IMO 1982 SL 8

Problem

B2 (POL 4) A convex, closed figure lies inside a given circle. The figure is seen from every point of the circumference at a right angle (that is, the two rays drawn from the point and supporting the convex figure are perpendicular). Prove that the center of the circle is a center of symmetry of the figure.

Solution

Let F be the given figure. Consider any chord AB of the circumcircle \gamma that supports F. The other supporting lines to F from A and B intersect \gamma again at D and C respectively so that \angleDAB = \angleABC = 90◦. Then ABCD is a rectangle, and hence CD must support F as well, from which it follows that F is inscribed in the rectangle ABCD touching each of its sides. We easily conclude that F is the intersection of all such rectan- gles. Now, since the center O of \gamma is the center of symmetry of all these rectangles, it must be so for their intersection F as well.