Question

If the incircle of the triangle $ABC(AB\ne BC\ne CA)$, passes through it's circumcentre, then the $(\cos A+\cos B+\cos C{)}^2$ is

Answer 1

From the given information it is evident that distance between $I$ (incentre) and $O$ (circumcentre) should be equal to inradius of triangle.

If ${}^{\prime }{r}^{\prime }$ and ${}^{\prime }{R}^{\prime }$ be the inradius and circumradius respectivley. then $AI=r\text{cosec}\frac{A}{2}=4R\sin \frac{B}{2}\sin \frac{C}{2}$
and $AO=R,\angle OAI=\frac{A}{2}-(\frac{\pi }{2}-C)=\frac{C-B}{2}$
Now, $I{O}^2=O{A}^2+A{I}^2-2\left(OA\right)\left(AI\right)\cos \frac{C-B}{2}$
$\Rightarrow r^2=R^2+16R^2{\sin }^2\frac{B}{2}{\sin }^2\frac{C}{2}-8R^2\sin \frac{B}{2}\sin \frac{C}{2}\cos \left(\frac{C-B}{2}\right)$
$\Rightarrow r^2=R^2(1+8\sin \frac{B}{2}\sin \frac{C}{2}(2\sin \frac{B}{2}-\cos \left(\frac{C-B}{2}\right)))$
$=R^2(1+8\sin \frac{B}{2}\sin \frac{C}{2}(\sin \frac{B}{2}\sin \frac{C}{2}-\cos \frac{B}{2}\cos \frac{C}{2}))$
$=R^2(1-8\sin \frac{B}{2}\sin \frac{C}{2}\cos \left(\frac{B+C}{2}\right))$
$=R^2(1-8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})$
$=R^2(1-8\frac{r}{4R})=R^2-2rR$
$\Rightarrow {\left(\frac{r}{r}\right)}^2+2\left(\frac{r}{R}\right)-1=0$
$\Rightarrow \frac{r}{R}=\frac{-2\pm \sqrt{4+4}}{2}=\sqrt{2}-1,\text{as}\frac{r}{R}>0$
$\Rightarrow 1+\frac{r}{R}=\sqrt{2}\Rightarrow \cos A+\cos B+\cos C=\sqrt{2}$