当三角形PHG为等腰三角形时
设NGH;PGM;OGQ分别是OP;OH;PH的中线,
(1)PH=GH时
HG=2/3*HN=2/3*1/2OP=3=PH
(2)PG=PH时
设PH=PG=X
PM=3X/2
MH=√(9X^2/4-X^2)=√5X/2
OH=2MH=√5X
OP^2=PH^2+OH^2
81=X^2+5X^2
X=3√6/2
(3)PG=GH时
HN=PM=1/2OP=9/2
设PH=X
MH=√(PM^2-PH^2)=√(81/4-X^2)
OH=2MH=√(81-4X^2)=√(OP^2-PH^2)=√(81-X^2)
X=0
显然不满足题意,应该舍去.
所以PH=3或3√6/2