一个圆内接六边形有三边均为3,另三边长均为5,圆的一条弦将它分成二个四边形,其中一个四边形三边长为3,

设半径为R
则6×arcsin[3/(2R)]+6×arcsin[5/(2R)] = 2π
arcsin[3/(2R)]+arcsin[5/(2R)] = π/3
cos{arcsin[3/(2R)]+arcsin[5/(2R)] }=1/2
√{1-[3/(2R)]^2}*√{1-[5/(2R)]^2} - 3/(2R)*5/(2R) = 1/2
√{1-[3/(2R)]^2}*√{1-[5/(2R)]^2} = 15/(4R^2) + 1/2
{1-[3/(2R)]^2}*{1-[5/(2R)]^2} = [15/(4R^2)]^2 + 15/(4R^2) + 1/4
1 - [3/(2R)]^2 - [5/(2R)]^2 + [15/(4R^2)]^2= [15/(4R^2)]^2 + 15/(4R^2) + 1/4
(15+9+25)/(4R^2)= 3/4
7/(2R)=√3/2
R=7√3/3
M/N = 2Rsin{3arcsin[3/(2R)]}
= 2R{3sinarcsin[3/(2R)]-4[sinarcsin[3/(2R)]]^3}
= 2R{3*3/(2R)-4[3/(2R)]^3}
= 9 - 27/R^2
= 9 - 81/49
= 360/49
M+N=409
A正确