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1. A necklace consists of 50 blue and a certain number of red beads. It is known that in any segment of the necklace containing 8 blue beads, there are at least 4 red ones. What is the minimum number of red beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is a... | Answer: 29.
Solution. Note that any segment of the necklace consisting of 11 beads contains no more than 7 blue and no fewer than 4 red beads (otherwise, it would contain 8 blue beads and no more than 3 red ones). Fix a red bead in the necklace. The 7 consecutive segments of 11 beads adjacent to it do not cover the en... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,665 |
2. A 100-digit natural number had one of its digits erased (not the leading one). As a result, the number decreased by 13 times. Find all numbers for which this is possible. | Answer: $1625 \cdot 10^{96}, 195 \cdot 10^{97}, 2925 \cdot 10^{96}$, and $13 b \cdot 10^{98}$ for $b=1,2,3$.
Solution. Represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $k, m, n$ are non-negative integers, and $a$ is a decimal digit. By erasing the digit $a$, we obtain the number $m+10^{k} n$. A... | 1625\cdot10^{96},195\cdot10^{97},2925\cdot10^{96},13b\cdot10^{98}forb=1,2,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,666 |
3. Find the minimum value of the expression
$$
A=\left(2\left(\sin x_{1}+\ldots+\sin x_{n}\right)+\cos x_{1}+\ldots+\cos x_{n}\right) \cdot\left(\sin x_{1}+\ldots+\sin x_{n}-2\left(\cos x_{1}+\ldots+\cos x_{n}\right)\right)
$$ | Answer: $-\frac{5 n^{2}}{2}$.
Solution. For $k=1, \ldots, n$ let $t_{k}=3 \sin x_{k}-\cos x_{k}, s_{k}=\sin x_{k}+3 \cos x_{k}$. Then
$$
\begin{aligned}
4 A=\left(\sum_{k=1}^{n}\left(4 \sin x_{k}+2 \cos x_{k}\right)\right) \cdot\left(\sum_{k=1}^{n}\left(2 \sin x_{k}-4 \cos x_{k}\right)\right)= \\
=\left(\sum_{k=1}^{n... | -\frac{5n^{2}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,667 |
4. The bisector $A L$ and the median $B M$ of triangle $A B C$ intersect at point $X$. Line $C X$ intersects side $A B$ at point $Y$. Find the area of triangle $C Y L$, given that $\angle B A C=60^{\circ}$ and $A L=x$. | Answer: $\frac{x^{2}}{4 \sqrt{3}}$.
Solution. Let $P$ and $Q$ be the midpoints of segments $A Y$ and $C L$. Then $M P$ and $M Q$ are the midlines of triangles $A C Y$ and $A C L$, from which... | \frac{x^{2}}{4\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,668 |
5. Each cell of an $m \times n$ board is painted either black or white. It is known that for any cell of the board, the number of cells that have the same color and share at least one vertex is odd. Find all pairs of natural numbers $m$ and $n$ for which this is possible. | Answer: $n$ or $m$ is even.
Solution. We will call cells neighboring if they share a vertex. Let's provide an example of coloring that satisfies the condition if $m$ or $n$ is even. For definiteness, let the number of rows be even. We will color the first and second rows in black, the third and fourth in white, the fi... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,669 |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 6, 24, and 24. A truncated cone is placed on the table with its smaller base down, and it shares a generatrix with each of the other cones. Find the radius of the smaller base of the truncated cone. | Answer: 2.
Solution. Let $C$ be the center of the smaller base of the truncated cone, and $r$ be the radius of this base. Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other c... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,670 |
1. A necklace consists of 100 red and a certain number of blue beads. It is known that in any segment of the necklace containing 10 red beads, there are at least 7 blue ones. What is the minimum number of blue beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one i... | Answer: 78.
Solution. Note that any segment of the necklace containing 16 beads has no more than 9 red and no fewer than 7 blue beads (otherwise, it would contain 10 red beads and no more than 6 blue ones). Fix a blue bead in the necklace. The 11 consecutive segments of 16 beads adjacent to it do not cover the entire ... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,671 |
2. A 100-digit natural number had one of its digits replaced by zero (if it was the leading digit, it was simply erased). As a result, the number decreased by 13 times. Find all numbers for which this is possible. | Answer: $325 b \cdot 10^{97}$ for $b=1,2,3$.
Solution. Represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, and $k, m, n$ are non-negative integers, with $m<10^{k}$. By replacing the digit $a$ with zero, we obtain the number $m+10^{k+1} n$. According to the condition,
$$
m+... | 325b\cdot10^{97}forb=1,2,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,672 |
3. Find the minimum value of the expression
$$
A=\left(5\left(\cos x_{1}+\ldots+\cos x_{n}\right)+\sin x_{1}+\ldots+\sin x_{n}\right) \cdot\left(\cos x_{1}+\ldots+\cos x_{n}-5\left(\sin x_{1}+\ldots+\sin x_{n}\right)\right)
$$ | Answer: $-13 n^{2}$.
Solution. For $k=1, \ldots, n$ let $t_{k}=3 \cos x_{k}-2 \sin x_{k}, s_{k}=2 \cos x_{k}+3 \sin x_{k}$. Then
$$
\begin{aligned}
A=\left(\sum_{k=1}^{n}\left(5 \cos x_{k}+\sin x_{k}\right)\right) \cdot\left(\sum_{k=1}^{n}\left(\cos x_{k}-5 \sin x_{k}\right)\right)= \\
=\left(\sum_{k=1}^{n}\left(3 \c... | -13n^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,673 |
4. In a right-angled triangle $A B C$ with a right angle at $B$, the bisector $B L$ and the median $C M$ are drawn, intersecting at point $D$. The line $A D$ intersects side $B C$ at point $E$. Find the area of triangle $A E L$, given that $E L=x$. | Answer: $\frac{x^{2}}{2}$.
Solution. Let $P$ and $Q$ be the midpoints of segments $A L$ and $B E$. Then $M P$ and $M Q$ are the midlines of triangles $A B L$ and $A B E$, from which $M P \| ... | \frac{x^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,674 |
5. Each cell of an $m \times n$ board is painted either black or white. It is known that for any cell of the board, the number of cells that share a side with it and have the same color is odd. Find all pairs of natural numbers $m$ and $n$ for which this is possible. | Answer: $n$ or $m$ is even.
Solution. We will call cells that share a side neighboring. Let's provide an example of a coloring that satisfies the condition if $m$ or $n$ is even. For definiteness, let the number of rows be even. We will color the odd-numbered columns of the board as follows: the two top cells in black... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,675 |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 10, 15, and 15. A truncated cone is placed on the table with its smaller base down, and it shares a generatrix with each of the other cones. Find the area of the smaller base of the truncated cone. | Answer: $4 \pi$.
Solution. Let $C$ be the center of the smaller base of the truncated cone, and $r$ be the radius of this base. Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the o... | 4\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,676 |
1. A necklace consists of 50 blue, 100 red, and 100 green beads. We will call a sequence of four consecutive beads good if it contains exactly 2 blue beads and one each of red and green. What is the maximum number of good quartets that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning... | Answer: 99.
Solution. Blue beads make up one fifth of the total. Therefore, there will be two consecutive blue beads (let's call them $a$ and $b$), separated by at least three beads. Note that $a$ and $b$ are part of no more than three good quartets, while the other blue beads are part of no more than four. If we add ... | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,677 |
2. A 200-digit natural number had one of its digits erased. As a result, the number decreased by 5 times. Find all numbers for which this is possible. | Answer: $125 a \cdot 10^{197}$ for $a=1,2,3$.
Solution. Represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, $k, m$, $n$ are non-negative integers, and $m<10^{k}$. Therefore, $a$ will be the leading digit of the original number, from which $k=199$. Then
$$
4 m=10^{k} a=10^{... | 125\cdot10^{197}for=1,2,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,678 |
3. Given numbers $x_{1}, \ldots, x_{n} \in\left(0, \frac{\pi}{2}\right)$. Find the maximum value of the expression
$$
A=\frac{\sin x_{1}+\ldots+\sin x_{n}}{\sqrt{\operatorname{tg}^{2} x_{1}+\ldots+\operatorname{tg}^{2} x_{n}+n}}
$$ | Answer: $\frac{\sqrt{n}}{2}$.
Solution. Note that for any $a_{1}, \ldots, a_{n}$
$$
\left(\sum_{k=1}^{n} a_{k}\right)^{2} \leqslant n \sum_{k=1}^{n} a_{k}^{2}
$$
From this, by the inequality for the harmonic mean and the arithmetic mean,
$$
\frac{1}{\sqrt{\operatorname{tg}^{2} x_{1}+\ldots+\operatorname{tg}^{2} x_{... | \frac{\sqrt{n}}{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,679 |
4. A perpendicular bisector is drawn to the lateral side $AC$ of an isosceles triangle $ABC$. It intersects the lateral side $AB$ at point $L$ and the extension of the base at point $K$. It turns out that the areas of triangles $ALC$ and $KBL$ are equal. Find the angles of the triangle. | Answer: $36^{\circ}, 72^{\circ}, 72^{\circ}$.
Solution 1. Let $M$ and $N$ be the midpoints of segments $A C$ and $B C$ respectively, $\alpha=\angle A B C$. Then $M N$ is the midline of trian... | 36,72,72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,680 |
5. From $n^{2}$ bulbs, an $n \times n$ display was assembled. Each bulb has two states - on and off. When pressing any bulb, its state remains unchanged, while all bulbs in the same row or column change their state to the opposite. Initially, all bulbs on the display are off. Petya sequentially pressed several bulbs, a... | Answer: $2 n-2$.
Solution. Let's call the reversal of a set of bulbs the change of the state of all bulbs in this set to the opposite. Note two simple facts.
1) Pressing a bulb is equivalent to the reversal of the row and column in which this bulb is located. Indeed, with such reversals, the pressed bulb changes its ... | 2n-2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,681 |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 23, 46, and 69. A truncated cone is placed on the table with its smaller base down, and it shares a generatrix with each of the other cones. Find the radius of the smaller base of the truncated cone.
# | # Answer: 6.
Solution. Let $C$ be the center of the smaller base of the truncated cone, $R$ be the radius of this base, and $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other c... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,682 |
1. A thread is strung with 75 blue, 75 red, and 75 green beads. We will call a sequence of five consecutive beads good if it contains exactly 3 green beads and one each of red and blue. What is the maximum number of good quintets that can be on this thread? | Answer: 123.
Solution. Note that the first and last green beads are included in no more than three good fives, the second and second-to-last - in no more than four fives, and the rest - in no more than five fives. If we add these inequalities, we get $2 \cdot 3 + 2 \cdot 4 + 71 \cdot 5 = 369$ on the right side, and th... | 123 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,683 |
2. A 200-digit natural number had one of its digits replaced by zero (if it was the leading digit, it was simply erased). As a result, the number decreased by a factor of 5. Find all numbers for which this is possible. | Answer: $125 a \cdot 10^{97}$ for $a=1,2,3$.
Solution. Represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, and $k, m, n$ are non-negative integers, with $m<10^{k}$. By replacing the digit $a$ with zero, we obtain the number $m+10^{k+1} n$. According to the condition,
$$
m+... | 125\cdot10^{97}for=1,2,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,684 |
3. Given numbers $x_{1}, \ldots, x_{n} \in\left(0, \frac{\pi}{2}\right)$. Find the maximum value of the expression
$$
A=\frac{\cos ^{2} x_{1}+\ldots+\cos ^{2} x_{n}}{\sqrt{n}+\sqrt{\operatorname{ctg}^{4} x_{1}+\ldots+\operatorname{ctg}^{4} x_{n}}}
$$ | Answer: $\frac{\sqrt{n}}{4}$.
Solution. Note that for any $a_{1}, \ldots, a_{n}$
$$
\left(\sum_{k=1}^{n} a_{k}\right)^{2} \leqslant n \sum_{k=1}^{n} a_{k}^{2}
$$
From this, by the inequality for the harmonic mean and the arithmetic mean,
$$
\frac{1}{\sqrt{n}+\sqrt{\operatorname{ctg}^{4} x_{1}+\ldots+\operatorname{c... | \frac{\sqrt{n}}{4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,685 |
4. From the base $BC$ of the isosceles triangle $ABC$, the height $AH$ is dropped. On the side $AB$, a point $P$ is marked such that $CP = BC$. The segment $CP$ intersects the height $AH$ at point $Q$. It turns out that the area of triangle $BHQ$ is 4 times smaller than the area of triangle $APQ$. Find the angles of tr... | Answer: $30^{\circ}, 75^{\circ}, 75^{\circ}$.
Solution 1. Since triangle $A B C$ is isosceles, its height $A H$ is also the bisector and median, hence $B H=C H$ and $\angle C A H=\angle B A ... | 30,75,75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,686 |
5. From $n^{2}$ bulbs, an $n \times n$ display was assembled. Each bulb has two states - on and off. When pressing any bulb, its state remains the same, while all bulbs in the same row or column change their state to the opposite. Initially, all bulbs on the display are off. Vasya sequentially pressed several bulbs, no... | Answer: $\frac{n^{2}}{2}$ for even $n$ and $\frac{n^{2}-1}{2}$ for odd $n$.
Solution. Let's call the reversal of a set of bulbs the change of the state of all bulbs in this set to the opposite. Note two simple facts.
1) Pressing a bulb is equivalent to the reversal of the row and column in which this bulb is located.... | \frac{n^{2}}{2}forevenn\frac{n^{2}-1}{2}foroddn | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,687 |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are $2 r, 3 r$, and $10 r$. A frustum of a cone is placed on the table with its smaller base down, and it shares a common generatrix with each of the other cones. Find $r$ if the radius of the smaller base of t... | Answer: 29.
Solution. Let $C$ be the center of the smaller base of the truncated cone, and $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other cones, with $R=15$. Denote by $\math... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,688 |
1. A necklace consists of 80 beads of red, blue, and green colors. It is known that on any segment of the necklace between two blue beads, there is at least one red bead, and on any segment of the necklace between two red beads, there is at least one green bead. What is the minimum number of green beads that can be in ... | Answer: 27.
Solution. If the blue beads are arranged in a circle, the number of pairs of adjacent beads is equal to the number of beads. Since there is a red bead between any two blue beads, there are no fewer red beads in the necklace than blue ones. Similarly, it can be proven that there are no fewer green beads tha... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,689 |
2. A 100-digit natural number had two adjacent digits erased. As a result, the number decreased by 87 times. Find all numbers for which this is possible. | Answer: $435 \cdot 10^{97}, 1305 \cdot 10^{96}, 2175 \cdot 10^{96}, 3045 \cdot 10^{96}$.
Solution. Let's represent the original number in the form $m+10^{k} a+10^{k+2} n$, where $a, k, m, n$ are non-negative integers, and $m<10^{k}$ and $a<100$. By erasing the digits in the $k$-th and $(k+1)$-th positions, we get the ... | 435\cdot10^{97},1305\cdot10^{96},2175\cdot10^{96},3045\cdot10^{96} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,690 |
3. Given numbers $x_{1}, \ldots, x_{n} \in(0,1)$. Find the maximum value of the expression
$$
A=\frac{\sqrt{1-x_{1}}+\ldots+\sqrt{1-x_{n}}}{\sqrt{\frac{1}{x_{1}}+\ldots+\frac{1}{x_{n}}}}
$$ | Answer: $\frac{\sqrt{n}}{2}$.
Solution. Note that for any $a_{1}, \ldots, a_{n}$
$$
\left(\sum_{k=1}^{n} a_{k}\right)^{2} \leqslant n \sum_{k=1}^{n} a_{k}^{2}
$$
From this, by the inequality for the harmonic mean and the arithmetic mean,
$$
\frac{1}{\sqrt{\frac{1}{x_{1}}+\ldots+\frac{1}{x_{n}}}} \leqslant \frac{\sq... | \frac{\sqrt{n}}{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,691 |
4. Inside an angle of $30^{\circ}$ with vertex $A$, a point $K$ is chosen, the distances from which to the sides of the angle are 1 and 2. Through point $K$, all possible lines are drawn, intersecting the sides of the angle. Find the minimum perimeter of the triangle cut off by the line from the angle. | Answer: $4(2+\sqrt{3})$.
Solution. Draw a circle $\omega$ that is tangent to the sides of the angle at points $M$ and $N$, such that $K$ lies on the minor arc $MN$ (see the left figure). Let... | 4(2+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,692 |
5. In the cells of an $80 \times 80$ table, pairwise distinct natural numbers are placed. Each of them is either a prime number or a product of two prime numbers (possibly the same). It is known that for any number $a$ in the table, there is a number $b$ in the same row or column such that $a$ and $b$ are not coprime. ... | Answer: 4266.
Solution. We will say that a composite number $a$ serves a prime number $p$ if $a$ and $p$ are not coprime (i.e., $a$ is divisible by $p$). For each prime number in the table, there is a composite number that serves it. Since each composite number has no more than two distinct prime divisors, it serves n... | 4266 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,693 |
6. On the table, there are three cones standing on their bases, touching each other. The heights of the cones are the same, and the radii of their bases are 1, 2, and 3. A sphere is placed on the table, touching all the cones. It turns out that the center of the sphere is equidistant from all points of contact with the... | Answer: 1.
Solution: Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the cones, $O$ be the center of the sphere, $R$ be the radius of the sphere, $C$ be the point of contact of the... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,694 |
1. Around a round table, 50 schoolchildren are sitting: blondes, brunettes, and redheads. It is known that in any group of schoolchildren sitting in a row, between any two blondes there is at least one brunette, and between any two brunettes - at least one redhead. What is the minimum number of redheads that can sit at... | Answer: 17.
Solution. If only blondes were sitting at the table, the number of pairs of neighbors would be equal to the number of blondes. Since there is a brunette between any two blondes, there are no fewer brunettes than blondes sitting at the table. Similarly, it can be proven that there are no fewer redheads than... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,695 |
2. A 200-digit natural number had its leading digit and the digit standing one place after it erased. As a result, the number decreased by 44 times. Find all numbers for which this is possible. | Answer: $132 c \cdot 10^{197}$ for $c=1,2,3$.
Solution. Represent the original number in the form $m+10^{k} \cdot \overline{c b a}$, where $a, b, c$ are decimal digits, $k, m$ are non-negative integers, with $c>0$ and $m<10^{k}$. By erasing the digits $a$ and $c$, we obtain the number $m+10^{k} b$. According to the co... | 132\cdot10^{197}for=1,2,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,696 |
3. Given numbers $x_{1}, \ldots, x_{n} \in(0,1)$. Find the maximum value of the expression
$$
A=\frac{\sqrt[4]{1-x_{1}}+\ldots+\sqrt[4]{1-x_{n}}}{\frac{1}{\sqrt[4]{x_{1}}}+\ldots+\frac{1}{\sqrt[4]{x_{n}}}}
$$ | Answer: $\frac{\sqrt{2}}{2}$.
Solution. By the inequality for the harmonic mean and the arithmetic mean
$$
\frac{1}{\frac{1}{\sqrt[4]{x_{1}}}+\ldots+\frac{1}{\sqrt[4]{x_{n}}}} \leqslant \frac{\sqrt[4]{x_{1}}+\ldots+\sqrt[4]{x_{n}}}{n^{2}}
$$
Note that for any $a_{1}, \ldots, a_{n}$
$$
\left(\sum_{k=1}^{n} a_{k}\rig... | \frac{\sqrt{2}}{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,697 |
4. Inside an angle of $30^{\circ}$ with vertex $A$, a point $K$ is chosen, the distances from which to the sides of the angle are 1 and 2. Through point $K$, all possible lines are drawn, intersecting the sides of the angle. Find the minimum area of the triangle cut off by the line from the angle. | Answer: 8.
Solution 1. Choose points $P$ and $Q$ on the sides of the angle such that point $K$ is the midpoint of segment $P Q$. We will show that line $P Q$ cuts off a triangle of the small... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,698 |
5. In the cells of a $75 \times 75$ table, pairwise distinct natural numbers are placed. Each of them has no more than three distinct prime divisors. It is known that for any number $a$ in the table, there is a number $b$ in the same row or column such that $a$ and $b$ are not coprime. What is the maximum number of pri... | Answer: 4218.
Solution. We will say that a composite number $a$ serves a prime number $p$ if the numbers $a$ and $p$ are not coprime (that is, $a$ is divisible by $p$). For each prime number in the table, there is a composite number that serves it. Since each composite number has no more than three distinct prime divi... | 4218 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,699 |
6. On the table, there are three cones standing on their bases, touching each other. The heights of the cones are the same, and the radii of their bases are $2r$, $3r$, and $10r$. A sphere with a radius of 2 is placed on the table, touching all the cones. It turns out that the center of the sphere is equidistant from a... | Answer: 1.
Solution: Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the cones, $O$ be the center of the sphere, $R$ be the radius of the sphere, $C$ be the point of contact of the... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,700 |
1. (10 points) From the parabolas listed below, select those for which points $A(-1,-1)$ and $B(0,2)$ lie on the same side.
a) $y=2 x^{2}+4 x$
б) $y=x^{2} / 2-x-3 / 2$;
в) $y=-x^{2}+2 x-1$;
2) $y=-x^{2}-4 x-3$;
д) $y=-x^{2}+3$
е) none of the listed answers are correct.
Mark the correct answer, no need to provide t... | Answer: a), c) and d).
Solution. Transform the equations of the parabolas by completing the square:
1) $2 x^{2}+4 x=2(x+1)^{2}-2-$ the vertex of the parabola is the point with coordinates $(-1,-2)$, the branches of this parabola are directed upwards; it is easy to see that both given points lie inside the parabola;
2... | ),)) | Algebra | MCQ | Yes | Yes | olympiads | false | 7,701 |
2. (10 points) In triangle $A B C$, medians $A M$ and $B K$ are drawn. It is known that $A M=3, B K=5$. Determine which of the following statements is true:
a) the length of side $A B$ can be 6;
b) the perimeter of triangle $A B C$ can be 22;
c) based on the given data, it is impossible to estimate either the perimet... | Answer: g).
Solution. Let $O$ be the point of intersection of the medians. As is known, the medians intersect in the ratio $2: 1$, counting from the vertex, so $B O=10 / 3$, $O K=5 / 3$, $A O=2$, $O M=1$. Let's now analyze the statements formulated in the problem.
Case a). We will use the triangle inequality for $A O... | g | Geometry | MCQ | Yes | Yes | olympiads | false | 7,702 |
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second must be different. For example, 101 and 292 have this form, while 222 and 123 do not. Similarly, we define the form of the number $\overline{a b c a b d}$. How many odd numbers of the form $\o... | Answer: 448.
Solution. Odd numbers divisible by 5 are numbers ending in 5, so for $d$ we have only one option. For $a$ we have 8 options, as the number cannot start with zero, and $a$ cannot be equal to $d$. The digit $b$ cannot be equal to $a$ or $d$, and there are no other restrictions on it - we get 8 possible valu... | 448 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,703 |
5. (30 points) The diagonals of trapezoid $R S Q T$ with bases $R S$ and $Q T$ intersect at point $A$ at a right angle. It is known that base $R S$ is larger than base $Q T$ and angle $R$ is a right angle. The bisector of angle $R A T$ intersects $R T$ at point $U$, and the line passing through point $U$ parallel to $R... | The first solution. Let's find an expression for $U W$, using the equality of the areas of the trapezoids:
$$
\begin{gathered}
S_{R T Q S}=S_{R U W S}+S_{U T Q W} \Leftrightarrow \\
(R S+T Q) \cdot(T U+U R)=(R S+U W) \cdot U R+(U W+T Q) \cdot T U \Leftrightarrow \\
R S \cdot T U+T Q \cdot U R=U W \cdot(U R+T U) \Leftr... | UW=RT | Geometry | proof | Yes | Yes | olympiads | false | 7,705 |
6. (30 points) In triangle $X Y Z$, side $Y Z$ is twice as long as side $X Z$. A point $W$ is chosen on side $Y Z$ such that angles $Z X W$ and $Z Y X$ are equal. Line $X W$ intersects the bisector of the external angle at vertex $Z$ at point $A$. Prove that angle $Y A Z$ is a right angle. | Solution. Note that triangles $Z W X$ and $Z X Y$ are similar by two angles (angle $Z$ is common, and angles $\alpha$ are equal by condition). From this, we get
$$
\frac{W Z}{Z X}=\frac{Z X}{Z Y}=\frac{X W}{X Y}=\frac{1}{2}
$$
i.e., $Z W=\frac{1}{4} Z Y$.
Let $X K$ be the bisector of angle $W X Y$, with point $K$ ly... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,706 |
7. (30 points) In triangle $K I A$, side $K A$ is less than side $K I$, and points $R$ and $E$ are the feet of the perpendiculars dropped from points $I$ and $A$ to the angle bisector of angle $K$, respectively. Prove that the lines $IE$, $R A$, and the perpendicular to $K R$ raised at point $K$ intersect at one point.... | Solution. Let $M$ be the point of intersection of the lines $I E$ and $R A$; it exists because, by the condition, $K A < K I$ and, consequently, $A E \neq R I$ and $A E I R$ is not a parallelogram. Let $O$ be the point of intersection of the bisector of angle $K$ with side $A I$. It is easy to see that triangles $K A E... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,707 |
8. (40 points) On a strip, numbers from 1 to 1598 were written in order and the strip was cut into several parts. It turned out that the arithmetic mean of all numbers in the first part is some natural number $n$, the second part - the number $2 n$, the third part - the number $3 n$, and so on. Explain for which $n$ th... | Solution. Consider several consecutive natural numbers. If their quantity is even, then they can be paired with equal odd sums, hence their arithmetic mean is not an integer. If their quantity is odd, then their arithmetic mean is the middle number in this set. Therefore, all segments of the strip consist of an odd num... | 1,17,47,799 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,708 |
9. (40 points) The numbers $s_{1}, s_{2}, \ldots, s_{1008}$ are such that their sum is $2016^{2}$. It is known that
$$
\frac{s_{1}}{s_{1}+1}=\frac{s_{2}}{s_{2}+3}=\frac{s_{3}}{s_{3}+5}=\ldots=\frac{s_{1008}}{s_{1008}+2015}
$$
Find $s_{17}$. | Answer: 132.
Solution. Note that none of the $s_{i}$ is equal to zero (otherwise, all the fractions $\frac{s_{i}}{s_{i}+2 i-1}$ would be equal to zero, and, consequently, all $s_{i}$ would have to be equal to zero, which contradicts the fact that their sum is $2016^{2}$). Therefore, the original condition is equivalen... | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,709 |
5. (30 points) The diagonals of trapezoid $R S Q T$ with bases $R S$ and $Q T$ intersect at point $A$ at a right angle. It is known that base $R S$ is larger than base $Q T$ and angle $R$ is a right angle. The bisector of angle $R A T$ intersects $R T$ at point $U$, and the line passing through point $U$ parallel to $R... | The first solution. Let's find an expression for $U W$, using the equality of the areas of the trapezoids:
$$
\begin{gathered}
S_{R T Q S}=S_{R U W S}+S_{U T Q W} \Leftrightarrow \\
(R S+T Q) \cdot(T U+U R)=(R S+U W) \cdot U R+(U W+T Q) \cdot T U \Leftrightarrow \\
R S \cdot T U+T Q \cdot U R=U W \cdot(U R+T U) \Leftr... | UW=RT | Geometry | proof | Yes | Yes | olympiads | false | 7,710 |
7. (30 points) In triangle $K I A$, side $K A$ is less than side $K I$, and points $R$ and $E$ are the feet of the perpendiculars dropped from points $I$ and $A$ to the angle bisector of angle $K$, respectively. Prove that the lines $IE$, $R A$, and the perpendicular to $K R$ raised at point $K$ intersect at one point.... | Solution. Let $M$ be the point of intersection of the lines $I E$ and $R A$; it exists because, by the condition, $K A < K I$ and, consequently, $A E \neq R I$ and $A E I R$ is not a parallelogram. Let $O$ be the point of intersection of the bisector of angle $K$ with side $A I$. It is easy to see that triangles $K A E... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,711 |
1. Can all natural numbers from 1 to 2018 be arranged in a circle in such a way that the sum of any three consecutive numbers is an odd number? | Answer: No.
Solution. In each sum of three consecutive numbers, there is one or three odd addends. It is impossible for there to be only one odd addend in all sums. Indeed, the entire circle is covered by 673 triples, while it contains 1009 odd numbers in total. Therefore, there must be at least one triple with no few... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,712 |
2. Given non-zero real numbers $a, b$ and $c$. The parabola $y=a x^{2}+b x+c$ is located above the line $y=c x$. Prove that the parabola $y=c x^{2}-b x+a$ is located above the line $y=c x-b$. | Solution. The fact that the parabola $y=a x^{2}+b x+c$ is located above the line $y=$ $c x$ is equivalent to two conditions: the branches of the parabola are directed upwards, i.e., $a>0$, and the parabola does not intersect the line, i.e., the equation $a x^{2}+b x+c=c x$ has no solutions. Therefore, the discriminant ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 7,713 |
3. The product of positive numbers $a, b, c$ and $d$ is 1. Prove the inequality
$$
\frac{a^{4}+b^{4}}{a^{2}+b^{2}}+\frac{b^{4}+c^{4}}{b^{2}+c^{2}}+\frac{c^{4}+d^{4}}{c^{2}+d^{2}}+\frac{d^{4}+a^{4}}{d^{2}+a^{2}} \geqslant 4
$$ | First solution. Since $2\left(a^{4}+b^{4}\right) \geqslant\left(a^{2}+b^{2}\right)^{2}$, the inequality holds:
$$
\frac{a^{4}+b^{4}}{a^{2}+b^{2}} \geqslant \frac{\frac{1}{2}\left(a^{2}+b^{2}\right)^{2}}{a^{2}+b^{2}}=\frac{a^{2}+b^{2}}{2}
$$
By adding it to three similar inequalities, we get that the left side of the ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,714 |
4. Cells of an infinite grid paper are painted in $k$ colors (each cell is painted entirely in one color). What is the largest $k$ such that in every grid rectangle with sides 3 and 4, cells of all these colors will be found? | Answer: 10.
Solution. Divide the infinite grid paper into ten-cell figures as shown in the figure.
The required coloring with 10 colors can be achieved by coloring one such figure in 10 co... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,715 |
5. Given a triangle $A B C$. On its sides $B C, C A$, and $A B$, points $A_{1}, B_{1}$, and $C_{1}$ are chosen respectively such that the quadrilateral $A B_{1} A_{1} C_{1}$ is cyclic. Prove that
$$
\frac{S_{A_{1} B_{1} C_{1}}}{S_{A B C}} \leqslant\left(\frac{B_{1} C_{1}}{A A_{1}}\right)^{2}
$$
$$
Therefore, it is sufficient to prove that the expression in parentheses is not less than $\frac{9}{4}$. T... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,720 |
4. What is the minimum number of cells that need to be marked in a $50 \times 50$ table so that each vertical or horizontal strip of $1 \times 6$ contains at least one marked cell. | Answer: 416.
Solution. A $50 \times 50$ square can easily be cut into four rectangles of $24 \times 26$ and a central square of $2 \times 2$. Each rectangle can be cut into $4 \cdot 26=104$ strips of $1 \times 6$. Each such strip must have its own marked cell, so there will be no fewer than 416 such cells.
We will sh... | 416 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,721 |
5. Let $M$ be the point of intersection of the medians of triangle $ABC$. A line passing through $M$ intersects segments $BC$ and $CA$ at points $A_1$ and $B_1$ respectively. Point $K$ is the midpoint of side $AB$. Prove that $9 S_{K A_1 B_1} \geqslant 2 S_{A B C}$.
(n+2)(n+3)+m^{5}=5000 ?
$$ | Answer: none.
Solution. Let for some natural numbers $m$ and $n$ the equality $n(n+1)(n+2)(n+3)+m^{5}=5000$ holds. Since $n(n+1)(n+2)(n+3)=\left(n^{2}+3 n+1\right)^{2}-1$,
the equality can be rewritten as $k^{2}+m^{5}=5001$, where $k=n^{2}+3 n+1$. Consider the remainders upon division by 11 of the left and right parts... | none | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,723 |
1. Can all natural numbers from 1 to 2017 be arranged in a circle so that any two adjacent numbers differ by exactly 17 or 21? | Answer: No.
Solution. Since adjacent numbers differ by an odd number, they have different parity. Therefore, the parity of the numbers around the circle must alternate. But this is impossible, as there are more odd numbers than even ones. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,724 |
2. Given various real numbers $a, b$ and $c$. The quadratic trinomial $f(x)$ satisfies the relations $f(a)=b c, f(b)=c a$ and $f(c)=a b$. Find $f(a+b+c)$. | Answer: $a b+b c+c a$.
Solution. Let $f(x)=p x^{2}+q x+r$. Then from the equalities
$$
b c=f(a)=p a^{2}+q a+r \quad \text { and } \quad c a=f(b)=p b^{2}+q b+r
$$
it follows that $c(b-a)=p\left(a^{2}-b^{2}\right)+q(a-b)$, hence $c=-p(a+b)-q$, since $a \neq b$. Similarly, we obtain the relation $b=-p(a+c)-q$. Subtract... | ++ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,725 |
3. The sum of positive numbers $a, b$ and $c$ is 1. Prove the inequality
$$
\frac{a^{4}+b^{4}}{a^{6}+b^{6}}+\frac{b^{4}+c^{4}}{b^{6}+c^{6}}+\frac{c^{4}+a^{4}}{c^{6}+a^{6}} \leqslant \frac{1}{a b c}
$$ | Solution. Note that $\frac{a^{4}+b^{4}}{a^{6}+b^{6}} \leqslant \frac{1}{a b}$. Indeed, this inequality, after multiplying by the denominator, will take the form $a^{6}+b^{6}-a^{5} b-a b^{5} \geqslant 0$, which is equivalent to the inequality $\left(a^{5}-b^{5}\right)(a-b) \geqslant 0$. In this form, it is obvious, sinc... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,726 |
4. On a $50 \times 50$ chessboard, Petya places 50 non-attacking rooks, and Vasya selects a $k \times k$ square ($k \leqslant 50$) on the board. For which $k$ can Vasya always choose a square that does not contain any rooks, regardless of Petya's actions? | Answer: For $k \leqslant 7$.
Solution. We will show how Vasya can choose a $7 \times 7$ square. Notice that at least one corner of the $50 \times 50$ board is not occupied by a rook. Let's assume, for definiteness, that this is the top right corner. Then in the topmost row and the rightmost column, there is a rook. We... | k\leqslant7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,727 |
5. Given trapezoid $A B C D$. Points $Q$ and $S$ are chosen on the bases $B C$ and $A D$ respectively. Segments $A Q$ and $B S$ intersect at point $P$, and segments $C S$ and $D Q$ intersect at point $R$. Prove that $S_{P Q R S} \leqslant \frac{1}{4} S_{A B C D}$.
^{6}+b$ is divisible by $n^{2}+n+1$. | Answer: $a=2, b=27$.
Solution. Note that $n^{3}-1$ is divisible by $n^{2}+n+1$. Therefore, the number $n^{3}$ gives a remainder of 1 when divided by $n^{2}+n+1$. We will consider the remainder of the division of the number $(a n+1)^{6}+b$ by $n^{2}+n+1$. For this, we will expand the brackets:
$$
(a n+1)^{6}+b=a^{6} n... | =2,b=27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,729 |
1. Can all natural numbers from 1 to 2018 be arranged in a circle in such a way that the sum of any four consecutive numbers is an odd number? | Answer: No.
Solution. In each sum of four consecutive numbers, there is one or three odd addends. It is impossible for all sums to have exactly one even (odd) addend. Indeed, the entire circle is covered by 505 quartets, and it contains a total of 1009 even numbers. Therefore, there must be at least one quartet with n... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,730 |
2. Find all quadratic trinomials $a x^{2}+b x+c$ with real coefficients $a, b$, and $c$, such that if any one of the three coefficients in the trinomial is increased by 1, the resulting quadratic trinomial will have exactly one root. | Answer: $\frac{1}{8} \cdot x^{2}-\frac{3}{4} \cdot x+\frac{1}{8}$.
Solution. According to the condition, the quadratic polynomials $(a+1) x^{2}+b x+c, a x^{2}+(b+1) x+c$ and $a x^{2}+b x+(c+1)$ each have exactly one root. Therefore, their discriminants are zero. Thus,
$$
b^{2}-4(a+1) c=(b+1)^{2}-4 a c=b^{2}-4 a(c+1)=... | \frac{1}{8}\cdotx^{2}-\frac{3}{4}\cdotx+\frac{1}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,731 |
3. Positive numbers $a, b$, and $c$ satisfy the condition $a^{2}+b^{2}+c^{2}=3$. Prove the inequality
$$
\frac{a}{a+5}+\frac{b}{b+5}+\frac{c}{c+5} \leqslant \frac{1}{2}
$$ | Solution. Note that $\frac{a}{a+5}=1-\frac{5}{a+5}$, therefore
$$
\frac{a}{a+5}+\frac{b}{b+5}+\frac{c}{c+5}=3-\left(\frac{5}{a+5}+\frac{5}{b+5}+\frac{5}{c+5}\right)
$$
Thus, it is sufficient to prove that the expression in parentheses is not less than $\frac{5}{2}$. This is equivalent to the inequality
$$
\frac{1}{a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,732 |
4. Given a grid board $m \times n(m, n \geqslant 2)$. A corner is a three-cell figure consisting of a cell (let's call it the central cell) and two adjacent cells (which we will call the side cells). A cardboard corner covers the top-left corner and the two adjacent cells of the board. In one operation, you can choose ... | Answer: for odd $m$ and $n$.
Solution. We will show how to achieve the required movement for odd $m$ and $n$. For $m=n=3$, it is possible (on the diagram, the cell relative to which the rotation occurs is marked with a cross).
=(1,1),(1,2)$ and $(3,2)$.
Solution. If $n=1$, then $m=1$, and this pair obviously works. We will assume that $n \geqslant 2$. Let $m+1=k n$. The numbers $m$ and $n$ are coprime, since if $d$ is their common divisor, then $1=(m+1)-m=k n-m$ is divisible by $d$. Moreover, $m \leqslant n^{2}-n+1 \leqslant ... | (,n)=(1,1),(1,2),(3,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,735 |
1. In some cells of a $1 \times 2021$ strip, one chip is placed in each. For each empty cell, the number equal to the absolute difference between the number of chips to the left and to the right of this cell is written. It is known that all the written numbers are distinct and non-zero. What is the minimum number of ch... | Answer: 1347.
Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}... | 1347 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,736 |
2. Positive numbers $a, b, c$ are such that $a^{2}+b^{2}+c^{2}=3$. Find the minimum value of the expression
$$
A=\frac{a^{4}+b^{4}}{c^{2}+4 a b}+\frac{b^{4}+c^{4}}{a^{2}+4 b c}+\frac{c^{4}+a^{4}}{b^{2}+4 c a}
$$ | Answer: $\frac{6}{5}$.
Solution. By the inequalities for means
$$
\frac{a^{4}+b^{4}}{c^{2}+4 a b} \geqslant \frac{a^{4}+b^{4}}{c^{2}+2 a^{2}+2 b^{2}}=\frac{a^{4}+b^{4}}{3+a^{2}+b^{2}} \geqslant \frac{\left(a^{2}+b^{2}\right)^{2}}{2\left(3+a^{2}+b^{2}\right)}=\frac{\left(3-c^{2}\right)^{2}}{2\left(6-c^{2}\right)}=\fra... | \frac{6}{5} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,737 |
3. Given an isosceles triangle \(ABC\) with base \(BC\). On the extension of side \(AC\) beyond point \(C\), point \(K\) is marked, and a circle is inscribed in triangle \(ABK\) with center at point \(I\). A circle passing through points \(B\) and \(I\) is tangent to line \(AB\) at point \(B\). This circle intersects s... | Solution. Let $I M$ and $I N$ be the perpendiculars dropped from point $I$ to $A K$ and $B K$ respectively. Note that $\angle A B I = \angle B L I$, since line $A B$ is tangent to the circumcircle of triangle $B L I$. Line $A I$ is the angle bisector of angle $A$, and $A B = A C$ by the given condition. Therefore, tria... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,738 |
4. The number 1200 is written on the board. Petl appended $10 n+2$ fives to it on the right, where $n-$ is a non-negative integer. Vasya thought that this was the base-6 representation of a natural number $x$, and he factored $x$ into prime factors. It turned out that there were exactly two distinct primes among them. ... | Answer: $n=0$.
Solution. We agree to write the numbers in base 6 in parentheses to distinguish them from decimal numbers. Then
$$
\begin{aligned}
& x=(1200 \underbrace{55 \ldots 5}_{10 n+2})=(1201 \underbrace{00 \ldots 0}_{10 n+2})-1=289 \cdot 6^{10 n+2}-1= \\
& =\left(17 \cdot 6 \cdot 6^{5 n}-1\right)\left(17 \cdot ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,739 |
5. In the country of Lemonia, coins in circulation have denominations of $6n+1$, $6n+3$, $6n+5$, and $6n+7$ piastres, where $n$ is a natural number. A resident of the country went to the bank without any cash on hand. What is the largest amount that the bank will not be able to give him? | Answer: $12 n^{2}+8 n-1$.
Solution. Let's call a natural number $s$ decomposable if the bank can pay the sum $s$ using only coins of value $6 n+3, 6 n+5$, and $6 n+7$ piastres. For any $r=1, \ldots, 6 n$, denote by $m_{r}$ the smallest decomposable number that gives a remainder of $r$ when divided by $6 n+1$. We will ... | 12n^{2}+8n-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,740 |
1. In some cells of a $1 \times 2100$ strip, one chip is placed. In each of the empty cells, a number is written that is equal to the absolute difference between the number of chips to the left and to the right of this cell. It is known that all the written numbers are distinct and non-zero. What is the minimum number ... | Answer: 1400.
Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}... | 1400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,741 |
2. Positive numbers $a, b, c$ are such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{a^{3}+b^{3}}{8 a b+9-c^{2}}+\frac{b^{3}+c^{3}}{8 b c+9-a^{2}}+\frac{c^{3}+a^{3}}{8 c a+9-b^{2}}
$$ | Answer: $\frac{3}{8}$.
Solution. Note that
$$
4\left(a^{3}+b^{3}\right) \geqslant(a+b)^{3} \quad \text { and } \quad 9-c^{2}=(3-c)(3+c)=(a+b)(3+c) .
$$
Then, by the inequalities for means,
$$
\frac{4\left(a^{3}+b^{3}\right)}{8 a b+9-c^{2}} \geqslant \frac{(a+b)^{3}}{2(a+b)^{2}+9-c^{2}}=\frac{(a+b)^{2}}{2(a+b)+3+c}=... | \frac{3}{8} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,742 |
3. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $O$. Let $K$ and $L$ be the points of intersection of the circumcircle of triangle $A O B$ with the lines $A D$ and $B C$ respectively. Find the ratio $O K: O L$, given that $\angle B C A = \angle B D C$. | Answer: $1: 1$.
Solution. Since quadrilateral $A B C D$ is cyclic, the equalities $\angle B A C=\angle B D C$ and $\angle B C A=\angle B D A$ hold. Also note that $\angle B C A=\angle B D C$... | 1:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,743 |
4. The number 12320 is written on the board. Petya appended $10 n+1$ threes to it on the right, where $n-$ is a non-negative integer. Vasya thought that this was the quaternary (base-4) representation of a natural number $x$, and he factored $x$ into prime factors. It turned out that there were exactly two distinct pri... | Answer: $n=0$.
Solution. We agree to write quaternary numbers in parentheses to distinguish them from decimal numbers. Then
$$
x=(12320 \underbrace{33 \ldots 3}_{10 n+1})=(12321 \underbrace{00 \ldots 0}_{10 n+1})-1=441 \cdot 4 \cdot 4^{10 n}-1=\left(42 \cdot 1024^{n}-1\right)\left(42 \cdot 1024^{n}+1\right) .
$$
If ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,744 |
5. In the country of Lemonia, coins of denominations $6n+1$, $6n+4$, $6n+7$, and $6n+10$ piasters are in circulation, where $n$ is a natural number. A resident of the country went to the bank without any cash on hand. What is the largest amount that the bank will not be able to give him? | Answer: $12 n^{2}+14 n-1$.
Solution. Let's call a natural number $s$ decomposable if the bank can pay the sum $s$ using only coins of value $6 n+4, 6 n+7$, and $6 n+10$ piastres. For any $r=1, \ldots, 6 n$, let $m_{r}$ be the smallest decomposable number that gives a remainder of $r$ when divided by $6 n+1$. We will p... | 12n^{2}+14n-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,745 |
1. For which natural numbers $n$ can a square grid $n \times n$ be cut into $1 \times 1$ and $2 \times 2$ squares such that there are an equal number of squares of each size? | Answer: all $n>5$, multiples of 5.
Solution. Let the board can be cut into $m$ squares $1 \times 1$ and $m$ squares $2 \times 2$. Then $n^{2}=m+4 m=5 m$, from which $n^{2} \vdots 5$ and $n \vdots 5$, and $m=\frac{n^{2}}{5}$. Therefore, $n$ that are not divisible by 5 do not fit.
Let $n \vdots 5$. Notice that for $n \... | all\n>5,\multiples\of\5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,746 |
2. Positive numbers $a, b, c$ are such that $a^{2} b+b^{2} c+c^{2} a=3$. Find the minimum value of the expression
$$
A=a^{7} b+b^{7} c+c^{7} a+a b^{3}+b c^{3}+c a^{3}
$$ | Answer: 6.
Solution. By the Cauchy-Schwarz inequality
$$
A=\left(a^{7} b+a b^{3}\right)+\left(b^{7} c+b c^{3}\right)+\left(c^{7} a+c a^{3}\right) \geqslant 2\left(a^{4} b^{2}+b^{4} c^{2}+c^{4} a^{2}\right) \geqslant \frac{2}{3}\left(a^{2} b+b^{2} c+c^{2} a\right)^{2}=6
$$
Equality is achieved when $a=b=c=1$. | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,747 |
3. Given an acute scalene triangle $A B C$. In it, the altitudes $B B_{1}$ and $C C_{1}$ are drawn, intersecting at point $H$. Circles $\omega_{1}$ and $\omega_{2}$ with centers $H$ and $C$ respectively touch the line $A B$. From point $A$ to $\omega_{1}$ and $\omega_{2}$, tangents other than $A B$ are drawn. Denote th... | Answer: $180^{\circ}$.
Solution. Since right triangles $A C C_{1}$ and $A C E$ are equal by leg and hypotenuse, line $C A$ is the bisector of isosceles triangle $C_{1} C E$. Therefore, it i... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,748 |
4. Petya wrote down $n$ consecutive two-digit numbers $(n \geqslant 2)$ in a row on the board, the first of which does not contain the digit 4, and the last one does not contain the digit 7. Vasya thought that this was the decimal representation of a natural number $x$, and he factored $x$ into prime factors. It turned... | Answer: 2021.
Solution. Let $a$ and $b$ be the prime divisors of $x$, with $b=a+4$. The pairs of the last digits of $a$ and $b$ can be $(3,7),(7,1),(9,3)$. Since $x$ cannot end in 7, only the first case is possible. Then the number $y=\frac{a+b}{2}$ ends in 5 and, therefore, has the form $y=10 m+5$. From this,
$$
y^{... | 2021 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,749 |
5. In the country of Lemonia, coins in circulation have denominations of $3^{n}, 3^{n-1} \cdot 5, 3^{n-2} \cdot 5^{2}$, $3^{n-3} \cdot 5^{3}, \ldots, 3 \cdot 5^{n-1}, 5^{n}$ piastres, where $n-$ is a natural number. A resident of the country went to the bank without any cash on hand. What is the largest amount that the... | Answer: $5^{n+1}-2 \cdot 3^{n+1}$.
Solution. A natural number $s$ is called $n$-decomposable if the sum $s$ can be paid using coins of denominations $3^{n}, 3^{n-1} \cdot 5, \ldots, 5^{n}$ piastres, and $n$-indecomposable otherwise. For $n=1$, only coins of 3 and 5 piastres are available. If $s \geqslant 10$, then the... | 5^{n+1}-2\cdot3^{n+1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,750 |
1. For which natural numbers $n$ can a square grid $n \times n$ be cut into rectangles $1 \times 1$ and $2 \times 3$ such that there are an equal number of rectangles of each size? | Answer: all $n>7$, multiples of 7.
Solution. Let the board can be cut into $m$ squares $1 \times 1$ and $m$ rectangles $2 \times 3$. Then $n^{2}=m+6 m=7 m$, from which $n^{2} \vdots 7$ and $n \vdots 7$, and $m=\frac{n^{2}}{7}$. Therefore, $n$ that are not divisible by 7 do not fit.
Let $n \vdots$ 7. Notice that for $... | n>7, | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,751 |
2. Positive numbers $a, b, c$ are such that $a^{2} b+b^{2} c+c^{2} a=3$. Find the minimum value of the expression
$$
A=\frac{\sqrt{a^{6}+b^{4} c^{6}}}{b}+\frac{\sqrt{b^{6}+c^{4} a^{6}}}{c}+\frac{\sqrt{c^{6}+a^{4} b^{6}}}{a} .
$$ | Answer: $3 \sqrt{2}$.
Solution. We will use the inequality $\sqrt{x+y} \geqslant \frac{1}{\sqrt{2}}(\sqrt{x}+\sqrt{y})$ for $x, y \geqslant 0$. Taking into account the Cauchy inequality, we get
$$
\begin{aligned}
& A \geqslant \frac{1}{\sqrt{2}}\left(\frac{a^{3}+b^{2} c^{3}}{b}+\frac{b^{3}+c^{2} a^{3}}{c}+\frac{c^{3}... | 3\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,752 |
3. A circle is circumscribed around an acute-angled triangle $A B C$. Point $K$ is the midpoint of the smaller arc $A C$ of this circle, and point $L$ is the midpoint of the smaller arc $A K$ of this circle. Segments $B K$ and $A C$ intersect at point $P$. Find the angle between the lines $B C$ and $L P$, given that $B... | Answer: $90^{\circ}$.
Solution. Let $\angle A B L=\varphi$. Then $\angle K B L=\varphi$ and $\angle K B C=2 \varphi$. Since $B K=B C$, we get $\angle B K C=\angle B C K=90^{\circ}-\varphi$.... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,753 |
4. Petya wrote on the board in descending order $n$ consecutive two-digit numbers, the last of which does not contain the digit 7. Vasya thought that this was the decimal representation of a natural number $x$, and he factored $x$ into prime factors. It turned out that there were only two of them and they differ by 4. ... | Answer: 91.
Solution. Let $a$ and $b$ be the prime divisors of $x$, with $b=a+6$. The pairs of the last digits of $a$ and $b$ can be $(1,7),(3,9),(7,3)$. Since $x$ cannot end in 7, only the last case is possible. Then the number $y=\frac{a+b}{2}$ ends in 5 or 0. Note that the number $x$ is odd and $x=y^{2}-9$. Therefo... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,754 |
5. In the country of Lemonia, coins in circulation have denominations of $2^{n}, 2^{n-1} \cdot 3, 2^{n-2} \cdot 3^{2}$, $2^{n-3} \cdot 3^{3}, \ldots, 2 \cdot 3^{n-1}, 3^{n}$ piastres, where $n$ is a natural number. A resident of the country went to the bank without any cash. What is the largest amount that the bank wil... | Answer: $3^{n+1}-2^{n+2}$.
Solution. A natural number $s$ is called $n$-decomposable if the sum $s$ can be paid with coins of denominations $2^{n}, 2^{n-1} \cdot 3, \ldots, 3^{n}$ piastres, and $n$-indecomposable otherwise. For $n=1$, only coins of 2 and 3 piastres are available. If $s \geqslant 3$, there exists an $r... | 3^{n+1}-2^{n+2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,755 |
1. The natural numbers from 1 to 2021 are written in a row in some order. It turns out that for any number, its left and right neighbors have different parity. Which number can be in the first place? | Answer: any odd number.
Solution. First, let's prove that in any sequence of four consecutive numbers, there are two even and two odd numbers. Indeed, suppose in some quartet there are three numbers of the same parity. Then among them, there will be two numbers that are two positions apart. But the number between them... | any\odd\ | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,756 |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{3}+b^{3}+c^{3}}{(a+b+c)^{3}-26 a b c}
$$ | Answer: 3.
Solution. Note that $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+B$, where
$$
B=3\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right) \geqslant 18 \sqrt[6]{a^{6} b^{6} c^{6}}=18 a b c
$$
(we used the Cauchy inequality). Then
$$
a^{3}+b^{3}+c^{3} \leqslant(a+b+c)^{3}-24 a b c
$$
Let $t=\frac{(a+b+c)^{3... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,757 |
3. Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$, and a circle with center at point $O$ encompasses circles $\omega_{1}$ and $\omega_{2}$, touching them at points $C$ and $D$ respectively. It turns out that points $A, C$, and $D$ lie on the same line. Find the angle $A B O$. | Answer: $90^{\circ}$.
Solution. Let $O_{1}$ and $O_{2}$ be the centers of circles $\omega_{1}$ and $\omega_{2}$. Triangles $C O D$, $C O_{1} A$, and $A O_{2} D$ are isosceles. Since points ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,758 |
4. Petl wrote on the board in a row $n$ two-digit octal numbers ( $n \geqslant 2$ ), forming an arithmetic progression with a difference of -8. Vasya thought that this was the octal representation of a natural number $x$, and he factored $x$ into prime factors. It turned out that there were only two of them, and they d... | Answer: 7767.
Solution. Let $y, z$ be the prime divisors of $x, z=y+6$. Note that the number $x+9$ equals $\left(\frac{y+z}{2}\right)^{2}$, meaning it is a perfect square. Write the permissible pairs $(y, z)$ in octal form:
$$
(\overline{\ldots a 1}, \overline{\ldots a 7}), \quad(\overline{\ldots a 3}, \overline{\ldo... | 7767 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,759 |
5. In the country of Lemonia, coins in circulation have denominations of $3n-1$, $6n+1$, $6n+4$, and $6n+7$ piastres, where $n$ is a natural number. A resident of the country went to the bank without any cash on hand. What is the largest amount that the bank will not be able to give him? | Answer: $6 n^{2}+4 n-5$.
Solution. Let's call a natural number $s$ decomposable if the bank can pay the sum $s$ using only coins of value $6 n+1, 6 n+4$, and $6 n+7$ piastres. For any $r=1, \ldots, 3 n-2$, let $m_{r}$ be the smallest decomposable number that gives a remainder of $r$ when divided by $3 n-1$. We will pr... | 6n^{2}+4n-5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,760 |
1. Natural numbers from 1 to 2023 are written in a row in some order. It turned out that any three numbers, located every other one, give different remainders when divided by 3. Which number can be in the first place? | Answer: any number that gives a remainder of 1 when divided by 3.
Solution. Let's divide the range from 1 to 2023 into groups of numbers that give remainders of 0, 1, and 2 when divided by 3. The first and third groups contain 674 numbers each, while the second group contains 675 numbers. In any sequence of six consec... | anythatgivesremainderof1whendivided3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,761 |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{4}+b^{4}+c^{4}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}
$$ | Answer: 3.
Solution. Note that $(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+B$, where
\[
\begin{aligned}
& B=4\left(a^{3} b+a b^{3}+b^{3} c+b c^{3}+c^{3} a+c a^{3}\right)+6\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+12\left(a b c^{2}+b c a^{2}+c a b^{2}\right) \geqslant \\
& \geqslant 24 \sqrt[6]{(a b c)^{8}}+18 \sqrt[3]{(a b... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,762 |
3. Circle $\omega_{1}$ with center $O$ intersects circle $\omega_{2}$, which passes through point $O$, at points $K$ and $L$. A line through point $O$ intersects circle $\omega_{2}$ again at point $A$. Segment $OA$ intersects circle $\omega_{1}$ at point $B$. Find the ratio of the distances from point $B$ to the lines ... | Answer: $1: 1$.
Solution. Notice that $\angle O A K = \angle O A L$ as angles subtended by equal chords $O K$ and $O L$. Therefore, $A B$ is the angle bisector of $\angle L A K$. Moreover, $... | 1:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,763 |
4. Petl wrote on the board in a row $n$ two-digit octal numbers $(n \geqslant 2)$, forming an arithmetic progression with a common difference of 8, and the first number does not contain the digit 2. Vasya thought that this was the octal representation of a natural number $x$, and he factored $x$ into prime factors. It ... | Answer: 3343.
Solution. Let $y, z$ be the prime divisors of $x, z=y+2$. Note that the number $x+1$ equals $\left(\frac{y+z}{2}\right)^{2}$, meaning it is a perfect square. Let's write down the possible pairs $(y, z)$ in octal form:
$$
(\overline{\ldots a 1}, \overline{\ldots a 3}), \quad(\overline{\ldots a 3}, \overl... | 3343 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,764 |
5. In the country of Lemonia, coins in circulation have denominations of $3n-2$, $6n-1$, $6n+2$, and $6n+5$ piastres, where $n$ is a natural number. A resident of the country went to the bank without any cash on hand. What is the largest amount that the bank will not be able to give him? | Answer: $6 n^{2}-4 n-3$.
Solution. Let's call a natural number $s$ decomposable if the bank can pay the sum $s$ using only coins of value $6 n-1, 6 n+2$, and $6 n+5$ piastres. For any $r=1, \ldots, 3 n-3$, let $m_{r}$ be the smallest decomposable number that gives a remainder of $r$ when divided by $3 n-2$. We will pr... | 6n^{2}-4n-3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,765 |
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