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Variant 9.2.4. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=9$, $NC=39$, $AM=10$, $MB=5$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 675. (b) 247.5. | 675 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,258 |
Variant 9.3.1. Natural numbers $a$ and $b$ are such that $a$ is divisible by $b+1$ and 43 is divisible by $a+b$.
(a) (1 point) Indicate any possible value of $a$.
(b) (3 points) What can $b$ be? Indicate all possible options. | Answer: (a) any of the numbers $22,33,40,42$. (b) $1,3,10,21$. | 22,33,40,42 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,259 |
Variant 9.3.2. Natural numbers $a$ and $b$ are such that $a$ is divisible by $b+1$ and 67 is divisible by $a+b$.
(a) (1 point) Indicate any possible value of $a$.
(b) (3 points) What can $b$ be? Indicate all possible options. | Answer: (a) any of the numbers $34,51,64,66$. (b) $1,3,16,33$. | 34,51,64,66 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,260 |
Variant 9.4.1. An empty $3 \times 51$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1, 2, 3, \ldots, 153$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 152$, the n... | Answer: (a) 152. (b) 76. | 152 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,261 |
Variant 9.4.2. An empty $3 \times 53$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1,2,3, \ldots, 159$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 158$, the num... | Answer: (a) 158. (b) 79. | 158 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,262 |
Variant 9.4.3. An empty $3 \times 55$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1,2,3, \ldots, 165$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 164$, the num... | Answer: (a) 164 . (b) 82. | 82 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,263 |
Variant 9.4.4. An empty $3 \times 57$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1, 2, 3, \ldots, 171$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 170$, the n... | Answer: (a) 170 . (b) 85. | 170 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,264 |
Variant 9.5.1. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 34^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees d... | Answer: (a) $107^{\circ}$. (b) $47^{\circ}$. | 107 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,265 |
Variant 9.5.2. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 32^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees do... | Answer: (a) $106^{\circ}$. (b) $48^{\circ}$. | 106 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,266 |
Variant 9.5.3. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 36^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees d... | Answer: (a) $108^{\circ}$. (b) $46^{\circ}$. | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,267 |
Variant 9.5.4. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 38^{\circ}$ and $\angle X D C = 54^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees d... | Answer: (a) $109^{\circ}$. (b) $44^{\circ}$. | 109 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,268 |
9.1. The numbers $a$ and $b$ are such that each of the two quadratic trinomials $x^{2} + a x + b$ and $x^{2} + b x + a$ has two distinct roots, and the product of these trinomials has exactly three distinct roots. Find all possible values of the sum of these three roots.
(S. Berlov) | Answer: 0.
Solution: From the condition, it follows that the quadratic polynomials $x^{2}+a x+b$ and $x^{2}+b x+a$ have a common root $x_{0}$, as well as roots $x_{1}$ and $x_{2}$ different from it; in particular, $a \neq b$. Then $0=\left(x_{0}^{2}+a x_{0}+b\right)-\left(x_{0}^{2}+b x_{0}+a\right)=(a-b)\left(x_{0}-1\... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,269 |
9.2. Parallelogram $A B C D$ is such that $\angle B<90^{\circ}$ and $A B<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$. It turns out that $\angle E D A=\angle F D C$. Find the angle $A B C$.
(A. Yakub... | Answer: $60^{\circ}$.
Solution. Let $\ell$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents to $\omega$, the line $\ell$ passes through the center $O$ of the circle $\omega$.
Perform a reflection about $\ell$. Since $\angle E D A = \angle F D C$, the ray $D C$ will be transformed into the ray $D A... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,270 |
9.3. Natural numbers $a, x$ and $y$, greater than 100, are such that $y^{2}-1=$ $=a^{2}\left(x^{2}-1\right)$. What is the smallest value that the fraction $a / x$ can take? | # Answer. 2.
First solution. Rewrite the condition of the problem as $y^{2}=a^{2} x^{2}-a^{2}+1$. Notice that $y100$, if we set $a=2 x, y=a x-1=2 x^{2}-1$.
Second solution. We provide another proof of the estimate $a / x \geqslant 2$. Rewrite the equality from the condition as
$$
(a x-y)(a x+y)=a^{2} x^{2}-y^{2}=a^{... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,271 |
9.4. In a volleyball tournament, 110 teams participated, each playing exactly one game with each of the others (there are no ties in volleyball). It turned out that in any group of 55 teams, there is one that lost to no more than four of the other 54 teams in this group. Prove that in the entire tournament, there is a ... | Solution. Lemma. Let $k \geqslant 55, u$ be such that among any $k$ teams, there is one that lost to no more than four of the remaining $k-1$ teams. Then among any $k+1$ teams, there is one that lost to no more than four of the remaining $k$ teams.
Proof. Assuming the contrary, consider a set of $k+1$ teams $M=\left\{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,272 |
8.1. Find any four different natural numbers that have the following property: if the product of any two of them is added to the product of the other two numbers, the result is a prime number. | Solution. For example, the numbers 1, 2, 3, and 5 will work. Indeed, the values of all three expressions $1 \cdot 2 + 3 \cdot 5 = 17$, $1 \cdot 3 + 2 \cdot 5 = 13$, and $1 \cdot 5 + 2 \cdot 3 = 11$ are prime numbers.
Remark. There are other examples, such as $(1,2,3,13)$, $(2,5,11,17)$, or $(4,5,7,11)$. It is not hard... | 1,2,3,5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,273 |
8.2. In a convex quadrilateral $A B C D$, point $K$ is the midpoint of $A B$, point $L$ is the midpoint of $B C$, point $M$ is the midpoint of $C D$, and point $N$ is the midpoint of $D A$. For some point $S$ lying inside the quadrilateral $A B C D$, it turns out that $K S=L S$ and $N S=M S$. Prove that $\angle K S N=\... | Solution. Note that segment $K N$ is the midline of triangle $B A D$. Therefore, $K N=\frac{B D}{2}$. Similarly, from triangle $B C D$ we get $L M=\frac{B D}{2}$. Thus, triangles $K S N$ and $M S L$ are equal by the three sides. Hence, the required equality of the corresponding angles follows.
Remark. It follows from ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,274 |
8.3. Workers were laying a floor of size $n \times n$ using two types of tiles: $2 \times 2$ and $3 \times 1$. It turned out that they managed to completely cover the floor such that the same number of tiles of each type was used. For which $n$ could this have been possible? (Cutting tiles or overlapping them is not al... | Answer. For $n$ divisible by 7.
Solution. Let the workers use $x$ tiles of each type. Then the area covered by the tiles is $4x + 3x = 7x = n^2$. Therefore, $n$ must be divisible by 7.
If $n$ is divisible by 7, then the floor can be laid. It is sufficient to note that a $7 \times 7$ square can be laid using 7 tiles o... | ndivisible7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,275 |
8.4. The sum of the numbers $a, b$, and $c$ is zero, and their product is negative. Prove that the number $\frac{a^{2}+b^{2}}{c}+\frac{b^{2}+c^{2}}{a}+\frac{c^{2}+a^{2}}{b}$ is positive. | First solution. Since $a b c < 0$, $b > 0$, and $c < 0$. Or $\frac{b^{2}+c^{2}}{a}+\frac{c^{2}+a^{2}}{b}>-\frac{a^{2}+b^{2}}{c}=$ $=\frac{a^{2}+b^{2}}{a+b}$. Since $a > 0$ and $b > 0$, then $\frac{b^{2}+c^{2}}{a}>\frac{b^{2}}{a}>\frac{b^{2}}{a+b}$. Similarly, $\frac{c^{2}+a^{2}}{b}>\frac{a^{2}}{b}>\frac{a^{2}}{a+b}$. A... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,276 |
8.5. There are 300 coins on the table. Petya, Vasya, and Tolya are playing the following game. They take turns in the following order: Petya, Vasya, Tolya, Petya, Vasya, Tolya, and so on. In one move, Petya can take 1, 2, 3, or 4 coins, Vasya can take 1 or 2 coins, and Tolya can also take 1 or 2 coins. Can Vasya and To... | Answer: No.
Solution. We will show how Pete can play to take the last coin from the table regardless of Vasya and Tolya's moves. Let Pete's first move be to take 4 coins. Note that Vasya and Tolya can take a total of 2 to 4 coins in their turns. This means that after Tolya's first move, there will be between 292 and 2... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,277 |
9.1. The parabola $y=20 x^{2}+19 x$ and the line $y=20 x+19$ intersect at two points. Is it true that the graph of the function $y=20 x^{3}+19 x^{2}$ passes through these same two points? | Answer: Yes.
Solution. First method. If the parabola and the line intersect at two points, then the equation $20 x^{2}+19 x=20 x+19$ has two distinct roots. Multiplying both sides of the equation by $x$, we get the equation $20 x^{3}+19 x^{2}=20 x^{2}+19 x$, which has the same roots and an additional root $x=0$. There... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,278 |
9.2. Given an isosceles trapezoid with bases 4 and 12 and height 4. Can it be cut into three pieces and these pieces be assembled into a square? | Answer: It is possible.
Solution. Let in trapezoid $ABCD: BC=4, AD$ $=12$ and height $BH=4$ (see Fig. 9.2a). We also draw segment $CH$. Since $AH=\frac{AD-BC}{2}=4=BC$, then $ABCH$ is a parallelogram, composed of two
 / 2 * 45 = 2025 > 2019. Therefore, there are fewer than 45 addends, but the sum of 44 odd addends is an even number, so there are no more than 43 addends. Example. 2019 = 1 + 3 + ... + 81 ... | 43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,280 |
9.4. On a semicircle with diameter $A D$, points $B$ and $C$ are marked. Point $M$ is the midpoint of segment $B C$. Point $N$ is such that $M$ is the midpoint of segment $A N$. Prove that lines $B C$ and $D N$ are perpendicular. | Solution. First method. Let $O$ be the center of the circle. Since $M$ is the midpoint of $BC$, then $OM \perp BC$ (see Fig. 9.4a). Moreover, $OM$ is the midline of triangle $AND$, so $OM \| DN$. Therefore, $DN \perp BC$.
Second method. From the problem statement, it follows that $ABNC$ is a parallelogram (see Fig. 9.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,281 |
9.5. In a round-robin chess tournament, two boys and several girls participated. The boys scored a total of 8 points, while all the girls scored an equal number of points. How many girls could have participated in the tournament? (Win - 1 point, draw - 0.5 points, loss - 0 points.) | Answer: 7 or 14.
Solution. Let $n$ be the number of girls participating in the tournament and each of them scored $x$ points, then the total number of points scored is $n x+8$. On the other hand, each of the $n+2$ chess players played $n+1$ games, so the total number of games played in the tournament is $\frac{(n+2)(n... | 7or14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,282 |
9.6. Find the largest $n$ such that the sum of the fourth powers of any $n$ prime numbers greater than 10 is divisible by $n$. | Answer: $n=240$.
Solution. In fact, it is required that the fourth powers of all prime numbers greater than 10 give the same remainder when divided by $n$. Indeed, if the fourth powers of some two numbers give different remainders, then one can take $n-1$ of the first number and one of the second, then the sum of the ... | 240 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,283 |
2. A six-digit number $A$ is divisible by 19. The number obtained by removing its last digit is divisible by 17, and the number obtained by removing the last two digits of $A$ is divisible by 13. Find the largest $A$ that satisfies these conditions. | 2. Answer: 998412.
Solution. The largest four-digit number divisible by 13 is 9997. Among the numbers from 99970 to 99979, there is a number 99977 divisible by 17, but among the numbers from 999770 to 999779, there is no number divisible by 19. However, for the next number 9984, which is divisible by 13, the number 99... | 998412 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,285 |
3. The numbers $x$, $y$, $z$, and $t$ are such that $x>y^{3}$, $y>z^{3}$, $z>t^{3}$, $t>x^{3}$. Prove that $x y z t>0$. | 3. Solution. If $x>0$ then $x^{3}>0$. Then $t>x^{3}>0, z>t^{3}>0$ and $y>z^{3}>0$, that is, all numbers are positive. If $x<0$, then $x^{3}<0$. Then $t<x^{3}<0, z<t^{3}<0$ and $y<z^{3}<0$, that is, all numbers are negative. In both cases, we have a contradiction because if $x>0$, then $y>0, z>0, t>0$ and $x>0$, or, as ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,286 |
4. We will call a non-empty set of distinct natural numbers from 1 to 13 good if the sum of all the numbers in it is even. How many good sets are there in total? | 4. Answer: $2^{12}-1$.
Solution. There are a total of $2^{13}$ different subsets (including the empty set) of 13 numbers. We will show that exactly half of them are good. Indeed, since the sum of all numbers from 1 to 13, which is 91, is an odd number, for each good subset of numbers, the remaining numbers form a bad ... | 2^{12}-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,287 |
5. Quadrilateral $ABCD$ is inscribed in a circle. On the extension of diagonal $BD$ beyond point $D$, a point $F$ is chosen such that $\mathrm{AF}|| \mathrm{BC}$. Prove that the circle circumscribed around triangle $ADF$ is tangent to the line $AC$. | 5. Solution. We have $\angle A F B=\angle F B C=\angle D B C=\angle D A C$. The first equality follows from the parallelism of lines $A F$ and $B C$, the second equality - from the fact that point $D$ lies on segment $FB$, finally, the third equality follows from the equality of inscribed angles subtending the same arc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,288 |
1. A row of 50 people, all of different heights, stands. Exactly 15 of them are taller than their left neighbor. How many people can be taller than their right neighbor? (Provide all possible answers and prove that there are no others) | Answer: 34.
Solution. Let's call 15 people who are taller than their left neighbor - tall people.
If someone has a tall person to their right, then they are shorter than them, otherwise, they are taller.
There are a total of 49 pairs of adjacent people.
Tall people are on the right in exactly 15 pairs. In the remai... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,289 |
2. A line intersects the graph of the function $y=x^{2}$ at points with abscissas $x_{1}$ and $x_{2}$, and the x-axis at a point with abscissa $x_{3}$. Prove that $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{1}{x_{3}} \cdot\left(x_{1}, x_{2}, x_{3}\right.$ - are non-zero.) | Solution. Let the equation of the line be $y=a x+b$.
Then the graphs intersect at points with abscissas satisfying the equation $x^{2}-a x-$ $b=0$. By Vieta's theorem, $x_{1}+x_{2}=a, x_{1} x_{2}=-b$, then $-\frac{a}{b}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=\frac{1}{x_{1}}+\frac{1}{x_{2}}$. From $a x_{3}+b=0$, we get that $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,290 |
3. Prove that if $a^{2}+b^{2}+a b+b c+c a<0$, then $a^{2}+b^{2}<c^{2}$. | Solution.
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca \geq 0>a^{2}+b^{2}+ab+bc+ca=>$
$c^{2}+ab+bc+ca>0>a^{2}+b^{2}+ab+bc+ca=>c^{2}>a^{2}+b^{2}$ which was to be proved. | ^{2}>^{2}+b^{2} | Inequalities | proof | Yes | Yes | olympiads | false | 17,291 |
4. On the lateral sides $A B$ and $C D$ of trapezoid $A B C D$, points $X$ and $Z$ are chosen respectively. Segments $C X$ and $B Z$ intersect at point $Y$. It turns out that the pentagon $A X Y Z D$ is inscribed.
Prove that $A Y=D Y$. | Solution.
Since $A B C D$ is a trapezoid, then $\angle B A D + \angle A B C = 180$.
Since points $A, X, Z, D$ lie on the same circle, then $\angle B A D = \angle X A D = \angle X Z C$.
Then $\angle A B C + \angle X Z C = 180$, which means points $X, B, C, Z$ lie on the same circle.
Then the angles (as subtended by ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,292 |
5. Maxim wrote down 2021 different natural numbers and found their product. It turned out that this product is divisible by exactly 2020 different prime numbers. Prove that among the numbers written by Maxim, one can choose several numbers whose product is a perfect square, or one number which is a perfect square. | Solution.
Consider all non-empty subsets of a set of 2021 listed numbers. There are $2^{2021}-1$ of them.
Let's number the prime numbers that divide the product of the numbers: $\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots, \mathrm{p}_{2020}$.
We will associate each subset with a code consisting of 2020 numbers ( $\mathrm... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,293 |
1. Answer: $19890+8716=28606$. | Solution: Obvious remarks: $\Gamma=0, \mathrm{M}=Д+1$.
In the thousands place, the sum $\mathrm{E}+\mathrm{H}$ can end in $\mathrm{H}$ only if $\mathrm{E}=9$ and there is a carry of 1 from the hundreds place.
There is no carry from the units place to the tens place, so Д=1. From this, М=2.
In the word ДЕНЕГ, 4 digit... | 28606 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,294 |
11.5. Given a polynomial $P(x)$ and numbers $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ such that $a_{1} a_{2} a_{3} \neq 0$. It turns out that for any real $x$ the equality
$$
P\left(a_{1} x+b_{1}\right)+P\left(a_{2} x+b_{2}\right)=P\left(a_{3} x+b_{3}\right)
$$
holds. Prove that $P(x)$ has at least one real root.
(... | First solution. Suppose that $P(x)$ has no real roots. Then $P(x)$ has an even degree, not less than 2. Indeed, any polynomial of odd degree has at least one real root, and if $P(x)=$ const, then from the condition we get that $P(x) \equiv 0$.
Since $P(x)$ has no real roots, it takes values of one sign. Let's assume t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,295 |
11.6. Points $A_{1}, B_{1}, C_{1}$ are chosen on the sides $B C, C A$, and $A B$ of triangle $A B C$ respectively. It turns out that $A B_{1} - A C_{1} = C A_{1} - C B_{1} = B C_{1} - B A_{1}$. Let $O_{A}, O_{B}$, and $O_{C}$ be the centers of the circumcircles of triangles $A B_{1} C_{1}, A_{1} B C_{1}$, and $A_{1} B_... | Solution. Let $I$ be the center of the circle inscribed in triangle $ABC$, and let $A_{0}, B_{0}, C_{0}$ be the points of tangency with sides $BC$, $CA$, and $AB$, respectively. We will assume that point $A_{1}$ lies on the segment $A_{0}B$. Note that $C A_{0} + A C_{0} = C B_{0} + A B_{0} = C A$. From the condition, i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,296 |
11.7. On a circle, $2 n+1$ points are marked, dividing it into equal arcs $(n \geqslant 2)$. Two players take turns erasing one point each. If after a player's move, all triangles with vertices at the remaining marked points are obtuse, he immediately wins, and the game ends. Who will win with correct play: the player ... | Answer. The second player.
Solution. We will present a strategy for the second player that allows him to win. For this, he will strive to meet the following condition: before each move of the first player, if there are $2 k+1 \geqslant 5$ points left, then on any semicircle, there are at least $k$ marked points left.
... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,297 |
11.8. For a natural number $n$, denote $S_{n}=1!+2!+\ldots+n!$. Prove that for some $n$, the number $S_{n}$ has a prime divisor greater than $10^{2012}$.
( $\Phi$. Petrov $)$ | Solution. For a prime $p$ and a natural number $n$, denote by $\nu_{p}(n)$ the power to which $p$ is raised in the prime factorization of $n$. Note that if $\nu_{p}(n) \neq \nu_{p}(k)$, then $\nu_{p}(n \pm k) = \min \left(\nu_{p}(n), \nu_{p}(k)\right)$.
Assume the contrary; let $P=10^{2012}$. Then all prime divisors o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,298 |
11.2. First Solution. First, let's prove the following well-known fact.
Lemma. The orthocenter of a triangle, after reflection over a side, lies on the circumcircle.
Proof. Consider the case of an acute triangle (see Fig. 4; other cases are analogous). We have \(\angle A H^{\prime} C = \angle A H C = \angle A_{1} H C... | The second solution. Again, note that $O_{1} O_{2}$ and $O_{2} O_{3}$ are the perpendicular bisectors of segments $A T$ and $D T$. Therefore, the altitudes from vertices $O_{1}$ and $O_{3}$ of triangle $O_{1} O_{2} O_{3}$ are parallel to the lines $D T$ and $A T$, respectively. Let $H$ be the point of intersection of t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,299 |
1.1. In a cross-country skiing race, Petya and Vasya started simultaneously. Vasya ran the entire race at a constant speed of 12 km/h. Petya ran the first half of the distance at a speed of 9 km/h and fell behind Vasya. What should Petya's speed be on the second half of the distance to catch up with Vasya and finish at... | Answer: 18
Solution. Let $t$ be the time it takes for Petya to run half the distance, then the length of half the distance is $9t$, and the entire distance is $18t$. Vasya runs $12t$ in time $t$, and he has $6t$ left to the finish, which will take him time $t/2$. For Petya to run the second half of the distance and ca... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,300 |
2.1. On 40 cells of an $8 \times 8$ chessboard, stones were placed. The product of the number of stones lying on white cells and the number of stones lying on black cells was calculated. Find the minimum possible value of this product. | Answer: 256
Solution. Let $b$ and $w=40-b$ be the number of stones on black and white cells, respectively. Without loss of generality, assume that $b \geqslant w$, so $b$ can take any integer value from 20 to 32. We are interested in the minimum value of the product $b(40-b)$. The quadratic function $x(40-x)=-x^{2}+40... | 256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,301 |
3.1. Kolya had 10 sheets of paper. On the first step, he chooses one sheet and divides it into two parts. On the second step, he chooses one sheet from the available ones and divides it into 3 parts, on the third step, he chooses one sheet from the available ones and divides it into 4, and so on. After which step will ... | # Answer: 31
Solution. On the $k$-th step, the number of leaves increases by $k$, so after $k$ steps, the number of leaves will be $10+(1+2+\ldots+k)=10+\frac{k(k+1)}{2}$. When $k=30$, this number is less than 500 (it equals 475), and when $k=31$, it is already more than 500 (it equals 506), so the answer to the probl... | 31 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,302 |
4.1. A two-digit natural number $\overline{a b}$ is randomly selected from 21 to 45 (the probability of selecting each number is the same). The probability that the number $\overline{a 8573 b}$ is divisible by 6 is $n$ percent. Find $n$. | Answer: 16
Solution. There are a total of 25 options. The number $\overline{a 8573 b}$ is divisible by 6 if and only if both conditions are met simultaneously: $b$ is even and the sum of the digits $a+8+5+7+3+b$ is divisible by 3. Among the numbers from 21 to 45, the numbers $22,28,34,40$ (every sixth number) fit - th... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,303 |
5.1. $B$ of trapezoid $A B C D: \angle A=\angle B=90^{\circ}, A D=2 \sqrt{7}, A B=\sqrt{21}, B C=2$. What is the minimum value that the sum of the lengths $X A+X B+X C+X D$ can take, where $X-$ is an arbitrary point in the plane?
, B\left(x_{b}, y_{b}\right), C\left(x_{c}, y_{c}\right)$ are taken. It is known that $x_{c}=5$ and $y_{a}=y_{b}$. Find the abscissa of the point of intersection of the medians of triangle $A B C$.
. Find the number of rational numbers among the listed ones. | # Answer: 13
Solution. Multiply the numbers by 100, we get $\sqrt{73010}, \sqrt{73020}, \sqrt{73030}, \ldots, \sqrt{160030}$ (in this case, rational numbers will remain rational, and irrational numbers will remain irrational). The square root of a natural number $n$ is a rational number if and only if $n$ is a perfect... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,306 |
8.1. 72 vertices of a regular 3600-gon are painted red such that the painted vertices are the vertices of a regular 72-gon. In how many ways can 40 vertices of this 3600-gon be chosen so that they are the vertices of a regular 40-gon and none of them are red? | Answer: 81
Solution. Number the vertices in order $1,2, \ldots, 3600$, so that the painted vertices are those with numbers divisible by $3600 / 72=50$. 40 vertices form the vertices of a regular 40-gon if their numbers give the same remainder when divided by $3600 / 40=90$. There are 90 ways to choose a regular 40-gon... | 81 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,307 |
7.1. In the record $\frac{1}{4} \frac{1}{4} \frac{1}{4} \frac{1}{4}$, insert operation signs and, if necessary, parentheses so that the value of the resulting expression equals 2. | Answer: for example, $\frac{1}{4}: \frac{1}{4}+\frac{1}{4}: \frac{1}{4}=2$ or $\frac{1}{4}:\left(\frac{1}{4}+\frac{1}{4}\right): \frac{1}{4}=2$.
There are other solutions.
+ correct answer (or more than one correct answer) provided
$\pm$ correct and incorrect answers are provided together
- incorrect answer provide... | \frac{1}{4}:(\frac{1}{4}+\frac{1}{4}):\frac{1}{4}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,308 |
7.2. Getting ready for school, Misha found under the pillow, under the sofa, on the table, and under the table everything he needed: a notebook, a cheat sheet, a player, and sneakers. Under the table, he found neither the notebook nor the player. Misha's cheat sheets never lie on the floor. The player was not on the ta... | Answer: the notebook was under the sofa, the crib - on the table, the player - under the pillow, the sneakers - under the table.
The solution to the problem can be presented in various ways.
The first way. According to the condition, the player was not found under the table, not on the table, and not under the sofa. ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,309 |
7.4. Little One gave Karlson 111 candies. They immediately ate some of them together, $45\%$ of the remaining candies went to Karlson for lunch, and a third of the candies left after lunch were found by Fräulein Bok during cleaning. How many candies did she find? | Answer: 11.
First method. Let Carlson have had $n$ candies before lunch. Then, after lunch, there were $\frac{55}{100} n$ left, and Fräulein Bock found $\frac{1}{3} \cdot \frac{55}{100} n=\frac{11 n}{60}$ candies. Since the number of candies must be an integer, $11 n$ is divisible by 60. Since the numbers 11 and 60 ar... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,311 |
7.5. In the cells of a $3 \times 3$ square, numbers are arranged (see the left figure).
It is allowed to add the same number, not necessarily positive, simultaneously to the numbers in two adjacent cells. Can one obtain a square with numbers as shown in the right figure at some point? (Cells are considered adjacent if... | Answer: No, it cannot.
Let's divide all the cells of the square into two groups. One group will include the central cell and the 4 corner cells, while the other group will include the remaining 4 cells. Then, from any two adjacent cells of the square, one will fall into the first group, and the other into the second g... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,312 |
2. (7 points) The girl replaced each letter in her name with its number in the Russian alphabet. The result was the number 2011533. What is her name? | Answer: Tanya.
Other sequences of letters may also result, but only one of them is a name.
Criteria. Name Tanya obtained: 7 points.
Any other sequence of letters obtained: 0 points. | Tanya | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,313 |
3. (7 points) A seller bought a batch of pens and sold them. Some customers bought one pen for 10 rubles, while others bought 3 pens for 20 rubles. It turned out that the seller made the same profit from each sale. Find the price at which the seller bought the pens. | Answer: 5 rubles.
Solution. Let the purchase price of a pen be $x$. Then the profit from one pen is $10-x$, and from 3 pens is $20-3x$. Solving the equation $10-x=20-3x$, we get $x=5$.
Criteria. Correct solution by any method: 7 points.
If it is not justified that the purchase price of the pen should be 5 rubles, bu... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,314 |
11-1. Given three numbers. If each of them is decreased by 1, then their product also decreases by 1. If all the original numbers are decreased by 2, then their product also decreases by 2.
a) By how much will the product decrease if all the original numbers are decreased by 3?
b) Provide any three numbers that satis... | Answer: a) by 9; b) $1,1,1$.
Solution. a) Let the initial numbers be $a, b$, and $c$. Denote $ab + bc + ca = x$ and $a + b + c = y$. According to the condition, $(a-1)(b-1)(c-1) = abc - 1$ and $(a-2)(b-2)(c-2) = abc - 2$. Expanding the brackets and combining like terms, we get $x - y = 0$ and $-2x + 4y = 6$, from whic... | 1,1,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,318 |
11-2. In a certain coordinate system, the graph of the function $y=\cos ^{2} x$ was plotted. After that, the coordinate axes were erased. Construct a coordinate system so that the same line becomes the graph of the function $z=\cos t$ in the new coordinate system. | Solution. We have $y=\cos ^{2} x=\frac{1+\cos 2 x}{2}$. We can now set $2 x=t, y=(z+1) / 2$, then the variables $z$ and $t$ are related by $z=\cos t$, meaning the pair $(t, z)$ defines the position of a point in the desired coordinate system.
We see that $t=0$ when $x=0$, and $t=1$ when $x=2$. This means that the scal... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,319 |
11-3. Real numbers $a, b$ and $c$ are such that $|a| \geq|b+c|,|b| \geq|c+a|$ and $|c| \geq|a+b|$. Prove that $a+b+c=0$. | Solution. Let's square both sides of each inequality:
$$
\left\{\begin{array}{l}
a^{2} \geq(b+c)^{2} \\
b^{2} \geq(c+a)^{2} \\
c^{2} \geq(a+b)^{2}
\end{array}\right.
$$
Add all three inequalities and after rearranging the terms, we get $(a+b+c)^{2} \leq 0$. From this, it is clear that $a+b+c=0$.
Criteria. Full answe... | +b+=0 | Algebra | proof | Yes | Yes | olympiads | false | 17,320 |
11-4. In the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side 1, the diagonal $A C_{1}$ is drawn. For each point $N$ lying on this diagonal, a section of the cube is constructed by a plane perpendicular to $A C_{1}$ and passing through $N$. Let $P$ be the perimeter of this section. Construct the graph of the dependence... | Solution. Note that the sections $B D A_{l}$ and $B_{l} D_{l} A$ are perpendicular to the diagonal $A C_{l}$ and divide it into three equal parts, each of length $\frac{\sqrt{3}}{3}$. It is sufficient to consider the trace that the section leaves on two faces, for example, $A B B_{l} A_{l}$ and $B C C_{l} B_{l}$, the e... | \begin{cases}3\sqrt{6}x&for0\leqx\leq\frac{\sqrt{3}}{3}\\3\sqrt{2}&for\frac{\sqrt{3}}{3}\leqx\leq\frac{2\sqrt{3}}{3}\\3\sqrt{6}(\sqrt{3}-x | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,321 |
11-5. Santa Claus is preparing gifts. He has distributed 115 candies into bags, with each bag containing a different number of candies. In the three smallest gifts, there are 20 candies, and in the three largest - 50. How many bags are the candies distributed into? How many candies are in the smallest gift? | Answer: 10 packages, 5 candies.
Solution. Let's number the gifts from the smallest to the largest, from 1 to n. If the third gift has 7 or fewer candies, then the three smallest gifts have no more than $7+6+5=18$ candies. This contradicts the condition. Therefore, the third gift has at least 8 candies. Similarly, the ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,322 |
# 1. CONDITION
On a line, several points were marked. Then, between each pair of adjacent points, one more point was marked, and this operation was repeated once more. As a result, 101 points were obtained. How many points were marked initially? | Solution. Let there be $k$ points marked initially. Then $k-1$ more points were added to them (one between the first and second, second and third, $\ldots, (k-1)$-th and $k$-th marked points), and then another $(k+(k-1))-1=2k-2$ points. In total, the number of points became $4k-3$. Solving the equation $4k-3=101$, we f... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,323 |
# 2. CONDITION
The messenger needs to run 24 miles. Two-thirds of this distance he ran at an average speed of 8 miles per hour. Will he be able to, by increasing his speed, run the remaining distance so that his average speed for the entire journey equals 12 miles per hour? | Solution. To achieve an average speed of 12 miles per hour for the entire journey, the messenger must run 24 miles in 2 hours. But he has already spent these 2 hours on the first 16 miles.
Answer: cannot. | cannot | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,324 |
# 3. CONDITION
Perpendiculars $B E$ and $D F$, dropped from vertices $B$ and $D$ of parallelogram $A B C D$ to sides $A D$ and $B C$ respectively, divide the parallelogram into three parts of equal area. On the extension of diagonal $B D$ beyond vertex $D$, segment $D G$ is laid off, equal to segment $B D$. Line $B E$... | Solution. According to the condition $(\mathrm{AE} \cdot \mathrm{BE}): 2=\mathrm{ED} \cdot \mathrm{BE}$, hence $\mathrm{AE}=2$ ED. Note that AD is the median of triangle ABG. Therefore, the segment $\mathrm{BH}$, dividing the median $A D$ in the ratio $\mathrm{AE}: \mathrm{ED}=2$, is also a median of triangle ABG.
Ans... | 1:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,325 |
# 4. CONDITION
The sum of three non-negative numbers $x_{1}, x_{2}$, and $x_{3}$ does not exceed $1 / 2$. Prove that $\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right) \geq 1 / 2$. | Solution. Transform the left part $A$ of the inequality to be proven:
$\mathrm{A}=\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)=1-\left(x_{1}+x_{2}+x_{3}\right)+x_{1} x_{2}\left(1-x_{3}\right)+x_{2} x_{3}+x_{1} x_{3}$. From the condition, it follows that $x_{1} x_{2}\left(1-x_{3}\right) \geq 0, x_{2} x_{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,326 |
# 5. CONDITION
$n$ integers are written in a circle, the sum of which is 94. It is known that any number is equal to the absolute value of the difference of the two numbers following it. Find all possible values of $n$. | Solution. From the condition, it follows that all the recorded numbers are non-negative. Let $a$ be the largest of these numbers (if there are several, then we choose any one of them), and $b, c, d, e$ be the next numbers in the circle. Then $a=|b-c|$, which is possible only if one of the numbers $b$ or $c$ equals $a$,... | n=3orn=141 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,327 |
8.1. Find ten natural numbers whose sum and product are equal to 20. | Answer: $1,1,1,1,1,1,1,1,2,10$.
Comment. The given answer - 7 points. | 1,1,1,1,1,1,1,1,2,10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,328 |
8.2. If a class of 30 people is seated in a cinema hall, then in any case, at least two classmates will end up in the same row. If the same is done with a class of 26 people, then at least three rows will be empty. How many rows are there in the hall? | Answer: 29.
There are no more than 29 rows in the hall. Otherwise, a class of 30 students can be seated with one student per row. On the other hand, if a class of 26 students is seated with one student per row, at least three rows will be empty, which means there are no fewer than 29 rows.
Comment. An answer without ... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,329 |
8.3. With three integers, the following was done: one number was decreased by one, one was increased by one, and the third number was squared. As a result, the same set of numbers was obtained. Find these numbers, given that the sum of the numbers is 2013. | Answer: $1007,1006,0$.
Let the numbers in the original set be $a, b, c$. The resulting set will be $a-1, b+1, c^{2}$. From the fact that the sets coincide, we get the equality $a+b+c=(a-1)+(b+1)+c^{2}$. This simplifies to $c=c^{2}$. Therefore, $c=0$ or $c=1$. Since $c^{2}=c$, we get $a-1=b, b+1=a$. This means there ar... | 1007,1006,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,330 |
8.4. In triangle $A B C$, points $E$ and $D$ are on sides $A B$ and $B C$ respectively, such that segments $A D$ and $C E$ are equal, $\angle B A D=\angle E C A$ and $\angle A D C=\angle B E C$. Find the angles of the triangle. | Answer: All angles are $60^{\circ}$.
Triangles $C E A$ and $A D B$ are equal (by side and adjacent angles). Therefore, sides $A C$ and $A B$ are equal, and angles $C A E$ and $A B D$ are equal, which means angles $C A B$ and $A B C$ are equal. This implies the equality of sides $A C$ and $B C$. Thus, triangle $A B C$ ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,331 |
8.5. The numbers from 1 to 10 were written in some order and resulted in the numbers \(a_{1}, a_{2}, a_{3}, \ldots, a_{10}\), and then the sums \(S_{1}=a_{1}\), \(S_{2}=a_{1}+a_{2}\), \(S_{3}=a_{1}+a_{2}+a_{3}\), \ldots, \(S_{10}=a_{1}+a_{2}+a_{3}+\ldots+a_{10}\) were calculated. What is the maximum number of prime num... | Answer: 7.
Among the numbers from 1 to 10, there are five odd numbers. Adding an odd number changes the parity of the sum. Let $y_{1}^{\prime}, y_{2}, y_{3}, y_{4}, y_{5}$ be the odd numbers, in the order they appear on the board. After adding $y_{2}$, the sum will become even and greater than 2, as it will after addi... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,332 |
11.5. Eleven integers (not necessarily distinct) are written on a board. Can it happen that the product of any five of them is greater than the product of the remaining six?
(I. Bogdanov) | Answer: It can.
Solution: Let one of the numbers be 10, and each of the others be -1. Then the product of any five of them is greater than the product of the remaining six. Indeed, if the number 10 is included in the product of five numbers, then this product is 10, and the product of the remaining six numbers is 1, a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,333 |
11.6. Given a natural number $n$. Sasha claims that for any $n$ rays in space, no two of which have common points, he can mark $k$ points on these rays lying on the same sphere. For what largest $k$ is his statement true?
(A. Kuznetsov) | Answer. $k=n$ for even $n$, $k=n+1$ for odd $n$, that is, $2 \cdot\left\lceil\frac{n}{2}\right\rceil$.
Solution. Example. For even $n=2 m$, consider $m$ parallel lines and on each select a pair of non-intersecting rays. Note that in each pair of rays, there are no more than two intersections with the sphere, since a l... | 2\cdot\lceil\frac{n}{2}\rceil | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,334 |
11.8. From each vertex of triangle $ABC$, two rays, a red one and a blue one, symmetric with respect to the bisector of the corresponding angle, were drawn inside the triangle. Circles were circumscribed around the triangles formed by the intersections of the rays of the same color. Prove that if the circumcircle of tr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,335 | |
2. 50 businessmen - Japanese, Koreans, and Chinese - are sitting at a round table. It is known that between any two nearest Japanese, there are as many Chinese as there are Koreans at the table. How many Chinese can there be at the table?
## Answer: 32. | Solution. Let there be $-x$ Koreans and $y$ Japanese at the table, then there are $-x y$ Chinese. According to the condition: $x+y+x y=50$. By adding 1 to both sides of the equation and factoring both sides, we get: $(x+1)(y+1)=17 \times 3$. Since $x \neq 0$ and $y \neq 0$, one of the factors is 3, and the other is 17.... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,336 |
3. Prove that for any positive numbers $x$ and $y$ the inequality
$$
\frac{x}{x+5 y}+\frac{y}{y+5 x} \leq 1
$$
holds. | # Solution.
We use the lemma.
Lemma. If $a>0, b>c>0$, then $\frac{c}{b}<\frac{c+a}{b+a}$.
Then we get
$$
\begin{aligned}
& \frac{x}{x+5 y} \leq \frac{5 x}{5 x+5 y} \\
& \frac{y}{y+5 x} \leq \frac{5 y}{5 y+5 x}
\end{aligned}
$$
Adding these inequalities, we get
$$
\frac{x}{x+5 y}+\frac{y}{y+5 x} \leq \frac{5 x+5 y... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,337 |
4. Given a parallelogram $A B C D$ with an acute angle $A$. On the perpendicular bisector of $A D$, points $E$ and $F$ are chosen such that $\angle A B E=\angle D C F=90^{\circ}$. Point $M$ is the midpoint of $B C$. Prove that $\angle A M D=\angle E M F$.
# | # Solution.
First method. Let $N$ be the midpoint of $A D$, point $K$ be the intersection of $F N$ and $B C$, and line $M H$ be perpendicular to $A D$. Denote $A N = N D = a$. Then, by the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,338 |
5. Any two strangers in the company have exactly two common acquaintances. Dina and Tolya are acquainted with each other, but they do not have any common acquaintances. Prove that Dina and Tolya have the same number of acquaintances in this company. | Solution. Due to the symmetry of the concept of acquaintance, we can assume that Dina has no fewer acquaintances than Toly. Let $Y$ be an acquaintance of Dina. Since Dina and Toly have no common acquaintances, $Y$ and Toly are not acquainted with each other, and they have two common acquaintances. One common acquaintan... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,339 |
7.1. The sum of the minuend, subtrahend, and difference is 25. Find the minuend. | Answer: 12.5.
Since the sum of the subtrahend and the difference equals the minuend, the condition means that twice the minuend equals 25, hence the minuend is 12.5. | 12.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,340 |
7.2. On Monday morning, several fishermen arrived at the lake. Later, one more joined them. Each day, each of the fishermen caught 10 fish. From Monday to Friday inclusive, they caught a total of 370 fish. On which day of the week did the late-arriving fisherman arrive at the lake? | Answer: on Thursday.
Fishermen who were fishing from Monday to Friday inclusive caught 50 fish each. Since $370=50 \cdot 7+20$, the late-arriving fisherman caught twenty fish.
 | Thursday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,341 |
7.4. On the island, there live knights and liars. Knights always tell the truth, and liars always lie. Some of the inhabitants claimed that there is an even number of knights on the island, while the others claimed that there is an odd number of liars on the island. Is the number of inhabitants on the island even or od... | Answer: even.
Clearly, if two people made the same statement, then they are either both liars or both knights. Since the island has at least one liar and at least one knight, either all the knights made the first statement and all the liars made the second, or vice versa. In the first case, both the number of knights ... | even | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,342 |
7.5. In a row, there are 20 free chairs. From time to time, a person approaches and sits on one of the free chairs, at which point one of their neighbors, if any, immediately stands up and leaves (the two of them do not sit together). What is the maximum number of chairs that can be occupied? | Answer: 19.
Note that all chairs cannot be occupied, otherwise, when the 20th person sat down, no one would leave, and the neighboring chair would be occupied. We will show how 19 chairs can be occupied. The first person takes the first chair, the next person takes the third chair, the third person takes the second ch... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,343 |
# 3. Let the intersection point of segments
SK and AL be O. Note that CO is the median drawn to the hypotenuse of the right triangle ACL. Therefore, $\mathrm{AO}=\mathrm{OC}=\mathrm{OL}$ and
$\angle \mathrm{OCA}=\angle \mathrm{OAC}=\angle \mathrm{OAK}$
. These cards were laid out in a row such that there are no three consecutive cards with numbers in ascending order, and there are no three consecutive cards with numbers in descending order. Then, three cards were flipped over, as show... | Answer: The number 5 is written on card $A$, the number -2 on card $B$, and the number -9 on card $C$.
Solution. The missing numbers are $-2, 5$, and 9.
If the number 5 is on card $B$, then we get the consecutive numbers 3, 4, 5. If the number 5 is on card $C$, then we get the consecutive numbers 8, 7, 5. Therefore, ... | 5,-2,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,345 |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,346 |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
![](https://cd... | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,347 |
Problem 4.8. The figure shows a road map between the houses of five children. The shortest distance by road from Asya to Galia is 12 km, from Galia to Borya - 10 km, from Asya to Borya 8 km, from Dasha to Galia - 15 km, from Vasya to Galia - 17 km. How many kilometers is the shortest distance by road from Dasha to Vasy... | Answer: 18.
Solution. Add the distances from Dasha to Gala and from Vasya to Gala: $15+17=32$. This will include the "main" road (from Dasha to Vasya) and twice the "branch" from it to Gala.
Add the distance from Asey to Bory to the obtained sum: $32+8=40$. Now all three branches (to Gala - twice) and the main road w... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,348 |
Problem 5.3. The figure shows a plan of the road system of a certain city. In this city, there are 8 straight streets, and 11 intersections are named with Latin letters $A, B, C, \ldots, J, K$.
Three police officers need to be placed at some intersections so that at least one police officer is on each of the 8 streets... | Answer: $B, G, H$.
Solution. The option will work if police officers are placed at intersections $B, G, H$. It can be shown that this is the only possible option.
Since there are only three vertical streets, and each must have one police officer, there are definitely no police officers at intersections $C$ and $K$. T... | B,G,H | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,349 |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,350 |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,351 |
Problem 6.4. Misha made himself a homemade dartboard during the summer at the cottage. The round board is divided by circles into several sectors - darts can be thrown into it. Points are awarded for hitting a sector as indicated on the board.
Misha threw 8 darts 3 times. The second time he scored twice as many points... | Answer: 48.
Solution. The smallest possible score that can be achieved with eight darts is $3 \cdot 8=24$. Then, the second time, Misha scored no less than $24 \cdot 2=48$ points, and the third time, no less than $48 \cdot 1.5=72$.
On the other hand, $72=9 \cdot 8$ is the highest possible score that can be achieved w... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,353 |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,354 |
Problem 7.8. On a rectangular sheet of paper, a picture in the shape of a "cross" was drawn from two rectangles $A B C D$ and $E F G H$, the sides of which are parallel to the edges of the sheet. It is known that $A B=9, B C=5, E F=3, F G=10$. Find the area of the quadrilateral $A F C H$.
![](https://cdn.mathpix.com/c... | Answer: $52.5$.
Solution. The intersection of the two original rectangles forms a "small" rectangle with sides 5 and 3. Its area is 15.
Extend the segments $D A, G H, B C, E F$ to form lines. They form a "large" rectangle with sides 9 and 10, containing the "cross" (Fig. 2). Its area is 90.
![](https://cdn.mathpix.c... | 52.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,356 |
Problem 8.2. In the chat of students from one of the schools, a poll was being held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the poll.
Then, 80 more people participated in the poll, voting only for October 22. After that, th... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,357 |
Problem 8.7. In the figure, two equal triangles are depicted: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,358 |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), the right triangles $X A B$ and $Y D A$ are similar by the acut... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,359 |
Problem 9.5. Leonid has a white checkered rectangle. First, he painted every other column gray, starting with the leftmost one, and then every other row, starting with the topmost one. All cells adjacent to the border of the rectangle ended up being painted.
How many painted cells could there be in the rectangle if 74... | Answer: 301 or 373.
Solution. From the condition, it follows that the rectangle has an odd number of both rows and columns. Let's number the rows from top to bottom with the numbers $1,2, \ldots, 2 k+1$, and the columns from left to right with the numbers $1,2, \ldots, 2 l+1$ (for non-negative integers $k$ and $l$). W... | 301or373 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,360 |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=3... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,361 |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,363 |
Problem 11.1. The product of nine consecutive natural numbers is divisible by 1111. What is the smallest possible value that the arithmetic mean of these nine numbers can take? | Answer: 97.
Solution. Let these nine numbers be $-n, n+1, \ldots, n+8$ for some natural number $n$. It is clear that their arithmetic mean is $n+4$.
For the product to be divisible by $1111=11 \cdot 101$, it is necessary and sufficient that at least one of the factors is divisible by 11, and at least one of the facto... | 97 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,364 |
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