BMO 2017

Fix a direction in the plane. We cannot have three points in the same line parallel to the direction so suppose that in that direction there are $k$ pairs of points, each pair belonging to a parallel line to the fixed direction. Then there are at most $k-1$ parallelograms of area $1$ formed by these $k$ pairs of points.

Summing over all directions we get that the number of parallelograms of area $1$ are at most $\left(\genfrac{}{}{0ex}{}{n}{2}\right)-s$ where $s$ is the number of different directions. But in that way we count every parallelogram two times, so the number of parallelograms of area $1$ is at most $\frac{\left(\genfrac{}{}{0ex}{}{n}{2}\right)-s}{2}$.

We will prove that $s\ge n$. Indeed, taking the convex hull of the $n$ points, let $x$ be a point on the boundary of the convex hull. Because the convex hull has at least three points on its boundary, we can take two points which are neighbors of $x$ in the convex hull, say $y,z$ these points. Then every segment starting from $x$ has different direction from $yz$. So we have at least $n-1+1=n$ different directions. So the number of parallelograms is at most $$\frac{\left(\genfrac{}{}{0ex}{}{n}{2}\right)-n}{2}=\frac{n^2-3n}{4}$$