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4. In the maze diagram in Fig. 7, 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, 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 | 8,421 |
5. Numbers $0,1,2,3 \ldots$ are placed at the nodes of a grid in a spiral pattern. Then, in the center of each cell, the sum of the numbers at its nodes is written (see Fig. 10). Is it true that in the centers of the cells, numbers divisible by 68 will appear infinitely many times? Is it true that the numbers in the ce... | Answer: Yes, it is correct.
Solution. The cells of the plane form a spiral consisting of straight "corridors" that turn at right angles.
Fig. 11: Moving along the spiral
 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 | 8,423 |
6. (30 points) In triangle $K I A$, where side $K I$ is less than side $K A$, the bisector of angle $K$ intersects side $IA$ at point $O$. Let $N$ be the midpoint of $IA$, and $H$ be the foot of the perpendicular dropped from vertex $I$ to segment $KO$. Line $IH$ intersects segment $K N$ at point $Q$. Prove that $O Q$ ... | Solution: Let $I Q$ intersect side $K A$ at point $V$. In triangle $K I V$, the height $K H$ is also the bisector, and therefore the median, so $I H = H V$. Therefore, $H N$ is the midline of triangle $I V A$ and $H N \| A K$.
Let $W$ be the point of intersection of $N H$ and $K I$. We get that $N W$ is the midline of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,425 |
7. (40 points) Among the pensioners of one of the stars in the Tau Ceti system, the following pastime is popular: a board with a grid of 2016 by 2017 is painted in gold and silver colors in a checkerboard pattern, after which the numbers 0 or 1 are written at the vertices of each cell in such a way that the sum of the ... | Answer: 0, 2 or 4.
Solution: By adding the sums of the numbers at the vertices of all cells, we get an even number, since the entire board consists of an even number of cells. Since all vertices, except for the corner ones, are counted an even number of times, and the corner ones are counted once, the sum of the numbe... | 0,2or4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,426 |
7. (30 points) In an acute isosceles triangle \(ABC (AB = BC)\), altitudes \(AD\) and \(CE\) are drawn, intersecting at point \(H\). The circumcircle of triangle \(ACH\) intersects segment \(BC\) at point \(F\). Prove that \(\angle ABH = \angle BFH\).
=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 | 8,431 |
6. (30 points) $S M$ is the angle bisector in triangle $S Q T$. Point $O$ on side $S T$ is such that angle $O Q T$ is equal to the sum of angles $Q T S$ and $Q S T$. Prove that $O M$ is the angle bisector of angle $Q O T$.
 Let $R E$ be the bisector of triangle $R S T$. Point $D$ on side $R S$ is such that $E D \| R T, F$ is the intersection point of $T D$ and $R E$. Prove that if $S D = R T$, then $T E = T F$.
 $S M$ is the angle bisector in triangle $S Q T$. Point $O$ on side $S T$ is such that angle $O Q T$ is equal to the sum of angles $Q T S$ and $Q S T$. Prove that $O M$ is the angle bisector of angle $Q O T$.
 Let $R E$ be the bisector of triangle $R S T$. Point $D$ on side $R S$ is such that $E D \| R T, F$ is the intersection point of $T D$ and $R E$. Prove that if $S D = R T$, then $T E = T F$.
 with base \(BC\). On the extension of side \(AC\) beyond point \(C\), point \(K\) is marked, and a circle is inscribed in triangle \(ABK\) with center at point \(I\). A circle passing through points \(B\) and \(I\) is tangent to line \(AB\) at point \(B\). This circle intersects s... | Solution. Let $I M$ and $I N$ be the perpendiculars dropped from point $I$ to $A K$ and $B K$ respectively. Note that $\angle A B I = \angle B L I$, since line $A B$ is tangent to the circumcircle of triangle $B L I$. Line $A I$ is the angle bisector of angle $A$, and $A B = A C$ by the given condition. Therefore, tria... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,436 |
4. The number 1200 is written on the board. Petl appended $10 n+2$ fives to it on the right, where $n-$ is a non-negative integer. Vasya thought that this was the base-6 representation of a natural number $x$, and factored $x$ into prime factors. It turned out that there were exactly two distinct primes among them. For... | Answer: $n=0$.
Solution. We agree to write the numbers in base 6 in parentheses to distinguish them from decimal numbers. Then
$$
\begin{aligned}
& x=(1200 \underbrace{55 \ldots 5}_{10 n+2})=(1201 \underbrace{00 \ldots 0}_{10 n+2})-1=289 \cdot 6^{10 n+2}-1= \\
& =\left(17 \cdot 6 \cdot 6^{5 n}-1\right)\left(17 \cdot ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,437 |
2. Kostya "combined" the addition and multiplication tables of natural numbers: he constructed a table whose rows and columns correspond to natural numbers, and filled in all the cells: in the cell at the intersection of the $r$-th row and the $s$-th column, he placed the number $r s + (s + r)$.
| | 1 | 2 | 3 | $\ldo... | Solution. If the number $n+1$ is not prime, then it can be factored into a product of two factors, both greater than 1. Let's write them as $r+1$ and $s+1$, where $r$ and $s$ are natural numbers. Then
$$
n=n+1-1=(r+1)(s+1)-1=r s+s+r
$$
which means the number $n$ is contained in the table. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,438 |
4. Given an isosceles triangle $A B C$ with base $B C$. Point $M$ is the midpoint of side $A C$, point $P$ is the midpoint of $A M$, and point $Q$ is marked on side $A B$ such that $A Q=3 B Q$. Prove that $B P+M Q>A C$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9b3c9b6b3121d7fd8c29g-3.jpg?height=436&width=1120&top... | Solution 1. By the triangle inequality
$$
M Q+M A>A Q \Longleftrightarrow M Q+\frac{1}{2} A C>\frac{3}{4} A B \Longleftrightarrow M Q>\frac{1}{4} A C=P A
$$
Then
$$
B P+M Q>B P+P A>A B=A C
$$
Solution 2. Let point $P^{\prime}$ be symmetric to $P$ with respect to the axis of symmetry of the triangle. Obviously, $B P... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,440 |
2. Kostya "combined" the addition table and the multiplication table of natural numbers: he constructed a table whose rows and columns correspond to natural numbers starting from 3, and filled in all the cells: in the cell at the intersection of the $r$-th row and $s$-th column, he placed the number $r s - (s + r)$.
!... | Solution. If the number $n+1$ is not prime, then it can be factored into a product of two factors, both greater than 1. Let's write them as $r-1$ and $s-1$, where $r$ and $s$ are natural numbers, $r, s \geqslant 3$. Then
$$
n=n+1-1=(r-1)(s-1)-1=r s-s-r
$$
which means the number $n$ is contained in the table. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,441 |
4. Given an isosceles triangle $ABC$ with base $BC$. Point $M$ is the midpoint of side $AB$, point $Q$ is the midpoint of $AM$, and point $P$ is marked on side $AC$ such that $AP = 3PC$. Prove that $PQ + CM > AB$.
$ for which the function
$$
f(x)=\frac{(a+5 b) x+a+b}{a x+b}
$$
is constant over its entire domain of definition.
(7 points) | Answer: $a= \pm \sqrt{5} b, b \neq 0$.
Solution:
First, note that when $a=b=0$, the function $f(x)$ is not defined for any value of $x$. If $a=0, b \neq 0$, then we get $f(x)=5 x+1$ and $f(x)$ is not a constant, so $a \neq 0$.
Now, suppose that for all $x$ in the domain $D(f)$ of the function $f(x)$, that is, for al... | \\sqrt{5}b,b\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,444 |
3. Solve the equation in integers:
$$
x^{2}-x y-2 y^{2}=7
$$ | Answer: $\{(3 ;-2),(5 ; 2),(-3 ; 2),(-5 ;-2)\}$
## Solution:
Let's factor the left side of the equation, for example, by grouping:
$x^{2}-x y-2 y^{2}=x^{2}-y^{2}-x y-y^{2}=(x+y)(x-y)-y(x+y)=(x+y)(x-2 y)$.
From this, we get the following form of the original equation:
$$
(x+y)(x-2 y)=7
$$
Considering that $x$ and ... | (3,-2),(5,2),(-3,2),(-5,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,445 |
4. From the town "U $" to the town "A x" at $10^{00}$ AM, Ivan set off on his bicycle. After traveling two-thirds of the way, he passed the town "Ox," from which at that moment Peter set off on foot towards the town "U x". At the moment Ivan arrived in the town "A x", Nikolai set off from there in the opposite directio... | Answer: 6 km.
## Solution:
We will solve the problem using a graphical-geometric method. Let's represent Ivan's movement as segment $K L$, Nikolai's movement as segment $L M$, and Petr's movement as segment $N P$ in a coordinate system $(t ; s)$, where $t$ is time in hours and $s$ is distance in kilometers from point... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,446 |
5. One side of a certain triangle is twice as large as another, and the perimeter of this triangle is 60. The largest side of the triangle, when added to four times the smallest side, equals 71. Find the sides of this triangle.
(7 points) | Answer: 11, 22, 27.
## Solution:
Let $a, b$, and $c$ be the sides of the triangle, without loss of generality, we assume that $a \leq b \leq c$. Considering the conditions of the problem, we write the system of equations:
$$
\left\{\begin{array}{c}
a+b+c=60 \\
4 a+c=71
\end{array}\right.
$$
Since one of the sides o... | 11,22,27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,447 |
1. Calculate:
$$
\frac{2 \cdot 2018}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+\cdots+2018}}
$$ | Answer: 2019.
Solution: a similar solution to this problem is present in variant 1 under the same number. | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,448 |
2. Find all pairs of numbers $(a, b)$ for which the function
$$
f(x)=\frac{(2 a+3 b) x+a+2 b}{a x+b}
$$
is constant over its entire domain of definition.
(7 points) | Answer: $a= \pm \sqrt{3} b, b \neq 0$.
Solution: a similar solution to this problem is present in Variant 1 under the same number. | \\sqrt{3}b,b\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,449 |
3. Solve the equation in integers: $\quad x^{2}-x y-2 y^{2}=7$. | Answer: $\{(3 ;-2),(5 ; 2),(-3 ; 2),(-5 ;-2)\}$
Solution: the solution to this problem is identical to the solution of the problem in variant 1 under the same number.
Note: 1 point for each correct solution found by trial and error. | {(3;-2),(5;2),(-3;2),(-5;-2)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,450 |
4. From the town "U ${ }^{\prime}$ " to the town " $A x$ ", Ivan set off on his bicycle at $11^{00}$ AM, having traveled two fifths of the distance, he passed the town " $O x$ ", from which at that moment Peter set off on foot towards the town "Ux". At the moment when Ivan arrived in the town " $A x$ ", from there in t... | Answer: 5 km.
Solution: a similar solution to this problem is present in option 1 under the same number. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,451 |
5. One side of a certain triangle is twice as large as another, and the perimeter of this triangle is 56. The quadrupled smallest side is 21 units longer than the largest side. Find the sides of this triangle. | Answer: 11, 22, 23.
Solution: A similar solution to this problem is present in Variant 1 under the same number.
Remark: 1 point for a correct solution found by trial and error.
Evaluation criteria are provided in the table:
| Points | $\quad$ Evaluation Criteria |
| :---: | :--- |
| $\mathbf{7}$ | Fully justified s... | 11,22,23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,452 |
1. Find all triples of real numbers $(u ; v ; w)$, satisfying the system of equations:
$$
\left\{\begin{array}{l}
u+v \cdot w=12 \\
v+w \cdot u=12 \\
w+u \cdot v=12
\end{array}\right.
$$ | Answer: $(\mathbf{3} ; \mathbf{3} ; \mathbf{3}),(-\mathbf{4} ;-\mathbf{4} ;-\mathbf{4}),(\mathbf{1} ; \mathbf{1} ; \mathbf{1 1}),(\mathbf{1} ; \mathbf{1 1} ; \mathbf{1}),(\mathbf{1 1} ; \mathbf{1} ; \mathbf{1})$.
Solution: Subtract the first equation from the second
$$
\begin{gathered}
(v-u)-w(v-u)=0 \\
(v-u)(1-w)=0
... | (3,3,3),(-4,-4,-4),(1,1,11),(1,11,1),(11,1,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,453 |
2. Three integers are written in a row $a, b, c$. Below these numbers, in the second row, the numbers $a-b, b-c, c-a$ are written. The numbers in the third row and subsequent rows are formed from the numbers of the previous row according to the same rule. Is it possible to encounter the number 2021 among the numbers in... | Answer: impossible.
Solution.
Let's write down several rows according to the condition of the problem:
| $a$ | $b$ | $c$ |
| :---: | :---: | :---: |
| $a-b$ | $b-c$ | $c-a$ |
| $a-2 b+c$ | $b-2 c+a$ | $c-2 a+b$ |
| $3(c-b)$ | $3(a-c)$ | $3(b-a)$ |
| $3(c-b)-3(a-c)$ | $3(a-c)-3(b-a)$ | $3(b-a)-3(c-b)$ |
| $\cdots$ | ... | impossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,454 |
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 | 8,455 |
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 | 8,456 |
5. In the convex polygon $M N K L P$, the segment $N L$ bisects each of the angles $K N P$ and $K L M$, and the segment $K P$ bisects each of the angles $M K L$ and $N P L$. The diagonal $N P$ intersects the diagonals $M K$ and $M L$ at points $F$ and $E$. Is it true that $K F = L E$? | Answer: Yes, correct.
## Solution.
Triangles $F P K$ and $L P K$ are congruent by side and the angles adjacent to it ( $P K$ - common). Therefore, $K F = L K$. Similarly,
$L E = L K$.
Th... | KF=LE | Geometry | proof | Yes | Yes | olympiads | false | 8,457 |
1. Find all triples of real numbers $(u ; v ; w)$, satisfying the system of equations:
$$
\left\{\begin{array}{l}
u+v \cdot w=20 \\
v+w \cdot u=20 \\
w+u \cdot v=20
\end{array}\right.
$$ | Answer: $(4 ; 4 ; 4),(-5 ;-5 ;-5),(1 ; 1 ; 19),(1 ; 19 ; 1),(19 ; 1 ; 1)$.
Solution: Subtract the first equation from the second
$$
\begin{gathered}
(v-u)-w(v-u)=0 \\
(v-u)(1-w)=0
\end{gathered}
$$
1) $v=u \Rightarrow\left\{\begin{array}{c}v+w \cdot v=20, \\ w+v^{2}=20 .\end{array} \Rightarrow(w-v)-v(w-v)=0 \Rightar... | (4;4;4),(-5;-5;-5),(1;1;19),(1;19;1),(19;1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,458 |
2. Three integers are written in a row $a, b, c$. Below these numbers, in the second row, the numbers $a-b, b-c, c-a$ are written. The numbers in the third row and subsequent rows are formed from the numbers of the previous row according to the same rule. Is it possible to encounter the number 2023 among the numbers in... | Answer: impossible.
Solution.
Let's write down several rows according to the condition of the problem:
| $a$ | $b$ | $c$ |
| :---: | :---: | :---: |
| $a-b$ | $b-c$ | $c-a$ |
| $a-2 b+c$ | $b-2 c+a$ | $c-2 a+b$ |
| $3(c-b)$ | $3(a-c)$ | $3(b-a)$ |
| $3(c-b)-3(a-c)$ | $3(a-c)-3(b-a)$ | $3(b-a)-3(c-b)$ |
| $\cdots$ | ... | impossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,459 |
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 | 8,460 |
5. In the convex polygon $M N K L P$, the segment $N L$ bisects each of the angles $K N P$ and $K L M$, and the segment $K P$ bisects each of the angles $M K L$ and $N P L$. The diagonal $N P$ intersects the diagonals $M K$ and $M L$ at points $F$ and $E$. Is it true that $K F > L E$? | # Answer: No, it is not correct.
Solution.
Triangles $F P K$ and $L P K$ are congruent by side and the angles adjacent to it ( $Р K-$ common). Therefore, $K F=L K$. Similarly,
$L E=L K$.
... | KF=LE | Geometry | proof | Yes | Yes | olympiads | false | 8,461 |
1. Solve the equation in natural numbers
$$
2 y^{2}-x y-x^{2}+2 y+7 x-84=0
$$ | Answer: $(1 ; 6),(14 ; 13)$.
Solution: The original equation can be represented as
$$
2 y^{2}-x y-x^{2}+2 y+7 x-12=72
$$
Next, we factorize the left side of the equation using the discriminant:
$$
(x+2 y-4)(y-x+3)=72
$$
Since $72=1 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$, we can reduce the number of cases to chec... | (1;6),(14;13) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,462 |
2. Given the sequence $x_{n}=1+2^{n}+3^{n}+4^{n}+5^{n}$. Is it possible to find five consecutive terms in this sequence, each of which is divisible by 2025? Explain your answer.
# | # Answer: No, it does not exist.
Solution: We will prove that when $n=4k, k \in \mathbb{N}$, the term of the sequence with number $n$ is not divisible by 5. Let's find
$$
x_{4k}=1+2^{4k}+3^{4k}+4^{4k}+5^{4k}=1+16^{k}+81^{k}+256^{k}+625^{k}
$$
The numbers $16^{k}, 81^{k}, 256^{k}$ can be represented as $(5m+1)^{k}$, ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,463 |
5. In a right triangle $ABC$ on the hypotenuse $AB$, a point $M$ is taken. From point $M$, two angle bisectors $MK$ and $MN$ of angles $BMC$ and $AMC$ are drawn respectively, with points $K$ and $N$ lying on the legs and $CM = KN$. Prove that point $M$ is the midpoint of the hypotenuse $AB$. | # Solution:
Since $M K$ and $M N$ are the bisectors of adjacent angles $B M C$ and $A M C$, the angle $N M K$ is a right angle. Therefore, points $C$ and $M$ lie on a circle with diameter $K... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,464 |
1. Find all $x$ for which $2[x]+\{3 x\}=\frac{7}{3}$, where $[x]$ is the integer part of $x$, and $\{x\}$ is the fractional part of $x$, that is, $\{x\}=x-[x]$. | Answer: $\left\{1 \frac{1}{9} ; 1 \frac{4}{9} ; 1 \frac{7}{9}\right\}$.
## Solution:
From the equation and definitions, it follows that $[x]=1$, and $\{3 x\}=\frac{1}{3}$.
Consider the equation $\{3 x\}=\frac{1}{3}$:
1) if $0 \leq\{x\}<\frac{1}{3}$, then $\{3 x\}=3\{x\} \Rightarrow\{x\}=\frac{1}{9}$;
2) if $\frac{1... | {1\frac{1}{9};1\frac{4}{9};1\frac{7}{9}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,465 |
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 | 8,466 |
3. Regarding the quadratic trinomial $f(x)$, it is known that it has two distinct roots and satisfies the condition $f\left(x^{2}+y^{2}\right) \geq f(2 x y)$ for any $x$ and $y$. Is it possible for at least one of the roots of $f(x)$ to be negative? | # Answer: No.
## Solution: (proof by contradiction)
Assume that both roots of the quadratic trinomial $f(x)$ are non-negative. Let $y=0$ and substitute into the inequality, thus we obtain that
$$
f\left(x^{2}\right) \geq f(0) \text { for any } x \text { . }
$$
Then $f(t) \geq f(0)$ for any $t \geq 0$. Note that thi... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,467 |
5. A student constructed a quadrilateral $M N K L$ and measured the distances from the vertices to a point $P$, which the teacher indicated. It turned out that $M P^{2}+N P^{2}+K P^{2}+L P^{2}=2 S$, where $S-$ is the area of the quadrilateral. What kind of quadrilateral did the student construct, and what point did the... | Answer: square, point $P$ is the intersection point of the diagonals.
## Solution:
Let $M P=a, N P=b, K P=c, L P=d$, the lengths of the diagonals be denoted by $d_{1}$ and $d_{2}$, and $\alpha$ be the angle between the diagonals. Then $\mathrm{S}=\frac{1}{2} d_{1} \cdot d_{2} \cdot \sin \alpha \leq \frac{1}{2} d_{1} ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,468 |
1. Find all $x$ for which $x^{2}-10[x]+9=0$, where $[x]$ is the integer part of $x$. | Answer: $\{1 ; \sqrt{61} ; \sqrt{71} ; 9\}$.
## Solution:
Estimate the left side of the equation from below:
$$
x^{2}-10 x+9 \leq x^{2}-10[x]+9=0
$$
Therefore, $1 \leq x \leq 9$. On the other hand, $x^{2}+9=10[x]$, which means $x^{2}+9$ is an integer and divisible by 10. By checking the values of $x^{2}+9$ from 10 ... | {1;\sqrt{61};\sqrt{71};9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,469 |
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 | 8,470 |
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 | 8,471 |
5. A student constructed a quadrilateral $M N K L$ and measured the distances from the vertices to a point $Q$, which the teacher indicated. It turned out that $M Q^{2}+N Q^{2}+K Q^{2}+L Q^{2}=2 S$, where $S-$ is the area of the quadrilateral. What kind of quadrilateral did the student construct, and what point did the... | # Answer: square, point $-Q$ is the intersection point of the diagonals.
## Solution:
Let $M Q=a, N Q=b, K Q=c, L Q=d$, the lengths of the diagonals be denoted by $d_{1}$ and $d_{2}$, and $\alpha$ be the angle between the diagonals. Then $\mathrm{S}=\frac{1}{2} d_{1} \cdot d_{2} \cdot \sin \alpha \leq \frac{1}{2} d_{... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,472 |
3. Solve the equation $\sqrt{6 \cos 4 x+15 \sin 2 x}=2 \cos 2 x$. | (10 points) Answer: $-\frac{1}{2} \arcsin \frac{1}{8}+\pi n$ | -\frac{1}{2}\arcsin\frac{1}{8}+\pin | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,473 |
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
$$
C D^{2}=D K^{2}+K C^{2}=x^{2}+1, A B^{2... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,474 |
6. Can the number $\left(x^{2}+3 x+1\right)^{2}+\left(y^{2}+3 y+1\right)^{2}$ be a perfect square for some integers $x$ and $y$
(15 points) | Solution. Since $x^{2}+3 x+1=x(x+3)+1$, for any integers $x$ and $y$, the value of each expression in parentheses is an odd number. The square of an odd number, when divided by 4, gives a remainder of 1, because $(2 n+1)^{2}=4 n^{2}+4 n+1=4 n(n+1)+1$. Therefore, the value of the given expression is an even number and g... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,475 |
3. Solve the equation
$$
\sqrt{-9 \cos 2 x-19 \cos x}=-2 \sqrt{3} \sin x
$$ | (10 points) Answer: $\pi+\arccos \frac{1}{6}+2 \pi n, n \in Z$. | \pi+\arccos\frac{1}{6}+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,476 |
5. In an isosceles trapezoid $A B C D$ with lateral sides $A B$ and $C D$, the lengths of which are 10, perpendiculars $B H$ and $D K$ are drawn from vertices $B$ and $D$ to the diagonal $A C$. It is known that the bases of the perpendiculars lie on segment $A C$ and $A H: A K: A C=5: 14: 15$. Find the area of trapezoi... | Solution. Let $x = BH$, $y = DK$. From the similarity of right triangles $DKA$ and $BHC$, since $\angle BHC = \angle DAK$, we get
$$
\frac{x}{y} = \frac{CH}{AK} = \frac{10}{14} = \frac{5}{7}, \quad 5y = 7x, \quad y = \frac{7x}{5}.
$$
By the condition $AH : AK : AC = 5 : 14 : 15$, therefore $AH : CK = 5 : 1$. By the P... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,477 |
1. Do there exist integers $x$ and $y$ such that
$(x+2019)(x+2020)+(x+2020)(x+2021)+(x+2019)(x+2021)=y^{2} ?$ | Answer: No, they do not exist.
Solution. $\quad$ Let $k=x+2019$, then the equation will take the form
$$
k(k+1)+(k+1)(k+2)+k(k+2)=y^{2} \quad \text { or } \quad 3 k^{2}+6 k+2=y^{2} .
$$
The left side of the equation, when divided by 3, gives a remainder of 2, while the square of an integer, when divided by 3, can gi... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,478 |
2. In a sports store, over two days, thirteen pairs of sneakers, two sports suits, and one T-shirt were sold, with the same amount of money earned on the first day as on the second day (from the sale of the aforementioned items). One pair of sneakers is cheaper than a sports suit and more expensive than a T-shirt by th... | Answer: 8 pairs of sneakers and no sports suits.
Solution. Let in one day $x$ suits and $y$ pairs of sneakers were sold with a T-shirt. Then in the other day, $(2-x)$ suits and $(13-y)$ pairs of sneakers were sold.
Let $c$ be the price of one pair of sneakers, and $s$ be the price difference.
Then, from the conditio... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,479 |
3. Find $g(2022)$, if for any real $x, y$ the equality holds
$$
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 | 8,480 |
5. In an isosceles triangle $M N K$, the sides $M N = N K = 8$, and $M K = 4$. A point $F$ is chosen on side $N K$ such that the circles inscribed in triangles $M N F$ and $M K F$ touch each other. Find the areas of triangles $M N F$ and $M K F$. | Answer: $S_{M N F}=3 \sqrt{15}, S_{M K F}=\sqrt{15}$.
## Solution.
Let $R, L, E, P, Q$ be the points of tangency, and $x, y, u, v$ be the lengths of the tangent segments as indicated in the diagram.
## Then
$$
\left\{\begin{array}{c}
\begin{array}{r}
x+y=8 \\
y+u=4, \\
x+2 v+u=8
\end{array} \Rightarrow\left\{\begin... | S_{MNF}=3\sqrt{15},S_{MKF}=\sqrt{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,481 |
1. Do there exist integers $x$ and $y$ such that
$(x+2020)(x+2021)+(x+2021)(x+2022)+(x+2020)(x+2022)=y^{2} ?$ | Answer: No, they do not exist.
Solution. $\quad$ Let $k=x+2020$, then the equation will take the form
$$
k(k+1)+(k+1)(k+2)+k(k+2)=y^{2} \text { or } 3 k^{2}+6 k+2=y^{2} .
$$
The left side of the equation, when divided by 3, gives a remainder of 2, while the square of an integer, when divided by 3, can give a remaind... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,482 |
2. Misha invited eighteen friends from his sports club and two of his brothers to celebrate his birthday, a total of twenty guests. All the guests and Misha himself, seated at two tables, ate all the hot dogs served equally on both tables, and everyone ate only from their own table. Each friend from the sports club ate... | # Answer: 9 friends from the sports club and no brothers.
Solution. Let $x$ be the number of brothers and $y$ be the number of friends from the sports club sitting at the same table with Misha. Then, at the other table, there were $2-x$ brothers and $18-y$ friends from the sports club.
Let $c$ be the number of hot do... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,483 |
3. Find $g(2021)$, if for any real $x, y$ the equality holds
$$
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 | 8,484 |
5. In an isosceles triangle $M N K$, the sides $M N = N K = 12$, and $M K = 8$. A point $F$ is chosen on side $N K$ such that the circles inscribed in triangles $M N F$ and $M K F$ touch each other. Find the areas of triangles $M N F$ and $M K F$. | Answer: $S_{M N F}=\frac{32 \sqrt{2}}{3}, S_{M K F}=\frac{64 \sqrt{2}}{3}$.
## Solution.
Let $R, L, E, P, Q$ be the points of tangency, and $x, y, u, v$ be the lengths of the tangent segments as indicated in the diagram.
Then
$$
\left\{\begin{array} { c }
{ x + y = 1 2 , } \\
{ y + u = 8 } \\
{ x + 2 v + u = 1 2 .... | S_{MNF}=\frac{32\sqrt{2}}{3},S_{MKF}=\frac{64\sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,485 |
1. Solve the equation in integers
$$
2 x^{2}+2 x^{2} z^{2}+z^{2}+7 y^{2}-42 y+33=0
$$ | Answer: $(\mathbf{1} ; \mathbf{5} ; \mathbf{0}),(-\mathbf{1} ; \mathbf{5} ; \mathbf{0}),(\mathbf{1} ; \mathbf{1} ; \mathbf{0}),(-\mathbf{1} ; \mathbf{1} ; \mathbf{0})$.
Solution: Transform the equation to the form:
$$
2 x^{2}+2 x^{2} z^{2}+z^{2}+7(y-3)^{2}=30
$$
There are 3 cases:
1) $|y-3|=0$, 2) $|y-3|=1$, 3) $|y... | (1;5;0),(-1;5;0),(1;1;0),(-1;1;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,486 |
2. Find the number of roots of the equation: $2^{\lg \left(x^{2}-2023\right)}-\lg 2^{x^{2}-2022}=0$. | # Answer: 4 roots.
Solution: Using the properties of logarithms, rewrite the equations in the following form
$$
\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 | 8,487 |
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 | 8,488 |
1. Find all solutions to the equation $(x-|x|)^{2}+x+|x|=2020$. | Answer: $\{1010 ;-\sqrt{505}\}$.
## Solution:
1) Let $x \geq 0$, then $(x-x)^{2}+x+x=2020$, we get $x=1010$.
2) Let $x<0$, then $(x+x)^{2}+x-x=2020$, we get $4 x^{2}=2020, x= \pm \sqrt{505}$, but since $x<0$, then $x=-\sqrt{505}$. | 1010;-\sqrt{505} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,489 |
2. It is known that a two-digit number, when divided by 4, gives a remainder of 3, and when divided by 3, gives a remainder of 2. Find all such numbers. | Answer: $\{11 ; 23 ; 35 ; 47 ; 59 ; 71 ; 83 ; 95\}$.
## Solution:
Let the desired number be $A$, then we have $A=4 m+3=3 n+2$, from which we obtain the linear Diophantine equation $3 n-4 m=1$. Easily selecting a particular solution $m=n=-1$, we get the general solution in the form
$$
\left\{\begin{array}{l}
m=-1+3 t... | {11;23;35;47;59;71;83;95} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,490 |
5. An internal point $P$ of an acute-angled triangle $A B C$ satisfies the condition
$$
A B^{2}+P C^{2}=B C^{2}+A P^{2}=A C^{2}+B P^{2}
$$
What is the point $P$ for triangle $A B C$? | # Answer: the intersection point of the altitudes of triangle $A B C$.
## Solution:
Draw the perpendicular $P H$ to side $A C$ and the altitude $B K$.
By the Pythagorean theorem, $A B^{2}-A K^{2}=B C^{2}-C K^{2}$ or $A B^{2}-B C^{2}=A K^{2}-C K^{2}$. But according to the condition, $A B^{2}-B C^{2}=A P^{2}-P C^{2}$.... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,492 |
1. Find all solutions to the equation $(x-|x|)^{2}+2020(x+|x|)=2020$. | Answer: $\{0.5 ;-\sqrt{505}\}$.
## Solution:
1) Let $x \geq 0$, then $(x-x)^{2}+2020(x+x)=2020$, we get $x=0.5$.
2) Let $x<0, \quad$ then $(x+x)^{2}+2020(x-x)=2020$, we get $4 x^{2}=2020, \quad x=$ $\pm \sqrt{505}$, but since $x<0$, then $x=-\sqrt{505}$. | 0.5;-\sqrt{505} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,493 |
2. It is known that a two-digit number, when divided by 3, gives a remainder of 1, and when divided by 5, gives a remainder of 3. Find all such numbers. | Answer: $\{13 ; 23 ; 43 ; 58 ; 73 ; 88\}$.
## Solution:
Let the required number be $A$, then we have $A=3 m+1=5 n+3$, from which we obtain the linear Diophantine equation $3 m-5 n=2$. Easily selecting a particular solution $m=n=-1$, we get the general solution in the form
$$
\left\{\begin{array}{l}
m=-1+5 t, \\
n=-1... | {13;23;43;58;73;88} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,494 |
3. The coefficients of the quadratic trinomials $f(x)=x^{2}+m x+n$ and $p(x)=x^{2}+k x+l$ satisfy the condition $k>m>n>l>0$. Is it possible for $f(x)$ and $g(x)$ to have a common root
# | # Answer: No.
## Solution: (by contradiction)
Assume that $f(x)$ and $g(x)$ have a common root $x_{0}$. Since all coefficients of the polynomials are positive, all roots (if they exist) are negative. $\Rightarrow$ $x_{0}m>n>l>0$, we get that $m-k0$. We have reached a contradiction. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,495 |
5. An internal point $Q$ of an acute-angled triangle $M N K$ satisfies the condition
$$
M N^{2}+Q K^{2}=N K^{2}+M Q^{2}=M K^{2}+N Q^{2}
$$
What is the point $Q$ for the triangle $M N K$? | # Answer: the intersection point of the altitudes of triangle $M N K$.
## Solution:
Draw the perpendicular $Q H$ to side $M K$ and the altitude $B L$.
By the Pythagorean theorem, $M N^{2}-M L^{2}=N K^{2}-K L^{2}$ or $M N^{2}-N K^{2}=M L^{2}-K L^{2}$. But by the condition, $M N^{2}-N K^{2}=M Q^{2}-Q K^{2}$.
On the o... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,496 |
1. Solve the equation in integers
$$
y^{2}(y-x+2)-y(x+4)+5 x+7=0
$$ | Answer: $(15 ; 2),(10 ;-3),(-2 ; 1),(-5 ;-2)$.
Solution: Express $x$ from this equation:
$$
x=\frac{y^{3}+2 y^{2}-4 y+7}{y^{2}+y-5}=y+1+\frac{12}{y^{2}+y-5}
$$
Therefore, the number $y^{2}+y-5$ is a divisor of the number 12 and at the same time $x, y \in \mathbb{Z}$.
For the number 12, the divisors are $\pm 1, \pm ... | (15,2),(10,-3),(-2,1),(-5,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,497 |
5. Triangle $M N K$ is inscribed in a circle of radius $R$. The bisectors of the internal and external angles at $K$ intersect the line $M N$ at points $E$ and $F$ respectively, and $K E=K F$. Prove that $M K^{2}+N K^{2}=4 R^{2}$. | # Solution:
Let points $M, E, N, F$ be located on the line $MN$ in the given order (the case of $F, M, E, N$ is considered similarly), then $\angle E K F=90^{\circ}$ and $\angle K E F=45^{\c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,498 |
1. Find the sum of the numbers:
$$
6+66+666+6666+\cdots+\underbrace{66666 \ldots 6}_{2018 \text { of them }}
$$ | Answer: $\frac{2\left(10^{2019}-18172\right)}{27}$.
## Solution:
$$
\begin{gathered}
6+66+666+6666+\cdots+\underbrace{66666 \ldots 6}_{2018 \text { items }}=\frac{2}{3}(9+99+999+9999+\cdots+\underbrace{99999 \ldots 9}_{2018 \text { items }})= \\
=\frac{2}{3}\left(10-1+10^{2}-1+10^{3}-1+\cdots+10^{2018}-1\right)=\frac... | \frac{2(10^{2019}-18172)}{27} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,499 |
2. The function $f$ satisfies the equation $(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$ for each value of $x$, not equal to 0 and 1. Find $f\left(\frac{2018}{2019}\right)$.
(7 points). | Answer: 2019.
## Solution:
Substitute $\frac{1}{x}$ for $x$ in the original equation. Together with the original equation, we get a system of linear equations in terms of $f(x)$ and $f\left(\frac{1}{x}\right)$.
$$
\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 | 8,500 |
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 | 8,501 |
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 | 8,503 |
1. Find the sum of the numbers:
$$
3+33+333+3333+\cdots+\underbrace{33333 \ldots 3}_{2018 \text { of them }} .
$$ | Answer: $\frac{10^{2019}-18172}{27}$.
Solution: a similar solution to this problem is present in Variant 1 under the same number. | \frac{10^{2019}-18172}{27} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,504 |
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 | 8,506 |
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 | 8,507 |
5. The angle between the lateral sides $A B$ and $C D$ of trapezoid $A B C D$ is $30^{\circ}$. A circle is inscribed in the trapezoid, and the side $A B$ is divided by the point of tangency into segments of length $\sqrt{3}$ and $3 \sqrt{3}$. Determine the length of the lateral side $C D$ of the trapezoid. | Answer: 6 or 12.
Solution: A similar solution to this problem is present in Variant 1 under the same number.
## Evaluation criteria are provided in the table:
| Points | Evaluation Criteria |
| :---: | :--- |
| **7** | Fully justified solution. |
| **6** | Justified solution with minor flaws. |
| **5 - 6** | The sol... | 6or12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,508 |
1. Find all $x$ for which $[x]+\{2 x\}=2.5$, where $[x]$ is the integer part of the number $x$, and $\{x\}$ is the fractional part of the number $x$, that is, $\{x\}=x-[x]$. | Answer: $\{2.25 ; 2.75\}$.
Solution:
From the equation and definitions, it follows that $[x]=2$, and $\{2 x\}=0.5$.
Consider the equation $\{2 x\}=0.5$:
1) if $0 \leq\{x\}<\frac{1}{2}$, then $\{2 x\}=2\{x\} \Rightarrow\{x\}=0.25$.
2) if $\frac{1}{2} \leq\{x\}<1$, then $\{2 x\}=2\{x\}-1 \Rightarrow\{x\}=0.75$.
Sinc... | 2.25;2.75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,509 |
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 | 8,510 |
3. For the quadratic polynomial $g(x)=m x^{2}+n x+k$, it is known that the values $g(k)$ and $g\left(\frac{1}{m}\right)$ have different signs. Can the roots of the polynomial $g(x)$ have the same sign?
# | # Answer: No.
## Solution:
According to the condition $g(k) \cdot g\left(\frac{1}{m}\right)<0$, on the other hand, we have
$$
g(k) \cdot g\left(\frac{1}{m}\right)=\left(m k^{2}+n k+k\right)\left(m \frac{1}{m^{2}}+n \frac{1}{m} m+k\right)=\frac{k}{m}(m k+n+1)^{2}
$$
Therefore, $\frac{k}{m}<0$, and by Vieta's theorem... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,511 |
5. In an isosceles trapezoid $M N K L$ with bases $M L$, $N K$ the diagonals are perpendicular to the sides $M N, \quad K L$ and intersect at an angle of $22.5^{\circ}$. Find the height of the trapezoid if the length $N Q=3$, where $Q-$ is the midpoint of the larger base. | Answer: $\frac{3 \sqrt{2-\sqrt{2}}}{2}\left(3 \sin 22.5^{\circ}\right)$.
## Solution:
Let $M L$ be the larger base of the trapezoid $M N K L$.
Consider the triangle $M N L: \angle M N L=90^{\circ}, Q$ is the midpoint of $M L$ (by condition) $\Rightarrow$ $Q$ is the midpoint of the hypotenuse $M L \Rightarrow N Q=M Q... | \frac{3\sqrt{2-\sqrt{2}}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,512 |
1. Find all $x$ for which $\left[\frac{8 x+19}{7}\right]=\frac{16(x+1)}{11}$, where $[t]$ is the integer part of $t$. Answer: $\left\{1 \frac{1}{16} ; 1 \frac{3}{4} ; 2 \frac{7}{16} ; 3 \frac{1}{8} ; 3 \frac{13}{16}\right\}$. | # Solution:
Let $t=\frac{16(x+1)}{11}$, where $t \in Z$ and express $x=\frac{11 t-16}{16}$. Then the equation becomes: $\left[\frac{8 \frac{11 t-16}{16}+19}{7}\right]=t$ or $\left[\frac{11 t+22}{14}\right]=t$. Therefore, $0 \leq \frac{11 t+22}{14}-t\frac{8}{3}, \\ t \leq \frac{22}{3},\end{array} \stackrel{t \in Z}{\Ri... | x_{1}=1\frac{1}{16},x_{2}=1\frac{3}{4},x_{3}=2\frac{7}{16},x_{4}=3\frac{1}{8},x_{5}=3\frac{13}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,513 |
2. Usually, Dima leaves home at $8:10$ AM, gets into Uncle Vanya's car, who delivers him to school by a certain time. But on Thursday, Dima left home at 7:20 and ran in the opposite direction. Uncle Vanya waited for him and at $8:20$ drove after him, caught up with Dima, turned around, and delivered him to school 26 mi... | # Answer: 8.5 times.
## Solution:
The car was on the road for 16 minutes longer than usual, due to spending 8 minutes catching up to Dima and 8 minutes returning home. The car caught up with Dima at 8:28, and during the 68 minutes from 7:20 to 8:28, Dima ran the same distance that the car traveled in 8 minutes, i.e.,... | 8.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,514 |
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 | 8,515 |
5. In an isosceles trapezoid $M N K L$ with bases $M L, N K$, the diagonals are perpendicular to the sides $M N$, $K L$ and intersect at an angle of $15^{\circ}$. Find the height of the trapezoid if the length $N Q=5$, where $Q-$ is the midpoint of the larger base. | Answer: $\frac{5 \sqrt{2-\sqrt{3}}}{2}\left(5 \sin 15^{\circ}\right)$.
## Solution:
Let $M L-$ be the larger base of the trapezoid $M N K L$.
Consider the triangle $M N L: \angle M N L=90^{\circ}, Q$ - the midpoint of $M L$ (by condition) $\Rightarrow$ $Q$ is the midpoint of the hypotenuse $M L \Rightarrow N Q=M Q=Q... | \frac{5\sqrt{2-\sqrt{3}}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,516 |
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 | 8,517 |
2. For each value of the parameter $k$, solve the equation
$$
4-\sin ^{2} x+\cos 4 x+\cos 2 x+2 \sin 3 x \cdot \sin 7 x-\cos ^{2} 7 x=\cos ^{2}\left(\frac{\pi k}{2021}\right)
$$ | Answer: $P$ When $k=2021 \cdot m, m \in Z \quad x=\frac{\pi}{4}+\frac{\pi n}{2}, n \in Z$;
for other values of the parameter $k$ there are no solutions.
## Solution.
The original equation is equivalent to the following equation:
$$
\begin{gathered}
(\sin 3 x+\sin 7 x)^{2}+(\cos 3 x+\cos x)^{2}+\sin ^{2}\left(\frac{... | \frac{\pi}{4}+\frac{\pin}{2},n\inZwhenk=2021\cdot,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,518 |
3. Let $p(x)=x^{2}+3 x+2$. Compute the product
$$
\left(1-\frac{2}{p(1)}\right) \cdot\left(1-\frac{2}{p(2)}\right) \cdot\left(1-\frac{2}{p(3)}\right) \cdots\left(1-\frac{2}{p(2021)}\right)
$$ | Answer: $\frac{1012}{3033}$.
Solution.
Consider the $n-$th factor in the given product
$$
1-\frac{2}{p(n)}=\frac{p(n)-2}{p(n)}=\frac{n^{2}+3 n}{n^{2}+3 n+2}=\frac{n(n+3)}{(n+1)(n+2)}
$$
## Compute
$$
\left(1-\frac{2}{p(1)}\right) \cdot\left(1-\frac{2}{p(2)}\right) \cdot\left(1-\frac{2}{p(3)}\right) \cdots\left(1-\... | \frac{1012}{3033} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,519 |
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 | 8,520 |
5. The base of the quadrilateral pyramid $S M N K L$ is a rectangle $M N K L$. The lengths of four edges of the pyramid are known: $M N=5, N K=2, S M=3, S N=4$. Determine the values of the lengths of the remaining two edges $S K$ and $S L$ for which the volume of the pyramid is maximized, and calculate this volume. | Answer: $V_{\max }=8, S K=2 \sqrt{5}, S L=\sqrt{13}$.
## Solution.
Drop perpendiculars $S P$ from point $S$ to edge $M N$ and $S H$ to the plane of the base $M N K L$.
All sides of triangle $S M N$ are known, so the length of $S P$ is uniquely determined. The base of the pyramid is fixed, so the volume reaches its m... | V_{\max}=8,SK=2\sqrt{5},SL=\sqrt{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,521 |
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 | 8,522 |
2. For each value of the parameter $k$, solve the equation
$$
2+\cos ^{2} x+\cos 4 x+\cos 2 x+2 \sin 3 x \cdot \sin 7 x+\sin ^{2} 7 x=\cos ^{2}\left(\frac{\pi k}{2022}\right)
$$ | Answer: $P r i=2022 \cdot m, m \in Z \quad x=\frac{\pi}{4}+\frac{\pi n}{2}, n \in Z$;
for other values of the parameter $k$ there are no solutions.
## Solution.
The original equation is equivalent to the following equation:
$$
\begin{gathered}
(\sin 3 x+\sin 7 x)^{2}+(\cos 3 x+\cos x)^{2}+\sin ^{2}\left(\frac{\pi k... | \frac{\pi}{4}+\frac{\pin}{2},n\inZ\quadfor\quadk=2022\cdot,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,523 |
3. Let $p(x)=x^{2}+3 x+2$. Compute the product
$$
\left(1-\frac{2}{p(1)}\right) \cdot\left(1-\frac{2}{p(2)}\right) \cdot\left(1-\frac{2}{p(3)}\right) \cdots\left(1-\frac{2}{p(2022)}\right)
$$ | Answer: $\frac{675}{2023}$
Solution.
$$
1-\frac{2}{p(n)}=\frac{p(n)-2}{p(n)}=\frac{n^{2}+3 n}{n^{2}+3 n+2}=\frac{n(n+3)}{(n+1)(n+2)}
$$
## Calculate
$$
\left(1-\frac{2}{p(1)}\right) \cdot\... | \frac{675}{2023} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,524 |
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 | 8,525 |
5. At the base of the quadrilateral pyramid $S M N K L$ lies a rhombus $M N K L$ with a side length of 4 and an acute angle $N M K$ of $60^{\circ}$. It is known that $S M=2, S N=4$. Determine the values of the lengths of the remaining two edges $S K$ and $S L$ for which the volume of the pyramid reaches its maximum, an... | Answer: $V_{\max }=4 \sqrt{5}, S K=3 \sqrt{2}, S L=\sqrt{46}$.
## Solution.
Drop perpendiculars $S P$ from point $S$ to edge $M N$ and $S H$ to the plane of the base $M N K L$.
All sides of triangle $S M N$ are known, so the length of $S P$ is uniquely determined. The base of the pyramid is fixed, so the volume reac... | V_{\max}=4\sqrt{5},SK=3\sqrt{2},SL=\sqrt{46} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,526 |
1. Solve the equation in integers
$$
2 y^{2}-2 x y+x+9 y-2=0
$$ | Answer: $(9 ; 1),(2 ; 0),(8 ; 2),(3 ;-1)$.
Solution: Express $x$ from this equation:
$$
x=\frac{2 y^{2}+9 y-2}{2 y-1}=y+5+\frac{3}{2 y-1} .
$$
Therefore, the number $2 y-1$ is a divisor of the number 3:
$$
\left[\begin{array} { c }
{ 2 y - 1 = 1 } \\
{ 2 y - 1 = - 1 , } \\
{ 2 y - 1 = 3 } \\
{ 2 y - 1 = - 3 }
\end... | (9;1),(2;0),(8;2),(3;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,527 |
2. Masha and Vanya found a pack of 11-ruble bills each along the road and decided to go to the store. In the store, Vanya bought 3 chocolates, 4 sodas, and 5 packs of cookies. Masha bought 9 chocolates, 1 soda, and 4 packs of cookies. A chocolate, a soda, and a pack of cookies each cost a whole number of rubles. Vanya ... | # Answer: Yes, she could.
Solution: Let's introduce the following variables: $\boldsymbol{m}$ - the cost of $\mathbf{1}$ chocolate bar,
$$
\boldsymbol{n} \text { - the cost of } \mathbf{1} \text { soda, } \boldsymbol{k} \text { - the cost of } \mathbf{1} \text { pack of cookies. }
$$
Since Vanya was able to pay with... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,528 |
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 | 8,529 |
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