Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. Let $r$ and $R$ be the inradius and the circumradius of the triangle, then the distance between the circumcenter and the incenter of $\Delta ABC$ is

• A. $\sqrt{R^2-2Rr}$
• B. $\sqrt{R^2-Rr}$
• C. $\sqrt{R^2-2R}$
• D. $\sqrt{R^2-2r}$

Answer 1

The correct option is A $\sqrt{R^2-2Rr}$

Let $O$ be the circumcenter and $OF$ be the perpendicular to $AB$

Let $I$ be the incenter and $IE$ be the perpendicular to $AC$

Then,$\angle OAF={90}^{\circ }-C$ $\Rightarrow \angle OAI=\angle IAF-\angle OAF$ $=\frac{A}{2}-({90}^{\circ }-C)=\frac{A}{2}+C-\frac{A+B+C}{2}=\frac{C-B}{2}$

Also,$AI=\frac{IE}{\sin \frac{A}{2}}=\frac{r}{\sin \frac{A}{2}}=4R\sin \frac{B}{2}\sin \frac{C}{2}$
Hence in $\Delta OAI,O{I}^2=O{A}^2+A{I}^2-2OAAI\cos \angle OAI$ $=R^2+16R^2{\sin }^2\frac{B}{2}{\sin }^2\frac{C}{2}-8R^2\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{C-B}{2}$ $\Rightarrow \frac{O{I}^2}{R^2}=1+16{\sin }^2\frac{B}{2}{\sin }^2\frac{C}{2}-8\sin \frac{B}{2}\sin \frac{C}{2}(\cos \frac{B}{2}\cos \frac{C}{2}+\sin \frac{B}{2}\sin \frac{C}{2})$ $=1-8\sin \frac{B}{2}\sin \frac{C}{2}(\cos \frac{B}{2}\cos \frac{C}{2}-\sin \frac{B}{2}\sin \frac{C}{2})$
$=1-8\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{B+C}{2}$ $=1-8\sin \frac{B}{2}\sin \frac{C}{2}\sin \frac{A}{2}$ $\therefore OI=R\sqrt{1-8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}=\sqrt{R^2-2Rr}$