20/08/02 21:30:44.28 XOQINjwE.net
>>917
AP^2 = BQ^2 = CR^2
= RR {3 - cos(2α) - cos(2β) - cos(2γ) + (√3)[sin(2α) + sin(2β) + sin(2γ)]}
= 4RR(1 + cosα cosβ cosγ + (√3)sinα sinβ sinγ)
= 4RR(1 + cosα cosβ cosγ) + 2(√3)S
ここで S = abc/4R = 2RR sinα sinβ sinγ. (正弦定理)
(AB + BC + CA)^2 = 4RR (sinα + sinβ + sinγ)^2
= 8RR(1+cosα)(1+cosβ)(1+cosγ),