solution
f is not identically zero, because f(-a1) = 1 + 1/2 cos(a2 - a1) + ... > 1 - 1/2 - 1/4 - ... - 1/2n-1 > 0.
Using the expression for cos(x + y) we obtain f(x) = b cos x + c sin x, where b = cos a1 + 1/2 cos a2 + ... + 1/2n-1 cos an, and c = - sin a1 - 1/2 sin a2 - ... - 1/2n-1 sin an. b and c are not both zero, since f is not identically zero, so f(x) = √(b2 + c2) cos(d + x), where cos d = b/√(b2 + c2), and sin d = c/√(b2 + c2). Hence the roots of f(x) = 0 are just mπ + π/2 - d.
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.
© John Scholes
jscholes@kalva.demon.co.uk
5 Oct 1998
Last corrected/updated 5 Oct 1998