Question

在△ABC中A+B=3C,其外接圆面积S=2π,求△ABC面积的最大值

Answer 1

R=√2
因为余弦定理：a^2+b^2-2abcosC=c^2
sinA^2+sinB^2-2sinAsinBcosC=sinC^2
(sinA+sinB)^2-2sinAsinB(1+cosC)=sinC^2
因为sinA+sinB=3sinC
sinAsinB=4sinC^2/(1+cosC)=8sin(C/2)^2
因为（sinA+sinB）^2>=4sinAsinB 得到cosC>=7/9,cos(C/2)^2>=8/9

故S=2R^2*sinAsinBsinC=4sinAsinBsinC=64sin(C/2)^3cos(C/2)
令f(C)=S/64
求导f’(C)=1/2*sin(C/2)^2*(3cos(C/2)^2-sin(C/2)^2)=0,在cos(C/2)^2>=8/9 单调递增
故cos(C/2)^2=8/9时取最大值，
Smax=128√2/81