Question

已知三角形ABC中，角A,B,C的对角分别为a,b,c,且a^4+b^4+c^4=2c^2(a^2+b^2),则角C等于

Answer 1

c^2=a^2+b^2-2abcosC
c^4=(a^2+b^2-2abcosC)^2
c^4=a^4+b^4+4a^2*b^2*(cosC)^2-4a^3*bcosC-4ab^3cosC+2a^2*b^2
将c^2和c^4分别带入原式
2a^4+2b^4+4a^2*b^2*(cosC)^2-4a^3*bcosC-4ab^3cosC+2a^2*b^2=2*(a^2+b^2-2abcosC)*(a^2+b^2)
2a^4+2b^4+4a^2*b^2*(cosC)^2-4a^3*bcosC-4ab^3cosC+2a^2*b^2=2a^4+2a^2*b^2+2b^4+2a^2*b^2-4a^3*bcosC-4ab^3*cosC
左右相消
4a^2*b^2*(cosC)^2=2a^2*b^2
(cosC)^2=1/2
所以角C等于45度或135度

Answer 2

(a²+b²-c²)²=2a²b²
即a²+b²-c²=±√2ab
亦即cosC=(a²+b²-c²)/2ab=±√2/2
解得C=π/4或3π/4