回答: 接着前面的思路想了一下 由 强生 于 2017-03-29 2:31

接上面，n=3，
等边三角形三点在圆上，用三角函数可得
G3=(2sin(pi/3))^3
若任意一点沿着圆周稍微一动一下，则有
G3'= (sin(pi/3+x)+sin(pi/3))*(sin(pi/3-x)+sin(pi/3)*2sin(pi/3),
where x is the change in angle resulted by moving of the corner of the circle， pi/3>x>-pi/3
G3' 化简得
(cos(pi/6-x)+sqrt(3)/2)(cos(pi/6+x)+3/2)
不难算的G3'的maximum在pi/3>x>-pi/3区间，x=0
同理，可以在G3->G3'基础上generalize到三点沿着圆周移动x，y，z
计算会复杂，但结果不难想象，max G3'''(x,y,z) happens at x=y=z=0;

for general n
Gn=(2sin(pi/n))^n

generalised Gn'(x1,x2...x_n) = (sin(pi/n+x1)+sin(pi/n-x_n))*(sin(pi/n+x2)+sin(pi/n-x1))+....(sin(pi/n+x_n)+sin(pi/n+x_{n-1}))
= prod_{i} (sin(pi/n+x_i)+sin(pi/n-x_j))
and the product uses periodic boundary condition, i.e. x_(n+1)=x1

max Gn' happens at x1=x2=...x_n=0