9. Circles $\omega_{a}, \omega_{b}, \omega_{c}$ have centers $A, B, C$, respectively and are pairwise externally tangent at points $D, E, F$ (with $D \in B C, E \in C A, F \in A B$ ). Lines $B E$ and $C F$ meet at $T$. Given that $\omega_{a}$ has radius 341 , there exists a line $\ell$ tangent to all three circles, and there exists a circle of radius 49 tangent to all three circles, compute the distance from $T$ to $\ell$.
Answer: 294
Solution 1: We will use the following notation: let $\omega$ be the circle of radius 49 tangent to each of $\omega_{a}, \omega_{b}, \omega_{c}$. Let $\omega_{a}, \omega_{b}, \omega_{c}$ have radii $r_{a}, r_{b}, r_{c}$ respectively. Let $\gamma$ be the incircle of $A B C$, with center $I$ and radius $r$. Note that $D E F$ is the intouch triangle of $A B C$ and $\gamma$ is orthogonal to $\omega_{a}, \omega_{b}, \omega_{c}$ (i.e. $I D, I E, I F$ are the common internal tangents). Since $A D, B E, C F$ are concurrent at $T$, we have $K=A B \cap D E$ satisfies $(A, B ; F, K)=-1$, so $K$ is the external center of homothety of $\omega_{a}$ and $\omega_{b}$. In particular, $K$ lies on $\ell$. Similarly, $B C \cap E F$ also lies on $\ell$, so $\ell$ is the polar of $T$ to $\gamma$. Hence $I T \perp \ell$ so if $L$ is the foot from $I$ to $\ell$, we have $I T \cdot I L=r^{2}$.
An inversion about $\gamma$ preserves $\omega_{a}, \omega_{b}, \omega_{c}$ and sends $\ell$ to the circle with diameter $I T$. Since inversion preserves tangency, the circle with diameter $I T$ must be $\omega$. Therefore $I T=98$ by the condition of the problem statement. Letting $a, b, c$ be the radii of $\omega_{a}, \omega_{b}, \omega_{c}$ respectively and invoking Heron's formula as well as $A=r s$ for triangle $A B C$, we see that $\gamma$ has radius
$$
r=\sqrt{\frac{r_{a} r_{b} r_{c}}{r_{a}+r_{b}+r_{c}}} .
$$

We will compute this quantity using Descartes' theorem. Note that there are two circles tangent to $\omega_{a}$, $\omega_{b}, \omega_{c}$, one with radius $I T / 2$ and one with radius $\infty$. By Descartes' circle theorem, we have (where $k_{a}:=1 / a$ is the curvature)
$$
k_{a}+k_{b}+k_{c}+2 \sqrt{k_{a} k_{b}+k_{b} k_{c}+k_{a} k_{c}}=\frac{1}{I T / 2}
$$
and
$$
k_{a}+k_{b}+k_{c}-2 \sqrt{k_{a} k_{b}+k_{b} k_{c}+k_{c} k_{a}}=0,
$$
which implies
$$
\sqrt{\frac{r_{a}+r_{b}+r_{c}}{r_{a} r_{b} r_{c}}}=\sqrt{k_{a} k_{b}+k_{b} k_{c}+k_{c} k_{a}}=\frac{1}{2 I T} .
$$

Therefore $r=2 I T$, which means $I L=\frac{r^{2}}{I T}=4 I T$ and $T L=3 I T=294$.
Solution 2: Using the same notation as the previous solution, note that the point $T$ can be expressed with un-normalized barycentric coordinates
$$
\left(\frac{1}{r_{a}}: \frac{1}{r_{b}}: \frac{1}{r_{c}}\right)
$$
with respect to $A B C$ because $T$ is the Gergonne point of triangle $A B C$. The distance from $T$ to $\ell$ can be expressed as a weighted average of the distances from each of the points $A, B, C$, which is
$$
\frac{1 / r_{a}}{1 / r_{a}+1 / r_{b}+1 / r_{c}} \cdot r_{a}+\frac{1 / r_{b}}{1 / r_{a}+1 / r_{b}+1 / r_{c}} \cdot r_{b}+\frac{1 / r_{c}}{1 / r_{a}+1 / r_{b}+1 / r_{c}} \cdot r_{c}=\frac{3}{1 / r_{a}+1 / r_{b}+1 / r_{c}} .
$$

Note that there are two circles tangent to $\omega_{a}, \omega_{b}, \omega_{c}$, one with radius 49 and one with radius $\infty$. By Descartes' circle theorem, we have (where $k_{a}:=1 / r_{a}$ is the curvature)
$$
k_{a}+k_{b}+k_{c}+2 \sqrt{k_{a} k_{b}+k_{b} k_{c}+k_{a} k_{c}}=\frac{1}{49}
$$
and
$$
k_{a}+k_{b}+k_{c}-2 \sqrt{k_{a} k_{b}+k_{b} k_{c}+k_{c} k_{a}}=0,
$$
so $k_{a}+k_{b}+k_{c}=1 / 98$. The distance from $T$ to $\ell$ is then $3 \cdot 98=294$.
Solution 3: As in the first solution, we deduce that $I T$ is a diameter of $\omega$, with $T$ being the point on $\omega$ closest to $\ell$. By Steiner's porism, we can hold $\omega$ and $\ell$ fixed while making $\omega_{c}$ into a line parallel to $\ell$, resulting in the following figure:
Let $R$ be the common radius of $\omega_{a}$ and $\omega_{b}$ and $r$ be the radius of $\omega$. Notice that $A B D E$ is a rectangle with center $T$, so $R=A E=2 I T=4 r$. The distance from $T$ to $\ell$ is $I L-I T=2 R-2 r=6 r=294$.