Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it Philippines

Problem:

Let $\triangle ABC$ be a right triangle with right angle at $B$ . Let the points $D$ , $E$ , and $F$ be on $AB$ , $BC$ , and $CA$ , respectively, such that $\triangle DEF$ is an equilateral triangle and $EC = FC$ . If $DB = 5\sqrt{3}$ , $BE = 3$ , and $\sin \angle ACB = 4\sqrt{3}/7$ , find the perimeter of $\triangle ADF$ .

Solution

Solution:

By Pythagorean Theorem, $DE = \sqrt{(5\sqrt{3})^{2} + 3^{2}} = \sqrt{84} = EF$ .

From $\sin \angle ACB = 4\sqrt{3}/7$ , we have $\cos \angle ACB = 1/7$ .

Let $EC = FC = x$ , then by Cosine Law on side $EF$ of $\triangle ECF$ , we have
$EF^{2} = x^{2} + x^{2} - 2x^{2} \cos \angle ACB.$
Solving for $x^{2}$ , we have
$x^{2} = 84 / [2(1 - 1/7)] = 49.$
Hence, $EC = FC = x = 7$ and $BC = BE + EC = 10$ .

Since $\cos \angle ACB = 1/7$ , then $AC = 70$ , which means $AF = 63$ .

Since $\sin \angle ACB = 4\sqrt{3}/7$ , then $AB = 40\sqrt{3}$ , which means $AD = 35\sqrt{3}$ .

Thus, the perimeter of $\triangle ADF$ is $35\sqrt{3} + 63 + 2\sqrt{21}$ .

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