problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle \(\alpha=45^{\circ}\). (15 points)
 | Solution. Let the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$ and the sector $A C B=a$. Then, if the area of the segment $A C$ is $b$, we have: $x+b=S_{0}=b+a \Rightarrow x=a$.
The entire shaded area is equal to $x+y=a+y$, but $a+y$ is the area of the ... | \frac{\piR^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,253 |
8. Solve the system $\left\{\begin{array}{l}x \log _{2} 3+y=\log _{2} 18 ; \\ 5^{x}=25^{y} .\end{array}\right.$
 | # Solution:
$\left\{\begin{array}{l}\log _{2} 3^{x}+y=\log _{2} 18, \\ 5^{x}=5^{2 y},\end{array}\left\{\begin{array}{l}2 y \log _{2} 3+y=18 \\ x=2 y,\end{array},\left\{\begin{array}{l}y\left(\log _{2} 3^{2}+1\right)=18, \\ x=2 y,\end{array}\left\{\begin{array}{l}y\left(\log _{2}\left(3^{2} \cdot 2\right)\right)=18, \\... | 2;1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,254 |
8. Solve the system $\left\{\begin{array}{l}x+y=13 \\ \log _{4} x+\log _{4} y=1+\log _{4} 10 .\end{array}\right.$
 | Solution: $\left\{\begin{array}{l}x+y=13, \\ \log _{4}(x y)=\log _{4} 40,\end{array}\left\{\begin{array}{l}x+y=13, \\ x y=40,\end{array}\left\{\begin{array}{l}x=5, y=8 \\ x=8, y=5\end{array}\right.\right.\right.$
Answer: $\{5,8\},\{8,5\}$ | {5,8},{8,5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,255 |
1. On the board, the numbers $a, b$, and $c$ are written. They were erased, and in their place, the numbers $a^{2}+2 b c, b^{2}+2 c a$, $c^{2}+2 a b$ were written. After this, it turned out that the numbers on the board are the same as those at the beginning (possibly in a different order). Find all possible values of ... | Answer: 0 or 1.
From the condition, it follows that the sum of the "new" numbers coincides with the sum $s=a+b+c$ of the original numbers, that is,
$$
a^{2}+2 b c+b^{2}+2 c a+c^{2}+2 a b=s
$$
The expression on the left side coincides with the number $(a+b+c)^{2}=s^{2}$, so this equality is equivalent to: $s^{2}=s$, ... | 0or1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,256 |
2. Find the smallest possible value of the function
$$
f(x)=|x+1|+|x+2|+\ldots+|x+100|
$$
$(25$ points. $)$ | Answer: 2500.
Using the known inequality $|a|+|b| \geqslant|a-b|$, then $|x+k|+|x+m| \geqslant|k-m|$. Grouping in the original sum the terms equally distant from the ends, we get
$$
\begin{aligned}
f(x) & =(|x+1|+|x+100|)+(|x+2|+|x+99|)+\ldots+(|x+50|+|x+51|) \geqslant \\
& \geqslant(100-1)+(99-2)+\ldots+(51-50)
\end... | 2500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,257 |
3. Find all values of $c$ for which the inequality $a+\sqrt{b+c}>b+\sqrt{a+c}$ holds for any positive $a, b$ and $a>b$.
(25 points.) | Answer: $c=\frac{1}{4}$.
Write the inequality in the form $a-b>\sqrt{a+c}-\sqrt{b+c}$. By multiplying and dividing the right part by its conjugate, we can represent the inequality as
$$
a-b>\frac{a-b}{\sqrt{a+c}+\sqrt{b+c}} \quad \Longleftrightarrow \quad \sqrt{a+c}+\sqrt{b+c}>1
$$
The last inequality must hold for ... | \frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,258 |
4. In triangle $A B C$, side $A C$ is greater than $A B$, line $l$ is the bisector of the external angle $C$. A line passing through the midpoint of $A B$ and parallel to $l$ intersects $A C$ at point $E$. Find $C E$, if $A C=7$ and $C B=4$. (The external angle of a triangle is the angle adjacent to the internal angle ... | Answer: $\frac{11}{2}$.
(Fig. 1.) Let $F$ be the intersection point of line $l$ and line $A B$. Since $A C>A B$, point $E$ lies on segment $A C$, and point $F$ lies on ray $A B$. Draw a line through point $B$ parallel to $A C$ until it intersects $C F$ at point $G$. Then $\angle B G C=\angle B C G$, so triangle $C B G... | \frac{11}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,259 |
1.
The answer to the task should be an integer. Enter all characters without spaces. Do not indicate units of measurement.
Doughnut eats a cake in 5 minutes, and Nosy eats it in 7 minutes. How many seconds will it take for the boys to eat the cake together if they do not conflict? | # 1. /2 points/
The answer to the task should be an integer. Enter all characters without spaces. Do not indicate units of measurement.
Doughnut eats a cake in 5 minutes, and Nosy eats it in 7 minutes. How many seconds will it take for the boys to eat the cake together if they do not conflict
Answer: 175
# | 175 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,260 |
# 3.
The answer to the task should be an integer. Enter all characters without spaces.
Find $n$, if $9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}=3^{2012}$. | # 3. /2 points/
The answer to the task should be an integer. Enter all characters without spaces.
Find $n$, if $9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}=3^{2012}$.
Answer: 1005
# | 1005 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,262 |
6.
Simplify the expression given below and indicate the number of the correct answer.
$\frac{x^{3}-9 x y^{2}}{9 y^{2}+x^{2}} \cdot\left(\frac{x+3 y}{x^{2}-3 x y}+\frac{x-3 y}{x^{2}+3 x y}\right)$
1) 0.5
2) $x-3 y$
3) 2
4) $x+3 y$
# | # 6. /2 points/
Simplify the expression given below and indicate the number of the correct answer option.
$\frac{x^{3}-9 x y^{2}}{9 y^{2}+x^{2}} \cdot\left(\frac{x+3 y}{x^{2}-3 x y}+\frac{x-3 y}{x^{2}+3 x y}\right)$
1) 0.5
2) $x-3 y$
3) 2
4) $x+3 y$
# | Algebra | MCQ | Yes | Yes | olympiads | false | 19,265 | |
# 7.
Among the statements given below, select the correct ones. In your answer, write down the numbers of these statements (in ascending order, without using spaces, commas, or other separators).
Example of answer format: 12345
1) Among the numbers $123,365,293,18$, exactly three numbers are divisible by 3.
2) If th... | # 7. /3 points/
Among the statements given below, select the correct ones. In your answer, write down the numbers of these statements (in ascending order, without using spaces, commas, or other separators). Example of answer format: $\quad 12345$
1) Among the numbers $123,365,293,18$, exactly three numbers are divisi... | 2456 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,266 |
# 8.
The answer to the task should be some integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all characters (comma, digits) without spaces. There is no need to indicate units of measurement.
The base and the side of the triangle are 30 and ... | # 8. /3 points/
The answer to the problem should be a certain integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all characters (comma, digits) without spaces. Do not specify units of measurement.
The base and the side of the triangle are 30... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,267 |
# 9.
The answer to the task should be some integer or a number written as a finite decimal fraction. If the answer contains a fractional number, use a comma when writing it. Enter all characters (minus sign, comma, digits) without spaces.
Find all distinct values of the parameter $p$, for each of which the equation g... | # 9. /3 points/
The answer to the task should be some integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all characters (minus sign, comma, digits) without spaces.
Find all distinct values of the parameter $p$ for which the equation given be... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,268 |
10.
The answer to the task should be some integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all symbols (comma, digits) without spaces.
Find the range of the function given below. In your answer, specify the value of the difference between... | # 10. /3 points/
The answer to the task should be some integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all symbols (comma, digits) without spaces.
Find the range of the function given below. In your answer, specify the value of the differ... | 4,5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,269 |
Task 1. Today, Vovochka started watching anime. If he watches 2 episodes per day, by April 1, 2023 (not inclusive), he will have 215 episodes left unwatched. If he starts watching 5 episodes per day, he will have only 50 left. What is the date today and how many episodes are in the series? | Answer: February 5, 2023, 325 episodes.
Solution. Let there be $n$ days left until April 1, and $N$ episodes in the anime. Then, according to the condition, $N-2n=215$ and $N-5n=50$. Subtracting the second equation from the first, we get $3n=215-50=165$, that is, $n=55$.
(We can also solve this without introducing va... | February5,2023,325episodes | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,276 |
Task 2. What is the smallest natural number that is divisible by 2022 and whose notation starts with 2023? | Answer: 20230110.
Solution. Let the number we are looking for have $n+4$ digits, then it has the form $2023 \cdot 10^{n}+a, a<10^{n}$. Subtract from it $2022 \cdot 10^{n}$, we get that $b=10^{n}+a$ is also divisible by 2022. That is, we need to find a number that starts with 1 and is divisible by 2022. Numbers divisib... | 20230110 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,277 |
Task 3. From six segments, two triangles with a perimeter of 2 each are formed. One segment from the first set of three was swapped with a segment from the second set of three. Now, it is impossible to form a triangle from the segments of the first set. Can we be sure that a triangle can still be formed from the segmen... | Answer: Yes, we can.
Solution. A triangle can be formed from segments of length $a \geq b \geq c$ if $a < b + c$. Given the condition $a + b + c = 2$, this inequality can be rewritten as $a < 2 - a$, which is equivalent to $a < 1$. Therefore, in the original set, all segments have lengths less than 1. Let the resultin... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,278 |
Task 4. Can the orthocenter (the point of intersection of the altitudes) of a triangle a) bisect one of its altitudes? b) bisect two altitudes? | Solution. a) Let's construct such a triangle. Let $P$ be the base of the altitude that is bisected. Construct an arbitrary right triangle $A P H$, and on the ray $P H$, mark off the segment $H B = P H$. Then the line $B R$ is drawn perpendicular to $A H$, and point $C$ is the intersection of $B R$ and $A P$. Triangle $... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,279 |
Task 5. The quadratic trinomial $P(x)=x^{2}+p x+q$ satisfies the condition $P(q)<0$. Prove that exactly one of its roots lies in the interval from 0 to 1. | Solution. The leading coefficient is positive, meaning the branches of the parabola are directed upwards, and at point $q$ the quadratic trinomial takes a negative value, so there are two roots. We have $P(q)=q^{2}+p q+q=q(p+q+1)=P(0) \cdot P(1)<0$. Therefore, at points 0 and 1, the quadratic trinomial takes different ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 19,280 |
1. Given three positive numbers, the product of any two of them is greater than the third by the same number $a$. Prove that $a \geq -1 / 4$. | 1. Let's denote the numbers by $x, y$ and $z$. We have $x y - z = a, x z - y = a, y z - x = a$. Then $x y - z = x z - y, x y - z - x z + y = 0$ or $(x + 1)(y - z) = 0$. Since $x > 0, x + 1 \neq 0$ and thus $y - z = 0$, i.e., $y = z$. Similarly, we obtain that $x = y$. Therefore, $a = x y - z = x^2 - x = \left(x - \frac... | \geq-\frac{1}{4} | Algebra | proof | Yes | Yes | olympiads | false | 19,281 |
2. There are 210 ninth-grade students in the school. In the mathematics classroom, there are 6 different algebra books and 5 different geometry books. Each student was asked to choose 6 titles: three in algebra and three in geometry. Prove that there will be at least two students whose lists of selected books coincide. | 2. Let's find the number of different ways to choose three titles of algebra books out of 6 possible ones. For the first book, we can choose any of the 6, for the second any of the 5 remaining, and for the third any of the 4 remaining after choosing the first two. Thus, we have $6 \cdot 5 \cdot 4=120$ different ways. H... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,282 |
3. On the sides $AB$ and $BC$ of triangle $ABC$, squares $BAMN$ and $BCKL$ are constructed outwardly with a common vertex $B$. Prove that the segment $NL$, connecting the ends of the sides of the squares originating from $B$, is twice as long as the median of the triangle originating from the same vertex $B$. | 3. Let's make an additional construction. Let $P$ be the point symmetric to point $B$ with respect to the midpoint $Q$ of side $A C$. Then $A B C P$ is a parallelogram.
Consider triangle $A B P$. We will prove that it is equal to triangle $B N L$. Indeed, $A B$ and $B N$ have the same lengths as the sides of the squar... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,283 |
4. Calculate the sum: $\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^{2}}{3}\right]+\ldots+\left[\frac{2^{1000}}{3}\right]$. (Here $[x]$ is the integer part of the number $x$, that is, the greatest integer not exceeding $x$.) | 4. Let's find the sum of the terms $S$ of the geometric progression
$$
S=\frac{1}{3}+\frac{2}{3}+\frac{2^{2}}{3}+\ldots+\frac{2^{1000}}{3}=\frac{1}{3}\left(1+2+2^{2}+\ldots+2^{1000}\right)=\frac{1}{3}\left(2^{1001}-1\right)
$$
Let $\{x\}$ denote the fractional part of the number $x$: by definition, $\{x\}=x-[x]$, whe... | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,284 | |
10. (3 points)
Find the smallest natural number that starts with the digit 3 and which, after erasing this digit, decreases by 25 times. | Answer: 3125
## Interregional Subject Olympiad KFU 2013-2014, MATHEMATICS 9th grade, 2nd variant, Internet round
The answer in each task should be an integer or a number written as a finite decimal. If the answer contains a fractional number, then when writing it, use a comma. All symbols (minus sign, comma, digits) ... | 3125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,285 |
Task 1. In the club, 36 schoolchildren are attending. If any 33 of them come to the lesson, then girls will always be more than half. But if 31 students come to the lesson, it may turn out that more than half of them are boys. How many girls are attending the club? (20 points) | Answer: 20.
Solution: Since among any 33 children, there are more girls than half, then boys among them are less than half, that is, no more than 16. Therefore, there are no more than 16 boys in total, because if there were at least 17, it could happen that 17 boys and another 16 students came, and the boys would alre... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,286 |
Task 2. Can the numbers from 1 to 1000 be divided into several groups such that each group contains at least two numbers, and in each group, the sum of any two numbers is divisible by 3? (20 points) | Answer. No, it is not possible.
Solution. Suppose the required is possible. Divide each number from 1 to 1000 by 3 with a remainder. It is easy to see that we have 333 numbers that are divisible by 3, 333 numbers that are divisible by 3 with a remainder of 2, and 334 numbers that are divisible by 3 with a remainder of... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,287 |
Task 3. Sasha chose five numbers from the numbers 1, 2, 3, 4, 5, 6, and 7 and told Anna their product. Based on this information, Anna realized that she could not uniquely determine the parity of the sum of the numbers chosen by Sasha. What number did Sasha tell Anna? (20 points) | Answer: 420.
Solution: Let's look at the two remaining numbers. Since Anya knows the sum of all numbers from 1 to 7 (which is 28), these two remaining numbers are such that their product cannot determine the parity of their sum. Therefore, their product can be represented in two ways: \( ab = xy \), where numbers \( a... | 420 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,288 |
Problem 4. Four points $A, B, C, D$ are such that $D$ and $C$ lie on the same side of the line $A B$ and the equality $A B = A D + B C$ holds. The bisectors of angles $A B C$ and $B A D$ intersect at point $E$. Prove that $C E = D E$. (20 points) | Solution. Mark point $F$ on segment $A B$ such that $A F=A D$, then $F B=$ $B C$. It remains to note that triangles $F A E$ and $D A E$ are equal by two sides and the angle between them, as are triangles $F B E$ and $C B E$, from which $E D=E F=E C$.
 without spaces. There is no need to specify units of measurement.
The bisector of angle $N$ of triang... | Answer: 70
The answer to the task should be some integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all symbols (minus sign, comma, digits) without spaces.
Find all different values of the parameter $p$ for which the equation given below has... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,291 |
1. Plot the graph of the function $y=x^{2}-\left|x-x^{2}\right|$ | Solution. The expression $x-x^{2}$ is positive for $0 \leq x \leq 1$. Therefore, $\left|x-x^{2}\right|=x-x^{2}$ if $0 \leq x \leq 1$ and $\left|x-x^{2}\right|=x^{2}-x$ if $x \in(-\infty, 0) \cup(1,+\infty)$. Therefore, $y=x$ for $x \in(-\infty, 0) \cup(1,+\infty)$ and $y=2 x^{2}-x$ for $0 \leq x \leq 1$.
$. Find $\frac{1}{\sin \alpha}+\frac{1}{\cos \alpha}$. | Solution. $\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha}=a \Rightarrow \frac{1}{\sin \alpha \cos \alpha}=a \Rightarrow \sin 2 \alpha=\frac{2}{a}$. Next,
$\frac{1}{\sin \alpha}+\frac{1}{\cos \alpha}=\frac{\sin \alpha+\cos \alpha}{\sin \alpha \cos \alpha} ;(\sin \alpha+\cos \alpha)^{2}=1+\sin 2 \alpha=... | \sqrt{^2+2a} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,297 |
2. In the circle $\omega$, a diameter $A B$ is drawn, and a point $X$ is fixed on this diameter. Point $P$ lies on the circle $\omega$ and does not coincide with points $A$ and $B$. Prove that the ratio
$$
\frac{\operatorname{tg} \angle A P X}{\operatorname{tg} \angle P A X}
$$
does not depend on the choice of point ... | Solution. Draw a perpendicular $X H$ from point $X$ to segment $A P$. Note that $\angle B P A=90^{\circ}$, since it subtends the diameter. Therefore, point $H$ lies within segment $A P$ and $X H \| B P$. Further,
$$
\frac{\operatorname{tg} \angle A P X}{\operatorname{tg} \angle P A X}=\frac{X H: H P}{X H: H A}=\frac{H... | \frac{AX}{XB} | Geometry | proof | Yes | Yes | olympiads | false | 19,298 |
3. Solve the equation
$$
2(x-6)=\frac{x^{2}}{(1+\sqrt{x+1})^{2}}
$$ | Solution 1. Rewrite the equation as $2(x-6)(1+\sqrt{x+1})^{2}=x^{2}$ and make the substitution $\sqrt{x+1}=y$. Then $x=y^{2}-1$ and $y \geq 0$. After the substitution, we get $2\left(y^{2}-7\right)(y+1)^{2}=$ $\left(y^{2}-1\right)^{2}$. Factor the right side as $(y+1)^{2}(y-1)^{2}$, move it to the left, and factor out ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,299 |
4. It is known that $a$ is an odd integer, and $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2} + a x - 1 = 0$. Prove that for any integer $n \geq 0$, the numbers $x_{1}^{n} + x_{2}^{n}$ and $x_{1}^{n+1} + x_{2}^{n+1}$ are integers and coprime. | Solution. Let's introduce the notation $x_{1}^{n}+x_{2}^{n}=A_{n}$. For $n=0$, we have: $A_{0}=2, A_{1}=-a$ - integers and coprime. Further, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=-1$. From this, $-a A_{n}=\left(x_{1}+x_{2}\right)\left(x_{1}^{n}+x_{2}^{n}\right)=x_{1}^{n+1}+x_{2}^{n+1}+x_{2} x_{1}^{n}+x_{1} x... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,300 |
1. How many positive numbers are there among the 2014 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}$, $\sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$ | Solution. For $n>3$ we have
$10^{n}-1000=10^{3}\left(10^{n-3}-1\right)=25 \cdot 40 \cdot\left(10^{n-3}-1\right)$.
Since $10^{n-3}-1$ is divisible by 9, it follows that $10^{n}-1000$ is divisible by 360. Therefore, all terms of the sequence, starting from the fourth, coincide with $\sin 1000^{\circ}=\sin \left(3 \cdot... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,301 |
2. Solve the inequality $\arccos x \geq \arccos \left(x^{2}\right)$. | Solution. Let's write down the domain of definition: $|x| \leq 1,\left|x^{2}\right| \leq 1$.
The solution to this system of inequalities is the interval $[-1 ; 1]$. Note that the function $y=\arccos x-$ is decreasing. Therefore, the inequality from the condition is equivalent to the system of inequalities $x \leq x^{2... | [-1;0]\cup{1} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,302 |
3. The area of a quadrilateral is 1. What is the smallest value that the sum of its diagonals can have? | Solution. First, let's find the minimum value for the product of the diagonals of a quadrilateral. For its area, we have the equality: \(2 S = x y \sin \alpha\), where \(x, y\) are the diagonals of the quadrilateral, and \(\alpha\) is the angle between them. Given \(S = 1\), we have \(2 = x y \sin \alpha \leq x y\).
N... | 2\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,303 |
4. What is the smallest natural number $\mathrm{N}$ by which 999,999 must be multiplied so that the result is a number written entirely with ones | Solution. Note that $999999=10^{6}-1$. A number written with all ones has the form $\frac{1}{9}\left(10^{n}-1\right)$. Therefore, the condition of the problem means that
$\left(10^{6}-1\right) \cdot N=\frac{10^{n}-1}{9}$, i.e., $N=\frac{1}{9} \cdot \frac{10^{n}-1}{10^{6}-1}$.
Now note that if $n=6 k+r$, then $10^{n}-... | 10^{48}+10^{42}+10^{36}+10^{30}+10^{24}+10^{18}+10^{12}+10^{6}+1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,304 |
Task 2. Let's call a year interesting if a person turns as many years old as the sum of the digits of their birth year in that year. A certain year turned out to be interesting for Ivan, who was born in the 20th century, and for Vovochka, who was born in the 21st century. What is the difference in their ages?
Note. Fo... | Answer: 18 years.
Solution. Let Ivan's year of birth be $\overline{19 x y}$, and Vovochka's $-\overline{20 z t}$. An interesting year for Ivan will be $1900+10 x+y+10+x+y$, and for Vovochka $-2000+10 z+t+2+z+t$, according to the condition these values are equal, that is, $2002+11 z+2 t=1910+11 x+2 y$, from which $11(x... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,308 |
Task 3. Find all solutions to the equation $x^{2}-[x]=1$. Here $[x]$ is the integer part of $x$, that is, the greatest integer not exceeding the given number. For example, $[2.9]=2 ;[-2.9]=-3$. | Solution. 1st method. Let $[x]=n$. Then the equation reduces to the form $x^{2}=n+1$ with the additional constraint $n \leq x<n+1$. We know the value of $x^{2}$, so it is desirable to square the inequality. This is possible if its terms are non-negative.
Since $x^{2}$ is non-negative, $n+1 \geq 0 ; n \geq-1$. The only... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,309 |
Problem 4. Given a triangle $ABC$ with sides $AB=5$; $BC=8$. The bisector of angle $B$ intersects the circumcircle of the triangle at point $D$. a) It is known that the area of triangle $ABD$ is 10. Find the area of triangle $BCD$. b) Can it be that $S_{ABD}$ equals 100? | Solution. a) The fact that point $D$ lies on the circle is irrelevant. The areas of triangles $A B O$ and $O B C$ are in the ratio $A O: O C$. The same applies to triangles $A O D$ and $D O C$. Therefore,
$$
S_{A B D}: S_{B C D}=A O: O C=A B: B C=5: 8
$$
by the property of the angle bisector.
 Vovochka lost some 11 cards. Can Lyudochka be sure that she can complete the task? b) The same question, if Vovochka lost 10 cards. | Answer: a) no; b) yes.
Solution. a) Suppose Vovochka lost cards numbered from 1 to 11. Among the remaining, the smallest four are 12, 13, 14, 15; the sum of the values on them is 54. Therefore, in this case, it is impossible to get 50 with any four cards.
b) We will select cards in pairs such that the sum in each pai... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,311 |
Task 1. A rectangular grid, both sides of which are even numbers, was cut into figures of the form $\square$ and $\square \square$ such that both types of figures are present. What is the smallest area that such a rectangle could have? Provide an example of the corresponding cutting and explain why a smaller area is im... | Answer: 40.
Solution: Since the area of both figures is 5, and the sum of numbers divisible by 5 is also divisible by 5, the area of the rectangle must be divisible by 5. Since 5 is a prime number, the length of one of the sides must be divisible by 5. Since the lengths of the sides are even numbers, it must be divisi... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,312 |
Task 2. Mom baked a cake for Anya's birthday, which weighs an integer number of grams. Before decorating it, Mom weighed the cake on digital scales that round the weight to the nearest ten grams (if the weight ends in 5, the scales round it down). The result was 1440 g. When Mom decorated the cake with identical candle... | Answer: 4 years.
Solution: From the condition, it follows that the cake weighed from 1436 to 1445 grams before decoration, and from 1606 to 1615 grams after decoration. Therefore, the total weight of the candles is between 161 and 179 grams. Since the weight of each candle is given as 40 grams, the weight of one candl... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,313 |
Task 3. Can the variables $a, b, c, d$ be replaced with any four consecutive natural numbers in some order so that the equation $(a+b)(b+c)(c+d)=(c+a)(a+d)(d+b)$ holds true? (20 points) | Answer: No.
Solution: Let's say we have chosen the numbers $n, n+1, n+2, n+3$. Then the set of their pairwise sums is $2n+1, 2n+2, 2n+3, 2n+3, 2n+4, 2n+5$. Among these sums, there are exactly two even numbers: $2(n+1)$ and $2(n+2)$, and since these are consecutive even numbers, exactly one of them is divisible by 4. T... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,314 |
Given trapezoid $A B C D$ with bases $B C$ and $A D$ such that $A D=3 B C$. Point $K$ is the midpoint of diagonal $B D$. It turns out that $A K$ is the bisector of angle $C A D$. Prove that $A C=2 B C .(20$ points $)$ | First solution. Extend the bisector $A K$ to intersect the line $B C$ at point $M$. Then $\angle B M A=\angle M A D=\angle C A M$, so $A C=C M$. Moreover, $\angle B D A=\angle D B M$, $\angle B K M=\angle D K A$ and $K B=K D$, so triangles $A K D$ and $M K B$ are equal, from which $B M=A D=3 B C$. Therefore, $A C=C M=B... | AC=2BC | Geometry | proof | Yes | Yes | olympiads | false | 19,315 |
Task 5. On a chessboard, there are 12 queens. Prove that it is possible to choose four rows and four columns such that none of the 16 cells at their intersections contain a queen. The board has dimensions $8 \times 8$. (20 points) | Solution. First, let's find four rows in which there are a total of no more than four queens. If there are four rows, each containing no more than one queen, then they fit. Otherwise, there will be four rows (actually at least 5), each containing at least two queens. Then, in these four rows, there are at least 8 queen... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,316 |
1. Prove that the function $f(x)=\sin x+\sin \sqrt{2} x$ is non-periodic. | 1. Suppose there exists $T>0$ such that for any $x$ the equality $\sin x+\sin \sqrt{2} x=\sin (x+T)+\sin \sqrt{2}(x+T)$ holds. Then $\sin (x+T)-\sin x+$ $+\sin \sqrt{2}(x+T)-\sin \sqrt{2} x=0$ or $2 \sin \frac{T}{2} \cos \left(x+\frac{T}{2}\right)+2 \sin \frac{\sqrt{2} T}{2} \cos \sqrt{2}\left(x+\frac{T}{2}\right)=0$, ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 19,318 |
2. Solve the system of equations: $\left\{\begin{array}{l}x^{2012}+y^{2012}=2, \\ 2012 x-y^{2011}=2012 y-x^{2011}\end{array}\right.$. | 2. From the second equation, we get $2012(x-y)=y^{2011}-x^{2011}$. Let's show that $x=y$. Indeed, if, for example, $x>y$, then $y^{2011}>x^{2011}$ and $y>x$ - contradiction. Therefore, $x=y$. From the first equation, we get $2 x^{2012}=2$, from which $x= \pm 1$. Answer: $x=y= \pm 1$. | \1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,319 |
3. Given several vectors. The length of the sum of any two of them does not exceed 2. Prove that the length of the sum of any three of these vectors does not exceed 3. | 3. Denote by $|\vec{v}|$ the length of the vector $\vec{v}$. For any two vectors $\overrightarrow{v_{1}}, \overrightarrow{v_{2}}$ we have
$$
\left|\overrightarrow{v_{1}}+\overrightarrow{v_{2}}\right| \leq\left|\overrightarrow{v_{1}}\right|+\left|\overrightarrow{v_{2}}\right|
$$
This inequality follows from the fact t... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 19,320 |
4. Prove that for any natural $n>1$ the number $n^{n-1}-1$ is divisible without remainder by $(n-1)^2$.
| 4. We have $x^{n-1}-1=(x-1)\left(x^{n-2}+x^{n-3}+\ldots+x+1\right)$, therefore
$$
n^{n-1}-1=(n-1)\left(n^{n-2}+n^{n-3}+\ldots+n+1\right)
$$
For any natural $k$ we have $n^{k}=(1+(n-1))^{k} \equiv 1(\bmod (n-1))$, therefore
$$
n^{n-2}+n^{n-3}+\ldots+1 \equiv \underbrace{1+1+\ldots+1}_{(n-1) \text { times }} \equiv 0 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,321 |
1. Can the parameters $a, b, c$ be chosen such that for all $x$ the equality
$$
(x+a)^{2}+(2 x+b)^{2}+(2 x+c)^{2}=(3 x+1)^{2} ?
$$ | Solution. Expanding the brackets: $x^{2}+2 a x+a^{2}+4 x^{2}+4 b x+b^{2}+4 x^{2}+4 c x+c^{2}=9 x^{2}+6 x+1$, from which $(2 a+4 b+4 c) x+a^{2}+b^{2}+c^{2}=6 x+1$. It is sufficient to choose numbers $a, b, c$ such that $a+2 b+2 c=3$, $a^{2}+b^{2}+c^{2}=1$. The numbers $a=1 / 3, b=c=2 / 3$ fit. | =\frac{1}{3},b==\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,322 |
3. 20 chess players played a round-robin tournament (each played one game with each other). A journalist reported that each participant won as many games as they drew. Prove that the journalist was wrong. | Solution. Each player played 19 games. If the $i$-th participant won $a_{i}$ games, then they also drew $a_{i}$ matches and lost $19-2 a_{i}$ games. Since a game won by one participant is a loss for another, the total number of all games won is equal to the total number of all games lost. Therefore, $a_{1}+\ldots+a_{20... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,324 |
4. In triangle $A B C$, the sides $C B=a, C A=b$. The bisector of angle $A C B$ intersects side $A B$ at point $K$, and the circumcircle of the triangle - at point $M$. The circumcircle of triangle $A M K$ intersects line $A C$ again at point $P$. Find the length of segment $A P$. | Solution. Triangle $C P K$ is equal to triangle $C B K$, since $C K$ is a common side, $\angle P C K=$ $\angle B C K, \angle K P C=\angle K M A=\angle C B K$. The last two equalities follow from the fact that inscribed angles subtending the same arc are equal. Therefore, $A P=|C P-C A|=|a-b|$. | |-b| | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,325 |
1. Given three integers. From the first number, the sum of the digits of the second number was subtracted; from the second number, the sum of the digits of the third number was subtracted; and from the third number, the sum of the digits of the first number was subtracted. Can these differences be equal to a) $2,3,4$? ... | Answer: a) can; b) cannot.
a) For example, the numbers $10, 8, 5$ work. Then the corresponding differences are $10-8=2$, $8-5=3$, $5-1=4$.
b) Let $a, b, c$ be the original numbers. Denote by $S(n)$ the sum of the digits of the number $n$. According to the divisibility rule by 9, the numbers $n$ and $S(n)$ have the sa... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,326 |
2. Does there exist a non-constant function $f(x)$ defined on the entire real line such that for all real $x$ the following equality holds:
a) $f(\sin x) + f(\cos x) = 1 ? \quad$ b) $f(\sin x) - f(\cos x) = 1 ?$
$(25$ points. $)$ | Answer: a) exists; b) does not exist.
a) For example, the function $f(t)=t^{2}$ works, since $f(\sin x)+f(\cos x)=\sin ^{2} x+\cos ^{2} x=1$.
b) Suppose such a function exists. Then for $x=0$ and $x=\frac{\pi}{2}$, we have the equalities
$$
\begin{aligned}
f(\sin 0) & =f(\cos 0)+1 \\
f\left(\sin \frac{\pi}{2}\right)... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,327 |
3. A natural number $n$ is called lucky if it can be uniquely decomposed into the sum of 10 different natural numbers (the order of the summands does not matter). Find all lucky numbers.
$(25$ points. $)$ | Answer: 55 and 56.
Let the number $n$ be lucky, $n=a_{1}+a_{2}+\ldots+a_{10}$, where $a_{1}1$, then $n$ can be split into a sum of different natural addends in another way: $n=\left(a_{1}-1\right)+a_{2}+\ldots+\left(a_{10}+1\right)$. Thus, $a_{1}=1$.
Next, if we assume that $a_{2}>2$, then for $n$ we can again provid... | 5556 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,328 |
4. A circle is inscribed in an angle with vertex $A$, touching the sides of the angle at points $B$ and $C$. A line passing through $A$ intersects the circle at points $D$ and $E$. Chord $B X$ is parallel to line $D E$. In what ratio does the line $X C$ divide the chord $D E ?$
$(25$ points. $)$ | Answer: $1: 1$.
First solution. (Fig. 2.) Let $O$ be the center of the circle, and $M$ be the intersection point of $C X$ and $D E$. We will prove that $O M \perp D E$, and thus, $D M = M E$.
First, the angle $\angle D M C$ is equal to half the sum of the arcs $C D$ and $E X$. Since the arcs between the parallel chor... | 1:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,329 |
1. Prove that the function
$$
f(x)=\frac{1+\sin x-\cos x}{1+\sin x+\cos x}, \quad x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
$$
is odd. | Solution. $f(x)=\frac{1-\cos x+\sin x}{1+\cos x+\sin x}=\frac{2 \sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}=\operatorname{tg} \frac{x}{2}$, and this function is odd. | proof | Algebra | proof | Yes | Yes | olympiads | false | 19,330 |
2. Prove that for any $n$ the polynomial $(x+1)^{2 n+1}+x^{n+2}$ is divisible by the polynomial $x^{2}+x+1$. | Solution. Induction on $n$. For $n=0$ we have: $(x+1)^{2 \cdot 0+1}+x^{0+2}=x^{2}+x+1$. For $n=k+1$ we have:
$$
\begin{aligned}
& (x+1)^{2(k+1)+1}+x^{(k+1)+2}=(x+1)^{2}(x+1)^{2 k+1}+x \cdot x^{k+2}= \\
& \quad\left(x^{2}+2 x+1\right)(x+1)^{2 k+1}+x \cdot x^{k+2}=\left(x^{2}+x+1\right)(x+1)^{2 k+1}+x\left((x+1)^{2 k+1}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 19,331 |
3. On the plane, a point $M$ and a circle $\omega$ are given. Prove that the sum of the squares of the distances from point $M$ to the two ends of a diameter of circle $\omega$ does not depend on the choice of the diameter. | Solution. Let $C$ be the center of the circle $\omega$, $R$ its radius, and $M=O$ the origin. Let $A$ and $B$ be the endpoints of some diameter of the circle $\omega$. Denote $\overrightarrow{O C}=\mathbf{c}, \overrightarrow{C A}=\mathbf{x}$, then $\overrightarrow{C B}=-\mathbf{x}, \overrightarrow{O A}=\mathbf{c}+\math... | 2(^{2}+R^{2}) | Geometry | proof | Yes | Yes | olympiads | false | 19,332 |
4. Prove that $(1+\sqrt{3})(1+\sqrt[3]{3})(1+\sqrt[4]{3}) \ldots(1+\sqrt[11]{3})>2013 \sqrt[3]{3}$. | Solution. We have: $1+\sqrt[n]{3}=1+3^{\frac{1}{n}}=1+\left(3^{\frac{1}{2 n}}\right)^{2}>2 \cdot 1 \cdot 3^{\frac{1}{2 n}}$. Therefore,
$$
(1+\sqrt{3})(1+\sqrt[3]{3})(1+\sqrt[4]{3}) \ldots(1+\sqrt[11]{3})>2^{10} \cdot 3^{\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\ldots+\frac{1}{22}\right)}>2^{10} \cdot 3=3072>2013 \cd... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 19,333 |
1. The product of four numbers - the roots of the equations $x^{2}+2 b x+c=0$ and $x^{2}+2 c x+b=0$, is equal to 1. It is known that the numbers $b$ and $c$ are positive. Find them. | Solution. Let the roots of the first equation be $x_{1}$ and $x_{2}$, and the roots of the second equation be $x_{3}$ and $x_{4}$. By Vieta's Theorem, $1=x_{1} x_{2} x_{3} x_{4}=b c$. Let's write down the discriminants of both equations: $4 b^{2}-4 c \geqslant 0,4 c^{2}-4 b \geqslant 0$. Therefore, $b^{2} \geqslant c=1... | b==1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,334 |
2. Two players take turns replacing one (any) of the coefficients $a_{7}, \ldots, a_{0}$ in the expression
$$
a_{7} \cdot 8^{7}+a_{6} \cdot 8^{6}+a_{5} \cdot 8^{5}+\ldots+a_{2} \cdot 8^{2}+a_{1} \cdot 8+a_{0} \cdot 1
$$
with +1 or -1 (at their choice). Prove that the second player can always ensure that the result af... | Solution. $8^{2}+1=65$ is divisible by 13. For the second player to win, it is sufficient to split all numbers into pairs: $\left(a_{7}, a_{5}\right),\left(a_{6}, a_{4}\right),\left(a_{3}, a_{1}\right)$ and $\left(a_{2}, a_{0}\right)$. Now, if the first player replaces one number from a certain pair, then the second pl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,335 |
3. Prove the inequality
$$
\frac{1}{1 \cdot \sqrt{2}+2 \cdot \sqrt{1}}+\frac{1}{2 \cdot \sqrt{3}+3 \cdot \sqrt{2}}+\ldots+\frac{1}{2012 \cdot \sqrt{2013}+2013 \cdot \sqrt{2012}}>0.97
$$ | Solution. We have
$$
\frac{1}{(k+1) \cdot \sqrt{k}+k \cdot \sqrt{k+1}}=\frac{(k+1) \cdot \sqrt{k}+k \cdot \sqrt{k+1}}{k(k+1)}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}
$$
Therefore, the entire sum is equal to
$$
1-\frac{1}{\sqrt{2013}}>0.97
$$ | 1-\frac{1}{\sqrt{2013}}>0.97 | Inequalities | proof | Yes | Yes | olympiads | false | 19,336 |
4. In triangle $ABC$, the bisector $BK$ is drawn, and points $M$ and $N$ are taken on sides $BA$ and $BC$ respectively such that $\angle AKM = \angle CKN = \frac{1}{2} \angle ABC$. Prove that line $AC$ is tangent to the circumcircle of triangle $MBN$. | Solution. Let $\angle A B C=2 \alpha$, then $\angle M K N=\pi-\angle A K M-\angle C K N=\pi-2 \alpha$, that is, $\angle M K N +$ $\angle M B N=\pi$. Therefore, the quadrilateral $M B N K$ is cyclic, that is, the circumcircle of triangle $M B N$ passes through point $K$. The line $C K$ forms an angle $\alpha$ with the c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,337 |
Task 1. From a $5 \times 7$ rectangle, three cells have been cut out (see the figure). Cut the remaining figure into 8 equal parts along the grid lines. Parts are considered equal if they can be superimposed by overlaying. It is sufficient to provide one correct example of cutting in the solution. (20 points)
Criteria. Any correct example -20 points. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,338 |
Task 2. The cat runs counterclockwise around eight squares, moving to a neighboring square each second. The mouse runs clockwise around the perimeter of the large square, moving to the next segment each second (see the figure). From left to right, the images show the starting position, and then the positions after 1, 2... | Answer. See the picture.
Solution. Every 8 seconds, the cat returns to the starting position, and every 12 seconds, the mouse returns to the starting position. Therefore, after 24 seconds, t... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,339 |
Task 3. At the party, there were several married couples (there were no other people present individually). Each man ate three more oranges and two fewer tangerines than his wife. It turned out that the number of oranges and tangerines eaten was the same. Prove that the total number of guests is divisible by 4. (20 poi... | Solution. Each couple together ate an even number of tangerines, since two numbers differing by 2 have the same parity, and thus their sum is even. Therefore, the total number of tangerines eaten by the guests is even, as the sum of several even numbers.
Similarly, each couple together ate an odd number of oranges, si... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,340 |
Task 4. Six people are standing in a row, each of whom is either a knight (always tells the truth) or a liar (always lies). The first says: "To the right of me, there are exactly five liars," the second: "To the left of me, there is exactly one liar," the third: "To the right of me, there are exactly three liars," the ... | Answer. From left to right: L, R, L, L, R, L.
Solution. We will call the people the natives of the island of knights and liars.
The first native cannot be telling the truth, because then all people standing to his right would be liars, but in this case, the fifth native would be telling the truth. Therefore, the firs... | L,R,L,L,R,L | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,341 |
Task 5. In class 5A, a survey was conducted on what fruits the students like. It turned out that 13 students like apples, 11 like plums, 15 like peaches, and 6 like melons. A student can like more than one fruit. Every student who likes plums also likes either apples or peaches (but not both at the same time). And ever... | Answer: 22 people.
Solution. Estimate. Since 15 people like peaches and 6 people like melons, there are at least 9 people who like peaches but do not like melons. Then all these people like plums but do not like apples. Therefore, in addition to these people, there are at least 13 people who like apples. Thus, the tot... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,342 |
Task 1. Masha has a piggy bank where she puts a 50 or 100 ruble bill every week. At the end of every 4 weeks, she picks the bill of the smallest denomination from the piggy bank and gives it to her little sister. After a year, it turned out that she had given her sister 1250 rubles. What is the minimum amount of money ... | Answer: 3750 rubles.
Solution. Let's call a 4-week period a "month", there are 13 such "months" in a year. If all the bills given by Masha were hundred-ruble bills, her sister would have received 1300 rubles. This means Masha gave 100 rubles twelve times and 50 rubles once. If in any "month" Masha gave 100 rubles, it ... | 3750 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,343 |
Task 2. a) Can it happen for some $a, b$ that $\log _{2} a \cdot \log _{2} b=\log _{2} a b$?
b) Can it happen for some $a, b$ that $\log _{2} a+\log _{2} b=\log _{2}(a+b)$?
c) Can both equalities hold for some $a, b$? | Answer: a) Yes; b) Yes; c) No.
Solution. It is clear that the numbers $a, b$ are positive.
a) The condition can be rewritten as $\log _{2} a \cdot \log _{2} b=\log _{2} a+\log _{2} b$. If $\log _{2} a \neq 1$, then $x=$ $\log _{2} b=\frac{\log _{2} a}{\log _{2} a-1}, b=2^{x}$. For example, when $a=4$ we have $\log _{... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,344 |
Task 3. Let $\min \frac{x-1}{x^{2}+1}=a ; \max \frac{x-1}{x^{2}+1}=b$. What are the minimum and maximum of the functions a) $\frac{x^{3}-1}{x^{6}+1}$; b) $\frac{x+1}{x^{2}+1}?$ | Answer: a) $\min \frac{x^{3}-1}{x^{6}+1}=a ; \max \frac{x^{3}-1}{x^{6}+1}=b$; b) $\min \frac{x+1}{x^{2}+1}=-b ; \max \frac{x+1}{x^{2}+1}=-a$.
Solution. Let $\frac{x-1}{x^{2}+1}=f(x)$.
a) We have $\frac{x^{3}-1}{x^{6}+1}=f\left(x^{3}\right)$. The value $x^{3}$ runs through all numerical values, so $f\left(x^{3}\right)... | )\\frac{x^{3}-1}{x^{6}+1}=\max\frac{x^{3}-1}{x^{6}+1}=b;\quadb)\\frac{x+1}{x^{2}+1}=-\max\frac{x+1}{x^{2}+1}=- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,345 |
Task 5. Consider the algebraic expression $F(a, \ldots, x)$, containing variables, parentheses, and multiplication and subtraction operations. Numerical constants are not used. Replace one of the operation signs with $\perp$, the other with $\bowtie$. We will call the resulting expression a "formula". For example, the ... | Answer: a) yes, for example, $(a \perp a) \bowtie(a \perp a)$; b) no.
Solution. a) Consider the formula $A=a \perp a$. If $\perp$ is subtraction, then the expression is identically equal to 0. If $\perp$ is multiplication, then $A=0$ when $a=0$. Therefore, the expression $N=(a \perp a) \bowtie(a \perp a)$ equals 0 for... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,347 |
Task 1. Cut the figure in the picture along the grid lines into six equal square parts and then assemble them into a $5 \times 6$ rectangle. Parts are considered equal if they can be superimposed. The solution must include two pictures - how to cut and how to assemble. (20 points)
Criteria. Any correct example - 20 points. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,348 |
Task 2. Amir is 8 kg heavier than Ilnur, and Daniyar is 4 kg heavier than Bulat. The sum of the weights of the heaviest and lightest boys is 2 kg less than the sum of the weights of the other two. The total weight of all four is 250 kg. How many kilograms does Amir weigh? (20 points) | Answer: 66 kg.
Solution. Let the weight of Bulat in kilograms be denoted by $x$, and the weight of Ilnur by $y$. Then Amir weighs $y+8$, and Daniyar weighs $-x+4$. Note that the heaviest is either Amir or Daniyar, and the lightest is either Ilnur or Bulat.
If the heaviest is Daniyar, then $x+4>y+8$, so $x>y$ and the ... | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,349 |
Task 3. During the break, one of the five students wrote an incorrect equation on the board. When asked by the class teacher who did it, the students gave the following answers.
Anton: "It was Borya or Vova."
Borya: "Neither Dima nor I did this."
Vova: "Anton and Borya are both lying."
Gosha: "One of Anton and Bory... | Answer: Vova.
Solution. Note that one of the two students - Gosha and Dima - is telling the truth, while the other is lying. If Vova is telling the truth, then already three children are lying - Anton, Borya, and one of Gosha and Dima. But according to the condition, there are only two liars, so Vova is lying.
In thi... | Vova | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,350 |
Problem 4. Let's call a natural number interesting if it can be represented both as the sum of two consecutive integers and as the sum of three consecutive integers. Kolya multiplied five different natural numbers. It turned out that the result is an interesting number. Prove that at least one of the factors was also a... | Solution. Note that the sum of two consecutive natural numbers is odd (since it is the sum of $k+(k+1)=2k+1$). Moreover, note that the sum of any three consecutive numbers is divisible by 3. Indeed, any three consecutive numbers can be represented as $n-1, n$, and $n+1$, and their sum is $3n$. Thus, the interesting num... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,351 |
Problem 5. Timofey placed 10 grid rectangles on a grid field, with areas of $1, 2, 3, \ldots, 10$ respectively. Some of the rectangles overlapped each other (possibly completely, or only partially). After this, he noticed that there is exactly one cell covered exactly once; there are exactly two cells covered exactly t... | Answer: 5 cells.
Solution. Estimation. There are a total of $1+2+\ldots+10=55$ cell coverings. The ten cells described in the condition are covered in total exactly $1 \cdot 1+2 \cdot 2+3 \cdot 3+4 \cdot 4=30$ times. There remain $55-30=25$ cell coverings. Therefore, the number of cells covered at least five times is ... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,352 |
1. A sequence of polynomials $P_{n}(x)$, where $n \geq 0$ is an integer, is defined recursively: $P_{0}(x)$ is identically equal to one (i.e., $P_{0}(x) \equiv 1$), and $P_{n+1}(x)=x^{7(n+1)}-P_{n}(x)$ for all $n \geq 0$. For each $n \geq 0$, find all real roots of $P_{n}(x)$.
A sequence of polynomials $P_{n}(x)$ (for... | 1. A sequence of polynomials $P_{n}(x)$, where $n \geq 0$ is an integer, is defined recursively: $P_{0}(x)$ is identically equal to one (i.e., $P_{0}(x) \equiv 1$), and $P_{n+1}(x)=x^{7(n+1)}-P_{n}(x)$ for all $n \geq 0$. For each $n \geq 0$, find all real roots of $P_{n}(x)$.
A sequence of polynomials $P_{n}(x)$ (for... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,353 |
2. The sequence of polynomials $P_{n}(x)$, where $n \geq 0$ is an integer, is defined recursively: $P_{0}(x)$ is identically equal to one (i.e., $P_{0}(x) \equiv 1$), and $P_{n+1}(x)=x^{11(n+1)}-P_{n}(x)$ for all $n \geq 0$. For each $n \geq 0$, find all real roots of $P_{n}(x)$.
A sequence of polynomials $P_{n}(x)$ (... | 2. The sequence of polynomials $P_{n}(x)$, where $n \geq 0$ is an integer, is defined recursively: $P_{0}(x)$ is identically equal to one (i.e., $P_{0}(x) \equiv 1$), and $P_{n+1}(x)=x^{11(n+1)}-P_{n}(x)$ for all $n \geq 0$. For each $n \geq 0$, find all real roots of $P_{n}(x)$.
A sequence of polynomials $P_{n}(x)$ (... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,354 |
3. The sequence of polynomials $P_{n}(x)$, where $n \geq 0$ is an integer, is defined recursively: $P_{0}(x)$ is identically equal to one (i.e., $P_{0}(x) \equiv 1$), and $P_{n+1}(x)=x^{17(n+1)}-P_{n}(x)$ for all $n \geq 0$. For each $n \geq 0$, find all real roots of $P_{n}(x)$.
A sequence of polynomials $P_{n}(x)$ (... | 3. The sequence of polynomials $P_{n}(x)$, where $n \geq 0$ is an integer, is defined recursively: $P_{0}(x)$ is identically equal to one (i.e., $P_{0}(x) \equiv 1$), and $P_{n+1}(x)=x^{17(n+1)}-P_{n}(x)$ for all $n \geq 0$. For each $n \geq 0$, find all real roots of $P_{n}(x)$.
A sequence of polynomials $P_{n}(x)$ (... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,355 |
4. The sequence of polynomials $P_{n}(x)$, where $n \geq 0$ is an integer, is defined recursively: $P_{0}(x)$ is identically equal to one (i.e., $P_{0}(x) \equiv 1$), and $P_{n+1}(x)=x^{5(n+1)}-P_{n}(x)$ for all $n \geq 0$. For each $n \geq 0$, find all real roots of $P_{n}(x)$.
A sequence of polynomials $P_{n}(x)$ (f... | 4. The sequence of polynomials $P_{n}(x)$, where $n \geq 0$ is an integer, is defined recursively: $P_{0}(x)$ is identically equal to one (i.e., $P_{0}(x) \equiv 1$), and $P_{n+1}(x) = x^{5(n+1)} - P_{n}(x)$ for all $n \geq 0$. For each $n \geq 0$, find all real roots of $P_{n}(x)$.
A sequence of polynomials $P_{n}(x)... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,356 |
2. (5 points) Given a polynomial $P(x)$ of the second degree. Moreover, $P\left(x_{1}\right)=y_{1}, P\left(x_{2}\right)=y_{2}$ and $P\left(x_{3}\right)=y_{3}$. Find the value of $P\left(x_{4}\right)$. | Solution: We will use the Lagrange interpolation polynomial and find
$$
P(x)=y_{1} \frac{\left(x-x_{2}\right)\left(x-x_{3}\right)}{\left(x_{1}-x_{2}\right)\left(x_{1}-x_{3}\right)}+y_{2} \frac{\left(x-x_{1}\right)\left(x-x_{3}\right)}{\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right)}+y_{3} \frac{\left(x-x_{1}\right)\l... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,357 | |
3. (7 points) Find all triples $(x, y, z)$ of pairwise coprime natural numbers satisfying the system
$$
\left\{\begin{array}{l}
(a+b) x+a y=b z ; \\
(a+b) x^{3}+a y^{3}=b z^{3},
\end{array} \quad \text { where } b>a\right. \text { are coprime natural numbers. }
$$
In your answer, write the smallest possible value of ... | Solution: Rewrite the system
$$
\left\{\begin{array} { l }
{ a ( x + y ) = b ( z - x ) ; } \\
{ a ( x ^ { 3 } + y ^ { 3 } ) = b ( z ^ { 3 } - x ^ { 3 } ) ; }
\end{array} \Longrightarrow \left\{\begin{array}{l}
a(x+y)=b(z-x) \\
a(x+y)\left(x^{2}-x y+y^{2}\right)=b(z-x)\left(z^{2}+z x+x^{2}\right)
\end{array}\right.\ri... | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,358 | |
4. (7 points) From the set $1,2, \ldots, 10$, three numbers (possibly the same) are chosen with equal probability. What is the probability that the sum of these numbers is 10? | Solution: We need to find how many ways there are to solve the equation
$$
x_{1}+x_{2}+x_{3}=10
$$
where $x_{1}, x_{2}, x_{3} \in\{1,2, \ldots, 10\}$. Let's write down a row of ten ones and place two partitions (in different places) between them. Then $x_{1}$ is the number of ones before the left partition, $x_{2}$ i... | 0.022 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,359 |
5. (8 points) The numbers $1,2, \ldots, n$ are arranged in some order at the vertices of a regular $n$-gon $A_{1} A_{2} \ldots A_{n}$. On each side of the $n$-gon, the absolute difference of the numbers at the endpoints of that side is written. What is the maximum value that the sum of all the numbers on the sides can ... | Solution: We will open all the modules, some numbers will open with a plus sign, and some with a minus sign. In the end, we will get some expression
$$
S= \pm 1 \pm 1 \pm 2 \pm 2 \pm \ldots \pm n \pm n
$$
where each number will be used twice and the number of pluses must be exactly the same as the number of minuses. ... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,360 | |
6. (8 points) At the University of Innopolis, $2 n$ applicants who were not acquainted with each other were admitted to the first year. How many pairs of freshmen need to be introduced to each other to ensure that there will definitely be a trio of mutually acquainted individuals? | Solution: The problem is solved by a simple application of Turán's theorem for $K_{3}$: the maximum number of edges in a graph without triangles is $n^{2}$. An example of such a graph is the complete bipartite graph $K_{n, n}$. In the end, the answer to the problem will be the number $n^{2}+1$.
Answers:
| Variant | 1... | n^2+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,361 |
7. (10 points) The bisector of angle $A L$ of triangle $A B C$ intersects the circumcircle of the triangle at point $D$. It turns out that $A L: L D=\alpha$. Find the perimeter of triangle $A B C$, if $B C=\beta$. | # Solution:
Since $AL$ is the angle bisector, then
$$
\frac{BA}{BL}=\frac{CA}{CL}=k \quad \text{and} \quad AL^2 = BA \cdot CA - BL \cdot CL
$$
We get
$$
\alpha^2 x^2 = k^2 bc - bc = (k^2 ... | \beta\sqrt{\alpha+1}+\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,362 |
9. (20 points) In triangle $ABC$, points $K$ and $L$ are marked on sides $AC$ and $BC$ such that $KL \parallel AB$. $X$ is the point of intersection of segments $KB$ and $AL$. Circle $\omega_{1}$ passes through point $X$ and is tangent to side $AB$ at point $A$. Circle $\omega_{2}$ passes through point $X$ and is tange... | # Solution:
Let $M$ be the midpoint of segment $AB$, and $N$ be the midpoint of segment $KL$. Since $ABLK$ is a trapezoid, $C, N, X$, and $M$ lie on the same line. We will prove that $Y$ al... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,364 |
1. Solve the system of inequalities:
$$
\left\{\begin{array}{l}
4 x^{2}+4 x y+19 y^{2} \leq 2 \\
x-y \leq-1
\end{array}\right.
$$
Solve the following system of inequalities:
$$
\left\{\begin{array}{l}
4 x^{2}+4 x y+19 y^{2} \leq 2 \\
x-y \leq-1
\end{array}\right.
$$ | 1. Solve the system of inequalities:
$$
\left\{\begin{array}{l}
4 x^{2}+4 x y+19 y^{2} \leq 2 \\
x-y \leq-1
\end{array}\right.
$$
Solve the following system of inequalities:
$$
\left\{\begin{array}{l}
4 x^{2}+4 x y+19 y^{2} \leq 2 \\
x-y \leq-1
\end{array}\right.
$$ | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,366 | |
2. Solve the system of inequalities:
$$
\left\{\begin{array}{l}
11 x^{2}-10 x y+3 y^{2} \leq 3 \\
5 x+y \leq-10
\end{array}\right.
$$
Solve the following system of inequalities:
$$
\left\{\begin{array}{l}
11 x^{2}-10 x y+3 y^{2} \leq 3 \\
5 x+y \leq-10
\end{array}\right.
$$ | 2. Solve the system of inequalities:
$$
\left\{\begin{array}{l}
11 x^{2}-10 x y+3 y^{2} \leq 3 \\
5 x+y \leq-10
\end{array}\right.
$$
Solve the following system of inequalities:
$$
\left\{\begin{array}{l}
11 x^{2}-10 x y+3 y^{2} \leq 3 \\
5 x+y \leq-10
\end{array}\right.
$$ | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,367 |
3. Solve the system of inequalities:
$$
\left\{\begin{array}{l}
11 x^{2}+8 x y+8 y^{2} \leq 3 \\
x-4 y \leq-3
\end{array}\right.
$$
Solve the following system of inequalities:
$$
\left\{\begin{array}{l}
11 x^{2}+8 x y+8 y^{2} \leq 3 \\
x-4 y \leq-3
\end{array}\right.
$$ | 3. Solve the system of inequalities:
$$
\left\{\begin{array}{l}
11 x^{2}+8 x y+8 y^{2} \leq 3 \\
x-4 y \leq-3
\end{array}\right.
$$
Solve the following system of inequalities:
$$
\left\{\begin{array}{l}
11 x^{2}+8 x y+8 y^{2} \leq 3 \\
x-4 y \leq-3
\end{array}\right.
$$ | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,368 |
4. Solve the system of inequalities:
$$
\left\{\begin{array}{l}
13 x^{2}-4 x y+4 y^{2} \leq 2 \\
2 x-4 y \leq-3
\end{array}\right.
$$
Solve the following system of inequalities:
$$
\left\{\begin{array}{l}
13 x^{2}-4 x y+4 y^{2} \leq 2 \\
2 x-4 y \leq-3
\end{array}\right.
$$ | 4. Solve the system of inequalities:
$$
\left\{\begin{array}{l}
13 x^{2}-4 x y+4 y^{2} \leq 2 \\
2 x-4 y \leq-3
\end{array}\right.
$$
Solution (RUS). (version 3) In the first inequality, group the complete squares: $(x+2 y)^{2} + 2(3 x)^{2} \leq 3$. Assume $u=x+2 y, v=3 x$. Then $2 u^{2} + v^{2} \leq 3$ and $v - 2 u ... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,369 |
1. Peter and Victor are playing a game, taking turns in painting out rectangles of size $1 \times 1, 1 \times 2$, and $2 \times 2$ in a checkered square of size $7 \times 7$. Each of the players paints in their own color (Peter's color is red, and Victor's green). Recoloring already colored cells is not allowed, initia... | 1. Peter and Victor are playing a game, taking turns in painting out rectangles of size $1 \times 1, 1 \times 2$, and $2 \times 2$ in a checkered square of size $7 \times 7$. Each of the players paints in their own color (Peter's color is red, and Victor's green). Recoloring already colored cells is not allowed, initia... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,374 |
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