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Problem 5.8. There are 7 completely identical cubes, on which 3 points are marked on one face, 2 points on two faces, and 1 point on the rest. These cubes were glued together to form a figure in the shape of the letter "P", as shown in the image, with the number of points on any two touching faces being the same.
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,103 |
Problem 7.2. Denis divided a triangle into nine smaller triangles, as shown in the figure, and placed numbers in them, with the numbers in the white triangles being equal to the sums of the numbers in the adjacent (by sides) gray triangles. After that, Lesha erased the numbers 1, 2, 3, 4, 5, and 6 and wrote the letters... | Answer: a1 b3 c2 d5 e6 f4.
Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is $6=1+2+3$ (or the same numbers in a different order).
Now let's look at the numbers $B, D$, and $E$. The maximum value of the sum $D+E$ is the sum $5+6=11$,... | a1b3c2d5e6f4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,105 |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,109 |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,110 |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,112 |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,113 |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles $1 \... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,114 |
Problem 10.6. The graph of the quadratic trinomial $y=\frac{2}{\sqrt{3}} x^{2}+b x+c$ intersects the coordinate axes at three points $K, L$ and $M$, as shown in the figure below. It turns out that $K L=K M$ and $\angle L K M=120^{\circ}$. Find the roots of the given trinomial.
. According to the problem, triangle $O M K$ is a right triangle with a $30^{\circ}$ angle at vertex $M$, so $O M=\sqrt{3} K O=\sqrt{3} p$ and $K M=2 K O=2 p$. Al... | 0.51.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,115 |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,116 |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,117 |
# 53.
## Solution
We will consider that the week starts on the day of the week on which January 1st of the year in question falls (i.e., if January 1st is a Wednesday, then we consider the week to start on Wednesday and end on Tuesday). Then, from the beginning of the year, [365/7] = 52 (in a leap year [366/7] = 52) ... | # Solution
Consider the calculations according to Malvina's plan. Adjacent operations "divide by 8" and "multiply by 6" can be replaced by a single multiplication operation by $6 / 8 = 3 / 4$. If after multiplying by $3 / 4$ we obtained a fractional number, then subtracting 9 from it would again result in a fractional... | 18\frac{1}{6} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,118 |
1. The sum of seven different natural numbers is $2 n$. Is it true that there will definitely be three numbers among these seven numbers whose sum is greater than $n ?$ | Solution. It is sufficient to provide a counterexample. One of the possible: $n=50$, seven numbers $11,12,13,14,15,16,19$. The sum of all seven numbers is 100. The sum $15+16+19=50$. Since these three numbers are the largest among the seven, any sum of three numbers does not exceed 50.
Supplement. To construct a count... | Notnecessarily.Acounterexampleexists | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,119 |
2. Find all positive solutions of the equation
$$
x^{101} + 100^{99} = x^{99} + 100^{101}
$$
Do not forget to justify your answer. | Solution. The equation is reduced to the form
$$
x^{99}\left(x^{2}-1\right)=100^{99}\left(100^{2}-1\right)
$$
Obviously, the solution is $x=100$. For $0<x<1$, the left side is less than the right side. For $x>1$, the left side is an increasing function, so the required value is taken no more than once. Therefore, the... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,120 |
3. Given trapezoid $ABCD$ with the larger base $AD$. Point $O$ is the midpoint of the midline of the trapezoid. Points $K, L, M, N$ are the images of vertices $A, B, C, D$ under central symmetry with respect to point $O$. Lines $KM$ and $NL$ intersect sides $CD$ and $AB$ at points $P$ and $Q$ respectively. Prove that e... | Solution.
All notations are shown in the figure. Let the length of the lower base $AD$ be $a$, and the upper base $BC$ be $b$. Since point $O$ is located at an equal distance from the paral... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,121 |
5. Prove that for any natural number $a$, all prime divisors of which are not less than 7, there exists a natural number $m$ such that the product $a m$ is a number whose decimal representation consists only of ones. | Solution. Let $a$ be a number, all prime divisors of which are different from $2,3,5$. The number $\frac{1}{a}$ can be expanded into a periodic decimal fraction with a period of $n$. This means that
$$
\frac{1}{a}=0, b_{0} b_{1} b_{2} \ldots b_{n} b_{1} b_{2} \ldots b_{n} \ldots
$$
Here, $b_{0}$ is the non-periodic b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,122 |
9.1 The sum of 100 numbers is 1000. The largest of these numbers was doubled, and some other number was decreased by 10. After these actions, the sum of all the numbers did not change. Find the smallest of the original numbers. | 9.1 The sum of 100 numbers is 1000. The largest of these numbers was doubled, and some other number was decreased by 10. After these actions, the sum of all the numbers did not change. Find the smallest of the original numbers.
Otvet: 10.
$\boldsymbol{P e s h e n i e : ~ P u s t ь ~} M$ - the largest number, $t$ - so... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,123 |
9.2 ABCD is a parallelogram. K is a point such that $\mathrm{AK}=\mathrm{BD}$. Point $\mathrm{M}$ is the midpoint of CK. Prove that $\angle \mathrm{BMD}=90^{\circ}$. | 9.2 ABCD is a parallelogram. K is a point such that $\mathrm{AK}=\mathrm{BD}$. Point $\mathrm{M}$ is the midpoint of CK. Prove that $\angle \mathrm{BMD}=90^{\circ}$.
Solution: We can base our proof on the following simple fact: if the median of a triangle is equal to half the side to which it is drawn, then the triang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,124 |
9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be? | 9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be equal to?
Answer: 0.
Solution: Let $\left(x_{1}, y_{1}\right)$ be the coordinates of the vertex of one para... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,125 |
9.4 In a company of $2 n+1$ people, for any $n$ people, there is a person different from them who is acquainted with each of them. Prove that there is a person in this company who knows everyone. | 9.4 In a company of $2 n+1$ people, for any $n$ people, there is a person different from them who is acquainted with each of them. Prove that there is a person in this company who knows everyone.
Proof: Consider in this company the largest group of people in which everyone is acquainted with everyone else. It is claim... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,126 |
9.5 Prove that every natural number is the difference of two natural numbers that have the same number of prime divisors (each prime divisor is counted once; for example, the number 12 has two prime divisors: 2 and 3). | 9.5 Prove that every natural number is the difference of two natural numbers that have the same number of prime divisors (each prime divisor is counted once; for example, the number 12 has two prime divisors: 2 and 3).
Solution: If \( n \) is an even number, \( n = 2m \), then \( n = 4m - 2m \), and the numbers \( 2m ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,127 |
1. Beginner seamstress Alyona decided to sew 8 dresses for dolls. In the store, she chose a piece of fabric 140 cm long and 75 cm wide. Each dress requires a whole piece of fabric at least 45 cm long and 26 cm wide. Will the purchased fabric be enough for everything? | Answer. Yes, enough.
Solution. Let's cut the fabric from the store into two pieces measuring $140 \times 30$ and $140 \times 45$. From the first piece, we can cut out 3 blanks measuring $45 \times 26$, and from the second piece - 5 blanks measuring $45 \times 26$.
Comment. A correct example of arranging 8 blanks (in ... | Yes,enough | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,128 |
3. In the box, there are balls of seven colors. One tenth of the balls are red, one eighth are orange, and one third are yellow. There are 9 more green balls than red ones, and 10 more blue balls than orange ones. There are 8 blue balls in the box. The remaining balls are purple. What is the smallest possible number of... | Answer: 25 balls.
Solution. Let the total number of balls be $x$, and the number of violet balls be $y$. Then
$$
\frac{x}{10}+\frac{x}{8}+\frac{x}{3}+\frac{x}{10}+9+\frac{x}{8}+10+8+y=x
$$
from which $\frac{47 x}{60}+27+y=x$, that is, $y=\frac{13 x}{60}-27$, and the smallest value of $y$ is achieved at the smallest ... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,130 |
4. Vanya and Petya decided to mow the lawn for a football field. Vanya could do it alone in 5 hours, and Petya in 6 hours. They started at 11 o'clock, and stopped mowing simultaneously when their parents called them, but Petya took a one-hour lunch break, and Vanya's lunch break lasted two hours. One tenth of the lawn ... | Answer: At 15 o'clock.
Solution: Let $t$ be the time from 11 o'clock to the end, then Petya worked $(t-1)$ hours, and Vanya worked $(t-2)$ hours. In one hour, Petya completed $\frac{1}{6}$ of the work, and Vanya completed $\frac{1}{5}$ of the work. We get the equation: $\frac{1}{6} \cdot(t-1)+\frac{1}{5} \cdot (t-2)=\... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,131 |
5. Masha and Alina are playing on a $5 \times 5$ board. Masha can place one chip in some cells. After that, Alina covers all these cells with L-shaped pieces consisting of three cells (non-overlapping and not extending beyond the boundaries of the square, L-shaped pieces can only be placed along the grid lines). If Ali... | Answer: 9 chips.
Solution. Example. Masha can place chips in the cells indicated in the figure (a). Then, Alina will need nine corners, as one corner cannot cover more than one cell with a chip. However, nine corners without overlapping cannot be placed on the board, since $27>25$.
Estimate. If Masha places fewer tha... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,132 |
1. Find such non-zero numbers $n$ and $m$ that the equation $\left(n \cdot 5^{n}\right)^{n}=m \cdot 5^{9}$ holds true. | # Solution.
There are infinitely many such pairs of numbers.
Let's show one of the most natural solutions. We need $n^{n} \cdot 5^{n^{2}}=m \cdot 5^{9}$. If $n=3$, then $5^{n^{2}}=5^{9}$. Now let's calculate $m$: $m=3^{3}=27$. The pair $n=3, m=27$ is a solution.
Let's show how to obtain other solutions. Take an arbi... | n=3,=27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,133 |
2. In the underwater kingdom, there live octopuses with seven and eight legs. Those with 7 legs always lie, while those with 8 legs always tell the truth. One day, three octopuses had the following conversation.
Green octopus: “We have 24 legs together.”
Blue octopus: “You are right!”
Red octopus: “Nonsense, Green i... | Answer. The Green Octopus has 7 legs, the Blue Octopus has 7 legs, the Red Octopus has 8 legs. Solution. From the condition of the problem, it follows that the number of legs and the truthfulness of the statements are uniquely related. The Blue and Red Octopuses made contradictory statements about the words of the Gree... | TheGreenOctopushas7legs,theBlueOctopushas7legs,theRedOctopushas8legs | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,134 |
3. A company produces a lemon drink by diluting lemon juice with water. Initially, the company produced a drink containing $15\%$ lemon juice. After some time, the CEO ordered to reduce the lemon juice content to $10\%$. By what percentage will the amount of lemon drink produced increase with the same volumes of lemon ... | Answer: By $50 \%$.
Solution. Method 1. The content of lemon juice in the drink after the CEO's directive decreased by one and a half times. This means that from the same lemons, one and a half times more lemon drink can be made. In other words, the amount of lemon drink produced will increase by one and a half times ... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,135 |
4. All natural numbers, the sum of the digits in the representation of which is divisible by 5, are listed in ascending order: $5,14,19,23,28,32, \ldots$ What is the smallest positive difference between consecutive numbers in this sequence? Provide an example and explain why it cannot be smaller.
---
The smallest pos... | Answer. The smallest difference is 1, for example, between the numbers 49999 and 50000.
Solution. The difference cannot be less than 1, as we are talking about the difference between different natural numbers.
Comment. How to guess the solution.
It is clear that if two adjacent numbers differ only in the units place... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,136 |
5. On a standard graph paper, an angle is drawn (see figure). Find its measure without using measuring instruments. Justify your answer. | Answer: $45^{\circ}$.
Solution. Connect the two "extreme" points with a segment (as shown in the figure). The resulting triangle is isosceles because two of its sides, $A B$ and $B C$, are diagonals of three-cell rectangles. Diagonal $A B$ divides the angle of the rectangle at vertex $B$ into two angles that complemen... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,137 |
6. On the coordinate plane, there are points with coordinates ( $x ; y$ ) that satisfy the equation $y(x+1)=x^{2}-1$. For example, one of them is the point with coordinates ( $1 ; 0$ ). Plot all points with coordinates ( $x ; y$ ) that satisfy this equation. | Answer. See the figure.
Solution. Transform the equation as follows: $y(x+1)=(x-1)(x+1) ;$ $y(x+1)-(x-1)(x+1)=0$
$(x+1)(y-x+1)=0$. From this, $x=-1$ or $y=x-1$. Thus, all points with coordinates satisfying the equation represent a combination of two lines: a line parallel to the y-axis passing through the point ($-1 ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,138 |
# Problem № 1
Two people take turns placing nickels on a round table so that they do not overlap each other and do not extend beyond the edge of the table. The one who cannot make a move loses. Who will win? | Answer: the first player will win.
## Solution.
We need to find such a sequence of moves that would allow, looking at the opponent's moves, to make moves that would lead to victory. So, how to move after the opponent's move? The table is round, so the first move is: place the coin in the center of the board. And then... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,139 |
# Task № 2
The captain's assistant, observing the loading of the ship, smoked one pipe after another from the start of the loading. When $2 / 3$ of the number of loaded containers became equal to $4/9$ of the number of unloaded containers, and the ship's bells struck noon, the old seafarer began to light another pipe.... | Answer: The assistant smoked 5 pipes.
## Solution.
Let $x$ be the part of containers that were loaded by noon, and $y$ be the remaining part of containers. Then from the conditions we get: $\left\{\begin{array}{l}\frac{2}{3} x=\frac{4}{9} y \\ x+y=1\end{array} \Rightarrow x=\frac{2}{5}, \quad y=\frac{3}{5}\right.$.
... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,140 |
# Problem 3
Find $g(x)$, if it is known that
$$
f(x+1)=3-2 x \text{ and } f(g(x))=6 x-3
$$ | Answer: $g(x)=4-3 x$.
## Solution.
Transform the first function to the form $f(x+1)=5-2(x+1)$. This makes it clear how this linear function "works" on its argument. It will act in the same way on the function $g(x)$. That is, $f(g(x))=5-2 g(x)=6 x-3$. From which $g(x)=4-3 x$. | (x)=4-3x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,141 |
# Problem 4
Find the area of the figure defined by the inequality
$$
|x|+|y|+|x-y| \leq \mathbf{2}
$$ | Answer: The area of the figure is 3 sq.units.
## Solution.
It is easy to see that if the point $(\boldsymbol{x}, \boldsymbol{y})$ satisfies the original inequality $|\boldsymbol{x}|+|\boldsymbol{y}|+|\boldsymbol{x} \boldsymbol{y}| \leq 2$, then the point $(-x,-\boldsymbol{y})$ will also satisfy this inequality, since... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 16,142 |
# Problem 5
The bisector of a right triangle divides a leg into segments of lengths 25 cm and 24 cm. Find the length of the circumscribed circle around this triangle. | Answer: The length of the circumscribed circle is $c=175 \pi$.
## Solution
Drop a perpendicular from point M to the hypotenuse. Triangles $A C M$ and $A M N$ are equal by hypotenuse and acu... | 175\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,143 |
2. (7 points) Petya is three times older than Anya, and Anya is 8 years younger than Petya. Determine how old each of them is. Justify your answer. | Answer: Petya is 12 years old, and Anya is 4 years old.
Solution: Petya's age is three times Anya's age. This means that the difference in age between Petya and Anya is twice Anya's age, and according to the problem, this difference is eight years. Therefore, Anya's age is half of this: \(8 \div 2 = 4\) years. Petya i... | Petyais12old,Anyais4old | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,144 |
4. (7 points) In a bag, there are 15 balls (see the figure). Color each ball in one of three colors: blue, green, or red - so that two of the statements are true, and one is false:
- there is one more blue ball than red balls;
- there are an equal number of red and green balls;
- there are 5 more blue balls than green... | # Answer.
7 blue balls, 6 red balls, 2 green balls.
Solution. We will prove that the second statement cannot be true. Indeed, if the first and second statements are true, then if we remove ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,146 |
5. (7 points) Four girls are singing songs, accompanying each other. Each time, one of them plays the piano while the other three sing. In the evening, they counted that Anya sang 8 songs, Tanya - 6 songs, Olya - 3 songs, and Katya - 7 songs. How many times did Tanya accompany? Justify your answer. | Answer. Twice.
Solution. If we add up the specified number of songs sung, each song will be counted 3 times (from the perspective of each of the three singing girls). Thus, we can find out how many songs were sung in total: $(8+6+3+7): 3=8$. It is known that Tanya sang 6 out of 8 songs, so she accompanied $8-6=2$ time... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,147 |
10.1. The numbers $a$ and $b$ are such that $|a| \neq|b|$ and $\frac{a+b}{a-b}+\frac{a-b}{a+b}=6$. Find the value of the expression $\frac{a^{3}+b^{3}}{a^{3}-b^{3}}+\frac{a^{3}-b^{3}}{a^{3}+b^{3}}$. | Answer: 18/7. From the ratio $\frac{a+b}{a-b}+\frac{a-b}{a+b}=6$, after bringing to a common denominator and transformations in the numerator, we get the ratio $\frac{2 a^{2}+2 b^{2}}{a^{2}-b^{2}}=6$ or $a^{2}=2 b^{2}$. The expression $\frac{a^{3}+b^{3}}{a^{3}-b^{3}}+\frac{a^{3}-b^{3}}{a^{3}+b^{3}}$ after bringing to a... | \frac{18}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,148 |
10.3. Numbers $1,2,3,4,5,6,7$ and 8 are placed at the vertices of a cube, such that the sum of any three numbers belonging to any face is not less than 10. Find the minimum possible sum of four numbers belonging to one face. | Answer: 16. Evaluation. Consider an arbitrary face. If the largest number written at the vertex of this face is no more than 5, then the sum of the remaining numbers is no more than $4+3+2=9$. Therefore, the largest number written at the vertex of any face is at least 6, and the minimum possible sum of four numbers bel... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,149 |
10.4. Initially, there is a pile with $n$ stones ( $n \geq 2$ ). Two players take turns dividing any of the existing piles (at the beginning of the game, there is only one) into two or three non-empty piles. The player who cannot make a move loses. Which player has a winning strategy? | Answer: the first player wins. The first move of the first player: if $n$ is even, divide the pile into two equal parts; if $n$ is odd, create a third pile with one stone. The strategy for the first player thereafter is as follows. Suppose that before the second player, there are "piles" with one stone, and all other p... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,150 |
# Task 11.1
For what values of $c$ are the numbers $\sin \alpha$ and $\cos \alpha$ the roots of the quadratic equation: $10 x^{2}-7 x-c=0$ (where $\alpha$ is some angle)?
## Number of points 7 | Answer:
when $c=2.55$
## Solution
By Vieta's theorem $\sin \alpha+\cos \alpha=0.7$.
Then $1+2 \sin \alpha \cos \alpha=(\sin \alpha+\cos \alpha)^{2}=0.49$.
Therefore, $\sin \alpha \cos \alpha=-0.255$.
Thus, $c=-10 \sin \alpha \cos \alpha=2.55$.
The roots of the obtained equation are indeed the sine and cosine of ... | 2.55 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,151 |
# Problem 11.2
It is known that among the members of a certain arithmetic progression $a_{1}, a_{2}, a_{3}, a_{4}, \ldots$ there are numbers $a_{1}^{2}, a_{2}^{2}, a_{3}^{2}$.
Prove that this progression consists of integers.
Number of points 7
# | # Solution
$a_{3}-a_{2}=a_{2}-a_{1}=d-$ the difference of the progression.
$a_{1}+a_{2}=\frac{a_{2}^{2}-a_{1}^{2}}{d}$ and $a_{2}+a_{3}=\frac{a_{3}^{2}-a_{2}^{2}}{d}$ - integers, which means $a_{3}-a_{1}=2d-$ is an integer, and $d$ is an integer or half-integer.
Since $2a_{1}+d=a_{1}+a_{2}$ is an integer, there are ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,152 |
# Task 11.3
Two identical gears with 14 teeth each are placed on top of each other so that the teeth align (thus the projection on a plane looks like a single gear). After this, four pairs of aligned teeth are cut out. Is it always possible to rotate these gears relative to each other so that the projection on a plane... | # Answer:
Always
## Solution
In the initial position, four cut-out pairs of teeth coincide (one tooth on the first gear and one on the second). There are 16 such pairs in total. Therefore, in one of the 13 remaining positions, no pair coincides.
## Additional Criteria
Only the idea of coloring is proposed - 2
poi... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,153 |
# Task 11.4
A rectangle of size $7 \times 9$ is divided into squares of size $1 \times 1$. The central square is shaded. Prove that it is impossible to draw a broken line through the centers of all unshaded squares such that each segment connects the centers of adjacent squares by side without passing through any of t... | Answer:
no.
## Hint
Use a chessboard coloring.
## Solution
We will color the cells of the $7 \times 9$ square in a checkerboard pattern, so that the corner cells are black. In this case, there will be one more black cell than white. It is easy to check that the central cell will be white. Let's assign black and wh... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,154 |
# Task 11.5
The incircle of triangle $ABC$ touches side $AB$ at point $C'$. The incircle of triangle $ACC'$ touches sides $AB$ and $AC$ at points $C_1, B_1$. The incircle of triangle $BCC'$ touches sides $AB$ and $BC$ at points $C_2, A_2$. Prove that the lines $B_1C_1$, $A_2C_2$, and $CC'$ intersect at one point.
## ... | # Solution
The inscribed circles of triangles $A C C^{\prime}$ and $B C C^{\prime}$ touch the side $C C^{\prime}$ at the same point. Therefore, $C B_{1}=C A_{2}$. Additionally, $A B_{1}=A C_{1}, B A_{2}=B C_{2}$. Thus,
$\angle B_{1} C_{1} C_{2}=\angle A+\angle A B_{1} C_{1}=90^{\circ}+1 / 2 \angle A$, $\angle B_{1} A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,155 |
1. Given a two-digit number, where the first digit is greater than the second. Its digits were rearranged, and a new two-digit number was obtained. The difference between the first and the second number turned out to be a perfect square. How many possible two-digit numbers satisfy this condition? | 1. Answer: 13 numbers.
Let the first number be $10a + b$. Then the second number is $10b + a$, where $a > b > 0$. The difference is $9(a - b)$, which implies that the difference $a - b$ is a perfect square. The difference can be 1 or 4. For the first case, we have 8 values for the larger number $(21, 32, \ldots, 98)$,... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,156 |
2. Given a $7 \times 3$ table. Each cell is filled with a number 0 or 1. If at the intersection of any two rows and any two columns all four numbers are the same, then the table is considered to be filled unsuccessfully. Is it possible to fill the table in such a way to avoid this phenomenon? | 2. Answer: no, it cannot.
Suppose there are two identical rows. Among the three numbers in the row, there are two identical ones. We select the columns where these numbers are located. Then at the intersection, we get four identical numbers.
Now let's assume there are no repeated rows. In total, there are 8 different... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,157 |
3. Between the digits of the number 987654321, signs + should be placed so that the sum is 99. In how many ways can this be achieved? | 3. Answer: in two ways.
It is clear that there cannot be three-digit addends, and there must be at least one two-digit addend, since the sum of all digits is 45. Let there be one two-digit addend, and we group the digits \(a+1\) and \(a\). Then the two-digit addend is \(10(a+1) + a = 11a + 10\), and the sum of the rem... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,158 |
4. On a plane, there are 10 different points. We considered the midpoints of all segments connecting all pairs of points. What is the smallest number of midpoints that could have arisen? | 4. Answer: 17 midpoints.
Consider the case when the points lie on a single line. Introduce coordinates on it, and let the points be $a_{1}<\cdots<a_{10}$. Consider the midpoints of segments involving $a_{1}$. There are nine of them, and they form an increasing sequence $\frac{a_{1}+a_{2}}{2}<\frac{a_{1}+a_{3}}{2}<\cdo... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,159 |
5. There is a $10 \times 10$ grid for playing "Battleship". In this case, the game follows unusual rules, and only ships of size $1 \times 2$ are allowed, which can be placed both horizontally and vertically. What is the maximum number of these ships that can be placed on such a field, if the ships must not extend beyo... | 5. Answer: 13.
One of the examples of placing 13 ships is shown in the figure:
| | | | | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | |
| | | | | | | | | | |... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,160 |
10.1. The roots of the equation $x^{2}+p x+q=0$ are integers, different from zero. Prove that the number $p^{2}+(q-1)^{2}$ is composite. | Solution. Let a and $b$ be the roots of the given equation. Then, using Vieta's theorem, we obtain: $p^{2}+(q-1)^{2}=(a+b)^{2}+(a b-1)^{2}=a^{2}+2 a b+b^{2}+a^{2} b^{2}-2 a b+1=\left(a^{2}+1\right)+\left(b^{2}+\right.$ $\left.a^{2} b^{2}\right)=\left(a^{2}+1\right)\left(b^{2}+1\right)$. Since $a \neq 0$ and $b \neq 0$,... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,161 |
10.2. Can a square with a side of 8 be completely covered by two circles with a diameter of 9?
# | # Answer: possible.
Solution. Divide the square into two equal rectangles of size $4 \times 8$. The diagonal of such a rectangle is $\sqrt{4^{2}+8^{2}}=4 \sqrt{5}<9$. Therefore, each of the resulting rectangles can be covered by a circle circumscribed around it, the diameter of which is equal to the length of the diag... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,162 |
10.3. Let $n$ - be a natural number. What digit stands immediately after the decimal point in the decimal representation of the number $\sqrt{n^{2}+n}$? | Answer: 4.
Solution. We will prove that $(n+0.4)^{2} < 0.8$. The last inequality is true for any natural $n$.
2) $n^{2}+n < (n+0.5)^{2} \Leftrightarrow n < n+0.25$, which is obvious.
Evaluation Criteria.
“+” A complete and justified solution is provided
“Ғ” The correct answer is provided, but only one of the two re... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,163 |
10.4. The incircle of triangle $A B C$ touches side $A C$ at point $D$. A second circle passes through point $D$, touches ray $B A$ at point $A$, and also touches the extension of side $B C$ beyond point $C$. Find the ratio $A D: D C$. | Answer: $3: 1$.
Solution. Let the second circle touch the ray $B C$ at point $K$, and the inscribed circle touch the sides $A B$ and $B C$ of the triangle at points $P$ and $Q$ respectively (see Fig. 10.4). We can reason in different ways.
First method. Let $C D = C Q = x$ and $A D = A P = k x$. Since $B A = B K$ and... | 3:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,164 |
10.5. The tsar has eight sons, and they are all fools. Every night the tsar sends three of them to guard the golden apples from the firebird. The tsareviches cannot catch the firebird, blame each other for it, and therefore no two of them agree to go on guard duty a second time. For how many nights can this continue at... | Answer: 8 nights.
Solution. Evaluation. Consider any of the sons. In each guard, he is with two brothers, so after his three appearances, there will be one brother with whom he has not been on guard and will not be able to go, as there will be no third for them. This situation is "symmetric," meaning if son A has not ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,165 |
10.6. Petya arranged cards with numbers from 1 to 10 in a row in some order, then for each pair of adjacent cards, he wrote down the number $\frac{1}{x+y}$, where $x$ and $y$ are the numbers on these cards. Prove that the sum of the numbers written by Petya is greater than 0.75. | Solution. We will use the inequality between the arithmetic mean and the harmonic mean, which holds for any sets of positive numbers: $\frac{a_{1}+a_{2}+\ldots+a_{n}}{n} \geq \frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}} \Leftrightarrow\left(a_{1}+a_{2}+\ldots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,166 |
11.1. Arkady, Boris, and Vasili decided to run the same distance, consisting of several laps. Arkady ran each lap 2 minutes faster than Boris, and Boris ran 3 minutes faster than Vasili, and all of them ran at a constant speed. When Arkady finished the distance, Boris had one lap left to run, and Vasili had two laps le... | Answer: 6.
Solution: Let the distance be $n$ laps, and Arkady takes $t$ minutes per lap. Then in $n t$ minutes, Arkady runs the entire distance. In this time, Boris runs one lap less, and Vasily runs two laps less. Therefore, $n t=(n-1)(t+2)$ and $n t=(n-2)(t+5)$. From this, it follows that $2 n=t+2$ and $5 n=2 t+10$,... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,167 |
11.2. Is there a natural number $n$ such that the number $6 n^{2}+5 n$ is some natural power of the number $2?$ | Answer. No.
Solution. Note that $6 n^{2}+5 n=n(6 n+5)$. Since $6 n+5$ will be an odd number greater than 1, it will have at least one odd prime divisor. Therefore, $6 n^{2}+5 n$ will not be a power of 2.
Comment. Answer only - 0 points. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,168 |
11.3. Circle $\omega$ touches the sides of the angle with vertex at point $A$ at points $B$ and $C$. Prove that the center of the circle inscribed in triangle $A B C$ lies on $\omega$. | First solution. Let $N$ be the midpoint of the smaller arc $BC$ of circle $\omega$. Then $\angle NBC = \angle NCB = \angle NBA$, where the last equality follows from the fact that the angle between a chord and a tangent is equal to half the arc cut off by the chord. Since $\angle NBC = \angle NBA$, $NB$ is the bisector... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,169 |
11.4. Polynomials $f(x)$ and $g(x)$ have real coefficients, with their leading coefficients equal to 1 and the degree of each being 10. It is known that the equation $f(x)=g(x)$ has no real roots. Prove that the equation $f(x+1)=$ $g(x-1)$ has at least one real root. | Solution. Let $f(x)=x^{10}+a_{9} x^{9}+\ldots+a_{0}, g(x)=x^{10}+$ $b_{9} x^{9}+\ldots+b_{0}$. The equation $f(x)=g(x)$ is equivalent to the equation $f(x)-g(x)=0$, and if $a_{9} \neq b_{9}$, then the left side of the last equation will be a polynomial of odd (ninth) degree, and thus this equation will have at least on... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,170 |
11.5. On each of ten cards, a real number is written. For each non-empty set of these cards, the sum of all numbers written on the cards in this set was found. It is known that not all of the obtained sums are integers. What is the maximum possible number of integer sums that could have resulted? | Answer: 511.
Solution: It is clear that at least one of the cards has a non-integer number, otherwise the sum of the numbers in any set would be an integer. We select one of these cards and call it $a$, and the number on it $-x$. For each set $C$ that does not contain $a$ (including the empty set), we associate the se... | 511 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,171 |
10.1. In the store, Sasha bought two pens, three notebooks, and one pencil and paid 33 rubles, Dima bought one pen, one notebook, and two pencils and paid 20 rubles. How much did Tanya pay for four pens, five notebooks, and five pencils? | Answer: 73 rubles.
Let $a, b, c$ be the price of a pen, a notebook, and a pencil, respectively. Then, according to the condition, we have $2a + 3b + c = 33$ and $a + b + 2c = 20$. Multiplying the second relation by 2 and adding it to the first equation, we get the answer. | 73 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,172 |
10.2. Does there exist an increasing geometric progression in which the first 100 terms are integers, and all subsequent terms are not integers? | Answer: Yes.
Let $a_{1}=2^{99}-$ be the first term of the geometric progression, $q=\frac{3}{2}-$ be the common ratio of the progression. Consider the general term of the progression $a_{n}=a_{1} q^{n-1}$. The fraction $\frac{3^{n-1}}{2^{n-100}}$ is irreducible if $n>100$, since 2 and 3 are prime numbers, and is an in... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,173 |
1. The quadratic trinomials $P(x), Q(x)$, and $R(x)$ are such that each of the equations $P(x)+Q(x)=0, P(x)+R(x)=0, Q(x)+R(x)=0$, and $P(x)+Q(x)+R(x)=0$ has two roots. For each equation, the product of its roots was found. For the first three equations, these products are equal to $r, q$, and $p$ respectively. What is ... | Answer: $\frac{p+q+r}{3}$.
Solution: Let $P(x)=x^{2}+a_{1} x+a_{0}, Q(x)=x^{2}+b_{1} x+b_{0}, R(x)=x^{2}+c_{1} x+c_{0}$. Since $P(x)+Q(x)=2 x^{2}+\left(a_{1}+b_{1}\right) x+a_{0}+b_{0}$, it follows from Vieta's theorem that the product of the roots of this quadratic trinomial is $\frac{a_{0}+b_{0}}{2}$, so $r=\frac{a_... | \frac{p+q+r}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,174 |
3. On the line $l$ there are three points $A, B$, and $F$, with $B$ lying between $A$ and $F$. Squares $ABCD$ and $BFNT$ lie on the same side of the line $l$. The circle passing through points $D, B$, and $N$ intersects the line $l$ again at point $S$, different from $B$. Prove that $DS = SN$. | Solution. Without loss of generality, we can assume that $AB > BF$, then point $S$ will be on the segment $AB$. Since $BN$ is the diagonal of the square $BFNT$, and $DB$ is the diagonal of the square $ABCD$, we have $\angle NBF = \angle DBS = 45^{\circ}$. The quadrilateral $DSBN$ is cyclic, so $\angle SDN = 180^{\circ}... | DS=SN | Geometry | proof | Yes | Yes | olympiads | false | 16,176 |
4. On his birthday, Mom baked a rectangular cake for Little One. Karlson divides the cake into 6 pieces with three cuts, where two cuts are parallel to one pair of sides of the cake, and the third is parallel to the other pair of sides. Then Little One takes the largest and the smallest piece, while Karlson takes every... | Answer: No.
First solution. We will assume that two cuts are vertical, and one is horizontal. Let the segments into which the horizontal side of the pie is divided be $a \geqslant b \geqslant c$, and the segments into which the vertical side is divided be $-x \geqslant y$. Note that Little One took pieces with areas $... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,177 |
5. In a $1000 \times 1000$ grid, 2000 cells are painted blue. Prove that there exist 4 blue cells such that the centers of these cells are the vertices of a parallelogram. | Solution. We will prove the problem for an $n \times n$ square and $2n$ painted cells. We will consider the distance between two cells as the distance between the centers of these cells. Consider a column with $k$ blue cells. There are $k-1$ distances from the bottom blue cell of this column to its other blue cells, an... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,178 |
Problem 4.1. Four arithmetic examples were written on the board. Vera erased one "plus" sign, one "minus" sign, one "multiply" sign, one "divide" sign, and four "equals" signs.
Instead of the same signs, she wrote the same letters, and instead of different signs, different letters. Restore the examples.
![](https://c... | Answer: $\mathrm{a} 4 \mathrm{~b} 5 \mathrm{c} 2 \mathrm{~d} 1 \mathrm{e} 3$.
Solution. Notice that the letter $B$ appears in all examples. It turns out that the letter $B$ must be replaced by the equals sign:
$$
\begin{aligned}
& 4 A 2=2 \\
& 8=4 C 2 \\
& 2 D 3=5 \\
& 4=5 E 1
\end{aligned}
$$
Now we will look at th... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,179 | |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,181 |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's numb... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,182 |
Problem 5.4. Arrange the digits from 1 to 6 (each must be used exactly once) so that the sum of the three numbers located on each of the 7 lines is equal to 15. In your answer, indicate which digits should be placed at positions $A-F$.
![](https://cdn.mathpix.com/cropped/2024_05_06_a7569897979ba1659de3g-09.jpg?height=... | Answer: $A=4, B=1, C=2, D=5, E=6, F=3$.
Solution. According to the condition, $A, D, E$ are distinct digits, not exceeding 6, the sum of which is 15. If these digits are taken to be the maximum possible, then their sum is $4+5+6=15$. Therefore, $A, D, E$ are $4,5,6$ in some order (if at least one of the digits is no m... | A=4,B=1,C=2,D=5,E=6,F=3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,183 |
Problem 5.8. There are 7 completely identical cubes, on which 3 points are marked on one face, 2 points on two faces, and 1 point on the rest. These cubes were glued together to form a figure in the shape of the letter "P", as shown in the image, with the number of points on any two touching faces being the same.
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,186 |
Problem 6.7. Eight paper squares $2 \times 2$ were laid out on the table one after another until a large square $4 \times 4$ was formed. The last square placed on the table was $E$. The image shows how the squares are visible: square $E$ is fully visible, and the other squares are partially visible. Which square was pl... | Answer: $G$.
Solution. It is clear that different squares were placed in different positions (otherwise, the one placed last would completely cover the one placed earlier, but we can see at least one cell from each square).
The positions of squares $A, C, G$, and $F$ can be determined immediately, as they contain cor... | G | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,187 |
Problem 7.2. Denis divided a triangle into nine smaller triangles, as shown in the figure, and placed numbers in them, with the numbers in the white triangles being equal to the sums of the numbers in the adjacent (by sides) gray triangles. After that, Lesha erased the numbers 1, 2, 3, 4, 5, and 6 and wrote the letters... | Answer: a1 b3 c2 d5 e6 f4.
Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is $6=1+2+3$ (or the same numbers in a different order).
Now let's look at the numbers $B, D$, and $E$. The maximum value of the sum $D+E$ is the sum $5+6=11$,... | a1b3c2d5e6f4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,188 |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,192 |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,193 |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,195 |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,196 |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles $1 \t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,197 |
Problem 10.6. The graph of the quadratic trinomial $y=\frac{2}{\sqrt{3}} x^{2}+b x+c$ intersects the coordinate axes at three points $K, L$, and $M$, as shown in the figure below. It turns out that $K L=K M$ and $\angle L K M=120^{\circ}$. Find the roots of the given trinomial.
. According to the problem, triangle $O M K$ is a right triangle with a $30^{\circ}$ angle at vertex $M$, so $O M=\sqrt{3} K O=\sqrt{3} p$ and $K M=2 K O=2 p$. Al... | 0.51.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,198 |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,199 |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,200 |
11.1 If a sequence of different numbers $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}$ from the interval $\left(0 ; \frac{\pi}{2}\right)$ is such that the numbers $\sin \mathrm{x}_{1}, \sin \mathrm{x}_{2}, \sin \mathrm{x}_{3}$ form an arithmetic progression, then the numbers $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x... | Solution. For example, by contradiction. Suppose that $x_{1}, x_{2}, x_{3}$ form an arithmetic progression, i.e., $x_{1} + x_{3} = 2 x_{2}$. Then we have
$2 \sin x_{2} = ($ by condition $) = \sin x_{1} + \sin x_{3} = ($ trigonometric transformation of the sum into a product $) =$
$$
= 2 \sin \frac{x_{1} + x_{3}}{2} \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,201 |
11.2 Prove that if the number $t+\sqrt{2}$ is rational, then the number $t^{3}+\sqrt{2}$ is irrational. | Solution. By the condition $\mathrm{t}=\mathrm{r}-\sqrt{2}$, where $\mathrm{r}$ is a rational number. Then
$$
\begin{aligned}
& \mathrm{t}^{3}+\sqrt{2}=(\mathrm{r}-\sqrt{2})^{3}=\mathrm{r}^{3}-3 \mathrm{r}^{2} \sqrt{2}+3 \mathrm{r}(\sqrt{2})^{2}-(\sqrt{2})^{3}+\sqrt{2}= \\
& =r^{3}+6 \mathrm{r}-\left(3 \mathrm{r}^{2}+... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,202 |
11.3 A plane is drawn through a point located at a distance $\mathrm{d}$ from the center of a parallelogram. Prove that the sum of the distances from the vertices of the parallelogram to this plane does not exceed 4d. The plane does not intersect the parallelogram. | Solution. See Fig.
$\mathrm{OP}=\mathrm{d} . \mathrm{AA}_{1}, \mathrm{OO}_{1}, \mathrm{CC}_{1}$ are perpendiculars to plane П, in particular, $\mathrm{OO}_{1} \leq \mathrm{OP}=\mathrm{d}$. $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,203 |
11.4 Prove that for numbers a and b belonging to the interval $[0 ; 1]$, the inequality $a^{5}+b^{3}+(a-b)^{2} \leq 2$ holds. | Solution. If $\mathrm{x} \in [0 ; 1]$, then $\mathrm{x}^{\mathrm{n}} \leq \mathrm{x}$ for any natural $\mathrm{n}$. Therefore, $\mathrm{a}^{5} \leq \mathrm{a}, \mathrm{b}^{3} \leq \mathrm{b}, \quad(\mathrm{a}-\mathrm{b})^{2}=|\mathrm{a}-\mathrm{b}|^{2} \leq|\mathrm{a}-\mathrm{b}| \quad$ (the last inequality holds becau... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,204 |
11.5 A square with a side of 1 was cut into rectangles, and one side was marked in each of them. Prove that the sum of the lengths of all marked sides is not less than 1. | Solution. Let $\mathrm{x}_{\mathrm{n}}$ denote the length of the marked side of the $\mathrm{n}$-th rectangle, and $\mathrm{y}_{\mathrm{n}}$ the length of its other side, so that $\mathrm{x}_{\mathrm{n}} \mathrm{y}_{\mathrm{n}}$ is its area. The sum of the areas of all rectangles is equal to the area of the square, i.e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,205 |
Problem 5.3. In five of the nine circles in the picture, the numbers 1, 2, 3, 4, 5 are written. Replace the digits $6, 7, 8, 9$ in the remaining circles $A, B, C, D$ so that the sums of the four numbers along each of the three sides of the triangle are the same.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,209 |
Problem 5.6. There are 4 absolutely identical cubes, each of which has 6 dots marked on one face, 5 dots on another, ..., and 1 dot on the remaining face. These cubes were glued together to form the figure shown in the image.
How many dots are on the four left faces?
![](https://cdn.mathpix.com/cropped/2024_05_06_8ce... | Answer: On face $A$ there are 3 points, on face $B-5$, on face $C-6$, on face $D-5$. Solution. Let's consider the arrangement of the faces on one die. We will denote the faces by numbers corresponding to the number of dots on them.
From the picture, it is clear that face 1 borders with faces $2,3,4$ and 5. Therefore, ... | 3,5,6,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,210 |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,211 |
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