4. In $\triangle ABC$, $AB=AC$, $\angle BAC=100^{\circ}$. Inside $\triangle ABC$, a point $M$ is taken such that $\angle MCB=20^{\circ}$, $\angle MBC=30^{\circ}$. Find $\angle BAM$.

![](https://cdn.mathpix.com/cropped/2024_06_04_b9e7a1c03f5c15132d5ag-03.jpg?height=754&width=1008&top_left_y=43&top_left_x=521)

Fig. 1: To the solution of problem 1

## Senior League
4. Construct point $M^{\prime}$, symmetric to $M$ with respect to line $B C$. (see fig. 4). $\angle B M^{\prime} C=\angle B M C=130^{\circ}=180^{\circ}-\frac{\angle B A C}{2}$, so point $M^{\prime}$ lies on the circle with center at $A$ and radius $A B$. $\angle B M^{\prime} M=\angle B M M^{\prime}=90^{\circ}-\angle M B C=60^{\circ}$, $\angle M B M^{\prime}=2 \angle M B C=60^{\circ}$. Then $\angle M M^{\prime} B=180^{\circ}-\angle M B M^{\prime}-\angle B M M^{\prime}=60^{\circ}=\angle M B M^{\prime}$. Therefore, point $M^{\prime}$ lies on the bisector of isosceles triangle $B A M^{\prime}$ and $\angle B A M=\angle M A M^{\prime}=\frac{\angle B A M^{\prime}}{2}=\frac{180^{\circ}-2 \angle A B M^{\prime}}{2}=20^{\circ}$.

## Senior League