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1. A bridge over a river connects two different regions of a country. Once, one of the regions repainted the part of the bridge that belonged to it. If the freshly painted part of the bridge had been $30 \%$ larger, then the unpainted part would have been $50 \%$ smaller. Could the painted part of the bridge be exactly... | # Solution
Let $x$ and $y$ be the fractions of the painted and unpainted parts, respectively. Clearly, $x + y = 1$. According to the condition, $1.3x + 0.5y = 1$. We obtain the equation
$$
x + y = 1.3x + 0.5y
$$
from which we can find the ratio
$$
\frac{x}{y} = \frac{5}{3}
$$
Now we can find the fraction of the pa... | 62.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,967 |
2. The truncated difference of numbers $x$ and $y$ is the operation $x-y$, the result of which is equal to the usual difference $x-y$ if $x \geqslant y$, and zero if $x<y$.
Solve the system of equations $\left\{\begin{array}{l}2 x-y=0 \\ x+2 y=1\end{array}\right.$. | Solution. The first equation of the system is equivalent to the inequality $2 x \leqslant y$. Express $y$ from the second equation ( $y=(1-x) / 2$ ) and substitute it into the inequality.
$$
\frac{1-x}{2} \geqslant 2 x \Longleftrightarrow 1 \geqslant 5 x
$$
Thus, the system has an infinite number of solutions lying o... | {\begin{pmatrix}x\leqslant1/5,\\(1-x)/2\end{pmatrix}. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,968 |
3. A row contains 2015 positive numbers. The product of all the numbers is 2015, and the product of any three consecutive numbers is 1. What is the 1008th number in the sequence?
# | # Solution
From the additional condition, it follows that if the sequence is divided into blocks of 3 numbers "end-to-end," moving simultaneously from left and right towards each other, the 336th block from the left and the 336th block from the right will overlap on the middle number. Let's denote this number as $C$.
... | \frac{1}{2015} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,969 |
4. Does there exist a convex polygon with 2015 diagonals?
# | # Solution.
The number of diagonals in a convex $n$-gon is
$$
\frac{n(n-1)}{2}-n=\frac{n^{2}-3 n}{2}
$$
Solve the equation
$$
\frac{n^{2}-3 n}{2}=2015 \quad \Longleftrightarrow \quad n^{2}-3 n-4030=0
$$
Its discriminant $D=9+4 \cdot 4030=16129$. Note that $120^{2}<16129<130^{2}$ and $D$ ends with the digit 9. If $... | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,970 |
# Problem 1.
A sequence of 100 non-zero numbers is written in a row. Each number, except the first and the last, is equal to the product of the two adjacent numbers. The first number is 2018. Find the last number in such a sequence.
# | # Solution.
Let's denote the hundred numbers as $a_{1}, \ldots, a_{100}$. According to the condition,
$$
a_{1}=\frac{a_{2}}{a_{3}}, \quad a_{2}=\frac{a_{3}}{a_{4}}, \ldots, \quad a_{98}=\frac{a_{99}}{a_{100}}
$$
Multiplying the first and second equations, we get
$$
a_{1} a_{2}=\frac{a_{2} a_{3}}{a_{3} a_{4}} \Right... | \frac{1}{2018} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,972 |
# Problem 2.
In a football tournament, each team is supposed to play one match against each of the others. But during the tournament, half of all the teams were disqualified and withdrew from further participation. As a result, 77 matches were played, and the teams that withdrew had managed to play all their matches a... | # Solution.
According to the condition, an even number of teams $2n$ started, of which $n$ were disqualified. The eliminated teams played $\frac{n(n-1)}{2}$ matches among themselves. The same number of matches were played by the teams that remained in the tournament.
Let each of the eliminated teams have played with ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,973 |
# Problem 3.
The number $b$ is the arithmetic mean of the numbers $a$ and $c$. Find all ordered triples $(a, b, c)$ of such numbers for which at least one of the numbers $1 / a, 1 / b, 1 / c$ is the arithmetic mean of the other two. | # Solution.
Obviously, $a, b, c$ should not be equal to 0. As indicated,
$$
2 b=a+c
$$
and one of the following conditions must be satisfied:
$$
\frac{2}{b}=\frac{1}{a}+\frac{1}{c} \quad \text { or } \quad \frac{2}{a}=\frac{1}{b}+\frac{1}{c} \quad \text { or } \quad \frac{2}{c}=\frac{1}{a}+\frac{1}{b} .
$$
Solving... | (x,x,x)wherex\neq0,or(-4x,-x,2x)wherex\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,974 |
# Problem 4.
At 9:00 AM, ships "Anin" and "Vinin" departed from port O to port E. At the same moment, ship "Sanin" departed from port E to port O. All three vessels are traveling on the same course ("Sanin" heading towards "Anin" and "Vinin") at constant but different speeds. At 10:00 AM, "Vinin" was at the same dista... | # Solution.
For simplicity, let's assume the distance between the ports is 1. Let $x$ be the speed of "Anina", $y$ be the speed of "Vanina", and $z$ be the speed of "Sanina". Then at 10 o'clock (i.e., one hour after departure),
$$
2 y = x + 1 - z,
$$
and one and a half hours after departure,
$$
2(1 - 1.5 z) = 1.5 x... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,975 |
# Problem 5.
Numbers $a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{n}, b_{n}$ satisfy the conditions
$$
\begin{gathered}
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=2018^{2}, \quad b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=2017^{2} \\
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}=2017 \cdot 2018
\end{gathered}
$$
Find the ratios $a_{1} ... | # Solution.
Let's perform the transformations.
$$
\begin{gathered}
\left(2017 a_{1}-2018 b_{1}\right)^{2}+\left(2017 a_{2}-2018 b_{2}\right)^{2}+\ldots+\left(2017 a_{n}-2018 b_{n}\right)^{2}= \\
=2017^{2}\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}\right)+2018^{2}\left(b_{1}^{2}+b_{2}^{2}+\ldots+b_{n}^{2}\right)- \\
-2... | \frac{2018}{2017} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,976 |
# Task 2.
Sasha, Pasha, and Arkasha are virtual businessmen. Sasha transferred to Pasha exactly the number of bitcoins Pasha had. After that, Sasha transferred to Arkasha exactly the number of bitcoins Arkasha had. Then, Pasha transferred to both Sasha and Arkasha the exact number of bitcoins each of them had before t... | # Solution
Since Sasha and Pasha ended up with 8 bitcoins each, before the last transfer, Sasha and Pasha each had 4 bitcoins. Therefore, before the second exchange, Sasha had 2 bitcoins, and Arkasha had 8. This means Pasha had 14 bitcoins (4+2+8=14). Consequently, Pasha originally had 7 bitcoins, Arkasha had 4 bitcoi... | Pasha:7,Arkasha:4,Sasha:13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,983 |
# Problem 4.
All numbers $1,2,3, \ldots, n^{2}$ are arranged in the form of an $n \times n$ square table. Find all $n$ for which the sum of the numbers in each subsequent row of such a table is 1 more than the sum of the numbers in the previous row. | # Solution.
First, we prove that the number $n$ cannot be even. We reason by contradiction. Suppose $n=2k (k \in \mathbb{N})$ and $S$ is the sum of the numbers in the first row. We calculate the sum of all numbers in the table in two ways. The first way is by adding all numbers from 1 to $4k^2$ inclusive, and the seco... | all\odd\natural\\n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,985 |
# Problem 5.
Given numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{2018}$. It is known that $x_{1}=1 / 2$ and
$$
x_{n}=\frac{x_{n-1}}{2 n x_{n-1}+1} \text { for } n=2, \ldots, 2018
$$
Find the sum $x_{1}+x_{2}+\cdots+x_{2018}$.
# | # Solution.
Let's introduce the sequence $y_{n}=1 / x_{n}$. For this sequence, the recurrence relation can be rewritten as
$$
y_{n}=y_{n-1}+2 n
$$
Thus,
$$
y_{n}=2+4+\ldots+2 n=n(n+1)
$$
from which we get
$$
x_{n}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}
$$
Summing such terms for all $\mathrm{n}$ from 1 to 201... | \frac{2018}{2019} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,986 |
1. It is known that $x+\frac{1}{x} \leqslant 4$. Find the range of the function
$$
f(x)=x^{3}+\frac{1}{x^{3}}
$$ | Solution. Let $g(x)=x+\frac{1}{x}$. If $x>0$, then $g(x) \geqslant 2$. Together with the condition of the problem, we get that $2 \leqslant g(x) \leqslant 4$ for $x>0$. Now
$$
f(x)=\left(x+\frac{1}{x}\right)\left(x^{2}-1+\frac{1}{x^{2}}\right)=g(x)\left(\left(x+\frac{1}{x}\right)^{2}-2-1\right)=g(x)\left(g^{2}(x)-3\ri... | f(x)\in[-\infty;-2]\cup[2;52] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,987 |
2. The truncated difference of numbers $x$ and $y$ is the operation $x-y$, the result of which is equal to the usual difference $x-y$ if $x \geqslant y$, and zero if $x<y$. Solve the system of equations $\left\{\begin{array}{l}5 x \cdot(y+6)=0 \\ 2 x+3 y=1\end{array}\right.$. | # Solution.
The first equation of the system is equivalent to the inequality $5 x \leqslant y+6$. Express $y$ from the second equation $(y=(1-2 x) / 3)$ and substitute it into the inequality.
$$
5 x \leqslant \frac{1-2 x}{3}+6 \Longleftrightarrow 17 x \leqslant 19
$$
Thus, the system has an infinite number of soluti... | {\begin{pmatrix}x\leqslant19/17,\\(1-2x)/3\end{pmatrix}. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,988 |
3. In a square table with 2015 rows and columns, positive numbers are arranged. The product of the numbers in each row and each column is 1, and the product of the numbers in any 1008 x 1008 square is 2. What number is in the center of the table?
# | # Solution
Consider the first 1008 rows of the table. From the additional condition, it follows that if these rows are covered by two squares of size $1008 \times 1008$, then these squares will overlap by one column. Denote the product of the numbers in this column (1008 numbers) by $M$.
Then the product of all the n... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,989 |
4. A little rascal ran a race with a mechanical horse on a carousel. After a seconds, she discovered that the horse, having made a full circle, had caught up with her. Instantly turning around, the little rascal ran towards the horse at the same speed and met it after \(\frac{a}{2}\) seconds. Determine how long it take... | # Solution.
Let $v$ be the angular velocity of the hedgehog, $U$ be the angular velocity of the carousel, and $x$ be the period of the carousel. Note that $\frac{a}{2}<x<a$. The distance the hedgehog ran in $a$ seconds is the same distance the carousel traveled in $a-x$ seconds. Therefore, we have
$$
v a=U(a-x)
$$
A... | \frac{2}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,990 |
5. Solve the equation $[\sin x]^{2}=\cos ^{2} x-1$, where $[a]$ denotes the integer part of the number $a$.
---
The original text has been translated into English while preserving the original formatting and line breaks. | Solution. The integer $[\sin x]$ can only take the values $0, 1, -1$.
1) If $[\sin x]=0$, then $\cos ^{2} x=0+1=1$, i.e., $\cos x= \pm 1$. Solving this equation, we find $x=\pi n$. At these points, $\sin ^{2} x=0$. The equality is true.
2) If $[\sin x]= \pm 1$, then $\cos ^{2} x=( \pm 1)^{2}+1=2$, which is impossible.... | \pin,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,991 |
# Problem 2.
Can the number $n^{2}+n+2$ be divisible by 2019 for any natural $n$? Either find the smallest such $n$, or prove that it is impossible.
# | # Solution.
Let's factorize the divisor: $2019=3 \cdot 673$.
The number $n^{2}$ gives remainders of 0 and 1 when divided by 3 (which is easy to verify, for example, by constructing a table of remainders)
| n | $n \bmod 3$ | $n^{2}$ | $\bmod 3$ |
| :---: | :---: | :---: | :---: |
| 0 | 0 | | 0 |
| 1 | 1 | | 1 |
| 乙... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,992 |
# Problem 3.
Two swimmers are training in a rectangular quarry. The first swimmer finds it more convenient to exit at a corner of the quarry, so he swims along the diagonal to the opposite corner and back. The second swimmer finds it more convenient to start from a point that divides one of the shores of the quarry in... | # Solution.
Let's draw a rectangular quarry $A D C D$ and the inscribed quadrilateral route of the second swimmer $N L M K$.
Reflect the drawing symmetrically first with respect to side $C D$, then with respect to side $C B^{\prime}$, and finally with respect to side $A^{\prime} B^{\prime}$.
 and two free pocket... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,995 |
3. Through a point lying inside a triangle, three lines parallel to its sides are drawn, which divide the triangle into six parts: three triangles and three quadrilaterals. The areas of all three inner triangles are equal. Determine the range in which the ratio of the area of each inner triangle to the area of the orig... | # Solution
Draw $K F\|A C, L Q\| B C, P E \| A B$.
Triangles $K L O, O E F, P O Q$ are similar to each other and to triangle $A B C$ (their corresponding angles are equal as angles formed by... | \frac{1}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,996 |
4. Four managers of responsibility shifting reported: "If they are arranged in pairs, there will be 1 left. If they are arranged in threes, there will also be 1 left. If they are arranged in fours, there will be 2 left, and if they are arranged in fives, there will also be 2 left." Should the head of the report recepti... | # Solution
Let $a$ be the desired quantity. If when dividing by 4, 2 remains, then $a$ is even. But then when dividing by two, 1 cannot remain. Therefore, these two statements are contradictory and the message as a whole is false.
Remove one of the contradictory statements.
1st var.: leave the division by 3, by 4, a... | 1042 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,997 |
5. Represent the number $\frac{3}{7}$ as the sum of several different common fractions, the numerators of which are equal to one.
# | # Solution
We use the fact that
$$
\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}
$$
Then
$$
\begin{aligned}
& \frac{3}{7}=\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{1}{7}+\frac{1}{8}+\frac{1}{56}+\frac{1}{8}+\frac{1}{56}= \\
& =\frac{1}{7}+\frac{1}{8}+\frac{1}{56}+\frac{1}{9}+\frac{1}{72}+\frac{1}{57}+\frac{1}{56 \cdot... | \frac{1}{7}+\frac{1}{8}+\frac{1}{56}+\frac{1}{9}+\frac{1}{72}+\frac{1}{57}+\frac{1}{56\cdot57} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,998 |
# Task 3.
Vintik and Shpuntik were designing a nanorover. Vintik drew a rectangle $9 \times 11$ cm and marked two hundred points in it - places for rivets. Shpuntik divided the rectangle into square compartments with a side of 1 cm. In doing so, none of Vintik's marks fell on Shpuntik's lines. Prove that there must be... | # Solution.
It is clear that Sputnik divided the original rectangle into $9 \times 11=99$ squares.
If each square has fewer than three holes, then the total number of holes will be less than 198, which contradicts the condition.
Olympiad for School Students "Hope of Energy". Final Stage. Solutions
# | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,999 |
Problem 8.1. (15 points) In a $5 \times 5$ grid, each cell is painted one of three colors: red, blue, or green. To the right of each row, the total number of blue and red cells in that row is written, and below each column, the total number of blue and green cells in that column is written.
To the right of the table, ... | Answer: Yes.
Solution. For example, the following coloring will work:
| c | K | k | k | K |
| :---: | :---: | :---: | :---: | :---: |
| 3 | C | K | K | K |
| 3 | 3 | c | K | K |
| 3 | 3 | 3 | c | k |
| 3 | 3 | 3 | 3 | c |
| | 4 | 3 | 2 | 1 |
## Criteria
The most suitable criterion is used:
15 p. A correct example... | Yes | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,000 |
Problem 8.2. (15 points) Real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ are such that
$$
\left\{\begin{array}{l}
x_{1}+x_{2} \geqslant 12 \\
x_{1}+x_{3} \geqslant 13 \\
x_{1}+x_{4} \geqslant 14 \\
x_{3}+x_{4} \geqslant 22 \\
x_{2}+x_{3} \geqslant 23 \\
x_{2}+x_{4} \geq 24
\end{array}\right.
$$
What is the smallest value t... | Answer: 37.
Solution. By adding the second equality to the last one, we get $x_{1}+x_{2}+x_{3}+x_{4} \geqslant 37$.
It is also worth noting that the value of the expression $x_{1}+x_{2}+x_{3}+x_{4}$ can be equal to 37, for example, when $x_{1}=1, x_{2}=11, x_{3}=12, x_{4}=13$. It is easy to verify that such numbers s... | 37 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 20,001 |
Problem 8.3. (15 points) In one move, you can choose a natural number $x$ and cross out all natural numbers $y$ such that $|x-y|$ is a natural composite number. At the same time, $x$ can be chosen from already crossed out numbers.
What is the minimum number of moves needed to cross out all numbers from the natural num... | Answer: 2.
Solution. First, note that one move is not enough, because if we choose some natural number $x$, then the number $y=x$ will remain unchecked. Let's prove that two moves are enough.
We will look for two suitable numbers $x_{1}$ and $x_{2}$ of different parity. Then one of the differences $\left|y-x_{1}\righ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,002 |
Problem 8.4. (15 points) In a class, there are an equal number of boys and girls. We will call a non-empty group of boys popular if every girl in the class is friends with at least one boy from this group (all friendships are mutual). It turned out that there are exactly 63 popular groups in the class. Prove that each ... | Solution. Suppose there exists a boy $x$ who does not befriend any girl. Let $M$ denote the set of all other boys in the class. Note that a group of boys $G \subset M$ is popular if and only if the group of boys $G \cup\{x\}$ is popular. Thus, all popular groups can be paired, where each pair differs only by the presen... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 20,003 |
Problem 8.5. (20 points) Triangle $ABC$ is such that $BC < AC < AB$. Point $M$ is the midpoint of side $AC$. On side $AB$, there is a point $K$ such that $CK = BC$ and $BK = AC$. Prove that $\angle BAC = 2 \angle ABM$.
. In the quadrilateral $AKCD$, the diagonals intersect at their midpoints, so it is a parallelogram. We obtain $AD = CK$ and $\angle BAD = \angle BKC$; and using the isosceles property of triangle $BCK$, we find $AD = BC$ and $\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,004 |
Problem 8.6. (20 points) On the table, there are 55 piles of candies. In one pile, there is 1 candy, in another - two, in the third - 3, ..., in the last - 55. Petya and Vasya are playing the following game, taking turns; Petya starts. In one move, a player takes one candy from any pile. If a player takes the last cand... | Answer: 1.
Solution. It is clear that Petya can eat 1 candy, for example, if he takes a candy from the pile with 1 candy on his very first move.
We will prove that Vasya can prevent Petya from eating more than 1 candy. For this, Vasya will act as follows. If there is exactly 1 candy left in any pile, he will take it ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,005 |
# 7. Problem 7.1*
Misha thought of a five-digit number, all digits of which are different, and Igor is trying to guess it. In one move, Igor can choose several digits of the number, and Misha, in any order, tells the digits that stand in these places. The order in which to tell the digits is chosen by Misha. For examp... | # 7. Problem $7.1^{*}$
Misha thought of a five-digit number, all digits of which are different, and Igor is trying to guess it. In one move, Igor can choose several digits of the number, and Misha reports the digits in these positions in any order. The order in which to report the digits is chosen by Misha. For exampl... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,012 |
# 9. Problem 9.1*
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fin... | # 9. Problem 9.1*
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fin... | 806 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,014 |
Problem 10.1. (15 points) Do there exist polynomials $P(x), Q(x)$, and $R(x)$ with real coefficients such that the polynomials $P(x) \cdot Q(x), Q(x) \cdot R(x)$, and $P(x) \cdot R(x)$ have the same degree, while the polynomials $P(x)+Q(x), P(x)+R(x)$, and $Q(x)+R(x)$ have pairwise distinct degrees? (We consider that t... | Answer: $: 2$.
Solution. It is sufficient to take, for example, $P(x)=x^{2}, Q(x)=-x^{2}+1, R(x)=x^{2}+x$. Then the polynomials $P(x) \cdot Q(x), Q(x) \cdot R(x)$, and $P(x) \cdot R(x)$ have degree 4 (these are products of two polynomials of degree 2); the polynomial $P(x)+Q(x)$ equals 1 and has degree 0; the polynomi... | 2 | Algebra | proof | Yes | Yes | olympiads | false | 20,016 |
Problem 10.2. (15 points) Given a cyclic quadrilateral $A B C D$, in which $A B=B C$. On the side $C D$ there is a point $N$ such that $\angle D N B=90^{\circ}$. Prove that $A D+N C=D N$.
Fi... | Solution. On the extension of segment $D A$ beyond point $A$, mark point $K$ such that $A K=N C$ (Fig. 1). Note that triangles $A K B$ and $C N B$ are equal ($A B=C B$ by condition, $A K=C N$ by construction, $\angle K A D=\angle N C B$ due to the inscribed quadrilateral $A B C D$). Therefore, $B K=B N$ and $\angle B K... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,017 |
Problem 10.3. (15 points) For real numbers $x>2$ and $y>2$, prove that
$$
\frac{x^{2}-x}{y^{2}+y}+\frac{y^{2}-y}{x^{2}+x}>\frac{2}{3}
$$ | Solution. Multiplying both sides by the product of the denominators, we get
$$
3\left(x^{4}-x^{2}+y^{4}-y^{2}\right)>2\left(x^{2}+x\right)\left(y^{2}+y\right) .
$$
Expanding the brackets on the right side and moving the negative terms to the right side, we get
$$
3 x^{4}+3 y^{4}>2 x^{2} y^{2}+2 x^{2} y+2 x y^{2}+2 x... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 20,018 |
Problem 10.4. (15 points) Once, 45 friends living in different parts of the world decided to exchange news with each other. To do this, they plan to arrange $k$ video meetings, at each of which every person will share their news, as well as all the news from other people they have learned previously.
For the video mee... | Answer: 5 days.
Solution. Let's provide an example of a situation where 4 days would not be enough. Suppose each of the 45 people has their own unique pair of days when they cannot participate in the meeting. Since the number of ways to choose a pair of days from the 10 offered is $\mathrm{C}_{10}^{2}=45$, for any pai... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,019 |
Task 10.5. (20 points) Find all composite natural numbers $n$ that have the following property: each natural divisor of the number $n$ (including $n$ itself), decreased by 1, is a square of an integer. | Answer: 10.
Solution. Suppose that $n$ is divisible by the square of some prime number $p$. Then it has a divisor $p^{2}=b^{2}+1$; but two squares of integers can only differ by 1 if they are 0 and 1.
Let $n$ be divisible by some two prime numbers $p$ and $q$. Without loss of generality, we can assume that $p>q$. Fro... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,020 |
Problem 10.6. (20 points) The altitudes of an acute triangle \(ABC\) intersect at point \(H\). It is known that \(AH^2 = BH^2 + CH^2\). On the circumcircle of triangle \(ABC\), points \(D\) and \(E\) were found such that \(CE \parallel AB\) and \(BD \parallel AC\). Prove that point \(H\) lies on the line \(DE\).
. From the conditions $C E \| A B$ and $B D \| A C$, it follows that $A D=A E=B C$. Moreover, from $\angle B C C_{1}=90^{\circ}-\angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,021 |
# 8. Problem $8^{*}$
A natural number $\mathrm{N}$ ends in 5. Ninth-grader Dima found all its divisors and discovered that the sum of the two largest proper divisors does not divide evenly by the sum of the two smallest proper divisors. Find the smallest possible value of the number N. A divisor of a natural number is... | # 8. Problem $\mathbf{8}^{2}$
A natural number $\mathrm{N}$ ends with 5. Ninth-grader Dima found all its divisors and noticed that the sum of the two largest proper divisors does not divide evenly by the sum of the two smallest proper divisors. Find the smallest possible value of the number N. A divisor of a natural n... | 725 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,029 |
11.1. Find all integer solutions $(x, y)$ of the equation $3 x^{2}-y^{2}=3^{x+y}$ and prove that there are no others. | Answer: $(1,0),(3,0),(-2,3),(-6,9)$.
CRITERIA
+/- 1 or 2 solutions are lost due to non-critical (computational) errors.
+/- all solutions are found. The justification for the absence of other solutions contains minor gaps.
+/2 3 solutions are found. The justification for the absence of other solutions contains mino... | (1,0),(3,0),(-2,3),(-6,9) | Number Theory | proof | Yes | Yes | olympiads | false | 20,032 |
11.2. A trinomial of degree $p$ is a function of the form $f(x)=x^{p}+a x^{q}+1$, where $p, q-$ are natural numbers, $q<p$, and $a$ is an arbitrary real number (possibly equal to 0). Find all pairs of trinomials that give a trinomial of degree 15 when multiplied.
ANSWER: $\left(1+x^{5}\right)\left(1-x^{5}+x^{10}\right... | Solution. Let
$$
\left(x^{p}+a x^{q}+1\right)\left(x^{r}+b x^{s}+1\right)=x^{15}+c x^{t}+1
$$
In particular, the monomials of the highest degree on the right and left sides are equal: $x^{p+r}=x^{15}$. Since $p+r=15$, then $p \neq r$.
First, we investigate how many of the coefficients $a, b$ can be zero. If $a=b=0$,... | (1+x^{5})(1-x^{5}+x^{10}),(1-x^{3}+x^{9})(1+x^{3}+x^{6}),(1-x^{6}+x^{9})(1+x^{3}+x^{6}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,033 |
11.4. Instead of crosses in the expression $x \cdot x + x \cdot x + \ldots + x \cdot x$ (50 terms), the numbers $1, \ldots, 100$ were placed, each exactly once. What is the maximum and minimum value that the resulting expression can have?
ANSWER: $169150, 85850$.
CRITERIA
The score was assigned based on the sum of a... | Solution. Let $aa c+b d>a d+b c$. The first inequality is obtained by expanding the brackets in the inequality $(d-a)(c-b)>0$, the second - in the inequality $(b-a)(d-c)>0$.
Then, for every arrangement of numbers with a term of the form $a b+c d$ or $a c+b d$, we can associate a smaller value by substituting $a d+b c$... | 169150,85850 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,034 |
1. Can 2010 rays be arranged on a plane such that no point on the plane is passed through by more than two rays, each ray intersects exactly two other rays, and any two points on any two rays can be connected by a broken line entirely contained within the union of these rays? | Solution. This can be done. Here is one of the possible constructions. If $A_{1} A_{2} \ldots A_{2010}$ is a regular 2010-gon, then the set of rays $A_{1} A_{2}, A_{2} A_{3}, \ldots A_{2010} A_{1}$ satisfies the condition of the problem. Indeed, for any pair of points $M \in A_{k} A_{k+1}, N \in A_{l} A_{l+1}$, where $... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,035 |
2. Among all quadruples of natural numbers $(k, l, m, n), k>l>m>n$, find the one such that the sum $\frac{1}{k}+\frac{1}{l}+\frac{1}{m}+\frac{1}{n}$ is less than one and as close to it as possible. | Solution. First, note that using a "greedy algorithm" (i.e., sequentially choosing the numbers $n, m, l, k$ to be as large as possible), we obtain the quadruple $(43,7,3,2):$
$$
1 / 43 + 1 / 7 + 1 / 3 + 1 / 2 = 1805 / 1806
$$
Thus, for the desired quadruple $(k, l, m, n)$, we have $1 / k + 1 / l + 1 / m + 1 / n \geq ... | (2,3,7,43) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,036 |
4. In an acute-angled triangle $A B C$, the altitude $A A_{1}$ is drawn. $H$ is the orthocenter of triangle $A B C$. It is known that $A H=3, A_{1} H=2$, and the radius of the circumcircle of triangle $A B C$ is 4. Find the distance from the center of this circle to $H$. | Solution. Draw the altitudes $B B_{1}$ and $C C_{1}$ in the triangle. Then the quadrilateral $A C_{1} H B_{1}$ is cyclic, since its opposite angles $C_{1}$ and $B_{1}$ are right angles. Therefore: $\angle B H C=$ $\angle C_{1} H B_{1}=180^{\circ}-\angle C_{1} A B_{1}$. Reflect point $H$ symmetrically with respect to th... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,038 |
5. Let $x$ be a number from the interval $(\pi / 2, \pi)$ such that
$$
\frac{4}{3}\left(\frac{1}{\sin x}+\frac{1}{\cos x}\right)=1
$$
Prove that the number
$$
\left(\frac{4}{3}\right)^{4}\left(\frac{1}{\sin ^{4} x}+\frac{1}{\cos ^{4} x}\right)
$$
is an integer and find it. | Solution. For brevity, let $s=\sin x, c=\cos x$. We need to find the number $t=$ $1 / c s$. For this, we square the relation $\left(s^{-1}+c^{-1}\right)=3 / 4$. We get $s^{2}+c^{2}+2 c s=$ $9(c s)^{2} / 16$. Considering that $s^{2}+c^{2}=1$, we obtain the following quadratic equation for $t$:
$$
t^{2}+2 t-9 / 16=0
$$
... | 49 | Algebra | proof | Yes | Yes | olympiads | false | 20,039 |
1. Given a cube, each face of which is a grid of 2015 by 2015
cells. In the center of one of the faces stands a pawn. Danil and Maxim move the pawn across the cells of the cube. Danil can only move to an adjacent cell (it is allowed to move to another face if the cells are adjacent), while Maxim can place the pawn in ... | Solution. Answer: Danil wins.
We will present a winning strategy for Danil.
The number of cells on the surface is even (equal to 2015*2015*6). We will divide the entire surface of the cube into dominoes; the dominoes do not overlap and cover the entire cube. We will not provide an example of the division. It is easy ... | Danilwins | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,041 |
2. Given a triangle ABC, points A_1, B_1, C_1 are the midpoints of sides BC, AC, AB respectively. Prove that the three lines passing through these points and parallel to the angle bisectors of the opposite angles intersect at one point. | Solution. Let these lines be $a, b, c$ respectively.
Consider the quadrilateral A C_1 A_1 B_1. It is a parallelogram, since C_1 A_1 and A_1 B_1 are midlines in
triangle ABC. Draw the angle bisector of angle A and line $a$. From the parallelism of these two lines and the fact that A C_1 A_1 B_1 is a parallelogram, it f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,042 |
3. In the gnome clan, some are acquainted with each other. Each gnome owns a certain number of coins. During the day, each gnome finds out how many coins each of his acquaintances has. In the evening, he gives one coin to each of his acquaintances who were richer than him during the day. A gnome cannot give more than h... | Solution. We will prove the required statement by induction on the number of gnomes.
Base case - one gnome; the statement is obvious.
Step - suppose the statement is true for any clan of p gnomes, let's prove it for any clan of $\mathrm{n}+1$ gnomes.
Consider an arbitrary clan of n+1 gnomes and how they exchanged co... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 20,043 |
4. On the board, the numbers $1 / 1, 1 / 2, 1 / 3, 1 / 4, \ldots 1 / 100$ are written.
It is allowed to erase any two numbers a, b and write $a b + a + b$ in their place,
then do the same with any two of the remaining numbers, and so on. What number can be left last? | Solution. If initially on the board there were not one hundred, but three numbers: a, b, c, then at the end on the board there would be the number abc + ab + bc + ac + a + b + c. That is, the sum of all possible monomials composed of the numbers a, b, c, taken no more than once each.
In the case when one hundred numbe... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,044 |
5. Into how many parts can 7 different tangents to a given circle divide the plane? Provide examples for all answers and prove that no others exist. | Solution. 26,27,28,29. To get the correct answer, one should notice that it is possible to consider cases of pairwise parallelism of several lines. For example, no parallel lines, one pair of parallel lines, two pairs of parallel lines, three pairs of parallel lines.
Criteria. -+ - incomplete enumeration, partially co... | 26,27,28,29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,045 |
6. The words of the language of the robots on the planet Shelyakia are sequences of arrows "up", "down", "left", and "right", with the condition that two opposite arrows cannot stand next to each other. The teacher wrote 1000000 words of this language on the board. Four students copy the words into their notebooks, mak... | Solution. Let's divide the words written on the board into 4 classes U, D, $\mathrm{L}, \mathrm{R}$ according to the first letter (we will use lowercase u, d, l, r to denote the number of elements in them, $u+d+l+r=1000000$). Then in the student's notebook, U will
## be written
u + l + r words starting with "up" (bec... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 20,046 |
# 1. Task 1
What number should the asterisk be replaced with so that the equation $(2 x-7)^{2}+(5 x-*)^{2}=0$ has a root?
# | # 1. Task 1
What number should the asterisk be replaced with so that the equation $(2 x-7)^{2}+(5 x-*)^{2}=0$ has a root?
Answer: 17.5 | 17.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,047 |
# 3. Problem 3
In an isosceles triangle $A B C$, the perpendicular bisector of the lateral side $B C$ intersects the base $A B$ at point $D$ such that $A C=A D$. Find the angle $A B C$.
Write the answer in degrees without the degree symbol. | # 3. Problem 3
In an isosceles triangle $ABC$, the perpendicular bisector of the lateral side $BC$ intersects the base $AB$ at point $D$ such that $AC = AD$. Find the angle $ABC$.
Write the answer in degrees without the degree symbol.
Answer: 36 | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,049 |
3. Spartak fans tell the truth when Spartak wins, and lie when it loses. Similarly, fans of Dynamo, Zenit, and Lokomotiv behave. After two matches involving these four teams, each of which ended with one team winning and not in a draw, out of the fans who watched the broadcast, 200 people answered positively to the que... | Solution: Let $x$ and $y$ be the number of people cheering for the winning teams, and $z$ and $t$ for the losing teams. Then the number of people answering questions about these teams are $x+z+t, y+z+t, t, z$ respectively. Therefore, the number of people cheering for Zenit is 0, for Lokomotiv 100, for Spartak 300, and ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,078 |
4. A triangle with sides 2, 3, and 3 was cut into four triangles similar to it. What could the similarity coefficients be? | Answer: $1 / 2$ and $6 / 13, 4 / 13, 9 / 13, 6 / 13$.
Solution: Note that the only arrangement of four triangles so that they are similar to the larger one (and thus to each other) is such that each triangle has one vertex at a vertex of the larger triangle and the other two on its sides, with one point marked on each... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,079 |
5. On the board, the numbers $1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{100}$ are written. It is allowed to erase any two numbers $a$ and $b$ and write $a+b+a b$ instead. After several such operations, only one number remains on the board. What can this number be? | Answer: $2^{99}(1+1)(1+1 / 2) \ldots(1+1 / 100)-1=2^{99} 101-1$. Solution: It is necessary to show the associativity of the operation for three arbitrary $a, b, c$, and then perform the calculation, noting that at each step, an expression of the form $2^{k}\left(1+a_{1}\right)\left(1+a_{2}\right) \cdot \ldots \cdot\lef... | 2^{99}\cdot101-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,080 |
6. The words of the language of the robots on the planet Shelyakia are sequences of arrows "up", "down", "left", and "right", with the condition that two opposite arrows cannot stand next to each other. The teacher wrote 1000000 words of this language on the board. Four students copy the words into their notebooks, mak... | Solution: Let's divide the words written on the board into 4 classes $U, D, L, R$ according to the first letter (we will use lowercase $u, d, l, r$ to denote the number of elements in them, $u+d+l+r=1000000$). Then in the notebook of student $U$, there will be $u+l+r$ words starting with "up" (since he will append "up"... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 20,081 |
10.1. Find all integer solutions $(x, y)$ of the equation $x^{2}-2 y^{2}=2^{x+y}$ and prove that there are no others. | Answer: $(2,0),(2,-1),(6,-4),(4,0),(12,-8)$.
CRITERIA
+/- losing 1 or 2 solutions due to non-critical (computational) errors.
+/- finding 3-5 solutions. The justification for the absence of other solutions has minor gaps.
+/2 finding 3-5 solutions. The justification for the absence of other solutions has minor gaps... | (2,0),(2,-1),(6,-4),(4,0),(12,-8) | Number Theory | proof | Yes | Yes | olympiads | false | 20,082 |
10.6. Two players play the following game. They have a chocolate bar, divided by grooves parallel to the sides of the bar into pieces. The grooves divide the bar into M vertical and N horizontal strips. The first player, on their turn, breaks the bar along one of the grooves into two rectangular parts and gives them to... | Answer: The maximum power of 2 that divides $M$ does not coincide with that for $\mathrm{N}$.
## CRITERIA
+ correct solution.
+/- correct answer and the beginning of the proof, including the case of a tile $2 n \times(2 m+1)$, but not limited to it.
+/2 cases of tiles $2 n \times(2 m+1)$ and $4 n \times(4 m+2)$ are... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,083 |
Problem 9-1. Calculate the sum $1^{2}+2^{2}-3^{2}-4^{2}+5^{2}+6^{2}-7^{2}-8^{2}+9^{2}+10^{2}-\ldots+2017^{2}+2018^{2}$. | Answer: 4074341.
Solution. Note that for any $k$ the equality $k^{2}-(k+1)^{2}-(k+2)^{2}+(k+3)^{2}=4$ holds. Therefore, the entire sum is equal to $1+504 \cdot 4+2018^{2}=4074341$.
| Rating | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| $+/-$ | 16 | Correct solution wit... | 4074341 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,094 |
Problem 9-2. Vovochka wants to pass a signed note to Natasha in class, with the envelope first passing through the entire rest of the class in a known order. Each student, except Natasha, may dislike one classmate, and if they pass an envelope signed by themselves, they change it to the one they dislike, and if it is s... | Answer: Any number.
Solution. Each student performs a permutation: if they dislike someone, then a permutation that swaps their name and the name of the person they dislike; otherwise, the permutation is trivial. In other words, no two names will become the same after passing through any student. And for each name in ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,095 |
Problem 9-3. In a right triangle $A B C$, angle $B$ is a right angle. On the leg $A B$, a point $M$ is chosen such that $A M=B C$, and on the leg $B C$, a point $N$ is chosen such that $C N=M B$. Find the acute angle between the lines $A N$ and $C M$.
Answer: $45^{\circ}$.
$.
Solution. One of two cases applies.
A) Suppose the two smallest divisors $p$ and $q$ are prime numbers. Then the number $r=(n / p+n / q)-(p+q)$ is also prime, and $p q r=(p+q)(n-p q)$. Since the numbers $p+q$ and $p q$ are coprime, we get $r=p+q$, from which $p=2$ and $n=4 q$. But then, due to ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,098 |
Problem 9-6. In a cubic chest with a side of $2^{n}$ dm, there are $8^{n}$ different spices: it contains eight closed cubic boxes with a side of $2^{n-1}$ dm, each of which contains eight closed cubic boxes with a side of $2^{n-2}$ dm, and so on down to boxes with a side of 1 dm, each containing its own spice.
In one ... | Solution. The conditions of the problem require that the mouse gnaws through each box in at least two places: to enter it and to leave it. Thus, the number of holes is no less than twice the number of boxes, that is, \(2 \cdot \left(8^{n+1}-1\right) / 7\). We will construct a path for the mouse with this number of hole... | 2\cdot(8^{n+1}-1)/7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,099 |
10.1. Can non-zero numbers $x, y$, and $z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x=y^{2}-y \\
y^{2}+y=z^{2}-z \\
z^{2}+z=x^{2}-x
\end{array}\right.
$$ | Answer: No.
Solution. Move all terms of the first equation to one side and factor the resulting expression:
$$
(x+y)(x-y+1)=0
$$
Also, add all three equations:
$$
x+y+z=0
$$
Since the numbers $x, y, z$ are not equal to zero, the numbers $x+y, y+z, z+x$ are also not equal to zero. Therefore, $x-y+1=0$, similarly $y... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,100 |
10.2. Real numbers $a, b$, and $c$ are such that the numbers $ab, bc, ca$ are rational. Prove that there exist integers $x, y, z$ such that $ax + by + cz = 0$. | Solution. Unfortunately, the phrase "not equal to zero simultaneously" was omitted in the condition, so the problem has a trivial solution $x=y=z=0$. In the problem that was intended, the numbers $x, y, z$ should not all be 0 at the same time. In this case, the problem has the following solution:
If none of the number... | proof | Algebra | proof | Yes | Yes | olympiads | false | 20,101 |
10.3. On the coordinate plane, a set of points defined by the equation $x=y^{2}$ is drawn. A circle of radius 5 with center at the point $(11 ; 1)$ intersects this set at points $A, B, C$ and $D$. Prove that all points $A, B, C, D$ lie on one parabola, i.e., on a curve defined by the equation $y=a x^{2}+b x+c$, and fin... | Answer. $y=\frac{1}{2} x^{2}-\frac{21}{2} x+\frac{97}{2}$.
Solution. The coordinates of points $A, B, C, D$ are solutions to the system
$$
\left\{\begin{array}{l}
y^{2}=x \\
(x-11)^{2}+(y-1)^{2}=25
\end{array}\right.
$$
Expanding the brackets in the second equation and substituting $y^{2}$ from the first, we obtain ... | \frac{1}{2}x^{2}-\frac{21}{2}x+\frac{97}{2} | Geometry | proof | Yes | Yes | olympiads | false | 20,102 |
10.4. Point $O$ is the center of the circumcircle of an acute-angled triangle $ABC$. The line $AO$ intersects side $BC$ at point $P$. Points $E$ and $F$ on sides $AB$ and $AC$ respectively are chosen such that a circle can be circumscribed around quadrilateral $AEPF$. Prove that the length of the projection of segment ... | Solution. Extend $E F$ to intersect side $B C$ at point $M$, and let $\gamma$ be the acute angle between the specified lines at point $M$. Then the projection of segment $E F$ onto side $B C$ is $|E F| \cos \gamma$ (if $E F \| B C$, then we set $\gamma=0$). Next, let $\angle B A O=\alpha$ and $\angle C A O=\beta$, and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,103 |
10.5. On a plane, there are eight different points. A numbering of these points with numbers from 1 to 8 is called good if the following condition is met:
there exists a line such that all points lie on one side of it and at different distances from it, and the distances from the points to this line increase with the ... | Answer. $2 C_{8}^{2}=56$.
Solution. Note that when numbering points relative to a line, we are only interested in its direction and which side of it the points are on. All this information can be uniquely restored from a unit normal to it, directed towards the points. Therefore, the numbering of the octet is derived f... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,104 |
10.6. Let $p>2$ be an integer not divisible by 3. Prove that there exist integers $a_{1}, a_{2}, \ldots, a_{k}$ such that
$$
-\frac{p}{2}<a_{1}<a_{2}<\cdots<a_{k}<\frac{p}{2}
$$
and the product
$$
\frac{p-a_{1}}{\left|a_{1}\right|} \cdot \frac{p-a_{2}}{\left|a_{2}\right|} \cdots \frac{p-a_{k}}{\left|a_{k}\right|}
$$... | Solution. FIRST METHOD. Let us take as numbers $a_{i}$ all numbers from the interval $\left(-\frac{p}{2}, \frac{p}{2}\right)$ such that $p \equiv a_{i}(\bmod 3)$. We will prove that they satisfy the condition of the problem. Note the following:
- All numbers $\left|a_{i}\right|$ belong to the interval $\left(0, \frac{... | 3^{3n} | Number Theory | proof | Yes | Yes | olympiads | false | 20,105 |
9.2. Is it possible to cut a circle into parts in such a way that a) the center of the circle is on the boundary of each part and b) from some of the parts obtained as a result of the cutting, a regular hexagon inscribed in this circle can be formed? If it is possible, describe the cutting and indicate how to form the ... | Answer: Inscribed in a circle with center $O$ and radius $r$ is a regular hexagon $A B C D E F$. Cut the circle along segments $A O, C O, E O$ and arcs of circles with centers $A, C$, and $E$, radius $r$, and endpoints at points $B, D$, and $F$ (in the left picture). Also, cut the regular hexagon along arcs of circles ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,106 |
9.3. A trinomial of degree $p$ is a function of the form $f(x)=x^{p}+a x^{q}+1$, where $p, q-$ are natural numbers, $q<p$, and $a$ is an arbitrary real number (possibly equal to 0). Find all factorizations of the polynomial $x^{12}+1$ into a product of two trinomials.
ANSWER: $\left(x^{4}+1\right)\left(x^{8}-x^{4}+1\r... | Solution. Let $\left(x^{p}+a x^{q}+1\right)\left(x^{r}+b x^{s}+1\right)=x^{12}+1(*)$.
Since the degrees of the monomials on both sides are equal, then $p+r=12$. The numbers $a$ and $b$ are not both zero (otherwise, after expanding the brackets and combining like terms, there would be at least three monomials in the le... | (x^{4}+1)(x^{8}-x^{4}+1),(x^{6}-\sqrt{2}x^{3}+1)(x^{6}+\sqrt{2}x^{3}+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,107 |
9.4. Along the shore of a circular lake with a perimeter of 1 km, two salmons are swimming - one at a constant speed of $500 \mathrm{m} /$ min clockwise, the other at a constant speed of 750 m/min counterclockwise. Along the edge of the shore, a bear is running, always moving along the shore at a speed of 200 m/min in ... | Solution. Note that the bear runs to the nearest salmon if and only if it runs from a point on the shore that is equidistant from the salmons and not separated from the bear by the salmons. Since both points equidistant from the salmons move counterclockwise at a speed of $|750-500| / 2=125<200$ m/min, they will never ... | 7 | Other | math-word-problem | Yes | Yes | olympiads | false | 20,108 |
9.6. Two players play the following game. They have a chocolate bar, divided by grooves parallel to the sides of the bar into pieces. The grooves divide the bar into M vertical and N horizontal strips. The first player, on their turn, breaks the bar along one of the grooves into two rectangular parts and gives them to ... | Answer: The maximum power of 2 that divides $M$ does not coincide with that for $\mathrm{N}$.
Criteria
+ correct solution.
+/- correct answer and the beginning of the proof, including the case of a tile $2 n \times(2 m+1)$, but not limited to it.
+/2 cases of tiles $2 n \times(2 m+1)$ and $4 n \times(4 m+2)$ are co... | Themaximumpowerof2thatdividesMdoesnotcoincidewiththatforN | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,109 |
2. Does there exist a quadrilateral that can be cut into three equal triangles in two different ways? If not, prove it; if it does, provide an example. | Solution: A trapezoid with bases $B C$ and $A D$ of lengths 1 and 2 respectively, and a lateral side $C D$ of length $\sqrt{3}$, perpendicular to the bases. The dissections are $\triangle A B M, \triangle B C M, \triangle C D M$ and $\triangle A B M, \triangle B D M, \triangle B C D$ where $M$ is the midpoint of $A D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,110 |
6. The words of the language of the robots on the planet Shelyakia are sequences of arrows "up", "down", "left", and "right", with the condition that two opposite arrows cannot stand next to each other. The teacher wrote 1000000 words of this language on the board. Four students copy the words into their notebooks, mak... | Solution: Let's divide the words written on the board into 4 classes $U, D, L, R$ according to the first letter (we will use lowercase $u, d, l, r$ to denote the number of elements in them, $u+d+l+r=1000000$). Then in the notebook of student $U$, there will be $u+l+r$ words starting with "up" (since he will append "up"... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 20,112 |
# 6. Problem 6
In triangle $A B C$, points $M$ and $N$ are chosen on sides $A B$ and $B C$ respectively, such that $A M=2 M B$ and $B N=N C$. Segments $A N$ and $C M$ intersect at point $P$. Find the area of quadrilateral $M B N P$, given that the area of triangle $A B C$ is 30. | # 6. Problem 6
In triangle $A B C$, points $M$ and $N$ are chosen on sides $A B$ and $B C$ respectively such that $A M=2 M B$ and $B N=N C$. Segments $A N$ and $C M$ intersect at point $P$. Find the area of quadrilateral $M B N P$, given that the area of triangle $A B C$ is 30.
## Answer: 7
# | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,118 |
Problem 10-1. For real numbers $a, b$, and $c$, it is known that $a b c + a + b + c = 10$, and $a b + b c + a c = 9$. For which numbers $x$ can it be asserted that at least one of the numbers $a, b, c$ is equal to $x$? (Find all such numbers $x$ and prove that there are no others.)
Answer: 1. | Solution. We will provide two solutions to the problem.
First Solution. Let $a+b+c=\lambda$. Vieta's theorem allows us to write a cubic equation depending on the parameter $\lambda$, whose roots are the set $a, b, c$ corresponding to the given $\lambda$:
$$
t^{3}-\lambda t^{2}+9 t-(10-\lambda)=0 \quad \Leftrightarrow... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,122 |
Problem 10-2. Vovochka wants to pass a signed note to Natasha in class, with the envelope first passing through the entire rest of the class in a known order. Each student, except Natasha, may dislike one classmate, and if they pass an envelope signed by themselves, they change it to the one they dislike, and if it is ... | Answer: Any number.
Solution. Each student performs a permutation: if they dislike someone, then a permutation that swaps their name and the name of the person they dislike; otherwise, the permutation is trivial. In other words, no two names will become the same after passing through any student. And for each name in ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,123 |
Problem 10-3. Grisha drew a convex 100-gon on the plane and drew all its diagonals, and, wonder of wonders, no more than two segments intersected at any point other than the vertices of the 100-gon. In how many ways can Grisha highlight some of the lines on the drawing with a marker to form a triangle (not necessarily ... | Solution. Let's find the ends of the diagonals outlined by Grisha. Each diagonal has two ends, but some ends can coincide, but no more than two in one point. Thus, we will get from 3 to 6 vertices of the hundred-gon. We will find the corresponding number of triangles.
For three vertices, the number of triangles is equ... | \binom{100}{3}+4\binom{100}{4}+5\binom{100}{5}+\binom{100}{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,124 |
Problem 10-4. In a cubic chest with a side of $2^{n}$ dm, $8^{n}$ different spices are stored: it is packed with eight closed cubic boxes with a side of $2^{n-1}$ dm, each of which contains eight closed cubic boxes with a side of $2^{n-2}$ dm, and so on down to boxes with a side of 1 dm, each containing its own spice.
... | Answer: $2 \cdot\left(8^{n+1}-1\right) / 7$.
Solution. The conditions of the problem require that the mouse must gnaw through each box in at least two places: to enter it and to leave it. Thus, the number of holes is not less than twice the number of boxes, that is, $2 \cdot\left(8^{n+1}-1\right) / 7$. We will constru... | 2\cdot(8^{n+1}-1)/7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,125 |
Problem 10-5. Consider all reduced quadratic trinomials $x^{2}+p x+$ $q$ with integer coefficients $p$ and $q$. Let's call the range of such a trinomial the set of its values at all integer points $x=0, \pm 1, \pm 2, \ldots$ What is the maximum number of such trinomials that can be chosen so that their ranges do not in... | Solution. Note that the substitution of the variable $x \rightarrow x+k$ for any integer $k$ does not change the range of the polynomial. Then, by making the substitution $x \rightarrow x-\left[\frac{p}{2}\right]$ (square brackets denote the integer part), we can assume that any polynomial has one of two forms: $x^{2}+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,126 |
Problem 10-6. In an acute-angled triangle $A B C$, altitudes $A D, B E, C F$ are drawn; $H-$ is the orthocenter. A circle with center at point $O$ passes through points $H$ and $A$, intersecting sides $A B$ and $A C$ at points $Q$ and $P$, respectively (point $O$ does not lie on sides $A B$ and $A C$). The circumcircle... | Solution. $\quad$ Let $\angle C A B=x, \angle A B C=y, \angle D C A=z$. Suppose point $Q$ is between $A$ and $F$, and point $P$ is between $C$ and $E$. $(\angle F Q H=\angle A P H)$. We will provide two solutions to the problem.
## First Solution.
(1+a b c) \ldots(1+a b c \ldots z)=0$. Prove that $(a+b)(a+b c) \ldots(a+$ $b c \ldots z)=0$. | Solution. Since the first product is zero, one of the factors is zero. Since all numbers are integers, this means that in some group $a, b, c, \ldots, x$ all numbers are $\pm 1$, and the number of -1s in this group is odd. But then in the sum $a+b c \ldots x$ one term is 1, and the other is -1.
| Evaluation | Score | ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 20,129 |
Problem 7-3. Midway between points A and B is a coffee shop. A cyclist first set off from A to B. When he was halfway to the coffee shop, a motorist set off from A. It is known that when the motorist arrived at the coffee shop, the cyclist was still on the way between A and B, and the distance between him and the motor... | Answer: simultaneously.
Solution. Let the distance $A C$ be 1. When the motorist set off from , the cyclist was at point $1 / 2$, and when he reached - at point $3 / 4$. Therefore, the speed of the car is four times the speed of the bicycle. At this moment, the distance between the cyclist and was four times less tha... | simultaneously | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,130 |
Problem 7-4. In triangle $A B C$, where all three sides are pairwise distinct, the angle bisectors of angles $A$ and $B$ are drawn, dividing it into a quadrilateral and three triangles, two of which are isosceles. Find the angles of the original triangle. | Answer: $4 \pi / 7, 2 \pi / 7, \pi / 7$.
Solution. Let in triangle $ABC$ the bisectors $AK$ and $BL$ intersect at point $O, \angle CAB = \alpha, \angle CBA = \beta$. Triangle $ABO$ cannot be isosceles, as none of the following conditions can be satisfied: $\alpha = \beta, \alpha + \beta / 2 = 180^{\circ}, \beta + \alp... | \frac{4\pi}{7},\frac{2\pi}{7},\frac{\pi}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,131 |
Problem 7-5. In the garden of the oracle, there live four turtles. A visitor can choose any subset of turtles in a move and ask the oracle how many of these turtles are males (the oracle's answers are always truthful). What is the minimum number of moves required to find out the gender of all the turtles?
Answer: 3. | Solution. In three questions, the answer can be obtained as follows. The first two questions are about turtles 1 and 2, and 2 and 3. If at least one of the answers is 0 or 2, we know who they are for the corresponding pair, and for the remaining one of the three, we know from the other question, leaving one question fo... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,132 |
Problem 7-6. There are several coins, each worth an integer number of tugriks. It is known that with these coins, you can make any other sum from 1 to 51 tugriks inclusive, except for the sum of 50 tugriks. Is it necessarily possible to make a sum of exactly 100 tugriks with these coins? | Answer: yes.
Solution. Let's take a set of coins $A$ with a total value of 51 tugriks. We will prove that $A$ consists of one coin worth 51 tugriks. Then it will remain to add it to the set worth 49 tugriks. Let $k$ tugriks be the value of the cheapest coin in the set $A$. If $k$ is less than or equal to 25, then ther... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,133 |
Problem 11.1. (15 points) Each natural number is painted in one of three colors: red, blue, or green, and all three colors are used. Can it happen that the sum of any two numbers of different colors is a number of the remaining color? | Answer: No.
Solution. Suppose such a coloring is possible. Without loss of generality, we can assume that the number 1 is colored in the first color. Choose an arbitrary number $x$ of the second color. Note that then $x+1$ must be of the third color, $x+2$ - of the second, $x+3$ - of the third, and so on. Thus, all nu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 20,134 |
Problem 11.2. (15 points) Different real numbers $x, y, z$ are such that among the three numbers
$$
\frac{x+y}{x^{2}+x y+y^{2}}, \quad \frac{y+z}{y^{2}+y z+z^{2}}, \quad \frac{z+x}{z^{2}+z x+x^{2}}
$$
some two are equal. Is it true that all three numbers are equal? | Answer: 2.
Solution. In the given expressions, multiply the numerators and denominators by $x-y, y-z, z-x$ respectively (according to the condition, these differences are non-zero). We get the same numbers in another form:
$$
\frac{x^{2}-y^{2}}{x^{3}-y^{3}}, \quad \frac{y^{2}-z^{2}}{y^{3}-z^{3}}, \quad \frac{z^{2}-x^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 20,135 |
Problem 11.3. (15 points) Natural numbers $a, b, c$ are such that $1 \leqslant a<b<c \leqslant 3000$. Find the greatest possible value of the quantity
$$
\text { GCD }(a, b)+\text { GCD }(b, c)+\text { GCD }(c, a)
$$ | Answer: 3000.
Solution. Note that GCD $(a, b)=$ GCD $(a, b-a) \leqslant b-a$, since the GCD of two natural numbers does not exceed either of them. Similarly, we obtain that GCD $(b, c) \leqslant c-b$, and also GCD $(c, a) \leqslant a$.
Adding these three inequalities, we get
$$
\text { GCD }(a, b)+\text { GCD }(b, c... | 3000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,136 |
Problem 11.4. (15 points) In a circle $\omega$, a triangle $ABC$ is inscribed such that $AB < BC$. The bisector of the external angle at $B$ intersects $\omega$ at point $M$. A line parallel to $BM$ intersects sides $BC$, $AB$, and the extension of side $CA$ beyond point $A$ at points $P$, $Q$, and $R$ respectively. Th... | Solution. Fig. 1. We will prove that points $R, X, P, C$ lie on the same circle $\Omega$. Indeed, $\angle X R P = \angle B M X$ as alternate interior angles when lines $B M$ and $R P$ are parallel, and $\angle B M X = \angle B C X$ as angles subtending the same arc in $\omega$, hence $\angle X R P = \angle X C P$.
Now... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,137 |
Problem 11.5. (20 points) Given a $100 \times 100$ grid. Each cell of the grid is painted in one of two colors: white or black. We call a coloring of the grid balanced if each row and each column contains 50 white and 50 black cells. In one operation, it is allowed to choose two rows and two columns such that out of th... | Solution. We will prove that from any balanced board, we can obtain a board with a checkerboard pattern, and at each step, the board will remain balanced. From this, it will follow that from any balanced board, we can obtain any other, since the operation is reversible.
We will obtain the checkerboard pattern as follo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 20,138 |
Problem 11.6. (20 points) Quadratic trinomials $P(x)$ and $Q(x)$ with real coefficients are such that together they have 4 distinct real roots, and each of the polynomials $P(Q(x))$ and $Q(P(x))$ has 4 distinct real roots. What is the smallest number of distinct real numbers that can be among the roots of the polynomia... | # Answer: 6.
Solution. Note that if among the roots of the polynomial $P(Q(x))$ there is a root of $Q(x)$, say, the number $x_{0}$, then $P\left(Q\left(x_{0}\right)\right)=P(0)=0$, from which 0 is a root of $P(x)$. Similarly, if among the roots of $Q(P(x))$ there is a root of $P(x)$, then 0 is a root of $Q(x)$. Howeve... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,139 |
Problem 8-2. Calculate the sum $1^{2}+2^{2}-3^{2}-4^{2}+5^{2}+6^{2}-7^{2}-8^{2}+9^{2}+10^{2}-\ldots+2017^{2}+2018^{2}$. | Solution. Note that for any $k$ the equality $k^{2}-(k+1)^{2}-(k+$ $2)^{2}+(k+3)^{2}=4$ holds. Therefore, the entire sum is equal to $1+504 \cdot 4+2018^{2}=4074341$.
| Grade | Score | Content criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Solution is correct, but there are minor flaws... | 4074341 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,141 |
Problem 8-3. $\quad$ In a right triangle $A B C$, angle $B$ is a right angle. On the leg $A B$, a point $M$ is chosen such that $A M=B C$, and on the leg $B C$, a point $N$ is chosen such that $C N=M B$. Find the acute angle between the lines $A N$ and $C M$.
Answer: $45^{\circ}$.
. What is the minimum number of moves required to find out the gender of all the turtles?
Answer: 3. | Solution. In three questions, the answer can be obtained as follows. The first two questions are about turtles 1 and 2, and 2 and 3. If at least one of the answers is 0 or 2, we know who they are for the corresponding pair, and for the remaining one of the three, we know from the other question, leaving one question fo... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,143 |
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