Gapoichkin A.i.  A paper triangle with angles $20^{\circ}, 20^{\circ}, 140^{\circ}$ is cut along one of its angle bisectors into two triangles, one of which is also cut along its angle bisector, and so on. Can a triangle similar to the original one be obtained after several cuts?
Let's call a triangle bad if one of its angles is not a multiple of $20 \|$. On the first step, both obtained triangles are bad. Note that in a bad triangle, there are at least two angles that are not multiples of $20 \|$. From this, it is clear that when cutting a bad triangle along its bisector, two bad triangles are obtained. Therefore, a triangle similar to the original, which is not bad, cannot be obtained at any step.

## [ Auxiliary Circle $\quad]$  Problem 108031 Topics: [ [Inscribed angle is half of the central angle] [ Angle between tangent and chord $\quad]$

Inside the square $A B C D$, a point $M$ is chosen such that $\angle M A C=\angle M C D=\alpha$. Find the measure of angle $A B M$.

## Hint

Prove that point $M$ lies on the circle centered at point $B$ with radius $B A$.

## Solution

Since $\angle D C M=\angle M A C<45^{\circ}$, point $M$ lies inside triangle $A C D$. Consider the circumcircle of triangle $A M C$. Due to the equality of angles, $D C$ is a tangent to this circle (points $D$ and $A$ lie on opposite sides of the chord $M C$). Therefore, the center of the circle is the intersection of line $C B$ and the perpendicular bisector of diagonal $A C$, which coincides with $B$. Consequently, $\angle A B M=2 \angle A C M=90^{\circ}-2 \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-43.jpg?height=500&width=506&top_left_y=796&top_left_x=799)

Answer

$90^{\circ}-2 \alpha$.