B4. $|A E|>|E B|,|E B|=|B C|=|A D|,|A B|=17 \text{ cm},|E D|=13 \text{ cm}$. Calculate the angle $\varphi$ to the nearest minute. (Points $A, E$ and $B$ are collinear).

![](https://cdn.mathpix.com/cropped/2024_06_07_c1f0ff7099009a973f5fg-05.jpg?height=297&width=777&top_left_y=937&top_left_x=682)

## PROBLEMS FOR THE THIRD YEAR

Before you are two sets of problems. Problems 1 to 6 of the first set are solved by selecting the correct answer from the five proposed answers and writing it in the table below the corresponding number. Only one answer is correct. A correct answer will be awarded two points, while a wrong answer will result in the deduction of one point.

Problems 1 to 4 of the second set are solved on the provided paper. The solution to each of these problems will be evaluated on a scale from 0 to 6 points. On the sheets where you solve the problems, do not sign, but write only your code. Write your solution clearly and legibly with a pen. If you solve a problem in multiple ways, it must be clearly marked which one should be considered.

The time for solving is 90 minutes.

THE NATIONAL COMPETITION COMMITTEE WISHES YOU GREAT SUCCESS.

## I. PART
B4. Let $|E B|=|B C|=|A D|=x$ and apply the Pythagorean theorem to triangle $A E D$, thus $13^{2}=x^{2}+(17-x)^{2}$. The equation has two solutions, $x_{1}=12$, which does not meet the first condition of the problem, and $x_{2}=5$. It follows that $\sin \alpha=\sin \angle A E D=\frac{5}{13}$ and $\alpha=22^{\circ} 38^{\prime}$ and $\varphi=180^{\circ}-\alpha-45^{\circ}=112^{\circ} 22^{\prime}$.

## Third Year

I. PART

| 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $\mathrm{B}$ | $\mathrm{E}$ | $\mathrm{A}$ | $\mathrm{B}$ | $\mathrm{C}$ | $\mathrm{B}$ |