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 3/4 步：译文：Sum over contestants
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通俗地说

The lower 且upper counts refer 到same N N .

N≥a(b−12)2N\ge a\left(\frac{b-1}{2}\right)^2
详细分析

Summing preceding lower bound over 所有 a a 译文： contestants gives N≥a((b−1)/2)2 N\ge a((b-1)/2)^2. Combining both estimates 对于 N N 译文： yields a(b−1)2/4≤kb(b−1)/2 a(b-1)^2/4\le kb(b-1)/2.