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# Task 5.
When establishing a connection between computers A and B, the following variant of the so-called "handshake procedure" is used: 1) A selects a natural number $x$, not greater than 5250, and sends B the value of the function $F(x)$, and then B sends A the number $F(x+1)$; 2) now B selects a natural number $y$... | Answer: $(x, y)=(102,72)$.
# | (x,y)=(102,72) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,991 |
2. To store a password written in a 32-letter alphabet ("е" is identified with "ё"), each letter of the password is represented by its ordinal number -- a pair of digits (i.e., А - 01, Б - 02, etc.). This results in a sequence of digits $y_{1}, y_{2}, \ldots$ Simultaneously, according to the rule $x_{i+1}=r_{10}\left(a... | 2. To store a password written in a 32-letter alphabet ("е" is identified with "ё"), each letter of the password is represented by its ordinal number -- a pair of digits (i.e., А - 01, Б - 02, etc.). This results in a sequence of digits $y_{1}, y_{2}, \ldots$ Simultaneously, according to the rule $x_{i+1}=r_{10}\left(a... | yacht | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,992 |
3. At the robot running competition, a certain number of mechanisms were presented. The robots were released on the same distance in pairs. The protocol recorded the differences in the finish times of the winner and the loser in each of the races. All of them turned out to be different: 1 sec., 2 sec., 3 sec., 4 sec., ... | 3. At the robot running competition, a certain number of mechanisms were presented. The robots were released on the same distance in pairs. The protocol recorded the differences in the finish times of the winner and the loser in each of the races. All of them turned out to be different: 1 sec., 2 sec., 3 sec., 4 sec., ... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,993 |
6. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The use... | 6. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The use... | 3212 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,996 |
2. Nезнayka and Knopochka developed the following encryption system. The original text, written without spaces, is divided into parts of 16 letters each. In each part, the letters are numbered from left to right from 1 to 16 and then rearranged according to the rule specified by Table 1. That is, the first letter of ea... | 2. Nезнayka and Knopochka developed the following encryption system. The original text, written without spaces, is divided into parts of 16 letters. In each part, the letters are numbered from left to right from 1 to 16 and then rearranged according to the rule specified by Table 1. That is, the first letter of each pa... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,997 | |
3. To store a password written in a $32$-letter alphabet ("i" is identified with "j"), each letter is represented by its ordinal number -- a pair of digits (i.e., A - 01, B - 02, etc.). This results in a sequence of digits $y_{1}, y_{2}, \ldots$. Simultaneously, according to the rule $x_{i+1}=r_{10}\left(a x_{i}+b\righ... | 3. To store a password written in a $32$-letter alphabet ("и" is identified with "й"), each letter is represented by its ordinal number -- a pair of digits (i.e., А - 01, Б - 02, etc.). This results in a sequence of digits $y_{1}, y_{2}, \ldots$. Simultaneously, according to the rule $x_{i+1}=r_{10}\left(a x_{i}+b\righ... | yachta | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,998 |
5. To transmit a three-letter word through a channel, the following method is used. Each letter of the word is assigned a pair of digits according to the rule: А -00, Б $-01, \mathrm{~B}-02, \ldots$, Я - 32. After that, the obtained sequence of digits $m_{1}, m_{2}, \ldots, m_{6}$. is transformed by the formula:
$c_{i... | 5. To transmit a three-letter word through a channel, the following method is used. Each letter of the word is assigned a pair of digits according to the rule: А -00, Б -01, В -02, ..., Я - 32. After that, the obtained sequence of digits $m_{1}, m_{2}, \ldots, m_{6}$. is transformed by the formula:
$c_{i}=f\left(m_{i}... | миф | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,000 |
8. To conduct an investigation, operational officers need to enter the gaming hall of an underground casino, which opens using electronic devices A and B, located in different rooms. One of the officers can access device A between 6:00 and 7:15, while another can access device B at the same time. Before the operation b... | Solution: Let's assume, for definiteness, that the notebook initially is with the operative working with device A. In 33 minutes, he goes through all combinations of the switches and records these combinations on the pages of the notebook. Each combination he writes on the page with the number that appears on the three... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,006 |
4. While waiting for the results of the "13 element. ALchemy of the future" competition, fifth-grader Vasya rolled a die, the sum of the points on opposite faces of which is 7, along the path shown in the figure. How many points
 until they had cards of all 4 suits among the drawn cards. What is the minimum number of cards that must be drawn to ensure that all 4 suits are among them?
a) 4
b) 9
c) 18
d) 28
e) 36 | 8. answer: g. There are 9 cards of each suit. If you draw 27 cards, they could all be from only three suits. Suppose you draw 28 cards and among the first 27 cards, they are only from three suits, then in the deck, only cards of the fourth suit will remain. And the 28th card will already be of the 4th suit. | 28 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 9,021 |
12. The director of an aluminum plant wants the advertisement of his plant, which lasts 1 min. 30 sec., to be shown exactly every 18 min. 30 sec. after the end of the previous advertising break. How much time will the advertisement of the aluminum plant be shown in a week?
a) 12 h.
b) 12 h. 12 min.
c) 12 h. 24 min.
... | 12. answer: g. In a week, there are $7 \cdot 24=168$ hours. For every hour, advertisements are shown for $1.5 \cdot 3=4.5$ minutes. Therefore, in a week, advertisements make up $4.5 \cdot 168=756=12$ hours and 36 minutes of airtime. In a week, there are $7 \cdot 24=168$ hours. For every hour, advertisements are shown f... | 12 | Algebra | MCQ | Yes | Yes | olympiads | false | 9,023 |
13. Substance A is a colorless liquid with a distinctive odor, lighter than water and well soluble in it. When this substance is heated in the presence of concentrated sulfuric acid, gas B, which is lighter than air, is formed. By reacting with hydrogen bromide, B forms a heavy liquid C. Provide the names of substances... | 13. A- ethanol, B - ethylene, C- bromoethane
## $\mathrm{H}_{2} \mathrm{SO}_{4}$
1) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightarrow \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \mathrm{O}$
$$
\text { 2) } \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{HBr} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}
$$ | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 9,024 |
8. During the lunch break on the day of the "13th Element. ALchemy of the Future" competition, a cafeteria worker mixed 2 liters of one juice with a $10 \%$ sugar content and 3 liters of another juice with a $15 \%$ sugar content. What is the sugar content of the resulting mixture?
a) $5 \%$
б) $12.5 \%$
в) $12.75 \%... | 8. Answer g. In the first juice, there is $0.1 \cdot 2=0.2$ liters of sugar, and in the second juice, there is $0.15 \cdot 3=0.45$ liters of sugar. Then, in the resulting mixture, there will be $0.2+0.45=0.65$ liters of sugar. The total volume is 5 liters. Then the percentage of sugar in the mixture is $\frac{0.65}{5} ... | 13 | Algebra | MCQ | Yes | Yes | olympiads | false | 9,031 |
14. A thin wooden rod is suspended from one end and can freely rotate relative to the point of suspension. The lower end of the rod is lowered into water. At a certain depth of immersion, the equilibrium position of the rod becomes unstable, and it begins to deviate from the vertical position. Find this depth and the a... | 14. Let the distance from the point of suspension of the rod to the water surface be \( h \). The length of the rod is denoted by \( l \). Suppose the rod is deflected from the vertical by an angle \( \boldsymbol{\alpha} \). Since the rod is in equilibrium, the algebraic sum of the moments of the forces of gravity and ... | \frac{}{}\leq\sqrt{\frac{\rho_{}-\rho_{i}}{\rho_{\mathrm{}}}}=\sqrt{0.4}=0.63 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 9,034 |
15. Hydrogen was passed over a heated powder (X1). The resulting red substance (X2) was dissolved in concentrated sulfuric acid. The resulting solution of the substance blue (X3) was neutralized with potassium hydroxide - a blue precipitate (X4) formed, which upon heating turned into a black powder (X1). What substance... | 15. 80 g/mol; $\mathrm{m}(\mathrm{CuO})=80$ g/mol; $\mathrm{X}_{1}-\mathrm{CuO} ; \mathrm{X}_{2}-\mathrm{Cu} ; \mathrm{X}_{3}-\mathrm{CuSO}_{4 ;} \mathrm{X}_{4}-\mathrm{Cu}(\mathrm{OH})_{2}$ | 80 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,035 |
16. At the quiz in the Museum of Entertaining Sciences of SFU, 10 schoolchildren are participating. In each round, the students are divided into pairs. Each participant meets every other participant exactly once. A win in a match earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. What is the minimum num... | 16. 7. Evaluation. After the sixth round, 30 points have been played, and the leader has no more than 6 points, while the other nine participants have collectively scored no less than 24 points. Therefore, among them, there is at least one participant with more than three points. Since there are still 3 rounds ahead, t... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,036 |
8. Young researcher Petya noticed that the iron at home heats up by $9^{\circ} \mathrm{C}$ every 20 seconds, and after being turned off, it cools down by $15^{\circ} \mathrm{C}$ every 30 seconds. Last evening, after turning off the iron, it cooled down for exactly 3 minutes. How long was the iron turned on?
a) 3 min 20... | 8. answer: a. 3 minutes $=180$ seconds. In 180 seconds, the iron cools down by $180 / 30 \cdot 15=90$ degrees. To heat up by 90 degrees, the iron takes $90 / 9 \cdot 20=200$ seconds. 200 seconds $=3$ minutes 20 seconds. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false | 9,040 |
12. The security guard of the aluminum plant works on Tuesdays, Fridays, and on odd-numbered days. What is the maximum number of consecutive days the security guard can work?
a) 3
b) 4
c) 5
d) 6
e) 7 | 12. answer: g. 6 days. An example of such a situation: 29th (odd), 30th (Tuesday), 31st (odd), 1st (odd), 2nd (Friday), 3rd (odd). | 6 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 9,043 |
13. Upon complete interaction of a solution containing 0.88 g of compound A and an equivalent amount of hydroiodic acid, only a suspension of iodine in water was formed. The iodine that precipitated reacted with an excess of sodium sulfide solution, resulting in a precipitate with a mass of 0.48 g. Determine the compos... | 13. Since only iodine and water were formed, A could only contain iodine, hydrogen, and oxygen atoms. A is an oxidizer. Possible compounds - $\mathrm{H}_{2} \mathrm{O}_{2}$ or oxygen-containing iodine acids. $\mathrm{I}_{2}+\mathrm{Na}_{2} \mathrm{~S}=2 \mathrm{NaI}+$ S. Precipitate - sulfur. Let's calculate the number... | \mathrm{HIO}_{3} | Other | math-word-problem | Yes | Yes | olympiads | false | 9,044 |
14. A wooden block floating in a cylindrical vessel with a base area of $\mathrm{S}=25 \mathrm{~cm}^2$, partially filled with water, had a small stone placed on it. As a result, the block remained afloat, and the water level in the vessel rose by $h_{1}=1.5 \mathrm{~cm}$. Then the stone was removed from the block and d... | 14. When the stone was placed on the block, the pressure on the bottom of the vessel increased by the amount
$$
p_{1}=\rho g h_{1}
$$
and the force of pressure on the bottom of the vessel increased, on the one hand, by the weight of the stone $\mathrm{mg}$, where $\mathrm{m}$ is the mass of the stone, and on the othe... | 3\frac{\mathrm{r}}{\mathrm{}^{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,045 |
16. A worker at an aluminum plant can produce 16 blanks or 10 parts from blanks in one shift. It is known that exactly one part is made from each blank. What is the maximum number of blanks the worker can produce in one shift to make parts from them in the same shift? | 16. 6. Let's denote by $x$ the duration of the worker's working day in hours. Then the worker makes one part from a blank in $\frac{x}{10}$ hours, one blank in $\frac{x}{16}$ hours, and a blank and a part from it in $\frac{x}{10}+\frac{x}{16}=\frac{13 x}{80}$ hours. Since $x: \frac{13 x}{80}=6 \frac{2}{13}$, the maximu... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,046 |
6. A bucket with a volume of 12 liters was filled to the top with gravel, and then water was added until it started to overflow over the edge of the bucket. Find the density of the material of the gravel particles if an additional 3 liters of water fit into the bucket, and the weight of the bucket became 28 kg. Assume ... | 6. $267 \mathrm{kr} / \mathrm{m} 3$.
The text above has been translated into English, maintaining the original formatting and line breaks. | 267\mathrm{kr}/\mathrm{}3 | Algebra | MCQ | Yes | Yes | olympiads | false | 9,048 |
7. What is the mass fraction (%) of oxygen in aluminum oxide $\mathrm{Al}_{2} \mathrm{O}_{3}$?
a) $53 \%$
b) $27 \%$
c) $16 \%$
d) $102 \%$ | 7. $53 \% . \operatorname{Mr}(\mathrm{Al} 2 \mathrm{O} 3)=27 * 2+16 * 3=102 \quad \mathrm{~W}(\mathrm{Al})=27 * 2 / 102 * 100 \%=53 \%$
7. $53 \% . \operatorname{Mr}(\mathrm{Al} 2 \mathrm{O} 3)=27 * 2+16 * 3=102 \quad \mathrm{~W}(\mathrm{Al})=27 * 2 / 102 * 100 \%=53 \%$
The text is already in a format that is a mix... | 53 | Other | MCQ | Yes | Yes | olympiads | false | 9,049 |
8. Pete, Kolya, and Vasya participated in the third competition task of the "13 Element. ALchemy of the Future" contest. Kolya guessed 3 times fewer logos than Pete, and Vasya guessed 5 times fewer logos than Pete. It is known that one point was given for each correctly guessed logo. How many points did the boys score ... | 8. 23 logos. Let the number of logos guessed by Petya be denoted by $x$. Then the number of logos guessed by Kolya and Vasya are $x / 3$ and $x / 5$ respectively. The total number of logos guessed by the three boys is $x + x / 3 + x / 5 = 23 \cdot x / 15$. Since the numbers 23 and 15 are coprime, and the number of gues... | 23 | Algebra | MCQ | Yes | Yes | olympiads | false | 9,050 |
12. An aluminum disk with a diameter of 20 cm weighs 2.4 kg. A worker cut out a disk with a diameter of 10 cm from it. What is the weight in kilograms of the remaining part?
a) 1.2
b) 1.6
c) 1.8
d) 1.9
e) 2 | 12. 1.8 kg. Area of the aluminum disk: $S=10^{2} \pi=100 \pi \mathrm{cm}^{2}$, then per 1 kg, $\frac{100 \pi}{2.4}=41 \frac{2}{3} \pi \mathrm{cm}^{2}$. Area of the cut-out disk: $S=5^{2} \pi=25 \pi$ $\mathrm{cm}^{2}$. We find the mass of the cut-out part $25 \pi: 41 \frac{2}{3} \pi=0.6$ kg, so the mass of the remaining... | 1.8 | Geometry | MCQ | Yes | Yes | olympiads | false | 9,053 |
13. To obtain phosphorus, calcium phosphate was previously treated with sulfuric acid, the resulting orthophosphoric acid was mixed with carbon and calcined. In this process, orthophosphoric acid turned into metaphosphoric acid, which, upon interaction with carbon, yielded phosphorus, hydrogen, and carbon monoxide. Ill... | 13. $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+3 \mathrm{H}_{2} \mathrm{SO}_{4}=2 \mathrm{H}_{3} \mathrm{PO}_{4}+3 \mathrm{CaSO}_{4} \cdot \mathrm{H}_{3} \mathrm{PO}_{4}=\mathrm{HPO}_{3}+\mathrm{H}_{2} \mathrm{O}$ (heating). $2 \mathrm{HPO}_{3}+6 \mathrm{C}=2 \mathrm{P}+\mathrm{H}_{2}+6 \mathrm{CO}$. Let's find t... | 23.4 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,054 |
14. A wooden block floating in a cylindrical vessel with a base area $\mathrm{S}=25$ cm², partially filled with water, had a small stone placed on it. As a result, the block remained afloat, and the water level in the vessel rose by $h_{1}=1.5$ cm. Then the stone was removed from the block and dropped into the water. T... | 14. When the stone was placed on the block, the pressure on the bottom of the vessel increased by the amount
$$
p_{1}=\rho g h_{1}
$$
and the force of pressure on the bottom of the vessel increased, on one side by the weight of the stone $m g$, where $m$ is the mass of the stone, and on the other by the amount $p_{1}... | 3\frac{\tau}{\mathrm{}^{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,055 |
16. Every sixth bus in the bus park of the aluminum plant is equipped with an air conditioner. After the plant director ordered to install it on 5 more buses, a quarter of the buses had an air conditioner. How many buses are there in the park of the plant, if each bus is equipped with only one air conditioner? | 16. 60. Let the number of buses in the park be denoted by $x$. Then, $\frac{1}{6} x + 5$ buses are equipped with air conditioning. This constitutes a quarter of all the buses in the park. We have the equation: $\left(\frac{1}{6} x + 5\right) \cdot 4 = x$ or $\frac{1}{3} x = 20$. From this, $x = 60$. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,056 |
Problem 6. The solution is generally correct but with an error in counting the number of addends by 1 "土".
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | The solution relies on incorrect trigonometric formulas - no higher than "G".
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 9,061 |
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three is to the sum of all terms except the last three as $4: 3$. Find the number of terms in this progressio... | Answer: 20.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all t... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,062 |
Task 2. On the island, every resident is either a knight (always tells the truth), a liar (always lies), or a commoner (can either tell the truth or lie). Knights are considered to be of the highest rank, commoners are of medium rank, and liars are of the lowest rank. A, B, and C are residents of this island. One of th... | Answer: B.
Solution. Let's go through the options for who A can be.
If A is a knight, then B is an ordinary person, and C is a liar, and C will answer: "B". If A is a liar, then C is a knight, and B is an ordinary person, and C will also answer: "B". Finally, if A is an ordinary person, then B cannot be a knight (oth... | B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,063 |
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number written with the same digits in reverse order is equal to $N$. It turns out that the number $N$ is divisible by 100. Find $N$. | Answer: 11000.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{d c b a}=1000 d+100 c+10 b+a$, where $a$, $b, c, d$ are digits and $a \neq 0$.
According to the condition, $X+Y$ is divisible by 100, i.e., $1001(a+d)+110(b+c) \vdots 100$.
We have $1001(a+d) \vdots 10$, i.e., $a+d \vdots 10$, from w... | 11000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,064 |
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | Answer: 2015.
Solution. Let $O$ be the intersection point of diagonals $AC$ and $BD$. From the similarity of triangles $AOB$ and $COD$, it follows that $\overrightarrow{OC} = \frac{31}{65} \overrightarrow{AO}$, and $\overrightarrow{OD} = \frac{31}{65} \overrightarrow{BO}$.
Let the vector $\overrightarrow{AO}$ be deno... | 2015 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,065 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
26 x^{2}+42 x y+17 y^{2}=10 \\
10 x^{2}+18 x y+8 y^{2}=6
\end{array}\right.
$$ | Answer: $(-1,2),(-11,14),(11,-14),(1,-2)$.
Solution. By adding and subtracting the two equations of the system, we obtain that the original system is equivalent to the following:
$$
\begin{cases}(6 x+5 y)^{2} & =16 \\ (4 x+3 y)^{2} & =4\end{cases}
$$
From which we get 4 possible cases
$$
\left\{\begin{array} { l } ... | (-1,2),(-11,14),(11,-14),(1,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,066 |
Problem 6. For $x=\frac{\pi}{2 n}$, find the value of the sum
$$
\cos ^{2}(x)+\cos ^{2}(2 x)+\cos ^{2}(3 x)+\ldots+\cos ^{2}(n x)
$$ | Answer: $\frac{n-1}{2}$.
Solution. Note that
$$
\cos ^{2}\left(\frac{k \pi}{2 n}\right)+\cos ^{2}\left(\frac{(n-k) \pi}{2 n}\right)=\cos ^{2}\left(\frac{k \pi}{2 n}\right)+\cos ^{2}\left(\frac{\pi}{2}-\frac{k \pi}{2 n}\right)=\cos ^{2}\left(\frac{k \pi}{2 n}\right)+\sin ^{2}\left(\frac{k \pi}{2 n}\right)=1
$$
If $n$... | \frac{n-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,067 |
Task 7. The line c is given by the equation $y=x+1$. Points $A$ and $B$ have coordinates $A(1 ; 0)$ and $B(3 ; 0)$. On the line $c$, find the point $C$ from which the segment $A B$ is seen at the largest angle. | Answer: $(1 ; 2)$.
Solution. The desired point is point $C(1 ; 2)$. Indeed, consider a circle with center at point $P(2 ; 1)$ and radius $P A = P B$. It touches the line $c$ at point $C$. The angle $A C B$ is half the arc $A B$. For any point $C^{\prime}$ on the line $c$, different from point $C$ and lying in the same... | (1;2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,068 |
Problem 8. For what values of the parameter $a$ does the equation $\ln (x+a)-4(x+a)^{2}+a=0$ have a unique root 3 | Answer: when $a=\frac{3 \ln 2+1}{2}$.
Solution. Let's make the substitution $t=x+a$ and write the equation in the form $\ln t=4 t^{2}-a$. We will plot the graph of the function $y=\ln t$ and the family of parabolas $y=4 t^{2}-a$ in the same coordinate system.
 h=63$. We also know that $\pi r^{2} h=9$. Dividing the corresponding part... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,071 |
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is to the sum of all terms except the last three as $5: 4$. Find the number of terms in this prog... | Answer: 22.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all t... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,072 |
Problem 2. On the island, every resident is either a knight (always tells the truth), a liar (always lies), or a regular person (can either tell the truth or lie). The residents of this island, A and B, said the following. A: “B is a knight.” B: “A is a liar.” Prove that either one of them is telling the truth, but it ... | Solution. Let's go through the options of who A can be.
If A is a knight, then B is also a knight, but then A is a liar. Therefore, A cannot be a knight. If A is a liar, then B is not a knight, but B told the truth, and we have found the required inhabitant. Finally, if A is an ordinary person, then B lied, meaning B ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 9,073 |
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number obtained from $X$ by swapping its second and third digits is divisible by 900. Find the remainder when the number $X$ is divided by 90. | Answer: 45.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{a c b d}=1000 a+100 c+10 b+d$, where $a$, $b, c, d$ are digits and $a \neq 0, d \neq 0$.
According to the condition, $X+Y$ is divisible by 900, i.e., $2000 a+110(b+c)+2 d \vdots 900$.
We have, $2 d \vdots 10$, i.e., $d \vdots 5$, so sin... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,074 |
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 155 and 13, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | Answer: 2015.
Solution. Let $O$ be the intersection point of diagonals $AC$ and $BD$. From the similarity of triangles $AOB$ and $COD$, it follows that $\overrightarrow{OC} = \frac{13}{155} \overrightarrow{AO}$, and $\overrightarrow{OD} = \frac{13}{155} \overrightarrow{BO}$.
Let the vector $\overrightarrow{AO}$ be de... | 2015 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,075 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
26 x^{2}-42 x y+17 y^{2}=10 \\
10 x^{2}-18 x y+8 y^{2}=6
\end{array}\right.
$$ | Answer: $(-1,-2),(-11,-14),(11,14),(1,2)$.
Solution. By adding and subtracting the two equations of the system, we obtain that the original system is equivalent to the following:
$$
\begin{cases}(6 x-5 y)^{2} & =16 \\ (4 x-3 y)^{2} & =4\end{cases}
$$
From which we get 4 possible cases
$$
\left\{\begin{array} { l } ... | (-1,-2),(-11,-14),(11,14),(1,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,076 |
Problem 6. For $x=\frac{\pi}{2 n}$, find the value of the sum
$$
\sin ^{2}(x)+\sin ^{2}(2 x)+\sin ^{2}(3 x)+\ldots+\sin ^{2}(n x)
$$ | Answer: $\frac{n+1}{2}$.
Solution. Note that
$$
\sin ^{2}\left(\frac{k \pi}{2 n}\right)+\sin ^{2}\left(\frac{(n-k) \pi}{2 n}\right)=\sin ^{2}\left(\frac{k \pi}{2 n}\right)+\sin ^{2}\left(\frac{\pi}{2}-\frac{k \pi}{2 n}\right)=\sin ^{2}\left(\frac{k \pi}{2 n}\right)+\cos ^{2}\left(\frac{k \pi}{2 n}\right)=1
$$
If $n$... | \frac{n+1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,077 |
Task 7. The line $c$ is given by the equation $y=2 x$. Points $A$ and $B$ have coordinates $A(2 ; 2)$ and $B(6 ; 2)$. On the line $c$, find the point $C$ from which the segment $A B$ is seen at the largest angle. | Answer: $(2 ; 4)$.
Solution. The desired point is point $C(2 ; 4)$. Indeed, consider a circle centered at point $P(4 ; 3)$ with radius $P A = P B$. It touches the line $c$ at point $C$. The angle $A C B$ is half the arc $A B$. For any point $C^{\prime}$ on the line $c$, different from point $C$, the angle $A C^{\prime... | (2;4) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,078 |
Problem 8. For what values of the parameter $a$ does the equation
$\ln (x-2 a)-3(x-2 a)^{2}+2 a=0$ have a unique root? | Answer: when $a=\frac{\ln 6+1}{4}$.
Solution. Let's make the substitution $t=x-2 a$ and write the equation in the form
$\ln t=3 t^{2}-2 a$. We will plot the graph of the function $y=\ln t$ and the family of parabolas $y=3 t^{2}-2 a$ in the same coordinate system.
)}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all t... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,081 |
Problem 2. On the island, every resident is either a knight (always tells the truth) or a liar (always lies). Two residents are called of the same type if they are either both knights or both liars. A, B, and C are residents of this island. A says: "B and C are of the same type." What will C answer to the question "Are... | # Answer: “Yes".
Solution. Let's go through the options of who A might be.
If A is a knight, then B and C are of the same type. If B is a knight in this case, then C is also a knight, and C will answer "Yes." If B is a liar, then C is also a liar, and will still answer "Yes." If A is a liar, then B and C are of diffe... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,082 |
Problem 3. If the sum of the digits of a four-digit number $X$ is subtracted from $X$, the result is a natural number $N=K^{2}$, where $K$ is a natural number that gives a remainder of 2 when divided by 10 and a remainder of 6 when divided by 11. Find the number $N$. | Answer: there are no solutions.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d$, where $a, b, c, d$ are digits and $a \neq 0$.
By the condition, $X-a-b-c-d=999 a+99 b+9 c=K^{2}$, where $K=10 u+2, K=11 v+6$.
Notice that $999 a+99 b+9 c \vdots 9$, i.e., $K^{2} \vdots 9$, so $K=3 M$, and $M^{2}=111 a+11 b+c \l... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,083 |
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$. | Answer: -2015.
Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{31}{65} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{31}{65} \overrightarrow{AO}$.
Let the v... | -2015 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,084 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
5 x^{2}+14 x y+10 y^{2}=17 \\
4 x^{2}+10 x y+6 y^{2}=8
\end{array}\right.
$$ | Answer: $(-1,2),(11,-7),(-11,7),(1,-2)$.
Solution. By adding and subtracting the two equations of the system, we obtain that the original system is equivalent to the following:
$$
\left\{\begin{aligned}
(3 x+4 y)^{2} & =25 \\
(x+2 y)^{2} & =9
\end{aligned}\right.
$$
From which we get 4 possible cases
$$
\left\{\beg... | (-1,2),(11,-7),(-11,7),(1,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,085 |
Problem 8. For what values of the parameter $a$ does the equation
$\ln (x+a)-4(x+a)^{2}+a=0$ have a unique root? | Answer: when $a=\frac{3 \ln 2+1}{2}$.
Solution. Let's make the substitution $t=x+a$ and write the equation in the form
$\ln t=4 t^{2}-a$. We will plot the graph of the function $y=\ln t$ and the family of parabolas $y=4 t^{2}-a$ in the same coordinate system.
 h=91$. We also know that $\pi r^{2} h=21$. Dividing the corresponding p... | 94.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,088 |
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is in the ratio $6: 5$ to the sum of all terms except the last three. Find the number of terms in... | Answer: 24.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all t... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,089 |
Problem 3. If the sum of the digits of a four-digit number $X$ is subtracted from $X$, the result is a natural number $N=K^{2}$, where $K$ is a natural number that gives a remainder of 5 when divided by 20 and a remainder of 3 when divided by 21. Find the number $N$. | Answer: 2025.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d$, where $a, b, c, d$ are digits and $a \neq 0$.
According to the condition, $X-a-b-c-d=999 a+99 b+9 c=K^{2}$, where $K=20 u+5, K=21 v+3$.
Notice that $999 a+99 b+9 c \vdots 9$, i.e., $K^{2} \vdots 9$, which means $K=3 M$, and $M^{2}=111 a+11 b+c \... | 2025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,091 |
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 155 and 13, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$. | Answer: -2015.
Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{13}{155} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{13}{155} \overrightarrow{AO}$.
Let the... | -2015 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,092 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
5 x^{2}-14 x y+10 y^{2}=17 \\
4 x^{2}-10 x y+6 y^{2}=8
\end{array}\right.
$$ | Answer: $(-1,-2),(11,7),(-11,-7),(1,2)$.
Solution. By adding and subtracting the two equations of the system, we obtain that the original system is equivalent to the following:
$$
\left\{\begin{aligned}
(3 x-4 y)^{2} & =25 \\
(x-2 y)^{2} & =9
\end{aligned}\right.
$$
From which we get 4 possible cases
$$
\left\{\beg... | (-1,-2),(11,7),(-11,-7),(1,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,093 |
Task 7. The line c is given by the equation $y=2x$. Points $A$ and $B$ have coordinates $A(2; 2)$ and $B(6; 2)$. On the line $c$, find the point $C$ from which the segment $AB$ is seen at the largest angle. | Answer: $(2 ; 4)$.
Solution. The desired point is point $C(2 ; 4)$. Indeed, consider a circle centered at point $P(4 ; 3)$ with radius $P A = P B$. It touches the line $c$ at point $C$. The angle $A C B$ is half the arc $A B$. For any point $C^{\prime}$ on the line $c$, different from point $C$, the angle $A C^{\prime... | (2;4) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,094 |
Problem 9. In a volleyball tournament, bets are accepted on four teams. Bets on the first team are accepted at a ratio of $1: 2$ (if the first team wins, the player receives the amount they bet on this team plus twice the amount, i.e., they receive three times the amount of money they bet, and if they lose, the money i... | Answer: Yes, it is possible.
Solution. If the first team wins, the bet is returned tripled, so more than $1 / 3$ of all the money should be placed on it. Similarly, more than $1 / 5$ of all the money should be placed on the second team, more than $1 / 6$ on the third, and more than $1 / 6$ on the fourth. Since the sum... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,095 |
Task 1. What is greater: 1 or $\frac{21}{64}+\frac{51}{154}+\frac{71}{214} ?$ | Answer: One is greater.
## First solution.
$$
\frac{21}{64}+\frac{51}{154}+\frac{71}{214}<\frac{21}{63}+\frac{51}{153}+\frac{71}{213}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1
$$
## Second solution.
$$
\begin{aligned}
\frac{21}{64}+\frac{51}{154}+\frac{71}{214} & =\frac{21 \cdot 154 \cdot 214+64 \cdot 51 \cdot 214+64 \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,096 |
Problem 2. In a football tournament, seven teams played: each team played once with each other. In the next round, teams that scored thirteen or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next rou... | Answer: 4.
Solution. In total, the teams played $\frac{7 \cdot 6}{2}=21$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $21 \cdot 3=63$. Hence, the number of teams $n$ advancing to the next stage satisfies the inequality $n \cdot 13 \leqslant 63$, f... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,097 |
Problem 3. For what least natural $k$ is the expression $2017 \cdot 2018 \cdot 2019 \cdot 2020+k$ a square of a natural number? | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2018$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,098 |
Problem 4. Point $O$ lies inside an isosceles right triangle $A B C$. The distance from it to vertex $A$ of the right angle is 6, to vertex $B$ is 4, and to vertex $C$ is 8. Find the area of triangle $A B C$. | Answer: $20+6 \sqrt{7}$.
First solution. Consider a $90^{\circ}$ rotation around point $A$ that maps point $C$ to point $B$. Let point $O$ be mapped to point $D$ during this rotation; then segment $BD$ is the image of segment $CO$; since the length of segments does not change during rotation, $BD=CO=8$. We obtain quad... | 20+6\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,099 |
Problem 5. Let $f(x)=9 x^{2}+8 x-2$. Solve the equation $f(f(x))=x$.
---
The provided text has been translated while preserving the original formatting and line breaks. | Answer: $-1, \frac{2}{9}, \frac{-3 \pm \sqrt{13}}{6}$.
First solution. The number $x$ satisfies the equation $f(f(x))=x$ if and only if there exists a number $y$ such that the system
$$
\left\{\begin{array}{l}
9 x^{2}+8 x-2=y \\
9 y^{2}+8 y-2=x
\end{array}\right.
$$
is satisfied. Subtracting the second equation from... | -1,\frac{2}{9},\frac{-3\\sqrt{13}}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,100 |
Task 6. Find all values that the expression
$$
3 \arcsin x - 2 \arccos y
$$
can take under the condition $x^{2} + y^{2} = 1$. | Answer: $\left[-\frac{5 \pi}{2} ; \frac{\pi}{2}\right]$.
Solution. Note that $x^{2}+y^{2}=1$ if and only if there exists some $\varphi \in[0 ; 2 \pi]$ such that $x=\sin \varphi, y=\cos \varphi$. Then the expression from the condition takes the form
$$
3 \arcsin \sin \varphi-2 \arccos \cos \varphi
$$
Let's consider s... | [-\frac{5\pi}{2};\frac{\pi}{2}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,101 |
Problem 7. Given a triangle $ABC$. Points $X$ and $Y$ are chosen on segments $AB$ and $BC$ respectively such that $AX = BY$. It turns out that points $A, X, Y$, and $C$ lie on the same circle. Let $BL$ be the angle bisector of triangle $ABC$ ($L$ on segment $AC$). Prove that $XL \parallel BC$. | Solution. From the fact that points $A, X, Y$ and $C$ lie on the same circle, it follows that $B X \cdot B A = B Y \cdot B C$, or $A B: B C = B Y: B X$. From the fact that $B L$ is the bisector of triangle $A B C$, it follows that $A L: L C = A B: B C$. Then
$$
A L: L C = A B: B C = B Y: B X = A X: X B
$$
from which,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,102 |
Problem 8. For what values of the parameter $a$ does the equation
$$
\log _{2}\left(2 x^{2}+(2 a+1) x-2 a\right)-2 \log _{4}\left(x^{2}+3 a x+2 a^{2}\right)=0
$$
have two distinct roots, the sum of whose squares is greater than 4? | Answer: $(-\infty ;-1) \cup\left(\frac{3}{5} ; 1\right)$.
Solution. Rewrite the original equation as
$$
\log _{2}\left(2 x^{2}+(2 a+1) x-2 a\right)=\log _{2}\left(x^{2}+3 a x+2 a^{2}\right)
$$
Notice that this equation is equivalent to the system
$$
\left\{\begin{array}{l}
2 x^{2}+(2 a+1) x-2 a=x^{2}+3 a x+2 a^{2} ... | (-\infty;-1)\cup(\frac{3}{5};1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,103 |
Problem 10. Let's call a mountain a truncated right circular cone with the circumference of the lower base being 8, and the upper base - 6. The slope of the mountain is inclined at an angle of $60^{\circ}$ to the base plane. On the circumference of the lower base lies point $A$. A tourist starts climbing up the slope f... | Answer: $\frac{4 \sqrt{3}}{\pi}$
Solution. Let's denote (see the left figure) the vertex of the cone by $Q$, the center of the smaller base by $T$, the point on the upper base closest to $A-B$ as the starting point of the return path $-C$. Let $D$ be the last point of the return path on the circumference of the upper ... | \frac{4\sqrt{3}}{\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,105 |
Task 1. What is greater: 1 or $\frac{27}{80}+\frac{46}{137}+\frac{63}{188}$? | Answer: The sum of the fractions is greater
## First solution.
$$
\frac{27}{80}+\frac{46}{137}+\frac{63}{188}>\frac{27}{81}+\frac{46}{138}+\frac{63}{189}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1
$$
## Second solution.
$$
\begin{aligned}
\frac{27}{80}+\frac{46}{137}+\frac{63}{188} & =\frac{27 \cdot 137 \cdot 188+80 \cd... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 9,106 |
Problem 2. In a football tournament, seven teams played: each team played once with each other. In the next round, teams that scored twelve or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round... | Answer: 5.
Solution. In total, the teams played $\frac{7 \cdot 6}{2}=21$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $21 \cdot 3=63$. Hence, the number of teams $n$ advancing to the next stage satisfies the inequality $n \cdot 12 \leqslant 63$, f... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,107 |
Problem 3. For what least natural $k$ is the expression $2019 \cdot 2020 \cdot 2021 \cdot 2022 + k$ a square of a natural number? | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2020$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,108 |
Problem 4. Point $O$ lies inside an isosceles right triangle $A B C$. The distance from it to vertex $A$ of the right angle is 5, to vertex $B$ is 6, and to vertex $C$ is 4. Find the area of triangle $A B C$. | Answer: $13+\frac{5}{2} \sqrt{23}$.
First solution. Consider a $90^{\circ}$ rotation around point $A$, which maps point $C$ to point $B$. Let point $O$ be mapped to point $D$ during this rotation; then segment $BD$ is the image of segment $CO$; since the length of segments does not change during rotation, $BD = CO = 4... | 13+\frac{5}{2}\sqrt{23} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,109 |
Problem 5. Let $f(x)=3 x^{2}-7 x-11$. Solve the equation $f(f(x))=x$.
---
The original text has been translated into English, maintaining the original formatting and line breaks. | Answer: $-1, \frac{11}{3}, \frac{3 \pm 4 \sqrt{3}}{3}$.
First solution. A number $x$ satisfies the equation $f(f(x))=x$ if and only if there exists a number $y$ such that the system
$$
\left\{\begin{array}{l}
3 x^{2}-7 x-11=y \\
3 y^{2}-7 y-11=x
\end{array}\right.
$$
is satisfied. Subtracting the second equation fro... | -1,\frac{11}{3},\frac{3\4\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,110 |
Task 6. Find all values that the expression
$$
4 \arcsin x - \arccos y
$$
can take under the condition $x^{2} + y^{2} = 1$. | Answer: $\left[-\frac{5 \pi}{2} ; \frac{3 \pi}{2}\right]$.
Solution. Note that $x^{2}+y^{2}=1$ if and only if there exists some $\varphi \in[0 ; 2 \pi]$ such that $x=\sin \varphi, y=\cos \varphi$. Then the expression from the condition takes the form
$$
4 \arcsin \sin \varphi-\arccos \cos \varphi
$$
Let's consider s... | [-\frac{5\pi}{2};\frac{3\pi}{2}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,111 |
Problem 7. Given a triangle $ABC$. Points $X$ and $Y$ are chosen on segments $AB$ and $BC$ respectively such that $AX = BY$. It turns out that points $A, X, Y$, and $C$ lie on the same circle. Let $L$ be a point on segment $AC$ such that $XL \parallel BC$. Prove that $BL$ is the bisector of triangle $ABC$. | Solution. From the fact that points $A, X, Y$ and $C$ lie on the same circle, it follows that $B X \cdot B A = B Y \cdot B C$, or $A B: B C = B Y: B X$. From the fact that $X L \| B C$ it follows that $A L: L C = A X: X B$. Then
$$
A L: L C = A X: X B = B Y: B X = A B: B C
$$
from which, by the angle bisector theorem... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,112 |
Problem 8. For what values of the parameter $a$ does the equation
$$
2 \log _{16}\left(2 x^{2}-x-2 a-4 a^{2}\right)-\log _{4}\left(x^{2}-a x-2 a^{2}\right)=0
$$
have two distinct roots, the sum of whose squares lies in the interval $(0 ; 4) ?$ | Answer: $\left(-\frac{1}{2} ;-\frac{1}{3}\right) \cup\left(-\frac{1}{3} ; 0\right) \cup\left(0 ; \frac{3}{5}\right)$.
Solution. Rewrite the original equation as
$$
\log _{4}\left(2 x^{2}-x-2 a-4 a^{2}\right)=\log _{4}\left(x^{2}-a x-2 a^{2}\right)
$$
Notice that this equation is equivalent to the system
$$
\left\{\... | (-\frac{1}{2};-\frac{1}{3})\cup(-\frac{1}{3};0)\cup(0;\frac{3}{5}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,113 |
Problem 10. Let's call a mountain a truncated right circular cone with the circumference of the lower base being 10, and the upper base - 9. The slope of the mountain is inclined at an angle of $60^{\circ}$ to the base plane. On the circumference of the lower base lies point $A$. A tourist starts climbing up the slope ... | Answer: $\frac{5 \sqrt{3}}{\pi}$
Solution. Let's denote (see the left figure) the vertex of the cone by $Q$, the center of the smaller base by $T$, the point on the upper base closest to $A-B$ as the starting point of the return path $-C$. Let $D$ be the last point of the return path on the circumference of the upper ... | \frac{5\sqrt{3}}{\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,115 |
Task 1. What is greater: 1 or $\frac{23}{93}+\frac{41}{165}+\frac{71}{143}$? | Answer: One is greater.
## First solution.
$$
\frac{23}{93}+\frac{41}{165}+\frac{71}{143}<\frac{23}{92}+\frac{41}{164}+\frac{71}{142}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=1 .
$$
## Second solution.
$$
\begin{aligned}
\frac{23}{93}+\frac{41}{165}+\frac{71}{143} & =\frac{23 \cdot 165 \cdot 143+93 \cdot 41 \cdot 143+93... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,116 |
Problem 2. In a football tournament, eight teams played: each team played once with each other. In the next round, teams that scored fifteen or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next roun... | Answer: 5.
Solution. In total, the teams played $\frac{8 \cdot 7}{2}=28$ games, in each of which 2 or 3 points were at stake. Therefore, the maximum total number of points that all teams can have is $28 \cdot 3=84$. Hence, the number of teams $n$ that advance to the next stage satisfies the inequality $n \cdot 15 \leq... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,117 |
Problem 3. For what least natural $k$ is the expression $2018 \cdot 2019 \cdot 2020 \cdot 2021+k$ a square of a natural number | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2019$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,118 |
Problem 4. Point $O$ lies inside an isosceles right triangle $A B C$. The distance from it to vertex $A$ of the right angle is 5, to vertex $B$ is 7, and to vertex $C$ is 3. Find the area of triangle $A B C$. | Answer: $\frac{29}{2}+\frac{5}{2} \sqrt{17}$.
First solution. Consider a rotation around point $A$ by an angle of $90^{\circ}$, which maps point $C$ to point $B$. Let point $O$ be mapped to point $D$ during this rotation; then segment $BD$ is the image of segment $CO$; since the length of segments does not change duri... | \frac{29}{2}+\frac{5}{2}\sqrt{17} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,119 |
Problem 5. Let $f(x)=7 x^{2}+6 x-2$. Solve the equation $f(f(x))=x$.
---
The provided text has been translated while preserving the original formatting and line breaks. | Answer: $-1, \frac{2}{7}, \frac{-7 \pm \sqrt{77}}{14}$.
First solution. The number $x$ satisfies the equation $f(f(x))=x$ if and only if there exists a number $y$ such that the system
$$
\left\{\begin{array}{l}
7 x^{2}+6 x-2=y \\
7 y^{2}+6 y-2=x
\end{array}\right.
$$
is satisfied. Subtracting the second equation fro... | -1,\frac{2}{7},\frac{-7\\sqrt{77}}{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,120 |
Task 6. Find all values that the expression
$$
2 \arcsin x - \arccos y
$$
can take under the condition $x^{2} + y^{2} = 1$. | Answer: $\left[-\frac{3 \pi}{2} ; \frac{\pi}{2}\right]$.
Solution. Note that $x^{2}+y^{2}=1$ if and only if there exists some $\varphi \in[0 ; 2 \pi]$ such that $x=\sin \varphi, y=\cos \varphi$. Then the expression from the condition takes the form
$$
2 \arcsin \sin \varphi-\arccos \cos \varphi
$$
Let's consider sev... | [-\frac{3\pi}{2};\frac{\pi}{2}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,121 |
Problem 7. Given a triangle $ABC$. Let $BL$ be the bisector of triangle $ABC$ ($L$ on segment $AC$), $X$ be a point on segment $AB$ such that $XL \parallel BC$. On segment $BC$, there is a point $Y$ such that $AX = BY$. Prove that points $A, X, Y$, and $C$ lie on the same circle. | Solution. From the fact that $BL$ is the bisector of triangle $ABC$, it follows that $AL: LC = AB: BC$. From $XL \| BC$, it follows that $AL: LC = AX: XB$. Therefore,
$$
BY: BX = AX: XB = AL: LC = AB: BC
$$
that is, $BX \cdot BA = BY \cdot BC$. Consequently, by one of the criteria for a cyclic quadrilateral, points $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,122 |
Problem 8. For what values of the parameter $a$ does the equation
$$
\log _{3}\left(2 x^{2}-x+2 a-4 a^{2}\right)+\log _{1 / 3}\left(x^{2}+a x-2 a^{2}\right)=0
$$
have two distinct roots, the sum of whose squares is less than 1? | Answer: $\left(0 ; \frac{1}{3}\right) \cup\left(\frac{1}{3} ; \frac{2}{5}\right)$.
Solution. Rewrite the original equation as
$$
\log _{3}\left(2 x^{2}-x+2 a-4 a^{2}\right)=\log _{3}\left(x^{2}+a x-2 a^{2}\right)
$$
Notice that this equation is equivalent to the system
$$
\left\{\begin{array}{l}
2 x^{2}-x+2 a-4 a^{... | (0;\frac{1}{3})\cup(\frac{1}{3};\frac{2}{5}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,123 |
Problem 9. At the conference, there were several scientists, some of whom speak English, some French, and some German. The organizers of the conference noticed that among those who speak English, exactly $1 / 5$ speak French, and exactly $1 / 3$ speak German; among those who speak French, exactly $1 / 8$ speak English,... | Answer: $2 / 5$.
Solution. Let $x$ be the number of scientists who speak German. Then $x / 6$ scientists speak both German and English; thus, $x / 2$ scientists speak English, and $x / 10$ scientists speak both English and French; therefore, $4 x / 5$ scientists speak French, and $2 x / 5$ scientists speak both French... | \frac{2}{5} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,124 |
Problem 10. Let's call a mountain a truncated right circular cone with the circumference of the lower base being 8, and the upper base - 6. The slope of the mountain is inclined at an angle of $60^{\circ}$ to the base plane. On the circumference of the lower base lies point $A$. A tourist starts climbing up the slope f... | Answer: $\frac{4 \sqrt{3}}{\pi}$
Solution. Let's denote (see the left figure) the vertex of the cone by $Q$, the center of the smaller base by $T$, the point on the upper base closest to $A-B$ as the starting point of the return path $-C$. Let $D$ be the last point of the return path on the circumference of the upper ... | \frac{4\sqrt{3}}{\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,125 |
Task 1. What is greater: 1 or $\frac{18}{71}+\frac{47}{187}+\frac{59}{117}$? | Answer: The sum of the fractions is greater.
## First solution.
$$
\frac{18}{71}+\frac{47}{187}+\frac{59}{117}>\frac{18}{72}+\frac{47}{188}+\frac{59}{118}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=1
$$
## Second solution.
$$
\begin{aligned}
\frac{18}{71}+\frac{47}{187}+\frac{59}{117} & =\frac{18 \cdot 187 \cdot 117+71 \c... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,126 |
Problem 2. In a football tournament, six teams played: each team played once with each other. In the next round, teams that scored twelve or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round? | Answer: 3
Solution. In total, the teams played $\frac{6 \cdot 5}{2}=15$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $15 \cdot 3=45$. Hence, the number of teams $n$ that advance to the next stage satisfies the inequality $n \cdot 12 \leqslant 45$,... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,127 |
Problem 3. For what least natural $k$ is the expression $2016 \cdot 2017 \cdot 2018 \cdot 2019 + k$ a square of a natural number? | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2017$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,128 |
Problem 4. Point $O$ lies inside an isosceles right triangle $A B C$. The distance from it to vertex $A$ of the right angle is 6, to vertex $B$ is 9, and to vertex $C$ is 3. Find the area of triangle $A B C$. | Answer: $\frac{45}{2}+9 \sqrt{2}$.
First solution. Consider a rotation around point $A$ by an angle of $90^{\circ}$, which maps point $C$ to point $B$. Let point $O$ be mapped to point $D$ during this rotation; then segment $BD$ is the image of segment $CO$; since the length of segments does not change during rotation... | \frac{45}{2}+9\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,129 |
Problem 5. Let $f(x)=4 x^{2}+7 x-10$. Solve the equation $f(f(x))=x$.
---
The original text has been translated into English, maintaining the original line breaks and format. | Answer: $1,-\frac{5}{2},-1 \pm \sqrt{3}$.
First solution. The number $x$ satisfies the equation $f(f(x))=x$ if and only if there exists a number $y$ such that the system
$$
\left\{\begin{array}{l}
4 x^{2}+7 x-10=y \\
4 y^{2}+7 y-10=x
\end{array}\right.
$$
is satisfied. Subtracting the second equation from the first,... | 1,-\frac{5}{2},-1\\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,130 |
Task 6. Find all values that the expression
$$
5 \arcsin x - 2 \arccos y
$$
can take under the condition $x^{2} + y^{2} = 1$. | Answer: $\left[-\frac{7 \pi}{2} ; \frac{3 \pi}{2}\right]$.
Solution. Note that $x^{2}+y^{2}=1$ if and only if there exists some $\varphi \in[0 ; 2 \pi]$ such that $x=\sin \varphi, y=\cos \varphi$. Then the expression from the condition takes the form
Let's consider several cases:
- $\varphi \in\left[0 ; \frac{\pi}{2... | [-\frac{7\pi}{2};\frac{3\pi}{2}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,131 |
Problem 7. Given a triangle $ABC$. Let $BL$ be the bisector of triangle $ABC$ ($L$ on segment $AC$), and $X$ be a point on segment $AB$ such that $XL \parallel BC$. The circumcircle of triangle $AXC$ intersects segment $BC$ at points $C$ and $Y$. Prove that $AX = BY$. | Solution. From the fact that $BL$ is the bisector of triangle $ABC$, it follows that $AL: LC = AB: BC$. From the fact that points $A, X, Y$ and $C$ lie on the same circle, it follows that $BX \cdot BA = BY \cdot BC$, or $AB: BC = BY: BX$. From the fact that $XL \| BC$ it follows that $AL: LC = AX: XB$. Then
$$
AL: LC ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,132 |
Problem 8. For what values of the parameter $a$ does the equation
$$
\log _{2}\left(2 x^{2}-x-2 a-4 a^{2}\right)+3 \log _{1 / 8}\left(x^{2}-a x-2 a^{2}\right)=0
$$
have two distinct roots, the sum of whose squares lies in the interval $(4 ; 8)$? | Answer: $\left(\frac{3}{5} ; 1\right)$.
Solution. Rewrite the original equation as
$$
\log _{2}\left(2 x^{2}-x-2 a-4 a^{2}\right)=\log _{4}\left(x^{2}-a x-2 a^{2}\right)
$$
Notice that this equation is equivalent to the system
$$
\left\{\begin{array}{l}
2 x^{2}-x-2 a-4 a^{2}=x^{2}-a x-2 a^{2} \\
x^{2}-a x-2 a^{2}>0... | (\frac{3}{5};1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,133 |
Problem 10. Let's call a mountain a truncated right circular cone with the circumference of the lower base being 10, and the upper base - 9. The slope of the mountain is inclined at an angle of $60^{\circ}$ to the base plane. On the circumference of the lower base lies point $A$. A tourist starts climbing the slope fro... | Answer: $\frac{5 \sqrt{3}}{\pi}$
Solution. Let's denote (see the left figure) the vertex of the cone by $Q$, the center of the smaller base by $T$, the point on the upper base closest to $A-B$ as the starting point of the return path $-C$. Let $D$ be the last point of the return path on the circumference of the upper ... | \frac{5\sqrt{3}}{\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,135 |
Problem 1. The sum of the first three terms of an arithmetic progression, as well as the sum of the first six terms, are natural numbers. In addition, its first term $d_{1}$ satisfies the inequality $d_{1} \geqslant \frac{1}{2}$. What is the smallest value that $d_{1}$ can take? | Answer: $5 / 9$
Solution. Let $d_{n}$ be the $n$-th term of the progression, $d$ be the common difference of the progression, and $S_{n}=d_{1}+d_{2}+\ldots+d_{n}$ be the sum of the first $n$ terms of the progression.
Express $d_{1}$ in terms of $S_{3}$ and $S_{6}$. Note that $S_{3}=3 d_{1}+3 d, S_{6}=6 d_{1}+15 d$, f... | \frac{5}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,136 |
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