Question

In $\Delta ABC$ let $R$ = circumradius, $r$= inradius. If $r$ is the distance between the circumcenter and the incenter, then ratio $R/r$ is equal to

• A. $\sqrt{2}-1$
• B. $\sqrt{3}-1$
• C. $\sqrt{2}+1$
• D. $\sqrt{3}+1$

Answer 1

The correct option is C $\sqrt{2}+1$

$r=4R\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}$
$\frac{r}{R}=4\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}$
$=2[\cos \left(\frac{A-B}{2}\right)-\cos (\frac{A+B}{2}\left)\right].\sin \frac{C}{2}$
$=2[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}]\sin \frac{C}{2}$
$=2\cos \left(\frac{A-B}{2}\right)\sin \frac{C}{2}-2{\sin }^2\frac{C}{2}$
$=2\cos \left(\frac{A-B}{2}\right)\cos \frac{A+B}{2}-2{\sin }^2\frac{C}{2}$
$=\cos A+\cos B-(1-\cos C)$
$=\cos A+\cos B+\cos C-1$
Hence it is entirely dependent on the angles of the triangle.
Considering $B={90}^0$ and $A=C$
We get
$\frac{r}{R}=2\cos {45}^0-1$
$=\sqrt{2}-1$
Thus
$\frac{R}{r}=\frac{1}{\sqrt{2}-1}$
$=\sqrt{2}+1$.