Competition problem — Geometry — Maths Olympiad Prep

Library / /298 of 377

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

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

Solution

Solution:

Using Heron's formula we arrive at $[ABC] = \frac{3 \sqrt{15}}{4}$ . Now invoking the relation $[ABC] = \frac{abc}{4R}$ where $R$ is the circumradius of $ABC$ , we compute $R^2 = \left(\frac{2 \cdot 3}{[ABC]^2}\right) = \frac{64}{15}$ . Now observe that $\angle ABD$ is right, so that $BDEF$ is a cyclic quadrilateral. Hence $AB \cdot AF = AD \cdot AE = 2R \cdot 4R = \frac{512}{15}$ . Similarly, $AC \cdot AG = \frac{512}{15}$ . It follows that $BCGF$ is a cyclic quadrilateral, so that triangles $ABC$ and $AGF$ are similar. Then $FG = BC \cdot \frac{AF}{AC} = 4 \cdot \frac{512}{2 \cdot 15 \cdot 3} = \frac{1024}{45}$

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