4  [ Proof by contradiction $]$

Inside an isosceles right triangle $ABC$ with hypotenuse $AB$, a point $M$ is taken such that angle $MAB$ is $15^{\circ}$ greater than angle $MAC$, and angle $MCB$ is $15^{\circ}$ greater than angle $MBC$. Find angle $BMC$.
Let the point $X$ of intersection of $A M$ and the altitude $C H$ of triangle $A B C$ lie on the segment $A M$ (see the figure; at the end of the solution, we will show that the other case is impossible).

![](https://cdn.mathpix.com/cropped/2024_05_06_8c2fdb390ca80df56784g-45.jpg?height=680&width=675&top_left_y=1129&top_left_x=677)

From the condition, it follows that $\angle B A X=30^{\circ}$. Therefore, $\angle C X M=\angle A X H=90^{\circ}-\angle X A H=60^{\circ}$. Since $C H$ is also the median of triangle $A B C$, triangle $A X B$ is isosceles, that is, $\angle B X H=60^{\circ}$.

Thus, $\angle B X M=60^{\circ}$.

Consider triangle $C X B$ separately. In it, $\angle X C B=45^{\circ}, \angle X B C=15^{\circ}, \angle C X B=120^{\circ}$, and $X M$ is the angle bisector of $\angle C X B$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_8c2fdb390ca80df56784g-45.jpg?height=366&width=1144&top_left_y=2121&top_left_x=457)

Let $\angle M B C=\alpha$, then $\angle M C B=15^{\circ}+\alpha$. Choose a point $Y$ on segment $X B$ such that $\angle Y C B=15^{\circ}$, then $\angle X C Y=30^{\circ}$. But $\angle X Y C=30^{\circ}$ (as the exterior angle of triangle $C Y B$), hence triangle $C X Y$ is isosceles.

Since $X M$ is the angle bisector of the isosceles triangle $C X Y$, it is also the median and altitude, hence $C M Y$ is also isosceles, so $\angle M Y C=\angle M C Y=\alpha$. On the other hand, $\angle M B C=\alpha$, so quadrilateral $C M Y B$ is cyclic. Then $\angle M B Y=\angle M C Y=\alpha$, hence $2 \alpha=15^{\circ}, \alpha=7.5^{\circ}$ and $\angle C M B=150^{\circ}$.

We will prove that point $X$ lies on segment $A M$. Suppose this is not the case (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_8c2fdb390ca80df56784g-46.jpg?height=494&width=1033&top_left_y=53&top_left_x=521)

Consider triangle $A X B$ separately and draw segment $C Y$ such that $\angle Y C B=15^{\circ}$. By the condition, $\angle M C B=15^{\circ}+\angle M B C$. Since

$\angle X C B=30^{\circ}+\angle X B C$, to satisfy the condition, angle $M B X$ must be $15^{\circ}$ greater than angle $M C X$.

Triangle $C M Y$ is isosceles, hence $\angle M C X=\angle M Y X>\angle M B Y$, so such a configuration of points is impossible.

## Answer

$150^{\circ}$.