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Rioplatense Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Argentina

Let $ABC$ be an acute triangle with $AB < AC$ . Let $D$ , $E$ , $F$ be the feet of the altitudes from $A$ , $B$ , $C$ respectively. The circumcircles of $AEF$ and $ABC$ meet again at $M$ . Suppose the line $BM$ is tangent to the circumcircle of $AEF$ . Prove that $M$ , $F$ , $D$ are collinear.

Solution

First, by the tangency condition and $AMBC$ being a cyclic quadrilateral we have that
$\angle AEM = 180^\circ - \angle AMB = \angle ACB$
and hence $ME$ is parallel to $BC$ . Next, we claim that $M$ and $E$ are symmetric with respect to $AD$ . This is because $AH$ is a diameter of the circumcircle of $\triangle AFE$ and it is also perpendicular to $ME$ .

To finish the proof observe that $\angle ADM = \angle ADE = 90^\circ - \angle BAC = \angle ADF$ , where the second and third equalities hold in any triangle, by angle chasing.

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