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Geometry Difficulty 4.6 AIME Find the answer United States

Problem:

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 $100 a + b$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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 $A X = 20 - \frac{133}{10} = \frac{67}{10}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.