18. (USA 5) Inside triangle $A B C$ there are three circles $k_{1}, k_{2}, k_{3}$ each of which is tangent to two sides of the triangle and to its incircle $k$. The radii of $k_{1}, k_{2}, k_{3}$ are 1,4 , and 9 . Determine the radius of $k$.
18. Suppose that circles $k_{1}\left(O_{1}, r_{1}\right), k_{2}\left(O_{2}, r_{2}\right)$, and $k_{3}\left(O_{3}, r_{3}\right)$ touch the edges of the angles $\angle B A C, \angle A B C$, and $\angle A C B$, respectively. Denote also by $O$ and $r$ the center and radius of the incircle. Let $P$ be the point of tangency of the incircle with $A B$ and let $F$ be the foot of the perpendicular from $Q_{1}$ to $O P$. From $\triangle O_{1} F O$ we obtain $\cot (\alpha / 2)=2 \sqrt{r r_{1}} /\left(r-r_{1}\right)$ and analogously $\cot (\beta / 2)=2 \sqrt{\mathrm{T}_{2}} /\left(r-r_{2}\right), \cot (\gamma / 2)=2 \sqrt{\Gamma_{3}} /\left(r-r_{3}\right)$. We will now use a well-known trigonometric identity for the angles of a triangle:
$$
\cot \frac{\alpha}{2}+\cot \frac{\beta}{2}+\cot \frac{\gamma}{2}=\cot \frac{\alpha}{2} \cdot \cot \frac{\beta}{2} \cdot \cot \frac{\gamma}{2} .
$$
(This identity follows from $\tan (\gamma / 2)=\cot (\alpha / 2+\beta / 2)$ and the formula for the cotangent of a sum.)
Plugging in the obtained cotangents, we get
$$
\begin{aligned}
\frac{2 \sqrt{r T_{1}}}{r-r_{1}}+\frac{2 \sqrt{r r_{2}}}{r-r_{2}}+\frac{2 \sqrt{T_{3}}}{r-r_{3}}= & \frac{2 \sqrt{r T_{1}}}{r-r_{1}} \cdot \frac{2 \sqrt{r r_{2}}}{r-r_{2}} \cdot \frac{2 \sqrt{r T_{3}}}{r-r_{3}} \Rightarrow \\
& \sqrt{r_{1}}\left(r-r_{2}\right)\left(r-r_{3}\right)+\sqrt{r_{2}}\left(r-r_{1}\right)\left(r-r_{3}\right) \\
& +\sqrt{r_{3}}\left(r-r_{1}\right)\left(r-r_{2}\right)=4 r \sqrt{r_{1} r_{2} r_{3}} .
\end{aligned}
$$

For $r_{1}=1, r_{2}=4$, and $r_{3}=9$ we get
$$
(r-4)(r-9)+2(r-1)(r-9)+3(r-1)(r-4)=24 r \Rightarrow 6(r-1)(r-11)=0 \text {. }
$$

Clearly, $r=11$ is the only viable value for $r$.