三角形ABC是圆O的内接正三角形,P是圆O上任一点,则PA＾＋PB＾＋PC＾是否为定值?为什么?谢谢

Question

三角形ABC是圆O的内接正三角形,P是圆O上任一点,则PA＾＋PB＾＋PC＾是否为定值?为什么?谢谢

Answer 1

原题是不是PA²＋PB²＋PC²
设o为原点,等边三角形A在y轴上,可得
P(rcosa.rsina)
A(0.r)
B(-√3r/2,-r/2)
C(√3r/2,-r/2)
AP=(rcosa,rsina-r)
BP=(rcosa+√3r/2,rsina+r/2)
CP=(rcosa-√3r/2,rsina+r/2)
AP²+BP²+CP²=(rcosa)²+(rsina-r)²+(rcosa+√3r/2)²+(rsina+r/2)²+(rcosa-√3r/2)²+(rsina+r/2)²
=(rcosa)²+(rsina)²-2r²sina+r²+2(rcosa)²+3r²/2+2(rsina)²+2r²sina+r²/2
=r²-2r²sina+r²+2r²+2r²+2r²sina
=6r²
r为定值,所以PA²+PB²+PC²为定值