Determine all integers n > 3 for which there exist n points A1, ... , An in the plane, no three collinear, and real numbers r1, ... , rn such that for any distinct i, j, k, the area of the triangle AiAjAk is ri + rj + rk.
Answer
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 A1A2A3A4 as made up of two triangles in two ways, we have that r1 + r3 = r2 + r4. Similarly, A5A1A2A3 gives r1 + r3 = r2 + r5, so r4 = r5.
We show that we cannot have two r's equal (whether or not the 4 points form a convex pentagon). For suppose r4 = r5. Then A1A2A4 and A1A2A5 have equal area. If A4 and A5 are on the same side of the line A1A2, then since they must be equal distances from it, A4A5 is parallel to A1A2. If they are on opposite sides, then the midpoint of A4A5 must lie on A1A2. The same argument can be applied to A1 and A3, and to A2 and A3. But we cannot have two of A1A2, A1A3 and A2A3 parallel to A4A5, because then A1, A2 and A3 would be collinear. We also cannot have the midpoint of A4A5 lying on two of A1A2, A1A3 and A2A3 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 A1A2A3A4. A5 must lie inside one of A1A2A4 and A2A3A4. Again without loss of generality we may take it to be the latter, so that A1A2A5A4 is also a convex quadrilateral. Then r2 + r4 = r1 + r3 and also = r1 + r5. So r3 = r5, giving a contradiction as before.
The final case is the convex hull a triangle, which we may suppose to be A1A2A3. Each of the other two points divides its area into three triangles, so we have: (r1 + r2 + r4) + (r2 + r3 + r4) + (r3 + r1 + r4) = (r1 + r2 + r5) + (r2 + r3 + r5) + (r3 + r1 + r5) and hence r4 = r5, giving a contradiction.
So the arrangement is not possible for 5 and hence not for any n > 5.
(C) John Scholes
jscholes@kalva.demon.co.uk
24 Oct 1998
Last corrected/updated 23 Aug 03