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Sure, here is the translated text:
```
II. (40 points) Given the set of points
$$
S=\{(x, y) \mid 1 \leqslant x, y \leqslant 2016, x, y \in \mathbf{Z}\} \text {. }
$$
Now, assign a positive integer to each point in the set $S$. Question: Is it possible to provide an assignment scheme such that three points in the set... | II. The following scheme can be given.
Since the set $S$ contains only a finite number of integer points, the lines determined by these integer points are also finite. Let there be $n$ lines, denoted as $l_{1}, l_{2}, \cdots, l_{n}$. Let the prime numbers in ascending order be $p_{1}$, $p_{2}, \cdots, p_{n}$, and assig... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,595 |
Three. (50 points) As shown in Figure 2, the chords $FP$ and $EA$ of circle $\Gamma$ intersect at the midpoint $M$ of chord $BC$. Take a point $D$ on line $AE$ such that $MD = MA$. Line $FD$ intersects circle $\Gamma$ again at point $G$. $EP$ intersects $CB$ at point $Q$. Connect $QG$ and $QD$. Prove: $QD = QG$.
---
... | Three, as shown in Figure 4, let $A F$ intersect $B C$ at point $N$, $G Q$ intersect the circle $\Gamma$ at another point $R$, and $A R$ intersect $F P$ at point $S$.
Since $M$ is the midpoint of $B C$, by the Butterfly Theorem, $M$ is also the midpoint of $N Q$.
Considering the hexagon AEPFGR, by Pascal's Theorem, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,596 |
Four. (50 points) Find all positive integers $a$ such that there exists an integer-coefficient polynomial $P(x)$ and
$$
Q(x)=a_{1} x+a_{2} x^{2}+\cdots+a_{2016} x^{2016},
$$
where, $a_{i} \in\{-1,1\}, i=1,2, \cdots, 2016$, satisfying
$$
\left(x^{2}+a x+1\right) P(x)=Q(x) \text {. }
$$ | First, prove that $Q(x)=0$ has no complex roots with modulus greater than or equal to 2.
Assume $Q(x)=0$ has a complex root with modulus greater than or equal to 2, denoted as $z(|z| \geqslant 2)$.
$$
\begin{array}{l}
\text { Then } a_{1} z+a_{2} z^{2}+\cdots+a_{2016} z^{2016}=0 \\
\Rightarrow a_{1} z+a_{2} z^{2}+\cdo... | 1 \text{ or } 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,597 |
1. The domain of the function $f(x)$ is $D$. If
(1) $f(x)$ is a monotonic function in $D$,
(2) There exists an interval $[a, b] \subseteq D$ such that the range of $f(x)$ on the interval $[a, b]$ is $[a, b]$, then $y=f(x)$ is called a closed function.
If $f(x)=\sqrt{x+2}+k$ is a closed function, then the range of valu... | $-1 .\left(-\frac{9}{4},-2\right]$.
Notice that, $\sqrt{x+2}+k=x$ has two distinct real roots in the interval $[-2,+\infty)$.
Let $\sqrt{x+2}=t \in[0,+\infty)$.
Then $g(t)=t^{2}-t-(k+2)=0$ has two distinct real roots in the interval $[0,+\infty)$.
Thus, $\Delta=1+4(k+2)>0$, and $g(0) \geqslant 0$.
Solving this, we get ... | \left(-\frac{9}{4},-2\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,598 |
2. Roll a die six times, let the number obtained on the $i$-th roll be $a_{i}$. If there exists a positive integer $k$, such that $\sum_{i=1}^{k} a_{i}=6$ has a probability $p=\frac{n}{m}$, where $m$ and $n$ are coprime positive integers. Then
$$
\log _{6} m-\log _{7} n=
$$ | 2. 1.
When $k=1$, the probability is $\frac{1}{6}$;
When $k=2$,
$$
6=1+5=2+4=3+3 \text {, }
$$
the probability is $5\left(\frac{1}{6}\right)^{2}$;
When $k=3$,
$$
6=1+1+4=1+2+3=2+2+2 \text {, }
$$
the probability is $(3+6+1)\left(\frac{1}{6}\right)^{3}=10\left(\frac{1}{6}\right)^{3}$;
When $k=4$,
$$
6=1+1+1+3=1+1+2+2... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,599 |
Example 3 Let real numbers $x, y, z$ satisfy $|xyz|=1$. Prove:
$$
\sum\left(\frac{1}{x^{2}+x+1}+\frac{1}{x^{2}-x+1}\right) \leqslant 4 .
$$ | Prove that,
$$
\frac{1}{x^{2}+x+1}+\frac{1}{x^{2}-x+1}=\frac{2\left(x^{2}+1\right)}{x^{4}+x^{2}+1} \text {. }
$$
Using equation (2) completes the proof. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,600 |
3. Let the vectors be
$$
\begin{array}{l}
\boldsymbol{\alpha}=(x+3, x), \\
\boldsymbol{\beta}=(2 \sin \theta \cdot \cos \theta, a \sin \theta+a \cos \theta)
\end{array}
$$
satisfy that for any $x \in \mathbf{R}$ and $\theta \in\left[0, \frac{\pi}{2}\right]$, we have
$$
|\boldsymbol{\alpha}+\boldsymbol{\beta}| \geqslan... | 3. $a \leqslant 1$ or $a \geqslant 5$.
Let $\sin \theta + \cos \theta = t$. Then
$t \in [1, \sqrt{2}]$,
$2 \sin \theta \cdot \cos \theta = t^2 - 1$,
$$
\boldsymbol{\alpha} + \boldsymbol{\beta} = (t^2 + x + 2, x + a t) \text{.}
$$
Notice that,
$$
\begin{array}{l}
\left(t^2 + x + 2\right)^2 + (x + a t)^2 \\
\geqslant \... | a \leqslant 1 \text{ or } a \geqslant 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,601 |
5. Let the set
$$
S=\{1,2, \cdots, 12\}, A=\left\{a_{1}, a_{2}, a_{3}\right\}
$$
satisfy $a_{1}<a_{2}<a_{3}, a_{3}-a_{2} \leqslant 5, A \subseteq S$. Then the number of sets $A$ that satisfy the conditions is | 5. 185 .
Notice that, the number of all three-element subsets of set $S$ is
$$
\mathrm{C}_{12}^{3}=220 \text {. }
$$
The number of subsets satisfying $1 \leqslant a_{1}<a_{2}<a_{3}-5 \leqslant 7$ is $\mathrm{C}_{7}^{3}=35$, so the number of sets $A$ that meet the conditions of the problem is $220-35=185$. | 185 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,602 |
6. Let $\left\{f_{n}\right\}(n \geqslant 0)$ be the Fibonacci sequence, defined as follows:
$$
\begin{array}{l}
f_{0}=1, f_{1}=1, \\
f_{n+1}=f_{n}+f_{n-1}(n=1,2, \cdots) .
\end{array}
$$
Then the solution set of the equation $n f_{n} f_{n+1}=\left(f_{n+2}-1\right)^{2}$ is | 6. $\varnothing$.
$$
\begin{array}{l}
\text { Clearly, } f_{n+2}=f_{n+1}+f_{n} \leqslant 2 f_{n+1} \text {. } \\
\text { Then } n f_{n} f_{n+1}=\left(f_{n+2}-1\right)^{2}<f_{n+2}^{2} \\
\leqslant 2 f_{n+1} \cdot 2 f_{n+1} \leqslant 8 f_{n} f_{n+1} \\
\Rightarrow n<8 . \\
\text { And } f_{0}=1, f_{1}=1, f_{2}=2, f_{3}=3... | \varnothing | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,603 |
11. (20 points) In the hyperbola \( C: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1 \), \( F_{1} \) and \( F_{2} \) are the left and right foci of the hyperbola \( C \), respectively. \( P \) is a point on the hyperbola and in the first quadrant, \( \triangle P F_{1} F_{2} \) has a centroid \( G \) and an incenter \( I \).
(1) Do... | 11. (1) Suppose there exists a point $P\left(x_{0}, y_{0}\right)\left(y_{0}>0\right)$.
Since $G$ is the centroid of $\triangle P F_{1} F_{2}$, we know $G\left(\frac{x_{0}}{3}, \frac{y_{0}}{3}\right)$.
Let the inradius of $\triangle P F_{1} F_{2}$ be $r$.
Then $S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\r... | y=-2(x-3) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,604 |
Example 4 Let $a, b, c > 0, abc = 1$. Prove:
$$
\sum \sqrt{16a^2 + 9} \geq 3 + 4 \sum a.
$$ | Proof (This solution belongs to Can-Hang)
Notice,
$$
\begin{array}{l}
\sum\left(\sqrt{16 a^{2}+9}-4 a\right) \geqslant 3 \\
\Leftrightarrow \sum \frac{1}{\sqrt{16 a^{2}+9}+4 a} \geqslant \frac{1}{3} .
\end{array}
$$
By the AM-GM inequality,
$$
\begin{array}{l}
2\left(\sqrt{16 a^{2}+9}+4 a\right) \\
\leqslant\left(\fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,605 |
Three. (50 points) Divide $N k\left(N, k \in \mathbf{Z}_{+}, N \geqslant 2\right)$ beads into $2 N-1$ piles. If it is possible to take one bead from each of $N$ piles at a time until all beads are taken, such a division is called "harmonious." Provide a necessary and sufficient condition for a harmonious division and p... | Three, the sufficient and necessary condition for a distribution to be harmonious is that the number of beads in the largest pile does not exceed $k$.
Below, let the number of beads in $2N-1$ piles be $a_{1}$, $a_{2}, \cdots, a_{2 N-1}$, and $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{2 N-1}$.
First, we prove... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,606 |
Four. (50 points) Let $s(x)$ denote the sum of the digits of the positive integer $x$ in decimal notation. Define
$$
f_{k}(n)=\frac{s\left(k n^{2}\right)}{s\left(n^{3}\right)}\left(k, n \in \mathbf{Z}_{+}\right) .
$$
Prove: For any $k \in \mathbf{Z}_{+}$, there exist infinitely many $n \in \mathbf{Z}_{+}, n \geqslant ... | Lemma 1 Let $n_{1}, n_{2} \in \mathbf{Z}_{+}$, then $s\left(n_{1} n_{2}\right) \leqslant s\left(n_{1}\right) s\left(n_{2}\right)$.
Proof of Lemma 1 For any $a, b \in \mathbf{Z}_{+}$, we have
$s(a)+s(b) \geqslant s(a+b)$.
Let $n_{1}=b_{r} 10^{r}+b_{r-1} 10^{-1}+\cdots+b_{1} 10+b_{0}$,
$n_{2}=a_{s} 10^{s}+a_{s-1} 10^{s-1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,607 |
498 Given
$$
A=\left\{a^{2}+4 a b+b^{2}-10 a-8 b+13 \mid a, b \in \mathbf{Z}\right\} \text {. }
$$
Prove: If $x, y \in A$, then $x y \in A$. | Proof of a lemma.
Lemma Given $A=\left\{a^{2}+4 a b+b^{2} \mid a, b \in \mathbf{Z}\right\}$. Prove: if $x, y \in A$, then $x y \in A$.
Proof Without loss of generality, let $a, b, c, d \in \mathbf{Z}$. Then
$$
\begin{array}{l}
\left(a^{2}+4 a b+b^{2}\right)\left(c^{2}+4 c d+d^{2}\right) \\
= {\left.\left[(a+2 b)^{2}+\s... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,608 |
Let $U$ be any finite set of points in the plane, and $U$ contains $k(k \geqslant 3)$ non-collinear points. Find the minimum value of $|U-U|$. | First, we provide the construction:
When $k=2 l+1\left(l \in \mathbf{Z}_{+}\right)$, let
$$
\begin{aligned}
U= & \{(i, 0) \mid 1 \leqslant i \leqslant l, i \in \mathbf{Z}\} \cup \\
& \{(i, 1) \mid 1 \leqslant i \leqslant l+1, i \in \mathbf{Z}\} .
\end{aligned}
$$
When $k=2 l(l \in \mathbf{Z}, l \geqslant 2)$, let
$$
U... | 3k-3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,609 |
Example 5 Let $a, b, c > 0, abc = 1$. Prove:
$$
\sum \frac{1+\sqrt{bc}}{1+a} \geqslant 3 .
$$ | To prove is equivalent to proving
$$
\sum \frac{1+\frac{1}{\sqrt{a}}}{1+a}=\sum \frac{x^{2}+1}{x^{6}+x^{2}}(x y z=1) .
$$
Locally, we have
$$
\frac{x^{2}+1}{x^{6}+x^{2}} \geqslant \frac{3}{x^{6}+x^{3}+1},
$$
which is actually equivalent to
$$
(x-1)^{2}\left(x^{6}+2 x^{5}+x^{4}+x^{3}+x^{2}+2 x+1\right) \geqslant 0 \te... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,610 |
Example 1 Given that the circumcircle of $\triangle ABC$ is $\odot O$, with radius $R$, and points $O_{1}, O_{2}, O_{3}$ are symmetric to point $O$ with respect to $BC, CA, AB$ respectively. Prove:
(1) The circles with centers $O_{1}, O_{2}, O_{3}$ and radius $R$ intersect at one point (denoted as $T$);
(2) Any line th... | Proof (1) Let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $BC$, $CA$, and $AB$ respectively. Then $\triangle O_{1} O_{2} O_{3}$ is similar to $\triangle A_{1} B_{1} C_{1}$, with the similarity center at $O$ and the similarity ratio being 2.
It is easy to see that $O$ is the orthocenter of $\triangle A_{1} B_{1} C_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,611 |
6. Given trapezoid $A B C D(A B / / C D)$, its incircle $\odot O$ touches $A B, B C, C D, D A$ at points $E, F,$ $G, H$ respectively, $F H$ intersects $A B, C D$ at points $P, Q$, the midpoints of $A B, C D$ are $I, J$ respectively. Prove: $P J, Q I$ are both tangents to $\odot O$. | Let the radius of $\odot O$ be $r$,
$$
A E=x, B E=y(x>y) \text {. }
$$
Obviously, $E, O, G$ are collinear.
Then $A D^{2}=(A E-D G)^{2}+G E^{2}$
$$
\Rightarrow D G=\frac{r^{2}}{A E}=\frac{r^{2}}{x} \text {. }
$$
Similarly, $C G=\frac{r^{2}}{y}, D Q=\frac{r^{2}(x+y)}{x(x-y)}$.
As shown in Figure 7, draw a tangent line ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,613 |
Example 1 Try to determine all triples $(p, q, n)$ that simultaneously satisfy
$$
\begin{array}{l}
q^{n+2} \equiv 3^{n+2}\left(\bmod p^{n}\right), \\
p^{n+2} \equiv 3^{n+2}\left(\bmod q^{n}\right)
\end{array}
$$
where $p, q$ are odd primes, and $n$ is an integer greater than 1. ${ }^{[1]}$
(2008, China Mathematical Ol... | 【Analysis】Obviously, $(3,3, n)(n=2,3, \cdots)$ are the arrays that meet the requirements. To prove that there are no other solutions, we can assume that there is another set of solutions that meet the requirements, and ultimately derive a contradiction.
Solution Assume $(p, q, n)$ is another set of solutions that meet... | (3,3, n)(n=2,3, \cdots) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,614 |
Example 2 Proof: In an arithmetic sequence composed of 40 different positive integers, at least one term cannot be expressed in the form $2^{k}+3^{l}$, where $k, l$ are non-negative integers. ${ }^{[2]}$
(2009, China National Training Team Selection Exam) | 【Analysis】It is not easy to directly prove the conclusion. Instead, we can "hide the truth and show the false," by taking its opposite and deriving a contradiction.
Proof Assume there exists a 40-term arithmetic sequence where all terms are distinct and can be expressed in the form $2^{k}+3^{l}$. Let this arithmetic s... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,615 |
The first question: Let real numbers $a_{1}, a_{2}, \cdots, a_{2016}$ satisfy $9 a_{i}>11 a_{i+1}^{2}(i=1,2, \cdots, 2015)$. Find
$$
\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)
$$
the maximum value. | Solution 1 First prove a lemma
Lemma Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n} \in \mathbf{R}$, $n \geqslant 2, a_{i}>b_{i}(1 \leqslant i \leqslant n)$. Then
$\prod_{i=1}^{n}\left(a_{i}-b_{i}\right) \leqslant\left(\sqrt[n]{\prod_{i=1}^{n} a_{i}}-\sqrt[n]{\prod_{i=1}^{n} b_{i}}\right)^{n}$.
Proof (1)... | \frac{1}{4^{2016}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,616 |
Second Question As shown in Figure 1, in $\triangle A B C$, $X$ and $Y$ are two points on the line $B C$ $(X, B, C, Y$ are arranged in order $)$, such that
$$
B X \cdot A C=C Y \cdot A B \text {. }
$$
Let the circumcenters of $\triangle A C X$ and $\triangle A B Y$ be $O_{1}$ and $O_{2}$, respectively. The line $O_{1}... | Proof 1 As shown in Figure 2, construct $\odot O_{1}$ and $\odot O_{2}$, let the intersection points be $A$ and $M$, connect $AM$, intersecting $O_{1}O_{2}$ at point $Q$, extend $AB$ and $AC$, intersecting $\odot O_{1}$ and $\odot O_{2}$ at points $R$ and $S$ respectively.
From $BX \cdot AC = CY \cdot AB \Rightarrow \f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,617 |
Example $\mathbf{3}$ Let the circumcenter of the scalene $\triangle A B C$ be $O$, and the centroid be $G, A_{1} 、 B_{1} 、 C_{1}$ be the midpoints of sides $B C 、 C A 、 A B$, respectively. Draw perpendiculars $l_{B} 、 l_{C}$ from points $B 、 C$ to $O G$, intersecting $A_{1} C_{1} 、 A_{1} B_{1}$ at points $E 、 F$, respe... | Proof As shown in Figure 1, let $A_{0}, B_{0}, C_{0}$ be the projections of the orthocenter $H$ of $\triangle ABC$ onto its three sides. Construct the circumcircle $\odot O_{1}$ of $\triangle B A_{1} C_{1}$, the nine-point circle $\Gamma$ of $\triangle ABC$, and the circumcircle $\odot H_{1}$ of $\triangle B A_{0} C_{0... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,620 |
2. Let positive integers $a, b$ satisfy $(a!)(b!)$ is a multiple of $a!+b!$. Prove: $3a \geqslant 2b+2$.
The text has been translated while preserving the original line breaks and format. | 2. If $a > b$, then $3a \geqslant 2b + 2$ is obviously true.
If $a = b$, then $3a \geqslant 2b + 2 \Leftrightarrow a \geqslant 2$.
When $a = b = 1$, $(a! + b!)$ । $(a!)(b!)$ does not hold. Therefore, $a \geqslant 2$.
If $a = 2c$, then $a + 1 = 2c + 1$.
Thus, $(c + 1) \mid M$.
From equation (1), we know $(1 + M)$ । $(c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,622 |
6. Consider the function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, for any positive integers $m, n$, denote
$$
f^{n}(m)=\underbrace{f(f(\cdots f}_{n \uparrow}(m) \cdots)) .
$$
If the function $f$ satisfies the following two properties:
(1) If $m, n \in \mathbf{Z}_{+}$, then $\frac{f^{n}(m)-m}{n} \in \mathbf{Z}_{... | 6. 【Step 1】First, prove that $f$ is injective.
For any positive integers $m$, $k$ satisfying $f(m)=f(k)$, by property (1), for each positive integer $n$,
$$
\frac{k-m}{n}=\frac{f^{n}(m)-m}{n} - \frac{f^{n}(k)-k}{n}
$$
is the difference of two integers. Therefore, it is also an integer.
For $n=|k-m|+1$, it must be that... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,623 |
2. The unit prices of three types of books are 10 yuan, 15 yuan, and 20 yuan, respectively. A school plans to spend exactly 500 yuan to buy 30 books of the above types. Then the number of different purchasing plans is ( ).
(A) 9
(B) 10
(C) 11
(D) 12 | 2. C.
Let the quantities of the three types of books purchased be $a$, $b$, and $c$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
a+b+c=30, \\
10 a+15 b+20 c=500
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
b=20-2 a, \\
c=10+a .
\end{array}\right.
\end{array}
$$
Thus, $a$ has 11 possible valu... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 729,625 |
3. If a positive integer can be expressed as the difference of cubes of two consecutive odd numbers, then the positive integer is called a "harmonious number" (for example, $2=1^{3}-(-1)^{3}, 26=3^{3}-1^{3}, 2, 26$ are both harmonious numbers). Among the positive integers not exceeding 2016, the sum of all harmonious n... | 3. B.
Notice that,
$$
(2 k+1)^{3}-(2 k-1)^{3}=2\left(12 k^{2}+1\right) \text {. }
$$
From $2\left(12 k^{2}+1\right) \leqslant 2016 \Rightarrow|k|<10$.
Taking $k=0,1, \cdots, 9$, we get all the harmonious numbers not exceeding 2016, and their sum is
$$
\begin{array}{l}
{\left[1^{3}-(-1)^{3}\right]+\left(3^{3}-1^{3}\ri... | 6860 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,626 |
Example 5 Given that $P, Q$ are two points inside an acute $\triangle ABC$, and satisfy
$$
\angle PAB = \angle QAC, \angle PBA = \angle QBC \text{, }
$$
$D$ is a point on side $BC$. Prove:
$$
\begin{array}{l}
\angle DPC + \angle APB = 180^{\circ} \\
\Leftrightarrow \angle DQB + \angle AQC = 180^{\circ} .{ }^{[3]}
\end{... | 【Analysis】In $\triangle A B C$, two points $P$ and $Q$ that satisfy condition (1) are called isogonal conjugate points.
By the trigonometric form of Ceva's theorem, we have
$$
\begin{array}{l}
\frac{\sin \angle P A B}{\sin \angle P A C} \cdot \frac{\sin \angle P C A}{\sin \angle P C B} \cdot \frac{\sin \angle P B C}{\s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,627 |
Three. (25 points) Given positive real numbers $x, y, z$ satisfying
$$
\begin{array}{l}
x y + y z + z x \neq 1, \\
\frac{\left(x^{2}-1\right)\left(y^{2}-1\right)}{x y} + \frac{\left(y^{2}-1\right)\left(z^{2}-1\right)}{y z} + \\
\frac{\left(z^{2}-1\right)\left(x^{2}-1\right)}{z x} = 4 .
\end{array}
$$
(1) Find the value... | (1) From the given equation, we have
$$
\begin{array}{l}
z\left(x^{2}-1\right)\left(y^{2}-1\right)+x\left(y^{2}-1\right)\left(z^{2}-1\right)+ \\
y\left(z^{2}-1\right)\left(x^{2}-1\right)=4 x y z \\
\Rightarrow \quad x y z(x y+y z+z x)-(x+y+z)(x y+y z+z x)+ \\
\quad(x+y+z)-x y z=0 \\
\Rightarrow \quad[x y z-(x+y+z)](x y... | 1 | Inequalities | proof | Yes | Yes | cn_contest | false | 729,628 |
$$
\begin{array}{l}
a+b+c=5, a^{2}+b^{2}+c^{2}=15, \\
a^{3}+b^{3}+c^{3}=47 . \\
\text { Find }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)
\end{array}
$$ | Given $a+b+c=5, a^{2}+b^{2}+c^{2}=15$, we know
$$
\begin{array}{l}
2(a b+b c+c a) \\
=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=10 \\
\Rightarrow a b+b c+c a=5 .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
Then $47-3... | 625 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,629 |
Three. (25 points) As shown in Figure 3, in isosceles $\triangle ABC$, $AB = AC = \sqrt{5}$, $D$ is a point on side $BC$ other than the midpoint, the symmetric point of $C$ with respect to line $AD$ is $E$, and the extension of $EB$ intersects the extension of $AD$ at point $F$. Find the value of $AD \cdot AF$. | Three, connect $A E$, $E D$, $C F$. From the given conditions, we know
$$
\angle A B C=\angle A C B=\angle A E D \text {. }
$$
Then $A$, $E$, $B$, $D$ are concyclic
$$
\Rightarrow \angle B E D=\angle B A D \text {. }
$$
Since points $C$, $E$ are symmetric with respect to line $A D$
$$
\begin{array}{l}
\Rightarrow \an... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,630 |
1. Let real numbers $a, b, c, d$ satisfy $abcd > 0$. Prove: there exists a permutation $x, y, z, w$ of $a, b, c, d$ such that
$$
2(xz + yw)^2 > (x^2 + y^2)(z^2 + w^2).
$$
(Liu Shixiong) | 1. Take $x, z$ as the largest two of $a, b, c, d$, and $y, w$ as the smallest two of $a, b, c, d$.
Below is the proof that such an arrangement satisfies the requirement.
In fact, by
$$
\left(x^{2}+y^{2}\right)\left(z^{2}+w^{2}\right)-(x z+y w)^{2}=(x w-y z)^{2} \text {, }
$$
it suffices to prove
$$
(x z+y w)^{2}>(x w-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,631 |
2. As shown in Figure 1, let $\odot O_{1}$ and $\odot O_{2}$ intersect at points $P$ and $Q$. One of their external common tangents touches $\odot O_{1}$ and $\odot O_{2}$ at points $A$ and $B$, respectively. The circle $\Gamma$ passing through points $A$ and $B$ intersects $\odot O_{1}$ and $\odot O_{2}$ at points $D$... | 2. As shown in Figure 3, connect $A D$, $P Q$, $B C$, $A P$, $A Q$, $B P$, and $B Q$.
By Monge's theorem, we know that $A D$, $Q P$, and $B C$ intersect at one point, denoted as $K$.
By $\triangle K P D \backsim \triangle K A Q \Rightarrow \frac{D P}{A Q}=\frac{K P}{K A}$.
By $\triangle K P A \backsim \triangle K D Q ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,632 |
3. Given positive integers $n, k(k \leqslant n-2)$. Let the set of real numbers $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be such that the absolute value of the sum of any $k$-element subset does not exceed 1. Prove: If $\left|a_{1}\right| \geqslant 1$, then for any $i(2 \leqslant i \leqslant n)$, we have
$$
\left|a... | 3. Suppose $a_{1} \geqslant 1$, to prove the conclusion, it suffices to show that for any $j(2 \leqslant j \leqslant n)$, we have
$$
\begin{array}{l}
a_{j} \geqslant a_{1}-2, \text { and } a_{j} \leqslant 2-a_{1} . \\
\text { Let }[n]=\{1,2, \cdots, n\} .
\end{array}
$$
First, prove $a_{j} \geqslant a_{1}-2$.
Let $2 \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,633 |
5. Prove: There exist infinitely many positive integer triples ( $a$, $b, c)$, such that $a$, $b$, and $c$ are pairwise coprime, and $ab+c$, $bc+a$, and $ca+b$ are pairwise coprime.
(Zhang Duanyang, problem contributor) | 5. Take a positive integer $k$ such that $k-1$ is not a multiple of 5.
The following proves that the positive integer array $(2 k-1, 2 k, 2 k+1)$ satisfies the requirements of the problem.
In fact, it is obvious that $2 k-1$, $2 k$, and $2 k+1$ are pairwise coprime,
$$
\begin{array}{l}
(2 k-1) 2 k+(2 k+1)=4 k^{2}+1, \... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,634 |
6. Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ non-negative real numbers, and let
$$
\begin{aligned}
S_{k}= & \sum_{i=1}^{k} a_{i}(1 \leqslant k \leqslant n) . \text { Prove: } \\
& \sum_{i=1}^{n}\left(a_{i} S_{i} \sum_{j=i}^{n} a_{j}^{2}\right) \leqslant \sum_{i=1}^{n}\left(a_{i} S_{i}\right)^{2} .
\end{aligned}
$$
(Supp... | 6. Notice that,
$$
\sum_{i=1}^{n}\left(a_{i} S_{i} \sum_{j=i}^{n} a_{j}^{2}\right)=\sum_{j=1}^{n}\left(a_{j}^{2} \sum_{i=1}^{j} a_{i} S_{i}\right) .
$$
Therefore, to prove inequality (1), it suffices to prove
$$
\sum_{j=1}^{n}\left(a_{j}^{2} \sum_{i=1}^{j} a_{i} S_{i}\right) \leqslant \sum_{j=1}^{n} a_{j}^{2} S_{j}^{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,635 |
7. As shown in Figure 2, in the cyclic quadrilateral $ABCD$, $\angle BAC = \angle DAC$. Let $\odot I_{1}$ and $\odot I_{2}$ be the incircles of $\triangle ABC$ and $\triangle ADC$, respectively. Prove that one of the external common tangents of $\odot I_{1}$ and $\odot I_{2}$ is parallel to $BD$.
（Yang Mingliang's prob... | 7. As shown in Figure 4, let $I$ be the incenter of $\triangle A B D$, and connect $B I$. Draw a tangent line to $\odot I_{1}$ through $I$, with the point of tangency being $E$, and intersecting $A B$ at point $M$.
By a well-known conclusion and the fact that the sums of the lengths of opposite sides of a circumscribe... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,636 |
8. Given integers $m, n (2 \leqslant m < n)$, if $n > mk$, then there exist $n$ real numbers $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, such that
$$
\frac{1}{m} \sum_{i \in 1} a_{i} > \frac{1}{n} \sum_{i=1}^{n} a_{i} \text{. }
$$
(Supplied by Lin Jin) | 8. The smallest integer that satisfies the condition is
$$
k=\frac{m n+m-n+1}{2} \text {. }
$$
First, prove that when $k=\frac{m n+m-n+1}{2}$, the condition is satisfied.
For the set $\{1,2, \cdots, n\}$, consider an $m$-element subset $I$ such that $\sum_{i \in I} i > k$. Let
$$
I=\left\{i_{1}, i_{2}, \cdots, i_{m}\r... | k=\frac{m n+m-n+1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 729,637 |
4. For a positive integer $k$, let the sum of all its digits in decimal representation be $S(k)$.
(1) Prove: For any positive integer $n$, there exists a positive integer arithmetic sequence $a_{1}, a_{2}, \cdots, a_{n}$, such that
$$
S\left(a_{1}\right)<S\left(a_{2}\right)<\cdots<S\left(a_{n}\right) \text {; }
$$
(2) ... | 4. (1) For $n=10^{k}$, construct the following sequence of positive integers: $a_{i}\left(i=1,2, \cdots, 10^{k}\right)$ is a $(k+1) 10^{k}$-digit number formed by concatenating the $10^{k}$ numbers from $10^{k}+i$ to $2 \times 10^{k}+i-1$ in their decimal representations.
In this sequence, the difference between corre... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,638 |
1. Let the function $f(x)=\left|\log _{2} x\right|$, and real numbers $a, b (a<b)$ satisfy
$$
\begin{array}{l}
f(a+1)=f(b+2), \\
f(10 a+6 b+22)=4 .
\end{array}
$$
Then $a b=$ . $\qquad$ | $\begin{array}{l}\text {-、1. } \frac{2}{15} . \\ \text { Given } f(a+1)=f(b+2) \\ \Rightarrow\left|\log _{2}(a+1)\right|=\left|\log _{2}(b+2)\right| \text {. } \\ \text { Given } a<b \Rightarrow a+1<b+2 \text {. } \\ \text { Then }(a+1)(b+2)=1 \text {, and. } \\ 0<a+1<1<b+2 \text {. } \\ \text { Given } f(10 a+6 b+22)=... | \frac{2}{15} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,639 |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=0,5 a_{n+1}=4 a_{n}+3 \sqrt{1-a_{n}^{2}}(n \in \mathbf{N}) \text {. }
$$
Let $S_{n}=\sum_{i=0}^{n} a_{i}$. Then $S_{51}=$ $\qquad$ | 2. 48 .
Let $a_{n}=\sin \alpha_{n}$.
Suppose $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$.
Then $30^{\circ}0 ; \\
\sin \left(\alpha_{n}-\theta\right), \cos \alpha_{n}<0 .
\end{array}\right.
$
Thus, $a_{n}=\left\{\begin{array}{l}\sin n \theta, n=0,1 ; \\ \sin 2 \theta, n=2 k\left(k \in \mathbf{Z}_{+}\right) ; \\... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,640 |
3. Given the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ with its right focus at $F$, and $P$ as a point on the ellipse, point $A(0,2 \sqrt{3})$. When the perimeter of $\triangle A P F$ is maximized, the area of $\triangle A P F$ is $\qquad$ | 3. $\frac{21 \sqrt{3}}{4}$.
Notice that, the perimeter of $\triangle A P F$ is
$$
\begin{array}{l}
|A F|+|A P|+|P F| \\
=|A F|+2 a-\left|P F^{\prime}\right|+|A P| .
\end{array}
$$
To maximize the perimeter, we need to maximize $|A P|-\left|P F^{\prime}\right|$. As shown in Figure 2, this is achieved when $A, P, F^{\p... | \frac{21 \sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,641 |
Example 7 Define a sequence of points $\left\{A_{i}\right\}(i=1,2, \cdots)$ in a plane based on three non-collinear points $A_{1}, A_{2}, A_{3}$. For $n \geqslant 3$, let $A_{n}$ be the centroid of $\triangle A_{n-3} A_{n-2} A_{n-1}$.
(1) Prove: There is exactly one point $S$ that lies in the interior of each $\triangl... | 【Analysis】The problem naturally leads to the use of limits for proof.
(1) Proof Since $A_{n}$ is the centroid of $\triangle A_{n-3} A_{n-2} A_{n-1}$, we have
$$
S_{\triangle A_{n-2} A_{n-1} A_{n}}=\frac{1}{3} S_{\triangle A_{n-3} A_{n-2} A_{n-1}} .
$$
Thus, $\lim _{n \rightarrow+\infty} S_{\Delta A_{n-A_{n} A_{n} A_{n... | \frac{A_{1} T}{T A_{2}}=2, \frac{T S}{S A_{3}}=1 | Geometry | proof | Yes | Yes | cn_contest | false | 729,642 |
4. Given that $AB$ is a diameter of the smallest circle with center $C(0,1)$ that has common points with the graph of the function $y=\frac{1}{|x|-1}$, and $O$ is the origin. Then $\overrightarrow{O A} \cdot \overrightarrow{O B}$ $=$ | 4. -2 .
For any point $P(x, y)$ on the function $y=\frac{1}{|x|-1}(x>1)$, we have
$$
R^{2}=x^{2}+(y-1)^{2}=x^{2}+\left(\frac{1}{x-1}-1\right)^{2} \text {. }
$$
Let $t=x-1$. Then $t>0$,
$$
\begin{array}{l}
R^{2}=(t+1)^{2}+\left(\frac{1}{t}-1\right)^{2} \\
=t^{2}+\frac{1}{t^{2}}+2\left(t-\frac{1}{t}\right)+2 .
\end{arr... | -2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,643 |
5. A drawer contains red and blue socks, with a total number not exceeding 2016. If two socks are randomly drawn, the probability that they are the same color is $\frac{1}{2}$. Then the maximum number of red socks in the drawer is . $\qquad$ | 5.990.
Let $x$ and $y$ be the number of red and blue socks in the drawer, respectively. Then
$$
\frac{x y}{\mathrm{C}_{x+y}^{2}}=\frac{1}{2} \Rightarrow (x-y)^{2}=x+y.
$$
Thus, the total number of socks is a perfect square.
Let $n=x-y$, i.e., $n^{2}=x+y$. Therefore, $x=\frac{n^{2}+n}{2}$.
Since $x+y \leqslant 2016$, ... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,644 |
7. Given that the center $O$ of a sphere with radius $\frac{\sqrt{22}}{2}$ is the center of a regular tetrahedron $\Gamma$, and the total length of the curves formed by the intersection of the sphere's surface with the four faces of the tetrahedron $\Gamma$ is $8 \pi$. Then the volume of the tetrahedron $\Gamma$ is $\q... | $7.18 \sqrt{2}$ or $54 \sqrt{6}$.
Let the edge length of the tetrahedron $\Gamma$ be $a$, and the radius of the small circle be $r$.
If a face of the tetrahedron $\Gamma$ intersects the sphere as shown in Figure 3, then the circumference of the small circle is $2 \pi$, and the radius is 1.
Therefore, the distance from ... | 18 \sqrt{2} \text{ or } 54 \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,645 |
10. (20 points) Given the parabola $C: y^{2}=4 x$ with focus $F$, and $M$ is a point on the circle $\Gamma:(x-3)^{2}+y^{2}=8$. A circle $\odot F$ is constructed with $F$ as the center and $F M$ as the radius. A line $l$ is tangent to $\odot F$ at point $M$, and intersects the parabola $C$ at points $A$ and $B$. Lines $... | 10. Let point \( M\left(x_{0}, y_{0}\right), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right) \).
According to the problem,
\[
\left(x_{0}-3\right)^{2}+y_{0}^{2}=8 \Rightarrow y_{0}^{2}=-x_{0}^{2}+6 x_{0}-1 \text{. }
\]
(1) When the slope of line \( l \) does not exist,
\[
M(3 \pm 2 \sqrt{2}, 0) \text{. }
\]
Le... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,646 |
II. (40 points) Given that there exist integers $x_{1}, x_{2}, \cdots, x_{n}$ $\left(n \in \mathbf{Z}_{+}\right)$, such that
$$
(4 n+1)\left(\sum_{i=1}^{n} x_{i}^{2}-1\right)=4\left(\sum_{i=1}^{n} x_{i}\right)^{2} \text {. }
$$
Find all possible values of $n$ and the corresponding integer solutions
$$
\left(x_{1}, x_{... | Let $X=\sum_{i=1}^{n} x_{i}$, and assume $X \geqslant 0$. Otherwise, replace with $-x_{1},-x_{2}, \cdots,-x_{n}$.
Since for all integers $x$, we have $x^{2} \geqslant x$. Then
$$
\sum_{i=1}^{n} x_{i}^{2} \geqslant X \text {. }
$$
(1) If $\sum_{i=1}^{n} x_{i}^{2}=X$, then
$$
x_{i} \in\{0,1\}(i=1,2, \cdots, n) \text {. }... | n=2 \text{ or } 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,647 |
Three. (50 points) Let $k$ be a positive integer, and $m$ be an odd number. Prove: there exists a positive integer $n$, such that
$$
2^{k} \mid \left(n^{n}-m\right) \text {. }
$$ | Three, prove by mathematical induction on $k$.
When $k=1$, if we let $n=1$, then from $n^{n}-m=1-m$ being divisible by 2, we know the proposition holds.
Assume when $k=t\left(t \in \mathbf{Z}_{+}\right)$, the proposition holds, i.e., there exists a positive integer $n_{0}$ such that
$$
2^{t} \mid \left(n_{0}^{n_{0}}-m... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,648 |
Given $\left\{a_{n}\right\}$ is a sequence with 1007 terms, and its general term formula is
$$
a_{n}=10^{n}+10^{2016-n}(1 \leqslant n \leqslant 1007) .
$$
In the sequence $\left\{a_{n}\right\}$, find the solutions to the following congruence equations (given that 2017 is a prime number, and 10 is a primitive root modu... | Notice that, if $10^{2016}$ is analogous to $\mathrm{e}^{2 \pi \mathrm{i}}$, then $10^{n} + 10^{2016-n}$ can be analogized to
$$
\mathrm{e}^{\frac{n \pi}{1008} \mathrm{i}} + \mathrm{e}^{-\frac{n \pi}{1008} \mathrm{i}} = 2 \cos \frac{n \pi}{1008};
$$
In the sequence $\left\{a_{n}\right\}$, any two terms cannot be congr... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,649 |
There are $n$ types of tasks $x_{1}, x_{2}, \cdots, x_{n}$ to be completed by $m$ soldiers $a_{1}, a_{2}, \cdots, a_{m}$. The set of tasks that soldier $a_{i}$ can complete is denoted as $A_{i}$, and $A_{1}, A_{2}, \cdots, A_{m}$ satisfy that for any $1 \leqslant k \leqslant m$ and $1 \leqslant i_{1}<i_{2}<\cdots<i_{k}... | Proof First, we prove a lemma.
Lemma (Marriage Theorem) Let $G$ be a bipartite graph with a bipartition $\{X, Y\}$. Then
$G$ has a matching that saturates $X$
$\Leftrightarrow\left|N_{G}(S)\right| \geqslant|S|$ (for any $S \subseteq X$), where $N_{G}(S)$ denotes the set of neighbors of $S$ in $G$.
Proof The necessity i... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,650 |
High $\mathbf{5 0 4}$ Given that $\triangle A B C$ is a right triangle, and the points of tangency of its three excircles with $B C, C A, A B$ are $P, Q, R$ respectively. Prove: the circumcenter of $\triangle P Q R$ lies on the circumcircle of $\triangle A B C$. | Prove that in $\triangle ABC$, let's assume $\angle C=90^{\circ}$.
Let $BC=a$, $CA=b$, $AB=c$.
It is easy to see that $a^{2}+b^{2}=c^{2}$.
As shown in Figure 1, let $O$ be the circumcenter of $\triangle PQR$, and connect $OP$, $OQ$, $OR$, $OA$, $OB$.
Draw $AE$ to bisect $\angle BAC$, and $BF$ to bisect $\angle ABC$. T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,651 |
2. In a non-isosceles $\triangle ABC$, the circumcenter is $O$. The lines $AB$ and $AC$ intersect the circumcircle of $\triangle OBC$ at points $B_{1}$ (different from $B$) and $C_{1}$ (different from $C$), respectively. The lines $BA$ and $BC$ intersect the circumcircle of $\triangle OAC$ at points $A_{2}$ (different ... | Hint: Assume $A B>A C$.
As shown in Figure 6, construct the perpendicular bisector $O D$ of $B C$, intersecting lines $A B$ and $A C$ at points $A_{2}^{\prime}$ and $A_{3}^{\prime}$, respectively. Then $A_{2}^{\prime}$ lies on segment $A B$, and $A_{3}^{\prime}$ lies on the extension of $C A$.
It is easy to prove that... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,652 |
Example 1 Given that the interval $[\alpha, \beta]$ does not contain any integers. Prove: There exists a positive integer $n$, such that the interval $[n \alpha, n \beta]$ does not contain any integers, and the interval length is no less than $\frac{1}{6}$.
【Analysis】Since an interval with a length greater than 1 must... | Let the set
$A=\left\{n \in \mathbf{Z}_{+}\right.$ such that the interval $[n \alpha, n \beta]$ contains no integers $\}$.
Then, by analysis, the set $A$ must have a maximum element, denoted as $n_{0}$.
Next, we prove: The length of the interval $\left[n_{0} \alpha, n_{0} \beta\right]$ is no less than $\frac{1}{6}$.
By... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,653 |
Example: 22016 players participate in a chess tournament. After the tournament, it is found that if player $A$ and player $B$ draw, then all other players either lose to player $A$ or lose to player $B$. Prove: If there are at least two draws, then all players can be arranged in a line such that each player defeats the... | Proof First, no player can draw with two other players (this is because if player $A$ draws with player $B$ and $C$, then by the given condition, player $B$ must beat player $C$, and player $C$ must also beat player $B$, which is a contradiction).
Second, let $A_{1}, A_{2}, \cdots, A_{k}$ be the longest sequence satis... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,654 |
Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $(a, b)$ denote the greatest common divisor of integers $a$ and $b$. Prove:
(1) There are infinitely many positive integers $n$ such that
$$
(n,[\sqrt{2} n])=1 \text{; }
$$
(2) There are infinitely many positive integers $n$ such that
$$
(n,[\... | Prove that taking $n=2^{k}\left(k \in \mathbf{Z}_{+}\right)$.
Then when $\left[2^{k} \sqrt{2}\right]$ is odd, $\left(2^{k},\left[2^{k} \sqrt{2}\right]\right)=1$;
when $\left[2^{k} \sqrt{2}\right]$ is even, $\left(2^{k},\left[2^{k} \sqrt{2}\right]\right) \geqslant 2$.
Therefore, the original proposition is equivalent to... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,655 |
Let $n$ be a positive integer. Given $n$ positive numbers $x_{1}$, $x_{2}, \cdots, x_{n}$ whose product is 1. Prove:
$$
\sum_{i=1}^{n}\left(x_{i} \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{i}^{2}}\right) \geqslant \frac{n+1}{2} \sqrt{n} .
$$
(13th China Southeast Mathematical Olympiad) When $n=2$, it is convenient to check:
$... | Proof when $n=2$, it has been proven.
When $n=3$, it needs to be proven that
$$
\begin{array}{l}
g\left(x_{1}, x_{2}\right) \\
=x_{1} \sqrt{x_{1}^{2}}+x_{2} \sqrt{x_{1}^{2}+x_{2}^{2}}+\frac{1}{x_{1} x_{2}} \sqrt{x_{1}^{2}+x_{2}^{2}+\frac{1}{x_{1}^{2} x_{2}^{2}}} \\
\geqslant 1+\sqrt{2}+\sqrt{3} .
\end{array}
$$
Withou... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,656 |
Proposition 2 As shown in Figure 3, in $\triangle A B C$, $\angle B$ is obtuse, and other conditions are the same as in Proposition 1. Then
$$
O I^{2}>R^{2}+2 R r \text {. }
$$ | Proof: From the proof of Proposition 1, we obtain points $D, E, F$. Let the extension of $AB$ intersect $\odot O$ at point $H$, and connect $HG, HF, CF, CI$; further, let $\odot I$ touch the extensions of $BA, BC$ at points $K, S$ respectively. Connect $IK, IS$. Similar to the proof of the proposition, we get
$$
A I \c... | O I^{2}>R^{2}+2 R r | Geometry | proof | Yes | Yes | cn_contest | false | 729,658 |
2. Equation
$$
|2 x-5|+|3 x-7|+|5 x-11|=\frac{2015}{2016}
$$
has ( ) real roots.
(A) 0
(B) 1
(C) 2
(D) greater than or equal to 3 | 2. A.
Notice,
$$
\begin{array}{l}
|2 x-5|+|3 x-7|+|5 x-11| \\
\geqslant|(2 x-5)+(3 x-7)+(11-5 x)|=1 \\
>\frac{2015}{2016} .
\end{array}
$$
Therefore, the equation has no solution. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,659 |
5. Let $a$ be a real number. To ensure that for any $x \in (-\infty, 1) \cup (5,+\infty)$, we have
$$
x^{2}-2(a-2) x+a>0 \text {. }
$$
Then the set of all $a$ that satisfy the above requirement is ( ).
(A) $(-\infty, 5]$
(B) $(1,4)$
(C) $(1,7]$
(D) $(1,5]$ | 5. D.
Let $f(x)=x^{2}-2(a-2) x+a$.
We discuss the following two cases.
(1) $f(x)$ has no zeros.
In this case, $f(x)$ is always greater than 0, which clearly meets the requirement. By $\Delta=4(a-2)^{2}-4 a<0 \Rightarrow a \in(1,4)$.
(2) $f(x)$ has zeros.
In this case, by the graph of the function, $f(x)$ meets the re... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,660 |
Example 3 Given vectors $a_{1}, a_{2}, \cdots, a_{n}$ whose magnitudes are all no greater than 1. Prove: these $n$ vectors can be divided into two groups (one of the two groups may be empty), such that
$$
|| S|-| T|| \leqslant 1,
$$
where $S$ and $T$ are the sums of the vectors in each of the two groups, respectively.... | Proof by analysis, without loss of generality, assume that when
$$
\begin{array}{l}
\boldsymbol{S}=\boldsymbol{a}_{1}+\boldsymbol{a}_{2}+\cdots+\boldsymbol{a}_{k}, \\
\boldsymbol{T}=\boldsymbol{a}_{k+1}+\boldsymbol{a}_{k+2}+\cdots+\boldsymbol{a}_{n}
\end{array}
$$
$||S|-| T||$ takes the minimum value, and without loss... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,661 |
1. The ellipse $x^{2}+k y^{2}=1$ and the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$ have the same directrix. Then $k=$ $\qquad$ . | $=1 \cdot \frac{16}{7}$.
The equations of the directrices of the hyperbola are
$$
x= \pm \frac{4}{\sqrt{4+5}}= \pm \frac{4}{3} \text {, }
$$
The equations of the directrices of the ellipse are $x= \pm \frac{1}{\sqrt{1-\frac{1}{k}}}$. Then $\frac{1}{\sqrt{1-\frac{1}{k}}}=\frac{4}{3} \Rightarrow 1-\frac{1}{k}=\frac{9}{1... | \frac{16}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,663 |
3. Given that the perimeter of $\triangle A B C$ is 20, the radius of the inscribed circle is $\sqrt{3}$, and $B C=7$. Then the value of $\tan A$ is | 3. $\sqrt{3}$.
Let $A B=x, A C=y$.
From the given information, we have
$$
x+y=13, S_{\triangle A B C}=10 \sqrt{3} \text {. }
$$
Using the triangle area formula and the cosine rule, we get
$$
\frac{1}{2} x y \sin A=10 \sqrt{3} \text {, }
$$
$$
x^{2}+y^{2}-2 x y \cos A=49 \text {. }
$$
From $x+y=13$, we obtain
$$
x^{2... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,664 |
4. Two students, A and B, are selecting courses from five subjects. They share exactly one course, and A selects more courses than B. The number of ways A and B can select courses to meet the above conditions is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, a... | 4. 155 .
For the courses that both A and B choose, there are $\mathrm{C}_{5}^{1}$ ways to select. For the remaining courses, at most one of A or B can choose each course. Let $(a, b)$ represent the scenario where A chooses $a$ courses and B chooses $b$ courses from the remaining four courses.
Given $a > b$, there are
... | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 729,665 |
6. Let the vectors be
$$
a_{1}=(1,5), a_{2}=(4,-1), a_{3}=(2,1) \text {, }
$$
$\lambda_{1} 、 \lambda_{2} 、 \lambda_{3}$ are non-negative real numbers, $\lambda_{1}+\frac{\lambda_{2}}{2}+\frac{\lambda_{3}}{3}=1$. Then the minimum value of $\left|\lambda_{1} a_{1}+\lambda_{2} a_{2}+\lambda_{3} a_{3}\right|$ is $\qquad$ | $6.3 \sqrt{2}$.
$$
\begin{array}{l}
\text { Let } b_{1}=a_{1}=(1,5), b_{2}=2 a_{2}=(8,-2), \\
b_{3}=3 a_{3}=(6,3), p_{1}=\lambda_{1}, p_{2}=\frac{\lambda_{2}}{2}, p_{3}=\frac{\lambda_{3}}{3} . \\
\text { Then } \lambda_{1} a_{1}+\lambda_{2} a_{2}+\lambda_{3} a_{3}=p_{1} b_{1}+p_{2} b_{2}+p_{3} b_{3} .
\end{array}
$$
T... | 3 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,666 |
3. Let $f(x)=\frac{x}{x+1}$, and
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{2}=\frac{3}{4} \\
a_{n+2}=f\left(a_{n}\right)+f\left(a_{n+1}\right)(n=1,2, \cdots) .
\end{array}
$$
Prove: For any positive integer $n$, we have
$$
f\left(3 \times 2^{n-1}\right) \leqslant a_{2 n} \leqslant f\left(3 \times 2^{2 n-2}\right) \tex... | 3. First, prove by mathematical induction: for any positive integer $n$, we have $a_{n}1$) when the proposition holds.
Then, for $n=k+1$, by the induction hypothesis,
$$
\begin{array}{l}
a_{k-1}2 f\left(a_{n}\right) \text {. }
\end{array}
$$
Next, prove by mathematical induction: for any positive integer $n$, we have
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,667 |
Example 4 Let integer $n \geqslant 3, a_{1}, a_{2}, \cdots, a_{n}$ be any $n$ distinct real numbers with a positive sum; $b_{1}, b_{2}, \cdots, b_{n}$ is a permutation of these $n$ numbers. If for $k=1,2, \cdots, n$, we have $\sum_{i=1}^{k} b_{i}>0$, then $b_{1}, b_{2}, \cdots, b_{n}$ is called a "good permutation". Fi... | Let $a_{1}>0, a_{2}, a_{3}, \cdots, a_{n}$ all be less than 0. In this case, it is easy to see that a good permutation must have $a_{1}$ in the first position, and the order of the last $n-1$ positions $a_{2}, a_{3}, \cdots, a_{n}$ is not required. Therefore, the number of good permutations is $(n-1)!$.
To prove: $(n-1... | (n-1)! | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,668 |
1. On the plane, $n$ circles intersect each other pairwise, the maximum number of intersection points is
$\qquad$. | 1. $n(n-1)$.
When two circles intersect, they can have at most 2 intersection points;
When three circles intersect, they can have at most $2+4=6$ intersection points;
When four circles intersect, they can have at most $2+4+6=12$ intersection points;
When $n$ circles intersect, they can have at most
$$
2+4+\cdots+2(n-1)... | n(n-1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,669 |
2. If the function $f(x)$ is an odd function with a period of 3, and when $x \in[0,1)$, $f(x)=3^{x}-1$, then $f\left(\log _{\frac{1}{3}} 54\right)=$ $\qquad$ . | 2. -1 .
Notice that,
$$
\log _{\frac{1}{3}} 54=\log _{\frac{1}{3}} 27+\log _{\frac{1}{3}} 2=-3-\log _{3} 2 \text {, }
$$
and $f(x)$ has a period of 3.
Therefore, $f\left(\log _{\frac{1}{3}} 54\right)=f\left(-\log _{3} 2\right)$.
Furthermore, since $f(x)$ is an odd function and $\log _{3} 2 \in[0,1)$, we have
$$
\begin... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,670 |
3. Let the three sides of a right-angled triangle be $a$, $b$, and $c$ ($c$ being the hypotenuse), with the perimeter being 6. Then the minimum value of $c$ is
保留源文本的换行和格式，直接输出翻译结果。 | 3. $6(\sqrt{2}-1)$.
Let the side $a$ of a right-angled triangle be opposite the angle $\alpha$. Then $0<\alpha<\frac{\pi}{2}$.
Thus, $a=c \sin \alpha, b=c \cos \alpha$.
Also, $a+b+c=6$
$\Rightarrow c(\sin \alpha+\cos \alpha+1)=6$
$\Rightarrow c\left[\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right)+1\right]=6$
$\Rightar... | 6(\sqrt{2}-1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,671 |
5. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
\left(2-a_{n+1}\right)\left(4+a_{n}\right)=8(n \geqslant 1), \text { and } a_{1}=2 . \\
\text { Then } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=
\end{array}
$$ | $\begin{array}{l}\text { 5. } 2^{n}-\frac{n}{2}-1 . \\ \text { From }\left(2-a_{n+1}\right)\left(4+a_{n}\right)=8 \\ \Rightarrow 2 a_{n}-4 a_{n+1}=a_{n} a_{n+1} \\ \Rightarrow \frac{2}{a_{n+1}}-\frac{4}{a_{n}}=1 \\ \Rightarrow \frac{2}{a_{n+1}}+1=2\left(\frac{2}{a_{n}}+1\right) . \\ \text { Then } \frac{2}{a_{n}}+1=\le... | 2^{n}-\frac{n}{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,672 |
6. Let the set $I=\{1,2, \cdots, n\}(n \geqslant 3)$. If two non-empty proper subsets $A$ and $B$ of $I$ satisfy $A \cap B=\varnothing, A \cup$ $B=I$, then $A$ and $B$ are called a partition of $I$. If for any partition $A, B$ of the set $I$, there exist two numbers in $A$ or $B$ such that their sum is a perfect square... | 6. 15 .
When $n=14$, take
$A=\{1,2,4,6,9,11,13\}$,
$B=\{3,5,7,8,10,12,14\}$.
It is easy to see that the sum of any two numbers in $A$ and $B$ is not a perfect square.
Thus, $n=14$ does not meet the requirement.
Therefore, $n<14$ also does not meet the requirement.
Now consider $n=15$.
Use proof by contradiction. Assum... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,673 |
7. In $\triangle A B C$, the sides opposite to $\angle A 、 \angle B 、 \angle C$ are $a 、 b 、 c$ respectively. If $\frac{\cos B}{\cos C}=-\frac{b}{2 a+c}$. Then $\angle B=$ $\qquad$ | 7. $\frac{2 \pi}{3}\left(\right.$ or $\left.120^{\circ}\right)$.
By the cosine rule, we have
$$
\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}, \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b} .
$$
Substituting the above two equations into the condition, we get
$$
\begin{array}{l}
\frac{a^{2}+c^{2}-b^{2}}{2 a c} \cdot \frac{2 a b}{a^{... | \frac{2 \pi}{3} \text{ or } 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,674 |
8. Given $x, y>0$, and $x+2 y=2$. Then the minimum value of $\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}$ is . $\qquad$ | 8. 2 .
$$
\begin{array}{l}
\text { Let } \boldsymbol{a}=\left(\sqrt{\frac{x^{2}}{2 y}}, \sqrt{\frac{4 y^{2}}{x}}\right), \boldsymbol{b}=(\sqrt{2 y}, \sqrt{x}) . \\
\text { Then }|\boldsymbol{a}|=\sqrt{\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}},|\boldsymbol{b}|=\sqrt{2 y+x}=\sqrt{2} \\
\boldsymbol{a} \cdot \boldsymbol{b}=x+2 ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,675 |
1. (16 points) Given $a, b \neq 0$, and $\frac{\sin ^{4} x}{a^{2}}+\frac{\cos ^{4} x}{b^{2}}=\frac{1}{a^{2}+b^{2}}$. Find the value of $\frac{\sin ^{100} x}{a^{100}}+\frac{\cos ^{100} x}{b^{100}}$. | $$
\begin{array}{l}
\text { II. 1. From } \frac{\sin ^{4} x}{a^{2}}+\frac{\cos ^{4} x}{b^{2}}=\frac{1}{a^{2}+b^{2}} \\
\Rightarrow\left(a^{2}+b^{2}\right)\left(\frac{\sin ^{4} x}{a^{2}}+\frac{\cos ^{4} x}{b^{2}}\right)=1 \\
=\left(\sin ^{2} x+\cos ^{2} x\right)^{2} \\
\Rightarrow \sin ^{4} x+\cos ^{4} x+\frac{b^{2}}{a^... | \frac{2}{\left(a^{2}+b^{2}\right)^{50}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,676 |
Example 5 Let $A=\{1,2, \cdots, 2012\}$,
$$
B=\{1,2, \cdots, 19\},
$$
and $S$ is the set of all subsets of $A$. Find the number of mappings $f: S \rightarrow B$ that satisfy the condition
$$
f\left(A_{1} \cap A_{2}\right)=\min \left\{f\left(A_{1}\right), f\left(A_{2}\right)\right\}
$$
for any $A_{1}, A_{2} \in S$. ${... | The number of functions $f$ that satisfy the conditions is
$$
1^{2012}+2^{2012}+\cdots+19^{2012} \text {. }
$$
For any finite sets of integers $U$ and $V$, $f(U \cap V)$ is the minimum of $f(U)$ and $f(V)$. Let $f(A)=n$. Then for all $X \in S$, we have
$$
f(X) \leqslant n \text {. }
$$
If $X \in S$ and $|X|=2011$, th... | 1^{2012}+2^{2012}+\cdots+19^{2012} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,677 |
3. (20 points) Let $A$ be a set of $n$ positive integers, and the sum of all elements in $A$ is less than $2^n-1$. Prove: the set $A$ has at least two non-empty subsets with no common elements, whose sums of elements are the same.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
3. (20 points) Let $A$ be a set of $n$ positive... | 3. Given that set $A$ contains $n$ elements, the number of non-empty subsets of set $A$ is $2^{n}-1$.
Let $B$ be any non-empty subset of set $A$, and let $S_{B}$ be the sum of the elements in set $B$.
Since the sum of all elements in set $A$ is less than $2^{n}-1$, it follows that $1 \leqslant S_{B} \leqslant 2^{n}-2$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,678 |
1. Let $\sin x+\cos x=\frac{1}{2}$. Then $\sin ^{3} x+\cos ^{3} x=$ $\qquad$ | $\begin{array}{l}\text { 1. } \frac{11}{16} \cdot \\ \text { Given } \sin x+\cos x=\frac{1}{2} \\ \Rightarrow 1+2 \sin x \cdot \cos x=\frac{1}{4} \\ \Rightarrow \sin x \cdot \cos x=-\frac{3}{8} \\ \Rightarrow \sin ^{3} x+\cos ^{3} x \\ =(\sin x+\cos x)\left(\sin ^{2} x-\sin x \cdot \cos x+\cos ^{2} x\right) \\ =\frac{1... | \frac{11}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,679 |
2. Simplify $(\mathrm{i}+1)^{2016}+(\mathrm{i}-1)^{2016}=$ | 2. $2^{1009}$.
From $(i+1)^{2}=2 i \Rightarrow(i+1)^{2016}=2^{1008}$.
Similarly, $(i-1)^{2016}=2^{1008}$.
Therefore, $(i+1)^{2016}+(i-1)^{2016}=2^{1009}$. | 2^{1009} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,680 |
3. Given an arithmetic sequence $a_{1}, a_{2}, \cdots, a_{1000}$, the sum of the first 100 terms is 100, and the sum of the last 100 terms is 1000. Then $a_{1}=$ $\qquad$ . | 3. 0. 505 .
Let the common difference of the arithmetic sequence be $d$. Then
$$
\begin{array}{l}
100 a_{1}+4950 d=100, \\
100 a_{1}+94950 d=1000 .
\end{array}
$$
Therefore, $d=0.01, a_{1}=1-0.495=0.505$ | 0.505 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,681 |
4. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the set
$$
\{[x]+[2 x]+[3 x] \mid x \in \mathbf{R}\} \cap\{1,2, \cdots, 100\}
$$
has elements. | 4. 67 .
Let $f(x)=[x]+[2 x]+[3 x]$.
Then $f(x+1)=f(x)+6$.
When $0 \leqslant x<1$, all possible values of $f(x)$ are 0, 1, 2, 3.
Therefore, the range of $f(x)$ is
$$
S=\{6 k, 6 k+1,6 k+2,6 k+3 । k \in \mathbf{Z}\} \text {. }
$$
Thus, $S \cap\{1,2, \cdots, 100\}$ has $4 \times 17-1=67$ elements. | 67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,682 |
7. Given that the parabola $C$ has the center of the ellipse $E$ as its focus, the parabola $C$ passes through the two foci of the ellipse $E$, and intersects the ellipse $E$ at exactly three points. Then the eccentricity of the ellipse $E$ is | 7. $\frac{2 \sqrt{5}}{5}$.
Assume the equation of the ellipse $E$ is
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$
Since the parabola $C$ passes through the two foci of the ellipse $E$, the equation of the parabola $C$ is
$$
x^{2}=2 c y+c^{2}\left(c=\sqrt{a^{2}-b^{2}}\right) \text {. }
$$
Given th... | \frac{2 \sqrt{5}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,683 |
8. A positive integer $x \in$ $\{1,2, \cdots, 2016\}$ is generated with equal probability. Then the probability that the sum of the digits of $x$ in binary is no more than 8 is $\qquad$ . | 8. $\frac{655}{672}$.
Let $x \in\{1,2, \cdots, 2016\}$ be represented in binary as $\left(x_{10} x_{9} \cdots x_{1} x_{0}\right)_{2}$,
i.e., $x=\sum_{i=0}^{10} x_{i} 2^{i}\left(x_{i} \in\{0,1\}\right)$.
In particular, the binary representation of 2016 is
(11111100000) $)_{2}$.
Next, consider the number of $x$ such tha... | \frac{655}{672} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,684 |
Example 6 If the family of sets $\mathscr{F}$ satisfies: for any three sets $X_{1}, X_{2}, X_{3} \in \mathscr{F}$,
$$
\left(X_{1} \backslash X_{2}\right) \cap X_{3},\left(X_{2} \backslash X_{1}\right) \cap X_{3}
$$
at least one of them is an empty set, then the family of sets $\mathscr{F}$ is called "perfect". Prove: ... | Proof: Let $\mathscr{F}$ be a perfect subset family of $U$. We prove by induction on $|U|$ that:
$$
|\mathscr{F}| \leqslant |U| + 1.
$$
If $|U| = 0$, i.e., $U = \varnothing$, then $|\mathscr{F}| \leqslant 1$, and (1) holds.
Assume the conclusion holds for $|U| = k-1$ ($k \geqslant 1$), i.e., (1) holds. Now consider t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,685 |
11. (22 points) Prove: For any real numbers $a, b,$ we have
$$
\begin{array}{l}
\sqrt{a^{2}+a b+b^{2}}+\sqrt{a^{2}+a c+c^{2}} \\
\geqslant \sqrt{3 a^{2}+(a+b+c)^{2}},
\end{array}
$$
and find the necessary and sufficient conditions for equality to hold. | 11. Let $\boldsymbol{\alpha}=\left(a+\frac{b}{2}, \frac{\sqrt{3}}{2} b\right), \boldsymbol{\beta}=\left(a+\frac{c}{2}, \frac{\sqrt{3}}{2} c\right)$.
Then $|\boldsymbol{\alpha}|+|\boldsymbol{\beta}|=\sqrt{a^{2}+a b+b^{2}}+\sqrt{a^{2}+a c+c^{2}}$,
$|\boldsymbol{\alpha}+\boldsymbol{\beta}|=\sqrt{\left(2 a+\frac{b+c}{2}\r... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,686 |
1. Let the sequence $\left\{\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}\right\}$ have the sum of its first $n$ terms as $S_{n}$. Then the number of rational terms in the first 2016 terms of the sequence $\left\{S_{n}\right\}$ is | - 1.43.
$$
\begin{array}{l}
\text { Given } \frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1} \\
\Rightarrow S_{n}=1-\frac{\sqrt{n+1}}{n+1} .
\end{array}
$$
Since $44<\sqrt{2016}<45$, the number of values for which $S$ is rational is 43. | 43 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,687 |
3. For the equation with respect to $x$
$$
\sqrt{2}|x-k|=k \sqrt{x}
$$
to have two distinct real roots in the interval $[k-1, k+1]$, the range of the real number $k$ is | 3. $00$.
$$
\text { Let } f(x)=|x-k|, g(x)=\frac{\sqrt{2}}{2} k \sqrt{x} \text {. }
$$
When $01$, $\left\{\begin{array}{l}f(k-1) \geqslant g(k-1), \\ f(k+1) \geqslant g(k+1) .\end{array}\right.$
But $f(k+1)<g(k+1)$, no solution. | 0<k<2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,689 |
4. Given the curve $C_{1}: y=\sqrt{-x^{2}+10 x-9}$ and point $A(1,0)$. If there exist two distinct points $B$ and $C$ on curve $C_{1}$, whose distances to the line $l: 3 x+1=0$ are $|A B|$ and $|A C|$, respectively, then $|A B|+|A C|=$ | 4.8.
Let points $B\left(x_{B}, y_{B}\right), C\left(x_{C}, y_{C}\right)$, the locus of points whose distance to point $A(1,0)$ is equal to the distance to the line $l: x=-\frac{1}{3}$ is given by the equation $y^{2}=\frac{8}{3} x-\frac{8}{9}$.
Solving the system of equations
$$
\begin{array}{l}
\left\{\begin{array}{l}... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,690 |
5. If the function $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a sum of its maximum and minimum values equal to 4, then $a+b=$ $\qquad$ | 5.3.
Given $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a maximum or minimum value, we know $b=0$.
Then $y=\frac{a+\sin x}{2+\cos x}$
$$
\begin{array}{l}
\Rightarrow \sin x-y \cos x=a-2 y \\
\Rightarrow \sin (x+\alpha)=\frac{a-2 y}{\sqrt{1+y^{2}}} \\
\Rightarrow|a-2 y| \leqslant \sqrt{1+y^{2}} .
\end{array}
$$
By Vi... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,691 |
6. Given planar vectors $\boldsymbol{a} \boldsymbol{b} 、 \boldsymbol{c}$ satisfy
$$
|a|=|b|=a \cdot b=c \cdot(a+2 b-2 c)=2 \text {. }
$$
If the maximum and minimum values of $|\boldsymbol{a}-\boldsymbol{c}|$ are $M$ and $m$, respectively, then $\frac{M}{m}=$ $\qquad$ | 6. $\frac{5+\sqrt{21}}{2}$.
Given $|\boldsymbol{a}|=|\boldsymbol{b}|=\boldsymbol{a} \cdot \boldsymbol{b}=2$, we know $\langle\boldsymbol{a}, \boldsymbol{b}\rangle=60^{\circ}$.
Establish a rectangular coordinate system with the origin at the starting point of $\boldsymbol{a}$ and the direction of $\boldsymbol{a}$ as th... | \frac{5+\sqrt{21}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,692 |
7. Given a square $A B C D$ with side length $a$, vertex $A$ lies in plane $\beta$, and the other vertices are on the same side of plane $\beta$. The distances from points $B$ and $D$ to plane $\beta$ are $1$ and $2$, respectively. If the angle between plane $A B C D$ and plane $\beta$ is $30^{\circ}$, then $a=$ | $7.2 \sqrt{5}$.
As shown in Figure 3, draw perpendiculars from points $B$ and $D$ to plane $\beta$, intersecting plane $\beta$ at points $B_{1}$ and $D_{1}$, respectively.
Then $B B_{1}=1$,
$D D_{1}=2$.
Let $A D=a$. Then
$$
\begin{array}{l}
A B_{1}=\sqrt{a^{2}-1}, A D_{1}=\sqrt{a^{2}-4}, \\
B_{1} D_{1}=\sqrt{2 a^{2}-1... | 2 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,693 |
The distances between points are all different. For each point, the 22 closest points to it are called its "neighbors". Question: What is the maximum number of points for which a given point can be a neighbor? ${ }^{[3]}$
(2015, Turkey National Team Selection Exam) | 【Analysis】This problem is a covering problem in combinatorial geometry. The solver needs to find a method to calculate the number of neighbors from the key conditions.
According to the definition of neighbors in the problem, if $M$ is a neighbor of $N$, then a circle with $N$ as the center and $NM$ as the radius conta... | 44 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,694 |
9. (16 points) In the complex plane, the complex numbers $-1, 1, z_{1}, z_{2}$ $\left(z_{1} z_{2}=1\right)$ correspond to points $A, B, C, D$ respectively. Circle $\Gamma$ passes through points $A, B$, but does not pass through points $C, D$. Prove: Exactly one of $C, D$ is inside circle $\Gamma$.
保留源文本的换行和格式，直接输出翻译结果... | 9. Let the circle $\Gamma: x^{2}+(y-n)^{2}=n^{2}+1$.
Let $z_{1}=a+b$ i. Then
$$
z_{2}=\frac{a}{a^{2}+b^{2}}-\frac{b}{a^{2}+b^{2}} \text { i. }
$$
Let $f(z)$ satisfy
$$
\begin{array}{l}
f(a+b \mathrm{i})=a^{2}+(b-n)^{2}-n^{2}-1 \\
=a^{2}+b^{2}-2 b n-1 .
\end{array}
$$
Then $f\left(z_{1}\right)=a^{2}+b^{2}-2 b n-1$,
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,696 |
10. (20 points) Given positive numbers $a, b, c$ satisfying $a b + b c + c a + 2 a b c = 1$.
Find the maximum value of $\sqrt{a b} + \sqrt{b c} + \sqrt{c a}$. | 10. Let $x=\sqrt{a b}, y=\sqrt{b c}, z=\sqrt{c a}$. Then
$$
x^{2}+y^{2}+z^{2}+2 x y z=1 \text {. }
$$
Let $x+y+z=t$, and combine with equation (1) to eliminate $x$, we get
$$
2 y^{2}(1-z)+2(z-t)(1-z) y+(t-z)^{2}+z^{2}-1=0 \text {. }
$$
Given $0<z<1$, the necessary and sufficient condition for the equation in $y$ to h... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,697 |
Example 8 One evening, 21 people made $n$ phone calls to each other. It is known that among them, there are $m$ ($m$ is an odd number) people $a_{1}, a_{2}$, $\cdots, a_{m}$ such that $a_{i}$ and $a_{i+1}\left(i=1,2, \cdots, m ; a_{m+1}=a_{1}\right)$ made phone calls. If no three people among these 21 people made phone... | Solve: Represent 21 people with 21 points.
If two people communicate by phone, connect the corresponding two points with an edge; otherwise, do not connect an edge, resulting in a graph $G$. It is known that graph $G$ contains an odd cycle, and there are no triangles in graph $G$. The task is to find the maximum number... | 101 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,698 |
3. Mahdi and Motaz each drew a 93-sided polygon inscribed in a circle, denoted as $A: A_{1} A_{2} \cdots A_{93}$ and $B: B_{1} B_{2} \cdots B_{93}$. It is known that $A_{i} A_{i+1} // B_{i} B_{i+1}\left(1 \leqslant i \leqslant 93, A_{94}=A_{1}, B_{94}\right.$ $\left.=B_{1}\right)$. Prove: $\frac{A_{i} A_{i+1}}{B_{B} B_... | 3. Construct a 93-sided polygon similar to 93-sided polygon $B$, such that it is inscribed in the circumcircle of 93-sided polygon $A$. Thus, the ratio of the corresponding sides of the newly constructed 93-sided polygon to 93-sided polygon $B$ is a constant $c$.
Therefore, it is sufficient to prove that two 93-sided ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,699 |
4. In $\triangle A B C$, $\angle C=\angle A+90^{\circ}$, point $D$ lies on the extension of segment $B C$, such that $A C=A D$, point $E$ is on the opposite side of $A$ with respect to side $B C$, and satisfies
$$
\angle E B C=\angle A, \angle E D C=\frac{1}{2} \angle A .
$$
Prove: $\angle C E D=\angle A B C$. | 4. As shown in Figure 3, let $M$ be the midpoint of segment $CD$. Then $AM$ is the perpendicular bisector of $CD$. Let the line $AM$ intersect $DE$ and $BE$ at points $P$ and $Q$, respectively. Then $PC = PD$.
From the given conditions, we have
$$
\begin{array}{l}
\angle EBA + \angle CAB = \angle A + \angle B + \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,700 |
5. Given that $X, Y$ are points on the arc $\overparen{B C}$ (excluding point $A$) of the circumcircle of $\triangle A B C$, satisfying $\angle B A X=\angle C A Y$. Let $M$ be the midpoint of segment $A X$. Prove:
$$
B M+C M>A Y \text {. }
$$ | 5. As shown in Figure 4, let $O$ be the circumcenter of $\triangle A B C$. Then $O M \perp A X$. Draw a perpendicular from point $B$ to $O M$, intersecting the circumcircle of $\triangle A B C$ at another point $Z$.
Since $O M \perp B Z$, $O M$ is the perpendicular bisector of segment $B Z$.
Thus, $M Z=M B$.
By the tr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,701 |
4. In the acute triangle $\triangle ABC$, $AC > AB$, the circumcircle $\odot O$ of $\triangle ABC$ is tangent to the line at point $A$ and intersects the extension of line segment $CB$ at point $P$. $X$ is a point on line segment $OP$ such that $\angle AXP = 90^{\circ}$. Points $E$ and $F$ are on sides $AB$ and $AC$ re... | 4. As shown in Figure 8, let the extensions of $X E$ and $X F$ intersect the circumcircle of $\triangle X K L$ at points $M$ and $N$ respectively.
Obviously, $X F \cdot F M = F L \cdot F K = A F \cdot F C$.
Therefore, $A, M, C, X$ are concyclic.
Thus, $\angle A M X = \angle A C X$.
Similarly, $A, N, B, X$ are concyclic... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,702 |
2. Connecting the intersection points of $x^{2}+y^{2}=10$ and $y=\frac{4}{x}$ in sequence, a convex quadrilateral is formed. The area of this quadrilateral is $\qquad$ | 2. 12 .
Let $A\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$ be the intersection point of the two curves in the first quadrant.
Since the two curves are symmetric with respect to the origin and the line $y=x$, the coordinates of the other three intersection points are
$$
B\left(y_{0}, x_{0}\right), C\left(-x... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,703 |
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