Question

已知三角形ABC的内切圆为r,角A=60度,BC=6,则r的取值范围是

Answer 1

如图

Answer 2

由均值不等式得AB+AC>=2√(AB*AC),AB*AC<=[(AB+AC)/2]²=(AB+AC)²/4,-3AB*AC>=-3(AB+AC)²/4,
由余弦定理得BC²=AB²+AC²-2AB*AC*cosA,36=AB²+AC²-2AB*AC*cos60°=AB²+AC²-AB*AC
=(AB+AC)²-3AB*AC>=(AB+AC)²-3(AB+AC)²/4=(AB+AC)²/4,
(AB+AC)²/4<=36,6=BC<AB+AC<=12
AB*AC=[(AB+AC)²-36]/3
S△=(1/2)AB*AC*sinA=(1/2)AB*AC*sin60°=(1/2)[(√3)/2]AB*AC=(1/2)(AB+AC+BC)r
r=[(√3)/2]AB*AC/(AB+AC+6)=[(√3)/2]{[(AB+AC)²-36]/3}/(AB+AC+6)=(√3)[(AB+AC)-6]/6,
6<AB+AC<=12,0<(AB+AC)-6<=6,0<(√3)[(AB+AC)-6]/6<=√3,
即r的取值范围是(0,√3]