IMO 1995 Problem A3 Determine all integers n > 3 for which there exist n points A 1 , ... , A n in the plane, no three collinear, and real numbers r 1 , ... , r n such that for any distinct i, j, k, the area of the triangle A i A j A k is r i + r j + r k .
n = 4. Solution The first point to notice is that if no arrangement is possible for n, then no arrangement is possible for any higher integer. Clearly the four points of a square work for n = 4, so we focus on n = 5. If the 5 points form a convex pentagon, then considering the quadrilateral A 1 A 2 A 3 A 4 as made up of two triangles in two ways, we have that r 1 + r 3 = r 2 + r 4 . Similarly, A 5 A 1 A 2 A 3 gives r 1 + r 3 = r 2 + r 5 , so r 4 = r 5 . We show that we cannot have two r's equal (whether or not the 4 points form a convex pentagon). For suppose r 4 = r 5 . Then A 1 A 2 A 4 and A 1 A 2 A 5 have equal area. If A 4 and A 5 are on the same side of the line A 1 A 2 , then since they must be equal distances from it, A 4 A 5 is parallel to A 1 A 2 . If they are on opposite sides, then the midpoint of A 4 A 5 must lie on A 1 A 2 . The same argument can be applied to A 1 and A 3 , and to A 2 and A 3 . But we cannot have two of A 1 A 2 , A 1 A 3 and A 2 A 3 parallel to A 4 A 5 , because then A 1 , A 2 and A 3 would be collinear. We also cannot have the midpoint of A 4 A 5 lying on two of A 1 A 2 , A 1 A 3 and A 2 A 3 for the same reason. So we have established a contradiction. hence no two of the r's can be equal. In particular, this shows that the 5 points cannot form a convex pentagon. Suppose the convex hull is a quadrilateral. Without loss of generality, we may take it to be A 1 A 2 A 3 A 4 . A 5 must lie inside one of A 1 A 2 A 4 and A 2 A 3 A 4 . Again without loss of generality we may take it to be the latter, so that A 1 A 2 A 5 A 4 is also a convex quadrilateral. Then r 2 + r 4 = r 1 + r 3 and also = r 1 + r 5 . So r 3 = r 5 , giving a contradiction as before. The final case is the convex hull a triangle, which we may suppose to be A 1 A 2 A 3 . Each of the other two points divides its area into three triangles, so we have: (r 1 + r 2 + r 4 ) + (r 2 + r 3 + r 4 ) + (r 3 + r 1 + r 4 ) = (r 1 + r 2 + r 5 ) + (r 2 + r 3 + r 5 ) + (r 3 + r 1 + r 5 ) and hence r 4 = r 5 , giving a contradiction. So the arrangement is not possible for 5 and hence not for any n > 5. 36th IMO 1995 (C) John Scholes jscholes@kalva.demon.co.uk 24 Oct 1998 Last corrected/updated 23 Aug 03