0
$\begingroup$

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 $OAB=90^{\circ}$.

$\endgroup$
0
$\begingroup$

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 $R_2+R_3=O_2B+O_3B=OO_3+O_3B=R$.

$\endgroup$
2
  • $\begingroup$ Do you mind showing how, @Alex Fok ? Thanks! $\endgroup$ – WilliamKin May 20 '15 at 11:26
  • $\begingroup$ Please see the updated version of my answer. $\endgroup$ – Alex Fok May 20 '15 at 11:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.