XXI OM - III - Task 1

Diameter $ \overline{AB} $ divides the circle into two semicircles. On one semicircle, n points $ P_1 P_2, \ldots, P_n $ are chosen such that $ P_1 $ lies between $ A $ and $ P_2 $, $ P_2 $ lies between $ P_1 $ and $ P_3 $, $ \ldots $, $ P_n $ lies between $ P_{n-1} $ and $ B $. How should point $ C $ be chosen on the other semicircle so that the sum of the areas of triangles $ CP_1P_2, CP_2P_3, CP_3P_4, \ldots, CP_{n-1}P_n $ is maximized?
Notice that the sum of the areas of triangles $ CP_1P_2, CP_2P_3, CP_3P_4, \ldots, CP_{n-1}P_n $ is equal to the sum of the areas of the polygon $ P_1P_2 \ldots P_n $ and triangle $ CP_1P_n $ (Fig. 13). The first area does not depend on the choice of point $ C $. The second area will be maximal when the distance from point $ C $ to the line $ P_1P_n $ is as large as possible. The point on the circle farthest from a given chord lies on the perpendicular bisector of that chord. Therefore, $ C $ should be the point of intersection of the perpendicular bisector of segment $ \overline{P_1P_n} $ with the other semicircle.