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31. Around a round table, 300 people are sitting: some of them are knights, and the rest are liars. Anton asked each of them: "How many liars are among your neighbors?" and added up the numbers he received. Then Anya did the same. When answering the question, knights always tell the truth, while liars always lie, but t... | 31. Answer: there are 200 liars at the table.
It is clear that the knights gave two identical answers, so the difference between the sums of Anton and Anya could have arisen because some liars gave Anton and Anya different answers. At the same time, the answers of any liar differ by only 1 or 2. The total difference o... | 200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,990 |
32. The number 2018 is written on the board. Players take turns, with Sasha starting. In one move, Sasha can append one digit to the number on the board, and Andrey can append two digits. If after Andrey's move, the number on the board becomes divisible by 112, he wins. If this does not happen, and a 2018-digit number ... | 32. Answer: Sasha will win.
We will show that Sasha can write such a digit on each move that, regardless of the two digits chosen by Andrey, the resulting number will not be divisible by 2018.
Let the number Sasha gets be \( N \). If Sasha appends the digit 0, then after Andrey's move, one of a hundred numbers from \... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,991 |
34. On a straight road running from north to south, there stands a cart driven by a Swan. Exactly at midnight, a Crab and a Pike chose natural numbers $m>n$. Every $n$ minutes (i.e., after $n, 2 n, 3 n \ldots$ minutes past midnight), the Pike commands "South!", and every $m$ minutes (after $m, 2 m, 3 m \ldots$ minutes ... | 34. Answer: the cart moved $(m-n) n$ meters.
It is not hard to see that the cart stopped when the Carp gave the $m$-th command, and the Crab gave the $n$-th command. For convenience, we will call the interval between successive commands of the Carp (as well as the interval between midnight and the first command) an ho... | (-n)n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,992 |
2. Finding himself on an uninhabited island, on the very first day Robinson Crusoe met a native named Friday. Robinson knows that Friday tells the truth only on Fridays, and lies on other days. Every day Robinson Crusoe asks Friday one question of the form "Is it true that today is such-and-such a day of the week?" Can... | Answer: he can. Let's ask Friday the question "Is it true that today is Monday?" The answer "No" can only be given on Friday or Monday. If Robinson asks this question for three consecutive days and hears at least one negative answer, then the day following this answer is Saturday or Tuesday, and it is easy to distingui... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,993 |
5. Inside an isosceles triangle $A B C (A B=A C)$, a point $K$ is marked. Point $L$ is the midpoint of segment $B K$. It turns out that $\angle A K B=\angle A L C=90^{\circ}, A K=C L$. Find the angles of triangle $A B C$. | Answer: the triangle is equilateral. Note that $\triangle A L C=\triangle B K A$ by the leg and hypotenuse. Therefore, $A L=B K=2 K L$, so $\angle K A L=30$. In addition, the sum of angles $B A K$ and $C A L$ in these triangles is 90, so $\angle B A C=90-30-60$. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,994 |
6. Find all triples of positive numbers $a, b, c$, satisfying the conditions $a+b+c=3, a^{2}-a \geqslant 1-b c$, $b^{2}-b \geqslant 1-a c, \quad c^{2}-c \geqslant 1-a b$. | Answer: $a=b=c=1$. If none of the numbers equals 1, then there are two on one side of 1, let these be $a$ and $b$. Adding the first two conditions and substituting $c=3-a-b$, we get $(a-1)(b-1) \leqslant 0$, a contradiction. If, however, $c=1$, then $a+b=2$ and from the third condition $ab \geqslant 1$, which only happ... | =b==1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 18,995 |
7. The figure shows a city plan. Nodes are intersections, and the 35 lines connecting them are streets. There are $N$ buses operating on these streets. All buses start simultaneously at intersections and move to adjacent intersections along the streets every minute. Each bus follows a closed, non-self-intersecting rout... | Answer: 35 (by the number of streets), i.e., it is possible to launch the minibuses so that at any moment in time, exactly one minibus is moving along each street. Obviously, $N$ cannot be greater than 35, because otherwise, in the first minute, some two minibuses will inevitably end up on the same street. Let's provid... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,996 |
35. In the cells of a $9 \times 9$ square, non-negative numbers are placed. The sum of the numbers in any two adjacent rows is at least 20, and the sum of the numbers in any two adjacent columns does not exceed 16. What can the sum of the numbers in the entire table be?
(A. Chukhnov) | 35. The total sum is no less than $4 \cdot 20=80$ (since the entire table is divided into 4 pairs of columns) and does not exceed $5 \cdot 16=80$ (since the table is covered by five pairs of rows with an overlap at the 8th row), therefore it is equal to 80. | 80 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 18,997 |
36. Many city residents engage in dancing, many in mathematics, and at least one in both. Those who engage only in dancing are exactly $p+1$ times more than those who engage only in mathematics, where $p-$ is some prime number. If you square the number of all mathematicians, you get the number of all dancers. How many ... | 36. Answer: 1 person is engaged in both dancing and mathematics.
Let $a$ people be engaged only in mathematics, and $b \geqslant 1$ people be engaged in both dancing and mathematics. Then, according to the condition, $(a+b)^{2}=(p+1) a+b$. Subtract $a+b$ from both sides: $(a+b)^{2}-(a+b)=p a$. Factor out the common te... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,998 |
38. Vasya placed numbers from 1 to $99^2$ once in each cell of a $99 \times 99$ board. Petya chooses a cell on the board, places a chess king on it, and wants to make as many moves as possible with the king so that the number under it keeps increasing. What is the maximum number of moves Petya can definitely make, no m... | 38. Answer: 3 moves (the king will visit 4 squares).
The rule by which a chess king moves allows it to visit the squares of any $2 \times 2$ square within the board in any order, so Petya will always be able to make three moves with the king: he will choose any such square, place the king on the smallest number in thi... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,000 |
39. The number 2018 is written on the board. In one move, Sasha appends one digit to the number on the board, and Andrey appends two digits. The players take turns, with Sasha starting. If after Andrey's move, the number on the board is divisible by 111, he wins. If this does not happen, and a 2018-digit number is writ... | 39. Answer: Sasha will win.
We will show that Sasha can write such a digit on each move so that, regardless of the two digits chosen by Andrey, the resulting number will not be divisible by 2018.
Let the number Sasha has be \( N \), and at this point, he has not lost yet (i.e., \( N \) is not divisible by 111). If Sa... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,001 |
63. A natural number is called hypotenuse if it can be represented as the sum of two squares of non-negative integers. Prove that any natural number greater than 10 is the difference of two hypotenuse numbers. | 63. The difference of two hypotenuse numbers has the form $\left(a^{2}+b^{2}\right)-\left(c^{2}+d^{2}\right)=\left(a^{2}-c^{2}\right)+\left(b^{2}-d^{2}\right)$. As is known, any number that gives a remainder other than 2 when divided by 4 can be represented as the difference of two squares. On the other hand, any natur... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,004 |
64. A nearsighted rook attacks all the cells in its row and column that can be reached in no more than 60 steps, moving from cell to adjacent cell by side. What is the maximum number of non-attacking nearsighted rooks that can be placed on a $100 \times 100$ square? | 64. Answer: 178 myopic rooks.
Evaluation. Divide the $100 \times 100$ square into a central $22 \times 22$ square and $4 \cdot 39=156$ rectangles of $1 \times 61$. In each rectangle of the partition and in each row of the $22 \times 22$ square, no more than one rook can be placed, so there are no more than 178 rooks.
... | 178 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,005 |
65. In a non-isosceles triangle \(ABC\), the angle bisector \(BB_1\) is drawn. Point \(I\) is the incenter of triangle \(ABC\). The perpendicular bisector of segment \(AC\) intersects the circumcircle of triangle \(AIC\) at points \(D\) and \(E\). Point \(F\) is chosen on segment \(B_1C\) such that \(AB_1 = CF\). Prove... | 65. Let $\Psi$ denote the midpoint of the arc $AC$ of the circumcircle of triangle $ABC$, not containing point $B$. Then $\Psi$ lies on the line $BB_{1}$. Moreover, by the trident lemma, point $\Psi$ is equidistant from points $I, A$, and $C$, so $\Psi$ is the center of the circumcircle of triangle $AIC$ and $\Psi$ lie... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,006 |
66. The sum
$$
\begin{gathered}
2 \\
3 \cdot 6
\end{gathered}+\begin{gathered}
2 \cdot 5 \\
3 \cdot 6 \cdot 9
\end{gathered}+\ldots+\begin{gathered}
2 \cdot 5 \cdot \ldots \cdot 2015 \\
3 \cdot 6 \cdot \ldots \cdot 2019
\end{gathered}
$$
was written as a decimal fraction. Find the first digit after the decimal point. | 66. Answer: the first digit after the decimal point is 5.
To start, let's simplify the given sum. Each term can be written as a difference
$$
\begin{aligned}
\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1)}{3 \cdot 6 \cdot 9 \cdot \ldots \cdot(3 k+3)}=\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1) \cdot(3 k+3)}{3 \cdot 6 \cdot ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,007 |
67. The altitudes $B B_{1}$ and $C C_{1}$ of an acute-angled triangle $A B C$ intersect at point $H$. A circle with center at point $O_{b}$ passes through points $A, C_{1}$, and the midpoint of segment $B H$. A circle with center at point $O_{c}$ passes through points $A, B_{1}$, and the midpoint of segment $C H$. Prov... | 67. First, let's denote the midpoint of segment $B M$ as $M$, and the circle passing through $A, C_{1}$, and $M$ as $w$.
Since $B C / 4 < B H / 4 + C H / 4$, to solve the problem, it is sufficient to prove the inequality $B_{1} O_{b} \geqslant B H / 4$ (and similarly, $C_{1} O_{c} \geqslant C H / 4$). This inequality ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,008 |
68. The sequence $a_{n}$ is defined by the conditions $a_{1}=1, a_{2}=2, a_{n+2}=a_{n}\left(a_{n+1}+1\right)$ for $n \geqslant 1$. Prove that $a_{a_{n}}$ is divisible by $\left(a_{n}\right)^{n}$ for $n \geqslant 100$. | 68. Let a prime number $p$ enter $a_{n}$ to the $k$-th power. We will prove that $a_{a_{n}}$ is divisible by $p^{k n}$. Then the statement of the problem will be satisfied.
Let $a_{i}$ be the first number in our sequence that is divisible by $p$. If $p \neq 2$, then $i>2$ and $a_{i}=a_{i-2}\left(a_{i-1}+1\right)$. The... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,009 |
69. $N$ oligarchs built a country with $N$ cities, each oligarch owns exactly one city. In addition, each oligarch built several roads between the cities: any pair of cities is connected by at most one road from each oligarch (between two cities there can be several roads belonging to different oligarchs). In total, $d... | 69. Answer: the maximum number of roads is $\frac{N(N-1)(N-2)}{6}$.
Let's number the oligarchs and their cities from 1 to $N$ respectively.
Estimation. We will say that a road is liked by an oligarch if it belongs to this oligarch or one of the cities at the ends of the road belongs to this oligarch. Note that a road... | \frac{N(N-1)(N-2)}{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,010 |
10. Autumn has arrived. Some of the green leaves on the tree are turning yellow, while some green leaves are turning red. After hanging for a while, the yellow and red leaves fall off. Yesterday, $1 / 9$ of all the leaves on the tree were green, another $1 / 9$ were red, and the rest were yellow. Today, $1 / 9$ of all ... | 10. Let's keep track of the total number of green and red leaves. Yesterday, green and red leaves together made up $2 / 9$ of the number of leaves on the tree yesterday. On the other hand, today, green and red leaves make up $8 / 9$ of the number of leaves on the tree today. During the night, the total number of green ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 19,012 |
11. The cellular figure "corner" consists of a central cell, to which horizontal and vertical rectangles $1 \times 10$ are attached (the figure shows one of the four possible types of corners, the side of each cell is 1, and there are 21 cells in total in the figure). Prove that for any coloring of the cells of a $2017... | 11. Consider an $11 \times 11$ square located "deep inside" a $2017 \times 2017$ square (for example, a $11 \times 11$ square whose central cell coincides with the central cell of the $2017 \times 2017$ square). Since the considered $11 \times 11$ square contains 121 cells, and there are only 120 colors, it must contai... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,013 |
12. In triangle $ABC$, $\angle B = 2 \angle C$. On the ray $BA$, a point $D$ is chosen such that $AC = BD$. Prove that $AB + BC > CD$.
(A. Kuznetsov) | 12. On the bisector of angle $B$, we mark off a segment $B K$ equal to segment $B C$. Then triangles $A C B$ and $D B K$ are congruent by two sides and the included angle. Therefore, $A B = D K$.
Let $\angle A C B = \alpha$. Then $\angle A C B + \angle C B A = 3 \alpha$ - this is the sum of two angles of triangle $A B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,014 |
13. Leshа wrote on the board in ascending order all natural divisors of a natural number $n$, and Dima erased several first and several last numbers of the resulting sequence, leaving 151 numbers. What is the maximum number of these 151 divisors that could be fifth powers of natural numbers?
(M. Achtipov) | 13. Answer: 31.
Lemma. If $n$ is divisible by $a^{5}$ and $b^{5}$, then $n$ is also divisible by $a^{4} b, a^{3} b^{2}, a^{2} b^{3}, a b^{4}$.
Proof of the lemma. Note that $n^{5}=n^{4} \cdot n$ is divisible by $a^{20} \cdot b^{5}$; taking the fifth root, we get that $n$ is divisible by $a^{4} b$. Similarly, $n^{5}=n... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,015 |
56. For a non-constant arithmetic progression $\left(a_{n}\right)$, there exists a natural number $n>1$ such that $a_{n}+a_{n+1}=a_{1}+\ldots+a_{3 n-1}$. Prove that this progression has no zero terms.
(S. Ivanov) | 56. Express the condition in terms of $a_{1}$ and $d: 2 a_{1}+(2 n-1) d=(3 n-1) a_{1}+(3 n-1)(3 n-2) d / 2$. From this, it is easy to find $a_{1}=(9 n-4) d / 6$. For the progression to contain a zero term, the quotient $a_{1} / d$ must be an integer (and negative). However, $(9 n-4) / 6$ is clearly not an integer! | proof | Algebra | proof | Yes | Yes | olympiads | false | 19,016 |
57. In Ruritania, every two of the $n$ cities are connected by a direct flight operated by one of two airlines, Alpha or Beta. The antimonopoly committee wants at least $k$ flights to be operated by Alpha. To achieve this, the committee can, every day, choose any three cities and change the ownership of the three fligh... | 57. Answer: when $k=\frac{n(n-1)}{2}-\left[\frac{n}{2}\right]$.
If from some city $A$ there are two flights of the Beta airline - to cities $B$ and $C$, then by applying the operation to the triplet of cities $(A, B, C)$, we will reduce the total number of flights operated by the Beta airline. Acting in this way, we c... | \frac{n(n-1)}{2}-[\frac{n}{2}] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,017 |
58. Given natural numbers $a, b$ and $c \geqslant b$. Prove that $a^{b}(a+b)^{c}>c^{b} a^{c}$. | 58. Simplify the inequality by $a^{b}:(a+b)^{c}>c^{b} a^{c-b}$. The left part, if expanded by the binomial, contains the term $C_{c}^{b} a^{c-b} b^{b}$. Let's check that this term alone is larger than the right part, i.e., that the inequality $C_{c}^{b} a^{c-b} b^{b}>c^{b} a^{c-b}$ holds. Cancel $a^{c-b}$, use the form... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 19,018 |
60. In a class, there are 25 students. The teacher wants to stock $N$ candies, conduct an olympiad, and distribute all $N$ candies for success in it (students who solve the same number of problems should receive the same number of candies, those who solve fewer should receive fewer, including possibly zero candies). Wh... | 60. Answer: $600=25 \cdot 24$ candies.
Let's show that a smaller number of candies might not be enough. If all participants solved the same number of problems, the number of candies must be a multiple of 25. Let $N=25 k$. Imagine that 24 people solved the same number of problems, while the 25th solved fewer. If each o... | 600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,019 |
43. Solution 1 (difference of squares). Notice that
$$
x^{2}+x y+y^{2}=(x+y)^{2}-x y=(x+y-\sqrt{x y})(x+y+\sqrt{x y}) .
$$
Since $x+y \geqslant 2 \sqrt{x y} \geqslant \sqrt{x y}$, it is sufficient to check the inequality
$$
x+y+\sqrt{x y} \leqslant 3(x+y-\sqrt{x y}) .
$$
After combining like terms, it can be writte... | Solution 2. Expand the brackets on the right side of the inequality:
$$
x^{2}+x y+y^{2} \leqslant 3 x^{2}+3 x y+3 y^{3}-6 x \sqrt{x y}-6 y \sqrt{x y}+6 x y .
$$
Combine like terms and move the radicals to the left side and divide by 2:
$$
3 x \sqrt{x y}+3 y \sqrt{x y} \leqslant x^{2}+4 x y+y^{2}
$$
Notice that the ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 19,020 |
5. From the edge of a large square sheet, a small square was cut off, as shown in the figure, and as a result, the perimeter of the sheet increased by $10 \%$. By what percent did the area of the sheet decrease? | 5. Answer: by $4 \%$.
Let the side of the larger square be denoted by $a$, and the side of the smaller square by $b$. As a result of cutting out the smaller square from the perimeter of the sheet, one segment of length $b$ disappears, and three such segments appear instead, meaning the perimeter increases by $2 b$. Th... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,024 |
1. Sasha went to bed at 10 PM and set the alarm clock (with hands and a 12-hour dial) for 7 AM. During the night, at some point, the alarm clock, which had been working properly, broke, and its hands started moving in the opposite direction (at the same speed). Nevertheless, the alarm rang exactly at the scheduled time... | 1. Answer: the alarm clock broke at 1 o'clock at night.
Let's imagine that the minute hand on the alarm clock is missing, and the hour hand, at the moment when the alarm clock broke, split into two halves, one of which (as stated in the condition) started moving in the opposite direction, while the other continued its... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,029 |
3. Autumn has arrived. On the tree, there are green, yellow, and red leaves. A green leaf can turn yellow or red. Yellow and red leaves, after hanging for a while, fall off. Yesterday, there were as many green leaves as red ones, and there were 7 times as many yellow leaves as red ones. Today, there are as many green l... | 3. Let the number of green leaves hanging on the tree yesterday be denoted by $x$. Then there were also $x$ red leaves, and there were $7 x$ yellow leaves. In total, there were $x+x+7 x=9 x$ leaves. Additionally, let the number of green leaves today be denoted by $y$. Then there are also $y$ yellow leaves, and $7 y$ re... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 19,031 |
4. At a large round table, 100 people are sitting: knights, who always tell the truth, and liars, who always lie, and it is known that among those present, there is at least one knight and at least one liar. Each person can see only 10 nearest neighbors to the right and 10 nearest neighbors to the left of themselves. E... | 4. Suppose that everyone sitting at the table answered "yes". Consider a knight $A$ and a liar $B$ sitting next to each other (such a pair must exist: there is at least one knight and at least one liar at the table, so we can move along the table from this knight to this liar...). Let's number the people sitting at the... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 19,032 |
42. In each cell of a $10 \times 10$ board, there are zeros. In one move, you can add one to all the numbers in one row or to all the numbers in one column. After several moves, it turned out that all the numbers on the diagonal from the top left corner to the bottom right corner are the same, and they are not less tha... | 42. Let the number in the cell with coordinates $(m, n)$ be $a_{m, n}$, and let all diagonal numbers be equal to $d$. Notice that $a_{m, n} + a_{n, m} = a_{m, m} + a_{n, n} = 2d$. On the other hand, by the condition $a_{m, n} \leqslant d$ and $a_{n, m} \leqslant d$, so $2d = a_{m, n} + a_{n, m} \leqslant 2d$, and equal... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,033 |
43. On the ray $(0,+\infty)$ of the number line, several (more than two) segments of length 1 are placed. For any two different segments, one can choose a number on each such that these numbers differ exactly by a factor of 2. The left endpoint of the leftmost segment is the number $a$, and the right endpoint of the ri... | 43. Answer: 5.5. Example: segments $[2.5,3.5],[4.5]$ and $[7,8]$.
Consider a number $x$ such that $x \in[a, a+1]$ and $2 x \in[b-1, b]$. Then $2 a+2 \geqslant 2 x \geqslant b-1$, from which $2 a+3 \geqslant b$. Next, consider any segment $[c, c+1]$, different from $[a, a+1]$ and $[b-1, b]$. Then $a<c<b-1$. Consider su... | 5.5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,034 |
44. Given an isosceles triangle \(ABC\). On the extensions of the base \(AC\) beyond points \(A\) and \(C\), points \(D\) and \(E\) are chosen, respectively. On the extension of \(CB\) beyond point \(B\), point \(F\) is chosen. It is known that \(AD = BF\) and \(CE = CF\). Prove that \(BD + CF > EF\). | 44. Extend $AC$ beyond point $C$ to point $D'$ such that $DA = CD'$. Then $BD + CF = BD' + CE = BD' + CD' + D'E = FB + BD' + D'E > EF$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,035 |
45. Given a 129-gon. Petya and Vasya are playing a game. They take turns marking the vertices of this polygon, with Petya making the first move. Petya can mark any unmarked vertex on each of his turns. Vasya can mark any unmarked vertex that is two positions away from the vertex Petya marked on his last turn. The game ... | 45. Answer: 7 moves. Draw all the diagonals of a 129-gon, connecting vertices that are two apart. This will form a cycle of length 129. We will call vertices adjacent if they are adjacent in this cycle. Thus, Vasya should mark a vertex adjacent to the one marked by Petya on each move. We will remove the marked vertices... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,036 |
46. Sasha has a very long strip of paper and a 2019-digit number $n$. He writes down consecutive natural numbers starting from $n$ on the strip in a row without spaces: $n, n+1, n+2, \ldots$ Prove that sooner or later, after writing down the next number, there will be a number on the strip that is divisible by 101. | 46. Let $k>100$ be a number of the form $4 \ell+2$, such that $10^{k}>n$. Note that $10^{k} \equiv-1(\bmod 101)$. From this, it is not difficult to deduce that after appending two consecutive $k$-digit numbers, the remainder modulo 101 will increase by 1. Since there will be more than 101 such pairs, in 101 such pair m... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,037 |
47. Inside an equilateral triangle $A B C$, points $P$ and $Q$ are chosen such that $P$ is inside triangle $A Q B$, $P Q = Q C$, and $\angle P A Q = \angle P B Q = 30$. Find $\angle A Q B$. | 47. We construct regular triangles $A P X$ and $B P Y$ on the sides $A P$ and $P B$ of triangle $A P B$. Triangle $A X C$ is obtained by rotating triangle $A P B$ by $60^{\circ}$ around point $A$, so these triangles are equal. Therefore, $X C = P B = P Y$. Similarly, $C Y = A P = X P$. Consequently, quadrilateral $P X ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,038 |
48. On the board, a convex 1000-gon is drawn. Dima wants to mark 500 points in it and connect each of them with broken lines to at least four vertices. All these broken lines must not intersect each other or the sides of the 1000-gon (but they can have common endpoints). Will he be able to do this? | 48. Answer: Dima will not be able to do this. Let's denote the given 1000-gon as $M$. Suppose point $B$ inside the polygon $M$ is connected to vertices $A_{1}, A_{2}, A_{3}, A_{4}$ (counterclockwise). If $A_{i} A_{i+1}$ is not a side of the 1000-gon $M$, then a broken line $A_{i} A_{i+1}$ can be drawn very close to the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,039 |
2. Find the number of all natural numbers in which each subsequent digit is less than the previous one. (8 points) | Solution. The largest possible number that satisfies the condition of the problem is 9876543210. In addition, the number must be at least two digits. All other such numbers can be obtained from 9876543210 by deleting one, two, three, four, five, six, seven, or eight digits out of ten. Then the total number
$$
\begin{g... | 1013 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,041 |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{5,8,11,14, \ldots\}$ and in the geometric progression $\{10,20,40,80, \ldots\} \cdot(10$ points $)$ | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{10,20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=10 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equali... | 6990500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,042 |
5. Find the maximum value of the expression $(\sin x + \sin 2y + \sin 3z)(\cos x + \cos 2y + \cos 3z)$. (15 points) | Solution. Note that for any $a, b$ and $c$, the following inequalities hold:
$$
a b \leq \frac{a^{2}+b^{2}}{2}
$$
(this inequality is equivalent to $2 a b \leq a^{2}+b^{2}$, or $0 \leq(a-b)^{2}$) and
$$
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right):
$$
$$
\begin{gathered}
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 a ... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,044 |
2. Find the sum of all four-digit numbers in which only the digits $1,2,3,4,5$ appear, and each digit appears no more than once. (8 points) | Solution. Any of these digits appears in any place as many times as the remaining four digits can be distributed among the remaining three places. This number is $4 \cdot 3 \cdot 2=24$. Therefore, the sum of the digits in each of the four places, taken over all four-digit numbers satisfying the conditions of the proble... | 399960 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,046 |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{4,7,10,13, \ldots\}$ and in the geometric progression $\{10,20,40,80, \ldots\} \cdot(10$ points $)$ | Solution. The members of the arithmetic progression $\{4,7,10,13,16,19, \ldots\}$ are given by the formula
$$
a_{n}=4+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{10,20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=10 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equality ... | 3495250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,047 |
5. Find the maximum value of the expression $(\sin 3 x+\sin 2 y+\sin z)(\cos 3 x+\cos 2 y+\cos z)$. $(15$ points) | Solution. Note that for any $a, b$ and $c$, the following inequalities hold:
$$
a b \leq \frac{a^{2}+b^{2}}{2}
$$
(this inequality is equivalent to $2 a b \leq a^{2}+b^{2}$, or $0 \leq(a-b)^{2}$) and
$$
\begin{gathered}
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right): \\
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 ... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,049 |
2. Find the number of all natural numbers in which each subsequent digit is greater than the previous one. (8 points) | Solution. The largest possible number satisfying the condition of the problem is 123456789. Moreover, the number must be at least two digits. All other such numbers can be obtained from 123456789 by erasing one, two, three, four, five, six, or seven digits out of nine. Then the total number
$1+C_{9}^{1}+C_{9}^{2}+C_{9... | 502 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,051 |
3. Find the sum of the first 10 elements that are found both among the members of the arithmetic progression $\{5,8,11,13, \ldots\}$, and among the members of the geometric progression $\{20,40,80,160, \ldots\} \cdot(10$ points) | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{20,40,80,160, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For common elements, the equality ... | 6990500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,052 |
4. Two adjacent faces of a tetrahedron, representing equilateral triangles with side length 3, form a dihedral angle of 30 degrees. The tetrahedron is rotated around the common edge of these faces. Find the maximum area of the projection of the rotating tetraheron onto the plane containing the given edge. (12 points) | Solution. Let the area of each of the given faces be $S$. If the face is located in the plane of projection, then the projection of the tetrahedron is equal to the area of this face $\Pi=S$.
When rotated by an angle $0<\varphi<60^{\circ}$, the area of the projection is $\Pi=S \cos \varphi<S$.
(\cos 2 x+\cos y+\cos 3 z)$. $(15$ points) | Solution. Note that for any $a, b$ and $c$, the following inequalities hold:
$$
a b \leq \frac{a^{2}+b^{2}}{2}
$$
(this inequality is equivalent to $2 a b \leq a^{2}+b^{2}$, or $0 \leq(a-b)^{2}$) and
$$
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right)
$$
$$
\begin{gathered}
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 a c... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,054 |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{4,7,10,13, \ldots\}$ and in the geometric progression $\{20,40,80,160, \ldots\} \cdot(10$ points $)$ | Solution. The members of the arithmetic progression $\{4,7,10,13,16,19, \ldots\}$ are given by the formula
$$
a_{n}=4+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For common elements, the equation $4+3 n=... | 13981000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,056 |
5. Find the maximum value of the expression $(\sin 2 x+\sin 3 y+\sin 4 z)(\cos 2 x+\cos 3 y+\cos 4 z)$. $(15$ points) | Solution. Note that for any $a, b$ and $c$, the following inequalities hold:
$$
a b \leq \frac{a^{2}+b^{2}}{2}
$$
(this inequality is equivalent to $2 a b \leq a^{2}+b^{2}$, or $0 \leq(a-b)^{2}$) and
$$
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right)
$$
$$
\begin{gathered}
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 a c... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,058 |
1. Solve the equation $\left(\sin ^{2} x+\frac{1}{\sin ^{2} x}\right)^{3}+\left(\sin ^{2} y+\frac{1}{\sin ^{2} y}\right)^{3}=16 \cos z \quad$ (5 points) | Solution. The domain of the inequality is limited by the conditions $\sin x \neq 0, \sin y \neq 0$. Note that $u+\frac{1}{u} \geq 2$ for any positive $u$, as this inequality is equivalent to
$$
\frac{u^{2}+1}{u} \geq 2 \Leftrightarrow u^{2}+1 \geq 2 u \Leftrightarrow u^{2}+1-2 u \geq 0 \Leftrightarrow(u-1)^{2} \geq 0
... | \frac{\pi}{2}+\pin,\frac{\pi}{2}+\pik,2\pi, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,059 |
2. A nine-digit natural number $A$, written in the decimal system, is obtained from the number $B$ by moving the last digit to the first place. It is known that the number $B$ is coprime with the number 18 and $B>222222222$. Find the largest and smallest among the numbers $A$ that satisfy these conditions. (Two natural... | Solution. Obviously, the last digit of the number $B$ cannot be zero, since when rearranged, it becomes the first digit of the number $A$. $18=2 \cdot 3^{2}$, so the number $B$ will be coprime with 18 if and only if it is not divisible by 2 or 3.
If $b$ is the last digit of the number $B$, then the number $A$ is expre... | A_{\}=122222224,A_{\max}=999999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,060 |
4. Plot the graph of the function $y=x^{3}+\frac{x^{3}}{1+x^{3}}+\frac{x^{3}}{\left(1+x^{3}\right)^{2}}+\frac{x^{3}}{\left(1+x^{3}\right)^{3}}+\ldots+\frac{x^{3}}{\left(1+x^{3}\right)^{n}}+\ldots$. (12 points) | Solution. The function is defined for $x \neq -1$.
for $x=0 \quad y=0$.
For other values of $x$, we transform the given expression in the condition:
$$
\begin{aligned}
& y=x^{3}+\frac{x^{3}}{1+x^{3}}+\frac{x^{3}}{\left(1+x^{3}\right)^{2}}+\frac{x^{3}}{\left(1+x^{3}\right)^{3}}+\ldots+\frac{x^{3}}{\left(1+x^{3}\right... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,061 | |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle \(\alpha=30^{\circ} .(15\) points)
 | Solution. Let the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$ and the sector $A C B=a$. Then, if the area of the segment $A C$ is $b$, we have: $x+b=S_{0}=b+a \Rightarrow x=a$.
The entire shaded area is equal to $x+y=a+y$, but $a+y$ is the area of the ... | \frac{\piR^{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,062 |
1. Solve the equation $\left(\sin ^{2} x+\frac{1}{\sin ^{2} x}\right)^{3}+\left(\cos ^{2} y+\frac{1}{\cos ^{2} y}\right)^{3}=16 \sin ^{2} z . \quad(5$ points) | Solution. The domain of the inequality is limited by the conditions $\sin x \neq 0, \cos y \neq 0$. Note that $u+\frac{1}{u} \geq 2$ for any positive $u$, as this inequality is equivalent to
$$
\frac{u^{2}+1}{u} \geq 2 \Leftrightarrow u^{2}+1 \geq 2 u \Leftrightarrow u^{2}+1-2 u \geq 0 \Leftrightarrow(u-1)^{2} \geq 0
... | \frac{\pi}{2}+\pin,\pi,\frac{\pi}{2}+\pik, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,063 |
2. An eight-digit natural number $A$, written in the decimal system, is obtained from the number $B$ by swapping the last digit to the first place. It is known that the number $B$ is coprime with the number 12 and $B>44444444$. Find the largest and smallest among the numbers $A$ that satisfy these conditions. (Two natu... | Solution. Obviously, the last digit of the number $B$ cannot be zero, since when rearranged, it becomes the first digit of the number $A$. $B=2^{2} \cdot 3$, so the number $B$ will be coprime with 12 if and only if it is not divisible by 2 or 3.
If $b$ is the last digit of the number $B$, then the number $A=10^{7} \cd... | A_{\}=14444446,A_{\max}=99999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,064 |
3. Solve the inequality $2021 \cdot \sqrt[202]{x^{2020}}-1 \geq 2020 x$ for $x \geq 0$. (10 points) | Solution. Transform the inequality into the form:
$$
\begin{aligned}
& \frac{2020 x+1}{2021} \leq \sqrt[202]{x^{2020}}, \text { from which } \\
& x+x+\ldots+x+1
\end{aligned}
$$
But by the r... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,065 |
4. Plot the graph of the function $y=\sqrt[3]{x}+\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}+\frac{\sqrt[3]{x}}{(1+\sqrt[3]{x})^{2}}+\frac{\sqrt[3]{x}}{(1+\sqrt[3]{x})^{3}}+\ldots+\frac{\sqrt[3]{x}}{(1+\sqrt[3]{x})^{n}}+\ldots$. (12 points) | Solution. The function is defined for $x \neq -1$.
for $x=0 \quad y=0$.
For other values of $x$, we transform the given expression in the condition:
$$
\begin{gathered}
y=\sqrt[3]{x}+\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}+\frac{\sqrt[3]{x}}{(1+\sqrt[3]{x})^{2}}+\frac{\sqrt[3]{x}}{(1+\sqrt[3]{x})^{3}}+\ldots+\frac{\sqrt[3... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,066 | |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle $\alpha=45^{\circ} .(15$ points $)$
 | Solution. Let the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$, and the sector $A C B=a$. Then, if the area of the segment $A C$ is $b$, we have: $x+b=S_{0}=b+a \Rightarrow x=a$.
The entire shaded area is equal to $x+y=a+y$, but $a+y$ is the area of the... | \frac{\piR^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,067 |
1. Solve the equation $\left(\sin ^{2} x+\frac{1}{\sin ^{2} x}\right)^{3}+\left(\cos ^{2} y+\frac{1}{\cos ^{2} y}\right)^{3}=16 \cos z$. (5 points) | Solution. The domain of the inequality is limited by the conditions $\sin x \neq 0, \cos y \neq 0$. Note that $u+\frac{1}{u} \geq 2$ for any positive $u$, as this inequality is equivalent to
$$
\frac{u^{2}+1}{u} \geq 2 \Leftrightarrow u^{2}+1 \geq 2 u \Leftrightarrow u^{2}+1-2 u \geq 0 \Leftrightarrow(u-1)^{2} \geq 0
... | \frac{\pi}{2}+\pi,\pik,2\pi, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,068 |
2. A nine-digit natural number $A$, written in the decimal system, is obtained from the number $B$ by moving the last digit to the first place. It is known that the number $B$ is coprime with the number 24 and $B>666666666$. Find the largest and smallest of the numbers $A$ that satisfy these conditions. (Two natural nu... | Solution. It is obvious that the last digit of the number $B$ cannot be zero, since when rearranged, it becomes the first digit of the number $A$. $24=2^{3} \cdot 3$, so the number $B$ will be coprime with 24 if and only if it is not divisible by 2 or 3.
If $b$ is the last digit of the number $B$, then the number $A$ ... | A_{\}=166666667,A_{\max}=999999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,069 |
3. Solve the equation $2021 x=2022 \cdot \sqrt[2022]{x^{2021}}-1 .(10$ points $)$ | Solution. $x \geq 0$. Transform the equation to the form:
But ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,070 |
4. Plot the graph of the function $\quad y=\sqrt[5]{x}+\frac{\sqrt[5]{x}}{1+\sqrt[5]{x}}+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{2}}+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{3}}+\ldots+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{n}}+\ldots$ $(12$ points) | Solution. The function is defined for $x \neq -1$.
for $x=0 \quad y=0$.
For other values of $x$, we transform the given expression in the condition:
$$
\begin{gathered}
y=\sqrt[5]{x}+\frac{\sqrt[5]{x}}{1+\sqrt[5]{x}}+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{2}}+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{3}}+\ldots+\frac{\sqrt[5... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,071 | |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle $\alpha=60^{\circ} .(15$ points) | Solution. $\quad$ Let
the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$, and the sector $A C B=a$. Then, if the area of the ... | \frac{2\piR^{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,072 |
1. Solve the equation $\left(\cos ^{2} x+\frac{1}{\cos ^{2} x}\right)^{3}+\left(\cos ^{2} y+\frac{1}{\cos ^{2} y}\right)^{3}=16 \sin z \cdot(5$ points $)$ | Solution. The domain of the inequality is limited by the conditions $\cos x \neq 0, \cos y \neq 0$.
Notice that $u+\frac{1}{u} \geq 2$ for any positive $u$, since this inequality is equivalent to
$$
\frac{u^{2}+1}{u} \geq 2 \Leftrightarrow u^{2}+1 \geq 2 u \Leftrightarrow u^{2}+1-2 u \geq 0 \Leftrightarrow(u-1)^{2} \... | \pin,\pik,\frac{\pi}{2}+2\pi, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,073 |
2. An eight-digit natural number $A$, written in the decimal system, is obtained from the number $B$ by swapping the last digit to the first place. It is known that the number $B$ is coprime with the number 36 and $B>77777777$. Find the largest and smallest of the numbers $A$ that satisfy these conditions. (Two natural... | Solution. It is obvious that the last digit of the number $B$ cannot be zero, since when rearranged, it becomes the first digit of the number $A$. In the prime factorization of 36, only the numbers 2 and 3 are present, so the number $B$ will be coprime with 36 if and only if it is not divisible by 2 or 3.
If $b-$ is t... | A_{\}=17777779,A_{\max}=99999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,074 |
3. Solve the equation $2021 \cdot \sqrt[202]{x^{2020}}-1=2020 x$ for $x \geq 0 \cdot$ (10 points) | Solution. Transform the equation to the form:
$$
\begin{aligned}
& \frac{2020 x+1}{2021}=\sqrt[202]{x^{2020}}, \text { from which } \\
& x+x+\ldots+x+1 \\
& \frac{(2020 \text { instances) }}{2021}=\sqrt[202]{x^{2020}}
\end{aligned}
$$
But by the relation for the arithmetic mean and the geometric mean
^{2}}+\frac{x^{5}}{\left(1+x^{5}\right)^{3}}+\ldots+\frac{x^{5}}{\left(1+x^{5}\right)^{n}}+\ldots$. (12 points) | Solution. The function is defined for $x \neq -1$.
for $x=0 \quad y=0$.
For other values of $x$, we transform the given expression in the condition:
$$
\begin{gathered}
y=x^{5}+\frac{x^{5}}{1+x^{5}}+\frac{x^{5}}{\left(1+x^{5}\right)^{2}}+\frac{x^{5}}{\left(1+x^{5}\right)^{3}}+\ldots+\frac{x^{5}}{\left(1+x^{5}\right)... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,076 | |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle $\alpha=20^{\circ} .(15$ points $)$ | Solution. Let the area of the semicircle be \( S_{0}=\frac{\pi R^{2}}{2} \), the area of the lune \( A B_{1} C \) be \( x \), the lune \( C B_{1} B \) be \( y \), and the sector \( A C B \) be \( a \). Then, if the area of the segment \( A C \) is \( b \), we have: \( x + b = S_{0} = b + a \Rightarrow x = a \).
The en... | \frac{2\piR^{2}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,077 |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{5,8,11,14, \ldots\}$ and in the geometric progression $\{10,20,40,80, \ldots\} \cdot(10$ points) | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{10,20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=10 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equati... | 6990500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,078 |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{4,7,10,13, \ldots\}$ and in the geometric progression $\{10,20,40,80, \ldots\} \cdot(10$ points) | Solution. The members of the arithmetic progression $\{4,7,10,13,16,19, \ldots\}$ are given by the formula
$$
a_{n}=4+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{10,20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=10 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equality ... | 3495250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,080 |
4. Two adjacent faces of a tetrahedron, which are isosceles right triangles with a hypotenuse of 2, form a dihedral angle of 60 degrees. The tetrahedron is rotated around the common edge of these faces. Find the maximum area of the projection of the rotating tetrahedron onto the plane containing
(\cos 3 x+\cos 2 y+\cos z)$. (15 points) | Solution. Note that for any $a, b$ and $c$, the following inequalities hold:
$$
a b \leq \frac{a^{2}+b^{2}}{2}
$$
(this inequality is equivalent to $2 a b \leq a^{2}+b^{2}$, or $0 \leq(a-b)^{2}$) and
$$
\begin{gathered}
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right): \\
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 ... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,082 |
3. Find the sum of the first 10 elements that are common to both the arithmetic progression $\{5,8,11,13, \ldots\}$ and the geometric progression $\{20,40,80,160, \ldots\}$. (10 points) | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{20,40,80,160, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equat... | 6990500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,083 |
5. Find the maximum value of the expression $(\sin 2 x+\sin y+\sin 3 z)(\cos 2 x+\cos y+\cos 3 z)$. (15 points) | Solution. Note that for any $a, b$ and $c$, the inequalities
$$
a b \leq \frac{a^{2}+b^{2}}{2}
$$
(this inequality is equivalent to $2 a b \leq a^{2}+b^{2}$, or $0 \leq(a-b)^{2}$) and
$$
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right)
$$
$$
\begin{gathered}
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 b c \leq \\
\... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,085 |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{4,7,10,13, \ldots\}$ and in the geometric progression $\{20,40,80,160, \ldots\} .(10$ points $)$ | Solution. The members of the arithmetic progression $\{4,7,10,13,16,19, \ldots\}$ are given by the formula
$$
a_{n}=4+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $(20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For common elements, the equality $4+3 n=2... | 13981000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,086 |
5. Find the maximum value of the expression $(\sin 2 x+\sin 3 y+\sin 4 z)(\cos 2 x+\cos 3 y+\cos 4 z)$. (15 points) | Solution. Note that for any $a, b$ and $c$, the following inequalities hold:
$$
a b \leq \frac{a^{2}+b^{2}}{2}
$$
(this inequality is equivalent to $2 a b \leq a^{2}+b^{2}$, or $0 \leq(a-b)^{2}$) and
$$
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right)
$$
$$
\begin{gathered}
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 a c... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,088 |
2. A nine-digit natural number $A$, written in the decimal system, is obtained from the number $B$ by moving the last digit to the first place. It is known that the number $B$ is coprime with the number 18 and $B>22222222$. Find the largest and smallest among the numbers $A$ that satisfy these conditions. (Two natural ... | Solution. Obviously, the last digit of the number $B$ cannot be zero, since when rearranged, it becomes the first digit of the number $A$. $18=2 \cdot 3^{2}$, so the number $B$ will be coprime with 18 if and only if it is not divisible by 2 or 3.
If $b$ is the last digit of the number $B$, then the number $A$ is expre... | A_{\}=122222224,A_{\max}=999999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,089 |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle $\alpha=30^{\circ}$. (15 points)
 | Solution. Let the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$ and the sector $A C B=a$. Then, if the area of the segment $A C$ is $b$, we have: $x+b=S_{0}=b+a \Rightarrow x=a$.
The entire shaded area is equal to $x+y=a+y$, but $a+y$ is the area of the ... | \frac{\piR^{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,090 |
2. An eight-digit natural number $A$, written in the decimal system, is obtained from the number $B$ by swapping the last digit to the first place. It is known that the number $B$ is coprime with the number 12 and $B>4444444$. Find the largest and smallest among the numbers $A$ that satisfy these conditions. (Two natur... | Solution. Obviously, the last digit of the number $B$ cannot be zero, since when rearranged, it becomes the first digit of the number $A$. $B=2^{2} \cdot 3$, so the number $B$ will be coprime with 12 if and only if it is not divisible by 2 or 3.
If $b$ is the last digit of the number $B$, then the number $A=10^{7} \cd... | A_{\}=14444446,A_{\max}=99999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,092 |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle $\alpha=45^{\circ}$. (15 points)
 | Solution. Let the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$ and the sector $A C B=a$. Then, if the area of the segment $A C$ is $b$, we have: $x+b=S_{0}=b+a \Rightarrow x=a$.
The entire shaded area is equal to $x+y=a+y$, but $a+y$ is the area of the ... | \frac{\piR^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,093 |
4. Plot the graph of the function $y=\sqrt[5]{x}+\frac{\sqrt[5]{x}}{1+\sqrt[5]{x}}+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{2}}+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{3}}+\ldots+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{n}}+\ldots$. (12 points) | Solution. The function is defined for $x \neq -1$.
for $x=0 \quad y=0$.
For other values of $x$, we transform the given expression in the condition:
$$
\begin{gathered}
y=\sqrt[5]{x}+\frac{\sqrt[5]{x}}{1+\sqrt[5]{x}}+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{2}}+\frac{\sqrt[5]{x}}{(1+\sqrt[5]{x})^{3}}+\ldots+\frac{\sqrt[5... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,095 | |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle $\alpha=60^{\circ}$. (15 points) | Solution.
Let
the area of the semicircle $S_{0}=\frac{\pi... | \frac{2\piR^{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,096 |
1. Solve the equation $\left(\cos ^{2} x+\frac{1}{\cos ^{2} x}\right)^{3}+\left(\cos ^{2} y+\frac{1}{\cos ^{2} y}\right)^{3}=16 \sin z \cdot(5$ points) | Solution. The domain of the inequality is limited by the conditions $\cos x \neq 0, \cos y \neq 0$.
Notice that $u+\frac{1}{u} \geq 2$ for any positive $u$, since this inequality is equivalent to
$$
\frac{u^{2}+1}{u} \geq 2 \Leftrightarrow u^{2}+1 \geq 2 u \Leftrightarrow u^{2}+1-2 u \geq 0 \Leftrightarrow(u-1)^{2} \... | \pin,\pik,\frac{\pi}{2}+2\pi, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,097 |
2. An eight-digit natural number $A$, written in the decimal system, is obtained from the number $B$ by swapping the last digit to the first place. It is known that the number $B$ is coprime with the number 36 and $B>7777777$. Find the largest and smallest of the numbers $A$ that satisfy these conditions. (Two natural ... | Solution. It is obvious that the last digit of the number $B$ cannot be zero, since when rearranged, it becomes the first digit of the number $A$. In the prime factorization of 36, only the numbers 2 and 3 are present, so the number $B$ will be coprime with 36 if and only if it is not divisible by 2 or 3.
If $b$ is th... | A_{\}=17777779,A_{\max}=999999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,098 |
3. Solve the equation $2021 \cdot \sqrt[202]{x^{2020}}-1=2020 x$ for $x \geq 0$. (10 points) | Solution. Transform the equation to the form:
$$
\begin{aligned}
& \frac{2020 x+1}{2021}=\sqrt[202]{x^{2020}}, \text { from which } \\
& \frac{\begin{array}{l}
x+x+\ldots+x+1 \\
(2020 \text { terms) }
\end{array}}{2021}=\sqrt[202]{x^{2020}} .
\end{aligned}
$$
But by the inequality between the arithmetic mean and the ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,099 |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle $\alpha=20^{\circ} .(15$ points) | Solution. Let the area of the
semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$, and the sector $A C B=a$. Then, if the area of the segment $... | \frac{2\piR^{2}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,101 |
5. The force with which the airflow acts on the sail can be calculated using the formula
$F=\frac{c s \rho\left(v_{0}-v\right)^{2}}{2}$, where $C$ is the aerodynamic force coefficient, $S$ is the area of the sail $S=5 \mathrm{m}^{2} ; \rho$ is the density of air, $v_{0}$ is the wind speed $v_{0}=6 \mathrm{~m} / \mathr... | Solution:
$$
\begin{aligned}
& \left.\begin{array}{l}
F=f(v) \\
N=F \cdot v
\end{array}\right\} \Rightarrow N=f(v) \\
& N=\frac{\operatorname{CS\rho }}{2}\left(v_{0}^{2}-2 v_{0} v+v^{2}\right) v=\frac{\operatorname{CS\rho }}{2}\left(v_{0}^{2} v-2 v_{0} v^{2}+v^{3}\right) \\
& N=N_{\max } \Rightarrow N^{\prime}(v)=0 \\... | 2\mathrm{~}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,102 |
5. The force with which the airflow acts on the sail can be calculated using the formula
$F=\frac{A S \rho\left(v_{0}-v\right)^{2}}{2}$, where $A-$ is the aerodynamic force coefficient, $S-$ is the area of the sail $S$ $=4 \mathrm{M}^{2} ; \rho-$ is the density of air, $v_{0}$ - is the wind speed $v_{0}=4.8 \mathrm{~m... | Solution:
$$
\begin{aligned}
& \left.\begin{array}{l}
F=f(v) \\
N=F \cdot v
\end{array}\right\} \Rightarrow N=f(v) \\
& N=\frac{A S \rho}{2}\left(v_{0}^{2}-2 v_{0} v+v^{2}\right) v=\frac{A S \rho}{2}\left(v_{0}^{2} v-2 v_{0} v^{2}+v^{3}\right) \\
& N=N_{\max } \Rightarrow N^{\prime}(v)=0 \\
& N^{\prime}(v)=\frac{A S \... | 1.6\mathrm{M}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,103 |
5. The force with which the airflow acts on the sail can be calculated using the formula
$F=\frac{B S \rho\left(v_{0}-v\right)^{2}}{2}$, where $B-$ is the aerodynamic force coefficient, $S-$ is the area of the sail $S$ $=7 \mathrm{M}^{2} ; \rho$ - air density, $v_{0}$ - wind speed $v_{0}=6.3 \mathrm{~m} / \mathrm{c}, ... | $$
\begin{aligned}
& \left.\begin{array}{l}
F=f(v) \\
N=F \cdot v
\end{array}\right\} \Rightarrow N=f(v) \\
& N=\frac{B S \rho}{2}\left(v_{0}^{2}-2 v_{0} v+v^{2}\right) v=\frac{B S \rho}{2}\left(v_{0}^{2} v-2 v_{0} v^{2}+v^{3}\right) \\
& N=N_{\max } \Rightarrow N^{\prime}(v)=0 \\
& N^{\prime}(v)=\frac{B S \rho}{2}\lef... | 3.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,104 |
5. The force with which the airflow acts on the sail can be calculated using the formula $F=\frac{c s \rho\left(v_{0}-v\right)^{2}}{2}$, where $C$ - the aerodynamic force coefficient, $S$ - the area of the sail; $\rho$ - the density of the air, $v_{0}$ - the wind speed, $v$ - the speed of the sailing vessel. At some po... | $$
\begin{aligned}
& \left.\begin{array}{l}
F=f(v) \\
N=F \cdot v
\end{array}\right\} \Rightarrow N=f(v) \\
& N=\frac{C S \rho}{2}\left(v_{0}^{2}-2 v_{0} v+v^{2}\right) v=\frac{C S \rho}{2}\left(v_{0}^{2} v-2 v_{0} v^{2}+v^{3}\right) \\
& N=N_{\max } \Rightarrow N^{\prime}(v)=0 \\
& N^{\prime}(v)=\frac{C S \rho}{2}\lef... | \frac{v_0}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,105 |
2. An athlete with a mass of 78.75 kg is testing a net used by firefighters to save people. The net sags by 100 cm when the athlete jumps from a height of 15 m. Assuming the net behaves elastically like a spring, calculate how much it will sag when a person with a mass of 45 kg jumps from a height of 29 m.
Given:
$m_... | Solution. The mechanical system "Earth-athlete-net" $x_{2}-?$ can be considered closed. According to the law of conservation of energy, when the athlete jumps, his potential energy should completely transform into the energy of the elastic deformation of the net: $m_{2} g\left(h_{2}+x_{2}\right)=\frac{k x_{2}^{2}}{2} ;... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,107 |
5. Consider the crank-slider mechanism. (1-crank, 2- connecting rod). They were first applied in antiquity, on Roman sawmills. There, the rotation of the wheel, driven by the force of falling water, was converted into reciprocating motion of the saw blade. The crank rotates at a constant angular velocity \(\omega=10 \m... | Solution. Let's write the equations of motion of point M in coordinate form. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,108 | |
5. Consider the crank-slider mechanism (1-crank, 2- connecting rod). They were first used in antiquity, in Roman sawmills. There, the rotation of the wheel, driven by the force of falling water, was converted into reciprocating motion of the saw blade. The crank rotates at a constant angular velocity $\omega=10$ $rad/s... | Solution. Let's write the equations of motion of point M in coordinate form. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,109 |
5. The force with which the airflow acts on the sail can be calculated using the formula
$F=\frac{C S \rho\left(v_{0}-v\right)^{2}}{2}$, where $C-$ is the aerodynamic force coefficient, $S-$ is the area of the sail $S=5 \mathrm{~m}^{2} ; \rho$ - density of air, $v_{0}$ - wind speed $v_{0}=6 \mathrm{M} / c, v$ - speed ... | Solution:
\[
\left.\begin{array}{l}
F=f(v) \\
N=F \cdot v
\end{array}\right\} \Rightarrow N=f(v)
\]
\[
N=\frac{\operatorname{CS\rho }}{2}\left(v_{0}^{2}-2 v_{0} v+v^{2}\right) v=\frac{\operatorname{CS\rho }}{2}\left(v_{0}^{2} v-2 v_{0} v^{2}+v^{3}\right)
\]
\[
N=N_{\max } \Rightarrow N^{\prime}(v)=0
\]
\[
N^{\prime... | v=\frac{v_0}{3}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,111 |
5. The force with which the airflow acts on the sail can be calculated using the formula
$F=\frac{A S \rho\left(v_{0}-v\right)^{2}}{2}$, where $A-$ is the aerodynamic force coefficient, $S-$ is the area of the sail $S$ $=4 \mathrm{m}^{2} ; \rho$ - density of air, $v_{0}$ - wind speed $v_{0}=4.8 \mu / c, v$ - speed of ... | Solution:
$$
\begin{aligned}
& \left.\begin{array}{l}
F=f(v) \\
N=F \cdot v
\end{array}\right\} \Rightarrow N=f(v) \\
& N=\frac{A S \rho}{2}\left(v_{0}^{2}-2 v_{0} v+v^{2}\right) v=\frac{A S \rho}{2}\left(v_{0}^{2} v-2 v_{0} v^{2}+v^{3}\right) \\
& N=N_{\max } \Rightarrow N^{\prime}(v)=0 \\
& N^{\prime}(v)=\frac{A S \... | 1.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,112 |
5. The force with which the airflow acts on the sail can be calculated using the formula
$F=\frac{B S \rho\left(v_{0}-v\right)^{2}}{2}$, where $B-$ is the aerodynamic force coefficient, $S-$ is the area of the sail $S$ $=7 \mathrm{M}^{2} ; \rho$ - air density, $v_{0}$ - wind speed $v_{0}=6.3 \mu / c, v-$ speed of the ... | Solution:
$\left.\begin{array}{l}F=f(v) \\ N=F \cdot v\end{array}\right\} \Rightarrow N=f(v)$
$N=\frac{B S \rho}{2}\left(v_{0}^{2}-2 v_{0} v+v^{2}\right) v=\frac{B S \rho}{2}\left(v_{0}^{2} v-2 v_{0} v^{2}+v^{3}\right)$
$N=N_{\max } \Rightarrow N^{\prime}(v)=0$
$N^{\prime}(v)=\frac{B S \rho}{2}\left(v_{0}^{2}-4 v_{... | 3.1\mathrm{M}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,113 |
5. The force with which the airflow acts on the sail can be calculated using the formula $F=\frac{c S \rho\left(v_{0}-v\right)^{2}}{2}$, where $C$ is the aerodynamic force coefficient, $S$ is the area of the sail; $\rho$ is the density of the air, $v_{0}$ is the wind speed, and $v$ is the speed of the sailboat. At some... | Solution:
\[\left.\begin{array}{l}F=f(v) \\ N=F \cdot v\end{array}\right\} \Rightarrow N=f(v)\]
\[N=\frac{C S \rho}{2}\left(v_{0}^{2}-2 v_{0} v+v^{2}\right) v=\frac{C S \rho}{2}\left(v_{0}^{2} v-2 v_{0} v^{2}+v^{3}\right)\]
\[N=N_{\max } \Rightarrow N^{\prime}(v)=0\]
\[N^{\prime}(v)=\frac{C S \rho}{2}\left(v_{0}^{2... | \frac{v_0}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,114 |
5. Consider the crank-slider mechanism (1-crank, 2-connecting rod). They were first used in antiquity, in Roman sawmills. There, the rotation of the wheel, driven by the force of falling water, was converted into reciprocating motion of the saw blade. The crank in question rotates at a constant
. They were first applied in antiquity, on Roman sawmills. There, the rotation of the wheel, driven by the force of falling water, was converted into reciprocating motion of the saw blade. The crank rotates at a constant angular velocity \(\omega=10 \m... | Solution. Let's write the equations of motion of point M in coordinate form. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,120 |
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