题目：
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设
Δn=maxA1,…,An∈R2maxi<jd(Ai,Aj)≤1(mini<jd(Ai,Aj))\Delta_n = \max_{\substack{A_1,\dots,A_n \in \mathbb{R}^2 \\ \max_{i<j} d(A_i, A_j) \le 1}} \left( \min_{i<j} d(A_i, A_j) \right) Δn =A1 ,…,An ∈R2maxi<j d(Ai ,Aj )≤1 max (i<jmin d(Ai ,Aj ))
即在任意两点间距离不超过 1 的条件下，nnn 个点的“最近距离”能取得的最大值。
构造：正 nnn 边形
将 nnn 个点构造成单位直径的正 nnn 边形，即取正 nnn 边形的顶点，使得其直径为 1，此时边长（即最小距离）为所求最大值。
* 若 nnn 为偶数：
2Rsin(⌊n/2⌋πn)=2Rsin(n/2⋅πn)=2Rsin(π2)=2R⋅1=2R2R \sin\left(\frac{\lfloor n/2 \rfloor \pi}{n}\right) = 2R \sin\left(\frac{n/2 \cdot \pi}{n}\right) = 2R \sin\left(\frac{\pi}{2}\right) = 2R \cdot 1 = 2R 2Rsin(n⌊n/2⌋π )=2Rsin(nn/2⋅π )=2Rsin(2π )=2R⋅1=2R
要求直径为 1，得 R=12R = \frac{1}{2}R=21 。此时边长为：
δn=2Rsin(πn)=sin(πn)\delta_n = 2R \sin\left(\frac{\pi}{n}\right) = \sin\left(\frac{\pi}{n}\right) δn =2Rsin(nπ )=sin(nπ )
* 若 nnn 为奇数：
两最远顶点间夹角为 (n−1)πn=π−πn\frac{(n-1)\pi}{n} = \pi - \frac{\pi}{n}n(n−1)π =π−nπ ，最大距离为：
2Rsin((n−1)π2n)=2Rcos(π2n)2R \sin\left( \frac{(n-1)\pi}{2n} \right) = 2R \cos\left( \frac{\pi}{2n} \right) 2Rsin(2n(n−1)π )=2Rcos(2nπ )
要求其为 1，得：
R=12cos(π/(2n))R = \frac{1}{2\cos(\pi/(2n))} R=2cos(π/(2n))1
此时边长为：
δn=2Rsin(πn)=2sin(π/n)2cos(π/(2n))=2sin(π2n)\delta_n = 2R \sin\left(\frac{\pi}{n}\right) = \frac{2\sin(\pi/n)}{2\cos(\pi/(2n))} = 2\sin\left(\frac{\pi}{2n}\right) δn =2Rsin(nπ )=2cos(π/(2n))2sin(π/n) =2sin(2nπ )
最终结论：
Δn={sin(πn),若 n 为偶数2sin(π2n),若 n 为奇数\boxed{ \Delta_n = \begin{cases} \sin\left( \frac{\pi}{n} \right), & \text{若 } n \text{ 为偶数} \\[1em] 2\sin\left( \frac{\pi}{2n} \right), & \text{若 } n \text{ 为奇数} \end{cases} } Δn =⎩⎨⎧ sin(nπ ),2sin(2nπ ), 若 n 为偶数若 n 为奇数
