problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
5. Given a square table of size $2023 \times 2023$. Each of its cells is colored in one of $n$ colors. It is known that for any six cells of the same color located in the same row, there are no cells of the same color above the leftmost of them and below the rightmost of them. What is the smallest $n$ for which this is... | Answer: 338.
Solution. Consider some specific color (say, blue). In each row, mark the five leftmost blue cells with a cross, and the rest with a zero (if there are fewer than six blue cells in the row, mark all cells with a cross). Note that in any column, there is no more than one blue cell with a zero (otherwise, f... | 338 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Nastya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the difference \( a - b \) must be copr... | Answer: $n=5 \cdot 7 \cdot 11=385$.
Solution. We will make two observations.
1) $n$ is not divisible by 2 and 3. Among seven numbers, there are always three numbers of the same parity. If $n$ is even, then they must be pairwise connected. Moreover, among seven numbers, there will be three numbers that give the same r... | 385 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the picture, several circles are drawn, connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be co... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among any five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) $n$ has at least two distinct pr... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the picture, several circles are drawn, connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be co... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. If there are three even and three odd numbers, then each of these triples forms a cycle. Suppose there are four numbers $a, b, c, d$ of the same parity. Then all of them are pairwise connected, and we again get two th... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers, the difference of whose square roots is less than 1? | Answer: 45.
Solution. We will show that $k=45$ works. Let's divide the numbers from 1 to 2016 into 44 groups:
\[
\begin{aligned}
& (1,2,3), \quad(4,5,6,7,8), \quad(9,10, \ldots, 15), \ldots, \quad\left(k^{2}, k^{2}+1, \ldots, k^{2}+2 k\right), \ldots, \\
& (1936,1937, \ldots, 2016)
\end{aligned}
\]
Since there are 4... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. What is the maximum number of different numbers from 1 to 1000 that can be chosen so that the difference between any two chosen numbers is not equal to any of the numbers 4, 5, 6. | Answer: 400.
Solution. Consider ten consecutive natural numbers. We will prove that no more than four of them are selected. If at least five numbers are selected, then three of them have the same parity, but then their pairwise differences cannot be only 2 and 8. Indeed, if $a<b<c$, then $b-a=2$ and $c-a=8$, but then ... | 400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. What is the maximum number of chess kings that can be placed on a $12 \times 12$ board so that each king attacks exactly one of the others? | Answer: 56.
Solution. Note that two kings attack each other if and only if their cells share at least one vertex. For each pair of attacking kings, mark the vertices of the cells they occupy. In this case, no fewer than six vertices are marked for each such pair. Since different pairs of kings mark different vertices ... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. From the numbers 1, 2, 3, ..., 2016, $k$ numbers are chosen. What is the smallest $k$ such that among the chosen numbers, there will definitely be two numbers whose difference is greater than 672 and less than 1344? | Answer: 674.
Solution. Let $n=672$. Then $2 n=1344$ and $3 n=2016$. Suppose it is possible to choose $674=n+2$ numbers such that no required pair of numbers can be found among them. Let $m$ be the smallest of the chosen numbers. Then the numbers $m+n+1, m+n+2$, $\ldots, m+2 n-1$ are not chosen. Remove them and the num... | 674 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. At a cactus lovers' meeting, 80 cactus enthusiasts presented their collections, each consisting of cacti of different species. It turned out that no species of cactus is present in all collections at once, but any 15 people have cacti of the same species. What is the smallest total number of cactus species that can ... | Answer: 16.
Solution. We will show that 16 cacti were possible. Number the cacti from 1 to 16. Let the 1st cactus lover have all cacti except the first; the 2nd cactus lover have all except the second cactus; the 15th cactus lover have all except the fifteenth cactus; and the cactus lovers from the 16th to the 80th ha... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers $a$ and $b$ such that $|\sqrt[3]{a}-\sqrt[3]{b}|<1$? | Answer: 13.
Solution. We will show that $k=13$ works. Divide the numbers from 1 to 2016 into 12 groups:
\[
\begin{aligned}
& (1,2,3,4,5,6,7), \quad(8,9, \ldots, 26), \quad(27,28, \ldots, 63), \ldots \\
& \left(k^{3}, k^{3}+1, \ldots,(k+1)^{3}-1\right), \ldots,(1728,1729, \ldots, 2016)
\end{aligned}
\]
Since there ar... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The cells of a $9 \times 9$ board are painted in black and white in a checkerboard pattern. How many ways are there to place 9 rooks on the same-colored cells of the board so that they do not attack each other? (A rook attacks any cell that is in the same row or column as itself.) | Answer: $4!5!=2880$.
Solution. Let the upper left corner of the board be black for definiteness. Note that black cells come in two types: cells with both coordinates even, and cells with both coordinates odd. If a rook is on a black cell with both even coordinates, then all black cells it attacks also have even coordi... | 2880 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. On an island, there live only knights, who always tell the truth, and liars, who always lie. One fine day, 30 islanders sat around a round table. Each of them can see everyone except themselves and their neighbors. Each person in turn said the phrase: "All I see are liars." How many liars were sitting at the table? | Answer: 28.
Solution. Not all of those sitting at the table are liars (otherwise they would all be telling the truth). Therefore, there is at least one knight sitting at the table. Everyone he sees is a liar. Let's determine who his neighbors are. Both of them cannot be liars (otherwise they would be telling the truth... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In the cells of a $2015 \times n$ table, non-negative numbers are arranged such that in each row and each column there is at least one positive number. It is known that if a cell contains a positive number, then the sum of all numbers in its column is equal to the sum of all numbers in its row. For which $n$ is this... | Answer: $n=2015$.
Solution. We will prove by induction on $m+n$ that for an $m \times n$ table, the specified arrangement is possible only when $m=n$. For $m+n=2$, the statement is obvious. Consider an $m \times n$ table. Let the sum of the numbers in the first row be $a$. Consider all rows and columns whose sum of nu... | 2015 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. How many ways are there to color the cells of a $10 \times 10$ board in blue and green so that in each $2 \times 2$ square there are two blue and two green cells? | Answer: $2^{11}-2=2046$.
Solution. Note that if the colors of three cells in a $2 \times 2$ square are known, the color of the remaining cell is uniquely determined. Also, the colors of the two remaining cells are uniquely determined if there are two adjacent cells of the same color in a $2 \times 2$ square.
Consider... | 2046 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. On an island, there live only knights, who always tell the truth, and liars, who always lie, and there are at least two knights and at least two liars. One fine day, each islander, in turn, pointed to each of the others and said one of two phrases: "You are a knight!" or "You are a liar!" The phrase "You are a liar!... | Answer: 526.
Solution. Let $r$ and $\ell$ denote the number of knights and liars, respectively. Note that a knight will say to another knight and a liar will say to another liar: "You are a knight!", while a knight will say to a liar and a liar will say to a knight: "You are a liar!" Therefore, the number of liar-knig... | 526 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In an $8 \times 8$ table, natural numbers are arranged. The numbers in cells symmetric with respect to both diagonals of the table are equal. It is known that the sum of all numbers in the table is 1000, and the sum of the numbers on the diagonals is 200. For what smallest number $M$ can we assert that the sum of th... | Answer: $M=288$.
Solution. Consider the upper half of the table. Let the numbers be arranged as shown in the figure. Due to symmetry, the marked numbers completely determine the placement of the remaining numbers in the table.
| $a_{1}$ | $b_{1}$ | $b_{2}$ | $b_{3}$ | $c_{3}$ | $c_{2}$ | $c_{1}$ | $d_{1}$ |
| :--- | ... | 288 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Real numbers $a, b, c$ and $d$ satisfy the condition $a^{6}+b^{6}+c^{6}+d^{6}=64$. Find the maximum value of the expression $a^{7}+b^{7}+c^{7}+d^{7}$. | Answer: 128
Solution. By the condition $a^{6} \leqslant a^{6}+b^{6}+c^{6}+d^{6}=64$, therefore $a \leqslant 2$. Similarly, we get that $b \leqslant 2, c \leqslant 2$ and $d \leqslant 2$. Consequently,
$$
a^{7}+b^{7}+c^{7}+d^{7}=a \cdot a^{6}+b \cdot b^{6}+c \cdot c^{6}+d \cdot d^{6} \leqslant 2\left(a^{6}+b^{6}+c^{6}... | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. On an island, there live liars and knights, a total of 2021 people. Knights always tell the truth, and liars always lie. Every resident of the island knows whether each person is a knight or a liar. One fine day, all the residents of the island lined up. After that, each resident of the island stated: "The number of... | Answer: 1010
Solution. The rightmost islander cannot be telling the truth, since there is no one to the left of him, and in particular, there are no liars. Therefore, he is a liar. The leftmost islander cannot be lying, since there is at least one liar to the left of him, and there is no one to the right of him, and i... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. (15 points) If all the trees on one hectare of forest are cut down, then 100 cubic meters of boards can be produced. Assuming that all the trees in the forest are the same, are evenly distributed, and that 0.4 m$^{3}$ of boards can be obtained from each tree, determine the area in square meters on which one tree gro... | Answer: 40.
Solution. Let's find out how many trees grow on one hectare of forest: $\frac{100 \mathrm{m}^{3}}{0.4 \mathrm{M}^{3}}=250$. Let's recall that 1 ha is $100 \mathrm{~m} \times 100 \mathrm{~m}=10000 \mathrm{~m}^{2}$. Thus, one tree grows on $\frac{10000 \mathrm{M}^{2}}{250}=40 \mathrm{M}^{2}$. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) Witch Gingema enchanted the wall clock so that the minute hand moves in the correct direction for five minutes, then three minutes in the opposite direction, then again five minutes in the correct direction, and so on. How many minutes will the hand show after 2022 minutes from the moment it pointed exac... | Answer: 28.
Solution. In 8 minutes of magical time, the hand will move 2 minutes in the clockwise direction. Therefore, in 2022 minutes, it will complete 252 full eight-minute cycles and have 6 minutes left. Since $252 \cdot 2=60 \cdot 8+24$, the hand will travel 8 full circles, plus 24 minutes, and then 5 minutes in ... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. (50 points) Two multiplication examples of two natural numbers were written on the board, and in one of the examples, the numbers were the same. The troublemaker Petya erased both the original numbers and almost all the digits of the multiplication results, leaving only the last three digits: ... 689 and ... 759. Ca... | Answer: It could be 689.
First solution. Statement: at least one of the two last digits of the square of a natural number is necessarily even. Then the second number is eliminated. Example for the first number: $133 \cdot 133=17689$.
Proof of the statement: the statement is true for even numbers, let's consider the s... | 689 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. How many ways are there to color the cells of a $10 \times 10$ board in blue and green such that in each $2 \times 2$ square there are two blue and two green cells? | Answer: $2^{11}-2=2046$.
Solution. Note that if the colors of three cells in a $2 \times 2$ square are known, the color of the remaining cell is uniquely determined. Also, the colors of the two remaining cells are uniquely determined if there are two adjacent cells of the same color in a $2 \times 2$ square.
Consider... | 2046 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second digit does not have to be different. For example, 101 and 222 have this form, while 220 and 123 do not. Similarly, we define the form of the number $\overline{\overline{b a b c}}$. How many nu... | Answer: 180.
Solution. Numbers divisible by $5$ are those ending in 0 or 5, so we have two options for $c$. For $a$, we have 9 options, as the number cannot start with zero, and the value of $b$ can be anything.
Thus, we get that the total number of such numbers is $2 \cdot 9 \cdot 10=180$. | 180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) Nезнayka came up with a password for his email consisting of five characters. Deciding to check the reliability of this password, he calculated all possible combinations that can be formed from these five characters. In the end, he got 20 different combinations. Is a password with this number of combinat... | For example, error.
Solution. The maximum number of different combinations that can be formed from 5 symbols equals $5!(n!=1 \cdot 2 \cdots n$ - the factorial of number $n$, i.e., the product of all natural numbers from 1 to $n$ inclusive). This number is obtained when all 5 symbols are distinct.
Suppose among the gi... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Masha has 1000 beads in 50 different colors, 20 beads of each color. What is the smallest $n$ such that for any way of making a necklace from all the beads, one can choose $n$ consecutive beads, among which there are beads of 25 different colors? | Answer: $n=462$.
First solution. Let's call a segment of the necklace of length $m$ a set of $m$ consecutive beads. If the beads in the necklace are arranged in groups of 20 of the same color, then a segment of length 461 cannot contain more than 24 different colors. Therefore, $n \geqslant 462$. Consider a segment of... | 462 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. What is the minimum number of chips that can be placed in the cells of a $99 \times 99$ table so that in each $4 \times 4$ square there are at least eight chips? | Answer: 4801.
Solution. Add a row and a column with number 100 to the table. Place a chip in each of their cells. Divide the expanded table into 625 squares $4 \times 4$. In each square, there can be no more than eight empty cells, so there are no more than 5000 empty cells in the entire table. Therefore, the total nu... | 4801 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Boy Tolya loves to swim. When he comes to one grandmother's cottage, he swims in the Volkhov and floats downstream from one beach to another in 18 minutes. On the way back, it takes him exactly 1 hour. When he came to another grandmother's cottage, he swam the same distance downstream in the Luga River in 20 minutes... | Answer: 45 minutes.
Solution. Let the distance between the beaches be $x$ km. Then in Volkhov, Tolya swims downstream at a speed of $x / 18$ km/min, and upstream at a speed of $x / 60$ km/min. Therefore, Tolya's own speed is $\frac{1}{2}(x / 60 + x / 18) = 13 x / 360$ km/min. In Luga, Tolya swims downstream at a speed... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The cells of a $20 \times 20$ table are painted in $n$ colors, and there are cells of each color. In each row and each column of the table, no more than six different colors are used. What is the largest $n$ for which this is possible? | Answer: 101.
Solution. Suppose there is a coloring with 102 colors. Then there are two rows that together use at least 12 colors (otherwise, the total number of colors does not exceed $11+5 \cdot 18=101$). Let's assume for definiteness that these are the first and second rows. By the condition, no more than six differ... | 101 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) Several positive numbers are written on the board. The sum of the five largest of them is 0.29 of the sum of all the numbers, and the sum of the five smallest is 0.26 of the sum of all the numbers. How many numbers are written on the board? | Answer: 18.
Solution. Let the total number of numbers be $k+10$. From the condition, it is clear that $k>0$. We can assume that the sum of all numbers is 1 (otherwise, we divide each number by this sum, and the condition of the problem remains). We order the numbers in ascending order:
$$
a_{1} \leqslant a_{2} \leqsl... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Students' written work is evaluated on a two-point grading system, i.e., a work will either be credited if it is done well, or not credited if it is done poorly. The works are first checked by a neural network, which gives an incorrect answer in $10 \%$ of cases, and then all works deemed uncredited are ... | Answer: 66.
Solution. We will represent the students' works in the diagram below.
Neural network errors can:
a) classify all 10% of good works as bad;
b) classify part of the good works as bad and part of the bad works as good;
c) classify all 10% of bad works as good.
 Point $M$ is the midpoint of the hypotenuse $A C$ of the right triangle $A B C$. Points $P$ and $Q$ on lines $A B$ and $B C$ respectively are such that $A P = P M$ and $C Q = Q M$. Find the measure of angle $\angle P Q M$, if $\angle B A C = 17^{\circ}$.
 Let's call the tail of a natural number any number that is obtained from it by discarding one or several of its first digits. For example, 234, 34, and 4 are tails of the number 1234. Does there exist a six-digit number without zeros in its decimal representation that is divisible by each of its tails? | Answer: Yes (721875 fits).
Solution. Suppose the required number exists. Let's write it as $A=$ $=\overline{a_{5} a_{4} \ldots a_{0}}$. Then $A$ is divisible by its five-digit tail, that is, by $\overline{a_{4} \ldots a_{0}}$. Therefore, $\overline{a_{4} \ldots a_{0}}$ divides the difference between $A$ and its tail, ... | 721875 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (15 points) In a bookstore, there is a rule for "summing" discounts: "if different types of discounts apply to an item, they are applied sequentially one after another." For example, if two discounts A% and B% apply to an item, the first discount is applied to the original price, and the second discount is applied t... | Answer: $50 \%$.
Solution. Note that the sequence of applying discounts is irrelevant, since applying a discount of A\% is equivalent to multiplying the cost of the item by ( $1-\frac{A}{100}$ ). Let $R$ be the magnitude of the "random" discount. Then, as a result of "summing" the three discounts, we get
$$
230 \cdot... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different four-letter words can be formed from the letters of the word КАША? And from the letters of the word ХЛЕБ? In your answer, indicate the sum of the found numbers. | Answer: 36.
Solution. In the word ХЛЕБ, all letters are different. Therefore, by rearranging the letters, we get $4 \cdot 3 \cdot 2 \cdot 1=24$ words. From the word КАША, we can form 12 words. Indeed, for the letters K and Ш, there are $4 \cdot 3=12$ positions, and we write the letters А in the remaining places. Thus,... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (35 points) Journalists have found out that
a) in the lineage of Tsar Pafnuty, all descendants are male: the tsar himself had 2 sons, 60 of his descendants also had 2 sons each, and 20 had 1 son each;
b) in the lineage of Tsar Zinovy, all descendants are female: the tsar himself had 4 daughters, 35 of his descenda... | Answer: At Zinovy.
Solution. We need to find the total number of children in the lineage of Pafnuty, including the children of the king himself. It is equal to $60 \cdot 2+20 \cdot 1+2=142$. The total number of children in the lineage of King Zinovy is $4+35 \cdot 3+35 \cdot 1=144$. | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. (50 points) From the numbers 1 to 200, one or several were selected into a separate group with the following property: if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5. What is the maximum number of numbers that can be in a group with this property? | Answer: 40.
Solution. Suppose that the group selected a number $A$, which gives a remainder $i \neq 0$ when divided by 5. If there is another number $B$ in the group, then it must give a remainder $5-i$ when divided by 5, so that $A+B$ is divisible by 5. We will show that there cannot be any other numbers in this grou... | 40 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On a $5 \times 7$ grid, 9 cells are marked. We will call a pair of cells with a common side interesting if exactly one cell in the pair is marked. What is the maximum number of interesting pairs that can be? | Answer: 35.
Solution. Let's call two cells adjacent if they share a side. The number of interesting pairs containing a given marked cell is no more than 4, and for a boundary cell, it is no more than 3. Then the total number of interesting pairs does not exceed $9 \cdot 4 = 36$. At the same time, if there are two adja... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a+b}{c}\right)^{4}+\left(\frac{b+c}{d}\right)^{4}+\left(\frac{c+d}{a}\right)^{4}+\left(\frac{d+a}{b}\right)^{4}
$$ | Answer: 64.
Solution. By Cauchy's inequality for means,
$$
A \geqslant 4 \cdot \frac{(a+b)(b+c)(c+d)(d+a)}{a b c d}=64 \cdot \frac{a+b}{2 \sqrt{a b}} \cdot \frac{b+c}{2 \sqrt{b c}} \cdot \frac{c+d}{2 \sqrt{c d}} \cdot \frac{d+a}{2 \sqrt{d a}} \geqslant 64
$$
Equality is achieved when $a=b=c=d=1$. | 64 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. On a $7 \times 7$ checkerboard, 14 cells are marked. We will call a pair of cells with a common side interesting if at least one cell in the pair is marked. What is the maximum number of interesting pairs that can be? | # Answer: 55.
Solution. Let's call two cells adjacent if they share a side. The number of interesting pairs containing a given marked cell is no more than 4, and for a boundary cell, no more than 3. Then the total number of interesting pairs does not exceed $14 \cdot 4 = 56$. At the same time, if there are two adjacen... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a^{2}+b^{2}}{c d}\right)^{4}+\left(\frac{b^{2}+c^{2}}{a d}\right)^{4}+\left(\frac{c^{2}+d^{2}}{a b}\right)^{4}+\left(\frac{d^{2}+a^{2}}{b c}\right)^{4}
$$ | Answer: 64.
Solution. By the Cauchy inequalities for means,
$$
A \geqslant 4 \cdot \frac{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+d^{2}\right)\left(d^{2}+a^{2}\right)}{c d \cdot a d \cdot a b \cdot b c}=64 \cdot \frac{a^{2}+b^{2}}{2 a b} \cdot \frac{b^{2}+c^{2}}{2 b c} \cdot \frac{c^{2}+d^{2}}{2 c ... | 64 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. A circle $\omega$ is circumscribed around triangle $A B C$. Tangents to the circle, drawn at points $A$ and $B$, intersect at point $K$. Point $M$ is the midpoint of side $A C$. A line passing through point $K$ parallel to $A C$ intersects side $B C$ at point $L$. Find the angle $A M L$. | Answer: $90^{\circ}$.
Solution. Let $\alpha=\angle A C B$. The angle between the tangent $A K$ and the chord $A B$ of the circle $\omega$ is equal to the inscribed angle that subtends $A B$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. On the table, there are three spheres and a cone (with its base $k$ on the table), touching each other externally. The radii of the spheres are 20, 40, and 40, and the radius of the base of the cone is 21. Find the height of the cone. | Answer: 28.
Solution. Let $O$ be the center of the base of the cone, $h$ be its height, and $2 \alpha$ be the angle of inclination of the cone's generators to the table. Consider the sectio... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Quadrilateral $ABCD$ is inscribed in circle $\omega$, the center of which lies on side $AB$. Circle $\omega_{1}$ is externally tangent to circle $\omega$ at point $C$. Circle $\omega_{2}$ is tangent to circles $\omega$ and $\omega_{1}$ at points $D$ and $E$ respectively. Line $BC$ intersects circle $\omega_{1}$ agai... | Answer: $180^{\circ}$.
Solution 1. Since $\angle B C O=\angle P C O_{1}$, the isosceles triangles $B O C$ and $P O_{1} C$ are similar, from which $\angle B O C=\angle P O_{1} C$. Similarly, it can be verified that $\angle A O D=\angle Q O_{2} D$. Then the segments $O_{1} P$ and $O_{2} Q$ are parallel to the line $A B$... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Quadrilateral $ABCD$ is inscribed in circle $\omega$, the center of which lies on side $AB$. Circle $\omega_{1}$ is externally tangent to circle $\omega$ at point $C$. Circle $\omega_{2}$ is tangent to circles $\omega$ and $\omega_{1}$ at points $D$ and $E$ respectively. Line $BD$ intersects circle $\omega_{2}$ agai... | Answer: $180^{\circ}$.
Solution 1. Since $\angle A C O=\angle Q C O_{1}$, the isosceles triangles $A O C$ and $Q O_{1} C$ are similar, from which $\angle A O C=\angle Q O_{1} C$. Similarly, ... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In the cells of a $5 \times 5$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table. | Answer: 48.
Solution. Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 5. Since all these sums are distinct, the minimal possible set of their values is $\{5,6, \ldots, 13,14\}$. By adding the sums of the rows and columns of the table, we get twice the... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a+\frac{1}{b}\right)^{3}+\left(b+\frac{1}{c}\right)^{3}+\left(c+\frac{1}{d}\right)^{3}+\left(d+\frac{1}{a}\right)^{3}
$$ | Answer: 32.
Solution 1. We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(2 \sqrt{\frac{a}{b}}\right)^{3}+\left(2 \sqrt{\frac{b}{c}}\right)^{3}+\left(2 \sqrt{\frac{c}{d}}\right)^{3}+\left(2 \sqrt{\frac{d}{a}}\right)^{3} \geqslant 32\left(\fr... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given an acute-angled triangle $A B C$. A circle with diameter $B C$ intersects sides $A B$ and $A C$ at points $D$ and $E$ respectively. Tangents drawn to the circle at points $D$ and $E$ intersect at point $K$. Find the angle between the lines $A K$ and $B C$. | Answer: $90^{\circ}$.
Solution 1. Let $\alpha=\angle D B E, \beta=\angle B C D, \gamma=\angle C B E, O$ - the intersection point of segments $C D$ and $B E$. The angle between the tangent $D... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In the cells of a $4 \times 6$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table. | Answer: 43.
Solution. Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 4. Since all these sums are different, the minimal possible set of their values is $\{4,5, \ldots, 12,13\}$. By adding the sums of the rows and columns of the table, we get twice th... | 43 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a^{2}+\frac{1}{b c}\right)^{3}+\left(b^{2}+\frac{1}{c d}\right)^{3}+\left(c^{2}+\frac{1}{d a}\right)^{3}+\left(d^{2}+\frac{1}{a b}\right)^{3}
$$ | Answer: 32.
Solution. We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(\frac{2 a}{\sqrt{b c}}\right)^{3}+\left(\frac{2 b}{\sqrt{c d}}\right)^{3}+\left(\frac{2 c}{\sqrt{d a}}\right)^{3}+\left(\frac{2 d}{\sqrt{a b}}\right)^{3} \geqslant 32\le... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Point $M$ is the midpoint of side $AB$ of triangle $ABC$. A circle $\omega_{1}$ is drawn through points $A$ and $M$, tangent to line $AC$, and a circle $\omega_{2}$ is drawn through points $B$ and $M$, tangent to line $BC$. Circles $\omega_{1}$ and $\omega_{2}$ intersect again at point $D$. Point $E$ lies inside tri... | Answer: $180^{\circ}$
Solution 1. The angle between the tangent $A C$ and the chord $A M$ of circle $\omega_{1}$ is equal to the inscribed angle in $\omega_{1}$ that subtends $A M$, i.e., $... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. What is the minimum number of cells that need to be marked in a $9 \times 9$ table so that each vertical or horizontal strip $1 \times 5$ contains at least one marked cell? | Answer: 16.
Solution. Consider a more general problem when the table has size $(2 n-1) \times(2 n-1)$, and the strip is $-1 \times n$. Let's call the $n$-th row and the $n$-th column central, and the marked cells on them, except for the center of the board, - axial. Suppose there are $k$ axial cells in the central row... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. A circle $\omega$ with center at point $O$ is circumscribed around triangle $ABC$. Circle $\omega_{1}$ touches the line $AB$ at point $A$ and passes through point $C$, while circle $\omega_{2}$ touches the line $AC$ at point $A$ and passes through point $B$. A line through point $A$ intersects circle $\omega_{1}$ ag... | Answer: $90^{\circ}$.
Solution 1. Let $\alpha=\angle B A C$. The angle between the tangent $A B$ and the chord $A C$ of circle $\omega_{1}$ is equal to the inscribed angle that subtends $A ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Given nine-digit numbers $m$ and $n$, obtained from each other by writing the digits in reverse order. It turned out that the product $mn$ consists of an odd number of digits and reads the same from left to right and from right to left. Find the largest number $m$ for which this is possible. | Answer: 220000001.
Solution. Let $m=\overline{a_{8} \ldots a_{0}}, n=\overline{a_{0} \ldots a_{8}}$. Since the number $m n$ contains an odd number of digits, it is a seventeen-digit number. Write $m n=\overline{b_{16} \ldots b_{0}}$. We will show by induction that
$$
b_{k}=a_{0} a_{8-k}+a_{1} a_{9-k}+\ldots+a_{k-1} a... | 220000001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Point $O$ is the center of the circumcircle of triangle $ABC$. On the circumcircle of triangle $BOC$, outside triangle $ABC$, point $X$ is chosen. On the rays $XB$ and $XC$ beyond points $B$ and $C$, points $Y$ and $Z$ are chosen respectively such that $XY = XZ$. The circumcircle of triangle $ABY$ intersects side $A... | Answer: $180^{\circ}$.
Solution. Note that $\angle B Y T=\angle B A T$ as inscribed angles subtending the same arc. Since quadrilateral $B O C X$ is cyclic, we get
$$
180^{\circ}-\angle B X C=\angle B O C=2 \angle B A C=2 \angle B Y T
$$
On the other hand, triangle $Y X Z$ is isosceles, so $180^{\circ}-\angle Y X Z=... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Point $O$ is the center of the circumcircle of triangle $A B C$. Points $Q$ and $R$ are chosen on sides $A B$ and $B C$ respectively. Line $Q R$ intersects the circumcircle of triangle $A B R$ again at point $P$ and intersects the circumcircle of triangle $B C Q$ again at point $S$. Lines $A P$ and $C S$ intersect a... | Answer: $90^{\circ}$.
Solution. Note that $\angle C S Q=\angle C B Q$ and $\angle A P R=\angle A B R$ as inscribed angles subtending the same arc. Therefore, $\angle K S P=\angle K P S$, whi... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (20 points) On the board, all two-digit numbers divisible by 5, where the number of tens is greater than the number of units, were written down. There turned out to be $A$ such numbers. Then, all two-digit numbers divisible by 5, where the number of tens is less than the number of units, were written down. There tur... | Answer: 413.
Solution: We will write two-digit numbers in the form $\overline{x y}$, where $x$ is the number of tens, and $y$ is the number of units.
Let's calculate what $A$ is. We need two-digit numbers divisible by 5, i.e., numbers where $y$ is either 0 or 5. Note that if $y=0$, then $x$ can take any value from 1 ... | 413 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) The pensioners of one of the planets in the Alpha Centauri system enjoy spending their free time solving cybrow solitaire puzzles: they choose natural numbers from a certain interval $[A, B]$ in such a way that the sum of any two of the chosen numbers is not divisible by a given number $N$. Last week, th... | Answer: 356.
Solution: For $k=0,1, \ldots, 10$, let $I_{k}$ be the set of all numbers in $[A, B)$ that give a remainder of $k$ when divided by 11. Since $A$ and $B$ are multiples of 11, all sets $I_{k}$ contain an equal number of elements. Therefore, all numbers in $[A, B)$ that are not multiples of 11 can be paired a... | 356 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) Before the geometry lesson, the teacher wrote on the board the values of all the angles (in degrees) of a certain convex polygon. However, the duty students erased one of the written numbers. When the lesson began, it turned out that the sum of the remaining numbers was 1703. What number did the duty stu... | Answer: 97.
Solution: Let the polygon have $n$ vertices. Since the $n$-gon is convex, each of its angles is less than $180^{\circ}$, and the sum of all angles is $(n-2) \cdot 180^{\circ}$. Therefore, the sum of all angles of the polygon minus one lies in the interval from $180(n-3)$ to $180(n-2)$. Then
$$
180(n-3)<17... | 97 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In each cell of a $100 \times 100$ table, a natural number is written. In each row, there are at least 10 different numbers, and in any four consecutive rows, there are no more than 15 different numbers. What is the maximum number of different numbers that can be in the table? | Answer: 175
Solution. In one line, there are no less than 10 different numbers, so in the next three lines together, there appear no more than 5 new numbers. Therefore, the first four lines contain no more than 15 different numbers, and each of the following three lines adds no more than 5 new numbers, making the tota... | 175 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In each cell of a $75 \times 75$ table, a natural number is written. In each row, there are at least 15 different numbers, and in any three consecutive rows, there are no more than 25 different numbers. What is the maximum number of different numbers that can be in the table? | # Answer: 385
Solution. In one line, there are no less than 15 different numbers, so in the next two lines together, there appear no more than 10 new numbers. Therefore, the first three lines contain no more than 25 different numbers, and each of the following two lines adds no more than 10 new numbers, making the tot... | 385 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In an acute-angled triangle $A B C$, the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are dropped. A point $T$ is chosen on the plane such that the lines $T A$ and $T B$ are tangents to the circumcircle of triangle $A B C$, and point $O$ is the center of this circle. The perpendicular dropped from point $T$ to the li... | Answer: $90^{\circ}$
Solution. Let $\angle B A C=\alpha$ and $\angle A C B=\gamma$. Triangles $A O T$ and $B O T$ are congruent by three sides, and the inscribed angle $\angle A C B$ subten... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. What is the minimum number of cells that need to be marked in a $15 \times 15$ table so that each vertical or horizontal strip $1 \times 10$ contains at least one marked cell. | Answer: 20
Solution. Let's cut the $15 \times 15$ table without the central $5 \times 5$ square into 20 rectangles of $1 \times 10$ (see the left figure). Therefore, we will need to mark at least 20 cells. An example with 20 cells: all cells of two parallel diagonals of length 10 are marked (see the right figure).
![]... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. What is the minimum number of cells that need to be marked in a $20 \times 20$ table so that each vertical or horizontal strip of $1 \times 12$ contains at least one marked cell. | Answer: 32
Solution. Let's cut the $20 \times 20$ table without the central $4 \times 4$ square into 32 rectangles of $1 \times 12$ (see the left figure). Therefore, we will need to mark at least 32 cells. An example with 32 cells: all cells of three parallel diagonals of lengths 4, 16, and 12 are marked (see the righ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the leg $AC$ of the right triangle $ABC$ with hypotenuse $AB$, a point $P$ is marked. Point $D$ is the foot of the perpendicular dropped from vertex $A$ to the line $BP$, and point $E$ is the foot of the perpendicular dropped from point $P$ to the side $AB$. On the plane, a point $T$ is chosen such that the lines... | Answer: $90^{\circ}$
Solution. Let $\angle A B P=\varphi$ and $\angle C B P=\psi$. Triangles $A O T$ and $P O T$ are equal by three sides, and the inscribed angle $\angle A B P$ subtends ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $P$, and triangle $A P D$ is acute-angled. Points $E$ and $F$ are the midpoints of sides $A B$ and $C D$ respectively. A perpendicular is drawn from point $E$ to line $A C$, and a perpendicular is drawn from point $F$ to line $B D$, these perp... | Answer: $90^{\circ}$
First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $D P$, respectively, and let $T$ be the point of inte... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $P$, and the angle $A P B$ is obtuse. Points $E$ and $F$ are the midpoints of sides $A D$ and $B C$ respectively. A perpendicular is drawn from point $E$ to the line $A C$, and a perpendicular is drawn from point $F$ to the line $B D$, these p... | Answer: $90^{\circ}$
First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $B P$, respectively, and let $T$ be the point of inte... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. (30 points) Point $M$ is the midpoint of the hypotenuse $A C$ of the right triangle $A B C$. Points $P$ and $Q$ on lines $A B$ and $B C$ respectively are such that $A P = P M$ and $C Q = Q M$. Find the measure of angle $\angle P Q M$, if $\angle B A C = 17^{\circ}$.
 On the sides $B C$ and $A C$ of the isosceles triangle $A B C (A B = A C)$, points $D$ and $E$ were found respectively such that $A E = A D, \angle E D C = 18^{\circ}$. Find the measure of the angle $\angle B A D$. | Answer: $36^{\circ}$.
Solution. Denote the angles as indicated in the figure. The angle $A D C$ is an exterior angle for triangle $A D B$; hence, $\beta+18^{\circ}=\alpha+x$. Angles $A B C$ ... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In the cells of an $11 \times 11$ square, zeros and ones are arranged in such a way that in any figure of four cells in the shape of $\square$, the sum of the numbers is odd. (The figure can be rotated and flipped). What is the minimum number of ones that can be in such an arrangement? | Answer: 25
Solution. Place 25 figures in the square without any common cells (see the left figure). Each of them contains at least one unit, so the total number of units is no less than 25. A suitable arrangement of 25 units: in all cells with even coordinates (see the right figure).
 and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime with... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among seven numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Sasha chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a+b$ must be coprime with $n$; if connected... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among eight numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Tanya chooses a natural number n and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers a and b are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with n; if connected... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, t... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Tanya chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with $n$; if $... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among six numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, th... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The board has the number 5555 written in an even base $r$ ($r \geqslant 18$). Petya found out that the $r$-ary representation of $x^{2}$ is an eight-digit palindrome, where the difference between the fourth and third digits is 2. (A palindrome is a number that reads the same from left to right and from right to left... | Answer: $r=24$.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. According to the condition, there exist such $r$-ary digits $a, b, c, d$ that $d-c=2$ and
$$
25(r+1)^{2}\left(r^{2}+1\right)^{2}=a\left(r^{7}+1\right)+b\left(r^{6}+r\right)+c\left(r^{5}+r^{2}\right)+d\left(r^{4}+r^{3}\right)
$$
... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Nastya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the difference \( a - b \) must be copr... | Answer: $n=5 \cdot 7 \cdot 11=385$.
Solution. We will make two observations.
1) $n$ is not divisible by 2 and 3. Among seven numbers, there are always three numbers of the same parity. If $n$ is even, then they must be pairwise connected. Moreover, among seven numbers, there will be three numbers that give the same r... | 385 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the picture, several circles are drawn, connected by segments. Nastl chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the difference \( a - b \) must... | Answer: $n=3 \cdot 5 \cdot 7=105$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among the six numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the diagram.
2) If $p$ is a pr... | 105 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime wit... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among any five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) $n$ has at least two distinct pr... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the picture, several circles are drawn, connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be co... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. If there are three even and three odd numbers, then each of these triples forms a cycle. Suppose there are four numbers $a, b, c, d$ of the same parity. Then all of them are pairwise connected, and we again get two th... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Triangle $A B C$ with angle $\angle A B C=135^{\circ}$ is inscribed in circle $\omega$. The lines tangent to $\omega$ at points $A$ and $C$ intersect at point $D$. Find $\angle A B D$, given that $A B$ bisects segment $C D$. Answer: $90^{\circ}$
=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secan... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (10 points) In each cell of a $50 \times 50$ square, a number is written that is equal to the number of $1 \times 16$ rectangles (both vertical and horizontal) in which this cell is an end cell. In how many cells are numbers greater than or equal to 3 written? | Answer: 1600.
Solution. We will denote the cells of the square by pairs $(i, j)$, where $i=1, \ldots, 50, j=$ $=1, \ldots, 50$. We will start the numbering from the bottom left corner of the square.
The cell $(i, j)$ is the rightmost for a horizontal rectangle if $16 \leqslant$ $\leqslant i-$ inequality (1), and the ... | 1600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. (40 points) What is the maximum number of numbers that can be chosen among the natural numbers from 1 to 3000 such that the difference between any two of them is different from 1, 4, and 5? | Answer: 1000.
Solution. Let's provide an example. We can choose all numbers divisible by 3. Then the difference between any two numbers will also be divisible by 3, while the numbers 1, 4, and 5 are not divisible by 3.
The estimate is based on the consideration that among 6 consecutive numbers, 3 numbers cannot be ch... | 1000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Triangle $A B C$ with angle $\angle A B C=135^{\circ}$ is inscribed in circle $\omega$. The lines tangent to $\omega$ at points $A$ and $C$ intersect at point $D$. Find $\angle A B D$, given that $A B$ bisects segment $C D$. Answer: $90^{\circ}$
=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secan... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate:
$$
\frac{2 \cdot 2018}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+\cdots+2018}}
$$ | Answer: 2019.
Solution: a similar solution to this problem is present in variant 1 under the same number. | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find $g(2022)$, if for any real $x, y$ the equality holds
$$
g(x-y)=g(x)+g(y)-2021(x+y)
$$
# | # Answer: 4086462.
Solution. Substitute $x=y=0$, we get
$$
g(0)=g(0)+g(0)-2021(0+0) \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=g(x)+g(x)-2021(x+x) \Rightarrow g(x)=2021 x \Rightarrow \\
g(2022)=2021 \cdot 2022=4086462
\end{gathered}
$$ | 4086462 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find $g$(2021), if for any real $x, y$ the equality holds
$$
g(x-y)=g(x)+g(y)-2022(x+y)
$$ | Answer: 4086462.
Solution. Substitute $x=y=0$, we get
$$
g(0)=g(0)+g(0)-2022(0+0) \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=g(x)+g(x)-2022(x+x) \Rightarrow g(x)=2022 x \Rightarrow \\
g(2021)=2022 \cdot 2021=4086462 .
\end{gathered}
$$ | 4086462 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Two math teachers are conducting a geometry test, checking each 10th-grade student's ability to solve problems and their knowledge of theory. The first teacher spends 5 and 7 minutes per student, respectively, while the second teacher spends 3 and 4 minutes per student. What is the minimum time they will need to int... | # Answer: 110 minutes.
## Solution: (estimation + example)
Let the first teacher accept the test on problems from $X$ students, and on theory from $Y$ students. Then the second teacher accepts the test on problems from (25-X) students, and on theory from (25-Y) students. Let $T$ be the minimum time required for them ... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Two math teachers are conducting a geometry test, checking each 10th-grade student's ability to solve problems and their knowledge of theory. The first teacher spends 5 and 7 minutes per student, respectively, while the second teacher spends 3 and 4 minutes per student. What is the minimum time required for them to ... | # Answer: 110 minutes.
Solution: fully corresponds to the solution of problem 2, option 1. | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In an isosceles trapezoid \(ABCD\) with bases \(AD\) and \(BC\), perpendiculars \(BH\) and \(DK\) are drawn from vertices \(B\) and \(D\) to the diagonal \(AC\). It is known that the feet of the perpendiculars lie on the segment \(AC\) and \(AC=20\), \(AK=19\), \(AH=3\). Find the area of trapezoid \(ABCD\).
(10 poi... | Solution. Note that right triangles $D K A$ and $B H C$ are similar, since
$\angle B C H=\angle D A K$. Let $D K=x, B H=y$. Due to similarity $\frac{D K}{K A}=\frac{B H}{H C}, \frac{x}{19}=\frac{y}{17}$. On the other hand, $C D$
$=\mathrm{AB}$ and by the Pythagorean theorem
$$
C D^{2}=D K^{2}+K C^{2}=x^{2}+1, A B^{2... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In an isosceles trapezoid $A B C D$ with lateral sides $A B$ and $C D$, the lengths of which are 10, perpendiculars $B H$ and $D K$ are drawn from vertices $B$ and $D$ to the diagonal $A C$. It is known that the bases of the perpendiculars lie on segment $A C$ and $A H: A K: A C=5: 14: 15$. Find the area of trapezoi... | Solution. Let $x = BH$, $y = DK$. From the similarity of right triangles $DKA$ and $BHC$, since $\angle BHC = \angle DAK$, we get
$$
\frac{x}{y} = \frac{CH}{AK} = \frac{10}{14} = \frac{5}{7}, \quad 5y = 7x, \quad y = \frac{7x}{5}.
$$
By the condition $AH : AK : AC = 5 : 14 : 15$, therefore $AH : CK = 5 : 1$. By the P... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find $g(2022)$, if for any real $x, y$ the equality holds
$$
g(x-y)=2022(g(x)+g(y))-2021 x y .
$$ | Answer: 2043231.
Solution. Substitute $x=y=0$, we get
$$
g(0)=2022(g(0)+g(0))-2021 \cdot 0 \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=2022(g(x)+g(x))-2021 \cdot x^{2} \Rightarrow g(x)=\frac{2021 x^{2}}{2 \cdot 2022} \Rightarrow \\
g(2022)=\frac{2021 \cdot 2022^{2}}{2 \cdot 2022}=\frac{2... | 2043231 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find $g(2021)$, if for any real $x, y$ the equality holds
$$
g(x-y)=2021(g(x)+g(y))-2022 x y
$$ | Answer: 2043231.
Solution. Substitute $x=y=0$, we get
$$
g(0)=2021(g(0)+g(0))-2022 \cdot 0 \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=2021(g(x)+g(x))-2022 \cdot x^{2} \Rightarrow g(x)=\frac{2022 x^{2}}{2 \cdot 2021}=\frac{1011 x^{2}}{2021} \Rightarrow \\
g(2021)=\frac{1011 \cdot 2021^{2... | 2043231 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. An equilateral triangle $M N K$ is inscribed in a circle. A point $F$ is taken on this circle. Prove that the value of $F M^{4}+F N^{4}+F K^{4}$ does not depend on the choice of point $F$.
# | # Solution:
Without loss of generality, we can assume that point \( \mathrm{M} \) lies on the arc \( M N \) of the circumscribed circle with center \( O \) and radius \( R \). Let \( \angle ... | 18 | Geometry | proof | Yes | Yes | olympiads | false |
2. The function $f$ satisfies the equation $(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$ for each value of $x$, not equal to 0 and 1. Find $f\left(\frac{2018}{2019}\right)$.
(7 points). | Answer: 2019.
## Solution:
Substitute $\frac{1}{x}$ for $x$ in the original equation. Together with the original equation, we get a system of linear equations in terms of $f(x)$ and $f\left(\frac{1}{x}\right)$.
$$
\left\{\begin{array}{l}
(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1} \\
\left(\frac{1}{x}-1\right... | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For any pair of numbers, a certain operation «*» is defined, satisfying the following properties: $a *(b * c)=(a * b) \cdot c$ and $a * a=1$, where the operation «$\cdot$» is the multiplication operation. Find the root $x$ of the equation: $\quad x * 2=2018$. | Answer: 4036.
## Solution:
Given the condition of the problem, we have $x * 1=x *(x * x)=(x * x) \cdot x=1 \cdot x=x$. Then
1) $(x * 2) \cdot 2=2018 \cdot 2=4036$,
2) $(x * 2) \cdot 2=x *(2 * 2)=x \cdot 1=x$.
Therefore, $x=4036$. | 4036 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For any pair of numbers, a certain operation «*» is defined, satisfying the following properties: $\quad a *(b * c)=(a * b) \cdot c$ and $a * a=1$, where the operation «$\cdot$» is the multiplication operation. Find the root $x$ of the equation: $\quad x * 3=2019$. | Answer: 6057.
Solution: a similar solution to this problem is present in variant 1 under the same number. | 6057 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. All passengers of the small cruise liner "Victory" can be accommodated in 7- and 11-seat lifeboats in case of emergency evacuation, with the number of 11-seat lifeboats being greater than the number of 7-seat lifeboats. If the number of 11-seat lifeboats is doubled, the total number of lifeboats will be more than 25... | Answer: 60 possible options are given in the table below.
## Solution:
Let $x, y$ be the number of 7-seater and 11-seater boats, respectively, and $z$ be the total number of passengers.
Then $z=7 x+11 y$, where $x, y$ satisfy the system of inequalities: $\left\{\begin{array}{c}2 y+x>25, \\ 2 x+yx\end{array}\right.$ ... | 159 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Usually, Nikita leaves home at 8:00 AM, gets into Uncle Vanya's car, who drives him to school by a certain time. But on Friday, Nikita left home at 7:10 and ran in the opposite direction. Uncle Vanya waited for him and at $8: 10$ drove after him, caught up with Nikita, turned around, and delivered him to school 20 m... | # Answer: 13 times.
Solution:
The car was on the road for 10 minutes longer than usual due to the 5 minutes spent catching up to Nikita and the 5 minutes spent returning home. The car caught up with Nikita at 8:15, and in 65 minutes (from 7:10 to 8:15), Nikita ran as far as the car traveled in 5 minutes, i.e., he spe... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In isosceles triangle $A B C$ with base $A B$, the angle bisectors $C L$ and $A K$ are drawn. Find $\angle A C B$ of triangle $A B C$, given that $A K = 2 C L$. | Answer: $108^{\circ}$
## Solution:
 | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. For the quadratic trinomial $p(x)=(a+1) x^{2}-(a+1) x+2022$, it is known that $-2022 \leq p(x) \leq 2022$ for $x \in[0 ; 1]$. Find the greatest possible value of $a$. | # Answer: 16175.
Solution. Since $p(0)=p(1)=2022$, the graph of the quadratic trinomial is a parabola symmetric about the line $x=\frac{1}{2}$. From the conditions that $-2022 \leq$ $p(x) \leq 2022$ for $x \in[0 ; 1]$ and $p(0)=p(1)=2022$, it follows that the branches of the parabola are directed upwards. Then the min... | 16175 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.