question stringlengths 37 4.33k | thinking stringlengths 2.04k 2.53k | answer stringlengths 1 261 | id int64 56 93.7k | content stringlengths 2.04k 2.77k | id_ stringlengths 3 6 |
|---|---|---|---|---|---|
How many types of convex pentahedra exist? | <think>If a polyhedron had a face that is a pentagon or has more than five sides, then the solid would have at least 6 faces. Therefore, the faces can only be triangles and quadrilaterals. Let $e$ be the number of edges, and let the number of triangular faces be $k$. Then the number of quadrilateral faces is $5-k$. Cou... | 2 | 27,080 | <think>If a polyhedron had a face that is a pentagon or has more than five sides, then the solid would have at least 6 faces. Therefore, the faces can only be triangles and quadrilaterals. Let $e$ be the number of edges, and let the number of triangular faces be $k$. Then the number of quadrilateral faces is $5-k$. Cou... | 1_100 |
A convex $n$-gon $P$, where $n > 3$, is dissected into equal triangles by diagonals non-intersecting inside it. Which values of $n$ are possible, if $P$ is circumscribed? | <think>1. **Claim**: The only possible value of \( n \) for which a convex \( n \)-gon \( P \) can be dissected into equal triangles by diagonals non-intersecting inside it, and \( P \) is circumscribed, is \( n = 4 \).
2. **Verification for \( n = 4 \)**:
- Consider a square (a convex quadrilateral) which is circu... | n = 4 | 37,396 | <think>1. **Claim**: The only possible value of \( n \) for which a convex \( n \)-gon \( P \) can be dissected into equal triangles by diagonals non-intersecting inside it, and \( P \) is circumscribed, is \( n = 4 \).
2. **Verification for \( n = 4 \)**:
- Consider a square (a convex quadrilateral) which is circu... | 1_101 |
In a triangle $ ABC$, let $ M_{a}$, $ M_{b}$, $ M_{c}$ be the midpoints of the sides $ BC$, $ CA$, $ AB$, respectively, and $ T_{a}$, $ T_{b}$, $ T_{c}$ be the midpoints of the arcs $ BC$, $ CA$, $ AB$ of the circumcircle of $ ABC$, not containing the vertices $ A$, $ B$, $ C$, respectively. For $ i \in \left\{a, b, c\... | <think>To solve this problem, we need to show that the lines \( p_a \), \( p_b \), and \( p_c \) form a triangle similar to \( \triangle ABC \) and find the ratio of similitude. Let's proceed step-by-step.
1. **Identify the Midpoints and Arcs:**
- \( M_a \), \( M_b \), and \( M_c \) are the midpoints of sides \( BC... | \frac{1}{2} | 8,144 | <think>To solve this problem, we need to show that the lines \( p_a \), \( p_b \), and \( p_c \) form a triangle similar to \( \triangle ABC \) and find the ratio of similitude. Let's proceed step-by-step.
1. **Identify the Midpoints and Arcs:**
- \( M_a \), \( M_b \), and \( M_c \) are the midpoints of sides \( BC... | 1_102 |
What are the conditions for two medians of a triangle to be perpendicular to each other? | <think>Let the medians from vertices $A$ and $B$ of triangle $ABC$ be perpendicular to each other. Thus, from their common point, the centroid $S$, the side $AB$ is seen at a right angle, meaning $S$ lies on the circumference of the Thales circle with diameter $AB$, so the distance from $S$ to the midpoint $C_1$ of sid... | 5c^2=^2+b^2 | 31,362 | <think>Let the medians from vertices $A$ and $B$ of triangle $ABC$ be perpendicular to each other. Thus, from their common point, the centroid $S$, the side $AB$ is seen at a right angle, meaning $S$ lies on the circumference of the Thales circle with diameter $AB$, so the distance from $S$ to the midpoint $C_1$ of sid... | 1_103 |
Rectangle $ABCD$ has side lengths $AB = 6\sqrt3$ and $BC = 8\sqrt3$. The probability that a randomly chosen point inside the rectangle is closer to the diagonal $\overline{AC}$ than to the outside of the rectangle is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | <think>1. **Simplify the Problem:**
Given the rectangle \(ABCD\) with side lengths \(AB = 6\sqrt{3}\) and \(BC = 8\sqrt{3}\), we need to find the probability that a randomly chosen point inside the rectangle is closer to the diagonal \(\overline{AC}\) than to the sides of the rectangle.
2. **Normalize the Rectangl... | 17 | 35,806 | <think>1. **Simplify the Problem:**
Given the rectangle \(ABCD\) with side lengths \(AB = 6\sqrt{3}\) and \(BC = 8\sqrt{3}\), we need to find the probability that a randomly chosen point inside the rectangle is closer to the diagonal \(\overline{AC}\) than to the sides of the rectangle.
2. **Normalize the Rectangl... | 1_104 |
What minimum number of colours is sufficient to colour all positive real numbers so that every two numbers whose ratio is 4 or 8 have different colours? | <think>1. **Understanding the Problem:**
We need to color all positive real numbers such that any two numbers whose ratio is 4 or 8 have different colors. This means if \( a \) and \( b \) are two numbers such that \( \frac{a}{b} = 4 \) or \( \frac{a}{b} = 8 \), then \( a \) and \( b \) must have different colors.
... | 3 | 81,283 | <think>1. **Understanding the Problem:**
We need to color all positive real numbers such that any two numbers whose ratio is 4 or 8 have different colors. This means if \( a \) and \( b \) are two numbers such that \( \frac{a}{b} = 4 \) or \( \frac{a}{b} = 8 \), then \( a \) and \( b \) must have different colors.
... | 1_105 |
In [trapezoid](https://artofproblemsolving.com/wiki/index.php/Trapezoid) $ABCD$, leg $\overline{BC}$ is [perpendicular](https://artofproblemsolving.com/wiki/index.php/Perpendicular) to bases $\overline{AB}$ and $\overline{CD}$, and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11... | <think>Solution 1
Let $x = BC$ be the height of the trapezoid, and let $y = CD$. Since $AC \perp BD$, it follows that $\triangle BAC \sim \triangle CBD$, so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$.
Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$. Then $AE = x$, and $ADE$ is ... | 110 | 92,803 | <think>Solution 1
Let $x = BC$ be the height of the trapezoid, and let $y = CD$. Since $AC \perp BD$, it follows that $\triangle BAC \sim \triangle CBD$, so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$.
Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$. Then $AE = x$, and $ADE$ is ... | 1_106 |
On two perpendicular axes $(O X$ and $O Y)$, we take two points, $A$ and $B$, such that $A O + B O$ is equal to a constant, $k$. From point $O$, we draw a line parallel to $A B$, which intersects the circumcircle of triangle $A O B$ at point $M$. What will be the geometric locus of point $M$? | <think>I. Solution. Let us project $k$ onto the legs of the right angle $O X$ and $O Y$, and denote the points thus obtained by $L$ and $N$ with the letters. The geometric place of point $M$ is the semicircle that can be circumscribed around the right-angled triangle $L O N$ with center $O_{1}$.
} d z
$$
where $L-$ is a rectangle bounded by the following lines: $x=2, x=-1, y=2, y=-1$. | <think>Solution. The special points of the integrand function, which in this example has the form
$$
f(z)=\frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}
$$
are the points $z=0$ and $z=\frac{\pi}{2}$. The point $z=0$ is a removable singularity of the integrand function, since at this point
$$
\lim _{z \rightarrow 0} f(... | 4i | 1,343 | <think>Solution. The special points of the integrand function, which in this example has the form
$$
f(z)=\frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}
$$
are the points $z=0$ and $z=\frac{\pi}{2}$. The point $z=0$ is a removable singularity of the integrand function, since at this point
$$
\lim _{z \rightarrow 0} f(... | 1_108 |
If $x$ and $y$ are positive integers such that $x y z=240, x y+z=46$ and $x+y z=64$, what is the value of $x+y+z$?
(a) 19
(b) 20
(c) 21
(d) 24
(e) 36 | <think>The correct option is (b).
Solution 1: From $x y z=240$, it follows that $x y=\frac{240}{z}$. Substituting in $x y+z=46$, we get $\frac{240}{z}+z=46$, that is, $z^{2}-46 z+240=0$. The roots of this equation are numbers whose sum is 46 and whose product is 240, and it is easy to verify that these roots are 6 and... | 20 | 22,327 | <think>The correct option is (b).
Solution 1: From $x y z=240$, it follows that $x y=\frac{240}{z}$. Substituting in $x y+z=46$, we get $\frac{240}{z}+z=46$, that is, $z^{2}-46 z+240=0$. The roots of this equation are numbers whose sum is 46 and whose product is 240, and it is easy to verify that these roots are 6 and... | 1_109 |
A positive integer $N$ is [i]interoceanic[/i] if its prime factorization
$$N=p_1^{x_1}p_2^{x_2}\cdots p_k^{x_k}$$
satisfies
$$x_1+x_2+\dots +x_k=p_1+p_2+\cdots +p_k.$$
Find all interoceanic numbers less than 2020. | <think>1. **Prove that \( k \leq 3 \):**
- If there are at least 4 prime numbers, we have \( p_1 + p_2 + p_3 + p_4 \geq 2 + 3 + 5 + 7 = 17 \).
- Since \( x_1 + x_2 + x_3 + x_4 = p_1 + p_2 + p_3 + p_4 \geq 17 \), it follows that \( N > 2^{x_1} \cdot 2^{x_2} \cdot 2^{x_3} \cdot 2^{x_4} = 2^{x_1 + x_2 + x_3 + x_4} =... | \{4, 27, 48, 72, 108, 162, 320, 800, 1792, 2000\} | 57,254 | <think>1. **Prove that \( k \leq 3 \):**
- If there are at least 4 prime numbers, we have \( p_1 + p_2 + p_3 + p_4 \geq 2 + 3 + 5 + 7 = 17 \).
- Since \( x_1 + x_2 + x_3 + x_4 = p_1 + p_2 + p_3 + p_4 \geq 17 \), it follows that \( N > 2^{x_1} \cdot 2^{x_2} \cdot 2^{x_3} \cdot 2^{x_4} = 2^{x_1 + x_2 + x_3 + x_4} =... | 1_110 |
6.30 There is a sports competition consisting of $M$ events, with athletes $A, B, C$ participating. In each event, the first, second, and third places receive $p_{1}, p_{2}, p_{3}$ points, respectively, where $p_{1}, p_{2}, p_{3}$ are positive integers, and $p_{1}>p_{2}>p_{3}$. In the end, $A$ scores 22 points, $B$ and... | <think>[Solution] Considering the total score of each person, we have
$$
M\left(p_{1}+p_{2}+p_{3}\right)=22+9+9=40.
$$
Given that $p_{1}, p_{2}, p_{3}$ are positive integers, and $p_{1}>p_{2}>p_{3}$, we get
$$
p_{1}+p_{2}+p_{3} \geqslant 1+2+3=6,
$$
thus
$$
\begin{array}{c}
6 M \leqslant 40, \\
M \leqslant 6.
\end{ar... | C | 32,197 | <think>[Solution] Considering the total score of each person, we have
$$
M\left(p_{1}+p_{2}+p_{3}\right)=22+9+9=40.
$$
Given that $p_{1}, p_{2}, p_{3}$ are positive integers, and $p_{1}>p_{2}>p_{3}$, we get
$$
p_{1}+p_{2}+p_{3} \geqslant 1+2+3=6,
$$
thus
$$
\begin{array}{c}
6 M \leqslant 40, \\
M \leqslant 6.
\end{ar... | 1_111 |
Let $ \left( x_n\right)_{n\ge 1} $ be a sequence having $ x_1=3 $ and defined as $ x_{n+1} =\left\lfloor \sqrt 2x_n\right\rfloor , $ for every natural number $ n. $ Find all values $ m $ for which the terms $ x_m,x_{m+1},x_{m+2} $ are in arithmetic progression, where $ \lfloor\rfloor $ denotes the integer part. | <think>1. **Initial Sequence Analysis:**
Given the sequence \( (x_n)_{n \ge 1} \) with \( x_1 = 3 \) and defined by \( x_{n+1} = \left\lfloor \sqrt{2} x_n \right\rfloor \), we start by calculating the first few terms:
\[
x_1 = 3
\]
\[
x_2 = \left\lfloor \sqrt{2} \cdot 3 \right\rfloor = \left\lfloor 4.... | m = 1, 3 | 22,011 | <think>1. **Initial Sequence Analysis:**
Given the sequence \( (x_n)_{n \ge 1} \) with \( x_1 = 3 \) and defined by \( x_{n+1} = \left\lfloor \sqrt{2} x_n \right\rfloor \), we start by calculating the first few terms:
\[
x_1 = 3
\]
\[
x_2 = \left\lfloor \sqrt{2} \cdot 3 \right\rfloor = \left\lfloor 4.... | 1_112 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{3}\left(x^{2}-3 x\right) \sin 2 x \, d x
$$ | <think>## Solution
$$
\int_{0}^{3}\left(x^{2}-3 x\right) \sin 2 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2}-3 x ; d u=(2 x-3) d x \\
& d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x
\end{aligned}
$$
Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get:
$$
\begin{aligned}
& =\left.\left... | \frac{3\sin6+\cos6-1}{4} | 3,463 | <think>## Solution
$$
\int_{0}^{3}\left(x^{2}-3 x\right) \sin 2 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2}-3 x ; d u=(2 x-3) d x \\
& d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x
\end{aligned}
$$
Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get:
$$
\begin{aligned}
& =\left.\left... | 1_113 |
Determine the minimum value of
$$x^{2014} + 2x^{2013} + 3x^{2012} + 4x^{2011} +\ldots + 2014x + 2015$$
where $x$ is a real number. | <think>To determine the minimum value of the function
\[ f(x) = x^{2014} + 2x^{2013} + 3x^{2012} + \cdots + 2014x + 2015, \]
we will analyze the behavior of \( f(x) \) and its derivative.
1. **Expression for \( f(x) \):**
The given function is:
\[ f(x) = x^{2014} + 2x^{2013} + 3x^{2012} + \cdots + 2014x + 2015.... | 1008 | 72,908 | <think>To determine the minimum value of the function
\[ f(x) = x^{2014} + 2x^{2013} + 3x^{2012} + \cdots + 2014x + 2015, \]
we will analyze the behavior of \( f(x) \) and its derivative.
1. **Expression for \( f(x) \):**
The given function is:
\[ f(x) = x^{2014} + 2x^{2013} + 3x^{2012} + \cdots + 2014x + 2015.... | 1_114 |
Task 3. Let $n \geq 2$ be a positive integer. Each cell of an $n \times n$ board is colored red or blue. We place dominoes on the board, each covering two cells. We call a domino plain if it lies on two red or two blue cells, and colorful if it lies on one red and one blue cell. Find the largest positive integer $k$ wi... | <think>Solution. We prove that $k=\left\lfloor\frac{n^{2}}{4}\right\rfloor$ is the largest possible value.
Suppose $n$ is even. Then it is possible to cover the board with $\frac{n^{2}}{2}$ dominoes (ignoring colors). Since there are $\frac{n^{2}}{2}$ dominoes, all of which are either colorful or plain, there are at l... | \lfloor\frac{n^{2}}{4}\rfloor | 14,444 | <think>Solution. We prove that $k=\left\lfloor\frac{n^{2}}{4}\right\rfloor$ is the largest possible value.
Suppose $n$ is even. Then it is possible to cover the board with $\frac{n^{2}}{2}$ dominoes (ignoring colors). Since there are $\frac{n^{2}}{2}$ dominoes, all of which are either colorful or plain, there are at l... | 1_115 |
In a round-robin tournament, 23 teams participated. Each team played exactly once against all the others. We say that 3 teams have cycled victories if, considering only their games against each other, each of them won exactly once. What is the maximum number of cycled victories that could have occurred during the tourn... | <think>Let's solve the problem more generally for $n$ teams, where $n$ is an odd number. We will represent each team with a point. After a match between two teams, an arrow should point from the losing team to the winning team. (In this context, we can assume that there has never been a tie.)
If the teams $A$, $B$, an... | 506 | 83,593 | <think>Let's solve the problem more generally for $n$ teams, where $n$ is an odd number. We will represent each team with a point. After a match between two teams, an arrow should point from the losing team to the winning team. (In this context, we can assume that there has never been a tie.)
If the teams $A$, $B$, an... | 1_116 |
Find all positive integers $A,B$ satisfying the following properties:
(i) $A$ and $B$ has same digit in decimal.
(ii) $2\cdot A\cdot B=\overline{AB}$ (Here $\cdot$ denotes multiplication, $\overline{AB}$ denotes we write $A$ and $B$ in turn. For example, if $A=12,B=34$, then $\overline{AB}=1234$) | <think>To find all positive integers \( A \) and \( B \) satisfying the given properties, we need to follow these steps:
1. **Express the given condition mathematically:**
The problem states that \( 2 \cdot A \cdot B = \overline{AB} \). Here, \(\overline{AB}\) denotes the concatenation of \( A \) and \( B \). If \(... | (3, 6), (13, 52) | 68,106 | <think>To find all positive integers \( A \) and \( B \) satisfying the given properties, we need to follow these steps:
1. **Express the given condition mathematically:**
The problem states that \( 2 \cdot A \cdot B = \overline{AB} \). Here, \(\overline{AB}\) denotes the concatenation of \( A \) and \( B \). If \(... | 1_117 |
A sphere with radius $r$ is inscribed in a cylinder, and the volume of the cylinder is $V_{1}$; the volume of a cone circumscribed around the sphere is $V_{2}$. What is the smallest value of the ratio $V_{2} / V_{1}$? | <think>A sphere can be circumscribed by a cylinder in only one way, and the volume of this cylinder is constant, specifically expressed by the sphere's radius as $V_{1}=2 r^{3} \pi$.
The ratio $V_{2} / V_{1}$ will be the smallest when $V_{2}$ takes its smallest value.
Thus, among the cones that can be circumscribed a... | \frac{4}{3} | 13,728 | <think>A sphere can be circumscribed by a cylinder in only one way, and the volume of this cylinder is constant, specifically expressed by the sphere's radius as $V_{1}=2 r^{3} \pi$.
The ratio $V_{2} / V_{1}$ will be the smallest when $V_{2}$ takes its smallest value.
Thus, among the cones that can be circumscribed a... | 1_118 |
A computer generates all pairs of real numbers $x, y \in (0, 1)$ for which the numbers $a = x+my$ and $b = y+mx$ are both integers, where $m$ is a given positive integer. Finding one such pair $(x, y)$ takes $5$ seconds. Find $m$ if the computer needs $595$ seconds to find all possible ordered pairs $(x, y)$. | <think>1. **Understanding the problem:**
We need to find pairs \((x, y) \in (0, 1)\) such that \(a = x + my\) and \(b = y + mx\) are both integers, where \(m\) is a given positive integer. Each pair takes 5 seconds to find, and the computer needs 595 seconds to find all possible pairs. We need to determine \(m\).
2... | m = 11 | 55,243 | <think>1. **Understanding the problem:**
We need to find pairs \((x, y) \in (0, 1)\) such that \(a = x + my\) and \(b = y + mx\) are both integers, where \(m\) is a given positive integer. Each pair takes 5 seconds to find, and the computer needs 595 seconds to find all possible pairs. We need to determine \(m\).
2... | 1_119 |
Let $ABCD$ be a quadrilateral with $AD = 20$ and $BC = 13$. The area of $\triangle ABC$ is $338$ and the area of $\triangle DBC$ is $212$. Compute the smallest possible perimeter of $ABCD$.
[i]Proposed by Evan Chen[/i] | <think>1. **Identify the given information and set up the problem:**
- We have a quadrilateral \(ABCD\) with \(AD = 20\) and \(BC = 13\).
- The area of \(\triangle ABC\) is \(338\) and the area of \(\triangle DBC\) is \(212\).
2. **Introduce the feet of the altitudes:**
- Let \(X\) and \(Y\) be the feet of th... | 118 | 70,436 | <think>1. **Identify the given information and set up the problem:**
- We have a quadrilateral \(ABCD\) with \(AD = 20\) and \(BC = 13\).
- The area of \(\triangle ABC\) is \(338\) and the area of \(\triangle DBC\) is \(212\).
2. **Introduce the feet of the altitudes:**
- Let \(X\) and \(Y\) be the feet of th... | 1_120 |
[color=darkred]Find all differentiable functions $f\colon [0,\infty)\to [0,\infty)$ for which $f(0)=0$ and $f^{\prime}(x^2)=f(x)$ for any $x\in [0,\infty)$ .[/color] | <think>1. Given the function \( f: [0, \infty) \to [0, \infty) \) with \( f(0) = 0 \) and \( f'(x^2) = f(x) \) for any \( x \in [0, \infty) \), we start by differentiating both sides of the equation \( f'(x^2) = f(x) \) with respect to \( x \).
2. Using the chain rule, we get:
\[
\frac{d}{dx} f'(x^2) = \frac{d}{... | f(x) = 0 | 24,117 | <think>1. Given the function \( f: [0, \infty) \to [0, \infty) \) with \( f(0) = 0 \) and \( f'(x^2) = f(x) \) for any \( x \in [0, \infty) \), we start by differentiating both sides of the equation \( f'(x^2) = f(x) \) with respect to \( x \).
2. Using the chain rule, we get:
\[
\frac{d}{dx} f'(x^2) = \frac{d}{... | 1_121 |
Given a real number $t\neq -1$. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[(t+1)f(1+xy)-f(x+y)=f(x+1)f(y+1)\]
for all $x,y\in\mathbb{R}$. | <think>1. Let \( P(x, y) \) be the assertion of the given functional equation:
\[
(t+1)f(1+xy) - f(x+y) = f(x+1)f(y+1)
\]
for all \( x, y \in \mathbb{R} \).
2. Consider the constant solutions first. If \( f(x) = 0 \) for all \( x \), then the equation becomes:
\[
(t+1) \cdot 0 - 0 = 0 \cdot 0 \implie... | f(x) = 0 \text{ or } f(x) = t \text{ or } f(x) = (t+1)x - (t+2) | 52,135 | <think>1. Let \( P(x, y) \) be the assertion of the given functional equation:
\[
(t+1)f(1+xy) - f(x+y) = f(x+1)f(y+1)
\]
for all \( x, y \in \mathbb{R} \).
2. Consider the constant solutions first. If \( f(x) = 0 \) for all \( x \), then the equation becomes:
\[
(t+1) \cdot 0 - 0 = 0 \cdot 0 \implie... | 1_122 |
61. Solve the matrix equation
$$
\left(\begin{array}{rrr}
3 & -1 & 0 \\
-2 & 1 & 1 \\
2 & -1 & 4
\end{array}\right) X=\left(\begin{array}{r}
5 \\
0 \\
15
\end{array}\right)
$$ | <think>Solution. $1^{0}$. Find the inverse matrix $A^{-1}$.
Calculate the determinant of matrix $A$:
$$
\begin{gathered}
D=\left|\begin{array}{rrr}
3 & -1 & 0 \\
-2 & 1 & 1 \\
2 & -1 & 4
\end{array}\right|=3 \cdot 1 \cdot 4+(-1) \cdot 1 \cdot 2+(-2) \cdot(-1) \cdot 0- \\
-2 \cdot 1 \cdot 0-(-1) \cdot 1 \cdot 3-(-2) \... | (\begin{pmatrix}2\\1\\3\end{pmatrix}) | 70,301 | <think>Solution. $1^{0}$. Find the inverse matrix $A^{-1}$.
Calculate the determinant of matrix $A$:
$$
\begin{gathered}
D=\left|\begin{array}{rrr}
3 & -1 & 0 \\
-2 & 1 & 1 \\
2 & -1 & 4
\end{array}\right|=3 \cdot 1 \cdot 4+(-1) \cdot 1 \cdot 2+(-2) \cdot(-1) \cdot 0- \\
-2 \cdot 1 \cdot 0-(-1) \cdot 1 \cdot 3-(-2) \... | 1_123 |
I am thinking of a four-digit number, each digit of which is different. When I strike out the last two digits in this number, I get a prime number. Similarly, I get a prime number if I strike out the second and fourth digits, and even if I strike out the middle two digits. However, the number I am thinking of is not a ... | <think>We are looking for a number in the form $\overline{a b c d}$ (digits $a, b, c, d$ are different). According to the problem, $\overline{a b}$ is a prime number, as well as $\overline{a c}$ and $\overline{a d}$. We are therefore looking for three different two-digit prime numbers that start with the same digit (i.... | 4731 | 21,254 | <think>We are looking for a number in the form $\overline{a b c d}$ (digits $a, b, c, d$ are different). According to the problem, $\overline{a b}$ is a prime number, as well as $\overline{a c}$ and $\overline{a d}$. We are therefore looking for three different two-digit prime numbers that start with the same digit (i.... | 1_124 |
36. Find all positive integer triples $(a, m, n)$ such that
$$\left(a^{m}+1\right) \mid(a+1)^{n}$$ | <think>36. Notice that when $a, m$ is equal to 1, it is easy to get the positive integer solutions $(a, m, n)=(1, m, n)$ or $(a, 1, n)$, where $a, m, n \in \mathbf{N}^{*}$.
Below, we discuss the case where $a, m$ are both greater than 1.
First, we prove a lemma: Let $u, v, l \in \mathbf{N}^{*}$, and $u \mid v^{l}$, th... | (a, m, n)=(1, m, n), (a, 1, n), (2,3, r) | 69,827 | <think>36. Notice that when $a, m$ is equal to 1, it is easy to get the positive integer solutions $(a, m, n)=(1, m, n)$ or $(a, 1, n)$, where $a, m, n \in \mathbf{N}^{*}$.
Below, we discuss the case where $a, m$ are both greater than 1.
First, we prove a lemma: Let $u, v, l \in \mathbf{N}^{*}$, and $u \mid v^{l}$, th... | 1_125 |
In the triangle $ABC$, the projection of the center of the circumscribed circle onto the line of the median from vertex $A$ is $P$. What relationship holds between the distances $PA$, $PB$, and $PC$? | <think>I. solution. Let the midpoint of side $BC$ be denoted by $F$. As long as $O, F$, and $P$ are distinct (and also distinct from $A$), from the (right-angled) triangles $OPA$ and $OPF$,
$$
P A^{2}=O A^{2}-O P^{2}=O B^{2}-\left(O F^{2}-P F^{2}\right)=B F^{2}+P F^{2}=\frac{B C^{2}}{4}+P F^{2}
$$
since $O A=O B$. No... | PA^{2}=\frac{1}{2}(PB^{2}+PC^{2}) | 31,245 | <think>I. solution. Let the midpoint of side $BC$ be denoted by $F$. As long as $O, F$, and $P$ are distinct (and also distinct from $A$), from the (right-angled) triangles $OPA$ and $OPF$,
$$
P A^{2}=O A^{2}-O P^{2}=O B^{2}-\left(O F^{2}-P F^{2}\right)=B F^{2}+P F^{2}=\frac{B C^{2}}{4}+P F^{2}
$$
since $O A=O B$. No... | 1_126 |
Let $\sigma(n)$ be the number of positive divisors of $n$, and let $\operatorname{rad} n$ be the product of the distinct prime divisors of $n$. By convention, $\operatorname{rad} 1 = 1$. Find the greatest integer not exceeding \[ 100\left(\sum_{n=1}^{\infty}\frac{\sigma(n)\sigma(n \operatorname{rad} n)}{n^2\sigma(\ope... | <think>1. **Define the function \( f(n) \):**
\[
f(n) = \frac{\sigma(n) \sigma(n \operatorname{rad} n)}{n^2 \sigma(\operatorname{rad} n)}
\]
where \(\sigma(n)\) is the sum of the divisors of \(n\) and \(\operatorname{rad}(n)\) is the product of the distinct prime divisors of \(n\).
2. **Identify the multip... | 164 | 14,627 | <think>1. **Define the function \( f(n) \):**
\[
f(n) = \frac{\sigma(n) \sigma(n \operatorname{rad} n)}{n^2 \sigma(\operatorname{rad} n)}
\]
where \(\sigma(n)\) is the sum of the divisors of \(n\) and \(\operatorname{rad}(n)\) is the product of the distinct prime divisors of \(n\).
2. **Identify the multip... | 1_127 |
Find the functions $ f:\mathbb{N}\longrightarrow\mathbb{N} $ that satisfy the following relation:
$$ \gcd\left( x,f(y)\right)\cdot\text{lcm}\left(f(x), y\right) = \gcd (x,y)\cdot\text{lcm}\left( f(x), f(y)\right) ,\quad\forall x,y\in\mathbb{N} . $$ | <think>1. **Initial Setup and Simplification:**
We start with the given functional equation:
\[
\gcd(x, f(y)) \cdot \text{lcm}(f(x), y) = \gcd(x, y) \cdot \text{lcm}(f(x), f(y)), \quad \forall x, y \in \mathbb{N}.
\]
Recall the relationship between gcd and lcm:
\[
\gcd(a, b) \cdot \text{lcm}(a, b) ... | f(x) = x | 73,760 | <think>1. **Initial Setup and Simplification:**
We start with the given functional equation:
\[
\gcd(x, f(y)) \cdot \text{lcm}(f(x), y) = \gcd(x, y) \cdot \text{lcm}(f(x), f(y)), \quad \forall x, y \in \mathbb{N}.
\]
Recall the relationship between gcd and lcm:
\[
\gcd(a, b) \cdot \text{lcm}(a, b) ... | 1_128 |
For an integer $n\ge 3$ and a permutation $\sigma=(p_{1},p_{2},\cdots ,p_{n})$ of $\{1,2,\cdots , n\}$, we say $p_{l}$ is a $landmark$ point if $2\le l\le n-1$ and $(p_{l-1}-p_{l})(p_{l+1}-p_{l})>0$. For example , for $n=7$,\\
the permutation $(2,7,6,4,5,1,3)$ has four landmark points: $p_{2}=7$, $p_{4}=4$, $p_{5}=5$ a... | <think>To solve the problem, we need to determine the number of permutations of $\{1, 2, \cdots, n\}$ with exactly one landmark point and find the maximum $n \geq 3$ for which this number is a perfect square.
1. **Definition of Landmark Point**:
A point $p_l$ in a permutation $\sigma = (p_1, p_2, \cdots, p_n)$ is a... | 3 | 87,732 | <think>To solve the problem, we need to determine the number of permutations of $\{1, 2, \cdots, n\}$ with exactly one landmark point and find the maximum $n \geq 3$ for which this number is a perfect square.
1. **Definition of Landmark Point**:
A point $p_l$ in a permutation $\sigma = (p_1, p_2, \cdots, p_n)$ is a... | 1_129 |
Determine all polynomials $p$ with real coefficients for which $p(0)=0$ and
$$f(f(n))+n=4f(n)\qquad\text{for all }n\in\mathbb N,$$where $f(n)=\lfloor p(n)\rfloor$. | <think>1. We start with the given conditions:
- \( p(0) = 0 \)
- \( f(f(n)) + n = 4f(n) \) for all \( n \in \mathbb{N} \), where \( f(n) = \lfloor p(n) \rfloor \).
2. Assume \( p(x) \) is a polynomial of degree \( d \). Since \( p(0) = 0 \), we can write \( p(x) \) as:
\[
p(x) = a_d x^d + a_{d-1} x^{d-1} +... | p(x) = (2 + \sqrt{3}) x | 60,097 | <think>1. We start with the given conditions:
- \( p(0) = 0 \)
- \( f(f(n)) + n = 4f(n) \) for all \( n \in \mathbb{N} \), where \( f(n) = \lfloor p(n) \rfloor \).
2. Assume \( p(x) \) is a polynomial of degree \( d \). Since \( p(0) = 0 \), we can write \( p(x) \) as:
\[
p(x) = a_d x^d + a_{d-1} x^{d-1} +... | 1_130 |
Let $ f:\mathbb{Z}_{>0}\rightarrow\mathbb{R} $ be a function such that for all $n > 1$ there is a prime divisor $p$ of $n$ such that \[ f(n)=f\left(\frac{n}{p}\right)-f(p). \]
Furthermore, it is given that $ f(2^{2014})+f(3^{2015})+f(5^{2016})=2013 $. Determine $ f(2014^2)+f(2015^3)+f(2016^5) $. | <think>1. **Understanding the given functional equation:**
\[
f(n) = f\left(\frac{n}{p}\right) - f(p)
\]
where \( p \) is a prime divisor of \( n \).
2. **Analyzing the function for prime numbers:**
Let \( p \) be a prime number. The only prime divisor of \( p \) is \( p \) itself. Thus, we have:
\[
... | \frac{49}{3} | 58,884 | <think>1. **Understanding the given functional equation:**
\[
f(n) = f\left(\frac{n}{p}\right) - f(p)
\]
where \( p \) is a prime divisor of \( n \).
2. **Analyzing the function for prime numbers:**
Let \( p \) be a prime number. The only prime divisor of \( p \) is \( p \) itself. Thus, we have:
\[
... | 1_131 |
Task 4. Determine all integers $n$ for which the polynomial $P(x)=3 x^{3}-n x-n-2$ can be written as the product of two non-constant polynomials with integer coefficients. | <think>Solution I. Suppose that $P(x)$ can be written as $P(x)=A(x) B(x)$ with $A$ and $B$ being non-constant polynomials with integer coefficients. Since $A$ and $B$ are not constant, each has a degree of at least 1. The sum of the two degrees is equal to the degree of $P$, which is 3. This means that the two degrees ... | n=38,n=-2,n=26,n=130 | 9,446 | <think>Solution I. Suppose that $P(x)$ can be written as $P(x)=A(x) B(x)$ with $A$ and $B$ being non-constant polynomials with integer coefficients. Since $A$ and $B$ are not constant, each has a degree of at least 1. The sum of the two degrees is equal to the degree of $P$, which is 3. This means that the two degrees ... | 1_132 |
A regular $n$-gon is inscribed in a unit circle. Compute the product from a fixed vertex to all the other vertices. | <think>1. **Move to the complex plane:**
Consider the vertices of the regular $n$-gon as the $n$th roots of unity. These roots are given by:
\[
z_k = e^{2\pi i k / n} \quad \text{for} \quad k = 0, 1, 2, \ldots, n-1
\]
These points lie on the unit circle centered at the origin in the complex plane.
2. **... | n | 50,306 | <think>1. **Move to the complex plane:**
Consider the vertices of the regular $n$-gon as the $n$th roots of unity. These roots are given by:
\[
z_k = e^{2\pi i k / n} \quad \text{for} \quad k = 0, 1, 2, \ldots, n-1
\]
These points lie on the unit circle centered at the origin in the complex plane.
2. **... | 1_133 |
Let $s(a)$ denote the sum of digits of a given positive integer a. The sequence $a_1, a_2,..., a_n, ...$ of positive integers is such that $a_{n+1} = a_n+s(a_n)$ for each positive integer $n$. Find the greatest possible n for which it is possible to have $a_n = 2008$. | <think>To find the greatest possible \( n \) for which it is possible to have \( a_n = 2008 \), we need to trace back the sequence \( a_n \) to see how far we can go. The sequence is defined by \( a_{n+1} = a_n + s(a_n) \), where \( s(a) \) denotes the sum of the digits of \( a \).
1. **Starting with \( a_n = 2008 \):... | 6 | 20,050 | <think>To find the greatest possible \( n \) for which it is possible to have \( a_n = 2008 \), we need to trace back the sequence \( a_n \) to see how far we can go. The sequence is defined by \( a_{n+1} = a_n + s(a_n) \), where \( s(a) \) denotes the sum of the digits of \( a \).
1. **Starting with \( a_n = 2008 \):... | 1_134 |
Let $a < b < c < d < e$ be real numbers. We calculate all possible sums in pairs of these 5 numbers. Of these 10 sums, the three smaller ones are 32, 36, 37, while the two larger ones are 48 and 51. Determine all possible values that $e$ can take. | <think>1. We start by noting that we have five real numbers \(a < b < c < d < e\) and we are given the sums of pairs of these numbers. Specifically, the three smallest sums are 32, 36, and 37, and the two largest sums are 48 and 51.
2. Let's denote the sums of pairs as follows:
\[
a + b = 32, \quad a + c = 36, \... | 27.5 | 32,336 | <think>1. We start by noting that we have five real numbers \(a < b < c < d < e\) and we are given the sums of pairs of these numbers. Specifically, the three smallest sums are 32, 36, and 37, and the two largest sums are 48 and 51.
2. Let's denote the sums of pairs as follows:
\[
a + b = 32, \quad a + c = 36, \... | 1_135 |
An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.
a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$;
b) Calculate the distance of $p$ from either base;
c) Determine under what conditions such points $P$ actu... | <think>Given an isosceles trapezoid with bases \( a \) and \( c \) and altitude \( h \), we need to find points \( P \) on the axis of symmetry such that both legs of the trapezoid subtend right angles at \( P \).
### Part (a)
1. **Identify the Trapezoid and Symmetry Axis:**
Let the trapezoid be \( ABCD \) with \(... | h^2 \leq ac | 5,430 | <think>Given an isosceles trapezoid with bases \( a \) and \( c \) and altitude \( h \), we need to find points \( P \) on the axis of symmetry such that both legs of the trapezoid subtend right angles at \( P \).
### Part (a)
1. **Identify the Trapezoid and Symmetry Axis:**
Let the trapezoid be \( ABCD \) with \(... | 1_136 |
Given an integer ${n>1}$, denote by $P_{n}$ the product of all positive integers $x$ less than $n$ and such that $n$ divides ${x^2-1}$. For each ${n>1}$, find the remainder of $P_{n}$ on division by $n$.
[i]Proposed by John Murray, Ireland[/i] | <think>Given an integer \( n > 1 \), denote by \( P_n \) the product of all positive integers \( x \) less than \( n \) such that \( n \) divides \( x^2 - 1 \). We need to find the remainder of \( P_n \) on division by \( n \).
1. **Identify the set \( S_n \):**
\[
S_n = \{ x \in \mathbb{N} \mid x < n \text{ and... | P_n \equiv -1 \pmod{n} \text{ if } n \text{ has a primitive root and 1 otherwise.} | 20,879 | <think>Given an integer \( n > 1 \), denote by \( P_n \) the product of all positive integers \( x \) less than \( n \) such that \( n \) divides \( x^2 - 1 \). We need to find the remainder of \( P_n \) on division by \( n \).
1. **Identify the set \( S_n \):**
\[
S_n = \{ x \in \mathbb{N} \mid x < n \text{ and... | 1_137 |
4. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{c}
(a y-a x+2)(4 y-3|x-a|-x+5 a)=0 \\
\left(\log _{a} x^{2}+\log _{a} y^{2}-2\right) \log _{2} a^{2}=8
\end{array}\right.
$$
has six distinct solutions. (16 points) | <think>Solution. Simplify the second equation of the system: $\left(\log _{a} x^{2}+\log _{a} y^{2}-2\right) \log _{2} a^{2}=8 \Leftrightarrow a>$ $0, a \neq 1, \log _{a} x^{2} y^{2}=2+4 \log _{a} 2,|x y|=4 a$. We have
\cup(4;32) | 84,068 | <think>Solution. Simplify the second equation of the system: $\left(\log _{a} x^{2}+\log _{a} y^{2}-2\right) \log _{2} a^{2}=8 \Leftrightarrow a>$ $0, a \neq 1, \log _{a} x^{2} y^{2}=2+4 \log _{a} 2,|x y|=4 a$. We have
}(n)=n$ for each $n \in S$.
Find all possible values for $f(2021)$.
(Here, $f^{(n)}(n) = \underbrace{f(f(f\dots f(}_{n \text{ times} }n)))\dots))$.)
[i]Proposed by Viktor Simjanoski[/i] | <think>1. Define \( \text{ord}(n) := \min \{k \in \mathbb{Z}^+ \mid f^k(n) = n\} \), which represents the minimal period of the sequence \( n, f(n), f^2(n), \ldots \) (its orbit). This is the smallest positive integer \( k \) such that applying the function \( f \) \( k \) times returns \( n \) to itself.
2. Given tha... | \{43, 86, 129, 172, 215, 258, 301, 344, 387, 430, 473, 516, 559, 602, 645, 688, 731, 774, 817, 860, 903, 946, 989, 1032, 1075, 1118, 1161, 1204, 1247, 1290, 1333, 1376, 1419, 1462, 1505, 1548, 1591, 1634, 1677, 1720, 1763, 1806, 1849, 1892, 1935, 1978, 2021\} | 23,508 | <think>1. Define \( \text{ord}(n) := \min \{k \in \mathbb{Z}^+ \mid f^k(n) = n\} \), which represents the minimal period of the sequence \( n, f(n), f^2(n), \ldots \) (its orbit). This is the smallest positive integer \( k \) such that applying the function \( f \) \( k \) times returns \( n \) to itself.
2. Given tha... | 1_139 |
Find all triplets of positive rational numbers $(m,n,p)$ such that the numbers $m+\frac 1{np}$, $n+\frac 1{pm}$, $p+\frac 1{mn}$ are integers.
[i]Valentin Vornicu, Romania[/i] | <think>1. Let \( m, n, p \) be positive rational numbers. Define \( t = mnp \). We are given that \( m + \frac{1}{np} \), \( n + \frac{1}{pm} \), and \( p + \frac{1}{mn} \) are integers. Let's denote these integers by \( a, b, \) and \( c \) respectively. Thus, we have:
\[
a = m + \frac{1}{np}, \quad b = n + \fra... | \left(\frac{1}{2}, \frac{1}{2}, 4\right), \left(\frac{1}{2}, 1, 2\right), (1, 1, 1) | 88,674 | <think>1. Let \( m, n, p \) be positive rational numbers. Define \( t = mnp \). We are given that \( m + \frac{1}{np} \), \( n + \frac{1}{pm} \), and \( p + \frac{1}{mn} \) are integers. Let's denote these integers by \( a, b, \) and \( c \) respectively. Thus, we have:
\[
a = m + \frac{1}{np}, \quad b = n + \fra... | 1_140 |
$10 \cdot 98$ Find the maximum positive integer $n$, such that there exist $n$ integer points in the coordinate plane, where any 3 points form a triangle, and the centroid of the triangle is not an integer point. | <think>[Solution] Let the points $\left(x_{1}, y_{1}\right) ;\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ be 3 integer points, then the coordinates of the centroid of the triangle formed by them are
$$
\left(\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right), \frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right)\right) .
$$
It is ... | 8 | 22,411 | <think>[Solution] Let the points $\left(x_{1}, y_{1}\right) ;\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ be 3 integer points, then the coordinates of the centroid of the triangle formed by them are
$$
\left(\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right), \frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right)\right) .
$$
It is ... | 1_141 |
Let $N$ and $M$ be positive integers, and $p$ and $q$ be distinct prime numbers. Suppose that $N+M$ is a five-digit number, $N$ is divisible by $p$, and the number of its divisors is $q$, while $M$ is divisible by $q$, and the number of its divisors is $p$. Determine the possible values of $N$ and $M$. | <think>To determine the number of divisors of a number, we multiply the numbers that are one greater than the exponents in its prime factorization. In our case, this product must be a prime number. If more than one prime factor were present in the prime factorization, the number of divisors would not be a prime number;... | \begin{pmatrix}\hlinep&N&M\\\hline2&17&65536&17\\17&2&17&65536\\3&11&59049&121\\11&3&121&590 | 871 | <think>To determine the number of divisors of a number, we multiply the numbers that are one greater than the exponents in its prime factorization. In our case, this product must be a prime number. If more than one prime factor were present in the prime factorization, the number of divisors would not be a prime number;... | 1_142 |
Let $ABC$ be an equilateral triangle, and let $P$ be some point in its circumcircle.
Determine all positive integers $n$, for which the value of the sum
$S_n (P) = |PA|^n + |PB|^n + |PC|^n$ is independent of the choice of point $P$. | <think>1. Let $ABC$ be an equilateral triangle with side length $a$, and let $P$ be a point on its circumcircle. We need to determine all positive integers $n$ for which the value of the sum $S_n(P) = |PA|^n + |PB|^n + |PC|^n$ is independent of the choice of point $P$.
2. Consider the midpoint $D$ of the arc $BC$ that... | 2, 4 | 35,343 | <think>1. Let $ABC$ be an equilateral triangle with side length $a$, and let $P$ be a point on its circumcircle. We need to determine all positive integers $n$ for which the value of the sum $S_n(P) = |PA|^n + |PB|^n + |PC|^n$ is independent of the choice of point $P$.
2. Consider the midpoint $D$ of the arc $BC$ that... | 1_143 |
Knight Milivoj was preparing to go to Veselín for a tournament. The tournament takes place on Wednesday. Since the journey from Knight's home, Rytířov, to Veselín would take two days, he set off on Monday. The road leads through two other towns, Kostín and Zubín. On the first day of travel, he covered 25 miles and spen... | <think>Let's first visualize the situation from the assignment.
The journey to the tournament:
The journey from the tournament:
The journey from the tournament:
 A shepherd uses 15 sections of fence, each 2 meters long, to enclose a rectangular or square sheep pen with one side against a wall. The maximum area of the sheep pen is ( ) square meters.
A. 100
B. 108
C. 112
D. 122 | <think>【Analysis】According to the problem, the length of these 15 sections of fence is the sum of one length and two widths of the rectangular sheep pen formed, from which we can list the values of length and width: when the width is 1 section, the length is $15-1 \times 2=13$ sections; when the width is 2 sections, th... | 112 | 84,341 | <think>【Analysis】According to the problem, the length of these 15 sections of fence is the sum of one length and two widths of the rectangular sheep pen formed, from which we can list the values of length and width: when the width is 1 section, the length is $15-1 \times 2=13$ sections; when the width is 2 sections, th... | 1_145 |
Let $a,b,c $ be the lengths of the three sides of a triangle and $a,b$ be the two roots of the equation $ax^2-bx+c=0 $$ (a<b) . $ Find the value range of $ a+b-c .$ | <think>Given the problem, we need to find the value range of \(a + b - c\) where \(a, b, c\) are the lengths of the sides of a triangle and \(a, b\) are the roots of the quadratic equation \(ax^2 - bx + c = 0\) with \(a < b\).
1. **Roots of the Quadratic Equation:**
Since \(a\) and \(b\) are the roots of the quadra... | \left(\frac{7}{8}, \sqrt{5} - 1\right) | 86,412 | <think>Given the problem, we need to find the value range of \(a + b - c\) where \(a, b, c\) are the lengths of the sides of a triangle and \(a, b\) are the roots of the quadratic equation \(ax^2 - bx + c = 0\) with \(a < b\).
1. **Roots of the Quadratic Equation:**
Since \(a\) and \(b\) are the roots of the quadra... | 1_146 |
From the midpoint of the height of a regular triangular pyramid, perpendiculars are dropped to the lateral edge and to the lateral face. These perpendiculars are equal to $a$ and $b$ respectively. Find the volume of the pyramid. Does the problem have a solution for any $a$ and $b$? | <think>Drop perpendiculars $H P$ and $H Q$ from the base $H$ of the height $D H$ of the given regular triangular pyramid $A B C D$ to the lateral edge $C D$ and the lateral face $A B D$, respectively. Then $H P=2 a, H Q=2 b$. Let $\alpha$ and $\beta$ be the angles between the lateral edge and the plane of the base $A B... | 18^{3}b^{3}/((^{2}-b^{2})\sqrt{4b^{2}-^{2}}) | 27,121 | <think>Drop perpendiculars $H P$ and $H Q$ from the base $H$ of the height $D H$ of the given regular triangular pyramid $A B C D$ to the lateral edge $C D$ and the lateral face $A B D$, respectively. Then $H P=2 a, H Q=2 b$. Let $\alpha$ and $\beta$ be the angles between the lateral edge and the plane of the base $A B... | 1_147 |
A certain frog that was placed on a vertex of a convex polygon chose to jump to another vertex, either clockwise skipping one vertex, either counterclockwise skipping two vertexes, and repeated the procedure.
If the number of jumps that the frog made is equal to the number of sides of the polygon, the frog has passed t... | <think>1. Let the polygon have \( n \) vertices. The frog can jump either clockwise skipping one vertex or counterclockwise skipping two vertices. The frog makes \( n \) jumps and ends up on the initial vertex after passing through all vertices.
2. We need to determine the set of possible values for \( n \) such that ... | \{ n \in \mathbb{N} \mid 6 \nmid n \text{ or } 30 \mid n \} | 47,045 | <think>1. Let the polygon have \( n \) vertices. The frog can jump either clockwise skipping one vertex or counterclockwise skipping two vertices. The frog makes \( n \) jumps and ends up on the initial vertex after passing through all vertices.
2. We need to determine the set of possible values for \( n \) such that ... | 1_148 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+11 x+7}{(x+1)(x+2)^{3}} d x
$$ | <think>## Solution
$$
\int \frac{x^{3}+6 x^{2}+11 x+7}{(x+1)(x+2)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+11 x+7}{(x+1)(x+2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_... | \ln|x+1|+\frac{1}{2(x+2)^{2}}+C | 86,603 | <think>## Solution
$$
\int \frac{x^{3}+6 x^{2}+11 x+7}{(x+1)(x+2)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+11 x+7}{(x+1)(x+2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_... | 1_149 |
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 $x$. | <think>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... | 3124 | 59,187 | <think>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... | 1_150 |
The eleventh question: Given a positive integer $n \geq 2$, find the smallest real number $c(n)$ such that for any positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$, the following inequality holds: $\frac{\sum_{i=1}^{n} a_{i}}{n}-\sqrt[n]{\prod_{i=1}^{n} a_{i}} \leq c(n) \cdot \max _{1 \leq i<j \leq n}\left\{\left(\s... | <think>The 11th question:
Solution: On one hand, we take $a_{1}=a_{2}=\ldots=a_{n-1}=1, a_{n}=\varepsilon$. Let $\varepsilon \rightarrow 0^{+}$, then we get $\frac{n-1}{n} \leq c(n)$.
Next, we prove that for any positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$, we have
$$
\frac{\sum_{i=1}^{n} a_{i}}{n}-\sqrt[n]{\pro... | (n)=\frac{n-1}{n} | 83,929 | <think>The 11th question:
Solution: On one hand, we take $a_{1}=a_{2}=\ldots=a_{n-1}=1, a_{n}=\varepsilon$. Let $\varepsilon \rightarrow 0^{+}$, then we get $\frac{n-1}{n} \leq c(n)$.
Next, we prove that for any positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$, we have
$$
\frac{\sum_{i=1}^{n} a_{i}}{n}-\sqrt[n]{\pro... | 1_151 |
18. Stirling's Problem (from 9th grade, 5 points). An incredible legend says that one day Stirling ${ }^{4}$ was contemplating Stirling numbers of the second kind and, in a pensive mood, was tossing 10 fair dice onto the table. After one of the tosses, he suddenly noticed that the combination of points that came up inc... | <think>Solution. Let's find the probability of event $A$ "some number is missing in the rolled combination". Denote $A_{k}$ as the event "number $k$ is missing". Then
$$
A=A_{1} \cup A_{2} \cup A_{3} \cup A_{4} \cup A_{5} \cup A_{6}
$$
Using the inclusion-exclusion principle, we get:
$$
\begin{gathered}
\mathrm{P}(A... | 0.272 | 7,836 | <think>Solution. Let's find the probability of event $A$ "some number is missing in the rolled combination". Denote $A_{k}$ as the event "number $k$ is missing". Then
$$
A=A_{1} \cup A_{2} \cup A_{3} \cup A_{4} \cup A_{5} \cup A_{6}
$$
Using the inclusion-exclusion principle, we get:
$$
\begin{gathered}
\mathrm{P}(A... | 1_152 |
Three, in a football round-robin tournament, the 1st to 4th places scored 11 points, 9 points, 7 points, and 5 points respectively, and the 5th place's score is different from the 4th place's score (the winner gets 2 points, the loser gets 0 points, and in a draw, both get 1 point). How many teams participated in the c... | <think>Three, suppose there are $k$ teams participating in the competition, a total of $C_{k}^{2}=\frac{1}{2} k(k-1)$ matches are played, so the total score is $2 C_{k}^{2}=k(k-1)$.
The top 4 teams scored $11+9+7+5=32$ points, the last $k-4$ teams can score at most $4(k-4)$ points, thus we have
Therefore, $k=7$.
The r... | 7teams,scores:11,9,7,5,4,4,2or11,9,7,5,4,3,3 | 85,912 | <think>Three, suppose there are $k$ teams participating in the competition, a total of $C_{k}^{2}=\frac{1}{2} k(k-1)$ matches are played, so the total score is $2 C_{k}^{2}=k(k-1)$.
The top 4 teams scored $11+9+7+5=32$ points, the last $k-4$ teams can score at most $4(k-4)$ points, thus we have
Therefore, $k=7$.
The r... | 1_153 |
In a triangle $ABC$ the area is $18$, the length $AB$ is $5$, and the medians from $A$ and $B$ are orthogonal. Find the lengths of the sides $BC,AC$. | <think>1. **Identify the given information and setup the problem:**
- The area of triangle \(ABC\) is \(18\).
- The length of side \(AB\) is \(5\).
- The medians from \(A\) and \(B\) are orthogonal.
2. **Use the properties of medians:**
- Let \(G\) be the centroid of the triangle, and \(D\), \(E\), and \(F... | BC = 2\sqrt{13} | 18,457 | <think>1. **Identify the given information and setup the problem:**
- The area of triangle \(ABC\) is \(18\).
- The length of side \(AB\) is \(5\).
- The medians from \(A\) and \(B\) are orthogonal.
2. **Use the properties of medians:**
- Let \(G\) be the centroid of the triangle, and \(D\), \(E\), and \(F... | 1_154 |
## Exercise 1
In the plane, we are given 2011 distinct points colored either blue or red.
(i) Suppose that for every blue point, the disk centered at this point with radius 1 contains exactly two red points. What is the maximum number of blue points possible?
(ii) Suppose that for every blue point, the circle center... | <think>## Solution.
(i) If there is a blue point, there must be at least two red points, so there cannot be more than 2009 blue points.
Conversely, if we consider 2009 disks of radius 1 having a common interior that is non-empty, it suffices to mark two red points in this common part and to mark in blue the centers o... | 2009 | 29,870 | <think>## Solution.
(i) If there is a blue point, there must be at least two red points, so there cannot be more than 2009 blue points.
Conversely, if we consider 2009 disks of radius 1 having a common interior that is non-empty, it suffices to mark two red points in this common part and to mark in blue the centers o... | 1_155 |
Determine all real numbers $x$ between $0$ and $180$ such that it is possible to partition an equilateral triangle into finitely many triangles, each of which has an angle of $x^{o}$.
| <think>1. **Show that \( x > 120^\circ \) is impossible:**
Suppose \( x > 120^\circ \) and such a partition into small triangles exists. Consider the set of vertices of the small triangles, which can be partitioned into 3 types:
- Let \( A \) be the 3 vertices of the original equilateral triangle.
- Let \( B ... | x \in (0, 120] | 39,234 | <think>1. **Show that \( x > 120^\circ \) is impossible:**
Suppose \( x > 120^\circ \) and such a partition into small triangles exists. Consider the set of vertices of the small triangles, which can be partitioned into 3 types:
- Let \( A \) be the 3 vertices of the original equilateral triangle.
- Let \( B ... | 1_156 |
Points $B$, $D$, and $J$ are midpoints of the sides of right triangle $ACG$. Points $K$, $E$, $I$ are midpoints of the sides of triangle $JDG$, etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$, then the total area of the shaded triangles is nearest
$\text{(A)}\ 6 \qq... | <think>Solution 1
Since $\triangle FGH$ is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram. All triangles are right-isosceles triangles.
$CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}$
$CB = CD = 3, DK = DE = \frac{3... | A | 77,798 | <think>Solution 1
Since $\triangle FGH$ is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram. All triangles are right-isosceles triangles.
$CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}$
$CB = CD = 3, DK = DE = \frac{3... | 1_157 |
Example 2. Consider the line integral
$$
\int_{L} \frac{-y d x}{x^{2}+y^{2}}+\frac{x d y}{x^{2}+y^{2}}
$$ | <think>Solution. The given expression makes sense everywhere except at the point $O(0,0)$; therefore, we exclude this point. In the remaining part of the plane (which will already be a disconnected region), the coefficients of $d x$ and $d y$ are continuous and have continuous partial derivatives, and they are identica... | 2\pi | 71,759 | <think>Solution. The given expression makes sense everywhere except at the point $O(0,0)$; therefore, we exclude this point. In the remaining part of the plane (which will already be a disconnected region), the coefficients of $d x$ and $d y$ are continuous and have continuous partial derivatives, and they are identica... | 1_158 |
9. A square has vertices at $(0,0),(1,0),(1,1)$ and $(0,1)$. Graphs of the following equations are drawn on the same set of axes as the square.
$$
x^{2}+y^{2}=1, \quad y=x+1, \quad y=-x^{2}+1, \quad y=x, \quad y=\frac{1}{x}
$$
How many of the graphs pass through exactly two of the vertices of the square?
A 1
B 2
C 3
D... | <think>SolUtion
C
Method 1
One method is to draw the graphs.
The graph of the equation $x^{2}+y^{2}=1$ is the circle with centre $(0,0)$ and radius 1. [By Pythagoras' Theorem, the distance, $d$ of the point with coordinates $(x, y)$ from the point $(0,0)$ is given by the equation $x^{2}+y^{2}=d^{2}$. Therefore the equa... | 3 | 76,737 | <think>SolUtion
C
Method 1
One method is to draw the graphs.
The graph of the equation $x^{2}+y^{2}=1$ is the circle with centre $(0,0)$ and radius 1. [By Pythagoras' Theorem, the distance, $d$ of the point with coordinates $(x, y)$ from the point $(0,0)$ is given by the equation $x^{2}+y^{2}=d^{2}$. Therefore the equa... | 1_159 |
174. $n^{2}$ integers from 1 to $n^{2}$ are written in a square table of size $n \times n$: in this case, the number 1 is placed at any position in the table; 2 belongs to the row whose ordinal number is equal to the column number containing 1; the number 3 belongs to the row whose number matches the column number cont... | <think>174. Let us agree to denote the number at the intersection of the $i$-th row and $j$-th column of the table by $a_{ij}$; in this case $a_{ij} = 1, \ldots, n^2$, where $i, j = 1, \ldots, n$. Let now $1 = a_{ji}$; then according to the problem's condition, $2 = a_{j_2 j_2}, \quad 3 = a_{j_3 j_4}$, and so on, up to... | n^2-n | 64,563 | <think>174. Let us agree to denote the number at the intersection of the $i$-th row and $j$-th column of the table by $a_{ij}$; in this case $a_{ij} = 1, \ldots, n^2$, where $i, j = 1, \ldots, n$. Let now $1 = a_{ji}$; then according to the problem's condition, $2 = a_{j_2 j_2}, \quad 3 = a_{j_3 j_4}$, and so on, up to... | 1_160 |
Example 3. Find the logarithmic residue of the function
$$
f(z)=\frac{1+z^{2}}{1-\cos 2 \pi z}
$$
with respect to the circle $|z|=\pi$. | <think>Solution. Setting $1+z^{2}=0$, we find two simple zeros of the function $f(z): a_{1}=-i, a_{2}=i$. Setting $1-\cos 2 \pi z=0$, we find the poles of the function $f(z): z_{n}=n, n=0, \pm 1, \pm 2, \ldots$. The multiplicity of the poles is $k=2$.
In the circle $|z|<\pi$, the function has two simple zeros $a_{1}=-... | -12 | 4,167 | <think>Solution. Setting $1+z^{2}=0$, we find two simple zeros of the function $f(z): a_{1}=-i, a_{2}=i$. Setting $1-\cos 2 \pi z=0$, we find the poles of the function $f(z): z_{n}=n, n=0, \pm 1, \pm 2, \ldots$. The multiplicity of the poles is $k=2$.
In the circle $|z|<\pi$, the function has two simple zeros $a_{1}=-... | 1_161 |
Find all functions $f:\mathbb{Z} \to \mathbb{Z}$ such that for all $m\in\mathbb{Z}$: [list][*] $f(m+8) \le f(m)+8$, [*] $f(m+11) \ge f(m)+11$.[/list] | <think>1. We start with the given conditions for the function \( f: \mathbb{Z} \to \mathbb{Z} \):
\[
f(m+8) \leq f(m) + 8
\]
\[
f(m+11) \geq f(m) + 11
\]
2. From the first condition, we have:
\[
f(m+8) \leq f(m) + 8
\]
Adding 3 to both sides, we get:
\[
f(m+8) + 3 \leq f(m) + 11
... | f(m) = m + a | 19,437 | <think>1. We start with the given conditions for the function \( f: \mathbb{Z} \to \mathbb{Z} \):
\[
f(m+8) \leq f(m) + 8
\]
\[
f(m+11) \geq f(m) + 11
\]
2. From the first condition, we have:
\[
f(m+8) \leq f(m) + 8
\]
Adding 3 to both sides, we get:
\[
f(m+8) + 3 \leq f(m) + 11
... | 1_162 |
Example 1.39 (Cai Jianxing) For non-negative real numbers $a, b, c$ where not two of them are zero simultaneously, determine the constant $k$ such that the following inequality always holds:
$$\sqrt{S+k} \leqslant \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}} \leqslant \sqrt{S+k+1}$$
where $\frac{a}{b... | <think>Proof: Let $a=b=c$, we have $2 \leqslant k \leqslant 3$, and let $a=0, b=c$, we have $1 \leqslant k \leqslant 2$, hence $k=2$. We first give two identities:
$$\begin{array}{c}
\sum(a+b)(a+c)=(a+b+c)^{2}+a b+b c+c a \\
\sum a(a+b)(a+c)=\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)+3 a b c
\end{array}$$
For the right sid... | 2 | 74,096 | <think>Proof: Let $a=b=c$, we have $2 \leqslant k \leqslant 3$, and let $a=0, b=c$, we have $1 \leqslant k \leqslant 2$, hence $k=2$. We first give two identities:
$$\begin{array}{c}
\sum(a+b)(a+c)=(a+b+c)^{2}+a b+b c+c a \\
\sum a(a+b)(a+c)=\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)+3 a b c
\end{array}$$
For the right sid... | 1_163 |
$23$ frat brothers are sitting in a circle. One, call him Alex, starts with a gallon of water. On the first turn, Alex gives each person in the circle some rational fraction of his water. On each subsequent turn, every person with water uses the same scheme as Alex did to distribute his water, but in relation to themse... | <think>1. **Understanding the Problem:**
- Alex starts with 1 gallon of water.
- On the first turn, Alex distributes the water to his neighbors and keeps some for himself.
- On each subsequent turn, every person with water distributes it in the same manner.
- After 23 turns, Alex has exactly 1 gallon of wat... | a = 0, b = 0, c = 1 | 26,070 | <think>1. **Understanding the Problem:**
- Alex starts with 1 gallon of water.
- On the first turn, Alex distributes the water to his neighbors and keeps some for himself.
- On each subsequent turn, every person with water distributes it in the same manner.
- After 23 turns, Alex has exactly 1 gallon of wat... | 1_164 |
Solve the following equation:
$$
\sqrt{4^{x}+\frac{17}{64}}-\sqrt{2 \cdot 4^{x}-\frac{7}{64}}=\sqrt{4^{x}-\frac{1}{16}}
$$ | <think>I. Solution: Square both sides of the equation:
$$
4^{x}+\frac{17}{64}-2 \sqrt{2 \cdot 4^{2 x}+\left(\frac{34}{64}-\frac{7}{64}\right) 4^{x}-\frac{7 \cdot 17}{64^{2}}}+2 \cdot 4^{x}-\frac{7}{64}=4^{x}-\frac{1}{16}
$$
After simplification and rearrangement:
$$
2 \cdot 4^{x}+\frac{14}{64}=2 \sqrt{2 \cdot 4^{2 x... | -\frac{3}{2} | 75,312 | <think>I. Solution: Square both sides of the equation:
$$
4^{x}+\frac{17}{64}-2 \sqrt{2 \cdot 4^{2 x}+\left(\frac{34}{64}-\frac{7}{64}\right) 4^{x}-\frac{7 \cdot 17}{64^{2}}}+2 \cdot 4^{x}-\frac{7}{64}=4^{x}-\frac{1}{16}
$$
After simplification and rearrangement:
$$
2 \cdot 4^{x}+\frac{14}{64}=2 \sqrt{2 \cdot 4^{2 x... | 1_165 |
When is the distance between the object and the real image produced by a convex lens the smallest? | <think>I. Solution. Speaking of a real image, the object and the image are on opposite sides of the lens. If the distances from the optical center of the lens are $t$ and $k$, then the quantity to be examined is
$$
y=t+k, \text { if } \frac{1}{t}+\frac{1}{k}=\frac{1}{f}
$$
From the latter law,
$$
k=\frac{t f}{t-f}
$... | 4f | 33,407 | <think>I. Solution. Speaking of a real image, the object and the image are on opposite sides of the lens. If the distances from the optical center of the lens are $t$ and $k$, then the quantity to be examined is
$$
y=t+k, \text { if } \frac{1}{t}+\frac{1}{k}=\frac{1}{f}
$$
From the latter law,
$$
k=\frac{t f}{t-f}
$... | 1_166 |
Example 1. Find the approximate value of the smallest characteristic number of the kernel by the Ritz method
$$
K(x, t)=x t ; \quad a=0, b=1
$$ | <think>Solution. As the coordinate system of functions $\psi_{n}(x)$, we choose the system of Legendre polynomials: $\psi_{n}(x)=P_{n}(2 x-1)$. In formula (1), we limit ourselves to two terms, so that
$$
\varphi_{2}(x)=a_{1} \cdot P_{0}(2 x-1)+a_{2} \cdot P_{1}(2 x-1) .
$$
Noting that
$$
\psi_{1} \equiv P_{0}(2 x-1)... | 3 | 36,139 | <think>Solution. As the coordinate system of functions $\psi_{n}(x)$, we choose the system of Legendre polynomials: $\psi_{n}(x)=P_{n}(2 x-1)$. In formula (1), we limit ourselves to two terms, so that
$$
\varphi_{2}(x)=a_{1} \cdot P_{0}(2 x-1)+a_{2} \cdot P_{1}(2 x-1) .
$$
Noting that
$$
\psi_{1} \equiv P_{0}(2 x-1)... | 1_167 |
257. a) How many roots does the equation
$$
\sin x=\frac{x}{100} ?
$$
b) How many roots does the equation
$$
\sin x=\lg x ?
$$ | <think>257. a) First of all, note that if $x_{0}$ is a root of the equation, then $-x_{0}$ is also a root. Therefore, the number of negative roots is the same as the number of positive roots. Furthermore, the number 0 is a root of the equation. Thus, it is sufficient to find the number of positive roots. Now, let's not... | 63 | 79,006 | <think>257. a) First of all, note that if $x_{0}$ is a root of the equation, then $-x_{0}$ is also a root. Therefore, the number of negative roots is the same as the number of positive roots. Furthermore, the number 0 is a root of the equation. Thus, it is sufficient to find the number of positive roots. Now, let's not... | 1_168 |
Suppose that points $A$ and $B$ lie on circle $\Omega$, and suppose that points $C$ and $D$ are the trisection points of major arc $AB$, with $C$ closer to $B$ than $A$. Let $E$ be the intersection of line $AB$ with the line tangent to $\Omega$ at $C$. Suppose that $DC = 8$ and $DB = 11$. If $DE = a\sqrt{b}$ for intege... | <think>1. **Identify the given points and their relationships:**
- Points \( A \) and \( B \) lie on circle \( \Omega \).
- Points \( C \) and \( D \) are the trisection points of the major arc \( AB \), with \( C \) closer to \( B \) than \( A \).
- \( E \) is the intersection of line \( AB \) with the line t... | 37 | 35,410 | <think>1. **Identify the given points and their relationships:**
- Points \( A \) and \( B \) lie on circle \( \Omega \).
- Points \( C \) and \( D \) are the trisection points of the major arc \( AB \), with \( C \) closer to \( B \) than \( A \).
- \( E \) is the intersection of line \( AB \) with the line t... | 1_169 |
11.195. The radius of the base of the cone is $R$, and the angle of the sector of its lateral surface is $90^{\circ}$. Determine the volume of the cone. | <think>## Solution.
Let $l$ be the slant height of the cone. Since the length of the arc of the lateral surface development of the cone is equal to the circumference of the base, we have $2 \pi R = \frac{2 \pi l}{4}$ (by the condition that the development is a quarter circle), from which $l = 4R$.
Next, we find the h... | \frac{\piR^3\sqrt{15}}{3} | 65,909 | <think>## Solution.
Let $l$ be the slant height of the cone. Since the length of the arc of the lateral surface development of the cone is equal to the circumference of the base, we have $2 \pi R = \frac{2 \pi l}{4}$ (by the condition that the development is a quarter circle), from which $l = 4R$.
Next, we find the h... | 1_170 |
Compute the $100^{\text{th}}$ smallest positive integer $n$ that satisfies the three congruences \[\begin{aligned} \left\lfloor \dfrac{n}{8} \right\rfloor &\equiv 3 \pmod{4}, \\ \left\lfloor \dfrac{n}{32} \right\rfloor &\equiv 2 \pmod{4}, \\ \left\lfloor \dfrac{n}{256} \right\rfloor &\equiv 1 \pmod{4}. \end{aligned}\] ... | <think>To solve the problem, we need to find the 100th smallest positive integer \( n \) that satisfies the following three congruences:
\[
\begin{aligned}
\left\lfloor \frac{n}{8} \right\rfloor &\equiv 3 \pmod{4}, \\
\left\lfloor \frac{n}{32} \right\rfloor &\equiv 2 \pmod{4}, \\
\left\lfloor \frac{n}{256} \right\rfloo... | 6491 | 89,710 | <think>To solve the problem, we need to find the 100th smallest positive integer \( n \) that satisfies the following three congruences:
\[
\begin{aligned}
\left\lfloor \frac{n}{8} \right\rfloor &\equiv 3 \pmod{4}, \\
\left\lfloor \frac{n}{32} \right\rfloor &\equiv 2 \pmod{4}, \\
\left\lfloor \frac{n}{256} \right\rfloo... | 1_171 |
In the fraction below and its decimal notation (with period of length $ 4$) every letter represents a digit, and different letters denote different digits. The numerator and denominator are coprime. Determine the value of the fraction:
$ \frac{ADA}{KOK}\equal{}0.SNELSNELSNELSNEL...$
$ Note.$ Ada Kok is a famous ... | <think>1. Given the fraction $\frac{ADA}{KOK} = 0.SNELSNELSNEL...$, we know that the decimal representation is periodic with a period of 4 digits. This implies that:
\[
\frac{ADA}{KOK} = \frac{SNEL}{9999}
\]
because $0.SNELSNELSNEL... = \frac{SNEL}{9999}$, where $9999 = 10^4 - 1$.
2. Since the numerator an... | \frac{242}{303} = 0.798679867986... | 59,898 | <think>1. Given the fraction $\frac{ADA}{KOK} = 0.SNELSNELSNEL...$, we know that the decimal representation is periodic with a period of 4 digits. This implies that:
\[
\frac{ADA}{KOK} = \frac{SNEL}{9999}
\]
because $0.SNELSNELSNEL... = \frac{SNEL}{9999}$, where $9999 = 10^4 - 1$.
2. Since the numerator an... | 1_172 |
10.86 Try to divide the seven digits $3,4,5,6,7,8,9$ into two groups, and arrange them into a three-digit number and a four-digit number, respectively, such that the product of these two numbers is maximized. How should the groups be divided and arranged, and prove your conclusion.
(China Beijing Junior High School Gra... | <think>[Solution] When forming each number, the larger digit should be placed at the highest position. Thus, 9 and 8 should be the leading digits of the two numbers, respectively.
Now, select one digit from the remaining digits to add to 9 and 8 to form two two-digit numbers, and make their product as large as possibl... | 964\cdot8753 | 16,538 | <think>[Solution] When forming each number, the larger digit should be placed at the highest position. Thus, 9 and 8 should be the leading digits of the two numbers, respectively.
Now, select one digit from the remaining digits to add to 9 and 8 to form two two-digit numbers, and make their product as large as possibl... | 1_173 |
Find all real numbers $a$ for which the equation $x^2a- 2x + 1 = 3 |x|$ has exactly three distinct real solutions in $x$. | <think>To find all real numbers \( a \) for which the equation \( x^2a - 2x + 1 = 3 |x| \) has exactly three distinct real solutions in \( x \), we will analyze the equation by considering the cases \( x = 0 \), \( x > 0 \), and \( x < 0 \).
1. **Case 1: \( x = 0 \)**
\[
a \cdot 0^2 - 2 \cdot 0 + 1 = 3 |0|
\]... | \frac{1}{4} | 81,506 | <think>To find all real numbers \( a \) for which the equation \( x^2a - 2x + 1 = 3 |x| \) has exactly three distinct real solutions in \( x \), we will analyze the equation by considering the cases \( x = 0 \), \( x > 0 \), and \( x < 0 \).
1. **Case 1: \( x = 0 \)**
\[
a \cdot 0^2 - 2 \cdot 0 + 1 = 3 |0|
\]... | 1_174 |
57. Find the sum of the squares of the first $n$ numbers of the natural series.
## Heron's Problem. From the treatise
Find the sum of the squares of the first $n$ numbers of the natural series. | <think>57. Archimedes solved this problem in the following form.
It goes as follows: if the excess is equal to the smallest of all, and if other lines, each equal to the largest of the line... | 1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2n+1)}{6} | 44,037 | <think>57. Archimedes solved this problem in the following form.
It goes as follows: if the excess is equal to the smallest of all, and if other lines, each equal to the largest of the line... | 1_175 |
Given a positive integer $n>1$. In the cells of an $n\times n$ board, marbles are placed one by one. Initially there are no marbles on the board. A marble could be placed in a free cell neighboring (by side) with at least two cells which are still free. Find the greatest possible number of marbles that could be placed ... | <think>1. **Initial Setup**:
- We start with an \( n \times n \) board where \( n > 1 \).
- Initially, there are no marbles on the board.
2. **Counting Neighboring Pairs**:
- Each cell on the board has up to 4 neighboring cells (up, down, left, right).
- We need to count the number of pairs of neighboring ... | n^2 - n | 51,041 | <think>1. **Initial Setup**:
- We start with an \( n \times n \) board where \( n > 1 \).
- Initially, there are no marbles on the board.
2. **Counting Neighboring Pairs**:
- Each cell on the board has up to 4 neighboring cells (up, down, left, right).
- We need to count the number of pairs of neighboring ... | 1_176 |
4. Determine all ordered pairs $(x, y)$ of nonnegative integers that satisfy the equation
$$
3 x^{2}+2 \cdot 9^{y}=x\left(4^{y+1}-1\right) .
$$
This is a modified version of Problem N2 from the 2010 IMO shortlist. | <think>Solution: The equation is equivalent to
$$
(3 x)^{2}+2 \cdot 3^{2 y+1}=3 x\left[2^{(2 y+1)+1}-1\right] .
$$
Letting $a=3 x$ and $b=2 y+1$, we have
$$
a^{2}+2 \cdot 3^{b}=a\left(2^{b+1}-1\right) \text {. }
$$
Case 1: $b=1$
We have $a^{2}-3 a+6=0$ which has no integer solution for $a$. Thus, there is no solution... | (2,1),(3,1),(3,2),(18,2) | 31,226 | <think>Solution: The equation is equivalent to
$$
(3 x)^{2}+2 \cdot 3^{2 y+1}=3 x\left[2^{(2 y+1)+1}-1\right] .
$$
Letting $a=3 x$ and $b=2 y+1$, we have
$$
a^{2}+2 \cdot 3^{b}=a\left(2^{b+1}-1\right) \text {. }
$$
Case 1: $b=1$
We have $a^{2}-3 a+6=0$ which has no integer solution for $a$. Thus, there is no solution... | 1_177 |
In $\triangle ABC$, $AB= 425$, $BC=450$, and $AC=510$. An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle. If these three segments are of an equal length $d$, find $d$. | <think>1. **Identify the given elements and the goal:**
- We are given a triangle \( \triangle ABC \) with sides \( AB = 425 \), \( BC = 450 \), and \( AC = 510 \).
- An interior point \( P \) is drawn, and segments through \( P \) parallel to the sides of the triangle are of equal length \( d \).
- We need to... | 306 | 65,176 | <think>1. **Identify the given elements and the goal:**
- We are given a triangle \( \triangle ABC \) with sides \( AB = 425 \), \( BC = 450 \), and \( AC = 510 \).
- An interior point \( P \) is drawn, and segments through \( P \) parallel to the sides of the triangle are of equal length \( d \).
- We need to... | 1_178 |
[b](i)[/b] Determine the smallest number of edges which a graph of $ n$ nodes may have given that adding an arbitrary new edge would give rise to a 3-clique (3 nodes joined pairwise by edges).
[b](ii)[/b] Determine the smallest number of edges which a graph of $ n$ nodes may have given that adding an arbitrary new edg... | <think>**(i)** Determine the smallest number of edges which a graph of \( n \) nodes may have given that adding an arbitrary new edge would give rise to a 3-clique (3 nodes joined pairwise by edges).
1. **Understanding the Problem:**
- We need to find the smallest number of edges in a graph with \( n \) vertices su... | n-1 | 19,374 | <think>**(i)** Determine the smallest number of edges which a graph of \( n \) nodes may have given that adding an arbitrary new edge would give rise to a 3-clique (3 nodes joined pairwise by edges).
1. **Understanding the Problem:**
- We need to find the smallest number of edges in a graph with \( n \) vertices su... | 1_179 |
Bogdanov I.I.
In a store, 21 white and 21 purple shirts are hanging in a row. Find the smallest $k$ such that for any initial order of the shirts, it is possible to remove $k$ white and $k$ purple shirts so that the remaining white shirts hang together and the remaining purple shirts also hang together. | <think>First, let's show that \( k \), equal to 10, is sufficient.
First method. We will walk along the row of shirts and count the white and purple shirts separately. As soon as we count 11 of one color - let's assume, without loss of generality, purple - shirts, we will stop. Now we will remove all the white shirts ... | 10 | 29,927 | <think>First, let's show that \( k \), equal to 10, is sufficient.
First method. We will walk along the row of shirts and count the white and purple shirts separately. As soon as we count 11 of one color - let's assume, without loss of generality, purple - shirts, we will stop. Now we will remove all the white shirts ... | 1_180 |
The sequence $(x_n)$ is determined by the conditions:
$x_0=1992,x_n=-\frac{1992}{n} \cdot \sum_{k=0}^{n-1} x_k$ for $n \geq 1$.
Find $\sum_{n=0}^{1992} 2^nx_n$. | <think>1. **Initial Conditions and Recurrence Relation:**
The sequence \((x_n)\) is defined by:
\[
x_0 = 1992
\]
and for \(n \geq 1\),
\[
x_n = -\frac{1992}{n} \sum_{k=0}^{n-1} x_k
\]
2. **Simplifying the Recurrence Relation:**
We start by examining the recurrence relation. For \(n \geq 1\),... | 1992 | 87,231 | <think>1. **Initial Conditions and Recurrence Relation:**
The sequence \((x_n)\) is defined by:
\[
x_0 = 1992
\]
and for \(n \geq 1\),
\[
x_n = -\frac{1992}{n} \sum_{k=0}^{n-1} x_k
\]
2. **Simplifying the Recurrence Relation:**
We start by examining the recurrence relation. For \(n \geq 1\),... | 1_181 |
Determine all functions $ f$ defined in the set of rational numbers and taking their values in the same set such that the equation $ f(x + y) + f(x - y) = 2f(x) + 2f(y)$ holds for all rational numbers $x$ and $y$. | <think>1. **Define the assertion**: Let \( P(x, y) \) be the assertion \( f(x + y) + f(x - y) = 2f(x) + 2f(y) \).
2. **Evaluate \( P(0, 0) \)**:
\[
P(0, 0) \implies f(0 + 0) + f(0 - 0) = 2f(0) + 2f(0) \implies 2f(0) = 4f(0) \implies f(0) = 0
\]
3. **Evaluate \( P(x, 0) \)**:
\[
P(x, 0) \implies f(x + 0... | f(x) = ax^2 \quad \forall x \in \mathbb{Q} | 54,292 | <think>1. **Define the assertion**: Let \( P(x, y) \) be the assertion \( f(x + y) + f(x - y) = 2f(x) + 2f(y) \).
2. **Evaluate \( P(0, 0) \)**:
\[
P(0, 0) \implies f(0 + 0) + f(0 - 0) = 2f(0) + 2f(0) \implies 2f(0) = 4f(0) \implies f(0) = 0
\]
3. **Evaluate \( P(x, 0) \)**:
\[
P(x, 0) \implies f(x + 0... | 1_182 |
p1. Determine the smallest natural number that has the property that it's cube ends in $888$.
p2. Triangle $ABC$ is isosceles with $AC = BC$. The angle bisector at $A$ intercepts side $BC$ at the point $D$ and the bisector of the angle at $C$ intercepts side $AB$ at $E$. If $AD = 2CE$, find the measure of the angles ... | <think>1. **Determine the smallest natural number that has the property that its cube ends in $888$.**
To solve this problem, we need to find the smallest natural number \( n \) such that \( n^3 \equiv 888 \pmod{1000} \).
First, note that \( n^3 \equiv 888 \pmod{1000} \) implies that the last three digits of \( n^3 \... | 192 | 14,605 | <think>1. **Determine the smallest natural number that has the property that its cube ends in $888$.**
To solve this problem, we need to find the smallest natural number \( n \) such that \( n^3 \equiv 888 \pmod{1000} \).
First, note that \( n^3 \equiv 888 \pmod{1000} \) implies that the last three digits of \( n^3 \... | 1_183 |
There are $27$ unit cubes. We are marking one point on each of the two opposing faces, two points on each of the other two opposing faces, and three points on each of the remaining two opposing faces of each cube. We are constructing a $3\times 3 \times 3$ cube with these $27$ cubes. What is the least number of marked ... | <think>1. **Understanding the Problem:**
We have 27 unit cubes, each with points marked on their faces. We need to construct a \(3 \times 3 \times 3\) cube using these unit cubes and determine the least number of marked points visible on the faces of the new cube.
2. **Classifying the Unit Cubes:**
- **Corner Cu... | 90 | 92,824 | <think>1. **Understanding the Problem:**
We have 27 unit cubes, each with points marked on their faces. We need to construct a \(3 \times 3 \times 3\) cube using these unit cubes and determine the least number of marked points visible on the faces of the new cube.
2. **Classifying the Unit Cubes:**
- **Corner Cu... | 1_184 |
12 Let $x_{1}, x_{2}, \cdots, x_{n}$ all be no less than 0, and $\sum_{i=1}^{n} x_{i}=1$, find the maximum value of the sum
$$\sum_{1 \leq i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$$ | <think>$$\begin{array}{l}
\text { 12. } \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)=\frac{1}{2} \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)+\frac{1}{2} \sum_{1 \leqslant i<j \leqslant n} x_{j} x_{i}\left(x_{j}+\right. \\
\left.x_{i}\right)=\frac{1}{2} \sum_{i \neq j... | \frac{1}{4} | 21,543 | <think>$$\begin{array}{l}
\text { 12. } \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)=\frac{1}{2} \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)+\frac{1}{2} \sum_{1 \leqslant i<j \leqslant n} x_{j} x_{i}\left(x_{j}+\right. \\
\left.x_{i}\right)=\frac{1}{2} \sum_{i \neq j... | 1_185 |
Find the least positive integer $n$ such that \[ \frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}. \] | <think>1. We start with the given equation:
\[
\frac{1}{\sin 45^\circ \sin 46^\circ} + \frac{1}{\sin 47^\circ \sin 48^\circ} + \cdots + \frac{1}{\sin 133^\circ \sin 134^\circ} = \frac{1}{\sin n^\circ}
\]
2. Notice that \(\sin(180^\circ - x) = \sin x\). Therefore, we can rewrite the sines of angles greater tha... | 1 | 27,490 | <think>1. We start with the given equation:
\[
\frac{1}{\sin 45^\circ \sin 46^\circ} + \frac{1}{\sin 47^\circ \sin 48^\circ} + \cdots + \frac{1}{\sin 133^\circ \sin 134^\circ} = \frac{1}{\sin n^\circ}
\]
2. Notice that \(\sin(180^\circ - x) = \sin x\). Therefore, we can rewrite the sines of angles greater tha... | 1_186 |
In a right-angled triangle, the radius of the circumscribed circle is $14.5 \mathrm{~cm}$ and the radius of the inscribed circle is $6 \mathrm{~cm}$.
Determine the perimeter of this triangle.
(L. Hozová) | <think>The center of the circumcircle of a right-angled triangle is at the midpoint of its hypotenuse; this point is marked as $O$ in the diagram. The center of the incircle is the common intersection of the angle bisectors of the triangle; this point is marked as $V$, and the points of tangency of the circle with the ... | 70\, | 47,016 | <think>The center of the circumcircle of a right-angled triangle is at the midpoint of its hypotenuse; this point is marked as $O$ in the diagram. The center of the incircle is the common intersection of the angle bisectors of the triangle; this point is marked as $V$, and the points of tangency of the circle with the ... | 1_187 |
4. Consider an $n \times n$ matrix, where each element is a real number with absolute value not exceeding 1, and the sum of all elements is 0. Let $n$ be a positive even number. Find the minimum value of $c$ such that for every such matrix, there exists a row or a column whose sum of elements has an absolute value not ... | <think>4. The minimum value of $c$ is $\frac{n}{2}$.
Define the matrix $\left(a_{i j}\right)_{1 \leqslant i, j \leqslant n}$ as follows:
$$
a_{i j}=\left\{\begin{array}{ll}
1, & i, j \leqslant \frac{n}{2} ; \\
-1, & i, j>\frac{n}{2} ; \\
0, & \text { otherwise. }
\end{array}\right.
$$
Then the sum of all elements is ... | \frac{n}{2} | 12,954 | <think>4. The minimum value of $c$ is $\frac{n}{2}$.
Define the matrix $\left(a_{i j}\right)_{1 \leqslant i, j \leqslant n}$ as follows:
$$
a_{i j}=\left\{\begin{array}{ll}
1, & i, j \leqslant \frac{n}{2} ; \\
-1, & i, j>\frac{n}{2} ; \\
0, & \text { otherwise. }
\end{array}\right.
$$
Then the sum of all elements is ... | 1_188 |
Let $f_1(x) = \frac{2}{3}-\frac{3}{3x+1}$, and for $n \ge 2$, define $f_n(x) = f_1(f_{n-1} (x))$. The value of x that satisfies $f_{1001}(x) = x - 3$ can be expressed in the form $\frac{m}{n}$,
where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | <think>1. **Define the function \( f_1(x) \):**
\[
f_1(x) = \frac{2}{3} - \frac{3}{3x+1}
\]
2. **Calculate \( f_2(x) \):**
\[
f_2(x) = f_1(f_1(x))
\]
Substitute \( f_1(x) \) into itself:
\[
f_2(x) = \frac{2}{3} - \frac{3}{3\left(\frac{2}{3} - \frac{3}{3x+1}\right) + 1}
\]
Simplify the ... | 8 | 66,196 | <think>1. **Define the function \( f_1(x) \):**
\[
f_1(x) = \frac{2}{3} - \frac{3}{3x+1}
\]
2. **Calculate \( f_2(x) \):**
\[
f_2(x) = f_1(f_1(x))
\]
Substitute \( f_1(x) \) into itself:
\[
f_2(x) = \frac{2}{3} - \frac{3}{3\left(\frac{2}{3} - \frac{3}{3x+1}\right) + 1}
\]
Simplify the ... | 1_189 |
5. (2002 Japan Mathematical Olympiad) 14 people participate in a Japanese chess round-robin tournament, where each person plays against the other 13 people. There are no ties in the matches. Find the maximum number of "triangular ties" (here, a "triangular tie" refers to a situation where 3 people each have one win and... | <think>5. Represent 14 people as $A_{1}, A_{2}, \cdots, A_{14}$, and let $A_{i}$ win $a_{i}$ games. If there is no "triangle" among any 3 people, then there must be one person who wins the other two, hence no "triangle" trio. The number of such trios is $\sum_{i=1}^{14} C_{a_{i}}^{2}$ (with the convention that $C_{0}^{... | 112 | 59,140 | <think>5. Represent 14 people as $A_{1}, A_{2}, \cdots, A_{14}$, and let $A_{i}$ win $a_{i}$ games. If there is no "triangle" among any 3 people, then there must be one person who wins the other two, hence no "triangle" trio. The number of such trios is $\sum_{i=1}^{14} C_{a_{i}}^{2}$ (with the convention that $C_{0}^{... | 1_190 |
Let $n\ge 4$ be a positive integer and let $M$ be a set of $n$ points in the plane, where no three points are collinear and not all of the $n$ points being concyclic. Find all real functions $f:M\to\mathbb{R}$ such that for any circle $\mathcal{C}$ containing at least three points from $M$, the following equality holds... | <think>1. **Define Interesting Circles:**
Call the circles which contain at least three of the points from \( M \) interesting. The problem's conditions tell us that every triplet of the points belongs to exactly one interesting circle.
2. **Lemma:**
\[
\sum_{P \in M} f(P) = 0
\]
**Proof:**
For a tr... | f(P) = 0 | 57,381 | <think>1. **Define Interesting Circles:**
Call the circles which contain at least three of the points from \( M \) interesting. The problem's conditions tell us that every triplet of the points belongs to exactly one interesting circle.
2. **Lemma:**
\[
\sum_{P \in M} f(P) = 0
\]
**Proof:**
For a tr... | 1_191 |
Calculate the exact value of the series $\sum _{n=2} ^\infty \log (n^3 +1) - \log (n^3 - 1)$ and provide justification. | <think>1. We start with the given series:
\[
\sum_{n=2}^\infty \left( \log(n^3 + 1) - \log(n^3 - 1) \right)
\]
Using the properties of logarithms, we can combine the terms inside the logarithm:
\[
\log(n^3 + 1) - \log(n^3 - 1) = \log \left( \frac{n^3 + 1}{n^3 - 1} \right)
\]
Thus, the series bec... | \log \left( \frac{3}{2} \right) | 28,386 | <think>1. We start with the given series:
\[
\sum_{n=2}^\infty \left( \log(n^3 + 1) - \log(n^3 - 1) \right)
\]
Using the properties of logarithms, we can combine the terms inside the logarithm:
\[
\log(n^3 + 1) - \log(n^3 - 1) = \log \left( \frac{n^3 + 1}{n^3 - 1} \right)
\]
Thus, the series bec... | 1_192 |
Let $[ABCD]$ be a convex quadrilateral with area $2014$, and let $P$ be a point on $[AB]$ and $Q$ a point on $[AD]$ such that triangles $[ABQ]$ and $[ADP]$ have area $1$. Let $R$ be the intersection of $[AC]$ and $[PQ]$. Determine $\frac{\overline{RC}}{\overline{RA}}$. | <think>1. **Identify the given information and set up the problem:**
- We have a convex quadrilateral \(ABCD\) with area \(2014\).
- Points \(P\) and \(Q\) are on segments \([AB]\) and \([AD]\) respectively, such that the areas of triangles \([ABQ]\) and \([ADP]\) are both \(1\).
- We need to determine the rat... | 2013 | 82,062 | <think>1. **Identify the given information and set up the problem:**
- We have a convex quadrilateral \(ABCD\) with area \(2014\).
- Points \(P\) and \(Q\) are on segments \([AB]\) and \([AD]\) respectively, such that the areas of triangles \([ABQ]\) and \([ADP]\) are both \(1\).
- We need to determine the rat... | 1_193 |
Petya and 9 other people are playing a game: each of them rolls a die. A player wins a prize if they roll a number that no one else rolls.
a) What is the probability that Petya will win a prize?
b) What is the probability that at least someone will win a prize? | <think>a) Suppose (for illustrative purposes only) that Petya throws first. He scores some number of points. The probability that each subsequent player will score a different number of points is 5/6.
Since the throws are independent, the probability that the nine remaining players will all score differently from Pety... | )(5/6)^{9}\approx0.194;b)\approx0.919 | 45,619 | <think>a) Suppose (for illustrative purposes only) that Petya throws first. He scores some number of points. The probability that each subsequent player will score a different number of points is 5/6.
Since the throws are independent, the probability that the nine remaining players will all score differently from Pety... | 1_194 |
Let $T_1$ and $T_2$ be the points of tangency of the excircles of a triangle $ABC$ with its sides $BC$ and $AC$ respectively. It is known that the reflection of the incenter of $ABC$ across the midpoint of $AB$ lies on the circumcircle of triangle $CT_1T_2$. Find $\angle BCA$. | <think>1. **Define the coordinates and parameters:**
- Let \( A = (1, 0, 0) \), \( B = (0, 1, 0) \), \( C = (0, 0, 1) \).
- Let \( a = BC \), \( b = CA \), \( c = AB \), and let \( s \) be the semiperimeter of \(\triangle ABC\).
- The incenter \( I \) of \(\triangle ABC\) has coordinates \( I = (a : b : c) \).... | 90^\circ | 84,865 | <think>1. **Define the coordinates and parameters:**
- Let \( A = (1, 0, 0) \), \( B = (0, 1, 0) \), \( C = (0, 0, 1) \).
- Let \( a = BC \), \( b = CA \), \( c = AB \), and let \( s \) be the semiperimeter of \(\triangle ABC\).
- The incenter \( I \) of \(\triangle ABC\) has coordinates \( I = (a : b : c) \).... | 1_195 |
\section*{Exercise 2 - 031212}
During the calibration of a dynamometer, the load \(P\) required to deflect the pointer to certain divisions on the scale was measured. The following values were obtained:
\begin{tabular}{cc}
Number of divisions & Load in kp \\
\(N\) & \(P\) \\
\hline 0 & 0 \\
5 & 4.87 \\
10 & 10.52 \\... | <think>}
a) It can be observed that the polynomial relationship set between the number of segments \(N\) and the load \(P(N)\) simplifies when one takes the differences between successive values in the value table. Specifically, the polynomial degree and thus the number of undetermined parameters are reduced by one.
T... | 35.15 | 24,224 | <think>}
a) It can be observed that the polynomial relationship set between the number of segments \(N\) and the load \(P(N)\) simplifies when one takes the differences between successive values in the value table. Specifically, the polynomial degree and thus the number of undetermined parameters are reduced by one.
T... | 1_196 |
II. (50 points $\}$
$a, b, c \in \mathbf{R}$. Satisfy $|a|>1,|b|>1,|c|>1$ and $b=\frac{a^{2}}{2-a^{2}}, c=\frac{b^{2}}{2-b^{2}}, a=$ $\frac{c^{2}}{2-c^{2}}$. Find all possible values of $a+b+c$. | <think>II. (50 points)
Make a trigonometric substitution. Since $|a|>1$, we can set
$$
a=\frac{1}{\cos \theta}(\theta \in(0,2 \pi)) \text {. }
$$
Then $b=\frac{a^{2}}{2-a^{2}}=\frac{\frac{1}{\cos ^{2} \theta}}{2-\frac{1}{\cos ^{2} \theta}}=\frac{1}{\cos ^{2} \theta}$,
Similarly, $c=\frac{1}{\cos ^{4} \theta}, a=\frac{... | 6,-4,-6 | 3,763 | <think>II. (50 points)
Make a trigonometric substitution. Since $|a|>1$, we can set
$$
a=\frac{1}{\cos \theta}(\theta \in(0,2 \pi)) \text {. }
$$
Then $b=\frac{a^{2}}{2-a^{2}}=\frac{\frac{1}{\cos ^{2} \theta}}{2-\frac{1}{\cos ^{2} \theta}}=\frac{1}{\cos ^{2} \theta}$,
Similarly, $c=\frac{1}{\cos ^{4} \theta}, a=\frac{... | 1_197 |
8. There are $n$ mathematicians attending a conference, and each mathematician has exactly 3 friends (friendship is mutual). Now, if we arrange some of the mathematicians to sit around a round table to discuss problems, it turns out that if everyone is seated next to their friends, then the number of people at the tabl... | <think>Analysis: Consider mathematicians as vertices, and if two mathematicians are friends, connect an edge between the corresponding vertices, thus obtaining a simple graph $G$. According to the problem, graph $G$ has $n$ vertices, each vertex has a degree of 3, and the length of each cycle $\geq 7$. Since each verte... | 24 | 90,426 | <think>Analysis: Consider mathematicians as vertices, and if two mathematicians are friends, connect an edge between the corresponding vertices, thus obtaining a simple graph $G$. According to the problem, graph $G$ has $n$ vertices, each vertex has a degree of 3, and the length of each cycle $\geq 7$. Since each verte... | 1_198 |
9.3. Given a triangle $A B C$. A line parallel to $A C$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively, and the median $A M$ at point $Q$. It is known that $P Q=3$, and $Q T=5$. Find the length of $A C$. | <think>Answer: $AC=11$.
Solution. First method. Draw a line through point $Q$ parallel to $BC$ (where $N$ and $L$ are the points of intersection of this line with sides $AB$ and $AC$ respectively, see Fig. 9.3a). Since $AM$ is the median of triangle $ABC$, then $LQ=NQ$. Moreover, $PT \| AC$, so $PQ$ is the midline of ... | 11 | 91,540 | <think>Answer: $AC=11$.
Solution. First method. Draw a line through point $Q$ parallel to $BC$ (where $N$ and $L$ are the points of intersection of this line with sides $AB$ and $AC$ respectively, see Fig. 9.3a). Since $AM$ is the median of triangle $ABC$, then $LQ=NQ$. Moreover, $PT \| AC$, so $PQ$ is the midline of ... | 1_199 |
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