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The Number of Halving Circles

For the rest of sections 1 and 2, n is a fixed positive integer and S signifies an arbitrary set of 2n + 1 points in general position in the plane. There are several solutions to Problem 1. One possible approach is the following. Let A and B be two consecutive vertices of the convex hull of S. We claim that some circle going through A and B is halving. All circles through A and B have their centers on the perpendicular bisector e of the segment A B. Pick a point O on f that lies on the same side of A B as S and is sufficiently far away from A B that the circle F with center O and passing through A and B completely contains S. This can clearly be done. Now slowly "push" O along e, moving it towards AB. The circle F changes continuously with O. As we do this, F stops containing some points of S. In fact, it loses the points of S one at a time: if it lost P and Q simultaneously, then points P, Q, A, and B would be concyclic. We can move O sufficiently far away past AB that, in the end, the circle does not contain any points of S. Originally, F contained all the points of S. Now, as it loses one point of S at a time in this process, we can decide how many points we want it to contain. In particular, if we stop moving O when the circle is about to lose the nth point P of S, then the