# How to solve geometry question on internal tangency

Let $\Gamma_1$ be a circle with centre at the Point $O$ and radius $R$. Two other circles $\Gamma_2$ and $\Gamma_3$ with centres $O_2$ and $O_3$ respectively are internally tangent to $\Gamma_1$ and meet each other at Points $A$ and $B$. Find the sum of the radii of $\Gamma_2$ and $\Gamma_3$, given that angle $OAB=90^{\circ}$.

## 1 Answer

The sum of radii of $\Gamma_2$ and $\Gamma_3$ is that of $\Gamma_1$. Let $C_2$ and $C_3$ be the points of intersection of $OA$ and the circles $O_2$ and $O_3$ respectively. We have that $C_2B$ and $C_3B$ are the diameters of circles $O_2$ and $O_3$ respectively, and $O_2$ and $O_3$ are their respective mid-points. So $O_2O_3$ is parallel to $C_2C_3$ and thus perpendicular to $AB$. Moreover $O_2O_3$ bisects $AB$. Consider the triangles $BO_2O_3$ and $OO_3O_2$. They share the common base $O_2O_3$ and by the above have the same heights. Moreover, by the internal tangency condition, $OO_3+R_3=OO_3+O_3B=OO_2+O_2B=OO_2+R_2$. So $O_2B-O_3B=OO_3-OO_2$. This forces $O_2B=OO_3$ and $O_3B=OO_2$, and $R_2+R_3=O_2B+O_3B=OO_3+O_3B=R$.

• Do you mind showing how, @Alex Fok ? Thanks! – WilliamKin May 20 '15 at 11:26
• Please see the updated version of my answer. – Alex Fok May 20 '15 at 11:45