Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Find the answer

Triangle $ABC$ has side lengths $AB=19, BC=20$ , and $CA=21$ . Points $X$ and $Y$ are selected on sides $AB$ and $AC$ , respectively, such that $AY=XY$ and $XY$ is tangent to the incircle of $\triangle ABC$ . If the length of segment $AX$ 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. Spacing and $ signs are ignored.

Solution

Note that the incircle of $\triangle ABC$ is the $A$ -excenter of $\triangle AXY$ . Let $r$ be the radius of this circle. We can compute the area of $\triangle AXY$ in two ways: $\begin{aligned} K_{AXY} & =\frac{1}{2} \cdot AX \cdot AY \sin A \\ & =r \cdot(AX+AY-XY) / 2 \\ \Longrightarrow AY & =\frac{r}{\sin A} \end{aligned}$ We also know that $\begin{aligned} K_{ABC} & =\frac{1}{2} \cdot 19 \cdot 21 \sin A \\ & =r \cdot(19+20+21) / 2 \\ \Longrightarrow \frac{r}{\sin A} & =\frac{19 \cdot 21}{60}=\frac{133}{20} \end{aligned}$ so $AY=133 / 20$ . Let the incircle of $\triangle ABC$ be tangent to $AB$ and $AC$ at $D$ and $E$ , respectively. We know that $AX+AY+XY=AD+AE=19+21-20$ , so $AX=20-\frac{133}{10}=\frac{67}{10}$ .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.