1. In triangle $\mathrm{ABC}$, angle $\mathrm{C}$ is a right angle, and $\mathrm{CD}$ is the altitude. Find the length of the radius of the circle inscribed in triangle $\mathrm{ABC}$, if the lengths of the radii of the circles inscribed in triangles $\mathrm{ACD}$ and $\mathrm{BCD}$ are 6 and 8, respectively.
Answer: 10

## №4: Progression.

The numbers $5 \mathrm{x}-\mathrm{y} ; 2 \mathrm{x}+3 \mathrm{y} ; \mathrm{x}+2 \mathrm{y}$ are consecutive terms of an arithmetic progression. The numbers $(\mathrm{y}+1)^{2} ; \mathrm{xy}+1 ;(\mathrm{x}-1)^{2}$ are consecutive terms of a geometric progression. Find the numbers x and y. If there are multiple values, write the negative sum $x+y$ rounded to the nearest integer in the answer.

## Answer: -1

## №5: Percentages.

A flask is completely filled with a 96% hydrochloric acid solution (by volume). From it, 12 liters of the solution were poured out, the flask was topped up to the brim with water and mixed. Then, 18 liters of the resulting solution were poured out again, the flask was topped up to the brim with water and mixed. After this, the concentration of acid in the flask was 32%. Find the volume of the flask (in liters)?

Answer: 36

## №6: Set

Solve the equation $\frac{\sqrt{x+3.3}+\sqrt{x+11.3}+\sqrt{x+27.3}}{\sqrt{x+2.3}+\sqrt{x+6.3}+\sqrt{x+18.3}}=\frac{3}{2}$. In the answer, specify the root or the sum of the roots if there are several.

## Solution:

$$
\begin{aligned}
& \text { Rewrite } \quad \text { the equation } \\
& 2(\sqrt{x+3.3}+\sqrt{x+11.3}+\sqrt{x+27.3})=3(\sqrt{x+2.3}+\sqrt{x+6.3}+\sqrt{x+18.3}) \text {, and then in } \\
& \text { the form } \\
& 2(\sqrt{x+3.3}-\sqrt{x+2.3}+\sqrt{x+11.3}-\sqrt{x+6.3}+\sqrt{x+27.3}-\sqrt{x+18.3})=\sqrt{x+2.3}+\sqrt{x+6.3}+\sqrt{x+18.3}
\end{aligned}
$$

Let the left side of the equation be $f(x)$, and the right side be $g(x)$. $g(x)$ is monotonically increasing on the interval $[-2.3 ;+\infty)$ as the sum of three increasing functions. Transform the left side - $f(x)$. $f(x)=2\left(\frac{1}{\sqrt{x+3.3}+\sqrt{x+2.3}}+\frac{5}{\sqrt{x+11.3}+\sqrt{x+6.3}}+\frac{9}{\sqrt{x+27.3}+\sqrt{x+18.3}}\right)$.

$f(x)$ is monotonically decreasing on $[-2.3 ;+\infty)$ as the sum of monotonically decreasing functions. Therefore, the equation $f(x)=g(x)$ has no more than one root. Notice that $f(-2.3)=g(-2.3)$, so $x=-2.3$ is the only root of the equation.

Answer: $-2.3$

## №7: Circles.

In triangle $A B C$, the height $B H$ is drawn. A circle with its center lying on the height $B H$ passes through the vertices of the triangle $B, C$ and intersects the side $A B$ at point $E$. Find the length of the segment $A E$, if $A B=8, B C=6$.

## Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_bae16753c49d7589df8eg-4.jpg?height=723&width=534&top_left_y=224&top_left_x=310)

Let the line $B H$ and the circle intersect at point $D$. Right triangles $B H A$ and $B E D$ are similar ( $\angle B$ is common), $\angle B C E=\angle B D E$ (as inscribed angles subtending the same arc), therefore, $\angle B A C=\angle B D E=\angle B C E$. Next, triangles $C B E$ and $A B C$ are similar (by two angles: $\angle B$ is common, $\angle B A C=\angle B C E)$.

From the equality of the ratios of corresponding sides, we find:

$$
\frac{B E}{B C}=\frac{B C}{A B^{\prime}}, B E=\frac{B C^{2}}{A B}=\frac{36}{8}=4.5
$$

Therefore, $A E=8-4.5=3.5$.

Answer: 3.5

## №8: Parameter.

Find all values of the parameter $a$ for which the equation $a^{2}+13|x-1|+3 \sqrt{4 x^{2}-8 x+13}=3 a+3|4 x-3 a-4|$ has at least one root. In the answer, specify the sum of all integer values of the parameter that satisfy the condition of the problem.

## Solution:

Rewrite the equation as $a^{2}-3 a+3 \sqrt{(2 x-2)^{2}+9}=3|4 x-3 a-4|-13|x-1|$ and consider the functions $f(x)=a^{2}-3 a+3 \sqrt{(2 x-2)^{2}+9}$ and $g(x)=3|4 x-3 a-4|-13|x-1|$, defined and continuous on the set of real numbers. The graph of the function $y=g(x)$ represents a piecewise-linear function consisting of segments of lines and rays. For $x \geq 1$ each segment of the graph is part of a line of the form $y=k x+b$, where $k<0$ (since for any expansion of the first modulus, the coefficient of $x$ will be negative). Therefore, on the interval $[1 ;+\infty)$ the function $y=g(x)$ decreases from $g(0)$ to $-\infty$. Similarly, it can be shown that on the interval $(-\infty ; 1]$ the function $g(x)$ increases from $-\infty$ to $g(0)$. Therefore, at the point $x=1$ this function reaches its maximum value. $g_{\text {max. }}=g(1)=3|3 a|=9|a|$.

Based on the properties of monotonic functions, the function $f(x)$ takes its minimum value at $x=1 . f_{\text {min. }}=f(1)=a^{2}-3 a+9$. On $[1 ;+\infty) f(x)$ increases from $f(0)$ to $+\infty$, and on the interval $(-\infty ; 0]$ it decreases from $+\infty$ to $f(0)$.

Therefore, the equation $f(x)=g(x)$ has at least one root only when $f_{\text {min. }} \leq g_{\text {max., }}$, that is, when the inequality $a^{2}-3 a+9 \leq 9|a|$ is satisfied. For $a \geq 0 \quad$ we get $a^{2}-3 a+9 \leq 9 a ; a^{2}-12 a+9 \leq 0 ; \frac{D}{4}=36-9=27$; $a_{1,2}=6 \pm 3 \sqrt{3} ; \quad a \in[6-3 \sqrt{3} ; 6+3 \sqrt{3}] . \quad$ For $\quad a<0 \quad$ we have $a^{2}+6 a+9 \leq 0 ;(a+3)^{2} \leq 0 ; a=-3$. Thus, the equation has at least one root for $a \in\{-3\} \cup[6-3 \sqrt{3} ; 6+3 \sqrt{3}]$. Estimate between which integers the boundaries of the interval lie: $0<6-3 \sqrt{3}<1 ; 11<6+3 \sqrt{3}<12$. Therefore, the integers on the considered interval are from 1 to 11 inclusive. Sum all the integers that satisfy the condition of the problem: $1+2+3+4+5+6+7+8+9+10+11-3=63$.

Answer: 63

## №9: Divisibility.

Find the smallest number among natural numbers exceeding 2022 that divides the number 2021!! + 2022!!. (The symbol n!! denotes the product of all natural numbers not exceeding $n$ and having the same parity: $n!!=n \cdot(n-2) \cdot(n-4) \ldots)$

## Solution.

We will prove that 2021!! + 2022!! is divisible by the number 2023. Indeed,

$$
2022!!=(2023-1)(2023-3)(2023-5) \ldots(2023-2021)
$$

the remainder of the division of this number by 2023 is equal to the remainder of the division of the product

$$
(-1)(-3)(-5) \ldots(-2021)=(-1)^{\frac{2022}{2}} \cdot 2021!!=-2021!!
$$

by 2023. Therefore, 2021!! + 2022!! is divisible by the number 2023. Since 2023 is the smallest number among natural numbers exceeding 2022, we get the answer.

Answer. 2023.