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.
I apologise for forgetting to put the 'to the power of' symbols in.