Notice middle-points of all diagonals and sides of a $2010$-gon. What's the maximum number of those points which can lie on a single circle?
The solution goes like this:
Note point $O$ which is the center of circumcircle. It's obvious that middle-points of all diagonals (or sides) with same length lie on a circle with center $O$. There are 1005 of these circles, and it's obvious that a circle with maximum number of points will be one of these or will contain at most two points from each of them, that is, $2\cdot1005=2010$ in total. Since every circle with center $O$ contains at most 2010 points, and the circle which contains middle-points contains exactly 2010 points, the answer is 2010.
Can someone explain me the part in bold and after that? Also, what would the answer be for $(2k+1)$-gon?