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1. From Central Square to the station, there is a straight street divided by 11 intersections into 12 equal blocks. At each intersection, there is a traffic light. All traffic lights simultaneously turn green for 3 minutes, then red for 1 minute. It takes the bus two minutes to travel one block (from intersection to in... | Solution. A car will travel three blocks without obstacles in the first three minutes. Upon approaching the intersection separating the third and fourth blocks, the car will stop at the traffic light for 1 minute. Thus, to travel three blocks and start moving on the fourth block, the car will need four minutes. The car... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,563 |
2. Find all solutions to the equation
$$
x^{4}=y^{2}+2 y+2
$$
where $x, y$ are integers. | Solution. Let's factor out the square of the sum on the right side, move the square to the left, and factor the left side.
$$
\begin{gathered}
x^{4}=y^{2}+2 y+1+1 \quad x^{4}=(y+1)^{2}+1 \\
x^{4}-(y+1)^{2}=1 \\
\left(x^{2}-y-1\right)\left(x^{2}+y+1\right)=1
\end{gathered}
$$
The product of two integer factors equals ... | (-1,-1)(1,-1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,564 |
3. Student Petya stores his files on USB drives and portable hard drives, and the number of files on his portable hard drives is greater than the number of files on USB drives. The number of photo files Petya has is greater than the number of text files. Is it necessarily true that Petya has photo files on his portable... | Answer. Portable disks necessarily contain files with photos. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 16,565 |
4. In quadrilateral $A B C D$, points $X, Y, Z$ are the midpoints of segments $A B, A D, B C$ respectively. It is known that $X Y$ is perpendicular to $A B$, $Y Z$ is perpendicular to $B C$, and the measure of angle $A B C$ is $100^{\circ}$. Find the measure of angle $A C D$.
. We are given that $x+y=100^{\circ}$. We need to find $z+v$.
Th... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,566 |
5. Do there exist six natural numbers such that none of them is divisible by any of the others, but the square of each of them is divisible by each of the others? Solution. Such numbers do exist. Take six different prime numbers $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}$. Construct six numbers
$$
\begin{aligned}
& a_{... | Answer. Such numbers exist. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,567 |
9.1. For which $p$ and $q$ does the equation $x^{2}+p x+q=0$ have two distinct solutions $2 p$ and $p+q$? Answer: $p=2 / 3, q=-8 / 3$. | Solution. By Vieta's theorem $3 p+q=-p, 2 p(p+q)=q$. From the first equation $q=-4 p$. Substituting into the second, we get $6 p^{2}=4 p$, from which $p=0$ or $p=2 / 3$. The first case does not satisfy the condition. | p=2/3,q=-8/3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,568 |
9.2. It is known that $5(a-1)=b+a^{2}$. Compare the numbers $a$ and $b$.
untranslated text remains unchanged. | Answer: $a>b$.
Solution: $a-b=a+a^{2}-5a+5=(a-2)^{2}+1>0$. | >b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,569 |
9.3. Three runners - Anton, Seryozha, and Tolya - are participating in a 100 m race. When Anton finished, Seryozha was 10 meters behind him, and when Seryozha finished, Tolya was 10 meters behind him. What distance was Tolya from Anton when Anton finished? (It is assumed that all boys run at constant, but of course, no... | Answer: 19 m.
Solution: Since Tolya's speed is $9 / 10$ of Seryozha's speed, by the time Anton finished, Tolya had run $9 / 10$ of the distance covered by Seryozha, which is $90 \cdot 9 / 10=81$ m. Thus, when Anton finished, Tolya and Anton were 19 m apart. | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,570 |
9.4. Two circles touch at point A. A common (external) tangent is drawn to them, touching the circles at points $C$ and $B$. Prove that $\angle C A B=90^{\circ}$. | # Solution
Let $M$ be the intersection point of line $C B$ and the tangent to the circles at point $A$. Then $M C = M A$ $= M B$ (equality of tangent segments). Therefore, point $A$ lies on the circle with diameter $C B$.
 chess pieces on an $8 \times 8$ board. Lena noticed that in each $2 \times 2$ square, the same number of pieces is placed. And Vlad noticed that in each $3 \times 1$ (or $1 \times 3$) rectangle, the same number of pieces is placed. How many pieces were placed on the board? (List al... | Answer: 0 or 64.
Solution. Suppose that in each $2 \times 2$ square there are $m$ figures, and in each $1 \times 3$ rectangle there are $n$ figures. Let's select a $2 \times 6$ rectangle from the board. On one hand, this rectangle can be divided into three $2 \times 2$ squares, and thus it contains $3m$ figures. On th... | 0or64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,572 |
1. The head of the fish weighs as much as the tail and half of the body, the body weighs as much as the head and the tail together. The tail weighs 1 kg. How much does the fish weigh? | Answer: 8 kg.
Solution 1. The body weighs as much as the head and tail, i.e., two tails and half the body. This means that half the body weighs as much as two tails, i.e., the body weighs 4 kg. Then the head weighs $1+2=3$ kg, and the whole fish weighs $4+3+1=8$ kg.
Solution 2. Let $\Gamma, T, X$ be the weight of the... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,573 |
2. The sum of the minuend, subtrahend, and difference is 555. Can the minuend be an integer? If yes, provide an example; if no, explain why. | # Answer. Hem.
Solution. Since the sum of the subtrahend and the difference is equal to the minuend, then the sum of the minuend, subtrahend, and difference is equal to twice the minuend, i.e., the minuend is 555/2 - a non-integer. | 555 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,574 |
3. In a psychiatric hospital, there is a chief doctor and many lunatics. During the week, each lunatic bit someone (possibly even themselves) once a day. At the end of the week, it turned out that each of the patients had two bites, and the chief doctor had a hundred bites. How many lunatics are there in the hospital | Answer: 20 madmen.
Solution: Let there be n madmen in the hospital. Then, by the end of the week, on the one hand, 7n bites were made, and on the other hand, $2 n+100$. i.e.
$7 n=2 n+100$, from which $n=20$. | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,575 |
5. Given a numerical riddle: TETA + BETA = GAMMA. (Different letters correspond to different digits, the same letters correspond to the same digits.) Find all its solutions and prove that there are no others. | Answer: $4940+5940=10880$
Solution. Since $\Gamma$ is the result of carrying over to the next digit, $\Gamma=1$. Since $A+A$ ends in $A$, then $A=0$. Therefore, there is no carry to the tens place, i.e., $T+T$ ends in $M$, and thus $M$ is even. There is also no carry to the hundreds place, otherwise the odd number Э+Э... | 4940+5940=10880 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,576 |
1. Real numbers $x, y, z$ (non-zero) satisfy the equation: $x+y=z$. Find all possible values that the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}$ can take. | 1. Answer: 3.
Solution. Transform the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}=$
$=\frac{z}{x}-\frac{y}{x}+\frac{z}{y}-\frac{x}{y}+\frac{y}{z}+\frac{x}{z}=\frac{z-y}{x}+\frac{z-x}{y}+\frac{y+x}{z}$.
Using the condition $x+y=z$ and the derived formulas $y=z-x$ and $x=$ $z-y$... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,578 |
3. In pentagon $\mathrm{ABCDE}$, all sides are equal: $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DE}=\mathrm{EA}$. Prove that the bisectors of angles C and E and the perpendicular bisector of segment AB intersect at one point | 3. Solution. In isosceles triangles AED and DCB, the bisectors of angles E and C are altitudes and medians of these triangles and, therefore, lie on the perpendicular bisectors of segments AD and BD. The perpendicular bisectors drawn to the sides of triangle ABD intersect at one point. Thus, the statement of the proble... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,580 |
5. How to arrange 12 ones and 52 zeros in the cells of an $8 \times 8$ board so that the sum of the numbers in any $2 \times 3$ rectangle is positive. | 5. Answer: possible.
An example is provided in the figure.
In the empty cells, there are zeros.
Evaluation criteria:
Correct example provided: 7 points.
Only the correct answer is provided without justifications.
(a+b)}{a b}=b-a$. Since the numbers $a$ and $b$ are different, we can divide both sides of the equation by $a-b$, after which we get: $\frac{a+b}{a b}=-1$. This is the requi... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,583 |
8.3. In parallelogram $A B C D$, from the vertex of the obtuse angle $B$, heights $B M$ and $B N$ are drawn, and from vertex $D$, heights $D P$ and $D Q$ are drawn. Prove that points $M, N, P$, and $Q$ are the vertices of a rectangle. | Solution. Let, for definiteness, point $N$ lie on line $A D$, and point $Q$ - on line $A B$ (see Fig. 8.3). Then the diagonals $B D$ and $P N$ of rectangle $P B N D$ are equal and intersect at their common midpoint $O$. Similarly, the diagonals $B D$ and $Q M$ of rectangle $Q B M D$ are equal and intersect at their com... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,584 |
8.4. The numbers 3, 9, and 15 were written on the board. It was allowed to add two of the written numbers, subtract the third from this sum, and write the result on the board instead of the number that was subtracted. After performing this operation multiple times, three numbers were left on the board, the smallest of ... | Answer: 2019 and 2025.
Solution. Note that $9-3=6$ and $15-9=6$. We will show that at any moment, one of the numbers on the board will be 6 less than the second and 6 more than the third.
Indeed, suppose this property is satisfied, and the numbers $x-6, x$, and $x+6$ are written on the board. If we add the two outer ... | 20192025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,585 |
8.5. On the sides $B C$ and $C D$ of the square $A B C D$, points $M$ and $K$ are marked respectively such that $\angle B A M = \angle C K M = 30^{\circ}$. Find $\angle A K D$. | Answer: $75^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle B M A=\angle C M K=60^{\circ}$, and then $\angle A M K=60^{\circ}$ (see Fig. 8.5a). Further reasoning can be done in different ways.
First method. Let $A H$ be the perpendicular from vertex $A$ to $M K$. Then the right triangle... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,586 |
8.6. Sasha drew a square of size $6 \times 6$ cells and alternately colors one cell at a time. After coloring the next cell, he writes down the number of colored cells adjacent to it. After coloring the entire square, Sasha adds up the numbers written in all the cells. Prove that no matter in what order Sasha colors th... | Solution. Consider all unit segments that are common sides for two cells. There are exactly sixty such segments - 30 vertical and 30 horizontal. If a segment separates two shaded cells, we will say that it is "painted." Note that when Sasha writes a number in a cell, he indicates the number of segments that were not pa... | 60 | Combinatorics | proof | Yes | Yes | olympiads | false | 16,587 |
Task 1. In a confectionery store, three types of candies are sold: caramels for 3 rubles, toffees for 5 rubles, and chocolates for 10 rubles. Varya wanted to buy exactly 8 candies of each type and took 200 rubles with her. In the morning, she saw announcements in the store: “When paying for three chocolates, get a free... | Answer: 72 rubles.
Solution. Since Varya will buy more than six but less than nine chocolates, she will get two free caramels. Then she will need to buy another 6 caramels, for which she will get two free toffees, after which she will need to buy another 6 toffees. Then she will spend a total of $8 \cdot 10 + 6 \cdot ... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,588 |
Problem 2. Find any solution to the puzzle
$$
A \cdot B + A + B = \overline{A B}
$$
$A$ and $B$ are two different digits; the notation $\overline{A B}$ represents a two-digit number (i.e., $A \neq 0$) formed by the digits $A$ and $B$. Write the number $\overline{A B}$ as your answer. | Answer: Accept any of the following answers: $19,29,39,49,59,69,79,89$.
Solution. Let's prove that $B=9$.
$$
\begin{gathered}
A \cdot B + A + B = \overline{A B} \\
A \cdot B + A = \overline{A 0} \\
(B + 1) \cdot A = \overline{A 0} \\
B + 1 = 10 \\
B = 9
\end{gathered}
$$
In this case, any $A$, not equal to 0 and 9, ... | 19,29,39,49,59,69,79,89 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,589 |
Problem 3. During a math test, Oleg was supposed to divide a given number by 2, and then add 6 to the result. But he hurried and instead multiplied the given number by 2, and then subtracted 6 from the result. Nevertheless, he got the correct answer. What number was given to Oleg | Answer: 8.
Solution. Since Oleg multiplied the number by 2 instead of dividing it, he got a result that is four times greater than the required one. This means that the difference between these two results is three times greater than the required result. However, according to the condition, this difference is equal to... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,590 |
Problem 4. In an album, a grid rectangle $3 \times 7$ is drawn. Robot Igor was asked to trace all the lines with a marker, and it took him 26 minutes (the robot draws lines at a constant speed). How many minutes will it take him to trace all the lines of a grid square $5 \times 5 ?$ | Answer: 30 minutes.
Solution. Note that the grid rectangle $3 \times 7$ consists of four horizontal lines of length 7 and eight vertical lines of length 3. Then the total length of all lines is $4 \cdot 7 + 8 \cdot 3 = 52$. It turns out that Igor spends $26 : 52 = \frac{1}{2}$ minutes on the side of one cell.
The rec... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,591 |
Problem 5. A messenger was riding a horse to deliver a message to Ilya Muromets. At some point, he noticed that Ilya Muromets had passed by him (and continued walking in the opposite direction). After 10 seconds (when the horse stopped), the messenger dismounted and ran to catch up with Ilya. How many seconds will it t... | Answer: 110.
Solution. Let Ilya Muromets walk at a speed of $x$ meters per second. Then the messenger's speed is $2x$, and the horse's speed is $10x$. Therefore, 10 seconds after the meeting, the distance between the messenger and the horse will be $10 \cdot x$ (Ilya Muromets has walked) $+10 \cdot 10 x$ (the horse ha... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,592 |
Problem 6. In the kingdom, there live counts, dukes, and marquises. One day, each count dueled with three dukes and several marquises. Each duke dueled with two counts and six marquises. Each marquise dueled with three dukes and two counts. It is known that all counts dueled with an equal number of marquises. How many ... | Answer: With 6 marquises.
Solution. Let there be $x$ counts, $y$ dukes, and $z$ marquises in the kingdom. Each count fought with three dukes, so there were $3 x$ duels between counts and dukes. But each duke fought with two counts, meaning there were $2 y$ such duels. Therefore, $3 x=2 y$.
Each duke fought with six m... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,593 |
Problem 7. On an island, there live knights who always tell the truth, and liars who always lie. One day, 15 natives, among whom were both knights and liars, stood in a circle, and each said: "Of the two people standing opposite me, one is a knight, and the other is a liar." How many of them are knights? | Answer: 10 knights.
Solution. Consider any knight. He tells the truth, which means that opposite him stand a knight and a liar. One of the people opposite the found liar is the initial knight, so next to him stands another knight. Opposite this new knight stands the previously found liar and another person, who must b... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,594 |
Problem 8. Candies are in the shape of $1 \times 1 \times 1$ cubes. The teacher arranged them into a parallelepiped $3 \times 4 \times 5$ and offered the children to enjoy. In the first minute, Petya took one of the corner candies (see figure). Each subsequent minute, the children took all the candies that had a neighb... | Solution. Let Petya take a candy from the left upper front corner of the parallelepiped. We will number the vertical layers of the parallelepiped from left to right, the transverse layers of the parallelepiped from front to back, and the horizontal layers from top to bottom. For each cube, we will calculate the sum of ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,595 |
1. Prove that the rebus $\mathrm{ABCD}+\mathrm{BCD}+\mathrm{CD}+\mathrm{D}=2019$ has no solutions. | Solution. Since the last digit in all the addends is the same, they are of the same parity, and the sum of four numbers of the same parity is even and cannot equal the odd number 2019. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,596 |
2. On four cards, four numbers are written, the sum of which is 360. You can choose three cards with the same number written on them. There are two cards, one of which has a number three times larger than the other. What numbers could be written on the cards? | Answer: $36,108,108,108$ or $60,60,60,180$.
Solution. Let the smallest of the four numbers be x. Then the largest is 3x. The three equal numbers can be either the three larger or the three smaller, from which we get two equations: $3 x+3 x+3 x+x=360$ and $x+x+x+3 x=360$. From the first, we find $x=36$, and the numbers... | 36,108,108,108or60,60,60,180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,597 |
3. Four boys, four girls, and a coach are arranged on a circular track. Each girl stands diametrically opposite one of the boys. The length of the track is 50m. On the coach's command, all of them run to him along the shortest path on the track. What is the total distance all the children will run together? | Answer: $100 \mathrm{m}$.
Solution. Consider a pair - a boy and a girl standing diametrically opposite each other - and the coach. It is clear that these boy and girl together will run half a circle, that is, 25 meters. Since there are four pairs, all the children together will run $4 \cdot 25=100$ m.
Criteria. Answer... | 100\mathrm{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,598 |
4. Some squares of the table are mined. Each number written in a square shows the number of mines in the squares adjacent to the given square. (See the figure. Adjacent are squares that have at least one common point; a square with a number is not mined). In how many ways can the mines be placed in the table? Justify y... | # Answer: 14.
Solution. The mine adjacent to the number 1 is in the second, third, or fourth column.
1 case. If the mine is in the second column, it has three possible positions, and it is adjacent to the number two in the first column, which means the second mine adjacent to this two is in the first column, and it h... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,599 |
5. Out of ten seventh graders, each is acquainted with exactly two others. Prove that among them, four can be chosen such that no two of them are acquainted. We assume that if A is acquainted with B, then B is also acquainted with A. | Solution. Let's take any of the seventh-graders and assign them number 1. Exclude their acquaintances and choose any of the remaining seven and assign them number 2. Exclude the acquaintances of the second from the remaining. We will have at least four left. Take any of them and assign them number 3. Exclude the acquai... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,600 |
6. Six coins of two types, differing only in weight - fake and real, are arranged in a circle. Among any three consecutive coins, there is no more than one fake. Find the fake coins in two weighings on a two-pan balance, given that the real coins weigh the same and are lighter than the fake ones, which also weigh the s... | Solution. There can be two or one counterfeit coins. Number the coins from 1 to 6 in the order of their placement. For the first weighing, place $1,2,3$ on the left pan and $4,5,6$ on the right pan. If equilibrium is achieved, then there are two counterfeit coins. If not, then there is one counterfeit coin and it lies ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,601 |
Task 1. On the board, in a row, the digits are written as follows:
$$
\begin{array}{lllllllll}
1 & 1 & 1 & 2 & 2 & 2 & 5 & 5 & 5 .
\end{array}
$$
Between them, several plus signs can be placed so that the resulting sum ends in the digit one:
$$
1+1+12+2+25+5+5=51
$$
How can several plus signs be placed so that the ... | Answer: $111+22+2+555=690$.
Note. There are many other correct examples.
## Criteria
## 4 p. A correct example is provided.
(Examples with only one plus sign are also accepted.) | 111+22+2+555=690 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,602 |
Problem 2. A family of four octopuses came to a shoe store (each octopus has 8 legs). The father octopus already had half of his legs booted, the mother octopus had only 3 legs booted, and their two sons had 6 legs booted each. How many boots did they buy if they left the store fully booted? | Answer: 13.
Solution. The daddy octopus had half of his legs booted, that is, 4 legs. Thus, 4 legs were not booted.
The mommy octopus had 3 legs booted, meaning 5 legs were not booted.
Each of the two sons had 6 legs booted, meaning 2 legs were not booted.
Thus, a total of $4+5+2+2=13$ boots were bought.
## Criter... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,603 |
Problem 4. Three empty baskets for fruits are standing in a row. Dasha puts apples one by one into the baskets in the following order: first, second, third, second, first, second, third, second, first, and so on. She will stop when the second basket has 13 apples. In which of the two baskets, the first or the third, wi... | Solution. Let's divide all of Dasha's actions into pairs as follows:
- she puts an apple in the first and second baskets;
- puts an apple in the third and second baskets;
- puts an apple in the first and second baskets;
- puts an apple in the third and second baskets;
- and so on
Each such pair of actions ends with D... | there\will\be\\\in\the\first\basket | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,604 |
Problem 5. A gardener planted saplings of ten trees: birches, oaks, and maples in a row. Sasha, Yasha, and Lesha walked by.
Sasha said: "There are the most birches!".
Yasha said: "There are exactly two oaks, and they are planted at the ends".
Lesha said: "And no identical trees grow next to each other". Could it be ... | Answer: it could not.
Solution. Suppose that none of the boys made a mistake.
From Yasha's words, we know that oaks are planted only at the edges, which means that the "central" eight trees are birches and maples.
From Lёsha's words, we know that two identical trees cannot grow next to each other, so birches and map... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,605 |
3. Find all pairs of natural numbers $m$ and $n$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{LCM}(m, n)+\operatorname{GCD}(m, n)=m+n \\
m-n=2015
\end{array}\right.
$$
Answer. Possible options for the pair $(m, n):(2016,1),(2020,5),(2028,13),(2046,31)$, $(2080,65),(2170,155),(2418,403... | Solution. Method I. The product of the LCM and GCD of any two natural numbers is equal to the product of these numbers. From this fact, the first equation, and Vieta's theorem, it follows that the LCM and GCD of the numbers $m$ and $n$ are the roots of the same quadratic trinomial as $m$ and $n$ themselves, which means... | (2016,1),(2020,5),(2028,13),(2046,31),(2080,65),(2170,155),(2418,403),(4030,2015) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,607 |
4. Given an acute triangle $A B C$. In it, the altitudes $A D$ and $B E$ are drawn, and the circumcenter $O$ is marked. Prove that $C O \perp D E$. | Solution. The figure is as follows:
Let $\alpha = \angle BAC$. Since both angles $AED$ and $BDE$ are right angles, quadrilateral $ABDE$ is cyclic. Therefore, $\angle BDE = 180^\circ - \angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,608 |
Task No. 1.1
## Condition:
On the potion-making exam, each student at Hogwarts School had to brew 4 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron took 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same s... | # Answer: 36
## Exact match of the answer -1 point
## Solution.
Let's calculate the potion-making speed of each of the three students, measured in potions per hour. Hermione made 4 potions in half an hour, so in one hour she would make twice as many potions, which is 8. Therefore, her speed is 8 potions per hour. Ha... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,609 |
# Task № 1.3
## Condition:
At the Potions exam, each student of Hogwarts School had to brew 8 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron needed 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same speed... | Answer: 72
Exact match of the answer -1 point
Solution by analogy with task №1.1
# | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,611 |
# Task № 1.4
## Condition:
At the potion-making exam, each student from Hogwarts School had to brew 10 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron took 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the sam... | Answer: 90
Exact match of the answer -1 point
Solution by analogy with task №1.1
# | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,612 |
# Task № 2.1
## Condition:
Masha was on vacation with her friends at the sea and drew a geometric figure in the sand.
Anya exclaimed: O, it's a triangle!
Rozalina objected: No, it's a quadrilateral!
Olya also spoke up: It's a square!
Alena did not stay out of it: It's a rectangle!
Only two of the four statements... | # Solution.
If Olya was right, then Alyona and Rozalina would also be right, but in that case, there would already be at least three correct statements, which is impossible. Therefore, Olya is wrong. If Anya were right, then Rozalina, Olya, and Alyona would all be wrong, but then there would only be one correct statem... | Rectangle\with\\perimeter\of\166\ | Geometry | MCQ | Yes | Yes | olympiads | false | 16,613 |
# Task No. 3.1
## Condition:
Solve the equation $6 x^{2}+10 x+25+5 y^{2}+10 x y=0$. | # Answer:
$\mathrm{x}=-5, \mathrm{y}=5$
## Exact match of the answer -1 point
Solution.
Let's represent $6 x^{2}$ as $5 x^{2}+x^{2}$. Then the left side, if we change the order of the terms, can be represented in the following form:
$$
\left(x^{2}+10 x+25\right)+\left(5 y^{2}+10 x y+5 x^{2}\right)=0
$$
Now notice... | -5,5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,617 |
# Task No. 4.1
## Condition:
Six friends live in a small town in southern France. One day they decided to go to Paris together for one day and see the attractions - the Eiffel Tower, the Louvre, Notre-Dame Cathedral, the Rodin Museum, the Bois de Boulogne, and the Champs-Élysées. The friends agreed to split up and se... | # Answer:
$\circ 5$
० 6
## Exact match of the answer -1 point
## Solution.
Five people said they visited Notre-Dame Cathedral, but in reality, only three did. This means that at least two of them were mistaken, and therefore the one who did not mention the Cathedral (friend 2) must have been correct. Similarly, th... | 5,6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,621 |
# Task № 4.2
## Condition:
Six friends live in a small town in southern France. One day they decided to go to Paris together for one day and see the sights - the Eiffel Tower, the Louvre, Notre-Dame Cathedral, the Rodin Museum, the Bois de Boulogne, and the Champs-Élysées. The friends agreed to split up and see them ... | # Answer:
$\circ 1$
$\circ 3$
## Exact match of the answer -1 point
Solution by analogy with task №4.1
# | 1,3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,622 |
# Task № 4.3
## Condition:
Six friends live in a small town in southern France. One day they decided to go to Paris together for one day and see attractions - the Eiffel Tower, the Louvre, Notre-Dame Cathedral, the Rodin Museum, the Bois de Boulogne, and the Champs-Élysées. The friends agreed to split up and see them... | # Answer:
$\circ 4$
० 6
## Exact match of the answer -1 point
Solution by analogy with task №4.1
# | 4,6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,623 |
# Task № 4.4
## Condition:
Six friends live in a small town in southern France. One day they decided to go to Paris together for one day and see the attractions - the Eiffel Tower, the Louvre, Notre-Dame Cathedral, the Rodin Museum, the Bois de Boulogne, and the Champs-Élysées. The friends agreed to split up and see ... | # Answer:
$\circ 2$
$\circ 5$
## Exact match of the answer -1 point
## Solution by analogy with problem №4.1
# | 2,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,624 |
# Task № 5.1
## Condition:
In a certain language $\mathrm{N}$, letters denote only 10 consonants and 5 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into syllables i... | Answer: 43750000
Exact match of the answer -1 point
## Solution.
Note that the number of vowels cannot be more than half of the total number of letters in the word, that is, more than 4, since in each syllable with a vowel there is at least one consonant. Also, the number of vowels cannot be less than $1 / 3$ of the... | 43750000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,625 |
# Task № 5.2
## Condition:
In a certain language $\mathrm{N}$, letters denote only 10 consonants and 8 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into syllables i... | Answer: 194560000
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 194560000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,626 |
# Task № 5.4
## Condition:
In a certain language $\mathrm{N}$, letters denote a total of 20 consonants and 3 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into sylla... | Answer: 272160000
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 272160000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,628 |
# Task № 6.1
## Condition:
Given a triangle $\mathrm{ABC}$, in which $\mathrm{AB}=5$. The median $\mathrm{BM}$ is perpendicular to the bisector $\mathrm{AL}$. Find $\mathrm{AC}$. | Answer: 10
Exact match of the answer -1 point
Solution.
Consider triangle ABM: since the bisector drawn from vertex A is perpendicular to side $\mathrm{BM}$, triangle ABM is isosceles.
Therefore, $\mathrm{AB}=\mathrm{AM}=\mathrm{MC}$. Hence, $\mathrm{AC}=10$.
# | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,629 |
# Task No. 7.1
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid ... | Answer: 6
Exact match of the answer -1 point
## Solution.
It is claimed that each painting by Ivan Konstantinovich was sold for $1000 \times 2^{\text {x }}$ rubles, where $\mathrm{x}$ is some natural number. Indeed, initially, the painting's price is 1000 rubles, and each participant doubles its price, meaning the p... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,633 |
# Task No. 7.2
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid ... | Answer: 4
Exact match of the answer - 1 point
Solution by analogy with task №7.1
# | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,634 |
# Task № 7.3
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid on... | Answer: 6
Exact match of the answer - 1 point
Solution by analogy with task №7.1
# | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,635 |
# Task No. 8.1
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=74^{\circ}$. On ray $\mathrm{BC}$, segment $\mathrm{CD}=\mathrm{CB}$ is laid out. Then, from point $\mathrm{D}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, dr... | Answer: 37
## Exact match of the answer -1 point
## Solution.
Let $\mathrm{F}^{\prime}$ be the midpoint of segment $\mathrm{AC}$. Then $\mathrm{F}^{\prime} \mathrm{C}=0.5 \mathrm{AC}=\mathrm{BC}=\mathrm{CD}=0.5 \mathrm{BD}$. From this, it follows that triangle BDF' is a right triangle, since its median is equal to h... | 37 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,637 |
# Problem № 8.2
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=106^{\circ}$. On the ray $\mathrm{BC}$, segment $\mathrm{CX}=$ CB is laid out. Then, from point $\mathrm{X}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, draw... | Answer: 53
Exact match of the answer -1 point
Solution by analogy with task №8.1
# | 53 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,638 |
# Problem 8.3
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=46^{\circ}$. On the ray $\mathrm{BC}$, segment $\mathrm{CM}=$ CB is marked. Then, from point M, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, drawn from vertex $\... | Answer: 23
Exact match of the answer - 1 point
Solution by analogy with task №8.1
# | 23 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,639 |
9.1. Fyodor thought of a natural number that is divisible by 300, and listed all its natural divisors, except the number itself. Prove that the sum of the odd numbers listed by Fyodor is less than the sum of the even numbers. | Solution: Note that for each odd divisor $A$ written on the board, an even divisor $2A$ is also written. (The number $2A$ is not equal to the chosen number, since the chosen number is divisible by 4, while $2A$ is not.) Therefore, the sum of the even divisors written is at least twice the sum of the odd divisors (in fa... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,641 |
9.2. In triangle $ABC$, the median $BM$ was drawn. It turned out that $\angle ABM=40^{\circ}$, $\angle MBC=70^{\circ}$. Find the ratio $AB: BM$. Justify your answer.
To solve problem 9.2 Sol... | Answer: $A B: B M=2$.
| in progress | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| An example of a triangle satisfying the conditions of the problem is provided | 0 points |
| Incorrect solution or its absence | 0 points | | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,642 |
9.3. The numbers x and y satisfy the equation:
$$
\sqrt{x y}+\sqrt{(1-x)(1-y)}=\sqrt{7 x(1-y)}+\frac{\sqrt{y(1-x)}}{\sqrt{7}}
$$
Find the maximum value of the expression $x+7 y$. Justify your answer. | # Solution:
Method 1. Perform equivalent transformations:
$$
\begin{aligned}
& \sqrt{x y}+\sqrt{(1-x)(1-y)}=\sqrt{7 x(1-y)}+\frac{\sqrt{y(1-x)}}{\sqrt{7}} \\
& \sqrt{7}(\sqrt{x y}-\sqrt{7 x(1-y)})+(\sqrt{7(1-x)(1-y)}-\sqrt{y(1-x)})=0 \\
& \sqrt{7}\left(\sqrt{x}(\sqrt{y}-\sqrt{7(1-y)})+\frac{\sqrt{1-x}}{\sqrt{7}}(\sqr... | \frac{57}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,643 |
9.4. After the bankruptcy of the company "Horns and Hooves," 17 horns, 2 hooves, and one weight remained. All this wealth was divided equally by weight between Pankovskiy and Balaganov, with the weight entirely going to Balaganov. The horns and hooves were not cut into pieces. Each horn is heavier than each hoof and li... | Solution: Let one hoof weigh $k$, and one horn weigh $k+\delta$ (all weights in the same units of measurement, for example, in puds). Then, according to the condition, the weight of the weight (the iron piece) is $k+2\delta$, and the total weight of the divided property is $20k+19\delta$. Each person received $10k+9.5\... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,644 |
9.5. On the exam, each of the three students was given the same test consisting of 40 questions. We will call a question unmanageable if all answered it incorrectly; difficult if only one student answered correctly, easy if exactly two students answered correctly, and trivial if all three answered correctly. It is know... | Solution: Let $x_{3}, x_{2}, x_{1}$, and $x_{0}$ be the quantities of trivial, easy, difficult, and insurmountable questions, respectively. Then the condition of the problem can be written as a system:
$$
\left\{\begin{aligned}
x_{3}+x_{2}+x_{1}+x_{0} & =40 \\
3 x_{3}+2 x_{2}+x_{1} & =64 \\
x_{2} & =2 x_{0}
\end{align... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,645 |
9.6. Archimedes' Problem. Let $D$ be the midpoint of arc $A C$, and $B$ be some point on this arc, different from $D$. Prove that the point $E$ - the foot of the perpendicular dropped from point $D$ to the broken line $A B C$, divides the length of this broken line in half.
![](https://cdn.mathpix.com/cropped/2024_05_... | # Solution:
Method 1. Without loss of generality, we assume that point $B$ lies on the arc $A D$ that does not contain point $C$. Then point $E$ lies on the chord $C B$. Mark a point $F$ on the extension of the line $C B$ beyond point $B$ such that $A B = B F$—see the figure. Then the length of the broken line $A B C$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,646 |
1. Variant 1.
Masha is distributing tennis balls into identical boxes. If she uses 4 boxes, there is still room for 8 more balls in the last box, and if she uses 3 boxes, 4 balls will not fit into the boxes. How many balls is one box designed to hold? | Answer: 12
Solution.
Method 1
Notice that if the balls are in 3 boxes, there are 4 tennis balls left. Take these 4 balls and put them in the fourth box. According to the condition, there will still be room for 8 more balls in this box. This means that the box can hold a total of 12 balls.
Method 2.
Let the box hol... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,647 |
3. Variant 1.
A natural number is written on the board. Five schoolchildren said the following about it:
- Pete: “This number is greater than $10$.”
- Kolya: “This number is not less than $11$.”
- Vasya: “This number is greater than $12$.”
- Dima: “This number is less than $12$.”
- Oleg: “This number is not greater t... | Answer: 4
Solution. All five answers could not have been correct, since Vasya's and Dima's answers contradict each other. Four answers could have been correct if the number 11 was written on the board. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,649 |
4. Variant 1.
In the number, two digits were swapped, and as a result, it increased by more than 3 times. The resulting number is 8453719. Find the original number. | Answer: 1453789.
Solution. When the number 8453719 is reduced by at least 3 times, the result is a number less than 3000000. Therefore, 8 was swapped with a digit less than 3, and the only such digit is 1. | 1453789 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,650 |
# 5. Option 1.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the café was open for 18 days" and "from April 10 to April 30, the café was also open for 18 days." It is known that he was wrong once. How many days was the café open from April 1 to ... | Answer: 23
Solution: In the period from April 10 to April 30, there are exactly 21 days. Dividing this period into three weeks: from April 10 to April 16, from April 17 to April 23, and from April 24 to April 30, we get exactly one weekend (Monday) in each of the three weeks. Therefore, in the second statement, Kolya ... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,651 |
Variant 2.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the café was open for 18 days" and "from April 10 to April 30, the café was also open for 18 days." It is known that he was wrong once. How many days was the café open from April 1 to Apri... | Answer: 11
Option 3.
The cafe "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "the cafe was open for 18 days from April 1 to April 20" and "the cafe was also open for 18 days from April 10 to April 30." It is known that he was wrong once. How many days was the cafe open from Ap... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,652 |
6. Variant 1.
On a grid paper, a right triangle with sides 6 and 10 is drawn (see figure). Find the total length of the horizontal grid lines inside this triangle.
 | Answer: 27.
Solution. Extend the triangle to form a rectangle.
It is divided into two identical triangles, so the total length of the horizontal lines in the rectangle is twice the desired ... | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,653 |
7. Variant 1.
Petya wrote 9 consecutive natural numbers on the board. Kolya calculated their sum and got the answer 43040102. It turned out that he made a mistake only in the first digit of the sum. What should the first digit be | Answer: 8.
Solution: Let the consecutive natural numbers be $a-4, a-3, \cdots, a+3, a+4$. The sum of nine consecutive numbers is divisible by 9, indeed, this sum is equal to $(a-4)+(a-3)+\cdots+(a+3)+(a+4)=9a$. Therefore, the desired sum must be divisible by 9. This means that the sum of the digits of the obtained ans... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,654 |
# 8. Variant 1.
101 natural numbers are written in a circle. It is known that among any 5 consecutive numbers, there will be at least two even numbers. What is the minimum number of even numbers that can be among the written numbers? | Answer: 41.
Solution. Consider any 5 consecutive numbers. Among them, there is an even number. Fix it, and divide the remaining 100 into 20 sets of 5 consecutive numbers. In each such set, there are at least two even numbers. Thus, the total number of even numbers is no less than $1+2 \cdot 20=41$. Such a situation is... | 41 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,655 |
8.1. If in the product of two natural numbers one factor is increased by 2, and the other is decreased by 2, the product of the numbers does not change. Prove that if 1 is added to this product, the result is a square of an integer. | Solution. Let the natural numbers be denoted as $x$ and $y$. Then $x y = (x + 2)(y - 2)$. From this, we get $y = x + 2$. Therefore, $x y + 1 = x(x + 2) + 1 = (x + 1)^2$.
Comment. It has been proven that the difference between the natural numbers is 2 - 2 points. | (x+1)^2 | Algebra | proof | Yes | Yes | olympiads | false | 16,656 |
8.2. On a line, there are blue and red points, with no fewer than 5 red points. It is known that on any segment with endpoints at red points, containing a red point inside, there are at least 3 blue points. And on any segment with endpoints at blue points, containing 2 blue points inside, there are at least 2 red point... | # Answer. 3.
Solution. Note that on a segment with endpoints at red points, not containing other red points, there cannot be 4 blue points. Indeed, in this case, between the outermost blue points there will be 2 blue points, which means there will be at least 2 more red points. Therefore, there are no more than 3 blue... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,657 |
8.3. In an isosceles triangle \(ABC (AB = BC)\), the altitude \(AH\) is drawn, and from the midpoint \(M\) of side \(AB\), a perpendicular \(MK\) is dropped to side \(AC\). It turns out that \(AH = MK\). Find the perimeter of triangle \(ABC\) if \(AK = a\). | Answer: 20a.
Solution. Triangle $ABC$ is isosceles, so $\angle MAK = \angle ACH$. Therefore, triangles $MAK$ and $ACH$ are congruent (right triangles with equal legs $MK = AH$ and equal acute angles opposite these legs). This means $AC = AM = \frac{1}{2} AB$. Draw the altitude $BO$ of triangle $ABC$. By Thales' theore... | 20a | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,658 |
8.4. In the castle, there are 25 identical square rooms arranged in a $5 \times 5$ square. In these rooms, 25 people—liars and knights (liars always lie, knights always tell the truth)—have settled, one person per room. Each of these 25 people said: "At least one of the rooms adjacent to mine is occupied by a liar." Wh... | Answer: 13 liars.
Solution. Note that liars cannot live in adjacent rooms (otherwise, they would be telling the truth). Let's select one corner room, and divide the remaining rooms into 12 pairs of adjacent rooms. Then, in each pair of rooms, there can be no more than one liar. Therefore, the total number of liars can... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,659 |
8.5. There are 90 cards - 10 with the digit 1, 10 with the digit $2, \ldots, 10$ with the digit 9. From all these cards, two numbers were formed, one of which is three times the other. Prove that one of these numbers can be factored into four not necessarily distinct natural factors, greater than one. | Solution. Let these numbers be $A$ and $B=3A$. Then the sum of the digits of the number $B$ is divisible by 3. But the sum of the digits on all cards is divisible by 9 (and therefore by 3), so the sum of the digits of the number $A$ is divisible by 3. This means that the number $A$ is divisible by 3. But then the numbe... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,660 |
10.1. The quadratic trinomial $y=a x^{2}+b x+c$ has no roots and $a+b+c>0$. Determine the sign of the coefficient $c$. | Answer: $c>0$.
Solution. The quadratic polynomial has no roots, which means its graph does not intersect the x-axis. Since $y(1)=a+b+c>0$, the graph is located in the upper half-plane (see figure), hence $c=y(0)>0$.
 games. Therefore, a total of 21 games wer... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,663 |
10.4. Let x be some natural number. Among the statements:
$2 \times$ greater than 70
x less than 100
$3 x$ greater than 25
x not less than 10
x greater than 5
three are true and two are false. What is x | Answer: 9.
Solution
The first statement is equivalent to x > 35, and the third statement is equivalent to $\mathrm{x}>8$, since x is a natural number. The first statement is false, because if it were true, then the third, fourth, and fifth statements would also be true, i.e., there would be at least 4 true statements... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 16,664 |
10.5. The diagonals of trapezoid $ABCD$ intersect at point $O$. The circumcircles of triangles $AOB$ and $COD$ intersect at point $M$ on the base $AD$. Prove that triangle $BMC$ is isosceles. | Solution
The first method. According to the theorem about angles inscribed in a circle, $\angle O B M=\angle O A M$ (see fig.). By the definition of a trapezoid, $\angle O A M=\angle O C B$. Thus, $\angle O B M=\angle O C B$. Similarly, it can be proven that $\angle O C M=\angle O B C$. Adding the obtained equalities,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,665 |
8.1. A natural number $n$ is called "good" if, when appended to the right of any natural number, the resulting number is divisible by $n$. Write down ten "good" numbers that are less than 1000. (It is sufficient to provide the answer.) | Answer: any 10 numbers from the set: $1,2,5,10,20,25,50,100,125,200,250,500$.
The answer can be obtained based on the following considerations. Appending a number p to the right is equivalent to multiplying the original number by $10^{p}$, where $p-$ is the number of digits in the number $n$, and adding the number p t... | 1,2,5,10,20,25,50,100,125,200,250,500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,666 |
8.2. The magpie-crow was cooking porridge and feeding her chicks. The third chick received as much porridge as the first two combined. The fourth one received as much as the second and third. The fifth one received as much as the third and fourth. The sixth one received as much as the fourth and fifth. And the seventh ... | Answer: 40 g.
Solution. Let the first chick receive $m$ g of porridge, and the second chick $-n$ g. Then the third chick received $m+n$ (g), the fourth chick $-m+2n$ (g), the fifth chick $-2m+3n$ (g), and the sixth chick $-3m+5n$ (g). Therefore, the total amount of porridge was: $m+n+(m+n)+(m+2n)+(2m+3n)+(3m+5n)=8m+12... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,667 |
8.3. $A B C D$ - a convex quadrilateral. It is known that $\angle C A D=\angle D B A=40^{\circ}, \angle C A B=60^{\circ}$, $\angle C B D=20^{\circ}$. Find the angle $C D B$. | Answer: $30^{\circ}$.
Solution. Since $\angle C A B=60^{\circ}, \angle A B C=\angle A B D+\angle D B C=60^{\circ}$, triangle $A B C$ is equilateral (see Fig. 8.3a). We can reason in different ways from here.
 stands up and leaves. What is the maximum number of people that can be sitting simultaneously, if initially all the chairs were empty? | Answer: 11.
Solution. Evaluation. Note that it is impossible to occupy all chairs simultaneously, as at the moment when a person sits on the last unoccupied chair, one of his neighbors will stand up. Therefore, the number of people sitting simultaneously cannot exceed 11.
Example. We will show how to seat 11 people. ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,669 |
6. In each cell of a $13 \times 13$ table, one of the natural numbers from 1 to 25 is written. A cell is called "good" if among the twenty-five numbers written in it and in all cells of the same row and the same column, there are no identical ones. Can all the cells of one of the main diagonals be "good"? | Answer: No, they cannot.
Solution. For each cell of one of the main diagonals, we will consider a set of twenty-five cells: the cell itself and all cells that are in the same row or column with it. We will call such a set a "cross."
Consider all "crosses" formed by the cells of the selected main diagonal. Note that a... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,670 |
1. (7 points) In a three-digit number, the first digit (hundreds place) was increased by 3, the second digit by 2, and the third digit by 1. As a result, the number increased by 4 times. Provide an example of such an original number.
Answer: 107. | Solution. The answer can be found in the following way. Let $x$ be the desired number. Then the condition of the problem immediately leads to the equation $x+321=4x$, the only solution of which is $x=107$.
Criteria. Correct answer, even without any comments: 7 points.
Incorrect answer: 0 points. | 107 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,671 |
2. (7 points) A movie ticket cost 300 rubles. When the price was reduced, the number of visitors increased by 50 percent, and the cinema's revenue increased by 35 percent. How many rubles does one ticket cost now? | Answer: 270.
Solution. Let the price of the new ticket be $s$ rubles. Suppose the initial number of visitors was $N$, and after increasing by $50 \%$ it became $1.5 N$. Then, according to the condition, the current revenue of the cinema $1.5 N \cdot s$ is $35 \%$ more than $N \cdot 300$, from which we have $1.5 N s = ... | 270 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,672 |
3. (7 points) Given an arithmetic progression. The sum of its first 10 terms is 60, and the sum of its first 20 terms is 320. What can the 15th term of this progression be? | Answer: 25.
Solution. Let the first term of the sequence be $a$, and the common difference be $b$. Then the sum of the first 10 terms is $a+(a+b)+\ldots+(a+9b)=$ $10a+45b$. The sum of the first 20 terms is $a+(a+b)+\ldots+(a+19b)=$ $20a+190b$. According to the problem, $10a+45b=60, 20a+190b=320$. Solving the system, w... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,673 |
4. (7 points) On the plane, there is a square $A B C D$ with side length 1 and a point $X$ (see the figure). It is known that $X A=\sqrt{5}, X C=\sqrt{7}$. What is $X B$?
 | Answer: $\sqrt{6-\sqrt{10}}$.
Solution. Drop perpendiculars $X M$ and $X N$ from point $X$ to lines $C B$ and $A B$ respectively. Quadrilateral $B M X N$ is a rectangle; denote the lengths of its sides $B M$ and $N X$ by $a$, and $B N$ and $M X$ by $b$.
 Consider the equation $\sin ^{3}(x)+\cos ^{3}(x)=-1$. How many solutions does it have on the interval $[0 ; 6 \pi]$?
Answer: 6. | Solution. From the main trigonometric identity, we have $\sin ^{2}(x)+$ $\cos ^{2}(x)=1$. Adding this to the given equality, we get
$$
0=\sin ^{2}(x) \cdot(1+\sin (x))+\cos ^{2}(x) \cdot(1+\cos (x))
$$
In this expression, all factors are non-negative, so both terms $\sin ^{2}(x)(1+\sin (x))$ and $\cos ^{2}(x)(1+\cos ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,675 |
6. (7 points) For the tetrahedron $ABCD$, it is known that $AB \cdot CD = AC \cdot BD = AD \cdot BC$. Let $I_{A}, I_{B}, I_{C}, I_{D}$ be the centers of the inscribed circles of triangles $BCD, CDA, DAB$, and $ABC$ respectively. Prove that the segments $AI_{A}, BI_{B}, CI_{C}, DI_{D}$ intersect at one point. | Solution. Let point $L$ be the foot of the bisector $AL$ of triangle $ABC$. By the property of the bisector, $BL / LC = BA / AC$. From the ratio of the lengths of the edges of the tetrahedron, it follows that $BA / AC = BD / DC$. By comparing these two equalities, we conclude that $BL / LC = BD / DC$. By the converse p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,676 |
9.1. Let $a_{1}, \ldots, a_{11}$ be distinct natural numbers, each not less than 2, whose sum is 407. Could it be that the sum of the remainders of some natural number $n$ when divided by the 22 numbers $a_{1}, a_{2}, \ldots, a_{11}, 4 a_{1}, 4 a_{2}, \ldots, 4 a_{11}$ equals 2012? (N. Agakhanov) | Answer: It cannot.
Solution: Suppose such a number $n$ exists.
Note that the maximum possible remainder when dividing by a natural number $m$ is $m-1$. Therefore, the sum of the remainders when dividing any number by $a_{1}, \ldots, a_{11}$ is no more than $407-11=396$, and the sum of the remainders when dividing it ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,677 |
9.2. On a circle, 2012 points are marked, dividing it into equal arcs. From these, $k$ points are chosen and a convex $k$-gon is constructed with vertices at the chosen points. What is the largest $k$ for which it could turn out that this polygon has no parallel sides? | Answer. For $k=1509$.
Solution. Let $A_{1}, A_{2}, \ldots, A_{2012}$ be the marked points in the order of traversal (we will assume that $A_{2013}=A_{1}, A_{2014}=$ $A_{2}$). We will divide them into quadruples of points $\left(A_{1}, A_{2}, A_{1007}, A_{1008}\right)$, $\left(A_{3}, A_{4}, A_{1009}, A_{1010}\right), \... | 1509 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,678 |
9.3. Given a parallelogram $A B C D$ with an obtuse angle $A$. Point $H$ is the foot of the perpendicular dropped from point $A$ to $B C$. The extension of the median of triangle $A B C$, drawn from vertex $C$, intersects the circumcircle of the triangle at point $K$. Prove that points $K, H, C$, and $D$ lie on the sam... | Solution. Let $E$ be the foot of the perpendicular dropped from point $B$ to $AD$. Then the quadrilateral $AHBE$ is a rectangle. Therefore, $\angle HED = \angle ABC = 180^\circ - \angle BCD$, which means that points $D, C, H, E$ lie on some circle $\omega$ (see Fig. 1).
 | Solution. Multiplying the equality
$$
a_{1}+a_{2}+\ldots+a_{n}=3 k
$$
and the inequality
$$
a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}>3 k^{3}+k
$$
we obtain the inequality
$$
\begin{gathered}
a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+a_{1}^{3} a_{2}+a_{1} a_{2}^{3}+a_{1}^{3} a_{3}+a_{1} a_{3}^{3}+\ldots+a_{n-1}^{3} a_{n}+a_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,680 |
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