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1. Given two groups of numerical sequences, each consisting of 15 arithmetic progressions containing 10 terms each. The first terms of the progressions in the first group are $1, 2, 3, \ldots, 15$, and their differences are respectively $2, 4, 6, \ldots, 30$. The second group of progressions has the same first terms $1... | Solution. Let $a_{1}, \ldots, a_{15}$ be the first terms of the progressions, $d_{1}, \ldots, d_{15}$ be the differences of the first group, and $p_{1}, \ldots, p_{15}$ be the differences of the second group. Then the sums of the first group are
$S_{1}=\frac{2 a_{1}+9 d_{1}}{2} \cdot 10=10 a_{1}+45 d_{1}, \ldots, S_{1... | \frac{160}{151} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,818 |
2. Two telephone companies agreed to release sets of all possible three-digit telephone numbers so that the first company would have all numbers consisting of odd digits, and the second company only even digits, except for 0. The first company sold its entire first set for $X$ rubles per number, the second - its entire... | Solution. a) There are 5 odd digits: $1,3,5,7,9$ and any of them can stand on each of the 3 positions, the total number of combinations is $N_{\text {odd }}=5^{3}=125$,
b) Even digits are 2,4,6,8, so 4 digits can stand on the first position, and any even digit out of 4 can stand further, the total number of combinatio... | (41;80),(105;205) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,819 |
3. Solve the equation $3 \sqrt{(x+1)^{2}}+\sqrt{x^{2}-4 x+4}=|1+6 x|-4|x-1|$. | Solution: Rewrite the equation as $|3 x+3|+|x-2|+|4 x-4|=|1+6 x|$. Notice that $3 x+3-x+2+4 x-4=1+6 x$, so the solution to the equation is found from the system $\left\{\begin{array}{l}3 x+3 \geq 0, \\ -x+2 \geq 0, \\ 4 x-4 \geq 0, \\ 1+6 x \geq 0 .\end{array} \Rightarrow\left\{\begin{array}{l}x \geq-1, \\ 2 \geq x, \\... | x\in[1,2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,820 |
4. Solve the inequality $4+x^{2}+2 x \sqrt{2-x^{2}}<8 \sqrt{2-x^{2}}+5 x$. points) | Solution. Rewrite it as $4+x^{2}-5 x+2 x \sqrt{2-x^{2}}-8 \sqrt{2-x^{2}}=0$. Solve the second inequality of the system
$(x-1)+2 \sqrt{2-x^{2}}>0 \Rightarrow 2 \sqrt{2-x^{2}}>1-x \Rightarrow\left[\begin{array}{l}\left\{\begin{array}{l}1-x \leq 0 \\ 2 \sqrt{2-x^{2}}>0\end{array}\right. \\ 1-x>0, \\ 2 \sqrt{2-x^{2}}>1-x,... | (-1;\sqrt{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,821 |
1. A telephone company decided to organize the procedure for issuing telephone numbers in a new microdistrict as follows: residents of apartments with even numbers will be given seven-digit numbers consisting only of even digits, and residents of apartments with odd numbers will be given seven-digit numbers consisting ... | Solution. a) Even digits are $0,2,4,6,8$, a seven-digit number cannot start with 0, so there can be 4 digits in the first place, and any of the 5 even digits can be used afterwards, the total number of combinations is $N_{\text {even }}=4 \cdot 5^{6}$,
b) there are 5 odd digits: $1,3,5,7,9$ and any of them can stand o... | )62500,b)78125,)1.25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,823 |
2. Find all values of $n, n \in N$, for which the sum of the first terms of the sequence $a_{k}=3 k^{2}-3 k+1, \quad k \in N$, is equal to the sum of the first $n$ terms of the sequence $b_{k}=2 k+89, k \in N . \quad(12$ points) | Solution. Note that $a_{k}=3 k^{2}-3 k+1=k^{3}-(k-1)^{3}$, and the sum is $S_{n}=n^{3}$. For the second sequence, $b_{k}=2 k+89=(k+45)^{2}-(k+44)^{2}$, the sum is $S_{n}=(n+45)^{2}-45^{2}=n(n+90)$.
We get the equation $n^{3}=n(n+90) \Rightarrow n^{2}-n-90=0 \Rightarrow n=10$.
Answer: 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,824 |
3. Solve the equation
$3 \sqrt{6 x^{2}+13 x+5}-6 \sqrt{2 x+1}-\sqrt{3 x+5}+2=0$ | Solution. Note that $6 x^{2}+13 x+5=(2 x+1)(3 x+5)$. Let's introduce the following notations:
$$
\left\{\begin{array}{l}
m=\sqrt{2 x+1} ; n=\sqrt{3 x+5} \\
m \geq 0 \\
n \geq 0
\end{array}\right.
$$
Then the equation can be represented as $3 m n-6 m-n+2=0$. Considering the domain of definition, we solve this equation... | -1/3;-4/9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,825 |
4. Solve the inequality
$$
\frac{x(x-2)(x-3)-(x-2)^{2}+1}{(|x-1|-|x-2|) \sqrt{16-x^{2}}} \geq 0
$$
points) | Solution. Domain of definition: $\left\{\begin{array}{l}|x-1| \neq|x-2|, \\ 16-x^{2}>0,\end{array} \Rightarrow\left\{\begin{array}{l}x \neq 3 / 2, \\ x \in(-4,4) .\end{array}\right.\right.$
Transform the numerator
$x(x-2)(x-3)+1-(x-2)^{2}=x(x-2)(x-3)+(1-x+2)(1+x-2)=$
$=x(x-2)(x-3)+(3-x)(x-1)=(x-3)\left(x^{2}-2 x+1-x... | x\in(-4;\frac{3-\sqrt{5}}{2}]\cup(\frac{3}{2};\frac{3+\sqrt{5}}{2}]\cup[3;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,826 |
5. Determine all values of the parameter $p$ for which the system
$$
\left\{\begin{array}{l}
(x-p)^{2}=16(y-3+p) \\
y^{2}+\left(\frac{x-3}{|x|-3}\right)^{2}=1
\end{array}\right.
$$
has solutions and solve it for each of the found values of $p$.
points) | Solution. Domain of definition: $|x| \neq 3$. For $x \geq 0 \quad y^{2}+1=1 \Rightarrow y=0$. If $x0, \\ x_{1} x_{2}=p^{2}-16 p+48=(p-12)(p-4)0, \\ x_{1} x_{2}=p^{2}-16 p+48=(p-12)(p-4) \geq 0, \Rightarrow \\ x_{1}+x_{2}>0\end{array}\right.$
Thus, we get the answer:
$x_{1,2}=p \pm 4 \sqrt{p-3}, p \in(3 ; 4] \cup[12 ;... | p\in(3;4]\cup[12;19)\cup(19;\infty),x_{1,2}=p\4\sqrt{p-3},\quad0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,827 |
6. A circle with center $O_{1}$ and radius 2, inscribed in triangle $A B C$, touches side $B C$ at point $D$. A second circle with center $O_{2}$ and radius 4 touches the extensions of sides $A B$ and $A C$, as well as side $B C$ at point $E$. Find the area of quadrilateral $O_{1} D O_{2} E$, if the measure of angle $A... | # Solution.
$S=S_{\square O_{1} E D}+S_{\square O_{2} E D}=1 / 2 \cdot E D \cdot\left(O_{1} D+O_{2} E\right)=3 E D$
$E D=E C-D C=O_{2} E / \operatorname{tg} 30-O_{1} D / \operatorname{tg} 60=4 \sqrt{3}-2 / \sqrt{3}=10 / \sqrt{3}$ $S=30 / \sqrt{3}$
Answer: $30 / \sqrt{3}$.
^{3}-x^{3}-y^{3}-z^{3}=18 .\end{array}\right.\right.$
$(x+y+z)^{3}-x^{3}-y^{3}-z^{3}=18 \Leftrightarrow\left((x+y+z)^{3}-x^{3}\right)-\left(y^{3}+z^{3}\right)=18 \Leftright... | (3,3,-4),(3,-4,3),(-4,3,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,829 |
3. A student wrote a program for recoloring a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 21, green - 45, blue - 75, indigo - 87, violet - 91, black - 12... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$ where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text { times }}(n) \ldots)-k$-fold composition of the function $f(n)$, which is equal to $n+4$ when $n \leq 19$, and equal to $|129-2 n|$ when $n \geq 20$. Let's compute and write down the first few valu... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,831 |
4. Find all pairs of natural numbers $a$ and $b$, for which out of the four statements
5) $a^{2}+6 a+8$ is divisible by $b$;
6) $\quad a^{2}+a b-6 b^{2}-15 b-9=0$
7) $a+2 b+2$ is divisible by 4;
8) $a+6 b+2$ is a prime number
three are true, and one is false. | Solution. Statements 3) and 4) cannot both be true. If 3) is true, then $a+6b+2=(a+2b+2)+4b$ is divisible by 4 and is not a prime number. Therefore, one of the statements 3) or 4) is false.
Let's consider statement 2): $a^{2}+a b-6 b^{2}-15 b-9=0$; Consider this equation as a quadratic equation in terms of $a$. Calcul... | =5,b=1;=17,b=7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,832 |
5. Find all values of the parameter $a$ for which the equation $\left(\left(1-x^{2}\right)^{2}+2 a^{2}+5 a\right)^{7}-\left((3 a+2)\left(1-x^{2}\right)+3\right)^{7}=5-2 a-(3 a+2) x^{2}-2 a^{2}-\left(1-x^{2}\right)^{2}$ has two distinct solutions on the interval $\left[-\frac{\sqrt{6}}{2} ; \sqrt{2}\right]$. Specify the... | Solution: $\left(\left(1-x^{2}\right)^{2}+2 a^{2}+5 a\right)^{7}-\left((3 a+2)\left(1-x^{2}\right)+3\right)^{7}=5-2 a-(3 a+2) x^{2}-2 a^{2}-\left(1-x^{2}\right)^{2} \Leftrightarrow$ $\left(\left(1-x^{2}\right)^{2}+2 a^{2}+5 a\right)^{7}+\left(1-x^{2}\right)^{2}+2 a^{2}+5 a=\left((3 a+2)\left(1-x^{2}\right)+3\right)^{7}... | \in[0.25;1),x_{1}=\sqrt{2-2},x_{2}=-\sqrt{2-2}\\\in[-3.5;-2),x_{1}=\sqrt{--2},x_{2}=-\sqrt{--2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,833 |
1. (10 points). Two scientific and production enterprises supply the market with substrates for growing orchids. In the substrate "Orchid-1," pine bark is three times more than sand; peat is twice as much as sand. In the substrate "Orchid-2," bark is half as much as peat; sand is one and a half times more than peat. In... | Solution. In the first substrate, bark, peat, and sand are in the ratio 3:2:1, in the second substrate, they are in the ratio 1:2:3. Let's take a parts of the first mixture and b parts of the second. Then $3a + b = 2a + 2b = a + 3b$. We get $a = b$.
Answer: $1: 1$.
| Points | Grading Criteria |
| :--- | :--- |
| 10 p... | 1:1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,834 |
2. (15 points). Form a quadratic equation whose roots are twice the roots of the equation $2 x^{2}-5 x-8=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: According to Vieta's theorem, $x_{1}+x_{2}=\frac{5}{2}, x_{1} \cdot x_{2}=-4$. Let $t_{1}$ and $t_{2}$ be the roots of the desired quadratic equation, then $t_{1}+t_{2}=2 x_{1}+2 x_{2}=5, t_{1} \cdot t_{2}=2 x_{1} \cdot 2 x_{2}=-16$ and by the inverse Vieta's theorem we get the equation: $x^{2}-5 x-16=0$. The... | x^{2}-5x-16=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,835 |
3. (15 points). On side $A B$ of parallelogram $A B C D$, a point $P$ is taken, and on side $B C$ - a point $Q$ such that $3 A B=7 B P, 3 B C=4 B Q$. Find the ratio $D O: O P$, where point $O$ is the intersection point of segments $A Q$ and $D P$. | # Solution
From the equality $3 A B=7 B P$, it follows that $A B=7 x, B P=3 x$, where $x$ is the coefficient of proportionality. Similarly, from $3 B C=4 B Q$, it follows that $B C=4 y, B Q=3 y$. Draw lines through points $P$ and $D$ parallel to $A Q$, and let points $M$ and $N$ be the points of intersection of these ... | \frac{7}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,836 |
4. (20 points). For which non-negative values of the parameter $a$ does the equation $\left|\frac{x^{3}-10 x^{2}+31 x-30}{x^{2}-8 x+15}\right|=(\sqrt{2 x-a})^{2}+2-2 x \quad$ have one solution? | # Solution.
Transform $\left|\frac{x^{3}-10 x^{2}+31 x-30}{x^{2}-8 x+15}\right|=(\sqrt{2 x-a})^{2}+2-2 x$
$\left|\frac{x\left(x^{2}-10 x+25\right)+6(x-5)}{(x-5)(x-3)}\right|=2 x-a+2-2 x$
$\left|\frac{(x-5)(x-3)(x-2)}{(x-5)(x-3)}\right|=2-a$
Solve the equation $|x-2|=2-a, \quad x \neq 5, x \neq 3, a \leq 2 x$ graphi... | =1,=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,837 |
5. (20 points). In an isosceles trapezoid $\mathrm{ABCD}$, the bases $\mathrm{BC}$ and $\mathrm{AD}$ are 8 cm and 20 cm respectively, and the angle $\mathrm{BAD}$ is $60^{\circ}$. Find the area of the quadrilateral connecting the midpoints of the sides of the trapezoid.
# | # Solution.
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,838 | |
6. (20 points). Solve the equation: $\sqrt{4+2 x}+\sqrt{6+3 x}+\sqrt{8+4 x}=9+\frac{3 x}{2}$ | Solution: Notice that $4+2x=2(2+x); 6+3x=3(2+x); 8+4x=4(2+x)$, by the inequality of means, the left side does not exceed $(2+2+x+3+2+x+4+2+x) / 2=15 / 2+3x / 2$
Answer: no solutions.
## Criteria:
| Points | Conditions for awarding |
| :---: | :--- |
| 20 points | Any complete and correct solution |
| 15 points | The... | nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,840 |
1. (10 points) On a tourist base, the number of two-room cottages is twice the number of one-room cottages. The number of three-room cottages is a multiple of the number of one-room cottages. If the number of three-room cottages is tripled, it will be 25 more than the number of two-room cottages. How many cottages are ... | Solution. Let the number of one-bedroom houses be $x$, two-bedroom houses $2 \mathrm{x}$, and three-bedroom houses nx.
$3 n x-25=2 x ; x(3 n-2)=25 ; \Rightarrow 25:(3 n-2) \Rightarrow 3 n-2=1 ; 5 ; 25$. If
$3 n-2=1 ; n=1 ; x=25 ; 25+50+25=100$ total houses.
If $3 n-2=5 ; 3 n=7$, which is impossible. If $3 n-2=25 ; 3... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,841 |
2. (15 points) For what values of the parameter $a$ is the sum of the squares of the roots of the equation $x^{2}+a x+2 a=0$ equal to $21?$ | Solution: By Vieta's theorem: $x_{1}+x_{2}=-a, x_{1} \cdot x_{2}=2 a, \quad$ therefore, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=a^{2}-4 a$. From the condition, we get that $a^{2}-4 a=21$ or $a \in\{-3 ; 7\}$.
The discriminant of the equation is $D=a^{2}-8 a$, when $a=7$ it is negative, meaning ... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,842 |
3. (15 points) In parallelogram $A B C D$, a point $P$ is taken on side $B C$ such that $3 P B=2 P C$, and a point $Q$ is taken on side $C D$ such that $4 C Q=5 Q D$. Find the ratio of the area of triangle $A P Q$ to the area of triangle $P Q C$.
# | # Solution:
From the equality $3 P B=2 P C$, it follows that $P B=2 x, P C=3 x$, where $x$ is the proportionality coefficient. Similarly, from $4 C Q=5 Q D$, it follows that $C Q=5 y, D Q=4 y... | \frac{37}{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,843 |
6. (20 points) Dasha added 158 numbers and got 1580. Then Seryozha tripled the largest of these numbers and decreased another number by 20. The resulting sum did not change. Find the smallest of the original numbers.
# | # Solution:
Let x be the largest of the original numbers, and y be the number that Seryozha decreased. Then: $x+y=3 x+y-20$, i.e., $x=10$.
Since the arithmetic mean of the original numbers is 10, and the largest of these numbers is also 10, each of the given numbers is 10.
Answer: 10.
## Criteria:
| 20 points | An... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,846 |
1. $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$, but according to Vieta's theorem $\left\{\begin{array}{l}D \geq 0 \\ x_{1}+x_{2}=-(m+1) . \text { Then, } \mathrm{c} \\ x_{1} x_{2}=2 m-2\end{array}\right.$ considering that $D=(m+3)^{2} \geq 0$, we have $x_{1}^{2}+x_{2}^{2}=$
$(-(m+1))^{2}-2(2 m-2)=... | Answer: For the equation $x^{2}+(m+1) x+2 m-2=0$, the smallest sum of the squares of its roots is 4 when $m=1$.
## Grading Criteria.
| 15 points | Correct and justified solution. |
| :--- | :--- |
| 10 points | Using Vieta's theorem, the expression for the sum of the squares of the roots is correctly written, but the... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,847 |
5. Solution: Let Masha have $x$ rubles, and Petya have $y$ rubles, then
$n(x-3)=y+3$
$x+n=3(y-n)$
Express $x$ from the second equation and substitute it into the first:
$n(3 y-4 n-3)=y+3$,
$3 n y-y=4 n^{2}+3 n+3$,
$y=\left(4 n^{2}+3 n+3\right):(3 n-1)$
For $y$ to be an integer, $\left(4 n^{2}+3 n+3\right)$ must b... | Answer: $1 ; 2 ; 3 ; 7$.
| 20 points | Complete justified solution |
| :--- | :--- |
| 15 points | Values of n are found, it is proven that there are no other values, but it is not shown why these values meet the condition of the problem. |
| 12 points | Correct approach to the solution, one value of n is lost. |
| 10... | 1;2;3;7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,851 |
1. Note that when $a=0$, the system is undefined. Let the solution of the system be of the form
$\left(x_{0} ; x_{0}\right)$. Then we get $\left\{\begin{array}{l}3 x_{0}+2 x_{0}=15 a \\ \frac{1}{a} x_{0}+x_{0}=9\end{array}\right.$, which is $\left\{\begin{array}{l}5 x_{0}=15 a \\ \frac{1+a}{a} \cdot x_{0}=9\end{array}... | Answer: when $a=2$ the solution to the system is $(6 ; 6)$.
| Points | Conditions for Awarding |
| :--- | :--- |
| 15 points | The problem is completely correctly solved. |
| 10 points | The value of the parameter is correctly found, but an error is made in calculating the coordinates of the point - the solution to th... | (6;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,852 |
1. According to the theorem inverse to Vieta's theorem, we form a quadratic equation. We get $x^{2}-\sqrt{2019} x+248.75=0$.
Next, solving it, we find the roots $a$ and $b$: $a=\frac{\sqrt{2019}}{2}+\frac{32}{2}$ and $b=\frac{\sqrt{2019}}{2}-\frac{32}{2}$, and consequently, the distance between the points $a$ and $b$:... | Answer: 32
| 15 points | The correct answer is obtained justifiably |
| :---: | :---: |
| 10 points | The quadratic equation is solved, but an arithmetic error is made or the distance between the points is not found |
| 5 points | The quadratic equation is correctly formulated according to the problem statement. |
| 0... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,857 |
6. The expression contains the following products:
$(2 \cdot 5) \cdot(10) \cdot(12 \cdot 15) \cdot(25 \cdot 4) \cdot(20 \cdot 30) \cdot(35 \cdot 14) \ldots \Rightarrow$ since each of these grouped products ends in 0, the number will end in 8 zeros $\Rightarrow B=0$
Let's break down the expression into the following p... | Answer: $\mathbf{A}=\mathbf{2}, \mathbf{B}=\mathbf{0}, \mathbf{C}=\mathbf{1}, \mathrm{D}=\mathbf{4}$.
| 20 points | Correct answer obtained with justification |
| :---: | :--- |
| 15 points | Three out of four digits are correctly found. |
| 10 points | Two out of four digits are correctly found. |
| 5 points | One ou... | A=2,B=0,C=1,D=4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,861 |
1. Initially, the farmer 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 the produce was finally placed i... | Solution. Let $x$ kg be the weight of the farmer's produce. Then $\quad 8(n-1)<x<8 n$, $6(n+7)<x<6(n+8), \quad 5(n+13)=x, \Rightarrow 8(n-1)<5(n+13)<8 n, \quad 6(n+7)<5(n+13)<6(n+8)$, $\Rightarrow 21 \frac{2}{3}<n<23, \quad n=22, \quad x=35 \cdot 5=175$.
Answer: 175. | 175 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,862 |
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 $[-1 ; 2]$.
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 | 21,863 |
3. Find the smallest distance from the point with coordinates $(10 ; 5 ; 10)$ to a point with positive coordinates that 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.
points) | 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 $x+y+z \geq 3 \sqrt[3]{x y z}$, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq 3 \sqrt[3]{\frac{1}{x y z}}$. Since all parts of the inequalities are positive, then $(x+y+z)\left(\frac{1}{x}+\frac{1}{y... | 115.2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,864 |
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 annual victories, 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 | 21,865 |
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 $N$.
(points) | Solution. The number $N$ can be represented as
$$
\begin{aligned}
& N=\frac{(2 \cdot 2018)!}{2018!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036}{2018!}=\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot(2 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4036)}{2018!}= \\
& =\frac{(1 \cdot 3 \cdot \... | 2018 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,866 |
6. What is the smallest area that a right triangle can have, if its hypotenuse lies on the tangent to the graph of the function $y=\sqrt{x-3}$, one of its legs lies on the $y$-axis, and one of its vertices coincides with the point of tangency
points) | Solution. $\quad f(x)=\sqrt{x-3}, \quad f^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x-3}}$
$S_{A B C}=\frac{1}{2} A B \cdot B C, x_{0}-$ abscissa of the point of tangency $A$,
$A\left(x_{0}, f\left(x_{0}\right)\right), \quad B\left(0, f\left(x_{0}\right)\right), \quad C \quad$ - the point of intersection of the tan... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 21,867 |
7. In triangle $A B C$, altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $1+\sqrt{3}$. The distances from the center of the inscribed circle in triangle $D E F$ to points $A$ and $C$ are $\sqrt{2}$ and 2, respectively. Find the length of side $A B$.
points) | Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=\sqrt{2}, C O=2$. Le... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,868 |
8. Find all values of the parameter $a$ for which the system
$$
\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$.
points | 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 | 21,869 |
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$.... | 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 | 21,870 |
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 an equal volume of argon was added. Then, the same volume of the mixture as the first time was pumped out and again topped up with the same volume of argon. In the new mixture, ... | Solution. Let $x$ be the volume of the mixture 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... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,872 |
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}}
$$
In your answer, write the square of the found distance.
(points) | Solution. Using the Cauchy inequality $a+b+c \geq 3 \sqrt[3]{a b c}$, which holds for all positive values of $a, b, c$. Then $((x+y)+(y+z)+(z+x)) \geq 3 \sqrt[3]{(x+y)(y+z)(z+x)}$, $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \geq 3 \sqrt[3]{\frac{1}{(x+y)(y+z)(z+x)}}$. Since all parts of the inequalities are positive, t... | 115.2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,873 |
5. Determine the smallest natural number $N$, among the divisors of which are all numbers of the form $x+y$, where $x$ and $y$ are natural solutions to the equation $6 x y-y^{2}-5 x^{2}=7$. points) | Solution. Transform the equation by factoring the right-hand side
$6 x y-y^{2}-5 x^{2}-x^{2}+x^{2}=7 \Rightarrow 6 x(y-x)-(y+x)(y-x)=7 \Rightarrow(y-x)(6 x-y-x)=7 \Rightarrow$ $(y-x)(5 x-y)=7$.
Considering that the variables are natural numbers, and 7 is a prime number, we get
$$
\left\{\begin{array} { l }
{ y - x ... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,874 |
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.
points) | 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 | 21,875 |
8. Find all values of the parameter $a$ for which the system $\left\{\begin{array}{l}y=\frac{x+1}{|x|-1}, \\ |x+y|+|x-y-2 a|=1 .\end{array}\right.$
has a unique solution. In your answer, specify the smallest of all the obtained values of the parameter $a$.
points | 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) $|x+y|+|x-y-2a|=1$.
a) $y=\frac{x+1}{|x|-1}=\left\{... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,876 |
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 $O$.... | 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, \qu... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,877 |
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 km/h, and after $3 / 8$ hours... | Solution. Let $x$ (km/h) be the speed of the motorcyclist, $x$ (km/h) be the speed of the car, $S$ (km) be the distance the motorcyclist travels before turning around, then the total length of the track is $2 S+5.25$. We have $\frac{S}{x}=\frac{3 S+5.25}{y}, \frac{3 x}{8}+6=S, \frac{3 y}{8}=S+5.25$. This leads to the q... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,879 |
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 $x^{2}+y^{2}+z^{2} \geq 3 \sqrt[3]{x^{2} y^{2} z^{2}}$, $\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \geq 3 \sqrt[3]{\frac{1}{x^{2} y^{2} z^{2}}}$. Since all parts of the inequalities are positive,... | 39.2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,881 |
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? (12 points) | Solution. 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 true fo... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,882 |
5. Find the sum of all numbers of the form $x+y$, where $x$ and $y$ are natural number solutions to the equation $5 x+17 y=307$.
points) | Solution. We solve the auxiliary equation $5 x+17 y=1$. For example, its solutions can be 7 and 2. Multiplying them by 307 and considering 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 ... | 164 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,883 |
6. What is the minimum length that segment $AB$ can have if point $A$ lies on the curve $10\left(x^{2}+y^{2}\right)+60 x-80 y+249=0$, and point $B$ lies on the graph of the function $y=\frac{1}{3}|x|$? Write the square of the found length in your answer.
points) | 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 | 21,884 |
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 $a$.
points | 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 | 21,885 |
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 $O$.... | 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 | 21,886 |
1. In 100 containers of three types with capacities of $1 \pi$, 10 liters, and 50 liters, 500 liters of oil have been distributed. How many containers of each type were used, if the amount of oil in each container corresponds to its capacity?
(12 points) | Solution. Let $n, m, k$ be the number of containers with capacities of 1 liter, 10 liters, and 50 liters, respectively. Then $\left\{\begin{array}{c}n+10 m+50 k=500, \\ n+m+k=100 .\end{array}\right.$ Since $n=500-10 m-50 k=10(50-m-5 k)$, $n$ is divisible by 10, i.e., $n=10 l, l \in N, n \leq 100, l \leq 10 .\left\{\beg... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,887 |
2. Solve the inequality $\quad \frac{x^{2}}{3}+\frac{40}{x}+\frac{48}{x^{2}} \geq \frac{10|x|}{3}$.
(12 points) | Solution. 1) $x>0, \frac{x^{2}}{3}+\frac{40}{x}+\frac{48}{x^{2}} \geq \frac{10 x}{3}, \quad x^{2}+\frac{144}{x^{2}} \geq 10\left(x-\frac{12}{x}\right) . \quad$ Let's make the substitution $\quad t=x-\frac{12}{x}$, $t^{2}-10 t+24 \geq 0, \quad t \leq 4$, or $t \geq 6$. Returning to the variable $x: x-\frac{12}{x} \leq 4... | x\in(-\infty;-6-2\sqrt{6}]\cup[-6+2\sqrt{6};0)\cup(0;6]\cup[3+\sqrt{21};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,888 |
3. The greatest common divisor of two natural numbers $a$ and $b$ is $d$. Determine the greatest common divisor of the numbers $5a + 3b$ and $13a + 8b$. | Solution. $\operatorname{GCD}(5 a+3 b, 13 a+8 b)=d_{1}, \operatorname{GCD}(a, b)=d$, $8(5 a+3 b): d_{1}, 3(13 a+8 b): d_{1} \Rightarrow(8(5 a+3 b)-3(13 a+8 b)): d_{1} \Rightarrow a: d_{1}$. $13(5 a+3 b): d_{1}, 5(13 a+8 b): d_{1} \Rightarrow(13(5 a+3 b)-5(13 a+8 b)): d_{1} \Rightarrow b: d_{1}$. $\Rightarrow d: d_{1}$.... | d | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,889 |
4. Find all natural numbers $n \geq 2$, for which the equality $4 x_{n}+2 y_{n}=55 n^{2}+61 n-116$ holds, where $x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n, y_{n}=1^{2}+2^{2}+3^{2}+\cdots+(n-1)^{2}$.
(20 points) | Solution: Let $z_{n}=1+2+\cdots+n=\frac{(n+1) n}{2}=\frac{n^{2}+n}{2}$. We have
$$
\begin{aligned}
& x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n=(2+3+\cdots n)+(3+4+\cdots n)+\cdots((n-1)+n)+n= \\
& =\left(z_{n}-z_{1}\right)+\left(z_{n}-z_{2}\right)+\cdots+\left(z_{n}-z_{n-1}\right)=(n-1) z_{n}-\left(z_{1}+z_{2}+\c... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,890 |
5. Specify all values of the parameter $a$ for which the system of equations
$$
\left\{\begin{array}{l}
\left(\frac{2 y}{3}-\frac{|x-1|}{x-1}-1\right)(y-1)=0, \\
y=a+\frac{|x-1|}{(x-1)(x-y)^{2}}
\end{array}\right. \text { has exactly four distinct solutions. }
$$
Find these solutions for each specified $a$.
(20 poin... | Solution: 1) $x>1,\left\{\begin{array}{l}\left(\frac{2 y}{3}-2\right)(y-1)=0, \\ y=a+\frac{1}{(x-y)^{2}}\end{array} \Leftrightarrow\left[\begin{array}{l}y=3,3=a+\frac{1}{(x-3)^{2}} \\ y=1,1=a+\frac{1}{(x-1)^{2}}\end{array} \Leftrightarrow\left[\begin{array}{l}y=3, a=3-\frac{1}{(x-3)^{2}} \\ y=1, a=1-\frac{1}{(x-1)^{2}}... | \in(0;1)\cup[11/4;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,891 |
6. In triangle $ABC$, lines parallel to the sides of the triangle are drawn through an arbitrary point $O$. As a result, triangle $ABC$ is divided into three parallelograms and three triangles. The areas of the resulting triangles are $6 \text{ cm}^2$, $24 \text{ cm}^2$, and $54 \text{ cm}^2$. Find the area of triangle... | # Solution:
The triangles are similar. Let the first and third be similar with a similarity coefficient $k_{1}$, and the second and third with a similarity coefficient $k_{2}$. Then for the areas of these triangles, we have the ratios $\frac{S_{1}}{S_{3}}=k_{1}^{2}, \frac{S_{2}}{S_{3}}=k_{2}^{2}$.
}_{n})(2+\underbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}... | A>\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,894 |
3. Schoolchildren were given 7 sheets of paper and asked to cut some of them into 7 pieces. The resulting pieces of paper were mixed and they were asked again to cut some of them into 7 pieces. This was repeated several times. How many pieces of paper need to be cut so that in the end there are 331 pieces. | Solution. After $k$ cuts, there remain $6 k+7$ pieces. Therefore, to determine the number of cuts made, we solve the equation $6 k+7=331 \Rightarrow k=54$.
Answer: 54. | 54 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,895 |
4. Solve the equation $10 x-6+x^{2}=\sqrt{2 x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}}$ | Solution. Transform the equation $10 x-6+x^{2}=\sqrt{2 x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}}$
$10 x-6+x^{2}=\sqrt{x^{2}+x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}} \Rightarrow 6 x-6+x^{2}+4 x=\sqrt{x^{2}+(x-1)^{2}} \cdot \sqrt{6^{2}+(x+4)^{2}}$,
This equation can be interpreted as the scalar product of vectors $\vec{a}=(... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,896 |
5. Given an arbitrary convex quadrilateral $A B C D$. Points $K$ and $M$ are the midpoints of sides $A B$ and $C D$ respectively. Point $P$ is the intersection of lines $K C$ and $B M$, point $N$ is the intersection of $A M$ and $K D$. Find the area of quadrilateral $K P N M$, if angles $C B P$ and $N D A$ are 30 degre... | # Solution and Answer.
$S_{K P M N}=S_{A M B}-S_{A N K}-S_{K P B}=S_{A D K}+S_{K C B}-S_{A N K}-S_{K P B}=S_{A D N}+S_{P C B}$
By the Law of Sines
$$
\begin{aligned}
& \frac{A N}{\sin 30}=\frac{N D}{\sin 15} \Rightarrow A N=\sqrt{3} \cdot \frac{1}{2} \cdot \frac{2 \sqrt{2}}{\sqrt{3}-1}=\frac{\sqrt{6}(\sqrt{3}+1)}{2}... | \frac{(\sqrt{3}+1)(4\sqrt{2}+3)}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,897 |
6. For what values of the parameter $a$ do the roots $x_{1}, x_{2}$ of the equation $x^{2}-2 a x-\frac{1}{a^{2}}=0$ satisfy the equality $x_{1}^{4}+x_{2}^{4}=16+8 \sqrt{2}$? | Solution. We will use Vieta's theorem: $x_{1} x_{2}=-\frac{1}{2 a^{2}}, x_{1}+x_{2}=2 a$, and express the sum of the fourth powers:
$$
\begin{aligned}
& x_{1}^{4}+x_{2}^{4}=\left(x_{1}+x_{2}\right)^{4}-4 x_{1} x_{2}\left(\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right)-6 x_{1}^{2} x_{2}^{2}=16 a^{4}-4\left(-\frac{1}{... | \\sqrt[8]{1/8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,898 |
1. (10 points) Solve the inequality: $3 \sqrt{x+4} \leq 5-2|x+2|$. | Solution. Consider the function $f(x)=3 \sqrt{x+4}-5+2|x+2|$. Domain: $\mathrm{x} \geq-4$. Let's find the zeros of this function, i.e., solve the equation $3 \sqrt{x+4}=5-2|x+2|$. Denote $\mathrm{y}=\sqrt{x+4}$. Consider the obtained system: $\left\{\begin{array}{c}3 y=5-2\left|y^{2}-2\right| \\ y \geq 0\end{array}\rig... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,899 | |
2. (15 points) Two medians of a triangle, equal to 18 and 24, are perpendicular to each other. Find the length of the third median of this triangle. | Solution. Let $\mathrm{AP}=18$ and $\mathrm{BH}=24$. We will find the length of the median $\mathrm{CM}$. By the property of medians in $\triangle A B C$ we have:
$A O: O P=B O: O H=2: 1 \Rightarrow A O=$
$=\frac{2}{3} A P=\frac{2}{3} \cdot 18=12 ; B O=\frac{2}{3} B H=\frac{2}{3} \cdot 24=16$.
Then, by the Pythagore... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,900 |
3. (15 points) Stepan and Pavel prepared feed for horses from oats, corn, and hay. It is known that Stepan prepared feed containing $40 \%$ hay, and Pavel - $26 \%$ oats. The percentage of corn in the first and second mixtures is the same. By mixing 150 kg of the mixture prepared by Stepan and 250 kg prepared by Pavel,... | Solution.
| | oats | corn | hay | weight |
| :--- | :--- | :--- | :--- | :--- |
| Stepan | | $x \%$ | $40 \%$ | 150 kg |
| Pavel | $26 \%$ | $x \%$ | $74 \% - x \%$ | 250 kg |
| New set | | $30 \%$ | | 400 kg |
1. Determine the percentage content of corn: $\frac{x}{100} \cdot 150 + \frac{x}{100} \cdot 250 = 400 \... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,901 |
4. (20 points) Determine the values of the parameter $\boldsymbol{a}$ for which the equation $(a+1)\left(x^{2}+1\right)^{2}-(2 a+3)\left(x^{2}+1\right) x+(a+2) x^{2}=0$ has exactly two distinct real roots. | Solution. 1. Note that $x=0$ is a solution when $a=-1$. Substituting this value of the parameter into the original equation, we get the equation $-x\left(x^{2}-x+1\right)=0$, which has no other roots besides zero. Therefore, if $x=0$ is a solution to the equation, then $a=-1$ and, moreover, $x=0$ is the only solution t... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,903 |
5. (20 points) Given triangle $K L M$. A circle is drawn passing through point $M$, touching segment $L K$ at point $A$, which is its midpoint, and intersecting sides $M L$ and $M K$ at points $C$ and $B$, respectively, such that $C B=4$, point $C$ is equidistant from points $A$ and $L$, and $\cos \angle K=\frac{\sqrt{... | # Solution.
$\angle C L A = \angle C A L = \angle C M A = \alpha, L A = M A = A K \Rightarrow \triangle L M K$ is a right triangle with $\angle L M K = 90^{\circ}$.
Then $\angle A M K = 90... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,905 |
6. (20 points) Calculate the value of the expression:
$$
1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots+2019 \cdot 2020 \cdot(2019+2020)
$$ | Solution. We will prove by mathematical induction for natural n that:
$$
\begin{gathered}
-0 \cdot 1 \cdot(0+1)+1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots- \\
-(2 n-2) \cdot(2 n-1) \cdot(4 n-3)+(2 n-1) \cdot 2 n \cdot(4 n-1)= \\
=(2 n-1) \cdot 2 n \cdot(2 n+1) .
\end{gathered}
$$
Base: \( n... | 8242405980 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,906 |
1. (10 points) Solve the inequality:
$$
2 \sqrt{(4 x-9)^{2}}+\sqrt{3 \sqrt{x}-5+2|x-2|} \leq 18-8 x
$$ | Solution. Since $\sqrt{(4 x-9)^{2}}=|4 x-9|$, we have
$$
2|4 x-9|+\sqrt{3 \sqrt{x}-5+2|x-2|} \leq 18-8 x
$$
The left side (1) is greater than or equal to 0 as the sum of a non-negative modulus and a square root.
Then the right side (1) will also be greater than or equal to 0, i.e.
$$
18-8 x \geq 0, x \geq 0 \Righta... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,907 |
2. (15 points) The area of an isosceles trapezoid is 100, and its diagonals are perpendicular to each other. Find the height of this trapezoid. | Solution. Let $ABCD$ be an isosceles trapezoid, the diagonals $AC$ and $BD$ of which are perpendicular to each other. Denote $O = AC \cap BD$, point $M$ is the midpoint of $BC$, and point $H$ is the midpoint of $AD$. Draw the segment $MH: O \in MH, MH \perp AD$. Since $MH$ is the axis of symmetry of the given trapezoid... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,909 |
3. (15 points). In a hat box, there are 21 hats and 18 caps. They were distributed on two shelves: the first should accommodate 20 items, and the second - 19. After the distribution, the percentage of hats on each shelf was calculated and the resulting numbers were added together. What should be the distribution of hat... | Solution. Solution 1. Instead of the total percentage, we will calculate the total share of hats, which are obviously 100 times different and reach their maximum simultaneously. Each hat on a shelf of 20 items constitutes $1 / 20$ of the total number of items on that shelf, and on a shelf of 19 items, it constitutes $1... | Ontheshelf-19hats,onthefirstshelf-2hats18caps | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,910 |
4. (20 points) Determine the number of solutions of the equation $a(x-|x|+2)=|x-3|-2$ for each value of the parameter $a$. | Solution. We will open the modules and investigate three cases separately:
1) $x \in(-\infty ; 0) a(2 x+2)=1-x ; x(2 a+1)=1-2 a$. When $a=-0.5$, the equation has no solutions for these x. Divide by the coefficient, $x=\frac{1-2 a}{1+2 a}$. Solve the inequality $\frac{1-2 a}{1+2 a} < 0 ; a > -1 ; a \in(-1 ;+\infty)$ - ... | For\\\in(-\infty;-1)\-\one\solution;\for\=-1\-\two\solutions;\for\\\in(-1;-0.5)\-\three\solutions;\for\\\in[-0.5;+\infty)\-\two\solutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,911 |
5. (20 points) In a right triangle $A B C\left(\angle C=90^{\circ}\right)$ with side $B C=a$, points $M$ and $N$ are the midpoints of sides $A B$ and $B C$, respectively. The bisector of $\angle A$ intersects line $M N$ at point $L$. Find the radius of the circumcircle of triangle $A C L$, if angle $\angle C B L=\alpha... | Solution.
Draw the line $B L$, intersecting $A C$ at point $X$. Then, in $\triangle A B X$, the bisector $A L$ is also a median, and therefore, an altitude. Thus, $\angle A L B=90^{\circ}$ ... | \frac{}{2\sin2\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,912 |
6. (20 points) Is the number 390629 prime? Provide a justified answer (without using a calculator). | Solution. Since $25^{4}=390625$, then
$390629=25^{4}+4=25^{4}+4+4 \cdot 25^{2}-4 \cdot 25^{2}=\left(25^{2}+2\right)^{2}-50^{2}=577 \cdot 677$.
Second method.
$390629 \cdot 256=10001024=10^{8}+2^{10}=2^{8}\left(5^{8}+4\right) ; 390629=\left(5^{4}+2\right)^{2}-50^{2}=577 \cdot 677$.
Answer: No.
| Content of the crit... | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,913 |
4. In the box, there are 3 white cups, 3 red cups, and 2 black cups. Sonya took out 5 cups at random. What is the probability that she took out 2 white, 2 red, and 1 black cup? (Round your answer to the nearest hundredth).
# | # Solution:
Let white cups be denoted as W, red cups as R, and black cups as B.
1) The probability of drawing 2 white, then 2 red, and then 1 black cup, in the order WWR RB: $\frac{3}{8} \cdot \frac{2}{7} \cdot \frac{3}{6} \cdot \frac{2}{5} \cdot \frac{2}{4}=\frac{3}{280}$
2) The number of ways to rearrange the lett... | 0.32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,914 |
5. Find all values of the parameter $a$ for which the equation
$$
\left(\left|\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}\right|-a\right) \cdot|4 a-3 x-19|=0
$$
has one solution. In your answer, write the largest value of the parameter $a$ | Solution:
Simplify $\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}=\frac{a x^{2}-x^{2}+(-a x+x)+(-12 a+12)}{a(x+3)-(x+3)}=$
$=\frac{x^{2}(a-1)-x(a-1)-12(a-1)}{(x+3)(a-1)}=\frac{(a-1)\left(x^{2}-x-12\right)}{(x+3)(a-1)}=$
$=\frac{x^{2}+3 x-4 x-12}{x+3}=\frac{(x+3)(x-4)}{x+3}=x-4$
We get $\left\{\begin{array}{c}(|x-... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,915 |
2. When adding two natural numbers, Alyosha accidentally added an extra digit, so instead of the number 23456, he got the number 55555.
Determine which digit Alyosha accidentally added.
## solution.
Let Alyosha be adding numbers $x$ and $y$. When he appended an extra digit $a$ to one of them (let's say to $x$), the ... | Answer: 36 liters
## №6: Plane Geometry.
№2. From points B and C of the acute angle BAC, perpendiculars CM and BK are drawn, intersecting at point E, such that point M lies on AB, and K lies on AC. A circle is drawn through points A, M, and C, intersecting BK at point O, and $BE=17, OK=72$. AK: $KC=1:4$. Find the tan... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,917 |
2. Determine for which $a$ the system $x+y-144=0, xy-5184-0.1a^2=0$ has one solution. Answer: 0.
Solution. $\left\{\begin{array}{c}x+y=144 \\ xy=5184+0.1a^2\end{array}\right.$
This means that $x$ and $y$ are equal to the roots of the quadratic equation
$p^2 - 144p + (5184 + 0.1a^2) = 0$. Therefore, $D = 72^2 - (5184 ... | Answer. 2023.
## Solution.
We will prove that $2021!! + 2022!!$ is divisible by the number 2023. Indeed,
$$
2022!! = (2023-1)(2023-3)(2023-5) \ldots (2023-2021)
$$
the remainder of this number when divided by 2023 is the same as the remainder of the product
$$
(-1)(-3)(-5) \ldots (-2021) = (-1)^{\frac{2022}{2}} \c... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,918 |
# Variant №1
№1. The number of male students in the university is $35 \%$ more than the number of female students. All students are distributed between two buildings, with $\frac{2}{5}$ of all male students in the first building and $\frac{4}{7}$ of all female students in the second building. How many students are the... | # Solution:
Let $n$ be the number of girls studying at the university, then the number of boys is $1.35 n = \frac{27}{20} n$, from which we can conclude that $n$ is a multiple of 20, i.e., $n = 20 m$.
Thus, the number of girls is $20 m$, and the number of boys is $27 m$.
In the first building, $\frac{2}{5}$ of all b... | 4935 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,919 |
# Variant №3
1. The number of ninth-graders participating in the math olympiad is $45\%$ more than the number of eighth-graders. The students were distributed between two buildings: in the first building, there were fewer than 1000 students, and in the second building, there were more than 1000. How many students took... | Solution: Let the number of eighth graders be $n$, then the number of ninth graders is $n\left(1+\frac{45}{100}\right)=\frac{29}{20} n$.
The number of students in the first building is $\frac{2}{3} \cdot \frac{29}{20} \cdot n + \frac{1}{7} \cdot n = \frac{233}{210} n \cdot 1000$, therefore, the number $n$ is a multipl... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,920 |
2. Given the vertices of a regular 100-gon $A_{1}, A_{2}, A_{3}, \ldots, A_{100}$. In how many ways can three vertices be chosen from them to form an obtuse triangle? (10 points) | Solution. Let the vertices be numbered clockwise.
Denote the selected vertices clockwise as $K, L, M$, where angle $K L M$ is obtuse. If $K=A_{k}, L=A_{l}, M=A_{m}$, then $\alpha=\angle K L M=\frac{180^{\circ}}{100}(100-(m-k))>90^{\circ}, 0<m-k<50$.
The difference $m-k$ is taken modulo 100
(for example, $15-\left.70... | 117600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,921 |
3. Specify all values of $a$ for which the system of equations $\left\{\begin{array}{c}x-4=a\left(y^{3}-2\right), \\ \frac{2 x}{\left|y^{3}\right|+y^{3}}=\sqrt{x}\end{array}\right.$ has at least one solution, and solve it for each $a$. (12 points) | Solution. Domain of definition: $y>0, x \geq 0$.
In the domain of definition, the second equation of the system takes the form: $x=y^{3} \sqrt{x}$.
I. $x=0, y=\sqrt[3]{2-\frac{4}{a}}=\sqrt[3]{\frac{2 a-4}{a}}>0 \Leftrightarrow\left[\begin{array}{l}a2\end{array}\right.$
II. $x>0, y^{3}=\sqrt{x} ;(\sqrt{x}-2)(\sqrt{x}+... | 50\sqrt{17} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,922 |
2. Given the vertices of a regular 120-gon $A_{1}, A_{2}, A_{3}, \ldots, A_{120}$. In how many ways can three vertices be chosen from them to form an obtuse triangle? (10 points) | Solution. Let the vertices be numbered clockwise.
Denote the selected vertices clockwise as $K, L, M$, where angle $K L M$ is obtuse. If $K=A_{k}, L=A_{l}, M=A_{m}$, then $\alpha=\angle K L M=\frac{180^{\circ}}{120}(120-(m-k))>90^{\circ}, 0<m-k<60$.
The difference $m-k$ is taken modulo 120
(for example, $15-\left.90... | 205320 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,923 |
3. Specify all values of $a$ for which the system of equations $\left\{\begin{array}{c}x-1=a\left(y^{3}-1\right) \\ \frac{2 x}{\left|y^{3}\right|+y^{3}}=\sqrt{x}\end{array}\right.$ has at least one solution, and solve it for each $a$.
# | # Solution.
Domain of definition: $y>0, x \geq 0$.
In the domain of definition, the second equation of the system takes the form: $x=y^{3} \sqrt{x}$.
I. $x=0, y=\sqrt[3]{1-\frac{1}{a}}=\sqrt[3]{\frac{a-1}{a}}>0 \Leftrightarrow\left[\begin{array}{l}a1\end{array}\right.$
II. $x>0, y=\sqrt{x} ;(\sqrt{x}-1)(\sqrt{x}+1)=... | 10\sqrt{2}+60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,924 |
1. The sequence of real numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{100}$ satisfies the inequalities $a_{n}-2022 a_{n+1}+2021 a_{n+2} \geq 0 \quad$ for $\quad n=1,2,3, \ldots, 98$, and $\quad a_{99}-2022 a_{100}+2021 a_{1} \geq 0$, $a_{100}-2022 a_{1}+2021 a_{2} \geq 0$. Find $a_{22}$, if $a_{10}=10$. (10 points) | # Solution.
Let $b=a_{1}+a_{2}+a_{3}+\cdots+a_{100}, b_{n}=a_{n}-2022 a_{n+1}+2021 a_{n+2}, n=1,2,3, \ldots, 98$, $b_{99}=a_{99}-2022 a_{100}+2021 a_{1}, b_{100}=a_{100}-2022 a_{1}+2021 a_{2}$. By the condition $b_{n} \geq 0$, for $n=$ $1,2,3, \ldots, 100$. We have $b_{1}+b_{2}+b_{3}+\cdots+b_{100}=b-2022 b+2021 b=0 .... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,925 |
2. Find the probability that a randomly selected five-digit natural number with non-repeating digits, formed from the digits $1,2,3,4,5,6,7,8$, is divisible by 8 without a remainder. (10 points) | Solution. A natural number $n$ is divisible by 8 without a remainder if the number formed by the last three digits in the number $n$ (in the order of their occurrence) is divisible by 8. If the number formed by the last three digits is not divisible by 8, then the number $n$ is not divisible by 8.
Let's find the numbe... | \frac{1}{8} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,926 |
1. There are two types of containers: 27 kg and 65 kg. How many containers of the first and second type were there in total, if the cargo in the first type of containers exceeds the cargo in the second type of containers by 34 kg, and the number of 65 kg containers does not exceed 44 units? | Solution. Let $x$ be the number of containers weighing 27 kg, and $y$ be the number of containers weighing 65 kg. We obtain the equation $27 x - 65 y = 34$.
$27(x - 2y) - 11y = 34$, let $x - 2y = k$.
$27k - 11y = 34$,
$11(2k - y) + 5k = 34$, let $2k - y = t$.
$11t + 5k = 34$,
$5(2t + k) + t = 34$, let $2t + k = n$... | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,928 |
4. Vasya and his father are collecting mushrooms in the forest. Father told Vasya: “Go ahead along this straight path. I will quickly check the glade and catch up with you.” The father's step is $20 \%$ longer than Vasya's. They both walk at a constant speed and do not change the length of their steps. There is a perio... | Solution. Let $\chi$ be the length of Vasya's step, $y$ be the number of steps he takes in the specified unit of time. Then $x y$ is the distance Vasya covers in this time, $1.2 x y\left(1-\frac{t}{100}\right)$ is the distance covered by his father. For Vasya and his father to start getting closer, the inequality $1.2 ... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,931 |
5. The clock shows 00:00, at which the hour and minute hands of the clock coincide. Counting this coincidence as number 0, determine how much time (in minutes) will pass before they coincide for the 23rd time. Round your answer to the nearest hundredth. | Solution. The minute hand passes 1 circle in an hour, while the hour hand passes $1 / 12$ of a circle, so their closing speed is $11 / 12$ of a circle per hour, one closing takes $1 /(11 / 12)=12 / 11$ hours or $720 / 11$ minutes. The 23rd closing will occur after $23 \cdot \frac{720}{11}=\frac{16560}{11}$ minutes.
An... | 1505.45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,932 |
6. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Find the degree measure of angle $C_{1} B_{1} A_{1}$. | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$. Considerin... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,933 |
7. Given a rectangular trapezoid $\mathrm{ABCE}$, the bases of which $\mathrm{BC}$ and $\mathrm{AE}$ are equal to 5 and 7, respectively. The smaller lateral side $\mathrm{AB}$ is equal to $\mathrm{BC}$. On $\mathrm{AB}$, a point $\mathrm{F}$ is marked such that $\mathrm{BF}: \mathrm{FA}=2: 3$, on $\mathrm{AC}$, a point... | Solution. Construct perpendiculars GI and GH.
1) $\Delta G I D = \Delta G F H$ - by two legs, since $G I = G H = 4; F H = I D = 1$, therefore $F G = G D$, $\angle F G H = \angle D G I = \alp... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,934 |
8. A circle with center $O$ is inscribed in a rectangular trapezoid with bases $B C$ and $A D$, where $B C: A D=2: 3$. The lines $D O$ and $B C$ intersect at point $E$. In what ratio does the tangent to the circle, drawn from point $E$, divide the side $A D$ of the trapezoid, counting from point A. | # Solution.
Let $BC = 2a, AD = 3a$. We need to find the radius of the inscribed circle in the trapezoid. The center of the circle, point $O$, is the intersection of the angle bisectors of t... | 2:13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,935 |
9. A chemistry student conducted an experiment: from a tank filled with syrup solution, he poured out several liters of liquid, refilled the tank with water, then poured out twice as much liquid and refilled the tank with water again. As a result, the amount of syrup in the tank decreased by $\frac{8}{3}$ times. Determ... | # Solution.
1) Let the syrup content in the original solution be $p \%$ and let $\mathcal{X}$ liters of the solution be poured out the first time.
2) Then after pouring out the liquid, there are $(1000-x)$ liters of solution left, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p... | 250 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,936 |
3. During the shooting practice, each soldier fired 10 times. One of them completed the task successfully and scored 90 points. How many times did he score 8 points, if there were 4 tens, and the results of the hits were sevens, eights, and nines. There were no misses at all. | Solution. Since the soldier scored 90 points and 40 of them were scored in 4 attempts, he scored 50 points with the remaining 6 shots. Since the soldier only hit the seven, eight, and nine, let's assume that for three shots (once in the seven, once in the eight, and once in the nine), he scored 24 points. Then, for the... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,938 |
4. The grandmother and granddaughter were picking strawberries. The granddaughter was using a children's bucket with a capacity of 2.5 liters, while the grandmother was using a two-liter jar. The grandmother had poor eyesight and found it hard to bend down, so the granddaughter was always picking berries faster than he... | Solution. Let the granddaughter collect $X$ liters of berries in the time it takes the grandmother to collect 1 liter, $x>1$ according to the problem. After swapping containers, the granddaughter collected 1 liter of berries, while the grandmother collected $(2.5-x)$ liters. Since their work efficiency does not change,... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,939 |
5. Elvira takes a triplet of numbers and transforms it according to the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after 2019 applications of this rule, if the initial triplet was $\{100 ; 89 ; 60\}$? If the... | Solution. Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b$.
$b=100-60=40$.
Answer: 40. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,940 |
7. On the sides $\mathrm{AB}$ and $\mathrm{AC}$ of the right triangle $\mathrm{ABC}\left(\angle B A C=90^{\circ}\right)$, right triangles АВТ and АСК are constructed externally such that $\angle A T B=\angle A K C=90^{\circ}$, $\angle A B T=\angle A C K=30^{\circ}$. On the side $\mathrm{BC}$, a point М is chosen such t... | Solution. Mark points P and O at the midpoints of sides AB and AC, respectively. Connect point P with points M and T, and point O with points K and M.
Then: 1) $\triangle T P M = \Delta K O... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,942 |
8. In rectangle $A B C D$, the ratio of the lengths of the sides $A B: A D=1: 3$ is known. Point $K$ lies on $A D$ and divides it in the ratio $2: 1$, counting from point $A$. Find the sum of the degree measures of angles $\angle A K B$ and $\angle A D B$. | Solution. Extend the rectangle $A B C D$ to a square $A E F D$ with side $A D$.
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,943 | |
9. A chemistry student conducted an experiment: from a bottle filled with syrup solution, he poured out one liter of liquid, refilled the bottle with water, then poured out one liter of liquid again and refilled the bottle with water. As a result, the percentage of syrup decreased from 9 to 4 percent. Determine the vol... | # Solution.
1) Let $X$ be the volume of the bottle in liters.
2) Then after pouring out one liter of liquid, there are $(x-1)$ liters of solution left, and in it $(x-1) \cdot \frac{9}{100}$ liters of syrup and $(x-1) \cdot \frac{91}{100}$ liters of water.
3) After adding one liter of water to the bottle, there are $X$... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,944 |
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