30th IMO 1989 shortlist Problem 15 Given a convex n-gon A 1 A 2 ... A n with area A and a point P, we rotate P through an angle x about A i to get the point P i . Find the area of the polygon P 1 P 2 ... P n .
Consider the triangle PA i P i . The angle A i PP i is 90 o - x/2, which is independent of i, and the ratio PP i /PA i = 2 sin x/2, which is also independent of i. So we can obtain P i by rotating A i through an angle 90 o - x/2 about P and then expanding by a factor 2. Thus the area of the polygon P 1 P 2 ... P n is 4 sin 2 (x/2) times the area of the polygon A 1 A 2 ... A n . 30th IMO shortlist 1989 © John Scholes jscholes@kalva.demon.co.uk 18 Dec 2002 Last corrected/updated 28 Dec 02