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Mongolian Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Mongolia

Let $M$ be an arbitrary point on the side $AB$ of regular triangle $ABC$ . If one erects regular triangle $AMN$ outwardly, lines $BN$ and $AC$ intersect at point $D$ . Denote $K$ the intersection point of lines $AN$ and $CM$ . Find the angle $ADK$ .

Solution

Since triangles $\triangle ABC$ and $\triangle MAN$ are regular, $\angle BAC = 60^\circ = \angle MAN \Rightarrow \angle NAD = 60^\circ$ .
$\left. \begin{array}{l} AC = AB \\ AM = AN \\ \angle CAM = \angle MAN = 60^\circ \\ \triangle CMA = \triangle BAN \Rightarrow \angle KCA = \angle DBA. \end{array} \right\} \Rightarrow$

$\left. \begin{array}{l} AC = AB \\ AM = AN \\ \angle CAM = \angle MAN = 60^\circ \end{array} \right\} \Rightarrow \text{we get } \triangle CAK = \triangle BAD.$
So $AK = AD$ and $\angle KAD = 60^\circ$ . It follows the triangle $\triangle AKD$ is regular and hence we have $\angle KDA = 60^\circ$ .

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