problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
1. Can different integers be arranged in a $35 \times 35$ table so that the values in cells sharing a side differ by no more than 18? | Answer: No.
Solution. Let $m$ be the smallest number in the table. By moving one cell at a time in a horizontal or vertical direction, we can reach any cell in the table in no more than $34 \cdot 2=68$ steps. According to the condition, the value in each cell can increase by no more than 18 on each step. Therefore, al... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,766 |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)}{a^{3}+b^{3}+c^{3}-2 a b c}
$$ | Answer: 6.
Solution. By the Cauchy-Schwarz inequality,
$$
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)=\frac{1}{3}\left((a+b+c)^{3}-\left(a^{3}+b^{3}+c^{3}+6 a b c\right)\right) \leqslant \frac{1}{3}\left((a+b+c)^{3}-9 a b c\right)
$$
Note that
$$
a^{3}+b^{3}+c^{3}-2 a b c \geqslant \frac{1}{9}(a+b+c)^{3}-2 a b c=\frac{1}{9}\l... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,767 |
3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $A$. Segment $O_{2} A$ intersects circle $\omega_{1}$ again at point $K$, and segment $O_{1} A$ intersects circle $\omega_{2}$ again at point $L$. The line passing through point $A$ parallel to $K L$ intersects cir... | Answer: $90^{\circ}$.
Solution. Triangles $O_{1} A K$ and $O_{2} A L$ are isosceles and have a common base angle. Therefore, their vertex angles are equal. Let the common value of these ang... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,768 |
4. In the file, 2021 two-digit quinary numbers are written in a row, and the numbers at odd positions are equal and one more than the numbers at even positions. The computer read the data from the file as one quinary number and factored it into prime factors. It turned out that there were exactly two such factors and t... | Answer: No.
Solution. We agree to write quinary numbers in parentheses to distinguish them from decimal numbers. Let $x$ be the read number, $u$ and $v$ be its prime divisors, $v=u+2$. Note that the number $x+1$ is even and equals $\left(\frac{u+v}{2}\right)^{2}$, that is, it is a perfect square. In particular, $x+1$ ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,769 |
5. In the country of Lemonia, coins in circulation have denominations of $5^{n}, 5^{n-1} \cdot 7, 5^{n-2} \cdot 7^{2}$, $5^{n-3} \cdot 7^{3}, \ldots, 5 \cdot 7^{n-1}, 7^{n}$ piasters, 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: $2 \cdot 7^{n+1}-3 \cdot 5^{n+1}$.
Solution. A natural number $s$ is called $n$-decomposable if the sum $s$ can be paid using coins of denominations $5^{n}, 5^{n-1} \cdot 7, \ldots, 7^{n}$ piastres, and $n$-indecomposable otherwise. For $n=1$, only coins of 5 and 7 piastres are available. If $s \geqslant 28$, ... | 2\cdot7^{n+1}-3\cdot5^{n+1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,770 |
1. Can different integers be arranged in a $25 \times 41$ table so that the numbers in cells sharing a side differ by no more than 16? | Answer: No.
Solution. Let $m$ be the smallest number in the table. By moving one cell at a time in a horizontal or vertical direction, we can reach any cell in the table in no more than $24+40=64$ steps. According to the condition, the value in a cell can increase by no more than 16 on each step. Therefore, all number... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,771 |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b)}{(a+b+c)^{4}-79(a b c)^{4 / 3}}
$$ | Answer: 3.
Solution. By the Cauchy-Schwarz inequality,
$a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b)=(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-\left(a^{4}+b^{4}+c^{4}\right) \leqslant(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-3(a b c)^{4 / 3}$.
Note that $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+B$, where
$$
B=3\left(a^{2} b+a b^{2}+b^{2}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,772 |
3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $B$. The extension of segment $O_{2} B$ beyond point $B$ intersects circle $\omega_{1}$ at point $K$, and the extension of segment $O_{1} B$ beyond point $B$ intersects circle $\omega_{2}$ at point $L$. The line pa... | Answer: $90^{\circ}$.
Solution. Since $\angle O_{1} B K = \angle O_{2} B L$, the isosceles triangles $O_{1} B K$ and $O_{2} B L$ are similar. Therefore, their vertex angles are equal. Let th... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,773 |
4. In the file, 2023 two-digit quinary numbers are recorded in a row, and the numbers at odd positions are equal and one more than the numbers at even positions. The computer read the data from the file as one quinary number and factored it into prime factors. It turned out that there were exactly two such factors and ... | Answer: No.
Solution. We agree to write quinary numbers in parentheses to distinguish them from decimal numbers. Let $x$ be the read number, $u$ and $v$ be its prime divisors, $v=u+2$. Note that the number $x+1$ is even and equals $\left(\frac{u+v}{2}\right)^{2}$, that is, it is a perfect square. In particular, $x+1$ ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,774 |
5. In the country of Lemonia, coins of denominations $3^{n}, 3^{n-1} \cdot 4, 3^{n-2} \cdot 4^{2}, 3^{n-3} \cdot 4^{3}, \ldots, 3 \cdot 4^{n-1}, 4^{n}$ piastres 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 ba... | Answer: $2 \cdot 4^{n+1}-3^{n+2}$.
Solution. A natural number $s$ is called $n$-decomposable if the sum $s$ can be paid using coins of denominations $3^{n}, 3^{n-1} \cdot 4, \ldots, 4^{n}$ piastres, and $n$-indecomposable otherwise. For $n=1$, only coins of 3 and 4 piastres are available. If $s \geqslant 8$, then ther... | 2\cdot4^{n+1}-3^{n+2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,775 |
1. For what smallest $k$ can $k$ cells be marked on a $10 \times 11$ board such that any placement of a three-cell corner on the board touches at least one marked cell? | Answer: 50.
Solution. It is not hard to notice that in any $2 \times 2$ square, there are at least two marked cells. Since 25 such squares can be cut out from a $10 \times 11$ board, there must be no fewer than 50 marked cells in it. An example with 50 marked cells is obtained if the cells with an even first coordinat... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,776 |
3. Given a right triangle $A B C$. On the extension of the hypotenuse $B C$, a point $D$ is chosen such that the line $A D$ is tangent to the circumcircle $\omega$ of triangle $A B C$. The line $A C$ intersects the circumcircle of triangle $A B D$ at point $E$. It turns out that the bisector of $\angle A D E$ is tangen... | Answer: $A C: C E=1: 2$.
Solution. Let $\alpha=\angle A B D, K$ and $L$ be the points of intersection of the angle bisector of $\angle A D E$ with $A E$ and $\omega$ respectively. The angle ... | AC:CE=1:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,778 |
4. A warehouse stores 400 tons of cargo, with the weight of each being a multiple of a centner and not exceeding 10 tons. It is known that any two cargos have different weights. What is the minimum number of trips that need to be made with a 10-ton truck to guarantee the transportation of these cargos from the warehous... | Answer: 51.
Solution. We will show that it is always possible to transport the goods in 51 trips, even if the warehouse contains all weights from 1 to 100 tons. Indeed, we can divide all the goods, except for the 50-ton and 100-ton ones, into 49 pairs as follows:
$$
(1,99), \quad(2,98), \quad(3,97), \quad \ldots, \qu... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,779 |
6. Three cones with vertex $A$ and slant height $\sqrt{8}$ touch each other externally. For two of the cones, the angle between the slant height and the axis of symmetry is $\frac{\pi}{6}$, while for the third cone, it is $\frac{\pi}{4}$. Find the volume of the pyramid $O_{1} O_{2} O_{3} A$, where $O_{1}, O_{2}, O_{3}$... | Answer: $\sqrt{\sqrt{3}+1}$.
Solution. Let $\alpha=\frac{\pi}{6}, \beta=\frac{\pi}{4}, l=\sqrt{8}, B-$ be the midpoint of segment $O_{1} O_{2}, A D-$ be the common generatrix of the second ... | \sqrt{\sqrt{3}+1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,781 |
1. For what smallest $k$ can $k$ cells be marked on a $9 \times 9$ board such that any placement of a three-cell corner piece will touch at least two marked cells? | Answer: 56.
Solution. It is not hard to notice that in any $2 \times 2$ square, at least three cells must be marked, and in each $1 \times 2$ rectangle, at least two cells must be marked. Since from a $9 \times 9$ board, 16 $2 \times 2$ squares and 8 $1 \times 2$ rectangles can be cut out, at least $16 \cdot 3 + 8 = 5... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,782 |
3. On the extension of side $B C$ of triangle $A B C$, a point $D$ is taken such that the line $A D$ is tangent to the circumcircle $\omega$ of triangle $A B C$. The line $A C$ intersects the circumcircle of triangle $A B D$ at point $E$, and it is given that $A C: C E=1: 2$. It turns out that the bisector of angle $A ... | Answer: $30^{\circ}, 60^{\circ}, 90^{\circ}$.
Solution 1. Let $M$ be the midpoint of side $A C$, and $K$ and $L$ be the points of intersection of the angle bisector of $\angle A D E$ with $A... | 30,60,90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,784 |
4. A cinema was visited by 50 viewers, the total age of whom is 1555 years, and among them, there are no viewers of the same age. For what largest $k$ can we guarantee to select 16 viewers whose total age is not less than $k$ years? | Answer: 776.
Solution. We will show that $k \geqslant 776$, that is, the total age of the 16 oldest viewers is always not less than 776. Arrange the viewers in order of increasing age, and let $a_{i}$ be the age of the $i$-th viewer. Since there are no viewers of the same age, we get
$$
a_{1} \leqslant a_{2}-1 \leqsl... | 776 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,785 |
5. Given a natural number $x=2^{n}-32$, where $n-$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 3. Find $x$. | Answer: 480 or 2016.
Solution. Let's write $x$ in the form $32 \cdot N$, where $N=2^{n-5}-1$. One prime divisor of $x$ is 2. Therefore, we need to find all $N$ that have exactly two odd prime divisors, one of which is 3. The divisibility of $N$ by 3 means that $n-5$ is even, i.e., $N=2^{2 m}-1$. If $m=1$, then $N=3$, ... | 480or2016 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,786 |
6. Three cones with vertex $A$ and slant height 6 touch each other externally. For two of the cones, the angle between the slant height and the axis of symmetry is $\frac{\pi}{8}$, while for the third cone, it is $\frac{\pi}{4}$. Find the volume of the pyramid $O_{1} O_{2} O_{3} A$, where $O_{1}, O_{2}, O_{3}$ are the ... | Answer: $9 \sqrt{\sqrt{2}+1}$.
Solution. Let $\alpha=\frac{\pi}{8}, \beta=\frac{\pi}{4}, l=6, B$ be the midpoint of segment $O_{1} O_{2}, A D$ be the common generatrix of the second and thir... | 9\sqrt{\sqrt{2}+1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,787 |
1. For what smallest $k$ can $k$ cells be marked on a $12 \times 12$ board such that any placement of a four-cell figure $\square \square$ on the board touches at least one marked cell? (The figure can be rotated and flipped.) | Answer: 48.
Solution. It is not hard to notice that in any $2 \times 3$ rectangle, there are at least two marked cells. Since the $12 \times 12$ board is divided into 24 such rectangles, there must be no fewer than 48 marked cells. An example with 48 marked cells can be obtained by marking the cells where the sum of t... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,788 |
3. On the height $B H$ of triangle $A B C$, a point $D$ is marked. Line $A D$ intersects side $B C$ at point $E$, and line $C D$ intersects side $A B$ at point $F$. Points $G$ and $J$ are the projections of points $F$ and $E$ onto side $A C$, respectively. The area of triangle $H E J$ is twice the area of triangle $H F... | Answer: $\sqrt{2}: 1$, measured from point $E$.
Solution 1. Let $T$ be the intersection of $A E$ and $F G$, $M$ be the intersection of $C F$ and $E J$, and $K$ be the intersection of $B H$ and $F E$ (see figure). Since triangles $F D T$ and $M D E$ are similar, by Thales' theorem
$$
\frac{E M}{F T}=\frac{D E}{D T}=\f... | \sqrt{2}:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,790 |
4. In the stands of the hockey arena, there are several rows with 168 seats in each row. For the final match, 2016 students from several sports schools were invited as spectators, with no more than 40 from each school. Students from any school must be seated in one row. What is the minimum number of rows that must be i... | Answer: 15.
Solution. Suppose that 57 schools sent 35 students each to the final match, and one school sent 21 students. Since only four groups of 35 students can fit in one row, the number of rows required to accommodate the students from 57 schools should be no less than $\frac{57}{4}=14 \frac{1}{4}$, which means at... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,791 |
5. Given a natural number $x=9^{n}-1$, where $n$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 13. Find $x$. | Answer: 728.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 13. The remainders of the powers of 9 when divided by 13 are $9, 3, 1$ and then repeat cyclically. Thus, the divisibility of $x$ by 13 means ... | 728 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,792 |
6. Three identical cones with vertex $A$ touch each other externally. Each of them touches internally a fourth cone with vertex at point $A$ and vertex angle $\frac{2 \pi}{3}$. Find the vertex angle of the identical cones. (The vertex angle of a cone is the angle between its generators in the axial section.) | Answer: $2 \operatorname{arcctg} \frac{4+\sqrt{3}}{3}$.
Solution. Let the desired angle be $2 \alpha$. Inscribed in the equal cones are spheres with centers $O_{1}, O_{2}, O_{3}$ of the sam... | 2\operatorname{arcctg}\frac{4+\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,793 |
1. For what smallest $k$ can $k$ cells be marked on an $8 \times 9$ board such that for any placement of a four-cell figure on the board, it can be rotated and flipped. | Answer: 16.
Solution. It is not hard to notice that in any $2 \times 4$ rectangle, there are at least two marked cells. Since from an $8 \times 9$ board, 8 non-overlapping $2 \times 4$ rectangles can be cut out, there must be at least 16 marked cells in it. An example with 16 marked cells is shown in the figure.
^{3} \leqslant 4\left(a^{3}+b^{3}\right) \quad \text { for all } a, b \geqslant 0
$$
Indeed, when $a=b=0$ it is obvious. In the opposite case, after dividing by $a+b$ we get $a^{2}+2 a b+b^{2} \leqslant 4\left(a^{2}-a b... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,795 |
3. On the height $B H$ of triangle $A B C$, a point $D$ is marked. The line $A D$ intersects side $B C$ at point $E$, and the line $C D$ intersects side $A B$ at point $F$. It is known that $B H$ divides segment $F E$ in the ratio $1: 3$, counting from point $F$. Find the ratio $F H: H E$. | Answer: $1: 3$.
Solution 1. Let $G$ and $H$ be the projections of points $F$ and $E$ onto $AC$, respectively, $T$ be the intersection of $AE$ and $FG$, $M$ be the intersection of $CF$ and $... | 1:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,796 |
4. In the stands of the hockey arena, there are several rows with 168 seats in each row. For the final match, 2016 students from several sports schools were invited as spectators, with no more than 45 from each school. Students from any school need to be seated in one row. What is the minimum number of rows that must b... | Answer: 16.
Solution. Suppose that 46 schools sent 43 students each to the final match, and one school sent 34 students. Since only three groups of 43 students can fit in one row, the number of rows required to accommodate the students from 46 schools should be no less than $\frac{46}{3}=15 \frac{1}{3}$, which means a... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,797 |
5. Given a natural number $x=9^{n}-1$, where $n$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 7. Find $x$. | Answer: 728.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 7. The remainders of the powers of 9 when divided by 7 are $2, 4, 1$ and then repeat cyclically. Thus, the divisibility of $x$ by 7 means tha... | 728 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,798 |
6. Three identical cones with vertex $A$ and vertex angle $\frac{\pi}{3}$ touch each other externally. Each of them touches internally a fourth cone with vertex at point $A$. Find the vertex angle of the fourth cone. (The vertex angle of a cone is the angle between its generators in the axial section.) | Answer: $\frac{\pi}{3}+2 \arcsin \frac{1}{\sqrt{3}}$.
Solution. Let the desired angle be $2 \gamma$. Inscribed in the equal cones are spheres with centers $O_{1}, O_{2}, O_{3}$ of the same ... | \frac{\pi}{3}+2\arcsin\frac{1}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,799 |
1. In the cells of a $10 \times 10$ table, the numbers $1,2,3, \ldots, 100$ are arranged such that the sum of the numbers in any $2 \times 2$ square does not exceed $S$. Find the smallest possible value of $S$. | Answer: 202.
Solution. Divide the $10 \times 10$ table into 25 squares of $2 \times 2$. Since the sum of the numbers in the entire table is
$$
1+2+\cdots+100=\frac{100 \cdot 101}{2}=5050
$$
the arithmetic mean of the sums of the numbers in these 25 squares is 202. Therefore, in at least one square, the sum of the nu... | 202 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,800 |
2. Given numbers $a, b, c, d, e, f$, and $b^{2} \leqslant a c$. Prove the inequality
$$
(a f-c d)^{2} \geqslant(a e-b d)(b f-c e)
$$ | Solution. If $a=0$ or $c=0$, then $b=0$, and the inequality is obvious. Let $a$ and $c$ be non-zero. The right-hand side of the inequality is a quadratic trinomial in $e$
$$
\varphi(e)=-a c e^{2}+e b(a f+c d)-b^{2} d f
$$
with a negative leading coefficient. The maximum of $\varphi$ is achieved at $e=\frac{b(a f+d c)... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,801 |
3. Given an acute-angled triangle $A B C$ with angle $\angle A B C=\alpha$. On the extension of side $В C$ a point $D$ is taken such that the line $A D$ is tangent to the circumcircle $\omega$ of triangle $A B C$. The line $A C$ intersects the circumcircle of triangle $A B D$ at point $E$. It turns out that the bisecto... | Answer: $A C: C E=\sin \alpha$.
Solution. Let $r$ be the radius of $\omega$, and $K$ and $L$ be the points of intersection of the angle bisector of $\angle A D E$ with $A E$ and $\omega$ respectively (see figure). The angle between the tangent $A D$ to the circle $\omega$ and the chord $A C$ is equal to the inscribed ... | AC:CE=\sin\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,802 |
4. In a school, there are 920 students, and in each class, there are no more than $k$ students. All students need to go on a bus tour. For this, 16 buses, each with 71 seats, have been booked. The students need to be seated on the buses so that students from each class end up in the same bus. For what largest $k$ is th... | Answer: 17.
Solution. If $k=18$, then it is not always possible to seat the students in the buses. Indeed, suppose the school has 50 classes of 18 students each and two classes of 10 students each. Only three classes of 18 students can fit in one bus. Therefore, it would require at least 17 buses to transport students... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,803 |
5. Given a natural number $x=5^{n}-1$, where $n-$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$. | Answer: 3124.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 11. The remainders of the powers of 5 when divided by 11 are $5, 3, 4, 9, 1$ and then repeat cyclically. Thus, the divisibility of $x$ by 11... | 3124 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,804 |
6. Three cones with vertex $A$ touch each other externally, and the angles at the vertex of the first two are $\frac{\pi}{6}$. Each of the cones touches internally a fourth cone with vertex at point $A$ and an angle at the vertex of $\frac{\pi}{3}$. Find the angle at the vertex of the third cone. (The angle at the vert... | Answer: $2 \operatorname{arcctg}(\sqrt{3}+4)$.
Solution 1. Let $2 \beta$ be the desired angle, $\alpha=\frac{\pi}{12}$. Inscribing spheres with centers $O_{1}, O_{2}, O_{3}$ in the first th... | 2\operatorname{arcctg}(\sqrt{3}+4) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,805 |
1. In the black cells of an $8 \times 8$ chessboard, the numbers $1,2,3, \ldots, 32$ are placed such that the sum of the numbers in any $2 \times 2$ square does not exceed $S$. Find the smallest possible value of $S$. | Answer: 33.
Solution. Divide the chessboard into 16 squares of $2 \times 2$. Since the sum of the numbers in all the black cells is
$$
1+2+\cdots+32=\frac{32 \cdot 33}{2}=16 \cdot 33
$$
the arithmetic mean of the sums of the numbers in these 16 squares is 33. Therefore, in at least one square, the sum of the numbers... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,806 |
2. Given numbers $a, b, c, d, e, f$, and $b^{2} \geqslant a^{2}+c^{2}$. Prove the inequality
$$
(a f-c d)^{2} \leqslant(a e-b d)^{2}+(b f-c e)^{2}
$$ | Solution. If $a=c=0$, the inequality is obvious. Otherwise, the right-hand side of the inequality is a quadratic trinomial in $e$
$$
\varphi(e)=\left(a^{2}+c^{2}\right) e^{2}-2 e b(a d+c f)+b^{2}\left(d^{2}+f^{2}\right)
$$
with a positive leading coefficient. The minimum of $\varphi$ is
$$
-\frac{D}{4\left(a^{2}+c^{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,807 |
3. Triangle $A B C$ with angle $\angle A B C=135^{\circ}$ is inscribed in circle $\omega$. The lines tangent to $\omega$ at points $A$ and $C$ intersect at point $D$. Find $\angle A B D$, given that $A B$ bisects segment $C D$. Answer: $90^{\circ}$
=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secan... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,808 |
4. A warehouse stores 1500 tons of various goods in containers. The weight of any container is a multiple of a ton and does not exceed $k$ tons. A train consisting of 25 platforms, each with a load capacity of 80 tons, has been dispatched to the warehouse. For what maximum $k$ can this train guarantee the transportatio... | Answer: 26.
Solution. If $k=27$, then it is not always possible to transport all the goods. Indeed, suppose there are 55 containers, each weighing 27 tons, and one weighing 15 tons. Only two containers of 27 tons each can fit on one platform. Therefore, to transport 55 such containers, at least 28 platforms are requir... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,809 |
5. Given a natural number $x=8^{n}-1$, where $n$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 31. Find $x$. | Answer: 32767.
Solution. One of the simple divisors of $x$ is obviously 7. First, let's prove that a number of the form $y=8^{k}-1$ is a power of 7 only in the case $k=1$. Indeed, for $k>1$, write $y=a b$, where $a=2^{k}-1, b=2^{2 k}+2^{k}+1$. Since
$$
b=2^{2 k}-1+2^{k}-1+3=a\left(2^{k}+2\right)+3
$$
the numbers $a$... | 32767 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,810 |
6. Three cones with vertex $A$ touch each other externally, and the first two are identical, while the angle at the vertex of the third cone is $2 \arcsin \frac{1}{4}$. Each of the cones touches the fourth cone internally, with the vertex at point $A$. Find the angle at the vertex of the first two cones, if it is half ... | Answer: $\frac{\pi}{6}+\arcsin \frac{1}{4}$.
Solution. Let $2 \alpha$ be the desired angle, $\beta=\arcsin \frac{1}{4}$. Inscribing spheres with centers $O_{1}, O_{2}, O_{3}$ in the first th... | \frac{\pi}{6}+\arcsin\frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,811 |
1. Pete and Vasya are playing the following game. Pete marks $k$ cells on a $9 \times 9$ board, after which Vasya places a $1 \times 4$ rectangle on the board and tells Pete which of the marked cells he has covered (the rectangle can be rotated). Vasya wins if Pete cannot uniquely determine the position of the rectangl... | Answer: 40.
Solution. We will show that it is necessary to mark at least 40 cells. Consider 5 consecutive cells: $A|B| C|D| E$. Petya must mark at least one of the cells $A$ and $E$. Indeed, if Petya does not mark cells $A$ and $E$, then he will not be able to distinguish the placement of the rectangle on cells $A B C... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,812 |
2. Given numbers $x, y$, satisfying the condition $x^{4}+y^{4} \geqslant 2$. Prove the inequality $\left|x^{12}-y^{12}\right|+2 x^{6} y^{6} \geqslant 2$. | Solution 1. Note that
$$
(u+v)^{3} \leqslant 4\left(u^{3}+v^{3}\right) \quad \text { for all } u, v \geqslant 0
$$
Indeed, for $u=v=0$ everything is obvious. In the opposite case, after dividing by $u+v$, we get $u^{2}+2 u v+v^{2} \leqslant 4\left(u^{2}-u v+v^{2}\right)$, which is equivalent to $3(u-v)^{2} \geqslant ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,813 |
3. A circle with center $O$ and radius $r$ is tangent to the sides $BA$ and $BC$ of the acute angle $ABC$ at points $M$ and $N$ respectively. A line passing through point $M$ parallel to $BC$ intersects the ray $BO$ at point $K$. On the ray $MN$, a point $T$ is chosen such that $\angle MTK = \frac{1}{2} \angle ABC$. Fi... | Answer: $\sqrt{r(a+r)}$.
Solution. Let $L$ be the intersection point of $M N$ with the ray $B O$. Note that triangle $M B N$ is isosceles, and $B L$ is the bisector of its angle $B$. Therefo... | \sqrt{r(+r)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,814 |
4. In the theater, there are $k$ rows of seats. 770 spectators came to the theater and sat down (possibly not occupying all the seats). After the intermission, all the spectators forgot which seats they had and sat down differently. For what largest $k$ will there definitely be 4 spectators who sat in the same row both... | Answer: 16
Solution. If the audience is seated on 16 rows, then on some row there are no fewer than 49 spectators (otherwise, there would be no more than 48 spectators on each row, and the total would not exceed $16 \cdot 48=7683$, making it impossible to seat the spectators of this row after the intermission such tha... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,815 |
5. Given a natural number $x=9^{n}-1$, where $n$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$. | Answer: 59048.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 11. First, let's prove that the number $3^{s}+1$ cannot be a power of two when $s>1$, and the number $3^{s}-1$ cannot be a power of two whe... | 59048 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,816 |
6. Three cones with vertex $A$ touch each other externally, with the first two being identical, and the third having an angle at the vertex equal to $\frac{\pi}{2}$. All cones touch the same plane passing through point $A$ and lie on the same side of it. Find the angle at the vertex of the first two cones. (The angle a... | Answer: $2 \operatorname{arctg} \frac{4}{5}$.
Solution. Let $2 \alpha$ be the desired angle, $\beta=\frac{\pi}{4}, \Pi$ be the plane tangent to the cones. Inscribed in the cones are spheres... | 2\operatorname{arctg}\frac{4}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,817 |
1. Pete and Vasya are playing the following game. Pete marks $k$ cells on a $13 \times 13$ board, after which Vasya places a $1 \times 6$ rectangle on the board and tells Pete which of the marked cells are covered by the rectangle (the rectangle can be rotated). Vasya wins if Pete cannot uniquely determine the position... | Answer: 84.
Solution. We will show that it is necessary to mark at least 84 cells. Consider 7 consecutive cells: $A|B| C|D| E|F| G$. Petya must mark at least one of the cells $A$ and $G$. Indeed, if Petya does not mark cells $A$ and $G$, then he will not be able to distinguish the placement of the rectangle on cells $... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,818 |
2. Given numbers $x, y$, satisfying the condition $x^{4}+y^{4} \geqslant 2$. Prove the inequality $\left|x^{16}-y^{16}\right|+4 x^{8} y^{8} \geqslant 4$. | Solution. Note that $u^{2}+v^{2} \geqslant \frac{1}{2}(u+v)^{2}$ for any $u, v \geqslant 0$, since this inequality is equivalent to $(u-v)^{2} \geqslant 0$. Due to the symmetry of the inequality, we can assume that $|x| \geqslant|y|$. Then $x^{8} y^{8} \geqslant y^{16}$, from which
$$
x^{16}-y^{16}+4 x^{8} y^{8} \geqs... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,819 |
3. Circle $\omega$ with center $O$ touches the sides $B A$ and $B C$ of the acute angle $A B C$ at points $M$ and $N$ respectively. The line passing through point $M$ parallel to $B C$ intersects the ray $B O$ at point $K$. On the ray $M N$, a point $T$ is chosen such that $\angle M T K=\frac{1}{2} \angle A B C$. It tu... | Answer: $\frac{a^{2}}{2}$.
Solution. Let $L$ be the point of tangency of $K T$ with $\omega$, and $S$ be the intersection point of $M N$ with the ray $B O$. Note that triangle $M B N$ is is... | \frac{^2}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,820 |
4. Vasya, a bibliophile, has 1300 books arranged on $k$ bookshelves in his cabinets. One day, Vasya moved the cabinets to a new place, taking all the books out beforehand. In the process, he completely forgot how the books were arranged before and placed them in the cabinets in a different order. For what maximum $k$ w... | Answer: 18.
Solution. If Vasya has 18 shelves, then on some shelf there are at least 73 books (otherwise, there would be no more than 72 books on each shelf, and the total would not exceed $18 \cdot 72=12964$, making it impossible to rearrange the books on this shelf so that there are no more than four books on each s... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,821 |
5. Given a natural number $x=9^{n}-1$, where $n-$ is an odd natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 61. Find $x$. | Answer: 59048.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 61. The remainders of the powers of 9 when divided by 61 are $9, 20, 58, 34, 1$ and then repeat cyclically. Thus, the divisibility of $x$ b... | 59048 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,822 |
6. Three cones with vertex $A$ touch each other externally, and the angles at the vertex of the first two are $\frac{\pi}{3}$. All cones also touch a plane passing through point $A$, and they lie on the same side of it. Find the angle at the vertex of the third cone. (The angle at the vertex of a cone is defined as the... | Answer: $2 \operatorname{arcctg} 2(\sqrt{3} \pm \sqrt{2})$.
Solution. Let $2 \beta$ be the desired angle, $\alpha=\frac{\pi}{6}, \Pi$ be the plane that the cones touch. Inscribed in the con... | 2\operatorname{arcctg}2(\sqrt{3}\\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,823 |
1. Pete and Vasya are playing the following game. Pete marks $k$ cells on a $9 \times 9$ board, after which Vasya places a corner of three cells on the board and tells Pete which of the marked cells he has covered. Vasya wins if Pete cannot uniquely determine the position of the corner. For what smallest $k$ can Pete m... | Answer: 68.
Solution. Let's note a few simple facts at the beginning.
1) In any $2 \times 3$ rectangle, there is no more than one unmarked cell. This is verified by direct enumeration.
2) In any $3 \times 3$ square, there are no more than two unmarked cells, and if there are two, they are located in opposite corners ... | 68 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,824 |
2. Given positive numbers $a, b, c, d, e, f$, and $|\sqrt{a b}-\sqrt{c d}| \leqslant 2$. Prove the inequality
$$
\left(\frac{e}{a}+\frac{b}{e}\right)\left(\frac{e}{c}+\frac{d}{e}\right) \geqslant\left(\frac{f}{a}-b\right)\left(d-\frac{f}{c}\right)
$$ | Solution. Let's denote the left side of the inequality by $\varphi(e)$, and the right side by $\psi(f)$. By the Cauchy inequality,
$$
\varphi(e)=\frac{d}{a}+\frac{b}{c}+\frac{e^{2}}{a c}+\frac{b d}{e^{2}} \geqslant \frac{d}{a}+\frac{b}{c}+2 \sqrt{\frac{b d}{a c}}=\left(\sqrt{\frac{d}{a}}+\sqrt{\frac{b}{c}}\right)^{2}
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,825 |
3. In triangle $ABC$, the median $AM$ is drawn. Circle $\alpha$ passes through point $A$, touches line $BC$ at point $M$, and intersects sides $AB$ and $AC$ at points $D$ and $E$ respectively. On the arc $AD$ that does not contain point $E$, a point $F$ is chosen such that $\angle BFE = 72^{\circ}$. It turns out that $... | Answer: $36^{\circ}$.
Solution. Inscribed angles $D A F$ and $D E F$ subtend the arc $D F$ and are therefore equal. Given that $\angle F A B = \angle A B C$, it follows that $B C \parallel ... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,826 |
4. In a box, there is a large batch of flowers of six types mixed together. Vasya randomly takes flowers one by one from the box. As soon as he collects 5 flowers of the same type, he makes a bouquet and sells it. What is the minimum number of flowers he needs to take to guarantee selling 10 bouquets? | Answer: 70.
Solution. Note that 69 flowers are not enough. Indeed, if Vasya pulled out 49 flowers of the first type and 4 flowers of each of the other types, then in total he took $49+5 \cdot 4=69$ flowers, but from them, he can only make 9 bouquets.
Suppose Vasya pulled out the 70th flower and possibly sold one bouq... | 70 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,827 |
5. Given a natural number $x=7^{n}+1$, where $n$ is an odd natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$. | Answer: 16808.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 11. The divisibility of $x$ by 11 is equivalent to $n$ being an odd multiple of 5, that is, $x=7^{5m}+1$ for some odd $m$. From here, using... | 16808 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,828 |
6. Three cones with vertex $A$ touch each other externally, and the first two are identical, while the angle at the vertex of the third cone is $\frac{\pi}{3}$. Each of the cones touches internally a fourth cone with vertex at point $A$ and an angle at the vertex of $\frac{5 \pi}{6}$. Find the angle at the vertex $y$ o... | Answer: $2 \operatorname{arctg}(\sqrt{3}-1)$.
Solution. Let $2 \alpha$ be the desired angle, $\beta=\frac{\pi}{6}, \gamma=\frac{5 \pi}{12}$. Inscribing spheres with centers $O_{1}, O_{2}, O... | 2\operatorname{arctg}(\sqrt{3}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,829 |
1. Pete and Vasya are playing the following game. Pete marks $k$ cells on an $8 \times 8$ board, after which Vasya places a four-cell figure $\square$ on the board and tells Pete which of the marked cells he has covered (the figure can be rotated and flipped). Vasya wins if Pete cannot uniquely determine the position o... | Answer: 48.
Solution. We will show that at least 48 cells are marked. First, let's check the following statement: among any two cells that are two cells apart horizontally, vertically, or diagonally, at least one is marked. Indeed, consider a $3 \times 3$ square:
\left(c e+\frac{d}{e}\right) \geqslant\left(a^{2} f^{2}-\frac{b^{2}}{f^{2}}\right)\left(\frac{d^{2}}{f^{2}}-c^{2} f^{2}\right)
$$ | Solution. Let the left-hand side of the inequality be denoted by $\varphi(e)$, and the right-hand side by $\psi(f)$. By the Cauchy inequality,
$$
\varphi(e)=a d+b c+a c e^{2}+\frac{b d}{e^{2}} \geqslant a d+b c+2 \sqrt{a c b d}=(\sqrt{a d}+\sqrt{b c})^{2}
$$
and also
$$
\psi(f)=a^{2} d^{2}+b^{2} c^{2}-\left(a^{2} c^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,831 |
3. A circle $\omega$ is circumscribed around triangle $ABC$. A line tangent to $\omega$ at point $C$ intersects the ray $BA$ at point $P$. On the ray $PC$ beyond point $C$, a point $Q$ is marked such that $PC = QC$. The segment $BQ$ intersects the circle $\omega$ again at point $K$. On the smaller arc $BK$ of the circl... | Answer: $30^{\circ}$.
Solution. Inscribed angles $L A K$ and $L B K$ subtend the arc $L K$ and are therefore equal. Given that $\angle L B Q = \angle B Q C$, it follows that $B L \| P Q$. Th... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,832 |
4. What is the maximum number of vertices of a regular 2016-gon that can be marked so that no four marked vertices are the vertices of any rectangle? | Answer: 1009.
Solution. Note that an inscribed quadrilateral is a rectangle if and only if its diagonals are diameters of the circumscribed circle. The 2016-gon has exactly 1008 pairs of diametrically opposite vertices. If no rectangle can be formed from the marked vertices, then only in one pair can both vertices be ... | 1009 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,833 |
5. Given a natural number $x=6^{n}+1$, where $n-$ is an odd natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$. | Answer: 7777.
Solution. The divisibility of $x$ by 11 is equivalent to $n$ being an odd multiple of 5, that is, $x=6^{5m}+1$ for some odd $m$. From here, using the formula for the sum of a geometric progression,
$$
x=p \cdot q, \quad \text{where } p=6^{m}+1, q=1-6^{m}+6^{2m}-6^{3m}+6^{4m}
$$
Note that $p$ and $q$ ar... | 7777 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,834 |
6. Three cones with vertex $A$ touch each other externally, and the first two are identical, while the third has an angle at the vertex of $\frac{\pi}{4}$. Each of the cones touches internally a fourth cone with vertex at point $A$ and an angle at the vertex of $\frac{3 \pi}{4}$. Find the angle at the vertex of the fir... | Answer: $2 \operatorname{arctg} \frac{2}{3}$.
Solution. Let $2 \alpha$ be the desired angle, $\beta=\frac{\pi}{8}, \gamma=\frac{3 \pi}{8}$. Inscribing spheres with centers $O_{1}, O_{2}, O_... | 2\operatorname{arctg}\frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,835 |
1. (10 points) The rules of the game are as follows: from 64 different items, on each turn, the player must form a set of items that has not been mentioned in the game before, with the number of items equal to the player's age in years. Players take turns; either player can start the game; the player who cannot make a ... | Answer: 34 years
Solution: The game move is the selection of 64 elements from a certain subset, which contains as many elements as the player's age. Thus, the problem requires comparing the number of ways to make such a selection: the player with a greater number of ways to choose their "own" subset, i.e., $C_{64}^{x}... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,836 |
2. (10 points) Minister K. issued an order that citizens will only be received if the number of ways to choose a group of four people from those who have come is less than the number of ways to choose a group of two people from them. Determine the maximum number of citizens that the minister can receive? | Answer: 5.
Solution: The number of ways to choose a group of four people from $n$ people is $C_{n}^{4}$; a group of two people is $C_{n}^{2}$. We are interested in the maximum natural number $n$ such that $C_{n}^{4}<C_{n}^{2}$. Transforming:
$$
\frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4}<\frac{n(n-1)}{2} \Left... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,837 |
3. (20 points) In math class, each dwarf needs to find a three-digit number without zero digits, divisible by 3, such that when 297 is added to it, the result is a number with the same digits but in reverse order. What is the minimum number of dwarfs that should be in the class so that among the numbers they find, ther... | Answer: 19.
Solution: Let the number of numbers that satisfy the condition of the problem be $N$. Then, by the Pigeonhole Principle, the minimum number of gnomes must be $N+1$.
Let's write a three-digit number as $\overline{x y z}$, where $x$ is the number of hundreds, $y$ is the number of tens, and $z$ is the number... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,838 |
4. (20 points) Let the set $\boldsymbol{A}$ consist of all three-, five-, seven-, and nine-digit numbers, in the representation of which the decimal digits $1,2, \ldots, n$ (not necessarily distinct) are used, and the set $\boldsymbol{B}$ - of all two-, four-, six-, and eight-digit numbers, in the representation of whi... | Answer: $m=8$ or $m=9$.
Solution: The number of numbers in set $\boldsymbol{A}$ is $6^{3}+6^{5}+6^{7}+6^{9}=1036624$. The number of numbers in set $\boldsymbol{B}$ is $N_{m}=m^{2}+m^{4}+m^{6}+m^{8}$ and $N_{m}$ increases with the increase of $m$.
By direct verification, we find that
$$
N_{7}=58849001036624 .
$$
Thu... | =8or=9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,839 |
5. (30 points) In triangle $K I A$, a point $V$ is marked on side $K I$ such that $K I = V A$. Then, inside the triangle, a point $X$ is marked such that angle $X K I$ is half of angle $A V I$, and angle $X I K$ is half of angle $K V A$. Let $O$ be the intersection point of line $A X$ and side $K I$. Is it true that $K... | Answer: Yes, correct.
Solution: Note that angle $K X I$ is a right angle (its sides are parallel to the bisectors of angles that sum up to a straight angle). Let $S$ be the midpoint of $K I$.... | KO=VI | Geometry | proof | Yes | Yes | olympiads | false | 7,840 |
6. (30 points) In triangle $A B C$ with sides $A B=5, B C=6$ and $A C=4$, the angle bisector $A A_{1}$ of angle $B A C$ is drawn. Then, in triangle $A A_{1} C$, the angle bisector $A_{1} C_{1}$ of angle $A A_{1} C$ is drawn; in triangle $C_{1} A_{1} C$, the angle bisector $C_{1} A_{2}$ of angle $A_{1} C_{1} C$ is drawn... | Answer: The similarity coefficient is $\left(\frac{4}{9}\right)^{2021}$.
Solution: It is sufficient to prove that triangle $C_{1} A_{1} C$ is similar to triangle $A B C$ with a similarity coe... | (\frac{4}{9})^{2021} | Geometry | proof | Yes | Yes | olympiads | false | 7,841 |
7. (40 points) Positive numbers $x, y, z$ are such that $x y + y z + z x = 6$. Is it true that
$$
\frac{1}{2 \sqrt{2} + x^{2}(y+z)} + \frac{1}{2 \sqrt{2} + y^{2}(x+z)} + \frac{1}{2 \sqrt{2} + z^{2}(x+y)} \leq \frac{1}{x y z} ?
$$ | Answer: Yes, correct.
Solution: By the inequality between the arithmetic mean and the geometric mean $2=(x y+y z+z x) / 3 \geq \sqrt[3]{x^{2} y^{2} z^{2}}$, from which $2 \sqrt{2} \geq x y z$ Let's estimate each term separately:
$$
\begin{gathered}
\frac{1}{2 \sqrt{2}+x^{2}(y+z)}=\frac{1}{2 \sqrt{2}+x(x y+x z)}= \\
=... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,842 |
8. (40 points) $B$ In the 8-A class, there are $n$ students ( $n \geq 2$ ). For them, clubs are organized, each of which is attended by at least two students. Any two clubs that have at least two students in common differ in the number of participants. Prove that the number of clubs is no more than $(n-1)^{2}$. | Solution: It is sufficient to prove that for each $2 \leq k \leq n$ there are no more than $\frac{n(n-1)}{k(k-1)}$ circles attended by exactly $k$ people. Indeed, since there cannot be fewer than 2 or more than $n$ people in a circle, the number of circles does not exceed
$$
n(n-1)\left(\frac{1}{1 \cdot 2}+\frac{1}{2 ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 7,843 |
9. (40 points) Find the number of pairs of natural numbers $m$ and $n$, satisfying the equation $\frac{1}{m}+\frac{1}{n}=\frac{1}{2020}$. | Answer: 45.
Solution: Transform the given equation
\[
\begin{gathered}
\frac{1}{m}+\frac{1}{n}=\frac{1}{2020} \Leftrightarrow \frac{n+m}{m \cdot n}=\frac{1}{2020} \Leftrightarrow m n=2020(n+m) \Leftrightarrow \\
\Leftrightarrow m n-2020 n-2020 m+2020^{2}-2020^{2}=0 \Leftrightarrow(n-2020)(m-2020)=2020^{2}
\end{gather... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,844 |
1. (10 points) Two neighbors ordered sectional fences for their plots. Ivan Ivanovich bought a fence of traditional shape, each section of which is a rectangle with sides 5 by 2. Petr Petrovich, on the other hand, took a fence with slanted sections in the form of parallelograms with sides 5 by 2. The number of sections... | Answer: b).
Solution: From the condition, it is clear that the sections of the fence of Pyotr Petrovich have the shape of a parallelogram, in which there are no right angles. This means that if $\alpha$ is any angle between the sides of this parallelogram, then $\sin \alpha < 1$. Therefore, the area of a section of Py... | b) | Geometry | MCQ | Yes | Yes | olympiads | false | 7,845 |
3. (20 points) From village $A$ to village $B$, the Indians take three times as long to travel by canoe as from village $B$ to village $A$. How many times longer than usual can the Indians take to travel from $B$ to $A$ by canoe without paddles? | Answer: 3 times.
Solution: It is not hard to see that village V is upstream from village A. Let's denote the speed of the river current as $v_{r}$, and the average speed of the canoe in still water as $v$. If $S$ is the distance along the river from A to V, then from the condition we have
$$
3 \frac{S}{v+v_{r}}=\frac... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,847 |
4. (20 points) The ballroom in the palace of the thirtieth kingdom is a region on the plane, the coordinates of which satisfy the conditions $|x| \leqslant 4,|y| \leqslant 6$. How many identical parquet tiles, having the shape of a rectangle with sides 1.5 and 2, are needed to tile the floor of the room? Tiling is cons... | Answer: 32.
Solution: From the condition, we find that the magic room is a rectangle with sides parallel to the coordinate axes, and the lengths of the sides parallel to the x-axis are 8, while the lengths of the sides parallel to the y-axis are 12. It is not hard to see that a tiling exists (along the side of the roo... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,848 |
5. (30 points) Let in triangle KIA, point $O$ be the foot of the median from vertex K; $Y$ be the foot of the perpendicular dropped from point I to the bisector of angle IOK; $Z$ be the foot of the perpendicular dropped from point A to the bisector of angle AOK. X is the point of intersection of segments $K O$ and $Y Z... | # Solution:
Extend $A Z$ to intersect with $O K$ and denote the intersection point as $R_{1}$. Right triangles $Z O A$ and $Z O R_{1}$ are congruent (the leg $Z O$ is common, $\angle Z O A = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,849 |
6. (30 points) In triangle $B M W$, where $B M < B W < M W$, $B O$ is the altitude, and $B H$ is the median. Point $K$ is symmetric to point $M$ with respect to point $O$. The perpendicular to $M W$, drawn through point $K$, intersects segment $B W$ at point $P$. Prove that if $M P$ and $B H$ are perpendicular, then an... | Solution: Let $R$ be the intersection point of $M P$ and $B O$, and $E$ be the intersection point of $M P$ and $B H$. Since $P K \perp M W$, then $P K \| B O$. And since $O M=O K$, then $R M=R P$ (by Thales' theorem). Therefore, $H R$ is the midline in triangle $M P W$ and thus $H R \| B W$.
 To enter Ali Baba's cave, it is necessary to zero out 28 counters, each set to a natural number in the range from 1 to 2017. Treasure hunters are allowed, in one move, to decrease the values of some of the counters by the same number, which they can change from move to move. Indicate the minimum number o... | Answer: In 11 moves.
## Solution:
Estimation. Suppose the counters are set to all powers of two from 1 to 1024 and some 17 other numbers. Since the order of moves does not matter, we can arrange them in descending order of the numbers being subtracted. We will show by induction that after the $k$-th move, there is a ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,851 |
8. (40 points) On one of the planets in the Alpha Centauri system, elderly women love to paint the cells of $2016 \times 2016$ boards with gold and silver paints. One day, it turned out that in all the boards painted that day, each $3 \times 3$ square had exactly $A$ gold cells, and each $2 \times 4$ or $4 \times 2$ re... | Answer: $A=Z=0$ or $A=9, Z=8$.
Solution: Divide the board into $2 \times 2$ squares. If a $2 \times 2$ square contains $u$ gold cells, then an adjacent $2 \times 2$ square must contain $v=Z-u$ gold cells. Therefore, $2 \times 2$ squares containing $u$ and $v$ gold cells alternate in a checkerboard pattern. Consider a ... | A=Z=0orA=9,Z=8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,852 |
1. On an island, there are two tribes: the tribe of knights, who always tell the truth, and the tribe of liars, who always lie. On the main holiday, 2017 islanders sat around a large round table. Each islander said the phrase: "my neighbors are from the same tribe." It turned out that two liars made a mistake and accid... | Answer: 1344 liars
Solution. Note that no two knights can sit next to each other, meaning the neighbors of each knight are liars. Indeed, consider a chain of knights sitting in a row, surrounded by liars. Suppose there are at least two knights in this chain. The neighbors of the extreme knight are a knight and a liar,... | 1344 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,853 |
3. In triangle $A B C$ with angle $\angle C=60^{\circ}$, the angle bisectors $A A_{1}$ and $B B_{1}$ are drawn. Prove that $A B_{1}+B A_{1}=A B$. | The first solution. Let $I$ be the point of intersection of the angle bisectors in the triangle. Let $\alpha=\angle B A A_{1}$ and $\beta=\angle A B B_{1}$. Then $2 \alpha+2 \beta+60^{\circ}=180^{\circ}$.
Therefore, $\beta=60^{\circ}-\alpha$. Then
$$
\begin{aligned}
\angle A A_{1} C & =\angle A B C+\angle A_{1} A B=2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,855 |
4. Given integers $a$ and $b$, not equal to -1. The quadratic trinomial $x^{2} + abx + (a+b)$ has two integer roots. Prove that $a+b \leqslant 6$. | Solution. We will assume that $a+b>0$, otherwise the proof is already complete. Since the trinomial has integer roots, its discriminant $(ab)^2-4(a+b)$ is a perfect square, i.e., $(ab)^2-4(a+b)=k^2$ for some non-negative integer $k$. Then the numbers $ab$ and $k$ have the same parity. From the relation $00$ and even, s... | +b\leqslant6 | Algebra | proof | Yes | Yes | olympiads | false | 7,856 |
5. In each cell of a $2017 \times 2017$ board, there is a chip. In one operation, you can remove a chip that has a non-zero even number of neighbors (neighbors are chips located in cells that share a side or a corner). What is the minimum number of chips that can be left on the board using such operations?
# | # Answer: 2
Solution. If there are only two chips on the board, then each of them has no more than one neighbor, and according to the rules, they cannot be removed. Therefore, at least two chips will remain. We will show how to leave exactly two chips on the board. For this, we will provide an algorithm that allows cl... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,857 |
6. We will call a divisor $d$ of a natural number $n>1$ good if $d+1$ is also a divisor of $n$. Find all natural $n$ for which at least half of the divisors are good. | Answer: $2,6,12$
Solution. Consider such an $n$. If $d$ is a good divisor of $n$, then $d+1$ is also a divisor of $n$. However, the numbers $d$ and $d+1$ are coprime, so $n$ is divisible by $d(d+1)$ and, in particular, $n \geqslant d(d+1)>d^{2}$. Therefore, $dk^{3}$ and, thus, $k \leqslant 4$. It remains to check the ... | 2,6,12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,858 |
1. On an island, there live knights who always tell the truth, and liars who always lie. In a room, 15 islanders gathered. Each of those in the room said two phrases: "Among my acquaintances in this room, there are exactly six liars" and "among my acquaintances in this room, there are no more than seven knights." How m... | Answer: 9 knights
Solution. If there is a liar in the room, then his statement "among my acquaintances in this room, there are no more than seven knights" is a lie, and there are no fewer than 8 knights in the room. If there is a knight in the room, then his statement "among my acquaintances in this room, there are ex... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,859 |
2. Given various natural numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ and $a_{7}$. Prove that $\left(a_{1}-a_{2}\right)^{4}+\left(a_{2}-a_{3}\right)^{4}+\left(a_{3}-a_{4}\right)^{4}+\left(a_{4}-a_{5}\right)^{4}+\left(a_{5}-a_{6}\right)^{4}+\left(a_{6}-\right.$ $\left.a_{7}\right)^{4}+\left(a_{7}-a_{1}\right)^{4} ... | Solution. Note that all the terms are no less than 1. If one of the brackets is at least 3, then the entire sum is certainly greater than $3^{4}+1=82$. Therefore, we will assume from now on that the absolute value of any bracket does not exceed 2. Mark the numbers $a_{1}, a_{2}, \ldots, a_{7}$ on a line. Let $a_{j}$ be... | 82 | Inequalities | proof | Yes | Yes | olympiads | false | 7,860 |
3. On the side $AB$ of triangle $ABC$, a point $O$ is marked. The circle $\omega$ with center at point $O$ intersects segments $AO$ and $OB$ at points $K$ and $L$ respectively, and touches sides $AC$ and $BC$ at points $M$ and $N$ respectively. Prove that the intersection point of segments $KN$ and $LM$ lies on the alt... | The first solution. Let segments $K N$ and $L M$ intersect at point $P$, and let $Q$ be the foot of the perpendicular dropped from point $P$ to $A B$. We can assume that point $Q$ lies on segment $O K$. Right triangles $O C N$ and $O C M$ have equal legs $O M = O N$ and a common hypotenuse $O C$, so they are
$ with real coefficients. Prove that there exists a natural number $n$, such that the equation
$p(x)=1 / n$ has no rational solutions. | Solution. Let $p(x)=a x^{2}+b x+c$. Suppose the opposite. Let for each natural $n$ there exists a rational number $x_{n}$ such that $p\left(x_{n}\right)=\frac{1}{n}$. Clearly, the numbers $x_{n}$ are distinct. We will show that the numbers $a$ and $c$ are rational. Since
$\left(x_{k}-x_{n}\right)\left(a\left(x_{k}+x_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 7,862 |
5. What is the maximum number of rooks that can be placed on the cells of a $300 \times 300$ board so that each rook attacks no more than one other rook? (A rook attacks all cells it can reach according to chess rules, without passing through other pieces.)
# | # Answer: 400
First solution. We will prove that no more than 400 rooks can be placed on the board. In each row or column, there are no more than two rooks; otherwise, the rook that is not at the edge will attack at least two other rooks. Suppose there are $k$ columns with two rooks each. Consider one such pair. They ... | 400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,863 |
6. Find all such triples of distinct prime numbers $p, q, r$ and a natural number $k$, such that $p q - k$ is divisible by $r$, $q r - k$ is divisible by $p$, $r p - k$ is divisible by $q$, and the number $p q - k$ is positive. | Answer: $2,3,5$ and all permutations, $k=1$
Solution. Since $p q-k$ is divisible by $r$, the number $n=p q+q r+r p-k$ is also divisible by $r$. Note that $n \geqslant p q-k>0$. Similarly, the number $n$ is divisible by $p$ and by $q$. Since $p, q$ and $r$ are different prime numbers, they are coprime, and thus $n$ is ... | 2,3,5allpermutations,k=1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,864 |
1. On the island, there live knights who always tell the truth, and liars who always lie. At a celebration for the opening of the new football season, 50 football fans sat around a round table: 25 fans of the "Super Eagles" team and 25 fans of the "Super Lions" team. Each of them stated: "to my right is a fan of the 'S... | Answer: No
Solution. Let there be $k$ liars among the fans of the "Super Eagles" and $k$ liars among the fans of the "Super Lions." Then the total number of liars is exactly $2k$, i.e., the statement "to my right is a fan of the 'Super Eagles'" was a lie an even number of times. Thus, an even number of times, the righ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,865 |
2. For any numbers $x, y$ and $z$, prove the inequality
$$
x^{2}+y^{2}+z^{2} \geqslant \sqrt{2}(x y+y z)
$$ | First solution. By expanding the brackets, it is easy to verify that
$$
x^{2}+y^{2}+z^{2}-\sqrt{2}(x y+y z)=\left(x-\frac{y}{\sqrt{2}}\right)^{2}+\left(z-\frac{y}{\sqrt{2}}\right)^{2} \geqslant 0
$$
Second solution. Note that
$$
2\left(x^{2}+z^{2}\right) \geqslant x^{2}+2|x||z|+z^{2}=(|x|+|z|)^{2}
$$
Therefore,
$$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,866 |
3. A circle $\omega_{1}$ is circumscribed around quadrilateral $A B C D$. A circle $\omega_{2}$ is drawn through points $A$ and $B$, intersecting ray $D B$ at point $E \neq B$. Ray $C A$ intersects circle $\omega_{2}$ at point $F \neq A$. Prove that if the tangent to circle $\omega_{1}$ at point $C$ is parallel to line... | Solution. Let $G \neq A$ be the intersection point of line $D A$ with circle $\omega_{2}$. Mark a point $X$ on the tangent to $\omega_{1}$ at point $C$ such that $\angle D C X \leqslant 90^{\circ}$, and a point $Y$ on the tangent to $\omega_{2}$ at point $F$ such that $\angle G F Y \leqslant 90^{\circ}$.
By the proper... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,867 |
5. What is the maximum number of chips that can be placed in the cells of a chessboard (no more than one chip per cell) so that no more than three chips are placed on each diagonal? | Answer: 38
Solution. Consider the eight-cell diagonal consisting of black cells, and the parallel diagonals consisting of black cells. These diagonals have $2, 4, 6, 8, 6, 4,$ and 2 cells. On each of the outermost diagonals, no more than two chips can be placed, and on each of the middle diagonals, no more than three.... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,869 |
6. Find all pairs of natural numbers $a$ and $b$, for which $a^{3} b-1$ is divisible by $a+1$ and $a b^{3}+1$ is divisible by $b-1$. | Answer: $a=b=2, a=b=3$ and $a=1, b=3$.
First solution. Since $a^{3} b-1$ is divisible by $a+1$, the number $b+1=$ $b\left(a^{3}+1\right)-\left(a^{3} b-1\right)$ is also divisible by $a+1$. Similarly, from the fact that $a b^{3}+1$ is divisible by $b-1$, we conclude that the number $a+1=\left(a b^{3}+1\right)-a\left(b^... | =b=2,=b=3,=1,b=3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,870 |
1. On an island, there live knights who always tell the truth, and liars who always lie. On the main holiday, 100 islanders sat around a large round table. Half of those present said: "both my neighbors are liars," and the rest said: "one of my neighbors is a liar." What is the maximum number of knights that can sit at... | # Answer: 67
Solution. Note that three knights cannot sit in a row, since then the middle knight would not be able to utter any of the required phrases. Let $k$ be the number of pairs of neighboring knights. Then each of the knights said the phrase "among my neighbors there is exactly one liar," so $2 k \leqslant 50$.... | 67 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,871 |
2. Positive numbers $a, b$ and $c$ satisfy the condition $c^{2} + ab = a^{2} + b^{2}$. Prove the inequality $c^{2} + ab \leqslant ac + bc$. | First solution. It is sufficient to check the inequality
$\left(a^{2}+b^{2}\right)^{2}=\left(c^{2}+a b\right)^{2} \leqslant(a+b)^{2} c^{2}=(a+b)^{2}\left(a^{2}-a b+b^{2}\right)=(a+b)\left(a^{3}+b^{3}\right)$ or, equivalently, $a^{4}+2 a^{2} b^{2}+b^{4} \leqslant a^{4}+a^{3} b+a b^{3}+b^{4}$. The latter is obvious, sin... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 7,872 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.