Suppose there be $2n+3$ points in a plane so that no 4 lie on a circle. Then there exists a circle through 3 points such that $n$ points lie inside the circle and the rest $n$ points lie outside that circle.
I solved it for five points. But can anyone suggest any idea to solve this problem by pigeonhole principle or recurrence? Thanks in advance.
In $2n+3$ points, we can always choose two points say $A$ and $B$ so that all the 2n+1 points lie on one side of AB. Label the other $2n+1$ points as $P_1, P_2, ..., P_{2n+1}$. Without loss of generality, we can assume $$\angle AP_1B < \angle AP_2B < \angle AP_3B ...<\angle AP_nB < \angle AP_{n+1}B <\angle AP_{n+2}B <...\angle AP_{2n+1}B.$$ We choose $A, B, P_{n+1}$ as three points through which the circle passes. $P_1, P_2, ...P_n$ lie inside the circle and $P_{n+2}, P_{n+3}, ...P_{2n+1}$ lie outside the circle.