Question

已知ΔABC各条边的长分别为a,b,c,外接圆半径为R,求证：(a^2+b^2+c^2)(1/sin^2A+1/sin^

Answer 1

由正弦定理a/sinA=b/sinB=c/sinC=2R

(a^2+b^2+c^2)(1/sin^2A+1/sin^2B+1/sin^2C)

=(a/sinA)^2+(a/sinB)^2+(a/sinC)^2+(b/sinA)^2+(b/sinB)^2+(b/sinC)^2+(c/sinA)^2+(c/sinB)^2+(c/sinC)^2

=(2R)^2+(a/sinB)^2+(a/sinC)^2+(b/sinA)^2+(2R)^2+(b/sinC)^2+(c/sinA)^2+(c/sinB)^2+(2R)^2

=12R^2+(a/sinB)^2+(a/sinC)^2+(b/sinA)^2+(b/sinC)^2+(c/sinA)^2+(c/sinB)^2

又A+B+C=∏ sinA=sin(B+C) 即sinA>sinB,sinA>sinC 所以a/sinB>2R,a/sinC>2R

同理b/sinA>2R,b/sinC>2R,c/sinA>2R,c/sinB>2R

所以：(a^2+b^2+c^2)(1/sin^2A+1/sin^2B+1/sin^2C)>=36R^2

>=36R^2