边长为a的正三角形ABC的中心为
设角M’MO=x,角N’NO=y,则有x+60度=y,30度≤x≤60度。
由正弦定理得
b^2[(1/OM)^2+(1/ON)^2]=(OM’/OM)^2+(ON’/ON)^2
=[Sinx/sin(120度)]^2+[Sin(x+60度)/sin(60度)]^2
=4/3[(Sinx)^2+Sin(x+60度)^2]
=4/3[3/4+(1/2)(Sinx)^2+Sin(60度)SinxCosx]
=4/3[3/4+(Sinx)Sin(x+60度)]=4/3[3/4+(Sinx)Sin(x+60度)]
=1+(2/3)[Cos(60度)-Cos(60度+2x)]=4/3-(2/3)Co...全部
设角M’MO=x,角N’NO=y,则有x+60度=y,30度≤x≤60度。
由正弦定理得
b^2[(1/OM)^2+(1/ON)^2]=(OM’/OM)^2+(ON’/ON)^2
=[Sinx/sin(120度)]^2+[Sin(x+60度)/sin(60度)]^2
=4/3[(Sinx)^2+Sin(x+60度)^2]
=4/3[3/4+(1/2)(Sinx)^2+Sin(60度)SinxCosx]
=4/3[3/4+(Sinx)Sin(x+60度)]=4/3[3/4+(Sinx)Sin(x+60度)]
=1+(2/3)[Cos(60度)-Cos(60度+2x)]=4/3-(2/3)Cos(60度+2x)
x=60度时,b^2[(1/OM)^2+(1/ON)^2]取最大值=2,
x+30度时,b^2[(1/OM)^2+(1/ON)^2]取最小值=5/3。
x=60度时,[(1/OM)^2+(1/ON)^2]取最大值=2/b^2,
x+30度时,[(1/OM)^2+(1/ON)^2]取最小值=5/(3b^2)。收起