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Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Equilateral $\triangle ABC$ has side length $6$ . Let $\omega$ be the circle through $A$ and $B$ such that $CA$ and $CB$ are both tangent to $\omega$ . A point $D$ on $\omega$ satisfies $CD = 4$ . Let $E$ be the intersection of line $CD$ with segment $AB$ . What is the length of segment $DE$ ?

Solution

Solution:

Let $F$ be the second intersection of line $CD$ with $\omega$ . By power of a point, we have $CF = 9$ , so $DF = 5$ . This means that $\frac{[ADB]}{[AFB]} = \frac{DE}{EF} = \frac{DE}{5 - DE}$ .

Now, note that triangle $CAD$ is similar to triangle $CFA$ , so $\frac{FA}{AD} = \frac{CA}{CD} = \frac{3}{2}$ . Likewise, $\frac{FB}{BD} = \frac{CB}{CD} = \frac{3}{2}$ .

Also, note that $\angle ADB = 180^\circ - \angle DAB - \angle DBA = 180^\circ - \angle CAB = 120^\circ$ , and $\angle AFB = 180^\circ - \angle ADB = 60^\circ$ .

This means that $\frac{[ADB]}{[AFB]} = \frac{AD \cdot BD \cdot \sin 120^\circ}{FA \cdot FB \cdot \sin 60^\circ} = \frac{4}{9}$ .

Therefore, we have that $\frac{DE}{5 - DE} = \frac{4}{9}$ . Solving yields $DE = \frac{20}{13}$ .

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