IMO 1977 LL BUL1

Origin: BUL

Problem

A pentagon ABCDE inscribed in a circle for which BC < CD and AB < DE is the base of a pyramid with vertex S. If AS is the longest edge starting from S, prove that BS > CS.

Solution

Let P be the projection of S onto the plane ABCDE. Obviously BS > CS is equivalent to BP > CP. The conditions of the problem imply that PA > PB and PA > PE. The locus of such points P is the region of the plane that is determined by the perpendicular bisectors of segments AB and AE and that contains the point diametrically opposite A. But since AB < DE, the whole of this region lies on one side of the perpendicular bisector of BC. The result follows immediately. Remark. The assumption BC < CD is redundant.