problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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6. The region is bounded by two cones with a common base, the height of which is twice less than the slant height. Three spheres are placed in the region, touching each other externally. Two spheres are identical and touch both cones, while the third touches the boundary of the region. What is the maximum ratio of the ... | Answer: $\frac{7-\sqrt{22}}{3}$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $BO$ be the height of one of the cones, $R$ be the radius of the base of the cones, $r$ a... | \frac{7-\sqrt{22}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,988 |
6. (40 points) On a circle with center $O$, points $A$ and $B$ are taken such that the angle $A O B$ is $60^{\circ}$. From an arbitrary point $R$ on the smaller arc $A B$, segments $R X$ and $R Y$ are drawn such that point $X$ lies on segment $O A$ and point $Y$ lies on segment $O B$. It turns out that the angle $R X O... | First solution: Let $r_{1}$ be the radius of the circle centered at point $O$. Since $\angle R X O + \angle R Y O = 180^{\circ}$, the quadrilateral $R X O Y$ is cyclic; denote the radius of its circumscribed circle by $r_{2}$. The diagonal $O R$ equals $r_{1}$ and subtends an arc on which the inscribed angle $R X O$ is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,989 |
1. The picture shows several circles connected by segments. Sasha chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime with... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among seven numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,990 |
2. For $x, y, z \in (0,2]$, find the maximum value of the expression
$$
A=\frac{\left(x^{3}-6\right) \sqrt[3]{x+6}+\left(y^{3}-6\right) \sqrt[3]{y+6}+\left(z^{3}-6\right) \sqrt[3]{z+6}}{x^{2}+y^{2}+z^{2}}
$$ | Answer: 1.
Solution. For $x \in(0,2]$, the inequalities $\sqrt[3]{x+6} \leqslant 2$ and $x^{3} \leqslant 2 x^{2}$ hold, from which
$$
\left(x^{3}-6\right) \sqrt[3]{x+6} \leqslant 2\left(2 x^{2}-6\right)
$$
Similarly, the other two terms in the numerator of $A$ are estimated. Therefore,
$$
A \leqslant 2 \cdot \frac{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,991 |
3. The diagonals of the inscribed quadrilateral $ABCD$ intersect at point $O$. Inside triangle $AOB$, a point $K$ is chosen such that line $KO$ is the bisector of angle $CK$. Ray $DK$ intersects the circumcircle of triangle $COK$ again at point $L$, and ray $CK$ intersects the circumcircle of triangle $DOK$ again at po... | Answer: 1.
Solution. Let $r_{1}$ and $r_{2}$ be the radii of the circumcircles of triangles $C O K$ and $D O K$, respectively. Note that
$$
\angle L K O=180^{\circ}-\angle D K O=180^{\cir... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,992 |
4. A natural number $x$ in a base $r$ system ( $r \leqslant 36$ ) has the form $\overline{p p q q}$, and $2 q=5 p$. It turns out that the $r$-ary representation of the number $x^{2}$ is a seven-digit palindrome with a zero middle digit. (A palindrome is a number that reads the same from left to right and from right to ... | Answer: 36.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. Let $p=2 s, q=5 s$. Then $x=\overline{p p q q}_{r}=\left(p r^{2}+q\right)(r+1)=s\left(2 r^{2}+5\right)(r+1)$. From the condition on $x^{2}$, we get the equality
$$
s^{2}\left(2 r^{2}+5\right)^{2}\left(r^{2}+2 r+1\right)=a\left(1+r^{... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,993 |
5. For which $n$ can a square grid $n \times n$ be divided into one $2 \times 2$ square and some number of strips of five cells, such that the square is adjacent to the side of the board? | Answer: for all $n$ that give a remainder of 2 when divided by 5.
Solution. If a board $n \times n$ can be cut into one $2 \times 2$ square and some number of strips of five cells, then $n^{2}=2^{2}+5 m$, from which $n$ gives a remainder of 2 or 3 when divided by 5. Suppose that $n=5 k+3$ and the board can be cut in t... | 5k+2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,994 |
1. In the picture, several circles are drawn, connected by segments. Sasha chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a+b$ must be coprime with $n$; if ... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among eight numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,995 |
2. Given $x, y, z \geqslant 3$ find the minimum value of the expression
$$
A=\frac{\left(x^{3}-24\right) \sqrt[3]{x+24}+\left(y^{3}-24\right) \sqrt[3]{y+24}+\left(z^{3}-24\right) \sqrt[3]{z+24}}{x y+y z+z x}
$$ | Answer: 1.
Solution. For $x \geqslant 3$, the inequalities $\sqrt[3]{x+24} \geqslant 3$ and $x^{3} \geqslant 3 x^{2}$ hold, from which
$$
\left(x^{3}-24\right) \sqrt[3]{x+24} \geqslant 3\left(3 x^{2}-24\right)
$$
Similarly, the other two terms in the numerator of $A$ are estimated. Therefore,
$$
A \geqslant 3 \cdot... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,996 |
3. On the sides $AB$ and $AC$ of an acute-angled triangle $ABC$, points $K$ and $L$ are marked such that the quadrilateral $B K L C$ is cyclic. Inside this quadrilateral, a point $M$ is chosen so that the line $AM$ is the bisector of angle $BMC$. The ray $BM$ intersects the circumcircle of triangle $AMC$ again at point... | Answer: 1.
Solution. Let $r_{1}$ and $r_{2}$ be the radii of the circumcircles of triangles $A M B$ and $A M C$, respectively. Since $A M$ is the bisector of angle $B M C$, the following eq... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,997 |
4. A natural number $x$ in a base $r(r \leqslant 400)$ numeral system has the form $\overline{p p q q}$, and $7 q=17 p$. It turns out that the $r$-ary representation of the number $x^{2}$ is a seven-digit palindrome with a zero middle digit. (A palindrome is a number that reads the same from left to right and from righ... | Answer: 400.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. Let $p=7 s, q=17 s$. Then $x=\overline{p p q q}_{r}=\left(p r^{2}+q\right)(r+1)=s\left(7 r^{2}+17\right)(r+1)$. From the condition on $x^{2}$, we get the equality
$$
s^{2}\left(7 r^{2}+17\right)^{2}\left(r^{2}+2 r+1\right)=a\left(1... | 400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,998 |
5. For which $n$ can a square grid $n \times n$ be divided into one $3 \times 3$ square and some number of strips of seven cells, such that the square is adjacent to a side of the board? | Answer: for all $n$ that give a remainder of 3 when divided by 7.
Solution. If an $n \times n$ board can be cut into one $3 \times 3$ square and some number of strips of seven cells, then $n^{2}=3^{2}+7 m$, from which it follows that $n$ gives a remainder of 3 or 4 when divided by 7. Suppose $n=7 k+4$ and the board ca... | 7k+3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,999 |
1. In the picture, several circles are drawn, connected by segments. Tanya chooses a natural number n and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers a and b are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with n; if ... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, t... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,000 |
2. Given $x, y, z \geqslant 1$, find the minimum value of the expression
$$
A=\frac{\sqrt{3 x^{4}+y}+\sqrt{3 y^{4}+z}+\sqrt{3 z^{4}+x}-3}{x y+y z+z x}
$$ | Answer: 1.
Solution. Note that for $x, y \geqslant 1$
$$
\sqrt{3 x^{4}+y}-1=\sqrt{x^{4}+2 x^{4}+y}-1 \geqslant \sqrt{x^{4}+2 x^{2}+1}-1=x^{2}
$$
Similarly, it can be shown that
$$
\sqrt{3 y^{4}+z}-1 \geqslant y^{2}, \quad \sqrt{3 z^{4}+x}-1 \geqslant z^{2}
$$
Therefore,
$$
A \geqslant \frac{x^{2}+y^{2}+z^{2}}{x y... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,001 |
3. Given a cyclic quadrilateral $A B C D$ with perpendicular diagonals. On the circle circumscribed around it, a point $E$ is marked, diametrically opposite to $D$, and the segments $A B$ and $D E$ do not intersect. Find the ratio of the areas of triangle $B C D$ and quadrilateral $A B E D$. | Answer: 1.
Solution. Let $N$ be the intersection point of $A C$ and $B D$, and $H$ be the foot of the perpendicular dropped from point $E$ to line $A C$. By the problem's condition, $D E$ i... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,002 |
4. The number $x=9999$ is written on the board in an even base $r$. Vasya found out that the $r$-ary representation of $x^{2}$ is an eight-digit palindrome, where the sum of the second and third digits is 24. (A palindrome is a number that reads the same from left to right and from right to left). For which $r$ is this... | Answer: $r=26$.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. According to the condition, there exist such $r$-ary digits $a, b, c, d$ that $b+c=24$ and
$$
81(r+1)^{2}\left(r^{2}+1\right)^{2}=a\left(r^{7}+1\right)+b\left(r^{6}+r\right)+c\left(r^{5}+r^{2}\right)+d\left(r^{4}+r^{3}\right)
$$... | 26 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,003 |
5. Given a square table of size $2021 \times 2021$. Each of its cells is colored in one of $n$ colors. It is known that for any four cells of the same color located in the same column, there are no cells of the same color to the right of the top one and to the left of the bottom one. What is the smallest $n$ for which ... | Answer: 506.
Solution. Consider some specific color (say, blue). In each column, mark the three lower blue cells with a cross, and the rest with a zero (if there are fewer than four blue cells in the column, mark all cells with a cross). Note that in any row, there is no more than one blue cell with a zero (otherwise,... | 506 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,004 |
1. The picture shows several circles connected by segments. Tanya chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with $n$; if $... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among six numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, th... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,005 |
2. For $x, y, z \in(0,1]$ find the maximum value of the expression
$$
A=\frac{\sqrt{8 x^{4}+y}+\sqrt{8 y^{4}+z}+\sqrt{8 z^{4}+x}-3}{x+y+z}
$$ | Answer: 2.
Solution. Note that for $x, y \in(0,1]$
$$
\sqrt{8 x^{4}+y}-1=\sqrt{4 x^{4}+4 x^{4}+y}-1 \leqslant \sqrt{4 x^{2}+4 x+1}-1=2 x .
$$
Similarly, it can be shown that
$$
\sqrt{3 y^{4}+z}-1 \leqslant 2 y, \quad \sqrt{3 z^{4}+x}-1 \leqslant 2 z
$$
Therefore,
$$
A \leqslant \frac{2 x+2 y+2 z}{x+y+z}=2
$$
Equ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,006 |
3. The diagonal $AC$ of the inscribed quadrilateral $ABCD$ is the diameter of the circumscribed circle $\omega$ around it. A line perpendicular to the segment $BC$ was drawn from point $D$, intersecting the circle $\omega$ again at point $E$. Find the ratio of the areas of triangle $BCD$ and quadrilateral $ABEC$. | Answer: 1.
Solution. Let $N$ be the intersection point of $B C$ and $D E$, and $H$ be the foot of the perpendicular dropped from point $A$ to line $D E$. By the problem's condition, $A C$ i... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,007 |
4. The board has the number 5555 written in an even base $r$ ($r \geqslant 18$). Petya found out that the $r$-ary representation of $x^2$ is an eight-digit palindrome, where the difference between the fourth and third digits is 2. (A palindrome is a number that reads the same from left to right and from right to left).... | Answer: $r=24$.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. According to the condition, there exist such $r$-ary digits $a, b, c, d$ that $d-c=2$ and
$$
25(r+1)^{2}\left(r^{2}+1\right)^{2}=a\left(r^{7}+1\right)+b\left(r^{6}+r\right)+c\left(r^{5}+r^{2}\right)+d\left(r^{4}+r^{3}\right)
$$
... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,008 |
5. Given a square table of size $2023 \times 2023$. Each of its cells is colored in one of $n$ colors. It is known that for any six cells of the same color located in the same row, there are no cells of the same color above the leftmost of them and below the rightmost of them. What is the smallest $n$ for which this is... | Answer: 338.
Solution. Consider some specific color (say, blue). In each row, mark the five leftmost blue cells with a cross, and the rest with a zero (if there are fewer than six blue cells in the row, mark all cells with a cross). Note that in any column, there is no more than one blue cell with a zero (otherwise, f... | 338 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,009 |
1. The picture shows several circles connected by segments. Nastya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the difference \( a - b \) must be copr... | Answer: $n=5 \cdot 7 \cdot 11=385$.
Solution. We will make two observations.
1) $n$ is not divisible by 2 and 3. Among seven numbers, there are always three numbers of the same parity. If $n$ is even, then they must be pairwise connected. Moreover, among seven numbers, there will be three numbers that give the same r... | 385 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,010 |
2. For $x, y \in(0,1]$, find the maximum value of the expression
$$
A=\frac{\left(x^{2}-y\right) \sqrt{y+x^{3}-x y}+\left(y^{2}-x\right) \sqrt{x+y^{3}-x y}+1}{(x-y)^{2}+1}
$$ | Answer: 1.
Solution. Note that
$$
y+x^{3}-x y-x^{2}=\left(y-x^{2}\right)(1-x)
$$
If $y \geqslant x^{2}$, then
$$
\sqrt{y+x^{3}-x y} \geqslant x \quad \text { and } \quad\left(x^{2}-y\right)\left(\sqrt{y+x^{3}-x y}-x\right) \leqslant 0
$$
and when $y<x^{2}$
$$
\sqrt{y+x^{3}-x y} \leqslant x \quad \text { and } \qu... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,011 |
3. On the side $AB$ of an acute-angled triangle $ABC$, a point $M$ is marked. A point $D$ is chosen inside the triangle. Circles $\omega_{A}$ and $\omega_{B}$ are circumscribed around triangles $AMD$ and $BMD$ respectively. The side $AC$ intersects the circle $\omega_{A}$ again at point $P$, and the side $BC$ intersect... | # Answer: 1.
Solution. Notice that
$$
\angle P A M=180^{\circ}-\angle P D M=\angle R D M=\angle R B M
$$
from which it follows that $B R \| A C$. Therefore, the distances from points $B$ ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,012 |
4. In a number system with base $r$ ( $r \leqslant 100$ ), a natural number $x$ is a two-digit number with identical digits. It turns out that the $r$-ary representation of the number $x^{2}$ is four-digit, and the extreme digits are the same, while the middle ones are zero. For which $r$ is this possible? | Answer: $r=2$ or $r=23$.
Solution. Let's write the condition as $a^{2}(1+r)^{2}=b\left(1+r^{3}\right)$, where $a, b-r$ are digits. By canceling out $1+r$, we get
$$
a^{2}(1+r)=b\left(1-r+r^{2}\right)=b((1+r)(r-2)+3)
$$
Note that the GCD $(1+r,(1+r)(r-2)+3)$ is 1 or 3. In the first case, $b \vdots (r+1)$, which is im... | r=2orr=23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,013 |
5. For a parallelepiped $a \times b \times c$ with faces divided into unit cells, there is also a large number of three-cell strips, which can be bent along the cell boundaries. For which $a, b$, and $c$ can the three faces of the parallelepiped, sharing a common vertex, be completely covered by the strips without over... | Answer: Among the numbers $a, b, c$, at least two are divisible by 3.
Solution. Suppose at least two dimensions of the parallelepiped are divisible by 3. Then any face of the parallelepiped has a side that is a multiple of 3. This means we can completely cover the face with strips, laying them parallel to this side wi... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,014 |
2. For $x, y \in[1,3]$ find the minimum value of the expression
$$
A=\frac{\left(3 x y+x^{2}\right) \sqrt{3 x y+x-3 y}+\left(3 x y+y^{2}\right) \sqrt{3 x y+y-3 x}}{x^{2} y+y^{2} x}
$$ | Answer: 4.
Solution. Note that
$$
3 x y+x-3 y-x^{2}=(3 y-x)(x-1) \geqslant 0
$$
since $x \geqslant 1$ and $x \leqslant 3 \leqslant 3 y$. Then
$$
\left(3 x y+x^{2}\right) \sqrt{3 x y+x-3 y} \geqslant 3 x^{2} y+x^{3} \quad \text { and, similarly, } \quad\left(3 x y+y^{2}\right) \sqrt{3 x y+y-3 x} \geqslant 3 y^{2} x+... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,016 |
3. On the side $B C$ of triangle $A B C$ with obtuse angle $C$, a point $M$ is marked. Point $D$ is chosen such that triangle $B C D$ is acute, and points $A$ and $D$ lie on opposite sides of line $B C$. Circles $\omega_{B}$ and $\omega_{C}$ are circumscribed around triangles $B M D$ and $C M D$ respectively. Side $A B... | # Answer: 1.
Solution. Notice that
$$
\angle S B M=180^{\circ}-\angle S D M=\angle M D Q=180^{\circ}-\angle M C Q
$$
from which $S B \| A C$. Therefore, the distances from points $B$ and $... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,017 |
4. The record of a natural number x in a base-23 numeral system consists of $2 m$ identical digits. It turned out that in the 23-ary representation of the number $x^{2}$, the extreme digits are the same, and the other $4 m-2$ digits are zero. Find all such numbers x. Provide the answer in the 23-ary numeral system. (Di... | Answer: $x=D D_{23}$.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. Let $n=2 m$. According to the condition, for some 23-ary digits $a$ and $b$, the equality
$$
a^{2}\left(1+23+23^{2}+\ldots+23^{n-1}\right)^{2}=b\left(1+23^{2 n-1}\right)
$$
or, in another form,
$$
a^{2}\left(23^{n}-1\rig... | DD_{23} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,018 |
5. For a parallelepiped $a \times b \times c$ with faces divided into unit cells, there is also a large number of five-cell strips, which can be bent along the cell boundaries. For which $a, b$, and $c$ can the three faces of the parallelepiped, sharing a common vertex, be completely covered by the strips without overl... | Answer: Among the numbers $a, b, c$, at least two are divisible by 5.
Solution. Suppose at least two dimensions of the parallelepiped are divisible by 5. Then any face of the parallelepiped has a side that is a multiple of 5. Therefore, we can completely cover the face with strips, laying them parallel to this side wi... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,019 |
1. In the picture, several circles are drawn, connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be co... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among any five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) $n$ has at least two distinct pr... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,020 |
2. For $x, y, z \in(0,1]$ find the minimum value of the expression
$$
A=\frac{(x+2 y) \sqrt{x+y-x y}+(y+2 z) \sqrt{y+z-y z}+(z+2 x) \sqrt{z+x-z x}}{x y+y z+z x}
$$ | Answer: 3.
Solution. Note that for $x, y \in(0,1]$
$$
\sqrt{x+y-x y}=\sqrt{x+y(1-x)} \geqslant \sqrt{x} \geqslant x
$$
Then $(x+2 y) \sqrt{x+y-x y} \geqslant x^{2}+2 x y$ and similarly,
$$
(y+2 z) \sqrt{y+z-y z} \geqslant y^{2}+2 y z, \quad(z+2 x) \sqrt{z+x-z x} \geqslant z^{2}+2 z x
$$
Therefore,
$$
A \geqslant ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,021 |
3. On the sides $B C$ and $A D$ of the inscribed quadrilateral $A B C D$, points $K$ and $M$ are marked respectively, such that $B K: K C = A M: M D$. On the segment $K M$, a point $L$ is chosen such that $K L: L M = B C: A D$. Find the ratio of the areas of triangles $A C L$ and $B D L$, given that $A C = p$ and $B D ... | Answer: $\frac{p}{q}$.
Solution. Let $O$ be the point of intersection of $A C$ and $B D$. Since $A B C D$ is a cyclic quadrilateral, triangles $A O D$ and $B O C$ are similar by two angles, ... | \frac{p}{q} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,022 |
4. A natural number $x$ in a base $r$ numeral system (where $r>3$) has the form $\overline{p p q q}$, and $q=2p$. It turns out that the $r$-ary representation of the number $x^2$ is a seven-digit palindrome $c$ with three identical middle digits. (A palindrome is a number that reads the same from left to right and from... | Answer: $r=3 n^{2}$, where $n-$ is a natural number, $n>1$.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. Notice that
$$
x=\overline{p p q q}_{r}=\left(p r^{2}+q\right)(r+1)=p\left(r^{2}+2\right)(r+1)
$$
From the condition on $x^{2}$, we get the equality
$$
p^{2}\left(r^{2}+2\right)^{2}\... | 3n^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,023 |
1. In the picture, several circles are drawn, connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be co... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. If there are three even and three odd numbers, then each of these triples forms a cycle. Suppose there are four numbers $a, b, c, d$ of the same parity. Then all of them are pairwise connected, and we again get two th... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,025 |
2. Given $x, y, z > 0$, find the maximum value of the expression
$$
A=\frac{(x-y) \sqrt{x^{2}+y^{2}}+(y-z) \sqrt{y^{2}+z^{2}}+(z-x) \sqrt{z^{2}+x^{2}}+\sqrt{2}}{(x-y)^{2}+(y-z)^{2}+(z-x)^{2}+2}
$$ | Answer: $\frac{1}{\sqrt{2}}$.
Solution. If $x \geqslant y$, then
$$
\sqrt{x^{2}+y^{2}} \leqslant \sqrt{2 x^{2}}=x \sqrt{2} \quad \text { and } \quad(x-y) \sqrt{x^{2}+y^{2}} \leqslant \sqrt{2}\left(x^{2}-x y\right),
$$
and when $x < y$,
$$
\sqrt{x^{2}+y^{2}} < \sqrt{2 x^{2}}=x \sqrt{2} \quad \text { and } \quad(x-y)... | \frac{1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,026 |
3. On the sides $AB$ and $CD$ of the inscribed quadrilateral $ABCD$, which is not a trapezoid, points $P$ and $R$ are marked, respectively, such that $AP: PB = CR: RD$. On the segment $PR$, a point $Q$ is chosen such that $PQ: QR = AB: CD$. Find the ratio of the areas of triangles $AQD$ and $BQC$, given that $AD = x$ a... | Answer: $\frac{x}{y}$.
Solution. Let $O$ be the point of intersection of the lines $D A$ and $C B$. For definiteness, we assume that it lies on the rays $D A$ and $C B$. Since $A B C D$ is c... | \frac{x}{y} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,027 |
4. The record of a natural number $x$ in a base $r$ system ( $r \leqslant 70$ ) is obtained by $n$ repeated occurrences of a certain pair of digits, where $n$ is a natural number. It turns out that the $r$-ary representation of $x^{2}$ consists of $4 n$ ones. Find all pairs $(r, x)$ for which this is possible. | Answer: $\left(7,26_{7}\right)$.
Solution. Let $x=\overline{a b \ldots a b_{r}}$. Set $y=\overline{a b}_{r}$ and write the condition as
$$
\begin{aligned}
y^{2}\left(1+r^{2}+\ldots+r^{2(n-1)}\right)^{2}=1+r+\ldots+r^{4 n-1} & \Longleftrightarrow y^{2} \cdot \frac{\left(r^{2 n}-1\right)^{2}}{\left(r^{2}-1\right)^{2}}=... | (7,26_{7}) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,028 |
5. A board $m \times n(m, n>15)$ is cut into figures consisting of ten unit squares (the figures can be rotated and flipped). For which $m$ and $n$ is this possible? | Answer: for all $m$ and $n$, among which there is a multiple of 4 and a multiple of 5.
Solution. Suppose that the board $m \times n$ can be cut in the required way. Since each figure consists of ten cells, the total number of cells on the board is a multiple of 10. Then among $m$ and $n$ there is an even number (let t... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,029 |
1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-f(x)=0$ has exactly three solutions. Find the ordinate of the vertex of the trinomial $f(x)$. | # Answer: 0.
Solution. Suppose the leading coefficient of the trinomial is positive. Note that $(f(x))^{3}-f(x)=f(x) \cdot(f(x)-1) \cdot(f(x)+1)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-1$, and fewer roots than the equation $f(x)=1$. It is also clear that no two equations have common roots. Ther... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,030 |
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers, the difference of whose square roots is less than 1? | Answer: 45.
Solution. We will show that $k=45$ works. Let's divide the numbers from 1 to 2016 into 44 groups:
\[
\begin{aligned}
& (1,2,3), \quad(4,5,6,7,8), \quad(9,10, \ldots, 15), \ldots, \quad\left(k^{2}, k^{2}+1, \ldots, k^{2}+2 k\right), \ldots, \\
& (1936,1937, \ldots, 2016)
\end{aligned}
\]
Since there are 4... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,031 |
3. Real numbers $x, y$ and $z$ satisfy the conditions $x+y+z=0$ and $|x|+|y|+|z| \leqslant 1$. Prove the inequality
$$
x+\frac{y}{2}+\frac{z}{3} \leqslant \frac{1}{3}
$$ | Solution. Indeed,
$$
3\left(x+\frac{y}{2}+\frac{z}{3}\right)=2(x+y+z)+x-\frac{y}{2}-z \leqslant 2(x+y+z)+|x|+|y|+|z|=1
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,032 |
4. Is it possible to arrange the numbers 1 and -1 in a $300 \times 300$ table such that the absolute value of the sum of all numbers in the table is less than 30000, and in each of the $3 \times 5$ or $5 \times 3$ rectangles, the absolute value of the sum of the numbers is greater than 3? | Answer: No.
Solution. Since the sum of the numbers in a $3 \times 5$ rectangle is odd, if its modulus is greater than three, it must be at least five. Suppose such an arrangement exists. Note that in this arrangement, either there is no row consisting entirely of +1, or there is no column consisting entirely of -1 (if... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,033 |
5. Given a parallelogram $A B C D$. The circumcircle $\omega$ of triangle $A B C$ intersects side $A D$ again and the extension of side $D C$ at points $P$ and $Q$ respectively. Prove that the center of the circumcircle of triangle $P D Q$ lies on $\omega$. | Solution. Trapezoids $A B C P$ and $A C Q D$ are inscribed in the circle $\omega$, so they are isosceles and, in particular, their diagonals are equal. Then $P B=A C=B Q$,
and point $B$ lies on the perpendicular bisector of $P Q$. Let the point of its intersection with the circle $\omega$ be denoted by $O$. Then points... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,034 |
6. The numbers $a, b$ and $c$ satisfy the equation $(a+3)^{2}+(b+4)^{2}-(c+5)^{2}=a^{2}+b^{2}-c^{2}$. Prove that both sides of the equation are perfect squares. | Solution. Expanding the brackets in the condition, we get $6 a+8 b-10 c=0$. Then
$$
5^{2}\left(a^{2}+b^{2}-c^{2}\right)=(5 a)^{2}+(5 b)^{2}-(3 a+4 b)^{2}=(4 a-3 b)^{2} .
$$
Therefore, $4 a-3 b$ is divisible by 5, i.e., $4 a-3 b=5 k$ for some integer $k$ and
$$
a^{2}+b^{2}-c^{2}=\frac{(4 a-3 b)^{2}}{5^{2}}=k^{2}
$$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,035 |
1. Find all polynomials $f(x)$ of degree not higher than two such that for any real $x$ and $y$, the difference of which is rational, the difference $f(x)-f(y)$ is also rational. | Answer: linear with a rational leading coefficient or constants.
Solution. Let $f(x)=a x^{2}+b x+c$. Then for any $x$ the difference
$$
f(x+1)-f(x)=\left(a(x+1)^{2}+b(x+1)+c\right)-\left(a x^{2}+b x+c\right)=2 a x+a+b
$$
is rational. If $a \neq 0$ this is impossible, since the equation $2 a x+a+b=\sqrt{2}$ has a sol... | f(x)=bx+ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,036 |
2. What is the maximum number of different numbers from 1 to 1000 that can be chosen so that the difference between any two chosen numbers is not equal to any of the numbers 4, 5, 6. | Answer: 400.
Solution. Consider ten consecutive natural numbers. We will prove that no more than four of them are selected. If at least five numbers are selected, then three of them have the same parity, but then their pairwise differences cannot be only 2 and 8. Indeed, if $a<b<c$, then $b-a=2$ and $c-a=8$, but then ... | 400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,037 |
3. Find the maximum value of the expression
$$
\frac{a}{x}+\frac{a+b}{x+y}+\frac{a+b+c}{x+y+z}
$$
where $a, b, c \in[2,3]$, and the triplet of numbers $x$, $y$, and $z$ is some permutation of the triplet of numbers $a, b, c$. | Answer: 3.75.
Solution. We will prove that
$$
\frac{a}{x}+\frac{a+b}{x+y}+\frac{a+b+c}{x+y+z} \leqslant \frac{15}{4}
$$
Notice that the last fraction equals 1 and $a / x \leqslant 3 / 2$. It remains to show that $\frac{a+b}{x+y} \leqslant \frac{5}{4}$. Since the numbers $x$ and $y$ are some pair of numbers from $a, ... | \frac{15}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,038 |
4. What is the maximum number of chess kings that can be placed on a $12 \times 12$ board so that each king attacks exactly one of the others? | Answer: 56.
Solution. Note that two kings attack each other if and only if their cells share at least one vertex. For each pair of attacking kings, mark the vertices of the cells they occupy. In this case, no fewer than six vertices are marked for each such pair. Since different pairs of kings mark different vertices ... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,039 |
5. Given a parallelogram $A B C D$. A perpendicular $B O$ is dropped from vertex $B$ to side $A D$. Circle $\omega$ with center at point $O$ passes through points $A, B$ and intersects the extension of side $A D$ at point $K$. Segment $B K$ intersects side $C D$ at point $L$, and ray $O L$ intersects circle $\omega$ at... | Solution. Since $O A=O B=O K$ and $\angle B O A=90^{\circ}$, the right triangles $A B O$ and $B O K$ are isosceles, and thus $\angle B A K=\angle B K A=45^{\circ}$.
By the property of angles in a parallelogram, $\angle K B C=\angle C D K=45^{\circ}$. Therefore, quadrilateral $B D K C$ is cyclic. Hence, $\angle B K C=\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,040 |
6. Natural numbers \(a\) and \(b\) are such that \(a^2 + b^2 + a\) is divisible by \(ab\). Prove that \(a\) is a perfect square. | Solution. If $n=a^{2}+b^{2}+a$ is divisible by $ab$, then $n$ is also divisible by $a$. Therefore, $b^{2}$ is divisible by $a$. So, $b^{2}=ka$. We will show that $a$ and $k$ are coprime. Suppose the opposite. Then there exists a prime number $p$ such that $a$ and $k$ are divisible by $p$. In this case, $b^{2}$ is divis... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,041 |
1. Does there exist a quadratic trinomial $f(x)$ with integer coefficients such that $f(f(\sqrt{3}))=0$? | Answer: Yes.
Solution. For example, the polynomial $f(x)=2 x^{2}-8$ works. Indeed, $f(\sqrt{3})=-2$, which is a root of $f(x)$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,042 |
2. From the numbers 1, 2, 3, ..., 2016, $k$ numbers are chosen. What is the smallest $k$ such that among the chosen numbers, there will definitely be two numbers whose difference is greater than 672 and less than 1344? | Answer: 674.
Solution. Let $n=672$. Then $2 n=1344$ and $3 n=2016$. Suppose it is possible to choose $674=n+2$ numbers such that no required pair of numbers can be found among them. Let $m$ be the smallest of the chosen numbers. Then the numbers $m+n+1, m+n+2$, $\ldots, m+2 n-1$ are not chosen. Remove them and the num... | 674 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,043 |
3. Non-negative numbers $x$ and $y$ satisfy the condition $x+y \leqslant 1$. Prove that $12 x y \leqslant 4 x(1-y)+9 y(1-x)$. | Solution. From the inequality $x+y \leqslant 1$ we conclude that $y \leqslant 1-x$ and $x \leqslant 1-y$. Therefore, by the inequality of means for two numbers
$$
4 x(1-y)+9 y(1-x) \geqslant 4 x^{2}+9 y^{2} \geqslant 12 x y
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,044 |
4. Given a board $2016 \times 2016$. For what smallest $k$ can the cells of the board be colored in $k$ colors such that
1) one of the diagonals is painted in the first color;
2) cells symmetric with respect to this diagonal are painted in the same color;
3) any two cells located in the same row on opposite sides of th... | Answer: 11.
Solution. Let the cells of the diagonal running from the top left corner to the bottom right corner be painted in the first color. Denote by $C_{i}$ the set of colors in which the cells of the $i$-th row, located to the left of the diagonal of a single color, are painted. We will prove that $C_{i} \neq C_{... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,045 |
5. Given a parallelogram $A B C D$. A circle is tangent to side $A C$ of triangle $A B C$, as well as the extensions of sides $B A$ and $B C$ at points $P$ and $S$ respectively. Segment $P S$ intersects sides $D A$ and $D C$ at points $Q$ and $R$. Prove that the incircle of triangle $C D A$ touches sides $A D$ and $D C... | Solution. We will prove that the incircle of triangle $A D C$ touches side $A D$ at point $Q$. Since $B P$ and $B S$ are equal tangent segments, triangle $B P S$ is isosceles, and therefore, triangle $A P Q$, cut off from it by the line $A Q$ parallel to $B S$, is also isosceles. Thus, $A P=A Q$. Using the equality of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,046 |
6. For which natural numbers $n$ is the number
$$
\frac{1! \cdot 2! \cdot 3! \cdot \ldots \cdot (2n)!}{(n+1)!}
$$
a perfect square? (As usual, $n!$ denotes the product of all natural numbers not exceeding $n$. For example, $4! = 1 \cdot 2 \cdot 3 \cdot 4$.) | Answer: for numbers of the form $n=4 k(k+1)$ and $n=2 k^{2}-1$.
Solution. It is sufficient to understand for which $n$ the number $A=(n+1)!\cdot 1!\cdot 2!\cdot 3!\cdot \ldots \cdot(2 n)!$ is a perfect square. Note that $(2 k-1)!(2 k)!=2 k \cdot((2 k-1)!)^{2}$. Therefore,
$$
A=(n+1)!\cdot 2 \cdot 4 \cdot 6 \cdot \ldo... | n=4k(k+1)n=2k^{2}-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,047 |
1. Quadratic trinomials $f(x)$ and $g(x)$ are such that $[f(x)]=[g(x)]$ for all $x$. Prove that $f(x)=g(x)$ for all $x$. (Here $[a]$ denotes the integer part of $a$, that is, the greatest integer not exceeding $a$). | Solution. Let $h(x)=f(x)-g(x)=a x^{2}+b x+c$. From the equality $[f(x)]=[g(x)]$ it follows that $|h(x)|=|f(x)-g(x)| \leqslant 1$. However, a quadratic trinomial can take arbitrarily large values in modulus. Therefore, $a=0$. On the other hand, for $b \neq 0$ the value of $|b x+c|$ at the point $-c / b+2 / b$ is two, so... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,048 |
2. At a cactus lovers' meeting, 80 cactus enthusiasts presented their collections, each consisting of cacti of different species. It turned out that no species of cactus is present in all collections at once, but any 15 people have cacti of the same species. What is the smallest total number of cactus species that can ... | Answer: 16.
Solution. We will show that 16 cacti were possible. Number the cacti from 1 to 16. Let the 1st cactus lover have all cacti except the first; the 2nd cactus lover have all except the second cactus; the 15th cactus lover have all except the fifteenth cactus; and the cactus lovers from the 16th to the 80th ha... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,049 |
3. For positive numbers $x$ and $y$, prove the inequality
$$
1 \leqslant \frac{(x+y)\left(x^{3}+y^{3}\right)}{\left(x^{2}+y^{2}\right)^{2}} \leqslant \frac{9}{8}
$$ | Solution. The left inequality is equivalent to the inequality $\left(x^{2}+y^{2}\right)^{2} \leqslant(x+y)\left(x^{3}+y^{3}\right)$, which after expanding the brackets, combining like terms, and dividing by $x y$
reduces to the inequality $2 x y \leqslant x^{2}+y^{2}$. Multiply both sides of the right inequality by $\l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,050 |
4. The cells of a $2015 \times 2015$ board are colored in a checkerboard pattern such that the corner cells are black. A token is placed on one of the black cells, and another black cell is marked. In one move, it is allowed to move the token to an adjacent cell. Is it always possible to traverse all the cells of the b... | Answer: Yes.
Solution. By induction on $n$, we will prove that for any two black cells $A$ and $B$ on the board $(2 n+1) \times(2 n+1)$, there exists a path for the chip starting at $A$, passing through all cells of the board, and ending at $B$.
Base case $n=1$. There are two possible placements of black cells $A$ an... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 8,051 |
5. Given a parallelogram $A B C D$ with angle $\angle B$ equal to $60^{\circ}$. Point $O$ is the center of the circumscribed circle of triangle $A B C$. Line $B O$ intersects the bisector of the exterior angle $\angle D$ at point $E$. Find the ratio $\frac{B O}{O E}$. | Answer: $1 / 2$.
Solution. Let $P$ be the point of intersection of the circumcircle of triangle $ABC$ with line $CD$, distinct from $C$, and let $F$ be the point diametrically opposite to point $B$. Then $ABCP$ is an inscribed trapezoid. Therefore, it is isosceles and $\angle APD = \angle BAP = \angle ABC = 60^{\circ}... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,052 |
6. Natural numbers $a$ and $b$ are such that $a^{3}+b^{3}+ab$ is divisible by $ab(a-b)$. Prove that $\operatorname{LCM}(a, b)$ is a perfect square. (LCM - least common multiple). | Solution. Let $d$ be the greatest common divisor of the numbers $a$ and $b$. Set $a = d a_{1}$ and $b = d b_{1}$. Then the numbers $a_{1}$ and $b_{1}$ are coprime, $\operatorname{HOK}(a, b) = d a_{1} b_{1}$, and the divisibility of $a^{3} + b^{3} + ab$ by $ab(a - b)$ implies the divisibility of the number $n = d a_{1}^... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,053 |
1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-4 f(x)=0$ has exactly three solutions. How many solutions does the equation $(f(x))^{2}=1$ have? | Answer: 2.
Solution. Suppose the leading coefficient of the polynomial is positive. Note that $(f(x))^{3}-4 f(x)=f(x) \cdot(f(x)-2) \cdot(f(x)+2)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-2$, and fewer roots than the equation $f(x)=2$. It is also clear that no two equations have common roots. The... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,054 |
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers $a$ and $b$ such that $|\sqrt[3]{a}-\sqrt[3]{b}|<1$? | Answer: 13.
Solution. We will show that $k=13$ works. Divide the numbers from 1 to 2016 into 12 groups:
\[
\begin{aligned}
& (1,2,3,4,5,6,7), \quad(8,9, \ldots, 26), \quad(27,28, \ldots, 63), \ldots \\
& \left(k^{3}, k^{3}+1, \ldots,(k+1)^{3}-1\right), \ldots,(1728,1729, \ldots, 2016)
\end{aligned}
\]
Since there ar... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,055 |
3. Real numbers $x$, $y$, and $z$ satisfy the conditions $x+y+z=0$ and $|x|+|y|+|z| \leqslant 1$. Prove the inequality
$$
x+\frac{y}{3}+\frac{z}{5} \leqslant \frac{2}{5}
$$ | # Solution. Indeed,
$$
5\left(x+\frac{y}{3}+\frac{z}{5}\right)=3(x+y+z)+2 x-\frac{4 y}{3}-2 z \leqslant 3(x+y+z)+2(|x|+|y|+|z|)=2
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,056 |
4. Is it possible to arrange the numbers 1 and -1 in a $600 \times 600$ table such that the absolute value of the sum of all numbers in the table is less than 90,000, and in each of the rectangles $4 \times 6$ and $6 \times 4$, the absolute value of the sum of the numbers is greater than 4? | Answer: No.
Solution. Since the sum of the numbers in a $4 \times 6$ rectangle is even, if its modulus is greater than four, it must be at least six. Suppose such an arrangement exists. Note that in this arrangement, either there is no row consisting entirely of +1, or there is no column consisting entirely of -1 (if ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,057 |
5. Trapezoid $ABCD (AB \| CD)$ is inscribed in circle $\omega$. On the ray $DC$ beyond point $C$, a point $E$ is marked such that $BC = BE$. Line $BE$ intersects circle $\omega$ again at point $F$, which lies outside segment $BE$. Prove that the center of the circumcircle of triangle $CEF$ lies on $\omega$. | Solution. Trapezoid $ABCD$ is inscribed in circle $\omega$, which means it is isosceles, and in particular, its diagonals are equal. Thus, $AC = BD$ and $\angle BCD = \angle ADC$. Since triangle $BCE$ is isosceles,
$$
\angle BEC = \angle BCE = 180^{\circ} - \angle BCD = 180^{\circ} - \angle ADC
$$
Therefore, lines $B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,058 |
6. The numbers $a, b$ and $c$ satisfy the equation $(a-5)^{2}+(b-12)^{2}-(c-13)^{2}=a^{2}+b^{2}-c^{2}$. Prove that both sides of the equation are perfect squares. | Solution. Expanding the brackets in the condition, we get $10 a+24 b-26 c=0$. Then
$$
13^{2}\left(a^{2}+b^{2}-c^{2}\right)=(13 a)^{2}+(13 b)^{2}-(5 a+12 b)^{2}=(12 a-5 b)^{2}
$$
Therefore, $12 a-5 b$ is divisible by 13, i.e., $12 a-5 b=13 k$ for some integer $k$, and
$$
a^{2}+b^{2}-c^{2}=\frac{(12 a-5 b)^{2}}{13^{2}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,059 |
4. Is it possible to arrange the numbers 1 and -1 in a $600 \times 600$ table such that the absolute value of the sum of all numbers in the table is less than 90,000, and in each of the rectangles $4 \times 6$ and $6 \times 4$ the absolute value of the sum of the numbers is greater than 4? | Answer: No.
Solution. Since the sum of the numbers in a $4 \times 6$ rectangle is even, if its modulus is greater than four, it must be at least six. Suppose such an arrangement exists. Note that in this arrangement, either there is no row consisting entirely of +1, or there is no column consisting entirely of -1 (if ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,060 |
1. On an island, only knights, who always tell the truth, and liars, who always lie, live. One fine day, 450 islanders sat around a round table. Each person at the table said: "Of the person sitting to my right and the person immediately after them, exactly one is a liar." How many liars can be sitting at the table? | Answer: 150 or 450.
Solution. All the liars can sit at the table, then there are 450 of them. Suppose there is at least one knight sitting at the table. Then there is a knight who has a liar sitting to their right (otherwise, they are all knights, which is impossible). The person to the right of this liar must be a kn... | 150or450 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,061 |
2. The values of the quadratic trinomial $a x^{2}+b x+c$ are negative for all $x$. Prove that $\frac{b}{a}<\frac{c}{a}+1$. | Solution. Since the values of the quadratic trinomial $a x^{2}+b x+c$ are negative for all $x$, its leading coefficient and discriminant are negative: $a<0$ and $b^{2}-4ac<0$. If $b>0$, then the required inequality is obvious, since its left side is negative, and the right side is positive. Let $b \leqslant 0$. Then we... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,062 |
3. On the median $C M$ of triangle $A B C$, a point $K$ is chosen. The line $A K$ intersects side $B C$ at point $A_{1}$, and the line $B K$ intersects side $A C$ at point $B_{1}$. It turns out that the quadrilateral $A B_{1} A_{1} B$ is cyclic. Prove that triangle $A B C$ is isosceles. | First solution. Draw lines $M A_{2}$ and $M B_{2}$ through point $M$, parallel to $A A_{1}$ and $B B_{1}$ respectively. They will be the midlines in triangles $A A_{1} B$ and $B B_{1} A$. Therefore, $A_{2}$ is the midpoint of $B A_{1}$ and $B_{2}$ is the midpoint of $A B_{1}$. Then, by Thales' theorem for parallel line... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,063 |
4. The numbers $a, b$, and $c$ satisfy the condition $a^{2} b c+a b^{2} c+a b c^{2}=1$. Prove the inequality $a^{2}+b^{2}+c^{2} \geqslant \sqrt{3}$. | First solution. Note that $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$. Therefore, it is sufficient to prove the inequality $(a b+b c+c a)^{2} \geqslant 3$. Expanding the brackets, we get
$$
\begin{aligned}
(a b+b c+c a)^{2} & =a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2\left(a^{2} b c+a b^{2} c+a b c^{2}\right) \geqslant \\
&... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,064 |
5. The cells of a $9 \times 9$ board are painted in black and white in a checkerboard pattern. How many ways are there to place 9 rooks on the same-colored cells of the board so that they do not attack each other? (A rook attacks any cell that is in the same row or column as itself.) | Answer: $4!5!=2880$.
Solution. Let the upper left corner of the board be black for definiteness. Note that black cells come in two types: cells with both coordinates even, and cells with both coordinates odd. If a rook is on a black cell with both even coordinates, then all black cells it attacks also have even coordi... | 2880 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,065 |
6. Solve the equation $2^{x}+3^{y}=z^{2}$ in natural numbers. | Answer: $(4,2,5)$.
Solution. Since the left side of the equality is odd and not divisible by 3, the number $z$ is also odd and not divisible by 3. Then $z=6u \pm 1$ and $z^{2}=36u^{2} \pm 12u + 1$, which means $z^{2}$ gives a remainder of 1 when divided by 3 and a remainder of 1 when divided by 4. Note that the remain... | (4,2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,066 |
1. On an island, there live only knights, who always tell the truth, and liars, who always lie. One fine day, 30 islanders sat around a round table. Each of them can see everyone except themselves and their neighbors. Each person in turn said the phrase: "All I see are liars." How many liars were sitting at the table? | Answer: 28.
Solution. Not all of those sitting at the table are liars (otherwise they would all be telling the truth). Therefore, there is at least one knight sitting at the table. Everyone he sees is a liar. Let's determine who his neighbors are. Both of them cannot be liars (otherwise they would be telling the truth... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,067 |
2. The quadratic trinomial $a x^{2}+b x+c$ has two roots. Prove that the trinomial $3 a x^{2}+2(a+b) x+(b+c)$ also has two roots. | Solution. According to the condition, the discriminant of the first quadratic is positive: $b^{2}-4 a c>0$. Consider the discriminant of the second quadratic:
$$
\begin{aligned}
4(a+b)^{2}-4 \cdot & 3 a(b+c)=4\left(a^{2}+2 a b+b^{2}-3 a b-3 a c\right)= \\
\quad & 4 a^{2}-4 a b+4 b^{2}-12 a c>4 a^{2}-4 a b+b^{2}=(2 a-b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,068 |
3. In triangle $ABC$, the bisector $AL$ is drawn. A perpendicular is dropped from point $B$ to $AL$. It intersects side $AL$ at point $H$, and the circumcircle of triangle $ABL$ at point $K$. Prove that the center of the circumcircle of triangle $ABC$ lies on the line $AK$. | Solution. Let $\angle B A C=\alpha, \angle A B C=\beta, \angle B C A=\gamma$ and $O-$ the center of the circumcircle of triangle $A B C$. Then $\angle A O B=2 \gamma$. From the isosceles property of triangle $A O B$ we have $\angle O A B=90^{\circ}-\gamma$. To prove that $O$ lies on $A K$, it is sufficient to check tha... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,069 |
4. For positive numbers $a, b$ and $c$, prove the inequality
$$
1+\frac{3}{a b+b c+c a} \geqslant \frac{6}{a+b+c} .
$$ | Solution. Note that $(a+b+c)^{2} \geqslant 3(a b+b c+c a)$. Therefore,
$$
1+\frac{3}{a b+b c+c a} \geqslant 1+\frac{9}{(a+b+c)^{2}} \geqslant 2 \frac{3}{a+b+c}=\frac{6}{a+b+c}
$$
In the last transition, we used the inequality of means for two numbers. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,070 |
5. In the cells of a $2015 \times n$ table, non-negative numbers are arranged such that in each row and each column there is at least one positive number. It is known that if a cell contains a positive number, then the sum of all numbers in its column is equal to the sum of all numbers in its row. For which $n$ is this... | Answer: $n=2015$.
Solution. We will prove by induction on $m+n$ that for an $m \times n$ table, the specified arrangement is possible only when $m=n$. For $m+n=2$, the statement is obvious. Consider an $m \times n$ table. Let the sum of the numbers in the first row be $a$. Consider all rows and columns whose sum of nu... | 2015 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,071 |
6. For which prime numbers $p$ and $q$ does the equation $p^{2 x}+q^{2 y}=z^{2}$ have a solution in natural numbers $x, y$ and $z ?$ | Answer: $p=2, q=3$ or $p=3$ and $q=2$.
Solution. Due to symmetry, we can assume that $p \leqslant q$. For $p=2$ and $q=3$, the equation has a solution in natural numbers: $2^{4}+3^{2}=5^{2}$.
First, consider the case $p=q$. Let for definiteness $x \leqslant y$. Then $p^{2 x}\left(p^{2(y-x)}+1\right)=z^{2}$. Therefore... | p=2,q=3orp=3,q=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,072 |
1. On an island, there are two tribes: liars, who always lie, and knights, who always tell the truth. Each inhabitant of the island is friends with all members of their own tribe and with some other islanders. One day, every inhabitant of the island said the phrase: "More than half of my friends are from my tribe." Who... | Answer: knights.
Solution. Let $r$ and $\ell$ denote the number of knights and liars. Suppose $r \leqslant \ell$. Then each knight is friends with $r-1$ knights and, therefore, with fewer than $r-1$ liars. Consequently, the number of knight-liar friend pairs is less than $r(r-1) \leqslant \ell(\ell-1)$. Since each lia... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,073 |
2. Non-zero numbers $a, b$ and $c$ are such that the doubled roots of the quadratic polynomial $x^{2}+a x+b$ are the roots of the polynomial $x^{2}+b x+c$. What can the ratio $a / c$ be? | Answer: $1 / 8$.
Solution. Let $x_{1}$ and $x_{2}$ be the roots of the first equation. Then, by Vieta's theorem, $a=-x_{1}-x_{2}$ and $b=x_{1} x_{2}$. Since $2 x_{1}$ and $2 x_{2}$ are the roots of the second equation, $b=-2 x_{1}-2 x_{2}=2 a$ and $c=4 x_{1} x_{2}=4 b$. Therefore, $\frac{a}{c}=$ $\frac{b / 2}{4 b}=\fr... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,074 |
3. Sides $A B$ and $A D$ of the inscribed quadrilateral $A B C D$ are equal. A point $K$ is chosen on side $C D$ such that $\angle D A K=\angle A B D$. Prove that $A K^{2}=K D^{2}+B C \cdot K D$. | Solution. On the extension of side $C B$ beyond point $B$, mark a point $E$ such that $B E=D K$. Since the quadrilateral $A B C D$ is inscribed, $\angle A B E=180^{\circ}-\angle A B C=\angle A D C$. Triangles $A B E$ and $A D K$ are equal by two sides and the included angle $(A B=A D, \angle A B E=\angle A D C$ and $B ... | AK^{2}=KD^{2}+BC\cdotKD | Geometry | proof | Yes | Yes | olympiads | false | 8,075 |
4. Positive numbers $a, b$ and $c$ satisfy the condition $a+b+c=$ $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove the inequality $a+b+c \geqslant \frac{3}{a b c}$. | Solution. By the condition
$$
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+c a}{a b c}
$$
Therefore, it is sufficient to prove that $a b+b c+c a \geqslant 3$ under the condition that $a^{2} b c+$ $a b^{2} c+a b c^{2}=a b+b c+c a$. By the inequality $(x+y+z)^{2} \geqslant 3(x y+y z+z x)$
for the numbers $a ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,076 |
5. How many ways are there to color the cells of a $10 \times 10$ board in blue and green so that in each $2 \times 2$ square there are two blue and two green cells? | Answer: $2^{11}-2=2046$.
Solution. Note that if the colors of three cells in a $2 \times 2$ square are known, the color of the remaining cell is uniquely determined. Also, the colors of the two remaining cells are uniquely determined if there are two adjacent cells of the same color in a $2 \times 2$ square.
Consider... | 2046 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,077 |
6. Solve the equation $2^{x}+15^{y}=z^{2}$ in natural numbers. | Answer: $(6,2,17)$.
Solution. Since the left side of the equality is odd and not divisible by 3, the number $z$ is also odd and not divisible by 3. Then $z=6u \pm 1$ and $z^{2}=36u^{2} \pm 12u + 1$, which means $z^{2}$ gives a remainder of 1 when divided by 3 and a remainder of 1 when divided by 4. Note that the remai... | (6,2,17) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,078 |
1. On an island, there live only knights, who always tell the truth, and liars, who always lie, and there are at least two knights and at least two liars. One fine day, each islander, in turn, pointed to each of the others and said one of two phrases: "You are a knight!" or "You are a liar!" The phrase "You are a liar!... | Answer: 526.
Solution. Let $r$ and $\ell$ denote the number of knights and liars, respectively. Note that a knight will say to another knight and a liar will say to another liar: "You are a knight!", while a knight will say to a liar and a liar will say to a knight: "You are a liar!" Therefore, the number of liar-knig... | 526 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,079 |
2. The numbers $a, b, c$ and $d$ satisfy the relation $a c=2 b+2 d$. Prove that at least one of the quadratic polynomials $x^{2}+a x+b$ and $x^{2}+c x+d$ has a root. | Solution. Suppose the opposite. Then the discriminants of both quadratic trinomials are negative: $a^{2}-4 b<0$ and $c^{2}-4 d<0$. Consequently, $a^{2}+c^{2}<4 b+4 d=2(2 b+2 d)=2 a c$. But then $(a-c)^{2}=$ $a^{2}-2 a c+c^{2}<0$. Contradiction. | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,080 |
3. The center $O$ of the circle circumscribed around quadrilateral $A B C D$ lies on side $A B$. Point $E$ is symmetric to $D$ with respect to line $A B$. Segments $A C$ and $D O$ intersect at point $P$, and segments $B D$ and $C E$ intersect at point $Q$. Prove that $P Q$ is parallel to $A B$. | Solution. Since side $AB$ passes through the center of the circle, it is its diameter. Points $D$ and $E$ are symmetric with respect to the diameter, so the arcs subtended by segments $AD$ and $AE$ are equal. Therefore, $\angle ACD = \angle ACE$. Angles $\angle ABD$ and $\angle ACD$ are also equal, as they subtend the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,081 |
4. Positive numbers $a, b$ and $c$ satisfy the condition $a^{2}+b^{2}+c^{2}=1$. Prove the inequality $\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b} \geqslant \sqrt{3}$. | Solution. Let $x=\frac{b c}{a}, y=\frac{c a}{b}$ and $z=\frac{a b}{c}$. Then $a^{2}=y z, b^{2}=z x$ and $c^{2}=x y$. By the condition $x y+y z+z x=1$. Therefore, $(x+y+z)^{2} \geqslant$ $3(x y+y z+z x)=3$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,082 |
5. In an $8 \times 8$ table, natural numbers are arranged. The numbers in cells symmetric with respect to both diagonals of the table are equal. It is known that the sum of all numbers in the table is 1000, and the sum of the numbers on the diagonals is 200. For what smallest number $M$ can we assert that the sum of th... | Answer: $M=288$.
Solution. Consider the upper half of the table. Let the numbers be arranged as shown in the figure. Due to symmetry, the marked numbers completely determine the placement of the remaining numbers in the table.
| $a_{1}$ | $b_{1}$ | $b_{2}$ | $b_{3}$ | $c_{3}$ | $c_{2}$ | $c_{1}$ | $d_{1}$ |
| :--- | ... | 288 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,083 |
6. Solve the equation $2^{x}+63^{y}=z^{2}$ in natural numbers. | Answer: $(8,2,65)$.
Solution. Since the left side of the equality is odd and not divisible by 3, the number $z$ is also odd and not divisible by 3. Then $z=6u \pm 1$ and $z^{2}=36u^{2} \pm 12u + 1$, which means $z^{2}$ gives a remainder of 1 when divided by 3 and a remainder of 1 when divided by 4. Note that the remai... | (8,2,65) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,084 |
1. Prove that for any real numbers $a$ and $b$, the equation
$$
\left(a^{2}-b^{2}\right) x^{2}+2\left(a^{3}-b^{3}\right) x+\left(a^{4}-b^{4}\right)=0
$$
has a solution. | Solution. If $a=b$, then any $x$ is a solution to the equation. If $a \neq b$, we need to prove that the quadratic trinomial has a root. For this, it is sufficient to check that its discriminant is non-negative. The discriminant, divided by 4, is
$$
\left(a^{3}-b^{3}\right)^{2}-\left(a^{2}-b^{2}\right)\left(a^{4}-b^{4... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,085 |
2. On an island, there live liars and knights. Knights always tell the truth, and liars always lie. Each inhabitant of the island knows whether each of the others is a knight or a liar. One day, 19 islanders met. Three of them said: "Exactly three of us are liars," then six of the others said: "Exactly six of us are li... | Answer: 9, 18 or 19.
Solution. Let the third statement be true. Then all who said it are knights, and there must be exactly nine liars. Therefore, all who said the other statements are liars. There are nine of them, so the one who remained silent is a knight, and such a situation is possible. Let the third statement b... | 9,18,19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,086 |
3. Real numbers $a, b, c$ and $d$ satisfy the condition $a^{6}+b^{6}+c^{6}+d^{6}=64$. Find the maximum value of the expression $a^{7}+b^{7}+c^{7}+d^{7}$. | Answer: 128
Solution. By the condition $a^{6} \leqslant a^{6}+b^{6}+c^{6}+d^{6}=64$, therefore $a \leqslant 2$. Similarly, we get that $b \leqslant 2, c \leqslant 2$ and $d \leqslant 2$. Consequently,
$$
a^{7}+b^{7}+c^{7}+d^{7}=a \cdot a^{6}+b \cdot b^{6}+c \cdot c^{6}+d \cdot d^{6} \leqslant 2\left(a^{6}+b^{6}+c^{6}... | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,087 |
4. On the side $B C$ of triangle $A B C$, a point $K$ is marked. A tangent $\ell_{1}$ to the circumcircle of triangle $A K C$ is drawn, parallel to the line $A B$ and closest to it. It touches the circle at point $L$. The line $A L$ intersects the circumcircle of triangle $A B K$ at point $M (M \neq A)$. A tangent $\el... | Solution. Let $N$ be the intersection point of lines $B C$ and $\ell_{1}$. If we prove that $N M$ is a tangent to the circumcircle of triangle $A B K$, then point $N$ will lie on line $\ell_{2}$, which means that lines $B K, \ell_{1}$, and $\ell_{2}$ intersect at one point.
Since lines $A B$ and $L N$ are parallel, an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,088 |
5. Given a grid board $2020 \times 2021$. Petya and Vasya are playing the following game. They take turns placing chips in free cells of the board. The player who, after their move, ensures that every $4 \times 4$ square contains a chip wins. Petya starts. Which player can guarantee a win regardless of the opponent's a... | Answer: Vasya
Solution. Vasya should move in such a way that Petya cannot win on his next move. We will show that he can always achieve this. If there is a pair of non-overlapping empty $4 \times 4$ squares on the board, Petya must place a chip so that they remain even after his move. If such a move is impossible, the... | Vasya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,089 |
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