5. If $x=5 m+4$, then $(5 m+4)^{2}+1=25 m^{2}+40 m+17$ is not divisible by 5.
Answer: $\left\{\begin{array}{c}x=5 m+2, \\ y=5 m^{2}+4 m+1 .\end{array}\right.$ or $\left\{\begin{array}{c}x=5 m+3, \\ y=5 m^{2}+6 m+2 .\end{array}\right.$ where $m \in \mathbb{Z}$.

5 points - the solution is correct, complete, and contains no errors.

4 points - if the answer is incorrect, the solution method is correct, but a computational error or typo has been made, and the solution is completed with this error.

3 points - the problem is solved to the correct answer by trial and error, but not all possible cases are considered or the sequence of trials is not completed.

1 point - the solution is incorrect, but contains reasonable considerations, has some progress in solving the problem.

0 points - the solution is incorrect or missing.

## №3. (6 points)

In an isosceles triangle $ABC$, a circle is inscribed. A tangent to the circle, parallel to the base $AC$, intersects the lateral sides at points $D$ and $E$. Find the radius of the circle if $DE=8$ and $AC=18$.

Solution: Drop perpendiculars from points $D$ and $E$ to side $AC$ - we get a rectangle $DEMK$, in which $KM=DE=8$.

The diameter of the circle is equal to the perpendicular $DK$, to calculate the length of which we first need to find the length of segment $AD$.

Consider the right triangles $ADK$ and $CEM$. They are equal by the leg ($DK=EM$) and the acute angle ($\angle A = \angle C$ since $\triangle ABC$ is isosceles). From the equality of the triangles, it follows that the segments $AK=MC=5$.

Since in a quadrilateral circumscribed around a circle, the sums of the lengths of opposite sides are equal,

$AD + EC = DE + AC = 8 + 18 = 26 \Rightarrow AD = EC = \frac{26}{2} = 13$. The diameter of the circle is $DK = \sqrt{AD^2 - AK^2} = \sqrt{13^2 - 5^2} = 12$. Therefore, the radius is $\frac{12}{2} = 6$.

Answer: 6.

6 points - the solution is correct, complete, and contains no errors.

5 points - the solution is correct, but insufficiently justified or contains inaccuracies.

2 points - the solution is incorrect, but contains reasonable considerations, has some progress in solving the problem.

1 point - only the answer is recorded.

0 points - the solution is incorrect or missing.

## №4. (7 points)

Solve the equation: $x \sqrt{1-y^{2}} + y \sqrt{2-z^{2}} + z \sqrt{3-x^{2}} = 3$.

Solution: Applying the inequality between the arithmetic mean and the geometric mean to the numbers $x^2$ and $1-y^2$, we get: $|x| \sqrt{1-y^2} \leq \frac{x^2 + 1 - y^2}{2}$.

Similarly, $|y| \sqrt{2-z^2} \leq \frac{y^2 + 2 - z^2}{2}$ and $|z| \sqrt{3-x^2} \leq \frac{z^2 + 3 - x^2}{2}$. Adding these inequalities and considering that $U \leq |l|$, we have

$$
x \sqrt{1-y^2} + y \sqrt{2-z^2} + z \sqrt{3-x^2} \leq 3
$$

Therefore, equality is achieved in all the above inequalities. This is possible only if $x^2 = 1 - y^2$, $y^2 = 2 - z^2$, and $z^2 = 3 - x^2$. Solving the system, we get:
$x = \pm 1, y = 0, z = \pm \sqrt{2}$. Note that for negative values of $x$ and $z$, the inequality $(*)$ is strict. Therefore, the equation has a unique solution when: $x = 1, y = 0, z = \sqrt{2}$.

Answer: $x = 1, y = 0, z = \sqrt{2}$

7 points - the solution is correct, complete, and contains no errors.

6 points - the solution is correct, but insufficiently justified or contains inaccuracies.

4 points - the solution is found, but the uniqueness of the solution is not proven.

3 points - the solution is constructed without justification, assuming each term equals one.

2 points - the solution is incorrect, but contains reasonable considerations, has some progress in solving the problem.

1 point - only the answer is recorded.

0 points - the solution is incorrect or missing.

## №5. (7 points)

How many different pairs can be formed from 28 domino tiles so that the tiles in the pair can be placed next to each other? (Tiles can be placed next to each other if some number of dots appears on both tiles).

## Solution

Choosing a pair of tiles is simpler to consider as choosing two cards, but it is clear that there are half as many unordered pairs as ordered ones. It is easier to find the number of ordered pairs, as in this case, the rule of product can be applied.

Choose the first tile - this can be done in 28 ways, of which in 7 cases the tile will be a double, i.e., a tile of the form

![](https://cdn.mathpix.com/cropped/2024_05_06_c99a48608c8f12e2d26eg-3.jpg?height=54&width=123&top_left_y=1515&top_left_x=812)

In 21 cases, it will be a tile of the form $a < b$. In the first case, the second tile can be chosen in 6 ways, and the number of ways to choose a pair of tiles by the rule of product is $7 \cdot 6 = 42$.

In the second case, the second tile can be chosen in 12 ways - 6

![](https://cdn.mathpix.com/cropped/2024_05_06_c99a48608c8f12e2d26eg-3.jpg?height=63&width=1350&top_left_y=1802&top_left_x=290)

pairs, so the number of ways is $21 \cdot 12 = 252$.

Therefore, by the rule of sum, the total number of ways to choose an ordered pair is $42 + 252 = 294$.

Answer: 147 pairs.

7 points - the task is completed correctly.

5 points - the solution contains minor errors, gaps in justification, but is generally correct and can become fully correct after minor corrections or additions.

2 points - important cases are considered in the absence of a solution (or an incorrect solution).

1 point - only the answer is recorded.

0 points - the solution is incorrect or missing.