IMO 1975 SL 12

Origin: GRE

Problem

Consider on the first quadrant of the trigonometric circle the arcs AM1 = x1, AM2 = x2, AM3 = x3, . . . , AM\nu = x\nu, such that x1 < x2 < x3 < \cdot \cdot \cdot < x\nu. Prove that \nu−1  i=0 sin 2xi − \nu−1  i=0 sin(xi −xi+1) < \pi 2 + \nu−1  i=0 sin(xi + xi+1).

Solution

Since sin 2xi = 2 sin xi cos xi and sin(xi + xi+1) + sin(xi −xi+1) = 2 sin xi cos xi+1, the inequality from the problem is equivalent to (cos x1 −cos x2) sin x1 + (cos x2 −cos x3) sin x2 + \cdot \cdot \cdot \cdot \cdot \cdot + (cos x\nu−1 −cos x\nu) sin x\nu−1 < \pi 4 . (1) Consider the unit circle with center at O(0, 0) and points Mi(cos xi, sin xi) on it. Also, choose the points Ni(cos xi, 0) and M ′ i(cos xi+1, sin xi). It is clear that (cos xi −cos xi+1) sin xi is equal to the area of the rectangle MiNiNi+1M ′ i. Since all these rectangles are disjoint and lie inside the quarter circle in the first quadrant whose area is \pi 4 , inequality (1) follows.