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 |
|---|---|---|---|---|---|---|---|---|---|
Example 3 Let the coordinates of the three vertices of the equilateral $\triangle A B C$ be $\left(x_{A}, y_{A}\right),\left(x_{B}, y_{B}\right),\left(x_{C}, y_{C}\right)$. Please express $S_{\triangle A B C}$ using $x_{A}, x_{B}, x_{C}$. | 【Analysis】Let $\triangle A B C$ be counterclockwise, and $O$ be the center of the equilateral $\triangle A B C$, with $R$ being the radius of the circumcircle. Then
$$
x_{0}=\frac{x_{A}+x_{B}+x_{C}}{3} .
$$
Therefore, the x-coordinate of vector $\overrightarrow{O A}$ is
$$
x_{A}-x_{o}=\frac{2 x_{A}-x_{B}-x_{C}}{3}=R \... | \frac{\sqrt{3}}{3} \sum\left(x_{A}^{2}-x_{B} x_{C}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,750 |
Example 4 Given 60 points $P_{1}, P_{2}$, $\cdots, P_{60}$ inside the unit circle. Then there must exist a point $X$ on the unit circle such that
$$
\sum_{i=1}^{60} X P_{i}<80 .
$$ | 【Analysis】First, prove a lemma.
Lemma Let $\triangle A B C$ be an equilateral triangle inscribed in a unit circle, and point $P$ be inside the circle. Then
$$
P A+P B+P C \leqslant 2.
$$
For the cyclic quadrilateral $A B C P^{\prime}$, applying Ptolemy's theorem yields
$$
2=A P^{\prime}=C P^{\prime}+B P^{\prime} \text ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,751 |
Example 5 As shown in Figure 1, let $E$ be a certain ellipse. Draw a tangent line $l_{1}$ of the ellipse $E$ through a point $B_{1}$ outside the ellipse $E$, which is tangent to the ellipse $E$ at point $C_{1}$. Take a point $B_{2}$ on the extension of $B_{1} C_{1}$ such that $B_{1} C_{1}=B_{2} C_{1}$; draw a tangent l... | 【Analysis】Establish a Cartesian coordinate system, and let the equation of the ellipse $E$ be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>0, b>0)$.
Perform a coordinate scaling transformation $X=\frac{x}{a}, Y=\frac{y}{b}$, transforming the line $l_{1}: A x+B y+C=0$ into $l_{1}^{\prime}: a A X+b B Y+c=0$, and the ell... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,752 |
Example 6: In space, 24 points (no three points are collinear), is it possible to determine exactly 2013 different planes? | 【Analysis】If no four points are coplanar, then the number of planes determined by 24 points is
$$
\mathrm{C}_{24}^{3}=2024=2013+11 .
$$
If there is a case where six points are coplanar, then the number of planes determined will be less than or equal to
$$
\mathrm{C}_{24}^{3}-\mathrm{C}_{6}^{3}+1=2005 .
$$
Assume that... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,753 |
Example 7 On a plane, there exist $n$ points, no three of which are collinear, and when these $n$ points are arbitrarily labeled as $A_{1}, A_{2}, \cdots, A_{n}$, the broken line $A_{1} A_{2} \cdots A_{n}$ does not intersect itself. Find the maximum value of $n$.
| 【Analysis】When $n=2,3$, it is obviously true.
When $n=4$, if the convex hull of the four points is a triangle, then it satisfies the condition.
Next, we show that when $n \geqslant 5$, it is impossible to satisfy the condition.
Consider only five points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ among these $n$ points.
If th... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,754 |
Example 3 In $\triangle A B C$, it is known that $A B=A C$, $\angle B A C=100^{\circ}, P$ is a point on the angle bisector of $\angle A C B$, and $\angle P B C=10^{\circ}$. Find $\angle A P B$. | 【Analysis 1】First, identify which angles are known.
As shown in Figure 3, given $A B=A C, \angle B A C=100^{\circ}$, we have
$$
\angle A B C=\angle A C B=\frac{180^{\circ}-100^{\circ}}{2}=40^{\circ} \text {. }
$$
Thus, $\angle 1=\angle 2=20^{\circ}, \angle 3=10^{\circ}, \angle 4=30^{\circ}$.
Furthermore, since $\angle... | 70^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,755 |
Example 8 The vertices of Rt $\triangle ABC$ are all lattice points, and the side lengths are all integers. Prove: its incenter $I$ is a lattice point. | 【Analysis】Let $\angle C=\frac{\pi}{2}$, and let $C$ be the origin, with $P$ and $Q$ being the points where the incircle touches sides $AC$ and $BC$, respectively, as shown in Figure 2.
Let $A_{1}$ and $B_{1}$ be the grid points on $CA$ and $CB$ closest to point $C$, and let $A_{1}=(r, s)$, where $r$ and $s$ are coprim... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,756 |
Example 9 Proof: A convex $4n+2$-sided polygon with an area of 1 can certainly be cut along a diagonal to produce a triangle with an area less than or equal to $\frac{1}{6n}$. | 【Analysis】When $n=1$, as shown in Figure 3, in the convex hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ with an area of 1, the diagonals $A_{k} A_{k+3}$ $(k=1,2,3)$ intersect with $A_{k+1} A_{k+4}$ at point $B_{k}$. The hexagon is divided into three convex quadrilaterals $A_{1} A_{2} A_{3} B_{3}$, convex quadrilateral ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,757 |
2. Let $P$ be a point inside a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, and let the line $A_{i} P$ intersect the boundary of the polygon at another point denoted as $B_{i}$. Then $\sum_{i=1}^{n} A_{i} P \geqslant \sum_{i=1}^{n} P B_{i}$. | Let the longest diagonal length of the regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ be $l$.
Let point $B_{i}$ be on the side $A_{j} A_{j+1}$. Then
$$
\begin{array}{l}
A_{i} B_{i} \leqslant \max \left\{A_{i} A_{j}, A_{i} A_{j+1}\right\} \leqslant l . \\
\text { Hence } \sum_{i=1}^{n} A_{i} P+\sum_{i=1}^{n} P B_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,759 |
3. There are $n$ unit circles in the plane. Prove: There must exist a unit circle, on which there is a circular arc $\overparen{A B}$, such that $\overparen{A B} \geqslant \frac{2 \pi}{n}$, and the arc $\overparen{A B}$ does not intersect with any other unit circle. | When the centers of these $n$ unit circles are collinear, consider the rightmost unit circle. Let the right semicircle (closed) of the rightmost unit circle be $\overparen{A B}=\pi$, then the condition of the problem is satisfied.
When the centers of these $n$ unit circles are not collinear, assume the convex hull of ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,760 |
4. Find the smallest $n \in \mathbf{N}_{+}$, such that for any finite point set $M$ in the plane, if any $n$ points in $M$ can be covered by two lines, then there must exist two lines that can cover the point set $M$. | The required minimum value is 6.
When $|M| \leqslant 6$, the proposition is obviously true.
If $|M|>6$, for any six points $A_{1}, A_{2}, \cdots, A_{6}$ in $M$, they can be covered by two lines $l_{1}$ and $l_{2}$. Without loss of generality, assume that $l_{1}$ contains at least three points $A_{1}, A_{2}, A_{3}$. Let... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,761 |
For example, a 1000-bit digital string $a_{1} a_{2} \cdots a_{1000}$, where $a_{i} \in\{0,1\}(i=1,2, \cdots, 1000)$. If two adjacent digits $a_{k} 、 a_{k+1}$ satisfy $a_{k}=1, a_{k+1}=0$, then $a_{k} a_{k+1}$ is called an "inversion" in the digital string. Question: How many 1000-bit digital strings have exactly 300 in... | 【Analysis】For any thousand-digit number string $x=$ $\overline{a_{1} a_{2} \cdots a_{1000}}$, add the digit 0 at the very beginning and the digit 1 at the very end, to get a 1002-digit number string $y=$, generating 1001 gaps.
Now, add “=” or “$\neq$” in each gap according to the following requirements: if the numbers... | \mathrm{C}_{1000}^{600} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,763 |
Example 3 Let $A B$ be a line segment, and for any point $P$ outside the line, the angle $\angle A P B$ is called the "viewing angle" of point $P$ with respect to the line segment $A B$. Prove: For any point $P$ inside the tetrahedron $A B C D$, among the six viewing angles of point $P$ with respect to the six edges of... | 【Analysis】Suppose for a tetrahedron, the cosine of the angle subtended by each edge from an internal point $P$ is greater than $-\frac{1}{3}$.
Note that, for the center of a regular tetrahedron, the cosine of the angle subtended by each edge is $-\frac{1}{3}$. We can consider appropriately transforming any point $P'$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,764 |
1. $n$ positive integers are written in a row, Alice selects two adjacent numbers $x, y (x>y$, and $x$ is to the left of $y)$. She replaces the pair $(x, y)$ with $(y+1, x)$ or $(x-1, x)$. Prove: Alice can only perform the above operation a finite number of times. | 1. Let the largest number among $n$ numbers be $M$. Then, after each operation, the maximum value of these numbers remains unchanged. Suppose after a certain operation, these $n$ numbers are $a_{1}, a_{2}, \cdots, a_{n}$. Define
$$
S=a_{1}+2 a_{2}+\cdots+n a_{n} \text {. }
$$
Select a pair $\left(a_{i}, a_{i+1}\right)... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,765 |
Example 4 In the right triangle $\triangle ABC$, it is known that $AB=AC$, points $D$ and $E$ are on sides $AB$ and $AC$ respectively, and $AB=3BD$, $AC=3AE$. Prove: $\angle ADE=\angle EBC$.
---
The translation maintains the original text's formatting and line breaks. | $$
\begin{array}{l}
\text { Given } 3 B D=A B=A C=3 A E \Rightarrow B D=A E \\
\Rightarrow A D=E C=2 A E . \\
\text { Also, } E C^{2}=E F^{2}+F C^{2}=2 E F^{2} \\
\Rightarrow E F^{2}=\frac{1}{2} E C^{2}=2 A E^{2} \\
\Rightarrow C F=E F=\sqrt{2} A E . \\
\text { Since } B C^{2}=A B^{2}+A C^{2}=2 A C^{2}=18 A E^{2} \\
\R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,766 |
2. Given that $n$ is a positive integer. Find the maximum number of disjoint element pairs in the set $\{1,2, \cdots, n\}$, such that for any two different element pairs $(a, b)$ and $(c, d)$, both $a+b$ and $c+d$ are distinct integers not exceeding $n$. | 2. Let $\{1,2, \cdots, n\}$ contain $x$ pairs of elements that satisfy the condition.
Since these $x$ pairs of elements are pairwise disjoint, the sum $S$ of the $2 x$ numbers in these pairs is at least
$$
1+2+\cdots+2 x \text {. }
$$
On the other hand, since the sum of the two numbers in each pair does not exceed $n... | \left[\frac{2 n-1}{5}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,767 |
3. In a $999 \times 999$ grid, some cells are white, and the others are red. Let $T$ be the number of cell groups $\left(C_{1}, C_{2}, C_{3}\right)$ such that $C_{1}$ and $C_{2}$ are in the same row, $C_{2}$ and $C_{3}$ are in the same column, and $C_{1}$ and $C_{3}$ are white, while $C_{2}$ is red. Find the maximum va... | 3. First, prove: For an $n \times n$ grid, we have
$$
T \leqslant \frac{4 n^{4}}{27} \text {. }
$$
Let the $i$-th row have $a_{i}$ white cells, and the $j$-th column have $b_{j}$ white cells. Let $R$ be the set of red cells. For each red cell $(i, j)$, there are $a_{i} b_{j}$ valid quadruples $\left(C_{1}, C_{2}, C_{3... | 148 \times 999^{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,768 |
$5.3 n \times 3 n$ grid is numbered by rows and columns as $1,2, \cdots, 3 n$. For each cell $(x, y)(1 \leqslant x, y \leqslant 3 n)$, if $x+y$ modulo 3 leaves a remainder of 0, 1, or 2, then the cell $(x, y)$ is colored with color $a$, $b$, or $c$ respectively. Each cell contains a coin of color $a$, $b$, or $c$, and ... | 5. Let the coins of colors $a, b, c$ be denoted as $A$ coins, $B$ coins, $C$ coins, and the squares of colors $a, b, c$ be denoted as $A$ squares, $B$ squares, $C$ squares, respectively.
Without loss of generality, it suffices to prove that all $A$ coins can be placed in different $A$ squares such that the distance ea... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,770 |
7. Given that there are $2^{500}$ points on a circle, numbered in some order as $1,2, \cdots, 2^{500}$. Prove: It is possible to select 100 pairwise non-intersecting chords connecting these points, such that the sum of the numbers of the endpoints of all these chords is equal. | 7. First, prove a lemma.
Lemma Given a graph $G$ where each vertex $v$ has a degree of $d_{v}$. Then, graph $G$ contains an independent set of vertices $S$ such that $|S| \geqslant f(G)$, where $f(G)=\sum_{v \in G} \frac{1}{d_{v}+1}$.
Proof Use mathematical induction on $n=|G|$.
When $n=1$, the conclusion is obviously... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,771 |
1. (50 points) As shown in Figure 1, given $\triangle ABC$ satisfies $AB < AC$, $M$ is the midpoint of side $BC$, the circle $\odot O$ passing through point $A$ is tangent to side $BC$ at point $B$, and intersects line segment $AM$ at point $D$, and intersects the extension of $CA$ at point $E$. The line through point ... | 1. Since $\angle D A C=\angle E B D=\angle D F C$, therefore, points $A, D, C, F$ are concyclic, let this circle be $\odot O_{1}$.
From $M C^{2}=M B^{2}=M A \cdot M D$, we know that $M C$ is tangent to $\odot O_{1}$ at point $C$.
Since $B E / / C F$, therefore, $G$ is one of the homothetic centers of $\odot O$ and $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,772 |
2. (50 points) Given that $a$, $b$, and $c$ are three distinct real numbers. If any two of the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$
have exactly one common root, find the value of $a^{2}+b^{2}+$ $c^{2}$. | 2. From equations (1) and (2), we know their common root is $p=\frac{b-c}{b-a}$.
Similarly, the common roots of equations (2) and (3), and equations (1) and (3) are $q=\frac{c-a}{c-b}$ and $r=\frac{a-b}{a-c}$, respectively.
Thus, $p q r=-1$.
If any two of $p, q, r$ are equal, assume $p=q$, then the three equations hav... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,773 |
3. (50 points) For any $a \in\{0,1, \cdots, 9\}$, prove: the set $X=\left\{[n \sqrt{2}] \mid n \in \mathbf{N}_{+}\right\}$ contains infinitely many elements whose last digit is $a$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
---
The translation preserves the original text's formatting... | 3. Let $x_{n}=[n \sqrt{2}]$. Clearly,
$$
x_{n+1}=[n \sqrt{2}+\sqrt{2}] \geqslant[n \sqrt{2}+1]=x_{n}+1 \text {, }
$$
and $x_{n+1}=[n \sqrt{2}+\sqrt{2}] \leqslant[n \sqrt{2}+2]=x_{n}+2$,
i.e., $x_{n+1}-x_{n}=1$ or 2.
Assume the set $X=\left\{[n \sqrt{2}] \mid n \in \mathbf{N}_{+}\right\}$ contains only a finite number ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,774 |
4. (50 points) Given 2013 people, each holding $1, 2, \cdots, 2013$ cards respectively, sitting around a round table in any order. A pass refers to a person giving one of their cards to one of their two adjacent neighbors. Prove: if after $m$ passes, everyone holds the same number of cards, and $m$ is the smallest, the... | 4. Label 2013 people around a round table in a clockwise direction as $A_{0}, A_{1}, \cdots, A_{2 n}(n=1006)$. Suppose $A_{i}$ holds $a_{i}+(n+1)$ cards. Then $\left(a_{0}, a_{1}, \cdots, a_{2 n}\right)$ is a permutation of $(-n,-n+1, \cdots,-1,0,1, \cdots, n)$, and $\sum_{i=0}^{2 n} a_{i}=0$.
Let $x_{i}$ be the numbe... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,775 |
5. (50 points) As shown in Figure 2, given an acute triangle $\triangle ABC$ with the circumcircle $\odot O$, $AD$ is the diameter of $\odot O$. The lines through points $B$ and $C$ and perpendicular to $BC$ intersect the extensions of $CA$ and $BA$ at points $E$ and $F$ respectively. Prove:
$$
\angle ADF = \angle BED.... | 5. As shown in Figure 3, connect $B D$ and $C D$.
Then
$$
\angle D B F
=
\angle D C E=90^{\circ} \text {, }
$$
and
$$
\begin{array}{l}
\frac{B F}{C E}=\frac{\frac{B C}{\cos \angle C B F}}{\frac{B C}{\cos \angle B C E}} \\
=\frac{\sin \angle B C D}{\sin \angle C B D}=\frac{B D}{C D} . \\
\text { Therefore, } \triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,776 |
Example 5 In $\triangle A B C$, it is known that $A C=B C$, $\angle A C B=90^{\circ}$, and $D$ is a point on side $A C$. $A E \perp B D$ at point $E$. If $A E=\frac{1}{2} B D$, prove: $B D$ is the angle bisector of $\angle A B C$. | 【Analysis】This problem essentially requires proving that $\angle A B D = \angle D B C$, which can be seen as an angle equality. However, it can be proven using the "three lines coincide" theorem.
Notice that the angle bisector is the axis of symmetry of the angle. Therefore, consider the reflection transformation with... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,777 |
6. (50 points) Let the function $f$ :
$\mathbf{R}_{+} \rightarrow \mathbf{R}$. If $f\left(x^{2}\right)-x^{3}$ and $f\left(x^{3}\right)-x^{4}$ are both strictly increasing, prove:
$$
f\left(x^{5}\right)+100\left(f\left(x^{5}\right)-x^{7}\right)
$$
is strictly increasing. | 6. Given that $f\left(x^{2}\right)-x^{3}$ and $f\left(x^{3}\right)-x^{4}$ are strictly increasing, we know that $f\left(x^{30}\right)-x^{45}$ and $f\left(x^{30}\right)-x^{40}$ are also strictly increasing. Therefore, for a positive number $\lambda$ (to be determined), we have
$$
\begin{array}{l}
\frac{\left(f\left(x^{3... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,778 |
7. (50 points) Prove: There does not exist a prime number $p$ greater than 7 such that the number of positive divisors of $p^{12} + 5039 \times 5041$ is less than 120. | 7. Let $d(n)$ be the number of all positive divisors of $n$. For a prime $p$ greater than 7, by Fermat's Little Theorem, we have
$$
5 \mid (p^4 - 1), \quad 7 \mid (p^6 - 1) \Rightarrow 35 \mid (p^{12} - 1).
$$
Let $p = 2k + 1$. By the binomial theorem, we get
$$
\begin{array}{l}
(2k + 1)^{12} - 1 \\
\equiv \mathrm{C}_... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,779 |
8. (50 points) In a $4 n \times 4 n\left(n \in \mathbf{N}_{+}\right)$ real number table, the absolute value of each number does not exceed 1, and the sum of all numbers is 0. It is known that the absolute value of the sum of each row and the absolute value of the sum of each column are not less than $c$. Find the maxim... | 8. $c_{\max }=2 n$.
Example: Take the number at the $i$-th row and $j$-th column of the number table as
$$
a_{i j}=\left\{\begin{array}{ll}
0, & 1 \leqslant i, j \leqslant 2 n \text { or } 2 n+1 \leqslant i, j \leqslant 4 n ; \\
1, & 1 \leqslant i \leqslant 2 n, 2 n+1 \leqslant j \leqslant 4 n ; \\
-1, & 2 n+1 \leqsla... | c_{\max }=2 n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,780 |
1. Real numbers $a, b$ make the equation
$$
x^{3}-a x^{2}+b x-a=0
$$
have three positive real roots. Find the minimum value of $\frac{2 a^{3}-3 a b+3 a}{b+1}$. | 1. Let the three positive real roots of equation (1) be $x_{1}, x_{2}, x_{3}$. Then, by the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=a, \\
x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}=b, \\
x_{1} x_{2} x_{3}=a .
\end{array}
$$
Thus, $a>0, b>0$.
$$
\begin{array}{l}
\text { By }\... | 9 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,781 |
2. As shown in Figure 1, in $\triangle ABC$, $AB > AC$, the incircle $\odot I$ touches side $BC$ at point $D$, and $AD$ intersects $\odot I$ again at point $E$. The tangent line $EP$ of $\odot I$ intersects the extension of $BC$ at point $P$, and $CF \parallel PE$ intersects $AD$ at point $F$. The line $BF$ intersects ... | 2. Proof 1 As shown in Figure 5, let $\odot I$ be tangent to $AC$ and $AB$ at points $S$ and $T$ respectively. Connect $ST$, $AI$, and $IT$, and let $ST$ intersect $AI$ at point $G$.
Then $IT \perp AT$, $TG \perp AI$.
Thus, $AG \cdot AI = AT^2 = AD \cdot AE$.
Therefore, points $I$, $G$, $E$, and $D$ are concyclic.
Also... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,782 |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, \\
a_{n+1}=\frac{a_{n}^{2}+(-1)^{n}}{a_{n-1}}(n=2,3, \cdots) .
\end{array}
$$
Prove: the sum of the squares of any two consecutive terms of this sequence is still a term in the sequence.
(Tao Pingsheng) | 3. From $a_{n+1}=\frac{a_{n}^{2}+(-1)^{n}}{a_{n-1}}$, we know
$$
\begin{array}{l}
a_{n+1} a_{n-1}=a_{n}^{2}+(-1)^{n}(n=2,3, \cdots) . \\
\text { Then } \frac{a_{n}-a_{n-2}}{a_{n-1}}=\frac{a_{n} a_{n-2}-a_{n-2}^{2}}{a_{n-1} a_{n-2}} \\
=\frac{a_{n-1}^{2}+(-1)^{n-1}-a_{n-2}^{2}}{a_{n-1} a_{n-2}} \\
=\frac{a_{n-1}^{2}-a_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,783 |
4.12 The acrobats are numbered $1, 2, \cdots$, 12. They are to be arranged in two circles, $A$ and $B$, each with six people. In circle $B$, each acrobat stands on the shoulders of two adjacent acrobats in circle $A$. If the number of each acrobat in circle $B$ is equal to the sum of the numbers of the two acrobats bel... | 4. Let the sums of the elements in circles $A$ and $B$ be $x$ and $y$ respectively. Then $y=2x$. Therefore,
$$
3x = x + y = 1 + 2 + \cdots + 12 = 78.
$$
Solving for $x$ gives $x = 26$.
Clearly, $1, 2 \in A$ and $11, 12 \in B$.
Let $A = \{1, 2, a, b, c, d\} (a < b < c < d)$. Then $a + b + c + d = 23$, and $a \geq 3, 8 ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,784 |
5. Let $f(x)=\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{2013!}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. For an integer $n$, if the equation $f(x)=n$ has a real solution, then $n$ is called a "good number". Find the number of good numbers in the ... | 5. First, give two obvious conclusions:
(1) If $m$ is a positive integer and $x$ is a real number, then
$$
\left[\frac{x}{m}\right]=\left[\frac{[x]}{m}\right] ;
$$
(2) For any integer $l$ and positive even number $m$, we have
$$
\left[\frac{2 l+1}{m}\right]=\left[\frac{2 l}{m}\right] \text {. }
$$
Returning to the ori... | 587 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,785 |
6. Let $n$ be an integer greater than 1. The first $n$ prime numbers, in ascending order, are denoted as $p_{1}, p_{2}, \cdots, p_{n}$ (i.e., $p_{1}=2$, $p_{2}=3, \cdots$), and let $A=p_{1}^{p_{1}} p_{2}^{p_{2}} \cdots p_{n}^{p_{n}}$. Find all positive integers $x$ such that $\frac{A}{x}$ is even and $\frac{A}{x}$ has ... | 6. From the given, we have $2 x \mid A$.
Notice that, $A=4 p_{2}^{p_{2}} p_{3}^{p_{3}} \cdots p_{n}^{p_{n}}$.
Therefore, we can set $x=2^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}} \cdots p_{n}^{\alpha_{n}}$, where $0 \leqslant \alpha_{1} \leqslant 1, 0 \leqslant \alpha_{i} \leqslant p_{i} (i=2,3, \cdots, n)$.
... | 2 p_{1} p_{2} \cdots p_{n} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,786 |
7. Remove any $2 \times 2$ small square from the corner of a $3 \times 3$ grid, and the remaining shape is called a "corner shape" (Figure 3 is an example of a corner shape). Now, place some non-overlapping corner shapes in a $10 \times 10$ grid (Figure 4). The boundaries of the corner shapes must coincide with the bou... | 7. First, $k_{\max }$
$<8$. This is because, if eight corner shapes are placed in the manner shown in Figure 9, it is impossible to place another corner shape in the grid.
Next, we prove that after placing seven corner shapes arbitrarily, it is still possible to place another complete corner shape.
Cover the 5th and ... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,787 |
Example 6 In the convex hexagon $A B C D E F$, it is known that $A B$
$$
\begin{array}{l}
/ / D E, B C / / E F, C D / / A F \text {, and } \\
\quad A B-E D=C D-A F=E F-B C>0 .
\end{array}
$$ | Prove: Hexagon $A B C D E F$ is an equiangular hexagon.
【Analysis】To construct the difference $A B-E D$, we can translate segments $A B, E D$ to the same line. Therefore, construct $\square A B R F, \square D E F Q$, as shown in Figure 7. Similarly, translate other segments to concentrate three interior angles of the h... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,788 |
8. Let integer $n \geqslant 3, \alpha, \beta, \gamma \in (0,1), a_{k}, b_{k}, c_{k} \geqslant 0 (k=1,2, \cdots, n)$ satisfy
$$
\begin{array}{l}
\sum_{k=1}^{n}(k+\alpha) a_{k} \leqslant \alpha, \\
\sum_{k=1}^{n}(k+\beta) b_{k} \leqslant \beta, \\
\sum_{k=1}^{n}(k+\gamma) c_{k} \leqslant \gamma .
\end{array}
$$
If for a... | 8. Let $a_{1}=\frac{\alpha}{1+\alpha}, b_{1}=\frac{\beta}{1+\beta}, c_{1}=\frac{\gamma}{1+\gamma}$, where $a_{i}, b_{i}, c_{i}=0(i=2,3, \cdots, n)$, at this point, the condition holds. Therefore, $\lambda$ must satisfy
$$
\begin{array}{l}
(1+\lambda) \frac{\alpha}{1+\alpha} \cdot \frac{\beta}{1+\beta} \cdot \frac{\gamm... | \frac{\alpha \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha \beta \gamma} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,789 |
3. Find all positive integers $k$ such that the product of the first $k$ primes minus 1 is a positive integer that is a perfect power greater than 1. | 3. $k=1$.
Let $2=p_{1}1, k>1$.
Since $a$ is odd and coprime with $p_{i}(1 \leqslant i \leqslant k)$, we know $a>p_{k}$.
Without loss of generality, we can assume $n$ is a prime (if $n=s t$, then we can replace $n$ with $t$ and $a$ with $a^{2}$).
Since $a^{2}+1$ is not a multiple of 3, we have $n>2$.
If there exists $2... | k=1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,791 |
4. Given a cyclic quadrilateral $A B C D$ and points $P$ , $Q$ inside it such that
$$
\begin{array}{l}
\angle P D C + \angle P C B = \angle P A B + \angle P B C \\
= \angle Q C D + \angle Q D A \\
= \angle Q B A + \angle Q A D = 90^{\circ} .
\end{array}
$$
Prove: The angle between line $P Q$ and $A D$ is equal to the ... | 4. Let the circumcircle of quadrilateral $ABCD$ be circle $\Gamma$, and the circumcircles of $\triangle ABP$, $\triangle CDP$, $\triangle ABQ$, $\triangle CDQ$ be circles $\Gamma_{1}$, $\Gamma_{2}$, $\Gamma_{3}$, $\Gamma_{4}$, respectively. Let the projection of point $P$ on side $BC$ be $X$, and denote the line $PX$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,792 |
5. Does there exist a positive integer $n$ (in decimal representation) such that for any $a, b \in\{1,2, \cdots, 9\}$, we have $\overline{a b} \mid \overline{a n b}$, where $\overline{x \cdots y}$ represents the decimal positive integer formed by the digits $x, \cdots, y$ in sequence.
For example, for any $a, b \in\{1,... | 5. Does not exist. $4112,12 \mid \overline{1 n 2}$. Therefore, $4 \mid \overline{n_{1} 2}$.
Similarly, $4|24,24| \overline{2 n 4}$, we get $4 \mid \overline{n_{1} 4}$.
Thus $41\left(\overline{n_{1} 4}-\overline{n_{1} 2}\right)=2$, a contradiction. | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,793 |
6. Let $A$ be a set of ten real-coefficient polynomials of degree five. It is known that there exist $k$ consecutive positive integers $n+1$, $n+2, \cdots, n+k$, and $f_{i}(x) \in A(1 \leqslant i \leqslant k)$, such that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence. Find the maximum possible... | 6. Given that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence, we know there exist real numbers $a$ and $b$ such that
$$
f_{i}(n+i)=a i+b .
$$
Notice that, for any fifth-degree polynomial $f$, the equation
$$
f(n+x)=a x+b
$$
has at most five real roots. Therefore, each polynomial in $A$ appea... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,794 |
7. Given that the incircle $\odot I$ of $\triangle ABC$ touches sides $BC$, $CA$, and $AB$ at points $A_{1}$, $B_{1}$, and $C_{1}$ respectively. Let $I_{a}$, $I_{b}$, and $I_{c}$ be the centers of the excircles opposite to sides $BC$, $CA$, and $AB$ respectively. The line segments $I_{a} B_{1}$ and $I_{b} A_{1}$ inters... | 7. Since $B_{1} C_{1}$ and $I_{b} I_{c}$ are both perpendicular to the line $A I$, we have $B_{1} C_{1} \parallel I_{b} I_{c}$.
Similarly, $C_{1} A_{1} \parallel I_{c} I_{a}, A_{1} B_{1} \parallel I_{a} I_{b}$. Therefore, $\triangle A_{1} B_{1} C_{1}$ is homothetic to $\triangle I_{a} I_{b} I_{c}$.
Similarly, $\triangl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,795 |
8. There is a square on the plane, whose sides are parallel to the horizontal and vertical directions. Draw several line segments parallel to the sides of the square such that: no two line segments belong to the same straight line, and any two intersecting line segments intersect at an endpoint of one of the line segme... | 8. The number of rectangles is $k l$.
Prove this conclusion by induction on $k$.
When $k=1$, the conclusion is obviously true.
Assume $k>1$. Since the line segments divide the square into several rectangles, the intersection of any two intersecting line segments is precisely one of the endpoints of one of the line seg... | k l | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,796 |
1. If the cube root of $\sqrt{125}$ is $A$, and the square root of $\sqrt{25}$ is $B$. Then $A+B=(\quad)$.
(A) $2 \sqrt{5}$ or 0
(B) 10 or 0
(C) $2 \sqrt{5}$
(D) 10 | -1. A.
Since the cube root of $\sqrt{125}$ is $\sqrt{5}$, and the square root of $\sqrt{25}$ is $\pm \sqrt{5}$, therefore, $A+B=2 \sqrt{5}$ or 0. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,797 |
2. Let $x+y+z=2010$, and
$$
\frac{x}{1999}=\frac{y}{2010}=\frac{z}{2011} \text {. }
$$
Then $x^{3}+y^{3}+z^{3}-3 x y z=(\quad)$.
(A) $\frac{1999}{3}$
(B) 670
(C) $\frac{2011}{3}$
(D) $\frac{2012}{3}$ | 2. B.
Let $1 \frac{x}{1999}=\frac{y}{2010}=\frac{z}{2011}=k$. Then $x+y+z=3 \times 2010 k=2010$.
Thus, $k=\frac{1}{3}$.
Therefore, $x^{3}+y^{3}+z^{3}-3 x y z$
$$
\begin{array}{l}
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \\
=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \\
=\frac{1}{2} \times 2... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,798 |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If the real number $a$ satisfies
$$
a-\frac{4}{a}+\frac{3 \sqrt{a(a-2)}}{a}=2,
$$
then $[a]=$ ( ).
(A) 0 or 2
(B) -1 or 2
(C) 0 or 3
(D) 2 or -3 | 3. B.
Equation (1) $\Leftrightarrow a^{2}-4+3 \sqrt{a(a-2)}=2 a$. Let $\sqrt{a(a-2)}=x$. Then $x^{2}+3 x-4=0$. Solving, we get $x_{1}=1, x_{2}=-4$ (discard). Thus, $\sqrt{a(a-2)}=1$.
Rearranging gives $a^{2}-2 a-1=0 \Rightarrow a_{1,2}=1 \pm \sqrt{2}$. Therefore, $[a]=-1$ or 2. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,800 |
4. As shown in Figure 1, in Rt $\triangle A B C$, $\angle A C B=90^{\circ}$, $C D$ is the angle bisector, $D E \perp$ $A C$ at point $E, D F \perp B C$ at point $F$. If $A E=6, B F=3$, then $C D=(\quad)$.
(A) 6
(B) $3 \sqrt{2}$
(C) $\frac{9}{2}$
(D) $2 \sqrt{5}$ | 4. A.
$$
\begin{array}{l}
\text { It is easy to know that } C D=\sqrt{2} D E=\sqrt{2} D F . \\
\text { By } \triangle A E D \backsim \triangle D F B \\
\Rightarrow \frac{A E}{D F}=\frac{D E}{B F} \\
\Rightarrow C D^{2}=2 D E \cdot D F=2 A E \cdot B F=36 \\
\Rightarrow C D=6 .
\end{array}
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,801 |
5. As shown in Figure 2, in the plane, there are four line segments $A B, B C, C D, D A$ connected head to tail in sequence. Point $E$ is on the extension of $B A$, and the angle bisector of $\angle D A E$ intersects the angle bisector of $\angle B C D$ at point $F$. If $\angle B=26^{\circ}, \angle D=62^{\circ}$, then ... | 5. D.
According to the problem, let $\angle D A F=\angle E A F=\alpha$,
$$
\begin{array}{l}
\angle D C F=\angle B C F=\theta . \\
\text { Then } \angle B+180^{\circ}-2 \alpha=\angle D+2 \theta \\
\Rightarrow \alpha+\theta=90^{\circ}-\frac{1}{2}(\angle D-\angle B) .
\end{array}
$$
Therefore, $\angle A F C=\angle D+(\a... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,802 |
6. As shown in Figure 3, with the right angle side $AB$ of the right triangle $\triangle ABC$ as the diameter, a semicircle $\odot O$ is drawn intersecting side $BC$ at point $D$. A tangent line is drawn through $D$ intersecting side $AC$ at point $E$. A line $EF$ parallel to $AB$ is drawn through $E$, intersecting $BC... | 6. C.
From the problem, we know $A D \perp B C, D E=A E$. Therefore, $E$ is the midpoint of the hypotenuse $A C$ of the right triangle $\triangle A C D$.
$$
\begin{array}{l}
\text { Also, } E F / / A B \Rightarrow C F=B F \\
\Rightarrow A C=2 O F=15 \\
\Rightarrow B C=\sqrt{20^{2}+15^{2}}=25 .
\end{array}
$$
Thus, by... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,803 |
1. As shown in Figure 4, in the Cartesian coordinate system, $O$ is the origin, the diagonals of $\square A B O C$ intersect at point $M$, and the hyperbola $y=\frac{k}{x}(x<0)$ passes through points $B$ and $M$. If the area of $\square A B O C$ is 24, then $k=$ . $\qquad$ | 2, 1. -8.
Let $M\left(\frac{k}{y}, y\right)$. Then $B\left(\frac{k}{2 y}, 2 y\right), C\left(\frac{3 k}{2 y}, 0\right)$.
From $S_{\text {OABOC }}=4 S_{\triangle O C M}=2\left|x_{c} y_{M}\right|$
$\Rightarrow 24=2\left|\frac{3 k}{2}\right| \Rightarrow|k|=8 \Rightarrow k=-8$. | -8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,804 |
2. Given that $m$, $n$, and $p$ are real numbers. If $x-1$ and $x+4$ are both factors of the polynomial $x^{3}+m x^{2}+n x+p$, then
$$
2 m-2 n-p+86=
$$
$\qquad$. | 2. 100 .
From the divisibility property of polynomials, we know
$$
\left\{\begin{array} { l }
{ 1 + m + n + p = 0 , } \\
{ - 6 4 + 1 6 m - 4 n + p = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
p=12-4 m, \\
n=3 m-13 .
\end{array}\right.\right.
$$
Therefore, $2 m-2 n-p+86$
$$
\begin{array}{l}
=2 m-2(3 m-13)-(1... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,805 |
3. Let $a-b-c=17, a^{2}+b^{2}+c^{2}=17$. Then $bc-ca-ab=$ $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 136 .
$$
\begin{array}{l}
\text { Given } a-b-c=17 \\
\Rightarrow(a-b-c)^{2}=289 \\
\Rightarrow a^{2}+b^{2}+c^{2}+2(b c-c a-a b)=289 \\
\Rightarrow b c-c a-a b=\frac{289-17}{2}=136 .
\end{array}
$$ | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,806 |
4. As shown in Figure 5, in $\triangle A B C$, it is known that $\tan \angle B A C=\frac{3}{4}$, $A D \perp B C$, with the foot of the perpendicular being $D$, and $B D$ $=9, C D=3$. Then
$$
S_{\triangle A B C}=
$$
$\qquad$ | $4.48+6 \sqrt{91}$.
As shown in Figure 7, let O be
the circumcenter of $\triangle ABC$, and
draw $OE \perp BC$ at point $D$, $OF$
$\perp AD$ at point $F$, and connect $OA, OB, OC$.
Given $\tan \angle BAC = \frac{3}{4}$
$\Rightarrow \cos \angle BAC = \frac{4}{5}, \sin \angle BAC = \frac{3}{5}$.
It is easy to see that $B... | 48 + 6 \sqrt{91} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,807 |
II. (25 points) As shown in Figure 6, with the vertices of an equilateral triangle $\triangle ABC$ with side length 2 as centers and the side length as the radius, draw three equal circles, obtaining intersection points $D, E, F$. Connect $CF$, intersecting $\odot C$ at point $G$. With $E$ as the center and $EG$ as the... | II. As shown in Figure 8, draw $EP \perp AB$, intersecting the extension of $BA$ at point $P$, and connect $EA$, $EC$, $EG$, and $EM$.
It is easy to see that $\triangle EAC$ is an equilateral triangle. Therefore, $EC \parallel AB$.
Since $CG \perp AB \Rightarrow EC \perp CG$
$\Rightarrow EM = EG = 2\sqrt{2}$.
Given $\a... | \frac{3\sqrt{3} - \sqrt{15}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,809 |
Example 8 In an acute triangle $\triangle ABC$, it is known that $AP$ is the altitude on side $BC$, and $O$ is the circumcenter. If $\angle C \geqslant \angle B + 30^{\circ}$, prove: $\angle A + \angle COP < 90^{\circ}$. | Prove as shown in Figure 9, construct $OM \perp AB$ at point $M$. Let the symmetric points of $A, P$ about $OM$ be $A', P'$.
Since $A, A'$ and $B, C$ are symmetric about $OM$, then $\angle ABA' = \angle ACB - \angle ABC \geqslant 30^{\circ}$.
Since $O$ is the circumcenter of $\triangle ABA'$, we have $\angle AOA' \geqs... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,810 |
Three. (25 points) Given the equation about $x$
$$
4 x^{2}-8 n x-3 n-2=0
$$
and $x^{2}-(n+3) x-2 n^{2}+2=0$.
Question: Is there such a value of $n$ that the square of the difference of the two real roots of equation (1) equals an integer root of equation (2)? If it exists, find such $n$ values; if not, explain the rea... | Three, because the discriminant of equation (1) is
$$
\begin{array}{l}
\Delta_{1}=(-8 n)^{2}-4 \times 4(-3 n-2) . \\
=(8 n+3)^{2}+23>0,
\end{array}
$$
Therefore, equation (1) has two distinct real roots $\alpha_{1} 、 \beta_{1}$.
By Vieta's formulas, we get
$$
\begin{array}{l}
\left(\alpha_{1}-\beta_{1}\right)^{2}=\lef... | n=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,811 |
1. Given
$$
\begin{aligned}
A= & \left\{x \mid x^{2}-4 x+3<0, x \in \mathbf{R}\right\}, \\
B= & \left\{x \mid 2^{1-x}+a \leqslant 0, x^{2}-2(a+7) x+5 \leqslant\right. \\
& 0, x \in \mathbf{R}\} .
\end{aligned}
$$
If $A \subseteq B$, then the range of real number $a$ is | $-1 .-4 \leqslant a \leqslant-1$.
From the problem, we know $A=\{x \mid 1<x<3\}$.
Let $f(x)=2^{1-x}+a$,
$$
g(x)=x^{2}-2(a+7) x+5 \text {. }
$$
To make $A \subseteq B$, it is only necessary that the graphs of $f(x)$ and $g(x)$ are below the x-axis in the interval $(1,3)$, then
$$
f(1) \leqslant 0, f(3) \leqslant 0, g(1... | -4 \leqslant a \leqslant -1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,812 |
2. Given that the center of a hyperbola is at the origin, and its foci are on the coordinate axes, the distance from point $P(-2,0)$ to its asymptotes is $\frac{2 \sqrt{6}}{3}$. A line with a slope of $\frac{\sqrt{2}}{2}$ passing through point $P$ intersects the hyperbola at points $A$ and $B$, and intersects the $y$-a... | 2. $\sqrt{3}$ or $\sqrt{21}$.
Let the equation of the asymptote be $y=k x$.
From the given condition, we have $\frac{|-2 k|}{\sqrt{1+k^{2}}}=\frac{2 \sqrt{6}}{3} \Rightarrow k= \pm \sqrt{2}$.
Thus, the equations of the asymptotes of the hyperbola are $y= \pm \sqrt{2} x$. Therefore, we can assume the equation of the hy... | \sqrt{3} \text{ or } \sqrt{21} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,813 |
3. In a cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with edge length 1, it is known that $O_{1}$ is the center of the base $A_{1} B_{1} C_{1} D_{1}$, $M$ is a point on the edge $B B_{1}$, and $S_{\triangle D B M}: S_{\triangle O_{1} B_{1} M}=$ $2: 3$. Then the volume of the tetrahedron $O_{1} A D M$ is $\qquad$ . | 3. $\frac{7}{48}$.
As shown in Figure 2, let $O$ be the center of the base $ABCD$. Then $AO \perp$ plane $DO_1M$.
From $\frac{S_{\triangle DBM}}{S_{\triangle O_1B_1M}}=\frac{2BM}{B_1M}=\frac{2}{3} \Rightarrow \frac{BM}{B_1M}=\frac{1}{3}$.
Thus, $BM=\frac{1}{4}, B_1M=\frac{3}{4}$.
Therefore, $S_{\triangle DO_1M}$
$$
\b... | \frac{7}{48} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,814 |
4. Given that $f(x)$ satisfies
$$
f(x)+f\left(\frac{x-1}{x}\right)=1+x \quad (x \neq 0,1) \text {. }
$$
then the range of $g(x)=x-2 f(x)$ is $\qquad$ | 4. $(-\infty,-4] \cup(0,+\infty)$.
Let $\varphi(x)=\frac{x-1}{x}$. Then
$$
\begin{array}{l}
\varphi^{2}(x)=\varphi(\varphi(x))=\varphi\left(\frac{x-1}{x}\right)=\frac{1}{1-x}, \\
\varphi^{3}(x)=\varphi\left(\varphi^{2}(x)\right)=\varphi\left(\frac{1}{1-x}\right)=x .
\end{array}
$$
At this point, the condition is expr... | (-\infty,-4] \cup(0,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,815 |
5. Given positive numbers $a, b, c$ satisfy
$$
a \leqslant b+c \leqslant 3 a, 3 b^{2} \leqslant a(a+c) \leqslant 5 b^{2} \text {. }
$$
Then the minimum value of $\frac{b-2 c}{a}$ is $\qquad$ | 5. $-\frac{18}{5}$.
Transform the given inequality into $1 \leqslant \frac{b}{a}+\frac{c}{a} \leqslant 3, 3\left(\frac{b}{a}\right)^{2} \leqslant 1+\frac{c}{a} \leqslant 5\left(\frac{b}{a}\right)^{2}$. Let $\frac{b}{a}=x, \frac{c}{a}=y$. Then
$$
\left\{\begin{array}{l}
1 \leqslant x+y \leqslant 3, \\
3 x^{2} \leqslant... | -\frac{18}{5} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,816 |
9. (16 points) In a $(2 n+1) \times(2 n+1)$ number table, each row is an arithmetic sequence, and the squares of the numbers in each column form an arithmetic sequence. Prove: top left $\times$ bottom right $=$ bottom left $\times$ top right. | 8. 201.
Let the number of positive integer sequences $\left\{a_{n}\right\}$ with the first term $m$ and satisfying the given conditions be $b_{m}$. Then
$$
b_{m}=b_{1}+b_{2}+\cdots+b_{k}\left(k^{2}+1 \leqslant m \leqslant(k+1)^{2}\right) .
$$
Notice that,
$$
44^{2}+1=1937<2013<2025=45^{2} \text {. }
$$
It is easy to... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,820 |
Given a cyclic convex quadrilateral $A B C D$ satisfying $\angle B A D=60^{\circ}$, the ray $A B$ intersects $D C$ at point $E$, the ray $A D$ intersects $B C$ at point $F$, the point $A$ is symmetric to point $N$ with respect to the diagonal $B D$, and points $E, N, F$ are collinear. Prove: $A C$ bisects $\angle B C D... | Prove (1) As shown in Figure 2, $N$ is not the Miquel point $M$ of the complete quadrilateral $A B E C F D$. Connect $B M, D M, B N, D N$.
Then $\angle E M B = \angle B A D = 60^{\circ}$,
$\angle F M D = \angle B A D = 60^{\circ}$.
Thus, $\angle B M D = 60^{\circ} = \angle B N D$
$\Rightarrow B, N, M, D$ are concyclic
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,828 |
During the Teachers' Day, 200 teachers at a school sent text messages to greet each other, with each teacher sending exactly 1 text message to another teacher. Now, from them, the maximum number of $k$ teachers can be selected to attend an award ceremony, such that none of them has sent a text message to any of the oth... | Let the 200 teachers be denoted as $A_{1}, A_{2}, \cdots, A_{200}$, and let $G=\left\{A_{1}, A_{2}, \cdots, A_{200}\right\}$. If $A_{i}$ sends a message to $A_{j}$, then we mark an arrow from $A_{i}$ to $A_{j}$ as
$A_{i} \rightarrow A_{j}(1 \leqslant i \neq j \leqslant 200)$.
First, construct an instance for $k=67$.
In... | 67 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,830 |
Given positive integers $a, b$ and a positive number $\alpha$. Does
there exist infinitely many positive integers $n$, such that the greatest prime factor of $a n^{2}+b$ is less than $\alpha n$? | Prove that taking a sufficiently large positive integer $k$, for any positive integer $n$, we have
$$
\sqrt{\frac{a n^{2}+b}{a k^{2}+b}}<\alpha n \text {. }
$$
Since $a n^{2}+b=\left(a k^{2}+b\right) m^{2}$ has a positive integer solution $(k, 1)$, there are infinitely many positive integer solutions $(n, m)$ such tha... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,831 |
Example 1 In $\triangle ABC$, if $AB=5, BC=6$, $CA=7, H$ is the orthocenter, then the length of $AH$ is $\qquad$
$(2000$, Shanghai Junior High School Mathematics Competition) | Extend $A H$, intersecting $B C$ at point $D$, and connect $B H$ and extend it to intersect $A C$ at point $E$.
By the properties of the orthocenter, we know
$A D \perp B C$,
$B E \perp A C$.
In Rt $\triangle A B E$ and Rt $\triangle C B E$, by the Pythagorean theorem, we get
$$
\begin{array}{l}
B E^{2}=A B^{2}-A E^{2}... | \frac{19 \sqrt{6}}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,833 |
Example 2 In $\triangle A B C$, $\angle A B C=60^{\circ}, O, H$ are the circumcenter and orthocenter of $\triangle A B C$ respectively. Points $D, E$ lie on sides $B C, A B$ respectively, such that $B D=B H, B E=B O$. Given that $B O=1$. Find the area of $\triangle B D E$. ${ }^{[1]}$
(2008, International Youth Math Ci... | Solve As shown in Figure 2, draw the diameter $A M$ of the circumcircle of $\triangle A B C$, and connect $C M, B M, C H$.
Since $H$ is the orthocenter of $\triangle A B C$, we have
$$
\begin{array}{l}
B H \perp A C, \\
C H \perp A B .
\end{array}
$$
Since $A M$ is the diameter, we get $M C \perp A C, M B \perp A B$.... | \frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,834 |
Example 3 Lucas' Theorem Let $p$ be a prime, $a, b \in \mathbf{N}_{+}$, integers $a_{i}, b_{i}$ satisfy $0 \leqslant a_{i}, b_{i} \leqslant p-1$ $(i=0,1, \cdots, k)$, and
$$
\begin{array}{l}
a=a_{k} p^{k}+a_{k-1} p^{k-1}+\cdots+a_{1} p+a_{0}, \\
b=b_{k} p^{k}+b_{k-1} p^{k-1}+\cdots+b_{1} p+b_{0} .
\end{array}
$$
Then ... | Prove that for a prime $p$, for $j(1 \leqslant j \leqslant p-1)$, we have $\mathrm{C}_{\mathrm{p}}^{j}=\frac{p}{j} \mathrm{C}_{p-1}^{j-1} \equiv 0(\bmod p)$.
Thus, $(1+x)^{p}=1+\mathrm{C}_{p}^{1} x+\cdots+\mathrm{C}_{p}^{p-1} x^{p-1}+x^{p}$ $\equiv 1+x^{p}(\bmod p)$.
From this, we get
$$
\begin{array}{l}
(1+x)^{a}=(1+x... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,835 |
11. (20 points) Let the function
$$
f_{n}(\theta)=\sin ^{n} \theta+(-1)^{n} \cos ^{n} \theta\left(0 \leqslant \theta \leqslant \frac{\pi}{4}, n \in \mathbf{N}_{+}\right) \text {. }
$$
For any given positive integer $n$, find the maximum and minimum values of the function $f_{n}(\theta)$. | 11. Notice that, when $n$ is odd, $\sin ^{n} \theta$ and $(-\cos \theta)^{n}$ are both monotonically increasing functions on $\left[0, \frac{\pi}{4}\right]$.
Therefore, $f_{n}(\theta)$ is monotonically increasing on $\left[0, \frac{\pi}{4}\right]$, with its maximum value being $f_{n}\left(\frac{\pi}{4}\right)=0$ and i... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,836 |
一、(40 points) As shown in Figure 1, in the convex quadrilateral $ABCD$, it is known that $BA=AD=DC$, $AC \neq BD$, and $AC$ intersects $BD$ at point $P$. Prove that $\angle ABC + \angle BCD = 120^{\circ}$ if and only if $PB = PC$. | Necessity.
Given $\angle A B C+\angle B C D=120^{\circ}$.
As shown in Figure 2, extend $B A$ and $C D$ to intersect at point $Q$, and connect $P Q$. Let the two base angles of isosceles $\triangle A B D$ be $\alpha$, and the two base angles of isosceles $\triangle D A C$ be $\beta$.
Since $\angle A B C+\angle B C D=12... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,837 |
II. (40 points) Given that $p$ is a prime number, and $m, n$ are natural numbers. Find all tuples $(p, m, n)$ such that $p^{m}+p^{n}$ is a perfect square. | (1) When $m=n$, $p^{m}+p^{n}=2 p^{m}=t^{2}$, then $p=t=2$, and $m=2 k+1(k \in \mathbf{N})$.
Thus, the array
$$
(p, m, n)=(2,2 k+1,2 k+1)(k \in \mathbf{N}) \text {. }
$$
(2) When $m \neq n$, assume without loss of generality that $m>n$. Then
$$
p^{m}+p^{n}=p^{n}\left(p^{d}+1\right) \text {, }
$$
where $d=m-n>0(d \in \m... | (2,2 k+1,2 k+1),(2,2 k+3,2 k),(2,2 k, 2 k+3),(3,2 k+1,2 k),(3,2 k, 2 k+1)(k \in \mathbf{N}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,838 |
Three. (50 points) There are 2014 points distributed on a circle, which are arbitrarily colored red and yellow. If starting from a certain point and moving around the circle in any direction to any position, the number of red points (including the starting point) is always greater than the number of yellow points, then... | Three, the advantages must be red points.
First, consider the simple case.
When there are $1, 2, 3, 4, 5, 6, 7$ points on the circumference, if there is at least one advantage point on the circumference, then the maximum number of yellow points on the circumference are $0, 0, 0, 1, 1, 1, 2$. From this, we can derive a ... | 671 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,839 |
Four, (50 points) Find all real-coefficient polynomials $f(x)$ such that for all real numbers $a, b, c$ satisfying $a b+b c+c a=0$, we have
$$
f(a-b)+f(b-c)+f(c-a)=2 f(a+b+c) .
$$ | Let $a=b=c=0$ in equation (1).
Then $3 f(0)=2 f(0) \Rightarrow f(0)=0$.
Let $b=c=0$.
Then $f(a)+f(-a)=2 f(a) \Rightarrow f(-a)=f(a)$.
Therefore, $f(x)$ is an even function.
Assume $f(x)=a_{n} x^{2 n}+a_{n-1} x^{2 n-2}+\cdots+a_{1} x^{2}$,
where $a_{n} \neq 0$.
Let $a=x, b=2 x, c=-\frac{2}{3} x$. Then
$$
\begin{array}{l... | f(x)=\alpha x^{4}+\beta x^{2}(\alpha, \beta \in \mathbf{R}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,840 |
Given $x_{1}, x_{2}, \cdots, x_{n}>0$, satisfying $x_{1}+x_{2}+\cdots+x_{n}=n$. Prove:
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{i}{1+x_{i}+x_{i}^{2}+\cdots+x_{i}^{i-1}} \\
\leqslant \sum_{i=1}^{n} \frac{i+1}{1+x_{i}+x_{i}^{2}+\cdots+x_{i}^{i}} .
\end{array}
$$ | Notice,
$$
\begin{array}{l}
2 \sum_{i=1}^{n} \frac{i}{1+x_{i}+x_{i}^{2}+\cdots+x_{i}^{i-1}} \\
=\sum_{i=1}^{n} \frac{i\left(x_{i}^{i}+1\right)}{1+x_{i}+x_{i}^{2}+\cdots+x_{i}^{i-1}}-\sum_{i=1}^{n} i\left(x_{i}-1\right), \\
2 \sum_{i=1}^{n} \frac{i+1}{1+x_{i}+x_{i}^{2}+\cdots+x_{i}^{i}} \\
=\sum_{i=1}^{n} \frac{(i+1)\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,841 |
As shown in Figure 2, given that $\odot O$ is the incircle of $\triangle ABC$, and $D, E, F$ are the points of tangency on the three sides, while $X, Y, Z$ are the points of tangency of the three excircles of $\triangle ABC$ on the respective sides. Compare the sizes of $S_{\triangle DEF}$ and $S_{\triangle XYZ}$, and ... | Proof: Let the circumradius of $\triangle ABC$ be $R$ and the inradius be $r$. As shown in Figure 2, connect $OD$, $OE$, and $OF$. Then $OD = OE = OF = r$, and $OD \perp BC$, $OE \perp CA$, $OF \perp AB$.
It is easy to see that $A, F, O, E$, $B, D, O, F$, and $C, E, O, D$ are concyclic, respectively.
$$
\begin{array}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,842 |
Given 361, find the smallest positive integer $n$, such that 16 numbers can be selected from $1,2, \cdots, n$ and filled into a $4 \times 4$ grid so that the product of the numbers in each row and each column is equal. | The smallest value of the positive integer $n$ is 27.
First, when $n=27$, a valid arrangement is shown in Figure 3.
Next, we prove that the smallest value of $n$ is 27.
Consider the prime factorization of the product $P$ of each row or column.
(1) If $P$ has only one prime factor, then $n \geqslant 2^{15} > 27$.
(2) If... | 27 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,843 |
For any positive odd integer $n$, prove: there do not exist positive integers $j, k$ such that $\csc \frac{j \pi}{n}-\csc \frac{k \pi}{n}=2$.
---
The translation maintains the original format and line breaks as requested. | Proof: Let $\varepsilon_{m}=\mathrm{e}^{\frac{2 \pi i}{m}}$ be an $m$-th primitive root of unity (all $m$-th primitive roots of unity are $\varepsilon_{m}^{k}$, where $0 \leqslant k < m$ and $(k, m)=1$, and there are $\varphi(m)$ (Euler's totient function) such roots).
Using proof by contradiction:
Assume there exists... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,844 |
Example 4 Find the last three digits of $2013^{2013^{2013}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Notice,
$$
\begin{array}{l}
2013 \equiv 3(\bmod 5), 2013^{2} \equiv-1(\bmod 5), \\
2013^{4} \equiv 1(\bmod 5), 2013 \equiv 13(\bmod 25), \\
2013^{2} \equiv-6(\bmod 25), 2013^{4} \equiv 11(\bmod 25), \\
2013^{8} \equiv-4(\bmod 25), 2013^{12} \equiv 6(\bmod 25), \\
2013^{20} \equiv 1(\bmod 25) .
\end{array}
$$
Since $20... | 053 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,845 |
Example 5 Find all odd prime numbers $p$ such that $p \mid \sum_{k=1}^{103} k^{p-1}$.
untranslated text remains the same as requested. | If $p>103$, then for $1 \leqslant k \leqslant 103$, we have
$$
\begin{array}{l}
k^{p-1} \equiv 1(\bmod p), \\
\sum_{k=1}^{103} k^{p-1} \equiv 103(\bmod p) .
\end{array}
$$
Therefore, $p \leqslant 103$.
Let $103=p q+r(0 \leqslant r < q)$, then $r=q, 103=p q+r=(p+1) r$. Since 103 is a prime number, we get $p=102, r=1$, ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,846 |
Example 6 Find all prime numbers $p$ such that
$$
p^{3} \mid \sum_{k=1}^{p-1}\left(\mathrm{C}_{p}^{k}\right)^{2} .
$$ | Obviously, $p \neq 2,3$.
When $p \geqslant 5$, note that,
$$
\begin{array}{l}
p-1 \equiv-1(\bmod p), \\
p-2 \equiv-2(\bmod p), \\
\cdots \cdots \\
k \equiv k-p(\bmod p)(k=1,2, \cdots, p) .
\end{array}
$$
Multiplying all the equations, we get
$$
\frac{(p-1)!}{(k-1)!} \pm(p-k)!(\bmod p),
$$
which means $\mathrm{C}_{p-1... | p \geqslant 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,847 |
Example 7 Find all prime pairs $(p, q)$ such that $p q \mid\left(p^{p}+q^{q}+1\right)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | If $p=q$, then $p^{2} \nmid\left(2 p^{p}+1\right)$.
Without loss of generality, assume $p<q$.
If $p=2$, then
$$
2 q\left|\left(q^{q}+5\right) \Rightarrow q\right| 5 \Rightarrow q=5 \text {. }
$$
Therefore, $(p, q)=(2,5)$.
When $p$ and $q$ are both odd primes, we have
$$
\begin{array}{l}
p q \mid\left(p^{p}+q^{q}+1\rig... | (p, q)=(2,5),(5,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,848 |
Example 8 Proof: The equation
$$
x^{2014}=4 y^{2013}+4 y^{2012}+2011 y+2010
$$
has no integer solutions. | Proof of the lemma first.
Lemma If $x$ is any integer, then each odd prime factor of $x^{2}+1$ is of the form $4 k+1$.
Proof Let $p$ be an odd prime factor of $x^{2}+1$, i.e., $p \mid\left(x^{2}+1\right)$. Then $(p, x)=1$, and $x^{2}=-1(\bmod p)$.
Therefore, $\left(x^{2}\right)^{\frac{p-1}{2}} \equiv(-1)^{\frac{p-1}{2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,849 |
Example 9 Find all positive integers that are coprime with all terms of the sequence $\left\{a_{n}\right\}$ satisfying
$$
a_{n}=2^{n}+3^{n}+6^{n}-1\left(n \in Z_{+}\right)
$$ | Solution: Clearly, $\left(1, a_{n}\right)=1$.
Let $m(m>1)$ be a positive integer that is coprime with all terms in $\left\{a_{n}\right\}$, and let $p$ be a prime factor of $m$.
If $p>3$, then by Fermat's Little Theorem,
$$
\begin{array}{l}
2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), \\
6^{p-1} \equiv 1(\bmod ... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,850 |
Example 10 Proof: There do not exist positive integers $k, m$, such that
$$
k!+48=48(k+1)^{m} .
$$ | Prove (1) If $k+1$ is a composite number, then
$$
(k+1)|k!,(k+1)| 48 \text {. }
$$
From $48 \mid k!\Rightarrow k \geqslant 6 \Rightarrow k=7,11,23,47$, inspection shows none of these are solutions.
(2) If $k+1$ is a prime number, by Wilson's theorem we know
$$
k!\equiv-1(\bmod k+1) \Rightarrow(k+1) \mid(k!+1) \text {.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,851 |
1. Prove: For any positive integer $n, 3^{n}+2 \times 17^{n}$ is not a multiple of 5, and find the smallest positive integer $n$, such that
$$
11 \mid\left(3^{n}+2 \times 17^{n}\right) .
$$ | $$
\begin{array}{l}
3^{2 k}+2 \times 17^{2 k} \equiv(-1)^{k}+2 \times 2^{2 k} \\
\equiv 3(-1)^{k}(\bmod 5) \\
3^{2 k+1}+2 \times 17^{2 k+1} \equiv 3(-1)^{k}+4(-1)^{k} \\
\equiv 2(-1)^{k}(\bmod 5)
\end{array}
$$
Therefore, $3^{n}+2 \times 17^{n}$ is not a multiple of 5.
$$
\begin{array}{l}
\text { Also, } 3^{n}+2 \time... | 4 | Number Theory | proof | Yes | Yes | cn_contest | false | 726,852 |
2. For any positive integer $n(n \geqslant 1), b$ has prime factors all greater than $n$. Prove:
$$
n! \mid a(a+b)(a+2 b) \cdots[a+(n-1) b] .
$$ | ```
Because the prime factors of $b$ are all greater than $n$, we have
$$
\begin{array}{l}
(b, n!)=1, b b^{-1} \equiv 1(\bmod n!), \\
\left(b^{-1}\right)^{n} a(a+b)(a+2 b) \cdots[a+(n-1) b] \\
\equiv\left(a b^{-1}\right)\left(a b^{-1}+1\right)\left(a b^{-1}+2\right) \cdots\left[a b^{-1}+(n-1)\right] \\
\equiv 0(\bmod n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,853 |
Example 3 As shown in Figure 3, given $H$ is the orthocenter of $\triangle A B C$, the circle $\odot O_{1}$ with $A B$ as its diameter intersects with the circumcircle $\odot O_{2}$ of $\triangle B C H$ at point $D$. Extend $A D$ to intersect $C H$ at point $P$. Prove: $P$ is the midpoint of $\mathrm{CH}$.
[2] | Prove as shown in Figure 3, extend $AP$, intersecting $\odot O_{2}$ at point $E$, and connect $AH$, $HE$, $EC$, $BD$, $BE$.
Since $AB$ is the diameter of $\odot O_{1}$, we know
$\angle ADB = \angle BDE = 90^{\circ}$.
Therefore, $BE$ is the diameter of $\odot O_{2}$.
Thus, $CE \perp BC$, $HE \perp BH$.
Since $H$ is the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,854 |
4. Given that $p$ is an odd prime. Prove:
$$
\sum_{k=1}^{p-1} k^{2 p-1} \equiv \frac{p(p+1)}{2}\left(\bmod p^{2}\right) .
$$ | Given that $p-1$ is even, we have
$$
\sum_{k=1}^{p-1} k^{2 p-1}=\sum_{k=1}^{\frac{p-1}{2}}\left[k^{2 p-1}+(p-k)^{2 p-1}\right] \text {. }
$$
Notice that,
$$
\begin{array}{l}
k^{2 p-1}+(p-k)^{2 p-1} \\
=p^{2 p-1}-\mathrm{C}_{2 p-1}^{1} p^{2 p-2} k+\cdots+\mathrm{C}_{2 p-1}^{2 p-2} p k^{2 p-2} \\
\equiv \mathrm{C}_{2 p-... | \frac{p(p+1)}{2}\left(\bmod p^{2}\right) | Number Theory | proof | Yes | Yes | cn_contest | false | 726,856 |
Example 1 Let $n \geqslant 3$ be an integer, and let $a_{2}, a_{3}, \cdots, a_{n}$ be positive real numbers satisfying $a_{2} a_{3} \cdots a_{n}=1$. Prove:
$$
\left(1+a_{2}\right)^{2}\left(1+a_{3}\right)^{3} \cdots\left(1+a_{n}\right)^{n}>n^{n} \text {. }
$$
(53rd IMO) | Proof for $k=2,3, \cdots, n$, by the AM-GM inequality we have
$$
\begin{array}{l}
\left(1+a_{k}\right)^{k}=(\frac{1}{k \underbrace{-1}_{k-1 \uparrow}+\frac{1}{k-1}+\cdots+\frac{1}{k-1}}+a_{k})^{k} \\
\geqslant k^{k}\left(\frac{1}{k-1}\right)^{k-1} a_{k} . \\
\text { Hence }\left(1+a_{2}\right)^{2}\left(1+a_{3}\right)^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,857 |
Example 2 Given the sequence $\left\{a_{n}\right\}$:
$1<a_{i}<\sqrt{7}(i=1,2, \cdots, n$, positive integer $n \geqslant 2)$. Prove: (1) For all positive integers $i$, we have $\frac{1}{a_{i}^{2}-1}+\frac{1}{7-a_{i}^{2}} \geqslant \frac{2}{3}$;
(2) Let $a_{n+1}=a_{1}$. Then
$$
\sum_{i=1}^{n} \frac{1}{\sqrt{\left(a_{i}^{... | Prove (1) For all positive integers $i$ we have
$$
\begin{array}{l}
\frac{1}{a_{i}^{2}-1}+\frac{1}{7-a_{i}^{2}}=\frac{6}{\left(a_{i}^{2}-1\right)\left(7-a_{i}^{2}\right)} \\
\geqslant \frac{6}{\left(\frac{a_{i}^{2}-1+7-a_{i}^{2}}{2}\right)^{2}}=\frac{2}{3} .
\end{array}
$$
(2) By the Cauchy-Schwarz inequality, we know
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,858 |
Example 3 Proof:
\[
\begin{array}{l}
(1+1)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{7}\right) \cdots\left(1+\frac{1}{3 n-2}\right) \\
>\sqrt[3]{3 n+1}\left(n \in \mathbf{N}_{+}\right) .
\end{array}
\] | Proof: By Bernoulli's inequality, we know
$$
(1+x)^{n}>1+n x \text {, }
$$
where, $n \in \mathbf{N}_{+}, n \geqslant 2, x>-1, x \neq 0$.
Let $n=3, x=\frac{1}{3 k-2}$. Then
$$
\begin{array}{l}
\left(1+\frac{1}{3 k-2}\right)^{3}>1+3 \times \frac{1}{3 k-2} \\
\Rightarrow 1+\frac{1}{3 k-2}>\sqrt[3]{1+\frac{3}{3 k-2}}=\sqr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,859 |
Example 4 Given an integer $n(n>2)$, let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{k} \leqslant 1(k=1,2, \cdots, n)$. Let
$$
A_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, n) .
$$
Prove: $\left|\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} A_{k}\right|<\frac{n-1}{2}$.
(2010, National High School M... | Prove that when $00$, we have
$$
|x-y|<\max \{x, y\} \text {. }
$$
Thus, for $1 \leqslant k \leqslant n-1$, we have
$$
\begin{array}{l}
\left|A_{n}-A_{k}\right|=\left|\frac{1}{n} \sum_{i=k+1}^{n} a_{i}-\left(\frac{1}{k}-\frac{1}{n}\right) \sum_{i=1}^{k} a_{i}\right| \\
<\max \left\{\frac{1}{n} \sum_{i=k+1}^{n} a_{i},\... | \frac{n-1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 726,860 |
Example 6 Given an integer $n(n>2012)$, positive real numbers $x_{1}$, $x_{2}, \cdots, x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n}=1$. Prove:
$$
\left(\frac{x_{1}^{2}}{1+x_{1}}+\frac{x_{2}^{2}}{1+x_{2}}+\cdots+\frac{x_{n}^{2}}{1+x_{n}}\right)^{\frac{1}{n}}>\left(\frac{1}{2013}\right)^{\frac{1}{2012}} .
$$ | Proof: First, prove that when $n > m > 0$, $(1+n)^{m} < (1+m)^{n}$.
Let $g(x) = \frac{\ln (1+x)}{x}$, then $g^{\prime}(x) = \frac{x - (1+x) \ln (1+x)}{x^{2}(1+x)}$.
From Example 5, we know $\frac{x}{1+x} < \ln (1+x) < x$ for $x > 0$, thus
\[ g^{\prime}(x) < 0. \]
Therefore, $g(x)$ is a decreasing function for $x > 0$.
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,862 |
Example 7 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1} \in(1,2), a_{n+1}=a_{n}^{3}-3 a_{n}^{2}+3 a_{n}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Prove: $\sum_{k=1}^{n}\left(a_{k}-a_{k+1}\right)\left(a_{k+2}-1\right)<\frac{1}{4}$.
(2010, National High School Mathematics League, Jiangsu Province Re-... | Prove: From $a_{n+1}=a_{n}^{3}-3 a_{n}^{2}+3 a_{n}$
$$
\begin{array}{l}
\Rightarrow a_{n+1}-1=\left(a_{n}-1\right)^{3} . \\
\text { Let } b_{n}=a_{n}-1, \text { then } \\
0<b_{1}<1, b_{n+1}=b_{n}^{3}<b_{n}, 0<b_{n}<1 . \\
\text { Therefore, }\left(a_{k}-a_{k+1}\right)\left(a_{k+2}-1\right) \\
=\left(b_{k}-b_{k+1}\right... | \frac{1}{4} | Algebra | proof | Yes | Yes | cn_contest | false | 726,863 |
Example 8 Let $\left\{a_{n}\right\}(n \in \mathrm{N})$ satisfy
$$
a_{0}=\frac{1}{2} \text {, and } a_{k+1}=a_{k}+\frac{1}{n} a_{k}^{2}(k=0,1, \cdots, n-1) \text {. }
$$
Prove: $1-\frac{1}{n}<a_{n}<1$.
(1980, Finland, United Kingdom, Hungary, Sweden Mathematical Olympiad) | $$
\begin{array}{l}
a_{n}>a_{n-1}>\cdots>a_{1}>a_{0}>0 \\
\Rightarrow a_{k+1}=a_{k}+\frac{1}{n} a_{k}^{2}\frac{1}{a_{0}}-1=1 \Rightarrow a_{n}\frac{n}{n+1} a_{k+1} \\
\Rightarrow a_{k+1}=a_{k}+\frac{1}{n} a_{k}^{2}>a_{k}+\frac{1}{n} a_{k} \cdot \frac{n}{n+1} a_{k+1} \\
\quad=a_{k}+\frac{1}{n+1} a_{k} a_{k+1} \\
\Righta... | 1-\frac{1}{n}<a_{n}<1 | Algebra | proof | Yes | Yes | cn_contest | false | 726,864 |
Example 4 Let the diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ be perpendicular to each other. Through the midpoint $M$ of $AB$, draw a perpendicular to $CD$, and through the midpoint $N$ of $AD$, draw a perpendicular to $CB$. Prove: The intersection of these two perpendiculars lies on the line $AC$. ${ }^{... | Prove as shown in Figure 4, take the midpoint $P$ of $AC$, and connect $MN$, $MP$, $NP$.
By the Midline Theorem of a triangle, we get $MN \parallel BD$, $MP \parallel BC$, $NP \parallel CD$.
Since $PA \perp BD$, $NE \perp BC$, $MF \perp CD$, it follows that $PA \perp MN$, $NE \perp MP$, $MF \perp NP$.
Thus, accordin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,865 |
Example 9 Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{n+1}=x_{n}+\frac{x_{n}^{2}}{n^{2}}$.
Prove: $x_{2013}<1007$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
(Note: The provided text is al... | Prove that the proposition can be strengthened to $x_{n} \leqslant \frac{n}{2}$.
Below, we prove this using mathematical induction.
(1) When $n=1$, $x_{1}=\frac{1}{2}$, the conclusion holds.
(2) Assume that when $n=k$ the conclusion holds, i.e., $x_{k} \leqslant \frac{k}{2}$. Then when $n=k+1$, we have
$x_{k+1}=x_{k}+\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,866 |
4. As shown in Figure 9, line segments $A B$, $B C$, and $C D$ are tangent to the circle at points $E$, $F$, and $G$, respectively, and $A B=B C=C D$. Connect $A C$ and $B D$ to intersect at point $P$, and connect $P F$. Prove: $P F \perp B C$. | Let the center of the circle be $O$. First, prove that $B O \perp A C, C O \perp$ $B D$, then $O$ is the orthocenter of $\triangle P B C$, hence, $P O \perp$ $B C$. Also, it can be proven that $O F \perp B C$, so $P, O, F$ are collinear. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,867 |
Example 1 The terms of the sequence $\left\{a_{n}\right\}$ are all positive, and for any $n \in \mathbf{N}_{+}$, it satisfies $a_{n+1}=a_{n}+c a_{n}^{2}($ constant $c>0)$. Prove:
(1) For any positive number $M$, there exists $N \in \mathbf{N}_{+}$, such that when $n>N$, $a_{n}>M$;
(2) Let $b_{n}=\frac{1}{1+c a_{n}}, S_... | Prove (1) Since $a_{n+1}=a_{n}+c a_{n}^{2}>a_{n}$, therefore, $\left\{a_{n}\right\}$ is a monotonically increasing sequence.
Assume the conclusion is not true, i.e., there exists an $M>0$, for any $n \in \mathbf{N}_{+}$, we have $a_{n} \leqslant M$.
Thus, $\left\{a_{n}\right\}$ is monotonically increasing and bounded... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,868 |
Example 3 (1) Does there exist an infinite sequence of positive integers $\left\{a_{n}\right\}$, such that for any positive integer $n$ we have
$$
a_{n+1}^{2} \geqslant 2 a_{n} a_{n+2} \text { ? }
$$
(2) Does there exist an infinite sequence of positive irrational numbers $\left\{a_{n}\right\}$, such that for any posit... | (1) Assume there exists a sequence of positive integers $\left\{a_{n}\right\}$ satisfying the condition.
$$
\begin{array}{l}
\text { By } a_{n+1}^{2} \geqslant 2 a_{n} a_{n+2}\left(a_{n}>0\right) \\
\Rightarrow \frac{a_{n}}{a_{n-1}} \leqslant \frac{1}{2} \cdot \frac{a_{n-1}}{a_{n-2}} \leqslant \frac{1}{2^{2}} \cdot \fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,870 |
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