IMO 1995 - Problem 3

Problem: Determine all integers $n>3$ such that there are $n$ points $A_1,A_2,...,A_n$ in the plane which satisfy the following two conditions simultaneously.
(a) No three lie on the same line
(b) There exist real numbers $p_1,p_2,...,p_n$ such that the area of $\triangle{A_iA_jA_k}$ is equal to $p_i+p_j+p_k$ for $1\le i<j<k\le n$.
Solution: We claim $n=4$ is the only answer. For $n=4$, let $A_1A_2A_3A_4$ be a unit square and let $p_1=p_2=p_3=p_4=\frac{1}{6}$. It remains to show no solution exists for $n=5$ which implies that there no solution for any $n\ge 5$.
 Suppose to the contrary that there is solution with $n=5$. Denote the area of $\triangle{A_iA_jA_k}$ by $[ijk]=p_i+p_j+p_k$. We cannot have $p_i=p_j$. If for example $p_4=p_5,$ then $[124]=[125]$ and $[234]=[235]$, which implies that $A_4A_5$ is parallel to both $A_1A_2$ and $A_2A_3$. This is impossible since $A_1,A_2,A_3$ are not collinear. We also observe that if $A_iA_jA_kA_l$ is convex, then $p_i+p_k=p_j+p_l$. This follows from $[ijk]+[kli]=[jkl]+[lij]$.
Consider the convex hull of $A_1,...,A_5$, we have $3$ cases. First, suppose that the convex hull is a pentagon $A_1A_2A_3A_4A_5$. Since $A_1A_2A_3A_4$ and $A_1A_2A_3A_5$ are convex, our observation yields $p_1+p_3=p_2+p_4$ and $p_1+p_3=p_2+p_5$. Hence $p_4=p_5$, a contradiction.
Next we suppose that the convex hull is a quadrilateral $A_1A_2A_3A_4$. We may assume that $A_5$ is within $A_3A_4A_1$. Then $A_1A_2A_3A_5$ is convex and we have the same contradiction as before.
Finally suppose that the convex hull is a triangle $A_1A_2A_3$. Since $[124]+[234]+[314]=[125]+[235]+[315]$ we have $p_4=p_5$, a contradiction.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf