4. [4] Circle $\omega$ has radius 5 and is centered at $O$. Point $A$ lies outside $\omega$ such that $O A=13$. The two tangents to $\omega$ passing through $A$ are drawn, and points $B$ and $C$ are chosen on them (one on each tangent), such that line $B C$ is tangent to $\omega$ and $\omega$ lies outside triangle $A B C$. Compute $A B+A C$ given that $B C=7$.
Answer: 17. Let $T_{1}, T_{2}$, and $T_{3}$ denote the points of tangency of $A B, A C$, and $B C$ with $\omega$, respectively. Then $7=B C=B T_{3}+T_{3} C=B T_{1}+C T_{2}$. By Pythagoras, $A T_{1}=A T_{2}=\sqrt{13^{2}-5^{2}}=12$. Now note that $24=A T_{1}+A T_{2}=A B+B T_{1}+A C+C T_{2}=A B+A C+7$.