09/04/11 16:35:09
>>686 2), >>859 の略証･･･
X = {p+q+(p-q)x}/2, Y = {p+q+(p-q)y}/2, Z = {p+q+(p-q)z}/2,
とおくと
X^2 + Y^2 + Z^2 = (X+Y+Z)^2 -2(XY+YZ+ZX)
= (X+Y+Z-p-q)^2 -(p+q)^2 +2(p+q)(X+Y+Z) -2(XY+YZ+ZX)
= (X+Y+Z-p-q)^2 -(p+q)^2 +3(p+q)^2 -(3/2)(p+q)^2 -(1/2)(p-q)^2･(xy+yz+zx)
= p^2 + q^2 + (X+Y+Z-p-q)^2 -(1/2)(p-q)^2･(xy+yz+zx+1),
ところで、題意から
x = (a+b)(a-b), y = (b+c)/(b-c), z = (c+a)/(c-a),
∴ xy + yz + zx + 1 = 0,