Petr drew a trapezoid $ABCD$, whose base $AB$ was twice as long as the base $CD$, and sides $AD, DC, CB$ were congruent. Then he drew a square that had one side in common with the shorter base of the trapezoid. The new vertex of the square, which was closer to $B$ than to $A$, he labeled $N$.

What could be the measure of angle $ABN$? Determine all possibilities. (A. Bohiniková)
According to the assumptions in the problem, trapezoid $A B C D$ can be divided into three equilateral triangles. Therefore, the interior angle at vertex $B$ is $60^{\circ}$ and at vertex $C$ is $120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_eae8f615c6c75fc598beg-3.jpg?height=351&width=694&top_left_y=2052&top_left_x=681)

Depending on which half-plane bounded by the line $C D$ the remaining vertices of Peter's square were in, we distinguish two cases:
![](https://cdn.mathpix.com/cropped/2024_04_17_eae8f615c6c75fc598beg-4.jpg?height=686&width=1520&top_left_y=196&top_left_x=268)

In both cases, triangle $B C N$ is isosceles, with equal sides $B C, C N$ and equal angles at the base $B N$. The relevant angles are as follows:

- In the first case, angle $B C N$ is $30^{\circ}(120-90=30)$, so angle $C B N$ is $75^{\circ}(30+2 \cdot 75=180)$ and angle $A B N$ is $15^{\circ}(75-60=15)$.
- In the second case, angle $B C N$ is $150^{\circ}(360-120-90=150)$, so angle $C B N$ is $15^{\circ}(150+2 \cdot 15=180)$ and angle $A B N$ is $75^{\circ}(15+60=75)$.

The measure of angle $A B N$ could be either $15^{\circ}$ or $75^{\circ}$.

Evaluation. 1 point for the interior angles of trapezoid $A B C D$; 2 points for further observations and intermediate results; 3 points for solving and quality of the commentary. Solutions that include only one possibility should be rated no higher than 4 points.