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 |
|---|---|---|---|---|---|---|---|---|---|
3. The solution set of the inequality $\frac{4 x^{2}}{(1-\sqrt{1+2 x})^{2}}<2 x+9$ is | 3. $-\frac{1}{2} \leqslant x<0$ or $0<x<\frac{45}{8}$.
The original inequality can be simplified to
$$
x^{2}(8 x-45)<0 \Rightarrow x<\frac{45}{8} \text {. }
$$
From $1+2 x \geqslant 0$, and $1-\sqrt{1+2 x} \neq 0$ $\Rightarrow x \geqslant-\frac{1}{2}$, and $x \neq 0$.
Therefore, $-\frac{1}{2} \leqslant x<0$ or $0<x<\... | -\frac{1}{2} \leqslant x<0 \text{ or } 0<x<\frac{45}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,277 |
4. As shown in Figure 1, in the Cartesian coordinate system, the vertices $A, B$ of rectangle $A O B C$ have coordinates $A(0,4), B(4 \sqrt{3}, 0)$, respectively. Construct the symmetric point $P$ of point $A$ with respect to the line $y=k x (k>0)$. If $\triangle P O B$ is an isosceles triangle, then the coordinates of... | 4. $\left(\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{33}}{3}\right)$ or $(2 \sqrt{3}, 2)$.
From the problem, we know $O P=O A=4, O B=4 \sqrt{3}$.
Since $\triangle P O B$ is an isosceles triangle, we have
$B P=B O$ or $O P=P B$.
When $B P=B O=4 \sqrt{3}$,
As shown in Figure 3, draw $P F \perp O B$.
Let $P(x, y)$. Then
$$
\be... | \left(\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{33}}{3}\right) \text{ or } (2 \sqrt{3}, 2) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,278 |
One, (20 points) Given real numbers $x$, $y$, and $a$ satisfy
$$
x+y=x^{3}+y^{3}=x^{5}+y^{5}=a \text {. }
$$
Find all possible values of $a$. | If $x=-y$, then $a=0$.
Let $x, y$ be the two roots of the quadratic equation $z^{2}-a z+p=0$.
Thus, $x+y=a$.
Then $x^{2}+y^{2}=(x+y)^{2}-2 x y=a^{2}-2 p$, $x^{3}+y^{3}=a^{3}-3 a p$,
$x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}$
$=a^{4}-4 a^{2} p+2 p^{2}$,
$$
x^{5}+y^{5}=a^{5}-5 a^{3} p+5 a p^{2} \text {. }
$... | \pm 2, \pm 1, 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,279 |
II. (25 points) From a point $P$ outside a circle $\odot O$, draw two tangents $P A$ and $P B$ and a secant $P D C$. Let $M$ be the midpoint of $P A$, and connect $C M$, which intersects $A B$ at point $E$. Prove: $D E \parallel P A$.
| $$
\begin{array}{l}
\text { Given } \angle P B D=\angle P C B, \angle B P D=\angle C P D \\
\Rightarrow \triangle P D B \sim \triangle P B C \\
\Rightarrow \frac{B D}{C B}=\frac{P D}{P B}=\frac{P B}{P C} . \\
\text { Also, } \angle P A D=\angle P C A, \angle A P D=\angle C P A \\
\Rightarrow \triangle P D A \sim \trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,280 |
Three. (25 points) Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find all possible values of $k$.
Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find ... | Three, let the equation $x^{2}+p x+k p-1=0$ have integer roots $x_{1}$ and another root $x_{2}$.
By the relationship between roots and coefficients, we know
$$
x_{1}+x_{2}=-p, x_{1} x_{2}=k p-1 \text {. }
$$
Thus, $x_{2}$ must also be an integer.
Assume $k>1$.
Notice,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,281 |
1. The function $f(x)(x \neq 1)$ defined on $\mathbf{R}$ satisfies $f(x)+2 f\left(\frac{x+2002}{x-1}\right)=4015-x$. Then $f(2004)=(\quad)$. | Let $x=2, x=2004$, we get
$$
f(2004)=2005 .
$$ | 2005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,282 |
1. People numbered $1,2, \cdots, 2015$ are arranged in a line, and a position-swapping game is played among them, with the rule that each swap can only occur between adjacent individuals. Now, the person numbered 100 and the person numbered 1000 are to swap positions, with the minimum number of swaps required being $\q... | $$
-, 1.1799 .
$$
Using the formula, the minimum number of swaps required is
$$
(1000-100) \times 2-1=1799
$$
times.
| 1799 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,283 |
2. Given a rectangular cuboid $A B C D-A_{1} B_{1} C_{1} D_{1}$ with length, width, and height of $1, 2, 3$ respectively, and $P$ is a point within the plane $A_{1} B D$. Then the minimum length of $A P$ is
| 2. $\frac{6}{7}$.
Notice that, $A P$ is shortest if and only if $A P \perp$ plane $A_{1} B D$. At this time, by the Pythagorean theorem, we get $A_{1} D=\sqrt{5}, A_{1} B=\sqrt{10}, B D=\sqrt{13}$. Then $\cos \angle B A_{1} D=\frac{\sqrt{2}}{10} \Rightarrow \sin \angle B A_{1} D=\frac{7 \sqrt{2}}{10}$. Thus, $S_{\tria... | \frac{6}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,284 |
3. The solution set of the inequality $\sin x \cdot|\sin x|>\cos x \cdot|\cos x|$ is . $\qquad$ | 3. $\left\{x \left\lvert\, \frac{\pi}{4}+2 k \pi\cos x \cdot|\cos x|$
$$
\Leftrightarrow \sin x>\cos x \text {. }
$$
Therefore, the solution set is
$$
\left\{x \left\lvert\, \frac{\pi}{4}+2 k \pi<x<\frac{5 \pi}{4}+2 k \pi(k \in \mathbf{Z})\right.\right\} .
$$ | \left\{x \left\lvert\, \frac{\pi}{4}+2 k \pi<x<\frac{5 \pi}{4}+2 k \pi(k \in \mathbf{Z})\right.\right\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,285 |
4. Let $I$ be the incenter of $\triangle A B C$, and
$$
3 \overrightarrow{I A}+4 \overrightarrow{I B}+5 \overrightarrow{I C}=0 \text {. }
$$
Then the size of $\angle C$ is $\qquad$ | 4. $\frac{\pi}{2}$.
Given that $I$ is the incenter of $\triangle ABC$, we have $a \overrightarrow{I A}+b \overrightarrow{I B}+c \overrightarrow{I C}=\mathbf{0}$. Also, $3 \overrightarrow{I A}+4 \overrightarrow{I B}+5 \overrightarrow{I C}=\mathbf{0}$, hence $a: b: c=3: 4: 5$.
Therefore, $\angle C=\frac{\pi}{2}$. | \frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,286 |
5. In the Cartesian coordinate system, let $O$ be the origin, point $A(-1,0), B(0, \sqrt{3})$, and the moving point $C$ lies on the circle
$$
(x-3)^{2}+y^{2}=4
$$
Then the maximum value of $|\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}|$ is $\qquad$. | 5. $\sqrt{7}+2$.
Let $C(3+2 \cos \theta, 2 \sin \theta)(\theta \in[0,2 \pi))$. Then
$$
\begin{array}{l}
|\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}| \\
=\sqrt{(2+2 \cos \theta)^{2}+(\sqrt{3}+2 \sin \theta)^{2}} \\
\leqslant \sqrt{2^{2}+(\sqrt{3})^{2}}+\sqrt{(2 \cos \theta)^{2}+(2 \sin \theta)^{2}} \... | \sqrt{7}+2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,287 |
6. Given 2015 positive integers $a_{1}, a_{2}, \cdots, a_{2015}$ satisfying
$$
\begin{array}{l}
a_{1}=1, a_{2}=8, \\
a_{n+1}=3 a_{n}-2 a_{n-1}(n \geqslant 2, \text { and } n \in \mathbf{N}) .
\end{array}
$$
Then the sum of all positive divisors of $a_{2015}-a_{2014}$ is $\qquad$ | 6. $8\left(2^{2014}-1\right)$.
$$
\begin{array}{l}
\text { Given } a_{n+1}=3 a_{n}-2 a_{n-1} \\
\Rightarrow a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right) . \\
\text { Then } a_{2015}-a_{2014}=2\left(a_{2014}-a_{2013}\right) \\
=2^{2}\left(a_{2013}-a_{2012}\right)=\cdots \\
=2^{2013}\left(a_{2}-a_{1}\right)=7 \times 2^{2013... | 8\left(2^{2014}-1\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,288 |
7. Let $n$ be a positive integer. From the set $\{1,2, \cdots, 2015\}$, the probability that a randomly chosen positive integer $n$ is a solution to the equation
$$
\left[\frac{n}{2}\right]=\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]
$$
is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real numb... | 7. $\frac{1007}{2015}$.
When $n=6 k\left(k \in \mathbf{Z}_{+}\right)$,
$$
\begin{array}{l}
{\left[\frac{n}{2}\right]=\left[\frac{6 k}{2}\right]=3 k,} \\
{\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\left[\frac{6 k}{3}\right]+\left[\frac{6 k}{6}\right]=2 k+k=3 k,}
\end{array}
$$
The $n$ that satisfies the equati... | \frac{1007}{2015} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,289 |
8. Given that $\alpha, \beta, \gamma$ are the three distinct roots of the equation
$$
5 x^{3}-6 x^{2}+7 x-8=0
$$
then
$$
\left(\alpha^{2}+\alpha \beta+\beta^{2}\right)\left(\beta^{2}+\beta \gamma+\gamma^{2}\right)\left(\gamma^{2}+\gamma \alpha+\alpha^{2}\right)
$$
is . $\qquad$ | 8. $-\frac{1679}{625}$.
Notice,
$$
\begin{array}{l}
\left(\alpha^{2}+\alpha \beta+\beta^{2}\right)\left(\beta^{2}+\beta \gamma+\gamma^{2}\right)\left(\gamma^{2}+\gamma \alpha+\alpha^{2}\right) \\
=\frac{5\left(\alpha^{3}-\beta^{3}\right)}{5(\alpha-\beta)} \cdot \frac{5\left(\beta^{3}-\gamma^{3}\right)}{5(\beta-\gamma)... | -\frac{1679}{625} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,290 |
9. (16 points) Given $x, y, z > 0$. Find
$$
f(x, y, z)=\frac{\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+4 z^{2}}+\sqrt{z^{2}+16 x^{2}}}{9 x+3 y+5 z}
$$
the minimum value. | 9. Introduce the parameter $\alpha$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\sqrt{x^{2}+y^{2}} \sqrt{1+\alpha^{2}} \geqslant x+\alpha y \\
\sqrt{y^{2}+4 z^{2}} \sqrt{1+\alpha^{2}} \geqslant y+2 \alpha z \\
\sqrt{z^{2}+16 x^{2}} \sqrt{1+\alpha^{2}} \geqslant z+4 \alpha x .
\end{array}
$$
Then $f(... | \frac{\sqrt{5}}{5} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,291 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=2, a_{1} a_{2} \cdots a_{n-1}=a_{n}(n \geqslant 2), \\
T_{n}=\sum_{k=1}^{n}\left[k(k-1) \log _{a_{k}} 2\right]=16-f(n) .
\end{array}
$$
Find the expression for $f(n)$. | 10. From the given, we know $a_{n} \neq 0$, and $a_{n}=\frac{a_{n+1}}{a_{n}}$.
Thus, $a_{n}=a_{n-1}^{2}=a_{n-2}^{2^{2}}=\cdots=a_{2}^{2^{n-2}}=2^{2^{n-2}}(n \geqslant 2)$.
Therefore, $a_{n}=\left\{\begin{array}{ll}2, & n=1 ; \\ 2^{2^{n-2}}, & n \geqslant 2 .\end{array}\right.$
Hence, $T_{n}=\sum_{k=1}^{n} k(k-1) \log ... | \frac{n^{2}+3 n+4}{2^{n-2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,292 |
Example 1 Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c=-1, a+b+c=4, \\
\frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}=1 .
\end{array}
$$
Find the value of $a^{2}+b^{2}+c^{2}$. | From the conditions given in the problem, we have
$$
\frac{1}{a}=-b c, \quad a=4-b-c \text {. }
$$
Notice that,
$$
\begin{array}{l}
\frac{a}{a^{2}-3 a-1}=\frac{1}{a-3-\frac{1}{a}}=\frac{1}{b c-b-c+1} \\
=\frac{1}{(b-1)(c-1)} .
\end{array}
$$
Similarly, $\frac{b}{b^{2}-3 b-1}=\frac{1}{(c-1)(a-1)}$,
$\frac{c}{c^{2}-3 c... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,293 |
Example 2 If positive numbers $a, b, c$ satisfy
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=3 \text {, }
$$
find the value of the algebraic expression
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 ... | Notice,
$$
\begin{array}{l}
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}-1 \\
=\left(\frac{b^{2}+c^{2}-a^{2}+2 b c}{2 b c}\right)\left(\frac{b^{2}+c^{2}-a^{2}-2 b c}{2 b c}\right) \\
=\frac{(b+c+a)(b+c-a)(b-c+a)(b-c-a)}{4 b^{2} c^{2}} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\left(\frac{c^{2}+a^{2}-b^{2}}{2... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,294 |
2. Given real numbers $x, y, z$ satisfy
$$
x+\frac{1}{y}=4, y+\frac{1}{z}=1, z+\frac{1}{x}=\frac{7}{3} \text {. }
$$
Find the value of $x y z$. | Multiplying the three conditional expressions yields
$$
\frac{28}{3}=x y z+\frac{1}{x y z}+\frac{22}{3} \Rightarrow x y z=1 .
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,295 |
3. Given distinct real numbers $a$, $b$, $c$ satisfying $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=t$.
Find the value of $t$. | By eliminating $b$ and $c$ from the given equation, we get
$$
\left(t^{2}-1\right)\left(a^{2}-t a+1\right)=0 \text {. }
$$
If $a^{2}-t a+1=0$, then
$$
a+\frac{1}{a}=t=a+\frac{1}{b} \Rightarrow a=b \text {, }
$$
which is a contradiction.
Thus, $t^{2}=1 \Rightarrow t= \pm 1$. | t = \pm 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,296 |
4. Given that $x, y, z$ satisfy
$$
\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}=0 \text {. }
$$
Find the value of the algebraic expression $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$. | Notice,
$$
\begin{array}{l}
x+y+z \\
=\left(\frac{x^{2}}{y+z}+x\right)+\left(\frac{y^{2}}{z+x}+y\right)+\left(\frac{z^{2}}{x+y}+z\right) \\
=(x+y+z)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right) .
\end{array}
$$
Thus, $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=-3$ or 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,297 |
5. Let non-zero real numbers $a, b, c$ be not all equal and satisfy
$$
\frac{b c}{2 a^{2}+b c}+\frac{a c}{2 b^{2}+a c}+\frac{a b}{2 c^{2}+a b}=1 \text {. }
$$
Find the value of $a+b+c$. | Let $x=\frac{2 a^{2}}{b c}, y=\frac{2 b^{2}}{a c}, z=\frac{2 c^{2}}{a b}$. Then $x y z=8$.
Therefore, the original expression can be transformed as
$$
\begin{array}{l}
\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1 \\
\Rightarrow x y z=x+y+z+2 \\
\Rightarrow x+y+z=6 .
\end{array}
$$
Substituting $a, b, c$ into the expand... | a+b+c=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,298 |
Example $1^{\prime}$ Let $n$ be a positive integer, $v_{1}, v_{2}, \cdots, v_{n+1}$ be some vectors in an $n$-dimensional linear space $V$ over the field $F_{2}$, and each $v_{i}$ satisfies $v_{i} \cdot v_{i}=1$. Prove: There exist $1 \leqslant i<j \leqslant n+1$ such that $\boldsymbol{v}_{i} \cdot \boldsymbol{v}_{j}=1... | Prove that since $V$ is an $n$-dimensional linear space, and any $n+1$ vectors in it are necessarily linearly dependent, there exist non-zero elements $\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n+1}$ in the field $F_{2}$ such that
$$
\lambda_{1} v_{1}+\lambda_{2} v_{2}+\cdots+\lambda_{n+1} v_{n+1}=0.
$$
Since the fie... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,299 |
Example 2 Find the maximum value of the positive integer $r$ that satisfies the following condition: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. ${ }^{\text {[2] }}$
(2013, Romanian National Team Selection Exam) | 【Analysis】Similarly, map the five subsets of 500 elements each to five vectors in a 1000-dimensional linear space. Since the requirement is the number of elements rather than their parity, we can consider the Euclidean space.
Let $v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$ be the five vectors after transformation.
Notice,
$$
\... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,300 |
Example 3: Color an $n \times n$ (where $n$ is even) grid in black and white such that any two different rows have exactly $\frac{n}{2}$ pairs of cells in the same column that are the same color, and $\frac{n}{2}$ pairs of cells in the same column that are different colors. Prove: In any $a$ rows and $b$ columns, the n... | To prove that by symmetry, it is sufficient to prove that the number of black squares in the $ab$ intersections of the first $a$ rows and the first $b$ columns does not exceed $\frac{ab + \sqrt{nab}}{2}$.
Map the coloring of each row of squares to a vector in $n$-dimensional Euclidean space. For each $i=1,2, \cdots, n... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,301 |
Example 4 After a single round-robin ice hockey tournament (i.e., each pair of teams played one game), for any group of teams, there exists a team (which may be in the group), whose total score against all teams in the group is odd (each game's loser gets zero points, a draw gives each team one point, and the winner ge... | 【Analysis】Obviously, the teams that have a win or loss will not change the parity of their scores. Therefore, we only need to consider which teams have drawn.
Construct a simple graph with teams as vertices and draw relationships as edges
$$
G=G(V, E) \text {. }
$$
The problem condition is: for any set of vertices $V_... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,302 |
Example 1 If all the interior angles of a convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ are equal, then it is called an equiangular $n$-sided polygon. Prove: The sum of the distances from any point $P$ inside an equiangular $n$-sided polygon to each side is a constant. | Prove that, as shown in Figure 1, for an equiangular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ (simply referred to as $n$-sided polygon $A$), construct a sufficiently large regular $n$-sided polygon $B_{1} B_{2} \cdots B_{n}$ (simply referred to as regular $n$-sided polygon $B$) to cover the $n$-sided polygon $A$, a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,303 |
Example 2 Given any 16 points in a plane, the distance between any two points does not exceed 1. Prove: there must be two points among them whose distance does not exceed $\frac{\sqrt{2}}{4}$.
| Proof: Since the distance between any two points does not exceed 1, all 16 points can be covered by a square with a side length of 1 (this is because, if $A$ and $B$ are the points with the maximum distance, draw perpendicular lines $l_{1}$ and $l_{2}$ from points $A$ and $B$ to $AB$, respectively, then any point in th... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,304 |
Example 3 Let $a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2$. Find the value of the algebraic expression $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}$. | Solve: Regarding $a$ and $b$ as the main elements, then
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b=2-c, \\
a^{2}+b^{2}=2-c^{2} .
\end{array}\right. \\
\text { Hence } a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2} \\
=c^{2}-2 c+1=(c-1)^{2} \\
\Rightarrow \frac{(1-c)^{2}}{a b}=1 .
\end{array}
$$
Similarly, $\fra... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,305 |
Example 3 A $30 \times 67$ rectangle is divided into 2010 unit square cells by lines parallel to the boundaries of the rectangle. Now, 140 crosses (each covering exactly 5 cells, and no two crosses overlap) are placed arbitrarily in the given rectangular grid. Prove: there are at least 5 empty cells in the rectangular ... | Prove that for any cross (with its center cell being $A$) placed in a rectangular grid, there are 8 cells that share a common edge with it. Together with the cross itself, there are a total of 13 cells, forming a shape called the "block $A'$" of the cross $A$, as shown in Figure 4.
For another cross $B$, as long as it... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,306 |
Example 4 Place 120 unit squares arbitrarily inside a $20 \times 25$ rectangle. Prove: there must be a gap inside the rectangle such that a small circle with a diameter of 1 can be completely placed without intersecting any of the previously placed squares. | Proof by contradiction.
Suppose that after placing 120 unit squares in a rectangle in some way, it is no longer possible to fit a small circle with a diameter of 1 completely within the rectangle. This implies that no matter how the small circle is placed, it must intersect with one of the previously placed unit square... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,307 |
Example 5: On a rectangular table, $n$ coins of the same size are placed such that there is no more space on the table to place another such coin without overlapping with the previously placed coins. Prove: If overlapping of coins is allowed and they are placed appropriately, then only $4 n$ coins are needed to complet... | Proof Let the radius of the coin be $r$.
Initially, the placed coins are denoted as $\odot O_{1}, \odot O_{2}, \cdots$, $\odot O_{n}$.
For any point $P$, if a coin $\odot P$ is placed with $P$ as its center, according to the condition, it must intersect with some previous coin $\odot O_{i}$.
Then $P O_{i} \leqslant 2 ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,308 |
Example 6 As shown in Figure 5, a rhombus $A B C D$ with side length $n$, where the vertex angle $A$ is $60^{\circ}$. Three sets of equally spaced parallel lines, each parallel to $A B$, $A D$, and $B D$, respectively, divide the rhombus into $2 n^{2}$ equilateral triangles with side length 1. Try to find the number of... | Given that any two line segments in the figure are either parallel or intersect at a $60^{\circ}$ acute angle, all trapezoids formed by the line segments in the figure are isosceles trapezoids with base angles of $60^{\circ}$.
For such trapezoids, if the intersection point of the extensions of the two non-parallel sid... | \frac{n\left(n^{2}-1\right)(2 n+1)}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,309 |
Example 1 As shown in Figure 1, in the acute $\triangle A B C$, $\angle B A C \neq 60^{\circ}$. Through points $B$ and $C$, draw the tangents $B D$ and $C E$ to the circumcircle of $\triangle A B C$, and satisfy $B D=C E=B C$. Line $D E$ intersects the extensions of $A B$ and $A C$ at points $F$ and $G$, respectively. ... | Let $\triangle A B C$ have side lengths $B C=a$, $C A=b$, and $A B=c$, with the circumcircle $\Gamma$.
Given that $B D$ and $E C$ are both tangents to circle $\Gamma$, we have
$$
\angle D B C=\angle B A C=\angle E C B.
$$
Since $B D=C E$, quadrilateral $B C E D$ is an isosceles trapezoid.
Thus, $D E \parallel B C$.
G... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,310 |
Example 2 As shown in Figure $2, A B$ is a chord of circle $\Gamma$, $P$ is a point on arc $\overparen{A B}$, and $E, F$ are points on segment $A B$ such that $A E=E F=F B$. Connecting $P E$ and $P F$ and extending them, they intersect circle $\Gamma$ at points $C$ and $D$ respectively. Prove:
$$
E F \cdot C D=A C \cdo... | Given $\triangle P F A \backsim \triangle B F D$, we know
$$
\frac{B D}{E F}=\frac{B D}{B F}=\frac{P A}{P F} \text {. }
$$
By the Law of Sines, we have
$$
\begin{array}{l}
\frac{A C}{C D}=\frac{\sin \angle A P C}{\sin \angle C P D}=\frac{\sin \angle A P E}{\sin \angle E P F} . \\
\text { Also, } 1=\frac{S_{\triangle P... | E F \cdot C D=A C \cdot B D | Geometry | proof | Yes | Yes | cn_contest | false | 728,311 |
Example 3 As shown in Figure $3, \odot O_{1} 、 \odot O_{2}$ are tangent to the three sides of $\triangle A B C$, $E 、 F 、 G 、 H$ are the points of tangency, and the extensions of $E G$ and $F H$ intersect at point $P$. Prove: $P A \perp B C$. | Prove as shown in Figure 3, construct $A D \perp B C$ at point $D$, extend $A D$ in the opposite direction, intersecting $E G$ and $F H$ at points $P_{1}$ and $P_{2}$, respectively.
It is sufficient to prove that point $P_{1}$ coincides with $P_{2}$, i.e., to prove
$P_{1} D=P_{2} D$
$\Leftarrow E D \tan \angle G E C=D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,312 |
Example 4 Quadrilateral $A B C D$ is inscribed in $\odot O$, with the lines containing sides $A B$ and $D C$ intersecting at point $P$, and the lines containing sides $A D$ and $B C$ intersecting at point $Q$. Two tangents $Q E$ and $Q F$ are drawn from point $Q$ to $\odot O$, with points of tangency $E$ and $F$ respec... | Prove as shown in Figure 4, connect $A E, C E, D E, D F$.
Since $Q E, Q F$ are both tangents to $\odot O$, we have,
$$
\begin{array}{l}
\angle A E F=\angle A D F=180^{\circ}-\angle Q D F, \\
\angle F E D=\angle Q F D. \\
\text { Also, } \angle P D A=180^{\circ}-\angle P D Q, \\
\angle D A P=\angle D C Q, \angle E D P=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,313 |
Example 5 Given that $P$ is a point inside $\triangle ABC$ and satisfies $\angle APB - \angle ACB = \angle APC - \angle ABC$, let $D$ and $E$ be the incenters of $\triangle APB$ and $\triangle APC$ respectively. Prove that $AP$, $BD$, and $CE$ are concurrent. ${ }^{[2]}$
(37th IMO) | Prove that, as shown in Figure 5, extend $A P$ to intersect $B C$ at point $K$, and intersect the circumcircle of $\triangle A B C$ at point $F$. Connect $B F$ and $C F$.
From the given, we have $\angle P B F=\angle P C F$.
By the Law of Sines, we have
\[
\frac{P F}{\sin \angle P B F}=\frac{P B}{\sin \angle P F B},
\]
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,314 |
Example 6 As shown in Figure 6, given an acute triangle $\triangle ABC$ satisfying $AB > AC$, $O$ and $H$ are the circumcenter and orthocenter of $\triangle ABC$ respectively. Line $BH$ intersects $AC$ at point $B_{1}$, and line $CH$ intersects $AB$ at point $C_{1}$. If $OH \parallel B_{1}C_{1}$, prove:
$$
\cos 2B + \c... | Prove as shown in Figure 6, connect $A O$ and $A H$.
Since $A, B_{1}, H, C_{1}$ are concyclic, and $O H \parallel B_{1} C_{1}$, then
$$
\angle O H C_{1}=\angle H C_{1} B_{1}=\angle H A B_{1}=90^{\circ}-\angle C.
$$
Also, since $B, C, B_{1}, C_{1}$ are concyclic, thus,
$\angle A H C_{1}=\angle A B_{1} C_{1}=\angle A B ... | \cos 2B + \cos 2C + 1 = 0 | Geometry | proof | Yes | Yes | cn_contest | false | 728,315 |
Example 4 Given real numbers $x, y, z, u$ satisfy
$$
\frac{x}{y+z+u}=\frac{y}{z+u+x}=\frac{z}{u+x+y}=\frac{u}{x+y+z} \text {. }
$$
Find the value of $\frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{z+u}{x+y}+\frac{u+x}{y+z}$. | Let
$$
\frac{x}{y+z+u}=\frac{y}{z+u+x}=\frac{z}{u+x+y}=\frac{u}{x+y+z}=k \text {. }
$$
Then $x=k(y+z+u), y=k(z+u+x)$,
$$
z=k(u+x+y), u=k(x+y+z) \text {. }
$$
Adding the above four equations yields
$$
x+y+z+u=3 k(x+y+z+u) \text {. }
$$
(1) If $x+y+z+u=0$, then
$$
\begin{array}{l}
\frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{... | -4 \text{ or } 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,316 |
Example 7 Given seven circles, six smaller circles are inside a larger circle, each smaller circle is tangent to the larger circle, and is tangent to the two adjacent smaller circles. If the points of tangency of the six smaller circles with the larger circle are sequentially $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$,... | Proof As shown in Figure 7, let the radius of the large circle $\odot O$ be 1, and the radii of the small circles be $r_{1}, r_{2}, \cdots, r_{6}$. Denote $\angle A_{1} O A_{2}=\alpha$.
Then $A_{1} A_{2}=2 \sin \frac{\alpha}{2}$.
By the cosine rule, we have
$$
\begin{array}{l}
\cos \alpha=\frac{\left(1-r_{1}\right)^{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,317 |
Example 8 As shown in Figure 8, given that $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively, $AD$ is the altitude on side $BC$, and point $I$ lies on segment $OD$. Prove: the circumradius $R$ of $\triangle ABC$ equals the exradius $r_{a}$ on side $BC$.
(1998, National High School Mathematic... | $$
\begin{array}{l}
\text { Hence } \frac{A I}{I K}=\frac{A D}{O K}=\frac{c \sin B}{R}=2 \sin B \cdot \sin C \text {. } \\
\text { Also, } \angle A B I=\angle I B C=\frac{1}{2} \angle B \text {, } \\
\angle C B K=\angle C A K=\frac{1}{2} \angle A \text {, } \\
\angle A K B=\angle A C B=\angle C, \\
\angle B A K=\frac{1... | R = r_{a} | Geometry | proof | Yes | Yes | cn_contest | false | 728,318 |
Given that $a$, $b$, and $c$ are positive real numbers, and $abc = 1$. Prove:
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{3}{a + b + c} \geqslant 4.
$$
(2001, USA Mathematical Olympiad Summer Program) | Proof: Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$. Then $x y z=1$. Hence the original inequality is equivalent to
$$
x+y+z+\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \geqslant 4 \text {. }
$$
In fact,
$$
\begin{array}{l}
x+y+z+\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \\
=x+y+z+\frac{3}{x y+y z+z x} \\
\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,319 |
In a convex quadrilateral $ABCD$, it is given that $\angle ABC = \angle CDA = 90^{\circ}$. $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ is inside $\triangle SCT$, and
$$
\begin{array}{l}
\angle CHS - \angle CSB = 90^{\circ}, \\
\angle ... | Proof 1 As shown in Figure 1, extend $CB$ to point $E$ such that $CB = BE$; extend $CD$ to point $F$ such that $CD = DF$.
Let $X$ be the intersection of line $EC$ and $SH$, and $Y$ be the intersection of line $FC$ and $TH$. Then point $X$ lies on the extension of segment $EC$, and point $Y$ lies on the extension of se... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,320 |
1. Find all functions $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, such that for all positive integers $m, n$, we have
$$
\left(m^{2}+f(n)\right) \mid(m f(m)+n) .
$$ | 1. The function $f(n)=n$ satisfies the condition.
Let $m=n=2$. Then
$$
(4+f(2)) \mid(2 f(2)+2) \text {. }
$$
Since $2 f(2)+2<2(4+f(2))$, we have
$$
2 f(2)+2=4+f(2) \text {. }
$$
Thus, $f(2)=2$.
Let $m=2$. Then $(4+f(n)) \mid(4+n)$.
This indicates that for any positive integer $n$, we have $f(n) \leqslant n$.
Let $m=... | f(n)=n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,321 |
3. Prove: There exist infinitely many positive integers $n$, such that the greatest prime factor of $n^{4}+$ $n^{2}+1$ equals the greatest prime factor of $(n+1)^{4}+(n+1)^{2}+1$. | 3. Let $p_{n}$ be the largest prime factor of $n^{4}+n^{2}+1$, and $q_{n}$ be the largest prime factor of $n^{2}+n+1$. Then $p_{n}=q_{n^{2}}$.
For positive integers $n \geqslant 2$, we have
$$
\begin{array}{l}
n^{4}+n^{2}+1=\left(n^{2}+1\right)^{2}-n^{2} \\
=\left(n^{2}-n+1\right)\left(n^{2}+n+1\right) \\
=\left[(n-1)^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,322 |
4. Does there exist an infinite sequence of non-zero digits $a_{1}, a_{2}, \cdots$ and a positive integer $N$, such that for each integer $k>N$, $\overline{a_{k} a_{k-1} \cdots a_{1}}$ is a perfect square. | 4. Does not exist.
Assume there exists a sequence $a_{1}, a_{2}, \cdots$ and a positive integer $N$. For each positive integer $k$, let $y_{k}=\overline{a_{k} a_{k-1} \cdots a_{1}}$. Then for each integer $k>N$, there exists a positive integer $x_{k}$ such that $y_{k}=x_{k}^{2}$.
(1) For each positive integer $n$, let... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,323 |
5. Given a fixed integer $k \geqslant 2$. Two players, A and B, play the following number game. Before the game starts, an integer $n(n \geqslant k)$ is written on the blackboard, and the two players take turns to perform operations, with A making the first move. Each player erases the number $m$ written on the blackbo... | 5. Since the number on the blackboard strictly decreases with each operation, the game must stop after a finite number of operations. Therefore, there will always be a winning strategy for one of the players.
If an integer $n (n \geqslant k)$ is a bad number, then when the game starts with the number $n$ on the blackb... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,324 |
Example 5 Given positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{3}-x y z=-5, \\
y^{3}-x y z=2, \\
z^{3}-x y z=21 .
\end{array}\right.
$$
Find the value of $x+y+z$. | Let $x y z=k$. Then
$$
\left\{\begin{array}{l}
x^{3}=k-5, \\
y^{3}=k+2, \\
z^{3}=k+21 .
\end{array}\right.
$$
Multiplying the above three equations, we get
$$
\begin{array}{l}
k^{3}=(x y z)^{3}=(k-5)(k+2)(k+21) \\
\Rightarrow 18 k^{2}-73 k-210=0 \\
\Rightarrow k_{1}=6, k_{2}=-\frac{35}{18} \text { (not suitable, disca... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,327 |
1. The range of the function $y=\frac{x^{2}+2 x+3}{x^{2}+4 x+5}(x \in \mathbf{R})$ is | ,$- 1 . y \in[2-\sqrt{2}, 2+\sqrt{2}]$.
From the problem, we know
$$
y=\frac{x^{2}+2 x+3}{x^{2}+4 x+5}
$$
is equivalent to the equation in terms of $x$
$$
(y-1) x^{2}+(4 y-2) x+(5 y-3)=0
$$
having real roots, which means
$$
\begin{array}{l}
(2 y-1)^{2}-(y-1)(5 y-3) \geqslant 0 \\
\Rightarrow y^{2}-4 y+2 \leqslant 0 \... | y \in [2-\sqrt{2}, 2+\sqrt{2}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,328 |
2. Function
$$
y=\tan 2013 x-\tan 2014 x+\tan 2015 x
$$
The number of zeros of the function in the interval $[0, \pi]$ is $\qquad$. | 2. 2014 .
$$
\begin{aligned}
y & =\tan 2013 x-\tan 2014 x+\tan 2015 x \\
& =\frac{\sin (2013 x+2015 x)}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{\sin 4028 x}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{2 \sin 2014 x \cdot \cos 2014 x}{\cos 2013 x \cdot \c... | 2014 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,329 |
3. Let point $A(2,1)$, moving point $B$ on the $x$-axis, and moving point $C$ on the line $y=x$. Then the minimum value of the perimeter of $\triangle A B C$ is $\qquad$ . | 3. $\sqrt{10}$.
Let $P(2,-1), Q(1,2)$ be the points symmetric to point $A$ with respect to the $x$-axis and the line $y=x$, respectively. Then the perimeter of $\triangle A B C$ is
$$
P B+B C+C Q \geqslant P Q=\sqrt{10} .
$$ | \sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,330 |
4. Let $P_{1}$ and $P_{2}$ be two points on a plane, $P_{2 k+1}\left(k \in \mathbf{Z}_{+}\right)$ be the symmetric point of $P_{2 k}$ with respect to $P_{1}$, and $P_{2 k+2}$ be the symmetric point of $P_{2 k+1}$ with respect to $P_{2}$. If $\left|P_{1} P_{2}\right|=1$, then $\left|P_{2013} P_{2014}\right|=$ $\qquad$ . | 4.4024.
From the problem, we know
$$
\begin{array}{l}
\left\{\begin{array}{l}
P_{2 k+1}=2 P_{1}-P_{2 k}, \\
P_{2 k+2}=2 P_{2}-P_{2 k+1}
\end{array}\right. \\
\Rightarrow P_{2 k+2}=2\left(P_{2}-P_{1}\right)+P_{2 k} \\
\Rightarrow\left\{\begin{array}{l}
P_{2 k+2}=2 k\left(P_{2}-P_{1}\right)+P_{2}, \\
P_{2 k+1}=2 k\left(... | 4024 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,331 |
6. Let the complex number $z$ satisfy $\left|z+\frac{1}{z}\right| \leqslant 2$. Then the range of $|z|$ is $\qquad$ . | 6. $[\sqrt{2}-1, \sqrt{2}+1]$.
Let $|z|=r$. Then
$$
\begin{array}{l}
\left|1-r^{2}\right| \leqslant\left|1+z^{2}\right| \leqslant 2 r \\
\Rightarrow\left(1-r^{2}\right)^{2} \leqslant 4 r^{2} \\
\Rightarrow 3-2 \sqrt{2} \leqslant r^{2} \leqslant 3+2 \sqrt{2} \\
\Rightarrow \sqrt{2}-1 \leqslant r \leqslant \sqrt{2}+1 .
... | [\sqrt{2}-1, \sqrt{2}+1] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,333 |
7. Let the moving points be $P(t, 0), Q(1, t)(t \in[0,1])$. Then the area of the plane region swept by the line segment $P Q$ is $\qquad$ . | 7. $\frac{1}{6}$.
Let $l_{P Q}: y=\frac{t}{1-t}(x-t)$.
Fix $x \in[0,1]$, as $t$ varies in the interval $[0, x]$,
$$
y=2-x-\left(\frac{1-x}{1-t}+1-t\right)
$$
the range of values is $0 \leqslant y \leqslant 2-x-2 \sqrt{1-x}$.
Therefore, the area of the desired planar region is
$$
\int_{0}^{1}(2-x-2 \sqrt{1-x}) \mathr... | \frac{1}{6} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 728,334 |
8. The probability of selecting four vertices from a regular 12-sided polygon such that no two are adjacent is $\qquad$ . | 8. $\frac{10}{99}$.
Notice that, the total number of ways to choose four vertices from 12 vertices is $\mathrm{C}_{12}^{4}$.
Label the vertices of the regular 12-gon as $1 \sim 12$, and let the numbers of the four chosen vertices, in ascending order, be $a, b, c, d$.
If they are pairwise non-adjacent, then
$$
\begin{... | \frac{10}{99} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,335 |
9. (21 points) Given positive real numbers $x, y, z$ satisfying $x+y+z=1$.
Prove:
$$
\frac{z-y}{x+2 y}+\frac{x-z}{y+2 z}+\frac{y-x}{z+2 x} \geqslant 0 .
$$ | II, 9. According to the mean (or Cauchy) inequality, we have
$$
\begin{array}{l}
{[(x+2 y)+(y+2 z)+(z+2 x)]} \\
\left(\frac{1}{x+2 y}+\frac{1}{y+2 z}+\frac{1}{z+2 x}\right) \\
\geqslant 9 \sqrt[3]{(x+2 y)(y+2 z)(z+2 x)} . \\
\sqrt[3]{\frac{1}{x+2 y} \cdot \frac{1}{y+2 z} \cdot \frac{1}{z+2 x}} \\
=9 .
\end{array}
$$
T... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,336 |
10. (21 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{n+1}=\frac{a_{n}^{2}+3}{2 a_{n}}(n \geqslant 1) .
$$
Prove: (1) When $n \geqslant 2$, the sequence $\left\{a_{n}\right\}$ is strictly monotonically decreasing;
(2) When $n \geqslant 1$,
$$
\left|a_{n+1}-\sqrt{3}\right|=\frac{2 \sqrt{3} r^{2... | 10. (1) When $n \geqslant 2$, by the AM-GM inequality we have
$$
a_{n}=\frac{a_{n-1}^{2}+3}{2 a_{n-1}} \geqslant \sqrt{3} \text {. }
$$
Since $a_{n}$ is a rational number, we have $a_{n}>\sqrt{3}$.
Thus, $a_{n+1}-a_{n}=\frac{3-a_{n}^{2}}{2 a_{n}}<0$, which means the sequence $\left\{a_{n}\right\}$ is strictly decreasi... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,337 |
Example 6 Given real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{4}+y^{2}+4=5 y z, \\
y^{4}+z^{2}+4=5 z x, \\
z^{4}+x^{2}+4=5 x y .
\end{array}\right.
$$
Find the value of $x+y+z$. | $$
\begin{array}{l}
\left(x^{4}+y^{4}+z^{4}\right)+\left(x^{2}+y^{2}+z^{2}\right)-5(x y+y z+z x)+12 \\
=\left(x^{2}-2\right)^{2}+\left(y^{2}-2\right)^{2}+\left(z^{2}-2\right)^{2}+ \\
5\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x)\right] \\
=\left(x^{2}-2\right)^{2}+\left(y^{2}-2\right)^{2}+\left(z^{2}-2\right)^{2}... | x+y+z=3\sqrt{2} \text{ or } -3\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,338 |
11. (22 points) Given a convex quadrilateral $ABCD$ in the plane with an area of 1. Prove:
$$
\begin{array}{l}
|AB|+|AC|+|AD|+|BC|+|BD|+|CD| \\
\geqslant 4+2 \sqrt{2} .
\end{array}
$$ | 11. Suppose the convex quadrilateral $ABCD$ satisfies
$$
L=|AB|+|AC|+|AD|+|BC|+|BD|+|CD|
$$
is minimized. In this case, the quadrilateral $ABCD$ must be a rhombus. Otherwise, as shown in Figure 2, we can fix two diagonal points, say points $B$ and $D$. Draw lines through points $A$ and $C$ parallel to $BD$, and adjust... | 4+2\sqrt{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,339 |
12. (22 points) Prove: (1) The equation $x^{3}-x-1=0$ has exactly one real root $\omega$, and $\omega$ is an irrational number;
(2) $\omega$ is not a root of any quadratic equation with integer coefficients
$$
a x^{2}+b x+c=0(a, b, c \in \mathbf{Z}, a \neq 0)
$$ | 12. (1) Let $f(x)=x^{3}-x-1$. Then $f^{\prime}(x)=3 x^{2}-1$.
Therefore, $f(x)$ is monotonically increasing in the interval $\left(-\infty,-\frac{1}{\sqrt{3}}\right)$, and it attains a maximum value of $\frac{2}{3 \sqrt{3}}-10$ at $x=-\frac{1}{\sqrt{3}}$. It is known that the equation $f(x)=0$ has a unique real root $\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,340 |
1. Given a positive integer $n \geqslant 2$, let non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n}=n$. Find the maximum and minimum values of $\sum_{i=1}^{n-1}\left(x_{i}-1\right)\left(x_{i+1}-1\right)$. | 1. Note that,
$$
\begin{aligned}
f & =\sum_{i=1}^{n-1}\left(x_{i} x_{i+1}-x_{i}-x_{i+1}+1\right) \\
& =\sum_{i=1}^{n-1} x_{i} x_{i+1}+x_{1}+x_{n}-(n+1) .
\end{aligned}
$$
When $n=2$, $x_{1}+x_{2}=2, f=x_{1} x_{2}-1$.
Since $x_{1}, x_{2}$ are non-negative real numbers, we have $f \geqslant-1$.
When $x_{1}=0, x_{2}=2$, ... | \frac{(n+1)(n-3)}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,341 |
2. As shown in Figure 1, given that $\odot P$ and $\odot Q$ are internally tangent to $\odot O$ at points $A$ and $B$, respectively, and $\odot P$ and $\odot Q$ are externally tangent to each other. The internal common tangent of the two circles intersects $AB$ at point $C$. Prove: $OC$ bisects $PQ$. | 2. As shown in Figure 2, draw a diameter $M N$ through point $O$ parallel to $P Q$. Let $\odot P$ and $\odot Q$ be externally tangent at point $R$.
From $\triangle A P R \backsim \triangle A O N$, we know that points $A$, $R$, and $N$ are collinear.
Similarly, points $B$, $R$, and $M$ are collinear.
Extend $M A$ and $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,342 |
4. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=0, a_{2}=1$, and for all $n \geqslant 3, a_{n}$ is the smallest positive integer greater than $a_{n-1}$ such that there is no subsequence of $a_{1}, a_{2}, \cdots, a_{n}$ that forms an arithmetic sequence. Find $a_{2014}$. | 4. First, prove a lemma using mathematical induction.
Lemma A non-negative integer appears in the sequence if and only if its ternary expansion contains only 0 and 1.
Proof It is obvious that the proposition holds for 0.
Assume the proposition holds for all non-negative integers less than \( N \), and consider \( N \)... | 88327 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,344 |
5. Let real numbers $a, b, c, d$ be distinct. Prove:
$$
\left|\frac{a}{b-c}\right|+\left|\frac{b}{c-d}\right|+\left|\frac{c}{d-a}\right|+\left|\frac{d}{a-b}\right| \geqslant 2 .
$$ | 5. Note that,
$$
\begin{array}{l}
\frac{|a|}{|b-c|}+\frac{|b|}{|c-d|}+\frac{|c|}{|d-a|}+\frac{|d|}{|a-b|} \\
\geqslant \frac{|a|}{|b|+|c|}+\frac{|b|}{|c|+|d|}+\frac{|c|}{|d|+|a|}+\frac{|d|}{|a|+|b|} .
\end{array}
$$
Assume $a, b, c, d \geqslant 0$, where at most one of $a, b, c, d$ is 0. Prove:
$$
\frac{a}{b+c}+\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,345 |
6. Let $A, B$ be finite sets of real numbers, and define:
$$
A-B=\{a-b \mid a \in A, b \in B\} \text {. }
$$
Let $E(A, B)$ be the set of all ordered quadruples $\left(a_{1}, a_{2}, b_{1}, b_{2}\right)$ that satisfy the following conditions:
(1) $a_{1}, a_{2} \in A, b_{1}, b_{2} \in B$;
(2) $a_{1}+b_{1}=a_{2}+b_{2}$.
... | 6. Let $A-B=C=\left\{c_{1}, c_{2}, \cdots, c_{k}\right\}$,
$$
\begin{array}{l}
N_{i}=\left\{(a, b) \mid a-b=c_{i}, a \in A, b \in B\right\}, \\
n_{i}=\left|N_{i}\right|,
\end{array}
$$
where $i=1,2, \cdots, k$.
Then $n_{1}+n_{2}+\cdots+n_{k}=|A||B|,|A-B|=k$.
For $\left(a_{1}, a_{2}, b_{1}, b_{2}\right) \in E(A, B)$, w... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,346 |
7. Given a positive integer $n$. Find the smallest positive integer $k$, such that for any $d$ real numbers $a_{1}, a_{2}, \cdots, a_{d}$ satisfying
$$
a_{1}+a_{2}+\cdots+a_{d}=n\left(0 \leqslant a_{i} \leqslant 1, i=1,2, \cdots, n\right)
$$
they can be divided into no more than $k$ groups, and the sum of all numbers ... | 7. First prove: $k \geqslant 2 n-1$.
Take $d=2 n-1, a_{1}=a_{2}=\cdots=a_{2 n-1}=\frac{n}{2 n-1}$, it is easy to see that these numbers are all greater than $\frac{1}{2}$.
Therefore, any two of these numbers cannot be placed in the same group.
Thus, at least $2 n-1$ groups are needed.
Next, prove: $k \leqslant 2 n-1$.... | 2n-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,347 |
8. Let positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}<a_{2}<\cdots<a_{n}$. Prove,
$$
\sum_{i=1}^{n}\left(a_{i}, a_{i+1}\right)=a_{n} \Leftrightarrow \sum_{i=1}^{n} \frac{1}{\left[a_{i}, a_{i+1}\right]}=\frac{1}{a_{1}},
$$
where, $a_{n+1}=a_{1},(x, y) 、[x, y]$ represent the greatest common divisor and t... | For positive integers $m, n (m < n)$, we have
$$
(m, n) = (m, n-m) \leqslant \min \{m, n-m\}.
$$
Since $a_{1} < a_{2} < \cdots < a_{n}, a_{n+1} = a_{1} < a_{n}$, according to equation (1), we have
$$
\begin{array}{l}
\sum_{i=1}^{n} (a_{i}, a_{i+1}) = (a_{1}, a_{n}) + \sum_{i=1}^{n-1} (a_{i}, a_{i+1}) \\
\leqslant a_{1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,348 |
Example 7 Given positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=9, \\
y^{2}+y z+z^{2}=16, \\
z^{2}+z x+x^{2}=25 .
\end{array}\right.
$$
Find the value of $x y+y z+z x$. | As shown in Figure 1, with $P$ as the endpoint,
draw rays $P A, P B, P C$, such that
$$
\begin{array}{l}
\angle A P B=\angle B P C \\
=\angle C P A=120^{\circ} .
\end{array}
$$
Let the lengths of segments $P A, P B, P C$ be $x, y, z$ respectively.
By the Law of Cosines, we have
$$
\begin{array}{l}
A B=\sqrt{x^{2}+x y+... | 8 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,349 |
1. Prove: In the prime factorization of the product of any 10 consecutive three-digit numbers, there are at most 23 distinct prime factors. | First, in the prime factorization of each three-digit number, at most two prime factors greater than 10 can appear; otherwise, their product would exceed 1000, which is impossible.
Second, in any sequence of 10 consecutive three-digit numbers, there is one that is a multiple of 10, and in its prime factorization, at m... | 23 | Number Theory | proof | Yes | Yes | cn_contest | false | 728,350 |
2. In $\triangle A B C$, it is known that $\angle C=100^{\circ}$, points $P$ and $Q$ are both on side $A B$, such that $A P=B C, B Q=A C$, the midpoints of segments $A B$, $C P$, and $C Q$ are $M$, $N$, and $K$ respectively. Find $\angle N M K$. | 2. As shown in Figure 1, extend $\triangle A B C$ to form $\triangle C B D$.
Thus, $M$ is the midpoint of line segment $C D$.
Since $A P = B C = A D, B Q = A C = B D$, therefore,
$$
\begin{array}{c}
\angle Q D P = \angle A D P + \angle B D Q - \angle A D B \\
= \frac{180^{\circ} - \angle D A B}{2} + \frac{180^{\circ} ... | 40^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,351 |
3. In the 100th year of Besmiki's tenure as the President of the Currency Authority, he decided to issue new gold coins. In this year, he put into circulation an unlimited number of gold coins with a face value of $2^{100}-1$ yuan. In the following year, he put into circulation an unlimited number of gold coins with a ... | 3. It happens in the 200th year of Besmiki's presidency.
Assume that the described scenario occurs in the $k$-th year of Besmiki's presidency. Then,
$$
2^{k}-1=a_{1}+a_{2}+\cdots+a_{n}=N-n,
$$
where $N$ is the sum of some powers of 2, all of which are divisible by $2^{100}$.
Since $2^{k}$ is also divisible by $2^{10... | 200 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,352 |
4. There are 49 visually identical coins, among which 25 are genuine and 24 are counterfeit. There is a device available to identify the authenticity of the coins; you can place any number of coins into the device, and if the number of counterfeit coins exceeds half, the device will signal. How can you use this device ... | 4. If two counterfeit coins can be found from a pile of coins, then this pile of coins is called an "operational object".
For a pile containing no less than $M$ counterfeit coins and a total of $N$ coins, it is denoted as $N: M$.
Obviously, the initial operational object is 49: 24.
Next, for each operation, “$(+)$” in... | 2: 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 728,353 |
5. Betya and Vasya simultaneously enter the same non-zero integer into their calculators. At any moment, Betya either increases her number by 10 or multiplies it by 2014. At the same time, Vasya either decreases his number by 10 in the first case or divides it by 2014 in the second case. Can the numbers in their calcul... | 5. It is possible.
Assuming that before getting equal numbers again, Betia multiplies her number by 2014, and Vasia divides her number by 2014, this indicates that the numbers they have are negative, and in terms of absolute value, Vasia's number is $2014^{2}$ times Betia's number. Suppose these two numbers are obtain... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,354 |
6. If a positive integer, in its decimal representation, has no digit equal to 0, contains only one largest digit (this digit is called the peak), and this largest digit does not appear at the ends, and its digits first increase step by step (i.e., each digit is not less than the one before it) to the maximum, then dec... | 6. For two unimodal positive integers, if one number is obtained by arranging the digits of the other in reverse order, then these two numbers are called "friendly". Thus, for 100-digit mutually friendly numbers, the digits in the even positions of the former are all in the odd positions of the latter, and vice versa.
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,355 |
7. Let the decimal representation of the positive integer $N$ consist only of the digits 1 and 2. By deleting digits from $N$, one can obtain all 10000 different positive integers formed by 9999 digits 1 and 1 digit 2. Find the minimum possible number of digits in $N$. | 7. The minimum possible number of digits in $N$ is 10198.
For example, $N=\underbrace{1 \cdots 1}_{99 \uparrow} \underbrace{1 \cdots}_{100 \uparrow} 12 \underbrace{1 \cdots}_{100 \uparrow} \underbrace{2}_{98 \uparrow} \underbrace{1 \cdots 1}_{99 \uparrow}$.
For a number formed by 9999 digits 1 and 1 digit 2, if there ... | 10198 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,356 |
8. For a convex 101-gon, if a diagonal has 50 vertices on one side and 49 vertices on the other side, it is called a "main diagonal". Select several main diagonals that have no common endpoints. Prove: the sum of the lengths of these main diagonals is less than the sum of the lengths of the other diagonals. | 8. For a convex $(2n+1)$-gon $K=A_{1}A_{2}\cdots A_{2n+1}$, each diagonal $A_{i}A_{n+i}$ is called a main diagonal $\left(A_{j+2n+1}=A_{j}\right)$.
Next, we use mathematical induction to prove: For any positive integer $n$, the sum of the lengths of any set of diagonals in the $(2n+1)$-gon $K$ that have no common endp... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,357 |
2. The arithmetic sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}+a_{2}+\cdots+a_{14}=77 \text {, and } a_{1} 、 a_{11} \in \mathbf{Z}_{+} \text {. }
$$
Then $a_{18}=$ $\qquad$ | 2. -5 .
From the formula for the sum of an arithmetic sequence, we get
$$
\begin{array}{l}
a_{1}+a_{14}=11 \Rightarrow 2 a_{1}+13 \times \frac{a_{11}-a_{1}}{10}=11 \\
\Rightarrow 7 a_{1}+13 a_{11}=110 \\
\Rightarrow a_{1}=12(\bmod 13), a_{11}=2(\bmod 7) \\
\Rightarrow\left(a_{1}, a_{11}\right)=(12,2) \\
\Rightarrow a_... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,359 |
Example 8 If $\frac{a^{2}}{b+c-a}+\frac{b^{2}}{c+a-b}+\frac{c^{2}}{a+b-c}=0$, find the value of the algebraic expression $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { Let } \left\{\begin{array} { l }
{ b + c - a = 2 x , } \\
{ c + a - b = 2 y , } \\
{ a + b - c = 2 z . }
\end{array} \text { Then } \left\{\begin{array}{l}
a=y+z, \\
b=z+x, \\
c=x+y .
\end{array}\right.\right. \\
\text { Therefore } \frac{(y+z)^{2}}{2 x}+\frac{(z+x)^{2}}{2 y}+\frac{(x+y)^{2... | 1 \text{ or } -\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,360 |
3. If point $P\left(x_{0}, y_{0}\right)$ is such that the chord of contact of the ellipse $E: \frac{x}{4}+y^{2}=1$ and the hyperbola $H: x^{2}-\frac{y^{2}}{4}=1$ are perpendicular to each other, then $\frac{y_{0}}{x_{0}}=$ $\qquad$ | 3. $\pm 1$.
Point $P\left(x_{0}, y_{0}\right)$ has the tangent chord line equations for ellipse $E$ and hyperbola $H$ respectively as
$$
\frac{x x_{0}}{4}+y y_{0}=1, x x_{0}-\frac{y y_{0}}{4}=1 .
$$
From the fact that these two lines are perpendicular, we have
$$
\begin{array}{l}
\left(\frac{x_{0}}{4}, y_{0}\right) \... | \pm 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,361 |
4. Let the area of $\triangle ABC$ be 1, and the midpoints of sides $AB$ and $AC$ be $E$ and $F$, respectively. Let $P$ be a moving point on segment $EF$. Then
$$
f=\overrightarrow{P B} \cdot \overrightarrow{P C}+\overrightarrow{B C}^{2}
$$
the minimum value of $f$ is . $\qquad$ | 4. $\sqrt{3}$.
Construct $P D \perp B C$ at point $D$. Let $B C=a$.
As shown in Figure 2, when point $D$ is on the line segment $B C$ or the extension of $C B$,
$$
\begin{aligned}
f & =(\overrightarrow{P D}+\overrightarrow{D B}) \cdot(\overrightarrow{P D}+\overrightarrow{D C})+B C^{2} \\
& =\overrightarrow{P D}^{2}+\o... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,362 |
5. Let the function
$$
f(x)=x \log _{2} x+(a-x) \log _{2}(a-x)
$$
be symmetric about the line $x=\frac{1}{2}$. Then for any real numbers $x_{i} \in(0,1)(1 \leqslant i \leqslant 4)$ satisfying $\sum_{i=1}^{4} x_{i}=1$, the minimum value of $s=\sum_{i=1}^{4} x_{i} \log _{2} x_{i}$ is . $\qquad$ | 5. -2 .
From the problem, we know that the midpoint of the interval $(0, a)$ is $\frac{1}{2}$.
Thus, $a=1$.
Then, $f(x)=x \log _{2} x+(1-x) \log _{2}(1-x)$
$\Rightarrow f^{\prime}(x)=\log _{2} \frac{x}{1-x}$.
Let $f^{\prime}(x)=0$, we get $x=\frac{1}{2}$.
For any $x \in\left(0, \frac{1}{2}\right)$, $f^{\prime}(x) < 0$... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,363 |
6. The integer $n$ that satisfies $\left(1+\frac{1}{n}\right)^{n+1}=\left(1+\frac{1}{2014}\right)^{2014}$ is $=$. | 6. -2015 .
Notice that for any $x \in(-1,+\infty)$ we have
$$
\frac{x}{1+x} \leqslant \ln (1+x) \leqslant x \text {. }
$$
Then for $f(x)=\left(1+\frac{1}{x}\right)^{x+1}(x>0)$ and
$$
g(x)=\left(1+\frac{1}{x}\right)^{x}(x>0)
$$
the derivatives are respectively
$$
\begin{array}{l}
f^{\prime}(x)=\left(1+\frac{1}{x}\rig... | -2015 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,364 |
7. If $x, y, z > 0$ satisfy
$$
\left\{\begin{array}{l}
\frac{2}{5} \leqslant z \leqslant \min \{x, y\}, \\
x z \geqslant \frac{4}{15}, \\
y z \geqslant \frac{1}{5},
\end{array}\right.
$$ | 7.13.
From the problem, we have
$$
\frac{1}{\sqrt{x}} \leqslant \frac{\sqrt{15 z}}{2}, \frac{1}{z} \leqslant \frac{5}{2}, \frac{1}{\sqrt{y}} \leqslant \sqrt{5 z} \text {. }
$$
Then
$$
f=\frac{2}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}+\frac{1}{z}\left(1-\frac{z}{x}\right)+
$$
$$
\begin{aligned}
& 2\left[\frac{2}{\sqrt{y}... | 13 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,365 |
8. Rearrange the six-element array $(1,2,3,4,5,6)$ to
$$
\begin{array}{l}
A=\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right), \\
\text { and } \quad B=\left(b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\right) .
\end{array}
$$
Then the minimum value of $P=\sum_{i=1}^{6} i a_{i} b_{i}$ is | 8. 162 .
From the geometric mean
$$
G=\sqrt[6]{\prod_{i=1}^{6} i a_{i} b_{i}}=\sqrt[6]{(6!)^{3}}=12 \sqrt{5} \in(26,27),
$$
we know that there exists at least one term not less than 27, and at least one term not greater than 25.
Let $i_{1} a_{i_{1}} b_{i_{1}} \leqslant 25, i_{2} a_{i_{2}} b_{i_{2}} \geqslant 27$.
$$
... | 162 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,367 |
9. (16 points) For any $n \in \mathbf{Z}_{+}$, prove:
$$
\left(1-\frac{1}{3}\right)\left(1-\frac{1}{3^{2}}\right) \cdots\left(1-\frac{1}{3^{n}}\right)>\frac{1}{2} \text {. }
$$ | II. 9. Introducing a constant $a$, such that for all $k \in \mathbf{Z}_{+}$ we have
$$
\begin{array}{l}
1-\frac{1}{3^{k}}=\frac{1}{3} \times \frac{3^{k}-1}{3^{k-1}} \geqslant \frac{1}{3} \times \frac{3^{k}+a}{3^{k-1}+a} \\
\Rightarrow a \geqslant \frac{1}{2-\frac{3}{3^{k}}} \geqslant \frac{1}{2-\frac{3}{3^{1}}}=1 .
\en... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,368 |
10. (20 points) The function $f(x)$ defined on $\mathbf{R}$ satisfies:
(i) For any real numbers $x, y$,
$$
f(x+y+1)=f(x-y+1)-f(x) f(y) \text {; }
$$
(ii) $f(1)=2$;
(iii) $f(x)$ is an increasing function on the interval $[0,1]$.
(1) Find the values of $f(0), f(-1), f(2)$;
(2) Solve the inequality $f(x)>1$.
| 10. (1) Let $x=y=0$, we get
$$
\begin{array}{l}
f(1)=f(1)-f(0) f(0) \\
\Rightarrow f(0)=0 ;
\end{array}
$$
Let $x=-1, y=1$, we get
$$
\begin{array}{l}
f(1)=f(-1)-f(-1) f(1)=-f(-1) \\
\Rightarrow f(-1)=-2 ;
\end{array}
$$
Let $x=0, y=1$, we get
$$
f(2)=f(0)-f(0) f(1)=0 \text {. }
$$
(2) First, study a few properties o... | \left\{x \left\lvert\, \frac{1}{3}+4 k<x<\frac{5}{3}+4 k\right., k \in \mathbf{Z}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,369 |
11. (20 points) Given an arithmetic sequence where the first term is less than 0, the 100th term is not less than 74, the 200th term is less than 200, and the number of terms in the interval $\left(\frac{1}{2}, 5\right)$ is 2 less than the number of terms in the interval $\left[20, \frac{49}{2}\right]$. Find the genera... | $$
\begin{array}{l}
\text{Let the first term of the arithmetic sequence be } a\frac{2(l+1)}{9}>\frac{199}{250.74} \\
\Rightarrow \frac{891}{148}-1>l>\frac{1791}{501.48}-1 \\
\Rightarrow 5.03>l>2.57 .
\end{array}
$$
Combining with $3 \mid(l+1)$, we get $l=5$.
Thus, $d=\frac{3}{4}$.
Substituting $l=5$ into equation (7) ... | a_{n}=\frac{3}{4} n-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,370 |
Example 9 Given $a b c \neq 0$, and
$$
\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}=1 \text {. }
$$
Find the value of $a b c$. | From the given problem, we have
$$
\frac{1}{\frac{1}{a}+b+1}+\frac{1}{\frac{1}{b}+c+1}+\frac{1}{\frac{1}{c}+a+1}=1 \text {. }
$$
$$
\text { Let } \frac{1}{a}+b=x, \frac{1}{b}+c=y, \frac{1}{c}+a=z \text {. }
$$
Then equation (1) can be transformed into
$$
\begin{array}{l}
\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1 \\
... | a b c=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,371 |
One. (40 points) As shown in Figure 1, let $L$, $M$, and $N$ be points inside $\triangle ABC$ at $\angle BAC$, $\angle CBA$, and $\angle ACB$, respectively, and
$$
\begin{array}{l}
\angle BAL = \angle ACL, \\
\angle LBA = \angle LAC, \\
\angle CBM = \angle BAM, \\
\angle MCB = \angle MBA, \\
\angle ACN = \angle CBN, \\... | (1) As shown in Figure 4, let $A L$ intersect $B C$ at point $D$, $B M$ intersect $C A$ at point $E$, and $C N$ intersect $A B$ at point $F$.
From $\angle B A L = \angle A C L$, $\angle A B L = \angle C A L$, we get
$\triangle A B L \backsim \triangle C A L \Rightarrow \frac{L B}{L A} = \frac{L A}{L C} = \frac{A B}{A C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,372 |
$$
\begin{array}{l}
\text { II. (40 points) Let } f(x)=x^{2}+a, \text { and define } \\
f^{1}(x)=f(x), \\
f^{n}(x)=f\left(f^{n-1}(x)\right)(n=2,3, \cdots) .
\end{array}
$$
Find the set
$$
M=\left\{a \in \mathbf{R}|| f^{n}(0) \mid \leqslant 2, n \in \mathbf{Z}_{+}\right\}
$$ | Define the sequence $\left\{a_{n}\right\}$ satisfying
$$
a_{1}=a, a_{n}=a_{n-1}^{2}+a(n \geqslant 2) \text {. }
$$
First, from $\left|a_{1}\right|=|a| \leqslant 2$, we get $-2 \leqslant a \leqslant 2$.
Thus, $M \subseteq[-2,2]$.
Second, by
$$
\begin{array}{l}
a_{k}-a_{k-1}=\left(a_{k-1}-\frac{1}{2}\right)^{2}+a-\frac{... | M=\left[-2, \frac{1}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,373 |
Three, (50 points) Find all positive integer triples $(x, y, z)$ that satisfy $5^{x}+12^{y}=13^{z}$.
---
The translation maintains the original text's format and line breaks. | Three, taking the given equation modulo 3 yields $(-1)^{x}=1(\bmod 3)$.
Thus, $2 \mid x$.
Let $x=2 x_{1}\left(x_{1} \in \mathbf{Z}_{+}\right)$.
(1) When $y=1$, the original equation becomes
$13^{z}-5^{2 x_{1}}=12$.
Taking both sides modulo 13 yields $(-1)^{x_{1}}=1(\bmod 13)$.
Thus, $2 \mid x_{1}$.
Let $x_{1}=2 x_{2}\l... | (x, y, z)=(2,2,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,374 |
(1) Prove: $\left|\bigcap_{i=1}^{m} A_{i}\right|+\left|\bigcup_{i=1}^{m} A_{i}\right| \geqslant 2 k$;
(2) If adding one more edge to the graph $G$ results in the existence of a $k+1$ clique, find the minimum number of central vertices in the graph $G$. | (1) Let $\left|\bigcup_{i=1}^{m} A_{i}\right|=l_{m}\left(l_{m} \leqslant n\right)$, i.e., to prove
$$
\left|\bigcap_{i=1}^{m} A_{i}\right| \geqslant 2 k-l_{m} \text {. }
$$
When $m=1$,
$$
\left|\bigcap_{i=1}^{m} A_{i}\right|=\left|A_{1}\right|=k=2 k-k \geqslant 2 k-n .
$$
Assume that (1) holds for $m-1(m>1)$.
For $m$... | 2k-n | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,375 |
In $\triangle A B C$, prove:
$$
\frac{\sin ^{2} A}{1+\sin A}+\frac{\sin ^{2} B}{1+\sin B}+\frac{\sin ^{2} C}{1+\sin C} \leqslant \frac{9}{2}(2-\sqrt{3}) .
$$ | Prove in $\triangle ABC$:
$$
\begin{array}{l}
\sin A+\sin B+\sin C \leqslant \frac{3 \sqrt{3}}{2}, \\
\sin ^{2} A+\sin ^{2} B+\sin ^{2} C \leqslant \frac{9}{4} .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
\sin A+\sin B=2 \sin \frac{A+B}{2} \cdot \cos \frac{A-B}{2} \\
\leqslant 2 \sin \frac{A+B}{2} .
\end{array}
$... | \frac{9}{2}(2-\sqrt{3}) | Inequalities | proof | Yes | Yes | cn_contest | false | 728,376 |
407 From $1,2, \cdots, n$ choose any $k$ numbers, among which there must be three numbers that are pairwise not coprime. Find the minimum value of $k$, $f(n)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let the number of primes in $1,2, \cdots, m$ be denoted as $a_{m}$. Let $A=\left\{x \mid x=1\right.$ or $p$ or $p^{2}$, where $p$ is a prime, $\left.x \leqslant n\right\}$.
Then $|A|=a_{n}+1+a_{[\sqrt{n}]}$, where $[x]$ represents the greatest integer not exceeding the real number $x$.
Obviously, any three numbers in t... | f(n)=a_{n}+2+a_{[\sqrt{n}]} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 728,377 |
Given a unit circle $\odot O$. A side $AB$ of the square $ABCD$ is a chord of $\odot O$. Try to find $OD_{\max}$ and $OD_{\min}$. | Solve as shown in Figure 2, draw $O M \perp A B$ at point $M$, and let
$$
\angle O A B=\alpha \text {. }
$$
Then $A D=A B=2 A M=2 \cos \alpha$.
By the cosine rule, we have
$$
\begin{array}{l}
O D^{2}= 1^{2}+(2 \cos \alpha)^{2}- \\
2 \times 1(2 \cos \alpha) \cos \left(90^{\circ}+\alpha\right) \\
= 1+4 \cos ^{2} \alpha... | OD_{\max} = \sqrt{2} + 1, \quad OD_{\min} = \sqrt{2} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,378 |
1. Given
$$
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}, \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}, \frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} \text {. }
$$
Find the value of $\frac{2}{x}+\frac{3}{y}+\frac{4}{z}$. | $$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{x+y+z}{x(y+z)}=\frac{1}{2} \\
\Rightarrow \frac{2}{x}=\frac{y+z}{x+y+z} .
\end{array}
$$
Similarly, $\frac{3}{y}=\frac{z+x}{x+y+z}, \frac{4}{z}=\frac{x+y}{x+y+z}$.
Adding the three equations yields $\frac{2}{x}+\frac{3}{y}+\frac{4}{z}=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,379 |
Example 2 As shown in Figure 4, given that $I$ is the incenter of $\triangle A B C$, $I D \perp B C$ at point $D$, $A I$ intersects the circumcircle of $\triangle A B C$ at point $S$, and extending $S D$ intersects the circumcircle again at point $P$. Prove:
$$
\angle A P I=90^{\circ} \text {. }
$$ | Prove as shown in Figure 4, draw $I E \perp A C$ at point $E$, $I F \perp A B$ at point $F$. Connect $P B, P C, P E, P F$.
It is easy to know that $P S$ bisects $\angle B P C$.
In $\triangle P B C$, by the Angle Bisector Theorem, we have
$$
\begin{array}{l}
\frac{P B}{P C}=\frac{B D}{C D}=\frac{B F}{C E}, \text { and }... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,380 |
Three. (50 points) If a function $f$ satisfies: for any real number $a$, the number of solutions to the equation $f(x)=a$ is even (it can be 0, but not infinitely many), then $f$ is called an "even function". Prove:
(1) No polynomial $f$ is an even function;
(2) There exists a continuous function $f: \mathbf{R} \righta... | (1) Note that the domain of the polynomial $f$ is $\mathbf{R}$, which can be divided into alternating intervals of increase and decrease as follows:
$$
\begin{array}{l}
I_{0}=\left(-\infty, x_{1}\right], I_{1}=\left[x_{1}, x_{2}\right], \cdots, \\
I_{k}=\left[x_{k},+\infty\right),
\end{array}
$$
where $x_{1}, x_{2}, \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,381 |
In $\triangle A B C$, it is known that $a+b=3 c$. The incircle $\odot I$ touches sides $B C$ and $C A$ at points $D$ and $E$, respectively. $M$ is the point symmetric to $D$ with respect to $I$, and $N$ is the point symmetric to $E$ with respect to $I$. The line $A M$ intersects $B N$ at point $P$. Prove: Point $P$ lie... | Proof As shown in Figure 1, let $\odot I$ be tangent to side $AB$ at point $F$, with $AF=x, BF=y$. Draw a line through point $M$ parallel to $BC$, intersecting $AC$ and $AB$ at points $T$ and $S$ respectively. Let the line $AM$ intersect $\odot I$ at point $P'$ and $BC$ at point $Q$. Let the line $BN$ intersect $\odot ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,382 |
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