MathLabs

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 4 of 5: Count all pairs
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∑{A,B}qAB=3G,(2n+12)=n(2n+1)≡n(mod2)\sum_{\{A,B\}}q_{AB}=3G,\qquad \binom{2n+1}{2}=n(2n+1)\equiv n\pmod2
Detailed analysis

Let qAB be the number of dividing circles through A and B, and let G be the total number of dividing circles. Each qAB is odd by the preceding step. Summing qAB over all pairs gives 3G because every dividing circle has exactly three pairs of its points. There are binomial(2n+1,2)=n(2n+1) pairs, whose parity is n. Since 3 is odd, G has parity n.