Thread: Considering a set 'G' proove the number of possible parallelograms

Hello, struggling sooo much with this question
I hav made an attempt but honestly dont know how to prove the actually 2011^2


Consider the set

G of 2011^2 points (p, q) in the plane where p and q are both integers

between 1 and 2011 inclusive. Let

A be any subset of G containing at least 4×2011×square root of 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.

Help would b great.
Cheers
I apologise for forgetting to put the 'to the power of' symbols in.

i sat down to try this question again, ive seen similar equations but just can't apply the methods, this one seems slightly different. Any help?