.已知m是正三角形ABC外接圆上的任意已点,求证,|MA|^2+|MB|^2+|MC|^2为定值.

Question

.已知m是正三角形ABC外接圆上的任意已点,求证,|MA|^2+|MB|^2+|MC|^2为定值.

Answer 1

设圆方程为x^2+y^2=R^2,A:(R,0)
则B为(-0.5R,sin(pi/3)Ri),C为(-0.5R,sin(2pi/3)Ri)
设圆上任意一点M为(x0,y0)
有x0^2+y0^2=R^2
|MA|^2+|MB|^2+|MC|^2
=(x0-R)^2+y0^2+(x0+0.5R)^2+(y0-sin(pi/3)R)^2+(x0+0.5R)^2+(y0+sin(pi/3)R)^2
=x0^2+R^2-2Rx0+y0^2+x0^2+0.25R^2+Rx0+y0^2+0.75R^2-sqrt(3)Ry0+x0^2+0.25R^2+Rx0+y0^2+0.75R^2+sqrt(3)Ry0
=3x0^2+3R^2+3y0^2
=6R^2为定值