OJBM

Geometry Difficulty 4.8 AIME Prove it JBMO

Problem:
Let $ABCDEF$ be a regular hexagon. The points $M$ and $N$ are internal points of the sides $DE$ and $DC$ respectively, such that $\angle AMN = 90^{\circ}$ and $AN = \sqrt{2} \cdot CM$ . Find the measure of the angle $\angle BAM$ .

Solution

Solution:
Since $AC \perp CD$ and $AM \perp MN$ the quadrilateral $AMNC$ is inscribed. So, we have
$\angle MAN = \angle MCN$
Let $P$ be the projection of the point $M$ on the line $CD$ . The triangles $AMN$ and $CPM$ are similar implying
$\frac{AM}{CP} = \frac{MN}{PM} = \frac{AN}{CM} = \sqrt{2}$
So, we have
$\frac{MP}{MN} = \frac{1}{\sqrt{2}} \Rightarrow \angle MNP = 45^{\circ}$

Figure 4

Hence we have
$\angle CAM = \angle MNP = 45^{\circ}$
and finally, we obtain
$\angle BAM = \angle BAC + \angle CAM = 75^{\circ}$

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