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5. On the sides $B C$ and $C D$ of the square $A B C D$, points $E$ and $F$ are chosen such that the angle $E A F$ is $45^{\circ}$. The length of the side of the square is 1. Find the perimeter of triangle $C E F$. | Solution. If point $D$ is reflected across line $A F$, and then across line $A E$, it will transition to point $B$. Indeed, the composition of two axial symmetries relative to intersecting lines is a rotation by twice the angle between the lines. In our case, these two symmetries are equivalent to a $90^{\circ}$ rotati... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,756 |
6. Let $x_{1}$ and $x_{2}$ be the largest roots of the polynomials $f(x)=1-x-4 x^{2}+x^{4}$ and $g(x)=16-8 x-$ $16 x^{2}+x^{4}$ respectively. Find $\frac{x_{1}}{x_{2}}$. | # Solution.
## 1st method
Notice that $g(2 x)=16 f(x)$. Then $x_{1}-$ is a root of $f(x)$ if and only if $2 x_{1}-$ is a root of $g(x)$. Therefore, $\frac{x_{1}}{x_{2}}=\frac{1}{2}$.
2nd method
Comparing the coefficients of the polynomials
$$
f(x)=1-x-4 x^{2}+x^{4} \text { and } g(x)=16-8 x-16 x^{2}+x^{4}
$$
show... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,757 |
8. It is known that the number $\cos 6^{0}$ is a root of the equation $32 t^{5}-40 t^{3}+10 t-\sqrt{3}=0$. Find the other four roots of this equation. (Answers in the problem should be compact expressions, not containing summation signs, ellipses, etc.) | Solution: Substitution $t=\cos \varphi$. The equation will take the form: $32 \cos ^{5} \varphi-40 \cos ^{3} \varphi+10 \cos \varphi=\sqrt{3}$. Transform the left side:
$$
\begin{aligned}
& 2 \cos \varphi\left(16 \cos ^{4} \varphi-20 \cos ^{2} \varphi+5\right)=[\text { [reduction formulas }]= \\
& =2 \cos \varphi\left... | \cos78,\cos150,\cos222,\cos294 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,758 |
1. (2 points) In a puddle, there live three types of amoebas: red, blue, and yellow. From time to time, any two amoebas of different types can merge into one amoeba of the third type. It is known that in the morning, there were 26 red, 31 blue, and 16 yellow amoebas in the puddle, and by evening, only one amoeba remain... | # Solution:
It is not hard to notice that as a result of the changes that the original population of amoebas can undergo, the parities of the absolute differences of the numbers of different types of amoebas do not change.
Let $n_{1}, n_{2}$ and $n_{3}$ be the numbers of red, blue, and yellow amoebas in the initial p... | blue | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,759 |
2. (3 points) Solve the equation:
$$
\left[\frac{5+6 x}{8}\right]=\frac{15 x-7}{5}
$$ | # Solution:
By the definition of the integer part of a number, we have the system of inequalities
$$
\left\{\begin{array}{c}
\frac{5+6 x}{8} \geq \frac{15 x-7}{5} \\
\frac{5+6 x}{8} > \frac{41}{90}
\end{array}\right. \\
& x \in\left(\frac{41}{90} ; \frac{9}{10}\right]
$$
Now we need to find those values of \( x \) i... | \frac{7}{15};\frac{4}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,760 |
3. (3 points) Solve the equation of the form $f(f(x))=x$, given that $f(x)=x^{2}-4 x-5$
# | # Solution:
It is clear that some solutions to the equation $f(f(x))=x$ can be found by solving the simpler equation $f(x)=x$, which in this case is:
$$
x^{2}-4 x-5=x \Leftrightarrow x^{2}-5 x-5=0 \Leftrightarrow x=\frac{1}{2}(5 \pm 3 \sqrt{5})
$$
Since the roots of the polynomial $f(x)-x=x^{2}-5 x-5$ are also roots... | \frac{1}{2}(5\3\sqrt{5}),\frac{1}{2}(3\\sqrt{41}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,761 |
4. (4 points) Solve the system of equations:
$$
\left\{\begin{array}{c}
\log _{2}(y-x)=\log _{8}(9 y-15 x) \\
x^{2}+y^{2}=15
\end{array}\right.
$$ | # Solution:
The following equivalent transformations are valid:
$$
\begin{aligned}
& \left\{\begin{array} { c }
{ y - x > 0 } \\
{ 9 y - 1 5 x > 0 } \\
{ ( y - x ) ^ { 3 } = 9 y - 1 5 x } \\
{ x ^ { 2 } + y ^ { 2 } = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { c }
{ y - x > 0 } \\
{ ( y - x ) ^ { 3 } =... | {(-\sqrt{15};0);(\sqrt{\frac{3}{2}};3\sqrt{\frac{3}{2}});(\sqrt{3};2\sqrt{3})} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,762 |
5. (5 points) The sides of a parallelogram are 2 and 3, and the angle between them is $\arccos \frac{5}{16}$. Two mutually perpendicular lines divide this parallelogram into four equal-area quadrilaterals. Find the lengths of the segments into which these lines divide the sides of the parallelogram.
# | # Solution:
Let in parallelogram $ABCD$, $|AB| = |CD| = a$, $|AD| = |BC| = b$, and $\angle BAC = \alpha = \arccos c$.
Obviously, $\cos \alpha = c$, and $\cos (\pi - \alpha) = -c$.
Let $|D... | \frac{4}{3}\frac{2}{3},21 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,763 |
6. (5 points) Let $a_{n}$ be the first (leading) digit in the decimal expansion of $n^{2}$ for $n=1,2,3, \ldots\left(a_{1}=1, a_{2}=4, a_{3}=9, a_{4}=1, a_{5}=2, \ldots\right)$. Prove that this sequence is not periodic. | # Solution:
Let $b$ be any digit from 1 to 9 and $k$ be any natural number. Then for all numbers $n$ satisfying the inequality
$$
10^{k} \sqrt{b} \leq n < 10^{k+1} \sqrt{b} \quad (n \in \mathbb{N})
$$
it follows that the sequence of digits in the decimal representation of $n$ will contain a block of $k$ consecutive ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,764 |
1. (2 points) In a puddle, there live three types of amoebas: red, blue, and yellow. From time to time, any two amoebas of different types can merge into one amoeba of the third type. It is known that in the morning, there were 47 red, 40 blue, and 53 yellow amoebas in the puddle, and by evening, only one amoeba remain... | # Solution:
It is not hard to notice that as a result of the changes that the original population of amoebas can undergo, the parities of the absolute differences of the numbers of different types of amoebas do not change.
Let $n_{1}, n_{2}$ and $n_{3}$ be the numbers of red, blue, and yellow amoebas in the initial p... | blue | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,765 |
2. (3 points) Solve the equation:
$$
\left[\frac{9 x-4}{6}\right]=\frac{12 x+7}{4}
$$ | # Solution:
By the definition of the integer part of a number, we have the system of inequalities
$$
\left\{\begin{array}{c}
\frac{9 x-4}{6} \geq \frac{12 x+7}{4} \\
\frac{9 x-4}{6}-\frac{41}{18}
\end{array}\right. \\
& x \in\left(-\frac{41}{18} ;-\frac{29}{18}\right]
\end{aligned}
$$
Now we need to find those value... | -\frac{9}{4};-\frac{23}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,766 |
3. (3 points) Solve the equation of the form $f(f(x))=x$, given that $f(x)=x^{2}+2 x-4$
# | # Solution:
It is clear that some solutions to the equation $f(f(x))=x$ can be found by solving the simpler equation $f(x)=x$, which in this case is:
$$
x^{2}+2 x-4=x \Leftrightarrow x^{2}+2 x-5=0 \Leftrightarrow x=\frac{1}{2}(-1 \pm \sqrt{17})
$$
Since the roots of the polynomial $f(x)-x=x^{2}+2 x-5$ are also roots... | \frac{1}{2}(-1\\sqrt{17}),\frac{1}{2}(-3\\sqrt{13}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,767 |
4. (4 points) Solve the system of equations:
$$
\left\{\begin{aligned}
& \log _{2}(y-x)=\log _{8}(19 y-13 x) \\
& x^{2}+y^{2}=13
\end{aligned}\right.
$$ | # Solution:
The following equivalent transformations are valid:
$$
\left\{\begin{array} { c }
{ y - x > 0 } \\
{ 19 y - 13 x > 0 } \\
{ ( y - x ) ^ { 3 } = 19 y - 13 x } \\
{ x ^ { 2 } + y ^ { 2 } = 13 }
\end{array} \Leftrightarrow \left\{\begin{array} { c }
{ y - x > 0 } \\
{ ( y - x ) ^ { 3 } = 19 y - 13 x } \\
{... | {(-\sqrt{13};0);(-3;-2);(-\frac{1}{\sqrt{2}};\frac{5}{\sqrt{2}})} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,768 |
5. (5 points) The sides of a parallelogram are 5 and 13, and the angle between them is $\arccos \frac{6}{13}$. Two mutually perpendicular lines divide this parallelogram into four equal-area quadrilaterals. Find the lengths of the segments into which these lines divide the sides of the parallelogram.
# | # Solution:
Let in parallelogram $ABCD$, $|AB| = |CD| = a$, $|AD| = |BC| = b$, $\angle BAC = \alpha = \arccos c$. Clearly, $\cos \alpha = c$, and $\cos (\pi - \alpha) = -c$.
Denote $|DN| =... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,769 |
6. (5 points) Let $a_{n}$ be the first (leading) digit in the decimal expansion of $[\sqrt{n}]$ for $n=1,2,3, \ldots\left(a_{1}=1, a_{2}=1, a_{3}=1, a_{4}=2, a_{5}=2, \ldots\right)$. Prove that this sequence is not periodic. | # Solution:
Let $b$ be any digit from 1 to 9 and $k$ be any natural number. Then for all numbers $n$ satisfying the inequality
$$
10^{2 k} \cdot b^{2} \leq n < 10^{2 k+2} \cdot b^{2} \left(10^{2}-1\right)$.
Therefore, in our sequence, there will be arbitrarily long consecutive sequences of ones, twos, ..., nines. He... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,770 |
1. (2 points) For a natural number $x$, five statements are made:
$$
3 x>91
$$
$$
\begin{aligned}
& x37 \\
& 2 x \geq 21 \\
& x>7
\end{aligned}
$$
It is known that only three of them are true, and two are false. Find $x$. | Solution. Let's transform the original system of inequalities
$$
\begin{aligned}
& x>\frac{91}{3} \\
& x \geq \frac{21}{2} \\
& x>\frac{37}{4} \\
& x>7 \\
& x<120
\end{aligned}
$$
The first inequality does not hold, since otherwise, four inequalities would be satisfied immediately. Therefore, $x \leq \frac{91}{3}$, a... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,771 |
2. (2 points) For which integers $p$ and $q$ is the value
of the polynomial $f(x)=x^{2}+p x+q$ divisible by 2 for any integer $x ?$ | Solution: According to the condition
$f(x+1)-f(x)=2 x+1+p \vdots 2 \forall x \in Z \Leftrightarrow(p+1) \vdots 2 \Leftrightarrow p-$ is odd. Therefore, the numbers $x^{2}$ and $p x$ are either both even or both odd, which means $x^{2}+p x \vdots 2$. Consequently, $q$ is even.
Answer: $p$ is odd, $q$ is even. | p | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,772 |
3. (3 points) Solve the system of equations
$$
\left\{\begin{array}{c}
x + xy + y = 2 + 3 \sqrt{2} \\
x^{2} + y^{2} = 6
\end{array}\right.
$$ | Solution: Substitution
$$
u=x+y, v=x \cdot y
$$
The original system will take the form
$$
\left\{\begin{array}{l}
u+v=2+3 \sqrt{2} \\
u^{2}-2 v=6
\end{array}\right.
$$
Multiplying equation (2) by 2 and adding to (3), we get $u^{2}+2 u=10+6 \sqrt{2} \Leftrightarrow(u+1)^{2}=11+6 \sqrt{2}=(3+\sqrt{2})^{2}$. $u_{1}=2+... | [\begin{pmatrix}x_{1}=2,y_{1}=\sqrt{2}\\x_{2}=\sqrt{2},y_{2}=2\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,773 |
4. (4 points) The population of the city increased by $50\%$ annually for $x$ years, then decreased by $\frac{388}{9}\%$ annually for $y$ years, and then decreased by $\frac{50}{3}\%$ annually for $z$ years. As a result, the population doubled. It is known that $x, y, z$ are integers. Find them. | Solution: Let $a=50, b=-\frac{388}{9}, c=-\frac{50}{3}$. By the condition $\left(1+\frac{a}{100}\right)^{x} \cdot\left(1+\frac{b}{100}\right)^{y} \cdot\left(1+\frac{c}{100}\right)^{z}=2$.
Adding the fractions in the parentheses and factoring the numerator and denominator into prime factors:
$$
1+\frac{a}{100}=2^{-1} ... | 4,1,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,774 |
6. (5 points) Find all pairs of integers $(x, y)$ for which the following equality holds:
$$
x(x+1)(x+7)(x+8)=y^{2}
$$
# | # Solution:
$$
\begin{array}{ll}
x(x+1)(x+7)(x+8)=y^{2} \\
\left(x^{2}+8 x\right)\left(x^{2}+8 x+7\right)=y^{2}
\end{array}
$$
Let $z=x^{2}+8 x+\frac{7}{2} \Rightarrow 2 z=2 x^{2}+16 x+7 \in Z$. From (2) we find
$$
\left(z-\frac{7}{2}\right)\left(z+\frac{7}{2}\right)=y^{2} \Leftrightarrow z^{2}-\frac{49}{4}=y^{2}
$$... | (1;12),(1;-12),(-9;12),(-9;-12),(0;0),(-8;0),(-4;-12),(-4;12),(-1;0),(-7;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,776 |
3. On an 8x8 board, 9 checkers are placed. Movements of the checkers (moves) are carried out as follows: a first (movable) and a second checker are chosen. Then the first is placed on such a cell that the initial and resulting positions are symmetric relative to the second checker. Can such moves move the checkers from... | Solution: We will color the cells in white and black. As a result of a move, the checker that was on a white cell will again land on a white cell, and the one that was on a black cell will land on a black cell. This means that the number of white cells occupied by checkers is always the same. It remains to note that in... | impossible | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 8,778 |
4. Prove that for each natural number $n$ the equality $[\sqrt{4 n+1}]=[\sqrt{4 n+3}]$ holds. Here, the brackets [ ] denote the integer part of a number. (Recall that the integer part of a number $x$ is the greatest integer not exceeding $x$. For example, $[3.7]=3$.) | Solution: The integer parts of numbers $a$ and $b$ are equal if and only if the half-interval $(a, b]$ does not contain integers. Suppose the contrary: let for some natural $n$ the inequality
$$
[\sqrt{4 n+1}] \neq[\sqrt{4 n+3}]
$$
holds. Then
$\exists m \in \mathbb{N}: \sqrt{4 n+1}<m \leq \sqrt{4 n+3} \Leftrightarr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,779 |
5. The equations $x^{4}+2 x^{3}-x^{2}-2 x-3=0 \quad$ and $x^{4}+3 x^{3}+x^{2}-4 x-6=0$ have two common roots. Find them. | Answer: $\frac{-1 \pm \sqrt{13}}{2}$. | \frac{-1\\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,780 |
8. We have $n$ integers $0,1,2, \ldots, n-1$. By rearranging these numbers in a random order, we obtain some permutation $\left(i_{1}, i_{2}, \ldots, i_{n}\right)$. From the original set of numbers $(0,1,2, \ldots, n-1)$ and this permutation $\left(i_{1}, i_{2}, \ldots, i_{n}\right)$, we obtain a new set of numbers $\l... | Solution: a) For example, $\left(i_{1}, i_{2}, \ldots, i_{5}\right)=(3,4,0,1,2)$;
b) By the condition $(0+1+\ldots+5)+\left(i_{1}+\ldots+i_{6}\right)=a_{1}+\ldots+a_{6}(\bmod 6)$. If all $a_{i}$ were distinct, then the sum of the numbers on the right side would be 15. But the equality $15+15=15(\bmod 6)$ is not true. ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,781 |
1. Find the value of $f(2019)$, given that $f(x)$ simultaneously satisfies the following three conditions:
1) $f(x)>0$ for any $x>0$
2) $f(1)=1$;
3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}$ for any $a, b \in \mathbb{R}$. | Solution. In the identity given in the problem
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0
$$
Next, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Fina... | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,782 |
3. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent ... | Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be t... | 4900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,784 |
1. Real numbers $x, y, z$ satisfy the relations:
$$
4 x^{2}-2 x-30 y z=25 y^{2}+5 y+12 x z=9 z^{2}-3 z-20 x y .
$$
Find all possible triples of numbers $(a, b, c)$, where $a=2 x+5 y, b=3 z+5 y, c=3 z-2 x$. | Solution. Note that
$$
a-b+c=0
$$
Let $A=4 x^{2}-2 x-30 y z, B=25 y^{2}+5 y+12 x z$ and $C=9 z^{2}-3 z-20 x y$. Subtracting these equations from each other, we get
$$
\begin{aligned}
& A-B=a \cdot(2 x-6 z-5 y-1)=0 \\
& B-C=b \cdot(5 y+4 x-3 z+1)=0 \\
& A-C=c \cdot(1-2 x-10 y-3 z)=0
\end{aligned}
$$
Assume that all ... | (0,0,0),(0,1,1),(1,0,-1),(-1,-1,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,785 |
2. Find all functions $f(x)$ that simultaneously satisfy the following three conditions: 1) $f(x)>0$ for any $x>0$; 2) $f(1)=1$; 3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b) + a^{2} + b^{2}$ for any $a, b \in \mathbb{R}$. | Solution. In the identity from the condition
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0
$$
Then, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Finall... | f(x)=x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,786 |
3. In quadrilateral $ABCD$, the diagonals intersect at point $O$. It is known that $S_{ABO}=S_{CDO}=\frac{3}{2} BC=$ $3 \sqrt{2}, \cos \angle ADC=\frac{3}{\sqrt{10}}$. Find the sine of the angle between the diagonals of this quadrilateral, if its area takes the smallest possible value under the given conditions. | Solution. We will prove that quadrilateral $ABCD$ is a parallelogram. Let $x_{1}, x_{2}, y_{1}, y_{2}$ be the segments into which the diagonals are divided by their point of intersection. Denote the angle between the diagonals as $\alpha$. By the given condition, the areas of triangles $ABO$ and $CDO$ are equal, that i... | \frac{6}{\sqrt{37}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,787 |
4. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent ... | Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be t... | 4900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,788 |
5. Find all prime numbers whose decimal representation has the form 101010 ... 101 (ones and zeros alternate). | Solution: Let $2n+1$ be the number of digits in the number $A=101010$...101 under investigation. Let $q=$ 10 be the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2n}=\frac{q^{2n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2k \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k+1}-1}{q-1} ... | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,789 |
6. Find some integers $A$ and $B$, for which the inequality $0.999 < A + B \cdot \sqrt{2} < 1$ holds. | Solution: Note that if a number of the form $x+y \cdot \sqrt{2}$, where $x, y$ are integers, is raised to a non-negative integer power $n$, we will again obtain a number of the same form, i.e., $(x+y \cdot \sqrt{2})^{n}=x_{1}+y_{1} \cdot \sqrt{2}$, where $x_{1}$ and $y_{1}$ are also integers. The positive number $\sqrt... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,790 |
8. Anya and Borya are playing "battleship" with the following rules: 29 different points are chosen on a circle, numbered clockwise with natural numbers from 1 to 29. Anya draws a ship - an arbitrary triangle with vertices at these points. We will call a "shot" the selection of two different natural numbers $k$ and $m$... | Solution. The vertices of Anya's triangle divide the circle into three arcs. Let $x, y$ and $26-x-y$ be the number of points on these arcs (see figure), excluding the vertices of the triangle itself. For a "shot" with endpoints at points $k$ and $m$ to not hit the ship, both these points must lie on one of the arcs. Cl... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,791 |
3. A square table consists of 2014 rows and 2014 columns. In each cell at the intersection of the row with number $i$ and the column with number $j$, the number $a_{i, j}=(-1)^{i}(2015-i-j)^{2}$ is written. Find the sum of all the numbers in the table | Problem 3. Answer: $\mathbf{0 .}$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,792 |
4. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{l}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$.
... | Solution. We will prove that ${ }^{1}$
Multiply the equation (a) of the original system
$$
a_{2} b_{3}=a_{3} b_{2}
$$
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & (\mathrm{a}) \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1 & \text { (d) }\end{cas... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,793 |
6. At the vertices of a square with a side length of 4, there are four cities. These cities need to be connected by roads such that it is possible to travel from any city to any other. Propose at least one variant of such roads with a total length of less than 11.
Hint. The following statement may be useful in solving... | Solution. Suppose our road system has no intersections, that is, there is one road connecting the vertices of the square $E, F, G$ and $H$ sequentially. Then its length will be no less than three sides of the square (the letter "П" in Fig. (a)), that is, no less than 12 (a road connecting adjacent vertices of the squar... | 4\cdot(\sqrt{3}+1)<11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,794 |
8. In an acute-angled triangle $ABC$, a point $Q$ is chosen on side $AC$ such that $AQ: QC = 1: 2$.
From point $Q$, perpendiculars $QM$ and $QK$ are dropped to sides $AB$ and $BC$ respectively. It is given that $BM: MA = 4: 1, BK = KC$. Find $MK: AC$. | Solution. Draw the heights $A K_{1}$ and $C M_{1}$. The idea of the solution is as follows: we will show that triangles $M_{1} B K_{1}$, $M B K$, and $A B C$ are similar to each other; from this, it will be easy to find the required ratio.
Let's denote the lengths:
^{2018}$ and $\left(10^{2018}+10^{2017}+\cdots+10+1\right)^{2017}$. | Solution. Let
$$
A=\left(10^{2017}+10^{2016}+\cdots+10+1\right)^{2018}, \quad B=\left(10^{2018}+10^{2017}+\cdots+10+1\right)^{2017}
$$
Using the formula for the sum of terms of a geometric progression, we find:
$$
A=\frac{\left(10^{2018}-1\right)^{2018}}{9^{2018}}, \quad B=\frac{\left(10^{2019}-1\right)^{2017}}{9^{2... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,796 |
2. Find all even natural numbers $n$ for which the number of divisors (including 1 and $n$ itself) is equal to $\frac{n}{2}$. (For example, the number 12 has 6 divisors: $\left.1,2,3,4,6,12.\right)$ | Solution. Let the canonical decomposition of the number $n$ be: $n=2^{t_{1}} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdots p^{t_{k}}$. Then the number of divisors of the number $n$ is $\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)$. From the condition of the problem, we have th... | {8,12} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,797 |
5. For natural numbers $a, b, c$, the following is known:
- $a^{b}$ is divisible by $c$;
- $\quad b^{c}$ is divisible by $a$;
- $\quad c^{a}$ is divisible by $b$.
Prove that $(a+b+c)^{a+b+c}$ is divisible by the product $a b c$.
# | # Solution.
Let $p$ be a prime divisor of the number $a$. Then, from the 2nd condition, it follows that $b$ is divisible by $p$, and from the 3rd condition, that $c$ is divisible by $p$. Therefore, $(a+b+c)^{a+b+c}$ is divisible by $p^{a+b+c}$. Let $n, k, t$ be the largest natural numbers such that $a$ is divisible by... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,799 |
6. $\quad$ In trapezoid $ABCD$ with bases $BC$ and $AD$, angle $DAB$ is a right angle. It is known that there exists a unique point $M$ on side $CD$ such that angle $BMA$ is a right angle.
Prove that $BC = CM$ and $AD = MD$. | Solution. Construct a circle with side $AB$ as its diameter. Since angle $BMA$ is a right angle, point $M$ lies on this circle. Since such a point $M$ is unique on side $CD$, $CD$ is tangent to the circle. Therefore, $BC = CM$ and $AD = MD$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,800 |
7. It is known that there exists a natural number $N$ such that
$$
(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}
$$
Find $N$.
 | Solution. Suppose that raising the number $a+b \sqrt{3}$ to the power $N$ results in the number $A+B \sqrt{3}$ (here $a, b, A, B$ are integers). Expanding the expression $(a+b \sqrt{3})^{N}$, we get a sum of monomials (with integer coefficients that are not important to us now) of the form $a^{N-n}(b \sqrt{3})^{n}$. Th... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,801 |
8. A circle inscribed in a trapezoid intersects its diagonals at points $A, B, C, D$. Prove that the sum of the lengths of arcs $\widetilde{B A}+\overline{D C}$ is greater than the sum of the lengths of arcs $\overline{A D}+\overline{C B}$. | Solution. Let $O$ be the point of intersection of the diagonals. It is known that the measure of angle $A O D$ is half the sum of the angular measures of arcs $\overline{C B}$ and $\overline{A D}$. The problem essentially requires proving that the sum of the lengths of arcs
 For the polynomial $\left(x^{2}-x+1\right)^{100}$, find the sum of the coefficients of the even powers of $x$. | Solution: Let $P(x)=\left(x^{2}-x+1\right)^{100}$. The desired sum is $\frac{P(1)+P(-1)}{2}=\frac{1+3^{100}}{2}$
Answer: $\frac{1+3^{100}}{2}$ | \frac{1+3^{100}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,803 |
2. (2 points) It is known about the numbers $x_{1}$ and $x_{2}$ that $x_{1}+x_{2}=2 \sqrt{1703}$ and $\left|x_{1}-x_{2}\right|=90$. Find $x_{1} \cdot x_{2}$. | Solution: Let $A=x_{1}+x_{2}, B=x_{1} \cdot x_{2}, C=\left|x_{1}-x_{2}\right|$. Since $C^{2}=A^{2}-4 \cdot B$, we find $B=-322$.
Answer: -322 . | -322 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,804 |
3. (3 points) Find the number of integers from 1 to 1000 inclusive that give the same remainder when divided by 11 and by 12. | Solution. Let $r_{n}(a)$ be the remainder of the division of the number $a$ by the number $n$. Suppose $a \in [1 ; 1000]$ and $r_{11}(a)=r_{12}(a)=t$. Then $t \in \{0, \ldots, 10\}$ and the following equality holds:
$$
t+11 k=t+12 m=a, \quad k, m \in N_{0}
$$
From the last equality, it follows that $k$ is divisible b... | 87 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,805 |
5. (5 points) Solve the inequality $\log _{\cos x}\left(x^{2}-6 x-2\right)>\frac{2}{\log _{5} \cos x}$. | # Solution:
o.D.Z
\[
\left\{\begin{array}{c}
x^{2}-6 x-2>0 \Leftrightarrow x \in(-\infty, 3-\sqrt{11}) \cup(3+\sqrt{11},+\infty) \\
0 < \frac{2}{\log _{5} \cos x} \Leftrightarrow \log _{\cos x}\left(x^{2}-6 x-2\right)>\log _{\cos x} 25 \Leftrightarrow(c \text{ considering O.D.Z.})
\end{array}\right.
\]
\cup(3+\sqrt{11},\frac{5\pi}{2}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,806 |
6. (5 points). Point $\boldsymbol{D}$ lies on the extension of side $\boldsymbol{A} \boldsymbol{C}$ of triangle $\boldsymbol{A B C}$, the area of which is equal to $\boldsymbol{S}$; point $\boldsymbol{A}$ is located between $\boldsymbol{D}$ and $\boldsymbol{C}$. Let $\boldsymbol{O}$ be the point of intersection of the ... | Solution: We will show that
$$
S_{\triangle D O C} + S_{\triangle D O A} = S_{\triangle D O B}
$$
To do this (since these triangles share the side $D O$), it is sufficient to show that the l... | 2S_{1}-\frac{S}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,807 |
1. Find the sum of all natural numbers $n$ that are multiples of three and for which the number of divisors (including 1 and $n$ itself) is equal to $\frac{n}{3}$. (For example, the number 12 has 6 divisors: $1,2,3,4,6,12$.) | Solution. Let the canonical decomposition of the number $n$ be: $n=2^{t_{1}} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdot p^{t_{k}}$. Then the number of divisors of the number $n$ is $\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)$. From the condition of the problem, we have the... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,808 |
1. (2 points) Prove that in the expression $2012^{2} * 2011^{2} * 2010^{2} * \ldots * 2^{2} * 1^{2}$, the sign «*» can be replaced by the signs «+» and «-» so that the resulting expression equals 2012.
# | # Solution:
For any natural $n$
$$
n^{2}-(n+1)^{2}-(n+2)^{2}+(n+3)^{2}=-1 \cdot(2 n+1)+1 \cdot(2 n+5)=4 \text{. }
$$
From this and the fact that 2012 is divisible by 4, it is clear that the given problem can be solved by periodically repeating the combination of signs: “+”, “-”, “-”, “+”. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,810 |
2. (3 points) In a puddle, there live three types of amoebas: red, blue, and yellow. From time to time, any two amoebas of different types can merge into one amoeba of the third type. It is known that in the morning, there were 26 red, 31 blue, and 16 yellow amoebas in the puddle, and by evening, only one amoeba remain... | # Solution:
It is not hard to notice that as a result of the changes that the original population of amoebas can undergo, the parities of the absolute differences of the numbers of different types of amoebas do not change.
Let $n_{1}, n_{2}$ and $n_{3}$ be the numbers of red, blue, and yellow amoebas in the initial p... | blue | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,811 |
3. (4 points) Find all integer solutions of the equation:
$$
y^{2}=x^{2}+x+1
$$
# | # Solution:
For $|x| \leq 3$, solutions are easily found by enumeration: $\{(0 ; \pm 1) ;(-1 ; \pm 1)\}$. Now let $|x| \geq 4$. Consider two cases.
Case 1: $x=2 t-$ is even. Then
$$
x^{2}+x+1=4 t^{2}+2 t+1=(2 t+1)^{2}-2 t
$$
From the last equality, it follows that $x^{2}+x+1$ differs from the perfect square $(2 t+1... | {(0;\1);(-1;\1)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,812 |
4. (4 points) Solve the equation:
$$
\left[\frac{5+6 x}{8}\right]=\frac{15 x-7}{5}
$$
where the symbol $[a]$ denotes the integer part of the number $a$.
# | # Solution:
By the definition of the integer part of a number, we have the system of inequalities
$$
\left\{\begin{array}{c}
\frac{5+6 x}{8} \geq \frac{15 x-7}{5} \\
\frac{5+6 x}{8} > \frac{41}{90}
\end{array}\right. \\
& x \in\left(\frac{41}{90} ; \frac{9}{10}\right]
$$
Now we need to find those values of $\mathrm{... | \frac{7}{15};\frac{4}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,813 |
5. (4 points) Find the area of a triangle if two of its medians are equal to $\frac{15}{7}$ and $\sqrt{21}$, and the cosine of the angle between them is $\frac{2}{5}$.
# | # Solution:
Let $|A D|=a$ and $|B E|=b$ be the medians of the triangle, and $\cos (\angle A O F)=c$.
It is known that the area
$$
\begin{aligned}
& S_{A O E}=\frac{1}{6} \cdot S_{A B C} \\... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,814 |
6. (5 points) Numbers are marked on a chessboard (see Fig. 1). How many arrangements of 8 rooks are there such that no rook attacks another and the numbers on the squares occupied by the rooks include all numbers from 0 to 7?
\(\left(\begin{array}{llllllll}0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7... | Answer: 3456.
## Variant 2 | 3456 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,815 |
2. (3 points) In a puddle, there live three types of amoebas: red, blue, and yellow. From time to time, any two amoebas of different types can merge into one amoeba of the third type. It is known that in the morning, there were 47 red, 40 blue, and 53 yellow amoebas in the puddle, and by evening, only one amoeba remain... | # Solution:
It is not hard to notice that as a result of the changes that the original population of amoebas can undergo, the parities of the absolute differences of the numbers of different types of amoebas do not change.
Let $n_{1}, n_{2}$, and $n_{3}$ be the numbers of red, blue, and yellow amoebas in the initial ... | blue | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,816 |
3. (4 points) Find all integer solutions of the equation:
$$
y^{2}=x^{2}-x+2
$$
# | # Solution:
For $|x| \leq 3$, solutions are easily found by enumeration: $\{(2 ; \pm 2) ;(-1 ; \pm 2)\}$.
Now let $|x| \geq 4$. Consider two cases.
Case 1: $x=2 t-$ is even. Then
$$
x^{2}-x+2=4 t^{2}-2 t+2=(2 t-1)^{2}+2 t+1
$$
From the last equality, it follows that $x^{2}-x+2$ differs from the perfect square $(2 ... | {(2;\2);(-1;\2)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,817 |
4. (4 points) Solve the equation:
$$
\left[\frac{9 x-4}{6}\right]=\frac{12 x+7}{4}
$$
where the symbol $[a]$ denotes the integer part of the number $a$.
# | # Solution:
By the definition of the integer part of a number, we have the system of inequalities
$$
\left\{\begin{array}{l}
\frac{9 x-4}{6} \geq \frac{12 x+7}{4} \\
\frac{9 x-4}{6}-\frac{41}{18}
\end{array}\right. \\
& x \in\left(-\frac{41}{18} ;-\frac{29}{18}\right]
$$
Now we need to find those values of \( x \) i... | -\frac{9}{4};-\frac{23}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,818 |
5. (4 points) Find the area of a triangle if two of its medians are equal to 3 and $2 \sqrt{7}$, and the cosine of the angle between them is $-\frac{3}{4}$.
# | # Solution:
Let $|A D|=a$ and $|B E|=b$ be the medians of the triangle, and $\cos (\angle A O F)=c$.
It is known that the area
$$
\begin{aligned}
& S_{A O E}=\frac{1}{6} \cdot S_{A B C} \c... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,819 |
2. Solve the equation $\sqrt{\frac{2 t}{1+t^{2}}}+\sqrt[3]{\frac{1-t^{2}}{1+t^{2}}}=1$. | Solution. Let's make the substitution
$$
t=\operatorname{tg} \frac{\alpha}{2}, \alpha \in(-\pi, \pi)
$$
Then
$$
\frac{2 t}{1+t^{2}}=\sin \alpha, \frac{1-t^{2}}{1+t^{2}}=\cos \alpha
$$
and the original equation will take the form
$$
\sqrt{\sin \alpha}+\sqrt[3]{\cos \alpha}=1
$$
If $\cos \alpha<0$, then the left si... | \in{0,1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,821 |
3. Consider all possible 100-digit natural numbers, in the decimal representation of which only the digits 1 and 2 appear. How many of them are divisible by 3? | Solution: Each 100-digit natural number can be obtained by appending two digits to the right of a 98-digit number. Let $x$ be some 98-digit number. We will examine which two digits (each equal to 1 or 2) need to be appended to the number $x$ so that the resulting 100-digit number is divisible by 3. We will use the fact... | \frac{4^{50}+2}{3} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,822 |
5. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{l}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$. | Solution. We will prove that ${ }^{1}$
$$
a_{2} b_{3}=a_{3} b_{2}
$$
Multiply equation (a) of the original system
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & (\mathrm{a}) \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1 & \text { (d) }\end{cases}
... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,823 |
1. [4] In the room, there are several children and a pile of 1000 candies. The children take turns approaching the pile. Each child who approaches divides the number of candies in the pile by the number of children in the room, rounds (if the result is not an integer), takes the resulting number of candies, and leaves ... | Solution. Dividing a pile of candies with a remainder among $k$ children can be imagined as follows: we distribute the candies into $k$ piles, which are either all the same (if the remainder is 0), or in some of the piles there is one more candy than in the others (the number of such piles is equal to the remainder).
... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 8,824 |
2. [5] Does there exist a natural number $n$ such that for any real numbers $x$ and $y$, there are real numbers $a_{1}, \ldots, a_{n}$ satisfying the equations
$$
x=a_{1}+\ldots+a_{n} \quad \text { and } \quad y=\frac{1}{a_{1}}+\ldots+\frac{1}{a_{n}} ?
$$
Artemy Sokolov | Answer: it exists.
Solution 1. We will prove that $n=6$ works. First, note that any pair $(0, y)$ with a non-zero $y$ can be obtained as follows: $0=\frac{3}{2 y}+\frac{3}{2 y}-\frac{3}{y}, y=\frac{2 y}{3}+\frac{2 y}{3}-\frac{y}{3}$. Similarly, any pair $(x, 0)$ with a non-zero $x$ can be obtained. Then, any pair $(x,... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,825 |
3. [5] Point $M$ is the midpoint of side $BC$ of triangle $ABC$. Circle $\omega$ passes through point $A$, is tangent to line $BC$ at point $M$, and intersects side $AB$ at point $D$ and side $AC$ at point $E$. Let $X$ and $Y$ be the midpoints of segments $BE$ and $CD$ respectively. Prove that the circumcircle of trian... | Solution. Note that $M X$ and $M Y$ are the midlines of triangles $C B E$ and $B C D$ respectively. By the given condition
$$
B D \cdot B A=B M^{2}=C M^{2}=C E \cdot C A
$$
from which we have
$$
M X: M Y=C E: B D=B A: C A
$$
Since $M X \| A C$ and $M Y \| A B$, triangles $M X Y$ and $A B C$ are similar. Therefore, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,826 |
4. [8] There are $100 \mathrm{~N}$ sausages in a row. Uncle Fyodor and the cat Matroskin are playing a game. Uncle Fyodor, in one move, eats one of the end sausages. Matroskin, in one move, can remove the sausage from one of the sausages (or do nothing). Uncle Fyodor makes 100 moves in a row, while Matroskin makes only... | Solution. We will prove that when $N=3^{100}$, the cat Matroskin will win. For this, it is sufficient that at the last step of Uncle Fyodor, all the remaining 100 sandwiches are without sausage.
Let's number the sandwiches in order. We will divide the cat Matroskin's strategy into several stages. First, we will show t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,827 |
5. [8] At night, 100 tourists arrived at a hotel. They know that the hotel has single rooms 1, 2, ..., n, of which k are under repair (but it is unknown which ones), and the rest are free. The tourists can agree in advance on their actions, after which they go to check in one by one: each checks the rooms in any order,... | Answer: $n=100(m+1)$ for $k=2 m$ and $n=100(m+1)+1$ for $k=2 m+1$.
Solution. Let $k=2 m$ or $k=2 m+1$.
Algorithm. Mentally divide the room numbers into 100 segments, each containing $m+1$ numbers, and in the case of an odd $k$, declare the remaining number as a spare. Let the $i$-th tourist first check all the rooms ... | n=100(+1)fork=2mn=100(+1)+1fork=2m+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,828 |
6. [10] Find at least one real number $A$ with the property: for any natural number $n$, the distance from the ceiling of the number $A^{n}$ to the nearest square of an integer is 2. (The ceiling of a number $x$ is the smallest integer not less than $x$.)
Dmitry Krykov | Solution. Consider any quadratic equation with integer coefficients and the leading coefficient 1, which has two positive roots, the product of which is 1. For example, the equation $x^{2}-4 x+1$ has roots $2+\sqrt{3}$ and $2-\sqrt{3}$. Note that the sum and product of these roots are integers, and then the sum $(2+\sq... | (2+\sqrt{3})^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,829 |
7. Given an integer $n>2$. On a sphere of radius 1, it is required to place $n$ non-intersecting arcs of great circles, all of equal length $\alpha$. Prove that a) [6] for any $\alpha\pi+\frac{2 \pi}{n}$ this is impossible.
Ilya Bogdanov | Solution. a) Let the vertical line $\ell$ pass through the center of the sphere $O$. Let two parallel horizontal planes cut out two equal (not large!) circles $\gamma_{+}$ and $\gamma_{-}$ on the sphere. Then there exists a great circle $\Omega_{0}$ that is tangent to $\gamma_{+}$ and $\gamma_{-}$ at (diametrically opp... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,830 |
1. [4] A hundred friends, including Petya and Vasya, live in several cities. Petya learned the distance from his city to the city of each of the remaining 99 friends and added these 99 numbers. Vasya did the same. Petya got 1000 km. What is the largest number Vasya could have obtained? (Consider the cities as points on... | Answer: 99000 km.
Solution. Estimate. Let Vasya live in city $V$, and Petya - in city $P$. Consider an arbitrary friend of Vasya (this could be Petya), let's say he lives in city $X$. By the triangle inequality, $V X \leq P V + P X$, and this sum is no more than Petya's total, i.e., 1000 km. Therefore, the sum of the ... | 99000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,831 |
3. [6] Baron Munchausen claims that he drew a polygon and a point inside it such that any line passing through this point divides this polygon into three polygons. Can the baron be right? Tatyana Kazitsyna | Answer: it can. Solution. See examples on the right. In the second figure, it is essential that the triple of points $A, O, A^{\prime}$, the triple $B$, $O, B^{\prime}$, and the triple $C, O, C^{\prime}$ lie on one straight line.
(the inscribed angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,834 |
6. [10] Petya added a natural number $M$ to a natural number $N$ and noticed that the sum of the digits of the result is the same as that of $N$. Then he added $M$ to the result again, then once more, and so on. Will he necessarily get a number with the same sum of digits as $N$ again at some point?
Alexander Shapoval... | Answer: not necessarily. Solution. Example. Let $N=2, M=1008$. The number $M$ is divisible by 16, so all numbers obtained by Petya give a remainder of $2 \bmod 16$. A number with a digit sum of 2 can be represented as the sum of two powers of ten. The powers $1, 10, 10^{2}$, and $10^{3}$ give remainders of 1, 10, 4, an... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,835 |
1. [4] The number $2021=43 \cdot 47$ is composite. Prove that if you insert any number of eights between 20 and 21 in the number 2021, the resulting number will also be composite.
Mikhail Evdokimov | Solution. The difference between two such numbers, where the number of eights differs by 1, has the form $1880 \ldots 0$. But $188=47 \cdot 4$, so it is divisible by 47, as is 2021. Therefore, by adding eights one by one, we will get numbers that are divisible by 47. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,837 |
2. [5] In the room, there are several children and a pile of 1000 candies. The children take turns approaching the pile. Each child who approaches divides the number of candies in the pile by the number of children in the room, rounds (if the result is not an integer), takes the resulting number of candies, and leaves ... | Solution. Dividing a pile of candies with a remainder among $k$ children can be imagined as follows: we distribute the candies into $k$ piles, which are either all the same (if the remainder is 0), or in some of the piles there is one more candy than in the others (the number of such piles is equal to the remainder).
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,838 |
3. [6] Triangle $A B C$ is equilateral. Points $E$ and $F$ are chosen on sides $A B$ and $A C$, and point $K$ is chosen on the extension of side $A B$ such that $A E=C F=B K$. Point $P$ is the midpoint of $E F$. Prove that angle $K P C$ is a right angle. | Solution 1. On the extension of segment $C P$ beyond point $P$, mark a point $T$ such that $C P = P T$. Then $F C E T$ is a parallelogram, from which $T E$ is equal and parallel to $F C$. But then triangles $T E K$ and $K B C$ are congruent by the first criterion: the obtuse angles are equal to $120^{\circ}$ and the co... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,839 |
4. [7] A traveler arrived on an island where 50 natives live, each of whom is either a knight or a liar. All the natives stood in a circle, and each named the age of their left neighbor first, and then the age of their right neighbor. It is known that each knight named both numbers correctly, while each liar increased ... | Answer: always.
Solution 1. Choose any native - let's call him Petya, - and show how to find his age. Mentally put a hat on every second native, starting with Petya. Number the natives without hats, going clockwise after Petya: $1,2, \ldots, 24,25$.
Note that each native correctly reports the sum of the ages of his n... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,840 |
5. In the center of each cell of a grid rectangle $M$, there is a point light bulb, all of which are initially turned off. In one move, it is allowed to draw any straight line that does not touch the light bulbs and turn on all the light bulbs on one side of this line, if all of them are off. Each move must turn on at ... | Solution. Instead of the original rectangle $M$, we will consider rectangle $N$ with vertices at the corner lamps.
Estimates. Note that with each move, at least one of the four corner lamps is lit. Therefore, there are no more than 4 moves. In part a), note also that we must light the central lamp at some move. Togeth... | )3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,841 |
6. [10] At night, 100 tourists arrived at a hotel. They know that the hotel has single rooms 1, 2, ..., n, of which k are under repair (but it is unknown which ones), and the rest are free. The tourists can agree in advance on their actions, after which they go to check in one by one: each checks the rooms in any order... | Answer: $n=100(m+1)$ for $k=2 m$ and $n=100(m+1)+1$ for $k=2 m+1$.
Solution. Let $k=2 m$ or $k=2 m+1$.
Algorithm. Mentally divide the room numbers into 100 segments, each containing $m+1$ numbers, and in the case of an odd $k$, declare the remaining number as a spare. Let the $i$-th tourist first check all the rooms ... | n=100(+1)fork=2mn=100(+1)+1fork=2m+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,842 |
1. [5] In triangle $ABC$, point $M$ is the midpoint of side $BC$, and point $E$ is an arbitrary point inside side $AC$. It is known that $BE \geq 2AM$. Prove that triangle $ABC$ is obtuse. (N. Sedrakyan) | Solution 1. Let $X$ be the midpoint of segment $EC$. Then $MX = \frac{BE}{2}$. As is known, a cevian of a triangle is less than at least one of the sides emanating from the same vertex (this follows, for example, from the properties of oblique and projections). By the condition, $MX \geq MA$, so $MX < BE$. Contradictio... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,844 |
2. [6] On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 residents were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The residents answered in turn, and everyone could hear their answers. Knights answered tr... | Answer: 1009 yes-men. Solution: Let's call knights and liars principled people.
Estimate. First method. (Buchaev Abdulqadyr) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values ... | 1009 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,845 |
3. [8] It is required to write a number of the form 77...7 using only sevens (they can be written individually or in a row), and only addition, subtraction, multiplication, division, and exponentiation, as well as parentheses are allowed. For the number 77, the shortest notation is simply 77. Is there a number of the f... | Answer. There is. Solution 1. Note that $\underbrace{7 \ldots 7}_{n}=\frac{10^{n}-1}{9} \cdot 7=\frac{7 \cdot 10^{n}-7}{9}$. The number 10 can be written as ( $77-7): 7$, and 9 as $7+(7+7): 7$. For $n$, we can take 77 or $14=7+7$.
Remark. This solution uses 12 sevens. By replacing ( $77-7): 7$ with $7+(7+7+7): 7$, we ... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 8,846 |
4. [8] A $7 \times 7$ board is either empty or has an invisible $2 \times 2$ ship placed on it "by cells." It is allowed to place detectors in some cells of the board and then turn them on simultaneously. An activated detector signals if its cell is occupied by the ship. What is the smallest number of detectors needed ... | Answer: 16 detectors. Solution: Evaluation. In each $2 \times 3$ rectangle, there should be at least two detectors: the rectangle consists of three $1 \times 2$ dominoes, and if a detector is in the outer domino, we cannot determine whether there is a ship on the other two dominoes, and if a detector is in the middle d... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,847 |
7. In a virtual computer state, there are no fewer than two cities. Some pairs of cities are connected by roads, and from each city, it is possible to travel by roads to any other city (here and below, switching from one road to another is only allowed in cities). If, at the same time, it is possible to start moving fr... | Solution. Consider a graph where the vertices are cities and the edges are roads.
a) The condition means that the graph is a tree. Petya chooses an arbitrary vertex. From each vertex, there is exactly one path to the chosen one. He orients all edges on this path towards the chosen vertex.
On the first move, Petya mov... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,848 |
7. A $2 \times 2$ square grid is covered by two triangles. Is it necessarily true that
a) at least one of the four cells is completely covered by one of these triangles;
b) one of these triangles can contain a $1 \times 1$ square? | 7. A $2 \times 2$ grid square is covered by two triangles. Is it necessarily true that a) (6 points) at least one of the four cells is completely covered by one of these triangles; b) (6 points) a $1 \times 1$ square can fit into one of these triangles?
(Alexander Shapovalov)
a) Answer: not necessarily. Inscribing a ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,849 |
2. A group of eight tennis players once a year played for a cup according to the Olympic system (players are divided into pairs by lot; the winners are divided by lot into two pairs playing in the semifinals; their winners play the final match). After several years, it turned out that everyone had played exactly once w... | Answer: when $n$ is not a multiple of 4.
Strategy: each time leave a multiple of 4 stones in the pile: when $n=4 k+1$ take one stone, when $n=4 k+2$ take two stones; when $n=4 k+3$ take $p$ stones, where $p$ is a prime divisor of the number $n$ of the form $4 q+3$ (such a divisor exists, otherwise all prime divisors o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,850 |
4. [7] A traveler arrived on an island where 50 natives live, each of whom is either a knight or a liar. All the natives stood in a circle, and each one first named the age of their left neighbor, and then the age of their right neighbor. It is known that each knight named both numbers correctly, while each liar increa... | Answer: always.
Solution 1. Choose any native - let's call him Petya, - and show how to find his age. Mentally put a hat on every second native, starting with Petya. Number the natives without hats, going clockwise after Petya: $1,2, \ldots, 24,25$.
Note that each native correctly reports the sum of the ages of his n... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,851 |
1. [5] The polynomial $P(x, y)$ is such that for every integer $n \geq 0$, each of the polynomials $P(n, y)$ and $P(x, n)$ is either identically zero or has a degree not higher than $n$. Can the polynomial $P(x, x)$ have an odd degree?
(Boris Frenkin) | Answer. It cannot. Solution. Let the highest power in which $x$ appears be $m$, and the highest power in which $y$ appears be $n$. For definiteness, assume $n \geq m$. Write the polynomial $P(x, y)$ in the form $A(x) y^{n}+B(x) y^{n-1}+\ldots$, where $A(x), B(x), \ldots$ are polynomials in $x$. Since for all integers $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,852 |
2. [5] Segments $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$ with endpoints on the sides of an acute triangle $A B C$ intersect at a point $P$ inside the triangle. On each of these segments, a circle is constructed with the segment as its diameter, and a chord perpendicular to this diameter is drawn through the poi... | Solution. Let $2 x$ be the length of the specified chords. According to the theorem of the product of segments of chords, $x^{2}=A P \cdot A^{\prime} P=B P \cdot B^{\prime} P=C P \cdot C^{\prime} P$. By the converse theorem, points $A, A^{\prime}, B$ and $B^{\prime}$ lie on the same circle. Therefore, $\angle A A^{\pri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,853 |
3. [6] There are 100 indistinguishable coins of three types: gold, silver, and copper (each type appears at least once). It is known that gold coins weigh 3 g each, silver coins weigh 2 g each, and copper coins weigh 1 g each. How can you determine the type of all the coins using a balance scale without weights in no m... | Solution 1. (A. Shapovalov) Let's call a situation a winning one if no more than $k+1$ weighings have been conducted and the weights of $k$ coins have been determined, with at least one being silver or two being copper. In a winning situation, by comparing an unknown coin with a weight of 2 g, we can determine its weig... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,854 |
4. [10] Given an increasing sequence of positive numbers
$$
\ldots<a_{-2}<a_{-1}<a_{0}<a_{1}<a_{2}<\ldots
$$
infinite in both directions. Let $b_{k}$ be the smallest integer with the property: the ratio of the sum of any $k$ consecutive terms of the given sequence to the largest of these $k$ terms does not exceed $b_... | Solution. Obviously, $b_{1}=1$, and for $k>1$ the ratio from the condition is less than $k$, so $b_{k} \leq k$ for all natural $k$. If the sequence $b_{1}, b_{2}, b_{3}, \ldots$ does not coincide with the natural series, then $b_{k} \leq k-1$ for some $k$. Then $a_{i}+a_{i+1}+\ldots+a_{i+k-1} \leq(k-1) a_{i+k-1}$ for e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,855 |
5. Point $M$ lies inside the convex quadrilateral $ABCD$ at the same distance from the lines $AB$ and $CD$ and at the same distance from the lines $BC$ and $AD$. It turns out that the area of quadrilateral $ABCD$ is equal to $MA \cdot MC + MB \cdot MD$. Prove that quadrilateral $ABCD$
a) [6] is cyclic; b) [6] is tange... | Solution. a) Drop perpendiculars $M P, M Q, M R, M T$ to the lines $A B, B C, C D, D A$ respectively. Then $S_{A B C D}=S_{A M B}+S_{\text {BMC }}+S_{C M D}+S_{D M A} \leq\left(S_{\text {AMP }}+S_{\text {BMP }}\right)+\left(S_{\text {BMQ }}+S_{\text {CMQ }}\right)+$ $+\left(S_{C M R}+S_{D M R}\right)+\left(S_{D M S}+S_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,856 |
6. A cube consisting of $(2 n)^{3}$ unit cubes is pierced by several spokes parallel to the edges of the cube. Each spoke pierces exactly $2 n$ cubes, and each cube is pierced by at least one spoke.
a) [6] Prove that it is possible to choose such $2 n^{2}$ spokes, going in one or two directions in total, so that no tw... | Solution. (A. Shapovalov) Let the edges of the cube be parallel to the coordinate axes.
a) Divide the cube into layers of thickness 1, parallel to the plane $Oxy$. Consider only the directions $Ox$ and $Oy$. In each layer, find the maximum number of such spokes going in one direction. Similarly, find the maximum numbe... | 2n^2 | Combinatorics | proof | Yes | Yes | olympiads | false | 8,857 |
1. [3] The magician lays out a deck of 52 cards in a row and announces that 51 of them will be thrown off the table, leaving only the three of clubs. On each step, the spectator says which card to discard by its position from the edge, and the magician chooses whether to count from the left or the right edge and discar... | Answer: at the extreme positions. Solution. The three of clubs will have to be discarded only if it ends up in the center of the row at some point, otherwise, another card can be discarded. Since the row always contains more than one card, an end card can be kept until the end.
Let the three of clubs $T$ initially not... | attheextremepositions | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,858 |
2. [4] Given a convex pentagon $A B C D E$, in which $A E \| C D$ and $A B = B C$. The bisectors of its angles $A$ and $C$ intersect at point $K$. Prove that $B K \| A E$.
(Egor Bakaev) | Solution. Let the bisector of angle $C$ intersect line $A E$ at point $F$, and the line passing through $B$ parallel to $A E$ intersect segment $C F$ at point $X$. Then $\angle B X C = \angle D C X = \angle B C X$. Therefore, $B X = B C = B A$. Hence, $\angle B A X = \angle B X A = \angle F A X$. Consequently, $A X$ is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,859 |
4. [5] Given a polygon where each two adjacent sides are perpendicular. We will call two of its vertices not friendly if the bisectors of the polygon, emanating from these vertices, are perpendicular. Prove that for any vertex, the number of vertices not friendly with it is even.
(Mikhail Skopenkov) | Solution 1. Arrange the polygon so that its sides are horizontal and vertical. Let the number of vertical sides be $k$, then the number of horizontal sides is also $k$. All vertices of the polygon can be divided into 4 types: $\Gamma, \urcorner, \mathrm{L}$, $\lrcorner$. Suppose vertex $A$ has type 2 (without loss of g... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,861 |
5. [5] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly 4 chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order?
(Egor Bakaev) | Answer: For 61 rubles. Solution: Let's number the chips and cells in order from 0 to 99. The free operation does not change the remainder of the cell number when divided by 5.
Estimate. Let's mentally arrange the piles of chips in a circle. First, the pile of chips with a remainder of 0, then with 1, and so on up to 4... | 61 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,862 |
5. [8] Do there exist 100 such natural numbers, all distinct, such that the cube of one of them is equal to the sum of the cubes of the others?
(Mikhail Evdokimov) | Answer: they exist.
First solution. Note that $3^{3}+4^{3}+5^{3}=6^{3}$ (check it!). Multiplying this equality by $2^{3}$, we get: $6^{3}+8^{3}+10^{3}=12^{3}$. Replacing $6^{3}$ with the sum from the previous equality, we obtain five cubes that sum up to a cube: $3^{3}+4^{3}+5^{3}+8^{3}+10^{3}=12^{3}$. Multiplying the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,864 |
6. [10] Behind each of two round tables sit $n$ gnomes. Each gnome is friends only with their neighbors to the left and right at their table. A kind wizard wants to seat the gnomes at one round table so that every two neighboring gnomes are friends. He has the ability to make $2 n$ pairs of gnomes friends (the gnomes i... | Answer: for all odd $n>1$.
Let $A_{1}, A_{2}, \ldots, A_{n}$ be the gnomes sitting at the first table, and $B_{1}, B_{2}, \ldots, B_{n}$ be the gnomes sitting at the second table.
Case of odd $n$. Let $n=2 k-1$. The good wizard's strategy: to befriend pairs of gnomes $\left(A_{i}, B_{i}\right)$ and $\left(A_{i}, B_{i... | foralloddn>1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,865 |
7. A convex quadrilateral $ABCD$ has the property that no triangle can be formed from any three of its sides. Prove that
a) [6] one of the angles of this quadrilateral is not greater than $60^{\circ}$;
b) [6] one of the angles of this quadrilateral is not less than $120^{\circ}$.
(Maxim Diden) | The first solution. Let $AD$ be the largest side of the quadrilateral $ABCD$. By the condition, the sum of any two other sides is not greater than $AD$.
a) Suppose angles $A$ and $D$ are both greater than $60^{\circ}$. Construct an equilateral triangle $AKD$ on side $AD$ inside the quadrilateral. Its sides $AK$ and $D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,866 |
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