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5. Inside a regular quadrilateral pyramid $S A B C D$ with base $A B C D$, there is a regular quadrilateral prism $K L M N K_{1} L_{1} M_{1} N_{1}$, the base $K L M N$ of which lies in the plane $A B C$. The center of the base $K L M N$ of the prism is located on the segment $A C$, $K L\|A C$, $K N\|$ $B D$ (points $K$... | Solution. According to the problem, we have $AB = a = 8\sqrt{2}$, $SO = h = 12$, $KL = b = 2$, $KK_1 = h_0 = 3$.
$PR \parallel AC \parallel KL$, $RF \parallel SO$, $RF = h_0$.
$\triangle RCF \sim \triangle SCO$, $\frac{CF}{CO} = \frac{RF}{SO} = \frac{h_0}{h} = \frac{1}{4}$, $CF = \frac{CO}{4}$.
The plane $PQR$ is pa... | \frac{64\sqrt{34}}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,030 |
6. To improve living conditions for urban residents, a renovation program is being implemented in some cities - replacing housing stock that cannot be preserved by demolishing it and constructing new buildings on the freed-up land. Typically, the dismantling of old buildings and clearing the site for new construction o... | Solution. By the Law of Sines for triangle $O E F$, we have
$$
\begin{aligned}
& \frac{O E}{\sin \angle O F E}=\frac{E F}{\sin \angle E O F^{\prime}}, \frac{2}{\sin \angle O F E}=\frac{2 \sqrt{2}}{\sin 45^{\circ}}, \sin \angle O F E=\frac{1}{2^{\prime}} \\
& \angle O F E=30^{\circ}, \angle O E F=105^{\circ} \\
& \frac... | \frac{67\pi}{3}+2(1+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,031 |
# 2. Solution:
Domain of definition: $x>0, 6-x>0, x \neq 1, x \neq 2 \Rightarrow x \in(0 ; 1) \cup(1 ; 2) \cup(2 ; 6)$.
We have $\frac{\left(\log _{4}(6-x)-\log _{4} 1\right)(|x-5|-|x-1|)}{\left(3^{x}-9\right)\left(3^{x}-3\right)\left(\log _{3} x-\log _{3} 1\right)} \leq 0$.
On the domain of definition, the original... | Answer: $x \in (0; 1) \cup (1; 2) \cup [3; 5]$.
# | x\in(0;1)\cup(1;2)\cup[3;5] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,032 |
5. Solution: Consider the second equation of the system $3-x=\sqrt{x^{2}-6 x+y+8} \Leftrightarrow\left\{\begin{array}{l}x \leq 3, \\ y=1 .\end{array}\right.$
Therefore, $\left\{\begin{array}{c}\log _{|x-1|}(a x)=2 \log _{|x-1|}(x+y), \\ 3-x=\sqrt{x^{2}-6 x+y+8}\end{array} \Leftrightarrow\left\{\begin{array}{c}x^{2}+(2... | Answer: for $a=4.5 \quad x=0.5, y=1$, for $a \in(-\infty ; 0) \quad x=\frac{a-2+\sqrt{a^{2}-4 a}}{2}, y=1$, for $a \in(16 / 3 ;+\infty) \quad x=\frac{a-2-\sqrt{a^{2}-4 a}}{2}, y=1$. | \frac{4.5\quad0.5,1,\quad\in(-\infty;0)\quad\frac{-2+\sqrt{^{2}-4}}{2},1,\quad\in(16}{3;+\infty)\quad\frac{-2-\sqrt{^{2}-4}}{2},1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,034 |
# 2. Solution:
Domain of definition: $x+6 \geq 0, x+1>0,5-x>0, x \neq 0, x \neq 1, x \neq 4 \Rightarrow x \in(-1 ; 0) \cup(0 ; 1) \cup(1 ; 4) \cup(4 ; 5)$.
We have $\frac{(\sqrt{x+6}-1)(\sqrt{x+6}-3)\left(\log _{2}(x+1)-\log _{2} 1\right)}{\left(2^{x}-1\right)\left(2^{x}-2\right)\left(\log _{5}(5-x)-\log _{5} 1\right... | Answer: $x \in(-1 ; 0) \cup(0 ; 1) \cup[3 ; 4)$. | x\in(-1;0)\cup(0;1)\cup[3;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,036 |
1. Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). T... | Answer: 7200 ways.
# | 7200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,040 |
# 2. Solution:
$\sin ^{4} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=1 \Leftrightarrow\left(1-\cos ^{2} 2025 x\right)^{2}+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=1 \Leftrightarrow$ $\cos ^{4} 2025 x-2 \cos ^{2} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=0 \Leftrightarrow$ $\cos ^{2} 2025 x\le... | Answer: $x=\frac{\pi}{4050}+\frac{\pi n}{2025}, n \in Z, x=\frac{\pi k}{9}, k \in Z$. | \frac{\pi}{4050}+\frac{\pin}{2025},n\inZ,\frac{\pik}{9},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,041 |
1. Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). T... | Answer: 3720 ways.
# | 3720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,045 |
# 2. Solution:
$\sin ^{4} 2019 x+\cos ^{2019} 2022 x \cdot \cos ^{2018} 2019 x=1 \Leftrightarrow\left(1-\cos ^{2} 2019 x\right)^{2}+\cos ^{2019} 2022 x \cdot \cos ^{2018} 2019 x=1 \Leftrightarrow$ $\cos ^{4} 2019 x-2 \cos ^{2} 2019 x+\cos ^{2019} 2022 x \cdot \cos ^{2018} 2019 x=0 \Leftrightarrow$ $\cos ^{2} 2019 x\le... | Answer: $x=\frac{\pi}{4038}+\frac{\pi n}{2019}, n \in Z, x=\frac{\pi k}{3}, k \in Z$.
$O_{2}$ is the center of the inscribed circle in triangle $A D E$, and $r$ is its radius. Let $p_{1}$ be... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,046 |
1. Prove that the expression $3+3^{2}+3^{3}+\ldots+3^{2015}$ is divisible by 121. | Task 1.
| Points | |
| :---: | :--- |
| 12 | The correct answer is justified. |
| 10 | With the correct understanding of the problem and the correct answer, there are remarks about the clarity of the justification. |
| 0 | The solution does not meet the above requirements. | | proof | Number Theory | proof | Yes | Yes | olympiads | false | 10,049 |
3. Specify the smallest natural number, different from 1, which when divided by each of the numbers $2, 3, 5$ and 9, gives a remainder of 1. | Task 3.
| Points | |
| :---: | :--- |
| 12 | A justified and correctly executed solution to the problem (including by selection, if all numbers less than the sought one have been considered). |
| 10 | The solution is generally correct but has remarks regarding the clarity of its presentation. |
| 0 | The solution doe... | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,051 | |
4. Find all values of the parameter $a$, for which the area of the figure bounded on the coordinate plane by the lines $y=2 x+a, y=2 x-a, y=x+a, y=x-a$ is 36. | Task 4.
| Points | |
| :---: | :--- |
| 16 | A well-reasoned and correctly executed solution to the problem. |
| 12 | Only one of the two parameter values is found. |
| 5 | The geometric model of the problem is correctly constructed. |
| 0 | The solution does not meet the above requirements. |
| Points | |
| :---:... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,052 | |
6. Given a pentagon $A B C D E$, in which line $A B$ is parallel to line $C D$, line $D E$ is parallel to line $B C$, the length of diagonal $A C$ is 24, and the length of diagonal $C E$ is 20. Find the distance from point $D$ to line $A C$, if the distance from point $B$ to line $C E$ is 18. | Problem 6.
| Points | |
| :---: | :--- |
| 16 | A well-reasoned and correctly executed solution to the problem. |
| 14 | With the correct approach to the proof, there are remarks about the clarity of its presentation. |
| 6 | Intermediate results necessary for solving the problem have been obtained. |
| 0 | The solut... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,053 | |
7. On a plane, there are 1580 points such that the area of any triangle with vertices at these points does not exceed 1. Prove that all these points can be covered by one triangle of area 4.
| Task | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | $\mathbf{6}$ | $\mathbf{7}$ | Total |
| :---... | Problem 7.
| Points | |
| :---: | :--- |
| 16 | A well-reasoned and correctly executed solution to the problem. |
| 14 | With the correct approach to the proof, there are remarks about the clarity of its presentation. |
| 6 | Intermediate results necessary for solving the problem have been obtained. |
| 0 | The solut... | proof | Geometry | proof | Yes | Yes | olympiads | false | 10,054 |
1. Find the maximum value of the expression $x+y$, where $x, y-$ are integer solutions of the equation $3 x^{2}+5 y^{2}=345$ | # Solution
Notice that 345 and $5 y^{2}$ are divisible by 5, so $3 x^{2}$ must also be divisible by 5. Therefore, $\quad x=5 t, t \in Z$. Similarly, $y=3 n, n \in Z$. After simplification, the equation becomes $5 t^{2}+3 n^{2}=23$. Therefore, $t^{2} \leq \frac{23}{5}$, $n^{2} \leq \frac{23}{3}$ or $|t| \leq 2,|n| \leq... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,055 |
2. The car traveled half of the distance at a speed of 60 km/h, then one third of the remaining distance at a speed of 120 km/h, and the remaining distance at a speed of 80 km/h.
Find the average speed of the car during this trip. Give your answer in km/h. | Solution: Let x hours be the time the car traveled at a speed of 60 km/h, then $60 x=\frac{s}{2}$. Let y hours be the time the car traveled at a speed of 120 km/h, then $120 y=\frac{s}{6}$. Let z hours be the time the car traveled at a speed of 80 km/h, then $80 z=\frac{s}{3}$. By definition $v_{cp}=\frac{s}{t_{\text{t... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,056 |
4.
On a coordinate line, 16 points are marked and numbered from left to right. The coordinate of any point, except for the extreme points, is equal to half the sum of the coordinates of the two adjacent points. Find the coordinate of the fifth point if the first point has a coordinate of 2 and the sixteenth point has... | # Solution
Solution. Let $a, b$ and $c$ be the coordinates of three consecutive points (from left to right). Then $b=\frac{a+c}{2}$, which means the second point is the midpoint of the segment with endpoints at the neighboring points. This condition holds for any triple of consecutive points, meaning the distances bet... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,058 |
5.
In a 6-liter vessel, 4 liters of a 70% (by volume) sulfuric acid solution are poured, and in a second vessel of the same capacity, 3 liters of a 90% solution of sulfuric acid are poured. A certain amount of the solution is transferred from the second vessel to the first so that it becomes an $r-\%$ solution of sul... | # Solution.
Let $x$ liters of the solution be transferred from the second vessel to the first. Since it follows from the condition that $0 \leq x \leq 2$, to find the amount of pure acid in the new solution, we obtain the equation $2.8 + 0.9x = (4 + x) \frac{r}{100}$, from which $x = \frac{4r - 280}{90 - r}$. Now, con... | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,059 |
6.
Find the roots of the equation $f(x)=8$, if $4 f(3-x)-f(x)=3 x^{2}-4 x-3$ for any real value of $x$. In your answer, specify the product of the found roots. # | # Solution:
Notice that when $x$ is replaced by $3-x$, the expression $3-x$ changes to $x$. That is, the pair $f(x)$ and $f(3-x)$ is invariant under this substitution. Replace $x$ with $3-x$ in the equation given in the problem. We get:
$4 f(x) - f(3-x) = 3(3-x)^2 - 4(3-x) - 3 = 3x^2 - 14x + 12$. Express $f(x)$ from ... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,060 |
7.
In triangle $A B C$, the bisector $A L$ ( $L \in B C$ ) is drawn, and $M$ and $N$ are points on the other two bisectors (or their extensions) such that $M A=M L$ and $N A=N L, \angle B A C=50^{\circ}$.
Find the measure of $\angle M A N$ in degrees. | # Solution
We will use the auxiliary statements.
If the bisector $B K$ in triangle $A B C$ intersects the circumscribed circle at point $W$, then:
1) $A W=C W$ (since $\angle C A W=\angle C B W=\angle A B W=\angle A C W$, that is, triangle $A W C$ is isosceles and $A W=C W$).
 x+a}{x^{2}+5 x+4} \geq 0$ is the union of three non-overlapping intervals. In your answer, specify the sum of the three smallest integer values of $a$ from the obtained interval. | # Solution
Let's factorize the numerator and the denominator of the left part of the inequality, it will take the form: $\frac{(x+1)(x+a)}{(x+1)(x+4)} \geq 0$. There are five possible cases for the placement of the number ($-a$) relative to the numbers (-4) and (-1). In each case, the inequality is solved using the in... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,062 |
9.
Calculate the number $8^{2021}$, find the sum of the digits in this number, and write down the result. Then, in the newly written number, find the sum of the digits and write down the result again. These actions were repeated until a single-digit number was obtained. Find this number. | # Solution.
Consider the natural powers of 8. Notice that even powers of the number 8 give a remainder of 1 when divided by 9, while odd powers (including the number \(8^{2021}\)) give a remainder of 8. Indeed, let's analyze the powers of 8:
\[
\begin{gathered}
8^{2}=(9-1)^{2}=9 n+1, n \in N \\
8^{3}=(9 n+1) \cdot 8=... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,063 |
1. Which of the numbers is greater $2^{5^{4^{3}}}$ or $3^{4^{2^{5}}}$? | Solution:
$$
2^{5^{4^{3}}} \vee 3^{4^{2^{5}}} \Leftrightarrow 2^{5^{64}} \vee 3^{4^{32}} \Leftrightarrow 2^{5^{64}}>2^{4^{64}}=2^{4^{63} \cdot 4}=\left(2^{4}\right)^{4^{63}}=16^{4^{63}}>3^{4^{63}}>3^{4^{32}}
$$
Answer: $2^{5^{4^{3}}}>3^{4^{2^{5}}}$ | 2^{5^{4^{3}}}>3^{4^{2^{5}}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,064 |
# 2. Find the value of the expression
$$
\sqrt{\frac{1 \cdot 2 \cdot 3+2 \cdot 4 \cdot 6+3 \cdot 6 \cdot 9+\ldots+n \cdot 2 n \cdot 3 n}{2 \cdot 3 \cdot 4+4 \cdot 6 \cdot 8+6 \cdot 9 \cdot 12+\ldots+2 n \cdot 3 n \cdot 4 n}}
$$ | # Solution:
$$
\sqrt{\frac{1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 6+\ldots+n \cdot 2 n \cdot 3 n}{2 \cdot 3 \cdot 4+4 \cdot 6 \cdot 8+\ldots+2 n \cdot 3 n \cdot 4 n}}=\sqrt{\frac{6\left(1+8+27+\ldots+n^{3}\right)}{24\left(1+8+27+\ldots+n^{3}\right)}}=\sqrt{\frac{1}{4}}=\frac{1}{2}
$$
Answer: $\frac{1}{2}$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,065 |
3. Prove that if an integer $n$ is not divisible by 5, then the number $n^{5}-n$ is divisible by 30. | Solution: $n^{5}-n=n\left(n^{4}-1\right)=n\left(n^{2}-1\right)\left(n^{2}+1\right)$.
If the integer $n$ is not divisible by 5, then $n=5 k \pm 1$ or $n=5 k \pm 2$.
a) $\quad n^{2}-1=(5 k \pm 1)^{2}-1=25 k^{2} \pm 10 k: 5 \quad$ or $\quad n^{2}+1=(5 k \pm 2)^{2}+1=25 k^{2} \pm 20 k+5: 5$, therefore, $n^{2}-n: 5$
b) $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 10,066 |
4. We will call a ticket with a number from 0001 to 2014 excellent if the difference between some two adjacent digits of its number is 5. Find the number of excellent tickets. | Solution. The number of excellent tickets from 0001 to 2014 is equal to the number of excellent tickets from 0000 to 2014. First, let's calculate the number of non-excellent tickets from 0000 to 2014.
The number of non-excellent tickets from 0000 to 1999 can be found as follows. Let $a_{1} a_{2} a_{3} a_{4}$ be the nu... | 543 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,067 |
5. Find all values of the parameter $a$, for which the area of the quadrilateral bounded on the coordinate plane by the x-axis and the lines $x=a+5, x=a+1$, $y=(a-3)x$, is 28. | Solution. The lines $x=a+5, x=a+1$ are vertical and the distance between them is 4 for any value of the parameter $a$.
The line $y=(a-3) x$ passes through the origin. These lines and the x-... | \sqrt{2};4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,068 |
1. (15 points) Pavel caught 32 crayfish and decided to sell them at the market. When part of his catch was bought, it turned out that the buyer paid 4.5 rubles less for each one than the number of crayfish that remained on the counter. At the same time, the boy earned the maximum amount of money possible. How much mone... | Solution: Let $x$ be the number of crayfish left on the counter, then (32-x) crayfish were bought, ($x-4.5$) - the cost of one crayfish.
$(32-x)(x-4.5)$ - the cost of all crayfish.
$y=(32-x)(x-4.5)$
$x_{B}=18.25$.
$x_{1}=19, y_{1}=(32-19)(19-4.5)=13 * 14.5=188.5$.
$x_{2}=18, y_{2}=(32-18)(18-4.5)=14 * 13.5=189$. T... | 189 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,070 |
2. (15 points) Find the range of the similarity coefficient of triangles with side lengths $x, y, z$ and $y, z, p$. In your answer, specify the nearest integers between which the found interval lies. | Solution: Let's write down the similarity condition for the triangles.
$\frac{x}{y}=\frac{y}{z}=\frac{z}{p}=k ; y^{2}=x z ; z^{2}=y p \quad$ or $\quad x=y k, y=z k$. Since the triangles are similar, it is sufficient to write the triangle existence conditions (triangle inequality) for only one of them $\left\{\begin{ar... | 02 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,071 |
3. Solve the inequality:
$$
\frac{2|2 x-1|+2}{3}+\frac{6}{1+|2 x-1|} \leq 4-\sqrt{16 x^{4}-8 x^{2}+1}
$$ | # Solution:
The domain of the inequality is all real numbers. Rewriting the left side of the inequality as $2\left(\frac{|2 x-1|+1}{3}+\frac{3}{1+|2 x-1|}\right)$, we notice that it is not less than 4, as it is twice the sum of two positive reciprocals, and it equals 4 only when $1+|2 x-1|=3$.
At the same time, the r... | -0.5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,072 |
4. (15 points) Find all values of the parameter $a$ for which the equation $\left(x^{2}+(2 a-1) x-4 a-2\right) \cdot\left(x^{2}+x+a\right)=0$ has three distinct roots. | Solution. The given equation is equivalent to the system $\left[\begin{array}{l}x^{2}+(2 a-1) x-4 a-2=0 \\ x^{2}+x+a=0\end{array}\right.$ and can have from zero to four roots. It has three distinct roots if one of the equations in the system has two distinct roots ( $\mathrm{D}>0$ ), and the other has one ( $\mathrm{D}... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,073 | |
5. (20 points) In an acute-angled triangle $\mathrm{ABC}$, a point $\mathrm{D}$ is chosen on side $\mathrm{BC}$ such that $\mathrm{CD}: \mathrm{DB}=2: 1$, and a point $\mathrm{K}$ is chosen on segment $\mathrm{AD}$ such that $\mathrm{AK}=\mathrm{CD}+\mathrm{DK}$. A line is drawn through point $\mathrm{K}$ and vertex $\... | Solution: Extend SV beyond point V so that BN = BD. Draw NM || BE. NM intersects AD at point L. Draw segment $\mathrm{CH} \perp \mathrm{AD}$. Extend it to $\mathrm{P}$ such that $\mathrm{HP}=\mathrm{HC}, \mathrm{PN}|| \mathrm{AD}$.
In triangle DLN, segment $\mathrm{BK}$ is the midline, therefore, $\mathrm{DK}=\mathrm{... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,074 |
6. (20 points) Each of the two baskets contains white and black balls, and the total number of balls in both baskets is 25. A ball is randomly drawn from each basket. It is known that the probability that both drawn balls will be white is 0.54. Find the probability that both drawn balls will be black. | Answer: 0.04.
Solution. Let in the $i$-th basket there be $n_{i}$ balls, among which $k_{i}$ are white, $i=1,2$. Then $\frac{k_{1}}{n_{1}} \cdot \frac{k_{2}}{n_{2}}=0.54=\frac{27}{50}$. Therefore, for some natural number $m$, the equalities $k_{1} \cdot k_{2}=27 m, n_{1} \cdot n_{2}=50 m$ hold. One of the numbers $n_{... | 0.04 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,075 |
1. A farmer initially placed his produce in boxes with a capacity of 8 kg each, but one box was not fully loaded. Then the farmer repackaged all the produce into boxes with a capacity of 6 kg each, which required 8 more boxes, but in this case, one box was also not fully loaded. When all the produce was placed in boxes... | Solution. Let $x$ kg be the weight of the farmer's produce. Then $\quad 8(n-1)<x<8 n, \quad 6(n+7)<x<6(n+8)$, $5(n+13)=x, \Rightarrow 8(n-1)<5(n+13)<8 n, \quad 6(n+7)<5(n+13)<6(n+8)$,
$\Rightarrow 21 \frac{2}{3}<n<23, \quad n=22, \quad x=35 \cdot 5=175$.
Answer: 175. | 175 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,076 |
2. Solve the equation $8 \sin ^{4}(\pi x)-\sin ^{2} x=\cos ^{2} x-\cos (4 \pi x)$. In your answer, specify the sum of the roots that belong to the interval $[-1 ; 2]$.
(5 points) | Solution. Considering the basic trigonometric identity, we get
$8 \sin ^{4}(\pi x)-1+\cos (4 \pi x)=0 \quad \Rightarrow \quad 8 \sin ^{4}(\pi x)-2 \sin ^{2}(2 \pi x)=0 \quad \Rightarrow$
$\left(2 \sin ^{2}(\pi x)-2 \sin (\pi x) \cos (\pi x)\right)\left(2 \sin ^{2}(\pi x)+2 \sin (\pi x) \cos (\pi x)\right)=0 \Rightarr... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,077 |
3. Find the smallest distance from the point with coordinates $(10 ; 5 ; 10)$ to a point whose coordinates are positive and satisfy the inequality $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 9 \sqrt{1-(2 x+y)^{2}}$. In your answer, write the square of the found distance. | Solution. We use the Cauchy inequality $a+b+c \geq 3 \sqrt[3]{a b c}$, which holds for all positive values of $a, b, c$. Then, $x+y+z \geq 3 \sqrt[3]{x y z}, \quad \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq 3 \sqrt[3]{\frac{1}{x y z}}$. Since all parts of the inequalities are positive, we have
$$
(x+y+z)\left(\frac{1}{x... | 115.2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,078 |
4. In the country of Landia, which breeds an elite breed of horses, an annual festival is held to test their speed, in which only one-year-old, two-year-old, three-year-old, and four-year-old horses can participate. For each horse that meets the speed standard, the festival organizers pay a fixed amount of money to the... | Solution. A four-year-old horse can earn a maximum of 4 landricks over its entire participation in festivals. If the horse starts participating in festivals at 1 year old, it can participate for another 3 years after that. In the case of winning every year, it will earn 1+2+3+4=10 landricks over 4 years. If the horse s... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,079 |
5. The number $N$ is written as the product of consecutive natural numbers from 2019 to 4036: $N=2019 \cdot 2020 \cdot 2021 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036$. Determine the power of two in the prime factorization of the number $N$. | Solution. The number $N$ can be represented as
$$
\begin{aligned}
& N=\frac{(2 \cdot 2018)!}{2018!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036}{2018!}=\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot(2 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4036)}{2018!}= \\
& =\frac{(1 \cdot 3 \cdot \... | 2018 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,080 |
6. What is the smallest area that a right triangle can have, if its hypotenuse lies on the tangent to the graph of the function $y=\sqrt{x-3}$, one of its legs lies on the $y$-axis, and one of its vertices coincides with the point of tangency | Solution. $\quad f(x)=\sqrt{x-3}, \quad f^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x-3}}$
$$
\begin{aligned}
& S_{A B C}=\frac{1}{2} A B \cdot B C, x_{0}-\text { abscissa of the point of tangency } A, \\
& A\left(x_{0}, f\left(x_{0}\right)\right), \quad B\left(0, f\left(x_{0}\right)\right), \quad C \quad \text { - ... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 10,081 |
7. In triangle $A B C$, altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $1+\sqrt{3}$. The distances from the center of the inscribed circle in triangle $D E F$ to points $A$ and $C$ are $\sqrt{2}$ and 2, respectively. Find the length of side $A B$. | Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=\sqrt{2}, C O=2$. Le... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,082 |
8. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
y=\frac{|x|-1}{|x+1|}-2 \\
4|x-1.5-a|+|y-1-2a|=1
\end{array}\right.
$$
has a unique solution. In your answer, specify the smallest of all the obtained values of the parameter $a$. | Solution. We will construct the graphs of the functions:
a) $y=\frac{|x|-1}{|x+1|}-2=\left\{\begin{array}{l}-1, \text { for } x<-1, \\ -3, \text { for }-1<x \leq 0, \\ \frac{x-1}{x+1}-2=\frac{-x-3}{x+1}=-1-\frac{2}{x+1}, x \geq 0,\end{array}\right.$
b) $4|x-1.5-a|+|y-1-2a|=1$ - a rhombus, the center of which moves al... | -1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,083 |
9. Given a regular quadrilateral pyramid $T A B C D$ with base $A B C D$, where $A B=9 / 2$. On its height $T O$, a point $T_{1}$ is chosen such that $T T_{1}=T O / 3$. Points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ divide the segments $O A, O B, O C$, and $O D$, respectively, in the ratio $1: 2$, counting from point $O$. F... | Solution. Let $\alpha$ be the plane of the section, $\rho$ be the distance from point $C$ to this plane of the section, $a$ be the side of the base of the pyramid $T A B C D, \varphi$ be the angle between the plane of the section and the base of the pyramid. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,084 |
1. A vessel with a capacity of 10 liters is filled with air containing $24\%$ oxygen. A certain volume of air was pumped out of the vessel and the same volume of argon was added. Then, the same volume of the mixture as the first time was pumped out and again the same volume of argon was added. In the new mixture, $11.7... | Solution. Let $x$ liters of the mixture be released each time from the vessel. Then, the first time, the amount of oxygen left in the vessel is $2.4 - 0.24x$. The percentage of oxygen in the mixture after adding argon is $(2.4 - 0.24x) \times 10$. The second time, the amount of oxygen left in the vessel is $2.4 - 0.24x... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,086 |
2. Solve the equation $\cos \left(\pi x^{2}\right)-\cos ^{2}\left(\frac{\pi x^{2}}{2}\right)+1+\cos \left(\pi x^{2}-4 \pi x\right)=\sin ^{2}\left(\frac{\pi x^{2}}{2}\right)$. In your answer, specify the third term of the increasing sequence of all positive roots of the equation. | Solution. Considering the basic trigonometric identity, we get $\cos \left(\pi x^{2}\right)+\cos \left(\pi x^{2}-4 \pi x\right)=0 \quad \Rightarrow \quad 2 \cos \left(\pi x^{2}-2 \pi x\right) \cos (2 \pi x)=0$.
Therefore, $\left[\begin{array}{l}\cos \left(\pi x^{2}-2 \pi x\right)=0, \\ \cos (2 \pi x)=0,\end{array} \Ri... | 0.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,087 |
3. Find the smallest distance from the point with coordinates $(5 ; 10 ; 13)$ to a point whose coordinates are positive and satisfy the inequality
$$
(x+y+z)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right) \geq 4.5 \sqrt[4]{1-(2 y+x)^{2}}
$$
In your answer, write the square of the found distance. | Solution. Using the Cauchy inequality $a+b+c \geq 3 \sqrt[3]{a b c}$, which holds for all positive values of $a, b, c$. Then $((x+y)+(y+z)+(z+x)) \geq 3 \sqrt[3]{(x+y)(y+z)(z+x)}$, $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \geq 3 \sqrt[3]{\frac{1}{(x+y)(y+z)(z+x)}}$. Since all parts of the inequalities are positive, t... | 115.2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,088 |
4. In how many ways can a rectangular board of size $2 \times 18$ be covered with identical rectangular tiles of size $1 \times 2$? The tiles must be placed so that they fit entirely on the board and do not overlap.
(12 points) | Solution. Let there be a board of size $2 \times$ n. Denote the number of ways to tile it with tiles of size $1 \times 2$ by $P_{n}$. Then the following recurrence relation holds: $P_{n}=P_{n-1}+P_{n-2}$. Since $P_{1}=1, P_{2}=2$, performing sequential calculations using the recurrence relation, we arrive at the answer... | 4181 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,089 |
5. Determine the smallest natural number $N$, among the divisors of which are all numbers of the form $x+y$, where $x$ and $y$ are natural solutions to the equation $6 x y-y^{2}-5 x^{2}=7$. | Solution. Transform the equation by factoring the right-hand side
$6 x y-y^{2}-5 x^{2}-x^{2}+x^{2}=7 \Rightarrow 6 x(y-x)-(y+x)(y-x)=7 \Rightarrow(y-x)(6 x-y-x)=7 \Rightarrow$ $(y-x)(5 x-y)=7$.
Considering that the variables are natural numbers, and 7 is a prime number, we get
$$
\left\{\begin{array} { l }
{ y - x ... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,090 |
6. What is the greatest value that the area of a right triangle can take, one vertex of which coincides with the origin, another lies on the curve $x^{2}+y^{2}=2(x+y)$, and the vertex of the right angle is located on the line $y=x$? In the answer, write the square of the found area. | Solution. $\quad x^{2}+y^{2}-2 x-2 y=0, \quad(x-1)^{2}+(y-1)^{2}=2$.
We have the equation of a circle with center at point $(1 ; 1)$ and radius $R=\sqrt{2}$. Let's transition to the coordinate system Ouv while maintaining the scale (see figure). The equation of the circle in this coordinate system is $(u-\sqrt{2})^{2}... | 1.6875 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,091 |
7. In triangle $A B C$, the altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $\sqrt{6}+\sqrt{2}$. The distances from the center of the inscribed circle of triangle $D E F$ to points $A$ and $C$ are 2 and $2 \sqrt{2}$, respectively. Find the radius of the circumscribed circle around triangle $D E F$. (16... | Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=2, C O=2 \sqrt{2}$. ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,092 |
9. Given a regular quadrilateral pyramid $T A B C D$ with base $A B C D$, where $A B=9 / 2$. On its height $T O$, a point $T_{1}$ is chosen such that $T T_{1}=T O / 3$. Points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ divide the segments $O A, O B, O C$, and $O D$, respectively, in the ratio $1: 2$, counting from point $O$. F... | Solution. Let $\alpha$ be the plane of the section, $\rho$ be the distance from point $C$ to this section plane, $a$ be the side of the base of the pyramid $T A B C D, \varphi$ be the angle between the section plane and the base of the pyramid. Note that the distance from point $B$ to the plane $\alpha$ is
$2 \rho, \q... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,094 |
1. From point $A$ of a circular track, a car and a motorcycle started simultaneously and in the same direction. The car drove two laps without stopping in one direction. At the moment when the car caught up with the motorcyclist, the motorcyclist turned around and increased his speed by $16 \mathrm{~km} / \mathrm{u}$, ... | Solution. Let $x$ (km/h) be the speed of the motorcyclist, $y$ (km/h) be the speed of the car, and $S$ (km) be the distance the motorcyclist travels before turning around. Then the total length of the track is $2 S + 5.25$. We have $\frac{S}{x} = \frac{3 S + 5.25}{y}$, $\frac{3 x}{8} + 6 = S$, $\frac{3 y}{8} = S + 5.25... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,096 |
3. Find the smallest distance from the point with coordinates $(7 ; 3 ; 6)$ to a point whose coordinates are positive and satisfy the inequality
$$
\left(x^{2}+y^{2}+z^{2}\right)\left(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{x z}\right) \geq 9 \sqrt{1-(2 z+y)^{2}}
$$
In your answer, write the square of the found distance... | Solution. Using the Cauchy inequality $a+b+c \geq 3 \sqrt[3]{a b c}$, which holds for all positive values of $a, b, c$. Then $x^{2}+y^{2}+z^{2} \geq 3 \sqrt[3]{x^{2} y^{2} z^{2}}$, $\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \geq 3 \sqrt[3]{\frac{1}{x^{2} y^{2} z^{2}}}$.
$\left(x^{2}+y^{2}+z^{2}\right)\left(\frac{1}{x y... | 39.2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,098 |
4. Given 2019 indistinguishable coins. All coins have the same weight, except for one, which is lighter. What is the minimum number of weighings required to guarantee finding the lighter coin using a balance scale without weights? | Solution. $\quad$ We will prove the following statement by induction on $k$: if there are $N$ visually identical coins, with $3^{k-1}<N \leq 3^{k}$, and one of them is lighter, then it can be found in $k$ weighings. Base case: $k=0, N=1$, no weighing is needed for a single coin. Inductive step: suppose the statement is... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,099 |
5. Find the sum of all numbers of the form $x+y$, where $x$ and $y$ are natural number solutions to the equation $5 x+17 y=307$. | Solution. We solve the auxiliary equation $5 x+17 y=1$. For example, its solutions can be 7 and 2. Multiply them by 307, and consider linear combinations for integer $t$, we get values in natural numbers
$\left\{\begin{array}{l}x=7 \cdot 307-17 t, \\ y=-2 \cdot 307+5 t,\end{array} t \in Z, x>0, y>0 \Rightarrow t \in\{... | 164 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,100 |
6. What is the minimum length that segment $AB$ can have if point $A$ belongs to the curve $10\left(x^{2}+y^{2}\right)+60 x-80 y+249=0$, and point $B$ belongs to the graph of the function $y=\frac{1}{3}|x|$? In the answer, write the square of the found length. | Solution. The curve $10(x+3)^{2}+10(y-4)^{2}=1$ is a circle with radius $R=1 / \sqrt{10}$ and center at point $O(-3 ; 4)$. OB is the segment of the perpendicular from point $O$ to the line $\quad y=-x / 3$. Point $A$ is the intersection of $O B$ with the circle. The smallest length is the length of segment $A B$. Angle... | 6.4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,101 |
7. In triangle $A B C$, the altitudes $A D, B E, C F$ are drawn. The length of side $B C$ is 6. The distances from the center of the inscribed circle of triangle $D E F$ to points $B$ and $C$ are $3 \sqrt{2}-\sqrt{6}$ and $2 \sqrt{6}$, respectively. Find the height of triangle $D E F$ drawn to side $D E$. (16 points) | Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $B O=3 \sqrt{2}-\sqrt{6},... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,102 |
8. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
y=\frac{x+1}{|x|-1} \\
|x+y+a|+|x-y-a|=1
\end{array}\right.
$$
has a unique solution. In your answer, specify the smallest of all the obtained values of the parameter $a$. | Solution. We will solve the given system of equations graphically.
Let's plot the graphs of the functions a) $y=\frac{x+1}{|x|-1}$ and b) $|x+y+a|+|x-y-a|=1$.
a) $y=\frac{x+1}{|x|-1}=\left\{\begin{array}{l}\frac{x+1}{x-1}=1+\frac{2}{x-1}, \\ \frac{x+1}{-x-1}=-1, \text { for } x<0\end{array}\right.$
For $x>1$ there a... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,103 |
9. Given a regular quadrilateral pyramid $T A B C D$ with base $A B C D$, where $A B=9 / 2$. On its height $T O$, a point $T_{1}$ is chosen such that $T T_{1}=T O / 3$. Points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ divide the segments $O A, O B, O C$, and $O D$, respectively, in the ratio $1: 2$, counting from point $O$. F... | Solution. Let $\alpha$ be the plane of the section, $\rho$ be the distance from point $C$ to this plane of the section, $a$ be the side of the base of the pyramid $T A B C D, \varphi$ be the angle between the plane of the section and the base of the pyramid. $\rho=4 \sqrt{\frac{5}{13}}$. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,104 |
# 4. Solution:
Note that $\sin x \neq 0, \cos x \neq 0$, and multiply both sides of the equation by $\operatorname{tg} x$. We get $\sqrt{1+\operatorname{tg} x}=\sin x+\cos x$. Given that $\sin x+\cos x \geq 0$, both sides of this equation can be squared. Since $\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$... | Answer: $\pm \frac{\pi}{4}+2 \pi n, \frac{3 \pi}{4}+2 \pi n, n \in Z$. | \\frac{\pi}{4}+2\pin,\frac{3\pi}{4}+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,106 |
# 5. Solution:
$$
\begin{aligned}
& \frac{8\left(x^{3}+27\right) \sqrt{x^{2}+8 x+16}}{\left(x^{2}-3 x+9\right)\left(x^{2}+7 x+12\right)} \geq x+3 \Leftrightarrow \frac{8(x+3)\left(x^{2}-3 x+9\right) \sqrt{(x+4)^{2}}}{\left(x^{2}-3 x+9\right)(x+3)(x+4)} \geq x+3 \Leftrightarrow \\
& x \neq-3, \frac{8 | x+4|}{(x+4)} \ge... | Answer: $x \in (-\infty; -11] \cup (-4; -3) \cup (-3; 5]$. | x\in(-\infty;-11]\cup(-4;-3)\cup(-3;5] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,107 |
# 6. Solution:
Let's find the range of the function $z=g(x)=14 \cos 2 x+28 \sin x+15$. The function $g(x)$ is defined for all real numbers. We will make a substitution. Let $t=\sin x$. Then $z=14\left(1-2 t^{2}\right)+28 t+15=29-28\left(t^{2}-t\right)=36-28(t-0.5)^{2}$ for $t \in[-1 ; 1]$, and $E_{g}=[-27 ; 36]$. The ... | Answer: $E_{f}=[0,5 ; 1]$. | E_{f}=[0.5;1] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,108 |
10. The base of the pyramid $TABC$ is the triangle $ABC$, all sides of which are equal to $a$, and the height of the pyramid coincides with the lateral edge $TA$. Find the area of the section of the pyramid by a plane that passes through the midpoint of the base side $AC$, is parallel to the median $AM$ of the lateral ... | Solution: Draw
$P N \| A M, S=(P N) \cap(A B)$. Let $H-$ be the midpoint
$A B, \quad M H=1 / 2 \cdot A T$,
$A N=1 / 4 \cdot A T=1 / 2 \cdot M H$.
$\Rightarrow \quad S A=1 / 2 \cdot A H=1 / 4 \cdot A B$.
Since $P N=3 / 4 \cdot A M$, $S N=1 / 2 A M \Rightarrow S N=2 / 3 \cdot P N$, or $S N=2 / 5 \cdot S P$.
Draw $S... | \frac{19^{3}}{64\sqrt{3^{2}-112^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,111 |
4. Solution: Note that $\sin x \neq 0, \cos x \neq 0$, and multiply both sides of the equation by $\operatorname{ctg} x$. We get $\sqrt{1+\operatorname{ctg} x}=\sin x+\cos x$. Given that $\sin x+\cos x \geq 0$, both sides of this equation can be squared. Since $\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$,... | Answer: $x= \pm \frac{\pi}{4}+2 \pi n, \quad x=\frac{3 \pi}{4}+2 \pi n, n \in Z$. | \\frac{\pi}{4}+2\pin,\quad\frac{3\pi}{4}+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,112 |
5. Solution: $\frac{6\left(x^{3}-8\right) \sqrt{x^{2}+6 x+9}}{\left(x^{2}+2 x+4\right)\left(x^{2}+x-6\right)} \geq x-2 . \Leftrightarrow \frac{6(x-2)\left(x^{2}+2 x+4\right) \sqrt{(x+3)^{2}}}{\left(x^{2}+2 x+4\right)(x-2)(x+3)} \geq x-2 \Leftrightarrow$ $x \neq 2, \quad \frac{6|x+3|}{(x+3)} \geq x-2 \Leftrightarrow\lef... | Answer: $x \in (-\infty; -4] \cup (-3; 2) \cup (2; 8]$. | x\in(-\infty;-4]\cup(-3;2)\cup(2;8] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,113 |
6. Solution: Let's find the range of the function $z=g(x)=\cos 2x - 2 \sin x$. The function $g(x)$ is defined for all real numbers. We will make a substitution. Let $t=\sin x$. Then
$z=(1-2t^2)-2t=1-2(t^2+t)=1.5-2(t+0.5)^2$ for $t \in [-1; 1]$, and $E_g=[-3; 1.5]$. The function $u=\frac{\pi}{9} z$ takes all values from... | Answer: $E_{f}=[0,5 ; 1]$. | E_{f}=[0.5;1] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,114 |
4. Solution: Note that $\sin x \neq 0, \cos x \neq 0$, and multiply both sides of the equation by $\operatorname{tg} x$. We get $\sqrt{2+\cos 2 x+\sqrt{3} \operatorname{tg} x}=\sin x+\sqrt{3} \cos x$. Under the condition $\sin x+\sqrt{3} \cos x \geq 0$, both sides of this equation can be squared. Since $\sin x+\sqrt{3}... | Answer: $x= \pm \frac{\pi}{4}+2 \pi n, \quad n \in Z$. | \\frac{\pi}{4}+2\pin,\quadn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,118 |
# 5. Solution:
$$
\frac{-3\left(x^{2}-4 x+16\right)\left(x^{2}+6 x+8\right)}{\left(x^{3}+64\right) \sqrt{x^{2}+4 x+4}} \leq x^{2}+x-3 \Leftrightarrow \frac{-3\left(x^{2}-4 x+16\right)(x+2)(x+4)}{(x+4)\left(x^{2}-4 x+16\right) \sqrt{(x+2)^{2}}} \leq x^{2}+x-3 \Leftrightarrow
$$
$x \neq-4, \frac{-3(x+2)}{|x+2|} \leq x^... | Answer: $x \in(-\infty ;-4) \cup(-4 ;-3] \cup(-2 ;-1] \cup[0 ;+\infty)$.
# | x\in(-\infty;-4)\cup(-4;-3]\cup(-2;-1]\cup[0;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,119 |
# 6. Solution:
Let's find the range of the function $z=g(x)=30+14 \cos x-7 \cos 2 x$. The function $g(x)$ is defined for all real numbers. We will make a substitution. Let $t=\cos x$. Then $z=30+14 t-7\left(2 t^{2}-1\right)=37-14\left(t^{2}-t\right)=40.5-14(t-0.5)^{2}$ for $t \in[-1 ; 1]$, and $E_{g}=[9 ; 40.5]$. The ... | Answer: $E_{f}=[0,5 ; 1]$. | E_{f}=[0.5;1] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,120 |
# 5. Solution:
$\frac{-2\left(x^{2}-5 x+25\right)\left(2 x^{2}+17 x+35\right)}{\left(x^{3}+125\right) \sqrt{4 x^{2}+28 x+49}} \leq x^{2}+3 x-2 . \Leftrightarrow \frac{-2\left(x^{2}-5 x+25\right)(x+5)(2 x+7)}{(x+5)\left(x^{2}-5 x+25\right) \sqrt{(2 x+7)^{2}}} \leq x^{2}+3 x-2 \Leftrightarrow$
$x \neq-5, \frac{-2(2 x+7... | Answer: $x \in(-\infty ;-5) \cup(-5 ;-4] \cup(-3.5 ;-3] \cup[0 ;+\infty)$. | x\in(-\infty;-5)\cup(-5;-4]\cup(-3.5;-3]\cup[0;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,123 |
# 6. Solution:
Let's find the range of the function $z=g(x)=4.5-2 \cos x+\cos 2 x$. The function $g(x)$ is defined for all real numbers. We will make a substitution. Let $t=\cos x$. Then $z=4.5-2 t+(2 t^{2}-1)=3.5+2(t^{2}-t)=3+2(t-0.5)^{2}$ for $t \in[-1 ; 1]$, and $E_{g}=[3 ; 7.5]$. The function $u=\frac{\pi}{9} z$ t... | Answer: $E_{f}=[0,5 ; 1]$.
# | E_{f}=[0.5;1] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,124 |
# Variant №1
№1: (15 points). Solve the inequality:
$3 \sqrt{(2 x-3)^{2}}+\sqrt[6]{\sqrt{x^{3}-x}+\sqrt[4]{x-x^{2}}-x^{3}+3 x-2} \leq 9-6 x$ | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,127 | |
# Variant №3
1. (15 points). Solve the inequality:
$2 \sqrt{(4 x-9)^{2}}+\sqrt[4]{\sqrt{3 x^{2}+6 x+7}+\sqrt{5 x^{2}+10 x+14}+x^{2}+2 x-4} \leq 18-8 x$ | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,128 | |
1. (10 points) The creative competition at the institute consisted of four tasks. In total, there were 70 applicants. The first test was successfully passed by 35, the second by 48, the third by 64, and the fourth by 63 people, with no one failing all 4 tasks. Those who passed both the third and fourth tests were admit... | Solution. 1st and 2nd tasks were solved by at least $35+48-70=13$ people. 3rd and 4th - at least $64+63-70=57$ people. No one failed all tasks, so 1st and 2nd were solved by 13 people, 3rd and 4th - 57 people.
Answer: 57 people.
Criteria.
| Points | Conditions for awarding |
| :--- | :--- |
| 10 points | Justified s... | 57 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,129 |
2. (15 points) Let $f(x)=x^{2}-5 x+2020$. Solve the equation $f(3-x)=f(3 x-1)$ | Solution: Let the equation be of the form $f(a)=f(b)$, we get:
$a^{2}-5 a+2020=b^{2}-5 b+2020$
$a^{2}-b^{2}-5 a+5 b=0$
$(a-b)(a+b-5)=0$
The product of two factors is zero if at least one of the factors is zero. Therefore, $a=b$ or $a+b=5$.
$\left[\begin{array}{l}a=b \\ a+b=5\end{array} \Leftrightarrow\left[\begin{... | 1.51 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,130 |
3. (15 points) In a convex quadrilateral $A B C D, A B=10, C D=15$. Diagonals $A C$ and $B D$ intersect at point $O, A C=20$, triangles $A O D$ and $B O C$ have equal areas. Find $A O$.
# | # Solution:
From the equality of the areas of triangles $A O D$ and $B O C$ and the equality of angles $\angle A O D=\angle B O C$, it follows that $\frac{A O \cdot O D}{B O \cdot O C}=1$ (by the theorem on the ratio of areas of triangles with one equal angle). From this, we get $\frac{A O}{O C}=\frac{B O}{O D}$. Addi... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,131 |
4. (20 points) For what values of the parameter \( a \) does the equation \(\left|\frac{-4 x^{4}-(6 a+10) x^{3}+(16-4 a) x^{2}-\left(6 a^{2}-14 a-40\right) x}{\left(4-x^{2}-a\right)(3 a+2 x+5)}\right|=\sqrt{a^{2}-2 a+1}\) have one solution?
# | # Solution
Transform
$$
\begin{gathered}
\left|\frac{-4 x^{4}-(6 a+10) x^{3}+(16-4 a) x^{2}-\left(6 a^{2}-14 a-40\right) x}{\left(4-x^{2}-a\right)(3 a+2 x+5)}\right|= \\
=\sqrt{a^{2}-2 a+1}
\end{gathered}
$$
 Given 10 natural numbers, the sum of any four of them is even. Can the product of all ten numbers end in 1580? Justify your answer.
# | # Solution:
It is clear that the parity of all numbers is the same. Otherwise, if there are numbers with different parities, then there are at least 3 numbers of at least one of these parities (by the pigeonhole principle). In such a case, we can take these 3 numbers and 1 number of the other parity, and the sum of th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,134 |
1. (10 points) In one of the regions on the planet, seismic activity was studied. 80 percent of all days were quiet. The instrument predictions promised a calm situation in 64 out of 100 cases; moreover, in 70 percent of all cases when the day was quiet, the instrument predictions came true. What percentage of days wit... | Solution. Let the total number of observed days be x. The number of actually quiet days was $0.8x$, and seismically active days were $0.2x$. The predictions of quiet days matched the actually quiet days: $0.7 \cdot 0.8x = 0.56x$. Then the number of active days that did not match the predictions was $0.64x - 0.56x = 0.0... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,135 |
2. (15 points) Solve the system of equations: $\left\{\begin{array}{l}y^{2}+x y=15 \\ x^{2}+x y=10\end{array}\right.$ | Solution: By adding the equations, we get: $(x+y)^{2}=25$ or $x+y= \pm 5$.
Therefore, the system of equations splits into two:
$\left[\begin{array}{l}\left\{\begin{array}{l}x+y=5 \\ x^{2}+x y=10\end{array}\right. \\ \left\{\begin{array}{l}x+y=-5 \\ x^{2}+x y=10\end{array}\right.\end{array} \Leftrightarrow\left[\begin... | (2,3),(-2,-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,136 |
3. (15 points) In triangle $P Q R$, the median $P A$ and the angle bisector $Q B$ (where $A$ and $B$ are the points of intersection with the corresponding sides of the triangle) intersect at point $O$. It is known that $3 P Q = 5 Q R$. Find the ratio of the area of triangle $P Q R$ to the area of triangle $P Q O$.
# | # Solution:
From the equality $3 P Q=5 Q R$ and the fact that $P A$ is a median, it follows that $P Q=10 x, \quad Q A=A R=3 x$, where $x$ is a proportionality coefficient. By the property of... | 2.6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,137 |
4. (20 points) For what values of the parameter \( a \) does the equation \(\frac{\sqrt{x-1} \cdot\left(\left|x^{2}-10 x+16\right|-a\right)}{a x^{2}-7 x^{2}-10 a x+70 x+21 a-147}=0\) have three solutions? | # Solution.
Transform $\frac{\sqrt{x-1} \cdot\left(\left|x^{2}-10 x+16\right|-a\right)}{a x^{2}-7 x^{2}-10 a x+70 x+21 a-147}=$
$=\frac{\sqrt{x-1} \cdot\left(\left|x^{2}-10 x+16\right|-a\right)}{x^{2}(a-7)-10 x(a-7)+21(a-7)}=$
$=\frac{\sqrt{x-1} \cdot\left(\left|x^{2}-10 x+16\right|-a\right)}{(a-7)(x-7)(x-3)}$
Solv... | =0,=5,=9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,138 |
5. (20 points) The quadrilateral connecting the midpoints of the sides of trapezoid $\mathrm{ABCD}$ is a rhombus. Find its area if the height of the trapezoid $\mathrm{BH}=5 \mathrm{~cm}$, the smaller base $\mathrm{BC}=6 \mathrm{~cm}$, and the angle $\mathrm{ABC}$ is $120^{\circ}$.
# | # Solution.
$\mathrm{H} \quad \mathrm{P}$ | 15\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,139 |
6. (20 points) The average life expectancy in the country of Gondor is 64 years. And the average life expectancy in the country of Númenor is 92 years. The average life expectancy in these two countries is 85 years. How many times does the population of Gondor differ from the population of Númenor
# | # Solution:
Let the population of Númenor be n, and the population of Gondor be $g$, then:
$64 \mathrm{~g}+92 \mathrm{n}=85(\mathrm{~g}+\mathrm{n})$
$64 g+92 n=85 g+85 n$
$7 \mathrm{n}=21 \mathrm{~g}$
$\mathrm{n}=3 \mathrm{~g}$
Answer: The population of Gondor is 3 times smaller
## Criteria:
| Points | Conditio... | \mathrm{n}=3\mathrm{~} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,141 |
1. A car left point A for point B, and a second car left with some delay. When the first car had traveled half the distance, the second car had traveled $26 \frac{1}{4}$ km, and when the second car had traveled half the distance, the first car had traveled $31 \frac{1}{5}$ km. After overtaking the first car, the second... | Solution. S - the distance between points A and B.
$$
\frac{S-2-S / 2}{S+2-26.25}=\frac{S-2-31.2}{S+2-S / 2}, \quad 5 S^{2}-383 S+5394=0, \quad \sqrt{D}=197, \quad S=58
$$
Answer: 58. | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,143 |
2. Solve the equation $\sqrt{8 x+5}+2\{x\}=2 x+2$. Here $\{x\}$ is the fractional part of the number $\mathrm{x}$, i.e., $\{x\}=x-[x]$. In the answer, write the sum of all solutions. | Solution. The equation is equivalent to the following $\sqrt{8 x+5}-2[x]-2=0, \quad[x]=n \in \mathbb{N}$, $n \leq x<n+1,2 n+2=\sqrt{8 x+5}$, for $n \geq-1$ we have $4 n^{2}+8 n-1=8 x$. Substituting into the double inequality
$$
8 n \leq 4 n^{2}+8 n-1<8 n+8, \quad 1 \leq 4 n^{2}<9, \quad \text { we get } \quad n=1 \qua... | 0.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,144 |
3. Find the largest integer \( a \) such that the expression
\[
a^{2}-15 a-(\tan x-1)(\tan x+2)(\tan x+5)(\tan x+8)
\]
is less than 35 for any value of \( x \in (-\pi / 2, \pi / 2) \).
(6 points) | Solution. Let's make the substitution $t=\operatorname{tg} x$. We need to determine for which values of $a$ the inequality $a^{2}-15 a-(t-1)(t+2)(t+5)(t+8)a^{2}-15 a-35,\left(t^{2}+7 t-8\right)\left(t^{2}+7 t+10\right)>a^{2}-15 a-35$
$z=t^{2}+7 t+1,(z-9)(z+9)>a^{2}-15 a-35, z^{2}>a^{2}-15 a+46$, $0>a^{2}-15 a+46, \sqr... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,145 |
4. Given six socks, all of different colors and easily stretchable. You cannot turn them inside out. In how many ways can you put on 3 socks on each foot, considering which one to put on earlier and which one later? | Solution. There is a sequence of 6 sock puttings on: $\mathrm{C}_{6}^{3}=20$ ways to choose which puttings are for the right foot. For each such choice, there are $6!=720$ ways to choose which sock to take for each putting.
Answer: 14400. | 14400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,146 |
5. Let $x, y, z$ be the roots of the equation $t^{3}-2 t^{2}-9 t-1=0$. Find $\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z}$.
(12 points) | Solution. Let's bring the desired expression to a common denominator: $\frac{y^{2} z^{2}+x^{2} z^{2}+x^{2} y^{2}}{x y z}$. The polynomial has 3 different real roots, since $\mathrm{P}(-100)0, \mathrm{P}(0)0$. By Vieta's theorem $x+y+z=2, x y+x z+y z=-9, x y z=1$.
$$
\begin{aligned}
& x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2... | 77 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,147 |
6. On the plane $x O y$, the lines $y=3 x-3$ and $x=-1$ intersect at point $\mathrm{B}$, and the line passing through point $M(1 ; 2)$ intersects the given lines at points A and C respectively. For what positive value of the abscissa of point A will the area of triangle $\mathrm{ABC}$ be the smallest?
(12 points) | # Solution.
$A C: \quad y=k x+d, \quad M \in A C \Rightarrow d=2-k$
$A(a ; 3 a-3) \in A C \Rightarrow 3 a-3=k a+2-k \Rightarrow a=\frac{5-k}{3-k}$,
$C(-1 ; c) \in A C \Rightarrow c=-2 k+2$,
$S_{A B C}=\frac{1}{2}(c+6) \cdot(a+1)=\frac{2(k-4)^{2}}{3-k}$
$S^{\prime}=\frac{2(k-4)(2-k)}{(3-k)^{2}}=0, k_{\min }=2, \qua... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,148 |
8. Specify the smallest integer value of \(a\) for which the system has a unique solution
\[
\left\{\begin{array}{l}
\frac{y}{a-\sqrt{x}-1}=4 \\
y=\frac{\sqrt{x}+5}{\sqrt{x}+1}
\end{array}\right.
\] | # Solution.
Solving the system by substitution, we arrive at an equation with constraints on the unknown quantity ${ }^{x}$.
$$
\left\{\begin{array} { l }
{ \frac { y } { a - \sqrt { x } - 1 } = 4 } \\
{ y = \frac { \sqrt { x } + 5 } { \sqrt { x } + 1 } }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ x \geq ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,149 |
9. The base of the pyramid $\mathrm{TABCD}$ is an isosceles trapezoid $\mathrm{ABCD}$, the midline of which is equal to $5 \sqrt{3}$. The ratio of the areas of the parts of the trapezoid $\mathrm{ABCD}$, into which it is divided by the midline, is $7: 13$. All lateral faces of the pyramid $\mathrm{TABCD}$ are inclined ... | # Solution.
Let $TO$ be the height of the pyramid. Since all lateral faces are inclined to the base at the same angle, $O$ is the center of the circle inscribed in the base. Let $MP$ be the midline of the trapezoid, $\quad AD=a, BC=b. \quad$ According to the problem, we have $S_{MB CP}=7x, S_{\text{AMPD}}=13x, \quad \... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,150 |
1. A group of schoolchildren heading to a school camp was to be seated in buses so that each bus had the same number of passengers. Initially, 22 people were to be seated in each bus, but it turned out that three schoolchildren could not be seated this way. When one bus left empty, all the schoolchildren were able to s... | Answer in the form of a number without indicating the unit.
(5 points) | 396 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,151 |
1. A group of schoolchildren heading to a school camp was to be seated in buses so that each bus had the same number of passengers. Initially, 22 people were to be seated in each bus, but it turned out that three schoolchildren could not be seated. When one bus left empty, all the schoolchildren were able to sit evenly... | Solution. Let $\mathrm{n}$ be the number of buses, $\mathrm{m}$ be the number of schoolchildren in each bus, and $\mathrm{S}$ be the total number of schoolchildren.
We have
$$
S=22 n+3, \quad S=(n-1) m, n \leq 10, m \leq 36, \quad 22 n+3=(n-1) m, \quad n=1+\frac{25}{m-22}
$$
Considering the constraints on $\mathrm{n... | 135 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,152 |
2. Find all pairs of integers $(x, y)$ that satisfy the equation $x^{2}-x y-6 y^{2}-11=0$. For each pair $(x, y)$ found, calculate the product $x y$. In the answer, write the sum of these products. | Solution. $x^{2}-x y-6 y^{2}-11=0,(x-3 y)(x+2 y)=11$. Since $x$ and $y$ are integers, we have four cases:
$\left\{\begin{array}{c}x-3 y=11 \\ x+2 y=1\end{array} \Leftrightarrow\left\{\begin{array}{c}y=-2 \\ x=5\end{array}\right.\right.$
2)
$\left\{\begin{array}{c}x-3 y=-11 \\ x+2 y=-1,\end{array} \Leftrightarrow\left\{... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,153 |
3. Let $g(x)=\frac{2}{x^{2}-8x+17}$. Find all possible values of the parameter $a$ such that the inequality holds for all real $x$. In your answer, write the difference between the largest and smallest possible values of the parameter $a$.
(6 points) | Solution. The function
$$
g(x)=\frac{2}{x^{2}-8 x+17}=\frac{2}{(x-4)^{2}+1}
$$
is defined on the entire number line and takes all values from the interval $(0 ; 2]$. The function $g(x)$ reaches its maximum value at the point $x=4, g_{\max }=g(4)=2$, on the interval $(-\infty ; 4)$ the function $g(x)$ is increasing, o... | 2.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,154 |
4. In how many ways can the line $x \sin \sqrt{16-x^{2}-y^{2}}=0$ be drawn without lifting the pencil and without retracing any part of the line? (12 points) | Solution. Since $\pi^{2}<16<(2 \pi)^{2}$, the given line consists of 2 circles with radii 4 and $\sqrt{16-\pi^{2}}$ and a vertical segment.
This line is unicursal, as it has only 2 odd poin... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,155 |
5. For how many two-digit natural numbers n are exactly two of these three statements true: (A) n is odd; (B) n is not divisible by $3 ;$ (C) n is divisible by 5? | Solution. We can consider the first 30 two-digit numbers (from 10 to 39), and then multiply the result by 3, since the remainders when dividing by 2, 3, and 5 do not change when shifted by 30 or 60. There are three mutually exclusive cases.
1) (A) and (B) are satisfied, and (C) is not. From (A) and (B), it follows tha... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,156 |
6. What is the smallest area that triangle $O A B$ can have, if its sides $O A_{\text {and }} O B$ lie on the graph of the function $y=2|x|-x+1$, and the line $A B_{\text {passes through the point }} M(0 ; 2)$? | Solution.
$S_{\text {AOB }}=S_{\text {AOM }}+S_{\text {BOM }}, \quad A(a ; a+1), B(b ;-3 b+1)$,
$S_{AOM}=\frac{1}{2} OM \cdot a, \quad S_{BOM}=\frac{1}{2} OM \cdot(-b), \quad OM=1$,
$S_{AOB}=\frac{a-b}{2}$.
The line $AB_{\text {passes through }}$ point $\mathrm{M}$, its equation is $y=k x+2$. Express the variables ... | 0.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,157 |
8. Find all integer values of the parameter \(a\) for which the system has at least one solution
\[
\left\{\begin{array}{l}
y-2=x(x+2) \\
x^{2}+a^{2}+2 x=y(2 a-y)
\end{array}\right.
\]
In the answer, specify the sum of the found values of the parameter \(a\). | Solution. Transform the system
$$
\left\{\begin{array} { l }
{ y - 1 = ( x + 1 ) ^ { 2 } , } \\
{ ( x + 1 ) ^ { 2 } + ( y - a ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
y-1=(x+1)^{2} \\
y-2+(y-a)^{2}=0
\end{array}\right.\right.
$$
Consider the second equation of the system
$$
y^{2}-y(2 a-1)+a^{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,158 |
9. The base of the pyramid $\mathrm{TABCD}$ is an isosceles trapezoid $\mathrm{ABCD}$, the length of the larger base $A D$ of which is $12 \sqrt{3}$. The ratio of the areas of the parts of the trapezoid $A B C D$, into which it is divided by the midline, is $5: 7$. All lateral faces of the pyramid TABCD are inclined to... | # Solution.
Let TO be the height of the pyramid. Since all lateral faces are inclined to the base at the same angle, O is the center of the circle inscribed in the base. Let MP be the midline of the trapezoid, $A D=a=12 \sqrt{3}, B C=b$.
According to the problem, we have
$S_{\text {MBCP }}=5 x, S_{\text {AMPD }}=7 x... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,159 |
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