Suppose there are $2n+3$ points such that no $3$ points are collinear and no $4$ points are concyclic. It is required to show that one can find at least one circle passing through $3$ points such that it has $n$ points inside it and $n$ points outside it. How to prove this?
I could do nothing significant in this problem. So any help will be appreciated.
HINT: Pick two points, $P$ and $Q$, such that all of the other $2n+1$ points are on the same side of the line $\overline{PQ}$. Let $\ell$ be the perpendicular bisector of the segment $PQ$, and let $R$ be the point of intersection of $\ell$ and that segment. Now imagine $R$ moving slowly away from the segment along $\ell$ on the side of $\overline{PQ}$ that does not contain the other $2n+1$ points. As soon as $R$ moves off $PQ$, the three points $P,Q$, and $R$ determine a circle. Initially that circle contains all $2n+1$ of the other points. (Why?) When $R$ has moved far enough, it contains none of them. (Why?) It never passes through more than one of them at a time. (Why?)