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Geometry Difficulty 4.8 AIME Prove it Hong Kong

There are 212212 points inside or on a circle with radius 11. Prove that there are at least 20012001 pairs of these points having distances at most 11.

Solution

Let A,B,C,D,E,FA, B, C, D, E, F be 66 distinct points on the circle such that
AB=BC=CD=DE=EF=FA. \overline{AB} = \overline{BC} = \overline{CD} = \overline{DE} = \overline{EF} = \overline{FA}.
Let OO be the centre of the circle. Note that any two points in the same sector among AOB,BOC,COD,DOE,EOF,FOAAOB, BOC, COD, DOE, EOF, FOA have distance at most 11. Let n1,n2,,n6n_1, n_2, \dots, n_6 be the number of points in the 66 sectors respectively. It suffices to show
(n12)+(n22)++(n62)2001. \binom{n_1}{2} + \binom{n_2}{2} + \dots + \binom{n_6}{2} \ge 2001.
Indeed, since the binomial function (x2)\binom{x}{2} is convex, by Jensen's inequality, we have
k=16(nk2)6(n1+n2++n66)212=6(352)=3570>2001. \sum_{k=1}^{6} \binom{n_k}{2} \ge 6 \left( \frac{n_1+n_2+\dots+n_6}{6} \right)^2 \frac{1}{2} = 6 \binom{35}{2} = 3570 > 2001.

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