Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Find the answer

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

Your answer

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

Solution

Let $F$ be the second intersection of line $C D$ with $\omega$ . By power of a point, we have $C F=9$ , so $D F=5$ . This means that $\frac{[A D B]}{[A F B]}=\frac{D E}{E F}=\frac{D E}{5-D E}$ . Now, note that triangle $C A D$ is similar to triangle $C F A$ , so $\frac{F A}{A D}=\frac{C A}{C D}=\frac{3}{2}$ . Likewise, $\frac{F B}{B D}=\frac{C B}{C D}=\frac{3}{2}$ . Also, note that $\angle A D B=180-\angle D A B-\angle D B A=180-\angle C A B=120$ , and $\angle A F B=180-\angle A D B=60$ . This means that $\frac{[A D B]}{[A F B]}=\frac{A D \cdot B D \cdot \sin 120}{F A \cdot F B \cdot \sin 60}=\frac{4}{9}$ . Therefore, we have that $\frac{D E}{5-D E}=\frac{4}{9}$ . Solving yields $D E=\frac{20}{13}$ .

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