Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it Ireland

Given a set of 2016 distinct points in the plane, show that we can choose a "circle of evil" CC in the plane such that exactly 666 of these points lie strictly inside CC, and none of them lies on CC.

Solution

Let SS denote the set of the 2016 given points. Consider the collection of perpendicular bisectors of pairs of distinct points in SS. This is a finite collection of lines, so we can pick a point PP not on any one of these lines, and also not in SS.

For r0r \ge 0, let N(r)N(r) be the number of points in SS whose distance from PP is at most rr. By construction, the circle C(r)C(r) of radius rr about PP contains at most one point of SS for each rr, so N(r)N(r) increases from 0 for small rr to 2016 for very big rr, and each increase occurs as a jump of 1. Thus, there are numbers
0=r0<r1<r2<<r2016<r2017= 0 = r_0 < r_1 < r_2 < \dots < r_{2016} < r_{2017} = \infty
such that if 0n20160 \le n \le 2016, then N(r)=nN(r) = n exactly when rnr<rn+1r_n \le r < r_{n+1}. The circle C(r)C(r) has the desired properties if we pick r666<r<r667r_{666} < r < r_{667}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.