Question

在△ABC中，∠A＝120°，BC＝6，若△ABC的内接圆半径为r，求r的最大值。

Answer 1

r(a+b+c)/2=S
a^2=b^2+c^2-bccosA=b^2+c^2+bc
S=bcsinA
a/sinA=b/sinB=c/sinC=k
S=bcsinA/2=k^2sinBsinCsinA/2
a+b+c=(sinA+sinB+sinC)k
r=k(sinAsinBsinC)/(sinA+sinB+sinC)=5sinBsinC/(sinA+sinB+sinC)
sinB+sinC=2sin[(B+C)/2]cos[(B-C)/2]=cos[(B-C)/2]
sinBsinC= [-cos(B+C)+cos(B-C)]/2=[cos(B-C)-1/2]/2
r=5*[cos(B-C)-1/2]/[√3+cos[(B-C)/2]],
B=C时，r最大值=5*(1/2)/(√3+1)=5(√3-1)/4