problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4. Let $\left\{a_{n}\right\}$ be a convex sequence. Prove that for odd $n$ greater than 1,
$$
\sum_{i=0}^{n} \frac{a_{i}}{\mathrm{C}_{n}^{i}} \geqslant \frac{1}{n+1}\left(\sum_{i=0}^{n} a_{i}\right)\left(\sum_{i=0}^{n} \frac{1}{\mathrm{C}_{n}^{i}}\right) .
$$ | Prompt: As in Example 6, use Chebyshev's inequality.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | proof | Yes | Yes | cn_contest | false | 722,729 |
4. Given that $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a+b+c=0, \\
a b c=1 .
\end{array}\right.
$$
Prove: $a, b, c$ contain at least one number greater than $\frac{3}{2}$. | Hint: As in Example 7, we can set \(a=-\frac{c}{2}+t, b=-\frac{c}{2}-t\). | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,730 |
5. The system of equations
$$
\left\{\begin{array}{l}
\sqrt{x+\frac{1}{y}}-\sqrt{x+y-3}=\sqrt{3}, \\
2 x+y+\frac{1}{y}=6
\end{array}\right.
$$
has the solution set $\qquad$ | The answer: $(x, y)=(2,1)$ or $(4,-1)$. | (x, y)=(2,1) \text{ or } (4,-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,731 |
Example 1 (E. J. Nanson Inequality) Let the sequence $\left\{a_{n}\right\}$ be a convex sequence. Prove: For any $n \in \mathbf{N}_{+}$, we have
$$
\frac{\sum_{i=0}^{n} a_{2 i+1}}{n+1} \geqslant \frac{\sum_{i=1}^{n} a_{2 i}}{n} .
$$ | Note that $a_{2 i+1}$ and $a_{2 i}$ are the odd and even terms of the sequence $\left\{a_{n}\right\}$, respectively. We can start from the definition of a convex sequence.
By the convexity of the sequence $\left\{a_{n}\right\}$, we have
$a_{2 i-1}+a_{2 i+1} \geqslant 2 a_{2 i}(i=1,2, \cdots)$.
Thus, $\sum_{i=1}^{n} a_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,733 |
Example 2 Let $\left\{a_{k}\right\}$ be a convex sequence, $0 \leqslant k \leqslant n, n \geqslant 2$.
Prove:
(1) If $a_{0}=a_{n}=0$, then there are no positive numbers in $a_{0}, a_{1}, \cdots, a_{n}$;
(18th Polish Mathematical Competition)
(2) If there exists $m(1 \leqslant m<n)$, such that $a_{0} \leqslant a_{m}$, $... | Prove (1) For $0 \leqslant i \leqslant n$, by Corollary 1 we get
$$
a_{i} \leqslant\left(1-\frac{i}{n}\right) a_{0}+\frac{i}{n} a_{n}=0 .
$$
Therefore, there are no positive numbers in the sequence $a_{0}, a_{1}, \cdots, a_{n}$.
(2) On one hand, for $0 \leqslant i \leqslant n$, by Corollary 2 we get
$$
a_{i} \leqslant... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,734 |
Example 3 Let $\left\{a_{k}\right\}_{k=1}^{\infty}$ be a sequence of non-negative real numbers such that for $k=1,2, \cdots$, we have
$$
a_{k}-2 a_{k+1}+a_{k+2} \geqslant 0, \sum_{j=1}^{k} a_{j} \leqslant 1 .
$$
Prove: $0 \leqslant a_{k}-a_{k+1}<\frac{2}{k^{2}}$. | To prove the left inequality of (1).
By property 2(1) and the given conditions, for any $n, k \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
a_{n} \geqslant a_{k}+(n-k)\left(a_{k+1}-a_{k}\right) \\
\geqslant(n-k)\left(a_{k+1}-a_{k}\right).
\end{array}
$$
Thus, when $n > k$,
$$
a_{k+1}-a_{k} \leqslant \frac{a_{n}}{n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,735 |
Example 3 Let $a$ and $b$ be positive real numbers, and
$$
\begin{array}{l}
\frac{1}{a}-\frac{1}{b}-\frac{1}{a+b}=0 . \\
\text { Then }\left(\frac{b}{a}\right)^{3}+\left(\frac{a}{b}\right)^{3}=
\end{array}
$$ | Given that $\frac{b}{a}-\frac{a}{b}=1$.
$$
\text { Let } \frac{b}{a}=\frac{1}{2}+t, -\frac{a}{b}=\frac{1}{2}-t \text {. }
$$
Multiplying the two equations gives $\frac{1}{4}-t^{2}=-1$.
Solving for $t$ yields $t= \pm \frac{\sqrt{5}}{2}$.
Since $a$ and $b$ are both positive real numbers, then
$$
\frac{b}{a}=\frac{1}{2}+... | 2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,736 |
Example 4 Let the sequence $a_{1}, a_{2}, \cdots, a_{n}$ be a convex sequence. Prove: when $n \geqslant 4$,
$$
\sum_{i=2}^{n} \frac{2 a_{i}-a_{i-1}}{i} \leqslant \max \left\{a_{1}, a_{n}\right\} \sum_{i=2}^{n} \frac{1}{i} .
$$ | Prove that by Abel's identity and property 4, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{a_{i}}{i}=\frac{1}{n} \sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n-1}\left[\left(\sum_{j=1}^{i} a_{j}\right)\left(\frac{1}{i}-\frac{1}{i+1}\right)\right] \\
\leqslant \frac{1}{2}\left(a_{1}+a_{n}\right)+\sum_{i=1}^{n-1} \frac{i}{2}\lef... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,737 |
Example 5 The sequence of real numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfies
$$
a_{1}+a_{2}+\cdots+a_{n}=0 \text {, }
$$
and $2 a_{k} \leqslant a_{k-1}+a_{k+1}(k=2,3, \cdots, n-1)$.
Find the smallest $\lambda(n)$, such that for all $k \in\{1$, $2, \cdots, n\}$, we have
$$
\left|a_{k}\right| \leqslant ... | Let $a_{1}=1, a_{2}=-\frac{n+1}{n-1}$,
$$
a_{k}=-\frac{n+1}{n-1}+\frac{2 n(k-2)}{(n-1)(n-2)}(k=3,4, \cdots, n),
$$
then it satisfies $a_{1}+a_{2}+\cdots+a_{n}=0$, and
$$
2 a_{k} \leqslant a_{k-1}+a_{k+1} \text {. }
$$
At this point, $\lambda(n) \geqslant \frac{n+1}{n-1}$.
Next, we prove:
When $\lambda(n)=\frac{n+1}{n... | \frac{n+1}{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,738 |
Example 6 Let $\left\{a_{n}\right\}$ be a convex sequence. Prove:
$$
\frac{1}{2^{n}} \sum_{i=0}^{n} a_{i} \mathrm{C}_{n}^{i} \leqslant \frac{1}{n+1} \sum_{i=0}^{n} a_{i} \leqslant \frac{1}{2}\left(a_{0}+a_{n}\right) .
$$ | To prove that by property 4, the right-hand side inequality of (1) holds.
Next, we prove the left-hand side inequality of (1). It suffices to prove:
$$
(n+1) \sum_{i=0}^{n} a_{i} \mathrm{C}_{n}^{i} \leqslant 2^{n} \sum_{i=0}^{n} a_{i} \text {. }
$$
When $n$ is odd, i.e., $k=\frac{n-1}{2}$, we have
$$
\begin{array}{l}
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,739 |
Find all complex-coefficient polynomials $P(x)$ such that for any three integers $a, b, c$ whose sum is not zero, the following holds:
$$
\frac{P(a)+P(b)+P(c)}{a+b+c}
$$
is an integer. | Analysis and Solution The condition given in the problem, "for any three integers $a, b, c$ whose sum is not 0, the algebraic expression (1) is an integer," is extremely useful and should be fully utilized.
First, take integers $a \neq 0, b=c=0$, we get
$$
a \mid(P(a)+2 P(0)) \text {. }
$$
Next, take integers $a=b=c \... | P(x)=kx(k \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,740 |
Question 2: Prove that for any prime $p$, the number of positive integers $n$ satisfying $p!(n!+1)$ is at most $c p^{\frac{2}{3}}$, where $c$ is a constant independent of $p$.
The above text is translated into English, preserving the original text's line breaks and format. | Analysis and Solution: Clearly, when $n \geqslant p$, $n!+1$ is not divisible by $p$, so the $n$ that satisfy $p \mid(n!+1)$ are in the set $\{1, 2, \cdots, p-1\}$. Represent these $n$ on the number line, and let there be $k$ of them, i.e., $a_{1}<a_{2}<\cdots<a_{k}$ in Figure 1.
Consider the $k-1$ subintervals of the... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,741 |
Question 3 Let $n$ be a composite number. Prove: there exists a positive integer $m$, satisfying $m \mid n (m \leqslant \sqrt{n})$, and $d(n) \leqslant d^{3}(m)$, where $d(k)$ denotes the number of positive divisors of the positive integer $k$. | Analysis and Solution: We can first consider some special cases. (1) $n=p^{\alpha}, p$ is a prime number, $\alpha$ is an integer greater than 1. In this case, take $m={ }_{p}\left[\frac{\alpha}{2}\right]$. Then $m \mid n(m \leqslant \sqrt{n})$, and
$$
\begin{array}{l}
d^{3}(m)=\left(1+\left[\frac{\alpha}{2}\right]\righ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,742 |
Question 4 Let $a>b>1, b$ be an odd number, and $n$ be a positive integer. If $b^{n} \mid \left(a^{n}-1\right)$, prove that: $a^{b}>\frac{3^{n}}{n}$.
保留源文本的换行和格式，直接输出翻译结果。 | Analysis and Solution Since $b$ is an odd number greater than 1, $b$ must have an odd prime factor $p$. Without loss of generality, let $b$ be $p$, i.e., assume
$$
p^{n} \mid\left(a^{n}-1\right) \text {. }
$$
To prove: $a^{p}>\frac{p^{n}}{n}$.
Equation (1) $\Leftrightarrow a^{n}=1\left(\bmod p^{n}\right)$.
Therefore, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,743 |
Question 5 Let non-negative real numbers $a_{1}, a_{2}, a_{3}, a_{4}$, satisfy $a_{1}+a_{2}+a_{3}+a_{4}=1$. Prove:
$$
\begin{array}{l}
\max \left\{\sum_{i=1}^{4} \sqrt{a_{i}^{2}+a_{i} a_{i-1}+a_{i-1}^{2}+a_{i-1} a_{i-2}},\right. \\
\left.\sum_{i=1}^{4} \sqrt{a_{i}^{2}+a_{i} a_{i+1}+a_{i+1}^{2}+a_{i+1} a_{i+2}}\right\} ... | Analysis and Solution: Not fond of using too many subscripts, let's change $a_{1}, a_{2}, a_{3}, a_{4}$ to $a, b, c, d$. The problem is then:
Given non-negative real numbers $a, b, c, d$, satisfying
$$
a+b+c+d=1 \text {. }
$$
Prove: $\max \{A, B\} \geqslant 2$,
where, $A=\sum \sqrt{a^{2}+a d+d^{2}+d c}$,
$$
B=\sum \sq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,744 |
Question 6 Let $m$ be an integer greater than 1, $n$ be an odd number, and $3 \leqslant n<2 m$. The numbers $a_{i, j}(i, j \in \mathbf{N}, 1 \leqslant i \leqslant m, 1 \leqslant j \leqslant n)$ satisfy:
(1) For any $1 \leqslant j \leqslant n, a_{1, j}, a_{2, j}, \cdots, a_{m, j}$ is a permutation of $1,2, \cdots, m$.
(... | Let's analyze and solve the problem. Set $n=2k+1(k \geqslant 1)$. Without loss of generality, assume that the $m$ in the $(k+1)$-th column is in the 1st row.
If $a_{1, k}=m$, then by condition (1) we have
\[
\begin{array}{l}
a_{1, k-1} \geqslant m-1, a_{1, k-2} \geqslant m-2, \cdots \cdots \\
a_{1,1} \geqslant m-(k-1),... | mn - k^2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,745 |
Example 4 Given real numbers $a, b$ satisfy
$$
a^{2}+a b+b^{2}=1 \text {, and } t=a b-a^{2}-b^{2} \text {. }
$$
Then the range of values for $t$ is | From the given conditions, we have
$$
\begin{array}{l}
a^{2}+b^{2}=\frac{1-t}{2}, \\
4 a^{2} b^{2}=(1+t)^{2} .
\end{array}
$$
From equation (1), we can set
$$
a^{2}=\frac{1-t}{4}+p, b^{2}=\frac{1-t}{4}-p \text {. }
$$
Substituting into equation (2) and simplifying, we get
$$
(t+3)(3 t+1)=-16 p^{2} \text {. }
$$
Ther... | -3 \leqslant t \leqslant -\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,747 |
Example 2 As shown in Figure 2 of the hopscotch game: a person can only enter the first
square from outside; in the squares, each time they can
jump forward 1 square or 2 squares.
Then, the number of ways a person can jump
from outside to the sixth square is $\qquad$. | Solution 1 (Enumeration Method) Transform the problem as follows:
Enter the 6th grid from the 1st grid, walking 5 grids. Represent the number 5 as the sum of several 1s or 2s, where different orders of 1s and 2s represent different methods (e.g., $5=1+1+1+1+1$). For convenience, abbreviate the expression to $5=11111$. ... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,748 |
Example 3 The condition for three line segments to form a triangle is: the sum of the lengths of any two line segments is greater than the length of the third line segment. There is a wire of length $144 \mathrm{~cm}$, which is to be cut into $n$ $(n>2)$ small segments, with each segment being no less than $1 \mathrm{~... | Since the necessary and sufficient condition for forming a triangle is that the sum of any two sides is greater than the third side, the condition for not forming a triangle is that the sum of any two sides is less than or equal to the largest side. The shortest piece of wire is 1, so we can place 2 ones, and the third... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,749 |
Example 4 Perform the following operations on a positive integer: if it is even, divide it by 2; if it is odd, add 1. Continue doing this until the number becomes 1 and the operation stops. Find how many numbers will become 1 after 9 operations. | The number that becomes 1 after 1 operation is only 1, which is 2;
The number that becomes 1 after 2 operations is also only 1, which is 4;
The numbers that become 1 after 3 operations are 2, which are 3, 8; ......
It can be observed that if a number $a$ is odd, there is only 1 way to get $a$; if it is even, there are... | F_9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,750 |
Example 5 Let $\alpha, \beta$ be the roots of the equation $x^{2}-x-1=0$, and the sequence $\left\{a_{n}\right\}: a_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}(n=1,2, \cdots)$. Prove:
(1) For any positive integer $n$, we have $a_{n+2}=a_{n+1}+a_{n}$;
(2) All terms in the sequence $\left\{a_{n}\right\}$ are integers, ... | Proof (1) omitted.
(2) The conclusion of this problem is actually a basic property of the F-sequence. A brief proof is as follows:
Let $\left(a_{n+2}, a_{n+1}\right)=d$.
Using the Euclidean algorithm, we get
$$
\begin{array}{l}
d=\left(a_{n+2}, a_{n+1}\right)=\left(a_{n+1}+a_{n}, a_{n+1}\right) \\
=\left(a_{n+1}, a_{n}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,751 |
Example 6 Let $\alpha, \beta$ be the roots of the equation $x^{2}-x-1=0$, and let $a_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}(n=1,2, \cdots)$.
(1) Prove: For any positive integer $n$, we have
$$
a_{n+2}=a_{n+1}+a_{n} \text {; }
$$
(2) Find all positive integers $a, b(a<b)$, such that for any positive integer $n$, ... | (1) Omitted.
(2) Solution According to the condition, $b \mid (a_1 - 2a)$.
Since $a_1 = a_2 = 1$, we have $b \mid (1 - 2a)$.
From $1 \leqslant 2a - 1 < 2b - 1 < 2b$, we get $b = 2a - 1$.
Also, $b \mid (a_3 - 6a^3) = 2 - 6a^3$,
$6a^3 - 2 = 3a^2(2a - 1) + (3a^2 - 2)$,
so $(2a - 1) \mid 2(3a^2 - 2)$.
And $2(3a^2 - 2) = 6... | a = 3, b = 5 | Algebra | proof | Yes | Yes | cn_contest | false | 722,752 |
Example 7 Proof: The sequence
$$
y_{0}=1, y_{n+1}=\frac{1}{2}\left(3 y_{n}+\sqrt{5 y_{n}^{2}-4}\right)(n \geqslant 0)
$$
consists of integers. | Prove that from $y_{n+1}=\frac{1}{2}\left(3 y_{n}+\sqrt{5 y_{n}^{2}-4}\right)$, we get $\left(2 y_{n+1}-3 y_{n}\right)^{2}=5 y_{n}^{2}-4$.
Then $y_{n+1}^{2}-3 y_{n+1} y_{n}+y_{n}^{2}=-1$.
Thus, $y_{n}^{2}-3 y_{n} y_{n-1}+y_{n-1}^{2}=-1$.
Subtracting these, we have
$$
y_{n+1}^{2}-3 y_{n}\left(y_{n+1}-y_{n-1}\right)-y_{n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,753 |
Example 8 The general term formula of the sequence $\left\{f_{n}\right\}$ is
$$
\begin{array}{l}
f_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]\left(n \in \mathbf{Z}_{+}\right) . \\
\text {Let } S_{n}=\mathrm{C}_{n}^{1} f_{1}+\mathrm{C}_{n}^{2} f_{2}+\cd... | Solve: This problem is based on a property of the F-sequence:
$$
\mathrm{C}_{n}^{1} f_{1}+\mathrm{C}_{n}^{2} f_{2}+\cdots+\mathrm{C}_{n}^{n} f_{n}=f_{2 n} \text {. }
$$
Therefore, $8\left|S_{n} \Leftrightarrow 8\right| f_{2 n}$.
According to the recurrence relation,
$$
f_{n}=8 f_{n-5}+5 f_{n-6} \text {, and } 81 f_{6}... | 3 \mid n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,754 |
2. Let positive real numbers $a, b, c$ satisfy
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c \text {. }
$$
Prove:
$$
\frac{1}{(2 a+b+c)^{2}}+\frac{1}{(2 b+c+a)^{2}}+\frac{1}{(2 c+a+b)^{2}} \leqslant \frac{3}{16} .
$$ | 2. Note that for any positive real numbers $x, y, z$, by the AM-GM inequality we have
$$
\begin{array}{l}
2 x+y+z=(x+y)+(x+z) \\
\geqslant 2 \sqrt{(x+y)(x+z)} .
\end{array}
$$
Thus, $\frac{1}{(2 x+y+z)^{2}} \leqslant \frac{1}{4(x+y)(x+z)}$.
Applying this to the left side of the original inequality, we get
$$
\begin{ar... | \frac{3}{16} | Inequalities | proof | Yes | Yes | cn_contest | false | 722,756 |
4. Given positive real numbers $a, b, c$ satisfying $a b + b c + c a \leqslant 3 a b c$.
Prove:
$$
\begin{array}{l}
\sqrt{\frac{a^{2}+b^{2}}{a+b}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}}+3 \\
\leqslant \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}) .
\end{array}
$$ | 4. By the power mean inequality, we have
$$
\begin{array}{l}
\sqrt{2} \cdot \sqrt{a+b}=2 \sqrt{\frac{a b}{a+b}} \cdot \sqrt{\frac{1}{2}\left(2+\frac{a^{2}+b^{2}}{a b}\right)} \\
\geqslant 2 \sqrt{\frac{a b}{a+b}} \cdot \frac{1}{2}\left(\sqrt{2}+\sqrt{\frac{a^{2}+b^{2}}{a b}}\right) \\
=\sqrt{\frac{2 a b}{a+b}}+\sqrt{\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,757 |
Example 5 Given real numbers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a^{2}-b c-8 a+7=0, \\
b^{2}+c^{2}+b c-6 a+6=0 .
\end{array}\right.
$$
Try to find the range of values for $a$. | From the given conditions, we have
$$
\left\{\begin{array}{l}
b c=a^{2}-8 a+7, \\
(b+c)^{2}=(a-1)^{2} .
\end{array}\right.
$$
Therefore, from equation (1), we get $b+c= \pm(a-1)$.
Assume $b=\frac{ \pm(a-1)}{2}+t, c=\frac{ \pm(a-1)}{2}-t$.
Multiplying the two equations, we get
$$
b c=a^{2}-8 a+7=\frac{(a-1)^{2}}{4}-t^{... | 1 \leqslant a \leqslant 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,758 |
5. Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$. Prove: There exist real numbers $x, y$, such that
$$
f(x-f(y))>y f(x)+x .
$$ | 5. Suppose for all real numbers $x, y$, we have
$$
f(x-f(y)) \leqslant y f(x)+x \text {. }
$$
Let $a=f(0)$.
By substituting $y=0$ in equation (1), we get
$$
f(x-a) \leqslant x
$$
for all real numbers $x$.
Thus, for all real numbers $y$, we have
$$
f(y) \leqslant y+a \text {. }
$$
By substituting $x=f(y)$ in equation... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,759 |
7. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for all real numbers $x, y$, we have
$$
f(x f(x+y))=f(y f(x))+x^{2} .
$$ | 7. It is not hard to see that $f(x)=x$ and $f(x)=-x$ are solutions to the original functional equation.
Below, we prove that the original functional equation has no other solutions.
Let $f$ be a function that satisfies the original functional equation. Then $f$ is not a constant.
First, we prove that $f(0)=0$.
Assume $... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,760 |
1. Given the polynomial in $x$ and $y$: $x^{8} y - x^{m^{2}-m} y^{2} + x y^{4}$, which is arranged in descending powers of $x$. Then the number of integer values of $m$ is ( . )
(A) 2
(B) 4
(C) 6
(D) 8 | 一、1. B.
According to the problem, $1<m^{2}-m<8$, and $m^{2}-m$ is the product of two consecutive integers, so $m^{2}-m=2,6$.
Solving this, we get $m=2,-1,3,-2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,761 |
2. If $a \neq b$, the simplified result of $\sqrt{2 \sqrt{a b}-a-b}$ is ( ).
(A) $\sqrt{a}-\sqrt{b}$
(B) $-\sqrt{a}-\sqrt{b}$
(C) $\sqrt{-a}+\sqrt{-b}$
(D) 0 | 2. C.
According to the problem, we have $a \leqslant 0, b \leqslant 0$. Then
$$
\begin{array}{l}
\sqrt{2 \sqrt{a b}-a-b}=\sqrt{(\sqrt{-a}+\sqrt{-b})^{2}} \\
=\sqrt{-a}+\sqrt{-b} .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,762 |
4. There are 25 cards numbered $1, 3, \cdots, 49$. If a card with number $a$ is drawn, the next card to be drawn is the card with the largest odd divisor of $99-a$. This process is repeated until no more cards can be drawn. Xiao Hua drew the card numbered 1. When the operation ends, how many cards are left?
(A) 9
(B) 1... | 4. C.
The card numbers drawn are
$$
1 \rightarrow 49 \rightarrow 25 \rightarrow 37 \rightarrow 31 \rightarrow 17 \rightarrow 41 \rightarrow 29 \rightarrow 35 \rightarrow 1 \text {. }
$$
A total of 9 cards were drawn, leaving 16 cards. | 16 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,764 |
5. Given $\angle M O N=40^{\circ}, P$ is a fixed point inside $\angle M O N$, and points $A, B$ move on rays $O M, O N$ respectively. When the perimeter of $\triangle P A B$ is minimized, the value of $\angle A P B$ is ( ).
(A) $80^{\circ}$
(B) $100^{\circ}$
(C) $120^{\circ}$
(D) $140^{\circ}$ | 5. B.
As shown in Figure 3, construct the symmetric points $C$ and $D$ of point $P$ with respect to lines $O M$ and $O N$, respectively, and connect $C D$, intersecting $O M$ and $O N$ at points $A$ and $B$. At this point, the perimeter of $\triangle P A B$ is minimized.
$$
\begin{array}{l}
\text { Given } P C \perp O... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,765 |
6. For the equation in terms of $x$
$$
\frac{1}{x^{2}-x}+\frac{k-5}{x^{2}+x}=\frac{k-1}{x^{2}-1}
$$
if it has an extraneous root $x=1$. Then the value of $k$ is ( ).
(A) 2
(B) 3
(C) 6
(D) 10 | 6. B.
Eliminating the denominator, we get
$$
x+1+(k-5)(x-1)=(k-1) x \text {. }
$$
Substituting $x=1$ yields $k=3$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,766 |
7. In a Cartesian coordinate system, given points $A(2,2)$, $B(-2,-3)$, point $P$ lies on the coordinate axes, and $\triangle A B P$ is a right-angled triangle. Then the number of points $P$ is ( ).
(A) 3
(B) 4
(C) 6
(D) 8 | 7. D.
Draw a circle with $AB$ as the diameter, and draw perpendicular lines to $AB$ through points $A$ and $B$. The circle and the perpendicular lines intersect the coordinate axes at 8 points. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,767 |
8. In Rt $\triangle A B C$, $\angle A C B=90^{\circ}$, the incircle $\odot I$ touches $A C$, $B C$ at points $E$, $F$, and the rays $B I$, $A I$ intersect the line $E F$ at points $M$, $N$. Let $S_{\triangle A I B}=S_{1}$, $S_{\triangle M I N}=S_{2}$. Then the value of $\frac{S_{1}}{S_{2}}$ is $(\quad)$.
(A) $\frac{3}{... | 8. B.
As shown in Figure 4, connect $A M$, $I E$, and $B N$.
It is easy to see that $\angle A E M = \angle C E F = \angle A I M = 45^{\circ}$
$\Rightarrow A$, $M$, $E$, $I$ are concyclic
$$
\Rightarrow \angle A M I = \angle A E I = 90^{\circ}
$$
$\Rightarrow \triangle A I M$ is an isosceles right triangle
$$
\Rightarr... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,768 |
Example 6 If $a+b+c=0, a^{3}+b^{3}+c^{3}=0$, prove: $a^{19}+b^{19}+c^{19}=0$. | Prove that given $a+b=-c$.
Let $a=-\frac{c}{2}+t, b=-\frac{c}{2}-t$.
Substitute into $a^{3}+b^{3}+c^{3}=0$ and simplify to get $c\left(\frac{c^{2}}{4}-t^{2}\right)=0$.
Therefore, $c=0$ or $t= \pm \frac{c}{2}$.
If $c=0$, then $a=-b$, and we have $a^{19}+b^{19}+c^{19}=0$;
If $t= \pm \frac{c}{2}$, then
$$
a=0, b=-c \text ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,769 |
9. The circumradius of $\triangle A B C$ is $R$, and the lines obtained by reflecting $A C$ about $A B$ and $B C$ intersect at point $K$. The radius of the incircle $\odot I$ of $\triangle A C K$ is $r$. Then ( ).
(A) $B I>2 R$
(B) $B I<2 R$
(C) $B I=2 R$
(D) $B I=R+r$ | 9. C.
As shown in Figure 5, connect $B I$ and $C I$. By the given conditions, it is easy to prove that $\angle B A I = \angle B C I = 90^{\circ}$. Therefore, $B I$ is the diameter of the circumcircle of $\triangle A B C$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,770 |
10. In a Cartesian coordinate system, given points $A(-1,2)$, $B(1,2)$, $C(-3,1)$, $D(3,1)$, $E(-2,-2)$, $F(2,-2)$, $G(-4,3)$, $H(4,3)$. Using these eight points as vertices to form triangles, and ensuring that no two vertices of the triangle are symmetric with respect to the $y$-axis. Then the number of such triangles... | 10. C.
These eight points consist of four pairs of points that are symmetric about the $y$-axis.
The vertices of the triangles that meet the condition cannot be symmetric about the $y$-axis, so the total number of such triangles that can be formed is
$$
\frac{8 \times 6 \times 4}{3 \times 2 \times 1}=32 \text{ (triang... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,771 |
11. Observe the sequence of arrays $(1),(3,5),(7,9,11)$,
$(13,15,17,19), \cdots$. Then 2009 is in the $\qquad$ group. | $$
\begin{array}{l}
\text { Notice that 2009 is the 1005th positive odd number. } \\
1+2+\cdots+44<1005 \\
<1+2+\cdots+45,
\end{array}
$$
Then 2009 is in the 45th group. | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,772 |
12. Given a positive integer $m$ that satisfies $m^{2}+5 m+30$ is a perfect square. Then the value of $m$ is $\qquad$ . | 12.21 or 1 .
By $(m+2)^{2}<m^{2}+5 m+30<(m+6)^{2}$, then
$$
\begin{array}{l}
m^{2}+5 m+30 \\
=(m+3)^{2},(m+4)^{2},(m+5)^{2} .
\end{array}
$$
Solving yields $m=21, \frac{14}{3}$ (discard) ,1. | 21 \text{ or } 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,773 |
13. $(x, y)=\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\left(x_{1}<x_{2}\right)$ are two integer solutions of the equation $x^{2}-y^{2}-2 x+6 y-8=0$. In the Cartesian coordinate system, point $A\left(x_{1}, y_{1}\right)$ and point $B\left(x_{2}, y_{2}\right)$ are symmetric about $P(1,3)$, and point $C(5,-1)$. I... | 13. $(-1,1)$.
According to the problem, we have
$$
\begin{array}{l}
(x-y+2)(x+y-4)=0 \\
\Rightarrow y=x+2 \text { or } y=-x+4 .
\end{array}
$$
Since point $P(1,3)$ is exactly the intersection of the two lines, and point $C(5,-1)$ lies on the line $y=-x+4$, therefore, points $A$ and $B$ lie on the line $y=x+2$.
There... | (-1,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,774 |
14. $[x]$ represents the greatest integer not exceeding the real number $x$. The equation $[2 x]+[3 x]=9 x-\frac{7}{4}$ has all real solutions as $\qquad$ | 14. $-\frac{1}{36}, \frac{7}{36}$.
From $2 x-1<[2 x] \leqslant 2 x, 3 x-1<[3 x] \leqslant 3 x$, then $5 x-2<9 x-\frac{7}{4} \leqslant 5 x \Rightarrow-\frac{1}{16}<x \leqslant \frac{7}{16}$. Therefore, $-\frac{37}{16}<9 x-\frac{7}{4} \leqslant \frac{35}{16}$.
Since $9 x-\frac{7}{4}$ is an integer, we have,
$$
9 x-\frac... | -\frac{1}{36}, \frac{7}{36} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,775 |
15. (25 points) Given a square $A B C D$ with side length 1, points $E, F, G$ are on sides $A D, A B, D C$ respectively (they can coincide with the vertices). If $\triangle E F G$ is an equilateral triangle, find the maximum and minimum values of the area of $\triangle E F G$. | Three, 15. As shown in Figure 6, draw $E M \perp F G$ at point $M$, and connect $A M, D M$.
Since $\angle E D G=\angle E M G=90^{\circ}$
$\Rightarrow E, D, G, M$ are concyclic
$\Rightarrow \angle E D M=\angle E G M=60^{\circ}$.
Similarly, $\angle E A M=60^{\circ}$.
Therefore, $\triangle A M D$ is an equilateral triang... | \frac{\sqrt{3}}{4}, 2 \sqrt{3}-3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,776 |
16. (25 points) Given that $a$, $b$, and $c$ are integers, satisfying
$$
c>0, a+b=3, c^{2}-2 c-a b=-2 \text {. }
$$
If the equation in $x$
$$
d x^{2}+(x+d) x+a b+d=0
$$
has only one solution, find the value of $d$. | 16. It is known that $a$ and $b$ are the roots of the equation
$$
x^{2}-3 x+c^{2}-2 c+2=0
$$
with $\Delta=9-4\left(c^{2}-2 c+2\right) \geqslant 0$, then $4 c^{2}-8 c-1 \leqslant 0$.
Given that $c$ is a positive integer, solving yields $c=1,2$.
When $c=1$, $a$ and $b$ have no integer solutions;
When $c=2$, $a=1, b=2$ o... | 0, \frac{2}{3}, -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,777 |
2. Let $a<b, f(x)=|x-a|-|x-b|$. Then the range of $f(x)$ is $ـ$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. $a-b \leqslant f(x) \leqslant b-a$.
If $x \leqslant a$, then
$$
f(x)=a-x-b+x=a-b \text{; }
$$
If $a<x<b$, then
$$
f(x)=-a-b+2 x, a-b<f(x)<b-a \text{; }
$$
If $x \geqslant b$, then $f(x)=x-a-x+b=b-a$.
In summary, $a-b \leqslant f(x) \leqslant b-a$. | null | Number Theory | proof | Yes | Yes | cn_contest | false | 722,779 |
Example 7 Let $x, y, z$ be real numbers satisfying the conditions
$$
x+y+z=a, x^{2}+y^{2}+z^{2}=\frac{1}{2} a^{2}(a>0) \text {. }
$$
Prove: $0 \leqslant x \leqslant \frac{2}{3} a, 0 \leqslant y \leqslant \frac{2}{3} a, 0 \leqslant z \leqslant \frac{2}{3} a$. | Prove that from $x+y+z=a$, we get $y+z=a-x$.
Let $y=\frac{a-x}{2}+t, z=\frac{a-x}{2}-t$.
Substitute into $x^{2}+y^{2}+z^{2}=\frac{1}{2} a^{2}$ and simplify to get $3 x^{2}-2 a x=-4 t^{2}$.
Therefore, $3 x^{2}-2 a x \leqslant 0$.
Solving gives $0 \leqslant x \leqslant \frac{2}{3} a$.
By symmetry, similarly we get
$$
0 \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,780 |
3. A certain linear function graph is parallel to the line $y=\frac{5}{4} x+\frac{95}{4}$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and passes through the point $(-1,-25)$. Then on the line segment $AB$ (including $A$ and $B$), the number of points with both integer coordinates is $... | 3.5.
Let the linear function be $y=\frac{5}{4} x+b$.
Given that it passes through the point $(-1,-25)$, we get $b=-\frac{95}{4}$.
Therefore, $y=\frac{5}{4} x-\frac{95}{4}$.
Thus, $A(19,0)$ and $B\left(0,-\frac{95}{4}\right)$.
From $y=\frac{5}{4} x-\frac{95}{4}(0 \leqslant x \leqslant 19)$, taking $x=3,7,11$,
15,19, $y$... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,781 |
4. Given the function $f(x)$ satisfies for all real numbers $x, y$,
$$
\begin{array}{l}
f(x)+f(2 x+y)=f(3 x-y)+x-2010 \text {. } \\
\text { Then } f(2010)=
\end{array}
$$ | 4.0.
Let $x=2010, y=1005$, then
$$
\begin{array}{l}
f(2010)+f(5025) \\
=f(5025)+2010-2010 .
\end{array}
$$
Therefore, $f(2010)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,782 |
5. As shown in Figure 1, given $A(-2,0)$, $B(0,-4)$, and $P$ as any point on the hyperbola $y=\frac{8}{x}$ $(x>0)$. A perpendicular line from $P$ to the x-axis meets at point $C$, and a perpendicular line from $P$ to the y-axis meets at point $D$. The minimum value of the area of quadrilateral $A B C D$ is $\qquad$ | 5.16.
Let point \( P\left(x_{0}, y_{0}\right) \). Then \( y_{0}=\frac{8}{x_{0}}\left(x_{0}>0\right) \). Hence \( C\left(x_{0}, 0\right) \) and \( D\left(0, \frac{8}{x_{0}}\right) \).
From the given conditions,
\[
\begin{array}{l}
|C A|=x_{0}+2,|D B|=\frac{8}{x_{0}}+4 . \\
\text { Therefore, } S=\frac{1}{2}\left(x_{0}+... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,783 |
6. Given a regular triangular prism with a base edge length of 1, the diagonals of the two lateral faces are perpendicular to each other. Then the length of the lateral edge of the prism is $\qquad$ | 6. $\frac{\sqrt{2}}{2}$.
Let the triangular prism $A B C-A_{1} B_{1} C_{1}$, with side edge length $a$, and the skew diagonals of the side faces $A B_{1}$ and $B C_{1}$ are perpendicular to each other. Then
$$
\begin{array}{l}
\overrightarrow{A B_{1}} \cdot \overrightarrow{B C_{1}}=0 \\
\Rightarrow\left(\overrightarro... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,784 |
7. If the parabola $y=\frac{1}{2} x^{2}-m x+m-1$ intersects the $x$-axis at integer points, then the equation of the axis of symmetry of the parabola is $\qquad$. | 7. $x=1$.
From the problem, we know that the equation $\frac{1}{2} x^{2}-m x+m-1=0$ has two integer roots $x_{1} 、 x_{2}\left(x_{1} \leqslant x_{2}\right)$. Therefore,
$$
\frac{1}{2} x^{2}-m x+m-1=\frac{1}{2}\left(x-x_{1}\right)\left(x-x_{2}\right) \text {. }
$$
Substituting $x=1$, we get $\left(1-x_{1}\right)\left(1... | x=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,785 |
8. Given real numbers $a, b$ satisfy
$$
a^{2}+a b+b^{2}=1 \text {, and } t=a b-a^{2}-b^{2} \text {. }
$$
Then the product of the maximum and minimum values of $t$ is $\qquad$ | 8. 1.
Let $a=x+y, b=x-y$. Then
$$
(x+y)^{2}+(x+y)(x-y)+(x-y)^{2}=1 \text {. }
$$
Simplifying, we get $y^{2}=1-3 x^{2}$.
Since $y^{2} \geqslant 0$, we have $0 \leqslant x^{2} \leqslant \frac{1}{3}$.
Thus, $t=a b-a^{2}-b^{2}$
$$
\begin{array}{l}
=(x+y)(x-y)-(x+y)^{2}-(x-y)^{2} \\
=-x^{2}-3 y^{2}=8 x^{2}-3 .
\end{array}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,786 |
10. Let two non-zero and non-collinear vectors $a, b$ in the same plane satisfy $b \perp (a-b)$. Then for any $x \in \mathbf{R}$, the range of $|a-b x|$ is $\qquad$ (expressed in terms of vectors $a, b$). | 10. $[|a-b|,+\infty)$.
Let $\overrightarrow{O A}=a, \overrightarrow{O B}=b, b x$ represent any vector collinear with $\overrightarrow{O B}$, $|a-b x|$ represents the distance from point $A$ to any point on line $O B$, and $|\boldsymbol{a}-\boldsymbol{b}|$ represents the distance from point $A$ to $B$.
When $b \perp(a-... | [|a-b|,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,788 |
II. (12 points) As shown in Figure 3, in $\triangle ABC$, it is given that $AB=9, BC=8, AC=7$, and $AD$ is the angle bisector. A circle is drawn with $AD$ as a chord, tangent to $BC$, and intersecting $AB$ and $AC$ at points $M$ and $N$, respectively. Find the length of $MN$. | II. As shown in Figure 3, connect DM. Given
$\angle B D M=\angle B A D=\angle C A D=\angle D M N$
$\Rightarrow M N / / B C \Rightarrow \triangle A M N \backsim \triangle A B C$.
It is easy to know that $B D=\frac{9}{2}$.
Then $B M \cdot B A=B D^{2} \Rightarrow B M \cdot 9=\left(\frac{9}{2}\right)^{2}$
$\Rightarrow B M=... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,789 |
Three, (13 points) As shown in Figure 4, two planes $m \parallel n$, line segment $AD$ intersects planes $m$ and $n$ at points $B$ and $C$, respectively. Another line through point $A$ intersects planes $m$ and $n$ at points $M$ and $P$, respectively. Another line through point $D$ intersects planes $m$ and $n$ at poin... | Three, it is known that $B M / / C P, B N / / C Q$. Then $\sin \angle M B N=\sin \angle P C Q$,
and
$$
\begin{array}{l}
\frac{B M}{C P}=\frac{A B}{A C}, \frac{B N}{C Q}=\frac{B D}{C D}. \\
\text { Also } S_{\triangle B M N}=\frac{1}{2} B M \cdot B N \sin \angle M B N, \\
S_{\triangle C P Q}=\frac{1}{2} C P \cdot C Q \... | 3+\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,790 |
Example 8 Solve the equation $\frac{2\left(x^{2}+1\right)}{x+1}+\frac{6(x+1)}{x^{2}+1}=7$. | Solving, we assume
$$
\begin{array}{l}
\frac{2\left(x^{2}+1\right)}{x+1}=\frac{7}{2}+t, \\
\frac{6(x+1)}{x^{2}+1}=\frac{7}{2}-t .
\end{array}
$$
Multiplying the two equations, we get $12=\frac{49}{4}-t^{2}$.
Therefore, $t= \pm \frac{1}{2}$.
When $t=\frac{1}{2}$, $\frac{2\left(x^{2}+1\right)}{x+1}=\frac{7}{2}+\frac{1}{... | x=1 \pm \sqrt{2}, x=\frac{3 \pm \sqrt{17}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,791 |
1. (Compulsory 3) A bag contains 6 balls of the same shape and size, including 2 black balls, 2 white balls, and 2 red balls. It is stipulated that drawing a black ball scores 0 points, drawing a white ball scores 1 point, and drawing a red ball scores 2 points; when drawing these balls, no one can see the color of the... | Four, 1. Let black balls be numbered 1, 2; white balls be numbered 3, 4; and red balls be numbered 5, 6. Then all possible outcomes of player A drawing the balls are as follows:
$$
\begin{array}{l}
123,124,125,126,134,135,136, \\
145,146,156,234,235,236,245, \\
246,256,345,346,356,456 .
\end{array}
$$
(1) In the event ... | \frac{3}{10} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,792 |
2. (Compulsory 4) Given vectors
$$
\boldsymbol{a}=(\sqrt{3} \sin \omega x, 1), \boldsymbol{b}=(\cos \omega x, 0)(\omega>0) \text {, }
$$
and the function $f(x)=\boldsymbol{b} \cdot(\boldsymbol{a}-k \boldsymbol{b})$ is a periodic function with the smallest positive period $\frac{\pi}{2}$. If the maximum value of the fu... | 2. Notice that
$$
\begin{array}{l}
f(x)=\boldsymbol{b} \cdot(\boldsymbol{a}-k \boldsymbol{b}) \\
=\sqrt{3} \sin \omega x \cdot \cos \omega x-k \cos ^{2} \omega x \\
=\frac{\sqrt{3}}{2} \sin 2 \omega x-\frac{1}{2} k \cos 2 \omega x-\frac{1}{2} k \\
=\frac{1}{2} \sqrt{k^{2}+3} \sin (2 \omega x+\theta)-\frac{1}{2} k .
\en... | t=\frac{2 \sqrt{3}}{3}, \boldsymbol{m}=\left(\frac{\pi}{24},-\frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,793 |
Five. (13 points) Let the quadratic function
$$
f(x)=x^{2}+b x+c(b, c \in \mathbf{R})
$$
intersect the $x$-axis. If for all $x \in \mathbf{R}$, we have
$$
f\left(x+\frac{1}{x}\right) \geqslant 0 \text {, and } f\left(\frac{2 x^{2}+3}{x^{2}+1}\right) \leqslant 1 .
$$
Find the values of $b$ and $c$. | Five, because $\left|x+\frac{1}{x}\right|=|x|+\left|\frac{1}{x}\right| \geqslant 2$, so for all real numbers $x$ satisfying $|x| \geqslant 2$ we have
$$
f(x) \geqslant 0 \text {. }
$$
Thus, the real roots of $f(x)=x^{2}+b x+c=0$ lie within the interval $[-2,2]$.
Therefore, the quadratic function $f(x)=x^{2}+b x+c(b, ... | b=-4, c=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,794 |
2. Given $f(1,1)=1, f(m, n) \in \mathbf{N}_{+}(m, n$ $\in \mathbf{N}_{+}$), and for any $m, n \in \mathbf{N}_{+}$, we have
(1) $f(m, n+1)=f(m, n)+2$;
(2) $f(m+1,1)=2 f(m, 1)$.
Then the value of $f(2010,2008)$ is ( ).
(A) $2^{2009}+2007$
(B) $2^{2009}+4014$
(C) $2^{2010}+2007$
(D) $2^{2010}+4014$ | 2. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,796 |
3. Given the function $f(x)=x^{2}-4 x+3$, the sets are
$$
\begin{array}{l}
M=\{(x, y) \mid f(x)+f(y) \leqslant 0\}, \\
N=\{(x, y) \mid f(x)-f(y) \geqslant 0\} .
\end{array}
$$
Then in the Cartesian coordinate system, the area of the region represented by the set $M \cap N$ is ( ).
(A) $\frac{\pi}{4}$
(B) $\frac{\pi}{2... | 3. C.
From the given information, we have
$$
\begin{array}{l}
M=\{(x, y) \mid f(x)+f(y) \leqslant 0\} \\
=\left\{(x, y) \mid(x-2)^{2}+(y-2)^{2} \leqslant 2\right\}, \\
N=\{(x, y) \mid f(x)-f(y) \geqslant 0\} \\
=\{(x, y) \mid(x-y)(x+y-4) \geqslant 0\} .
\end{array}
$$
Then, $M \cap N=\left\{\begin{array}{l}(x-2)^{2}+... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,797 |
4. The faces of a regular hexahedron and a regular octahedron are all equilateral triangles with side length $a$. The ratio of the radii of the inscribed spheres of these two polyhedra is a simplest fraction $\frac{m}{n}$. Then, the product $m n$ is ( ).
(A) 3
(B) 4
(C) 6
(D) 12 | 4. C.
Using the equal volume method, we can find that $\frac{m}{n}=\frac{2}{3}$. Therefore, $m n=6$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,798 |
6. There are ten equally spaced points on a circle. Then, among the convex quadrilaterals with four of these ten equally spaced points as vertices, the ratio of trapezoids is ( ).
(A) $\frac{8}{21}$
(B) $\frac{4}{21}$
(C) $\frac{1}{126}$
(D) $\frac{2}{7}$ | 6. D.
Choose any four points, there are $\mathrm{C}_{10}^{4}=210$ convex quadrilaterals, among which the two parallel sides of a trapezoid can be selected from five sets of 5 parallel chords parallel to the diameter, or from five sets of 4 parallel chords not parallel to the diameter, excluding rectangles, there are 6... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,800 |
Example 9 Let $x, y, z$ all be real numbers. Solve the system of equations
$$
\left\{\begin{array}{l}
2 x+3 y+z=13 \\
4 x^{2}+9 y^{2}+z^{2}-2 x+15 y+3 z=82
\end{array}\right.
$$ | Solve: From $2 x+3 y+z=13$, we get
$$
2 x+3 y=13-z \text {. }
$$
Let $2 x=\frac{13-z}{2}+t, 3 y=\frac{13-z}{2}-t$.
Substituting into equation (1) and simplifying, we get
$$
3(z-4)^{2}+4\left(t-\frac{3}{2}\right)^{2}=0 \text {. }
$$
Therefore, $z=4, t=\frac{3}{2}$.
Thus, $x=3, y=1$.
Hence, the solution to the original... | (3,1,4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,802 |
4. If $B x \geqslant \sin x \geqslant A x (A, B$ are constants $)$ holds for $0 \leqslant x \leqslant \frac{\pi}{2}$, then the minimum value of the constant $B-A$ is $\qquad$; for any acute triangle $\triangle ABC$, it is always true that
$$
\sin A+\sin B+\sin C>M
$$
then the maximum value of $M$ is $\qquad$ | 4. $1-\frac{2}{\pi}, 2$ | 1-\frac{2}{\pi}, 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,805 |
5. Given the circle $\odot O$ with radius 1, the angle between radii $O A$ and $O B$ is $\varphi(0<\varphi<\pi, \varphi$ is a constant $), C$ is a moving point on the circle. If $\overrightarrow{O C}=x \overrightarrow{O A}+y \overrightarrow{O B}(x, y \in \mathbf{R})$, then the maximum value of $x+y$ is $\qquad$. | $\frac{1}{\cos \frac{\varphi}{2}}$ | \frac{1}{\cos \frac{\varphi}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,806 |
6. The following is a mathematical model of a pastry chef's work process:
As shown in Figure 2, on the number line, take the line segment corresponding to the closed interval $[0,4]$, fold it in half (the point corresponding to coordinate 4 overlaps with the origin), and then uniformly stretch it to a line segment of ... | 6. $\frac{j}{2^{n-2}}\left(j\right.$ is $\left[1,2^{n}\right]$ of all odd numbers $)$ | \frac{j}{2^{n-2}}\left(j\right. is \left[1,2^{n}\right] \text{ of all odd numbers } ) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,807 |
1. (1) If $x>0, y>0$, prove:
$$
\frac{x^{2}}{x+y} \geqslant \frac{3 x-y}{4} \text {; }
$$
(2) If $x>0, y>0, z>0$, prove:
$$
\frac{x^{3}}{x+y}+\frac{y^{3}}{y+z}+\frac{z^{3}}{z+x} \geqslant \frac{x y+y z+z x}{2} .
$$ | Three, 1. (1) From $\frac{x^{2}}{x+y}-\frac{3 x-y}{4}=\frac{(x-y)^{2}}{4(x+y)} \geqslant 0$, we get
$$
\frac{x^{2}}{x+y} \geqslant \frac{3 x-y}{4} \text {. }
$$
(2) From (1) we get
$$
\frac{x^{3}}{x+y} \geqslant \frac{3 x^{2}-x y}{4} \text {. }
$$
Similarly, $\frac{y^{3}}{y+z} \geqslant \frac{3 y^{2}-y z}{4}, \frac{z^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,808 |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a(a \neq 0$, and $a \neq 1)$, the sum of the first $n$ terms is $S_{n}$, and $S_{n}=\frac{a}{1-a}\left(1-a_{n}\right)$. Let $b_{n}=a_{n} \lg \left|a_{n}\right|\left(n \in \mathbf{N}_{+}\right)$. When $a=-\frac{\sqrt{7}}{3}$, does there exist a positive integ... | 2. When $n \geqslant 2$,
$$
S_{n}=\frac{a}{1-a}\left(1-a_{n}\right), S_{n-1}=\frac{a}{1-a}\left(1-a_{n-1}\right) .
$$
Then $a_{n}=S_{n}-S_{n-1}$
$$
\begin{array}{l}
=\frac{a}{1-a}\left[\left(1-a_{n}\right)-\left(1-a_{n-1}\right)\right] \\
=\frac{a}{1-a}\left(a_{n-1}-a_{n}\right),
\end{array}
$$
i.e., $a_{n}=a a_{n-1}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,809 |
3. As shown in Figure 3, the diagonals of quadrilateral $ABCD$ intersect at point $O$, the angle bisector of $\angle DCO$ intersects line segment $OD$ at point $Q$, connect $AQ$, draw $OM \perp BC$ at point $M$, $ON \perp AQ$ at point $N$, and $P$ is the midpoint of side $AB$, $OA=\frac{OB \cdot OD}{OC+CD}$. Prove: $PM... | 3. Given $C Q$ bisects $\angle D C O$
$$
\begin{array}{l}
\Rightarrow \frac{D Q}{Q O}=\frac{D C}{C O} \Rightarrow C O=\frac{Q O}{D Q} \cdot D C . \\
\text { Also } O A=\frac{O B \cdot O D}{O C+C D} \\
\Rightarrow O A(O C+C D)=O B \cdot O D .
\end{array}
$$
Substituting equation (1) into equation (2) we get
$$
\begin{a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,810 |
4. In a Cartesian coordinate system, draw all rectangles that simultaneously satisfy the following conditions:
(1) The sides of these rectangles are parallel or coincide with the coordinate axes;
(2) All vertices of these rectangles (repeated vertices are counted only once) are exactly 100 integer points (points with b... | 4. First, prove that the number of such rectangles does not exceed 2025.
Take any 100 integer points. Let $O$ be one of the 100 integer points we have chosen. We call a rectangle "good" if $O$ is one of its vertices, the other three vertices are also chosen from these 100 integer points, and the sides are parallel or ... | 2025 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,811 |
9.1. Distribute 24 pencils of four colors (6 pencils of each color) to 6 students, with each student getting 4 pencils. It is known that no matter how the pencils are distributed, there will always be $n$ students such that the $4 n$ pencils they have are of four colors. Find the minimum value of $n$.
| 9.1. The minimum value of $n$ is 3.
First, we prove: There are always 3 students who have pencils of 4 different colors.
In fact, there are 6 pencils of each color, and each student has 4 pencils, so there exists a student who has at least two colors of pencils. Clearly, any other color must be owned by at least one ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,812 |
Example 10 Let real numbers $x, y, z$ satisfy
$$
x+y+z=5, \quad xy+yz+zx=3 \text{.}
$$
Then the maximum value of $z$ is $\qquad$. | Solve: From $x+y+z=5$, we get $x+y=5-z$.
Let $x=\frac{5-z}{2}+t, y=\frac{5-z}{2}-t$.
Substituting into $x y+y z+z x=3$ and simplifying, we get $3 z^{2}-10 z-13=-4 t^{2}$.
Therefore, $3 z^{2}-10 z-13 \leqslant 0$.
Solving, we get $-1 \leqslant z \leqslant \frac{13}{3}$.
Thus, the maximum value of $z$ is $\frac{13}{3}$. | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,813 |
9. 2. Place 100 distinct real numbers at different points on a circle. Prove that there must exist four consecutive numbers such that the sum of the two end numbers is greater than the sum of the two middle numbers. | 9.2. Assume the conclusion is not true.
Let 100 real numbers on a circle be $a_{1}, a_{2}, \cdots, a_{100}, a_{n}=a_{n+100}$. Then
$$
a_{n}+a_{n+3} \leqslant a_{n+1}+a_{n+2},
$$
which means $a_{n+3}-a_{n+2} \leqslant a_{n+1}-a_{n}(n=1,2, \cdots, 100)$. From this, we get
$$
a_{100}-a_{99} \leqslant a_{98}-a_{99} \leqs... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,814 |
9.3. From a point $O$ outside the circle $\omega$, draw tangents $OA$ and $OB$ to the circle $\omega$ ($A$ and $B$ are the points of tangency). $C$ is a point on the minor arc $\overparen{AB}$ such that $AC \neq BC$, and $I$ is the center of the circle $\omega$. The line $AC$ intersects $OB$ at point $D$, and the line ... | 9.3. Let $M$ be the second intersection point of the circumcircles of $\triangle A C E$ and $\triangle B C D$. Therefore, it suffices to prove that the circumcircle of $\triangle O C I$ also passes through $M$ (because in this case, the three circumcenters all lie on the perpendicular bisector of $C M$).
Let $\angle C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,815 |
9.4. In the kitchen, there are 100 apples with a total weight of $10 \mathrm{~kg}$, and each apple weighs no less than $25 \mathrm{~g}$. The chef hopes to cut these apples into several pieces, each weighing no less than $25 \mathrm{~g}$, and distribute them to 100 students, with each student receiving $100 \mathrm{~g}$... | 9.4. We call an apple (or a piece of an apple) with a weight not less than $25 \mathrm{~g}$ a large one.
Below, we use mathematical induction on $n$ to prove: a total weight of $100 n \mathrm{~g}$ of $n$ large apples can be divided into several large pieces to be evenly distributed among $n$ people.
When $n=1$, the co... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,816 |
9.5. Given the equation
$$
\left(x^{2}+20 a x+10 b\right)\left(x^{2}+20 b x+10 a\right)=0
$$
has no real roots, where $a, b$ are distinct real parameters. Prove: $20(b-a)$ is not an integer. | 9.5. Suppose the conclusion is not true, let's assume $b>a$. Then
$$
b-a \geqslant \frac{1}{20} \text {. }
$$
Since $x^{2}+20 b x+10 a=0$ has no real roots, we have
$$
100 b^{2}-10 a<0 \text {. }
$$
It is easy to see that $10 b^{2}<a \leqslant b-\frac{1}{20}$, which means
$$
10 b^{2}-b+\frac{1}{20}<0 \text {. }
$$
C... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,817 |
9.6. In a day, 1000 dwarfs wearing red or blue hats meet each other in pairs. Dwarfs wearing red hats tell lies, while those wearing blue hats tell the truth. Each dwarf may change the color of their hat several times (i.e., red to blue, blue to red). It is known that when any two dwarfs meet, they both say that the ot... | 9.6. Clearly, when two dwarfs meet, they say that the other is wearing a red hat if and only if their hat colors are different.
Thus, in a day, if three dwarfs never change the color of their hats, then two of them must have the same hat color, and they cannot both say that the other is wearing a red hat when they mee... | 998 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 722,818 |
9. 7. If a positive integer $n$ cannot be expressed as $n=\frac{x^{2}-1}{y^{2}-1}$ $(x, y>1$ are positive integers), then it is called an "unfortunate number". Question: Is the number of unfortunate numbers finite or infinite? | 9.7. Infinitely many.
In fact, it can be proven: the square of any odd prime $p$ is an unlucky number.
Assume, for contradiction, that $n=p^{2}$ is not an unlucky number. Then
$$
\left(y^{2}-1\right) p^{2}=x^{2}-1 \text {. }
$$
Thus, $p \mid(x+1)$ or $p \mid(x-1)$.
If $p \mid(x+1)$, since $x-1=(x+1)-2$ is not a multi... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,819 |
9.8. Given an acute triangle $\triangle A B C$ satisfying that the length of the median $A M$ is greater than the length of side $A B$. Prove: $\triangle A B C$ can be divided into three pieces that can be rearranged to form a rhombus. | 9.8. Let $N$ be the midpoint of $A C$, and $K$ be a point on line $M N$ such that $M$ is the midpoint of $K N$. Then $\triangle M N C$ and $\triangle M K B$ are symmetric about point $M$.
Cut the triangle along the midline $M N$ and place $\triangle M N C$ in the position of $\triangle M K B$ to get quadrilateral $A N... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,820 |
1. If the equation about $x$
$$
x^{2}+(1-2 \sqrt{2}) x+3 m-\sqrt{2}=0
$$
( $m$ is a rational number) has a rational root, then its other root is $(\quad)$.
(A) $\frac{3}{2}+2 \sqrt{2}$
(B) $\frac{3}{2}-2 \sqrt{2}$
(C) $-\frac{1}{2}+2 \sqrt{2}$
(D) $-\frac{1}{2}-2 \sqrt{2}$ | - 1. C.
Transform the original equation into
$$
x^{2}+x+3 m=\sqrt{2}(2 x+1) \text {. }
$$
Let one rational root of the original equation be $a$. Then
$$
a^{2}+a+3 m=\sqrt{2}(2 a+1) \text {. }
$$
Since the left side of the above equation is a rational number, the right side must also be a rational number. Therefore,
... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,821 |
2. As shown in Figure 1, two moving points, Jia and Yi, start moving along the sides of square $ABCD$ from vertices $A$ and $C$ simultaneously. Jia moves in a clockwise direction, while Yi moves in a counterclockwise direction. If Yi's speed is 5 times that of Jia, then they will meet for the 2010th time on side ( ).
(... | 2. A.
Let the side length of the square be $a$. Then, when they meet for the first time, A and B have traveled a total of $2a$. From the first meeting to the second meeting, A and B travel a total of $4a$, and for each subsequent meeting, A and B travel a total of $4a$. Therefore, when they meet for the 2010th time, A... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,822 |
3. Given points $A(1,2)$ and $B(2,1)$, the line segment $AB$ intersects with the line $y=3x+b$. Then the range of values for $b$ is ( ).
(A) $b>-5$
(B) $b<-1$
(C) $-5<b<-1$
(D) $-5 \leqslant b \leqslant-1$ | 3. D.
When the line $y=3 x+b$ passes through point $A$, we have $2=3+b$, solving for $b$ gives $b=-1$;
When the line $y=3 x+b$ passes through point $B$, we have $1=6+b$, solving for $b$ gives $b=-5$.
Therefore, options $(\mathrm{A})$, $(\mathrm{B})$, and $(\mathrm{C})$ are eliminated. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,823 |
Example 1 The side lengths of convex quadrilateral $ABCD$ are all positive integers, and the sum of any three sides is an integer multiple of the fourth side. Prove: Among the four sides of quadrilateral $ABCD$, there are always two sides of equal length. | Proof: Let the lengths of the four sides of a quadrilateral be $a_{1}, a_{2}, a_{3}, a_{4}$, and the perimeter be $s$, and
$$
\left\{\begin{array}{l}
a_{2}+a_{3}+a_{4}=p_{1} a_{1}, \\
a_{1}+a_{3}+a_{4}=p_{2} a_{2}, \\
a_{1}+a_{2}+a_{4}=p_{3} a_{3}, \\
a_{1}+a_{2}+a_{3}=p_{4} a_{4},
\end{array}\right.
$$
where all the ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,824 |
Example 2 Given that the lengths of the two legs of a right triangle are $l$ and $m$, and the hypotenuse is $n$, and $l$, $m$, and $n$ are all positive integers, with $l$ being a prime number. Prove: $2(l+m+1)$ is a perfect square.
$(1985$, Shanghai Junior High School Mathematics Competition) | Prove that because $l^{2}+m^{2}=n^{2}$, we have $l^{2}=n^{2}-m^{2}=(n+m)(n-m)$.
Since $l$ is a prime number, and $n+m>n-m>0$, then $n+m=l^{2}, n-m=1$.
Thus, $l^{2}=2 m+1,2 m=l^{2}-1$.
Therefore, $2(l+m+1)=2 l+2 m+2$ $=l^{2}+2 l+1=(l+1)^{2}$, which means $2(l+m+1)$ is a perfect square. | (l+1)^2 | Number Theory | proof | Yes | Yes | cn_contest | false | 722,825 |
5. As shown in Figure 6, in isosceles $\triangle ABC$, it is known that $AB = AC = k BC$ ($k$ is a natural number greater than 1), points $D$ and $E$ are on sides $AB$ and $AC$ respectively, and $DB = BC = CE$, $CD$ intersects $BE$ at point $O$. Find the smallest positive integer $k$ such that $\frac{OC}{BC}$ is a rati... | Connect $D E$. It is easy to see that quadrilateral $B C E D$ is an isosceles trapezoid.
By the given condition $\angle B D C=\angle B C D=\angle E B C$, hence
$$
O C \cdot C D=B C^{2} \text {. }
$$
On the other hand,
$$
\frac{O C}{O D}=\frac{B C}{D E}=\frac{A B}{A D}=\frac{A B}{A B-D B}=\frac{k}{k-1} .
$$
Therefore,... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,826 |
Example 1 Let $O(0,0), A(1,0), B(0,1)$, and $P$ be a moving point on the line segment $AB$, $\overrightarrow{A P}=\lambda \overrightarrow{A B}$. If
$$
\overrightarrow{O P} \cdot \overrightarrow{A B} \geqslant \overrightarrow{P A} \cdot \overrightarrow{P B} \text {, }
$$
then the range of the real number $\lambda$ is $... | From the given conditions, we have
$$
\begin{array}{l}
\overrightarrow{A P}=\overrightarrow{A B}=(-\lambda, \lambda), \text { and } 0 \leqslant \lambda \leqslant 1, \\
\overrightarrow{O P}=\overrightarrow{O A}+\overrightarrow{A P}=(1-\lambda, \lambda) .
\end{array}
$$
Therefore, $\overrightarrow{P A}=(\lambda,-\lambda... | \left[1-\frac{\sqrt{2}}{2}, 1\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,827 |
Example 2 defines the length of intervals $(c, d)$, $[c, d)$, $(c, d]$, and $[c, d]$ as $d-c$ ($d>c$). Given real numbers $a$, $b$ ($a>b$). The sum of the lengths of the intervals formed by $x$ that satisfy $\frac{1}{x-a}+\frac{1}{x-b} \geqslant 1$ is ( ).
(A) 1
(B) $a-b$
(C) $a+b$
(D) 2
(2007, National High School Mat... | Notice that
$$
\begin{array}{l}
\frac{1}{x-a}+\frac{1}{x-b} \geqslant 1 \\
\Leftrightarrow \frac{2 x-(a+b)}{(x-a)(x-b)} \geqslant 1 \\
\Leftrightarrow \frac{(x-a)(x-b)+a+b-2 x}{(x-a)(x-b)} \leqslant 0 .
\end{array}
$$
Let $g(x)=(x-a)(x-b)+a+b-2 x$
$$
=x^{2}-(a+b+2) x+a b+a+b \text {. }
$$
Notice that
$$
g(a)=b-a<0 .
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,828 |
Example 3 Solve the inequality
$$
\sqrt{x^{2}-a} \geqslant|x-1|-1 \text {. }
$$
$(2008$, Zhejiang Province High School Mathematics Competition) | Solution: Obviously, when $a=0$, the solution set of the inequality is $x \in \mathbf{R}$.
Next, assume $a \neq 0$. Then
$$
\begin{array}{l}
\sqrt{x^{2}-a} \geqslant|x-1|-1 \\
\Leftrightarrow \sqrt{x^{2}-a}+1 \geqslant|x-1| \\
\Leftrightarrow\left(\sqrt{x^{2}-a}+1\right)^{2} \geqslant(x-1)^{2} \\
\Leftrightarrow 2\left... | x \in \mathbf{R} \text{ when } a \leqslant 0; \, x \in [\sqrt{a}, +\infty) \text{ when } 0 < a \leqslant 4; \, x \in \left[\frac{a}{4}+1, +\infty\right) \text{ when } a > 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,829 |
Example 4 Inequality
$$
-2<\sqrt{x^{2}-2 x+4}-\sqrt{x^{2}-10 x+28}<2
$$
The solution set is $\qquad$ .
(2006, National High School Mathematics League, Heilongjiang Province Preliminary Contest) | Solve The original inequality can be transformed into
$\left|\sqrt{x^{2}-2 x+4}-\sqrt{x^{2}-10 x+28}\right|<2$.
Let $y=\sqrt{3}$. Then the inequality becomes
$\left|\sqrt{(x-1)^{2}+(y-0)^{2}}-\sqrt{(x-5)^{2}+(y-0)^{2}}\right|<2$
The above can be seen as the absolute value of the difference in distances from the point $... | 3-\sqrt{2}<x<3+\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,830 |
2. The solution set of the inequality $(x-2) \sqrt{x^{2}-2 x-3} \geqslant 0$ is $\qquad$ . $\qquad$ (2006, National High School Mathematics League Jiangsu Province Re-match)
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | From $\left\{\begin{array}{l}x-2 \geqslant 0, \\ x^{2}-2 x-3 \geqslant 0,\end{array}\right.$ we get $\{-1\} \cup[3,+\infty)$. | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,833 |
3. The solution set of the inequality $|x|^{3}-2 x^{2}-4|x|+3<0$ is $\qquad$
(2003, National High School Mathematics Competition) | Answer: $\left(-3,-\frac{\sqrt{5}-1}{2}\right) \cup\left(\frac{\sqrt{5}-1}{2}, 3\right)$. | \left(-3,-\frac{\sqrt{5}-1}{2}\right) \cup\left(\frac{\sqrt{5}-1}{2}, 3\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,834 |
Example 3 As shown in Figure 1, in $\triangle A B C$, it is known that $\angle B$ is an acute angle. A perpendicular line is drawn from vertex $A$ to side $B C$ or its extension, intersecting at point $H$; a perpendicular line is drawn from vertex $C$ to side $A B$ or its extension, intersecting at point $K$. Try to de... | Let $B C=a, A B=c$ (where $a, c$ are positive real numbers), $\frac{2 B H}{B C}=m, \frac{2 B K}{A B}=n$ (where $m, n$ are positive integers).
Thus, $B H=\frac{m a}{2}, B K=\frac{n c}{2}$.
Since $B H<c, B K<a$, we have $\frac{m a}{2}<c, \frac{n c}{2}<a$.
Therefore, $\frac{m n a c}{4}<a c$, which means $m n<4$.
Thus, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,836 |
Problem $\mathbf{B}$ Complete graph $K_{6}$ has its edges colored red or blue. Then the graph must contain a monochromatic triangle. | Prove that in the complete graph $K_{6}$, each vertex is incident to 5 edges, 3 of which are the same color.
Let $x_{1} x_{2}$, $x_{1} x_{3}$, $x_{1} x_{4}$ all be red.
If $\triangle x_{2} x_{3} x_{4}$ is not a monochromatic triangle, then there must be one edge that is red (let $x_{2} x_{3}$ be red). Hence, $\triangle... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,838 |
A1. Arbitrarily color the edges of the complete graph $K_{v}$ with two colors (red or blue). What is the minimum number of monochromatic triangles (triangles with all three sides the same color) in the graph? | The solution to $\mathrm{A1}$ adopts the method of calculating the number of bichromatic angles discussed in problem B. First, let $v$ be even $(v=2 n)$. For a vertex $x$ that connects to $r$ red edges and $b$ blue edges, the number of bichromatic angles with $x$ as the vertex is
$$
b r \leqslant n(n-1) \text {. }
$$
... | \left\{\begin{array}{ll}
2 \mathrm{C}_{n}^{3}, & v=2 n ; \\
\mathrm{C}_{n}^{3}+\mathrm{C}_{n+1}^{3}-\left[\frac{n}{2}\right], & v=2 n+1 .
\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,839 |
A2. For the complete graph $K_{8}$, if the edges are colored with two colors, there must be two independent (i.e., having no common vertex) monochromatic triangles. However, for the complete graph $K_{7}$, the corresponding conclusion does not hold. | A2 Solution: In the graph, there is a monochromatic triangle (let $\triangle x_{6} x_{7} x_{8}$ be a red triangle).
Assume the conclusion does not hold. Then, the complete graph $K_{5}$ formed by the other 5 points has no monochromatic triangle, and thus it is as shown in Figure 2.
When $x_{6}$ is added to the comple... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,840 |
A4. Find the smallest positive integer $v$, such that for any two-coloring of the edges of the complete graph $K_{v}$, there are two monochromatic triangles that share exactly one vertex. | The solution to A4 is $v=9$.
On one hand, take two red complete graphs $K_{4}$, and color all the edges between them blue. The resulting complete graph $K_{8}$ has no two monochromatic triangles with exactly one common vertex.
On the other hand, consider any two-coloring of the complete graph $K_{9}$. By problem $\mat... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,842 |
A5. Can the edges of the complete graph $K_{25}$ be four-colored without forming a monochromatic triangle? | The solution to A5 divides 25 points into five groups, each containing 5 points, forming five complete graphs $G_{i}(1 \leqslant i \leqslant 5)$, each colored as shown in Figure 2 (using colors 1 and 2).
Then, treat each complete graph $G_{i}$ as a single point, resulting in a complete graph $K_{5}$, which is still co... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,843 |
For example, prove: for any $k\left(k \in \mathbf{N}_{+}\right)$, there exists a positive integer $n$, such that $2^{k} \mid\left(3^{n}+5\right)$. | Prove that when $k=1,2$, the conclusion is obviously true. When $k \geqslant 3$, we first prove a lemma.
Lemma: $2^{k} \|\left(3^{2 k-2}-1\right)$, and $3^{3^{t}} \equiv 1\left(\bmod 2^{k}\right)$, where $t \in \mathbf{N}_{+}$, and $t<k-2$.
Proof: When $k=3$, the conclusion is obviously true.
Assume that when $k=m$, th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,846 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.