7. Let $\Gamma$ be a circle, and $\omega_{1}$ and $\omega_{2}$ be two non-intersecting circles inside $\Gamma$ that are internally tangent to $\Gamma$ at $X_{1}$ and $X_{2}$, respectively. Let one of the common internal tangents of $\omega_{1}$ and $\omega_{2}$ touch $\omega_{1}$ and $\omega_{2}$ at $T_{1}$ and $T_{2}$, respectively, while intersecting $\Gamma$ at two points $A$ and $B$. Given that $2 X_{1} T_{1}=X_{2} T_{2}$ and that $\omega_{1}, \omega_{2}$, and $\Gamma$ have radii 2,3 , and 12 , respectively, compute the length of $A B$.
Answer: $\frac{96 \sqrt{10}}{13}$

Solution 1: Let $\omega_{1}, \omega_{2}, \Gamma$ have centers $O_{1}, O_{2}, O$ and radii $r_{1}, r_{2}, R$ respectively. Let $d$ be the distance from $O$ to $A B$ (signed so that it is positive if $O$ and $O_{1}$ are on the same side of $A B$ ).

Note that
$$
\begin{array}{r}
O O_{i}=R-r_{i}, \\
\cos \angle T_{1} O_{1} O=\frac{O_{1} T_{1}-O M}{O O_{1}}=\frac{r_{1}-d}{R-r_{1}}, \\
\cos \angle T_{2} O_{2} O=\frac{O_{2} T_{2}+O M}{O O_{1}}=\frac{r_{2}+d}{R-r_{2}} .
\end{array}
$$

Then
$$
\begin{aligned}
X_{1} T_{1} & =r_{1} \sqrt{2-2 \cos \angle X_{1} O_{1} T_{1}} \\
& =r_{i} \sqrt{2+2 \cos \angle T_{1} O_{1} O} \\
& =r_{1} \sqrt{2+2 \frac{r_{1}-d}{R-r_{1}}} \\
& =r_{1} \sqrt{2 \frac{R-d}{R-r_{1}}} .
\end{aligned}
$$

Likewise,
$$
X_{2} T_{2}=r_{2} \sqrt{2 \frac{R+d}{R-r_{2}}} .
$$

From $2 X_{1} T_{1}=X_{2} T_{2}$ we have
$$
8 r_{1}^{2}\left(\frac{R-d}{R-r_{1}}\right)=4 X_{1} T_{1}^{2}=X_{2} T_{2}^{2}=2 r_{2}^{2}\left(\frac{R+d}{R-r_{2}}\right) .
$$

Plugging in $r_{1}=2, r_{2}=3, R=12$ and solving yields $d=\frac{36}{13}$. Hence $A B=2 \sqrt{R^{2}-d^{2}}=\frac{96 \sqrt{10}}{13}$.
Solution 2: We borrow the notation from the previous solution. Let $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect $\Gamma$ again at $M_{1}$ and $M_{2}$. Note that, if we orient $A B$ to be horizontal, then the circles $\omega_{1}$ and $\omega_{2}$ are on opposite sides of $A B$. In addition, for $i \in\{1,2\}$ there exist homotheties centered at $X_{i}$ with ratio $\frac{R}{r_{i}}$ which send $\omega_{i}$ to $\Gamma$. Since $T_{1}$ and $T_{2}$ are points of tangencies and thus top/bottom points, we see that $M_{1}$ and $M_{2}$ are the top and bottom points of $\Gamma$, and so $M_{1} M_{2}$ is a diameter perpendicular to $A B$.

Now, note that through power of a point and the aforementioned homotheties,
$$
P\left(M_{1}, \omega_{1}\right)=M_{1} T_{1} \cdot M_{1} X_{1}=X_{1} T_{1}^{2}\left(\frac{R}{r_{1}}\right)\left(\frac{R}{r_{1}}-1\right)=30 X_{1} T_{1}^{2},
$$
and similarly $P\left(M_{2}, \omega_{2}\right)=12 X_{2} T_{2}^{2}$. (Here $P$ is the power of a point with respect to a circle). Then
$$
\frac{P\left(M_{1}, \omega_{1}\right)}{P\left(M_{2}, \omega_{2}\right)}=\frac{30 X_{1} T_{1}^{2}}{12 X_{2} T_{2}^{2}}=\frac{30}{12(2)^{2}}=\frac{5}{8} .
$$

Let $M$ be the midpoint of $A B$, and suppose $M_{1} M=R+d$ (here $d$ may be negative). Noting that $M_{1}$ and $M_{2}$ are arc bisectors, we have $\angle A X_{1} M_{1}=\angle T_{1} A M_{1}$, so $\triangle M_{1} A T_{1} \sim \triangle M_{1} X_{1} A$, meaning that $M_{1} A^{2}=M_{1} T_{1} \cdot M_{1} X_{1}=P\left(M_{1}, \omega_{1}\right)$. Similarly, $\triangle M_{2} A T_{2} \sim \triangle M_{2} X_{2} A$, so $M_{2} A^{2}=P\left(M_{2}, \omega_{2}\right)$. Therefore,
$$
\frac{P\left(M_{1}, \omega_{1}\right)}{P\left(M_{2}, \omega_{2}\right)}=\frac{M_{1} A^{2}}{M_{2} A^{2}}=\frac{\left(R^{2}-d^{2}\right)+(R+d)^{2}}{\left(R^{2}-d^{2}\right)+(R-d)^{2}}=\frac{2 R^{2}+2 R d}{2 R^{2}-2 R d}=\frac{R+d}{R-d}=\frac{5}{8},
$$
giving $d=-\frac{3}{13} R$. Finally, we compute $A B=2 R \sqrt{1-\left(\frac{3}{13}\right)^{2}}=\frac{8 R \sqrt{10}}{13}=\frac{96 \sqrt{10}}{13}$.