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1. A and B take turns shooting at a target, with A starting first. Let the hit rates of the two people be $a, b \in (0,1)$. It is known that the probability of both hitting the target for the first time is the same, and once A hits the target, he stops shooting. Then the equation that the real numbers $a, b$ need to sa... | $-1 \cdot \frac{a}{1-a}=b$.
Since both miss on their first attempt, they are back to the starting point. Therefore, we only need to consider the relationship between the probability of A hitting the target on the first attempt, $a$, and the probability of A missing on the first attempt and B hitting on the first attemp... | \frac{a}{1-a}=b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,681 |
2. Let the real constant $k$ be such that the equation
$$
2 x^{2}+2 y^{2}-5 x y+x+y+k=0
$$
represents two intersecting lines in the plane coordinate system $x O y$, with the intersection point $P$. If points $A$ and $B$ lie on these two lines respectively, and $|\overrightarrow{P A}|=|\overrightarrow{P B}|=1$, then $\... | 2. $\pm \frac{4}{5}$.
Notice,
$$
\begin{array}{l}
2 x^{2}+2 y^{2}-5 x y+x+y+k \\
=(2 x-y)(x-2 y)+(2 x-y)-(x-2 y)+k \\
=(2 x-y-1)(x-2 y+1)+k+1=0 .
\end{array}
$$
Therefore, $k=-1$, the two lines are $2 x-y-1=0$ and $x-2 y+1=0$, their intersection point is $P(1,1)$.
The slopes of the two lines are $k_{1}=2, k_{2}=\fra... | \pm \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,682 |
3. In the tetrahedron $ABCD$, it is known that a plane parallel to $AB$ and $CD$ intersects the edge $BD$ in the ratio $k$. Then the ratio of the volumes of the two parts formed by the plane cutting the tetrahedron is $\qquad$ | 3. $\frac{k^{3}+3 k^{2}}{3 k+1}$.
As shown in Figure 3, complete the tetrahedron into a triangular prism, and let the volume of tetrahedron $ABCD$ be $V$.
The plane $PMNS$ is parallel to $AB$ and $CD$, dividing the tetrahedron $ABCD$ into two parts. Let the volume of the part containing point $B$ be $V_{1}$, and the ... | \frac{k^{3}+3 k^{2}}{3 k+1} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,683 |
4. Given $x, y \geqslant 0$, and $x+y \leqslant 2 \pi$. Then the function
$$
f(x, y)=\sin x+\sin y-\sin (x+y)
$$
has a maximum value of | 4. $\frac{3 \sqrt{3}}{2}$.
Notice,
$$
\begin{array}{l}
f(x, y)=\sin x+\sin y-\sin (x+y) \\
=2 \sin \frac{x+y}{2}\left(\cos \frac{x-y}{2}-\cos \frac{x+y}{2}\right) \\
\leqslant 2 \sin \frac{x+y}{2}\left(1-\cos \frac{x+y}{2}\right) \\
=8 \sin ^{3} \frac{x+y}{4} \cdot \cos \frac{x+y}{4} \\
=\frac{8}{\sqrt{3}} \sqrt{\sin ... | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,684 |
Example 9 Let real numbers $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ satisfy
$$
\left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}, \\
a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}=b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1}, \\
\min \left\{a_{1}, a_{2}, a_{3}\right\} \leqslant \min \left\{b_{1}, b_{2}, b_{3}\right\} .
\end{ar... | Proof: Without loss of generality, let $a_{1} \leqslant a_{2} \leqslant a_{3}, b_{1} \leqslant b_{2} \leqslant b_{3}$.
Then the given conditions are:
$$
\left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}, \\
a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}=b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1} \\
a_{1} \leqslant b_{1} .
\end... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,685 |
5. In the spatial quadrilateral $ABCD$, $\overrightarrow{AC}=a, \overrightarrow{BD}=$ $b, E, F$ are points on $AB, CD$ respectively, such that $\frac{AE}{EB}=\frac{CF}{FD}$ $=2012$. Then $\overrightarrow{EF}=$ $\qquad$ (express in terms of $\boldsymbol{a}, \boldsymbol{b}$). | 5. $\frac{a+2012 b}{2013}$.
As shown in Figure 4, $\overrightarrow{E F}=\overrightarrow{E B}+\overrightarrow{B C}+\overrightarrow{C F}$.
Let $\lambda=2012$. Then
$\frac{A E}{E B}=\lambda \Rightarrow \frac{E B}{A B}=\frac{1}{\lambda+1} \Rightarrow \overrightarrow{E B}=\frac{1}{\lambda+1} \overrightarrow{A B}$,
$\frac{C... | \frac{a+2012 b}{2013} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,686 |
6. Let $n=\sum_{a_{1}=0}^{2} \sum_{a_{2}=0}^{a_{1}} \cdots \sum_{a_{2} 012=0}^{a_{2} 011}\left(\prod_{i=1}^{2012} a_{i}\right)$. Then the remainder when $n$ is divided by 1000 is . $\qquad$ | 6. 191 .
It is evident that from $a_{1}$ to $a_{2012}$ forms a non-increasing sequence, and the maximum element does not exceed 2. Therefore, their product is a power of 2 or 0. Since each power of 2 can only be represented in one way (the sequence being non-increasing), we have
$$
\begin{array}{l}
n=1+2+4+\cdots+2^{2... | 191 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,687 |
7. Given $x, y, z \in \mathbf{R}$, then $\sum \frac{x^{2}}{(3 x-2 y-z)^{2}}$ has the minimum value of $\qquad$ ("sum" indicates cyclic sum).
| 7. $\frac{5}{49}$.
Let $a=\frac{4 x+2 y+z}{7}, b=\frac{4 y+2 z+x}{7}$, $c=\frac{4 z+2 x+y}{7}$.
Then the original expression $=\frac{1}{49} \sum\left(\frac{2 a-b}{a-b}\right)^{2}$ $=\frac{1}{49}\left[5+\left(\sum \frac{a}{a-b}\right)^{2}\right] \geqslant \frac{5}{49}$.
Furthermore, when $x=-1, y=0, z=4$, the equality ... | \frac{5}{49} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,688 |
8. If five vertices of a regular nonagon are colored red, then there are at least $\qquad$ pairs of congruent triangles (each pair of triangles has different vertex sets) whose vertices are all red. | 8. 4 .
A triangle with both vertices colored red is called a "red triangle". Thus, there are $\mathrm{C}_{5}^{3}=10$ red triangles. For a regular nonagon, the triangles formed by any three vertices are of only seven distinct types (the lengths of the minor arcs of the circumcircle of the regular nonagon corresponding ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,689 |
9. (16 points) Given a positive integer $n$, non-negative integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy that for $i(i=1,2, \cdots, n)$, $a_{i}+f(i)<n$, where $f(i)$ denotes the number of positive numbers in $a_{i+1}, a_{i+2}, \cdots, a_{n}$ (with the convention that $f(n)=0$). Try to find the maximum value of $\sum_{i... | Second, $a_{i}=i-1(1 \leqslant i \leqslant n)$ satisfies
$$
f(i)=n-i(1 \leqslant i \leqslant n) .
$$
Therefore, it meets the requirements. At this point,
$$
\sum_{i=1}^{n} a_{i}=\frac{n(n-1)}{2} \text {. }
$$
Next, assume $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the problem's requirements, and $a_{i_{1}}, a_{i_{2}}, \cd... | \frac{n(n-1)}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,690 |
10. (20 points) As shown in Figure 1, the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$ has an inscribed circle $x^{2}+y^{2}=b^{2}$. A tangent line to the circle intersects the ellipse at points $A$ and $B$, and the tangent point $Q$ of line $AB$ is on the left side of the $y$-axis. $F$ is the right foc... | 10. Note that,
$$
\begin{array}{l}
|A F|=\sqrt{(x-c)^{2}+y^{2}} \\
=\sqrt{x^{2}-2 c x+c^{2}+b^{2}-\frac{b^{2}}{a^{2}} \cdot x^{2}} \\
=\sqrt{\frac{c^{2}}{a^{2}} \cdot x^{2}-2 c x+a^{2}}=a-\frac{c}{a} \cdot x(\text { because } x<a), \\
|A Q|=\sqrt{A O^{2}-O Q^{2}}=\sqrt{x^{2}+y^{2}-b^{2}} \\
=\sqrt{x^{2}+b^{2}-\frac{b^{... | 2a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,691 |
11. (20 points) On a plane, there are an odd number of line segments. Two players, A and B, play the following game: They take turns (A first, then B) to assign a direction to any line segment that has not yet been assigned a direction, until after a certain turn (by A), all line segments have been assigned a direction... | 11. Player A has a winning strategy.
If there is only one line segment, obviously, A wins.
Below, let the total number of line segments be $2n+1$. Without loss of generality, assume that among the given line segments $l_{1}, l_{2}, \cdots, l_{2n+1}$, $l_{2n+1}$ is the longest. Let the projection length of $l_{i}$ $(1 ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,692 |
One. (40 points) As shown in Figure 2, given that circle $\Gamma$ is the circumcircle of $\triangle ABC$, $D$ is a point on the extension of $CB$, circle $\Gamma'$ is tangent to $\Gamma$ at point $S$, and is tangent to $AD$ and $BD$ at points $N$ and $M$ respectively. The extension of $MS$ intersects circle $\Gamma$ at... | (1) As shown in Figure 6, let the internal common tangent of circles $\Gamma$ and $\Gamma^{\prime}$ be $X S Y$, which intersects $C D$ and $A D$ at points $X$ and $Y$ respectively.
$$
\begin{array}{l}
\text { Then } \angle M N S=\angle M S X=\angle Y S T \\
=\angle S C T=\angle S A I_{a} .
\end{array}
$$
Thus, points ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,693 |
II. (40 points) Find the largest positive real number $\lambda$ such that for all positive integers $n$ and positive real numbers $a_{i} (i=1,2, \cdots, n)$, we have
$$
1+\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant \lambda\left[\sum_{k=1}^{n} \frac{1}{\left(1+\sum_{s=1}^{k} a_{s}\right)^{2}}\right] .
$$ | II. Define $S_{k}=\sum_{i=1}^{k} a_{i}+1$, and supplement the definition $S_{0}=1$. First, prove a lemma.
Lemma For any positive integer $k \geqslant 1$, we have $\frac{1}{a_{k}^{2}}+\frac{1}{S_{k-1}^{2}} \geqslant \frac{8}{S_{k}^{2}}$.
Proof Let $S_{k-1}=a_{k} t_{k}$. Then $S_{k}=a_{k}\left(t_{k}+1\right)$.
Thus, equa... | 7 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,694 |
Three, (50 points) Let $k, n$ be positive integers, $\sigma_{k}(n)$ denote the sum of the $k$-th powers of all positive divisors of $n$. Prove: For any $k \geqslant 2$, there exist infinitely many positive integers $n$ such that $n \mid \sigma_{k}(n)$. | Three, construct the sequence $\{n_i\}_{i>1}$ recursively, such that each term of this sequence of positive integers meets the requirements, and for any positive integer $n_i > 1$, $n_i$ strictly divides $n_{i+1}$.
First, assume $p$ is a prime factor of $2^k + 1$. Then $p$ is odd.
Thus, $2 \mid (p^k + 1)$.
Therefore, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,695 |
Example 1: For $n$ consecutive positive integers, if each number is written in its standard prime factorization form, and each prime factor is raised to an odd power, such a sequence of $n$ consecutive positive integers is called a "consecutive $n$ odd group" (for example, when $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{... | 【Analysis】Notice that, in a connected $n$-singular group, if there exists a multiple of 4, then by the definition of a connected $n$-singular group, it must be a multiple of 8.
Let this number be $2^{k} A\left(k, A \in \mathbf{N}_{+}, k \geqslant 3, A\right.$ is an odd number). Then $2^{k} A+4$ and $2^{k} A-4$ are bot... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,696 |
Example 2 The number of prime pairs \((a, b)\) that satisfy the equation
$$
a^{b} b^{a}=(2 a+b+1)(2 b+a+1)
$$
is \qquad (2]
(2011, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | 【Analysis】If $a$ and $b$ are both odd, then
the left side of equation (1) $\equiv 1 \times 1 \equiv 1(\bmod 2)$,
the right side of equation (1) $\equiv(2 \times 1+1+1)(2 \times 1+1+1)$ $\equiv 0(\bmod 2)$.
Clearly, the left side is not congruent to the right side $(\bmod 2)$, a contradiction.
Therefore, at least one of... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,697 |
Example 1 Let $X$ be the set of irreducible proper fractions with a denominator of 800, and $Y$ be the set of irreducible proper fractions with a denominator of 900, and let $A=\{x+y \mid x \in X, y \in Y\}$. Find the smallest denominator of the irreducible fractions in $A$. | 【Analysis】This problem is adapted from the 35th Russian Mathematical Olympiad question ${ }^{[1]}$.
Let $x=\frac{a}{800} \in X, y=\frac{b}{900} \in Y$, where,
$$
\begin{array}{l}
1 \leqslant a \leqslant 799, (a, 800)=1, \\
1 \leqslant b \leqslant 899, (b, 900)=1 .
\end{array}
$$
Then $x+y=\frac{9 a+8 b}{7200}$.
Since ... | 288 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,698 |
Example 2 Let $X$ be a subset of the set of real numbers, and satisfies the following conditions: $\frac{1}{2} \in X$, and if $x \in X$, then $\frac{1}{1+x} \in X$, $\frac{x}{1+x} \in X$. Prove: all rational numbers between $(0,1)$ belong to the set $X$. | 【Analysis】From $\frac{1}{2} \in X \Rightarrow \frac{1}{3}, \frac{2}{3} \in X$.
Similarly, from $\frac{1}{3} \in X \Rightarrow \frac{1}{4}, \frac{3}{4} \in X$,
$$
\begin{array}{l}
\frac{2}{3} \in X \Rightarrow \frac{2}{5}, \frac{3}{5} \in X, \\
\frac{1}{4} \in X \Rightarrow \frac{1}{5}, \frac{4}{5} \in X, \\
\cdots \cdo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,699 |
Example 3 Given six distinct non-zero real numbers, for any two numbers $x, y$, then $x+y, xy$ at least one is a rational number. Prove: the squares of these six numbers are all rational numbers. ${ }^{[2]}$
(2005, Russian Mathematical Olympiad) | 【Analysis】Let the six non-zero real numbers be $a_{1}, a_{2}, \cdots, a_{6}$.
(1) If one of them is a rational number, then the other five are all rational numbers.
When $x$ is a non-zero rational number,
if $x+y \in \mathbf{Q}$, then $y \in \mathbf{Q}$;
if $x y \in \mathbf{Q}$, then $y \in \mathbf{Q}$.
Therefore, we m... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,700 |
Example 4 For a positive integer $n \in \mathbf{N}_{+}$, let
$$
f_{n}=\left[2^{n} \sqrt{2008}\right]+\left[2^{n} \sqrt{2009}\right] \text {. }
$$
Prove: The sequence $\left\{f_{n}\right\}$ contains infinitely many odd numbers and infinitely many even numbers ([x] denotes the greatest integer not exceeding the real num... | 【Analysis】Let $x_{n}=\left[2^{n} \sqrt{2008}\right], y_{n}=\left[2^{n} \sqrt{2009}\right]$.
Then the parity of $f_{n}$ is closely related to the parity of $x_{n}$ and $y_{n}$. $\square$
Represent the irrational number $\sqrt{2008}$ in binary as
Since irrational numbers are infinite and non-repeating in any base represe... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,701 |
Example 5 Let $X=\left\{x_{n} \mid n \in \mathbf{N}_{+}\right\} \subseteq \mathbf{Q}, X$ contains any 2009 pairwise distinct elements whose product is an integer, and in the prime factorization of this integer, the exponents of all prime factors are less than 2009. Prove: All elements in $X$ are integers. ${ }^{[4]}$
(... | 【Analysis】Assume that not all elements in $X$ are integers, let's say $x_{1}=\frac{p_{1}}{q_{1}}$ is not an integer, where $\left(p_{1}, q_{1}\right)=1, p_{1}, q_{1} \in \mathbf{Z}, q_{1}>1$. Let $q^{*}$ be a prime factor of the denominator $q_{1}$, and other $x_{j}=\frac{p_{j}}{q_{j}}$, where $\left(p_{j}, q_{j}\right... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,702 |
Example 6 Find the smallest positive integer $n$, such that there exist rational-coefficient polynomials $f_{1}, f_{2}, \cdots, f_{n}$, satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+\cdots+f_{n}^{2}(x) .
$$
(51st IMO Shortlist) | 【Analysis】For the case $n=5$,
$$
x^{2}+7=x^{2}+2^{2}+1+1+1 \text {, }
$$
it meets the requirement.
Now we prove that $n \leqslant 4$ does not meet the requirement.
Assume there exist four rational coefficient polynomials $f_{1}(x)$, $f_{2}(x)$, $f_{3}(x)$, $f_{4}(x)$ (which may include the zero polynomial), such that ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,703 |
Example 7 Find a rational number $x$, such that $1+5 \times 2^{x}$ is the square of a rational number.
| 【Analysis】Obviously, when $x$ is a non-integer rational number, $1+5 \times 2^{x}$ is not a rational number, and thus does not satisfy the condition.
It is also obvious that $x=0,1,2$ do not satisfy the condition.
Let's start with the simpler cases.
When $x$ is a positive integer, $x \geqslant 3$.
Assume $1+5 \times 2^... | -2 \text{ or } 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,704 |
Example 8 Find the rational solutions of the equation
$$
x^{x+y}=(x+y)^{y}
$$ | 【Analysis】First, analyze several simple cases.
When $x=0$, $0^{y}=0=y^{y}$, it is only possible that $y=0$. Since $0^{0}$ is undefined, we have $x \neq 0$.
When $y=0$, $x^{x}=1 \Rightarrow x=1$, so $x=1, y=0$ is a solution.
Assume $x \neq 0, y \neq 0$ below. Let $z=\frac{y}{x}$. Then the original equation simplifies ... | x=(1+z)^{z}, y=z(1+z)^{z}(z \in \mathbf{Z}, z \neq -1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,705 |
Example 9 Find all real numbers $x$ such that $4 x^{5}-7$ and $4 x^{13}-7$ are both perfect squares. ${ }^{[6]}$
(2008, German Mathematical Olympiad) | 【Analysis】Let
$$
4 x^{5}-7=a^{2}, 4 x^{13}-7=b^{2}(a, b \in \mathbf{N}) \text {. }
$$
Then $x^{5}=\frac{a^{2}+7}{4}>1$ is a positive rational number, and $x^{13}=\frac{b^{2}+7}{4}$ is a positive rational number.
Therefore, $x=\frac{\left(x^{5}\right)^{8}}{\left(x^{13}\right)^{3}}$ is a positive rational number.
Let $x... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,706 |
Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 4)$ be given positive real numbers, $a_{1}<a_{2}<\cdots<a_{n}$, and for any positive real number $r$ satisfying $\frac{a_{j}-a_{i}}{a_{k}-a_{j}}=r(1 \leqslant i<j<k \leqslant n)$, the number of triples $(i, j, k)$ is denoted by $f_{n}(r)$. Prove: $f_{n}(r)<\frac{n^{2}}{4}$. | Prove using induction a stronger conclusion:
When $n \geqslant 4$, $f_{n}(r) \leqslant\left[\frac{(n-1)^{2}}{4}\right]$;
When $n=4$, among $(1,2,3)$ and $(1,2,4)$, at most one satisfies the condition, and among $(1,3,4)$ and $(2,3,4)$, at most one satisfies the condition, hence $f_{4}(r) \leqslant 2$.
Assume the propos... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,707 |
Example 3 Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $abcd$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$. ${ }^{[3]}$
(2011, Xin Zhi Cup Shanghai Junior High School Mathematics Competition) | 【Analysis】According to the problem, we set
$$
\begin{array}{l}
a b c d=k+(k+1)+\cdots+(k+34)\left(k \in \mathbf{N}_{+}\right) \\
\Rightarrow \frac{(2 k+34) \times 35}{2}=a b c d \\
\Rightarrow(k+17) \times 5 \times 7=a b c d .
\end{array}
$$
By symmetry, without loss of generality, let $c=5, d=7, a \leqslant b$.
We on... | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,708 |
Problem 1 If $f(x)=|x-a|+|x-b|(a<b)$, find the minimum value of this function. | Solution 1 (Geometric Method) By the geometric meaning of absolute value, the value of $|x-a|+|x-b|$ can be seen as the sum of distances from point $x$ to points $a$ and $b$ on the number line. Therefore, its minimum value is achieved when $x \in[a, b]$, and is $|a-b|$, which is $b-a$.
Solution 2 (Algebraic Method) Re... | b-a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,710 |
If
$$
f(x)=\sum_{i=1}^{n}\left|x-a_{i}\right|\left(a_{1}<a_{2}<\cdots<a_{n}\right),
$$
find the minimum value of this function. | Solve the geometric method of analogy to Problem 1, first formalize the function.
(1) When $n$ is even,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{n}\left|x-a_{i}\right| \\
=\sum_{i=1}^{\frac{n}{2}}\left(\left|x-a_{i}\right|+\left|x-a_{(n+1)-i}\right|\right) .
\end{array}
$$
Here, let $f_{i}=\left|x-a_{i}\right|+\left|x-a_... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,711 |
If
$$
f(x)=\sum_{i=1}^{n} N_{i}\left|x-a_{i}\right|\left(N_{i} \in \mathbf{N}_{+}, a_{1}<a_{2}<\cdots<a_{n}\right) \text {, }
$$
then the minimum value of $f(x)$ is attained at some segment point in the middle or between two adjacent segment points. | Solve: From the problem, we have
$$
\begin{array}{l}
f(x)=\sum_{k=1}^{2011} k\left|x-\frac{1}{k}\right| \\
=|x-1|+\left(\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{2}\right|\right)+\cdots+ \\
(\left\lvert\, \underbrace{\left|x-\frac{1}{2011}\right|+\cdots+\left|x-\frac{1}{2011}\right|}_{\text {2011 }} .\right.
\end{arr... | \frac{592043}{711} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,712 |
Problem 4 If
$$
f(x)=|x-a|-|x-b|(a<b) \text {, }
$$
find the minimum value of this function. | Solution 1 (Geometric Method): By the geometric meaning of absolute value, the value of $|x-a|-|x-b|$ can be regarded as the difference in distances from point $x$ to points $a$ and $b$ on the number line. Therefore,
$$
f(x)_{\text {min }}=a-b, f(x)_{\text {max }}=-a+b .
$$
Solution 2 (Algebraic Method): Remove the ab... | a-b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,713 |
Problem 5 If
$$
f(x)=\left|x-a_{1}\right|-\left|x-a_{2}\right|+\left|x-a_{3}\right|\left(a_{1}<a_{2}<a_{3}\right) \text {, }
$$
find the maximum and minimum values of this function. | Since there is no intuitive geometric meaning, let's use the algebraic method.
First, remove the absolute value and convert it into a piecewise function to find the maximum and minimum values. It is easy to get,
$$
f(x)=\left\{\begin{array}{ll}
-x+a_{1}-a_{2}+a_{3}, & x \leqslant a_{1} ; \\
x-a_{1}-a_{2}+a_{3}, & a_{1... | f(x)_{\text {min }}= \begin{cases} f\left(a_{1}\right) & \text{if } 2 a_{1} \geqslant a_{2}+a_{3} \\ f\left(a_{3}\right) & \text{if } 2 a_{2} < a_{1}+a_{3} \end{cases} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,715 |
Question 3 Discuss the function
$$
f(x)=\sum_{i=1}^{n}(-1)^{n-1}\left|x-a_{i}\right|\left(a_{1}<a_{2}<\cdots<a_{n}\right)
$$
the maximum and minimum values. | From the lead-in questions 4, 5, and question 2, the terms of the function can be grouped and reduced.
(1) When $n$ is even, by
$$
k=\sum_{i=1}^{n}(-1)^{n-1}=0 \text {, }
$$
we know that $f(x)$ has a minimum value and a maximum value, i.e.,
$$
f(x)_{\text {min }}=f\left(a_{1}\right), f(x)_{\max }=f\left(a_{n}\right) .... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,716 |
Question 4 Discuss the existence of the extremum of the function
$$
f(x)=\sum_{i=1}^{n} k_{i}\left|x-a_{i}\right|
$$
where $k_{i} 、 a_{i} \in \mathbf{R}, i=1,2$, $\cdots, n$ | Let $k=\sum_{i=1}^{n} k_{i}$.
From the above analysis, we know that the extremum of the function $f(x)$ is related to the sign of $k$.
(1) When $k>0$, let
$$
a=\min \left\{a_{i}\right\}, b=\max \left\{a_{i}\right\}(i=1,2, \cdots, n) .
$$
When $x<a$,
$$
\begin{array}{r}
f(x)=\sum_{i=1}^{n} k_{i}\left(a_{i}-x\right) \\
... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,717 |
Question 1 Given that the two interior angles $\angle A, \angle B$, and $\angle C$ of $\triangle ABC$ are $\frac{\pi}{7}, \frac{2\pi}{7}, \frac{4\pi}{7}$ respectively, and the three angle bisectors intersect the opposite sides at points $A', B', C'$. Prove: $\triangle A'B'C'$ is an isosceles triangle.
(2009-2010 Hungar... | Prove as shown in Figure 1, let $A A^{\prime} 、 B B^{\prime} 、 C C^{\prime}$ intersect at the incenter $I$.
In $\triangle C B^{\prime} I$ and $\triangle A B C$, by the Law of Sines, we get
$\frac{C B^{\prime}}{C I}=\frac{\sin \angle C I B^{\prime}}{\sin \angle C B^{\prime} I}=\frac{\sin \frac{3 \pi}{7}}{\sin \frac{2 \p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,718 |
Example 4 Find all prime numbers $p$ and positive integers $m$ that satisfy $2 p^{2}+p+8=m^{2}-2 m$. ${ }^{[4]}$
(2010, "Mathematics Weekly Cup" National Junior High School Mathematics Competition). | If $p=2$, then $18=m^{2}-2 m$, that is,
$$
m^{2}-2 m-18=0,
$$
which has no integer solutions, a contradiction.
Therefore, $p \geqslant 3$.
$$
\begin{array}{l}
\text { By } 2 p^{2}+p+8=m^{2}-2 m \\
\Rightarrow 2 p^{2}+p=m^{2}-2 m-8 \\
\Rightarrow p(2 p+1)=(m-4)(m+2) .
\end{array}
$$
Since $p, 2 p+1, m+2$ are all posit... | p=5, m=9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,719 |
Question 2 Let $O$ be the circumcenter of acute $\triangle ABC$, and the extensions of $AO$, $BO$, and $CO$ intersect $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. If $\triangle ABC \sim \triangle DEF$, prove: $\triangle ABC$ is an equilateral triangle. ${ }^{[3]}$
(2nd Chern Shiing-Shen Cup National H... | Prove as shown in Figure 2, extend $A D$, $B E$, and $C F$ to intersect the circumcircle $\odot O$ of $\triangle A B C$ at points $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$, and connect $A^{\prime} B^{\prime}$, $B^{\prime} C^{\prime}$, and $A^{\prime} C^{\prime}$.
Then $\triangle A B C$ and $\triangle A^{\prime} B^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,720 |
1. Given an acute triangle $\triangle A B C, L$ is a point on side $B C$, and the circle $\omega$ with center $L$ is tangent to sides $A B$ and $A C$ at points $B^{\prime}$ and $C^{\prime}$, respectively. If the circumcenter $O$ of $\triangle A B C$ lies on the minor arc $\overparen{B^{\prime} C^{\prime}}$ of circle $\... | 1. Auxiliary lines as shown in Figure 1.
Since $B^{\prime}$ is the projection of point $L$ on $A B$, point $B^{\prime}$ lies on the interior of segment $A B$.
Similarly, point $C^{\prime}$ lies on the interior of segment $A C$.
Since point $O$ is inside $\triangle A B^{\prime} C^{\prime}$, then
$$
\begin{array}{l}
\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,721 |
2. Let quadrilateral $A_{1} A_{2} A_{3} A_{4}$ not be a cyclic quadrilateral, $O_{1}$ be the circumcenter of $\triangle A_{2} A_{3} A_{4}$, with circumradius $r_{1}$. Similarly, define $O_{2}, O_{3}, O_{4}$ and $r_{2}, r_{3}, r_{4}$. Prove:
$$
\begin{array}{l}
\frac{1}{O_{1} A_{1}^{2}-r_{1}^{2}}+\frac{1}{O_{2} A_{2}^{2... | 2. Let the diagonals $A_{1} A_{3}$ and $A_{2} A_{4}$ intersect at point $M$. On each diagonal, choose a direction, and let the directed distances from point $M$ to points $A_{1}$, $A_{2}$, $A_{3}$, and $A_{4}$ be $x$, $y$, $z$, and $w$, respectively. The circumcircle $\Gamma_{1}$ of $\triangle A_{2} A_{3} A_{4}$ inters... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,722 |
3. Given a convex quadrilateral $ABCD$ with sides $AD$ and $BC$ not parallel, the circles with diameters $AB$ and $CD$ intersect at points $E$ and $F$, and $E$ and $F$ are inside the quadrilateral $ABCD$. Let $\omega_{E}$ be the circle passing through the projections of point $E$ onto lines $AB$, $BC$, and $CD$, and $\... | 3. As shown in Figure 2, let point $E$ have projections $P, Q, R, S$ on lines $DA, AB, BC, CD$ respectively.
Then points $P, Q$ lie on the circle with $AE$ as its diameter.
Thus, $\angle QPE = \angle QAE$.
Similarly, $\angle QRE = \angle QBE$.
Therefore, $\angle QPE + \angle QRE$
$$
= \angle QAE + \angle QBE = 90^{\cir... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,723 |
4. Given an acute triangle $\triangle A B C$ with circumcircle $\Omega$, the midpoints of sides $A C$ and $A B$ are $B_{0}$ and $C_{0}$, respectively. The projection of point $A$ onto side $B C$ is $D$, and $G$ is the centroid of $\triangle A B C$. Let the circle $\omega$ passing through points $B_{0}$ and $C_{0}$ be t... | 4. If $A B=A C$, then the conclusion is trivial.
Without loss of generality, assume $A B<A C$.
As shown in Figure 4, let the midpoint of side $B C$ be $A_{0}$, and $O$ be the center of circle $\Omega$, and the circumcircle of $\triangle A B_{0} C_{0}$ be $\Omega_{1}$.
Since $B_{0} C_{0} / / B C$, the circle $\Omega$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,724 |
5. Let $I$ be the incenter of $\triangle ABC$, and its circumcircle be circle $\omega$. The second intersection points of lines $AI, BI$ with circle $\omega$ are points $D, E$ respectively. Chord $DE$ intersects $AC, BC$ at points $F, G$ respectively. The line through $F$ parallel to $AD$ and the line through $G$ paral... | 5. As shown in Figure 5, let the tangents to circle $\omega$ at points $D$ and $E$ intersect at point $M$, and the extensions of $EA$ and $DB$ intersect at point $T$.
If $AE \parallel BD$, assume that $T$ is at the point at infinity.
For the degenerate hexagon $AADBBE$ inscribed in the circle, by Pascal's theorem, poin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,725 |
6. Given $\triangle A B C$ satisfies $A B=A C, D$ is the midpoint of side $A C$, the angle bisector of $\angle B A C$ intersects the circle passing through points $D, B, C$ at a point $E$ inside $\triangle A B C$, line $B D$ intersects the circle passing through points $A, E, B$ at two points $B, F$, line $A F$ interse... | 6. As shown in Figure 6, let the midpoints of sides $A B$ and $B C$ be $D'$ and $M$ respectively.
Since $A M$ is the axis of symmetry of $\triangle A B C$, point $D'$ lies on the circumcircle of $\triangle B C D$.
Because $\overparen{D' E} = \overparen{D E}$, we have
$$
\angle A B I = \angle D' B E = \angle E B D = \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,726 |
7. Given a convex hexagon $A B C D E F$ with an inscribed circle $\odot O$, and the circumcircle of $\triangle A C E$ also has the center $O$. Let the projection of point $B$ onto line $C D$ be $J$, and the line through $B$ perpendicular to $D F$ intersects line $O E$ at point $K$. The projection of $K$ onto line $D E$... | 7. As shown in Figure 7.
Since the circumcircle of $\triangle A C E$ and $\odot O$ are concentric, the lengths of the tangents drawn from points $A, C, E$ to $\odot O$ are all equal. This indicates that,
$$
A B=B C, C D=D E, E F=F A,
$$
and $\angle B C D=\angle D E F=\angle F A B$.
Consider a rotation transformation ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,727 |
In $\triangle A B C$, as shown in Figure 1, $\angle C=90^{\circ}, I$ is the incenter, line $B I$ intersects $A C$ at point $D$, through $D$ a line $D E / / A I$ intersects $B C$ at point $E$, line $E I$ intersects $A B$ at point $F$. Prove: $D F \perp A I$. | Since $\angle A I D$ is the exterior angle of $\triangle A B I$, we have
$\angle A I D=\angle B A I+\angle A B I$
$=\frac{1}{2} \angle B A C+\frac{1}{2} \angle A B C=45^{\circ}$.
Also, since $D E / / A I$, then $\angle E D I=\angle A I D=45^{\circ}$.
And $\angle E C I=\frac{1}{2} \angle A C B=45^{\circ}$, therefore, $E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,728 |
II. Positive integers $x_{1}, x_{2}, \cdots, x_{n}\left(n \in \mathbf{N}_{+}\right)$ satisfy $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=111$.
Find the maximum possible value of $S=\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}$. | Due to $111 \equiv 7(\bmod 8)$, and $x^{2} \equiv 0,1,4(\bmod 8)(x \in \mathbf{N})$, hence $n \geqslant 4$.
(1) When $n=4$,
$$
S \leqslant \sqrt{\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}{4}}=\frac{2 \sqrt{111}}{4}<\frac{22}{4} \text {. }
$$
Taking $(5,5,5,6)$ as a solution, at this time, $S=\frac{21}{4}$.
(2) Whe... | \frac{21}{4} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,729 |
Example 5 Find a prime number $p$ greater than 2, such that the parabola
$$
y=\left(x-\frac{1}{p}\right)\left(x-\frac{p}{2}\right)
$$
has a point $\left(x_{0}, y_{0}\right)$ where $x_{0}$ is a positive integer and $y_{0}$ is the square of a prime number. [5]
(2010, I Love Mathematics Junior High School Summer Camp Mat... | Let $y_{0}=t^{2}$ (where $t$ is a prime number). Then
$$
\left(x_{0}-\frac{1}{p}\right)\left(x-\frac{p}{2}\right)=t^{2} \text {, }
$$
which implies $\left(p x_{0}-1\right)\left(2 x_{0}-p\right)=2 p t^{2}$.
Since $p$ is a prime number and $p \times\left(p x_{0}-1\right)$, we have
$$
p\left|\left(2 x_{0}-p\right) \Right... | p=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,730 |
Three, let $S=\left\{x \mid x=a^{2}+a b+b^{2}, a, b \in \mathbf{Z}\right\}$. Prove:
(1) If $m \in S, 3 \mid m$, then $\frac{m}{3} \in S$;
(2) If $m, n \in S$, then $m n \in S$. | (1) Let $m=a^{2}+a b+b^{2}$.
From $3 \mid m \Rightarrow 3 \mid \left[(a-b)^{2}+3 a b\right]$
$\Rightarrow a \equiv b(\bmod 3)$.
Then $x=\frac{b-a}{3}, y=\frac{b+2 a}{3}$ are positive integers, and
$$
\begin{array}{l}
3\left(x^{2}+x y+y^{2}\right) \\
=\frac{(b-a)^{2}}{3}+\frac{(b-a)(b+2 a)}{3}+\frac{(b+2 a)^{2}}{3} \\
=... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,731 |
Four, on a plane there are $n(n \geqslant 4)$ lines. For lines $a$ and $b$, among the remaining $n-2$ lines, if at least two lines intersect with both lines $a$ and $b$, then lines $a$ and $b$ are called a "congruent line pair"; otherwise, they are called a "separated line pair". If the number of congruent line pairs a... | (1) Among these $n$ lines, if there exist four lines that are pairwise non-parallel, then any two lines are coincident line pairs. However, $\mathrm{C}_{n}^{2}=2012$ has no integer solution, so there does not exist an $n$ that satisfies the condition.
(2) If the $n$ lines have only three different directions, let the n... | 72 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,732 |
Five, given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=0, a_{n}=\frac{1}{a_{n-1}-2}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
In the sequence $\left\{a_{n}\right\}$, arbitrarily select a term $a_{k}$, and construct the sequence $\left\{b_{n}\right\}$ satisfying
$$
b_{0}=a_{k}, b_{n}=\frac{2 b_{n-1}+1}... | $$
\begin{array}{l}
a_{n-1}=\frac{1}{a_{n}}+2\left(n \in \mathbf{N}_{+}\right), \\
b_{n}=\frac{1}{b_{n-1}}+2\left(n \in \mathbf{N}_{+}\right), \\
b_{1}=\frac{1}{b_{0}}+2=\frac{1}{a_{k}}+2=a_{k-1}, \\
b_{2}=\frac{1}{b_{1}}+2=\frac{1}{a_{k-1}}+2=a_{k-2}, \\
\cdots \cdots \\
b_{k}=\frac{1}{b_{k-1}}+2=\frac{1}{a_{1}}+2=a_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,733 |
Six, let $n$ be a positive integer. Prove:
$$
\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^{2}}\right) \cdots\left(1+\frac{1}{3^{n}}\right)<2 \text {. }
$$
(Supplied by Yang Yunxin) | Let $f(n)=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^{2}}\right) \cdots\left(1+\frac{1}{3^{n}}\right)$.
When $n=1$,
$$
f(1)=1+\frac{1}{3}=2-\frac{2}{3}<2-\frac{1}{3} \text {. }
$$
When $n=2$,
$$
\begin{array}{l}
f(2)=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^{2}}\right) \\
=1+\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,734 |
Seven, As shown in Figure 2, in pentagon $A B C D E$, $B C = D E$, $C D \parallel B E$, $A B > A E$. If $\angle B A C = \angle D A E$, and $\frac{A B}{B D} = \frac{A E}{E D}$, prove: $A C$ bisects segment $B E$. | Seven, as shown in Figure 3, let $A C$ intersect $B E$ at point $M$, and the perpendicular bisector of $B E$ be $m$. Then $m$ is also the perpendicular bisector of $C D$.
Construct the symmetric point $F$ of $A$ with respect to $m$, then $\triangle A D E$ and $\triangle F C B$ are symmetric about $m$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,735 |
Let $p$ be an odd prime. If there exists a positive integer $a$ such that $p! \mid \left(a^{p}+1\right)$, prove:
(1) $\left(a+1, \frac{a^{p}+1}{a+1}\right)=p$;
(2) $\frac{a^{p}+1}{a+1}$ has no prime factors less than $p$; (3) $p! \mid (a+1)$. | Thus, $\angle B F C=\angle D A E=\angle B A C$.
Therefore, $A, B, C, F$ are concyclic.
Also, $A F \perp m, B E \perp m$, so
$A F \parallel B E$, and $F B=A E$.
Hence, quadrilateral $A E B F$ is an isosceles trapezoid.
Thus, $A, E, B, F$ are concyclic, i.e., $F, A, B, C, E$ are concyclic.
Since quadrilateral $B C D E$ ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,736 |
1. Find a triplet of integers $(l, m, n)(1<l<m<n)$, such that $\sum_{k=1}^{l} k 、 \sum_{k=l+1}^{m} k 、 \sum_{k=m+1}^{n} k$ form a geometric sequence. | Day 1
1. For $t \in \mathbf{N}_{+}$, let $S_{t}=\sum_{k=1}^{t} k=\frac{t(t+1)}{2}$. Given that $\sum_{k=1}^{l} k=S_{l}, \sum_{k=l+1}^{m} k=S_{m}-S_{l}, \sum_{k=m+1}^{n} k=S_{n}-S_{m}$ form a geometric sequence, then
$$
\begin{array}{l}
S_{l}\left(S_{n}-S_{m}\right)=\left(S_{m}-S_{l}\right)^{2} \\
\Rightarrow S_{l}\left... | (3,11,36) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,737 |
2. As shown in Figure $1, \triangle A B C$ has an incircle $\odot I$ that touches sides $A B$, $B C$, and $C A$ at points $D$, $E$, and $F$ respectively. Line $E F$ intersects $A I$, $B I$, and $D I$ at points $M$, $N$, and $K$. Prove:
$$
D M \cdot K E=D N \cdot K F \text {. }
$$ | 2. It is known that points $I, D, E, B$ are concyclic.
Also, $\angle A I D=90^{\circ}-\angle I A D$,
$\angle M E D=\angle F D A=90^{\circ}-\angle I A D$,
thus $\angle A I D=\angle M E D$.
Therefore, points $I, D, E, M$ are concyclic.
Hence, points $I, D, B, E, M$ are concyclic.
So, $\angle I M B=\angle I E B=90^{\circ... | D M \cdot K E=D N \cdot K F | Geometry | proof | Yes | Yes | cn_contest | false | 725,738 |
3. For a composite number $n$, let $f(n)$ be the sum of its smallest three positive divisors, and $g(n)$ be the sum of its largest two positive divisors. Find all positive composite numbers $n$ such that $g(n)$ equals a positive integer power of $f(n)$.
(Provided by He Yijie) | 3. Solution 1 If $n$ is odd, then all divisors of $n$ are odd. Therefore, by the problem's condition, $f(n)$ is odd and $g(n)$ is even. Thus, $g(n)$ cannot be a positive integer power of $f(n)$.
Therefore, we only need to consider the case where $n$ is even. In this case, 1 and 2 are the smallest two positive divisors... | n=4 \times 6^{l}\left(l \in \mathbf{N}_{+}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,739 |
4. Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leq 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at that time.
(Supplied by Li Shenghong) | 4. Let $f(x)=a \cos x+b \cos 2 x+$ $c \cos 3 x+d \cos 4 x$.
From $f(0)=a+b+c+d$,
$f(\pi)=-a+b-c+d$,
$f\left(\frac{\pi}{3}\right)=\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2}$,
then $a+b-c+d$
$$
=f(0)+\frac{2}{3} f(\pi)+\frac{4}{3} f\left(\frac{\pi}{3}\right) \leqslant 3 .
$$
Equality holds if and only if $f(0)=f(\pi)=f\left... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,740 |
Example 6 Given positive integers $a, b$ satisfy that $a-b$ is a prime number, and $ab$ is a perfect square. When $a \geqslant 2012$, find the minimum value of $a$.
保留源文本的换行和格式，直接输出翻译结果。 | Given the problem, let's set
$a-b=p(p$ is a prime number $), ab=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
Then $a(a-p)=k^{2} \Rightarrow a^{2}-k^{2}=ap$
$$
\Rightarrow(a+k)(a-k)=ap \text {. }
$$
Since $a+k, a, p$ are all positive integers, we have
$$
a-k>0 \text {. }
$$
Given that $p$ is a prime number, from equation ... | 2025 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,741 |
5. If a non-negative integer $m$ and the sum of its digits are both multiples of 6, then $m$ is called a "Lucky Six Number". Find the number of Lucky Six Numbers among the non-negative integers less than 2012. | 5. Solution 1 It is easy to know that a non-negative integer is a hexagonal number if and only if its last digit is even and the sum of its digits is a multiple of 6.
For convenience, let
$$
M=\{0,1, \cdots, 2011\}
$$
write each number in $M$ as a four-digit number $\overline{a b c d}$ (when it is less than four digit... | 168 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,742 |
6. Find the smallest positive integer $n$ such that
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}} \\
<\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} .
\end{array}
$$ | 6. From the known, we must have $n \geqslant 2$ 013. At this time,
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}4023, \\
\sqrt[3]{\frac{n-2013}{2011}} \geqslant \sqrt[3]{\frac{n-2011}{2013}} \\
\Leftrightarrow 2013(n-2013) \geqslant 2011(n-2011) \\
\Leftrightarrow n \geqslant 4024 .
\end{array}
$$
From equations (1) ... | 4024 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,743 |
7. As shown in Figure 2, in $\triangle A B C$, $D$ is a point on side $A C$, and $\angle A B D = \angle C$. Point $E$ is on side $A B$, and $B E = D E$. $M$ is the midpoint of side $C D$, and $A H \perp D E$ at point $H$. Given that $A H = 2 - \sqrt{3}$ and $A B = 1$. Find the degree measure of $\angle A M E$. | 7. Solution 1 As shown in Figure 4, the circumcircle $\odot O$ of $\triangle BCD$ is tangent to line $AB$ at point $B$.
Draw $AN \perp BD$ at point $N$, and $EG \parallel BD$ intersects $AC$ at point $G$. Connect $OB, OE, OG, OD, OM$.
Since $OE \perp BD, OM \perp CD$, we have
$\angle GEO = 90^\circ = \angle GMO$.
Thus... | 15^\circ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,744 |
8. Let $m$ be a positive integer, $n=2^{m}-1$, and the set of $n$ points on the number line be $P_{n}=\{1,2, \cdots, n\}$.
A grasshopper jumps on these points, each step moving from one point to an adjacent point. Find the maximum value of $m$ such that for any $x, y \in P_{n}$, the number of ways to jump from point $... | 8. When $m \geqslant 11$, $n=2^{m}-1>2013$.
Since there is only one way to jump from point 1 to point 2013 in 2012 steps, this is a contradiction, so $m \leqslant 10$.
We will now prove that $m=10$ satisfies the condition.
We use mathematical induction on $m$ to prove a stronger proposition: $\square$
For any $k \geq... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,745 |
1. As shown in Figure $1, O$ is the circumcenter of $\triangle A B C$, points $D$, $E$, and $F$ lie on segments $B C$, $C A$, and $A B$ respectively, such that $D E \perp C O$ and $D F \perp B O$. Let $K$ be the circumcenter of $\triangle A F E$. Prove: $D K \perp B C$. | 1. Let $l_{c}$ be the tangent line through point $C$ to the circumcircle of $\triangle A B C$.
Since $C O \perp l_{c}$, it follows that $l_{c} / / D E$.
Thus, $\angle C D E=\measuredangle\left(B C, l_{c}\right)=\angle B A C$.
Therefore, $B 、 D 、 E 、 A$ are concyclic.
Similarly, $C 、 D 、 F 、 A$ are concyclic.
Hence $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,746 |
3. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for all $x, y \in \mathbf{R}$,
$$
f(y f(x+y)+f(x))=4 x+2 y f(x+y)
$$
holds. | 3. Let $y=0$, we get $f(f(x))=4 x$. Then $f$ is injective.
$$
\begin{array}{l}
\text { By } f(0)=f(4 \times 0)=f(f(f(0)))=4 f(0) \\
\Rightarrow f(0)=0 .
\end{array}
$$
Let $x=0, y=1$, we get
$$
\begin{array}{l}
2 f(1)=f(f(1))=4 \Rightarrow f(1)=2 \\
\Rightarrow f(2)=f(f(1))=4 .
\end{array}
$$
Let $y=1-x$, we get
$$
\... | f(x)=2x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,748 |
4. If the set of integers $A \subseteq A+A$, where,
$$
A+A=\{a+b \mid a \in A, b \in A\},
$$
then $A$ is called "saturated"; if all integers except 0 are the sum of all elements in some non-empty finite subset of the integer set $A$, then the set $A$ is called "free". Question: Does there exist an integer set that is ... | 4. There exists an integer set that is both saturated and free.
Take the integer set
$$
A=\left\{(-1)^{n} F_{n} \mid n \in \mathbf{N}, n \geqslant 2\right\},
$$
where, $\left\{F_{n}\right\}$ is the Fibonacci sequence, i.e.,
$$
F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
By $F_{n... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,749 |
5. Let $p, q$ be prime numbers, and $n$ be a positive integer, satisfying
$$
\frac{p}{p+1}+\frac{q+1}{q}=\frac{2 n}{n+2} \text {. }
$$
Find all possible values of $q-p$. | 5. Subtract 2 from both sides of the equation to get
$$
\frac{1}{p+1}-\frac{1}{q}=\frac{4}{n+2} \text {. }
$$
Since \( n \) is a positive integer, the left side of the equation is greater than 0.
Therefore, \( q > p + 1 \).
Since \( q \) is a prime number, then \( (q, p+1) = 1 \).
Simplifying equation (1) by finding a... | 2, 3, 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,750 |
6. An infinite number of people participate in a social network, some of whom are paired as friends, meaning each person has at least one friend and at most a finite number of friends (friendship is symmetric, if $A$ is a friend of $B$, then $B$ is also a friend of $A$). Each person must designate one of their friends ... | 6. (1) For any person $A$, let $f(A)$ denote $A$'s best friend.
Let $f^{0}(A)=A, f^{k+1}(A)=f\left(f^{k}(A)\right)$.
Therefore, any $k$-th best friend must belong to someone $A$'s $f^{k}(A)$.
Let $X$ be a popular person. For any positive integer $k$, let $x_{k}$ be a person satisfying $f^{k}\left(x_{k}\right)=X$.
Sin... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,751 |
1. If $a, b$ are both integers, the equation
$$
a x^{2}+b x-2008=0
$$
has two distinct roots that are prime numbers, then $3 a+b=$ $\qquad$ (2008, Taiyuan Junior High School Mathematics Competition) | Let the two prime roots of the equation be \(x_{1} 、 x_{2}\left(x_{1}<x_{2}\right)\). From the problem, we have
\[
x_{1} x_{2}=\frac{-2008}{a} \Rightarrow a x_{1} x_{2}=-2008 \text{. }
\]
It is easy to see that, \(2008=2^{3} \times 251\) (251 is a prime number).
Thus, \(x_{1}=2, x_{2}=251\).
Therefore, \(3 a+b=1000\). | 1000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,752 |
7. Circle $\Gamma$ is the circumcircle of acute $\triangle A B C$, $H$ is the orthocenter, $K$ is a point on the minor arc $\overparen{B C}$, $L$ and $M$ are the reflections of point $K$ over lines $A B$ and $B C$, respectively, and $E$ is the other intersection point (besides point $B$) of the circumcircle of $\triang... | 7. As shown in Figure 2.
From the fact that points $E, M, B, L$ are concyclic, we have
$$
\begin{array}{l}
\angle B E M=\angle B L M . \\
\text { Since } B K=B L=B M, \text { it follows that } \\
\angle B L M=90^{\circ}-\frac{1}{2} \angle M B L \\
\quad=90^{\circ}-\left(180^{\circ}-\frac{1}{2} \angle L B K-\frac{1}{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,753 |
8. A word is a finite sequence of letters from the alphabet. If a word is formed by concatenating at least two identical sub-words, it is called "repetitive" (for example, ababab and abcabc are repetitive, while $a b a b a$ and $a a b b$ are not repetitive). Prove: if a word becomes repetitive after swapping any two ad... | 8. A word is called "constant" if all its letters are the same.
Use proof by contradiction to derive a contradiction.
First, consider a non-constant word $W$, with length $|W|=\omega$.
Since $W$ has at least two different adjacent letters, assume without loss of generality that $W=A a b B (a \neq b)$. Further assume th... | proof | Other | proof | Yes | Yes | cn_contest | false | 725,754 |
1. If $(x, y, z)$ is a solution to the system of equations
$$
\left\{\begin{array}{l}
5 x-3 y+2 z=3, \\
2 x+4 y-z=7, \\
x-11 y+4 z=3
\end{array}\right.
$$
then the value of $z$ is ( ).
(A) 0
(B) -1
(C) 1
(D) does not exist. | - 1.D.
(1) - (3) yields
$4 x+8 y-2 z=0 \Rightarrow 2 x+4 y-z=0$, which contradicts equation (2). Therefore, the system of equations has no solution. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,755 |
2. Given positive integers $x, y$. Then $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5}$ has $(\quad)$ solutions $(x, y)$.
(A) 0
(B) 1
(C) 2
(D) More than 2, but finite | 2. C.
From $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5} \Rightarrow 50 y=x^{2}(y+5)$
$\Rightarrow x^{2}=\frac{50 y}{y+5}=50-\frac{250}{y+5}$.
If $(y, 5)=1$, then
$(5, y+5)=1 \Rightarrow(y+5) \mid 2$.
Also, $y+5>2$, which leads to a contradiction. Therefore, 5 । $y$.
In equation (1), from $y+5>10$, we have $25 \leqslant x... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,756 |
3. As shown in Figure 1, in
Rt $\triangle A B C$, it is
known that $\angle C=90^{\circ}, C D$
is the altitude. Let $B C=a$,
$$
\begin{array}{l}
C A=b(a \neq b), \\
C D=h, A D=m, \\
D B=n . \text { Let }
\end{array}
$$
$$
f=a m n+b m n, g=a h m+b h n .
$$
Then the relationship between $f$ and $g$ is ( ).
(A) $f>g$
(B)... | 3. C.
From the knowledge of similar triangles, we easily get
$$
\begin{array}{l}
\frac{a}{h}=\frac{b}{m}, \frac{b}{h}=\frac{a}{n} \\
\Rightarrow \frac{a+b}{h}=\frac{a}{n}+\frac{b}{m}=\frac{a m+b n}{m n} \\
\Rightarrow a m n+b m n=a h m+b h n .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,757 |
4. Let the average of $a, b, c$ be $M$, the average of $a, b$ be $N$, and the average of $N, c$ be $P$. If $a>b>c$, then the relationship between $M$ and $P$ is ( ).
(A) $M=P$
(B) $M>P$
(C) $M<P$
(D) Uncertain | 4. B.
Notice,
$$
\begin{array}{l}
M=\frac{a+b+c}{3}, N=\frac{a+b}{2}, \\
P=\frac{N+c}{2}=\frac{a+b+2 c}{4} .
\end{array}
$$
Then $M-P=\frac{a+b-2 c}{12}>0 \Rightarrow M>P$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,758 |
6. As shown in Figure 2, the convex quadrilateral $A B C D$ is inscribed in a circle, $A B$ and $D C$ intersect at point $P$, $B C$ and $A D$ intersect at point $Q$, $P E$ and $Q F$ are tangent to the circle at points $E$ and $F$ respectively. Then, $P Q$, $P E$, and $Q F$ can form a ( ) triangle.
(A) Acute
(B) Right
(... | 6. B.
Take a point $K$ on $P Q$ such that $P, B, C, K$ are concyclic. At this time,
$$
\begin{array}{l}
\angle C K P=\angle A B C=\angle C D Q \\
\Rightarrow Q, D, C, K \text{ are concyclic. }
\end{array}
$$
$$
\begin{array}{l}
\text{Then } P E^{2}+P F^{2}=P C \cdot P D+Q C \cdot Q B \\
=P K \cdot P Q+Q K \cdot Q P \\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,760 |
1. Given $[x]$ represents the greatest integer not exceeding the real number $x$. Then the solution to the equation $x^{2}-4[x]+3=0$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $\begin{array}{l}\text { 2 } 1.1, \sqrt{5}, 3 \text {. } \\ \text { From }[x] \leqslant x \Rightarrow x^{2}+3=4[x] \leqslant 4 x \\ \Rightarrow x^{2}-4 x+3 \leqslant 0 \Rightarrow 1 \leqslant x \leqslant 3 \text {. } \\ \text { When }[x]=1 \text {, } x^{2}+3=4 \Rightarrow x=1 \text {; } \\ \text { When }[x]=2 \text {, ... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,761 |
2. Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, find the maximum value of $k$; if not, please briefly explain the reason. | 提示: As the sum of the squares of the smallest 10 distinct prime numbers is
$$
\begin{array}{l}
4+9+25+49+121+169+289+361+529+841 \\
=2397>2010,
\end{array}
$$
thus, $k \leqslant 9$.
By analyzing the parity and the fact that the square of an odd number is congruent to 1 modulo 8, it is easy to prove that $k \neq 8, k \... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,763 |
4. Two circles are concentric, with radii $R$ and $r$ ($R>r$). The vertices of the convex quadrilateral $ABCD$ are such that vertex $A$ is on the smaller circle, and vertices $B$, $C$, and $D$ are on the larger circle. The maximum area of the convex quadrilateral $ABCD$ is $\qquad$ | 4. $R(R+r)$.
As shown in Figure 4, in the convex quadrilateral $ABCD$, $DB$ is the diameter of the larger circle, $AC$ passes through the center $O$, and $AC \perp BD$. | R(R+r) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,765 |
One, (20 points) Given the function
$$
y=(a+2) x^{2}-2\left(a^{2}-1\right) x+1 \text {, }
$$
where the independent variable $x$ is a positive integer, and $a$ is also a positive integer. Find the value of $x$ for which the function value is minimized. | One, the function is organized as
$$
y=(a+2)\left(x-\frac{a^{2}-1}{a+2}\right)^{2}+1-\frac{\left(a^{2}-1\right)^{2}}{a+2},
$$
its axis of symmetry is
$$
x=\frac{a^{2}-1}{a+2}=(a-2)+\frac{3}{a+2} \text {. }
$$
Since $a$ is a positive integer, hence
$$
0 < \frac{3}{a+2} \leq 1,
$$
we have
$$
a-2 < x \leq a-1.
$$
Let $x... | x=\left\{\begin{array}{ll}
1, & a=1 ； \\
a-1, & a=2 \text { or } 3 \text { ； } \\
2 \text { or } 3, & a=4 ； \\
a-2, & a>4
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,766 |
II. (25 points) Given 12 balls, one of which has a different weight from the other 11. How can you use a balance scale without weights to find this ball in three weighings?
| Second, define the ball to be found as the "bad ball," and the other 11 balls as "good balls."
Divide the 12 balls into two groups, each with 4 balls, and number the 12 balls from (1) to (12).
First weighing: Weigh (1)(2)(3)(4) against (5)(6)(7)(8). If they balance, the bad ball is in (9)(10)(11)(12), and we know tha... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 725,767 |
Three. (25 points) As shown in Figure 3, there is a semicircle inside $\triangle ABC$, with its diameter $PQ$ on side $BC$ and tangent to $AB$ and $AC$ at points $E$ and $F$, respectively. Let $PF$ and $QE$ intersect at point $S$. Prove: $AS \perp BC$.
保留源文本的换行和格式，直接输出翻译结果如下：
Three. (25 points) As shown in Figure 3, ... | Three, as shown in Figure 5, draw perpendiculars from $A$ to $Q E$ and $P F$, intersecting $P F$ and $Q E$ at points $X$ and $Y$ respectively. Connect $X Y$, $E F$, $P E$, and $Q F$.
Then $S$ is the orthocenter of $\triangle A X Y$.
Therefore, $A S \perp X Y$.
Given $\angle P E Q = \angle P F Q = 90^{\circ}$,
$A X \per... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,768 |
1. The maximum value of the function $y=\frac{\sin \alpha \cdot \cos \alpha}{1+\cos \alpha}\left(\alpha \in\left(0, \frac{\pi}{2}\right)\right)$ is . $\qquad$ | -1. $\sqrt{\frac{5 \sqrt{5}-11}{2}}$.
Let $x=\cos \alpha\left(\alpha \in\left(0, \frac{\pi}{2}\right)\right)$. Then $x \in(0,1)$.
Given $y=\frac{x \sqrt{1-x^{2}}}{1+x}$, we have
$$
y'=\frac{1-x-x^{2}}{\sqrt{1-x^{2}}(1+x)} \text {. }
$$
Let $y'=0$. Solving this, we get $x=\frac{\sqrt{5}-1}{2}$ (discard the negative roo... | \sqrt{\frac{5 \sqrt{5}-11}{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,769 |
2. Given a line $l$ passing through the point $(0,2)$ intersects the curve $C$: $y=x+\frac{1}{x}(x>0)$ at two distinct points $M$ and $N$. Then the locus of the intersection of the tangents to the curve $C$ at $M$ and $N$ is $\qquad$.
| 2. $x=1,10 \Rightarrow k>0 \text {. } \\
\text { and } x_{1}+x_{2}=\frac{2}{1-k}>0, x_{1} x_{2}=\frac{1}{1-k}>0 \\
\Rightarrow 0<k<1 \text {. } \\
\end{array}
$$
It is easy to know that the equation of $l_{1}$ is
$$
\begin{aligned}
& y-y_{1}=\left(1-\frac{1}{x_{1}^{2}}\right)\left(x-x_{1}\right) \\
\Rightarrow & y=\le... | x=1,1<y<2 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 725,770 |
3. Given that $a$ is a constant, and real numbers $x, y, z$ satisfy
$$
(x-1)^{2}+(y-\sqrt{5})^{2}+(z+1)^{2}=a
$$
when, $-8 \leqslant 4 x-\sqrt{5} y+2 z \leqslant 2$. Then $a=$ $\qquad$ | 3. 1.
Let $4 x-\sqrt{5} y+2 z=k$. Then $-8 \leqslant k \leqslant 2$.
From the given equation, we have
$$
\frac{(4 x-4)^{2}}{16 a}+\frac{(-\sqrt{5} y+5)^{2}}{5 a}+\frac{(2 z+2)^{2}}{4 a}=1 \text {. }
$$
Using the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
(16 a+5 a+4 a)\left[\frac{(4 x-4)^{2}}{16 a}+\frac{(... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,771 |
4. Given real numbers $x, y, z \in (0, \sqrt{2})$, and satisfying
$$
\left(2-x^{2}\right)\left(2-y^{2}\right)\left(2-z^{2}\right)=x^{2} y^{2} z^{2} \text{. }
$$
Then the maximum value of $x+y+z$ is | 4.3.
$$
\begin{array}{l}
\text { Let } x=\sqrt{2} \cos \alpha, y=\sqrt{2} \cos \beta, \\
z=\sqrt{2} \cos \gamma\left(\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)\right) \text {. }
\end{array}
$$
Then the given equation transforms to
$$
\tan \alpha \cdot \tan \beta \cdot \tan \gamma=1 \text {. }
$$
Assume wi... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,772 |
5. The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{1}=1, \\
x_{i+1}-x_{i}=\sqrt{x_{i+1}+x_{i}}(i=1,2, \cdots) .
\end{array}
$$
Then the general term formula $x_{n}=$ . $\qquad$ | 5. $\frac{n^{2}+n}{2}$.
From $x_{i+1}-x_{i}=\sqrt{x_{i+1}+x_{i}}$, we know $x_{i+1}-x_{i} \geqslant 0$.
Since $x_{1}=1$, we have
$$
x_{i+1} \geqslant x_{i} \geqslant 1(i=1,2, \cdots) \text {. }
$$
Squaring both sides of $x_{i+1}-x_{i}=\sqrt{x_{i+1}+x_{i}}$ and rearranging, we get
$$
x_{i+1}^{2}-\left(2 x_{i}+1\right)... | x_{n}=\frac{n^{2}+n}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,773 |
3. Let $p$ and $5 p^{2}-2$ both be prime numbers: Find the value of $p$.
(2012, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | It is easy to prove that when $3 \times p$, $3 \mid \left(5 p^{2}-2\right)$.
Since $5 p^{2}-2>3$, thus $5 p^{2}-2$ is not a prime number, which contradicts the given condition.
Therefore, $3 \mid p$. Then $p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,774 |
6. Given that $\alpha, \beta$ are acute angles, and
$$
\begin{array}{l}
(1+\sin \alpha-\cos \alpha)(1+\sin \beta-\cos \beta) \\
=2 \sin \alpha \cdot \sin \beta .
\end{array}
$$
Then $\alpha+\beta=$ $\qquad$ | 6. $\frac{\pi}{2}$.
From the given equation, we have
$$
\begin{array}{l}
\left(2 \sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2} + 2 \sin^2 \frac{\alpha}{2}\right)\left(2 \sin \frac{\beta}{2} \cdot \cos \frac{\beta}{2} + 2 \sin^2 \frac{\beta}{2}\right) \\
\quad=8 \sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2} \cd... | \frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,775 |
7. Given the real-coefficient equation $a x^{3}-x^{2}+b x-1=0$ has three positive real roots. Then
$$
P=\frac{5 a^{2}-6 a b+3}{a^{3}(b-a)}
$$
the minimum value of $P$ is | 7. 108.
Let the three positive real roots of $a x^{3}-x^{2}+b x-1=0$ be $v_{1}, v_{2}, v_{3}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
v_{1}+v_{2}+v_{3}=\frac{1}{a}, \\
v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}=\frac{b}{a}, \\
v_{1} v_{2} v_{3}=\frac{1}{a} .
\end{array}
$$
From (1) and (2), we get $a>0, b>0$.
From... | 108 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,776 |
8. Given $a, b \in [1,3], a+b=4$. Then
$$
\left|\sqrt{a+\frac{1}{a}}-\sqrt{b+\frac{1}{b}}\right|
$$
the maximum value is $\qquad$. | 8. $\sqrt{\frac{10}{3}}-\sqrt{2}$.
From the fact that $x+\frac{1}{x}$ is an increasing function on $[1,3]$, we know that
$$
\sqrt{a+\frac{1}{a}}-\sqrt{(4-a)-\frac{1}{4-a}}
$$
is an increasing function.
Therefore, $\left|\sqrt{a+\frac{1}{a}}-\sqrt{b+\frac{1}{b}}\right| \leqslant \sqrt{\frac{10}{3}}-\sqrt{2}$.
The equa... | \sqrt{\frac{10}{3}}-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,777 |
9. (16 points) Let $a, b, c$ be positive real numbers, and $-2 < \lambda < 2$. Prove:
$$
\begin{array}{l}
\sqrt{\left(a^{2}-\lambda a b+b^{2}\right)\left(b^{2}-\lambda b c+c^{2}\right)}+ \\
\sqrt{\left(b^{2}-\lambda b c+c^{2}\right)\left(c^{2}-\lambda c a+a^{2}\right)}+ \\
\sqrt{\left(c^{2}-\lambda c a+a^{2}\right)\lef... | $$
\begin{array}{l}
a^{2}-\lambda a b+b^{2} \\
=\frac{2-\lambda}{4}(a+b)^{2}+\frac{2+\lambda}{4}(a-b)^{2} .
\end{array}
$$
Thus, we can construct the complex numbers
$$
\begin{array}{l}
z_{1}=\frac{\sqrt{2-\lambda}}{2}(a+b)+\frac{\sqrt{2+\lambda}}{2}(a-b) \mathrm{i}, \\
z_{2}=\frac{\sqrt{2-\lambda}}{2}(b+c)+\frac{\sqr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,778 |
10. (20 points) The lengths of two adjacent sides of a rectangle are $a$ and $b$ $(a \leqslant b)$. A straight line cuts the rectangle to form a right-angled triangle with a perimeter of $l$. Find the minimum value of the remaining area of the rectangle. | 10. Let the segments intercepted by the line on the shorter and longer sides of the rectangle be $x$ and $y$, respectively. Then,
$$
\begin{array}{l}
x+y+\sqrt{x^{2}+y^{2}}=l \\
\Rightarrow \sqrt{x^{2}+y^{2}}=l-x-y \\
\Rightarrow(l-x)(l-y)=\frac{l^{2}}{2} .
\end{array}
$$
First, find the maximum area of the right tria... | a b-\left(\frac{3}{4}-\frac{\sqrt{2}}{2}\right) l^{2}, \quad a \geqslant\left(1-\frac{\sqrt{2}}{2}\right) l \\ a b-\frac{a l(l-2 a)}{4(l-a)}, \quad a<\left(1-\frac{\sqrt{2}}{2}\right) l | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,779 |
11. (20 points) Let $A\left(x_{0}, y_{0}\right)$ be any point on the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
other than the vertices. Prove: there do not exist two distinct points on the hyperbola that are symmetric with respect to the normal line (the line perpendicular to the tangent line... | 11. First, prove that the equation of the normal line $l$ passing through point $A$ of the hyperbola is
$$
a^{2} y_{0} x + b^{2} x_{0} y = c^{2} x_{0} y_{0},
$$
where $c$ is the semi-focal distance of the hyperbola.
It is known that the equation of the tangent line $l_{0}$ passing through point $A$ of the hyperbola is... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,780 |
One. (40 points) Given that the circumcircle of the equilateral $\triangle ABC$ is $\odot O, P$ is a point on the arc $\overparen{BC}$, the line $AB$ intersects $CP$ at point $M$, the line $AC$ intersects $BP$ at point $N$, $D$ and $E$ are the midpoints of $BM$ and $CN$ respectively, $DE$ intersects $BN$ and $CM$ at po... | (1) As shown in Figure 1, connect $ID, IF, IE, IG, OB$. Since $DI, EI$ are the midlines of $\triangle BCM, \triangle BCN$ respectively, we have
$DI \parallel CM, EI \parallel BN$.
Thus, $\angle M = \angle BDI, \angle CBN = \angle CIE$.
Also, $\angle ABC = \angle M + \angle BCM$,
$\angle BAC = \angle BPM = \angle CBP + ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,781 |
For positive integer $n$, if there exists a permutation $a_{1}, a_{2}, \cdots, a_{n}$ of 1, 2, $\cdots, n$ satisfying
$$
a_{i+1}=\left\{\begin{array}{ll}
2 a_{i}, & 2 a_{i} \leqslant n ; \\
2 n+1-2 a_{i}, & 2 a_{i}>n
\end{array}\right.
$$
$\left(i=1,2, \cdots, n, a_{n+1}=a_{1}\right)$, then $n$ is called a "cyclic numb... | (1) It is easy to see that “$1,2,4,8,3,6,7,5,9$” and “$1,2,4,8,7,9,5,10,3,6,11$” are permutations of “$1,2, \cdots, 9$” and “$1,2, \cdots, 11$” that satisfy the definition of a cyclic number. Therefore, $9$ and $11$ are both cyclic numbers.
(2) If $2 n+1$ is not a prime number, by $2 n+1 \geqslant 9$, we know there exi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,782 |
$$
\begin{array}{l}
\text { Three. (50 points) Given } a, b, c > 0 \text {, and } \\
a^{2}+b^{2}+c^{2}+a b c=4, p=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}, \\
s=a+b+c, t=2+a b c .
\end{array}
$$
Try to compare the sizes of $p$, $s$, and $t$, and find the maximum and minimum values of $p$, $s$, and $t$ respectively. | When $a=b=c=1$, $a^{2}+b^{2}+c^{2}+a b c=4, p=s=t=3$.
When $a=b=\frac{1}{2}$, according to the problem,
$\frac{1}{4}+\frac{1}{4}+c^{2}+\frac{1}{4} c=4$
$\Rightarrow 4 c^{2}+c-14=0$
$\Rightarrow c=\frac{7}{4}$ (discard the negative root).
At this time, $p>3>s>t$.
Guess: $p \geqslant 3 \geqslant s \geqslant t$.
Next, pro... | p_{\min }=3, s_{\max }=3, t_{\max }=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,783 |
Four, (50 points) Find the maximum real number $k$, such that for any simple graph $G$ of order $(n \geqslant 3)$, the inequality
$$
x^{3} \geqslant k y^{2} \text {, }
$$
holds, where $x$ is the number of edges in the graph $G$, and $y$ is the number of triangles in the graph $G$. | First, prove a lemma.
Lemma: Let a simple graph of order $n$ have $x$ edges and $y$ triangles. Then
$$
y \leqslant \frac{n-2}{3} x \text {. }
$$
Proof: In a simple graph of order $n$, $y$ triangles have a total of $3 y$ edges, and each edge can be part of at most $n-2$ triangles, so,
$$
\frac{3 y}{n-2} \leqslant x \Ri... | \frac{9}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,784 |
4. Let $a$ be a prime number, $b$ be a positive integer, and
$$
9(2 a+b)^{2}=509(4 a+511 b) \text {. }
$$
Find the values of $a$ and $b$.
$(2008$, National Junior High School Mathematics League) | Hint: $2a + b$ can be treated as a whole, then $9(2a + b)^2 = 509[509b + 2(2a + b)]$. It is easy to see that 509 is a prime number, so 509 and 9 are coprime.
Therefore, 509 divides $(2a + b)$.
Let $2a + b = 509k\left(k \in \mathbf{N}_{+}\right)$.
Thus, $9k^2 = b + 2k$.
So, $9k^2 = 511k - 2a$.
Hence, we can find $a = 2... | a = 251, b = 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,785 |
Example 2 As shown in Figure 3, in rectangle $A B C D$, $A B=20$, $B C=10$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value. | Solve as shown in Figure 4, construct the symmetric point $B'$ of point $B$ with respect to $AC$, and connect $AB'$. Construct $B'N \perp AB$, intersecting $AC$ at point $M$, and connect $BM$. Then $BM=B'M$.
Therefore, $BM+MN=B'M+MN$.
By the shortest distance of a perpendicular segment, we know that $B'M+MN$ is minimiz... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,787 |
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