problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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Question 3 In the convex quadrilateral $A B C D$,
$$
\angle A D B+\angle A C B=\angle C A B+\angle D B A=30^{\circ} \text {, }
$$
and $A D=B C$. Prove: The segments $D B$, $C A$, and $D C$ can form a right triangle. ${ }^{[3]}$ | Proof: Let $AC$ and $BD$ intersect at point $E$.
Let $\angle EAB = x$, $\angle EBA = y$,
$\angle EDA = \alpha$, $\angle ECB = \beta$.
Clearly, $x + y = \alpha + \beta = 30^\circ$,
$$
\angle AEB = \angle BEC = 30^\circ.
$$
Construct an equilateral $\triangle PAB$, and connect $PD$ and $PC$, as shown in Figure 3.
It is ... | DC^2 = CA^2 + DB^2 | Geometry | proof | Yes | Yes | cn_contest | false | 730,027 |
Question 4 In $\square A B C D$, $\angle A<90^{\circ}$, point $T$ is on side $B C$ such that $\triangle A T D$ is an acute triangle. Let the circumcenters of $\triangle A B T$, $\triangle A D T$, and $\triangle C D T$ be $O_{1}$, $O_{2}$, and $O_{3}$, respectively. Prove: The orthocenter of $\triangle O_{1} O_{2} O_{3}... | To facilitate the description, we agree to denote the circumcircle of $\triangle XYZ$ as $\odot XYZ$.
The auxiliary lines are shown in Figure 4. Using the properties of central angles and inscribed angles, we get
$$
\begin{array}{l}
\angle A O_{1} T = \angle A O_{1} B + \angle T O_{1} B \\
= 2 \angle A T B + 2 \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,028 |
Let $n$ be an integer greater than 1, and let positive real numbers $x_{1}$, $x_{2}, \cdots, x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n}=1$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}-x_{i+1}^{3}} \geqslant \frac{n^{3}}{n^{2}-1}\left(x_{n+1}=x_{1}\right) .
$$ | Proof From $\sum_{i=1}^{n} x_{i}^{2} \geqslant \frac{1}{n}\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\frac{1}{n}$, we know
$$
\sum_{i=1}^{n}\left(1-x_{i+1}^{2}\right)=n-\sum_{i=1}^{n} x_{i}^{2} \leqslant n-\frac{1}{n} \text {. }
$$
Using the AM-GM inequality, we get
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}-x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,029 |
Example 1 Fill the $n^{2}$ positive integers $1,2, \cdots, n^{2}$ into an $n \times n$ grid (one number per cell). Prove: there exist two adjacent cells (sharing a common edge) such that the difference between the two numbers is at least $n$. | Proof by contradiction.
Assume the conclusion is not true, i.e., the difference between the numbers in any two adjacent cells is at most $n-1$.
For $k=1,2, \cdots, n^{2}-n$, let $A_{k}$ be the set of cells containing the numbers $1,2, \cdots, k$; $B_{k}$ be the set of cells containing the numbers $k+n, k+n+1, \cdots, ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,030 |
Example 2 From a set of $n$ people, select $q$ different pairs and label them with numbers $1,2, \cdots, q$. Let
$$
m=\min \left\{t \in \mathbf{Z} \left\lvert\, t \geqslant \frac{2 q}{n}\right.\right\} .
$$
Prove: It is possible to find $m$ different pairs such that these pairs can be arranged in a sequence, satisfyin... | 【Analysis】Expressed in terms of graph theory: Let graph $A$ consist of $n$ vertices and $q$ edges, with the $q$ edges labeled 1, $2, \cdots, q$. Prove: There exists a chain of $m\left(m \geqslant \frac{2 q}{n}\right)$ edges, arranged in order of their labels.
Proof Let $L_{A}(v)$ denote the length of the longest direc... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,031 |
Example 3 Given that there are three beetles at the same point on a table, they start crawling, and after a while, they are at the three vertices of a triangle, and the radius of the incircle of this triangle is 2. Prove: At least one of the beetles has crawled a distance greater than 3. | Proof: Suppose at the beginning, three beetles are all at point $O$, and after a while, they are at the three vertices of $\triangle A B C$.
Let $a=BC, b=CA, c=AB$,
$$
x=OA, y=OB, z=OC \text{. }
$$
By the triangle inequality, we have
$$
\begin{array}{l}
x+y \geqslant c, y+z \geqslant a, y+x \geqslant b \\
\Rightarrow ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,032 |
$$
\begin{array}{l}
16\left(\frac{1}{5}-\frac{1}{3} \times \frac{1}{5^{3}}+\frac{1}{5} \times \frac{1}{5^{5}}-\frac{1}{7} \times \frac{1}{5^{7}}+ \\
\frac{1}{9} \times \frac{1}{5^{9}}-\frac{1}{11} \times \frac{1}{5^{11}}\right)-4\left(\frac{1}{239}-\frac{1}{3} \times \frac{1}{239^{3}}\right) \\
=
\end{array}
$$ | -、1.3. 14159265
The above text has been translated into English, but since it consists of numbers and symbols without specific meaning in a language, the format and content are retained as is. | 3.14159265 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,033 |
2. A four-digit number divided by 433 has a quotient of $a$ and a remainder of $r$ $(a 、 r \in \mathbf{N})$. Then the maximum value of $a+r$ is $\qquad$ . | 2.454.
Let the four-digit number be $433 a+r(0 \leqslant r \leqslant 432)$. Since $433 \times 24=10392>9999$, then $a \leqslant 23$.
When $a=23$, $433 \times 23+40=9999$. At this point, $a+r=23+40=63$. When $a=22$, $433 \times 22+432=9958$. At this point, $a+r=22+432=454$. In summary, the maximum value of $a+r$ is 454... | 454 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,034 |
Example 3 Snow White and the seven dwarfs live in the forest. In any consecutive 16 days, some of the dwarfs work in the diamond mine, while the remaining dwarfs pick strawberries in the forest. No dwarf does both jobs on the same day, and on any two days (not necessarily consecutive), at least three dwarfs do these tw... | Prove that constructing a $7 \times 16$ 0-1 matrix as follows: if the $i$-th dwarf works in the mine on the $j$-th day, then the cell in the $i$-th row and $j$-th column is filled with 0; otherwise, it is filled with 1.
The problem is transformed into: the $7 \times 16$ grid satisfies that the first column is all 0, a... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,035 |
3. Let points $A(-0.8,4.132), B(1.2,-1.948)$, $C(2.8,-3.932)$ be on the graph of the quadratic function
$$
y=a x^{2}+b x+c
$$
When the point $D$ on the graph has the x-coordinate $x=1.8$, its y-coordinate $y$ is $\qquad$ . | 3. -2.992 .
From the given, we have
$$
\left\{\begin{array}{l}
0.64 a-0.8 b+c=4.132, \\
1.44 a+1.2 b+c=-1.948, \\
7.84 a+2.8 b+c=-3.992 .
\end{array}\right.
$$
Solving, we get $a=0.5, b=-3.24, c=1.22$.
Thus, $y=0.5 \times 3.24-3.24 \times 1.8+1.22$ $=-2.992$. | -2.992 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,036 |
5. As shown in Figure $1, P$ is a point on the chord $AB$ of $\odot O$, $AP$ $=m, PB=n$, and $m>n$.
When $AB$ moves around $\odot O$ once,
the trajectory of point $P$ is curve
C. If the area between
$\odot O$ and curve $C$ is
$\left(m^{2}-n^{2}\right) \pi$, then the value of $\frac{m}{n}$ is | 5. $\frac{1+\sqrt{5}}{2}$.
As shown in Figure 5, let the midpoint of $A B$ be $M$, connect $O M$, $O P$, and $O B$, and let the radius of $\odot O$ be $R$.
$$
\begin{array}{l}
\text { By } O M \perp A B, M B=\frac{1}{2}(m+n), \\
M P=\frac{1}{2}(m+n)-n=\frac{1}{2}(m-n),
\end{array}
$$
we get $\quad R^{2}=O B^{2}=O M^{... | \frac{1+\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,037 |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
\begin{array}{c}
S=\left[\frac{1}{1}\right]+\left[\frac{2}{1}\right]+\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\left[\frac{3}{2}\right]+ \\
{\left[\frac{4}{2}\right]+\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{3}{3}... | 6.1078.
$$
2+4+\cdots+2 \times 44=1980 \text {. }
$$
For any integer \( k \) satisfying \( 1 \leqslant k \leqslant 44 \), the sum includes \( 2k \) terms with the denominator \( k \): \(\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]\), whose sum is \( k+2 \).
Also, \( 2016-1980=3... | 1078 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,038 |
7. If real numbers $a, b, c$ make the quadratic function $f(x) = a x^{2} + b x + c$ such that when $0 \leqslant x \leqslant 1$, always $|f(x)| \leqslant 1$. Then the maximum value of $|a| + |b| + |c|$ is $\qquad$ | 7. 17.
Take $x=0, \frac{1}{2}, 1$.
From the problem, we have
$$
\begin{array}{l}
|c| \leqslant 1,|a+2 b+4 c| \leqslant 4, \\
|a+b+c| \leqslant 1 .
\end{array}
$$
Let $m=a+2 b+4 c, n=a+b+c$.
Then $a=-m+2 n+2 c, b=m-n-3 c$
$$
\begin{aligned}
\Rightarrow & |a| \leqslant|m|+2|n|+2|c| \leqslant 8, \\
& |b| \leqslant|m|+|n... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,039 |
8. Given that $a$, $b$, $c$, and $d$ are four positive constants, when the real numbers $x$, $y$ satisfy $a x^{2}+b y^{2}=1$, the minimum value of $c x+d y^{2}$ is $\qquad$ . | 8. $-\frac{c}{\sqrt{a}}$.
Let $k=c x+d y^{2}$.
From $a x^{2}+b y^{2}=1$, we get $y^{2}=\frac{1-a x^{2}}{b}$.
Then $k=c x+\frac{d-a d x^{2}}{b}$
$$
=-\frac{a d}{b}\left(x-\frac{b c}{2 a d}\right)^{2}+\frac{d}{b}+\frac{b c^{2}}{4 a d} \text {. }
$$
From $a x^{2}+b y^{2}=1 \Rightarrow a x^{2} \leqslant 1$
$$
\Rightarrow... | -\frac{c}{\sqrt{a}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,040 |
Example 4 Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. | Let the ten points be $A_{1}, A_{2}, \cdots, A_{10}$. Using these ten points as vertices and the line segments connecting them as edges, we obtain a simple graph $G$ of order 10. Let the degree of point $A_{i} (i=1,2, \cdots, 10)$ be $d_{i}$. Then the total number of edges in graph $G$ is $\frac{1}{2} \sum_{i=1}^{10} d... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,041 |
1. Let the sum of the digits of the natural number $x$ be $S(x)$. Then the solution set of the equation $x+S(x)+S(S(x))+S(S(S(x)))=2016$ is $\qquad$ | $-1 .\{1980\}$.
It is easy to see that $x<2016$.
Note that, the sum of the digits of natural numbers less than 2016 is at most 28, for example, $S(1999)=28$, which indicates,
$$
S(x) \leqslant 28 \text {. }
$$
Furthermore, $S(S(x)) \leqslant S(19)=10$.
Finally, $S(S(S(x))) \leqslant 9$.
From the equation we get
$$
\be... | 1980 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,042 |
2. As shown in Figure $1, \odot O$ is tangent to the sides $A B$ and $A D$ of the square $A B C D$ at points $L$ and $K$, respectively, and intersects side $B C$ at points $M$ and $P$. Given that $B M=8$ cm and $M C=17$ cm, the area of $\odot O$ is ______ square centimeters. | 2. $169 \pi$.
As shown in Figure 5, connect the radii $O K, O L, O M$ of $\odot O$, and let the radius be $r$. Draw $M N \perp O L$ at point $N$.
Then $O N=r-8$.
Also, $B C=B M+M C=25$, so
$$
M N=B L=25-r \text {. }
$$
In the right triangle $\triangle O N M$, by the Pythagorean theorem, we have
$$
r^{2}=(25-r)^{2}+(r... | 169 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,043 |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set $A=\left[\frac{7}{8}\right]+\left[\frac{7^{2}}{8}\right]+\cdots+\left[\frac{7^{2016}}{8}\right]$. Then the remainder when $A$ is divided by 50 is $\qquad$ | 3. 42.
Since $\frac{7^{2 k-1}}{8}$ and $\frac{7^{2 k}}{8}$ are not integers, and
$$
\frac{7^{2 k-1}}{8}+\frac{7^{2 k}}{8}=7^{2 k-1},
$$
for any $k \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
{\left[\frac{7^{2 k-1}}{8}\right]+\left[\frac{7^{2 k}}{8}\right]=7^{2 k-1}-1} \\
\equiv 7(-1)^{k-1}-1(\bmod 50) .
\end{arr... | 42 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,044 |
4. As shown in Figure $2, P A$ is tangent to $\odot O$ at point $A, P C$ intersects $\odot O$ at points $B$ and $C, P O$ intersects $\odot O$ at point $D, A E \perp P O$ at point $E$. Connect $B E$ and extend it to intersect $\odot O$ at point $F$, connect $O C, O F, A D, A F$. If $\angle B C O=30^{\circ}, \angle B F O... | 4. $115^{\circ}$.
Connect $O A, C D, C F$.
Since $P A$ is tangent to $\odot O$ at point $A$, we have $O A \perp P A$.
By $A E \perp P O \Rightarrow P A^{2}=P E \cdot P O$.
Also, $P A^{2}=P B \cdot P C$, thus
$P E \cdot P O=P B \cdot P C$
$\Rightarrow C, B, E, O$ are concyclic
$\Rightarrow \angle F E O=\angle B C O=30^... | 115^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,045 |
Six. (15 points) The sequence given by $a_{0}=a_{1}=1, a_{n+2}=a_{n}+a_{n+1}(n=0,1, \cdots)$ is the famous Fibonacci sequence:
$$
1,1,2,3,5,8,13,21,34,55,89, \cdots \text{, }
$$
where each number is called a Fibonacci number. Prove: there exists a Fibonacci number that ends with three 0s. | Six, in the pairs of adjacent Fibonacci numbers $\left\{\left(a_{i}, a_{i+1}\right)\right\}_{i=0}^{106}$, there exist two such pairs
$$
\left(a_{k}, a_{k+1}\right) 、\left(a_{l}, a_{l+1}\right)(k>l) \text {, }
$$
such that $a_{k+1}-a_{l+1}$ and $a_{k}-a_{l}$ are both divisible by 1000.
According to the definition of th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,046 |
1. Let the plane point sets be
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(y-x)\left(y-\frac{1}{x}\right) \geqslant 0\right.\right\}, \\
B=\left\{(x, y) \mid(x-1)^{2}+(y-1)^{2} \leqslant 1\right\} .
\end{array}
$$
Then the area of the plane figure represented by $A \cap B$ is ).
(A) $\frac{4 \pi}{3}$
(B) $\frac{... | -,1.D.
$A \cap B$ represents the shaded area in Figure 5.
Given that the curve $y=\frac{1}{x}$ and the circle $(x-1)^{2}+(y-1)^{2}=1$ are both symmetric about the line $y=x$, the shaded area is half of the circle's area. Therefore, the required area is $\frac{\pi}{2}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,047 |
The objective function $z=x+m y$ has a maximum value less than 2. Then the range of the real number $m$ is ( ).
(A) $(1,1+\sqrt{2})$
(B) $(1+\sqrt{2},+\infty)$
(C) $(1, \sqrt{2})$
(D) $(3,+\infty)$ | 2. A.
It is easy to know that the maximum value is taken at the intersection of $y=m x$ and $x+y=1$, at this point, the intersection is $A\left(\frac{1}{1+m}, \frac{m}{1+m}\right)$.
$$
\begin{aligned}
& \text { Hence } z_{\max }=\frac{1}{1+m}+\frac{m^{2}}{1+m}<2 \\
\Rightarrow & m^{2}-2 m-1<0 \\
\Rightarrow & 1<m<1+\s... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,048 |
3. Let the function $f(x)=M \sin (\omega x+\varphi)(M \neq 0$, $\left.\omega>0,-\frac{\pi}{2}<\varphi<\frac{\pi}{2}\right)$ have a graph that is symmetric about the line $x=\frac{2 \pi}{3}$, and its period is $\pi$. Then ( ).
(A) The graph of $f(x)$ passes through the point $\left(0, \frac{1}{2}\right)$
(B) $f(x)$ is a... | 3. C.
From the problem, we get $\omega=2, \varphi=\frac{\pi}{6}$. Therefore, $f(x)=M \sin \left(2 x+\frac{\pi}{6}\right)$. At this point, $f\left(\frac{5 \pi}{12}\right)=0$ holds. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,049 |
8. Given an arithmetic sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$,
$$
\begin{array}{l}
\left(a_{6}-1\right)^{3}+2013\left(a_{6}-1\right)=1, \\
\left(a_{2008}-1\right)^{3}+2013\left(a_{2008}-1\right)=-1 .
\end{array}
$$
Then which of the following conclusions is correct? ( ).
(A) $S_... | 8. A.
Construct the function $f(x)=x^{3}+2013 x$, which is an odd and increasing function.
From $f\left(a_{6}-1\right)=1, f\left(a_{2008}-1\right)=-1$, we know $\left(a_{6}-1\right)+\left(a_{2008}-1\right)=0$,
and $a_{6}-1>a_{2008}-1$.
Therefore, $a_{6}+a_{2008}=2$, and $a_{6}>a_{2008}$.
Thus, $S_{2013}=2013$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,051 |
Example 6 Rectangle $R$ is divided into 2016 small rectangles, with each small rectangle's sides parallel to the sides of rectangle $R$. The vertices of the small rectangles are called "nodes". For a line segment on the side of a small rectangle, if both endpoints are nodes and its interior does not contain any other n... | Consider a graph $G$ with all nodes as vertices and basic segments as edges. Let the number of vertices in graph $G$ be $v$, and the number of edges be $e$. Treat the external region of rectangle $R$ as one face (region). Then, the total number of faces in graph $G$ is $f=2017$.
By Euler's formula, we have $v+f-e=2$.
T... | 4122 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,053 |
13. In $\triangle A B C$, $\angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively. Let
$$
\begin{array}{l}
f(x)=\boldsymbol{m} \cdot \boldsymbol{n}, \boldsymbol{m}=(2 \cos x, 1), \\
\boldsymbol{n}=(\cos x, \sqrt{3} \sin 2 x), \\
f(A)=2, b=1, S_{\triangle A B C}=\frac{\sqrt{3}}{2} . \\
\text { Then }... | $=13.2$.
It is easy to know, $f(x)=1+2 \sin \left(2 x+\frac{\pi}{6}\right)$.
Combining the conditions, we get $\angle A=\frac{\pi}{3}, c=2, a=\sqrt{3}$. Therefore, $\frac{a}{\sin A}=\frac{b+c}{\sin B+\sin C}=2$. | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,054 |
15. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}+a_{n+1}=n(-1)^{\frac{a(a+1)}{2}} \text {, }
$$
the sum of the first $n$ terms is $S_{n}, m+S_{2015}=-1007, a_{1} m>0$. Then the minimum value of $\frac{1}{a_{1}}+\frac{4}{m}$ is $\qquad$ . | 15.9.
$$
\begin{array}{l}
\text { Given } S_{2015}=a_{1}+\sum_{k=1}^{1007}\left(a_{2 k}+a_{2 k+1}\right) \\
=a_{1}-1008 \\
\Rightarrow m+a_{1}-1008=-1007 \\
\Rightarrow m+a_{1}=1 . \\
\text { Also, } m a_{1}>0 \text {, so } m>0, a_{1}>0 . \\
\text { Therefore, } \frac{1}{a_{1}}+\frac{4}{m}=\left(m+a_{1}\right)\left(\fr... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,055 |
16. Given a function $f(x)$ defined on $(0,+\infty)$ that satisfies: for any $x \in(0,+\infty)$, it always holds that
$$
f(2 x)=2 f(x) \text {; }
$$
when $x \in(1,2]$, $f(x)=2-x$.
The following conclusions are given:
(1) For any $m \in \mathbf{Z}$, $f\left(2^{m}\right)=0$;
(2) The range of the function $f(x)$ is $[0,+... | 16. (1)(2)(4).
From the graph of $f(x)$, we know (3) is incorrect. | (1)(2)(4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,056 |
19. (12 points) As shown in Figure 2, for the pyramid $P-ABCD$:
(1) Find a point $F$ on edge $PD$ such that $AF \parallel$ plane $PEC$;
(2) Find the cosine value of the dihedral angle $D-PE-A$. The base $ABCD$ is a rhombus with side length 2, $\angle ABC = 60^{\circ}$, $E$ is the midpoint of $AB$, $PA \perp$ plane $ABC... | 19. (1) Taking $BD$ as the $x$-axis, $CA$ as the $y$-axis, and the intersection of $AC$ and $BD$ as $O$, and drawing a perpendicular line to the plane $ABCD$ through point $O$ as the $z$-axis, we establish a spatial rectangular coordinate system.
Thus, $A(0,1,0), B(-\sqrt{3}, 0,0), C(0,-1,0)$, $D(\sqrt{3}, 0,0), P(0,1,... | \frac{4 \sqrt{31}}{31} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,057 |
Example 1 The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{0}=b_{0}=1, \\
a_{n+1}=\alpha a_{n}+\beta b_{n}, \\
b_{n+1}=\beta a_{n}+\gamma b_{n},
\end{array}
$$
where $\alpha, \beta, \gamma \in \mathbf{Z}_{+}, \alpha<\gamma$, and $\alpha \gamma=\beta^{2}+1$.
Prove: For any ... | To prove: For $0 \leqslant k \leqslant m+n$, we have
$$
a_{m+n}+b_{m+n}=a_{k} a_{m+n-k}+b_{k} b_{m+n-k} \text {. }
$$
We use induction on $k$.
When $k=0$, the conclusion is obviously true.
When $k=1$, since
$$
\begin{array}{l}
a_{m+n}=\alpha a_{m+n-1}+\beta a_{m+n-1}, \\
b_{m+n}=\beta a_{m+n-1}+\gamma b_{m+n-1},
\end{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,058 |
1. Define the length of intervals $(m, n)$, $[m, n)$, $(m, n]$, and $[m, n]$ to be $n-m$ (where $n, m \in \mathbf{R}$, and $n > m$). Then the sum of the lengths of the intervals of real numbers $x$ that satisfy
$$
\frac{1}{x-20}+\frac{1}{x-17} \geqslant \frac{1}{512}
$$
is $\qquad$ . | $-1.1024$.
Let $a=20, b=17, c=\frac{1}{512}$.
Then $a>b>c>0$.
The original inequality is equivalent to $\frac{2 x-(a+b)}{(x-a)(x-b)} \geqslant c$.
When $x>a$ or $x0, f(a)=b-a<0$.
Let the two real roots of $f(x)=0$ be $x_{1}$ and $x_{2}$ $\left(x_{1}<x_{2}\right)$. Then the interval of $x$ that satisfies $f(x) \leqslant... | 1024 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,059 |
2. The equation
$$
\sqrt[3]{(x+7)(x+8)}-\sqrt[3]{(x+5)(x+10)}=2
$$
has $\qquad$ real solutions $x$ that are not equal. | 2.4.
Let $a=\sqrt[3]{(x+7)(x+8)}$,
$$
b=\sqrt[3]{(x+5)(x+10)} \text {. }
$$
Then $a-b=2, a^{3}-b^{3}=6$.
Eliminating $a$ gives
$$
\begin{array}{l}
3 b^{2}+6 b+1=0 \\
\Rightarrow b_{1}=\frac{-3+\sqrt{6}}{3}, b_{2}=\frac{-3-\sqrt{6}}{3} .
\end{array}
$$
If $b_{1}=\frac{-3+\sqrt{6}}{3}$, then
$$
x^{2}+15 x+50+\left(\fr... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,060 |
3. Given the function
$$
\begin{aligned}
f(x)= & a \tan ^{2017} x+b x^{2017}+ \\
& c \ln \left(x+\sqrt{x^{2}+1}\right)+20,
\end{aligned}
$$
where $a$, $b$, and $c$ are real numbers. If $f\left(\ln \log _{5} 21\right)=17$, then $f\left(\ln \log _{21} 5\right)=$ $\qquad$ | 3. 23 .
Let $g(x)$
$$
=a \tan ^{2017} x+b x^{2017}+c \ln \left(x+\sqrt{x^{2}+1}\right) \text {. }
$$
Then $g(-x)=-g(x)$
$$
\begin{array}{l}
\Rightarrow f(-x)-20=-(f(x)-20) \\
\Rightarrow f(-x)=40-f(x) .
\end{array}
$$
Therefore, $f\left(\ln \log _{21} 5\right)=f\left(-\ln \log _{5} 21\right)$
$$
=40-f\left(\ln \log ... | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,061 |
6. Given the complex number $z$ satisfies
$$
(a-2) z^{2018}+a z^{2017} \mathrm{i}+a z \mathrm{i}+2-a=0 \text {, }
$$
where, $a<1, \mathrm{i}=\sqrt{-1}$. Then $|z|=$ $\qquad$ | 6. 1 .
Notice,
$$
z^{2017}((a-2) z+a \mathrm{i})=a-2-a z \mathrm{i} \text {. }
$$
Thus, $|z|^{2017}|(a-2) z+a \mathrm{i}|=|a-2-a z \mathrm{i}|$.
Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
$$
\begin{array}{l}
\text { Then }|(a-2) z+a \mathrm{i}|^{2}-|a-2-a z \mathrm{i}|^{2} \\
=|(a-2) x+((a-2) y+a) \mathrm{i}|^{2}-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,062 |
7. For any positive integer $n$, define
$$
S(n)=\left[\frac{n}{10^{[\lg n]}}\right]+10\left(n-10^{[\lg n]}\left[\frac{n}{10^{[\lg n]}}\right]\right) \text {. }
$$
Then among the positive integers $1,2, \cdots, 5000$, the number of positive integers $n$ that satisfy $S(S(n))=n$ is $\qquad$ . | 7. 135.
Let $t=10^{[18 n]}$, then
$$
S(n)=\left[\frac{n}{t}\right]+10\left(n-t\left[\frac{n}{t}\right]\right)
$$
Notice that, $n-t\left[\frac{n}{t}\right]$ is the remainder of $n$ modulo $t$, and $\left[\frac{n}{t}\right]$ is the leading digit of $n$.
We will discuss the cases separately.
(1) If $n$ is a one-digit nu... | 135 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,063 |
9. Given the ellipse $\Gamma: \frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, a line passing through the left focus $F(-2,0)$ of the ellipse $\Gamma$ with a slope of $k_{1}\left(k_{1} \notin\{0\right.$, $\infty\})$ intersects the ellipse $\Gamma$ at points $A$ and $B$. Let point $R(1,0)$, and extend $A R$ and $B R$ to intersect th... | 9. 305 .
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, $C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$,
$l_{A R}: x=\frac{x_{1}-1}{y_{1}} y+1$.
Substitute into the equation of the ellipse $\Gamma$, eliminate $x$ to get
$$
\frac{5-x_{1}}{y_{1}^{2}} y^{2}+\frac{x_{1}-1}{y_{1}} y-4=0 \text {. }
$... | 305 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,064 |
13. Given the hyperbola \( C: \frac{x^{2}}{4}-y^{2}=1 \), a line \( l \) passing through the point \( M(1,-1) \) intersects the right branch of the hyperbola \( C \) at points \( A \) and \( B \), and intersects the \( x \)-axis at point \( N \). Let
\[
\begin{array}{l}
\overrightarrow{M A}=\lambda_{1} \overrightarrow{... | 13. It is known that the slope of the line exists, let it be $k$.
Then the line $l: y+1=k(x-1), N\left(\frac{1}{k}+1,0\right)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Substituting the line $l$ into the equation of the hyperbola $C$ yields
$\left(1-4 k^{2}\right) x^{2}+8 k(k+1) x-$
$4(k+1)^{2}-4=... | \left(-\infty,-\frac{74}{35}\right) \cup(2,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,065 |
12. Given positive real numbers $a, b, c$ satisfying $a b + b c + c a = 1$. Prove: $\square$
$$
\frac{3}{\sqrt{a^{2}+1}}+\frac{4}{\sqrt{b^{2}+1}}+\frac{12}{\sqrt{c^{2}+1}}<\frac{39}{2} .
$$ | 12. Clearly,
$\frac{3}{\sqrt{a^{2}+1}} \leqslant 3, \frac{4}{\sqrt{b^{2}+1}} \leqslant 4, \frac{12}{\sqrt{c^{2}+1}} \leqslant 12$.
Then $\frac{3}{\sqrt{a^{2}+1}}+\frac{4}{\sqrt{b^{2}+1}}+\frac{12}{\sqrt{c^{2}+1}}$
$\leqslant 3+4+12=19<\frac{39}{2}$.
[Note] The conclusion of this problem can be strengthened to
$$
\frac{... | \frac{15 \sqrt{2}}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 730,066 |
3. In a cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with edge length 2, $M$ and $N$ are the midpoints of edges $B B_{1}$ and $B_{1} C_{1}$, respectively. If $P$ is a moving point in the plane $D M N$, when the distance from point $P$ to the plane $B C C_{1} B_{1}$ equals the length of $P D$, the eccentricity of the traje... | 3. $\frac{2 \sqrt{34}}{17}$ | \frac{2 \sqrt{34}}{17} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,067 |
4. The distance between the foci of the conic section
$$
(3 x+4 y-13)(7 x-24 y+3)=200
$$
is $\qquad$ . | 4. $2 \sqrt{10}$.
Let $x=X+3, y=Y+1$.
Then $21 X^{2}-44 X Y-96 Y^{2}=200$.
Let $X=\frac{11 \sqrt{5}}{25} a+\frac{2 \sqrt{5}}{25} b$,
$Y=-\frac{2 \sqrt{5}}{25} a+\frac{11 \sqrt{5}}{25} b$.
Thus, $\frac{a^{2}}{8}-\frac{b^{2}}{2}=1$.
Therefore, the distance between the foci is $2 \sqrt{10}$. | 2 \sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,068 |
5. Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=2, b_{0}=2$,
and $\left\{\begin{array}{l}a_{n+1}=a_{n} \sqrt{1+a_{n}^{2}+b_{n}^{2}}-b_{n} \\ b_{n+1}=b_{n} \sqrt{1+a_{n}^{2}+b_{n}^{2}}+a_{n} \text {. }\end{array}\right.$
Then $a_{2017}^{2}+b_{2017}^{2}=$ $\qquad$ | 5. $3^{2^{2018}}-1$.
Notice,
$$
\begin{array}{l}
1+a_{n+1}^{2}+b_{n+1}^{2} \\
= 1+a_{n}^{2}\left(1+a_{n}^{2}+b_{n}^{2}\right)+b_{n}^{2}+ \\
b_{n}^{2}\left(1+a_{n}^{2}+b_{n}^{2}\right)+a_{n}^{2} \\
=\left(1+a_{n}^{2}+b_{n}^{2}\right)^{2} .
\end{array}
$$
Then $1+a_{n}^{2}+b_{n}^{2}=3^{2^{n+1}}$
$$
\Rightarrow a_{2017... | 3^{2^{2018}}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,069 |
6. Let the non-real complex number $z$, satisfy $z^{23}=1$. Then $\sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}}=$ | 6. $\frac{46}{3}$.
Notice that,
$$
\begin{array}{l}
\sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}}=\frac{1}{3}+\sum_{k=1}^{22} \frac{1-z^{k}}{1-z^{3 k}} \\
=\frac{1}{3}+\sum_{k=1}^{22} \frac{1-\left(z^{24}\right)^{k}}{1-z^{3 k}}=\frac{1}{3}+\sum_{k=1}^{22} \sum_{l=0}^{7} z^{3 k l} \\
=\frac{1}{3}+(22-7)=\frac{46}{3} .
\end... | \frac{46}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,070 |
9. (16 points) Let $O$ be the circumcenter of acute $\triangle A B C$, and the areas of $\triangle B O C$, $\triangle C O A$, and $\triangle A O B$ form an arithmetic sequence. Find the minimum value of $\tan A + 3 \tan C$, and determine the values of the three interior angles at this point.
---
Translate the text ab... | Given $\triangle A B C$ with circumradius $R$.
From the problem, we have
$$
\begin{array}{l}
2 S_{\triangle C O A}=S_{\triangle B O C}+S_{\triangle A O B} \\
\Rightarrow R^{2} \sin 2 B=\frac{1}{2}\left(R^{2} \sin 2 A+R^{2} \sin 2 C\right) \\
\Rightarrow 2 \sin 2 B=\sin 2 A+\sin 2 C \\
\Rightarrow 2 \cos B=\cos (A-C) \\... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,072 |
Example 4 Let $a, b, c$ be distinct positive integers. Prove:
$$
\begin{array}{l}
(a+b+c) \mid\left(a^{3} b+b^{3} c+c^{3} a\right) \\
\Leftrightarrow(a+b+c) \mid\left(a b^{3}+b c^{3}+c a^{3}\right)
\end{array}
$$ | $$
\begin{array}{l}
\left(a b^{3}+b c^{3}+c a^{3}\right)-\left(a^{3} b+b^{3} c+c^{3} a\right) \\
=\left(a^{3} c-a^{3} b\right)+\left(b^{3} a-b^{3} c\right)+\left(c^{3} b-c^{3} a\right) \\
=\left(\left(a^{3} c-a^{3} b\right)+\left(a^{2} b^{2}-a^{2} c^{2}\right)+\right. \\
\left.\left(c^{2} a b-b^{2} a c\right)\right)+\l... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,073 |
Three, (50 points) Find the maximum value of the positive integer $r$ such that: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. | Three, first explain $r \leqslant 200$.
Take $k \in\{1,2, \cdots, 10\}$. Let
$$
A_{k}=\{100 k-99,100 k-98, \cdots, 100 k\} \text {. }
$$
Consider the set
$$
\begin{array}{l}
A_{1} \cup A_{5} \cup A_{6} \cup A_{7} \cup A_{9}, A_{1} \cup A_{2} \cup A_{7} \cup A_{8} \cup A_{10}, \\
A_{2} \cup A_{3} \cup A_{6} \cup A_{8} ... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,074 |
Given 533 As shown in Figure $1, \odot O_{1}$ and $\odot O_{2}$ are both inside $\odot O$, $\odot O_{1}$ and $\odot O_{2}$ are separate and both internally tangent to $\odot O$. An external common tangent of $\odot O_{1}$ and $\odot O_{2}$ intersects $\odot O$ at points $A$ and $B$. Line $O_{1} O_{2}$ intersects $A B$ ... | Proof of a lemma first.
Lemma As shown in Figure 2, points $A, B$ are on the circle $\odot O$ with radius $R$, and the radius of $\odot I$ is $r (R > r)$. $\odot I$ is internally tangent to $\odot O$ at point $T$. $AE, BF$ are the tangents from points $A, B$ to $\odot I$, with $E, F$ being the points of tangency. Then ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,075 |
Let points $D$, $E$, and $F$ be on the sides $BC$, $CA$, and $AB$ of $\triangle ABC$, respectively. If the circumcenters and orthocenters of $\triangle ABC$ and $\triangle DEF$ coincide, prove that $\triangle ABC$ is an equilateral triangle. | Proof As shown in Figure 4, let the circumcenter of $\triangle ABC$ be $O$, and the orthocenter be $H$.
By the property of the Euler line, we know that the centroids of $\triangle ABC$ and $\triangle DEF$ coincide, denoted as point $G$.
$$
\text { Let } \frac{B D}{B C}=\lambda_{1}, \frac{C E}{C A}=\lambda_{2}, \frac{A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,076 |
In Rt $\triangle A B C$, $A D$ is the altitude to the hypotenuse, points $E_{1}$ and $E_{2}$ are both on segment $B D$, and $\angle B A E_{1}=\angle D A E_{2}$. Through point $C$, draw lines parallel to $A E_{1}$ and $A E_{2}$, intersecting the extension of $A D$ at points $G_{1}$ and $G_{2}$, respectively. Lines $G_{1... | Proof: Given points $E_{1}$ and $E_{2}$ on segment $BD$, and
$$
\angle B A E_{1}=\angle D A E_{2},
$$
we know that $A E_{1}$ and $A E_{2}$ are the internal angle bisectors of $\angle B A D$.
Then, $\frac{A B^{2}}{A D^{2}}=\frac{B E_{1} \cdot B E_{2}}{D E_{1} \cdot D E_{2}}$.
Given $C G_{1} \parallel A E_{1}$
$\Rightar... | A F_{1}=B F_{2} | Geometry | proof | Yes | Yes | cn_contest | false | 730,077 |
Given 536 Let $P$ be a moving point on the arc $\overparen{A B}$ of the circumcircle of the regular pentagon $A B C D E$. Prove: $\frac{P A+P B}{P C+P D+P E}$ is always a constant. | Let \( P A=a, P B=b, P C=c \),
\[
P D=d, P E=e \text {. }
\]
As shown in Figure 6, draw all the diagonals of the regular pentagon \( A B C D E \).
Assume the side length of the regular pentagon \( A B C D E \) is 1, and the diagonal length is \( x \). Applying Ptolemy's theorem to the cyclic quadrilateral \( A B C D ... | \sqrt{5} - 2 | Geometry | proof | Yes | Yes | cn_contest | false | 730,078 |
Example 2 Find the number of integers in the set $\left\{\left.\frac{2015[a, b]}{a+b} \right\rvert\, a 、 b \in \mathbf{Z}_{+}\right\}$. | Let $d=(a, b)$, and $a=A d, b=B d$, where $A$ and $B$ are coprime positive integers.
Since $[a, b]=A B(a, b)$, then
$(a+b)|2015[a, b] \Leftrightarrow(A+B)| 2015 A B$.
Because $(A, B)=1$, we have
$(A+B, A)=1,(A+B, B)=1$,
$(A+B, A B)=1$.
Thus, $(A+B) \mid 2015$.
For a fixed divisor $k$ of 2015 greater than 1, and $k$ is ... | 1007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,079 |
Given five numbers $u_{0}, u_{1}, \cdots, u_{4}$. Prove: there always exist five real numbers $v_{0}, v_{1}, \cdots, v_{4}$, satisfying
(1) $u_{i}-v_{i} \in \mathbf{N}$;
(2) $\sum_{0 \leqslant i<j \leqslant 4}\left(v_{i}-v_{j}\right)^{2}<4$.
(28th IMO Preliminary Problem) | First, we point out that condition (1) can be replaced by the following condition:
(1') $u_{i}-v_{i} \in \mathbf{Z}$.
In fact, if a set $\left\{v_{i}^{\prime}\right\}$ is found that satisfies (1') and (2), but does not satisfy (1), then by setting
$$
v_{j}=v_{j}^{\prime}-\sum_{k=0}^{4}\left|u_{k}-v_{k}^{\prime}\right|(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,080 |
For each integer $a_{0}>1$, define the sequence $a_{0}, a_{1}, \cdots$ as follows: for any $n \geqslant 0$,
$a_{n+1}=\left\{\begin{array}{ll}\sqrt{a_{n}}, & \sqrt{a_{n}} \text { is an integer; } \\ a_{n}+3, & \text { otherwise. }\end{array}\right.$
Find all $a_{0}$ such that there exists a number $A$ for which $a_{n}=A... | 1. $a_{0}$ is all multiples of 3.
Since $a_{n+1}$ is determined solely by $a_{n}$, the sequence $\left\{a_{n}\right\}$ has infinitely many equal terms if and only if the sequence is eventually periodic, which is equivalent to the sequence being bounded.
Notice that perfect squares modulo 3 are congruent to 0 or 1. If... | a_{0} \text{ is all multiples of 3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,081 |
2. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$, we have
$$
f(f(x) f(y))+f(x+y)=f(x y) .
$$ | 2. Denote the given equation as $P(x, y)$.
From $P(0,0)$, we have
$f\left(f^{2}(0)\right)=0$.
For any real number $x \neq 1$, there exists a real number $y$ such that $x+y=xy$.
In fact, taking $y=\frac{x}{x-1}$ suffices.
From $P\left(x, \frac{x}{x-1}\right)$, we get
$f\left(f(x) f\left(\frac{x}{x-1}\right)\right)=0 \q... | f_{1}(x)=0, f_{2}(x)=x-1, f_{3}(x)=1-x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,082 |
3. A hunter and an invisible rabbit are playing a game on the Euclidean plane. It is known that the rabbit's starting position $A_{0}$ coincides with the hunter's starting position $B_{0}$. After $n-1$ rounds of the game, the rabbit is at point $A_{n-1}$, and the hunter is at point $B_{n-1}$. In the $n$-th round, the f... | 3. The hunter cannot ensure that the distance to the rabbit will be no more than 100 after $10^{9}$ rounds.
Let $d_{n}=A_{n} B_{n}$.
If for some $n \leqslant 10^{9}$, we already have $d_{n} \geqslant 100$, then after this, the rabbit only needs to jump in the opposite direction of the line connecting it and the hunter ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,083 |
Example 3 Prove: For any $l \in \mathbf{Z}_{+}$, there exists $n \in \mathbf{N}$ such that
$$
n^{n}+47 \equiv 0\left(\bmod 2^{l}\right) .
$$
(47th Mongolian Mathematical Olympiad)
[Analysis] First, conduct simple mathematical experiments for $l$.
When $l=1$, taking $n=1$ works.
Obviously, when $n=1$, $1^{1}+47=48$, $l=... | Prove by mathematical induction on $l$.
When $l=1$, $1^{1}+47 \equiv 0\left(\bmod 2^{1}\right)$, i.e., $n=1$ satisfies the condition.
Assume when $l=k$, there exists $n=n_{k}$ satisfying the condition.
Now consider the case when $l=k+1$.
From $n_{k}^{n_{k}}+47 \equiv 0\left(\bmod 2^{k}\right)$, set
$n_{k}^{n_{k}}+47=2^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,084 |
5. Given an integer $N \geqslant 2 . N(N+1)$ football players, all of different heights, stand in a row. The team coach wants to remove $N(N-1)$ players so that the remaining $2 N$ players satisfy the following $N$ conditions:
(1) The two tallest players among them have no other players between them;
(2) The third and ... | 5. Divide all $N(N+1)$ players into $N$ groups by height, the tallest $N+1$ people form the 1st group, the next $N+1$ people form the 2nd group, $\cdots \cdots$ the shortest $N+1$ people form the $N$th group. The coach observes these players from left to right according to the arrangement, until he finds two people in ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,085 |
2. Find the smallest real number $C$, such that for any positive real numbers $a_{1}, a_{2}, \cdots, a_{5}$ (allowing duplicates), we can always choose different indices $i, j, k, l$ such that $\left|\frac{a_{i}}{a_{j}}-\frac{a_{k}}{a_{l}}\right| \leqslant C$. | 2. The smallest real number $C=\frac{1}{2}$.
First, prove: $C \leqslant \frac{1}{2}$.
For positive real numbers $a_{1}, a_{2}, \cdots, a_{5}$, without loss of generality, assume $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5}$.
Consider the five ratios $\frac{a_{1}}{a_{2}}, \frac{a_{3}}{a_{4}}, ... | \frac{1}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,086 |
5. (1) Prove: For each positive integer $n$, there exists a fraction $\frac{a}{b} (a, b$ being integers $)$, such that
$$
0 < b \leqslant \sqrt{n} + 1 \text{, and } \sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1} \text{. }
$$
(2) Prove: There exist infinitely many positive integers $n$, such that there does not exi... | 5. (1) For each positive integer $n$, there exists a unique positive integer $r$, such that
$$
r^{2} \leqslant n<(r+1)^{2} \text {. }
$$
Let $n=r^{2}+s$. Then $0 \leqslant s \leqslant 2 r$.
Depending on the parity of $s$, we consider two cases.
(i) $s$ is even.
$$
\frac{r^{2}+\frac{s}{2}}{r}=r+\frac{s}{2 r} \text {. }... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,087 |
In $\triangle A B C$, prove:
\[
\begin{array}{l}
\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \\
\geqslant \frac{b+c-a}{a}+\frac{c+a-b}{b}+\frac{a+b-c}{c} \geqslant 3 .
\end{array}
\] | 2. Let “ $\sum$ ” denote the cyclic sum.
First, we prove the left inequality holds.
Let $b+c-a=2 x, c+a-b=2 y$,
$$
a+b-c=2 z \text {. }
$$
Thus, $x, y, z \in \mathbf{R}_{+}$, and
$$
a=y+z, b=z+x, c=x+y \text {. }
$$
Then the left inequality
$$
\begin{array}{l}
\Leftrightarrow \sum \frac{y+z}{2 x} \geqslant \sum \fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,088 |
4. Positive real numbers $\alpha, \beta, \gamma, \delta$, for any $n \in \mathbf{N}_{+}$, satisfy
$$
[\alpha n][\beta n]=[\gamma n][\delta n],
$$
and $\{\alpha, \beta\} \neq\{\gamma, \delta\}$. Prove: $\alpha \beta=\gamma \delta$, and $\alpha, \beta, \gamma, \delta$ are positive integers. | 4. Notice that, $\alpha n-1\gamma \geqslant \delta>\beta$.
Thus, there exists a positive integer $N$ satisfying
(1) $(\alpha-\gamma) N>1$;
(2) $(\delta-\beta) N>1$;
(3) $\beta N>1$.
For any $m>N$, the above three conditions hold. At this point,
$[\alpha m]-[\gamma m] \geqslant 1, [\delta m]-[\beta m] \geqslant 1$,
$[\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,089 |
1. Given the function $f(x)=a x^{2}+b x+c(a \neq 0, a$, $b$, $c$ are all constants), the graph of function $f_{1}(x)$ is symmetric to the graph of function $f(x)$ about the $y$-axis, and the graph of function $f_{2}(x)$ is symmetric to the graph of function $f_{1}(x)$ about the line $y=1$. Then the analytical expressio... | $$
-1 \cdot f_{2}(x)=-a x^{2}+b x+2-c \text {. }
$$
Let point $P(x, y)$ be on the graph of the function $f_{2}(x)$. Then the point $P^{\prime}(x, 2-y)$, which is symmetric to $P$ with respect to the line $y=1$, is on the graph of the function $f_{1}(x)$. Furthermore, the point $P^{\prime \prime}(-x, 2-y)$, which is sy... | f_{2}(x)=-a x^{2}+b x+2-c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,090 |
2. The complex number $z$ satisfies $|z|=1, w=3 z^{2}-\frac{2}{z^{2}}$. The ordinary equation of the curve represented by the moving point $W$ in the complex plane is $\qquad$ . | 2. $x^{2}+\frac{y^{2}}{25}=1$.
Since $|z|=1$, we have $\left|z^{2}\right|=1$.
Let $z^{2}=\cos \theta+\mathrm{i} \sin \theta$. Then
$$
\begin{array}{l}
w=3(\cos \theta+\mathrm{i} \sin \theta)-2(\cos \theta-\mathrm{i} \sin \theta) \\
=\cos \theta+5 \mathrm{i} \sin \theta \\
\Rightarrow x=\cos \theta, y=5 \sin \theta .
\... | x^{2}+\frac{y^{2}}{25}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,091 |
3. The solution to the equation $\arctan 2^{x}-\arctan 2^{-x}=\frac{\pi}{6}$ with respect to $x$ is $\qquad$ . | 3. $\log _{2} \sqrt{3}$.
Taking the tangent of both sides of the known equation, we get
$$
\begin{array}{l}
\frac{2^{x}-2^{-x}}{1+2^{x} \times 2^{-x}}=\frac{\sqrt{3}}{3} \Rightarrow 2^{x}-2^{-x}=\frac{2 \sqrt{3}}{3} \\
\Rightarrow\left(2^{x}\right)^{2}-\frac{2 \sqrt{3}}{3} \times 2^{x}-1=0 .
\end{array}
$$
Since $2^{... | \log _{2} \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,092 |
4. Red, blue, green, and white four dice, each die's six faces have numbers $1, 2, 3, 4, 5, 6$. Simultaneously roll these four dice so that the product of the numbers facing up on the four dice equals 36, there are $\qquad$ possible ways. | 4. 48 .
$$
\begin{array}{l}
36=6 \times 6 \times 1 \times 1=6 \times 3 \times 2 \times 1 \\
=4 \times 3 \times 3 \times 1=3 \times 3 \times 2 \times 2 .
\end{array}
$$
For each of the above cases, there are respectively
$$
\begin{array}{l}
\frac{4!}{(2!)(2!)}=6,4!=24, \\
\frac{4!}{2!}=12, \frac{4!}{(2!)(2!)}=6
\end{ar... | 48 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,093 |
5. Given the functions
$$
f(x)=\cos \pi x, g(x)=2^{x} a-\frac{1}{2}(a \neq 0) .
$$
If there exist $x_{1} 、 x_{2} \in[0,1]$, such that $f\left(x_{1}\right)=g\left(x_{2}\right)$ holds, then the range of the real number $a$ is | 5. $\left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{3}{2}\right]$.
Let $F$ and $G$ be the ranges of $f(x)$ and $g(x)$ defined on the interval $[0,1]$.
Then $F=[-1,1]$,
$$
g(x)=\left\{\begin{array}{ll}
{\left[a-\frac{1}{2}, 2 a-\frac{1}{2}\right],} & a>0 ; \\
{\left[2 a-\frac{1}{2}, a-\frac{1}{2}\right],} & a < 0 ;
\en... | \left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{3}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,094 |
6. As shown in Figure 1, there are 16 small triangular rooms. Two people, A and B, are randomly placed in different small triangular rooms. The probability that they are in non-adjacent (no common side) rooms is (expressed as a fraction).
保留源文本的换行和格式，翻译结果如下：
6. As shown in Figure 1, there are 16 small triangular room... | 6. $\frac{17}{20}$.
It is easy to see that the rooms at the vertices are each adjacent to one room, the rooms on the sides of the large triangle (but not at the vertices) are each adjacent to two rooms, and the remaining rooms are each adjacent to three rooms. Therefore, the number of possible adjacent room pairs is
$... | \frac{17}{20} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,095 |
7. In space, four non-collinear vectors $\overrightarrow{O A} 、 \overrightarrow{O B}$ $\overrightarrow{O C} 、 \overrightarrow{O D}$ have pairwise angles of $\alpha$. Then the size of $\alpha$ is | 7. $\arccos \left(-\frac{1}{3}\right)$.
Let's assume $|\overrightarrow{O A}|=|\overrightarrow{O B}|=|\overrightarrow{O C}|=|\overrightarrow{O D}|=1$ and
$$
\overrightarrow{O D}=a \overrightarrow{O A}+b \overrightarrow{O B}+c \overrightarrow{O C}
$$
Taking the dot product of both sides with $\overrightarrow{O D}$, we ... | \arccos \left(-\frac{1}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,096 |
8. Given $a>0, b>0, a^{3}+b^{3}=1$. Then the range of $a+b$ is $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 8. $(1, \sqrt[3]{4}]$.
Let $u=a+b$.
Then $u>0$, and $a=u-b(00
\end{array} \Leftrightarrow 1<u \leqslant \sqrt[3]{4}\right. \text {. }
$$
Therefore, the range of values for $a+b$ is $(1, \sqrt[3]{4}]$. | (1, \sqrt[3]{4}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,097 |
Example 6 Given positive integers $p, q$, define the sequence $\left\{a_{n}\right\}$:
$$
\begin{array}{l}
a_{1}=a_{2}=1, \\
a_{n+2}=p a_{n+1}+q a_{n}(n=1,2, \cdots) .
\end{array}
$$
Prove: For any positive integers $m, n$, $\left(a_{m}, a_{n}\right) = a_{(m, n)}$ if and only if $p=1$.
$(2016$, Peking University Mathem... | 【Analysis】Necessity.
Simple calculations show that
$$
\begin{array}{l}
a_{3}=p+q, a_{4}=p^{2}+p q+q, \\
\left(a_{3}, a_{4}\right)=a_{(3,4)}=1 .
\end{array}
$$
Thus, $(p, q)=1$.
$$
\begin{array}{l}
\text { Also, } a_{6}=p a_{5}+q a_{4}=\left(p^{2}+q\right) a_{4}+p q a_{3} \text {, then } \\
\left(a_{6}, a_{3}\right)=a_... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,098 |
2. Given a scalene $\triangle ABC$ with orthocenter $H$, circumcenter $O$, and $M, N$ being the midpoints of $AH$ and $BC$ respectively. The circle $\Gamma$ with diameter $AH$ intersects the circumcircle of $\triangle ABC$ at a point $G$ different from $A$, and intersects $AN$ at a point $Q$ different from $A$. A tange... | 2. As shown in Figure 1, let $A D$, $B E$, and $C F$ be the three altitudes of $\triangle A B C$. Then $B$, $C$, $E$, and $F$ are concyclic, with the circle center at $N$, denoted as $\odot N$.
Since circle $\Gamma$ passes through points $E$ and $F$, for circles $\odot O$, $\odot N$, and $\Gamma$, by Monge's theorem, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,099 |
4. If positive integers $n, k$ satisfy
$$
1=\underbrace{\varphi(\varphi(\cdots \varphi(n) \cdots)),}_{k \uparrow}
$$
where $\varphi(n)$ is the Euler's function, i.e., $\varphi(n)$ is the number of elements in $\{1,2, \cdots, n\}$ that are coprime with $n$. In particular, $\varphi(1) = 1$. Prove: $n \leqslant 3^{k}$. | 4. Since each application of $\varphi$ reduces the power of 2 by at most 1, this provides the required estimate.
Define the function $\omega$ for positive integers:
$\omega(2)=1$;
For any odd prime $p, \omega(p)=\omega(p-1)$;
For any positive integers $m, l$,
$\omega(m l)=\omega(m)+\omega(l)$.
By mathematical induction... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,100 |
2. The eccentricity of the conic section $2 x^{2}+3 x y+2 y^{2}=1$ is $\qquad$ . | 2. $\frac{\sqrt{42}}{7}$.
$$
\begin{array}{l}
\text { Given } 2 x^{2}+3 x y+2 y^{2}=1 \\
\Rightarrow \frac{1}{2}\left(x^{2}+y^{2}\right)+\frac{3}{2}(x+y)^{2}=1 \\
\Rightarrow x^{2}+y^{2} \leqslant 2 .
\end{array}
$$
It can be seen that the quadratic curve is an ellipse centered at the origin.
Furthermore, when $x+y=0$... | \frac{\sqrt{42}}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,102 |
3. In the complex number range, the number of common solutions of the equations $z^{4}+z=1$ and $|z|=1$ is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 0 .
From $|z|=1$, we know that $z$ lies on the unit circle centered at $O$.
Furthermore, from $|1-z|=\left|z^{4}\right|=1$, we know that $z$ also lies on the unit circle centered at 1.
It is easy to see that the two circles intersect at two points, which are
$$
\begin{array}{l}
z_{1}=\cos \frac{\pi}{3}+\mathrm{i} \s... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,103 |
4. Fold a triangle with side lengths of $10$, $12$, and $14$ along its three midlines to form a tetrahedron. Then the diameter of the circumscribed sphere of the tetrahedron is $\qquad$ . | 4. $\sqrt{55}$.
The tetrahedron has equal opposite edges, and the lengths of the three pairs of opposite edges are 5, 6, and 7.
As shown in Figure 4, let $A B=C D=5, A C=B D=7$,
$$
A D=B C=6 \text {. }
$$
Take the midpoints $M$ and $N$ of $B C$ and $A D$ respectively.
It is easy to see that $M N$ is the common perpen... | \sqrt{55} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,104 |
5. Let the sequence of positive numbers $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=\sqrt{2}-1, \\
a_{n+1}=\frac{2 n+1}{S_{n}+S_{n+1}+2}(n=1,2, \cdots),
\end{array}
$$
where $S_{n}$ is the sum of the first $n$ terms of $a_{n}$. Then the general term formula of the sequence is $\qquad$ | 5. $a_{n}=\sqrt{n^{2}+1}-\sqrt{(n-1)^{2}+1}$.
By $a_{n+1}=S_{n+1}-S_{n}=\left(S_{n+1}+1\right)-\left(S_{n}+1\right)$
$\Rightarrow\left(S_{n+1}+1\right)-\left(S_{n}+1\right)=\frac{2 n+1}{\left(S_{n+1}+1\right)+\left(S_{n}+1\right)}$
$\Rightarrow\left(S_{n+1}+1\right)^{2}-\left(S_{n}+1\right)^{2}=(n+1)^{2}-n^{2}$.
Let $... | a_{n}=\sqrt{n^{2}+1}-\sqrt{(n-1)^{2}+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,105 |
6. In $\triangle A B C$, $\angle A \leqslant \angle B \leqslant \angle C$, if $\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3}$,
then the value of $\sin B+\sin 2 B$ is $\qquad$ | 6. $\sqrt{3}$.
Let $\alpha=\angle A-\frac{\pi}{3}, \beta=\angle B-\frac{\pi}{3}, \gamma=\angle C-\frac{\pi}{3}$.
Obviously, $\alpha+\beta+\gamma=0$.
From the given information, we have
$$
\begin{array}{l}
(\sin A-\sqrt{3} \cos A)+(\sin B-\sqrt{3} \cos B)+ \\
(\sin C-\sqrt{3} \cos C)=0 \\
\Rightarrow 2 \sin \left(A-\fr... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,106 |
7. Zheng rolls four dice. The probability that the product of the four numbers is divisible by 6 is $\qquad$ | 7. $\frac{325}{432}$.
Let the product of the four numbers be $m$, i.e., find $P(61 m)$.
$$
\begin{array}{l}
\text { Then } P(2 \nmid m)=\left(\frac{1}{2}\right)^{4}=\frac{1}{16} \\
\Rightarrow P(2 \mid m)=\frac{15}{16}, \\
P(3 \nmid m)=\left(\frac{2}{3}\right)^{4}=\frac{16}{81} \Rightarrow P(3 \mid m)=\frac{65}{81}, \... | \frac{325}{432} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,107 |
One, (40 points) Find all positive real solutions of the equation
$$
17 x^{19}-4 x^{17}-17 x^{15}+4=0
$$ | The equation to be solved is
$$
17\left(x^{19}-x^{15}\right)=4\left(x^{17}-1\right) \text {. }
$$
Obviously, $x=1$ is a solution.
When $x-1 \neq 0$, we can divide both sides by the factor $x-1$ to get
$$
\begin{array}{l}
17 x^{15}\left(x^{3}+x^{2}+x+1\right)=4\left(x^{16}+\cdots+x+1\right) \\
\Rightarrow \frac{x^{18}+... | x=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,108 |
Four. (50 points) Divide an equilateral triangle with side length $n$ into $n^{2}$ smaller equilateral triangles with side length 1 using lines parallel to the sides. Figure 3 shows the case for $n=2$. Prove: there exists a positive integer $n$, such that among the vertices of the smaller triangles, 2000n points can be... | Lemma If $x_{m}$ points can be selected inside (excluding the boundary) a regular triangle with side length $m$ (such that no three points form a triangle), then $4 x_{m}$ points can be selected inside a regular triangle with side length $3 m$.
Proof In fact, a regular triangle with side length $3 m$ can be divided in... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,110 |
Given $a, b, x_{i}(i=1,2, \cdots, n) \in \mathbf{R}_{+}$, $b>1$. Try to determine the smallest real number $c$, such that
$$
c \sum_{i=1}^{n} x_{i}^{a b+n} \geqslant\left(\prod_{i=1}^{n} x_{i}\right)\left(\sum_{i=1}^{n} x_{i}^{a}\right)^{b} .
$$ | According to the problem, by homogeneity, we may assume $\prod_{i=1}^{n} x_{i}=1$.
Then the original inequality
$$
\Leftrightarrow c \geqslant \frac{\left(\prod_{i=1}^{n} x_{i}\right)\left(\sum_{i=1}^{n} x_{i}^{a}\right)^{b}}{\sum_{i=1}^{n} x_{i}^{a b+n}}=-c \geqslant \frac{\left(\sum_{i=1}^{n} x_{i}^{a}\right)^{b}}{\s... | n^{b-1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,111 |
Given $x_{i}(i=1,2, \cdots, n)$ and $\lambda$ are all positive real numbers, and satisfy $\prod_{i=1}^{n} x_{i}=1$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+\lambda x_{i}}} \geqslant \frac{n}{\sqrt{1+\lambda}} .
$$ | Prove the construction of the function
$$
f(x)=\frac{x}{\sqrt{1+\lambda x}}-\frac{1}{\sqrt{1+\lambda}}-\frac{2+\lambda}{2 \sqrt{(1+\lambda)^{3}}} \ln x \quad (x>0) \text {. }
$$
Then $f^{\prime}(x)$
$$
=\frac{1}{\sqrt{1+\lambda x}}-\frac{\lambda x}{2 \sqrt{(1+\lambda x)^{3}}}-\frac{2+\lambda}{2 x \sqrt{(1+\lambda)^{3}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,112 |
Given $\triangle A B C$ with the angles $\angle A, \angle B, \angle C$ corresponding to the sides $a, b, c$ respectively, and the altitudes and the radii of the excircles opposite to these sides are $h_{a}, h_{b}, h_{c}$ and $r_{a}, r_{b}, r_{c}$ respectively. Prove:
$$
\sum \frac{h_{a}}{r_{a}} \geqslant 4 \sum \sin ^{... | Proof:
$$
\begin{array}{l}
\sin ^{2} \frac{A}{2}=\frac{1-\cos A}{2}=\frac{1-\frac{b^{2}+c^{2}-a^{2}}{2 b c}}{2} \\
=\frac{(a+b-c)(a-b+c)}{4 b c} \\
\Rightarrow 4 \sum \sin ^{2} \frac{A}{2}=\sum \frac{(a+b-c)(a-b+c)}{b c} .
\end{array}
$$
Let the semi-perimeter of $\triangle ABC$ be $p=\frac{a+b+c}{2}$.
It is easy to s... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,113 |
In $\triangle A B C$, $D$ is the midpoint of side $A B$, point $E$ lies on segment $B C$, and $\angle B D E=2 \angle A$. Point $F$ lies on side $A B$, and $\angle F C A=\angle B+\angle A$. Point $G$ lies on segment $C F$, and $B F=2 E G$. Prove:
$$
\angle D G E+\angle B=180^{\circ}
$$ | Prove as shown in Figure 2, extend $BC$ to point $H$ such that $CH = CF$, connect $FH$, intersecting $AC$ at point $I$, draw $AJ \perp CH$ at point $J$, connect $JG$, $JD$, and $EI$.
Then $\angle ACH = \angle CAB + \angle B = \angle ACF$.
Therefore, $\triangle ACF \cong \triangle ACH$.
Thus, $AC$ is the perpendicular b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,114 |
3. Find all pairs of positive integers $(a, b)$, satisfying $a\left|b^{2}, b\right| a^{2},(a+1) \mid\left(b^{2}+1\right)$.
(2012, Serbian Mathematical Olympiad) | Given $b^{2}=c a$.
Then the original problem is transformed into
$$
\begin{array}{l}
b^{2}=c a \mid a^{4} \text { and }(a+1) \mid(c a+1) \\
\Leftrightarrow c \mid a^{3} \text { and }(a+1) \mid(c-1) .
\end{array}
$$
Let $c=d(a+1)+1\left(d \in \mathbf{Z}_{+}\right)$.
Since $a^{3} \equiv-1 \equiv-c(\bmod a+1)$,
$c \equiv... | (a, b) = \left(t^{2}, t\right),\left((t+1)^{2}\left(t^{2}+2 t\right), (t+1)\left(t^{2}+2 t\right)^{2}\right),\left(t^{2}, t^{3}\right) (t \in \mathbf{N}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,115 |
4. Try to prove: The set $A=\left\{2,2^{2}, 2^{3}, \cdots\right\}$ satisfies the following properties:
(1) For any $a \in A, b \in \mathbf{Z}_{+}$, if $b < 2a-1$, then $b(b+1)$ is not a multiple of $2a$;
(2) For any $a \in \mathbf{Z}_{+} \backslash A$, and $a \neq 1$, there exists $b \in \mathbf{Z}_{+}, b < 2a-1$, suc... | (1) For any $a \in A$, let $a=2^{k}\left(k \in \mathbf{Z}_{+}\right)$.
Assume $b(b+1)$ is a multiple of $2a$. Then $2a \mid b(b+1)$.
Since $b$ and $b+1$ are one odd and one even, and $2a$ is a power of 2, it follows that $2a \mid b$ or $2a \mid (b+1)$, i.e., $2a \leqslant b$ or $2a \leqslant b+1$ with $b1$. Then $2a=2^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,116 |
Example 1 Let $a, b$ be positive integers, and $(4 a b-1) \mid\left(4 a^{2}-1\right)^{2}$.
Prove: $a=b$. | Prove that,
$$
\begin{array}{l}
\left(4 a^{2}-1\right)^{2} b^{2}=(a(4 a b-1)+a-b)^{2} \\
=a^{2}(4 a b-1)^{2}+2 a(a-b)(4 a b-1)+ \\
\quad(a-b)^{2} .
\end{array}
$$
Then $(4 a b-1)$ divides $\left(4 a^{2}-1\right)^{2}$
$$
\Rightarrow(4 a b-1) \mid(a-b)^{2} \text {. }
$$
If $a \neq b$, and $(4 a b-1)$ divides $(a-b)^{2}... | a=b | Number Theory | proof | Yes | Yes | cn_contest | false | 730,117 |
Example 2 Given that $P$ is a point on side $AB$ of the convex quadrilateral $ABCD$, $\Gamma$ is the incircle of $\triangle CPD$, and $I$ is its incenter. The circle $\Gamma$ is tangent to the incircle of $\triangle APD$ and the incircle of $\triangle BPC$ at points $K$ and $L$, respectively. $AC$ and $BD$ intersect at... | Proof: As per the analysis, construct circle $\Gamma_{1}$, with center $J$. We need to prove that points $E, F$ lie on line $IJ$.
(1) Let the incircle of $\triangle A P D$ be $\Gamma_{A}$, and the incircle of $\triangle B P C$ be $\Gamma_{B}$.
On one hand, the internal homothetic center of circles $\Gamma_{A}$ and $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,119 |
Example 3 As shown in Figure 3, $G_{e}$ is the Gergonne point of $\triangle A B C$ (the common point of the lines joining the vertices of $\triangle A B C$ to the points where the incircle touches the opposite sides), and circle $\Gamma_{1}$ is the smaller circle passing through $G_{e}$ and tangent to both $A B$ and $A... | Let point $I$ have projections $I_{1}$, $I_{2}$, and $I_{3}$ on $BC$, $CA$, and $AB$ respectively. Let circle $\Gamma_{1}$ be tangent to $AB$ and $AC$ at points $A_{3}$ and $A_{2}$ respectively, circle $\Gamma_{2}$ be tangent to $AB$ and $BC$ at points $B_{3}$ and $B_{1}$ respectively, and circle $\Gamma_{3}$ be tangen... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,120 |
Example 1 Proof: If each point in the plane is arbitrarily colored one of three colors: red, yellow, or blue, then there must be a color such that for any $d>0$, there exist two points $A$ and $B$ of this color, such that $A B=d$.
---
Translate the above text into English, please retain the original text's line break... | Proof by contradiction.
If the distance between any two red points is not $a$, the distance between any two yellow points is not $b$, and the distance between any two blue points is not $c$, a construction similar to Figure 1 can yield points $A_{1}, A_{2}, \cdots, A_{7}$, such that among any three points, there are t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,121 |
Example 3 On the plane, there are $n(n \geqslant 5)$ distinct points, each point is exactly at a distance of 1 from four other points. Find the minimum value of such $n$. [2] | Keep points $A, B, D, E, G$ in Figure 1, construct $\square E A B C, \square D A G F, \square B A G I$ to get points $C, F, I$, and construct $\square C I F H$ to get point $H$.
$$
\begin{array}{l}
\text { From } B I=A G=A E=B C, \\
\angle I B C=\angle A B C-\angle A B I \\
=\left(180^{\circ}-\angle B A E\right)-\left(... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,122 |
2. Find all positive integers $n$, such that all positive divisors of $n$ can be placed in the cells of a rectangular grid, satisfying the following constraints:
(1) Each cell contains a different divisor;
(2) The sum of the numbers in each row of cells is equal;
(3) The sum of the numbers in each column of cells is eq... | 2. $n=1$.
Assume all positive divisors of $n$ can be placed in a $k \times l$ $(k \leqslant l)$ rectangular grid, and satisfy the conditions.
Let the sum of the numbers in each column of the grid be $s$.
Since $n$ is one of the numbers in a column of the grid, we have $s \geqslant n$, and the equality holds if and onl... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,123 |
6. Given a city with $n(n \geqslant 3)$ islands. Initially, the ferry company provides some routes between the islands, such that it is impossible to divide these $n$ islands into two subsets where there is no route between any two islands in different subsets. Each year, the ferry company will close a route between tw... | 6. Initially, select any two islands $A$ and $B$ that are connected by a route, and place island $A$ in set $\mathscr{A}$, and island $B$ in set $\mathscr{B}$. By the condition, there must be another island that is connected by a route to either island $A$ or $B$. Without loss of generality, assume this island $C$ is c... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,124 |
2. Prove: There exist infinitely many positive integers $n$, such that $2017^{2} \mid \left(1^{n}+2^{n}+\cdots+2016^{n}\right)$. | 2. Let $n=2017 k$ (where $k$ is an odd number).
It is easy to see that there are infinitely many such $n$.
At this point, for any $1 \leqslant a \leqslant 2016$, by the binomial theorem we get
$$
\begin{array}{l}
a^{n}+(2017-a)^{n} \\
\equiv a^{n}+\mathrm{C}_{n}^{1} 2017(-a)^{n-1}+\mathrm{C}_{n}^{0}(-a)^{n} \\
=2017 n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,125 |
4. Given $n \geqslant 3\left(n \in \mathbf{Z}_{+}\right), n$ pairwise coprime positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy: it is possible to appropriately add “+” or “-” to make their algebraic sum 0. Ask: Does there exist a set of positive integers $b_{1}, b_{2}, \cdots, b_{n}$ (allowing repetition), such ... | 4. (1) When $n \geqslant 4$.
(i) If $b_{1}, b_{2}, \cdots, b_{n}$ contain two even numbers, then when
$k$ is even,
$$
b_{1}+a_{1} k, b_{2}+a_{2} k, \cdots, b_{n}+a_{n} k
$$
contains two even terms, which are not coprime.
(ii) If $b_{1}, b_{2}, \cdots, b_{n}$ contain at most one even number, then there are at least thr... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,126 |
2. As shown in Figure 1, in trapezoid $ABCD$, $AD \parallel BC$, diagonals $AC$ and $BD$ intersect at point $E$, point $M$ lies on segment $BC$, and $MA = MD$. Let the circumcircle of $\triangle ABM$ intersect the circumcircle of $\triangle CDM$ at point $N$ (distinct from point $M$). Prove that points $M$, $E$, and $N... | 2. Let $A C$ intersect the circumcircle of $\triangle A B M$ at point $P$, and $B D$ intersect the circumcircle of $\triangle C D M$ at point $Q$. Connect $B P$ and $C Q$.
Since $M A = M D$, we have
$\angle M A D = \angle M D A$.
Thus, $\angle B P E = \angle B M A = \angle M A D$
$= \angle M D A = \angle C M D = \angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,127 |
5. Given $a_{1}, a_{2}, \cdots, a_{10}$ and $b_{1}, b_{2}, \cdots, b_{10}$ are permutations of $1,2, \cdots, 10$, i.e., the sets
$$
\begin{array}{l}
\left\{a_{1}, a_{2}, \cdots, a_{10}\right\} \\
=\left\{b_{1}, b_{2}, \cdots, b_{10}\right\} \\
=\{1,2, \cdots, 10\} .
\end{array}
$$
Let $c_{k}=a_{k}^{2}+b_{k}^{2}(k=1,2,... | 5. Let $a_{k}=k, b_{k}=11-a_{k}=11-k$, i.e., $b_{1}$, $b_{2}, \cdots, b_{10}$ are exactly in reverse order of $a_{1}, a_{2}, \cdots, a_{10}$, at this time, the sequence
$$
c_{k}=a_{k}^{2}+b_{k}^{2}=k^{2}+(11-k)^{2}
$$
has the maximum number $M=c_{1}=c_{10}=1^{2}+10^{2}=101$, reaching its minimum possible value; the mi... | 101, 61 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,128 |
6. There are $n(n \geqslant 2)$ cards, each with a real number written on it, and these $n$ numbers are all distinct. Now, these cards are arbitrarily divided into two piles (each pile has at least one card). It is always possible to take one card from the first pile and place it in the second pile, and then take one c... | 6. The maximum possible value of $n$ is 7.
If given seven cards, each written with $0, \pm 1, \pm 2, \pm 3$, it is easy to verify that they meet the requirements.
Below is the proof that the number of cards cannot be more.
Take any one card as the first pile, and the remaining cards as the second pile. After the oper... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,129 |
1. In a convex quadrilateral $ABCD$, $\angle DAB = \angle BCD = 90^{\circ}$, $\angle ABC > \angle CDA$, $Q$ and $R$ are points on segments $BC$ and $CD$ respectively, line $QR$ intersects $AB$ and $AD$ at points $P$ and $S$ respectively, and $PQ = RS$. Let $M$ and $N$ be the midpoints of segments $BD$ and $QR$ respecti... | 1. As shown in Figure 1.
Since $N$ is also the midpoint of line segment $PS$, in the right triangles $\triangle P A S$ and $\triangle C Q R$, we have
$$
\begin{array}{l}
\angle A N P=2 \angle A S P, \\
\angle C N Q=2 \angle C R Q. \\
\text{Then } \angle A N C=\angle A N P+\angle C N Q \\
=2(\angle A S P+\angle C R Q) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,130 |
2. Let $k$ be a positive integer. Suppose that all positive integers can be colored using $k$ colors, and there exists a function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, satisfying:
(1) For any positive integers $m, n$ of the same color (allowing $m = n$), we have $f(m+n)=f(m)+f(n)$;
(2) There exist positive in... | 2. The minimum value of $k$ is 3.
First, construct an example for $k=3$.
Let $f(n)=\left\{\begin{array}{ll}2 n, & n \equiv 0(\bmod 3) ; \\ n, & n \equiv 1,2(\bmod 3)\end{array}\right.$
Then $f(1)+f(2)=3 \neq f(3)$, satisfying condition (2).
At the same time, color the numbers that are congruent to $0, 1, 2 \pmod{3}$ wi... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,131 |
5. Let $n \geqslant 2$ be a positive integer. An $n$-tuple $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ (allowing repeated numbers) is called an "expensive tuple" if and only if there exists a positive integer $k$ such that
\[
\begin{array}{l}
\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots\left(a_{n-1}+a_{n}\rig... | 5. (1) The required $n$ is all odd numbers greater than 1.
Notice that, for any odd number $n \geqslant 3$, the $n$-tuple $(1,1, \cdots, 1)$ is an expensive tuple.
We now prove: For any even number $n \geqslant 4$, if there exists an $n$-tuple expensive array, then there also exists an $(n-2)$-tuple expensive array.
... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,132 |
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