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Slovenia — National Math Olympiad

Geometry Difficulty 4.8 AIME Prove it Slovenia

In a triangle $ABC$ we have $|AB| = 2|AC|$ . Let $D$ and $E$ be the two points on the segments $AB$ and $BC$ , such that $\angle BAE = \angle ACD$ . The segments $AE$ and $CD$ intersect at $F$ , and $CFE$ is an equilateral triangle. Find the angles of the triangle $ABC$ .

Solution

Since $CEF$ is an equilateral triangle we have $\angle EFC = 60^\circ$ . This implies that $\angle CFA = 120^\circ$ , so $\angle FAC = 180^\circ - \angle CFA - \angle ACF = 60^\circ - \angle ACF$ and $\angle BAC = \angle BAE + \angle FAC = \angle BAE + 60^\circ - \angle ACD = 60^\circ$ . We have $|AB| = 2|AC|$ and $\angle BAC = 60^\circ$ , so the triangle $ABC$ is one half of an equilateral triangle and we see that $\angle CBA = 30^\circ$ and $\angle ACB = 90^\circ$ .

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