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14. (15 points) Let
$$
A=x^{4}+2 x^{3}-x^{2}-5 x+34 \text {. }
$$
Find the integer values of \( x \) for which \( A \) is a perfect square. | 14. Notice that,
$$
A=\left(x^{2}+x-1\right)^{2}-3(x-11) \text {. }
$$
So, when $x=11$, $A=131^{2}$ is a perfect square.
Next, we prove: there are no other integer $x$ that satisfy the condition.
(1) When $x>11$, we have $A>0$, thus, $A>\left(x^{2}+x-2\right)^{2}$.
Therefore, $\left(x^{2}+x-2\right)^{2}<A<\left(x^{2}+... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
381 A chess piece starts from one corner of an $8 \times 8$ chessboard, and moving one square in the row direction or column direction is called a step. After several steps, it passes through every square without repetition and returns to the starting point.
(1) Prove: the number of steps in the row direction is differ... | (1) Construct an $8 \times 8$ grid, and connect every adjacent point with a line segment to form a graph. Assume the distance between adjacent points is 1, and call a square with an area of 1 and vertices on the grid points a "cell."
The original problem is equivalent to: Prove that in every Hamiltonian cycle of the g... | 4 | Combinatorics | proof | Yes | Yes | cn_contest | false |
1. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$.
Then the integer part of $\sum_{n=0}^{201} \frac{1}{a_{n}+1}$ is $\qquad$.
(2011, National High School Mathematics League Gansu Province Preliminary) | Given: $\frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}$
$$
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\frac{1}{a_{0}}-\frac{1}{a_{2012}}=4-\frac{1}{a_{2012}} \text {. }
$$
Obviously, the sequence $\left\{a_{n}\right\}$ is monotonically increasing.
$$
\begin{array}{l}
\text { Also, } a_{1}=\frac{5}{16}, a_{2}... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (18 points) Given
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) .
$$
(1) If there exist positive numbers $a, b, c$ such that inequality (1) holds, prove: $k>5$;
(2) If there exist positive numbers $a, b, c$ such that inequality (1) holds, and any set of positive numbers $a, b, c$ that satisfy inequality (1) are... | 11. (1) Since $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, therefore, from inequality (1) we know
$$
k(a b+b c+c a)>5(a b+b c+c a) .
$$
Noting that $a, b, c$ are positive numbers.
Thus, $a b+b c+c a>0$.
Hence $k>5$.
(2) From (1) we know $k>5$.
Since $k$ is an integer, then $k \geqslant 6$.
Let $a=1, b=1, c=2$. It is ea... | 6 | Inequalities | proof | Yes | Yes | cn_contest | false |
11. (20 points) In the sequence $\left\{x_{n}\right\}$,
$$
x_{n}=p^{n}+q^{n}, x_{1}=1, x_{3}=4 \text {. }
$$
(1) Prove: $x_{n+2}=x_{n+1}+x_{n}$;
(2) Determine the last digit of $x_{2011}$, but no proof is required. | 11. (1) Notice,
$$
\begin{array}{l}
p+q=1, p^{3}+q^{3}=4, \\
p^{3}+q^{3}=(p+q)\left(p^{2}-p q+q^{2}\right) .
\end{array}
$$
Thus, $p^{2}-p q+q^{2}=4, p^{2}+2 p q+q^{2}=1$.
$$
\begin{array}{l}
\text { Then } x_{2}=p^{2}+q^{2}=3 \\
\Rightarrow p^{2}+(1-p)^{2}=3 \\
\Rightarrow p^{2}=p+1, q^{2}=q+1 \text {. }
\end{array}
... | 6 | Algebra | proof | Yes | Yes | cn_contest | false |
2. Given that for any real number $x$ we have $a \cos x + b \cos 2x \geqslant -1$.
Then the maximum value of $a + b$ is $\qquad$ | 2. 2 .
Let $x=\frac{2 \pi}{3}$, then $a+b \leqslant 2$.
When $a=\frac{4}{3}, b=\frac{2}{3}$,
$$
\begin{array}{l}
a \cos x+b \cos 2 x=\frac{4}{3} \cos x+\frac{2}{3} \cos 2 x \\
=\frac{4}{3} \cos ^{2} x+\frac{4}{3} \cos x-\frac{2}{3} \\
=\frac{4}{3}\left(\cos x+\frac{1}{2}\right)^{2}-1 \\
\geqslant-1 .
\end{array}
$$ | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the function
$$
f(x)=x^{3}-6 x^{2}+17 x-5 \text {, }
$$
real numbers $a, b$ satisfy $f(a)=3, f(b)=23$. Then $a+b=$ $\qquad$ | 4.4 .
Notice,
$$
\begin{array}{l}
f(x)=x^{3}-6 x^{2}+17 x-5 \\
=(x-2)^{3}+5(x-2)+13 \text {. } \\
\text { Let } g(y)=y^{3}+5 y .
\end{array}
$$
Then $g(y)$ is an odd function and monotonically increasing.
And $f(a)=(a-2)^{3}+5(a-2)+13=3$, so, $g(a-2)=-10$.
Also, $f(b)=(b-2)^{3}+5(b-2)+13=23$, then $g(b-2)=10$.
Thus $... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. The positive integer solutions of the equation $x+y^{2}+(x, y)^{3}=x y(x, y)$ are $\qquad$ groups $((x, y)$ represents the greatest common divisor of integers $x, y)$.
| 7.4.
Let $(x, y)=d$. Then $d^{2} \mid x$.
Let $x=a d^{2}, y=b d$. Then $(a d, b)=1$.
Thus, the original equation becomes
$$
a+b^{2}+d=a b d^{2} \text {. }
$$
Therefore, $b \mid(a+d) \Rightarrow b \leqslant a+d$.
Then $a+b^{2}+d=a b d^{2}$
$$
\begin{array}{l}
=(a+d) b+(a+d) b+b\left(a d^{2}-2 a-2 d\right) \\
\geqslant... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 1, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$ respectively. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
t... | 3. 2 .
Solution 1 Notice that,
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} .
$$
Since points $M, O, N$ are collinear, we have,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. }
$$
Solution 2 Since points $M, ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. (15 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram. | 12. Given that $\left|F_{1} F_{2}\right|=2$.
As shown in Figure 5, the inscribed quadrilateral $\square A B C D$ of the ellipse has a pair of opposite sides $B C$ and $A D$ passing through the foci $F_{1}$ and $F_{2}$, respectively.
Obviously, the center of symmetry of $\square A B C D$ is the origin.
Figure 5
Let th... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that the arithmetic mean of $\sin \theta$ and $\cos \theta$ is $\sin \alpha$, and the geometric mean is $\sin \beta$. Then $\cos 2 \alpha-\frac{1}{2} \cos 2 \beta=$ $\qquad$ . | 4. 0 .
From the given, we have
$$
\begin{array}{l}
\sin \alpha=\frac{\sin \theta+\cos \theta}{2}, \sin ^{2} \beta=\sin \theta \cdot \cos \theta. \\
\text { Therefore, } \cos 2 \alpha-\frac{1}{2} \cos 2 \beta \\
=1-2 \sin ^{2} \alpha-\frac{1}{2}\left(1-2 \sin ^{2} \beta\right) \\
=\frac{1}{2}-2 \sin ^{2} \alpha+\sin ^{... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If the edge length of the cube $A_{1} A_{2} A_{3} A_{4}-B_{1} B_{2} B_{3} B_{4}$ is 1, then the number of elements in the set
$$
\left\{x \mid x=\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}, i, j \in\{1,2,3,4\}\right\}
$$
is $\qquad$ | 5. 1 .
Solution 1 Note that,
$$
\begin{array}{l}
\overrightarrow{A_{1} B_{1}} \perp \overrightarrow{A_{i} A_{1}}, \overrightarrow{A_{1} B_{1}} \perp \overrightarrow{B_{1} B_{j}}(i, j \in\{2,3,4\}) \text {. } \\
\text { Hence } \overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}=\overrightarrow{A_{1} B_{1}}... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}^{2}=a_{n+1} a_{n}-1\left(n \in \mathbf{Z}_{+}\right) \text {, and } a_{1}=\sqrt{2} \text {. }
$$
Then the natural number closest to $\sqrt{a_{2014}}$ is $\qquad$ | 6.8.
From the given, we have
$$
\begin{array}{l}
a_{n+1}=a_{n}+\frac{1}{a_{n}} \Rightarrow a_{n+1}^{2}-a_{n}^{2}=2+\frac{1}{a_{n}^{2}} \\
\Rightarrow a_{n+1}^{2}=a_{1}^{2}+2 n+\sum_{i=1}^{n} \frac{1}{a_{i}^{2}} .
\end{array}
$$
Thus, $a_{2014}^{2}=2+2 \times 2013+\sum_{i=1}^{2013} \frac{1}{a_{i}^{2}}$
$$
>2+2 \times ... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=6$. Then the maximum value of $x+\sqrt{x y}+\sqrt[3]{x y z}$ is $\qquad$ . | 7.8.
By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{1}{4} x+y+4 z \geqslant 3 \sqrt[3]{x y z}, \\
\frac{3}{4} x+3 y \geqslant 3 \sqrt{x y} .
\end{array}
$$
Adding the two inequalities, we get
$$
\begin{array}{l}
x+4 y+4 z \geqslant 3 \sqrt{x y}+3 \sqrt[3]{x y z} \\
\Rightarrow x+\sqrt{x y}+\sqrt[3]{x y z}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$, $b$, and $c$ are constants, and for any real number $x$ we have
$$
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \text {. }
$$
then $a b c=$ . $\qquad$ | $$
\text { Two, 1. }-9 \text {. }
$$
From the problem, we know
$$
\begin{array}{c}
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \\
=x^{3}+(a+1) x^{2}+(a+b) x+b .
\end{array}
$$
Thus, $a+1=0, a+b=2, b=c$
$$
\begin{array}{l}
\Rightarrow a=-1, b=c=3 \\
\Rightarrow a b c=-9
\end{array}
$$ | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $S_{\triangle M B C}=4, 3 A B=2 B C$, draw the perpendicular from point $C$ to the angle bisector $B E$ of $\angle A B C$, and let the foot of the perpendicular be $D$. Then $S_{\triangle B D C}=$ $\qquad$ | 4.3.
Construct the reflection of point $C$ about $BD$ as point $F$, thus, $BC=BF$. Connect $FA$ and $FD$, then point $F$ lies on the extension of $BA$, and points $C$, $D$, and $F$ are collinear.
$$
\text{Therefore, } \frac{S_{\triangle ABC}}{S_{\triangle FBC}}=\frac{AB}{FB}=\frac{AB}{BC}=\frac{2}{3} \text{.}
$$
Sinc... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the function $y=f(x)$ is defined on $\mathbf{R}$, with a period of 3, and Figure 1 shows the graph of the function in the interval $[-2,1]$. Then $\frac{f(2014)}{f(5) f(15)}=$ $\qquad$ . | $$
\text { II,6. }-2 \text {. }
$$
From the problem statement and combining with the graph, we know that
$$
\frac{f(2014)}{f(5) f(15)}=\frac{f(1)}{f(-1) f(0)}=\frac{2}{(-1) \times 1}=-2 \text {. }
$$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. In $\triangle A B C$, it is known that
$$
\begin{array}{l}
|\overrightarrow{A B}|=\sqrt{3},|\overrightarrow{B C}|=1, \\
|\overrightarrow{A C}| \cos B=|\overrightarrow{B C}| \cos A \text {. } \\
\text { Then } \overrightarrow{A C} \cdot \overrightarrow{A B}=
\end{array}
$$ | 7.2.
Let the three sides of $\triangle A B C$ be $a, b, c$. From the given condition and using the Law of Sines, we have
$$
\begin{array}{l}
\sin B \cdot \cos B=\sin A \cdot \cos A \\
\Rightarrow \sin 2 B=\sin 2 A \\
\Rightarrow \angle B=\angle A \text { or } \angle B+\angle A=90^{\circ} .
\end{array}
$$
We will disc... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Given integers $a, b, c, d$. If the roots of the equation
$$
z^{4}+a z^{3}+b z^{2}+c z+d=0
$$
correspond to four points $A, B, C, D$ forming the vertices of a square in the complex plane, then the minimum value of the area of square $A B C D$ is $\qquad$ | 7. 2 .
Let the complex number corresponding to the center of the square be $m$. Then, after translating the origin of the complex plane to $m$, the vertices of the square are distributed on a circle, i.e., they are the solutions to the equation $(z-m)^{4}=n$ (where $n$ is a complex number).
From $z^{4}+a z^{3}+b z^{2... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Let $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}\left(n \in \mathbf{Z}_{+}\right)$. Find the smallest positive real number $\lambda$, such that for any $n \geqslant 2$, we have
$$
a_{n}^{2}<\lambda \sum_{k=1}^{n} \frac{a_{k}}{k} .
$$ | Sure, here is the translated text:
```
9. Notice that,
\[
\begin{array}{l}
a_{k}^{2}-a_{k-1}^{2}=\left(a_{k}-a_{k-1}\right)\left(a_{k}+a_{k-1}\right) \\
=\frac{1}{k}\left(2 a_{k}-\frac{1}{k}\right)(k \geqslant 2) .
\end{array}
\]
Then \(a_{n}^{2}-a_{1}^{2}=\sum_{k=2}^{n}\left(a_{k}^{2}-a_{k-1}^{2}\right)=2 \sum_{k=2}... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given that $AB$ is the major axis of the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and $CD$ is a chord of the ellipse $\Gamma$. The tangents at points $C$ and $D$ intersect at point $P$, the extension of $AD$ intersects the extension of $CB$ at point $E$, and the extension of $AC$ int... | 10. As shown in Figure 3, let the center of the ellipse be $O$. Connect $O P$, intersecting $C D$ at point $M$.
Let $C\left(x_{1}, y_{1}\right), D\left(x_{2}, y_{2}\right), P\left(x_{0}, y_{0}\right)$.
Then the equation of line $C D$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$.
From $\left\{\begin{array}{l}\frac... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
For $I$ being the excenter of $\triangle ABC$ with respect to $\angle A$, prove:
$$
\frac{A I^{2}}{C A \cdot A B}-\frac{B I^{2}}{A B \cdot B C}-\frac{C I^{2}}{B C \cdot C A}=1 .
$$ | Proof As shown in Figure 5, let $\odot I$ be tangent to $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively.
$$
\begin{array}{l}
\text { Let } \angle A I F=\alpha, \\
\angle B I D=\beta, \\
\angle C I E=\gamma .
\end{array}
$$
Then $\alpha=\beta+\gamma$.
Assume $I D=I E=I F=1$. Then $A I^{2}=\frac{1}{\cos... | 1 | Geometry | proof | Yes | Yes | cn_contest | false |
4. Try to determine the largest integer not exceeding $\frac{\sqrt{14}+2}{\sqrt{14}-2}$ | 4. 3 .
Notice that,
$$
\begin{array}{l}
\frac{\sqrt{14}+2}{\sqrt{14}-2}=\frac{(\sqrt{14}+2)^{2}}{(\sqrt{14}-2)(\sqrt{14}+2)} \\
=\frac{14+4+4 \sqrt{14}}{14-4}=\frac{18+4 \sqrt{14}}{10} .
\end{array}
$$
And $3<\sqrt{14}<4$, thus, $30<18+4 \sqrt{14}<34$.
Therefore, $3<\frac{18+4 \sqrt{14}}{10}<3.4$.
Hence, the largest ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given that vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ are two mutually perpendicular unit vectors in a plane, and
$$
(3 \alpha-\gamma) \cdot(4 \beta-\gamma)=0 .
$$
Then the maximum value of $|\boldsymbol{\gamma}|$ is . $\qquad$ | 11.5.
As shown in Figure 2, let
$$
\begin{array}{l}
\overrightarrow{O A}=3 \alpha, \\
\overrightarrow{O B}=4 \beta, \\
\overrightarrow{O C}=\gamma .
\end{array}
$$
From the given information, $\overrightarrow{A C} \perp \overrightarrow{B C}$. Therefore, point $C$ lies on the circle with $A B$ as its diameter, and thi... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Given $k$ as a positive integer, the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=3, a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$.
Let $b_{n}=\frac{1}{n} \log _{3} a_{1}... | 15. From the problem, we know
Also, $a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3$,
$$
a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n-1}+3(n \geqslant 2) \text {, }
$$
Thus, $a_{n+1}-a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) a_{n}$
$\Rightarrow a_{n+1}=3^{\frac{2}{2 k-1}} a_{n}$
$$
\Rightarrow a_{n}=a_{2}\left(3^... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The value of the complex number $\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{6 n}\left(n \in \mathbf{Z}_{+}\right)$ is | 3. 1 .
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{6 n}=\left[\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{3}\right]^{2 n} \\
=\left(\frac{1}{8}+\frac{3}{4} \times \frac{\sqrt{3}}{2} i-\frac{3}{2} \times \frac{3}{4}-\frac{3 \sqrt{3}}{8} i\right) \\
=(-1)^{2 n}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $(\sqrt{2}+1)^{21}=a+b \sqrt{2}$, where $a, b$ are positive integers. Then $(b, 27)=$ $\qquad$ | 8. 1 .
Notice,
$$
\begin{array}{l}
(\sqrt{2}+1)^{21} \\
=(\sqrt{2})^{21}+\mathrm{C}_{21}^{1}(\sqrt{2})^{20}+\mathrm{C}_{21}^{2}(\sqrt{2})^{19}+\cdots+ \\
\mathrm{C}_{21}^{20} \sqrt{2}+\mathrm{C}_{21}^{21}, \\
(\sqrt{2}-1)^{21} \\
=(\sqrt{2})^{21}-\mathrm{C}_{21}^{1}(\sqrt{2})^{20}+\mathrm{C}_{21}^{2}(\sqrt{2})^{19}-\c... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If $a+b=\sqrt{5}$, then
$$
\frac{a^{4}+a^{2} b^{2}+b^{4}}{a^{2}+a b+b^{2}}+3 a b=(\quad) \text {. }
$$
(A) 5
(B) $\frac{3 \sqrt{5}}{2}$
(C) $2 \sqrt{5}$
(D) $\frac{5 \sqrt{5}}{2}$ | $\begin{array}{l}\text {-1. A. } \\ \frac{a^{4}+a^{2} b^{2}+b^{4}}{a^{2}+a b+b^{2}}+3 a b \\ =\frac{\left(a^{2}+a b+b^{2}\right)\left(a^{2}-a b+b^{2}\right)}{a^{2}+a b+b^{2}}+3 a b \\ =\left(a^{2}-a b+b^{2}\right)+3 a b \\ =(a+b)^{2}=5 .\end{array}$ | 5 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. Given that $M$ is the least common multiple of 15 consecutive natural numbers $1,2, \cdots, 15$. If a divisor of $M$ is divisible by exactly 14 of these 15 natural numbers, it is called a "good number" of $M$. Then the number of good numbers of $M$ is $\qquad$.
| 4.4.
It is known that $M=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13$.
Since $2 \times 11, 2 \times 13$ are both greater than 15, therefore,
$$
\begin{array}{l}
\frac{M}{11}=2^{3} \times 3^{2} \times 5 \times 7 \times 13, \\
\frac{M}{13}=2^{3} \times 3^{2} \times 5 \times 7 \times 11,
\end{array}
$$
each... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Find all positive integers $n$, such that there exists an integer-coefficient polynomial $P(x)$, satisfying $P(d)=\left(\frac{n}{d}\right)^{2}$, where, $d$ is each divisor of $n$. [3] | For a polynomial $P(x)$ with integer coefficients, and positive integers $a, b (a \neq b)$, we always have
$$
(a-b) \mid (P(a)-P(b)).
$$
This is an invariant.
If $n=1$, then $P(1)=1$. Hence, we can take the polynomial $P(x)=x$.
If $n$ is a prime number, then $n$ has only two factors, 1 and $n$. The polynomial $P(x)$ ... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If positive real numbers $a, b$ satisfy
$$
\log _{8} a+\log _{4} b^{2}=5, \log _{8} b+\log _{4} a^{2}=7 \text {, }
$$
then $\log _{4} a+\log _{8} b=$ $\qquad$ | $=1.4$.
Let $a=2^{x}, b=2^{y}$. Then $\frac{x}{3}+y=5, \frac{y}{3}+x=7$.
Thus, $x=6, y=3$.
Therefore, $\log _{4} a+\log _{8} b=\frac{x}{2}+\frac{y}{3}=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Find the unit digit of $(2+\sqrt{3})^{2013}$. | Notice,
$$
\begin{array}{l}
{\left[(2+\sqrt{3})^{2013}\right]=(2+\sqrt{3})^{2013}+(2-\sqrt{3})^{2013}-1 .} \\
\text { Let } a_{n}=(2+\sqrt{3})^{2013}+(2-\sqrt{3})^{2013}=4 a_{n-1}-a_{n-2}, \\
a_{0}=2, a_{1}=4 .
\end{array}
$$
Then $2 \mid a_{n}$,
$$
\begin{array}{l}
a_{n} \equiv 2,4,4,2,4,4,2, \cdots(\bmod 5) \\
\Righ... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $a$ and $b$ are two different positive integers. Ask:
$$
\begin{array}{l}
a(a+2), a b, a(b+2), (a+2) b, \\
(a+2)(b+2), b(b+2)
\end{array}
$$
Among these six numbers, what is the maximum number of perfect squares? | 2. At most two numbers are perfect squares (such as $a=2, b=16$). Notice that,
$$
\begin{array}{l}
a(a+2)=(a+1)^{2}-1, \\
b(b+2)=(b+1)^{2}-1
\end{array}
$$
cannot be perfect squares;
$$
a b \cdot a(b+2)=a^{2}\left(b^{2}+2 b\right)
$$
is not a perfect square, so at most one of $a b$ and $a(b+2)$ is a perfect square.
... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 There are three piles of stones. Each time, A moves one stone from one pile to another, and A can receive a reward from B for each move, which is equal to the difference between the number of stones in the pile to which A moves the stone and the number of stones in the pile from which A moves the stone. If th... | 【Analysis】Due to the uncertainty of A's operations, it is necessary to start from the whole and establish a "substitute" for the increase or decrease of A's reward each time, which must be simple to calculate.
Solution A's reward is 0.
In fact, the three piles of stones can be imagined as three complete graphs (each st... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $n(n \geqslant 4)$ be a positive integer. $n$ players each play a table tennis match against every other player (each match has a winner and a loser). Find the minimum value of $n$ such that after all the matches, there always exists an ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$, satisfying that w... | 2. First, prove: when $n=8$, there always exists an ordered quartet that satisfies the problem's conditions.
Since 8 players have played a total of $\mathrm{C}_{8}^{2}=28$ matches, there must be at least one player who has won at least $\left\lceil\frac{28}{8}\right\rceil=4$ matches (where $\left\lceil x \right\rceil$... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given distinct complex numbers $a$ and $b$ satisfying $a b \neq 0$, the set $\{a, b\}=\left\{a^{2}, b^{2}\right\}$. Then $a+b=$ $\qquad$ . | $-1 .-1$.
If $a=a^{2}, b=b^{2}$, by $a b \neq 0$, we get $a=b=1$, which is a contradiction. If $a=b^{2}, b=a^{2}$, by $a b \neq 0$, we get $a^{3}=1$.
Clearly, $a \neq 1$.
Thus, $a^{2}+a+1=0 \Rightarrow a=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$.
Similarly, $b=-\frac{1}{2} \mp \frac{\sqrt{3}}{2} \mathrm{i}$.
Ther... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive integer $a$ such that the function
$$
f(x)=x+\sqrt{13-2 a x}
$$
has a maximum value that is also a positive integer. Then the maximum value of the function is $\qquad$ . | 2.7.
Let $t=\sqrt{13-2 a x} \geqslant 0$. Then
$$
\begin{aligned}
y= & f(x)=\frac{13-t^{2}}{2 a}+t \\
& =-\frac{1}{2 a}(t-a)^{2}+\frac{1}{2}\left(a+\frac{13}{a}\right) .
\end{aligned}
$$
Since $a$ is a positive integer, $y_{\max }=\frac{1}{2}\left(a+\frac{13}{a}\right)$ is also a positive integer, so, $y_{\max }=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In the positive geometric sequence $\left\{a_{n}\right\}$,
$$
a_{5}=\frac{1}{2}, a_{6}+a_{7}=3 \text {. }
$$
Then the maximum positive integer $n$ that satisfies $a_{1}+a_{2}+\cdots+a_{n}>a_{1} a_{2} \cdots a_{n}$ is $\qquad$ | 3. 12 .
According to the problem, $\frac{a_{6}+a_{7}}{a_{5}}=q+q^{2}=6$.
Since $a_{n}>0$, we have $q=2, a_{n}=2^{n-6}$.
Thus, $2^{-5}\left(2^{n}-1\right)>2^{\frac{n(n-11)}{2}} \Rightarrow 2^{n}-1>2^{\frac{n(n-11)}{2}+5}$.
Estimating $n>\frac{n(n-11)}{2}+5$, we get $n_{\max }=12$.
Upon verification, $n=12$ meets the re... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. As shown in Figure 1, given a regular tetrahedron $P-A B C$ with all edge lengths equal to 4, points $D, E, F$ are on edges $P A, P B, P C$ respectively. Then the number of $\triangle D E F$ that satisfy $D E = E F = 3, D F = 2$ is $\qquad$. | 7.3.
Let $P D=x, P E=y, P F=z$. Then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-x y=9, \\
y^{2}+z^{2}-y z=9, \\
z^{2}+x^{2}-z x=4 .
\end{array}\right.
$$
(1) - (2) gives $x=z$ or $x+z=y$.
When $x=z$, we get $x=z=2, y=1+\sqrt{6}$;
When $x+z=y$, $x z=\frac{5}{2}, x^{2}+z^{2}=\frac{13}{2}$, there are two sets of solutions. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the function
$$
f(x)=a \sin x+b \cos x \quad(a, b \in \mathbf{Z}),
$$
and it satisfies
$$
\{x \mid f(x)=0\}=\{x \mid f(f(x))=0\} .
$$
Then the maximum value of $a$ is . $\qquad$ | 2.3.
Let $A=\{x \mid f(x)=0\}, B=\{x \mid f(f(x))=0\}$.
Obviously, set $A$ is non-empty.
Take $x_{0} \in A$, i.e., $x_{0} \in B$, hence
$$
b=f(0)=f\left(f\left(x_{0}\right)\right)=0 \text {. }
$$
Thus, $f(x)=a \sin x(a \in \mathbf{Z})$.
When $a=0$, obviously, $A=B$.
Now assume $a \neq 0$, in this case,
$$
\begin{array... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let any real numbers $a>b>c>d>0$. To make
$$
\begin{array}{l}
\log _{\frac{b}{a}} 2014+\log _{\frac{d}{b}} 2014+\log _{\frac{d}{c}} 2014 \\
\geqslant m \log _{\frac{d}{a}} 2014
\end{array}
$$
always hold, then the minimum value of $m$ is $\qquad$. | 5.9.
$$
\begin{array}{l}
\text { Let } x_{1}=-\log _{2014} \frac{b}{a}, x_{2}=-\log _{2014} \frac{c}{b}, \\
x_{3}=-\log _{2014} \frac{d}{c} .
\end{array}
$$
Since $a>b>c>d>0$, we have
$$
x_{1}>0, x_{2}>0, x_{3}>0 \text {. }
$$
Thus, the given inequality can be transformed into
$$
\begin{array}{l}
\frac{1}{x_{1}}+\fra... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let \( x, y, z \) all be positive real numbers, and
\[
x+y+z=1 \text{. }
\]
Find the minimum value of the function
\[
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}
\]
and provide a proof. | 11. Notice that, $\frac{3 x^{2}-x}{1+x^{2}}=\frac{x(3 x-1)}{1+x^{2}}$.
Consider the function $g(t)=\frac{t}{1+t^{2}}$.
It is easy to see that $g(t)$ is an odd function.
Since when $t>0$, $\frac{1}{t}+t$ is decreasing in the interval $(0,1)$, hence $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in the interval $(0,1)$.
T... | 0 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 2: Twelve acrobats numbered $1, 2, \cdots, 12$ are divided into two groups, $A$ and $B$, each with six people. Let the actors in group $A$ form a circle, and each actor in group $B$ stand on the shoulders of two adjacent actors in the $A$ circle. If the number of each actor in the $B$ circle is equal to the sum... | Let the sum of the elements in groups $A$ and $B$ be denoted as $x$ and $y$ respectively.
Then $y = 2x$.
Therefore, $3x = x + y = 1 + 2 + \cdots + 12 = 78 \Rightarrow x = 26$.
Clearly, $1, 2 \in A, 11, 12 \in B$.
Let $A = \{1, 2, a, b, c, d\} (a < b < c < d)$.
If $d \leq 7$, then
$$
a + b + c + d \leq 4 + 5 + 6 + 7 = 2... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x, y$ be positive real numbers. Find the minimum value of
$$
x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}
$$
(Gu Bin, Jin Aiguo) | 1. Let $f(x, y)=x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}$.
If $x \geqslant 1, y \geqslant 1$, then
$$
f(x, y) \geqslant x+y \geqslant 2 \text {; }
$$
If $0<x<1, 0<y<1$, then
$$
f(x, y) = x + y + \frac{1-x}{y} + \frac{1-y}{x} = \left(x + \frac{1-x}{y}\right) + \left(y + \frac{1-y}{x}\right) \geqslant 2.
$$
For $0<x<1, y \... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Four, in an election, there are 12 candidates, and each member of the electoral committee casts 6 votes. It is known that any two members' votes have at most 2 candidates in common. Find the maximum number of members in the committee.
(Proposed by the Problem Committee) | The maximum number of committee members is 4.
Let the number of committee members be $k$, and the candidates be represented by $1,2, \cdots, 12$. Each person's vote is a set $A_{i}(1 \leqslant i \leqslant k)$, and the number of votes each candidate receives is $m_{i}(1 \leqslant i \leqslant 12)$. Then,
$$
\sum_{i=1}^{1... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Rational numbers $a, b$ have decimal expansions that are repeating decimals with the smallest period of 30. It is known that the decimal expansions of $a-b$ and $a+kb$ have the smallest period of 15. Find the smallest possible value of the positive integer $k$.
| 3. $k_{\min }=6$.
Notice that, $a, b, a-b, a+k b$ can simultaneously become pure repeating decimals by multiplying them by a power of 10. Therefore, assume they are all pure repeating decimals.
Since the decimal part of a rational number is a pure repeating decimal with a period of $T$ if and only if it can be writte... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find all possible values of $k$.
Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find ... | Three, let the equation $x^{2}+p x+k p-1=0$ have integer roots $x_{1}$ and another root $x_{2}$.
By the relationship between roots and coefficients, we know
$$
x_{1}+x_{2}=-p, x_{1} x_{2}=k p-1 \text {. }
$$
Thus, $x_{2}$ must also be an integer.
Assume $k>1$.
Notice,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If positive numbers $a, b, c$ satisfy
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=3 \text {, }
$$
find the value of the algebraic expression
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 ... | Notice,
$$
\begin{array}{l}
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}-1 \\
=\left(\frac{b^{2}+c^{2}-a^{2}+2 b c}{2 b c}\right)\left(\frac{b^{2}+c^{2}-a^{2}-2 b c}{2 b c}\right) \\
=\frac{(b+c+a)(b+c-a)(b-c+a)(b-c-a)}{4 b^{2} c^{2}} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\left(\frac{c^{2}+a^{2}-b^{2}}{2... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given real numbers $x, y, z$ satisfy
$$
x+\frac{1}{y}=4, y+\frac{1}{z}=1, z+\frac{1}{x}=\frac{7}{3} \text {. }
$$
Find the value of $x y z$. | Multiplying the three conditional expressions yields
$$
\frac{28}{3}=x y z+\frac{1}{x y z}+\frac{22}{3} \Rightarrow x y z=1 .
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $x, y, z$ satisfy
$$
\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}=0 \text {. }
$$
Find the value of the algebraic expression $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$. | Notice,
$$
\begin{array}{l}
x+y+z \\
=\left(\frac{x^{2}}{y+z}+x\right)+\left(\frac{y^{2}}{z+x}+y\right)+\left(\frac{z^{2}}{x+y}+z\right) \\
=(x+y+z)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right) .
\end{array}
$$
Thus, $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=-3$ or 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let $a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2$. Find the value of the algebraic expression $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}$. | Solve: Regarding $a$ and $b$ as the main elements, then
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b=2-c, \\
a^{2}+b^{2}=2-c^{2} .
\end{array}\right. \\
\text { Hence } a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2} \\
=c^{2}-2 c+1=(c-1)^{2} \\
\Rightarrow \frac{(1-c)^{2}}{a b}=1 .
\end{array}
$$
Similarly, $\fra... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{3}-x y z=-5, \\
y^{3}-x y z=2, \\
z^{3}-x y z=21 .
\end{array}\right.
$$
Find the value of $x+y+z$. | Let $x y z=k$. Then
$$
\left\{\begin{array}{l}
x^{3}=k-5, \\
y^{3}=k+2, \\
z^{3}=k+21 .
\end{array}\right.
$$
Multiplying the above three equations, we get
$$
\begin{array}{l}
k^{3}=(x y z)^{3}=(k-5)(k+2)(k+21) \\
\Rightarrow 18 k^{2}-73 k-210=0 \\
\Rightarrow k_{1}=6, k_{2}=-\frac{35}{18} \text { (not suitable, disca... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The arithmetic sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}+a_{2}+\cdots+a_{14}=77 \text {, and } a_{1} 、 a_{11} \in \mathbf{Z}_{+} \text {. }
$$
Then $a_{18}=$ $\qquad$ | 2. -5 .
From the formula for the sum of an arithmetic sequence, we get
$$
\begin{array}{l}
a_{1}+a_{14}=11 \Rightarrow 2 a_{1}+13 \times \frac{a_{11}-a_{1}}{10}=11 \\
\Rightarrow 7 a_{1}+13 a_{11}=110 \\
\Rightarrow a_{1}=12(\bmod 13), a_{11}=2(\bmod 7) \\
\Rightarrow\left(a_{1}, a_{11}\right)=(12,2) \\
\Rightarrow a_... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the function
$$
f(x)=x \log _{2} x+(a-x) \log _{2}(a-x)
$$
be symmetric about the line $x=\frac{1}{2}$. Then for any real numbers $x_{i} \in(0,1)(1 \leqslant i \leqslant 4)$ satisfying $\sum_{i=1}^{4} x_{i}=1$, the minimum value of $s=\sum_{i=1}^{4} x_{i} \log _{2} x_{i}$ is . $\qquad$ | 5. -2 .
From the problem, we know that the midpoint of the interval $(0, a)$ is $\frac{1}{2}$.
Thus, $a=1$.
Then, $f(x)=x \log _{2} x+(1-x) \log _{2}(1-x)$
$\Rightarrow f^{\prime}(x)=\log _{2} \frac{x}{1-x}$.
Let $f^{\prime}(x)=0$, we get $x=\frac{1}{2}$.
For any $x \in\left(0, \frac{1}{2}\right)$, $f^{\prime}(x) < 0$... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given
$$
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}, \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}, \frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} \text {. }
$$
Find the value of $\frac{2}{x}+\frac{3}{y}+\frac{4}{z}$. | $$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{x+y+z}{x(y+z)}=\frac{1}{2} \\
\Rightarrow \frac{2}{x}=\frac{y+z}{x+y+z} .
\end{array}
$$
Similarly, $\frac{3}{y}=\frac{z+x}{x+y+z}, \frac{4}{z}=\frac{x+y}{x+y+z}$.
Adding the three equations yields $\frac{2}{x}+\frac{3}{y}+\frac{4}{z}=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b+c=1, \\
\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}=1 .
\end{array}
$$
Then $a b c=$ | $$
\begin{aligned}
& \frac{1}{1-2 c}+\frac{1}{1-2 a}+\frac{1}{1-2 b}=1 \\
\Rightarrow & (1-2 a)(1-2 b)+(1-2 b)(1-2 c)+(1-2 a)(1-2 c) \\
& =(1-2 a)(1-2 b)(1-2 c) \\
\Rightarrow & 2-2(a+b+c)=8 a b c \\
\Rightarrow & a b c=0 .
\end{aligned}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (16 points) Find all natural numbers $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square. | Let $N$ be the square of the desired natural number.
Below, we discuss different cases.
(1) When $n \leqslant 8$,
$$
N=2^{n}\left(2^{8-n}+2^{11-n}+1\right) \text {. }
$$
Since the result inside the parentheses is odd, for $N$ to be a square number, $n$ must be even.
By verifying $n=2,4,6,8$ one by one, we find that $... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. Given a regular tetrahedron $P-ABC$, points $P, A, B, C$ are all on a sphere with radius $\sqrt{3}$. If $PA, PB, PC$ are mutually perpendicular, then the distance from the center of the sphere to the plane $ABC$ is $\qquad$. | $=13 \cdot \frac{\sqrt{3}}{3}$.
Let $P A=P B=P C=x$. Then $A B=\sqrt{2} x$.
Let the centroid of $\triangle A B C$ be $M$, and the center of the circumscribed sphere of the regular tetrahedron be $O, O M=y$. Then
$$
\left.\begin{array}{l}
A M=\frac{2 \sqrt{3}}{3} \times \frac{\sqrt{2}}{2} x=\frac{\sqrt{6}}{3} x .
\end{a... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. In the right trapezoid $A B C D$, it is known that $A D \perp A B$, $A B / / D C, A B=4, A D=D C=2$. Let $N$ be the midpoint of side $D C$, and $M$ be a moving point within or on the boundary of trapezoid $A B C D$. Then the maximum value of $\overrightarrow{A M} \cdot \overrightarrow{A N}$ is | 8. 6 . | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram. ${ }^{[4]}$
(2013, National High School Mathematics League Shandong Province Preliminary... | Solve the general ellipse:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
Corresponding problems.
With $F_{1}$ as the pole and $F_{1} x$ as the polar axis, the equation of the ellipse is transformed into
$$
\rho=\frac{e p}{1+e \cos \theta}=\frac{b^{2}}{a-c \cos \theta} .
$$
Therefore, the length of the chord ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: x^{2}-\frac{y^{2}}{24}=1$, respectively, and let $P$ be a point on the hyperbola $C$ in the first quadrant. If $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{4}{3}$, then the radius of the incircle of $\triangle P F_{1} F_{2}$ is . $\q... | 4. 2 .
Let $\left|P F_{1}\right|=4 t$. Then $\left|P F_{2}\right|=3 t$.
Thus $4 t-3 t=\left|P F_{1}\right|-\left|P F_{2}\right|=2$
$$
\Rightarrow t=2,\left|P F_{1}\right|=8,\left|P F_{2}\right|=6 \text {. }
$$
Combining with $\left|F_{1} F_{2}\right|=10$, we know that $\triangle P F_{1} F_{2}$ is a right triangle, $P... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given points $A(1,-1), B(4,0), C(2,2)$, the plane region $D$ consists of all points $P(x, y)$ that satisfy
$$
\overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}(1<\lambda \leqslant a, 1<\mu \leqslant b)
$$
If the area of region $D$ is 8, then the minimum value of $a+b$ is . $\qquad$ | 8. 4 .
As shown in Figure 3, extend $A B$ to point $N$, and extend $A C$ to point $M$, such that
$$
|A N|=a|A B|,|A M|=b|A C| .
$$
Construct $\square A B E C$ and $\square A N G M$. Then quadrilateral $E H G F$ is a parallelogram.
From the conditions, the region $D$ composed of points $P(x, y)$ is the shaded area in... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. A certain meeting has 30 participants, each of whom knows at most five others; among any five people, at least two are not acquaintances. Find the largest positive integer $k$, such that in any group of 30 people satisfying the above conditions, there always exists a group of $k$ people, none of whom are acquaintanc... | 5, hence we have
$30-|X| \leqslant 5|X|$.
Therefore, $|X| \geqslant 5$.
If $|X|=5$, then by equation (1) the equality is achieved, which means the 25 edges are distributed among the five vertices in set $X$, i.e., the neighborhood of each vertex in set $X$ is a set of five points in set $V \backslash X$.
Since $|V \ba... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y>0 \text {. }
$$
Then the minimum value of $x+y$ is $\qquad$. | 2. 2 .
Notice,
$$
\begin{array}{l}
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right) \geqslant 1 \\
\Rightarrow 2 x+\sqrt{4 x^{2}+1} \\
\geqslant \frac{1}{\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}}=\sqrt{1+\frac{4}{y^{2}}}+\frac{2}{y} .
\end{array}
$$
When $x=\frac{1}{y}$,
$$
\left(2 x+\sq... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 2. If } \frac{y}{x}+\frac{x}{z}=a, \frac{z}{y}+\frac{y}{x}=b, \frac{x}{z}+\frac{z}{y}=c, \\ \text { then }(b+c-a)(c+a-b)(a+b-c)=\end{array}$ | 2.8.
Notice that,
$$
\begin{array}{l}
b+c-a=\left(\frac{z}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{y}\right)-\left(\frac{y}{x}+\frac{x}{z}\right) \\
=\frac{2 z}{y} .
\end{array}
$$
Similarly, $c+a-b=\frac{2 x}{z}, a+b-c=\frac{2 y}{x}$.
Therefore, $(b+c-a)(c+a-b)(a+b-c)$ $=\frac{2 z}{y} \cdot \frac{2 x}{z} \c... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. In the Cartesian coordinate plane $x O y$, the area of the region determined by the system of inequalities
$$
\left\{\begin{array}{l}
|x| \leqslant 2, \\
|y| \leqslant 2, \\
|| x|-| y|| \leqslant 1
\end{array}\right.
$$
is $\qquad$ | 6. 12 .
Obviously, the region is symmetric with respect to both the $x$-axis and the $y$-axis. Therefore, we can assume $x \geqslant 0, y \geqslant 0$.
From $\left\{\begin{array}{l}0 \leqslant x \leqslant 2, \\ 0 \leqslant y \leqslant 2, \\ -1 \leqslant x-y \leqslant 1\end{array}\right.$, draw the part of the region ... | 12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
7. Given
$$
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \text {. }
$$
Then $f(20)+f(14)=$ $\qquad$ . | $$
\begin{array}{l}
\text { II.7.0. } \\
\text { Let } g(x)=x^{2}-53 x+196 \\
=(x-4)(x-49)
\end{array}
$$
Then when $x<4$ or $x>49$, $g(x)>0$;
when $4 \leqslant x \leqslant 49$, $g(x) \leqslant 0$.
Thus, when $x=4,5, \cdots, 49$,
$$
f(x)=g(x)+|g(x)|=g(x)-g(x)=0 \text {. }
$$
Therefore, $f(20)+f(14)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { 4. If } x_{1}>x_{2}>x_{3}>x_{4}>0, \text { and the inequality } \\
\log _{\frac{x_{1}}{2}} 2014+\log _{\frac{x_{2}}{x_{3}}} 2014+\log _{x_{4} \frac{1}{4}} 2014 \\
\geqslant k \log _{\frac{x_{1}}{}} 2014
\end{array}
$$
always holds, then the maximum value of the real number $k$ is . $\qquad$ | 4.9.
From the given, we have
$$
\begin{array}{l}
\frac{\ln 2014}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln 2014}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln 2014}{\ln \frac{x_{3}}{x_{4}}} \\
\geqslant k \frac{\ln 2014}{\ln \frac{x_{1}}{x_{4}}}.
\end{array}
$$
Since $x_{1}>x_{2}>x_{3}>x_{4}>0$, we have
$$
\begin{array}{l}
\ln \frac{x... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. If $x, y, z$ are real numbers, satisfying
$$
x+\frac{1}{y}=2 y+\frac{2}{z}=3 z+\frac{3}{x}=k \text{, and } x y z=3 \text{, }
$$
then $k=$ | 4. 4 .
Multiplying the three expressions yields
$$
\begin{array}{l}
k^{3}=6\left[x y z+\left(x+\frac{1}{y}\right)+\left(y+\frac{1}{z}\right)+\left(z+\frac{1}{x}\right)+\frac{1}{x y z}\right] \\
=6\left(3+k+\frac{k}{2}+\frac{k}{3}+\frac{1}{3}\right) \\
\Rightarrow k^{3}-11 k-20=0 \\
\Rightarrow(k-4)\left(k^{2}+4 k+5\ri... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) If $x, y \in [0,1]$, try to find the maximum value of
$$
x \sqrt{1-y} + y \sqrt{1-x}
$$ | $$
\begin{array}{l}
\text { I. Let } s=\sqrt{1-x}, t=\sqrt{1-y}, \\
z=x \sqrt{1-y}+y \sqrt{1-x} .
\end{array}
$$
Then $z=\left(1-s^{2}\right) t+\left(1-t^{2}\right) s=(s+t)(1-s t)$.
Also, $(1-s)(1-t)=1-(s+t)+s t \geqslant 0$, which means $s+t \leqslant 1+s t$,
thus $z \leqslant(1+s t)(1-s t)=1-s^{2} t^{2} \leqslant 1$... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
7. If a non-zero complex number $x$ satisfies $x+\frac{1}{x}=1$, then $x^{2014}+\frac{1}{x^{2014}}=$ $\qquad$ | $=, 7 .-1$. Therefore $x^{2014}+\frac{1}{x^{2014}}=2 \cos \frac{4 \pi}{3}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{4028} x^{4028}$ is the expansion of $\left(x^{2}+x+2\right)^{2014}$, then
$$
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+2 a_{4026}-a_{4007}-a_{4028}
$$
is $\qquad$ | 6.2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$.
Then $x^{2}+x+2=1$.
Therefore, $a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3}+a_{4} \omega+\cdots+$ $a_{4026}+a_{4027} \omega+a_{4028} \omega^{2}=1$.
Taking the conjugate of the above equation, we get
$$
\begin{array}{l}
a_{0}+a_{1} \omega... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. A line $l$ is drawn through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersecting the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly three lines $l$, then $\lambda=$ $\qquad$. | 2.4.
Since there are an odd number of lines that satisfy the condition, by symmetry, the line perpendicular to the $x$-axis satisfies the condition, at this time,
$$
x=\sqrt{3}, y= \pm 2, \lambda=|A B|=4 \text {. }
$$ | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. In an All-Star basketball game, 27 players participate, each wearing a jersey with their favorite number, which is a non-negative integer. After the game, they line up in a 3-row, 9-column formation for fans to take photos. An eccentric fan only takes photos where the players in the frame form a rectangle of $a$ row... | Prompt: $s_{\text {min }}=2$.
Assume at most one player is photographed.
By symmetry, without loss of generality, assume no players in the first row are photographed.
For $i=1,2, \cdots, 9$, let the players in the $i$-th column of the 1st, 2nd, and 3rd rows have jersey numbers $a_{i}, b_{i}, c_{i}$, respectively, and ... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given
\[
\begin{aligned}
f(x, y)= & x^{3}+y^{3}+x^{2} y+x y^{2}- \\
& 3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{aligned}
\]
and \( x, y \geqslant \frac{1}{2} \). Find the minimum value of \( f(x, y) \). ${ }^{[4]}$
(2011, Hebei Province High School Mathematics Competition) | 【Analysis】This problem involves finding the extremum of a bivariate function, with a very complex expression, making it difficult to find a breakthrough. Observing the symmetry in the expression, we can attempt the following transformation.
First, when $x \neq y$, multiply both sides of the function by $x-y$, yielding... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given the sequence $\left\{a_{n}\right\}$:
$$
a_{1}=2, a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}(n \geqslant 1) \text {. }
$$
Determine the periodicity of the sequence $\left\{a_{n}\right\}$. | Solving, from $a_{1}=2$, we get
$$
a_{2}=\frac{5 a_{1}-13}{3 a_{1}-7}=3, a_{3}=\frac{5 a_{2}-13}{3 a_{2}-7}=1,
$$
which means $a_{1}, a_{2}, a_{3}$ are all distinct.
From the given conditions, we have
$$
\begin{array}{l}
a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}, \\
a_{n+2}=\frac{5 a_{n+1}-13}{3 a_{n+1}-7}=\frac{7 a_{n}-13... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2 . \\
\text { Then } \frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}=
\end{array}
$$ | 7.3.
From the given, we have $a b+b c+c a=1$.
Then $b c=1-a b-a c=1-a(b+c)$ $=1-a(2-a)=(1-a)^{2}$.
Thus, the desired value is 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
435 Given a convex polygon $F$, consider all the figures that are positively homothetic to the convex polygon $F$ and smaller than $F$. Let $n(F)$ be the minimum number of such figures (allowing translation but not rotation) needed to cover the convex polygon $F$. Find the value of $n(F)$. | (1) On the one hand, let the convex polygon $F$ be the parallelogram $ABCD$. It is easy to see that the convex polygon $F$ can be covered by four smaller parallelograms that are similar to the convex polygon $F$ (as shown in Figure 3).
On the other hand, for any smaller parallelogram $F_1$ that is similar to the conve... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Add a “+” or “-” in front of each number in $1,2, \cdots, 1989$. Find the minimum non-negative algebraic sum, and write down the equation. | First, prove that the algebraic sum is odd.
Consider the simplest case: all filled with " + ", at this time,
$$
1+2+\cdots+1989=995 \times 1989
$$
is odd.
For the general case, it is only necessary to adjust some " + " to " - ".
Since $a+b$ and $a-b$ have the same parity, each adjustment does not change the parity of... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\sum_{c y c} \frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}} \leqslant 8 \text {. }
$$ | To prove that since the inequality to be proved is homogeneous with respect to $a, b, c$, we may assume without loss of generality that $a+b+c=3$.
The original inequality is then $\sum_{\text {cyc }} \frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}} \leqslant 8$.
Consider the function $f(x)=\frac{(x+3)^{2}}{2 x^{2}+(3-x)^{2}}(0<x<3)... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false |
8. Given real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \\
\leqslant 60 .
\end{array}
$$
Then $x+y-z=$ . $\qquad$ | 8. 0 .
Original expression
$$
\begin{array}{l}
=\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \\
\leqslant 60 \\
\Rightarrow x=-2, y=5, z=3 \\
\Rightarrow x+y-z=0 .
\end{array}
$$ | 0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $S_{n}=1+2+\cdots+n$. Then among $S_{1}, S_{2}$, $\cdots, S_{2015}$, there are. $\qquad$ that are multiples of 2015. | 10.8 .
Obviously, $S_{n}=\frac{1}{2} n(n+1)$.
For any positive divisor $d$ of 2015, it is easy to see that in the range $1 \sim 2015$, there is exactly one $n$ that satisfies $n$ being a multiple of $d$ and $n+1$ being a multiple of $\frac{2015}{d}$. Therefore, each divisor of 2015 will generate one $n$ such that $S_{... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive number $x$ satisfies
$$
x^{10}+x^{5}+\frac{1}{x^{5}}+\frac{1}{x^{10}}=15250 \text {. }
$$
then the value of $x+\frac{1}{x}$ is | 2.3.
Let $a=x^{5}+\frac{1}{x^{5}}$. Then $x^{10}+\frac{1}{x^{10}}=a^{2}-2$.
Thus, the original equation becomes
$$
\begin{array}{l}
a^{2}+a-15252=0 \\
\Rightarrow(a-123)(a+124)=0
\end{array}
$$
$\Rightarrow a=123$ or $a=-124$ (discard).
Therefore, $x^{5}+\frac{1}{x^{5}}=123$.
Let $x+\frac{1}{x}=b>0$. Then
$$
\begin{ar... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $d$ be a positive divisor of 2015. Then the maximum value of the unit digit of $d^{\frac{2011}{d}}$ is $\qquad$ . | 9.7.
Notice that, $2015=5 \times 13 \times 31$.
Therefore, 2015 has eight positive divisors.
Let $G(n)$ denote the unit digit of $n$, then
$$
\begin{array}{l}
G\left(1^{2015}\right)=G\left(31^{165}\right)=1, \\
G\left(2015^{1}\right)=G\left(5^{403}\right)=G\left(65^{31}\right) \\
=G\left(155^{13}\right)=5 .
\end{array... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given sets
$$
A=\{1,3,5,7,9\}, B=\{2,4,6,8,10\} \text {. }
$$
If set $C=\{x \mid x=a+b, a \in A, b \in B\}$, then the number of elements in set $C$ is $\qquad$ . | ,- 1.9 .
Since $a$ is odd and $b$ is even, all elements in set $C$ are odd.
Also, the minimum value of $a+b$ is 3, and the maximum value is 19, and all odd numbers between 3 and 19 can be obtained, thus, set $C$ contains 9 elements. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the function
$$
f(x)=\left\{\begin{array}{ll}
0, & x<0, \\
1, & x \geqslant 0 .
\end{array}\right.
$$
Then $f(f(x))=$ $\qquad$ | 2. 1 .
Since for any $x \in \mathbf{R}$, we have $f(x) \geqslant 0$, therefore, $f(f(x))=1$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given
$$
\sin \alpha+\sqrt{3} \sin \beta=1, \cos \alpha+\sqrt{3} \cos \beta=\sqrt{3} \text {. }
$$
Then the value of $\cos (\alpha-\beta)$ is $\qquad$ . | 3. 0 .
Squaring and adding the two known equations, we get
$$
\begin{array}{l}
4+2 \sqrt{3}(\sin \alpha \cdot \sin \beta+\cos \alpha \cdot \cos \beta)=4 \\
\Rightarrow \cos (\alpha-\beta)=0
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. For a sequence of real numbers $x_{1}, x_{2}, \cdots, x_{n}$, define its "value" as $\max _{1 \leqslant i \leqslant n}\left\{\left|x_{1}+x_{2}+\cdots+x_{i}\right|\right\}$. Given $n$ real numbers, David and George want to arrange these $n$ numbers into a sequence with low value. On one hand, diligent David examines ... | 3. $c=2$.
If the initial numbers given are $1, -1, 2, -2$, then David arranges these four numbers as $1, -2, 2, -1$, and George can get the sequence $1, -1, 2, -2$.
Thus, $D=1, G=2$.
Therefore, $c \geqslant 2$.
We now prove: $G \leqslant 2 D$.
Let the $n$ real numbers be $x_{1}, x_{2}, \cdots, x_{n}$, and assume David... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. If the set $S=\{1,2,3, \cdots, 16\}$ is arbitrarily divided into $n$ subsets, then there must exist a subset in which there are elements $a, b, c$ (which can be the same), satisfying $a+b=c$. Find the maximum value of $n$.
【Note】If the subsets $A_{1}, A_{2}, \cdots, A_{n}$ of set $S$ satisfy the following condition... | 4. First, when $n=3$, assume there exists a partition of the set that does not satisfy the condition. Thus, there must be a subset with at least six elements, without loss of generality, let
$$
A=\left\{x_{1}, x_{2}, \cdots, x_{6}\right\}\left(x_{1}<x_{2}<\cdots<x_{6}\right) \text {. }
$$
Then $x_{6}-x_{1}, x_{6}-x_{2... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. A shape obtained by removing one unit square from a $2 \times 2$ grid is called an "L-shape". In an $8 \times 8$ grid, place $k$ L-shapes, each of which can be rotated, but each L-shape must cover exactly three unit squares of the grid, and any two L-shapes must not overlap. Additionally, no other L-shape can be pla... | 6. On one hand, using six straight lines to divide the large square grid into 16 smaller $2 \times 2$ grids, each $2 \times 2$ grid must have at least two cells covered by an L-shape. Therefore, the number of cells covered is no less than 32. Thus, the number of L-shapes is no less than $\left[\frac{32}{3}\right]+1$ $=... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$
and a fixed point $B(a, b)$, both satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$ | 2. 2 .
From the problem, we know that for any $A(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking
$$
(x, y)=(1,0),(0,1),(-1,-1) \text {, }
$$
we get that the fixed point $B(a, b)$ satisfies the necessary conditions
$$
\left\{\begin{array}{l}
a \leqslant 1, \\
b \leqslant 1, \\
a+b \geqslant-1
\end{array} \quad ... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Define the function on $\mathbf{R}$
$$
f(x)=\left\{\begin{array}{ll}
\log _{2}(1-x), & x \leqslant 0 ; \\
f(x-1)-f(x-2), & x>0 .
\end{array}\right.
$$
Then $f(2014)=$ | 3. 1.
First, consider the method for finding the function values on the interval $(0,+\infty)$.
If $x>3$, then
$$
\begin{array}{l}
f(x)=f(x-1)-f(x-2) \\
=f(x-1)-(f(x-1)+f(x-3)) \\
=-f(x-3) .
\end{array}
$$
Thus, for $x>6$, we have
$$
f(x)=-f(x-3)=f(x-6) \text {. }
$$
Therefore, $f(x)$ is a periodic function with a p... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The set $X \backslash Y=\{a \mid a \in X, a \notin Y\}$ is called the difference set of set $X$ and set $Y$. Define the symmetric difference of sets $A$ and $B$ as
$$
A \Delta B=(A \backslash B) \cup(B \backslash A) \text {. }
$$
If two non-empty finite sets $S$ and $T$ satisfy $|S \Delta T|=1$, then the minimum va... | 6. 3 .
If $|S \Delta T|=1$, without loss of generality, assume
$$
\left\{\begin{array}{l}
|S \backslash T|=1, \\
|T \backslash S|=0,
\end{array}\right.
$$
i.e., $T$ is a proper subset of $S$, thus,
$$
\begin{array}{l}
|S|>|T| \geqslant 1 . \\
\text { Hence } k \geqslant|T|+1+|T| \geqslant 3 .
\end{array}
$$
$$
\text ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. In an $m$ row by 10 column grid, fill each cell with either 0 or 1, such that each column contains exactly three 1s. Let the sum of the numbers in the $i$-th row ($i=1,2, \cdots, m$) be denoted as $x_{i}$, and for any two columns, there always exists a row where the cells intersecting with these two columns are both... | 8. 5 .
Construct an application model: "10 people go to a bookstore to buy $m$ kinds of books, each person buys 3 books, and any two of them buy at least one book in common. Find the sales volume of the book that is purchased by the most people."
Let the $i(i=1,2, \cdots, m)$-th kind of book be purchased by $x_{i}$ p... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a, b$ be the roots of the quadratic equation $x^{2}-x-1=0$. Then the value of $3 a^{2}+4 b+\frac{2}{a^{2}}$ is | From the given, we have $a b=-1, a+b=1$,
$$
\begin{array}{l}
a^{2}=a+1, b^{2}=b+1 . \\
\text { Therefore, } 3 a^{3}+4 b+\frac{2}{a^{2}}=3 a^{3}+4 b+2 b^{2} \\
=3 a^{2}+3 a+6 b+2=6(a+b)+5=11 .
\end{array}
$$ | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a$ and $b$ be distinct real numbers. If the quadratic function $f(x)=x^{2}+a x+b$ satisfies $f(a)=f(b)$, then the value of $f(2)$ is $\qquad$ . | $$
-, 1.4
$$
From the given conditions and the symmetry of the quadratic function graph, we get
$$
\begin{array}{l}
\frac{a+b}{2}=-\frac{a}{2} \Rightarrow 2 a+b=0 \\
\Rightarrow f(2)=4+2 a+b=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If the real number $\alpha$ satisfies $\cos \alpha=\tan \alpha$, then $\frac{1}{\sin \alpha}+\cos ^{4} \alpha=$ $\qquad$ | 2. 2 .
From the condition, we know that $\cos ^{2} \alpha=\sin \alpha$.
$$
\begin{array}{l}
\text { Then } \frac{1}{\sin \alpha}+\cos ^{4} \alpha=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\sin \alpha}+\sin ^{2} \alpha \\
=(1+\sin \alpha)+\left(1-\cos ^{2} \alpha\right) \\
=2+\sin \alpha-\cos ^{2} \alpha=2 .
\end{array}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $\{x\}=x-[x]$. Then the sum of the squares of all natural numbers $m$ that satisfy
$$
[(2 m+1)\{\sqrt{2 m+1}\}]=m
$$
is $\qquad$ . | 8.0.
Let $2 m+1=n$. Then $2[n\{\sqrt{n}\}]=n-1$. Note that,
$$
\begin{array}{l}
2 n\{\sqrt{n}\}-21$ when, by $n=2[n\{\sqrt{n}\}]+1$ being odd, we know $n \geqslant 3$.
Thus, $\left|4 n-(2[\sqrt{n}]+1)^{2}\right| \leqslant \frac{4}{\sqrt{n}}+\frac{1}{n}<3$.
Therefore, $4 n-(2[\sqrt{n}]+1)^{2}= \pm 1$.
This leads to
$$
... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Color each cell of a $12 \times 12$ grid either black or white, such that any $3 \times 4$ or $4 \times 3$ subgrid contains at least one black cell. Find the minimum number of black cells.
（Liang Yingde provided the problem） | 3. The minimum number of black cells required is $n=12$.
First, prove that $n \geqslant 12$.
Since a $12 \times 12$ grid can be divided into $\frac{12 \times 12}{3 \times 4}=12$ non-overlapping $3 \times 4$ sub-grids (excluding the boundaries), according to the problem statement, there must be at least 12 black cells.... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the 20 vertices of a regular 20-sided polygon inscribed in the unit circle in the complex plane correspond to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$. Then the number of distinct points corresponding to $z_{1}^{2015}, z_{2}^{2015}, \cdots, z_{20}^{2015}$ is $\qquad$ | 2. 4 .
Assume the vertices corresponding to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$ are arranged in a counterclockwise direction.
Let $z_{1}=\mathrm{e}^{\mathrm{i} \theta}$. Then
$$
\begin{array}{l}
z_{k+1}=\mathrm{e}^{\mathrm{i}\left(\theta+\frac{k}{10}\right)}(k=0,1, \cdots, 19), \\
z_{k+1}^{2015}=\mathrm... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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