Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME Prove it United States

Problem:

In equilateral triangle $ABC$ , a circle $\omega$ is drawn such that it is tangent to all three sides of the triangle. A line is drawn from $A$ to point $D$ on segment $BC$ such that $AD$ intersects $\omega$ at points $E$ and $F$ . If $EF = 4$ and $AB = 8$ , determine $|AE - FD|$ .

Solution

Solution:

Answer: $\frac{4}{\sqrt{5}}$ OR $\frac{4 \sqrt{5}}{5}$ Without loss of generality, $A, E, F, D$ lie in that order. Let $x = AE$ , $y = DF$ .

By power of a point, $x(x+4) = 4^{2} \Longrightarrow x = 2 \sqrt{5} - 2$ , and $y(y+4) = (x+4+y)^{2} - (4 \sqrt{3})^{2} \Longrightarrow y = \frac{48 - (x+4)^{2}}{2(x+2)} = \frac{12 - (1+\sqrt{5})^{2}}{\sqrt{5}}$ . It readily follows that $x - y = \frac{4}{\sqrt{5}} = \frac{4 \sqrt{5}}{5}$ .

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