problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4. The integers from 1 to $n$ are written in a row. Below them, the same numbers are written in a different order. Can it happen that the sum of each number and the one written below it is a perfect square a) for $n=9$, b) for $n=11$, c) for $n=1996$. | 4. a) The sum of the number 9 and the number written below it is between 10 and 18. Since there is only one perfect square in this range, the number below 9 must be 7. Similarly, the number 7 must be written above the number 9. It is also verified that the numbers below
4, 5, and 6 should be 5, 4, and 3, respectively.... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,819 |
4. In the corner of a chessboard of size $m \times n$ fields, a rook is placed. Two players take turns moving it vertically or horizontally by any number of fields; it is not allowed for the rook to land on or pass through a field it has already visited (or passed through). The player who cannot make a move loses. Who ... | 4. The case $m=n=1$ is obvious (the first player has nowhere to move). For definiteness, let's assume that the board consists of $m$ verticals and $n$ horizontals, with $m \geqslant n$, and the rook, initially, stands in the top-left corner.
It turns out that the first player can always make the longest moves. We will... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,820 |
4. Prove that there are infinitely many natural numbers $n$ such that the number $n$ can be represented as the sum of the squares of two natural numbers, while the numbers $n-1$ and $n+1$ cannot. | 4. First solution. First of all, note that the squares of integers, when divided by 4, can give remainders 0 or 1, and when divided by 9, they can give remainders 0, 1, 4, or 7 (see comment). Therefore, numbers of the form \(4k+3\) and \(9k+3\) cannot be represented as the sum of squares of two integers.
For any natur... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,821 |
5. In the corner of a chessboard of size $n \times n$ fields, a rook is placed. For which $n$, alternating between horizontal and vertical moves, can it visit all the fields of the board and return to its starting position in $n^{2}$ moves? (Only the fields where the rook stops are considered, not those it flies over d... | 5. For even $n$ - see Fig. 73. The rook first visits all the cells of the first two rows, then all the cells of the 3rd and 4th rows, and so on.
Let's show that for odd $n$ this is impossible. Indeed, consider any row other than the first. When the rook reaches this row, it needs to move to another cell in the same ro... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,822 |
5. Let $A$ and $B$ be points lying on a circle. They divide the circle into two arcs. Find the geometric locus of the midpoints of all possible chords whose endpoints lie on different arcs $AB$.
Comment. The condition does not clearly specify whether chords passing through points $A$ and $B$ are allowed. We will assum... | 5. First, let's formulate two simple statements:
1) The segment connecting the midpoint of a chord with the center of the circle is perpendicular to this chord;
2) The condition that the ends of the chord lie on different arcs $AB$ is equivalent to the chord intersecting the segment $AB$ (at an internal point).
Then t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,823 |
5. In a country where residents' homes are represented as points on a plane, two laws are in effect:
1) A person can play basketball only if they are taller than the majority of their neighbors.
2) A person is entitled to free public transportation only if they are shorter than the majority of their neighbors.
In each... | 5. Let's provide an example. Suppose in a country there are 10 people, and all their houses are located on a straight line, in the order of increasing height of these people. Let the distances between them be: 1 km, 2 km, 3 km, 4 km, 5 km, 4 km, 3 km, 2 km, 1 km (Fig. 80). Then, all except the tallest person can ride t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,824 |
5. Point $X$, lying outside two non-intersecting circles $\omega_{1}$ and $\omega_{2}$, is such that the segments of the tangents drawn from $X$ to $\omega_{1}$ and $\omega_{2}$ are equal. Prove that the point of intersection of the diagonals of the quadrilateral formed by the points of tangency coincides with the poin... | 5. Let the centers of circles $\omega_{1}$ and $\omega_{2}$ be denoted by $O_{1}$ and $O_{2}$, and their radii by $r_{1}$ and $r_{2}$, respectively.
Consider the point $D$ where the common internal tangent to the circles intersects the segment $O_{1} O_{2}$. Then (see fact 17)
$$
\frac{D O_{1}}{D O_{2}}=\frac{r_{1}}{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,825 |
6. a) Eight schoolchildren solved 8 problems. It turned out that each problem was solved by 5 schoolchildren. Prove that there will be such two schoolchildren that each problem was solved by at least one of them.
b) If each problem was solved by 4 students, it may happen that such a pair does not exist (provide an exa... | 6. The first method. a) Consider two cases.
1) Suppose there is a student who solved at least 6 problems. We will prove that there is a student who solved the remaining 2 problems. Indeed, each of these two problems was solved by 5 students, and since there are a total of 8 students, which is less than $5+5$, there is ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,826 |
6. Ali-Baba and the bandit are dividing a treasure consisting of 100 gold coins, arranged in 10 piles of 10 coins each. Ali-Baba chooses 4 piles, places a cup next to each, and sets aside several coins from each pile (at least one, but not the entire pile). The bandit must then rearrange the cups, changing their initia... | 6. We will show that Ali-Baba can achieve no more than 4 coins in 7 piles, while the robber can ensure that there are no piles with fewer than 4 coins. Therefore, Ali-Baba will take $100 - 7 \cdot 4 = 72$ coins.
First, we will prove that the robber can act in such a way that there are no piles with fewer than 4 coins.... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,827 |
6. Prove that for any polynomial $P(x)$ of degree $n$ with natural coefficients, there exists an integer $k$ such that the numbers $P(k), P(k+1), \ldots, P(k+1996)$ are composite, if a) $n=1$; b) $n$ is any natural number.
11 c l a s s
Translate the above text into English, please retain the original text's line brea... | 6. It is not hard to see that (since the coefficients are natural numbers) $P(N)>P(M)$ when $N>M>0$. Moreover, $P(N)>1$ when $N>0$.
Further, note that if $k \mid x-y$, then $k \mid P(x)-P(y)$ (see the comment). Let
$$
A=P(1) P(2) \ldots P(1996)
$$
Then $P(k) \mid P(A+k)-P(k)$ for $k=1, \ldots, 1996$ (since $P(k) \mi... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,828 |
6. In a $2^{n} \times n$ table, all possible rows of length $n$ consisting of the numbers 1 and -1 were written down. Then some of the numbers were replaced by zeros. Prove that it is possible to select several rows such that their sum is a row of zeros. (The sum of rows is a row whose elements are the sums of the corr... | 6. Let's proceed with the proof. We will choose the rows of the "spoiled" zero table one by one and keep track of the sum $S = S(m)$ of the selected rows. The first row we take is the "spoiled" row that was derived from $(1,1, \ldots, 1)$. The row $S(1)$ will consist of 0 and 1 (if it contains only zeros, then this sin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,829 |
1. Some cells of a chessboard are occupied by pieces. It is known that each row contains at least one piece, and different rows contain a different number of pieces. Prove that it is always possible to mark 8 pieces so that each column and each row contains exactly one marked piece. | 1. On the horizontal lines, there can be from one to eight figures. Since there are different numbers of figures on different horizontal lines, on some horizontal line there is exactly one figure, on some other - two figures, ..., finally, some horizontal line is filled with eight figures. Let's number the horizontal l... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,830 |
1*. Does there exist a convex body, different from a sphere, whose orthogonal projections onto some three mutually perpendicular planes are circles? | 1. The first method. Let's introduce coordinates in space and consider the coordinate planes $\alpha, \beta$, and $\gamma$, defined by the equations $x=0, y=0$, and $z=0$ respectively.
Consider the ball $B$ defined by the inequality
$$
x^{2}+y^{2}+z^{2} \leqslant 1
$$
Its projection onto the plane $\alpha$ is a circ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,832 |
1. On the sides $AB$, $BC$, and $CA$ of triangle $ABC$, points $C'$, $A'$, and $B'$ are taken respectively. Prove that the area of triangle $A'B'C'$ is
$$
\frac{AB' \cdot BC' \cdot CA' + AC' \cdot CB' \cdot BA'}{4 R}
$$
where $R$ is the radius of the circumscribed circle of triangle $ABC$ (Fig. 3).
 takes 16 hours. Therefore, if you start immediately after the eruption of the first crater, it will not be dangerous.
Movement along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the eruption of the s... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,834 |
2. Compute
$$
\int_{0}^{\pi / 2} \cos ^{2}(\cos x)+\sin ^{2}(\sin x) d x
$$ | 2. We have
$$
\begin{aligned}
\int_{0}^{\pi / 2} \cos ^{2}(\cos x)+\sin ^{2}(\sin x) d x & = \\
= & \int_{0}^{\pi / 2} \cos ^{2}(\cos x) d x+\int_{0}^{\pi / 2} \sin ^{2}(\sin x) d x .
\end{aligned}
$$
Perform the substitution (see fact 28) $y=\frac{\pi}{2}-x$ in the second integral, then $d y=-d x$, and
$\int_{0}^{\... | \frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 25,837 |
3. Inside the acute angle $X O Y$, points $M$ and $N$ are taken such that $\angle X O N = \angle Y O M$. On the segment $O X$, a point $Q$ is chosen such that $\angle N Q O = \angle M Q X$, and on the segment $O Y$, a point $P$ is chosen such that $\angle N P O = \angle M P Y$. Prove that the lengths of the broken line... | 3. Let points $L$ and $K$ be symmetric to point $M$ with respect to the lines $O X$ and $O Y$ respectively (Fig. 82, a). Then points $K, P$, and $N$ lie on the same line, and $N K = N P + P K = N P + P M$. Indeed, the segment $M K$ is perpendicular to the line $O Y$, and if $A$ is the point of intersection, then $M A =... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,838 |
3. In the convex hexagon $A C_{1} B A_{1} C B_{1}$, it is given: $A B_{1} = A C_{1}, B C_{1} = B A_{1}, C A_{1} = C B_{1}$ and $\angle A + \angle B + \angle C = \angle A_{1} + \angle B_{1} + \angle C_{1}$.
Prove that the area of triangle $A B C$ is half the area of the hexagon. | 3. First, note that the sum of all angles of a hexagon is $720^{\circ}$ (this can be seen, for example, by cutting it into two quadrilaterals).
Thus,
$$
\angle A+\angle B+\angle C=\angle A_{1}+\angle B_{1}+\angle C_{1}=360^{\circ} .
$$
The area of the hexagon is equal to the sum of the areas of triangles $A B_{1} C$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,839 |
4. Prove that there exists a natural number which remains composite after replacing any triple of consecutive digits with an arbitrary triple. Does such a $1997-$digit number exist? | 4. Consider any even number $N$ greater than 9992. By the divisibility rule for 2 (see fact 6), it ends in an even digit. Therefore, if we change any triplet of digits other than the last one, the number will remain even, and hence composite (from the number 9992, we can obtain the prime number 0002).
Therefore, we on... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,841 |
4. $n$ trains are circulating in one direction at equal distances around a circle. On this road, stations $A$, $B$, and $C$ are located at the vertices of an equilateral triangle (labeled in the direction of travel). Ira enters station $A$ and at the same time, Lesha enters station $B$ to catch the nearest trains. It i... | 4. For clarity, we will consider trains and stations as points.
It is clear that if $n$ is divisible by 3, then Lёsha and Ira always leave at the same time. Therefore, in this case, there is no forest. Let $n$ not be divisible by 3. Then, whenever they arrive at the station, either Ira leaves before Lёsha, or Lёsha le... | \frac{2}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,842 |
4. Given real numbers $a_{1} \leqslant a_{2} \leqslant a_{3}$ and $b_{1} \leqslant b_{2} \leqslant b_{3}$ such that
$$
\begin{aligned}
a_{1}+a_{2}+a_{3} & =b_{1}+b_{2}+b_{3} \\
a_{1} a_{2}+a_{2} a_{3}+a_{1} a_{3} & =b_{1} b_{2}+b_{2} b_{3}+b_{1} b_{3}
\end{aligned}
$$
Prove that if $a_{1} \leqslant b_{1}$, then $a_{3... | 4. These problems remind us of Vieta's theorem (see fact 20). Consider the polynomials $P(x)=\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)$ and $Q(x)=\left(x-b_{1}\right)\left(x-b_{2}\right)\left(x-b_{3}\right)$. From the problem statement, it follows that these polynomials differ only by the constant te... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,843 |
4. Can a regular tetrahedron with edge length 1 be divided into regular tetrahedrons and octahedrons, each with edge lengths less than $1 / 100$? | 4. By marking the midpoints of the edges of a regular tetrahedron, we easily obtain a partition of the regular tetrahedron into a regular octahedron and 4 regular tetrahedra. In other words, the small tetrahedra are obtained from the larger one by homotheties with a coefficient of $1 / 2$ and centers at the vertices of... | 1/128<1/100 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,844 |
5*. In rhombus $A B C D$ the measure of angle $B$ is $40^{\circ}, E-$ is the midpoint of $B C, F$ - is the foot of the perpendicular dropped from $A$ to $D E$. Find the measure of angle $D F C$. | 5. Let lines $D E$ and $A B$ intersect at point $G$ (Fig. 83). Then triangles $D E C$ and $B E G$ are congruent by the second criterion. Therefore, $B G=C D=B A$. Hence, points $A, G$, and $C$ lie on a circle with center at point $B$, and $A G$ is the diameter. Since $\angle A F G=90^{\circ}$, point $F$ lies on the sam... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,845 |
5. $2 n$ athletes conducted a round-robin tournament twice (in a round-robin tournament, everyone meets everyone, one point is awarded for a win, $-1 / 2$ for a draw, and 0 for a loss). Prove that if the sum of points of each changed by at least $n$, then it changed exactly by $n$. | 5. Let's divide the tournament participants into two groups: those who scored more points in the second tournament than in the first, and those who scored more points in the first tournament than in the second. At least one of these two groups includes no fewer than $n$ athletes. Suppose, for example, that the first gr... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,846 |
5. In a round-robin tournament, there were no draws; 1 point was awarded for a win, and 0 for a loss. Then, a coefficient for each participant was determined. It was equal to the sum of the points scored by those whom the athlete defeated. It turned out that all participants had equal coefficients. The number of partic... | 5. Suppose not all participants scored the same number of points. Let those who took first place (the "firsts") have scored $K$ points, and those who took last place (the "lasts") have scored $-L$ points. (Places are determined by points, not by coefficients.)
The coefficient of the "firsts" is the sum of $K$ numbers,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,847 |
5*. Positive numbers $a, b$ and $c$ are such that $a b c=1$. Prove the inequality
$$
\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leqslant 1
$$ | 5. Let $a=x^{3}, b=y^{3}, c=z^{3}$. Then $x y z=\sqrt[3]{a b c}=1$.
Obviously, for any $x$ and $y$, the inequality $x^{2}-x y + y^{2} \geqslant x y$ holds. Multiplying both sides of this inequality by $x+y$ (which is valid since $x>0, y>0$), we get $x^{3}+y^{3} \geqslant (x+y) x y$, from which
$$
\frac{1}{1+a+b}=\fra... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,848 |
6. A banker learned that among identical-looking coins, one is counterfeit (lighter). He asked an expert to identify this coin using a balance scale without weights, and required that each coin participate in weighings no more than twice. What is the maximum number of coins the banker can have so that the expert can de... | 6. Let's first solve a simpler problem. Suppose the banker allows placing coins on the scales no more than once. From what largest number of coins can a lighter one be identified in $k$ weighings?
If at some weighing there are more than one coin on the scale, it will be impossible to identify the fake coin (since we c... | 2n^{2}+1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,849 |
6. Let $1+x+x^{2}+\ldots+x^{n}=F(x) G(x)$, where $F$ and $G$ are polynomials with coefficients that are zeros and ones. Prove that one of the polynomials $F(x), G(x)$ can be represented
in the form $\left(1+x+x^{2}+\ldots+x^{k}\right) T(x)$, where $T$ is also a polynomial with coefficients 0 and $1 (k>0)$.
10 t h g ... | ## 6. Let
$$
\begin{aligned}
& F(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \\
& G(x)=b_{0}+b_{1} x+b_{2} x^{2}+\ldots
\end{aligned}
$$
Then, by the condition,
$$
\begin{aligned}
\left(a_{0}+a_{1} x+a_{2} x^{2}+\ldots\right)\left(b_{0}+b_{1} x+b_{2} x^{2}+\right. & \ldots)= \\
& =1+x+x^{2}+\ldots+x^{n}
\end{aligned}
$$
Th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,850 |
6. Let's consider the powers of five:
$$
1,5,25,125,625, \ldots
$$
We form a sequence of their first digits:
$$
1,5,2,1,6, \ldots
$$
Prove that any segment of this sequence, written in reverse order, will appear in the sequence of the first digits of the powers of two ( $1,2,4,8,1$, $3,6,1, \ldots)$.
11 k l a s s | 6. It is sufficient to prove that any initial segment of the sequence of the first digits of the powers of five appears (in reverse order) in the sequence of the first digits of the powers of two.
Consider the numbers: $1 / 2, 1 / 4, \ldots, 1 / 2^{n}$. The sequence of the first non-zero digits of their decimal repres... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,851 |
6. On a plane, there are a finite number of strips, the sum of whose widths is 100, and a circle of radius 1. Prove that each of the strips can be translated parallel to itself so that together they cover the circle. | 6. We will associate each strip with a vector perpendicular to its boundary, the length of which is equal to the width of this strip. We will lay them out from the same point $O$.
We will divide the plane into 12 angles of $30^{\circ}$ with the vertex at point $O$. For each of these angles, we will calculate the sum o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,852 |
1. Are there natural numbers $x, y$ and $z$ that satisfy the equation $28 x+30 y+31 z=365$? | 1. In a year, there are 12 months. One of them, February, consists of 28 days, four months (April, June, September, November) consist of 30 days, and the remaining 7 months consist of 31 days. Since there are 365 days in a year, we get
$$
28 \cdot 1 + 30 \cdot 4 + 31 \cdot 7 = 365
$$
Comments. $1^{\circ}$. Compare wi... | 28\cdot1+30\cdot4+31\cdot7=365 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,853 |
1. Is the number $4^{9}+6^{10}+3^{20}$ prime? | 1. We have
$$
4^{9}+6^{10}+3^{20}=\left(2^{9}\right)^{2}+2 \cdot 2^{9} \cdot 3^{10}+\left(3^{10}\right)^{2}=\left(2^{9}+3^{10}\right)^{2}
$$
Comment. One could try to compute the remainders of this number when divided by different prime numbers, hoping to find a small number that divides our number. However, this wou... | (2^{9}+3^{10})^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,854 |
1. Let $a, b, c$ be non-negative integers such that $28a + 30b + 31c = 365$. Prove that $a + b + c = 12$. | 1. Let $a+b+c \leqslant 11$. Then
$28 a+30 b+31 c \leqslant 31(a+b+c) \leqslant 11 \cdot 31=341365
$$
This cannot be true! It remains to prove that $a+b+c$ cannot equal 13. So, let $a+b+c=13$. The case $a=13, b=c=0$ does not satisfy the condition:
$$
28 \cdot 13+30 \cdot 0+31 \cdot 0=364 \neq 365
$$
The only remain... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,855 |
1. The numbers $x, y, z$ satisfy the equation
$$
x+y+z-2(x y+y z+x z)+4 x y z=\frac{1}{2}
$$
Prove that at least one of them is equal to $1 / 2$. | 1. Note that
$x+y+z-2(x y+y z+x z)+4 x y z-\frac{1}{2}=\frac{1}{2}(2 x-1)(2 y-1)(2 z-1)$.
If the left side of the equality is zero, then at least one of the factors on the right is zero. Therefore, one of the numbers $x, y, z$ is equal to $1 / 2$.
Comments. $1^{\circ}$. How to guess the factorization? Let this polyn... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,856 |
2. Can we find eight such natural numbers that none of them is divisible by any other, but the square of any of these numbers is divisible by each of the others? | 2. Let $p_{1}, p_{2}, \ldots, p_{8}$ be distinct prime numbers. Then the desired numbers are $n_{1}=p_{1}^{2} \cdot p_{2} \cdot \ldots \cdot p_{8}, n_{2}=$ $=p_{1} \cdot p_{2}^{2} \cdot p_{3} \cdot \ldots \cdot p_{8}, \ldots, n_{8}=p_{1} \cdot p_{2} \cdot \ldots \cdot p_{8}^{2}$.
Indeed, in the prime factorization of ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,857 |
2. In an acute-angled triangle $ABC$, altitudes $AD$ and $CE$ are drawn. A square $ACPQ$ is constructed, and rectangles $CDMN$ and $AEKL$ are constructed such that $AL = AB$ and $CN = CB$. Prove that the area of the square $ACPQ$ is equal to the sum of the areas of the rectangles $AEKL$ and $CDMN$.
In an acute-angled ... | 2. Let's draw the third altitude $BS$ and extend it to intersect $PQ$ at point $T$ (Fig. 94). We will prove that the areas of rectangles $ASTQ$ and $AEKL$ are equal. From the similarity of right triangles $ABS$ and $AEC$, we get: $AE / AC = AS / AB \Leftrightarrow AE \cdot AB = AS \cdot AC \Leftrightarrow AE \cdot AL =... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,858 |
2. For the continuous function $f$, it is known that:
1) $f$ is defined on the entire number line;
2) $f$ has a derivative at each point (and thus the graph of $f$ has a unique tangent at each point);
3) the graph of the function $f$ does not contain points where one of the coordinates is rational and the other is irra... | 2. Example:
$$
f(x)= \begin{cases}2-x, & \text { if } x \leqslant 1 \\ 1 / x, & \text { if } x>1\end{cases}
$$
(The graph of this function is shown in Fig. 100). The first property is obvious, the second property follows from the fact that the derivatives of the functions $2-x$ and $1 / x$ at the point 1 are equal.
... | proof | Calculus | proof | Yes | Yes | olympiads | false | 25,860 |
3. The diagonals $A C$ and $B D$ of parallelogram $A B C D$ intersect at point $O$. Point $M$ lies on line $A B$, such that $\angle A M O = \angle M A D$. Prove that point $M$ is equidistant from points $C$ and $D$. | 3. Let $P$ and $Q$ be the midpoints of sides $A B$ and $C D$ respectively. The segment $P Q$ is the midline of parallelogram $A B C D$, hence it passes through point $O$, and $P Q \| A D$.
This can be proven as follows: quadrilateral $A P Q D$ is a parallelogram because sides $A P$ and $Q D$ are parallel and equal. Th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,861 |
3. A traveler visited a village where every person either always tells the truth or always lies. The villagers stood in a circle, and each one told the traveler whether the neighbor to their right was truthful or a liar. Based on these statements, the traveler was able to determine unequivocally what fraction of the vi... | 3. Let $x$ be the fraction of truthful residents. Imagine that all truthful residents became liars, and all liars "reformed." Then the traveler would hear the same thing! Indeed, the truthfulness of any resident has changed, but the truthfulness of the neighbor they are talking about has also changed. But the fraction ... | \frac{1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,862 |
3. A road 1 km long is fully illuminated by street lamps, each of which illuminates a section of the road 1 m long. What is the maximum number of street lamps that can be on the road, given that after turning off any street lamp, the road will no longer be fully illuminated? | 3. Let's number the street lamps with natural numbers in the order of their placement along the road. If the segments illuminated by the $n$-th and $(n+2)$-th street lamps intersect (at least at one point), then the $(n+1)$-th street lamp can be turned off. Therefore, segments with different odd numbers do not intersec... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,863 |
3. In a non-isosceles triangle $A B C$, medians $A K$ and $B L$ are drawn. Angles $B A K$ and $C B L$ are equal to $30^{\circ}$. Find the angles of triangle $A B C$. | 3. $1^{\circ}$. Let $M$ be the point of intersection of the medians of triangle $ABC$. Consider the equilateral triangle $APQ$, for which segment $AK$ is a median, and point $P$ lies on line $AB$ (Fig. 101). Since the point of intersection of the medians always divides them in the ratio $2:1$, point $M$ is the point of... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,864 |
4. Some of the numbers $a_{1}, a_{2}, \ldots, a_{200}$ are written in blue pencil, and the rest in red. If all the red numbers are erased, the remaining numbers are all natural numbers from 1 to 100, written in ascending order. If all the blue numbers are erased, the remaining numbers are all natural numbers from 100 t... | 4. Suppose that among the numbers $a_{1}, a_{2}, \ldots, a_{100}$ there are $k$ blue and, respectively, $100-k$ red. Since the blue numbers are written in ascending order, these $k$ blue numbers are the numbers from 1 to $k$ inclusive. Similarly, the $100-k$ red numbers are the numbers $100,99, \ldots$ $\ldots, k+1$. T... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,865 |
4*. In the country of Nashii, there are military bases connected by roads. A set of roads is called important if after closing these roads, there will be two bases that are not connected by a path. An important set is called strategic if it does not contain a smaller important set. Prove that the set of roads, each of ... | 4. From any base, you can get to any other, otherwise an empty set would form an important set, but then the empty set would be the only strategic set, while the problem statement mentions two different strategic sets.
Consider some strategic set.
L e m m a. When all the roads of this set are closed, the set of bases... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,866 |
4. Does there exist a natural number divisible by 1998, the sum of whose digits is less than 27? | 4. $1^{\circ}$. If a number is divisible by 1998, then it is also divisible by 999. We will show that there does not exist a number divisible by 999 whose sum of digits is less than 27.
It is well known that a number is divisible by 9 if and only if the sum of the digits in its decimal representation is divisible by 9... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,867 |
4. Solve the equation $3^{x}+4^{y}=5^{z}$ in natural numbers. | 4. The right side of the equation, when divided by 3, should give the same remainder as the left side, i.e., 1 (see fact 7). Therefore, \( z \) is an even number (see comment). Similarly, the left side of the equation, when divided by 4, gives a remainder of 1, so the number \( x \) is also even. Thus, \( 4^{y} = 5^{z}... | x=y=z=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,868 |
5. At a round table, several guests are sitting. Some of them know each other; the acquaintance is mutual. All acquaintances of any guest (including the guest themselves) are seated around the table at equal intervals. (For another person, these intervals may be different.) It is known that any two people have at least... | 5. Note that if a person has acquaintances sitting next to each other (in particular, if he is acquainted with his neighbor), then this person is acquainted with everyone.
Let's prove that such a guest exists. Suppose $A$ and $B$ are two neighbors. If they are not acquainted with each other, then their common acquaint... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,869 |
5. Point $O$ lies inside rhombus $A B C D$. Angle $D A B$ is $110^{\circ}$. Angles $A O D$ and $B O C$ are $80^{\circ}$ and $100^{\circ}$ respectively. What can the measure of angle $A O B$ be? | 5. $1^{\circ}$. Note that the geometric locus of points $O$ such that $\angle A O D=80^{\circ}$ and point $O$ lies on the same side of line $A D$ as $B$, is the arc of a circle with endpoints at points $A$ and $D$, and the set of points $O$ for which $\angle B O C=100^{\circ}$, and point $O$ lies on the same side of li... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,870 |
5. A regular triangle $A B C$, cut out of plywood, was placed on the floor. Three nails were hammered into the floor (one next to each side of the triangle) so that the triangle cannot be rotated without lifting it off the floor. The first nail divides side $A B$ in the ratio $1: 3$, counting from vertex $A$, the secon... | 5. $1^{\circ}$. Let's first solve the "one nail problem". Suppose one nail is hammered in, touching the triangle at point $M$ on side $A C$. We fix the center of the supposed rotation (point $O$). Can the triangle be rotated around this point by a small angle, and if so, in which direction?
Let the triangle be positio... | 5:7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,871 |
5. Is it possible to form a closed chain of 61 identical synchronously rotating gears in space so that the angles between meshed gears are no less than $150^{\circ}$? At the same time:
1) for simplicity, the gears are considered as circles;
2) gears are meshed if the corresponding circles have a common tangent at the p... | 5. At first glance, it may seem that such a construction is impossible: if the first gear rotates clockwise, then the second counterclockwise, the third clockwise, and so on; therefore, the 61st should rotate clockwise, and then it cannot rotate in sync with the first.
However, we are working in space. And whether a g... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,872 |
6. A red square is covered by 100 white squares. In this case, all squares are the same, and the sides of each white square are parallel to the sides of the red square. Is it always possible to remove one of the white squares so that the remaining white squares still completely cover the red square?
Comment. In the ph... | 6. See Fig. 93 for 8 white squares.
Let the vertices of the red square be denoted by the letters $A, B$, $C$, and $D$. The diagonal $A C$ will be divided into 100 equal segments, the endpoints of which will be sequentially labeled with the numbers $1,2, \ldots, 101$ (point $A$ is labeled with the number 1, and point $... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,873 |
6*. On the segment $[0 ; 1]$, several distinct points are marked. Each marked point is either exactly in the middle between two other marked points (not necessarily adjacent to it), or exactly in the middle between a marked point and an endpoint of the segment. Prove that all marked points are rational.
10 t h g r a... | 6. First method. $1^{\circ}$. Let's denote the coordinates of the ends of the segment and the marked points as $x_{0}, x_{1}, \ldots, x_{n+1}$ ( $0=x_{0}i$ is not zero. Indeed, this is true for the initial set. Making the next substitution from the $j$-th equality ( $j0$. When dividing both sides of the equality by the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,874 |
6. Natural numbers from 1 to $n$ are arranged in a row in any order (the number $n$ is fixed). An arrangement is called bad if in it one can mark 10 numbers (not necessarily consecutive) in decreasing order. All other arrangements are called good. Prove that the number of good arrangements does not exceed $81^{n}$.
##... | 6. $1^{\circ}$. Let's prove that if the arrangement is good, then the numbers in the row can be colored in nine colors such that the numbers of each color are in increasing order. Indeed, we will color the numbers from left to right, each time using the color with the smallest number such that the last number painted i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,875 |
1. Comparing the fractions $x=\frac{111110}{111111}, y=\frac{222221}{222223}, z=\frac{333331}{333334}$, arrange them in ascending order. | 1. Consider the numbers
$$
1-x=\frac{1}{111111}, \quad 1-y=\frac{2}{222223}, \quad 1-z=\frac{3}{333334}
$$
and their reciprocals
$$
\frac{1}{1-x}=111111, \quad \frac{1}{1-y}=111111 \frac{1}{2}, \quad \frac{1}{1-z}=111111 \frac{1}{3}
$$
We see that $\frac{1}{1-x}1-z>1-y$. Therefore, $x<z<y$. | x<z<y | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,877 |
1. On the board in the laboratory, two numbers are written. Every day, senior researcher Petya erases both numbers from the board and writes down their arithmetic mean and harmonic mean instead. On the morning of the first day, the numbers 1 and 2 were written on the board. Find the product of the numbers written on th... | 1. The product of the numbers on the board does not change. Indeed,
$$
\frac{a+b}{2} \cdot \frac{2}{1 / a+1 / b}=\frac{a+b}{2} \cdot \frac{2 a b}{a+b}=a b
$$
Therefore, on the 1999th day, the product will be the same as it was on the first day. See also fact 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,878 |
1. $a, b, c$ - sides of a triangle. Prove the inequality
$$
\frac{a^{2}+2 b c}{b^{2}+c^{2}}+\frac{b^{2}+2 a c}{c^{2}+a^{2}}+\frac{c^{2}+2 a b}{a^{2}+b^{2}}>3
$$ | 1. From the triangle inequality $a>|b-c|$. Squaring both sides, we get: $a^{2}>(b-c)^{2}$. From this, it follows that $a^{2}+2 b c>b^{2}+c^{2}$. The right side is positive, so we can divide by it. This means the first term in the condition of the problem is greater than 1. The same is true for the other two terms. Ther... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,880 |
2. Two players play the following game: the first writes down letters A or B in a row as they wish (from left to right, one after another; one letter per move), and the second, after each move of the first, swaps any two of the written letters or does nothing (this also counts as a move). After both players have made 1... | 2. Let's outline the strategy of the second player. For the first 1000 moves, he skips. On move $k+1000$, he makes a move such that the last $2 k+1$ letters form a palindrome. We will prove that he can always do this.
To do this, we will use induction on $k$ (see fact 24). For $k=0$, this is obvious. Suppose that afte... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,882 |
3. Find any 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. | 3. If it suddenly turned out that $a b=c d$, then
$$
\begin{aligned}
& a^{2}+2 c d+b^{2}=a^{2}+2 a b+b^{2}=(a+b)^{2} \\
& c^{2}+2 a b+d^{2}=c^{2}+2 c d+d^{2}=(c+d)^{2}
\end{aligned}
$$
Thus, it is sufficient to find four different natural numbers $a, b, c$, and $d$ for which $a b=c d$. Or, equivalently, to find a num... | 6=1\cdot6=2\cdot3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,885 |
3. The diagonals of parallelogram $A B C D$ intersect at point $O$. The circle passing through points $A, O, B$ is tangent to line $B C$. Prove that the circle passing through points $B, O, C$ is tangent to line $C D$. | 3. By the theorem on the angle between a tangent and a chord, $\angle C B O=\angle B A C$ (see fact 15). On the other hand, by the theorem on alternate interior angles for parallel lines, $\angle B A C=\angle A C D$ (Fig. 106); therefore, $\angle C B O=\angle O C D$. Applying now the converse of the theorem on the angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,886 |
3 *. Find all pairs of natural numbers $x, y$ such that $x^{3}+y$ and $y^{3}+x$ are divisible by $x^{2}+y^{2}$. | 3. Let's first prove that $x$ and $y$ are coprime. Assume the opposite. Then $x$ and $y$ are divisible by some prime number $p$. Let $p$ enter the prime factorizations of $x$ and $y$ with powers $a \geqslant 1$ and $b \geqslant 1$ respectively (see fact 10). Without loss of generality, we can assume that $a \geqslant b... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,887 |
3. The faces of a regular octahedron are
Fig. 5 painted in white and black. At the same time, any two faces that share an edge are painted in different colors. Prove that for any point in... | 3. The first method. The planes to which the faces of each color belong form equal regular tetrahedra. This can be represented as follows. Consider the cube $A B C D E F G H$ (Fig. 112) and two tetrahedra: $A C F H$ and $B D E G$. The intersection of these tetrahedra is an octahedron. Indeed, the vertices of the inters... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,888 |
4. Pete's bank account contains 500 dollars. The bank allows only two types of transactions: withdrawing 300 or adding 198 dollars. What is the maximum amount Pete can withdraw from his account if he has no other money? | 4. Since 300 and 198 are divisible by 6, Petya will only be able to withdraw an amount that is a multiple of 6 dollars (see fact 5). The maximum number that is a multiple of 6 and does not exceed 500 is 498.
Let's show how to withdraw 498 dollars. We will perform the following operations: $500-300=200, 200+198=398, 39... | 498 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,889 |
4*. Find all such positive integers $k$ for which the number
$$
\underbrace{1 \ldots 1 \overbrace{2 \ldots 2}^{k}}_{2000}-\underbrace{2 \ldots 2}_{1001}
$$
is a perfect square. | 4. Let $n=1000$. Consider two cases.
$1^{\circ} . k>n$. Then
$$
\underbrace{1 \ldots 12 \ldots 2}_{2 n}-\underbrace{2 \ldots 2}_{n+1}=\underbrace{1 \ldots 1}_{2 n-k} \overbrace{2 \ldots 2}^{k-(n+1)} \underbrace{0 \ldots 0}_{n+1} .
$$
This number ends with $n+1=1001$ zeros. But if a number is a square of a natural nu... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,890 |
4. $2 n$ radii divide a circle into $2 n$ equal sectors: $n$ blue and $n$ red. In the blue sectors, starting from a certain one, the numbers from 1 to $n$ are written counterclockwise. In the red sectors, starting from a certain one, the same numbers are written in the same way, but clockwise. Prove that there will be ... | 4. Let's agree to call numbers in red sectors red, and numbers in blue sectors blue. The distance between two numbers $a$ and $b$ will be the number of numbers located on the smaller arc between $a$ and $b$. The numbers arranged in a circle are paired equally. We will choose the pair of equal numbers with the smallest ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,891 |
4. On a meadow in the shape of a square, there is a circular pit. A grasshopper is jumping around the meadow. Before each jump, it selects a vertex of the square and jumps in its direction. The length of the jump is half the distance to the vertex. Will the grasshopper be able to jump into the pit? | 4. Let the side of the square have a length of 1. Divide each side into $2^{n}$ equal parts, and draw lines through the division points parallel to the sides. In this way, the square is divided into squares with side length $2^{-n}$. If $n$ is sufficiently large, one of these squares will be entirely within the hole (f... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,892 |
5. In a right triangle $ABC$, point $O$ is the midpoint of the hypotenuse $AC$. On segment $AB$, point $M$ is taken, and on segment $BC$, point $N$ is taken such that angle $MON$ is a right angle. Prove that $AM^2 + CN^2 = MN^2$. | 5. Let $N^{\prime}$ be the point symmetric to point $N$ with respect to point $O$ (Fig. 105).
Triangles $O N C$ and $O N^{\prime} A$ are equal by two sides and the angle between them. Moreover, the angle $N^{\prime} A M$ is a right angle. Indeed, $\angle N^{\prime} A M=\angle N^{\prime} A O+\angle M A O=$
$$
=\angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,893 |
5. The incircle of triangle $ABC$ touches sides $AB$ and $AC (AB > BC)$ at points $P$ and $Q$ respectively, $RS$ is the midline parallel to $AB$, $T$ is the intersection point of lines $PQ$ and $RS$. Prove that $T$ lies on the bisector of angle $B$ of the triangle. | 5. Let the lengths of sides $AB, BC$, and $AC$ be denoted by $c, a$, and $b$ respectively. Let $p$ be the semiperimeter of the triangle. We will assume that $R$ lies on $AC$ and $S$ lies on $BC$ (Fig. 107).
$. A point $a$ is chosen on the segment $[0 ; 1]$. Prove that, starting from any point, the grasshopper can, after several jumps,... | 5. Let
$$
f, g:[0 ; 1] \rightarrow[0 ; 1], \quad f(x)=\frac{x}{\sqrt{3}}, \quad g(x)=\frac{x}{\sqrt{3}}+\left(1-\frac{1}{\sqrt{3}}\right)
$$
- be functions corresponding to the jumps of a grasshopper. The range of the function $f$ is the interval $[0 ; 1 / \sqrt{3}]$, and the range of the function $g$ is the interval... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,895 |
5. A graph is a set of vertices, some of which are connected by edges (each edge connects exactly two vertices of the graph). A vertex coloring of a graph is called proper if no two vertices of the same color are connected by an edge. Suppose a certain graph is properly colored with $k$ colors, and it cannot be properl... | 5. The colors in which the graph is painted are numbered from 1 to $k$. Those vertices of color 2 that do not neighbor any vertices of color 1 will be repainted in color 1. The new coloring will be correct, so it will have $k$ colors. Therefore, some vertices of color 2 are not repainted and thus neighbor vertices of c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,896 |
6. In a chess tournament, each participant played two games with each other: one with white pieces, the other with black. At the end of the tournament, it turned out that everyone had scored the same number of points (1 point for a win, $-1 / 2$ point for a draw, 0 points for a loss). Prove that there will be two parti... | 6. Let there be $n$ chess players in the tournament. Then a total of $n(n-1)$ games were played in the tournament, and 1 point was at stake in each game. Therefore, if all results were equal, each participant scored $n-1$ points. Each chess player played $n-1$ games with white pieces, and the number of games won by the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,897 |
6*. Solve the equation $\left(1+n^{k}\right)^{l}=$ $=1+n^{m}$ in natural numbers, where $l>1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 6. The first method. Let $p$ be a prime divisor of the number $l$. Since $n^{m}=\left(1+n^{k}\right)^{l}-1$, then $n^{m}$ is divisible by $\left(1+n^{k}\right)^{p}-1$ (see fact 8). However, by the binomial theorem (see comment),
$$
\left(1+n^{k}\right)^{p}-1=n^{k} \cdot p+n^{2 k} \cdot \frac{p(p-1)}{2}+n^{3 k} \cdot r... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,900 |
7. Prove that the leading digits of numbers of the form $2^{2^{n}}$ form a non-periodic sequence.
Of the last three problems for the 11th grade, two were counted towards the final score. | 7. Consider a circle of length 1 as the segment $[0; 1]$ with identified endpoints (see the comment to the solution of problem 6 for the 10th grade of the 1997 competition). Then the fractional part $f_{m}$ of the number $\lg \left(2^{m}\right)=m \cdot \lg 2$ can be considered as a point on this circle. Consider the po... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,901 |
1. Two different numbers $x$ and $y$ (not necessarily integers) are such that $x^{2}-2000 x=y^{2}-2000 y$. Find the sum of the numbers $x$ and $y$. | 1. Move $2000 x$ to the right side, and $y^{2}$ - to the left. We get:
$$
x^{2}-y^{2}=2000 x-2000 y
$$
Factor the left side using the difference of squares formula:
$$
(x-y)(x+y)=2000(x-y) .
$$
Since $x \neq y$, we can cancel out $(x-y)$. Thus, $x+y=$ $=2000$. | x+2000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,902 |
1. Points $A$ and $B$ are taken on the graph of the function $y=1 / x$, $x>0$. Perpendiculars are dropped from them to the x-axis, with the bases of the perpendiculars being $H_{A}$ and $H_{B} ; O$ - the origin. Prove that the area of the figure bounded by the lines $O A, O B$ and the arc $A B$ is equal to the area of ... | 1. We can assume that the abscissa of point $A$ is less than the abscissa of point $B$ (Fig. 124). Consider the point $K$ of intersection of the segments $A H_{A}$ and $O B$. The considered figures intersect along a curvilinear "triangle" $A K B$. Therefore, the difference of the considered areas is equal to the differ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 25,904 |
1. The greatest common divisor (GCD) of natural numbers $m$ and $n$ is 1. What is the greatest possible value of the GCD of the numbers $m+2000n$ and $n+2000m?$ | 1. Let $a=2000 m+n, b=2000 n+m, d$ be the greatest common divisor of $a$ and $b$. Then $d$ also divides the numbers
$$
2000 a-b=\left(2000^{2}-1\right) m \quad \text { and } \quad 2000 b-a=\left(2000^{2}-1\right) n .
$$
Since $m$ and $n$ are coprime, $d$ is also a divisor of the number $2000^{2}-1$ (see comment). On ... | 2000^2-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,905 |
2. In the elections to the 100-seat parliament, 12 parties participated. Parties that received strictly more than $5 \%$ of the voters' votes enter the parliament. Among the parties that entered the parliament, seats are distributed proportionally to the number of votes they received (i.e., if one party received $x$ ti... | 2. Idea of the solution: The Party of Mathematics Enthusiasts (PME) will receive the maximum number of seats in parliament if the total number of votes cast for non-qualifying (i.e., receiving no more than $5 \%$ of the votes) parties is maximized.
If 10 parties receive exactly $5 \%$ of the votes each, and two, inclu... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,906 |
3. The lengths of the bases of a trapezoid are $m$ cm and $n$ cm ( $m$ and $n$ are natural numbers, $m \neq n$ ). Prove that the trapezoid can be cut into equal triangles. | 3. Let $m>n$. Extend the lateral sides of the trapezoid until they intersect and divide each side of the resulting triangle into $m$ equal parts.
Fig. 114 Through the division points, draw ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,909 |
3. Given a circle and a point $A$ inside it. Find the geometric locus of the vertices $C$ of all possible rectangles $A B C D$, where points $B$ and $D$ lie on the circle. | 3. $1^{\circ}$. Let $O$ be the center of the circle in the problem's condition. We will prove that for point $C$, the equality $O C^{2}=2 R^{2}-O A^{2}$ holds, where $O$ is the center of the given circle, and $R$ is its radius. We introduce the notation as shown in Fig. 118. The segment $O C$ is the hypotenuse of a rig... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,910 |
3. On a sheet of "graph paper", a convex polygon is drawn such that all its vertices are at the vertices of the cells and none of its sides run vertically or horizontally. Prove that the sum of the lengths of the vertical segments of the grid lines enclosed within the polygon is equal to the sum of the lengths of the h... | 3. We will prove that each of the considered quantities is equal to the area of the polygon. Let's check this for the sum of the lengths of the vertical segments. Draw these segments. Let their lengths be $a_{1}, \ldots, a_{n}$ (Fig. 125). Then the polygon will be divided into two triangles and $n-1$ trapezoids, with t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,911 |
3. Chords $A C$ and $B D$ of a circle with center $O$ intersect at point $K$. Let $M$ and $N$ be the centers of the circumcircles of triangles $A K B$ and $C K D$, respectively. Prove that $O M=K N$. | 3. Let $X$ be the midpoint of $KB$ (Fig. $127, a$). Then $\angle KMX = \frac{1}{2} \angle KMB = \angle KAB = \angle KDC$ (in the penultimate equality, we use the fact that an angle subtending an arc is equal to half the arc, and in the last equality, that angles subtending equal arcs are equal). It is clear that $MX \p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,912 |
4. In triangle $ABC$, the length of median $BM$ is equal to the length of side $AC$. On the extensions of sides $BA$ and $AC$, points $D$ and $E$ are chosen respectively such that the equalities $AD = AB$ and $CE = CM$ are satisfied (Fig. 6). Prove that lines $DM$ and $BE$ are perpendicular. | 4. Let $C M=a$ (Fig. 115). Let $K$ be the point symmetric to $M$ with respect to point $A$. Then $K M=M B=$ $=M E=2 a$. Therefore, triangle $K B E$ is a right triangle (see fact 14$): K B \perp B E$.
Quadrilateral $D K B M$ is a parallelogram, since its diagonals bisect each other at the point of intersection. Therefo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,913 |
4. Grisha wrote the numbers 1, $2,3, \ldots, 63,64$ in the cells of a chessboard in some order. He told Lesha only the sum of the numbers in each rectangle of two cells and added that 1 and 64 lie on the same diagonal. Prove that Lesha can determine exactly which number is in which cell based on this information. | 4. Let's say Lёsha has many dominoes $1 \times 2$. If he places two dominoes on the board so that they overlap by exactly one cell
Fig. 120 (Fig. 120, left), then by subtracting the sum of... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,914 |
4. Fedia has three sticks. If it is impossible to form a triangle with them, Fedia shortens the longest stick by the sum of the lengths of the other two. If the length of the stick does not become zero and it is still impossible to form a triangle, Fedia repeats the operation, and so on. Can this process continue indef... | 4. Consider the polynomial $P(x)=x^{3}-x^{2}-x-1$. This polynomial has a root $t$ greater than 1, since $P(1) < 0$ and $P(2) > 0$ (see comment). Let's take the lengths of the sticks to be $t^{3}, t^{2}, t$. Recall that three segments can form a triangle if and only if the length of the largest one is less than the sum ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,916 |
5*. $A B C D$ is a convex quadrilateral. The circles constructed on segments $A B$ and $C D$ as diameters touch each other externally at point $M$, different from the intersection point of the diagonals of the quadrilateral. The circle passing through points $A, M$ and $C$ intersects the line connecting point $M$ and t... | 5. The first method. Let $P$ and $Q$ be the midpoints of segments $AB$ and $CD$ respectively, $O_1$ and $O_2$ be the centers of the circles passing through points $A, M, C$ and $B, M, D$ respectively, $H_1$ and $H_2$ be the projections of $O_1$ and $O_2$ onto the line $PQ$ (Fig. 121).
, it is allowed to obtain the sequences $\left\{b_{n}+c_{n}\right\},\left... | 5. Let's explain how to act in points "a" and "b". Removing the first term transforms the sequence $\left\{a_{n}\right\}$ into the sequence $\left\{a_{n+1}\right\}$. Subtracting the original sequence from this sequence yields the sequence $\left\{a_{n+1}-a_{n}\right\}$. Let's denote such a transformation by $T$, and th... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,919 |
5. In a round-robin chess tournament, each participant played against every other participant once. We will call a game incorrect if the winner of the game ended up scoring fewer points in total than the loser. (A win gives 1 point, a draw $1 / 2$, a loss -0.$)$
Can incorrect games make up
a) more than $75 \%$ of the... | 5. a) Let $N$ be the number of players, $M=\left[\frac{N}{2}\right]$. Players who occupy the first $M$ places are called strong, and the rest are called weak (places among players with the same score are distributed arbitrarily). Let $X$ be the number of correct matches between strong and weak players. The sum of point... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,920 |
6. What is the maximum number of knights that can be placed on a $5 \times 5$ board such that each one attacks exactly two others? (Provide an example and explain why it is not possible to place more knights.)
## 9 t h g r a d e | 6. Fig. 116 shows the arrangement of 16 knights that satisfies the problem's condition. We will show that it is impossible to place more knights. Let's color the cells of the board in black and white, as shown in Fig. 116. Note that the number of knights on black cells is equal to the number of knights on white cells. ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,921 |
6. Seven cards were drawn from the deck, shown to everyone, shuffled, and then dealt to Grisha and Lesha, three cards each, while the remaining card was a) hidden; b) given to Kolya.
Grisha and Lesha can take turns providing any information about their cards out loud. Can they convey their cards to each other in such ... | 6. a) Let Grisha name two sets of cards: his own and a set of three cards that he does not have, and say: “I have one of these sets.” In this case, Lesha learns Grisha’s cards (since Grisha’s set does not intersect with Lesha’s set, while the second set named by Grisha must intersect).
Now, two situations are possible... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,923 |
6. Can an infinite number of equal convex polyhedra be arranged in a layer bounded by two parallel planes so that no polyhedron can be removed from the layer without moving the others? | 6. We will place regular tetrahedra between the planes, the distance between the opposite edges of which is equal to the distance between the planes. Let one of the edges of each tetrahedron lie in one of the boundary planes, and the opposite edge - in the other. Two tetrahedra can be arranged so that the end of the "u... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,924 |
1. On a grid paper, a rectangle 200 cells wide and 100 cells tall is drawn. It is filled cell by cell, starting from the top-left corner and following a spiral path (turning right when reaching an edge or a filled part, see Fig. 8). Which cell will be filled last? (Specify the number of its row and column. For example,... | 1. First, the outer layer of cells will be shaded, after which a rectangle of $98 \times 198$ cells will remain. This rectangle will also be shaded in a spiral; after shading its outer layer, a rectangle of $96 \times 196$ cells will remain, and so on.
Let's see what will happen after shading 49 layers (Fig. 130). The... | (51,50) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,925 |
1. Is it possible to place four football players on a football field so that the pairwise distances between them are 1, 2, 3, 4, 5, and 6 meters? | 1. Arrange them on a straight line so that the distance between the first and second football players is 2 m, between the second and third - 3 m, between the third and fourth - 1 m (the football players are listed in the order of their arrangement on the line, see Fig. 134).
^{2}$, and $(x-2)^{2}$. The sum of any two of them is greater than zero for any $x$. In Fig. 137, the graphs of the trinomials and their sums are shown.
Comment. Compare with Problem 1 for 11th grade. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,927 |
1. Do there exist three quadratic trinomials such that each of them has two distinct real roots, and the sum of any two trinomials has no real roots? | 1. For example, the trinomials $(x-10)^{2}-1$, $x^{2}-1$, and $(x+10)^{2}-1$ are suitable. Indeed, $x^{2}+(x \pm 10)^{2}=2(x \pm 5)^{2}+50 \geqslant 50$ and $(x-10)^{2}+(x+10)^{2}=2 x^{2}+200 \geqslant 200$. Therefore, the sum of any two trinomials does not turn into zero.
Comments. $1^{\circ}$. Compare with problem 1... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,928 |
2. Is it possible to place 100 points on a plane (first one point, then a second, and so on up to the hundredth) such that no three points are collinear and at any moment the figure consisting of the already placed points has an axis of symmetry? | 2. For example, one can place points on a circle at equal and sufficiently small intervals (as in Fig. 131, only smaller). At each moment, the resulting figure will have an axis of symmetry. Indeed, if the number of points placed is odd, the axis of symmetry is the diameter of the circle passing through the middle poin... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,929 |
2. In a certain country, the total salary of the $10 \%$ highest-paid workers accounts for $90 \%$ of the salary of all workers. Could it be that in each of the regions into which this country is divided, the salary of any $10 \%$ of workers does not exceed $11 \%$ of the total salary paid in that region? | 2. Let's say, for example, everyone in each region receives the same salary, and there is a region where the very $10 \%$ of workers earn $90 \%$ of the total salary.
Let's consider a specific example. Suppose there are only 2 regions. In the first region, there are 1 thousand workers, and in the second region, there ... | 81y | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,930 |
2. A geometric progression is given. It is known that its first, tenth, and thirtieth terms are natural numbers. Is it true that its twentieth term is also a natural number? | 2. Let $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ be a given geometric progression, $q$ be its common ratio. By the condition, $a_{1}$, $a_{10}=a_{1} q^{9}$, and $a_{30}=a_{1} q^{29}$ are natural numbers. Therefore, $q^{9}$ and $q^{29}$ are positive rational numbers. It follows that $q^{2}=\frac{q^{29}}{\left(q^{9}\right)^{... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,932 |
3. Six words are given: ZANOZA, ZIPUNY, KAZINO, KEFAL', OTMEL', SHELEST. 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 ZKNOZA from ZANOZA). How many steps are needed to make all the words the same (nonsense words are allowed)? Provide... | 3. Let's write the words in a column:
|  | | | |
| :---: | :---: | :---: | :---: |
After all the letter replacements in each column, the letters should become the same. The number of rep... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,933 |
3. Inside an angle with vertex $M$, a point $A$ is marked. A ball was launched from this point, which reflected off one side of the angle at point $B$, then off the other side at point $C$, and returned to $A$ (the "angle of incidence" equals the "angle of reflection," see Fig. 9). Prove that the center $O$ of the circ... | 3. First method. Consider on the circle circumscribed around triangle $B C M$, the point $M^{\prime}$ diametrically opposite to point $M$
Fig. 135 (Fig. 135). Since the angle subtended by ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,934 |
3. Provide an example of a polynomial $P(x)$ of degree 2001 for which the identity
$$
P(x)+P(1-x)=1
$$
holds. | 3. Let, for example, $P(x)=\left(x-\frac{1}{2}\right)^{2001}+\frac{1}{2}$. Then
$$
P(1-x)=\left(\frac{1}{2}-x\right)^{2001}+\frac{1}{2}=-\left(x-\frac{1}{2}\right)^{2001}+\frac{1}{2}
$$
therefore $P(x)+P(1-x)=1$. | P(x)+P(1-x)=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,935 |
3. In triangle $ABC$, point $I$ is the center of the inscribed circle, $I'$ is the center of the circle that touches side $AB$ and the extensions of sides $CB$ and $CA$; $L$ and $L'$ are the points where side $AB$ touches these circles. Prove that the lines $IL'$, $I'L$, and the altitude $CH$ of triangle $ABC$ intersec... | 3. We will prove that the lines $I L^{\prime}$ and $I^{\prime} L$ pass through the midpoint of the height $C H$. Let $M$ be the point on the incircle diametrically opposite to point $L$,
Fi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,936 |
4. In triangle $A B C$, the bisector $A K$, the median $B L$, and the altitude $C M$ are drawn. Triangle $K L M$ is equilateral. Prove that triangle $A B C$ is equilateral. | 4. Let's recall that in a right-angled triangle, the median to the hypotenuse is equal to half of the hypotenuse, and
Fig. 132 conversely, if the median is equal to half of the side to whic... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,937 |
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