IMO 1989 SL 18

Origin: MON

Problem

Given a convex polygon A1A2 . . . An with area S, and a point M in the same plane, determine the area of polygon M1M2 . . . Mn, where Mi is the image of M under rotation R\alpha Ai around Ai by \alpha, i = 1, 2, . . . , n.

Solution

Consider the triangle MAiMi. Obviously, the point Mi is the image of Ai under the composition C of rotation R\alpha/2−90◦ M and homothety H2 sin(\alpha/2) M . Therefore, the polygon M1M2 . . . Mn is obtained as the im- age of A1A2 . . . An under the rotational homothety C with coefficient 2 sin(\alpha/2). Therefore SM1M2...Mn = 4 sin2 (\alpha/2) \cdot S.