Question

已知等边三角形ABC的边长为A,P是△ABC所在平面内一点,求|PA|^2+|PB|^2+|PC|^2的

Answer 1

设A(xa,ya),B(xb,yb),C(xc,yc),P(xp,yp)
|PA|^2 + |PB|^2 + |PC|^2
= (xa - xp)^2 + (ya - yp)^2 + (xb - xp)^2 + (yb - yp)^2 + (xc - xp)^2 + (yc - yp)^2
=3xp^2 - (xa + xb + xc)xp + 3yp^2 - (ya + yb + yc)yp + (xa^2 + xb^2 + xc^2 + ya^2 + yb^2 + yc^2)
=(xp - (xa + xb + xc)/3)^2 + (yp - (ya + yb + yc)/3)^2 + (xa^2 + xb^2 + xc^2 + ya^2 + yb^2 + yc^2 - (xa + xb + xc)^2/9 - (ya + yb + yc)^2/9)
所以xp = (xa + xb + xc)/3,yp = (ya + yb + yc)/3,P是ABC重心时
|PA|^2 + |PB|^2 + |PC|^2 有最小值
不妨令A(-A/2,0),B(A/2,0),C(0,√3A/2)
P(0,√3/6A)时
|PA|^2 + |PB|^2 + |PC|^2 = 3*(√3A/3)^2 = A^2