Question

The radii of the escribed circles of triangle $ABC$ are $r_a,$ $r_b$ and $r_c$ respectively. If $r_a+r_b=3R$ and $r_b+r_c=2R$, where $R$ is radius of the circumcircle of triangle $ABC$ and $\angle B=\frac{\pi }{k}$, then the value of $k$ is

Answer 1

We have, $r_a+r_b=3R$
$\Rightarrow \frac{\Delta }{s-a}+\frac{\Delta }{s-b}=\frac{3abc}{4\Delta }\Rightarrow \frac{4{\Delta }^2}{(s-a)(s-b)}=3ab\Rightarrow 4s(s-c)=3ab\Rightarrow (a+b+c)(a+b-c)=3ab\Rightarrow a^2+b^2-c^2=ab\Rightarrow \frac{a^2+b^2-c^2}{2ab}=\frac{1}{2}\Rightarrow \cos C=\frac{1}{2}\Rightarrow C=\frac{\pi }{3}$

Similarly, from $r_b+r_c=2R$
$b^2+c^2-a^2=0\Rightarrow \cos A=0$
$\Rightarrow A=\frac{\pi }{2}$
Hence, $\angle B=\frac{\pi }{6}\Rightarrow k=6$