Determine all integers $ n > 3$ for which there exist $ n$ points $ A_{1},\cdots ,A_{n}$ in the plane, no three collinear, and real numbers $ r_{1},\cdots ,r_{n}$ such that for $ 1\leq i < j < k\leq n$, the area of $ \triangle A_{i}A_{j}A_{k}$ is $ r_{i} + r_{j} + r_{k}$.
Problem
Source: IMO 1995, Problem 3, Day 1, IMO Shortlist 1995, N3
Tags: geometry, area of a triangle, combinatorial geometry, point set, IMO, imo 1995
grobber
11.11.2005 01:33
Let's prove that only $n=4$ works. Obviously, $n=4$ is Ok, since we can take $A_1A_2A_3A_4$ to be a square and all the $r_i$'s equal to one third of the area of a triangle formed by three of its vertices. Since if we have $n\ge 5$ such points we also have $5$, all we need to do is prove that $n=5$ is impossible. First notice that if $A_1A_2A_3A_4$ is a convex quadrilateral, then $r_1+r_3=r_2+r_4$ and if $A_4$ lies inside $A_1A_2A_3$, then $r_4=-\frac{r_1+r_2+r_3}3$. I'll use these without mentioning it from now on. Case 1 $A_1A_2A_3A_4A_5$ is a convex pentagon In this case $r_i+r_{i+2}$ is the same for all $i$ (the indices are modulo $5$), which, in turn, implies that all the $r_i$ are equal. This is impossible, since $[A_1A_2A_3]=[A_1A_2A_5]$ clearly implies $[A_1A_2A_4]>[A_1A_2A_3]$. Case 2 $A_5$ lies inside the convex quadrilateral $A_1A_2A_3A_4$ Suppose furthermore that $A_5$ lies inside the triangles $A_1A_2A_3,A_1A_2A_4$, and outside the other two formed by the vertices $A_1,A_2,A_3,A_4$. We then get $r_5=-\frac{r_1+r_2+r_3}3=-\frac{r_1+r_2+r_4}3\Rightarrow r_3=r_4$, which means that $A_3A_4\|A_1A_2\Rightarrow r_1=r_2$. We also have $r_5+r_3=r_2+r_4=r_2+r_3\Rightarrow r_1=r_2=r_5$. This is impossible, since it would mean that $A_5$ lies on the parallel $A_1A_2$ to $A_3A_4$. Case 3 $A_4,A_5$ lie inside $A_1A_2A_3$ We get $r_4=r_5=-\frac{r_1+r_2+r_3}3$, so the areas of the following pairs of triangles are equal: $(A_1A_2A_4,A_1A_2A_5),(A_2A_3A_4,A_2A_3A_5)$. This means that $A_4A_5$ is parallel to both $A_1A_2$ and $A_2A_3$, which is, of course, absurd.