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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 |
2. In each cell of a $15 \times 15$ square, there is a natural number not exceeding 4, and the sum of the numbers in each $2 \times 2$ square is 7. What is the maximum value that the sum of the numbers in the entire table can take? | Answer: 417
Solution. Note that the sum of the numbers in two adjacent cells does not exceed five, since otherwise the sum of the numbers in the $2 \times 2$ square containing these two cells would be at least eight, which is impossible according to the condition.
Divide the table into 49 squares of $2 \times 2$ and ... | 417 | Combinatorics | 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. The picture shows several circles connected by segments. Sasha chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime with... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among seven numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Sasha chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a+b$ must be coprime with $n$; if connected... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among eight numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Tanya chooses a natural number n and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers a and b are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with n; if connected... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, t... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Tanya chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with $n$; if $... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among six numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, th... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The board has the number 5555 written in an even base $r$ ($r \geqslant 18$). Petya found out that the $r$-ary representation of $x^{2}$ is an eight-digit palindrome, where the difference between the fourth and third digits is 2. (A palindrome is a number that reads the same from left to right and from right to left... | Answer: $r=24$.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. According to the condition, there exist such $r$-ary digits $a, b, c, d$ that $d-c=2$ and
$$
25(r+1)^{2}\left(r^{2}+1\right)^{2}=a\left(r^{7}+1\right)+b\left(r^{6}+r\right)+c\left(r^{5}+r^{2}\right)+d\left(r^{4}+r^{3}\right)
$$
... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Nastya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the difference \( a - b \) must be copr... | Answer: $n=5 \cdot 7 \cdot 11=385$.
Solution. We will make two observations.
1) $n$ is not divisible by 2 and 3. Among seven numbers, there are always three numbers of the same parity. If $n$ is even, then they must be pairwise connected. Moreover, among seven numbers, there will be three numbers that give the same r... | 385 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the picture, several circles are drawn, connected by segments. Nastl chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the difference \( a - b \) must... | Answer: $n=3 \cdot 5 \cdot 7=105$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among the six numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the diagram.
2) If $p$ is a pr... | 105 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime wit... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among any five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) $n$ has at least two distinct pr... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the picture, several circles are drawn, connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be co... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. If there are three even and three odd numbers, then each of these triples forms a cycle. Suppose there are four numbers $a, b, c, d$ of the same parity. Then all of them are pairwise connected, and we again get two th... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Triangle $A B C$ with angle $\angle A B C=135^{\circ}$ is inscribed in circle $\omega$. The lines tangent to $\omega$ at points $A$ and $C$ intersect at point $D$. Find $\angle A B D$, given that $A B$ bisects segment $C D$. Answer: $90^{\circ}$
=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secan... | 90 | 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 |
2. (10 points) In each cell of a $50 \times 50$ square, a number is written that is equal to the number of $1 \times 16$ rectangles (both vertical and horizontal) in which this cell is an end cell. In how many cells are numbers greater than or equal to 3 written? | Answer: 1600.
Solution. We will denote the cells of the square by pairs $(i, j)$, where $i=1, \ldots, 50, j=$ $=1, \ldots, 50$. We will start the numbering from the bottom left corner of the square.
The cell $(i, j)$ is the rightmost for a horizontal rectangle if $16 \leqslant$ $\leqslant i-$ inequality (1), and the ... | 1600 | 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 |
9. (40 points) What is the maximum number of numbers that can be chosen among the natural numbers from 1 to 3000 such that the difference between any two of them is different from 1, 4, and 5? | Answer: 1000.
Solution. Let's provide an example. We can choose all numbers divisible by 3. Then the difference between any two numbers will also be divisible by 3, while the numbers 1, 4, and 5 are not divisible by 3.
The estimate is based on the consideration that among 6 consecutive numbers, 3 numbers cannot be ch... | 1000 | Combinatorics | 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 |
3. Triangle $A B C$ with angle $\angle A B C=135^{\circ}$ is inscribed in circle $\omega$. The lines tangent to $\omega$ at points $A$ and $C$ intersect at point $D$. Find $\angle A B D$, given that $A B$ bisects segment $C D$. Answer: $90^{\circ}$
=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secan... | 90 | Geometry | 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 |
1. Calculate:
$$
\frac{2 \cdot 2018}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+\cdots+2018}}
$$ | Answer: 2019.
Solution: a similar solution to this problem is present in variant 1 under the same number. | 2019 | Algebra | 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 |
3. Find $g(2022)$, if for any real $x, y$ the equality holds
$$
g(x-y)=g(x)+g(y)-2021(x+y)
$$
# | # Answer: 4086462.
Solution. Substitute $x=y=0$, we get
$$
g(0)=g(0)+g(0)-2021(0+0) \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=g(x)+g(x)-2021(x+x) \Rightarrow g(x)=2021 x \Rightarrow \\
g(2022)=2021 \cdot 2022=4086462
\end{gathered}
$$ | 4086462 | Algebra | 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. Find $g$(2021), if for any real $x, y$ the equality holds
$$
g(x-y)=g(x)+g(y)-2022(x+y)
$$ | Answer: 4086462.
Solution. Substitute $x=y=0$, we get
$$
g(0)=g(0)+g(0)-2022(0+0) \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=g(x)+g(x)-2022(x+x) \Rightarrow g(x)=2022 x \Rightarrow \\
g(2021)=2022 \cdot 2021=4086462 .
\end{gathered}
$$ | 4086462 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Two math teachers are conducting a geometry test, checking each 10th-grade student's ability to solve problems and their knowledge of theory. The first teacher spends 5 and 7 minutes per student, respectively, while the second teacher spends 3 and 4 minutes per student. What is the minimum time they will need to int... | # Answer: 110 minutes.
## Solution: (estimation + example)
Let the first teacher accept the test on problems from $X$ students, and on theory from $Y$ students. Then the second teacher accepts the test on problems from (25-X) students, and on theory from (25-Y) students. Let $T$ be the minimum time required for them ... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Two math teachers are conducting a geometry test, checking each 10th-grade student's ability to solve problems and their knowledge of theory. The first teacher spends 5 and 7 minutes per student, respectively, while the second teacher spends 3 and 4 minutes per student. What is the minimum time required for them to ... | # Answer: 110 minutes.
Solution: fully corresponds to the solution of problem 2, option 1. | 110 | Logic and Puzzles | 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 |
5. In an isosceles trapezoid \(ABCD\) with bases \(AD\) and \(BC\), perpendiculars \(BH\) and \(DK\) are drawn from vertices \(B\) and \(D\) to the diagonal \(AC\). It is known that the feet of the perpendiculars lie on the segment \(AC\) and \(AC=20\), \(AK=19\), \(AH=3\). Find the area of trapezoid \(ABCD\).
(10 poi... | Solution. Note that right triangles $D K A$ and $B H C$ are similar, since
$\angle B C H=\angle D A K$. Let $D K=x, B H=y$. Due to similarity $\frac{D K}{K A}=\frac{B H}{H C}, \frac{x}{19}=\frac{y}{17}$. On the other hand, $C D$
$=\mathrm{AB}$ and by the Pythagorean theorem
$$
C D^{2}=D K^{2}+K C^{2}=x^{2}+1, A B^{2... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In an isosceles trapezoid $A B C D$ with lateral sides $A B$ and $C D$, the lengths of which are 10, perpendiculars $B H$ and $D K$ are drawn from vertices $B$ and $D$ to the diagonal $A C$. It is known that the bases of the perpendiculars lie on segment $A C$ and $A H: A K: A C=5: 14: 15$. Find the area of trapezoi... | Solution. Let $x = BH$, $y = DK$. From the similarity of right triangles $DKA$ and $BHC$, since $\angle BHC = \angle DAK$, we get
$$
\frac{x}{y} = \frac{CH}{AK} = \frac{10}{14} = \frac{5}{7}, \quad 5y = 7x, \quad y = \frac{7x}{5}.
$$
By the condition $AH : AK : AC = 5 : 14 : 15$, therefore $AH : CK = 5 : 1$. By the P... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In a sports store, over two days, thirteen pairs of sneakers, two sports suits, and one T-shirt were sold, with the same amount of money earned on the first day as on the second day (from the sale of the aforementioned items). One pair of sneakers is cheaper than a sports suit and more expensive than a T-shirt by th... | Answer: 8 pairs of sneakers and no sports suits.
Solution. Let in one day $x$ suits and $y$ pairs of sneakers were sold with a T-shirt. Then in the other day, $(2-x)$ suits and $(13-y)$ pairs of sneakers were sold.
Let $c$ be the price of one pair of sneakers, and $s$ be the price difference.
Then, from the conditio... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find $g(2022)$, if for any real $x, y$ the equality holds
$$
g(x-y)=2022(g(x)+g(y))-2021 x y .
$$ | Answer: 2043231.
Solution. Substitute $x=y=0$, we get
$$
g(0)=2022(g(0)+g(0))-2021 \cdot 0 \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=2022(g(x)+g(x))-2021 \cdot x^{2} \Rightarrow g(x)=\frac{2021 x^{2}}{2 \cdot 2022} \Rightarrow \\
g(2022)=\frac{2021 \cdot 2022^{2}}{2 \cdot 2022}=\frac{2... | 2043231 | 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 |
3. Find $g(2021)$, if for any real $x, y$ the equality holds
$$
g(x-y)=2021(g(x)+g(y))-2022 x y
$$ | Answer: 2043231.
Solution. Substitute $x=y=0$, we get
$$
g(0)=2021(g(0)+g(0))-2022 \cdot 0 \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=2021(g(x)+g(x))-2022 \cdot x^{2} \Rightarrow g(x)=\frac{2022 x^{2}}{2 \cdot 2021}=\frac{1011 x^{2}}{2021} \Rightarrow \\
g(2021)=\frac{1011 \cdot 2021^{2... | 2043231 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the number of roots of the equation: $2^{\lg \left(x^{2}-2023\right)}-\lg 2^{x^{2}-2022}=0$. | # Answer: 4 roots.
Solution: Using the properties of logarithms, rewrite the equations in the following form
$$
\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. An equilateral triangle $M N K$ is inscribed in a circle. A point $F$ is taken on this circle. Prove that the value of $F M^{4}+F N^{4}+F K^{4}$ does not depend on the choice of point $F$.
# | # Solution:
Without loss of generality, we can assume that point \( \mathrm{M} \) lies on the arc \( M N \) of the circumscribed circle with center \( O \) and radius \( R \). Let \( \angle ... | 18 | Geometry | proof | Yes | Yes | olympiads | false |
2. The function $f$ satisfies the equation $(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$ for each value of $x$, not equal to 0 and 1. Find $f\left(\frac{2018}{2019}\right)$.
(7 points). | Answer: 2019.
## Solution:
Substitute $\frac{1}{x}$ for $x$ in the original equation. Together with the original equation, we get a system of linear equations in terms of $f(x)$ and $f\left(\frac{1}{x}\right)$.
$$
\left\{\begin{array}{l}
(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1} \\
\left(\frac{1}{x}-1\right... | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For any pair of numbers, a certain operation «*» is defined, satisfying the following properties: $a *(b * c)=(a * b) \cdot c$ and $a * a=1$, where the operation «$\cdot$» is the multiplication operation. Find the root $x$ of the equation: $\quad x * 2=2018$. | Answer: 4036.
## Solution:
Given the condition of the problem, we have $x * 1=x *(x * x)=(x * x) \cdot x=1 \cdot x=x$. Then
1) $(x * 2) \cdot 2=2018 \cdot 2=4036$,
2) $(x * 2) \cdot 2=x *(2 * 2)=x \cdot 1=x$.
Therefore, $x=4036$. | 4036 | 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 any pair of numbers, a certain operation «*» is defined, satisfying the following properties: $\quad a *(b * c)=(a * b) \cdot c$ and $a * a=1$, where the operation «$\cdot$» is the multiplication operation. Find the root $x$ of the equation: $\quad x * 3=2019$. | Answer: 6057.
Solution: a similar solution to this problem is present in variant 1 under the same number. | 6057 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. All passengers of the small cruise liner "Victory" can be accommodated in 7- and 11-seat lifeboats in case of emergency evacuation, with the number of 11-seat lifeboats being greater than the number of 7-seat lifeboats. If the number of 11-seat lifeboats is doubled, the total number of lifeboats will be more than 25... | Answer: 60 possible options are given in the table below.
## Solution:
Let $x, y$ be the number of 7-seater and 11-seater boats, respectively, and $z$ be the total number of passengers.
Then $z=7 x+11 y$, where $x, y$ satisfy the system of inequalities: $\left\{\begin{array}{c}2 y+x>25, \\ 2 x+yx\end{array}\right.$ ... | 159 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Usually, Nikita leaves home at 8:00 AM, gets into Uncle Vanya's car, who drives him to school by a certain time. But on Friday, Nikita left home at 7:10 and ran in the opposite direction. Uncle Vanya waited for him and at $8: 10$ drove after him, caught up with Nikita, turned around, and delivered him to school 20 m... | # Answer: 13 times.
Solution:
The car was on the road for 10 minutes longer than usual due to the 5 minutes spent catching up to Nikita and the 5 minutes spent returning home. The car caught up with Nikita at 8:15, and in 65 minutes (from 7:10 to 8:15), Nikita ran as far as the car traveled in 5 minutes, i.e., he spe... | 13 | Algebra | 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.
$$... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In isosceles triangle $A B C$ with base $A B$, the angle bisectors $C L$ and $A K$ are drawn. Find $\angle A C B$ of triangle $A B C$, given that $A K = 2 C L$. | Answer: $108^{\circ}$
## Solution:
 | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. For the quadratic trinomial $p(x)=(a+1) x^{2}-(a+1) x+2022$, it is known that $-2022 \leq p(x) \leq 2022$ for $x \in[0 ; 1]$. Find the greatest possible value of $a$. | # Answer: 16175.
Solution. Since $p(0)=p(1)=2022$, the graph of the quadratic trinomial is a parabola symmetric about the line $x=\frac{1}{2}$. From the conditions that $-2022 \leq$ $p(x) \leq 2022$ for $x \in[0 ; 1]$ and $p(0)=p(1)=2022$, it follows that the branches of the parabola are directed upwards. Then the min... | 16175 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three different real numbers satisfying the conditions:
$a^{3}-2022 a+1011=0, \quad b^{3}-2022 b+1011=0, \quad c^{3}-2022 c+1011=0$. | Answer: 2.
## Solution.
The cubic equation $t^{3}-2022 t+1011=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2022 t+1011: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\left\{\begin{array}{l}
a+b+c=0 \\
a b+b c+a c=-2022 \\
a b c=-1011
\end{array}\right.
$$
We find ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. For the quadratic trinomial $p(x)=(a-1) x^{2}-(a-1) x+2022$, it is known that $-2022 \leq p(x) \leq 2022$ for $x \in[0 ; 1]$. Find the greatest possible value of $a$. | Answer: 16177.
Solution. Since $p(0)=p(1)=2022$, the graph of the quadratic trinomial is a parabola symmetric about the line $x=\frac{1}{2}$. From the conditions that $-2022 \leq$ $p(x) \leq 2022$ for $x \in[0 ; 1]$ and $p(0)=p(1)=2022$, it follows that the branches of the parabola are directed upwards. Then the minim... | 16177 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three distinct real numbers satisfying the conditions:
$a^{3}-2020 a+1010=0, \quad b^{3}-2020 b+1010=0, c^{3}-2020 c+1010=0$. | Answer: 2.
Solution.
The cubic equation $t^{3}-2020 t+1010=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2020 t+1010: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\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 |
4. In the store "Everything for School," chalk is sold in packages of three grades: regular, unusual, and superior. Initially, the ratio of the quantities by grade was 3:4:6. As a result of sales and deliveries from the warehouse, this ratio changed to 2:5:8. It is known that the number of packages of superior chalk in... | # Answer: 260 packages.
Solution: Let $x$ be the initial number of packages of regular chalk, then the number of packages of unusual chalk is $\frac{4 x}{3}$. Since the latter number is an integer, then $x=3 n$, where $n \in N$. Therefore, the initial quantities of all three types of packages are $3 n, 4 n, 6 n$ respe... | 260 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the store "Everything for School," chalk is sold in packages of three grades: regular, unusual, and superior. Initially, the ratio of the quantities by grade was 2:3:6. After a certain number of packages of regular and unusual chalk, totaling no more than 100 packages, were delivered to the store, and 40% of the ... | # Answer: 24 packs.
Solution: a similar solution to this problem is present in Variant 1 under the same number. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Determine how many times the number $((2014)^{2^{2014}}-1)$ is larger than the number written in the following form: $\left.\left.\left((2014)^{2^{0}}+1\right) \cdot\left((2014)^{2^{1}}+1\right) \cdot\left((2014)^{2^{2}}+1\right) \cdot \ldots \quad \cdot\right)^{2^{2013}}+1\right)$. | Justify the solution.
Answer: 2013. | 2013 | Number Theory | 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. Find all prime numbers whose decimal representation has the form 101010 ... 101 (ones and zeros alternate). | Solution. Let $2n+1$ be the number of digits in the number $A=101010 \ldots 101$.
Let $q=10$ be the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2n}=\frac{q^{2n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2k \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k+1}-1}{q-1} \cdot \frac{q^{... | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The ordinary fraction $\frac{1}{221}$ is represented as a periodic decimal fraction. Find the length of the period. (For example, the length of the period of the fraction $\frac{25687}{99900}=0.25712712712 \ldots=0.25(712)$ is 3.) | Solution. Let's consider an example. We will convert the common fraction $\frac{34}{275}$ to a decimal. For this, we will perform long division (fig.).
As a result, we find $\frac{34}{275}=0.123636363 \ldots=0.123(63)$.
## Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Ma... | 48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. It is known that the lengths of the sides of a convex quadrilateral are respectively $a=4, b=5, c=6, d=7$. Find the radius $R$ of the circle circumscribed around this quadrilateral. Provide the integer part of $R^{2}$ as the answer. | Solution. By the cosine theorem, we express the length of the diagonal:
$$
l^{2}=a^{2}+b^{2}-2 a b \cos \gamma, l^{2}=c^{2}+d^{2}-2 c d \cos (\pi-\gamma)
$$
From this, we get $\cos \gamma=\frac{a^{2}+b^{2}-c^{2}-d^{2}}{2(a b+c d)}$. Since $R=\frac{l}{2 \sin \gamma^{\prime}}$, we obtain
$$
R^{2}=\frac{l^{2}}{4\left(1... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that the equation $x^{4}-8 x^{3}+a x^{2}+b x+16=0$ has (taking into account multiplicity) four positive roots. Find $a-b$. | Solution: Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of our equation (some of them may be the same). Therefore, the polynomial on the left side of the equation can be factored as:
$$
x^{4}-8 x^{3}+a x^{2}+b x+16=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)
$$
Expanding the brack... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. There is an unlimited number of test tubes of three types - A, B, and C. Each test tube contains one gram of a solution of the same substance. Test tubes of type A contain a $10\%$ solution of this substance, type B $-20\%$ solution, and type C $-90\%$ solution. Sequentially, one after another, the contents of the t... | Solution: Let the number of test tubes of types A, B, and C be \(a\), \(b\), and \(c\) respectively. According to the problem, \(0.1a + 0.2b + 0.9c = 0.2017 \cdot (a + b + c) \Leftrightarrow 1000 \cdot (a + 2b + 9c) = 2017 \cdot (a + b + c)\). The left side of the last equation is divisible by 1000, so the right side m... | 73 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $1^{2}+2^{2}+4^{2}=21$ ). | Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Co... | 5035485 | Number Theory | 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 $a=2, b=3$.
=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial $\quad$ of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has $\quad$ roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coe... | Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$: $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}$.
Then, according to the problem, we have:
$f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{... | -1216 | 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 $C E F$. | 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^{\cir... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$ | Solution: Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness of ... | 43 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{c}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
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 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 |
3. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$
# | # Solution:
Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness ... | 43 | 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 |
1. (2 points) A square of 90 by 90 is divided by two horizontal and two vertical lines into 9 rectangles. The sides of the central rectangle are 34 and 42. Find the total area of the four corner rectangles. | 1. 2688 .
+(2 points) - the solution is correct
-(0 points) - there are errors in the solution, including arithmetic errors | 2688 | Geometry | 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 |
1. Oleg has 550 rubles, and he wants to give his mother tulips for March 8, and there must be an odd number of them, and no color shade should be repeated. In the store where Oleg came, one tulip costs 49 rubles, and there are eleven shades of flowers available. How many ways are there for Oleg to give his mother flowe... | Solution. From the condition, it is obvious that the maximum number of flowers in a bouquet is 11.
1st method
Using the property of binomial coefficients
$$
\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{5}+\cdots=2^{n-1}
$$
and also considering their combinatorial meaning, we get that the number of ways to ... | 1024 | Combinatorics | 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 |
7. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$ | Solution. Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness of ... | 43 | Algebra | 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 |
1. A 100-digit number has the form $a=1777 \ldots 76$ (with 98 sevens in the middle). The number $\frac{1}{a}$ is represented as an infinite periodic decimal. Find its period. Justify your answer. | # Solution:
Notice that $a=16 \cdot 111 \ldots 11$. The last number $b$ consists of 99 ones. According to the rules for converting a common fraction to a decimal, the number $\frac{1}{b}=0,(00 \ldots 09)$. Its period is 99. Then, when multiplying this fraction by the number $\frac{1}{16}=0.0625$, the period will not c... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent ... | Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be t... | 36 | Combinatorics | 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}
$$
Assume that all ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the value of $f(2019)$, given that $f(x)$ simultaneously satisfies the following three conditions:
1) $f(x)>0$ for any $x>0$
2) $f(1)=1$;
3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}$ for any $a, b \in \mathbb{R}$. | Solution. In the identity given in the problem
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0
$$
Next, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Fina... | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In quadrilateral $A B C D$, the diagonals intersect at point $O$. It is known that $S_{A B O}=S_{C D O}=\frac{3}{2}$, $B C=3 \sqrt{2}$, $\cos \angle A D C=\frac{3}{\sqrt{10}}$. Find the smallest area that such a quadrilateral can have. | Solution. We will prove that quadrilateral $ABCD$ is a parallelogram. Let $x_{1}, x_{2}, y_{1}, y_{2}$ be the segments into which the diagonals are divided by their point of intersection. Denote the angle between the diagonals as $\alpha$. By the condition, the areas of triangles $ABO$ and $CDO$ are equal, that is, $\f... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. In a circle, three chords $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. The angular measures of the arcs $A C_{1}, A B, C A_{1}$, and $A_{1} B_{1}$ are $150^{\circ}, 30^{\circ}, 60^{\circ}$, and $30^{\circ}$, respectively. Find the angular measure of the arc $B_{1} C_{1}$. | Solution: Let's formulate several auxiliary statements.
1) Let the angular measure of the arc $AB$ (Fig.1) be $\varphi$. (This means that $\varphi$ is equal to the corresponding central angle $AOB$.) Then the length of the chord
=1-x-4 x^{2}+x^{4}$ and $g(x)=16-8 x-$ $16 x^{2}+x^{4}$ respectively. Find $\frac{x_{2}}{x_{1}}$. | # Solution:
Notice that $f(-2)>0, f(-1)<0, f(0)>0, f(1)<0$. Therefore, the polynomial $f(x)$ has 4 real roots. Similarly, from the inequalities $g(-4)>0, g(-2)<0, g(0)>0, g(2)<0$, it follows that the polynomial $g(x)$ has 4 real roots.
Comparison of the coefficients of the polynomials
$$
f(x)=1-x-4 x^{2}+x^{4} \text... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$
# | # Solution:
Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness ... | 43 | 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 |
1. Find the sum of all even natural numbers $n$ for which the number of divisors (including 1 and $n$ itself) is equal to $\frac{n}{2}$. (For example, the number 12 has 6 divisors: $1,2,3,4,6,12$.) | Solution. Let the canonical decomposition of the number $n$ be: $n=2^{t_{1}} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdots \cdot p^{t_{k}}$. Then the number of divisors of the number $n$ is $\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)$. From the condition of the problem, we h... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In how many ways can 4 numbers be chosen from the first 1000 natural numbers $1,2, \ldots, 1000$ to form an increasing arithmetic progression? | Solution. Let's find the formula for calculating the number of ways to choose 4 numbers from the first $n$ natural numbers $1,2, \ldots, n$ that form an increasing arithmetic progression. The number of progressions with a difference of 1 is $n-3$ (the first term of the progression can take values from 1 to $n-3$ inclus... | 166167 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. It is known that there exists a natural number $N$ such that $(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}$. Find $N$. | Solution. Suppose that raising the number $a+b \sqrt{3}$ to the power $N$ results in the number $A+B \sqrt{3}$ (where $a, b, A, B$ are integers). Expanding the expression $(a+b \sqrt{3})^{N}$, we get a sum of monomials (with non-essential (for us now) integer coefficients) of the form $a^{N-n}(b \sqrt{3})^{n}$. The ter... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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