Problem 8.7. Given an isosceles triangle $ABC$, where $AB = AC$ and $\angle ABC = 53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $AK$. Point $M$ is chosen such that:

- $B$ and $M$ are on the same side of line $AC$;
- $KM = AB$
- angle $MAK$ is the maximum possible.

How many degrees does angle $BAM$ measure?

![](https://cdn.mathpix.com/cropped/2024_05_06_038330058549e9b1ba1cg-27.jpg?height=400&width=301&top_left_y=224&top_left_x=576)
Answer: 44.

Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does not exceed the angle $PAK$, and these angles are equal only if points $M$ and $P$ coincide. Therefore, $M$ is this point of tangency.

![](https://cdn.mathpix.com/cropped/2024_05_06_038330058549e9b1ba1cg-27.jpg?height=403&width=311&top_left_y=935&top_left_x=575)

The radius $KM$ of the circle is perpendicular to the tangent $AM$. Also, in the right triangle $AMK$, the leg $MK$ is half the hypotenuse $AK$, so $\angle MAK=30^{\circ}$. Additionally, from the condition, we get that $\angle BAC=180^{\circ}-2 \cdot 53^{\circ}=74^{\circ}$. Therefore,

$$
\angle BAM=\angle BAC-\angle MAK=74^{\circ}-30^{\circ}=44^{\circ}
$$