[This puzzle was given to me by Arvind Hariharan a long time back]
The puzzle is easy to state:
Given $2n+3$ points in the general position (no three collinear, no four concyclic) in the 2D plane, show that there are three points among those, such that the circle through those points has exactly $n$ of the remaining $2n$ points inside the circle (and exactly $n$ outside).
[Solution]
The puzzle is easy to state:
Given $2n+3$ points in the general position (no three collinear, no four concyclic) in the 2D plane, show that there are three points among those, such that the circle through those points has exactly $n$ of the remaining $2n$ points inside the circle (and exactly $n$ outside).
[Solution]
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