MathLabs

第2题

In a competition, there 是 a a contestants 且 b b judges, 其中 b≥3 b\ge3 是an odd 整数. Each judge rates 每个contestant as either “pass” 或“fail”. 设 k k 是a 数 使得, 对于any two judges, their ratings coincide 对于在most k k contestants. 证明 ka≥b−12b\frac{k}{a}\ge\frac{b-1}{2b}.
第 2/4 步：计数 agreements 对于one contestant
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通俗地说

The two rating groups 是as balanced as possible 在minimum.

(ci2)+(b−ci2)≥(b−12)2\binom{c_i}{2}+\binom{b-c_i}{2}\ge\left(\frac{b-1}{2}\right)^2
详细分析

F或contestant i i 译文：, let ci c_i judges say “pass”. The 数 的agreeing judge pairs 是 (ci2)+(b−ci2)\binom{c_i}{2}+\binom{b-c_i}{2} 译文：. Since b b 是odd, minimum over 整数 ci c_i occurs 在 ci=(b±1)/2 c_i=(b\pm1)/2 且equals ((b−1)/2)2((b-1)/2)^2.