problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
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6. Find all natural numbers $n$ such that the number $2^{n}+n^{2}+25$ is a cube of a prime number. | Answer: $n=6$
Solution. Let $2^{n}+n^{2}+25=p^{3}$ for some prime number $p$. Since $p>3$, $p$ is an odd prime number. Then $n$ is an even number, and $2^{n}$ gives a remainder of 1 when divided by three. If $n$ is not divisible by three, then $n^{2}$ gives a remainder of 1 when divided by three, and then $2^{n}+n^{2}... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find all values of $a$ for which the quadratic trinomials $x^{2}+2 x+a$ and $x^{2}+a x+2=0$ have two roots each, and the sum of the squares of the roots of the first trinomial is equal to the sum of the squares of the roots of the second trinomial. | Answer: $a=-4$
Solution. If $x_{1}$ and $x_{2}$ are the roots of the quadratic polynomial $x^{2}+p x+q$, then by Vieta's theorem
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$
Therefore, we need to find such numbers $a$ for which $2^{2}-2 a=a^{2}-2 \cdot 2$. Thus, we need to solve the... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find all values of $a$ for which the quadratic trinomials $x^{2}-6 x+4 a$ and $x^{2}+a x+6=0$ have two roots each, and the sum of the squares of the roots of the first trinomial is equal to the sum of the squares of the roots of the second trinomial. | Answer: $a=-12$
If $x_{1}$ and $x_{2}$ are the roots of the quadratic polynomial $x^{2}+p x+q$, then by Vieta's theorem
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$
Therefore, we need to find such numbers $a$ for which $6^{2}-8 a=a^{2}-2 \cdot 6$. Thus, we need to solve the equation... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) We will call a date diverse if in its representation in the form DD/MM/YY (day-month-year) all digits from 0 to 5 are present. How many diverse dates are there in the year 2013? | Answer: 2.
Solution. Note that in any date of 2013 in the specified format, the digits 1 and 3 are used, so for the day and month of a diverse date, the digits left are 0, 2, 4, and 5.
Let $Д_{1}$ and $Д_{2}$ be the first and second digits in the day's notation, and $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ be the first ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (10 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different six-letter words can be formed from the letters of the word СКАЛКА? And how many seven-letter words from the letters of the word ТЕФТЕЛЬ? In your answer, indicate the quotient of dividing the larger of the foun... | # Answer: 7.
Solution. The word СКАЛКА has six letters, but the letter А and the letter K each appear twice. Therefore, the number of different words will be $\frac{6!}{2!\cdot 2!}$. In
the word ТЕФТЕЛЬ, there are seven letters, and the letters $\mathrm{T}$ and $\mathrm{E}$ each appear twice, so the number of differen... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) Let a sequence of non-negative integers be given
$$
k, k+1, k+2, \ldots, k+n
$$
Find the smallest $k$, for which the sum of all numbers in the sequence is equal to 100.
# | # Answer: 9.
Solution. The given sequence of numbers is an arithmetic progression with $n+1$ terms. Its sum is $\frac{(n+1)(2 k+n)}{2}$. Therefore, the condition can be rewritten as $(n+1)(2 k+n)=200$, where $n$ is a non-negative integer. It is clear that $k$ decreases as $n$ increases. Thus, we need to find the large... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (50 points) A $5 \times 5$ square of cells was cut into several pieces of different areas, each consisting of an integer number of cells. What is the maximum number of pieces that could result from such a cutting? | Answer: 6.
Solution. We will show that there cannot be more than 6 parts. Indeed, the total area of 7 parts cannot be less than $1+2+3+4+5+6+7=28$, which exceeds the area of the square.
Now we will show how a $5 \times 5$ square can be cut into 6 parts of different sizes:
 Each of the 12 knights sitting around a round table has thought of a number, and all the numbers are different. Each knight claims that the number they thought of is greater than the numbers thought of by their neighbors to the right and left. What is the maximum number of these claims that can be true? | Answer: 6.
Solution. Let's renumber the knights in order with numbers from 1 to 12. In the pairs $(1,2),(3,4)$, $\ldots,(11,12)$, at least one of the knights is lying (specifically, the one who guessed the smaller number). Therefore, there can be no more than 6 true statements. Now let's provide an example where exact... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Quadrilateral $A B C D$ is inscribed in a circle. A tangent line $\ell$ is drawn to this circle at point $C$. Circle $\omega$ passes through points $A$ and $B$ and is tangent to line $\ell$ at point $P$. Line $P B$ intersects segment $C D$ at point $Q$. Find the ratio $\frac{B C}{C Q}$, given that $B$ is the point o... | Answer: 1.
Solution. The angle between the tangent $BD$ and the chord $AB$ of the circle $\omega$ is equal to the inscribed angle that subtends $AB$, so $\angle APB = \angle ABD = \angle AC... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Each cell of a $5 \times 6$ table is colored in one of three colors: blue, red, or yellow. In each row of the table, the number of red cells is not less than the number of blue cells and not less than the number of yellow cells, and in each column of the table, the number of blue cells is not less than the number of... | Answer: 6.
| $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{C}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Each cell of a $5 \times 5$ table is colored in one of three colors: blue, red, or yellow. In each row of the table, the number of yellow cells is not less than the number of red cells and not less than the number of blue cells, and in each column of the table, the number of red cells is not less than the number of ... | Answer: 5.
| $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| :---: | :---: | :---: | :---: | :---: |
| $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\m... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. What is the minimum number of cells that need to be marked in a $7 \times 7$ table so that every vertical or horizontal strip $1 \times 4$ contains at least one marked cell? | Answer: 12.
Solution. Consider a more general problem when the table has size $(2 n-1) \times(2 n-1)$, and the strip is $-1 \times n$. Let's call the $n$-th row and the $n$-th column central, and the marked cells on them, except for the center of the board, - axial. Suppose there are $k$ axial cells in the central row... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given numbers $x, y, z \in\left[0, \frac{\pi}{2}\right]$. Find the minimum value of the expression
$$
A=\cos (x-y)+\cos (y-z)+\cos (z-x)
$$ | Answer: 1.
Solution. We can assume that $x \leqslant y \leqslant z$, since the expression $A$ does not change under pairwise permutations of the variables. Notice that
$$
\cos (x-y)+\cos (z-x)=2 \cos \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The first cosine in the right-hand side is positive a... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Given numbers $x, y, z \in [0, \pi]$. Find the minimum value of the expression
$$
A=\cos (x-y)+\cos (y-z)+\cos (z-x)
$$ | Answer: -1.
Solution. We can assume that $x \leqslant y \leqslant z$, since the expression $A$ does not change under pairwise permutations of the variables. Notice that
$$
\cos (x-y)+\cos (z-x)=2 \cos \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The first cosine in the right-hand side is non-negat... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In a $3 \times 3$ table, 9 numbers are arranged such that the six products of these numbers in the rows and columns of the table are all different. What is the maximum number of these numbers that can equal one? | Answer: 5.
Solution. Let's call the index of the table the total number of its rows and columns consisting of ones. According to the condition, the index does not exceed 1. Let $n$ be an element of the table different from 1. Then in one row or one column with $n$ there is another number not equal to 1 (otherwise the ... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In a $3 \times 4$ table, 12 numbers are arranged such that all seven sums of these numbers in the rows and columns of the table are distinct. What is the maximum number of numbers in this table that can be zero? | Answer: 8.
Solution. Let's call the index of the table the total number of its zero rows and columns. According to the condition, the index does not exceed 1. Let $n$ be a non-zero element of the table. Then in the same row or in the same column with $n$ there is another non-zero number (otherwise the sums in the row ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given numbers $x, y, z \in [0, \pi]$. Find the maximum value of the expression
$$
A=\sin (x-y)+\sin (y-z)+\sin (z-x)
$$ | Answer: 2.
Solution 1. We can assume that $x \leqslant y$ and $x \leqslant z$, since the expression $A$ does not change under cyclic permutation of the variables. Notice that
$$
\sin (x-y)+\sin (z-x)=2 \sin \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The argument of the sine on the right-hand sid... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. (40 points) For the quadratic function $p(x)=a x^{2}+b x+c$, for some integers $n$, the equality $p(n)=p\left(n^{2}\right)$ holds. Provide an example of the function $p(x)$ for which the number of such numbers $n$ is the largest. What is this largest number of numbers $n$? | Answer: The maximum number of numbers $n$ is 4. An example of a function: $p(x)=x^{2}-6 x+1$.
Solution: Since $0=0^{2}$ and $1=1^{2}$, for any function $p(x)$ there are at least two such numbers $n$.
Write the equality $p(n)=p\left(n^{2}\right)$ and transform it:
$$
a n^{2}+b n+c=a n^{4}+b n^{2}+c \Leftrightarrow a ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (20 points) Find the number of different four-digit numbers that can be obtained by rearranging the digits of the number 2021 (including this number itself). | Answer: 9.
Solution: The number of options can be found by enumerating the permutations of the digits: 2021, 2012, 2201, 2210, 2102, 2120, 1022, 1202, 1220.
The number of options can also be calculated using combinatorial methods. The position of zero can be chosen in three ways, as it should not be the first. Then, ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. All three-digit numbers from 100 to 999 are written in a row without spaces. Kostya underlined \( k \) consecutive digits in this sequence, and Andrey underlined other \( k \) consecutive digits in this sequence. It turned out that the \( k \)-digit numbers underlined by the boys are equal. For what largest \( k \) ... | Answer: $k=5$.
Solution. An example of a five-digit number that satisfies the condition of the problem is 22923. Indeed, Kostya could underline the fragment «22923» at the junction of the numbers 229 and 230, and Andrey - at the junction of the numbers 922 and 923.
Now we will show that it is not possible to choose a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. All four-digit numbers from 1000 to 9999 are written in a row without spaces. Kostya underlined \( k \) consecutive digits in this sequence, and Andrey underlined other \( k \) consecutive digits in this sequence. It turned out that the \( k \)-digit numbers underlined by the boys are equal. For what largest \( k \)... | Answer: $k=7$.
Solution. An example of a seven-digit number that satisfies the condition of the problem is 2229223. Indeed, Kostya could underline the fragment «2229223» at the junction of the numbers 2229 and 2230, and Andrey - at the junction of the numbers 9222 and 9223.
Now we will show that it is impossible to c... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the maze diagram in Fig. 1, each segment (link) is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, d3, a5.
## Solution.
Fig. 2: Parts of the maze
 is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
Solution.
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 8): part $\mathcal{K}$ is room $K$, the long dead-end corridor that exits from it to the left, and the dead-en... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In the maze diagram in Fig. 13, each segment (link) is a corridor, and each circle is a small room. Some rooms have beacons that hum, each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the shortest ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
## Solution.
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 14): part $\mathcal{K}$ is room $K$ and the dead-end corridors that lead from it to the right and down, par... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Find all primes $p$ for which the numbers $p+1$ and $p^{2}+1$ are double the squares of natural numbers.
# | # Answer: $p=7$
First solution. Let $p+1=2 x^{2}$ and $p^{2}+1=2 y^{2}$, then $2\left(y^{2}-x^{2}\right)=p(p-1)$. Therefore, either $y-x$ or $y+x$ is divisible by $p$. From the inequality $xy-x \geqslant p$ and, thus, $2(y-x)(y+x) \geqslant 2 p^{2}>p(p-1)$, which is impossible. Therefore, $y+x$ is divisible by $p$. No... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find all primes $p$ for which the numbers $p+7$ and $p^{2}+7$ are double the squares of natural numbers. | Answer: $p=11$
First solution. Let $p+7=2 x^{2}$ and $p^{2}+7=2 y^{2}$, then $2\left(y^{2}-x^{2}\right)=p(p-1)$. Since $p-$ is odd, $p \geqslant 3$ and $2 p^{2}>p^{2}+7$. Therefore, $xy-x \geqslant p$ and, thus, $2(y-x)(y+x) \geqslant 2 p^{2}>p(p-1)$, which is impossible. Therefore, $y+x$ is divisible by $p$. Note tha... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The sum of positive numbers $a, b, c$ and $d$ is not less than 8. Find the minimum value of the expression
$\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)}$. | Answer: 1
First solution. By the inequality of means for four numbers, we have
$$
\begin{aligned}
\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{a+b}{16}+\frac{a+c}{16} & +\frac{a+d}{16} \geqslant \\
& \geqslant 4 \sqrt[4]{\frac{a^{4}}{(a+b)(a+c)(a+d)} \cdot \frac{a+b}{16} \cdot \frac{a+c}{16} \cdot \frac{a+d}{16}}=\frac{a}{2}
... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) At the Journalism Faculty of the University of Enchanted Commonwealth, 4 chickens are applying. The faculty has 2 places in the daytime program and 3 places in the evening program. Assuming all 4 chickens will be admitted to the faculty, determine the number of outcomes in which exactly two chickens will... | Answer: 6.
Solution. The faculty has a total of 5 places, to which 4 applicants are applying. Since exactly two chickens will be admitted to the evening department, the other two will be admitted to the daytime department. The number of ways to choose 2 out of 4 applicants to be admitted to the daytime department is $... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. (40 points) For real numbers $a, b$ and $c$ it is known that $a b + b c + c a = 3$. What values can the expression $\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}$ take? | Answer: 6.
Solution. Consider the first term of the desired expression. Using the condition that $a b+b c+c a=3$. Then
$$
b^{2}+3=b^{2}+a b+b c+c a=(b+a)(b+c)
$$
Therefore,
$$
\frac{a\left(b^{2}+3\right)}{a+b}=\frac{a(b+a)(b+c)}{a+b}=a(b+c) \text {. }
$$
Similarly for the second and third terms, we get
$$
\frac{b... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $2 f(2 x-3)-f(3 x+1)$ also has exactly one root. Find the root of the trinomial $f(x)$. | Answer: -11.
Solution. Since dividing all the coefficients of the quadratic polynomial $f(x)$ by $a$ does not change its roots or the roots of the polynomial $g(x)=2 f(2 x-3)-f(3 x+1)$, we can assume that $a=1$. A quadratic polynomial has exactly one root if and only if its discriminant is zero. Therefore, $b^{2}=4 c$... | -11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $f(3 x+2)-2 f(2 x-1)$ also has exactly one root. Find the root of the trinomial $f(x)$. | Answer: -7.
Solution. Since dividing all the coefficients of the quadratic polynomial $f(x)$ by $a$ does not change its roots or the roots of the polynomial $g(x)=f(3 x+2)-2 f(2 x-1)$, we can assume that $a=1$. A quadratic polynomial has exactly one root if and only if its discriminant is zero. Therefore, $b^{2}=4 c$ ... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given the quadratic trinomial $f(x)=a x^{2}-a x+1$. It is known that $|f(x)| \leqslant 1$ for all $x \in[0,1]$. What is the greatest value that $a$ can take? | Answer: 8
Solution. It is not difficult to check that $a=8$ works. Indeed, $|2 x-1| \leqslant 1$ for $x \in[0,1]$, so $f(x)=8 x^{2}-8 x+1=2(2 x-1)^{2}-1 \leqslant 1$, and the inequality $f(x) \geqslant-1$ holds for all $x$.
Suppose that $a>8$. Then
$$
f\left(\frac{1}{2}\right)=\frac{a}{4}-\frac{a}{2}+1=1-\frac{a}{4}... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Given trapezoid $A B C D$ with bases $A B$ and $C D$, angles $\angle C=30^{\circ}$ and $\angle D=80^{\circ}$. Find $\angle A C B$, if it is known that $D B$ is the bisector of angle $\angle D$. | Answer: $10^{\circ}$
Let $E$ be the intersection point of lines $A D$ and $B C$, and $D^{\prime}$ be the point symmetric to point $D$ with respect to line $B C$. Then $C D = C D^{\prime}$ a... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) Chess clubs from Moscow, Saint Petersburg, and Kazan agreed to hold a tournament. Each Muscovite played exactly 9 Saint Petersburg residents and $n$ Kazan residents. Each Saint Petersburg resident played exactly 6 Muscovites and 2 Kazan residents. Each Kazan resident played exactly 8 Muscovites and 6 Sai... | # Answer: 4.
Solution. Let the team from Moscow consist of $m$ participants, the team from Saint Petersburg - of $p$ participants, and the team from Kazan - of $k$ participants.
According to the problem, each Muscovite, i.e., each of the $m$ people, played exactly 9 games with the Saint Petersburg residents; and each... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) At a market in Egypt, a tourist is bargaining with a seller over a souvenir worth 10000 Egyptian pounds. The tourist first reduces the price by x percent $(0<x<100)$, then the seller increases the price by $x$ percent, and so on. The number $x$ remains constant throughout the bargaining, and the seller i... | Answer: 5.
Solution. The final cost of the souvenir can be found using one of two formulas (depending on who had the last word): $10000 \cdot\left(1-\frac{x}{100}\right)^{n} \cdot\left(1+\frac{x}{100}\right)^{n}$ or $10000 \cdot\left(1-\frac{x}{100}\right)^{n+1} \cdot\left(1+\frac{x}{100}\right)^{n}$.
After some tran... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (30 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are 15 and 19, respectively. $AH$ and $BG$ are heights to the line $DC$, and $CF$ is a height to the line $AB$. Points $K, L, M$, and $N$ are the midpoints of segments $AB, CF, CD$, and $AH$ respectively. Find the ratio of the area of trapezoid $ABCD$ to the ar... | Answer: 2 or $\frac{2}{3}$.
Solution. Let the trapezoid $ABCD$ be labeled in a clockwise direction, with the bases assumed to be horizontal. Since the problem does not specify where point $G... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In the maze diagram in Fig. 1, each segment (link) is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, d3, a5.
## Solution.
Fig. 2: Parts of the maze
 is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
Solution.
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 8): part $\mathcal{K}$ is room $K$, the long dead-end corridor that exits from it to the left, and the dead-en... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In the maze diagram in Fig. 13, each segment (link) is a corridor, and each circle is a small room. Some rooms have beacons that hum, each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the shortest ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
## Solution.
b) Estimation. We will prove that two beacons are insufficient. Consider three parts of our maze (Fig. 14): part $\mathcal{K}$ is room $K$ and the dead-end corridors that lead from it to the right and down, p... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. From the town "U $" to the town "A x" at $10^{00}$ AM, Ivan set off on his bicycle. After traveling two-thirds of the way, he passed the town "Ox," from which at that moment Peter set off on foot towards the town "U x". At the moment Ivan arrived in the town "A x", Nikolai set off from there in the opposite directio... | Answer: 6 km.
## Solution:
We will solve the problem using a graphical-geometric method. Let's represent Ivan's movement as segment $K L$, Nikolai's movement as segment $L M$, and Petr's movement as segment $N P$ in a coordinate system $(t ; s)$, where $t$ is time in hours and $s$ is distance in kilometers from point... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. From the town "U ${ }^{\prime}$ " to the town " $A x$ ", Ivan set off on his bicycle at $11^{00}$ AM, having traveled two fifths of the distance, he passed the town " $O x$ ", from which at that moment Peter set off on foot towards the town "Ux". At the moment when Ivan arrived in the town " $A x$ ", from there in t... | Answer: 5 km.
Solution: a similar solution to this problem is present in option 1 under the same number. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Find the value of the expression
$$
\frac{1}{1+m+m n}+\frac{1}{1+n+n k}+\frac{1}{1+k+k m} \text { given that } m=\frac{1}{n k} \text {. }
$$ | Answer: 1.
## Solution.
$$
\begin{gathered}
\frac{1}{1+m+m n}+\frac{1}{1+n+n k}+\frac{1}{1+k+k m}=\frac{k}{k+k m+k m n}+\frac{k m}{k m+k m n+k m n k}+\frac{1}{1+k+k m}= \\
=\frac{k}{k+k m+1}+\frac{k m}{k m+1+k}+\frac{1}{1+k+k m}=\frac{k+k m+1}{1+k+k m}=1
\end{gathered}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For the quadratic trinomials $f_{1}(x)=a x^{2}+b x+c_{1}, f_{2}(x)=a x^{2}+b x+c_{2}$, $\ldots, f_{2020}(x)=a x^{2}+b x+c_{2020}$, it is known that each of them has two roots.
Denote by $x_{i}$ one of the roots of $f_{i}(x)$, where $i=1,2, \ldots, 2020$. Find the value
$$
f_{2}\left(x_{1}\right)+f_{3}\left(x_{2}\r... | Answer: 0.
## Solution:
Since $f_{1}\left(x_{1}\right)=0$, then $f_{2}\left(x_{1}\right)=f_{1}\left(x_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$.
Similarly, we can obtain the following equalities:
$$
f_{3}\left(x_{2}\right)=c_{3}-c_{2}, \ldots, f_{2020}\left(x_{2019}\right)=c_{2020}-c_{2019}, f_{1}\left(x_{20... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In a sports store, over two days, thirteen pairs of sneakers, two sports suits, and one T-shirt were sold, with the same amount of money earned on the first day as on the second day (from the sale of the aforementioned items). One pair of sneakers is cheaper than a sports suit and more expensive than a T-shirt by th... | Answer: 8 pairs of sneakers and no sports suits.
Solution. Let in one day $x$ suits and $y$ pairs of sneakers were sold with a T-shirt. Then in the other day, $(2-x)$ suits and $(13-y)$ pairs of sneakers were sold.
Let $c$ be the price of one pair of sneakers, and $s$ be the price difference.
Then, from the conditio... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Misha invited eighteen friends from his sports club and two of his brothers to celebrate his birthday, a total of twenty guests. All the guests and Misha himself, seated at two tables, ate all the hot dogs served equally on both tables, and everyone ate only from their own table. Each friend from the sports club ate... | # Answer: 9 friends from the sports club and no brothers.
Solution. Let $x$ be the number of brothers and $y$ be the number of friends from the sports club sitting at the same table with Misha. Then, at the other table, there were $2-x$ brothers and $18-y$ friends from the sports club.
Let $c$ be the number of hot do... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Find the number of roots of the equation: $2^{\lg \left(x^{2}-2023\right)}-\lg 2^{x^{2}-2022}=0$. | # Answer: 4 roots.
Solution: Using the properties of logarithms, rewrite the equations in the following form
$$
\left(x^{2}-2023\right)^{\lg 2}-\lg 2^{x^{2}-2022}=0
$$
Introduce the notations $z=x^{2}-2023, a=1 \mathrm{~g} 2$, in this case $z>0, a \in(0,1)$. Then $z^{a}=(z+1) a$
Let $y_{1}(z)=z^{a}, y_{2}(z)=(z+1) ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The cosine of the angle between the lateral sides $A D$ and $B C$ of trapezoid $A B C D$ is 0.8. A circle is inscribed in the trapezoid, and the side $A D$ is divided by the point of tangency into segments of lengths 1 and 4. Determine the length of the lateral side $B C$ of the trapezoid. | Answer: 4 or $\frac{100}{7}$.
Solution: Let $S$ be the intersection point of lines $A D$ and $B C$; $K, L, M$ be the points of tangency of the inscribed circle with sides $A B, A D$, and $C D$ respectively, and $O$ be its center. Then $O K \perp A B, O M \perp C D$, as radii, and since $A B \| C D$, points $K, O, M$ l... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. For the quadratic trinomial $f(x)=a x^{2}+b x+c$, it is known that
$$
f\left(\frac{a-b-c}{2 a}\right)=f\left(\frac{c-a-b}{2 a}\right)=0
$$
Find the value of the product $f(-1) \cdot f(1)$. | Answer: 0.
## Solution:
$$
f\left(\frac{a-b-c}{2 a}\right)=\frac{a(a-b-c)^{2}}{4 a^{2}}+\frac{b(a-b-c)}{2 a}+c=\frac{(a-b+c)(a+b+c)}{4 a}=\frac{f(-1) \cdot f(1)}{4 a}=0
$$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Compute $2022!\cdot\left(S_{2021}-1\right)$, if $S_{n}=\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}$. | Answer: -1.
Solution.
Given that $\frac{n}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$, we get
$S_{2021}=\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{2021}{2022!}=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{2021!}-\frac{1}{2022!}\right)... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the value of the expression $\frac{1}{a b}+\frac{1}{b c}+\frac{1}{a c}$, if it is known that $a, b, c$ are three different real numbers satisfying the conditions $a^{3}-2022 a^{2}+1011=0, b^{3}-2022 b^{2}+1011=0, c^{3}-2022 c^{2}+1011=0$. | Answer: -2.
## Solution.
The cubic equation $t^{3}-2022 t^{2}+1011=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2022 t^{2}+1011: f(-3000)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's formulas:
$$
\left\{\begin{array}{l}
a+b+c=2022 \\
a b+b c+a c=0 \\
a b c=-1011
\end{array}\right.
$$... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Compute $2023!\cdot\left(S_{2022}-1\right)$, if $S_{n}=\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}$. | Answer: -1.
Solution.
Given that $\frac{n}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$, we get
$S_{2022}=\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{2022}{2023!}=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{2022!}-\frac{1}{2023!}\right)... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the value of the expression $\frac{1}{a b}+\frac{1}{b c}+\frac{1}{a c}$, if it is known that $a, b, c$ are three different real numbers satisfying the conditions $a^{3}-2020 a^{2}+1010=0, b^{3}-2020 b^{2}+1010=0, \quad c^{3}-2020 c^{2}+1020=0$. | Answer: -2.
## Solution.
The cubic equation $t^{3}-2020 t^{2}+1010=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2020 t^{2}+1010: f(-3000)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's formulas:
$$
\left\{\begin{array}{l}
a+b+c=2020 \\
a b+b c+a c=0 \\
a b c=-1010
\end{array}\right.
$$... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three different real numbers satisfying the conditions:
$a^{3}-2022 a+1011=0, \quad b^{3}-2022 b+1011=0, \quad c^{3}-2022 c+1011=0$. | Answer: 2.
## Solution.
The cubic equation $t^{3}-2022 t+1011=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2022 t+1011: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\left\{\begin{array}{l}
a+b+c=0 \\
a b+b c+a c=-2022 \\
a b c=-1011
\end{array}\right.
$$
We find ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three distinct real numbers satisfying the conditions:
$a^{3}-2020 a+1010=0, \quad b^{3}-2020 b+1010=0, c^{3}-2020 c+1010=0$. | Answer: 2.
Solution.
The cubic equation $t^{3}-2020 t+1010=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2020 t+1010: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\left\{\begin{array}{l}
a+b+c=0 \\
a b+b c+a c=-2020 \\
a b c=-1010
\end{array}\right.
$$
We find the... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. To walk 2 km, ride 3 km on a bicycle, and drive 20 km by car, Uncle Vanya needs 1 hour 6 minutes. If he needs to walk 5 km, ride 8 km on a bicycle, and drive 30 km by car, it will take him 2 hours 24 minutes. How much time will Uncle Vanya need to walk 4 km, ride 5 km on a bicycle, and drive 80 km by car? | # Answer: 2 hours 54 minutes. (2.9 hours.)
## Solution:
Let $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ be the walking speed, cycling speed, and driving speed, respectively. Then, according to the problem, we can set up the system:
$$
\left\{\begin{array} { c }
{ 2 x + 3 y + 2 0 z = 6 6 } \\
{ 5 x + 8 y + 3 0 z = 1 4 4... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The base of the pyramid $S A B C D$ is a rectangle $A B C D$, with the height being the edge $S A=25$. Point $P$ lies on the median $D M$ of the face $S C D$, point $Q$ lies on the diagonal $B D$, and the lines $A P$ and $S Q$ intersect. Find the length of $P Q$, if $B Q: Q D=3: 2$. | # Answer: 10.
## Solution:
Since lines $A P$ and $S Q$ intersect, points $A, P, S, Q$ lie in the same plane. Let $R$ be the intersection point of $S P$ and $A Q$. Then
$$
\frac{R Q}{A Q}=\frac{D Q}{B Q}=\frac{2}{3} \Rightarrow \frac{R Q}{R A}=\frac{2}{5}
$$
We will prove that $\frac{R M}{R S}=\frac{2}{5}$. Note tha... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Real
numbers
$x, y, z$
satisfy
$$
4 x^{2}-2 x-30 y z=25 y^{2}+5 y+12 x z=9 z^{2}-3 z-20 x y .
$$
relations:
Find the maximum of the sum $a+b+c$, where $a=2 x+5 y, b=3 z+5 y, c=3 z-2 x$. | Solution. Note that
$$
a-b+c=0
$$
Let $A=4 x^{2}-2 x-30 y z, B=25 y^{2}+5 y+12 x z$ and $C=9 z^{2}-3 z-20 x y$. Subtracting these equations from each other, we get
$$
\begin{aligned}
& A-B=a \cdot(2 x-6 z-5 y-1)=0 \\
& B-C=b \cdot(5 y+4 x-3 z+1)=0 \\
& A-C=c \cdot(1-2 x-10 y-3 z)=0
\end{aligned}
$$
Assume that all ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a triangle with sides $a, b, c$ and angles $\alpha, \beta, \gamma$, the equality $3 \alpha + 2 \beta = 180^{0}$ is satisfied. The sides $a, b, c$ lie opposite the angles $\alpha, \beta, \gamma$ respectively. Find the length of side $c$ when $a=2, b=3$.
 of the original system by $b_{2}$ and subtract from it equation (b) multiplied by $b_{1}$. The result is
$$
a_{2} \cdot \Delta=b_{2}
$$
Here $\Delta=b_{2} b_{3}-b_{1} b_{4}$. Similarly, from (c) and (d) we find that
$$
a_{3}... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The function $y=f(x)$ is defined on the set $(0,+\infty)$ and takes positive values on it. It is known that for any points $A$ and $B$ on the graph of the function, the areas of the triangle $A O B$ and the trapezoid $A B H_{B} H_{A}$ are equal to each other $\left(H_{A}, H_{B}\right.$ - the bases of the perpendicul... | # Solution:
Let $M$ be the intersection point of segments $O B$ and $A H_{A}$. Since the areas of triangle $A O B$ and trapezoid $A B H_{B} H_{A}$ are equal, then
the areas of triangles $A M... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Let $x_{1}$ and $x_{2}$ be the largest roots of the polynomials
$$
\begin{gathered}
f(x)=1-x-4 x^{2}+x^{4} \\
\text { and } \\
g(x)=16-8 x-16 x^{2}+x^{4}
\end{gathered}
$$
respectively. Find $\frac{x_{2}}{x_{1}}$. | Solution:
Notice that $f(-2)>0, f(-1)<0, f(0)>0, f(1)<0$. Therefore, the polynomial $f(x)$ has 4 real roots. Similarly, from the inequalities $g(-4)>0, g(-2)<0, g(0)>0, g(2)<0$ it follows that the polynomial $g(x)$ has 4 real roots.
Comparison of the coefficients of the polynomials
$$
f(x)=1-x-4 x^{2}+x^{4} \text { ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Points $F$ and $G$ are chosen on the sides $AB$ and $BC$ of rectangle $ABCD$, respectively. A perpendicular $FK$ is dropped from point $F$ to side $CD$. A perpendicular $GH$ is dropped from point $G$ to side $AD$. The intersection point of $FK$ and $GH$ is denoted as $E$. Find the area of triangle $DFG$, given that ... | # Solution:
Let $A D=a, D C=b, H D=x$, and $D K=y$.
$$
\begin{aligned}
S_{D F G}=S_{A B C D} & -S_{A F D}-S_{F G B}-S_{D G C}=a b-\frac{1}{2} a y-\frac{1}{2}(b-y)(a-x)-\frac{1}{2} b x \\
&... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{l}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$. | Solution. We will prove that ${ }^{1}$
$$
a_{2} b_{3}=a_{3} b_{2}
$$
Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Mathematics
$$
\begin{cases... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (4 points) Oleg usually arrives on a business trip by the 11 o'clock train. A car arrives at the station for this train. This time, the train arrived an hour earlier, and Oleg started walking towards the car. Meeting the car on the way, he got in, and as a result, arrived 10 minutes earlier than planned. Determine t... | 4. 10 hours 55 minutes.
$+(4$ points) - solution is correct (by any method)
+- (3 points) - solution is correct, but there are arithmetic errors
-+ (2 points) - there are reasonable ideas in solving the problem, but the problem is not solved in general | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (4 points) Let two sequences of numbers $\left(x_{0}, x_{1}, \ldots, x_{2009}\right)$, $\left(y_{0}, y_{1}, \ldots, y_{2009}\right)$ be constructed according to the following rules:
a) $x_{0}=12, x_{1}=\frac{1}{3}, y_{0}=4, y_{1}=\frac{1}{18}$,
b) $x_{i+1}=x_{i-1}+4 x_{i}$ and $y_{i+1}=y_{i-1}-4 y_{i}$ for $i=1, \l... | 5. 2 .
$+(4$ points) - solution is correct
+- (3 points) - the idea of the recurrence relation is proven, but there are arithmetic errors in the solution
-+ (2 points) - the idea of the recurrence relation is formulated, but the problem is not completed | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Non-zero numbers $a$ and $b$ are roots of the quadratic equation $x^{2}-5 p x+2 p^{3}=0$. The equation $x^{2}-a x+b=0$ has a unique root. Find $p$. Justify your solution. | Solution. Since the equation $x^{2}-a x+b=0$ has a unique root, then $b=\frac{a^{2}}{4}$. By Vieta's theorem, we have the equalities: $a+b=5 p ; a b=2 p^{3}$. Substituting $b=\frac{a^{2}}{4}$ into the last equality, we get: $a=2 p$. Considering that $a$ and $b$ are non-zero, we find $p=3$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the sides $B C$ and $C D$ of the square $A B C D$, points $E$ and $F$ are chosen such that the angle $E A F$ is $45^{\circ}$. The length of the side of the square is 1. Find the perimeter of triangle $C E F$. Justify your solution. | Solution. If point $D$ is reflected across line $A F$, and then across line $A E$, it will transition to point $B$. Indeed, the composition of two axial symmetries relative to intersecting lines is a rotation by twice the angle between the lines. In our case, these two symmetries are equivalent to a $90^{\circ}$ rotati... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{aligned}
a_{1} b_{1}+a_{2} b_{3} & =1 \\
a_{1} b_{2}+a_{2} b_{4} & =0 \\
a_{3} b_{1}+a_{4} b_{3} & =0 \\
a_{3} b_{2}+a_{4} b_{4} & =1
\end{aligned}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_... | Solution. We will prove that ${ }^{1}$
$$
a_{2} b_{3}=a_{3} b_{2}
$$
Multiply equation (a) of the original system
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & \text { (a) } \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1\end{cases}
$$
by $b_{2}$ ... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. On the sides $B C$ and $C D$ of the square $A B C D$, points $E$ and $F$ are chosen such that the angle $E A F$ is $45^{\circ}$. The length of the side of the square is 1. Find the perimeter of triangle $C E F$. Justify your solution. | # Solution:
If point $D$ is reflected relative to line $A F$, and then relative to line $A E$, it will transition to point $B$. Indeed, the composition of two axial symmetries relative to intersecting lines is a rotation by twice the angle between the lines. In our case, these two symmetries are equivalent to a $90^{\... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Points $F$ and $G$ are chosen on the sides $AB$ and $BC$ of rectangle $ABCD$, respectively. A perpendicular $FK$ is dropped from point $F$ to side $CD$. A perpendicular $GH$ is dropped from point $G$ to side $AD$. The intersection point of $FK$ and $GH$ is denoted as $E$. Find the area of triangle $DFG$, given that ... | # Solution:
Let $A D=a, D C=b, H D=x$, and $D K=y$.
$$
\begin{aligned}
S_{D F G}=S_{A B C D} & -S_{A F D}-S_{F G B}-S_{D G C}=a b-\frac{1}{2} a y-\frac{1}{2}(b-y)(a-x)-\frac{1}{2} b x \\
&... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (3 points) Let two sequences of numbers $\left(x_{0}, x_{1}, \ldots, x_{2009}\right)$, $\left(y_{0}, y_{1}, \ldots, y_{2009}\right)$ be constructed according to the following rules:
a) $x_{0}=12, x_{1}=\frac{1}{3}, y_{0}=4, y_{1}=\frac{1}{18}$
b) $x_{i+1}=x_{i-1}+4 x_{i}$ and $y_{i+1}=y_{i-1}-4 y_{i}$ for $i=1, \ld... | 2. 2 .
$+(3$ points) - the solution is correct
+- (2 points) - the idea of the recurrence relation is proven, but there are arithmetic errors in the solution
-+ (1 point) - the idea of the recurrence relation is formulated, but the problem is not completed | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Real numbers $x, y, z$ satisfy the relations:
$$
4 x^{2}-2 x-30 y z=25 y^{2}+5 y+12 x z=9 z^{2}-3 z-20 x y .
$$
Find the maximum of the sum $a+b+c$, where $a=2 x+5 y, b=3 z+5 y, c=3 z-2 x$. | Solution. Note that
$$
a-b+c=0
$$
Let $A=4 x^{2}-2 x-30 y z, B=25 y^{2}+5 y+12 x z$ and $C=9 z^{2}-3 z-20 x y$. Subtracting these equations from each other, we get
$$
\begin{aligned}
& A-B=a \cdot(2 x-6 z-5 y-1)=0 \\
& B-C=b \cdot(5 y+4 x-3 z+1)=0 \\
& A-C=c \cdot(1-2 x-10 y-3 z)=0
\end{aligned}
$$
Assume that all ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In quadrilateral $A B C D$, the diagonals intersect at point $O$. It is known that $S_{A B O}=S_{C D O}=\frac{3}{2}$, $B C=3 \sqrt{2}$, $\cos \angle A D C=\frac{3}{\sqrt{10}}$. Find the smallest area that such a quadrilateral can have. | Solution. We will prove that quadrilateral $ABCD$ is a parallelogram. Let $x_{1}, x_{2}, y_{1}, y_{2}$ be the segments into which the diagonals are divided by their point of intersection. Denote the angle between the diagonals as $\alpha$. By the condition, the areas of triangles $ABO$ and $CDO$ are equal, that is, $\f... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Let $x_{1}$ and $x_{2}$ be the largest roots of the polynomials $f(x)=1-x-4 x^{2}+x^{4}$ and $g(x)=16-8 x-$ $16 x^{2}+x^{4}$ respectively. Find $\frac{x_{2}}{x_{1}}$. | # Solution:
Notice that $f(-2)>0, f(-1)<0, f(0)>0, f(1)<0$. Therefore, the polynomial $f(x)$ has 4 real roots. Similarly, from the inequalities $g(-4)>0, g(-2)<0, g(0)>0, g(2)<0$, it follows that the polynomial $g(x)$ has 4 real roots.
Comparison of the coefficients of the polynomials
$$
f(x)=1-x-4 x^{2}+x^{4} \text... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (3 points) Oleg usually arrives on a business trip by the 11 AM train. A car arrives at the station for this train. This time, the train arrived an hour earlier, and Oleg started walking towards the car. Meeting the car on the way, he got in, and as a result, arrived 10 minutes earlier than planned. Determine the ti... | 1. 10 hours 55 minutes.
+ (3 points) - the solution is correct (by any method)
$+-(2$ points) - the solution is correct, but there are arithmetic errors
-+ (1 point) - there are reasonable ideas in solving the problem, but the problem is not solved overall | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $\mathrm{ABC}$, the sides $A B=4, B C=6$. Point $M$ lies on the perpendicular bisector of segment $A B$, and lines $A M$ and $A C$ are perpendicular. Find $M A$, if the radius of the circumscribed circle around triangle $A B C$ is 9. | Solution. Introduce a coordinate system with the origin at point A such that point C lies on the x-axis. From the problem statement, point M lies on the y-axis. Let's introduce notations for the unknown coordinates: $\mathrm{A}(0,0), \mathrm{B}\left(\mathrm{x}_{\mathrm{B}}, \mathrm{y}_{\mathrm{B}}\right), \mathrm{C}\le... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. How many solutions of the equation $x^{2}-2 x \cdot \sin (x \cdot y)+1=0$ fall within the circle $x^{2}+y^{2} \leq 100$? | Solution. We will interpret the left side of the equation as a quadratic trinomial in terms of $x$. For the roots to exist, the discriminant must be non-negative, i.e., $D=4 \sin ^{2}(x \cdot y)-4 \geq 0 \Leftrightarrow \sin ^{2}(x \cdot y)=1 \Leftrightarrow \cos 2 x y=-1 \Leftrightarrow x y=\frac{\pi}{2}+\pi n, n \in ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (5 points) From point $A$, lying on a circle of radius 3, chords $A B, A C$ and tangent $A D$ are drawn. The angle between the chords is $\frac{\pi}{4}$, and the angle between chord $A C$ and tangent $A D$, which does not contain chord $A B$, is $\frac{5 \pi}{12}$. Calculate the integer area of triangle $A B C$.
# | # Solution:
Let $\angle D A C=\alpha, \angle B A C=\beta$, and the radius of the circle be $R$. It is known that $\angle A C B=\angle D A C=\alpha$.
By the Law of Sines, $\quad \frac{|A B|}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (5 points) From point $A$, lying on a circle, chords $A B$, $A C$, and tangent $A D$ are drawn. The angle between the chords is $\frac{\pi}{6}$, and the angle between chord $A C$ and tangent $A D$, which does not contain chord $A B$, is $\frac{5 \pi}{12}$. Calculate the integer part of the radius of the circle if th... | # Solution:
Let $\angle D A C=\alpha, \angle B A C=\beta$, and the radius of the circle be $R$. It is known that $\angle A C B=\angle D A C=\alpha$.
By the Law of Sines, $\quad \frac{|A B|}... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. In a triangle with sides $a, b, c$ and angles $\alpha, \beta, \gamma$, the equality $3 \alpha + 2 \beta = 180^{0}$ is satisfied. The sides $a, b, c$ lie opposite the angles $\alpha, \beta, \gamma$ respectively. Find the length of side $c$ when $a=2, b=3$.
 The sides of a parallelogram are 5 and 13, and the angle between them is $\arccos \frac{6}{13}$. Two mutually perpendicular lines divide this parallelogram into four equal-area quadrilaterals. Find the lengths of the segments into which these lines divide the sides of the parallelogram.
# | # Solution:
Let in parallelogram $ABCD$, $|AB| = |CD| = a$, $|AD| = |BC| = b$, $\angle BAC = \alpha = \arccos c$. Clearly, $\cos \alpha = c$, and $\cos (\pi - \alpha) = -c$.
Denote $|DN| =... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) For a natural number $x$, five statements are made:
$$
3 x>91
$$
$$
\begin{aligned}
& x37 \\
& 2 x \geq 21 \\
& x>7
\end{aligned}
$$
It is known that only three of them are true, and two are false. Find $x$. | Solution. Let's transform the original system of inequalities
$$
\begin{aligned}
& x>\frac{91}{3} \\
& x \geq \frac{21}{2} \\
& x>\frac{37}{4} \\
& x>7 \\
& x<120
\end{aligned}
$$
The first inequality does not hold, since otherwise, four inequalities would be satisfied immediately. Therefore, $x \leq \frac{91}{3}$, a... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A square table consists of 2014 rows and 2014 columns. In each cell at the intersection of the row with number $i$ and the column with number $j$, the number $a_{i, j}=(-1)^{i}(2015-i-j)^{2}$ is written. Find the sum of all the numbers in the table | Problem 3. Answer: $\mathbf{0 .}$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{l}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$.
... | Solution. We will prove that ${ }^{1}$
Multiply the equation (a) of the original system
$$
a_{2} b_{3}=a_{3} b_{2}
$$
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & (\mathrm{a}) \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1 & \text { (d) }\end{cas... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (4 points) Find the area of a triangle if two of its medians are equal to $\frac{15}{7}$ and $\sqrt{21}$, and the cosine of the angle between them is $\frac{2}{5}$.
# | # Solution:
Let $|A D|=a$ and $|B E|=b$ be the medians of the triangle, and $\cos (\angle A O F)=c$.
It is known that the area
$$
\begin{aligned}
& S_{A O E}=\frac{1}{6} \cdot S_{A B C} \\... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (4 points) Find the area of a triangle if two of its medians are equal to 3 and $2 \sqrt{7}$, and the cosine of the angle between them is $-\frac{3}{4}$.
# | # Solution:
Let $|A D|=a$ and $|B E|=b$ be the medians of the triangle, and $\cos (\angle A O F)=c$.
It is known that the area
$$
\begin{aligned}
& S_{A O E}=\frac{1}{6} \cdot S_{A B C} \c... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{l}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$. | Solution. We will prove that ${ }^{1}$
$$
a_{2} b_{3}=a_{3} b_{2}
$$
Multiply equation (a) of the original system
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & (\mathrm{a}) \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1 & \text { (d) }\end{cases}
... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. On the first day, $2^{n}$ schoolchildren played ping-pong in a "knockout" format: first, two played, then the winner played with the third, the winner of this pair played with the fourth, and so on until the last schoolchild (there are no draws in ping-pong). On the second day, the same schoolchildren played for the... | Answer: 3.
Solution. Let's construct the following graph: vertices are players, edges are played matches. According to the condition, for both tournaments, this graph is the same. Consider the first tournament and select the matches where the winners had not won before (for example, the first match). Then the correspo... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [8] The three medians of a triangle divide its angle into six angles, of which exactly $k$ are greater than $30^{\circ}$. What is the greatest possible value of $k$?
(N. Sedrakyan) | Answer. $k=3$.
Evaluation. First method. Let $h_{1} \leq h_{2} \leq h_{3}$ be the heights of the triangle, and $m_{1}, m_{2}, m_{3}$ be the medians from the corresponding vertices (in fact, $m_{1} \leq m_{2} \leq m_{3}$, but this will not be used). Then we have $m_{3} \geq h_{i}$ for $i=1,2$, 3. Dropping perpendicular... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2.
In a table consisting of $n$ rows and $m$ columns, numbers (not necessarily integers) were written such that the sum of the elements in each row is 408, and the sum of the elements in each column is 340. After that, $k$ columns were added to the table, the sum of the elements in each of which is 476, and a col... | Answer: 4 when $n=65, m=78, k=18$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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