IMO 1967 LL USS55

Origin: USS

Problem

Find all x for which for all n, sin x + sin 2x + sin 3x + \cdot \cdot \cdot + sin nx \leq \sqrt 2 .

Solution

It is enough to find all x from (0, 2\pi] such that the given inequality holds for all n. Suppose 0 < x < 2\pi/3. If n is the maximum integer for which nx \leq 2\pi/3, we have \pi/3 < nx \leq2\pi/3, and consequently sin nx \geq \sqrt 3/2. Thus sin x + sin 2x + \cdot \cdot \cdot + sin nx > \sqrt 3/2. Suppose now that 2\pi/3 \leqx < 2\pi. We have sin x + \cdot \cdot \cdot + sin nx = cos x 2 −cos 2n+1 x 2 sin x \leqcos x 2 + 1 2 sin x = cot x \leq \sqrt 2 . For x = 2\pi the given inequality clearly holds for all n. Hence, the inequal- ity holds for all n if and only if 2\pi/3 + 2k\pi \leqx \leq2\pi + 2k\pi for some integer k.