Problem 7.4. On the sides $AB$ and $AC$ of triangle $ABC$, points $X$ and $Y$ are chosen such that $\angle A Y B = \angle A X C = 134^{\circ}$. On the ray $YB$ beyond point $B$, point $M$ is marked, and on the ray $XC$ beyond point $C$, point $N$ is marked. It turns out that $MB = AC$ and $AB = CN$. Find $\angle MAN$.

Answer: $46^{\circ}$.
Solution. Note that $\angle A C N=\angle C A X+\angle A X C=\angle B A Y+\angle A Y B=\angle A B M$. Adding

![](https://cdn.mathpix.com/cropped/2024_05_06_88259ba1cefbf4a4fee2g-2.jpg?height=599&width=713&top_left_y=2019&top_left_x=680)

Fig. 1: to the solution of problem $?$ ?

## Kurchatov School Olympiad in Mathematics - 2019 Final Round.

to the obtained equality of angles, and given the equalities of segments $A C=B M$ and $A B=C N$, we conclude that by the first criterion, triangles $A C N$ and $M B A$ are equal (Fig. ??), which means $\angle N A C = \angle A M B$. Then the following chain of equalities holds: $\angle M A N=\angle N A C+\angle C A B+\angle B A M=\angle A M B + \angle Y A B + \angle B A M = \angle Y A M + \angle A M Y = 180^{\circ} - \angle A Y B = 180^{\circ} - 134^{\circ} = 46^{\circ}$.

## Criteria

The largest suitable criterion is used:

76 . Any correct solution to the problem.

5 6. A correct solution with an incorrect answer due to an arithmetic error.

3 6. It is proven that triangles $A C N$ and $M B A$ are equal.

16 . The correct answer is provided.