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, 2018

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:
Let $A$ , $B$ , $C$ be points in that order along a line, such that $AB = 20$ and $BC = 18$ . 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 $\overline{KA}$ and $Y$ lie on segment $\overline{KC}$ such that $XY \parallel BC$ and $XY$ is tangent to $\omega$ . What is the largest possible integer length for $XY$ ?

Solution

Solution:
Note that $B$ is the $K$ -excenter of $KXY$ , so $XB$ is the angle bisector of $\angle AKY$ . As $AB$ and $XY$ are parallel, $\angle XAB + 2\angle AXB = 180^\circ$ , so $\angle XBA = 180^\circ - \angle AXB - \angle XAB$ . This means that $AXB$ is isosceles with $AX = AB = 20$ . Similarly, $YC = BC = 18$ .

As $KXY$ is similar to $KAC$ , we have that $\frac{KX}{KY} = \frac{XA}{YC} = \frac{20}{18}$ . Let $KA = 20x$ , $KC = 18x$ , so the Triangle Inequality applied to triangle $KAC$ gives $KA < KC + AC \Longrightarrow 20x < 18x + 38 \Longrightarrow x < 19$ .

Then, $XY = AC \cdot \frac{KX}{KA} = 38 \cdot \frac{x-1}{x} = 38 - \frac{38}{x} < 36$ , so the maximum possible integer length of $XY$ 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 $AC$ .

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