若abc是⊿ABC的三边长,且满足a=2a2/1+a2,c=2b2/1+b2,a=2c2/1+c2则的面积为

Question

若a b c 是⊿ABC的三边长,且满足a=2a2/1+a2 ,c=2b2/1+b2,a=2c2/1+c2则的面积为

Answer 1

全部取倒数得,1/b=1/2a^2+1/2,1/c=1/2b^2+1/2,1/a=1/2c^2+1/2,三试相加,配方得,1/2(1/a^2-2/a+1)+1/2(1/b^2-2/b+1)+1/2(1/c^2-2/c+1)=0,即1/2(1/a-1)^2+1/2(1/b-1)^2+1/2(1/c-1)^2=0,所以a=b=c=1,所以△ABC是等边三角形,所以△ABC的面积＝1/2*1*(√3/2)＝√3/4