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Problem 1454

AIME late

Geometry Difficulty 5.4 Find the answer HMMT February

Let $A B C$ be a triangle with $A B=2, C A=3, B C=4$ . Let $D$ be the point diametrically opposite $A$ on the circumcircle of $A B C$ , and let $E$ lie on line $A D$ such that $D$ is the midpoint of $\overline{A E}$ . Line $l$ passes through $E$ perpendicular to $\overline{A E}$ , and $F$ and $G$ are the intersections of the extensions of $\overline{A B}$ and $\overline{A C}$ with $l$ . Compute $F G$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Using Heron's formula we arrive at $[A B C]=\frac{3 \sqrt{15}}{4}$ . Now invoking the relation $[A B C]=\frac{a b c}{4 R}$ where $R$ is the circumradius of $A B C$ , we compute $R^{2}=\left(\frac{2 \cdot 3}{[A B C]^{2}}\right)=$ $\frac{64}{15}$ . Now observe that $\angle A B D$ is right, so that $B D E F$ is a cyclic quadrilateral. Hence $A B \cdot A F=A D \cdot A E=2 R \cdot 4 R=\frac{512}{15}$ . Similarly, $A C \cdot A G=\frac{512}{15}$ . It follows that $B C G F$ is a cyclic quadrilateral, so that triangles $A B C$ and $A G F$ are similar. Then $F G=B C \cdot \frac{A F}{A C}=4 \cdot \frac{512}{2 \cdot 15 \cdot 3}=\frac{1024}{45}$

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.