正多边形的最优分割问题

记平面边长为1的正m边形为S_m,将S_m剖分成n块:S_(m1),S_(m2),…,S_(mn),这样的剖分称S_m的n剖分,并以T(m,n)表示.以d_(mi)表示区域S_(mi)(i=1,2,…,n)的直径(即区域S_(mi)任意两点之间距离的最大者).记D(m,n)=max{d_(m1),d_(m2),…,d_(mn)}及Ψ(m,n)=■{D(m,n)}.本文将估计Ψ(m,n)的上下界.证明Ψ(6,3)=3/2,Ψ(6,4)=3-3~(1/2),Ψ(6.6)=1,Ψ(6,7)=3/2,估计Ψ(6,n)的渐进性.提出几个猜想.

The Problem of Best Cut of Regular Polygon

Let S_m be a regular polygon that has a unit side length.Divide S_m into n regions. denoted by S_(m1),S_(m2),…,S_(mn) respectively.This is called the n-division of S_m and denoted by as T(m,n).Define d_(mi) as the diameter of S_(mi)(i=1,2,…,n),that is,the maximum distance between two arbitrary points on S_(mi),let D(m,n)=max{d_(m1),d_(m2),…,d_(mn)} andΨ(m,n)= ■ {D(m,n)}.This paper estimates the upper and lower bound ofΨ(m,n),estimates asymptotic properties ofΨ(6,n),and prove thatΨ(6,3)=3/2,Ψ(6,4)=3-3~(1/2),Ψ(6,6)= 1,Ψ(6,7)=3/2.