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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Points $A$ , $B$ , and $C$ lie in that order on line $\ell$ , such that $AB = 3$ and $BC = 2$ . Point $H$ is such that $CH$ is perpendicular to $\ell$ . Determine the length $CH$ such that $\angle AHB$ is as large as possible.

Solution

Solution:

Let $\omega$ denote the circumcircle of triangle $ABH$ . Since $AB$ is fixed, the smaller the radius of $\omega$ , the bigger the angle $AHB$ . If $\omega$ crosses the line $CH$ in more than one point, then there exists a smaller circle that goes through $A$ and $B$ that crosses $CH$ at a point $H'$ . But angle $AH'B$ is greater than $AHB$ , contradicting our assumption that $H$ is the optimal spot. Thus the circle $\omega$ crosses the line $CH$ at exactly one spot: i.e., $\omega$ is tangent to $CH$ at $H$ .

By Power of a Point, $CH^2 = CA \cdot CB = 5 \cdot 2 = 10$ , so $CH = \sqrt{10}$ .

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