Question

In an isosceles triangle, the length of equal sides is $b$ and the base angle $\alpha$ is less than $\frac{\pi }{4}$. Then which of the following is/are true ?

• A. the circumradius of the triangle is $\frac{b}{2}\text{cosec}\alpha$
• B. the inradius of the triangle is $\frac{b\sin 2\alpha }{2(1+\cos \alpha )}$
• C. distance between circumcenter and incenter is $\left|\frac{b\cos \left(3\alpha /2\right)}{2\sin \left(\alpha /2\right)\cos \left(\alpha /2\right)}\right|$
• D. distance between circumcenter and incenter is $\left|\frac{b\cos \left(3\alpha /2\right)}{2\sin \alpha \cos \left(\alpha /2\right)}\right|$

Answer 1

Answer: (A), (B), (D)

The correct options are
A the circumradius of the triangle is $\frac{b}{2}\text{cosec}\alpha$
B the inradius of the triangle is $\frac{b\sin 2\alpha }{2(1+\cos \alpha )}$
D distance between circumcenter and incenter is $\left|\frac{b\cos \left(3\alpha /2\right)}{2\sin \alpha \cos \left(\alpha /2\right)}\right|$


Using sine rule in $△ABC$,
$\frac{b}{\sin \alpha }=2R\Rightarrow R=\frac{b}{2}\text{cosec}\alpha$

Let $r$ be the distance of incenter $\left(I\right)$ from the side $BC$.
$r=\frac{\Delta }{s}=\frac{\frac{1}{2}{b}^2\sin (\pi -2\alpha )}{\frac{1}{2}(b+b+2b\cos \alpha )}$
$=\frac{b\sin 2\alpha }{2(1+\cos \alpha )}$

Distance of circumcenter $\left(O\right)$ from the side $BC$ is,
$|R\cos A|=|R\cos (\pi -2\alpha )|=R\cos 2\alpha$

Hence, distance between circumcenter and incenter is given by
$OI=|r+R\cos 2\alpha |$

$=|\frac{b\sin 2\alpha }{2(1+\cos \alpha )}+\frac{b\text{cosec}\alpha \cos 2\alpha }{2}|$

$=b|\frac{\sin 2\alpha }{4{\cos }^2\left(\alpha /2\right)}+\frac{\cos 2\alpha }{4\sin \left(\alpha /2\right)\cos \left(\alpha /2\right)}|$

$=b\left|\frac{\cos 2\alpha \cos \left(\alpha /2\right)+\sin 2\alpha \sin \left(\alpha /2\right)}{4\sin \left(\alpha /2\right){\cos }^2\left(\alpha /2\right)}\right|$

$=\left|\frac{b\cos \left(3\alpha /2\right)}{2\sin \alpha \cos \left(\alpha /2\right)}\right|$