, 2010

Suppose that for some integer $n \ge 2$, there exist $n+1$ distinct points in the plane, $X_0, X_1, \dots, X_n$, such that the number of triangles $X_0 X_i X_j$ of area $A$ be greater than $4n\sqrt{n}$. Choose the minimal such $n$, notice that $n \ge 4$, and let $G$ be the graph whose vertices are $X_1, \dots, X_n$ and whose edges are the $X_i X_j$ such that area $X_0 X_i X_j = A$. Then every vertex $X_i$ of $G$ is adjacent to at least $\lfloor 4\sqrt{n} \rfloor$ other vertices, since otherwise, removing $X_i$ would reduce the number of triangles by at most $4\sqrt{n}$, and we would be left with a configuration of $n$ distinct points such that the number of triangles $X_0 X_j X_k$ of area $A$ is at least $4n\sqrt{n} - 4\sqrt{n} > 4(n-1)\sqrt{n}-1$, contradicting the minimal choice of $n$. Consequently, for each $X_i$, there are at least $\lfloor 4\sqrt{n} \rfloor$ points $X_j$ such that the triangle $X_0 X_i X_j$ has area $A$. These points lie on two parallel lines to the line $X_0 X_i$. One of these linear sets of points, say $S_i$, contains at least $\frac{1}{2}\lfloor 4\sqrt{n} \rfloor$ points. Notice that at least $n/2$ of the $S_i$ are pairwise distinct. Without loss of generality, we may (and will) assume that the first $n/2$ of the $S_i$ are among these. Finally, recall that $n \ge 4$, so $\sqrt{n} \le n/2$, and consider the points $X_j$ on the first $\lfloor \sqrt{n} \rfloor$ lines $S_i$, $i = 1, \dots, \lfloor \sqrt{n} \rfloor$, to get
$$\begin{aligned} n \ge |S_1 \cup \dots \cup S_{\lfloor \sqrt{n} \rfloor}| &\ge \sum_{i=1}^{\lfloor \sqrt{n} \rfloor} |S_i| - \sum_{1 \le i < j \le \lfloor \sqrt{n} \rfloor} |S_i \cap S_j| \\ &\ge \frac{1}{2} \lfloor \sqrt{n} \rfloor \lfloor 4\sqrt{n} \rfloor - \binom{\lfloor \sqrt{n} \rfloor}{2}, \end{aligned}$$
which is false for $n \ge 4$.