Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Slovenia

In the triangle $ABC$ let $D$ be the foot of the altitude to the side $AB$ . Given the points $E$ and $F$ on the sides $AD$ and $BC$ , such that $\angle BAF = \angle ACE$ , let the segments $AF$ and $CE$ intersect at $G$ and let the segments $AF$ and $CD$ intersect at $T$ . Find the angles of the triangle $ABC$ , given that $CGF$ is an equilateral triangle and the triangle $AET$ is isosceles with the apex at $E$ .

Solution

Denote $\angle BAF = \angle ACE = \varphi$ . Since $CGF$ is an equilateral triangle, we have $\angle CGA = 120^\circ$ . Thus, $\angle GAC = 180^\circ - \angle CGA - \angle ACG = 60^\circ - \varphi$ and $\angle BAC = \angle BAF + \angle GAC = \varphi + 60^\circ - \varphi = 60^\circ$ .

Since the triangle $AET$ is isosceles, we have $\angle ETA = \varphi = \angle ECA$ . So the points $A$ , $E$ , $T$ and $C$ are concyclic. Thus, $\angle ECT = \angle EAT = \varphi$
and $2\varphi = \angle ACE + \angle ECD = \angle ACD = \frac{\pi}{2} - \angle DAC = 30^\circ$ , which implies $\varphi = 15^\circ$ .

Finally, we can calculate $\angle CBA = \angle FBA = 60^\circ - \varphi = 45^\circ$ and $\angle ACB = 75^\circ$ .

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