
G3. Let $A B C D E F$ be a regular hexagon. The points $\mathrm{M}$ and $\mathrm{N}$ are internal points of the sides $\mathrm{DE}$ and $\mathrm{DC}$ respectively, such that $\angle A M N=90^{\circ}$ and $A N=\sqrt{2} \cdot C M$. Find the measure of the angle $\angle B A M$.


## Solution

Since $A C \perp C D$ and $A M \perp M N$ the quadrilateral $A M N C$ is inscribed. So, we have

$$
\angle M A N=\angle M C N
$$

Let $P$ be the projection of the point $M$ on the line $C D$. The triangles $A M N$ and $C P M$ are similar implying

$$
\frac{A M}{C P}=\frac{M N}{P M}=\frac{A N}{C M}=\sqrt{2}
$$

So, we have

$$
\frac{M P}{M N}=\frac{1}{\sqrt{2}} \Rightarrow \angle M N P=45^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-07.jpg?height=679&width=830&top_left_y=1110&top_left_x=606)

Figure 4

Hence we have

$$
\angle C A M=\angle M N P=45^{\circ}
$$

and finally, we obtain

$$
\angle B A M=\angle B A C+\angle C A M=75^{\circ}
$$
