Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

Let $A, B, C$ be points in that order along a line, such that $A B=20$ and $B C=18$ . Let $\omega$ be a circle of nonzero radius centered at $B$ , and let $\ell_{1}$ and $\ell_{2}$ be tangents to $\omega$ through $A$ and $C$ , respectively. Let $K$ be the intersection of $\ell_{1}$ and $\ell_{2}$ . Let $X$ lie on segment $\overline{K A}$ and $Y$ lie on segment $\overline{K C}$ such that $X Y \| B C$ and $X Y$ is tangent to $\omega$ . What is the largest possible integer length for $X Y$ ?

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Note that $B$ is the $K$ -excenter of $K X Y$ , so $X B$ is the angle bisector of $\angle A K Y$ . As $A B$ and $X Y$ are parallel, $\angle X A B+2 \angle A X B=180^{\circ}$ , so $\angle X B A=180^{\circ}-\angle A X B-\angle X A B$ . This means that $A X B$ is isosceles with $A X=A B=20$ . Similarly, $Y C=B C=18$ . As $K X Y$ is similar to $K A C$ , we have that $\frac{K X}{K Y}=\frac{X A}{Y C}=\frac{20}{18}$ . Let $K A=20 x, K C=18 x$ , so the Triangle Inequality applied to triangle $K A C$ gives $K A<K C+A C \Longrightarrow 20 x<18 x+38 \Longrightarrow x<19$ . Then, $X Y=A C \cdot \frac{K X}{K A}=38 \cdot \frac{x-1}{x}=38-\frac{38}{x}<36$ , so the maximum possible integer length of $X Y$ is 35. The optimal configuration is achieved when the radius of $\omega$ becomes arbitrarily small and $\ell_{1}$ and $\ell_{2}$ are on opposite sides of $A C$ .

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