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7. (4 points) Four out of six midpoints of the edges of a certain tetrahedron form a regular tetrahedron with an edge length of 2. Find the edges of the original tetrahedron. | Answer: Five edges of length 4, one edge of length $4 \sqrt{2}$.
Solution:
Let the vertices of the tetrahedron be $A, B, C, D$. The midpoint of edge $A B$ is denoted as $M_{A B}$, and similar notations are introduced for the other edges.
Suppose the midpoints of the skew edges of the original tetrahedron do not fall... | Five\edges\of\length\4,\one\edge\of\length\4\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,008 |
8. (5 points) Prove that for positive $x, y, z$ the inequality
$$
(x+8 y+2 z)(x+2 y+z)(x+4 y+4 z) \geqslant 256 x y z
$$
holds. | # Solution:
Consider the first two parentheses and note that
$$
(x+8 y+2 z)(x+2 y+z)=(x+4 y)^{2}+3 z(x+4 y)+2 z^{2}+2 x y
$$
Then we can rewrite the required inequality as
$$
\left(\frac{(x+4 y)^{2}+3 z(x+4 y)+2 z^{2}}{x y}+2\right) \cdot \frac{x+4 y+4 z}{z} \geqslant 256
$$
Now fix $z$ and $x+4 y$ and shift $x$ a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 21,009 |
1. (2 points) Boy Vasya wrote down the non-zero coefficients of the polynomial $P(x)$ of the seventh degree in his notebook. Then he calculated the derivative of the resulting polynomial and wrote down its non-zero coefficients, and so on, until he obtained a constant, which he also wrote down.
What is the smallest nu... | # Answer: 7
Solution: Evaluation: since the polynomial has a degree of 7, it definitely has a non-zero coefficient for $x^{7}$, let's call it $a$. Then the leading coefficient of the derivative of this polynomial is $7a$, the leading coefficient of the second derivative is $7 \cdot 6a$, and so on, the leading coeffici... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,010 |
3. (3 points) In a certain country, there are 120 cities. The Ministry of Aviation requires that every two cities be connected by a two-way flight operated by exactly one airline, and that it should be possible to travel from any city to any other using the flights of each airline (possibly with layovers). What is the ... | Solution: Evaluation: in order to connect all cities, the airline must have at least 119 flights. The total number of city pairs is $120 \cdot 119 / 2$, so there are no more than 60 airlines.
Example: let's denote the cities as $a_{1}, a_{2}, \ldots a_{60}, b_{1}, b_{2}, \ldots, b_{60}$ in some way. Then the airline w... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,012 |
4. (3 points) Prove that for $n=12002$ the sum of binomial coefficients with a step of 6, i.e., $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$, gives a remainder of 1 when divided by 3.
$C_{n}^{k}-$ the number of ways to choose $k$ items from $n$, which is $\frac{n!}{k!(n-k)!}$ if $0 \leqslant k \leqslant n$ and 0 in all ot... | # Solution:
It is easy to verify that $C_{n}^{k}=C_{n-3}^{k-3}+3 C_{n-3}^{k-2}+3 C_{n-3}^{k-1}+C_{n-3}^{k}$, therefore, $C_{n}^{k}$ gives the same remainder when divided by 3 as $C_{n-3}^{k-3}+C_{n-3}^{k}$.
Thus, $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$ gives the same remainder when divided by 3 as $C_{n-3}^{1}+C_{n-... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false | 21,013 |
5. (3 points) Three circles with radii $\frac{1}{2}, \frac{1}{2}$, and $\sqrt{\frac{13-6 \sqrt{3}}{13}}$ are arranged such that the triangle formed by their centers is an equilateral triangle with side length $\frac{\sqrt{3}}{2}$. Find the radius of the circumcircle of the triangle, each vertex of which is the point of... | Answer: $2 \sqrt{3}-3$
Solution:
Let $A, B, C$ be the points of intersection of the first and second, second and third, and first and third circles, respectively, and $O_{1}, O_{2}, O_{3}$ be the centers of these circles.
The radii of the first and second circles are equal, so point $A$ will lie on the perpendicular... | 2\sqrt{3}-3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,014 |
6. (3 points) Find the total length of intervals on the number line where the inequalities $|x|<1$ and $\operatorname{tg} \log _{5}|x|<0$ are satisfied. | Answer:
$$
\frac{2 \cdot 5^{\frac{\pi}{2}}}{1+5^{\frac{\pi}{2}}}
$$
Solution:
It is sufficient to solve the problem for positive $x$, and then multiply the result by 2.
If $\operatorname{tg} \log _{5} x<0$, then $\pi\left(k-\frac{1}{2}\right)<\log _{5} x<\pi k$ for some integer $k$, which means $5^{\pi\left(k-\frac... | \frac{2\cdot5^{\frac{\pi}{2}}}{1+5^{\frac{\pi}{2}}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,015 |
7. (4 points) Four out of six midpoints of the edges of a certain tetrahedron form a regular tetrahedron with an edge length of 3. Find the edges of the original tetrahedron. | Answer: Five edges of length 6, one edge of length $6 \sqrt{2}$.
Solution:
Let the vertices of the tetrahedron be $A, B, C, D$. The midpoint of edge $A B$ is denoted as $M_{A B}$, and similar notations are introduced for the other edges.
Suppose the midpoints of the skew edges of the original tetrahedron do not fall... | Five\edges\of\length\6,\one\edge\of\length\6\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,016 |
8. (5 points) Prove that for positive $x, y, z$ the inequality
$$
(x+9 y+3 z)(x+4 y+2 z)(2 x+12 y+9 z) \geqslant 1029 x y z
$$
holds. | Solution:
Consider the first two parentheses and note that
$$
(x+9 y+3 z)(x+4 y+2 z)=(x+6 y)^{2}+5 z(x+6 y)+6 z^{2}+x y
$$
Then we can rewrite the required inequality as
$$
\left(\frac{(x+6 y)^{2}+5 z(x+6 y)+6 z^{2}}{x y}+1\right) \cdot \frac{2 x+12 y+9 z}{z} \geqslant 1029
$$
Now fix $z$ and $x+6 y$ and shift $x$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 21,017 |
1. (2 points) In an infinite volleyball tournament, 65 teams participate. Every morning, the organizer schedules the matches for the day. Each day, 64 teams are paired to play one match each, while one team rests. Each match ends with one team winning, there are no draws.
Prove that the organizer can schedule the matc... | # Solution:
For the first six days, the organizer conducts a tournament among 64 teams using the Olympic system (knockout), with extra matches played arbitrarily. The winner of this tournament will win six consecutive matches.
For the next six days, the winner of the first Olympic tournament rests, while the remainin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,019 |
2. (2 points) Solve the puzzle $T O K=K O T+K T O$. (Different letters represent different digits, numbers cannot start with zero). | Answer: $954=459+495$.
Solution: Let's write the numbers from our puzzle as the sum of place value addends:
$$
100 T+10 O+K=100 K+10 T+O+100 K+10 O+T
$$
from which $89 T=199 K+O$. The left side of the equation does not exceed 801, so it makes sense to try $K$ from 1 to 4. When $K=4$, we get $T=9$ and $O=5$. | 954=459+495 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,020 |
3. (2 points) A chess piece called a quadraliner attacks two verticals and two horizontals adjacent to the square it occupies. What is the maximum number of non-attacking quadraliners that can be placed on a $10 \times 10$ board? | Answer: 25
## Solution:
Example: place the four-line rulers at the intersections of even columns and even rows.
Evaluation: In each pair of rows, the four-line rulers can only stand on one row. At the same time, they cannot stand on two adjacent cells. | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,021 |
4. (3 points) Vasya found a triangle with sides $a, b, c$ and formed the quadratic trinomial $a x^{2}-b x+c$. This trinomial has two roots. Prove that they cannot both be less than $\frac{1}{3}$.
# | # Solution:
For non-positive $x$, the quadratic polynomial takes positive values, so they cannot be roots. Let $x_{1}, x_{2}$ be the positive roots of our quadratic polynomial, and both are less than $\frac{1}{3}$. Then
$b+c=a\left(\frac{b}{a}+\frac{c}{a}\right)=a\left(x_{1}+x_{2}+x_{1} x_{2}\right) \leqslant \frac{7... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,022 |
6. (3 points) Points $X, Y, Z$ lie on a straight line in that order. $\omega_{1}, \omega_{2}, \omega_{3}$ are circles with diameters $X Z, X Y, Y Z$ respectively. Line $\ell$ intersects circle $\omega_{1}$ at point $A$, then circle $\omega_{2}$ at point $B$, then it passes through $Y$, then it intersects circle $\omega... | # Solution:
Angles $\angle X B D$ and $\angle X A Z$ are subtended by the diameters of circles $\omega_{2}$ and $\omega_{1}$ respectively, hence they are right angles. $\angle B D X=\angle A D X=\angle A Z X$. Therefore, triangles $X A Z$ and $X B D$ are similar.
Thus, $B D=\frac{A Z \cdot X D}{X Z}$. Similarly, from... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,024 |
7. (4 points) Prove that if a convex 102-gon can be divided into triangles by non-intersecting diagonals such that from each vertex either exactly three diagonals or none emanate, then the number of resulting triangles, all three sides of which are diagonals of the original polygon, will be exactly 34.
# | # Solution:
A 102-gon is divided into 100 triangles by 99 diagonals. Each diagonal has two ends, so the sum of the number of diagonals emanating from the vertices of the polygon is 198. Therefore, the number of vertices from which three diagonals emanate is 66, and the number of remaining vertices from which no diagon... | 34 | Combinatorics | proof | Yes | Yes | olympiads | false | 21,025 |
8. (4 points) Given a triangle $ABC$ with sides $AB=13$, $BC=14$, $AC=15$. On side $AB$, point $K$ is marked; on side $BC$, point $L$ is marked; on side $AC$, point $N$ is marked. It is known that $BK=\frac{14}{13}$, $AN=10$, $BL=1$. A line parallel to $KL$ is drawn through point $N$, intersecting side $BC$ at point $M... | Solution:
The area of triangle $ABC$ by Heron's formula is 84.
$$
\begin{gathered}
S_{KBL}=\frac{KB \cdot BL}{AB \cdot BC} \cdot S_{ABC}=\frac{\frac{14}{13} \cdot 1}{13 \cdot 14} \cdot 84=\frac{84}{169} \\
S_{AKN}=\frac{AK \cdot AN}{AB \cdot AC} \cdot S_{ABC}=\frac{\left(13-\frac{14}{13}\right) \cdot 10}{13 \cdot 15}... | \frac{36503}{1183}=30\frac{1013}{1183} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,026 |
8. (4 points) Given a triangle $ABC$ with sides $AB=14$, $BC=13$, and $AC=15$. On side $AB$, point $K$ is marked; on side $AC$, point $L$ is marked; and on side $BC$, point $N$ is marked. It is known that $AK=\frac{15}{14}$, $BN=9$, and $AL=1$. A line parallel to $KL$ is drawn through point $N$, intersecting side $AC$ ... | Answer: $\frac{36503}{1183}=30 \frac{1013}{1183}$. | 30\frac{1013}{1183} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,028 |
3. A grid rectangle $4 \times 11$ is divided into $2 \times 2$ squares and strips of three cells. What is the maximum number of strips that can participate in this division? | Answer: 12
## Examples of answer notation:
100
90
# | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,032 |
3. In September, Vasya's salary was increased by $x$ percent, and in October it was decreased by the same $x$ percent. After these changes, Vasya was surprised to find that his salary had decreased by $36\%$ compared to the original. Find $x$. | Answer: 60
## Examples of answer recording:
23
# | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,033 |
3. On the board, there was a two-digit number that is not divisible by 10. Alyosha increased its first digit by five times, and decreased the second digit by some integer factor. The result was a new two-digit number that is divisible by the original number. Find the original number. If there are multiple possible answ... | Answer: $17 ; 18|17,18| 18,17 \mid 18 ; 17$
## Examples of answer notation:
11
$22 ; 33$
# | 17;18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,036 |
2. Given a fractional-linear function of the form $f(x)=\frac{a x+b}{c x+d}$, where $a, b, c$ and $d-$ are different natural numbers from 1 to 15. What is the smallest value that $f(3)$ can take? | Write the answer in the form of a proper fraction, do not round.
Example of answer recording: $27 / 34$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,039 |
3. Point $O$ is the center of a certain circle, $A$ is a point outside the circle, $B$ is a point on the circle such that $AB$ is a tangent. $AO=6$. Find the greatest possible value of the area of triangle $AOB$.
| Answer: 9.
## Examples of answer notations:
17
$1 / 7$
1.7
# | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,040 |
3. In country Gamma, there are 101 cities. It is known that at least 2 roads lead out of each city. It is also known that if there is a road from city A to city B and from city B to city C, then there is also a road from city A to city C. What is the minimum number of roads that can be in the country? | Answer: 205
## Examples of answer notation:
17
# | 205 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,042 |
3. Solve the equation $\mathrm{p}^{3}-\mathrm{q}^{3}=2716$ in prime numbers. In your answer, specify the pair $(p ; q)$. If there are several such pairs, specify the one for which $p-q$ is maximal. | Answer: $17 ; 13 \mid(17 ; 13)$
## Examples of answer notation:
$(37 ; 11)$
# | (17;13) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,044 |
1. Given a rectangular parallelepiped with sides 6, 6, and \( x \). For what values of the number \( x \) is the product of the surface area and the perimeter (the sum of the lengths of all edges) of this parallelepiped no more than \( \frac{224}{3} \) of its volume? | Write the answer in the form of an interval.
Answer: $[4 ; 9] \|[4,9]$ | [4,9] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,045 |
2. Given a rectangular parallelepiped with sides 4, 6, and \( x \). For what values of the number \( x \) is the product of the surface area and the perimeter (the sum of the lengths of all edges) of this parallelepiped no more than 78 times its volume? | Write the answer in the form of an interval.
Answer: $[3 ; 8] \|[3,8]$ | [3,8] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,046 |
3. Given a rectangular parallelepiped with sides 2, 6, and \( x \). For what values of the number \( x \) is the product of the surface area and the perimeter (the sum of the lengths of all edges) of this parallelepiped no more than 88 times its volume? | Write the answer in the form of an interval.
Answer: $[3 ; 4] \|[3,4]$
## Examples of writing answers:
$[1 ; 5]$
# | [3,4] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,047 |
1. It is known that the number $\sqrt{3}+\sqrt{5}$ is a root of a polynomial of the fourth degree with integer coefficients, the leading coefficient of which is 1. What is the sum of the coefficients of this polynomial? | Answer: -11
2. It is known that the number $\sqrt{3}+\sqrt{7}$ is a root of a polynomial of the fourth degree with integer coefficients, the leading coefficient of which is 1. What is the sum of the coefficients of this polynomial?
Answer: -3 | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,048 |
3. It is known that the number $\sqrt{5}+\sqrt{7}$ is a root of a polynomial of the fourth degree with integer coefficients, the leading coefficient of which is 1. What is the sum of the coefficients of this polynomial? | Answer: -19
## Examples of how to write answers:
17
$1 / 7$
1.7
## Problem 8 | -19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,049 |
3. Given two geometric progressions $a_{n}$ and $b_{n}$. It is known that $a_{0}=4$ and the sum of the denominators of these two progressions is 5. Find what $b_{0}$ can be, if it is known that $a_{0}, b_{0}, a_{1}, b_{1}$ are consecutive terms of some arithmetic progression (in that order) | Answer: $-2 ; 8 \| 8 ; -2$
## Examples of answer notation:
17
$1 / 7$
1.7
$1 ; 7$
1, $7 ; 17$
# | 8;-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,052 |
3. It is known that the function $f(x)$ satisfies the equality for any $x$:
$3 \sin x+f(x)=f(x+2)$
Find the value of $\mathrm{f}(2017) \sin 1$, given that $f(1)=\frac{3 \cos ^{2} 1008}{\sin 1}$. | Answer: 3
## Examples of answer notations:
17
$1 / 7$
1.7 | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,053 |
2. (3 points) On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, $n$ islanders gathered in a room.
The first one said: "Exactly 1 percent of those present in this room are liars."
The second one said: "Exactly 2 percent of those present in this room are liars... | Answer: 100
## Solution:
There must be exactly one knight, as all other islanders in the room contradict each other. This means that the knights are $\frac{1}{n}$ of the total number of people. Let them be $k$ percent of the total number of people present, which means the person with the number $100-k$ is telling the... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,056 |
3. (3 points) Anya, Vanya, Danya, and Tanya were collecting apples. It turned out that each of them collected a whole percentage of the total number of apples collected, and all these numbers were different and greater than zero. Then Tanya, who collected the most apples, ate her apples. After this, it turned out that ... | # Solution:
Let Anya have collected the fewest apples, which is $k$, and the total number of apples is $n$. Then we get that the total number of apples collected is no less than $4k + 6$. For $k \geq 5$, we already get more than 20 apples. For $k < 5$, we need to consider the cases where $k$ is 1, 2, 3, or 4.
If $k ... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,057 |
4. (3 points) A number written on the board is allowed to be multiplied by 5 or to have its digits rearranged (it is not allowed to place zero in the first position). Is it possible to obtain the hundred-digit number $5222 \ldots 2221$ from the number 1 in this way?
# | # Answer: No
## Solution:
Consider the reverse process: we will try to get from the number $5222 \ldots 2221$ to 1 using the specified operations.
First, we need to get a number divisible by 5, specifically $222 \ldots 22122 \ldots 5 = 222 \ldots 2225 - 10^k$ (where $k$ is a natural number). Dividing it by 5, we get... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,058 |
6. (3 points) Anya counted all nine-digit numbers, all digits in each of which are different, divisible by 9. Kolya counted all ten-digit numbers, all digits in each of which are different, divisible by 5. Who of them counted more numbers? | Answer: both counted the same.
## Solution:
Let's establish a one-to-one correspondence between these sets of numbers. Consider some of Anya's numbers. Since it is divisible by 9, the sum of its digits is also divisible by 9. To satisfy this condition, the number must contain all digits except 0 or 9.
Append the mis... | bothcountedthesame | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,059 |
7. (3 points) Given a convex pentagon $A B C D E$. Point $P$ is the intersection of $B E$ and $A C$, point $Q$ is the intersection of $C E$ and $A D$, point $O$ is the intersection of $A D$ and $B E$. It turns out that triangles $A B P$ and $D E Q$ are isosceles triangles with the angle at the vertex (exactly at the ve... | Answer: $60^{\circ}$ or $105^{\circ}$
Solution: Consider triangle $A B P$. The angle $\angle A P B$ in it can be $80^{\circ}$ or $50^{\circ}$. Then the adjacent $\angle A P O$ is $100^{\circ}$ or $130^{\circ}$. Therefore, in triangle $A P O$, this is the angle at the vertex and $\angle A O P$ is $40^{\circ}$ or $25^{\... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,060 |
8. (4 points) On the board, all three-digit natural numbers are written, the first digits of which are odd and greater than 1. What is the maximum number of quadratic equations of the form $a x^{2}+b x+c=0$ that can be formed using these numbers as $a, b$ and $c$, each no more than once, such that all these equations h... | # Answer: 100
## Solution:
In each equation, there must be a coefficient less than five hundred. Otherwise, the discriminant $b^{2}-4 a c$ 2 variant | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,061 |
2. (3 points) On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, $n$ islanders gathered in a room.
The first one said: "Exactly 1 percent of those present in this room are knights."
The second one said: "Exactly 2 percent of those present in this room are kni... | Answer: $10,20,25,50$ or 100.
Solution:
There must be exactly one knight, as all other islanders in the room contradict each other. This means that the knights are $\frac{1}{n}$ of the total number of people. Let them be $k$ percent of the total number of people present, which means the person with number $k$ told th... | 10,20,25,50,100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,062 |
3. (3 points) Anya, Vanya, Danya, Sanya, and Tanya were collecting apples. It turned out that each of them collected a whole percentage of the total number of collected apples, and all these numbers were different and greater than zero. Then Tanya, who collected the most apples, ate her apples. After that, it turned ou... | Answer: 20, for example $1+2+3+4+10$
## Solution:
Let Anna have collected the fewest apples, and this number is $k$, and the total number of apples is $n$. Then we get that the total number of apples collected is no less than $4 k+10$. For $k \geq 3$, we already get more than 20 apples. For $k=2$, we get $n \geq 18=9... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,063 |
4. (3 points) A number written on the board is allowed to be multiplied by 5 or to have its digits rearranged (it is not allowed to place zero in the first position). Is it possible to obtain a 150-digit number $5222 \ldots 2223$ from the number 1 in this way?
# | # Answer: No
## Solution:
Consider the reverse process: we will try to obtain the number $5222 \ldots 2223$ from 1 using the specified operations.
First, we need to get a number divisible by 5, specifically $222 \ldots 22322 \ldots 5 = 222 \ldots 22225 + 10^k$ (where $k$ is a natural number). Dividing it by 5, we ge... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,064 |
5. (3 points) Given a rectangle $A B C D, A B=5, B C=6$. Point $K$ lies on side $C D$, point $L$ lies on side $A B$, point $M$ lies on side $B C$, point $N$ lies on side $C D$. Prove that the length of the closed broken line $A K L M N A$ is not greater than 26. | Solution:
Let's add several more rectangles equal to the given one, as shown in the figure. Take point $L^{\prime}$ such that $A L=A L^{\prime}$, point $M$ such that $B^{\prime} M^{\prime}=... | 26 | Geometry | proof | Yes | Yes | olympiads | false | 21,065 |
6. (3 points) Anya counted all nine-digit numbers, all digits in each of which are distinct, divisible by 9. Kolya counted all ten-digit numbers, all digits in each of which are distinct, divisible by 5. Who of them counted more numbers? | Answer: both counted the same.
## Solution:
Let's establish a one-to-one correspondence between these sets of numbers. Consider some of Anya's numbers. Since it is divisible by 3, the sum of its digits is also divisible by 9. To satisfy this condition, the number must contain all digits except $0, 3, 6$ or 9.
Append... | 33\cdot8! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,066 |
7. (3 points) Given a convex pentagon $A B C D E$. Point $P$ is the intersection of $B E$ and $A C$, point $Q$ is the intersection of $C E$ and $A D$, point $O$ is the intersection of $A D$ and $B E$. It turns out that triangles $A B P$ and $D E Q$ are isosceles triangles with the angle at the vertex (exactly at the ve... | Answer: $120^{\circ}$ or $75^{\circ}$
Solution: Consider triangle $A B P$. The angle $\angle A P B$ in it can be $40^{\circ}$ or $70^{\circ}$. Then the adjacent $\angle A P O$ is $140^{\circ}$ or $110^{\circ}$. Therefore, in triangle $A P O$, this is the angle at the vertex and $\angle A O P$ is $20^{\circ}$ or $35^{\... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,067 |
8. (4 points) On the board, all three-digit natural numbers are written, the first digits of which are odd and greater than 1. What is the maximum number of quadratic equations of the form $a x^{2}+b x+c=0$ that can be formed using these numbers as $a, b$, and $c$, each no more than once, such that all these equations ... | Answer: 100.
Solution:
In each equation, there must be a coefficient less than five hundred. Otherwise, the discriminant $b^{2}-4 a c<1000^{2}-4 \cdot 500^{2}=0$ and the equation has no roots. Thus, we get no more than 100 equations.
Now for an example: Take $b$ as numbers starting with 9, and for each equation, tak... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,068 |
# 1. (1 point)
Kolya had a polynomial of the hundredth degree written in his notebook. Kolya can take one of the polynomials written in the notebook, add $a$ to the coefficient of the $k$-th degree and subtract $2a$ from the coefficient of the $(k+1)$-th degree, after which he can write the resulting polynomial in the... | # Answer: No.
Solution: Note that the value of the polynomial at the point $\frac{1}{2}$ does not change with these operations, so all polynomials that Kolya ends up with in his notebook will coincide at this point, and therefore none of them will be strictly greater than the other. | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,069 |
2. (2 points) Solve the equation $f^{-1}(g(x))=h(x)$, where $f^{-1}(x)$ is the inverse function of $f(x)$, given that $f(x)=x^{3}+2 x^{2}+3 x+1, g(x)=x^{4}-x^{3}+4 x^{2}+8 x+8, h(x)=x+1$.
Answer: $\frac{3 \pm \sqrt{5}}{2}$ | Solution: The original equation can be rewritten as $g(x)=f(h(x))$, that is, as $(x+1)^{3}+2(x+1)^{2}+3(x+$ 1) $+1=x^{4}-x^{3}+4 x^{2}+8 x+8$.
Expand the brackets and move everything to the right side: $x^{4}-2 x^{3}-2 x^{2}-2 x+1=0$. Notice that the polynomial on the left side is divisible by $x^{2}+x+1$. This trinom... | \frac{3\\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,070 |
3. (3 points) Quadrilateral $A B C D$ is inscribed in circle $S$. Circles $S_{1}$ and $S_{2}$ of equal radius are internally tangent to circle $S$ at points $A$ and $C$ respectively. Circle $S_{1}$ intersects sides $A B$ and $A D$ at points $K$ and $N$ respectively, and circle $S_{2}$ intersects sides $B C$ and $C D$ a... | # Solution:
Notice that triangle $A K N$ is similar to triangle $A B D$ with a similarity coefficient of $\frac{r}{R}$, where $R$ is the radius of circle $S$, and $r$ is the radius of circles $S_{1}$ and $S_{2}$. Similarly, triangle $C M L$ is similar to triangle $C B D$ with the same coefficient.
From this, segments... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,071 |
4. (3 points) Prove that the sum of an infinite arithmetic progression and an infinite non-constant geometric progression will never be an arithmetic progression. | Solution:
The statement of the problem is equivalent to the fact that the difference of two arithmetic progressions cannot be a geometric progression. But the difference of two arithmetic progressions is an arithmetic progression, and an arithmetic progression with more than two terms cannot coincide with a geometric ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,072 |
6. (3 points) In space, there is a cube $1000 \times 1000 \times 1000$ with a vertex at the origin and faces parallel to the coordinate planes. Vectors are drawn from the origin to all integer points inside and on the boundary of this cube. Find the remainder when the sum of the squares of the lengths of these vectors ... | # Answer: 0
Solution:
The sum of the squares of the lengths of these vectors is the sum of the squares of all their coordinates, that is, $3 \cdot 1001^{2} \cdot\left(0^{2}+1^{2}+\right.$ $2^{2}+\ldots+1000^{2}$ ), which is divisible by 13, since 1001 is divisible by 13. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,074 |
7. (4 points) Given a triangle $ABC$, point $I$ is the center of its inscribed circle. On the rays $BI$ and $CI$ respectively, points $E$ and $F$ (different from $I$) are marked such that $AI = AE = AF$. Prove that the areas of triangles $BIF$ and $CIE$ are equal. | Solution:
Consider point $D$ - the midpoint of arc $BC$ of the circumcircle of triangle $ABC$. By the trident lemma, $DB = DC = DI$. Triangles $DBI$ and $AEI$ are similar, as they are isosceles triangles with equal base angles (angles at point $I$ are equal as vertical angles, because points $A, I$, and $D$ lie on the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,075 |
8. (5 points) In a certain country, there are 100 cities and 146 airlines. Any two cities are connected by bidirectional flights of one or several airlines. The cost of a flight between cities connected by flights of $k$ airlines is the same for all companies and is $\frac{1}{k}$. It turned out that there are no cities... | Solution:
In fact, we need to prove that there exist two flights of the same price from one city. Suppose the opposite. Then all flights from any city have different prices.
Consider the most expensive flight. It is the most expensive in each of its cities, so there are at least 98 cheaper flights. Then the price of ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,076 |
1. (2 points)
Kolya had a polynomial of two hundredth degree written in his notebook. Kolya can take one of the polynomials written in the notebook, add $2a$ to the coefficient of the $k$-th degree and subtract $a$ from the coefficient of the $(k+1)$-th degree, after which he can write the resulting polynomial in the ... | Answer: No.
Solution: Note that with these operations, the value of the polynomial at point 2 does not change, which means that all polynomials that Kolya ended up with in his notebook will coincide at this point, and therefore none of them will be strictly greater than the other. | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,077 |
2. (2 points) Solve the equation $f^{-1}(g(x))=h(x)$, where $f^{-1}(x)$ is the inverse function of $f(x)$, if it is known that $f(x)=x^{3}+2 x^{2}+3 x+4, g(x)=x^{4}+2 x^{3}+1 x^{2}+11 x+11, h(x)=x+1$. | Answer: $\frac{-3 \pm \sqrt{5}}{2}$ and 1.
Solution: The original equation can be rewritten as $f(h(x))=g(x)$, that is, $(x+1)^{3}+2(x+1)^{2}+3(x+1)+4=x^{4}+2 x^{3}+x^{2}+11 x+11$.
Expanding the brackets and moving everything to the right side, we get: $x^{4}+x^{3}-4 x^{2}+x+1=0$. Notice that the polynomial on the le... | \frac{-3\\sqrt{5}}{2}1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,078 |
3. (3 points) Quadrilateral $A B C D$ is inscribed in circle $S$. Circles $S_{1}$ and $S_{2}$ touch circle $S$ internally at points $A$ and $C$ respectively. Circle $S_{1}$ intersects sides $A B$ and $A D$ at points $K$ and $N$ respectively, and circle $S_{2}$ intersects sides $B C$ and $C D$ at points $L$ and $M$ resp... | Solution:
Notice that triangle $A K N$ is similar to triangle $A B D$ with a similarity coefficient of $\frac{r_{1}}{R}$, where $R$ is the radius of circle $S$, and $r_{1}$ is the radius of circle $S_{1}$. Similarly, triangle $C M L$ is similar to triangle $C B D$ with a similarity coefficient of $\frac{r_{2}}{R}$, wh... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,079 |
4. (3 points) Find all pairs of geometric progressions such that their sum is also a geometric progression. | Answer: the denominators of these progressions must be the same and the sum of the first terms is not equal to 0.
Moreover, any progressions of two elements with non-zero sums of the first and second terms are suitable.
## Solution:
 In space, there is a cube $1000 \times 1000 \times 1000$ with a vertex at the origin and faces parallel to the coordinate planes. Vectors are drawn from the origin to all integer points inside and on the boundary of this cube. Find the remainder when the sum of the squares of the lengths of these vectors ... | # Answer: 0
## Solution:
The sum of the squares of the lengths of these vectors is the sum of the squares of all their coordinates, that is, $3 \cdot 1001^{2} \cdot\left(0^{2}+1^{2}+\right.$ $\left.2^{2}+\ldots+1000^{2}\right)$, which is divisible by 11, since 1001 is divisible by 11. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,081 |
7. (4 points) Given a triangle $ABC$, point $I$ is the center of its inscribed circle. On the rays $BI$ and $CI$ respectively, points $E$ and $F$ (different from $I$) are marked such that $AI = AE = AF$. Prove that $EF \parallel BC$ | # Solution:
Consider point $D$ - the midpoint of arc $BC$ of the circumcircle of triangle $ABC$. By the trident lemma, $DB = DC = DI$. Triangles $DBI$ and $AEI$ are similar, as they are isosceles triangles with equal base angles (angles at point $I$ are equal as vertical angles, because points $A, I$, and $D$ lie on t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,082 |
8. (5 points) In a certain country, there are 50 cities and 71 airlines. Any two cities are connected by two-way flights of one or several airlines. The cost of a flight between cities connected by flights of $k$ airlines is the same for all companies and is $\frac{1}{k}$. It turned out that there are no cities between... | # Solution:
In fact, we need to prove that there exist two flights of the same price from one city. Suppose the opposite. Then all flights from any city have different prices.
Consider the most expensive flight. It is the most expensive in each of its cities, so there are at least 48 cheaper flights. Then this most e... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,083 |
2. The train was supposed to travel 300 km. Having traveled 212 km at a certain speed, it then reduced its speed by 50 km/h. As a result, the train arrived 40 minutes later than the scheduled time. Find the initial speed of the train. Answer in km/h. | Answer: 110
3rd option. The train was supposed to travel 300 km. After traveling 50 km at a certain speed, it then reduced its speed by 40 km/h. As a result, the train arrived 1 hour and 40 minutes later than the scheduled time. Find the initial speed of the train. Record your answer in km/h.
Answer: 100
# | 110 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,084 |
# Problem 2. (2 points)
1st variant: In a $10 \times 10$ table, 18 cells are colored, and at each vertex of a cell not lying on the boundary of the table, the number of colored cells of which it is a vertex is written. What is the minimum sum that could have been obtained? | Answer: 32
2nd variant: In a $9 \times 9$ table, 25 cells are colored, and at each vertex of a cell not lying on the boundary of the table, the number of colored cells that the vertex belongs to is written. What is the minimum sum that could be obtained?
Answer: 46
3rd variant: In an 11 x 11 table, 32 cells are colo... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,085 |
3. Four different numbers $a, b, c, d$, greater than one and not divisible by 5, are such that $\gcd(a, b) = \gcd(c, d)$ and $\operatorname{lcm}(a, b) = \operatorname{lcm}(c, d)$. What is the smallest possible value of $a + b + c + d$? | Answer: 24
## Examples of how to write the answer:
17
# | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,088 |
3. At a round table, raccoons, hedgehogs, and hamsters were sitting, a total of 134 animals. When asked: "Are there any animals of the same kind as you among your neighbors?", everyone answered "No." What is the maximum number of hedgehogs that could have been sitting at the table, given that hamsters and hedgehogs alw... | Answer: 44
## Examples of how to write the answer:
17
# | 44 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,089 |
1. In a computer game, a turtle moves across a grid on the computer screen, which contains 5 columns and 7 rows. Initially, it is located at the bottom-left corner of the screen - on the cell with coordinates $(0,0)$. If the program instructs the turtle to move off the screen, it reappears on the opposite side - for ex... | Answer: $(2 ; 6)\|(2,6)\| 2 ; 6 \| 2,6$ | (2,6) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,090 |
2. In a computer game, a turtle moves across a grid on the computer screen, which contains 11 columns and 5 rows. Initially, it is located at the bottom-left corner of the screen - on the cell with coordinates $(0,0)$. If the program instructs the turtle to move off the screen, it reappears on the opposite side - for e... | Answer: $(4 ; 1)\|(4,1)\| 4 ; 1 \| 4,1$ | (4,1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,091 |
3. 4. In a computer game, a turtle moves across a grid on the computer screen, which contains 5 columns and 9 rows. Initially, it is located in the bottom-left corner of the screen - on the cell with coordinates $(0,0)$. If the program instructs the turtle to move off the screen, it reappears on the opposite side - for... | Answer: $(2 ; 1)\|(2,1)\| 2 ; 1 \| 2,1$
Examples of answer notation:
$(1 ; 7)$
# | (2,1) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,092 |
1. Given 11 natural numbers with a sum of 40. What is the smallest value that the sum of all 55 numbers, which are the reciprocals of their pairwise products, can take? | Write the answer in the form of a proper or improper fraction.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,093 |
# Problem 1. (2 points)
Do there exist two natural numbers $x$ and $y$, having the same number of natural divisors, such that $x>y$, and the sum of all divisors of $x$ (including 1 and the number itself) is less than the sum of natural divisors of $y$ (including 1 and the number itself)? | Answer: Yes
Solution:
For example, the numbers 38 and $39: 1+2+19+38=60>56=1+3+13+39$. | Yes | Number Theory | proof | Yes | Yes | olympiads | false | 21,094 |
# Task 2. (2 points)
Knights and liars played "Chinese whispers" according to the following rules: the first in the chain whispers a number to the second, the second to the third, and so on. The last person in the chain announces the number out loud. At the same time, a knight repeats the same number that was told to ... | Answer: Knight
## Solution:
In the first game, the number does not change parity, which means that the number of liars in the chain from the 2nd to the last is even.
From the result of the second game, we get that the number of liars in the chain from the second-to-last to the first is odd.
This means that the firs... | Knight | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,095 |
# Problem 4. (3 points)
Come up with two different sets of eight natural numbers each, such that the sum of the numbers in each set equals their product.
The sum and product can be different for different sets. Numbers in the sets can repeat. | Answer: $(1,1,1,1,1,1,2,8)$ and ( $1,1,1,1,1,2,2,3$ up to permutation
# | (1,1,1,1,1,1,2,8)(1,1,1,1,1,2,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,096 |
# Problem 5. (3 points)
A natural number had its largest digit subtracted from it. Then, from the resulting number, its largest digit was also subtracted, and so on. After six such operations, the number 100 was obtained. What number could have been written initially? List all the options and prove that there are no o... | Answer: $130,131,132$
## Solution:
For numbers from 101 to 139, we will show which number they transition to. In the third column, it is indicated how many operations it takes to get to 100.
| $10 x, x>0$ | 100 | 1 |
| :---: | :---: | :---: |
| 110 | 109 | 2 |
| $11 x, x>0$ | 110 | 3 |
| 120 | 118 | 4 |
| 121 | 119 ... | 130,131,132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,097 |
# Problem 8. (5 points)
A figure called a "half-bishop" moves one square diagonally. What is the maximum number of non-attacking half-bishops that can be placed on a $7 \times 7$ chessboard?
Answer: 28 | Solution:
Consider all diagonals of one direction. On each diagonal of even length, no more than half of the cells can be occupied by semi-bishops, as it can be divided into pairs of cells that cannot be simultaneously occupied.
On each diagonal of odd length $2n+1$, no more than $n+1$ semi-bishops can be placed for ... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,098 |
# Problem 1. (2 points)
Do there exist two natural numbers $x$ and $y$, having the same number of natural divisors, such that $x<y$, and the sum of all divisors of $x$ (including 1 and the number itself) is greater than the sum of natural divisors of $y$ (including 1 and the number itself)? | Answer: Yes
Solution:
For example, the numbers 38 and $39: 1+2+19+38=60>56=1+3+13+39$. | Yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,099 |
# Task 2. (2 points)
Knights and liars played "Chinese whispers" according to the following rules: the first in the chain whispers a number to the second, the second to the third, and so on. The last person in the chain announces the number aloud. At the same time, a knight repeats the same number that was told to the... | Answer: Knight
Solution:
In the first game, the number does not change parity, which means the number of liars in the chain from the second to the last is odd.
From the result of the second game, we get that the number of liars in the chain from the second-to-last to the first is also odd.
This means that the first... | Knight | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,100 |
# Problem 4. (3 points)
Come up with two different sets of five natural numbers each, such that the sum of the numbers in each set equals their product.
The sum and product can be different for different sets. Numbers in the sets can repeat. | Answer: $(1,1,1,2,5)$ and $(1,1,2,2,2)$ up to permutation
# | (1,1,1,2,5)(1,1,2,2,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,101 |
# Problem 5. (3 points)
From a natural number, its largest digit was subtracted. Then, from the resulting number, its largest digit was also subtracted, and so on. After five such operations, the number 200 was obtained. What number could have been written initially? List all the options and prove that there are no ot... | Answer: All numbers from 222 to 229
Solution:
For numbers from 201 to 139, we will show which number they transition to. In the third column, it is indicated how many operations it takes to get to 100.
| 201 | 199 | - |
| :---: | :---: | :---: |
| $20 x, x>1$ | 200 | 1 |
| 210 | 108 | 2 |
| 211 | 109 | 2 |
| $21 x, ... | 222to229 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,102 |
# Problem 8. (5 points)
A figure called a "half-bishop" moves one square diagonally. What is the maximum number of non-attacking half-bishops that can be placed on a $9 \times 9$ chessboard? | Answer: 45
Solution:
Consider all diagonals of one direction. On each diagonal of even length, no more than half of the cells can be occupied by pseudo-bishops, as it can be divided into pairs of cells that cannot be simultaneously occupied.
On each diagonal of odd length $2n+1$, no more than $n+1$ pseudo-bishops ca... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,103 |
1. (2 points) In trapezoid $A B C D$ with bases $A D=16$ and $B C=10$, the circles constructed on sides $A B, B C$ and $C D$ as diameters intersect at one point. The length of diagonal $A C$ is 10. Find the length of $B D$. | Answer: 24
## Solution:
Let the intersection point of the three circles be $O$. Then, since the circles are constructed on the sides of the trapezoid $AB, BC$, and $CD$ as diameters, the angles $\angle AOB, \angle BOC$, and $\angle COD$ are right angles. Therefore, points $A, O, C$ lie on the same line and points $B,... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,105 |
2. (3 points) In a regular tetrahedron, the bases and midpoints of all four of its medians are marked. Each base of the tetrahedron's median is connected to the midpoints of the other three. Prove that the resulting polyhedron is a rectangular parallelepiped. | Solution: Let the position vectors of the vertices of the tetrahedron be $\vec{a}, \vec{b}, \vec{c}$, and $\vec{d}$. Then the bases of the medians are given by the formula $\frac{\vec{b}+\vec{c}+\vec{d}}{3}$ and similar ones, and the midpoints of the medians by the formula $\frac{\frac{\vec{a}+\vec{b}+\vec{c}}{3}+\vec{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,106 |
3. (3 points) Find the distance between the curves $y=e^{3 x+5}$ and $y=(\ln x-5) / 3$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: $\sqrt{2}\left(2+\frac{\ln 3}{3}\right)$.
Solution: The distance between the graphs of the functions is the distance between their nearest points. Note that the functions, between the graphs of which we need to find the distance, are inverses of each other, and their graphs are symmetric with respect to the li... | \sqrt{2}(2+\frac{\ln3}{3}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 21,107 |
4. (3 points) In a certain country, there are 450 cities and 6 airlines. Any two cities are connected by flights of one of the six airlines. Can we assert that there will be an airline and more than 150 cities, between any two of which it is possible to travel using flights of this airline (possibly with layovers) | Answer: No, this is not necessarily true.
Solution: Let's construct a counterexample. Divide the cities into 6 groups of 75 cities each. Let's call these groups $A, B, C, D, E, F$.
Within each group, connect the cities with flights from company number 1.
Company number 2 will connect cities in group $A$ with cities ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,108 |
5. (3 points) Find all functions continuous on the entire number line that satisfy the identity $4 f(x+y)=f(x) f(y)$ and the condition $f(1)=12$. | Answer: $f(x)=4 \cdot 3^{x}$.
Solution:
First, we prove that $f(x) \geqslant 0$. Indeed, $4 f(x)=f\left(\frac{x}{2}\right)^{2} \geqslant 0$.
Substituting $x=y=0$ into the condition of the problem, we get $4 f(0)=f(0) \cdot f(0)$, from which $f(0)=0$ or $f(0)=4$. If $f(0)=0$, then $4 f(x+0)=f(x) f(0)$, from which $f(... | f(x)=4\cdot3^{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,109 |
6. (3 points) Let $x, y, z$ and $t$ be non-negative numbers such that $x+y+z+t=5$. Prove the inequality
$$
\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+1}+\sqrt{z^{2}+y^{2}}+\sqrt{z^{2}+t^{2}}+\sqrt{t^{2}+9} \geqslant 10
$$ | # Solution:
Consider the following points on the plane: $A(0,0) ; B(x, 1) ; C(x+y, 1+x) ; D(x+y+z, 1+$ $x+y) ; E(x+y+z+t, 1+x+y+z) ; F(x+y+z+t+3,1+x+y+z+t)$. Then the length of the broken line $A B C D E F$ coincides with the expression that needs to be evaluated. By the triangle inequality, the length of the broken l... | 10 | Inequalities | proof | Yes | Yes | olympiads | false | 21,110 |
7. (4 points) Solve the equation:
$$
\sin x + \sin 3x + \ldots + \sin 2017x = \cos x + \cos 3x + \ldots + \cos 2017x
$$ | # Answer:
1) $x=\frac{\pi k}{1009}, k \in \mathbb{Z}, k$ does not divide 1009.
2) $x=\frac{\pi+4 \pi k}{4036}, k \in \mathbb{Z}$
## Solution:
First, let's prove the following formulas by induction in the case where $\sin x \neq 0$:
$$
\begin{gathered}
\sum_{k=1}^{n} \sin ((2 k-1) x)=\frac{\sin ^{2}(n x)}{\sin x} \\... | \frac{\pik}{1009},k\in\mathbb{Z},k | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,111 |
8. (4 points) With the number written on the board, the following operation is allowed: erase two adjacent digits whose sum does not exceed 9, and write this sum in their place. Initially, a 200-digit number $12341234 \ldots 1234$ was written. The specified operation was performed on the number on the board until it wa... | # Answer:
$64 \ldots 64$ - a 100-digit number.
## Solution:
The sum of the digits of the original number is $(1+2+3+4) \cdot 50 = 500$. The sum of the digits of the final number is the same.
We will divide the digits of the final number into pairs: the first with the second, the third with the fourth, and so on. Th... | 64\ldots64 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,112 |
1. (2 points) In trapezoid $A B C D$ with bases $A D=12$ and $B C=8$, the circles constructed on sides $A B, B C$, and $C D$ as diameters intersect at one point. The length of diagonal $A C$ is 12. Find the length of $B D$. | Answer: 16
## Solution:
Let the intersection point of the three circles be $O$. Then, since the circles are constructed on the sides of the trapezoid $AB, BC$, and $CD$ as diameters, the angles $\angle AOB, \angle BOC$, and $\angle COD$ are right angles. Therefore, points $A, O, C$ lie on the same line and points $B,... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,113 |
2. (3 points) In an orthocentric tetrahedron, the bases and midpoints of all four of its medians are marked. Each base of the median of the tetrahedron is connected to the midpoints of the other three. Prove that in the resulting polyhedron, all edges have equal length. | Solution: Let the position vectors of the vertices of the tetrahedron be $\vec{a}, \vec{b}, \vec{c}$, and $\vec{d}$. Then the bases of the medians are given by the formula $\frac{\vec{b}+\vec{c}+\vec{d}}{3}$ and similar ones, and the midpoints of the medians by the formula $\frac{\frac{\vec{a}+\vec{b}+\vec{c}}{3}+\vec{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,114 |
3. (3 points) Find the distance between the curves $y=e^{3 x+7}$ and $y=(\ln x-7) / 3$.
---
The text has been translated while preserving the original formatting and line breaks. | Solution: The distance between the graphs of the functions is equal to the distance between their nearest points. Note that the functions, between the graphs of which we need to find the distance, are inverses of each other, and their graphs are symmetric with respect to the line $y=x$, and, moreover, are on opposite s... | \sqrt{2}(\frac{8+\ln3}{3}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 21,115 |
4. (3 points) In a certain country, there are 200 cities and 8 airlines. Every two cities are connected by flights of one of the eight airlines. Can we assert that there will be an airline and more than 50 cities, between any two of which it is possible to travel using flights of this airline (possibly with layovers)? | Answer: No, this is not necessarily true.
Solution: Let's construct a counterexample. Divide the cities into 8 groups of 25 cities each. Let's call these groups $A, B, C, D, E, F, G$ and $H$.
 Find all functions continuous on the entire number line that satisfy the identity $5 f(x+y)=f(x) f(y)$ and the condition $f(1)=10$.
Answer: $f(x)=5 \cdot 2^{x}$. | Solution:
First, we prove that \( f(x) \geqslant 0 \). Indeed, \( 5 f(x) = f\left(\frac{x}{2}\right)^{2} \geqslant 0 \).
Substituting \( x = y = 0 \) into the given condition, we get \( 5 f(0) = f(0) \cdot f(0) \), from which \( f(0) = 0 \) or \( f(0) = 5 \). If \( f(0) = 0 \), then \( 5 f(x + 0) = f(x) f(0) \), whic... | f(x)=5\cdot2^{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,117 |
6. (3 points) Let $x, y, z$ and $t$ be non-negative numbers such that $x+y+z+t=4$. Prove the inequality
$$
\sqrt{x^{2}+t^{2}}+\sqrt{z^{2}+1}+\sqrt{z^{2}+t^{2}}+\sqrt{y^{2}+x^{2}}+\sqrt{y^{2}+64} \geqslant 13
$$ | # Solution:
Consider the following points on the plane: $A(0,0) ; B(x, t) ; C(x+z, t+1) ; D(x+z+t, t+1+z)$; $E(x+z+t+y, t+1+z+x) ; F(x+z+t+y+8, t+1+z+x+y)$. Then the length of the broken line $A B C D E F$ coincides with the expression that needs to be evaluated. By the triangle inequality, the length of the broken li... | 13 | Inequalities | proof | Yes | Yes | olympiads | false | 21,118 |
7. (4 points) Solve the equation:
$$
\sin x + \sin 3x + \ldots + \sin 2013x = \cos x + \cos 3x + \ldots + \cos 2013x
$$ | # Answer:
1) $x=\frac{\pi k}{1007}, k \in \mathbb{Z}, k$ does not divide 1007.
2) $x=\frac{\pi+4 \pi k}{4028}, k \in \mathbb{Z}$.
## Solution:
First, let's prove the following formulas by induction in the case where $\sin x \neq 0$:
$$
\begin{gathered}
\sum_{k=1}^{n} \sin ((2 k-1) x)=\frac{\sin ^{2}(n x)}{\sin x} \... | \frac{\pik}{1007},k\in\mathbb{Z},k | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,119 |
8. (4 points) With the number written on the board, the following operation is allowed: erase two adjacent digits whose sum does not exceed 9, and write this sum in their place. Initially, a 300-digit number $12251225 \ldots 1225$ was written. The operation was performed on the number on the board until it was no longe... | # Solution:
The sum of the digits of the original number $(1+2+2+5) \cdot 75=750$. The sum of the digits of the final number is the same.
Let's break down the digits of the final number into pairs: the first with the second, the third with the fourth, and so on. The last digit may remain unpaired. In each pair, the s... | 555\ldots55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,120 |
1. (2 points) In trapezoid $A B C D$ with bases $A D=20$ and $B C=14$, the circles constructed on sides $A B, B C$ and $C D$ as diameters intersect at one point. The length of diagonal $A C$ is 16. Find the length of $B D$. | Answer: 30
## Solution:
Let the intersection point of the three circles be $O$. Then, since the circles are constructed on the sides $AB$, $BC$, and $CD$ of the trapezoid as diameters, the angles $\angle AOB$, $\angle BOC$, and $\angle COD$ are right angles. Therefore, points $A$, $O$, and $C$ lie on the same line, a... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,121 |
3. (3 points) Find the distance between the curves $y=e^{5 x+7}$ and $y=(\ln x-7) / 5$.
---
The text has been translated while preserving the original formatting and line breaks. | Solution: The distance between the graphs of the functions is equal to the distance between their nearest points. Note that the functions, between the graphs of which we need to find the distance, are inverses of each other, and their graphs are symmetric relative to the line $y=13x$, and, moreover, are on opposite sid... | \sqrt{2}(\frac{8+\ln5}{5}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 21,122 |
4. (3 points) In a certain country, there are 600 cities and 6 airlines. Any two cities are connected by flights of one of the six airlines. Can we assert that there will be an airline and more than 200 cities, between any two of which it is possible to travel using flights of this airline (possibly with layovers) | Solution: We will construct a counterexample. Divide the cities into 6 groups of 100 cities each. Let's call these groups $A, B, C, D, E, F$.
Within each group, connect the cities with flig... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,123 |
5. (3 points) Find all functions continuous on the entire number line that satisfy the identity $3 f(x+y)=f(x) f(y)$ and the condition $f(1)=12$.
Answer: $f(x)=3 \cdot 4^{x}$. | Solution:
First, we prove that $f(x) \geqslant 0$. Indeed, $3 f(x)=f\left(\frac{x}{2}\right)^{2} \geqslant 0$.
Substituting $x=y=0$ into the condition, we get $3 f(0)=f(0) \cdot f(0)$, from which $f(0)=0$ or $f(0)=3$. If $f(0)=0$, then $3 f(x+0)=f(x) f(0)$, which implies $f(x)=0$ for any $x$, which is incorrect. Ther... | f(x)=3\cdot4^{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,124 |
6. (3 points) Let $x, y, z$ and $t$ be non-negative numbers such that $x+y+z+t=7$. Prove the inequality
$$
\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+1}+\sqrt{z^{2}+y^{2}}+\sqrt{t^{2}+64}+\sqrt{z^{2}+t^{2}} \geqslant 17
$$ | Solution:
Consider the following points on the plane: $A(0,0) ; B(x, y) ; C(x+1, y+x) ; D(x+1+y, y+x+z) ; E(x+1+y+t, y+x+z+8) ; F(x+1+y+t+z, y+x+z+8+t)$. Then the length of the broken line $A B C D E F$ coincides with the expression that needs to be evaluated. By the triangle inequality, the length of the broken line ... | 17 | Inequalities | proof | Yes | Yes | olympiads | false | 21,125 |
7. (4 points) Solve the equation:
$$
\sin x + \sin 3x + \ldots + \sin 2019x = \cos x + \cos 3x + \ldots + \cos 2019x
$$ | # Answer:
1) $x=\frac{\pi k}{1010}, k \in \mathbb{Z}, k$ does not divide 1010.
2) $x=\frac{\pi+4 \pi k}{4040}, k \in \mathbb{Z}$.
## Solution:
First, let's prove the following formulas by induction in the case where $\sin x \neq 0$:
$$
\begin{gathered}
\sum_{k=1}^{n} \sin ((2 k-1) x)=\frac{\sin ^{2}(n x)}{\sin x} \... | \frac{\pik}{1010},k\in\mathbb{Z},k | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,126 |
8. (4 points) With the number written on the board, the following operation is allowed: erase two adjacent digits whose sum does not exceed 9, and write this sum in their place. Initially, a 200-digit number $3112331123 \ldots 31123$ was written. The operation was performed on the number on the board until it was no lo... | Answer:
Solution:
The sum of the digits of the original number is $(3+1+1+2+3) \cdot 40=400$. The sum of the digits of the final number is the same.
We will divide the digits of the final ... | 737373\ldots73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,127 |
1. (2 points) In trapezoid $A B C D$ with bases $A D=20$ and $B C=10$, circles constructed on sides $A B, B C$, and $C D$ as diameters intersect at one point. The length of diagonal $A C$ is 18. Find the length of $B D$. | Answer: 24
## Solution:
Let the intersection point of the three circles be $O$. Then, since the circles are constructed on the sides of the trapezoid $AB, BC$, and $CD$ as diameters, the angles $\angle AOB, \angle BOC$, and $\angle COD$ are right angles. Therefore, points $A, O, C$ lie on the same line and points $B,... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,128 |
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