yz/(bz+cy)=xz/(cx+az)=xy/(ay+bx)=(x^2+y^2+z^2)/(a^2+b^2+c^2) (abc均不为零)

yz/(bz+cy)=xz/(cx+az)=xy/(ay+bx)=>
1/x(bz+cy)=1/y(cx+az)=/z(ay+bx)
bxz+cxy=cxy+ayz=ayz+bxz =>
bxz=ayz=cxy=>
bx=ay, bz=cy, az=cx
x=a/b*y
z=c/b*y
(x^2+y^2+z^2)/(a^2+b^2+c^2)=(a^2+b^2+c^2)/b^2*y^2/(a^2+b^2+c^2)=(y/b)^2
同理
(x^2+y^2+z^2)/(a^2+b^2+c^2)=(x/a)^2=(z/c)^2
xy/(ay+bx)=a/b*y*y/(ay+b*a/b*y)=y/2b=(y/b)^2
y=b/2
同理x=a/2,z=c/2
故,解为
x=a,y=b,z=c