35. Let three circles $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$ with centres $A_{1}, A_{2}, A_{3}$ and radii $r_{1}, r_{2}, r_{3}$ respectively be mutually tangent to each other externally. Suppose that the tangent to the circumcircle of the triangle $A_{1} A_{2} A_{3}$ at $A_{3}$ and the two external common tangents of $\Gamma_{1}$ and $\Gamma_{2}$ meet at a common point $P$, as shown in the figure below. Given that $r_{1}=18 \mathrm{~cm}, r_{2}=8 \mathrm{~cm}$ and $r_{3}=k \mathrm{~cm}$, find the value of $k$.
35. Answer: 12

First we shall show that $r_{3}=\sqrt{r_{1} r_{2}}$. Let $P$ be the point of concurrence of the tangent to the circumcircle of the triangle $A_{1} A_{2} A_{3}$ at $A_{3}$ and the two external common tangents of $\Gamma_{1}$ and $\Gamma_{2}$. Note that the line joining $A_{1}$ and $A_{2}$ also passes through $P$.

On the other hand,
$$
\frac{P A_{3}}{P A_{2}}=\frac{\sin \angle A_{3} A_{2} P}{\sin \angle A_{2} A_{3} P}=\frac{\sin A_{2}}{\sin A_{1}}=\frac{r_{1}+r_{3}}{r_{2}+r_{3}} .
$$

Thus $\sqrt{\frac{r_{1}}{r_{2}}}=\frac{r_{1}+r_{3}}{r_{2}+r_{3}}$. Solving for $r_{3}$, we obtain $r_{3}=\sqrt{r_{1} r_{2}}$. Substituting $r_{1}=18$, $r_{2}=8$, we get $r_{3}=12$.