problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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2. Let $A_{n}$ and $B_{n}$ be the sums of the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively. If $\frac{A_{n}}{B_{n}}=\frac{5 n-3}{n+9}$, then $\frac{a_{8}}{b_{8}}=$ $\qquad$ | 2.3.
Let $\left\{a_{n}\right\}, \left\{b_{n}\right\}$ have common differences $d_{1}, d_{2}$, respectively. Then $\frac{A_{n}}{B_{n}}=\frac{a_{1}+\frac{1}{2}(n-1) d_{1}}{b_{1}+\frac{1}{2}(n-1) d_{2}}=\frac{5 n-3}{n+9}$.
Let $d_{2}=d$. Then $d_{1}=5 d$.
Thus, $a_{1}=d, b_{1}=5 d$.
Therefore, $\frac{a_{8}}{b_{8}}=\frac{... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $O$ be the origin, $A$ be a moving point on the parabola $x=\frac{1}{4} y^{2}+1$, and $B$ be a moving point on the parabola $y=x^{2}+4$. Then the minimum value of the area of $\triangle O A B$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output... | 3.2.
Let $A\left(s^{2}+1,2 s\right), B\left(t, t^{2}+4\right)$. Then $l_{O B}:\left(t^{2}+4\right) x-t y=0$.
Let the distance from point $A$ to line $O B$ be $h$, we have
$$
h=\frac{\left|\left(t^{2}+4\right)\left(s^{2}+1\right)-t \cdot 2 s\right|}{\sqrt{\left(t^{2}+4\right)^{2}+t^{2}}} \text {. }
$$
Therefore, $S_{\... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the complex number $z=\cos \frac{4 \pi}{7}+\mathrm{i} \sin \frac{4 \pi}{7}$. Then
$$
\left|\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}\right|
$$
is equal to $\qquad$ (answer with a number). | 5. 2 .
Given that $z$ satisfies the equation $z^{7}-1=0$.
From $z^{7}-1=(z-1) \sum_{i=0}^{6} z^{i}$, and $z \neq 1$, we get $z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=0$. After combining the fractions of $\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}$, the denominator is
$$
\begin{array}{l}
\left(1+z^{2}\right... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a, b, c, d$ be real numbers, satisfying
$$
a+2 b+3 c+4 d=\sqrt{10} \text {. }
$$
Then the minimum value of $a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}$ is $\qquad$ | 6. 1 .
From the given equation, we have
$$
\begin{array}{l}
(1-t) a+(2-t) b+(3-t) c+(4-t) d+ \\
t(a+b+c+d)=\sqrt{10} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
{\left[(1-t)^{2}+(2-t)^{2}+(3-t)^{2}+(4-t)^{2}+t^{2}\right] .} \\
{\left[a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}\right] \geqs... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If $\left(x^{2}-x-2\right)^{3}=a_{0}+a_{1} x+\cdots+a_{6} x^{6}$, then $a_{1}+a_{3}+a_{5}=$ | 3. -4 .
Let $x=0, x=1$, we get
$$
\begin{array}{l}
a_{0}=-8, \\
a_{0}+a_{1}+\cdots+a_{6}=\left(1^{2}-1-2\right)^{3}=-8 .
\end{array}
$$
Thus, $a_{1}+a_{2}+\cdots+a_{6}=0$.
Let $x=-1$, we get
$$
\begin{array}{l}
a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6} \\
=\left[(-1)^{2}-(-1)-2\right]^{3}=0 .
\end{array}
$$
Thus, $-... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given an isosceles triangle with a vertex angle of $20^{\circ}$ and a base length of $a$, the length of the legs is $b$. Then the value of $\frac{a^{3}+b^{3}}{a b^{2}}$ is $\qquad$ | 4.3.
Given $a=2 b \sin 10^{\circ}$.
Thus $a^{3}+b^{3}=8 b^{3} \sin ^{3} 10^{\circ}+b^{3}$
$$
\begin{array}{l}
=8 b^{3} \cdot \frac{1}{4}\left(3 \sin 10^{\circ}-\sin 30^{\circ}\right)+b^{3}=6 b^{3} \sin 10^{\circ} \\
\Rightarrow \frac{a^{3}+b^{3}}{a b^{2}}=\frac{6 b^{3} \sin 10^{\circ}}{2 b \sin 10^{\circ} \cdot b^{2}}... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a regular tetrahedron $P-ABC$ with the side length of the base being 6 and the side length of the lateral edges being $\sqrt{21}$. Then the radius of the inscribed sphere of the tetrahedron is $\qquad$ | 7.1 .
Let $P O \perp$ plane $A B C$ at point $O$. Then $O$ is the center of the equilateral $\triangle A B C$. Connect $A O$ and extend it to intersect $B C$ at point $D$, and connect $P D$. Thus, $D$ is the midpoint of $B C$.
It is easy to find that $P D=2 \sqrt{3}, O D=\sqrt{3}, P O=3$.
Let the radius of the inscrib... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. The largest prime $p$ such that $\frac{p+1}{2}$ and $\frac{p^{2}+1}{2}$ are both perfect squares is $\qquad$. | 10.7.
Let $\frac{p+1}{2}=x^{2} ， \frac{p^{2}+1}{2}=y^{2}\left(x, y \in \mathbf{Z}_{+}\right)$.
Obviously, $p>y>x, p>2$.
From $p+1=2 x^{2}, p^{2}+1=2 y^{2}$, subtracting the two equations gives $p(p-1)=2(y-x)(y+x)$.
Since $p(p>2)$ is a prime number and $p>y-x$, then $p \mid(y+x)$.
Because $2 p>y+x$, so $p=y+x$.
Thus, $... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. On a plane, there is an $8 \times 8$ grid colored in a black and white checkerboard pattern. Basil arbitrarily selects one of the cells. Each turn, Peter draws a polygon (which can be concave but not self-intersecting) on the grid, and Basil will honestly inform Peter whether the selected cell is inside or outside t... | 2. If the polygon drawn by Peter includes only all the cells of a certain color, then this polygon must intersect itself. Therefore, Peter cannot determine the color of the cell chosen by Basil in just one round.
Next, two strategies are given that can determine the color of the selected cell in two rounds.
【Strategy... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the complex numbers be
$$
\begin{array}{l}
z_{1}=(6-a)+(4-b) \mathrm{i}, \\
z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, \\
z_{3}=(3-a)+(3-2 b) \mathrm{i},
\end{array}
$$
where, $a, b \in \mathbf{R}$.
When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ reaches its minimum value, $3 a+4 b$ $=$ | 2. 12 .
Notice that,
$$
\begin{array}{l}
\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right| \\
=|12+9 \mathrm{i}|=15 .
\end{array}
$$
Equality holds if and only if
$$
\frac{6-a}{4-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{12}{9} \text {, }
$$
i.e., when $a=\frac{7}{... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the line $l$ passing through the origin intersect the graph of the function $y=|\sin x|$ $(x \geqslant 0)$ at exactly three points, with $\alpha$ being the largest of the x-coordinates of these intersection points. Then
$$
\frac{\left(1+\alpha^{2}\right) \sin 2 \alpha}{2 \alpha}=
$$
$\qquad$ . | 3. 1 .
As shown in Figure 2, let the line $l$ be tangent to the function $y=|\sin x|(x \geqslant 0)$
at point $P(\alpha,-\sin \alpha)$, and $k_{l}=-\cos \alpha$.
Then the line $l: y+\sin \alpha=-(x-\alpha) \cos \alpha$.
Substituting $(0,0)$, we get $\alpha=\tan \alpha$.
Therefore, $\frac{\left(1+\alpha^{2}\right) \sin... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the number of solutions to the equation
$$
|| \cdots|||x|-1|-2| \cdots|-2011|=2011
$$
The number of solutions. ${ }^{[4]}$ | 【Analysis】Remove the absolute value symbols from outside to inside, step by step.
From the original equation, we get
$$
|| \cdots|||x|-1|-2| \cdots|-2010|=0
$$
or ||$\cdots|||x|-1|-2| \cdots|-2010|=4002$.
For equation (1), we have
$$
\begin{array}{l}
|| \cdots|||x|-1|-2| \cdots|-2009|=2010 \\
\Rightarrow|| \cdots|||x|... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let real numbers $x, y$ satisfy
$$
x^{2}+\sqrt{3} y=4, y^{2}+\sqrt{3} x=4, x \neq y \text {. }
$$
Then the value of $\frac{y}{x}+\frac{x}{y}$ is $\qquad$ | 3. -5 .
From the conditions, we have
$$
\left\{\begin{array}{l}
x^{2}-y^{2}+\sqrt{3} y-\sqrt{3} x=0, \\
x^{2}+y^{2}+\sqrt{3}(x+y)=8 .
\end{array}\right.
$$
From equation (1) and $x \neq y$, we know $x+y=\sqrt{3}$.
Substituting into equation (2) gives $x^{2}+y^{2}=5$.
$$
\begin{array}{l}
\text { Also, } 2 x y=(x+y)^{2... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups.
The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups. | 7.6 .
Rewrite the given equation as
$$
\begin{array}{l}
x^{2}-x(2+y)+y^{2}-2 y=0 \\
\Rightarrow \Delta=(2+y)^{2}-4\left(y^{2}-2 y\right)=-3 y^{2}+12 y+4 \\
\quad=-3(y-2)^{2}+16 \geqslant 0 \\
\Rightarrow|y-2| \leqslant \frac{4}{\sqrt{3}}<3 .
\end{array}
$$
Since $y$ is an integer, thus, $y=0,1,2,3,4$.
Upon calculatio... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) As shown in Figure 4, the three sides of $\triangle ABC$ are all positive integers, and the perimeter is 35. $G$ and $I$ are the centroid and incenter of $\triangle ABC$, respectively, and $\angle GIC = 90^{\circ}$. Find the length of side $AB$. | 11. Extend $G I$, intersecting $C B$ and $C A$ (or their extensions) at points $P$ and $Q$ respectively.
Since $C I$ is the angle bisector of $\angle C$ and $\angle G I C=90^{\circ}$, we know that $\triangle C P Q$ is an isosceles triangle.
Draw $G E \perp P C$ and $G F \perp C Q$ from point $G$, and draw $I R \perp ... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the function $f: \mathbf{N} \rightarrow \mathbf{N}$ defined as follows:
$$
f(x)=\left\{\begin{array}{ll}
\frac{x}{2}, & x \text { is even; } \\
\frac{x+7}{2}, & x \text { is odd. }
\end{array}\right.
$$
Then the number of elements in the set $A=\{x \in \mathbf{N} \mid f(f(f(x)))=x\}$ is | - 1.8.
On one hand, when $x \in \{0,1, \cdots, 7\}$, it is calculated that $f(x) \in \{0,1, \cdots, 7\}$,
and it can be verified that $\{0,1, \cdots, 7\} \subseteq A$.
On the other hand, when $x \geqslant 8$,
$$
\frac{x}{2}<x \text{, and } \frac{x+7}{2}<\frac{2 x}{2}<x \text{. }
$$
Then $f(f(f(x)))<8$, which does not... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. (16 points) As shown in Figure 4, $A$ and $B$ are the common vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. $P$ and $Q$ are moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and sati... | Prove: (1) $O, P, Q$ are collinear;
(2) If the slopes of the lines $AP, BP, AQ, BQ$ are respectively
14. (1) Note that,
$\overrightarrow{A P}+\overrightarrow{B P}=2 \overrightarrow{O P}, \overrightarrow{A Q}+\overrightarrow{B Q}=2 \overrightarrow{O Q}$.
Also, $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrigh... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\sqrt{a+b+c+d}+\sqrt{a^{2}-2 a+3-b}- \\
\sqrt{b-c^{2}+4 c-8}=3 .
\end{array}
$$
Then the value of $a-b+c-d$ is ( ).
(A) $-7 \quad \square \quad$ -
(B) -8
(C) -4
(D) -6 | 5. A.
$$
\begin{array}{l}
\text { Given } a^{2}-2 a+3-b \geqslant 0, \\
b-c^{2}+4 c-8 \geqslant 0,
\end{array}
$$
we know
$$
\begin{array}{l}
b \leqslant-(a+1)^{2}+4 \leqslant 4, \\
b \geqslant(c-2)^{2}+4 \geqslant 4 .
\end{array}
$$
Thus, $b=4, a=-1, c=2$.
Substituting these into the known equations gives $d=4$.
The... | -7 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. Given a positive real number $x$ satisfies
$$
x^{3}+x^{-3}+x^{6}+x^{-6}=2754 \text {. }
$$
then $x+\frac{1}{x}=$ | 3. 4 .
Transform the given equation into
$$
\begin{aligned}
& \left(x^{3}+x^{-3}\right)^{2}+\left(x^{3}+x^{-3}\right)-2756=0 \\
\Rightarrow & \left(x^{3}+x^{-3}+53\right)\left(x^{3}+x^{-3}-52\right)=0 .
\end{aligned}
$$
Notice that, $x^{3}+x^{-3}+53>0$.
Thus, $x^{3}+x^{-3}=52$.
Let $b=x+x^{-1}$.
Then $x^{3}+x^{-3}=\l... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) There are 288 sets of cards, totaling 2016 cards, each set consisting of $1,2, \cdots, 7$ and stacked in the order $1,2, \cdots, 7$ from top to bottom. Now, these 288 sets of cards are stacked together from top to bottom. First, discard the top five cards, then place the top card at the bottom, and c... | (1) For the first 42 cards (in 6 groups each), according to the operation rule, after discarding 5 cards, the cards placed at the bottom are $6,5, \cdots, 1,7$. Thus, each number discards 5 cards.
And $288 \div 6=48, 48 \times 5 \times 7=1680$ (cards), $48 \times 5=240$ (cards), $2016-1680=336$ (cards). That is, when ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the set $A=\left\{x \left\lvert\, \frac{x+4}{x-3} \leqslant 0\right., x \in \mathbf{Z}\right\}$, and from set $A$ a random element $x$ is drawn, denoted by $\xi=x^{2}$. Then the mathematical expectation of the random variable $\xi$ is $\mathrm{E} \xi=$ $\qquad$ | -1.5 .
From the conditions, we know that
$$
A=\{-4,-3,-2,-1,0,1,2\},
$$
The values of the random variable $\xi$ are $0, 1, 4, 9, 16$.
It is easy to obtain that the probability distribution of $\xi$ is shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline$\xi$ & 0 & 1 & 4 & 9 & 16 \\
\hline$P$ & $\frac{1}{7}$... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $f(x)=x+g(x)$, where $g(x)$ is a function defined on $\mathbf{R}$ with the smallest positive period of 2. If the maximum value of $f(x)$ in the interval $[2,4)$ is 1, then the maximum value of $f(x)$ in the interval $[10,12)$ is $\qquad$
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 2.9.
According to the problem, we have
$$
\begin{array}{l}
f(x+2)=(x+2)+g(x+2) \\
=x+g(x)+2=f(x)+2 .
\end{array}
$$
Given that the maximum value of $f(x)$ in the interval $[2,4)$ is 1, we know that the maximum value of $f(x)$ in the interval $[4,6)$ is 3, ... and the maximum value of $f(x)$ in the interval $[10,12)$ ... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
4. Given real numbers $x, y, z$ satisfy $x^{2}+2 y^{2}+3 z^{2}=24$.
Then the minimum value of $x+2 y+3 z$ is $\qquad$ . | 4. -12 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x+2 y+3 z)^{2} \\
=(1 \times x+\sqrt{2} \times \sqrt{2} y+\sqrt{3} \times \sqrt{3} z)^{2} \\
\leqslant\left[1^{2}+(\sqrt{2})^{2}+(\sqrt{3})^{2}\right]\left(x^{2}+2 y^{2}+3 z^{2}\right) \\
\quad=144 .
\end{array}
$$
Therefore, $x+2 y+3 z \geqslant-12$... | -12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. If $\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\cdots+\sin \frac{n \pi}{9}=\frac{1}{2} \tan \frac{4 \pi}{9}$, then the smallest positive integer $n$ is $\qquad$. | 10.4.
Notice,
$$
\begin{array}{l}
2 \sin \frac{\pi}{18}\left(\sum_{k=1}^{n} \sin \frac{k \pi}{9}\right) \\
=\sum_{k=1}^{n}\left(\cos \frac{(2 k-1) \pi}{18}-\cos \frac{(2 k+1) \pi}{18}\right) \\
=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \tan \frac{4 \pi}{9} \cdot \sin \frac{\pi}{18}=\cos \frac{\pi... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the function $f(x)=\frac{(x+1)^{2}+\sin x}{x^{2}+1}$ have the maximum value and minimum value as $M$ and $N$, respectively. Then $M+N=$ | 3. 2 .
From the given information, we have
$$
f(x)=\frac{(x+1)^{2}+\sin x}{x^{2}+1}=1+\frac{2 x+\sin x}{x^{2}+1} \text {. }
$$
Notice that the function $g(x)=\frac{2 x+\sin x}{x^{2}+1}$ is an odd function. Therefore, the maximum value $M_{0}$ and the minimum value $N_{0}$ of $g(x)$ satisfy
$$
\begin{array}{l}
M_{0}+N... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Four numbers
$$
\begin{array}{l}
\sqrt{2-\sqrt{3}} \cdot \sqrt{2-\sqrt{2-\sqrt{3}}} 、 \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}} \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}}
\end{array}
$$
The product of these is ( ).
(A) $2+\sqrt{3}$
(B) 2
(C) 1
(D) $2-\sqrt{3}$ | $-1 . \mathrm{C}$.
$$
\begin{array}{l}
\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}} \times \\
\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \\
\quad \sqrt{2^{2}-\left(\sqrt{2-\sqrt{2-\sqrt{3}})^{2}}\right.} \\
=\sqrt{... | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. Several different numbers are written on the blackboard, such that the sum of any three of them is a rational number, while the sum of any two is an irrational number. The maximum number of numbers that can be written on the blackboard is $\qquad$ | 4. 3 .
Assume that the numbers written on the blackboard are no less than four, denoted as $a, b, c, d$. Then, $a+b+c$ and $b+c+d$ are both rational numbers, which implies that their difference
$$
(b+c+d)-(a+b+c)=d-a
$$
is also a rational number.
Similarly, $b-a$ and $c-a$ are also rational numbers.
Therefore, $b=a+r_... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$1.2014^{2015}$ 的个位数为
The unit digit of $1.2014^{2015}$ is | -1.4 .
Let $g(n)$ denote the unit digit of a natural number $n$.
$$
\begin{array}{l}
\text { Then } g\left(2014^{2015}\right) \\
=g\left((201 \times 10+4)^{2015}\right) \\
=g\left(4^{2015}\right)=g\left(\left(4^{2}\right)^{1007} \times 4\right) \\
=g\left((10+6)^{1007} \times 4\right)=g\left(6^{1007} \times 4\right) \\... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $P$ be any point on the graph of the function $f(x)=x+\frac{2}{x}(x>0)$, and draw perpendiculars from point $P$ to the $x$-axis and $y$-axis, with the feet of the perpendiculars being $A$ and $B$ respectively. Then the minimum value of $|P A|+|P B|$ is $\qquad$ | 2. 4 .
Let $P(x, y)$. According to the problem,
$$
|P A|=|y|=y,|P B|=|x|=x,
$$
where, $y=x+\frac{2}{x}(x>0)$.
$$
\begin{array}{l}
\text { Hence }|P A|+|P B|=y+x=2\left(x+\frac{1}{x}\right) \\
\geqslant 2 \times 2 \sqrt{x \cdot \frac{1}{x}}=4,
\end{array}
$$
The equality holds if and only if $x=\frac{1}{x}$.
Since $x... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. As shown in Figure 3, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through point $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
then th... | 14. 2 .
Since $O$ is the midpoint of side $B C$, we have
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} \text {. }
$$
Since points $M, O, N$ are collinear, it follows that,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\frac{1}{2} \sin 2 y+a=0$. Then the value of $\cos (x+2 y)$ is $\qquad$ | 15. 1 .
Solving the system of equations and eliminating $a$ yields
$$
x=-2 y \Rightarrow \cos (x+2 y)=1
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. For any $\alpha, \beta \in\left(0, \frac{2 \pi}{3}\right)$, we have
$$
\begin{array}{l}
4 \cos ^{2} \alpha+2 \cos \alpha \cdot \cos \beta+4 \cos ^{2} \beta- \\
3 \cos \alpha-3 \cos \beta-k<0 .
\end{array}
$$
Then the minimum value of $k$ is | 5.6.
Substitution $\left(x_{1}, x_{2}\right)=(2 \cos \alpha, 2 \cos \beta)$.
Then the given inequality becomes
$$
2 x_{1}^{2}+x_{1} x_{2}+2 x_{2}^{2}-3 x_{1}-3 x_{2}-2 k<0 \text {. }
$$
Since $x_{1}, x_{2} \in(-1,2)$, we have
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{1}-2\right)+\left(x_{2}+1\right)\left(x_{2}... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let $a_{1} \in \mathbf{Z}_{+}$, and $a_{1} \leqslant 18$, define the sequence $\left\{a_{n}\right\}:$
$$
a_{n+1}=\left\{\begin{array}{ll}
2 a_{n}, & a_{n} \leqslant 18 ; \\
2 a_{n}-36, & a_{n}>18
\end{array}(n=1,2, \cdots) .\right.
$$
Find the maximum number of elements in the set $M=\left\{a_{n} \mid ... | 11. Given the positive integer $a_{1} \leqslant 18$ and the recursive formula for the sequence $\left\{a_{n}\right\}$, the following properties can be derived:
(1) $a_{n+1} \equiv 2 a_{n}(\bmod 36)$, i.e.,
$a_{n+1} \equiv 2 a_{n}(\bmod 4)$, and $a_{n+1} \equiv 2 a_{n}(\bmod 9)$;
(2) All terms are positive integers, and... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2011^{3}}$.
Then the integer part of $4 S$ is ( ). ${ }^{[2]}$
(A) 4
(B) 5
(C) 6
(D) 7
(2011, "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | When $k=2,3, \cdots, 2011$,
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {. }
$$
Then $1<S=1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\cdots+\frac{1}{2011^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2011 \times 2012}\right)<\frac{5}{4} \text {. }
$$
... | 4 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. The smallest positive odd number that cannot be expressed as $7^{x}-3 \times 2^{y}\left(x 、 y \in \mathbf{Z}_{+}\right)$ is $\qquad$ | -1.3 .
Since $x, y \in \mathbf{Z}_{+}$, therefore, $7^{x}-3 \times 2^{y}$ is always an odd number, and $7^{1}-3 \times 2^{1}=1$.
If $7^{x}-3 \times 2^{y}=3$, then $317^{x}$.
And $7^{x}=(1+6)^{x}=1(\bmod 3)$, thus, there do not exist positive integers $x, y$ such that
$$
7^{x}-3 \times 2^{y}=3 \text {. }
$$
Therefore, ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $f(x)=\frac{\sin \pi x}{x^{2}}(x \in(0,1))$. Then
$$
g(x)=f(x)+f(1-x)
$$
the minimum value of $g(x)$ is . $\qquad$ | 7.8.
From the given, we have
$$
\begin{aligned}
f^{\prime}(x) & =\frac{\pi x \cos \pi x-2 \sin \pi x}{x^{3}}, \\
f^{\prime \prime}(x) & =\frac{\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x}{x^{3}} .
\end{aligned}
$$
Next, we need to prove
$$
\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x>0.
$... | 8 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
4. If $2016+3^{n}$ is a perfect square, then the positive integer $n=$ . $\qquad$ | 4. 2 .
Obviously, $2016+3^{n}$ is an odd perfect square.
So $2016+3^{n} \equiv 1(\bmod 8)$
$$
\Rightarrow 3^{n} \equiv 1(\bmod 8)
$$
$\Rightarrow n$ must be even.
Let $n=2+2 k(k \in \mathbf{N})$. Then
$$
2016+3^{n}=9\left(224+3^{2 k}\right) \text {. }
$$
So $224+3^{2 k}=224+\left(3^{k}\right)^{2}$ is a perfect square... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) If the pair of positive integers $(a, x)$ satisfies
$$
\sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}} \neq x \text {, }
$$
find all positive integers $a$ that meet the requirement. | $$
\text { Three, let } \sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}}=t \text {. }
$$
Then $t^{2} x+x+t^{2}-a=0$,
$$
(t+1) x^{2}+t-a=0 \text {. }
$$
$t \times$ (2) $-x \times$ (1) gives
$t x^{2}-x^{2}-t^{2} x+t^{2}-a t+a x=0$
$\Rightarrow(t-x)(t+x-t x-a)=0$
$\Rightarrow t=x$ (discard) or $t=\frac{a-x}{1-x}$.
Thus $\f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$
and a fixed point $B(a, b)$ satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$ | 2. 2 .
According to the problem, for any $(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking $(x, y)=(1,0),(0,1)$, we get the fixed point $B(a, b)$ satisfying the necessary conditions $\left\{\begin{array}{l}a \leqslant 1, \\ b \leqslant 1,\end{array}\right.$, which implies $a+b \leqslant 2$.
Thus, $(a+b)_{\text ... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the incenter $I(-1,7)$ of a right-angled triangle $\triangle O A B$ with all three vertices as integer points, and the origin $O$ as the right-angle vertex. The number of such right-angled triangles $\triangle O A B$ is $\qquad$
Translate the above text into English, please retain the original text's line bre... | 8.2.
As shown in Figure 4.
Let $\angle x O A=\alpha, \angle x O I=\beta$.
Then $\alpha=\beta-\frac{\pi}{4}$.
Also, $\tan \beta=-7$, so
$$
\begin{array}{l}
\tan \alpha=\tan \left(\beta-\frac{\pi}{4}\right) \\
=\frac{\tan \beta-1}{1+\tan \beta}=\frac{4}{3}, \\
\tan \angle x O B=\tan \left(\frac{\pi}{2}+\alpha\right) \\
... | 2 | Other | math-word-problem | Yes | Yes | cn_contest | false |
8. Given real numbers $a, b$ satisfy
$$
a+\lg a=10, b+10^{b}=10 \text {. }
$$
Then $\lg (a+b)=$ $\qquad$ . | 8.1.
Since $\lg a=10-a, 10^{b}=10-b$, therefore, $a$ is the x-coordinate of the intersection point of $y=\lg x$ and $y=10-x$, and $b$ is the y-coordinate of the intersection point of $y=10^{x}$ and $y=10-x$.
Also, $y=\lg x$ and $y=10^{x}$ are symmetric about the line $y=x$, and $y=10-x$ is symmetric about the line $y... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. $6^{11}+C_{11}^{1} 6^{10}+C_{11}^{2} 6^{9}+\cdots+C_{11}^{10} 6-1$ when divided by 8 yields a remainder of $\qquad$ . | 3.5.
Notice that,
$$
\begin{array}{l}
6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\mathrm{C}_{11}^{2} 6^{9}+\cdots+\mathrm{C}_{11}^{10} 6-1 \\
=\mathrm{C}_{10}^{0} 6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\cdots+\mathrm{C}_{11}^{10} 6+\mathrm{C}_{11}^{11} 6^{0}-2 \\
=(6+1)^{11}-2=7^{11}-2 \\
\equiv(-1)^{11}-2 \equiv 5(\bmod 8) .
\end{... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 2, point $A$ is on the positive $y$-axis, point $B$ is on the positive $x$-axis, $S_{\triangle A O B}=9$, segment $A B$ intersects the graph of the inverse proportion function $y=\frac{k}{x}$ at points $C$ and $D$. If $C D=$ $\frac{1}{3} A B$, and $A C=B D$, then $k=$ . $\qquad$ | 8. 4 .
Let point $A\left(0, y_{A}\right), B\left(x_{B}, 0\right)$.
Given $C D=\frac{1}{3} A B, A C=B D$, we know that $C$ and $D$ are the trisection points of segment $A B$.
Thus, $x_{C}=\frac{1}{3} x_{B}, y_{C}=\frac{2}{3} y_{A}$.
Therefore, $k=x_{C} y_{C}=\frac{2}{9} x_{B} y_{A}=\frac{4}{9} S_{\triangle A O B}=4$. | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. If $4^{a}=6^{b}=9^{c}$, then
$$
\frac{1}{a}-\frac{2}{b}+\frac{1}{c}=
$$
$\qquad$ | 8. 0 .
Let $4^{a}=6^{b}=9^{c}=k$.
Then $a=\log _{k} 4, b=\log _{k} 6, c=\log _{k} 9$
$$
\begin{array}{l}
\Rightarrow \frac{1}{a}-\frac{2}{b}+\frac{1}{c}=\log _{k} 4-2 \log _{k} 6+\log _{k} 9 \\
\quad=\log _{k} 1=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. In rectangle $A B C D$, $A B=3, A D=4, P$ is a point on the plane of rectangle $A B C D$, satisfying $P A=2$, $P C=\sqrt{21}$. Then $\overrightarrow{P B} \cdot \overrightarrow{P D}=$ | 11.0.
As shown in Figure 3, let $A C$ and $B D$ intersect at point $E$, and connect $P E$. Then $E$ is the midpoint of $A C$ and $B D$.
Notice that,
$$
\begin{array}{l}
\overrightarrow{P B} \cdot \overrightarrow{P D}=\frac{1}{4}\left[(\overrightarrow{P B}+\overrightarrow{P D})^{2}-(\overrightarrow{P B}-\overrightarrow... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that all positive integers are in $n$ sets, satisfying that when $|i-j|$ is a prime number, $i$ and $j$ belong to two different sets. Then the minimum value of $n$ is $\qquad$ | 6. 4 .
It is easy to see that $n \geqslant 4$ (2, 4, 7, 9 must be in four different sets).
Also, when $n=4$, the sets
$$
A_{i}=\left\{m \in \mathbf{Z}_{+} \mid m \equiv i(\bmod 4)\right\}(i=0,1,2,3)
$$
satisfy the condition.
Therefore, the minimum value of $n$ is 4. | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that the ellipse $C$ passes through the point $M(1,2)$, with two foci at $(0, \pm \sqrt{6})$, and $O$ is the origin, a line $l$ parallel to $OM$ intersects the ellipse $C$ at points $A$ and $B$. Then the maximum value of the area of $\triangle OAB$ is $\qquad$ | 7. 2 .
According to the problem, let $l_{A B}: y=2 x+m$.
The ellipse equation is $\frac{y^{2}}{8}+\frac{x^{2}}{2}=1$. By solving the system and eliminating $y$, we get $16 x^{2}+8 m x+2\left(m^{2}-8\right)=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\fra... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given positive integers $a, b, c, x, y, z$ satisfying $a \geqslant b \geqslant c \geqslant 1, x \geqslant y \geqslant z \geqslant 1$,
and $\left\{\begin{array}{l}2 a+b+4 c=4 x y z, \\ 2 x+y+4 z=4 a b c .\end{array}\right.$
Then the number of six-tuples $(a, b, c, x, y, z)$ that satisfy the conditions is $\qquad$ gro... | 8. 0 .
Assume $x \geqslant a$. Then
$$
\begin{array}{l}
4 x y z=2 a+b+4 c \leqslant 7 a \leqslant 7 x \\
\Rightarrow y z \leqslant \frac{7}{4} \Rightarrow y z \leqslant 1 \Rightarrow(y, z)=(1,1) \\
\Rightarrow\left\{\begin{array}{l}
a a+b+4 c=4 x, \\
2 x+5=4 a b c
\end{array}\right. \\
\Rightarrow 2 a+b+4 c+10=8 a b c... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $0<a<1$, and satisfies
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a]=$ $\qquad$ | Solve:
$$
\begin{array}{l}
0<a<1 \\
\Rightarrow 0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2 \\
\Rightarrow\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right] \text { is either }
\end{array}
$$
0 or 1.
By the problem, we know that 18 of them are equal to 1, and 11 ... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $O$ is the circumcenter of acute $\triangle A B C$, $\angle B A C$ $=60^{\circ}$, extend $C O$ to intersect $A B$ at point $D$, extend $B O$ to intersect $A C$ at point $E$. Then $\frac{B D}{C E}=$ $\qquad$ | $=, 7.1$.
Connect $O A, D E$.
$$
\begin{array}{l}
\text { Since } \angle B O C=2 \angle B A C=120^{\circ} \\
\Rightarrow \angle C O E=60^{\circ}=\angle D A E \\
\Rightarrow A, D, O, E \text { are concyclic } \\
\Rightarrow \angle D E B=\angle D A O=\angle D B E \\
\Rightarrow D B=D E .
\end{array}
$$
Similarly, $C E=D... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $x$ and $y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+2015(x-1)=-1, \\
(y-1)^{3}+2015(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$ | 4. 2 .
Notice that the function $f(z)=z^{3}+2015 z$ is a monotonically increasing function on $(-\infty,+\infty)$.
From the given condition, we have $f(x-1)=f(1-y)$.
Therefore, $x-1=1-y \Rightarrow x+y=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x-y+1 \geqslant 0 \\
y+1 \geqslant 0 \\
x+y+1 \leqslant 0 .
\end{array}\right.
$$
Then the maximum value of $2 x-y$ is $\qquad$ | 8. 1 .
When $x=0, y=-1$, $2 x-y$ takes the maximum value 1. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Given that $a$, $b$, and $c$ are three distinct real numbers. If in the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$
any two of these equations have exactly one common root, find the value of $a^{2}+$ $b^{2}+c^{2}$. | Let the equations (1) and (3) have only one common root \( x_{1} \), equations (1) and (2) have only one common root \( x_{2} \), and equations (2) and (3) have only one common root \( x_{3} \). Therefore, the roots of equation (1) are \( x_{1} \) and \( x_{2} \), the roots of equation (2) are \( x_{2} \) and \( x_{3} ... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2: A rope of length 2009 is operated as follows: first, it is divided into two ropes of positive integer lengths, and the lengths of the two ropes are recorded, then the above operation is repeated on one of the ropes, ... until 2009 ropes of length 1 are obtained. If the lengths of the two ropes obtained in a ... | 【Analysis】(1) It is easy to know that a rope of length 2 can only be divided into two segments of length 1, meaning the last operation on the rope must not be a good operation. A rope of length 2009 can be divided into 2009 segments of length 1 after exactly 2008 operations, so the number of good operations is no more ... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. In a lottery with 100000000 tickets, each ticket number consists of eight digits. A ticket number is called "lucky" if and only if the sum of its first four digits equals the sum of its last four digits. Then the sum of all lucky ticket numbers, when divided by 101, leaves a remainder of $\qquad$ | 7.0.
If the eight-digit number $x=\overline{a b c d e f g h}$ is lucky, then $y=99999999-x$ is also lucky, and $x \neq y$.
$$
\begin{array}{l}
\text { and } x+y=99999999=9999 \times 10001 \\
=99 \times 101 \times 10001,
\end{array}
$$
Therefore, the sum of all lucky numbers must be a multiple of 101. | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sets
$$
A=\left\{n^{2}+1 \mid n \in \mathbf{Z}_{+}\right\}, B=\left\{n^{3}+1 \mid n \in \mathbf{Z}_{+}\right\} \text {. }
$$
Arrange all elements in $A \cap B$ in ascending order to form the sequence $a_{1}, a_{2}, \cdots$. Then the units digit of $a_{99}$ is | 2. 2 .
From the given, we know that $A \cap B=\left\{n^{6}+1 \mid n \in \mathbf{Z}_{+}\right\}$.
Therefore, $a_{99}=99^{6}+1$.
Thus, its unit digit is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$, respectively. Point $P$ is on the right branch of the hyperbola $C$, and the excenter of $\triangle P F_{1} F_{2}$ opposite to $\angle P F_{1} F_{2}$ is $I$. The line $P I$ intersects the $x$-axis at point $Q$... | 6. 4 .
Since $I F_{1}$ is the angle bisector of $\angle P F_{1} Q$, we have
$$
\frac{|P Q|}{|P I|}=1+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|} \text {. }
$$
Let $P\left(x_{0}, y_{0}\right)$. Then, $\left|P F_{1}\right|=\frac{3}{2} x_{0}+2$.
By the optical property of the hyperbola, the tangent line to the hyp... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. Answer Questions (Total 56 points)
9. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=4, \\
a_{n}=a_{n-1}+a_{n-2}-a_{n-3}+1(n \geqslant 4) .
\end{array}
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Prove: $\frac{1}{a_{1... | (1) For $n \geqslant 4$, summing up we get
$$
a_{n}-a_{n-2}=a_{3}-a_{1}+n-3=n \text {. }
$$
When $n=2 m\left(m \in \mathbf{Z}_{+}\right)$,
$$
\begin{aligned}
a_{n} & =a_{2}+\sum_{k=1}^{m-1}\left(a_{2 k+2}-a_{2 k}\right) \\
& =2+\sum_{k=1}^{m-1}(2 k+2)=\frac{1}{4} n(n+2) ;
\end{aligned}
$$
When $n=2 m+1\left(m \in \ma... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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 $k=4$,
$$
6=1+1+1+3=1+1+2+2... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given positive real numbers $x, y, z$ satisfying
$$
\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 |
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 $A D$
$$
\begin{array}{l}
\Rightarrow \an... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $AB$ is a diameter of the smallest circle with center $C(0,1)$ that has common points with the graph of the function $y=\frac{1}{|x|-1}$, and $O$ is the origin. Then $\overrightarrow{O A} \cdot \overrightarrow{O B}$ $=$ | 4. -2 .
For any point $P(x, y)$ on the function $y=\frac{1}{|x|-1}(x>1)$, we have
$$
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 |
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
$$
\begin... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $x, y>0$, and $x+2 y=2$. Then the minimum value of $\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}$ is . $\qquad$ | 8. 2 .
$$
\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 |
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
$$
\begin{array}{l}
\left\{\begin{array}{l}... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. If the function $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a sum of its maximum and minimum values equal to 4, then $a+b=$ $\qquad$ | 5.3.
Given $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a maximum or minimum value, we know $b=0$.
Then $y=\frac{a+\sin x}{2+\cos x}$
$$
\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}
$$
By Vi... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Connecting the intersection points of $x^{2}+y^{2}=10$ and $y=\frac{4}{x}$ in sequence, a convex quadrilateral is formed. The area of this quadrilateral is $\qquad$ | 2. 12 .
Let $A\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$ be the intersection point of the two curves in the first quadrant.
Since the two curves are symmetric with respect to the origin and the line $y=x$, the coordinates of the other three intersection points are
$$
B\left(y_{0}, x_{0}\right), C\left(-x... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. If the complex coefficient equation with respect to $x$
$$
(1+2 \mathrm{i}) x^{2}+m x+1-2 \mathrm{i}=0
$$
has real roots, then the minimum value of the modulus of the complex number $m$ is $\qquad$ | 4.2.
Let $\alpha$ be a real root of the original equation, $m=p+q$ i.
$$
\begin{array}{l}
\text { Then }(1+2 \mathrm{i}) \alpha^{2}+(p+q \mathrm{i}) \alpha+1-2 \mathrm{i}=0 \\
\Rightarrow\left\{\begin{array}{l}
\alpha^{2}+p \alpha+1=0, \\
2 \alpha^{2}+q \alpha-2=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Title: A quadruple of positive integers $(p, a, b, c)$ that satisfies the following conditions is called a "Leyden quadruple":
(i) $p$ is an odd prime;
(ii) $a, b, c$ are distinct;
(iii) $p \mid (ab + 1), p \mid (bc + 1), p \mid (ca + 1)$.
(1) Prove that for each Leyden quadruple $(p, a, b, c)$, we have $p + 2 \leq \fr... | From the form, the Leiden quadruples that satisfy the conditions have a certain rotational symmetry. To facilitate expression and discovery of its inherent laws, it is extended as follows:
Definition If the $(k+1)$-tuple of positive integers $\left(p, a_{1}, a_{2}, \cdots, a_{k}\right)$ satisfies the following three p... | 5 | Number Theory | proof | Yes | Yes | cn_contest | false |
Given that $a$, $b$, and $c$ are three distinct real numbers. If the quadratic equations:
$$
\begin{array}{l}
x^{2} + a x + b = 0, \\
x^{2} + b x + c = 0, \\
x^{2} + c x + a = 0
\end{array}
$$
each have exactly one common root with any other, find the value of $a^{2} + b^{2} + c^{2}$.
(The 4th Chen Shengshen Cup Natio... | From [1] we know
$$
x_{1}=\frac{a-b}{a-c}, x_{2}=\frac{b-c}{b-a}, x_{3}=\frac{c-a}{c-b},
$$
where $x_{1}$ is the common root of equations (1) and (3), $x_{2}$ is the common root of equations (1) and (2), and $x_{3}$ is the common root of equations (2) and (3).
Since $x_{1}$ and $x_{2}$ are the two roots of equation (1... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $\sqrt{24-t^{2}}-\sqrt{8-t^{2}}=2$, then
$$
\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=
$$
$\qquad$ | II. 1.8.
From the known equation and using the formula $a+b=\frac{a^{2}-b^{2}}{a-b}$, we get $\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=\frac{24-8}{2}=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the set $M=\{(a, b) \mid a \leqslant-1, b \leqslant m\}$. If for any $(a, b) \in M$, it always holds that $a \cdot 2^{b}-b-3 a \geqslant 0$, then the maximum value of the real number $m$ is $\qquad$. | 3. 1 .
Notice that,
$$
a \cdot 2^{b}-b-3 a \geqslant 0 \Leftrightarrow\left(2^{b}-3\right) a-b \geqslant 0
$$
holds for any $a \leqslant-1$.
$$
\text { Then }\left\{\begin{array}{l}
2^{b}-3 \leqslant 0, \\
2^{b}+b \leqslant 3
\end{array} \Rightarrow b \leqslant 1\right. \text {. }
$$ | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. If $P$ is the circumcenter of $\triangle A B C$, and $\overrightarrow{P A}+\overrightarrow{P B}+\lambda \overrightarrow{P C}=\mathbf{0}, \angle C=120^{\circ}$. Then the value of the real number $\lambda$ is $\qquad$. | 5. -1 .
Let the circumradius of $\triangle ABC$ be $R$.
From the given information, we have
$$
\begin{array}{l}
|\overrightarrow{P A}+\overrightarrow{P B}|^{2}=\lambda^{2}|\overrightarrow{P C}|^{2} \\
=|\overrightarrow{P A}|^{2}+|\overrightarrow{P B}|^{2}-2|\overrightarrow{P A}||\overrightarrow{P B}| \cos \frac{C}{2} ... | -1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $\alpha, \beta$ satisfy the equations respectively
$$
\begin{array}{l}
\alpha^{3}-3 \alpha^{2}+5 \alpha-4=0, \\
\beta^{3}-3 \beta^{2}+5 \beta-2=0
\end{array}
$$
then $\alpha+\beta=$ $\qquad$ | 9. 2 .
We have
$$
\begin{array}{l}
(\alpha-1)^{3}+2(\alpha-1)-1=0, \\
(1-\beta)^{3}+2(1-\beta)-1=0,
\end{array}
$$
which means $\alpha-1$ and $1-\beta$ are solutions to the equation $x^{3}+2 x-1=0$.
Since $x^{3}+2 x-1=0$ has only one solution, then
$$
\alpha-1=1-\beta \Rightarrow \alpha+\beta=2
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $n$ be a positive integer such that $\sqrt{3}$ lies between $\frac{n+3}{n}$ and $\frac{n+4}{n+1}$. Then $n=$ $\qquad$ | 3. 4 .
Notice that,
$$
\frac{n+4}{n+1}=1+\frac{3}{n+1}<1+\frac{3}{n}=\frac{n+3}{n} \text {. }
$$
Since $\sqrt{3}$ is an irrational number, then
$$
\begin{array}{l}
1+\frac{3}{n+1}<\sqrt{3}<1+\frac{3}{n} \\
\Rightarrow \frac{3}{n+1}<\sqrt{3}-1<\frac{3}{n} \\
\Rightarrow n<\frac{3}{\sqrt{3}-1}<n+1 .
\end{array}
$$
Thus... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
9. In the sequence $\left\{a_{n}\right\}$, $a_{4}=1, a_{11}=9$, and the sum of any three consecutive terms is 15. Then $a_{2016}=$ | Ni, 9.5.
According to the problem, for any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}+a_{n+2}=a_{n+1}+a_{n+2}+a_{n+3}=15 \\
\Rightarrow a_{n+3}=a_{n} .
\end{array}
$$
Then $a_{1}=a_{4}=1, a_{2}=a_{11}=9$,
$$
a_{3}=15-a_{1}-a_{2}=5 \text {. }
$$
Therefore, $a_{2016}=a_{3 \times 672}=a_{3}=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Let $x \in \mathbf{R}$. Then the function
$$
f(x)=|2 x-1|+|3 x-2|+|4 x-3|+|5 x-4|
$$
has a minimum value of $\qquad$ | 12. 1 .
Notice,
$$
\begin{array}{l}
f(x)=2\left|x-\frac{1}{2}\right|+3\left|x-\frac{2}{3}\right|+ \\
4\left|x-\frac{3}{4}\right|+5\left|x-\frac{4}{5}\right| \\
=2\left(\left|x-\frac{1}{2}\right|+\left|x-\frac{4}{5}\right|\right)+ \\
\quad 3\left(\left|x-\frac{2}{3}\right|+\left|x-\frac{4}{5}\right|\right)+4\left|x-\fr... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of zeros of the function $f(x)=x^{2} \ln x+x^{2}-2$ is . $\qquad$ | 3.1 .
From the condition, we have
$$
f^{\prime}(x)=2 x \ln x+x+2 x=x(2 \ln x+3) \text {. }
$$
When $0\mathrm{e}^{-\frac{3}{2}}$, $f^{\prime}(x)>0$.
Thus, $f(x)$ is a decreasing function on the interval $\left(0, \mathrm{e}^{-\frac{3}{2}}\right)$ and an increasing function on the interval $\left(\mathrm{e}^{-\frac{3}{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $z \in \mathbf{C}$. If the equation $x^{2}-2 z x+\frac{3}{4}+\mathrm{i}=0$ (where $\mathrm{i}$ is the imaginary unit) has real roots, then the minimum value of $|z|$ is $\qquad$ . | 7.1 .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be a real root of the original equation.
$$
\begin{array}{l}
\text { Then } x_{0}^{2}-2(a+b \mathrm{i}) x_{0}+\frac{3}{4}+\mathrm{i}=0 \\
\Rightarrow\left\{\begin{array}{l}
x_{0}^{2}-2 a x_{0}+\frac{3}{4}=0, \\
-2 b x_{0}+1=0
\end{array}\right. \\
\Rightarrow ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $f(x)$ is a periodic function on $\mathbf{R}$ with the smallest positive period of 2, and when $0 \leqslant x<2$, $f(x)=x^{3}-x$. Then the number of intersections between the graph of the function $y=f(x)$ and the $x$-axis in the interval $[0,6]$ is $\qquad$ . | 2.7.
When $0 \leqslant x<2$, let $f(x)=x^{3}-x=0$, we get $x=0$ or 1.
According to the properties of periodic functions, given that the smallest period of $f(x)$ is 2, we know that $y=f(x)$ has six zeros in the interval $[0,6)$.
Also, $f(6)=f(3 \times 2)=f(0)=0$, so $f(x)$ has 7 intersection points with the $x$-axis ... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. If $f(x)=\sum_{k=0}^{4034} a_{k} x^{k}$ is the expansion of $\left(x^{2}+x+2\right)^{2017}$, then $\sum_{k=0}^{1344}\left(2 a_{3 k}-a_{3 k+1}-a_{3 k+2}\right)=$ $\qquad$ | 8. 2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$. Then
$$
\begin{array}{l}
x^{2}+x+2=1 \\
\Rightarrow \sum_{k=0}^{1344}\left(a_{3 k}+a_{3 k+1} \omega+a_{3 k+2} \omega^{2}\right)=1
\end{array}
$$
Taking the conjugate of the above equation, we get
$$
\sum_{k=0}^{1344}\left(a_{3 k}+a_{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that $M N$ is a moving chord of the circumcircle of equilateral $\triangle A B C$ with side length $2 \sqrt{6}$, $M N=4$, and $P$ is a moving point on the sides of $\triangle A B C$. Then the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is | 9.4.
Let the circumcenter of $\triangle ABC$ be $O$.
It is easy to find that the radius of $\odot O$ is $r=2 \sqrt{2}$.
Also, $MN=4$, so $\triangle OMN$ is an isosceles right triangle, and
$$
\begin{aligned}
& \overrightarrow{O M} \cdot \overrightarrow{O N}=0,|\overrightarrow{O M}+\overrightarrow{O N}|=4 . \\
& \text ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. As shown in Figure 1, in $\triangle A B C$,
$$
\begin{array}{l}
\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}, \\
\overrightarrow{A H} \cdot \overrightarrow{B C}=0, \\
\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0 .
\end{array}
$$
Then the eccentricity of the hyperbola passing through point $C$ an... | 9. 2 .
Given $\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow(\overrightarrow{C B}-\overrightarrow{C A}) \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow A C=B C$.
From $\overrightarrow{A H} \cdot \overrightarrow{B C}=0 \Rightarrow A H \perp B C$.
Since $\cos \frac... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure $2, P$ is a point on the incircle of square $A B C D$, and let $\angle A P C=\alpha$, $\angle B P D=\beta$. Then $\tan ^{2} \alpha+\tan ^{2} \beta$ $=$ | 4. 8 .
As shown in Figure 5, establish a Cartesian coordinate system.
Let the equation of the circle be $x^{2}+y^{2}=r^{2}$.
Then the coordinates of the vertices of the square are
$$
A(-r,-r), B(r,-r), C(r, r), D(-r, r) \text {. }
$$
If $P(r \cos \theta, r \sin \theta)$, then the slopes of the lines $P A, P B, P C$, ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { Find the largest positive integer } n, \text { such that for positive real } \\
\text { numbers } \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}, \text { we have } \\
\quad \sum_{i=1}^{n} \frac{\alpha_{i}^{2}-\alpha_{i} \alpha_{i+1}}{\alpha_{i}^{2}+\alpha_{i+1}^{2}} \geqslant 0\left(\alpha_{n+1}... | Let $a_{i}=\frac{\alpha_{i+1}}{\alpha_{i}}(i=1,2, \cdots, n)$.
Then $\prod_{i=1}^{n} a_{i}=1$, and $a_{i}>0$.
The inequality to be proved becomes
$\sum_{i=1}^{n} \frac{1-a_{i}}{1+a_{i}^{2}} \geqslant 0$.
Let $x_{i}=\ln a_{i}(i=1,2, \cdots, n)$.
Then $\sum_{i=1}^{n} \frac{1-\mathrm{e}^{x_{i}}}{1+\left(\mathrm{e}^{x_{i}}... | 5 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. If $f(x)=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$, then the maximum value of $f(x)$ is $\qquad$ | $-、 1.11$.
From the fact that $f(x)$ is defined, we know
$$
0 \leqslant x \leqslant 13 \text {. }
$$
Then $\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
$$
\begin{array}{l}
=\sqrt{\left(6 \sqrt{\frac{x+27}{36}}+2 \sqrt{\frac{13-x}{4}}+3 \sqrt{\frac{x}{9}}\right)^{2}} \\
\leqslant \sqrt{(6+2+3)\left(6 \times \frac{x+27}{36}+2 \tim... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $a \in \mathbf{R}$, the equation ||$x-a|-a|=2$ has exactly three distinct roots. Then $a=$ $\qquad$ | 11. 2 .
The original equation can be transformed into $|x-a|=a \pm 2$.
For the equation to have exactly three distinct roots, then $a=2$. At this point, the equation has exactly three distinct roots:
$$
x_{1}=2, x_{2}=6, x_{3}=-2 \text {. }
$$
Thus, $a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Given that $a$, $b$, and $c$ are distinct integers. Then
$$
4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2}
$$
the minimum value is $\qquad$ | 15.8.
Notice,
$$
\begin{array}{l}
4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2} \\
=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}+a^{2}+b^{2}+c^{2},
\end{array}
$$
its minimum value is 8. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If three distinct real numbers $a$, $b$, and $c$ satisfy
$$
a^{3}+b^{3}+c^{3}=3 a b c \text {, }
$$
then $a+b+c=$ $\qquad$ | 2.0.
Notice,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\
=\frac{1}{2}(a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right) \\
=0 .
\end{array}
$$
Since $a, b, c$ are not all equal, we have
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \neq 0 \text {. }
$$
Therefore, $a+b+c=0... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a$ and $b$ are real numbers. If the quadratic function
$$
f(x)=x^{2}+a x+b
$$
satisfies $f(f(0))=f(f(1))=0$, and $f(0) \neq f(1)$, then the value of $f(2)$ is $\qquad$. | 3. 3 .
It is known that $f(0)=b, f(1)=1+a+b$ are both roots of the equation $f(x)=0$.
$$
\begin{array}{l}
\text { Then } x^{2}+a x+b \equiv(x-b)(x-(1+a+b)) \\
\Rightarrow a=-1-a-2 b, b=b(1+a+b) \\
\Rightarrow a=-\frac{1}{2}, b=0 \\
\Rightarrow f(2)=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Prove: The function
$$
f(x)=x^{x+2}-(x+2)^{x}-2 x(x+1)(x+2)+2
$$
has only one integer zero in the interval $[0,+\infty)$ | 9. In fact, it is only necessary to prove that there is only one positive integer satisfying $f(x)=0$.
When $x \in \mathbf{Z}_{+}, x \geqslant 5$,
$$
\begin{array}{l}
(x+3)^{x}=\left(\frac{x+3}{2}\right)^{4}(x+3)^{x-4} 4^{2} \\
& (x+3)^{x}-(x-2)^{x}-2 x(x+1)(x+2)+2 \\
> & (x+3)(x+2)^{x-1}-(x+2)^{x}- \\
& 2 x(x+1)(x+2)... | 3 | Algebra | proof | Yes | Yes | cn_contest | false |
13. In $\triangle A B C$, $\angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively. Let
$$
\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 |
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.
$$
\begin{array}{l}
\text { Given } S_{2015}=a_{1}+\sum_{k=1}^{1007}\left(a_{2 k}+a_{2 k+1}\right) \\
=a_{1}-1008 \\
\Rightarrow m+a_{1}-1008=-1007 \\
\Rightarrow m+a_{1}=1 . \\
\text { Also, } m a_{1}>0 \text {, so } m>0, a_{1}>0 . \\
\text { Therefore, } \frac{1}{a_{1}}+\frac{4}{m}=\left(m+a_{1}\right)\left(\fr... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The equation
$$
\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 |
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 $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
$$
\begin{array}{l}
\text { Then }|(a-2) z+a \mathrm{i}|^{2}-|a-2-a z \mathrm{i}|^{2} \\
=|(a-2) x+((a-2) y+a) \mathrm{i}|^{2}-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 On the plane, there are $n(n \geqslant 5)$ distinct points, each point is exactly at a distance of 1 from four other points. Find the minimum value of such $n$. [2] | Keep points $A, B, D, E, G$ in Figure 1, construct $\square E A B C, \square D A G F, \square B A G I$ to get points $C, F, I$, and construct $\square C I F H$ to get point $H$.
$$
\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 |
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 |
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 |
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 |
II. (25 points) Several boxes are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each box does not exceed 1 ton. To ensure that these boxes can be transported away in one go, the question is: what is the minimum number of trucks with a carrying capacity of 3 tons needed? | II. First, notice that the weight of each box does not exceed 1 ton. Therefore, the weight of boxes that each vehicle can transport at once will not be less than 2 tons. Otherwise, another box can be added.
Let $n$ be the number of vehicles needed, and the weights of the boxes transported by each vehicle be $a_{1}, a_... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
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