Consider the set $G$ of $2011^2$ points $(x, y)$ in the plane where $x$ and $y$ are both integers between $ 1$ and $2011$ inclusive. Let $A$ be any subset of $G$ containing at least $4\times 2011\times \sqrt{2011}$ points. Show that there are at least $2011^2$ parallelograms whose vertices lie in $A$ and all of whose diagonals meet at a single point.