Problem 5
A set of 2n+1 points in the plane has no three collinear and no four concyclic. A circle divides the set if it passes through 3 of the points and has exactly n-1 points inside it. Prove that the number of circles dividing the set is even if and only if n is even.
Step 3 of 5: Show an odd number of the circles divide the set
Detailed analysis
A circle Ci divides the set exactly when fi=n-1. As the ordered sequence moves, a step across the level n-1 changes the parity of the number of indices on that level only when it enters or leaves the level. If a constant run lies on that level, the side labels Xi alternate at its two ends, so the run has odd length. Comparing the first and last circle (and treating an endpoint already on the level by the same one-sided argument) shows that the sequence starts and ends on opposite sides in this parity sense. Hence the number of indices with fi=n-1 is odd. Thus, for every pair A,B, an odd number of dividing circles pass through it.