20th BMO 1984 Problem 3 Find the maximum and minimum values of cos x + cos y + cos z, where x, y, z are non-negative reals with sum 4π/3.
max 3/2 at 0,0,4π/3 min 0 at 0,2π/3,2π/3 Solution wlog x ≤ y ≤ z. So 4π/9 ≤ z ≤ 4π/3. Put z = 4π/3 - k, so 0 ≤ k ≤ 8π/9. Then cos x + cos y + cos z = cos(4π/3 - k) + 2 cos k/2 cos(y-x)/2. Since cos(k/2) ≥ 0, the expression is maximised by taking x = y so that cos(y-x)/2 = 1. That gives cos x + cos y + cos z = cos(4π/3 - k) + 2 cos k/2 = -1/2 cos k - (√3)/2 sin k + 2 cos k/2 = - cos2k/2 + 1/2 + 2 cos k/2 - (√3)/2 sin k = 3/2 - (cos k/2 - 1)2 - (√3)/2 sin k. Since k < π, sin k ≥ 0, so we maximise this expression by taking k = 0 which gives (cos k/2 - 1) = 0 and sin k = 0. We minimise cos(4π/3 - k) + 2 cos k/2 cos(y-x)/2 by taking cos(y-x)/2 as small as possible, in other words by taking x = 0, y = k, giving cos x + cos y + cos z = cos(4π/3 - k) + 2 cos 2 k/2 = cos(4π/3 - k) + cos k + 1 = 1 + 2 cos 2π/3 cos(2π/3 - k) = 1 - cos(2π/3 - k), which is minimised by taking k = 2π/3 to give 0. 20th BMO 1984 © John Scholes jscholes@kalva.demon.co.uk 10 February 2004 Last corrected/updated 10 Feb 04