problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Bakayev E.B.
Around a circular lake, 2019 trees grow at equal intervals: 1009 pines and 1010 spruces. Prove that there must be a tree next to which a pine grows, and on the other side, one tree away, another pine also grows. | Let's walk around the lake in a circle and write letters on the trees: A, B, C, then again A, B, C, and so on. There will be $2019: 3$ = 673 trees with each letter. If there were no more than 336 pines with each letter, then there would be no more than 336 with $3=1008$. But since there are 1009, there will be at least... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,970 |
Each side of an equilateral triangle is divided into n equal parts. Lines are drawn through the division points, parallel to the sides. As a result, the triangle is divided into $n^{2}$ small triangles. We will call a chain a sequence of small triangles in which no triangle appears more than once and each subsequent tr... | To prove what often happens in olympiads, it is enough to make one unexpected move: color the triangles in a checkerboard pattern, as shown in the figure. The rest is quite simple. In the entire triangle, there are
$-th lamps intersect (at least at one point), then the $(n+1)$-th lamp can be turned off.
Therefore, segments with different odd numbers do not intersect. On a segment ... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,975 |
Prove that any line in Cartesian coordinates $x$ Oy has an equation of the form $a x+b y+c=0$, where $a, b, c$ are some numbers, and at least one of the numbers $a, b$ is not zero.
# | Take two points $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)$ such that the given line is the perpendicular bisector of the segment $A B$, and apply the theorem of the perpendicular bisector of a segment.
## Solution
Let $l$ be an arbitrary line on the plane $X O Y$. Consider two distinct points $A\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,976 |
[Coordinate Method on the Plane]
Prove that the distance from the point $M\left(x_{0} ; y_{0}\right)$ to the line given by the equation $a x+b y+c=0$ is
$$
\frac{\left|a x_{0}+b y_{0}+c\right|}{\sqrt{a^{2}+b^{2}}}
$$ | Express through $a, b$ and the coordinates of the intersection point of the given line with the line passing through point $M$ perpendicular to the given line.
## Solution
Let the given line not be parallel to the coordinate axes. Write its equation in the form $y=-\frac{a}{b} x-\frac{c}{b}$. Then its slope $k_{1}=-\... | \frac{|x_{0}+by_{0}+|}{\sqrt{^{2}+b^{2}}} | Geometry | proof | Yes | Yes | olympiads | false | 53,977 |
Mitkin D:
Find all quadruples of real numbers, in each of which any number is equal to the product of any two other numbers. | $0,0,0,0,1,1,1,1,-1,-1,1,1,-1,-1,-1,1$ (up to permutation of the numbers in the quartet). Consider the absolute values of the desired numbers and order them in non-decreasing order: $a \leq b \leq c \leq d$. Note that $a \geq b c$, since either $a=b c$, or $a=b d \geq b c$, or $a=c d \geq b c$. Similarly, $d \leq b c$.... | 0,0,0,0,1,1,1,1,-1,-1,1,1,-1,-1,-1,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,978 |
Senderov V.A.
Do there exist three pairwise distinct non-zero integers, the sum of which is zero, and the sum of their thirteenth powers is a square of some natural number?
# | For a natural number $t$, the triplet of numbers $3 t, -t, -2 t$ satisfies all conditions, except possibly the last one. In order for the sum $(3 t) 13 + (-t) 13 + (-2 t) 13 = t 13 (313 - 1 - 213)$ to be a perfect square, it is sufficient to set, for example, $t = 313 - 1 - 213$.
Thus, the condition of the problem is ... | 3,-,-2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,980 |
7,8, |
| :---: | :---: | :---: |
| | [Sum of angles in a triangle. Theorem about the exterior angle.] | |
| | $\left[\begin{array}{l}\text { [ Equilateral (equiangular) triangle } \\ \text { [ Rectangles and squares. Properties and criteria }\end{array}\right]$ | |
A point inside a square is connected to the verti... | Let's denote our square as $A B C D$, and the given point inside it as $M$. Suppose $\angle M D C=\angle M C D=15^{\circ}$.
We will solve the inverse problem. Construct an equilateral triangle $A B N$ on the side $A B$ of the square such that the vertex $N$ lies inside the square (see the figure).
.
Consider the numbers $b_{0}=0, b_{1}, \ldots, b_{70}$,
$b_{0}+20, \ldots, b_{70}+20$. In total, there are 142 numbers, each of which does not exceed $10 \cdot 12+20=140$. Therefore, among them
t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,983 |
4[Mathematical logic (other).]
In a gang of 50 bandits. All of them have never participated in any showdown together, but every pair of them has met in a showdown exactly once. Prove that one of the bandits was in at least eight showdowns.
# | Show that in one of the shootouts R, at least eight bandits participated; each of them must meet with a bandit who did not participate in shootout R at different shootouts.
## Solution
Assume the opposite. Choose a bandit A. He participated in no more than 7 shootouts, and each of the remaining bandits was present at... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 53,985 |
Find the number of permutations $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{10}$ of the numbers $1,2, \ldots, 10$, such that $\mathrm{a}_{\mathrm{i}+1}$ is not less than $\mathrm{a}_{\mathrm{i}}-1$ (for $\mathrm{i}=1,2, \ldots, 9)$. | We can encode the required permutations as follows: each $\mathrm{a}_{\mathrm{i}}, \mathrm{i}=1,2, \ldots, 9$, will be marked with a "+" if $\mathrm{a}_{\mathrm{i}+1}>\mathrm{a}_{\mathrm{i}}$, and with a "-", if $\mathrm{a}_{\mathrm{i}+1}=\mathrm{a}_{\mathrm{i}}-1$. Prove that the permutation is uniquely restored by it... | 512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,986 |
Indicate a six-digit number $N$, consisting of different digits, such that the numbers $2 N, 3 N, 4 N, 5 N, 6 N$ differ from it by a permutation of digits. | Consider the period of the fraction $1 / 7$.
## Solution
Let's take $N=142857$. A direct check shows that $2N=285714, 3N=428571, 4N=571428, 5N$ $=714285, 6N=857142$.
## Answer
$N=142857$. | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,987 |
Does there exist a triangle, all altitudes of which are less than 1, and the area is greater than or equal to 10?
# | Provide an example of such a triangle.
## Solution
Consider an isosceles triangle with a base of 40 and a height dropped to the base of $\frac{1}{2}$. Then its area is 10, and the lateral side is
$$
\sqrt{20^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{1601}}{2}
$$
The height from the vertex of the base is equal t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,988 |
[ Evenness and Oddness ] [ Regular Polygons $]$
Given a convex $2 n$-gon $A_{1} \ldots A_{2 n}$. Inside it, a point $P$ is taken, which does not lie on any of the diagonals. Prove that the point $P$ belongs to an even number of triangles with vertices at points $A_{1}, \ldots, A_{2 n}$. | Diagonals divide a polygon into several parts. We will call those parts neighboring if they share a common side. It is clear that from any interior point of the polygon, one can reach any other by moving only to neighboring parts each time. The part of the plane lying outside the polygon can also be considered one of t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,989 |
There are $2 k+1$ cards numbered from 1 to $2 k+1$. What is the maximum number of cards that can be selected so that no one of the extracted numbers is equal to the sum of two other extracted numbers? | If we take all cards with odd numbers (there are $k+1$ of them), the condition will be satisfied. If we take $k+2$ cards, then by subtracting the largest number $N$ from all the others, we get $k+1$ different numbers. All these numbers do not exceed $2k$, so at least two of them will coincide with the numbers on the
$... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,990 |
Izmeystiev I.V.
The network of bus routes in the suburb of Amsterdam is organized in such a way that:
a) each route has exactly three stops;
b) any two routes either have no common stops at all or have only one common stop. What is the maximum number of routes that can be in this suburb if there are a total of 9 sto... | Evaluation. Consider some stop A. Determine the maximum number of routes passing through it. Besides A, there are 8 other stops in the city. On each route passing through A, there are two more stops. Since no two of these routes can have common stops other than A, a total of no more than 8 / 2 = 4 routes can pass throu... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,992 |
$4-$ $[$ Formulas for abbreviated multiplication (other) $]$
Twenty-five coins are distributed into piles as follows. First, they are arbitrarily divided into two groups. Then any of the existing groups is again divided into two groups, and so on until each group consists of one coin. With each division of any group i... | The first method. Let's represent the coins as points and connect each pair of points with a segment. We will get 25(25 - 1) : 2 $=300$ segments. Each time we divide one group of coins into two, we will erase all segments connecting points corresponding to coins that ended up in different groups. Suppose at some step w... | 300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,993 |
10,11 Find the minimum value of the function $f(x)=\sqrt{a^{2}+x^{2}}+\sqrt{(b-x)^{2}+c^{2}}(a>0, b>0, c>0)$. | Consider the segment $D E$ of length $b$. Construct perpendiculars to this segment at its ends: $D A=a, E B=c$ (points $A$ and $B$ lie in different half-planes relative to $D E$, see the figure). Let point $C$ lie on segment $D E$ and $D C=x$, then $E C=b-x$. From the right triangles $A C D$ and $B C E$ we get:
$A C=\... | \sqrt{(+)^{2}+b^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,994 |
Zassavssiy A. A.
Bisectors $A A_{1}$ and $B B_{1}$ of triangle $A B C$ intersect at point $I$. On segments $A_{1} I$ and $B_{1} I$, isosceles triangles are constructed with vertices $A_{2}$ and $B_{2}$ lying on line $A B$. It is known that line $C I$ bisects segment $A_{2} B_{2}$. Is it true that triangle $A B C$ is i... | Let's show that the condition is satisfied by any triangle with angle $C$ equal to $120^{\circ}$. Let $C C_{1}$ be the angle bisector of angle $C$. Then $C A_{1}$ is the external bisector of angle $A C C_{1}$, meaning point $A_{1}$ is equidistant from lines $A C$ and $C C_{1}$. But it is also equidistant from lines $A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,995 |
$[$ Processes and Operations $]$ [ Evenness and Oddness $]$
Dwarves sat down at a round table and voted on many issues. For each issue, they could vote "for", "against", or abstain. If both neighbors of a dwarf chose the same option on a particular issue, then for the next issue, the dwarf would choose the same option... | If the gnomes voted unanimously on any issue, they will always vote the same way after that. Therefore, the question about the Dragon was discussed before the question about the gold.
It is possible that before the question about the gold, the gnomes had already voted "for" several times unanimously. Consider the last... | 4k | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,996 |
[Coordinate method on the plane]
Find the coordinates of the vertices of a triangle, the sides of which lie on the lines $2 x+y-6=0, x-y+4=0$ and $y+1$ $=0$. | Solving the system of equations
$$
\left\{\begin{array}{l}
2 x+y-6=0 \\
x-y+4=0
\end{array}\right.
$$
we find the coordinates of the point $A\left(x_{1} ; y_{1}\right)$ of intersection of the given lines: $x_{1}=\frac{2}{3}, y_{1}=\frac{14}{3}$.
Similarly, we will find the other vertices of the triangle.
## Answer
... | (\frac{2}{3};\frac{14}{3}),(-5;-1),(\frac{7}{2};-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,997 |
## [Coordinate method on the plane] $[$ Circles (other) $\quad]$
Given points $A(0 ; 0), B(-2 ; 1), C(3 ; 3), D(2 ;-1)$ and the circle $(x-1)^{2}+(y+3)^{2}=25$. Determine the location of these points: on the circle, inside the circle, or outside the circle. | Substituting the coordinates of the given points into the left side of the equation of the given circle, we will find the squares of the distances from these points to the center $Q(1; -3)$ of the circle:
$$
\begin{gathered}
Q A^{2}=(0-1)^{2}+(0+3)^{2}=1025, \\
Q D^{2}=(2-1)^{2}+(-1+3)^{2}=5<25
\end{gathered}
$$
Ther... | Points\A\\D\\located\inside\the\circle,\point\B\is\on\the\circle,\\point\C\is\outside\the\circle | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,998 |
Petya has several coins in his pocket. If Petya randomly pulls out 3 coins from his pocket, there will definitely be a "1 ruble" coin among them. If Petya randomly pulls out 4 coins from his pocket, there will definitely be a "2 ruble" coin among them. Petya pulled out 5 coins from his pocket. Name these coins.
# | Since among any three coins there is definitely a coin of "1 ruble", it means that there are no more than two coins of other denominations. That is, all Petya's coins, except possibly two, are "1 ruble" coins. Since among any four coins there is definitely a coin of "2 rubles", it means that coins other than "2 rubles"... | "1ruble","1ruble","1ruble","2rubles","2rubles" | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,999 |
Akopyan A.V. Do there exist two quadrilaterals such that the sides of the first are less than the corresponding sides of the second, while the corresponding diagonals are greater? | For example, consider square $A B C D$ and quadrilateral $A_{1} B_{1} C_{1} D_{1}$, positioned as shown in the figure. Then the length of the diagonals of the square is 4, and the lengths of the diagonals of quadrilateral $A_{1} B_{1} C_{1} D_{1}$ are 2. Additionally, the lengths of the corresponding sides of quadrilat... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,000 |
Authors: Kazitsina T.V., Frenkin B.R.
Does there exist an arithmetic progression of 2011 natural numbers in which the number of numbers divisible by 8 is less than the number of numbers divisible by 9, and the latter, in turn, is less than the number of numbers divisible by 10? | Consider an arithmetic progression with the initial term 10 and the common difference 40:
$$
10,50,90,130,170,210,250,290, \ldots
$$
The initial term is not divisible by 8, while the common difference is. Therefore, no term in the sequence is divisible by 8. From the given segment of the sequence, it is clear that so... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,001 |
When two residents of Flower City meet, one gives the other a 10-kopeck coin, and the other gives him 2 coins of 5 kopecks. Could it happen that over the course of a day, each of the 1990 residents of the city gave exactly 10 coins? | No. The total number of coins given should be divisible by 3. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,002 |
$2+$
Beads move along a wire circle with the same angular velocity, some clockwise, some counterclockwise. Upon collision, two beads scatter in opposite directions with their previous speeds. Prove that at some point, the initial arrangement of the beads will repeat. | Can we consider the collision to be just a re-labeling of the beads.
## Solution
We will assume that at the moment of collision, the beads continue their motion, "not noticing each other." Thus, at the moment of collision, we simply re-label the beads. It is clear that after the time it takes for one bead to complete... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,003 |
$2+$
In a 2000*2000 square, numbers are arranged such that in any 2*2 square, the sum of the number in the top-left corner and the number in the bottom-right corner is equal to the sum of the number in the top-right corner and the number in the bottom-left corner. Prove that the sum of the numbers in the top-left and ... | Solve the equations corresponding to all $2 * 2$ squares.
## Solution
For each $2 * 2$ square, consider the equation $\mathrm{a}+\mathrm{c}=\mathrm{b}+\mathrm{d}$, where a and c are the numbers in the top-left and bottom-right corners of the square, and b and d are the numbers in the bottom-left and top-right corners... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,004 |
[ Dirichlet's Principle (continued).]
Given 8 different natural numbers, not greater than 15. Prove that among their positive pairwise differences there are three identical.
# | There can be 14 different differences - from 1 to 14 - these are the 14 cells in which we will place the rabbits. Who will be our rabbits? They, of course, must be the differences between pairs of the given natural numbers. However, there are 28 pairs, and they can be placed in 14 cells such that exactly two "rabbits" ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,009 |
[ Dirichlet's Principle (continued).]
10 schoolchildren at the Olympiad solved 35 problems, and it is known that among them there are schoolchildren who solved exactly one problem, schoolchildren who solved exactly two problems, and schoolchildren who solved exactly three problems. Prove that there is a schoolchild wh... | From the conditions, it follows that there will be 7 schoolchildren who have solved $35-6=29$ problems. Since $29=4 \cdot 7+1$, there will be a schoolchild who has solved no fewer than five problems.
## Problem
Translate the above text into English, please retain the original text's line breaks and format, and outp... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,010 |
In a square with a side of 1 meter, 51 points were thrown. Prove that some three of them can be covered by a square with a side of 20 cm.
# | Let's divide our square into 25 squares with a side length of 20 cm. According to the generalized pigeonhole principle, at least three of the 51 thrown points will fall into one of them. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,011 |
[ Invariants $]
A change machine exchanges one coin for five others. Can it be used to exchange a metal ruble for 26 coins?
# | Cannot. Track the remainders modulo 4. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,012 |
[ Invariants and semi-invariants ]
Natural numbers from 1 to 6 are arranged in a circle in order. It is allowed to add 1 to any three consecutive numbers or subtract 1 from any three numbers, standing every other one. Is it possible to make all the numbers equal using several such operations?
# | Consider the sums of diametrically opposite numbers.
## Solution
Consider three sums - the first and fourth, the second and fifth, the third and sixth. Initially, these sums were equal to $1+4=5, 2+5=7, 3+6=9$. Notice that when the first operation described in the condition is performed, each of these three sums incr... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,013 |
Prove that the distance from the point $\left(x_{0}, y_{0}\right)$ to the line $a x+b y+c=0$ is $\frac{\left|a x_{0}+b y_{0}+c\right|}{\sqrt{a^{2}+b^{2}}}$. | If $a x_{1}+b y_{1}+c=0$ and $a x_{2}+b y_{2}+c=0$, then $a\left(x_{1}-x_{2}\right)+b\left(y_{1}-y_{2}\right)=0$. Therefore, the vector $(a, b)$ is perpendicular to the considered line. Hence, the perpendicular dropped from the point ( $x_{0}, y_{0}$ ) to the considered line consists of points with coordinates $\left(x... | \frac{|x_{0}+by_{0}+|}{\sqrt{^{2}+b^{2}}} | Geometry | proof | Yes | Yes | olympiads | false | 54,014 |
In the language of the Ancient Tribe, the alphabet consists of only two letters: "M" and "O". Two words are synonyms if one can be obtained from the other by excluding or adding the letter combinations "MO" and "OOMM", repeated in any order and any number of times. Are the words "OMM" and "MOO" synonyms in the language... | Notice that the difference between the number of letters M and O does not change when adding or removing allowed letter combinations.
## Solution
Notice that with each addition or removal of allowed letter combinations, the difference between the number of letters "M" and "O" in the word does not change - it is alway... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,016 |
[Coordinate method on the plane $]$
Form the equation of the line passing through the point $M(-3 ; 2)$ parallel to the line $2 x-3 y+4=0$.
# | Let's write the equation of the given line as $y=\frac{2}{3} x+\frac{4}{3}$. Since the required line is parallel to the given one, its slope is equal to the slope of the given line, i.e., $\frac{2}{3}$. Therefore, the equation of the required line is $y=\frac{2}{3} x+l$.
Since the point $M(-3 ; 2)$ lies on this line, ... | 2x-3y+12=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,017 |
[ $\underline{\text { invariants }}$]
In the alphabet of the UYU tribe's language, there are only two letters: U and Y. It is known that the meaning of a word does not change if the adjacent lettersUY are removed and if the letter combinations YU or UUYU are added at any position in the word.
Can we assert that the w... | Note that with any allowed operation, the difference between the number of letters У and the number of letters Ы in the word does not change. But in the words УЫЫ and ЫУУ, these differences are different. Therefore, it is impossible to obtain the word ЫУУ from the word УЫЫ using allowed operations, and consequently, it... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,021 |
[ $[$ Invariants $]$
On the board, the numbers $1,2,3, \ldots, 19,20$ are written. It is allowed to erase any two numbers $a$ and $b$ and write the number $a+b-1$ instead.
What number can remain on the board after 19 such operations?
# | For any set of $n$ numbers on the board, consider the following quantity $X$: the sum of all numbers, decreased by $n$. It is not hard to verify that this is an invariant. In the set from the condition, $X=(1+2+\ldots+20)-20=190$. After 19 operations, when there is one number $p$ left on the board, $X=p-1$. Therefore, ... | 191 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,022 |
If a class of 30 people is seated in a movie theater, then in any case, at least two classmates will end up in the same row. If the same is done with a class of 26 people, then at least three rows will be empty. How many rows are there in the hall?
# | The first condition means that there are no more than 29 rows in the hall. Indeed, if the number of rows were not less than 30, then obviously, a class of 30 students could be seated with no more than one student per row. The second condition means that the number of rows in the hall is not less than 29. Indeed, if the... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,023 |
In a class of 25 people, it is known that among any three of them, there are two friends. Prove that there is a student who has at least 12 friends.
# | Let's consider two students in the class who do not get along with each other. (If there are no such students, then all students in the class are friends with each other, meaning each student has 24 friends, and the problem is solved.) Let these two be Vasya and Petya. Then, out of the remaining 23 students, each is fr... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,024 |
$3-$
Prove that on the coordinate plane, a circle can be drawn that contains exactly $n$ integer points inside it. | Draw a small circle and inflate it.
## Solution
We can choose a point $A$ on the plane such that on any circle centered at $A$, there is no more than one integer point. For this, it is sufficient that all distances from $A$ to integer points are distinct. Note that the perpendicular bisector of a segment connecting i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,025 |
Let $\mathrm{d}_{\mathrm{k}}$ denote the number of houses in Moscow in which at least $\mathrm{k}$ residents live, and let $\mathrm{c}_{\mathrm{m}}$ denote the number of residents in the m-th largest house by population. Prove the equality $\mathrm{c}_{1}+\mathrm{c}_{2}+\mathrm{c}_{3}+\ldots=\mathrm{d}_{1}+\mathrm{d}_{... | Calculate the number of residents in Moscow in two ways.
## Solution
It is clear that $\mathrm{c}_{1}+\mathrm{c}_{2}+\mathrm{c}_{3}+\ldots$ is equal to the number of residents in Moscow (we simply sum the number of residents in all houses). Now, for convenience, let's imagine that in a house where $\mathrm{k}$ people... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,026 |
[ Auxiliary coloring (other).]
On a grid paper, 2000 cells are marked arbitrarily. Prove that among them, it is always possible to select at least 500 cells that do not touch each other pairwise (cells are considered touching if they share at least one vertex). | Color the cells in 4 colors so that no two cells of the same color touch.
## Solution
Consider some (infinite) row and all rows that are one cell away from it. We will color the cells of each of these rows alternately red and yellow. Consider the remaining rows. We will color the cells of each of these rows alternate... | 500 | Combinatorics | proof | Yes | Yes | olympiads | false | 54,027 |
$3-$Prove that if $a, b, c$ are odd numbers, then at least one of the numbers $a b-1, b c-1, c a-1$ is divisible by 4.
# | Consider the remainders of the division of the given numbers by 4 and use the Pigeonhole Principle.
## Solution
Since the numbers $a, b, c$ are odd, when divided by 4, they can give remainders of 1 or 3. Therefore, at least two of them have the same remainder. Let these be $a, b$ with a remainder of 3 (other cases ar... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,028 |
Nikolai with his son and Peter with his son went fishing. Nikolai caught as many fish as his son, and Peter caught three times as many as his son. In total, 25 fish were caught. What is the name of Peter's son?
# | Either three or four people went fishing. The question of the problem should clarify the situation.
## Solution
There are three possibilities: 1) four different people went fishing - Nikolai, Petr, and their sons; 2)
Petr is Nikolai's son; 3) Nikolai is Petr's son.
In the first case, each pair of father and son cau... | Nikolai | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,029 |
There are many identical squares. In the vertices of each of them, the numbers 1, 2, 3, and 4 are written in random order. The squares are stacked, and the sum of the numbers that ended up in each of the four corners of the stack is written down. Can it be that
a) in each corner of the stack, the sum is 2004?
b) in e... | a) The sum of all numbers should be divisible by $1+2+3+4=10$. But 2004$\cdot$4 is not divisible by 10.
b) We can take 401 pairs of squares with such a number arrangement $(1,2,3,4)$ and ( $4,3,2,1$).
## Answer
## Problem | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,031 |
[ Completing the square. Sums of squares ]
In the expression $x^{6}+x^{4}+x A$, replace $A$ with a monomial so that a perfect square is obtained. Find as many solutions as possible
# | Since the given expression is a trinomial, it will become a perfect square in one of three cases.
1) $x^{6}$ and $x^{4}$ are squares of two numbers, and $x A$ is twice the product of these numbers. We get $x^{6}+x^{4} \pm 2 x^{5}=\left(x^{3} \pm\right.$ $\left.x^{2}\right)^{2}$, that is, $A= \pm 2 x^{4}$.
2) $x^{4}$ a... | \2x^{4},\frac{1}{4}x^{7},\frac{1}{4}x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,032 |
On the plane, there are 7 lines, no two of which are parallel. Prove that there will be two of them, the angle between which is less than $26^{\circ}$. | Let's take an arbitrary point $A$ on the plane and draw 7 lines through it, parallel to the lines of the given system. The angles between the constructed lines will be equal to the corresponding angles between the lines of the original system. But through point $A$ we have 7 lines that divide the angle of $360^{\circ}$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,033 |
Nazarov $\Phi$.
Positive numbers $a, b, c$ are such that $a \geq b \geq c$ and $a+b+c \leq 1$. Prove that $a^{2}+3 b^{2}+5 c^{2} \leq 1$.
# | $1 \geq(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c \geq a^{2}+b^{2}+c^{2}+2 b^{2}+2 c^{2}+2 c^{2}=a^{2}+3 b^{2}+5 c^{2}$.
Send a comment
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
$1 \geq(a+b+c)^{2}=a^{2}+b^{2}+c^{2}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 54,034 |
[The Pigeonhole Principle (continued).]
In a photo studio, 20 birds flew in - 8 sparrows, 7 wagtails, and 5 woodpeckers. Each time the photographer clicks the camera shutter, one of the birds flies away (permanently). How many shots can the photographer take to be sure: he will have at least four birds of one species ... | 8 snapshots are dangerous: in this time, 3 woodpeckers and 5 wagtails may fly away, leaving only 2 of each.
Let's show that 7 snapshots can be made. Then, in the studio, there will be $20-7=13$ birds left. This means that the number of birds of one species is at least $13: 3$, that is, at least 5. On the other hand, t... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,035 |
#
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | According to the formula for the distance between two points $A B^{2}=(-6-3)^{2}+(-2-5)^{2}=130, A C^{2}=(0-3)^{2}+(-6-5)^{2}=$ 130. Therefore, $A B=A C$. Hence, triangle $A B C-$ is isosceles. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,037 |
Given a point $M(x ; y)$. Find the coordinates of the point symmetric to point $M$ with respect to: a) the OX axis; b) the OY axis.
# | Let the point $M^{\prime}\left(x^{\prime} ; y\right)$ be symmetric to the point $M(x ; y)$ with respect to the axis $O X$. Then the points $M$ and $M^{\prime}$ lie on opposite sides of the axis $O X$ on a line perpendicular to this axis, at equal distances from the point $P$ where this line intersects the axis $O X$. T... | (x,-y);(-x,y) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,038 |
[Coordinate method on the plane]
Find the distance between point $A(1,7)$ and the intersection point of the lines $x-y-1=0$ and $x+3 y-12=0$.
# | Solving the system, we find the point of intersection $B\left({ }^{15} / 4,{ }^{11} / 4\right)$ of the given lines. Using the formula for the distance between two points
$A B^{2}=\left({ }^{15} / 4-1\right)^{2}+\left({ }^{11 / 4}-1\right)^{2}=(11 / 4)^{2}+\left({ }^{17 / 4}\right)^{2}=1 / 16+(121+289)=410 / 16$.
## A... | \frac{\sqrt{410}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,039 |
Given points $A(-2 ; 2), B(-2 ;-2)$ and $C(6 ; 6)$. Form the equations of the lines on which the sides of triangle $A B C$ lie.
# | Since the abscissas of points $A(-2 ; 2)$ and $B(-2 ;-2)$ are equal, the equation of the line $A B$ is $x=-2$, or $x+2=0$.
If $x_{1} \neq x_{2}$ and $y_{1} \neq y_{2}$, then the equation of the line passing through points $M_{1}\left(x_{1} ; y_{1}\right)$ and $M_{2}\left(x_{2} ; y_{2}\right)$ can be written as
$$
\fr... | AB:x+2=0,AC:x-2y+6=0,BC:x-0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,040 |
$\left[\begin{array}{l}{[\text { Method of coordinates on the plane }} \\ {[\quad \underline{\text { Circles (other) }}]}\end{array}\right]$
A circle with its center at point $M(3 ; 1)$ passes through the origin. Formulate the equation of the circle. | A circle with radius $R$ and center at point $A(a ; b)$ has the equation
$$
(x-a)^{2}+(y-b)^{2}=R^{2}
$$
In our case, $a=3, b=1$. Since the point $O(0 ; 0)$ lies on the circle $(x-3)^{2}+(y-1)^{2}=R^{2}$, the coordinates of this point satisfy the equation of the circle. From the equality $(0-3)^{2}+(0-1)^{2}=R^{2}$, ... | (x-3)^{2}+(y-1)^{2}=10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,041 |
[ Examples and counterexamples. Constructions ]
There are five links in a chain, each with three rings.
What is the minimum number of rings that need to be unbuckled and buckled to connect these links into one chain?
# | Let's unravel all the rings of one link. The remaining four links will be connected by the three unlinked rings.
## Answer
Three rings.
In this case, the first and last solutions correspond to the scenario where everyone drank only milk or only coffee, while the second solution corresponds to the scenario where they... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,042 |
Two pirates were playing for gold coins. First, the first pirate lost half of his coins (gave them to the second), then the second pirate lost half of his, then the first pirate lost half of his again. As a result, the first pirate ended up with 15 coins, and the second with 33. How many coins did the first pirate have... | Let's try to trace "from the end" how many coins each pirate had after each game. In the last game, the first pirate lost half of his coins to the second pirate, after which he was left with 15 coins. But this is exactly the amount he just gave to the second pirate! This means that before this, the first pirate had $15... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,044 |
3[ Identical Transformations $]$
Authors: Proizvolov V.V., Senderov V.A.
Find some four pairwise distinct natural numbers $a, b, c, d$, for which the numbers $a^{2}+2 c d+b^{2}$ and $c^{2}+2 a b+d^{2}$ are perfect squares. | It is sufficient to find such numbers that $a b=c d$. Then $a^{2}+2 c d+b^{2}=a^{2}+2 a b+b^{2}=(a+b)^{2}, c^{2}+2 a b+d^{2}=(c+$ $d)^{2}$. For this, we need to find a number $n$ that can be factored into the product of two factors in different ways.
For example, $6=1 \cdot 6=2 \cdot 3$.
## Answer
For example, $a=1,... | =1,b=6,=2,=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,045 |
[ Processes and Operations ]
In one step, you can replace any letter in any of these words with any other letter (for example, in one step, you can get the word ЗКНОЗА from the word ЗАНОЗА. What is the minimum number of steps needed to make all the words the same (nonsensical words are allowed)?
# | After all the letter replacements in each column, the letters should become the same. The number of replacements will be the smallest if the most frequent letter (any of them if there are several) is kept in each column. For example, in the first column, letters 3 or К can be kept, both requiring four replacements. The... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,046 |
Find the perimeter of triangle $ABC$, if the coordinates of its vertices $A(-3,5), B(3,-3)$ and point $M(6$, 1), which is the midpoint of side $BC$, are known.
# | Use the formulas for the coordinates of the midpoint of a segment and the formulas for the distance between two points.
## Solution
Let $(x, y)$ be the coordinates of vertex $C$. According to the condition, $1 / 2(x+3)=6, 1 / 2(y-3)=1$, from which $x=9, y=5$.
Using the distance formula between two points, we find th... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,047 |
Find the length of the chord that the line $y=3 x$ cuts off from the circle $(x+1)^{2}+(y-2)^{2}=25$.
# | Let's find the coordinates of the points of intersection \(A\left(x_{1} ; y_{1}\right)\) and \(B\left(x_{2} ; y_{2}\right)\) of the given line and circle. For this, we will solve the system of equations
\[
\left\{\begin{array}{l}
y=3 x \\
(x+1)^{2}+(y-2)^{2}=25
\end{array}\right.
\]
We obtain: \(x_{1}=-1, y_{1}=-3, x... | 3\sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,048 |
Do there exist natural numbers $m$ and $n$ for which the equality is true: $\left(-2 a^{n} b^{n}\right)^{m}+\left(3 a^{m} b^{m}\right)^{n}=a^{6} b^{6}$? | Indeed, for $m=3, n=2,\left(-2 a^{n} b^{n}\right)^{m}+\left(3 a^{m} b^{m}\right)^{n}+\left(-2 a^{2} b^{2}\right)^{3}+\left(3 a^{3} b^{3}\right)^{2}=-8 a^{6} b^{6}+9 a^{6} b^{6}=a^{6} b^{6}$.
## Answer
There exist. | ^{6}b^{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,049 |
Osipov I.i.
In a cinema, there are seven rows with 10 seats each. A group of 50 children went to the morning session and then to the evening session.
Prove that there will be two children who sat in the same row during the morning session and also sat in the same row during the evening session. | If in each of the seven rows there were no more than seven children, then there would be no more than 49 children in total.
Therefore, there is a row in which no fewer than 8 children were sitting. Since there are only seven rows, at least two of these children will end up in the same row during the evening session. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,050 |
$:$ Folkiore
While tidying up the children's room before the guests arrived, mom found nine socks. Among any four of these socks, at least two belonged to one child, and among any five, no more than three had the same owner. How many children could there have been, and how many socks could have belonged to each child? | Since among any four socks, at least two belonged to one child, there are no more than three children. No child can own more than three socks (otherwise, there would be five socks, among which more than three belong to one owner).
The mother found a total of nine socks, so there cannot be fewer than three children. Th... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,052 |
In the center of a $3 \times 3 \times 3$ cube, there is a beetle. Prove that it cannot visit all $1 \times 1 \times 1$ cubes exactly once by crawling over the edges.
# | Paint the cube in a checkerboard pattern. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,053 |
A road 1 km long is fully illuminated by street lamps, with each lamp lighting a section of the road 1 m long. What is the maximum number of lamps that can be on the road, given that after turning off any lamp, the road will no longer be fully illuminated? | If the segments illuminated by the n-th and (n+2)-th street lamps intersect, then the (n+1)-th street lamp can be turned off.
## Solution
We will number the street lamps with natural numbers in the order along the road. If the segments illuminated by the n-th and (n+2)-th street lamps intersect, then the (n+1)-th str... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,054 |
[ Coordinate method on the plane ] [ Vectors help solve the problem. ]
Prove that the points $A(-1; -2), B(2; -1)$, and $C(8; 1)$ lie on the same line. | Prove that $A B+B C=B C$ (or establish the collinearity of vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$).
## Solution
## First method.
Using the distance formula between two points,
$$
\begin{gathered}
A B=\sqrt{(2-(-1))^{2}+(-1-(-2))^{2}}=\sqrt{9+1}=\sqrt{10} \\
A C=\sqrt{(8-(-1))^{2}+(1-(-2))^{2}}=\s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,055 |
3Given points $A(-2 ; 1), B(2 ; 5)$ and $C(4 ;-1)$. Point $D$ lies on the extension of median $A M$ beyond point $M$, and quadrilateral $A B D C$ is a parallelogram. Find the coordinates of point $D$. | ## First method.
The coordinates of point $M\left(x_{0} ; y_{0}\right)$ are the arithmetic means of the corresponding coordinates of the endpoints of segment $B C$, i.e.,
$$
x_{0}=\frac{2+4}{2}=3, y_{0}=\frac{5-1}{2}=2
$$
Since the diagonals of a parallelogram are bisected by the point of intersection, $M\left(x_{0}... | (8;3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,056 |
Given points $A(0 ; 0), B(4 ; 0)$ and $C(0 ; 6)$. Formulate the equation of the circle circumscribed around triangle $A B C$.
# | Since the abscissas of points $A$ and $C$ are 0, these points lie on the line $x=0$, i.e., on the $O Y$ axis. Since the ordinates of points $A$ and $B$ are 0, these points lie on the line $y=0$, i.e., on the $O Y$ axis. Therefore, triangle $A B C$ is a right triangle, $\angle B A C=90^{\circ}$. Hence, the center of its... | (x-2)^{2}+(y-3)^{2}=13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,057 |
[ Coordinate Method on the Plane
Prove that the lines $y=k_{1} x+l_{1}$ and $y=k_{2} x+l_{2}$ are parallel if and only if $k_{1}=k_{2}$ and $l_{1} \neq l_{2}$.
# | Lines are parallel if and only if the system $\left\{\begin{array}{l}y=k_{1} x+l_{1}, \\ y=k_{2} x+l_{2}\end{array}\right.$ has no solutions.
Send a comment | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,058 |
[Properties and characteristics of a parallelogram] [Vectors help solve tasks]
Given points $A(-6; -1), B(1; 2)$, and $C(-3; -2)$. Find the coordinates of vertex $M$ of parallelogram $A B M C$. | ## First method.
The coordinates of the midpoint $K\left(x_{0} ; y_{0}\right)$ of the diagonal $B C$ of the parallelogram $A B M C$ are the arithmetic means of the corresponding coordinates of the endpoints of the segment $B C$, i.e.,
$$
x_{0}=\frac{1-3}{2}=-1, y_{0}=\frac{2-2}{2}=0
$$
Since the diagonals of a paral... | M(4;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,059 |
Prove that among any six people there are either three who are pairwise acquainted or three who are pairwise unacquainted.
# | Among these people, this person has either no fewer than three acquaintances or no fewer than three strangers among the other five. Let's consider, for example, the first case. Among these three people, there are either two who know each other - in which case they, together with the person we initially chose, form the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,062 |
6,7 |
On six fir trees, six goldfinches are sitting, one on each tree. The trees grow in a row with intervals of 10 meters. If one goldfinch flies from one tree to another, then another goldfinch must necessarily fly the same distance but in the opposite direction.
a) Can all the goldfinches gather on one tree?
b) A... | a) First method. Let's number the trees from 1 to 6 in order. Suppose each chick gets a number equal to the number of the tree it is sitting on (at the moment). Then the sum of the numbers of the chicks is an invariant. Initially, it is equal to \(1+2+\ldots+6=21\). Since 21 is not divisible by 6, the chicks will not b... | ) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,064 |
The numbers 1, 2, 3, ..., 1989 are written on a board. It is allowed to erase any two numbers and write their difference instead.
Can it be achieved that all numbers on the board become zeros?
# | Notice that the parity of the sum of the numbers on the board is unchanged.
## Otvet
It is impossible. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,065 |
[ $\underline{\text { Invariants }}$ ]
In an $m \times n$ table, numbers are arranged such that the sum of the numbers in any row or column is 1. Prove that $m=n$.
Note. Strangely enough, in some sense, this is also an invariant problem.
# | The sum of the numbers in the table does not depend on the method of its calculation. It is in this sense that this is a problem of invariant. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,066 |
3 [ Divisibility of numbers. General properties ] $[$ Decimal number system $]$
Bus tickets have numbers from 000000 to 999999. A ticket is called lucky if the sum of the first three digits of its number equals the sum of the last three digits. Prove that:
a) the number of all lucky tickets is even
b) the sum of the... | a) Let's specify two ways to pair happy tickets.
First method. We will swap the first three digits with the last three digits in the ticket number. The obtained ticket will be paired with the original one (for example, the pair for ticket 239671 will be 671239). This way, we have paired all tickets except those that a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,068 |
A square area of size $100 \times 100$ is paved with $1 \times 1$ square tiles of four colors: white, red, black, and gray - such that no two tiles of the same color touch each other (that is, they do not share a side or a vertex). How many red tiles can there be? | We will divide the area into 2,500 squares of $2 \times 2$, each of which consists of four tiles sharing a common vertex. Therefore, each of these squares can contain no more than one red tile, and thus the number of red tiles cannot exceed 2,500. The same is true for tiles of other colors, meaning there are exactly 2,... | 2500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,069 |
3 [ Point Systems $]
In the plane, four points are marked. Prove that they can be divided into two groups such that these groups of points cannot be separated from each other by any straight line.
# | Let's consider separately all possible ways of arranging four points on a plane.
1) All four points lie on the same straight line. Let's denote them as A, B, C, and D in the order in which they are located on the line. Then the first group should include points A and C, and the second group - B and D.
2) Only three of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,070 |
a) From a regular 8 by 8 chessboard, the squares c5 and g2 have been cut out. Can the remaining part be tiled with dominoes 1 by 2?
b) The same question, if the squares c6 and g2 have been cut out. | a) Yes, it is possible. It is easy to provide an example of such a tiling.
b) No, it is not possible. Indeed, we have cut out two white cells; but each domino covers one white and one black cell, so if we were to tile the entire board, the number of white and black cells covered would be the same. However, there are 3... | )Yes;b)no | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,071 |
[ Examples and counterexamples. Constructions ]
Can the numbers $1,2, \ldots, 50$ be placed at the vertices and midpoints of the sides of a regular 25-gon so that the sum of the three numbers at the ends and midpoint of each side is the same for all sides? | Arrange the numbers $1,2, \ldots, 25$ in order at the midpoints of the sides.
## Solution
Let the vertices of the 25-gon be $A_{1}, A_{2}, \ldots, A_{25}$ in a clockwise direction. Place the numbers $1,2, \ldots, 25$ in order at the midpoints of the sides: the number 1 is at the midpoint of side $A_{1} A_{2}$, the nu... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,072 |
[ Combinations and Permutations ] [ Directed Graphs ]
In the discussion, 15 deputies participated. Each of them, in their speech, criticized exactly $k$ of the remaining 14 deputies.
For what smallest $k$ can we assert that there will be two deputies who criticized each other? | Consider a directed graph where the vertices correspond to deputies, and an edge leading from $A$ to $B$ means that deputy $A$ has criticized deputy $B$.
## Solution
If each deputy has criticized 8 others, then the number of edges in the graph is $15 \cdot 8 = 120$, which is greater than the number of pairs $C_{15}^{... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,073 |
3 [ Inequality Problems. Case Analysis ]
Can 50 stones, weighing $370, 372, \ldots, 468$ kg, be transported from a quarry on seven three-ton trucks? | Some three-ton truck will have to carry 8 stones.
## Solution
Even the eight lightest stones weigh $370+372+\ldots+384=3016$ kg, which is more than 3 tons. Therefore, on one three-ton truck, you can carry
no more than seven, and on seven three-ton trucks - no more than 49 stones.
## Answer
It is impossible. | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 54,075 |
[ Pairing and Grouping; Bijections ]
a) On the segment $[0,1]$, a set $M$ is defined as the union of several segments such that the distance between any two points in $M$ is not equal to $1 / 10$. Prove that the sum of the lengths of the segments that make up $M$ is no more than $1 / 2$.
b) Is the same statement true... | . When shifted by ${ }^{1 / 10}$, a point of set $M$ cannot transition to another point of set $M$.
## Solution
a) Let's divide the interval $[0,1]$ into five intervals of length $1 / 5$. Let $A=M \cap[0,1 / 10], B=M \cap\left[1 / 10, { }^{1 / 5}\right]$. Denote by $A^{\prime}$ the set obtained from $A$ by shifting i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,076 |
| [ Divisibility of numbers. General properties $]$ |
Let $p-$ be a prime number greater than 2, and $m / n=1+1 / 2+1 / 3+\ldots+1 / p-1$. Prove that $m$ is divisible by $p$. | Group the first term with the last, the second with the second to last, and so on.
## Solution
It is sufficient to prove this statement for the irreducible fraction ${ }^{m} / n$. In the sum $1+1 / 2+1 / 3+\ldots+1 / p-1$, group the first term with the last, the second with the second to last, and so on. We get ${ }^... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,077 |
$\left[\begin{array}{l}\text { Coordinate method in space }] \\ {[\quad \underline{\text { equation of a plane }}]}\end{array}\right]$
Form the equation of the plane passing through the points $A(-3 ; 0 ; 1), B(2 ; 1 ;-1)$ and $C(-2 ; 2 ; 0)$. | Let's find the coordinates of vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(2-(-3) ; 1-0 ;-1-1)=(5 ; 1 ;-2) \\
& \overrightarrow{A C}=(-2-(-3) ; 2-0 ; 0-1)=(1 ; 2 ;-1)
\end{aligned}
$$
Let $\vec{n}=(a ; b ; c)$ be a non-zero vector perpendicular to the desired p... | x+y+30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,078 |
Find the distance from the point $D(1 ; 3 ; 2)$ to the plane passing through the points $A(-3 ; 0 ; 1), B(2 ; 1 ;-1)$ and $C(-2 ; 2 ; 0)$.
# | Let's find the coordinates of vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(2-(-3) ; 1-0 ;-1-1)=(5 ; 1 ;-2) \\
& \overrightarrow{A C}=(-2-(-3) ; 2-0 ; 0-1)=(1 ; 2 ;-1)
\end{aligned}
$$
Let $\vec{n}=(a ; b ; c)$ be a non-zero vector perpendicular to the desired p... | \frac{10}{\sqrt{11}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,079 |
\left.\begin{array}{l}{[\text { Coordinate Method in Space }]} \\ {[\text { Parametric Equations of a Line }]\end{array}\right]
Derive the parametric equations of the line of intersection of the planes $2 x-y-3 z+5=0$ and $x+y-2=0$. | Let $x=t$ and solve for $y$ and $z$ the system of equations
$$
\left\{\begin{array}{l}
y+3 z=2 t+5 \\
y=-t+2
\end{array}\right.
$$
We get:
$$
\left\{\begin{array}{l}
x=t \\
y=2-t \\
z=1+t
\end{array}\right.
$$
## Answer
$x=t, y=2-t, z=1+t$ | ,2-,1+ | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,080 |
Find the angle between the line of intersection of the planes $2 x-y-3 z+5=0$ and $x+y-2=0$ and the plane passing through the points $M(-2 ; 0 ; 3), N(0 ; 2 ; 2)$ and $K(3 ;-3 ; 1)$.
# | First, let's find the direction vector of the line of intersection of the given planes. For this, let $x=t$ and solve the system of equations for $y$ and $z$:
$$
\left\{\begin{array}{l}
y+3 z=2 t+5 \\
y=-t+2
\end{array}\right.
$$
We get
$$
x=t, y=2-t, z=1+t
$$
Thus, the vector $\overrightarrow{\boldsymbol{m}}=(1 ;-... | \arcsin\frac{22}{3\sqrt{102}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,082 |
[Coordinate method in space] [equation of a plane]
Given points $A(1 ; 0 ; 1), B(-2 ; 2 ; 1), C(2 ; 0 ; 3)$. Form the equation of the plane $ABC$.
# | Let's find the coordinates of vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(-2-1 ; 2-0 ; 1-1)=(-3 ; 2 ; 0) \\
& \overrightarrow{A C}=(2-1 ; 0-0 ; 3-1)=A C(1 ; 0 ; 2)
\end{aligned}
$$
Let $\vec{n}=(a ; b ; c)$ be a non-zero vector perpendicular to the desired pla... | 2x+3y-z-1=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,083 |
\[
\left[\begin{array}{l}
\text { Coordinate method in space } \\
{[\text { equation of a plane }]}
\end{array}\right]
\]
Given points \( A(1 ; 0 ; 1), B(-2 ; 2 ; 1), C(2 ; 0 ; 3) \) and \( D(0 ; 4 ;-2) \). Form the equation of the plane passing through point \( B \) and parallel to the plane \( ABC \). | Let's find the coordinates of vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(-2-1 ; 2-0 ; 1-1)=(-3 ; 2 ; 0) \\
& \overrightarrow{A C}=(2-1 ; 0-0 ; 3-1)=(1 ; 0 ; 2)
\end{aligned}
$$
Let $\vec{n}=(a ; b ; c)$ be a non-zero vector perpendicular to the plane $A B C$.... | 2x+3y-z-14=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,084 |
\left.\begin{array}{l}{[\text { Method of coordinates in space }} \\ \text { [ Parametric equations of a line }]\end{array}\right]
Given points $A(-3 ; 0 ; 1)$ and $D(1 ; 3 ; 2)$. Write the parametric equations of the line $A D$. | Point $M(x ; y ; z)$ lies on the line $A D$ if and only if the vector $\overrightarrow{A M}=(x+3 ; y ; z-1)$ is collinear with the vector
$$
\overrightarrow{A D}=(1-(-3) ; 3-0 ; 2-1)=(4 ; 3 ; 1)
$$
The vector $\overrightarrow{A M}$ is collinear with the vector $\overrightarrow{A D}$ if and only if $\overrightarrow{A ... | {\begin{pmatrix}-3+4\\3\\1+\end{pmatrix}.} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,086 |
10,11
In the cube $A B C D A 1 B 1 C 1 D 1$, where $A A 1, B B 1, C C 1$ and $D D 1$ are parallel edges, the plane $P$ passes through point $D$ and the midpoints of edges $A 1 D 1$ and $C 1 D 1$. Find the distance from the midpoint of edge $A A 1$ to the plane $P$, if the edge of the cube is 2. | Let $M, N, K$ and $L$ be the midpoints of edges $A1D1, C1D1$, $AA1$ and $CC1$ respectively (Fig.1). The line $KL$ is parallel to the line $MN$, so the line $KL$ is parallel to the plane $P$. Moreover, the line $KL$ passes through the center $O$ of the cube $ABCD A1B1C1D1$. Therefore, the distance from the midpoint $K$ ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,087 |
48 blacksmiths need to shoe 60 horses. Each blacksmith spends 5 minutes on one horseshoe. What is the least amount of time they should spend on the work? (Note, a horse cannot stand on two legs.)
# | Note that it is impossible to shoe all the horses in less than 25 minutes. Why?
## Solution
Let's show how to proceed. First, 48 blacksmiths take 48 horses and shoe each one with one shoe, which takes 5 minutes. At this point, 48 horses have one shoe, and 12 have none. Then, 12 blacksmiths shoe the horses that still ... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,088 |
On the b8 cell of the chessboard, the number -1 is written, and on all other cells, the number 1. It is allowed to simultaneously change the sign in all cells of one column or one row. Prove that no matter how many times we do this, it is impossible to achieve that all numbers in the table become positive.
# | Note: When changing the signs in a row or column, the product of all numbers in the table does not change.
## Solution
Since we change the signs each time in eight cells, the product of all numbers in the table does not change. And since it was negative at the beginning, it cannot become positive.
## [ Mean values $... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 54,089 |
Find all solutions of the system of equations: $(x+y)^{3}=z,(y+z)^{3}=x,(z+x)^{3}=y$.
# | Subtracting the second equation from the first, we get: $(x-z)\left((x+y)^{2}+(x+y)(y+z)+(y+z)^{2}\right)=-(x-z)$.
Since an expression of the form $a^{2}+a b+b^{2}$ cannot be negative, $x=z$. Similarly, $x=y$.
It remains to solve the equation $(2 x)^{3}=x$.
## Answer
$(0,0,0),\left( \pm \frac{\sqrt{2}}{4}, \pm \fra... | (0,0,0),(\\frac{\sqrt{2}}{4},\\frac{\sqrt{2}}{4},\\frac{\sqrt{2}}{4}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,092 |
Brisquin I.
In the lower left corner of an $8 \times 8$ chessboard, nine chips are placed in the form of a $3 \times 3$ square. A chip can jump to a free square over an adjacent chip, i.e., it can reflect symmetrically relative to the center of the adjacent chip (jumps can be made vertically, horizontally, and diagona... | Let's paint the board with stripes: odd horizontals - white, even ones - black. When making a move, the chip does not change the color of the field it is on. It remains to note that in the initial arrangement, the chips occupy six white fields and three black ones, while in the final arrangement - three black and six w... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,095 |
Folklore
In the center of a square pool is a boy, and at one of the corners on the shore stands a teacher. The boy's maximum speed in the water is three times slower than the teacher's maximum speed on land. The teacher cannot swim, but on the shore, the boy can run faster than the teacher. Can the boy escape? | Let's assume that the side of the square $ABCD$ is equal to 2. Let the teacher be at vertex $C$. The boy swims to $A$ until the teacher runs to the midpoint of side $CD(CB)$. After that, the boy swims perpendicular to $AB(AD)$. The boy has to swim a distance of no more than 1 to reach the shore, while the teacher will ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,096 |
Tolpygo A.K.
On a $4 \times 4$ chessboard, there is a figure called a "flying rook," which moves like a regular rook but cannot move to a field adjacent to the previous one in a single move. Can it, in 16 moves, visit every field exactly once and return to the starting field? | In the figure, the squares of the board are numbered in the required order.
| 12 | 16 | 9 | 13 |
| :---: | :---: | :---: | :---: |
| 6 | 2 | 7 | 3 |
| 11 | 15 | 10 | 14 |
| 5 | 1 | 8 | 4 |
## Answer
It is possible. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,098 |
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