■三角形の相似(その4)
cos^2α+cos^2β+cos^2γ
=a^2(b^2+c^2−a^2)^2/4a^2b^2c^2+b^2(c^2+a^2−b^2)^2/4b^2c^2a^2+c^2(a^2+b^2−c^2)^2/4c^2a^2b^2
a^2(b^2+c^2−a^2)^2
=a^2(b^2+c^2)^2−2a^4(b^2+c^2)+a^6
b^2(c^2+a^2−b^2)^2
=b^2(c^2+a^2)^2−2b^4(c^2+a^2)+b^6
=b^2(c^4+2c^2a^2+a^4)−2a^2b^4−2c^2b^4+b^6
=b^2a^4+2a^2(b^2c^2−b^4)+b^2c^4−2b^4c^2+b^6
c^2(a^2+b^2−c^2)^2
=c^2(a^2+b^2)^2−2c^4(a^2+b^2)+c^6
=c^2a^4+2a^2(b^2c^2−c^4)+b^4c^2−2b^2c^4+c^6
これらの和は
a^6−a^4(b^2+c^2)+a^2(b^2+c^2)^2+a^2(2b^2c^2−2b^4)+a^2(2b^2c^2−2c^4)−b^2c^2(b^2+c^2)+b^6+c^6
=a^6−a^4(b^2+c^2)+a^2(−b^4+−c^4+6b^2c^2)−b^2c^2(b^2+c^2)+b^6+c^6
{1−(cos^2α+cos^2β+cos^2γ)}・4a^2b^2c^2
=4a^2b^2c^2−a^2(b^2+c^2−a^2)^2−b^2(c^2+a^2−b^2)^2−c^2(a^2+b^2−c^2)^2
=4a^2b^2c^2−a^6+a^4(b^2+c^2)+a^2(b^4+c^4−6b^2c^2)+b^2c^2(b^2+c^2)−b^6−c^6
=−a^6+a^4(b^2+c^2)+a^2(b^4+c^4−2b^2c^2)+b^2c^2(b^2+c^2)−b^6−c^6
=−a^6−b^6−c^6+a^4b^2+a^4c^2+a^2b^4+a^2c^4+b^4c^2+b^2c^4−2a^2b^2c^2
=−2a^6−2b^6−2c^6+a^4(a^2+b^2+c^2)+b^4(a^2+b^2+c^2)+c^4(a^2+b^2+c^2)−2a^2b^2c^2
=−2a^6−2b^6−2c^6+(a^4+b^4+c^4)(a^2+b^2+c^2)−2a^2b^2c^2
もっときれいに整理できると思われるのであるが,・・・
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