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# Task 7.1
A sheet of paper has the shape of a circle. Is it possible to draw five segments on it, each connecting two points on the boundary of the sheet, such that among the parts into which these segments divide the sheet, there is a pentagon and two quadrilaterals?
## Number of points 7 Answer:
## possible
# | # Solution
For example, see the figure. Two quadrilaterals and a pentagon are shaded gray.
# | possible | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,755 |
# Task 7.2
There are 30 logs, the lengths of which are 3 or 4 meters, and their total length is one hundred meters. How many cuts are needed to saw all these logs into pieces 1 meter long? (Each cut saws exactly one log).
Points 7
# | # Answer:
70
## Solution
## First Method
The total length of the logs is 100 meters. If it were a single log, 99 cuts would be needed. Since there are 30 logs, 29 cuts have already been made. Therefore, another $99-29=70$ cuts are needed.
## Second Method
Let's find the number of logs of each type. If all the log... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,756 |
# Task 7.3
Does the rebus АПЕЛЬСИН - СПАНИЕЛЬ $=2018 \cdot 2019$ have a solution?
Number of points 7 | Answer:
does not have
## Solution
The numbers ORANGE and SPANIEL consist of the same digits, which means their difference is divisible by 9. However, $2018 \cdot 2019$ is not divisible by 9.
# | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,757 |
# Task 7.5
On a checkerboard of size $8 \times 8$, Petya colors several cells. Vasya wins if he can cover all these cells with non-overlapping and non-protruding L-shaped corners consisting of three cells. What is the minimum number of cells Petya should color so that Vasya cannot win?
## Number of points 7 Answer:
... | # Solution
Since 64 is not divisible by 3, the entire board (64 squares) cannot be covered by non-overlapping and non-overflowing L-shaped pieces of three squares. We will show that any 63 colored squares can be covered by such L-shaped pieces. Divide the $8 \times 8$ square into four $4 \times 4$ squares, and each $4... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,759 |
# 1. CONDITION
Is there such an $x$ that the values of the expressions $x+\sqrt{2}$ and $x^{3}+\sqrt{2}$ are rational numbers? | Solution. Suppose the required $x$ is found. Then $x+\sqrt{2}=a-$ a rational number. From this, it follows that $x=a-\sqrt{2}$. But then $x^{3}+\sqrt{2}=(a-\sqrt{2})^{3}+\sqrt{2}=a^{3}-3 \sqrt{2} a^{2}+6 a-2 \sqrt{2}+\sqrt{2}=a^{3}+6 a-\left(3 a^{2}+1\right) \sqrt{2}$. This number is rational only if $3 a^{2}+1=0-$ a c... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,760 |
# 2. CONDITION
Find all pairs $(a ; b)$ of real numbers $a$ and $b$ such that the equations $x^{2}+a x+b^{2}=0$ and $x^{2}+b x+a^{2}=0$ have at least one common root. | Solution. Suppose that $a \neq b$, and let $x_{0}$ be a common root of the equations. Substituting $x_{0}$ into the equations and subtracting one from the other, we get
$$
\begin{gathered}
(a-b) x_{0}+b^{2}-a^{2}=0, \\
\text { hence } x_{0}=a+b . \text { Therefore, } \\
(a+b)^{2}+a(a+b)+b^{2}=0, \text { which means, }... | (0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,761 |
# 3. CONDITION
In a plane, there are $\mathrm{n}$ identical gear wheels arranged such that the first is engaged with the second, the second with the third, and so on, finally, the last n-th wheel is engaged with the first. Can the wheels in such a system rotate? | Solution. Any two adjacent wheels rotate in opposite directions, meaning that the first and last wheels also have opposite directions of rotation, which is possible if and only if the number of wheels is even. Thus, the wheels of such a system can rotate if the number of wheels is even, and cannot in the opposite case. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,762 |
# 5. CONDITION
A right triangle $ABC$ with hypotenuse $AB$ is inscribed in a circle. A point $D$ is taken on the larger leg $BC$ such that $AC = BD$, and point $E$ is the midpoint of the arc $AB$ containing point $C$. Find the angle $DEC$. | Solution. Point $E$ is the midpoint of arc $AB$, so $AE = BE$. Moreover, inscribed angles $CAE$ and $EBD$, subtending the same arc, are equal. Given that $AC = BD$, triangles $ACE$ and $BDE$ are congruent, which implies that angle $CEA$ is equal to angle $BED$. Therefore, angle $DEC$ is equal to angle $BEA$ and both ar... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,763 |
# 6. CONDITION
Vasya has three cans of paint of different colors. In how many different ways can he paint a fence consisting of 10 planks so that any two adjacent planks are of different colors and he uses all three colors of paint? Justify your answer. | Solution. Let's calculate the number of ways to paint the fence so that any 2 adjacent boards are painted in different colors. The first board can be painted with any of the three colors, the second with one of the two remaining colors. The third board can be painted with one of the two colors that differ from the colo... | 1530 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,764 |
11.2. $\left(a_{n}\right)$ is an arithmetic progression with a common difference of 1. It is known that $\mathrm{S}_{2022}$ is the smallest among all $S_{n}$ (less than the sum of the first $n$ terms for any other value of $n$). What values can the first term of the progression take? | Answer: $a_{1}$ belongs to the interval (-2022; -2021).
Solution. Since the difference of the progression is positive, the progression is increasing. Therefore, the described situation is possible if and only if the members of the progression from the first to the 2022nd are negative, and starting from the 2023rd, the... | -2022<a_1<-2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,766 |
11.3. Petya and Vasya live in neighboring houses (see the plan in Figure 1). Vasya lives in the fourth entrance. It is known that for Petya to run to Vasya by the shortest path (not necessarily along the sides of the cells), it does not matter which side of his house he runs around. Determine which entrance Petya lives... | Answer: in the sixth entrance.
Solution. The shortest path from point $A$ to Vasya's entrance is the segment $A D$ (Fig. 2). The shortest path from point $B$ to Vasya's entrance is the path along segment $B C$, and then along segment $C D$. Since triangles $A E D$ and $C E B$ are equal, $A D = B C$. Therefore, the pat... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,767 |
11.4. In the product of seven natural numbers, each factor was decreased by 3. Could the product have increased exactly 13 times? (N. Agakhanov, I. Bogdanov) | Answer. Yes, it could.
Solution. As an example, the product $1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 16=32$ fits. After the specified operation, it becomes $(-2) \cdot(-2) \cdot(-2) \cdot(-2) \cdot(-2) \cdot(-1) \cdot 13=13 \cdot 32$.
Remark 1. Let's explain how to come up with this example. Suppose five of t... | 1,1,1,1,1,2,16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,768 |
11.5. Quadrilateral $ABCD$ is inscribed in a circle $\Gamma$ with center at point $O$. Its diagonals $AC$ and $BD$ are perpendicular and intersect at point $P$, and point $O$ lies inside triangle $BPC$. A point $H$ is chosen on segment $BO$ such that $\angle BHP=90^{\circ}$. The circle $\omega$, circumscribed around tr... | Solution. Draw the diameter $B T$ in the circle $\Gamma$ (see Fig. 1). Note that $\angle P D T = \angle B D T = 90^{\circ}$. Therefore, $\angle P H T + \angle P D T = 180^{\circ}$, which means that point $T$ lies on the circle $\omega$. Hence, $\angle P Q T = \angle P H T = 90^{\circ}$, and the quadrilateral $P Q T D$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,769 |
1. The numbers $\sqrt{2}$ and $\sqrt{5}$ are written on the board. You can add to the board the sum, difference, or product of any two different numbers already written on the board. Prove that you can write the number 1 on the board. | Solution: For example, we get $\sqrt{5}-\sqrt{2}$, then $\sqrt{5}+\sqrt{2}$ and $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=5-2=3$, then $\sqrt{2} \cdot \sqrt{5}=\sqrt{10}$, then $\sqrt{10}-3$ and $\sqrt{10}+3$ and finally $(\sqrt{10}-3)(\sqrt{10}+3)=10-9=1$.
Criteria. The goal is achieved if the same numbers are used in ... | 1 | Number Theory | proof | Yes | Yes | olympiads | false | 15,771 |
2. Let $a, b, c$ be positive numbers. Prove that
$$
\frac{a}{b+2 c}+\frac{b}{c+2 a}+\frac{c}{a+2 b}>\frac{1}{2}
$$ | Solution.
$$
\frac{a}{b+2 c}+\frac{b}{c+2 a}+\frac{c}{a+2 b}>\frac{a}{2 a+2 b+2 c}+\frac{b}{2 b+2 c+2 a}+\frac{c}{2 c+2 a+2 b}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}
$$
Remark: In fact, this sum is not less than 1. Let $a+2 b=x, b+2 c=y$ and $c+2 a=z$. Then $a=\frac{x-2 y+4 z}{9}, b=\frac{y-2 z+4 x}{9}$ and $c=\frac{z-2 ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,772 |
3. From 80 identical Lego parts, several figures were assembled, with the number of parts used in all figures being different. For the manufacture of the three smallest figures, 14 parts were used, and in the three largest, 43 were used in total. How many figures were assembled? How many parts are in the largest figure... | Answer: 8 figurines, 16 parts.
Solution. Let the number of parts in the figurines be denoted by $a_{1}43$, so $a_{n-2} \leqslant 13$.
Remove the three largest and three smallest figurines. In the remaining figurines, there will be $80-14-$ $43=23$ parts, and each will have between 7 and 12 parts. One figurine is clea... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,773 |
4. The diagonals of the circumscribed trapezoid $A B C D$ with bases $A D$ and $B C$ intersect at point O. The radii of the inscribed circles of triangles $A O D, A O B, B O C$ are 6, 2, and $3 / 2$ respectively. Find the radius of the inscribed circle of triangle $C O D$. | Answer: 3
Solution. We will prove a more general statement, that $\frac{1}{r_{1}}+\frac{1}{r_{3}}=\frac{1}{r_{2}}+\frac{1}{r_{4}}$, where $r_{1}, r_{2}, r_{3}$ and $r_{4}$ are the radii of the inscribed circles of triangles $A O D, A O B, B O C$ and $C O D$ respectively.
Let $A B=a, B C=b, C D=c, A D=d, O A=x, O D=y,... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,774 |
5. On a $10 \times 10$ board, 10 chips are placed such that each horizontal and vertical row contains exactly one chip. Can the remaining part of the board be tiled with $1 \times 2$ rectangles (in cells)? | Answer: It is impossible.
Solution. We will color the board in a checkerboard pattern and number the rows and columns. If the remaining part of the board can be tiled with $1 \times 2$ rectangles, then it contains an equal number of black and white cells, meaning 5 chips are on black cells, and 5 on white cells. Note ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,775 |
9.1. In the product of three natural numbers, each factor was decreased by 3. Could the product have increased by exactly $2016$?
(N. Agakhanov, I. Bogdanov) | Answer. Yes, it could.
Solution. The product $1 \cdot 1 \cdot 676$ serves as an example. After the specified operation, it becomes $(-2) \cdot(-2) \cdot 673 = 2692 = 676 + 2016$.
Remark. The given example is the only one. Here is how to come up with it. Suppose two of the factors were 1, and the third was $-a$. Their... | 676 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,776 |
9.2. Vasya has thought of 8 cells on a chessboard, none of which lie in the same row or column. On his turn, Petya places 8 rooks on the board, none of which attack each other, and then Vasya indicates all the rooks standing on the thought-of cells. If the number of rooks indicated by Vasya on this turn is even (i.e., ... | Answer. In 2 moves.
Solution. First, we will show how Pete can win in 2 moves. On his first move, he will place 8 rooks along the diagonal of the board. If he hasn't won yet, there is an odd number of cells on the diagonal that Vasya has thought of. In particular, there is both a cell $A$ that Vasya has thought of and... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,777 |
9.3. Does there exist a triangle with sides $x, y$ and $z$ such that $x^{3}+y^{3}+z^{3}=(x+y)(y+z)(z+x)$? | Answer. No, it does not exist.
First solution. Suppose such a triangle exists. We can assume that $x \geqslant y \geqslant z$. Then, by the triangle inequality, $y+z>x$, from which
$$
\begin{aligned}
(x+y)(x+z)(y+z) &>(x+y)(x+z) x=x^{3}+x^{2} y+x^{2} z+x y z> \\
&>x^{3}+x^{2} y+x^{2} z \geqslant x^{3}+y^{3}+z^{3}
\en... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,778 |
9.4. An equilateral triangle $ABC$ is inscribed in a circle $\Omega$ and circumscribed around a circle $\omega$. Points $P$ and $Q$ are chosen on sides $AC$ and $AB$ respectively such that segment $PQ$ is tangent to $\omega$. The circle $\Omega_{b}$ with center $P$ passes through $B$, and the circle $\Omega_{c}$ with c... | First solution. Let $O$ be the center of triangle $ABC$, and let $\omega$ touch segments $BQ$, $QP$, and $PC$ at points $K$, $L$, and $M$ respectively (see Fig. 2). Due to the symmetry of the equilateral triangle, lines $BO$ and $CO$ pass through points $M$ and $K$ respectively.
We lay off segment $OX$ on ray $LO$, eq... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,779 |
6.2. Two cyclists decided to travel from point $A$ to point $B$. The speed of the first one is 35 km/h, and the second one's speed is 25 km/h. It is known that each of them only cycled when the other was resting (standing still), and in total, they covered the same distance in 2 hours. Could they have reached point $B$... | Answer: Could not.
Solution: We will assume that the cyclists did not rest simultaneously. Since they covered the same distance, and the ratio of their speeds is 7:5, the time of movement for the first cyclist is 5 parts, and for the second cyclist - 7 parts of time. In total, they rode for 2 hours = 120 minutes. Ther... | Couldnot | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,780 |
Problem 4.3. How many rectangles exist on this picture with sides running along the grid lines? (A square is also a rectangle.)
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,783 |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,784 |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
 in the squares such that the following condition is met: if two squares are connected, the number in the higher square is greater. How many ways are there to do this?
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,788 |
Problem 7.4. In triangle $ABC$, the median $CM$ and the bisector $BL$ were drawn. Then, all segments and points were erased from the drawing, except for points $A(2 ; 8)$, $M(4 ; 11)$, and $L(6 ; 6)$. What were the coordinates of point $C$?
$.
Solution. Since $M$ is the midpoint of $A B$, point $B$ has coordinates (6;14). Since $\angle A B L=\angle C B L$, point $C$ lies on the line symmetric to the line $A M$ with respect to the vertical line $B L$. Also, point $C$ lies on the line $A L$. Carefully finding the intersection point of thes... | (14;2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,789 |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,791 |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,792 |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then,
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,793 |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,794 |
Problem 9.7. Through points $A(0 ; 14)$ and $B(0 ; 4)$, two parallel lines are drawn. The first line, passing through point $A$, intersects the hyperbola $y=\frac{1}{x}$ at points $K$ and $L$. The second line, passing through point $B$, intersects the hyperbola $y=\frac{1}{x}$ at points $M$ and $N$.
What is $\frac{A L... | Answer: 3.5.
Solution. Let the slope of the given parallel lines be denoted by $k$. Since the line $K L$ passes through the point ( $0 ; 14$ ), its equation is $y=k x+14$. Similarly, the equation of the line $M N$ is $y=k x+4$.
The abscissas of points $K$ and $L$ (denoted as $x_{K}$ and $x_{L}$, respectively) are the... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,796 |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,797 |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
^{2}-(10 b+a)^{2}=9 \cdot 11(a+b)(a-b)$. This difference is a square of an integer,... | Nikolayis65old,Anatolyis56old | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,801 |
8-3. Vika has been recording her grades since the beginning of the year. At the beginning of the second quarter, she received a five, after which the proportion of fives increased by 0.15. After another grade, the proportion of fives increased by another 0.1. How many more fives does she need to get to increase their p... | Answer: 4.
Solution: Let's say Vika had $n$ grades in the first quarter, of which $k$ were fives. Then, after the first five in the second quarter, the proportion of fives increased by $\frac{k+1}{n+1}-\frac{k}{n}=0.15$. Similarly, after the second five, the increase was $\frac{k+2}{n+2}-\frac{k+1}{n+1}=0.1$. Simplify... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,802 |
8-4. Given a parallelogram $A B C D$. The bisectors of angles $A$ and $B$ intersect at point $M$, which lies on side $C D$. Denote the point of intersection of the bisectors of angles $C$ and $D$ as $N$. Prove that $M N$ is parallel to $A D$. | Solution. Due to the parallelism of $AB$ and $CD$, the angle $\angle CMA = \angle MAB = \angle MAC$, which implies that triangle $ACM$ is isosceles and $AC = CM$. Similarly, it can be shown that $MD = BD$, which coincides with $AC$. Therefore, $M$ is the midpoint of $CD$. Similarly, it can be shown that the angle bisec... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,803 |
8-5. Diligent Asey multiplied two three-digit numbers, while lazy Petya simply wrote them down one after the other. Petya's result turned out to be 7 times larger than Asey's. What numbers did she multiply? | Answer: 143 and 143
Solution. Let the original numbers be denoted by $a$ and $b$. Asey's result is $a b$, and Petya's result is $-1000 a + b$. They are related by the equation $1000 a + b = 7 a b$, from which we get $b = \frac{1000 a}{7 a - 1}$. Note that the numbers $a$ and $7 a - 1$ have no common divisors, so $7 a ... | 143143 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,804 |
9.1. The bike path consists of two sections: first there is an asphalt section, and then a sandy one. Petya and Vasya started separately (first Petya, then Vasya), and each rode the entire path. The speed of each boy on each of the two sections was constant. It turned out that they met in the middle of the asphalt sect... | Answer. They spent the same amount of time.
First solution. Between the two moments of meeting, each boy rode half of the asphalt and half of the sandy sections, and they spent the same amount of time on this. Therefore, on the entire path, each of them spent twice as much time, that is, also the same amount of time.
... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,805 |
9.2. Given a paper triangle with side lengths of 5 cm, 12 cm, and 13 cm. Can it be cut into several (more than one) polygons, each of which has an area (measured in cm ${}^{2}$ ) numerically equal to the perimeter (measured in cm)? (D. Khramtsov) | Answer: No.
Solution. A polygon, the area of which (measured in cm ${ }^{2}$ ) is numerically equal to the perimeter (measured in cm), is called good.
Notice that the original triangle is good: it is a right triangle with legs of 5 cm and 12 cm, so its area is 30 cm $^{2}$ and numerically coincides with its perimeter... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,806 |
9.3. Given a natural number $n$. On a $2 n \times 2 n$ chessboard, $2 n$ rooks are placed such that no two rooks are in the same row or column. After that, the board is cut along the grid lines into two connected parts, symmetric to each other with respect to the center of the board. What is the maximum number of rooks... | Answer: $2 n-1$.
Solution: Note that in each column and each row, there is exactly one rook.
First, we show that all $2 n$ rooks could not have ended up in one part. Let $A, B, C, D$ be the corner cells of the board (in counterclockwise order, see Fig. 2). By symmetry, $A$ and $C$ must belong to different parts, as m... | 2n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,807 |
9.4. Given natural numbers $a, b$, and $c$. None of them is a multiple of another. It is known that the number $a b c + 1$ is divisible by $a b - b + 1$. Prove that $c \geqslant b$.
(M. Antipov) | The first solution. Assume the opposite: let $b \geqslant c+1$. From the divisibility of $a b c+1$ by $a b-b+1$, it follows that the number $b(a c-a+1)=(a b c+1)-(a b-b+1)$ is divisible by $a b-b+1$. Since the numbers $b$ and $a b-b+1=b(a-1)+1$ are coprime, we get that $a c-a+1$ is divisible by $a b-b+1$. Clearly, $a c... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,808 |
9.5. Quadrilateral $ABCD$ is inscribed in circle $\gamma$. It turns out that the circles constructed on segments $AB$ and $CD$ as diameters touch each other externally at point $S$. Let points $M$ and $N$ be the midpoints of segments $AB$ and $CD$ respectively. Prove that the perpendicular $\ell$ to line $MN$, erected ... | Solution. Let the circles with diameters $AB$ and $CD$ be denoted by $\omega_{1}$ and $\omega_{2}$, respectively. Note that point $S$ lies on segment $MN$.
Let lines $CS$ and $DS$ intersect $\ell$ at points $P$ and $Q$, respectively (see Fig. 3). Since $CD$ is the diameter of $\omega_{2}$, we have $\angle PSQ = \angle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,809 |
1. Does the equation $x^{4}-4 x^{3}-6 x^{2}-3 x+9=0$ have negative roots? | Answer: no.
Solution. Transform the given equation: $x^{4}-4 x^{3}-6 x^{2}-3 x+9=0 \Leftrightarrow\left(x^{2}-3\right)^{2}-4 x^{3}-3 x=0$, $\left(x^{2}-3\right)^{2}=4 x^{3}+3 x \Leftrightarrow\left(x^{2}-3\right)^{2}=x\left(4 x^{2}+3\right)$. If $x0$ for all values of $x$ )
“-” - only the answer is provided
“-" - th... | no | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,810 |
2. Vasya wrote natural numbers from 1 to 72 in the cells of a $4 \times 18$ table (4 rows, 18 columns) in some order known only to him. First, he found the product of the numbers in each column, and then for each of the eighteen products, he calculated the sum of the digits. Could all the resulting sums be the same? | Answer: No.
Solution. Suppose that each of the specified digit sums is equal to S. Since some of the products contain factors that are multiples of nine, such products are divisible by 9, meaning their digit sum is also divisible by 9. Therefore, the number $S$ must be a multiple of nine.
Thus, the product of the num... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,811 |
3. A regular pentagon and a regular icosagon are inscribed in the same circle. Which is greater: the sum of the squares of the lengths of all sides of the pentagon or the sum of the squares of the lengths of all sides of the icosagon? | Answer: the sum of the squares of the side lengths of a regular pentagon is greater.
Solution. Note that in a regular 20-sided polygon, the vertices taken every other one form a regular 10-sided polygon, and the vertices of this 10-sided polygon, taken every other one, form a regular 5-sided polygon. Therefore, it is ... | a_{5}^{2}>4a_{20}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,812 |
4. Given a triangular pyramid $A B C D$ with right dihedral angles at vertex $D$, where $C D=A D+D B$. Prove that the sum of the plane angles at vertex $C$ is $90^{\circ}$. | Solution. Let's introduce the notation: $\angle A C D=\alpha, \angle B C D=\beta$, $\angle B C A=\gamma, D A=a, D B=b$. According to the problem: $C D=a+b$, and we need to prove that $\alpha+\beta+\gamma=90^{\circ}$.
First method. Let $C A=m, C B=n$ (see Fig. 11.4a).
First, we will prove that the angles $\alpha+\beta... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,813 |
5. The function $f(x)$ is defined for all real numbers, and for any $x$, the equalities $f(x+2)=f(2-x)$ and $f(x+7)=f(7-x)$ hold. Prove that $f(x)$ is a periodic function. | Solution. We will prove that the given function has a period of 10. There are different ways to reason about this.
The first method. Considering that $f(x)$ is defined everywhere, it is sufficient to prove that for any real $x$, the equality $f(x+10)=f(x)$ holds.
Sequentially using the second and first equations from... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,814 |
6. Each integer on the coordinate line is painted in one of two colors - white or black, and the numbers 2016 and 2017 are painted in different colors. Is it necessarily possible to find three identically colored integers whose sum is zero? | Answer: necessarily.
Solution. Suppose there are no such integers. Let's assume, for definiteness, that the number 2016 is painted black, and the number 2017 is painted white. Consider two cases:
1) The number 0 is painted white. Since the numbers 0 and 2017 are white, the number -2017 is painted black. From the fact... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,815 |
1. A new series of "Kinder Surprises" - chocolate eggs, each containing a toy car - was delivered to the store. The seller told Pete that there are only five different types of cars in the new series, and it is impossible to determine which car is inside by the appearance of the egg. What is the minimum number of "Kind... | Solution. If Petya buys 10 "Kinder Surprises," in the least favorable situation for him, he will get two cars of each type. If he buys 11 "Kinder Surprises," he will get three cars of one type. Let's prove this. Suppose Petya bought 11 "Kinder Surprises" but did not get three cars of one type. This means the number of ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,816 |
2. Prove the inequality for any real $a, b, c$
$$
5 a^{2}+5 b^{2}+5 c^{2} \geq 4 a b+4 b c+4 a c
$$
For which $a, b, c$ does equality hold? | Solution. Transform the inequality by performing equivalent transformations.
$$
\begin{gathered}
5 a^{2}+5 b^{2}+5 c^{2}-4 a b-4 b c-4 a c \geq 0 \\
4 a^{2}-4 a b+b^{2}+4 b^{2}-4 b c+c^{2}+4 c^{2}-4 a c+a^{2} \geq 0 \\
(2 a-b)^{2}+(2 b-c)^{2}+(2 c-a)^{2} \geq 0
\end{gathered}
$$
This inequality is equivalent to the o... | =b==0 | Inequalities | proof | Yes | Yes | olympiads | false | 15,817 |
3. Quadrilateral $ABCD$ is circumscribed around a circle. The vertices $A, B, C, D$ are the centers of circles $S_{1}, S_{2}, S_{3}, S_{4}$ respectively. Circles $S_{1}$ and $S_{2}$ touch externally at point $X$, similarly $S_{2}$ and $S_{3}$ touch at point $Y$, $S_{3}$ and $S_{4}$ at point $Z$, and $S_{4}$ and $S_{1}$... | Solution. The condition that two circles with centers $A$ and $B$ touch each other at point $X$ implies that point $X$ lies on the segment $A B$ connecting the centers,
and $A X + X B = A B$.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,818 |
4. Let $S$ be the sum of the digits of a certain five-digit number, all of whose digits are distinct and non-zero. We listed all five-digit numbers obtained by permuting the digits of the original number, and then added all the listed numbers, including the original number. Prove that the resulting sum is divisible by ... | Solution. Let the given number have the form $a_{4} \cdot 10^{4}+a_{3} \cdot 10^{3}+a_{2} \cdot 10^{2}+a_{1} \cdot 10+a_{0}$. We will systematize all the written numbers. First, we will do it this way. We will group all the written numbers into five groups, in each group the first digit is the same. In these groups, ea... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,819 |
5. A $3 \times 3$ table is initially filled with zeros. In one move, any $2 \times 2$ square in the table is selected, and all zeros in it are replaced with crosses, and all crosses with zeros. We will call any arrangement of crosses and zeros in the table a "pattern." How many different patterns can be obtained as a r... | Solution. First method. Different drawings differ by at least one cell in the table, either plus or minus. The symbol in a given specific cell is determined by the parity (even or odd) of the number of swaps performed in each $2 \times 2$ square containing that cell. Each $2 \times 2$ square contains exactly one corner... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,820 |
1. Find the ratio $\frac{2023 a c}{b^{2}}$, if it is known that one of the roots of the equation $a x^{2}+b x+c=0$ is 2022 times the other. Here $a \cdot b \cdot c \neq 0$. | Solution. Let $x_{1}=2022 x_{2}$. Then, by Vieta's theorem,
$x_{1}+x_{2}=2023 x_{2}=-\frac{b}{a} \quad x_{1} \cdot x_{2}=2022 x_{2}^{2}=\frac{c}{a}$.
Then
$$
\begin{gathered}
x_{2}=-\frac{b}{2023 a} \\
\frac{c}{a}=\frac{2022 b^{2}}{2023^{2} \cdot a^{2}} \\
a c=\frac{2022 b^{2}}{2023^{2}} \\
2023 a c=\frac{2022 b^{2}... | \frac{2022}{2023} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,821 |
2. Find all solutions to the equation
$$
m^{2}-2 m n-3 n^{2}=5
$$
where $m, n$ are integers. | Solution.
Transform the left side of the equation
$$
\begin{gathered}
m^{2}-2 m n+n^{2}-4 n^{2}=5 \\
(m-n)^{2}-(2 n)^{2}=5 \\
(m-n-2 n)(m-n+2 n)=5 \\
(m-3 n)(m+n)=5
\end{gathered}
$$
The integer 5 can be factored in four ways: $5=1 \cdot 5=5 \cdot 1=$ $(-1) \cdot(-5)=(-5) \cdot(-1)$. We get four systems of equations... | =4,n=1;=-4,n=-1;=2,n=-1;=-2,n=1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,822 |
3. Prove the inequality for all positive $a, b$
$$
\frac{a}{a^{4}+b^{2}}+\frac{b}{a^{2}+b^{4}} \leq \frac{1}{a b}
$$ | Solution. Applying the formula for the square of a difference, we get
$$
\begin{array}{cl}
\left(a^{2}-b\right)^{2} \geq 0, & \left(a-b^{2}\right)^{2} \geq 0 \\
a^{4}+b^{2} \geq 2 a^{2} b, & a^{2}+b^{4} \geq 2 a b^{2}
\end{array}
$$
The terms on the left side of the inequality can be compared with $\frac{1}{2 a b}$
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,823 |
4. After the football match, Andrey said that he scored 3 goals, and Pasha only 1. Pasha said that he scored 4 goals, and Vanya scored 5 goals. Vanya said that he scored 6 goals, and Andrey only 2. Could it be that together they scored 10 goals, if it is known that each of them told the truth once and lied once? | Solution. The first method. There are two possible cases.
1) If Andrey told the truth about himself, then he scored 3 goals. Then Vanya lied about Andrey, and told the truth about himself, meaning he indeed scored 6 goals. Consequently, Pasha lied about Vanya and told the truth about himself, meaning he scored 4 goals... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,824 |
5. In triangle $A B C$, $A C>A B$. A height $A H$ is dropped from vertex $A$. Which difference is greater, $(A C-A B)$ or $(H C-H B)$? | Solution. We need to consider two cases, see the figure.
1st case. Point $H$ lies inside the segment $B C$. By the Pythagorean theorem,
$$
A B^{2}-B H^{2}=A H^{2}=A C^{2}-C H^{2}
$$
We tran... | CH-BH>AC-AB | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,825 |
1. Given a $3 \times 3$ square with numbers in the cells (see Fig. 1). In one operation, you can add 1 to all the numbers in one row or subtract 1 from all the numbers in one column. Transform the square into the one shown in Fig. 2.
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
Fig. 1
| 1 | 4 | 7... | 1. Solution: in the first line, add 4, in the second line, add 2. Then in the first column, subtract 4, and in the second column, subtract 2.
Criteria: correct solution - 7 points; in other cases - 0 points. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,826 |
2. Given three linear functions: $y=a x+b+3, y=-b x+a-2, y=2 x-8$. It is known that the graphs of the first and second intersect on the y-axis, and the graphs of the second and third - on the x-axis. Find the point of intersection of the graphs of the first and third functions. | 2. Answer: $(-3,-14)$. Solution: the first and second lines intersect at $x=0$, that is, $b+3=a-2$. The second and third lines intersect at $y=0$, therefore, $x=4$ and $-4b+a-2=0$. Thus, $b=1, a=6$, from which we get the answer.
Criteria: correct solution - 7 points; only the first relation between $a$ and $b$ found -... | (-3,-14) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,827 |
3. Let $m$ and $n$ be consecutive odd numbers. Prove that the expression $7 m^{2}-5 n^{2}-2$ is divisible by 8. | 3. Solution: Notice that $7 m^{2}-5 n^{2}-2=2(m-1)(m+1)+5(m-n)(m+n)$. In the first term, $m-1$ and $m+1$ are even, so this term is divisible by 8. In the second term, $m-n$ equals 2 (or -2), and $m+n$ is divisible by 4 (since consecutive odd numbers give remainders of 1 and 3 or 3 and 1 when divided by 4), thus this te... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,828 |
4. Two cyclists are training on a circular stadium. In the first two hours, Ivanov lapped Petrov by 3 laps. Then Ivanov increased his speed by 10 km/h, and as a result, after 3 hours from the start, he lapped Petrov by 7 laps. Find the length of the lap. | 4. Answer: 4 km. Solution: let I be Ivanov's speed (initial), Π be Petrov's speed, K be the length of the circle. Then 2I - 2Π = 3K and 2I - 2(Ι - Π) = 3K and 3(Ι - Π) = 3K - 10, express Ι - Π from both equations and equate, we get 14K - 20 = 9K, from which K = 4.
Criteria: correct solution - 7 points; the system is s... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,829 |
5. In triangle $ABC$, segment $AF$ is a median, point $D$ is its midpoint, and point $E$ is the intersection of lines $AB$ and $CD$. Given that $BD=BF$, prove that $AE=DE$. | 5. Solution: note that angles $A D B$ and $D F C$ are equal as adjacent to equal ones, therefore triangles $A B D$ and $D C F$ are equal by the first criterion. Then angles $B A D$ and $F D C$ are equal, but angles $E D A$ and $F D C$ are also equal as vertical angles. Thus, angles $E D A$ and $E A D$ are equal, from w... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,830 |
10.1. Form a quadratic equation with integer coefficients, one of whose roots is the number $2-\sqrt{7}$. | Answer. $x^{2}-4 x-3=0$.
Solution. If $a-\sqrt{b}$ is a root of the equation $x^{2}+p x+q$, then $a+\sqrt{b}$ is also a root of this equation. Indeed, by Vieta's theorem, the sum of the roots is equal to $-p$ (an integer), which means the second root has the form $c+\sqrt{b}$, and their product is also equal to the in... | x^{2}-4x-3=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,831 |
10.2. How many rectangular trapezoids $A B C D$ exist, where angles $A$ and $B$ are right angles, $A D=2$, $C D=B C$, the sides have integer lengths, and the perimeter is less than 100? | Answer: 5.
Solution. Let $C D=B C=a, A B=b$. Drop a perpendicular from point $D$ to $B C$, and apply the Pythagorean theorem: $(a-2)^{2}+b^{2}=a^{2}$. From this, $b^{2}=4(a-1), a=\frac{b^{2}}{4}+1$. Suppose $A D=2$ is the smaller base of the trapezoid. The perimeter $P=2 a+b+2a$, so $b>2$. Considering that $a=\frac{b^... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,832 |
10.3. In a bag, there are chips of two colors, blue and green. The probability of drawing two chips of the same color is equal to the probability of drawing two chips of different colors (drawing one chip after another without replacement). It is known that there are more blue chips than green chips, and the total numb... | Answer: $3,6,10,15,21,28$.
Solution. Let the number of blue chips be $n$, and the number of green chips be $-m$. Then $\frac{n m+m n}{(n+m)(n+m-1)}=\frac{n(n-1)+m(m-1)}{(n+m)(n+m-1)}$, or $2 n m=n(n-1)+m(m-1)$, from which $(n-m)^{2}=n+m$. Thus, the total number of chips (denote it as $a$) must be a perfect square, tha... | 3,6,10,15,21,28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,833 |
10.4. For various non-zero real numbers $a, b$, and $c$, the following equalities are satisfied:
$$
a+\frac{2}{b}=b+\frac{2}{c}=c+\frac{2}{a}=d
$$
Find all possible values of the number $d$. | Answer: $\pm \sqrt{2}$.
Solution. Express $b$ and $c$ from the equations, we get $b=\frac{2}{d-a}, c=\frac{d a-2}{a}$. Hence, $b+\frac{2}{c}=\frac{2}{d-a}+\frac{2 a}{d a-2}=d$. Bring to a common denominator: $2(d a-2)+2 a(d-a)=d(d-a)(d a-2)$, or $2 d a-4-2 a^{2}-d^{3} a+a^{2} d^{2}+2 d^{2}=0$. Group the first term wit... | \\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,834 |
Problem 7.1
It is known that $\frac{1}{a}-\frac{1}{b}=\frac{1}{a+b}$. Prove that $\frac{1}{a^{2}}-\frac{1}{b^{2}}=\frac{1}{a b}$.
## Number of points 7 | Solution
$\frac{1}{a^{2}}-\frac{1}{b^{2}}=\left(\frac{1}{a}-\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{a+b} \cdot \frac{a+b}{a b}=\frac{1}{a b}$
# | \frac{1}{} | Algebra | proof | Yes | Yes | olympiads | false | 15,835 |
# Task 7.3
A tile has the shape of a right-angled triangle with legs measuring 1 dm and 2 dm. Can 20 such tiles be used to form a square?
## Number of points 7
# | # Answer:
it is possible
## Solution
The diagram shows how this can be done.
## Additional Criteria
Any correct diagram - 7 points
 | itispossible | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,837 |
# Task 7.4
On an island, there lives an odd number of people, each of whom is either a knight, who always tells the truth, or a liar, who always lies. One day, all the knights declared: “I only befriend 1 liar,” while all the liars said: “I do not befriend knights.” Who is more numerous on the island, knights or liars... | Answer:
There are more knights
## Solution
Each liar is friends with at least one knight. However, since each knight is friends with exactly one liar, two liars cannot share the same knight friend. Therefore, each liar can be paired with their knight friend, which means there are at least as many knights as there ar... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,838 |
# Task 7.5
Consider the number
$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \ldots \cdot \frac{399}{400}$
Prove that it is greater than ${ }^{1 / 29}$ and less than ${ }^{1 / 20}$
Number of points 7
# | # Solution
Let
$A=\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \ldots \cdot \frac{399}{400}$
$B=\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \frac{8}{9} \cdot \ldots \cdot \frac{398}{399}$
It is clear that $B>A$ and $A B=1 / 400$.
Therefore, $A^{2}B$.
Therefore, $A^{2}>1 / 2 A B=1 / 800>(1 / 29)^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,839 |
10.1. Find the area of the figure defined on the coordinate plane by the inequalities $|x|-1 \leq y \leq \sqrt{1-x^{2}}$ | Answer: $\frac{\pi}{2}+1$. Solution: The figure has the appearance shown in the diagram: it is bounded from below by the graph $\quad y=|x|-1, \quad$ and from above
by the semicircle $y=\sqrt{1-x^{2}} \Leftrightarrow\left\{\begin{array}{l}x^{2}+y^{2}=1 \\ y \geq 0\end{array}\right.$.
The area of the figure consists o... | \frac{\pi}{2}+1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,840 |
10.4. Find all values of the parameter $a$ for which the equation $x^{4}-a x^{2}+1=0$ has 4 roots forming an arithmetic progression. | Answer: $\frac{10}{3}$. Solution. Let $t=x^{2}$. Then the equation $t^{2}-a t+1=0$ must have two positive roots $t_{1}, t_{2}\left(t_{1}<t_{2}\right)$, and the roots of the original equation will be $-\sqrt{t_{2}},-\sqrt{t_{1}}, \sqrt{t_{1}}, \sqrt{t_{2}}$. The condition of the problem about the arithmetic progression ... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,843 |
10.5. We have 10 sticks of lengths $1 ; 1.9 ; (1.9)^{2} ; \ldots ; (1.9)^{9}$. Can we use these sticks, not necessarily all, to form a) a triangle b) an isosceles triangle? | Answer: a) it is possible; b) it is not possible. Solution. a) Let's take as the sides of the triangle $a=1+1.9+(1.9)^{2}+\ldots+(1.9)^{7}, b=(1.9)^{8}, c=(1.9)^{9}$. First, we will show that $c>a$. By the formula for the sum of a geometric progression, we have $a=\frac{(1.9)^{8}-1}{1.9-1}c \Leftrightarrow \frac{(1.9)^... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,844 |
1. The numbers $x, y$ and $z$ are allowed to take only natural values. Prove that under this condition, the equation $x+y+z=100$ has an odd number of solutions. | First solution. First, we find the number of solutions to the equation \(x + y = a\) (where \(a\) is a natural number) in natural numbers. It is clear that \(x\) can take any value from 1 to \(a-1\), and \(y\) is automatically determined by the condition. Thus, the number of solutions to the equation \(x + y = a\) is \... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,845 |
2. The number $t$ is a root of the quadratic equation $a x^{2}+b x+c=0$ with integer coefficients ( $a \neq 0$ ). Prove that the number $t+2$ is also a root of a quadratic equation with integer coefficients. | Solution. The number $t+2$ is a root of the equation $a(x-2)^{2}+b(x-2)+c=0$. This equation can be rewritten as $a x^{2}+(b-4 a) x+(4 a-2 b+c)=0$. From this, it is clear that all its coefficients are integers. | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,846 |
3. Given a rhombus $A B C D$ with angle $B A D$ equal to $60^{\circ}$. On side $A B$, point $T$ is taken, and on side $B C$, point $N$ is taken such that $A T=B N$. Prove that $T N=D T$. | Solution. Triangle $A D B$ is isosceles, and angle $B A D$ is $60^{\circ}$, so $A D B$ is equilateral. From this, it follows that $A D=D B$. The diagonal $D B$ of the rhombus is the bisector of angle $A B C$, so $\angle D B N=60^{\circ}$. Since $A T=B N$ by the condition, triangles $D A T$ and $D B N$ are equal by two ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,847 |
4. Each point on the contour of a square is painted either blue or red. Is it necessarily true that there will be a right-angled triangle with vertices of the same color lying on the contour of the square
# | # Answer: Yes.
Solution. Suppose there exists a coloring without such triangles and we will arrive at a contradiction. The vertices of the square and the midpoints of its sides are denoted as $A_{1}, \ldots, A_{8}$ in a clockwise direction, starting from one of the vertices.
Assume $A_{1}$ is blue. If $A_{3}$ is also... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,848 |
5. On a rectangular table, several rectangular sheets of paper are laid out such that their sides are parallel to the edges of the table. It is known that any two sheets have a common point. Prove that there is a point that belongs to all the sheets. | Solution. Everywhere below, the word "rectangle" will mean one of the sheets on the table, unless otherwise specified. We introduce a coordinate system such that the axes are parallel to the sides of the table (axes are needed to introduce the concepts of top, bottom, right, and left). Among all the lower sides of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,849 |
9.1. Is it possible to place black and white pawns on the cells of a chessboard so that three pawns of the same color do not go in a row neither horizontally, vertically, nor diagonally? The chessboard is 8x8 in size. There is an unlimited number of pawns of both colors. | # Solution:
Example of pawn placement:
| $\mathbf{4}$ | B | $\mathbf{Y}$ | b | $\mathbf{4}$ | $B$ | $\mathbf{4}$ | $b$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{4}$ | B | 4 | D | $\mathbf{4}$ | $B$ | $\mathbf{4}$ | B |
| $\overline{5}$ | $\mathbf{4}$ | B | $\mathbf{4}$ | $\overli... | Yes,itispossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,850 |
9.2. Solve the system of equations: $\left\{\begin{array}{c}x^{4}+x^{2} y^{2}+y^{4}=481 \\ x^{2}+x y+y^{2}=37\end{array}\right.$. | Solution:
$$
\begin{gathered}
\left\{\begin{array} { c }
{ ( x ^ { 2 } + y ^ { 2 } ) ^ { 2 } - ( x y ) ^ { 2 } = 4 8 1 } \\
{ x ^ { 2 } + y ^ { 2 } + x y = 3 7 }
\end{array} \Rightarrow \left\{\begin{array}{c}
\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)=481 \\
x^{2}+y^{2}+x y=37
\end{array}=>\right.\righ... | (4;3),(3;4),(-3;-4),(-4;-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,851 |
9.3. Prove that there are no three-digit numbers for which the sum of all digits equals the number itself, decreased by the product of all its digits.
# | # Solution:
Assume that such a number exists and represent it as $\overline{a b c}$, where $a>0$. The following relationship holds: $a+b+c=\overline{a b c}-a b c$.
$$
\begin{gathered}
a+b+c=100 a+10 b+c-a b c \\
99 a+9 b=a b c
\end{gathered}
$$
From the obtained expression, we conclude that $99 a \leq a b c$, hence ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,852 |
9.4. In triangle $ABC$, $\angle ABC = 2 \angle ACB$, $K$ is the midpoint of side $BC$, and $AE$ is the altitude. Prove that $AB = 2EK$. | # Solution:
Let $\mathrm{N}$ be the midpoint of side AC. Draw segments KN and EN. KN is the midline of triangle $\mathrm{ABC}$, so $\mathrm{AB}=2 \mathrm{KN}$, and it is sufficient to prove that EK=KN.
EN is the median of a right triangle drawn to the hypotenuse, so $\mathrm{EN}=\mathrm{AN}=\mathrm{NC}$ and $\angle N... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,853 |
9.5. What is the maximum number of figures consisting of 4 1x1 squares, as shown in the diagram, that can be cut out from a $6 \times 6$ table, if cutting can only be done along the grid lines?
# | # Solution:
Example:
The diagram shows that 8 figures can be cut out.
Evaluation:
 gives a remainder of 0 or 1 when divided by 3. The number 2021 gives a remainder of 2 when divided by 3.
Answer: no. | no | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,855 |
2. When writing a decimal fraction, a schoolboy wrote all its digits correctly, but placed the comma in such a way that the resulting fraction was less than the required one by 18.189. What decimal fraction should the schoolboy have written? | Solution: Since the number decreased as a result of the error, the decimal point must have been moved to the left. This means the number was reduced by a factor of 10. Let the resulting number be x, then the original number is 10x. According to the problem: $10x - x = 18.189$, so $x = 2.021$, and $10x = 20.21$.
Answer... | 20.21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,856 |
4. Point E is the midpoint of side AB of parallelogram ABCD. On segment DE, there is a point F such that $\mathrm{AD}=\mathbf{B F}$. Find the measure of angle CFD. | Solution: Extend DE until it intersects line BC at point K (see figure). Since $\mathrm{BK} \| \mathrm{AD}$, then $\angle \mathrm{KBE}=\angle \mathrm{DAE}$. Moreover, $\angle \mathrm{KEB}=\angle \mathrm{DEA}$ and $\mathrm{AE}=\mathrm{BE}$, so triangles $\mathrm{BKE}$ and $\mathrm{ADE}$ are congruent. Therefore, $\mathr... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,857 |
5. We took 2021 arbitrary natural numbers and arranged them in a circle. Prove that there will always be two adjacent numbers whose sum is an even number. | Solution: The sum of two numbers will be even if they are both even or both odd. The sum of two numbers will be odd if one is even and the other is odd. Suppose the sum of any two adjacent numbers is odd, then even and odd numbers must alternate. This means the total number of numbers will be even, but according to the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,858 |
11.1. Solve the equation $2 \sin ^{2} x+1=\cos (\sqrt{2} x)$. | Answer: $x=0$. Solution. The left side of the equation $\geq 1$, and the right side $\leq 1$. Therefore, the equation is equivalent to the system: $\sin x=0, \cos \sqrt{2} x=1$. We have: $x=\pi n, \sqrt{2} x=2 \pi k$ ( $n, k-$ integers). From this, $n=k \cdot \sqrt{2}$. Since $\sqrt{2}$ is an irrational number, the las... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,859 |
11.2. Given a right triangle $A B C$ with legs $A C=a$ and $C B=b$. Find a) the side of the square with vertex $C$ of the largest area, entirely lying within the triangle $A B C$; б) the dimensions of the rectangle with vertex $C$ of the largest area, entirely lying within the triangle $A B C$. | Answer. a) $\frac{a b}{a+b} ;$ b) $\frac{a}{2}$ and $\frac{b}{2}$. Solution. See problem 10.2 | \frac{}{+b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,860 |
11.3. The sequence $a_{n}$ is defined by the relations $a_{n+1}=\frac{1}{2}+\frac{a_{n}^{2}}{2} ; a_{1}=\frac{1}{2}$. Prove that $a_{n}$ is monotonically increasing and $a_{n}<2$ for all $n$. | Solution. We will prove the monotonicity of $a_{n}$. The inequality $a_{n+1}>a_{n}$ takes the form $\frac{1}{2}+\frac{a_{n}^{2}}{2}>a_{n} \Leftrightarrow\left(a_{n}-1\right)^{2}>0$. Note that this inequality is strict, since in fact $a_{n}<1$: indeed, let's prove the last fact by mathematical induction. We have: $a_{1}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,861 |
11.4. On the coordinate plane, consider the family of all concentric circles centered at the point $M(\sqrt{2} ; \sqrt{3})$. Prove that there exists a circle in this family, inside which (i.e., inside the disk) there are exactly 2019 points with integer coordinates. | Solution. First, we will prove that on any circle of the given family, there is no more than one point with integer (and even rational) coordinates. By contradiction, suppose that the rational points $M_{1}\left(x_{1}, y_{1}\right)$ and $M_{2}\left(x_{2}, y_{2}\right)$ lie on a circle of the given family. Then $\left(x... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,862 |
1. At a round table, 12 people are sitting. Some of them are knights, who always tell the truth, and the rest are liars, who always lie. Each person declared their left neighbor to be a liar. Can we definitely state how many knights and how many liars are at the table? | 1. Answer: Yes, it is possible.
If a knight is sitting in some place, then he told the truth, and to his left should sit a liar. Conversely, if a liar is sitting in some place, then to his left sits the one who was incorrectly called a liar, that is, a knight. This means that knights and liars alternate around the tab... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,863 |
2. Find the smallest natural number that ends in 56, is divisible by 56, and has the sum of its digits equal to 56. | 2. Answer: 29899856.
The number has the form $100 A+56$, where $A$ is divisible by 7 and is even. The sum of the digits of $A$ is 45. It is sufficient to indicate the smallest number with these properties. Clearly, it must be at least six digits. The smallest even number with a digit sum of 45 is 199998, but it is not... | 29899856 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,864 |
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