18. [12] Triangle $A B C$ has side lengths $A B=19, B C=20$, and $C A=21$. Points $X$ and $Y$ are selected on sides $A B$ and $A C$, respectively, such that $A Y=X Y$ and $X Y$ is tangent to the incircle of $\triangle A B C$. If the length of segment $A X$ can be written as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers, compute $100 a+b$.
Answer: 6710
Solution: Note that the incircle of $\triangle A B C$ is the $A$-excenter of $\triangle A X Y$. Let $r$ be the radius of this circle. We can compute the area of $\triangle A X Y$ in two ways:
$$
\begin{aligned}
K_{A X Y} & =\frac{1}{2} \cdot A X \cdot A Y \sin A \\
& =r \cdot(A X+A Y-X Y) / 2 \\
\Longrightarrow A Y & =\frac{r}{\sin A}
\end{aligned}
$$

We also know that
$$
\begin{aligned}
K_{A B C} & =\frac{1}{2} \cdot 19 \cdot 21 \sin A \\
& =r \cdot(19+20+21) / 2 \\
\Longrightarrow \frac{r}{\sin A} & =\frac{19 \cdot 21}{60}=\frac{133}{20}
\end{aligned}
$$
so $A Y=133 / 20$.
Let the incircle of $\triangle A B C$ be tangent to $A B$ and $A C$ at $D$ and $E$, respectively. We know that $A X+A Y+X Y=A D+A E=19+21-20$, so $A X=20-\frac{133}{10}=\frac{67}{10}$