IMO 1995 SL N3

Origin: CZE | Category: Number Theory

Problem

Determine all integers n > 3 such that there are n points A1, A2, . . . , An in the plane that satisfy the following two conditions si- multaneously: (a) No three lie on the same line. (b) There exist real numbers p1, p2, . . . , pn such that the area of \triangleAiAjAk is equal to pi + pj + pk, for 1 \leqi < j < k \leqn.

Solution

For n = 4, the vertices of a unit square A1A2A3A4 and p1 = p2 = p3 = p4 = 1 6 satisfy the conditions. We claim that there are no solutions for n = 5 (and thus for any n \geq5). Suppose to the contrary that points Ai and pi, i = 1, . . . , 5, satisfy the conditions. Denote the area of \triangleAiAjAk by Sijk = pi + pj + pk, 1 \leqi < j < k \leq5. Observe that all the pi’s must be distinct. Indeed, if p4 = p5, then S124 = S125 and S234 = S235, which implies that A4A5 is parallel to A1A2 and A2A3, so A1, A2, A3 are collinear, which is impossible. Also note that if AiAjAkAl is convex, then Sijk + Sikl = Sijl + Sjkl gives pi + pk = pj + pl. Now consider the convex hull of A1, A2, A3, A4, A5. There are three cases. (i) The convex hull is the pentagon A1A2A3A4A5. Then A1A2A3A4 and A1A2A3A5 are convex, so we have p1 + p3 = p2 + p4 and p1 + p3 = p2 + p5. Hence p4 = p5, a contradiction. (ii) The convex hull is w.l.o.g. the quadrilateral A1A2A3A4. Assume that A5 lies within A1A3A4. Then A1A2A3A5 is also convex, so as in (1) we get p4 = p5. (iii) The convex hull is w.l.o.g. the triangle A1A2A3. Since S124 + S134 + S234 = S125 + S135 + S235, we conclude that again p4 = p5.