problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
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10. (14 points) As shown in Figure 1, in the triangular prism $A B C-A_{1} B_{1} C_{1}$, it is known that the side edge $A A_{1} \perp$ plane $A B C$, and $\triangle A B C$ is an equilateral triangle with a side length of 2. $M$ is a point on $A A_{1}$, $A A_{1}=4, A_{1} M=1, P$ is a point on the edge $B C$, and the sh... | 10. (1) From $A A_{1} \perp$ plane $A B C$ and $\triangle A B C$ being an equilateral triangle, we know that all the side faces are congruent rectangles.
As shown in Figure 3, rotate the side face $B C_{1}$ by $120^{\circ}$ so that it lies in the same plane as the side face $A C_{1}$. Point $P$ moves to the position o... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, it is known that $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of side $A B$, and $\triangle A C M$ is folded along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Then the distance from point $M$ to plane $A B C$ is $\qquad$ | 3. 1 .
From the problem, we know that $M A=M B=M C=2$.
Therefore, the projection $O$ of $M$ on the plane $A B C$ is the circumcenter of $\triangle A B C$.
Given $A B=2 \sqrt{2}, A C=2, B C=2 \sqrt{3}$, we know that $A B^{2}+A C^{2}=B C^{2}$.
Thus, $O$ is the midpoint of the hypotenuse $B C$ of the right triangle $\tri... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 In $\triangle A B C$, it is known that $\angle A: \angle B: \angle C = 4: 2: 1, \angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively.
(1) Prove: $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$;
(2) Find the value of $\frac{(a+b-c)^{2}}{a^{2}+b^{2}+c^{2}}$. | (1) Proof As shown in Figure 3, construct the angle bisector of $\angle ABC$, intersecting the circumcircle of $\triangle ABC$ at point $M$, and construct the angle bisector of $\angle BAC$, intersecting the circumcircle of $\triangle ABC$ at point $N$.
Then $AM=MC=AB=c, AM \parallel BC$;
$$
CN=NB=AC=b, AB \parallel CN... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the equation in $x$
$$
(a-1) x^{2}+2 x-a-1=0
$$
has roots that are all integers. Then the number of integer values of $a$ that satisfy this condition is
$\qquad$. | 4.5.
When $a=1$, $x=1$.
When $a \neq 1$, it is easy to see that $x=1$ is an integer root of the equation.
Furthermore, from $1+x=\frac{2}{1-a}$ and $x$ being an integer, we know
$$
1-a= \pm 1, \pm 2 \text {. }
$$
Therefore, $a=-1,0,2,3$.
In summary, there are 5 integer values of $a$ that satisfy the condition. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let the quadratic function
$$
f(x)=a x^{2}+b x+c
$$
satisfy the following conditions:
(i) When $x$ is a real number, its minimum value is 0, and
$$
f(x-1)=f(-x-1)
$$
holds;
(ii) There exists a real number $m (m>1)$, such that there exists a real number $t$, as long as $1 \leqslant x \leqslant m$, t... | (1) In condition (ii), let $x=1$, we get $f(1)=1$.
From condition (i), we know that the quadratic function opens upwards and is symmetric about $x=-1$, so we can assume the quadratic function to be
$$
f(x)=a(x+1)^{2}(a>0) .
$$
Substituting $f(1)=1$ into the above equation, we get $a=\frac{1}{4}$.
Therefore, $f(x)=\fra... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. From the arithmetic sequence $2,5,8, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$ Ω. | 7.8.
First, let's take $x_{1}, x_{2}, \cdots, x_{k}$ from the known sequence such that
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1 .
$$
Let $y_{i}=\frac{x_{1} x_{2} \cdots x_{k}}{x_{i}}$. Then
$$
y_{1}+y_{2}+\cdots+y_{k}=x_{1} x_{2} \cdots x_{k} \text {. }
$$
It is easy to see that for any $n$, $x_{n... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Can 2010 be written as the sum of squares of $k$ distinct positive integers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason. ${ }^{[2]}$
(2010, Beijing Middle School Mathematics Competition (Grade 8)) | Estimate the approximate range of $k$ first, then discuss by classification.
Let $p_{i}$ be a prime number.
If 2010 can be written as the sum of squares of $k$ prime numbers, then by
$$
\begin{array}{l}
2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}+17^{2}+19^{2}+23^{2}+29^{2} \\
=2397>2010,
\end{array}
$$
we know $k \leqslant... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A. Let $a=\sqrt[3]{3}, b$ be the fractional part of $a^{2}$. Then $(b+2)^{3}=$ $\qquad$ | γγ6. A.9.
From $2<a^{2}<3$, we know $b=a^{2}-2=\sqrt[3]{9}-2$. Therefore, $(b+2)^{3}=(\sqrt[3]{9})^{3}=9$. | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
B. In the Cartesian coordinate system $x O y$, it is known that $O$ is the origin, point $A(10,100), B\left(x_{0}, y_{0}\right)$, where $x_{0} γ y_{0}$ are integers, and points $O γ A γ B$ are not collinear. For all points $B$ that satisfy the above conditions, find the minimum area of $\triangle O A B$. | B. Draw a line parallel to the $x$-axis through point $B$, intersecting line $O A$ at point $C\left(\frac{y_{0}}{10}, y_{0}\right)$.
Thus, $B C=\left|x_{0}-\frac{y_{0}}{10}\right|$.
Since points $O$, $A$, and $B$ are not collinear, we have
$$
\begin{array}{l}
S_{\triangle O A B}=\frac{1}{2} B C \times 100=50\left|x_{0}... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
14. A. If placing the positive integer $M$ to the left of the positive integer $m$ results in a new number that is divisible by 7, then $M$ is called the "magic number" of $m$ (for example, placing 86 to the left of 415 results in the number 86415, which is divisible by 7, so 86 is called the magic number of 415). Find... | 14. A. If $n \leqslant 6$, take $m=1,2, \cdots, 7$. By the pigeonhole principle, there must be a positive integer $M$ among $a_{1}, a_{2}, \cdots, a_{n}$ that is a common magic number of $i$ and $j (1 \leqslant i<j \leqslant 7)$, i.e.,
$$
7 \mid(10 M+i), 7 \mid(10 M+j) \text {. }
$$
Then $7 \mid (j-i)$. But $0<j-i \le... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If real numbers $x, y, z$ satisfy the equation
$$
\sqrt{x+9+\sqrt{x-7}}+\frac{|x+y-z|}{4}=4 \text {, }
$$
then the units digit of $(5 x+3 y-3 z)^{2013}$ is $\qquad$ | 3. 4 .
It is known that $x \geqslant 7$, then
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}} \geqslant 4, \\
\frac{|x+y-z|}{4} \geqslant 0 .
\end{array}\right.
$$
Combining the given equations, we have
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}}=4, \\
\frac{|x+y-z|}{4}=0 .
\end{array}\right.
$$
Therefore, $x=... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $m, n$ satisfy $m-n=\sqrt{10}$, $m^{2}-3 n^{2}$ is a prime number. If the maximum value of $m^{2}-3 n^{2}$ is $a$, and the minimum value is $b$, then $a-b=$ $\qquad$ | 5.11 .
Let $m^{2}-3 n^{2}=p$ (where $p$ is a prime number).
From $m-n=\sqrt{10}$, we get
$$
m=\sqrt{10}+n \text {. }
$$
Substituting equation (2) into equation (1) and simplifying, we get
$$
\begin{array}{l}
2 n^{2}-2 \sqrt{10} n+p-10=0 \\
\Rightarrow \Delta=40-8 p+80 \geqslant 0 \\
\Rightarrow p \leqslant 15 \\
\Rig... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In the Cartesian coordinate system $x O y$, it is known that there are three points $A(a, 1), B(2, b), C(3,4)$.
If the projections of $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are the same, then $3 a-4 b=$ $\qquad$ | 3. 2 .
Solution 1 The projections of vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are $\frac{\overrightarrow{O A} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}, \frac{\overrightarrow{O B} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}$, respectively.
A... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, $8 n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$. [4]
(2009, "Mathematics Weekly Cup" National Junior High Sch... | Let $a_{1}, a_{2}, \cdots, a_{n}$ be such that removing $a_{i} (i=1, 2, \cdots, n)$ leaves the arithmetic mean of the remaining $n-1$ numbers as a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Thus, for any $1 \leqslant i<j \leqslant n$, we have
$$
b_{i}-b_{j}=\f... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. If for any $x \in\left(-\frac{1}{2}, 1\right)$, we have
$$
\frac{x}{1+x-2 x^{2}}=\sum_{k=0}^{\infty} a_{k} x^{k},
$$
then $a_{3}+a_{4}=$ . $\qquad$ | 9. -2 .
In equation (1), let $x=0$, we get $a_{0}=0$. Then $\frac{1}{1+x-2 x^{2}}=\sum_{k=1}^{\infty} a_{k} x^{k-1}$. Substituting $x=0$ into the above equation, we get $a_{1}=1$. Then $\frac{-1+2 x}{1+x-2 x^{2}}=\sum_{k=2}^{\infty} a_{k} x^{k-2}$. Substituting $x=0$ into the above equation, we get $a_{2}=-1$. Similar... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. If $0 \leqslant x_{i} \leqslant 1(i=1,2, \cdots, 5)$, then
$$
M=x_{1}-x_{2}^{3}+x_{2}-x_{3}^{3}+x_{3}-x_{4}^{3}+x_{4}-x_{5}^{3}+x_{5}-x_{1}^{3}
$$
the maximum value is $\qquad$ | 10.4.
If there exists a positive integer $j$, such that
$$
x_{j}=x_{j+1}\left(j=1,2, \cdots, 5, x_{6}=x_{1}\right) \text {, }
$$
then $M \leqslant 4$.
When $x_{1}=0, x_{2}=1, x_{3}=0, x_{4}=1, x_{5}=0$, the equality holds.
If for any positive integer $i$, we have $x_{i} \neq x_{i+1}$ $(i=1,2, \cdots 5)$, then either... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given a moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
2 x+y \leqslant 2, \\
x \geqslant 0, \\
\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1 .
\end{array}\right.
$$
Then the area of the figure formed by the moving point $P(x, y)$ is | 8. 2 .
From equation (1), we have
$$
\begin{array}{l}
x+\sqrt{x^{2}+1} \geqslant \sqrt{y^{2}+1}-y \\
\Rightarrow \ln \left(x+\sqrt{x^{2}+1}\right) \geqslant \ln \left(\sqrt{y^{2}+1}-y\right) .
\end{array}
$$
It is easy to see that $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)$ is a strictly increasing function.
$$
\begin{a... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. The positive integer $n=$ $\qquad$ that makes $2^{n}+256$ a perfect square. | When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$.
If it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a positive integer). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, hence $k=1$.
Thus, $n=11$. | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 (1) Find the minimum value of the sum of the digits of integers of the form $3 n^{2}+n+1\left(n \in \mathbf{Z}_{+}\right)$;
(2) Does there exist a number of this form whose sum of digits is 1999? | (1) Since
$$
3 n^{2}+n+1=n(3 n+1)+1
$$
is an odd number, the last digit must be odd.
If the sum of the digits is 1, then the number is 1, but $n$ is a positive integer, $3 n^{2}+n+1 \geqslant 5$, which is a contradiction.
If the sum of the digits is 2 and it is an odd number, then the first and last digits must both ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $p$ is a prime number greater than 3. Find
$$
\prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)
$$
the value. | Let $\omega=\mathrm{e}^{\frac{2 \pi i}{p}}$. Then
$$
\begin{array}{l}
\omega^{p}=1, \omega^{-\frac{p}{2}}=-1, 2 \cos \frac{2 k \pi}{p}=\omega^{k}+\omega^{-k} . \\
\text { Hence } \prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)=\prod_{k=1}^{p-1}\left(1+\omega^{k}+\omega^{-k}\right) \\
=\prod_{k=1}^{p-1} \omega... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The sum of all real numbers $m$ that satisfy $(2-m)^{m^{2}-m-2}=1$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. A.
When $2-m=1$, i.e., $m=1$, it satisfies the condition.
When $2-m=-1$, i.e., $m=3$,
$$
(2-m)^{m^{2}-m-2}=(-1)^{4}=1 \text{, }
$$
it satisfies the condition.
When $2-m \neq \pm 1$, i.e., $m \neq 1$ and $m \neq 3$, by the condition, $m^{2}-m-2=0$, and $2-m \neq 0$.
Solving gives $m=-1$.
Therefore, the sum is $1+3+... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given real numbers $x, y, z$ satisfy
$$
x+y=4,|z+1|=x y+2 y-9 \text {. }
$$
then $x+2 y+3 z=$ $\qquad$ | $=, 1.4$.
From $x+y=4$, we get $x=4-y$. Then
$$
\begin{array}{l}
|z+1|=x y+2 y-9 \\
=6 y-y^{2}-9=-(y-3)^{2} \\
\Rightarrow z=-1, y=3 \Rightarrow x=1 \\
\Rightarrow x+2 y+3 z=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. A cube is painted red on all its faces, then cut into $n^{3}(n>2)$ identical smaller cubes. If the number of smaller cubes with only one face painted red is the same as the number of smaller cubes with no faces painted red, then $n=$ $\qquad$ . | 2.8.
The total number of small cubes with only one face painted red is $6(n-2)^{2}$, and the total number of small cubes with no faces painted red is $(n-2)^{3}$. According to the problem,
$$
6(n-2)^{2}=(n-2)^{3} \Rightarrow n=8 \text {. }
$$ | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Given real numbers $a, b, c, d$ satisfy $2a^2 + 3c^2 = 2b^2 + 3d^2 = (ad - bc)^2 = 6$. Find the value of $\left(a^2 + \dot{b}^2\right)\left(c^2 + d^2\right)$. | Let $m=a^{2}+b^{2}, n=c^{2}+d^{2}$. Then
$$
\begin{array}{l}
2 m+3 n=2 a^{2}+2 b^{2}+3 c^{2}+3 d^{2}=12 . \\
\text { By }(2 m+3 n)^{2}=(2 m-3 n)^{2}+24 m n \geqslant 24 m n \\
\Rightarrow 12^{2} \geqslant 24 m n \\
\Rightarrow m n \leqslant 6 . \\
\text { Also } m n=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\
=... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) If positive numbers $a, b, c$ satisfy
$$
\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}-... | Three, since $a, b, c$ have cyclic symmetry, without loss of generality, assume $0 < a < b < c$, then
$$
c-a>b>0, c-b>a>0 \text {. }
$$
Thus $\frac{b^{2}+c^{2}-a^{2}}{2 b c}=1+\frac{(c-b)^{2}-a^{2}}{2 b c}>1$,
$$
\begin{array}{l}
\frac{c^{2}+a^{2}-b^{2}}{2 c a}=1+\frac{(c-a)^{2}-b^{2}}{2 c a}>1, \\
\frac{a^{2}+b^{2}-c... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $\mathrm{i}$ is the imaginary unit. If
$$
z=1+\mathrm{i}+\cdots+\mathrm{i}^{2013},
$$
denote the complex conjugate of $z$ as $\bar{z}$, then $z \cdot \bar{z}=$ $\qquad$ | 8. 2 .
Let $a_{n}=\mathrm{i}^{n}$. Then $a_{n+4}=a_{n}$, and
$$
1+\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}=0 \text {. }
$$
Thus $z=1+\mathrm{i} \Rightarrow \bar{z}=1-\mathrm{i}$
$$
\Rightarrow z \cdot \bar{z}=(1+\mathrm{i})(1-\mathrm{i})=1-\mathrm{i}^{2}=2 .
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Given a positive integer $n$. Find $\sum_{k=1}^{n}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
---
Please note that the format and line breaks have been preserved as requested. | $$
\text { Three, let } n=2^{m} a_{m}+2^{m-1} a_{m-1}+\cdots+2^{1} a_{1}+a_{0}
$$
where, $a_{m} \neq 0$.
At this point, $2^{m} \leqslant n<2^{m+1}$, so, $\left[\log _{2} n\right]=m$.
If $k \geqslant m+2$, then
$$
\frac{n}{2^{k}}-\frac{1}{2}<\frac{2^{m+1}}{2^{m+2}}-\frac{1}{2}=0 \text {, }
$$
at this time, $\left[\fra... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given $a, b, c > 0$, and $a^{2} + b^{2} + c^{2} + abc = 4$. Prove:
$$
\begin{array}{l}
\sqrt{\frac{(2-a)(2-b)}{(2+a)(2+b)}} + \sqrt{\frac{(2-b)(2-c)}{(2+b)(2+c)}} + \\
\sqrt{\frac{(2-c)(2-a)}{(2+c)(2+a)}} = 1 .
\end{array}
$$ | $$
\begin{array}{l}
16=4 a^{2}+4 b^{2}+4 c^{2}+4 a b c \text {. } \\
\text { Then } \sqrt{\frac{(2-a)(2-b)}{(2+a)(2+b)}}+\sqrt{\frac{(2-b)(2-c)}{(2+b)(2+c)}}+ \\
\sqrt{\frac{(2-c)(2-a)}{(2+c)(2+a)}} \\
=\frac{\sqrt{(4-a)^{2}\left(4-b^{2}\right)}}{(2+a)(2+b)}+ \\
\frac{\sqrt{\left(4-b^{2}\right)\left(4-c^{2}\right)}}{(2... | 1 | Algebra | proof | Yes | Yes | cn_contest | false |
11. (20 points) Given
$$
f(x)=\frac{1+\ln (x+1)}{x}, g(x)=\frac{k}{x+1} .
$$
Find the largest positive integer $k$, such that for any positive number $c$, there exist real numbers $a$ and $b$ satisfying $-1<a<b<c$, and
$$
f(c)=f(a)=g(b) .
$$ | 11. For positive integer $k$, it is clear that $g(x)=\frac{k}{x+1}$ is a decreasing function on the interval $(-1,+\infty)$.
Thus, for any positive number $c$,
$$
f(c)=g(b)>g(c) \text {. }
$$
When $x>0$, the inequality
$$
\begin{array}{l}
f(x)>g(x) \\
\Leftrightarrow k0) \text {. }
$$
Then $h^{\prime}(x)=\frac{x-1-\l... | 3 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
354 Divide the sides of the equilateral $\triangle A B C$ into four equal parts, and draw lines parallel to the other two sides through each division point. The 15 points formed by the intersections of the sides of $\triangle A B C$ and these parallel lines are called lattice points. Among these 15 lattice points, if $... | Solve for the minimum value of $n$ being 6.
Let the three equal division points from point $A$ to $B$ on side $AB$ be $L, F, W$; the three equal division points from point $B$ to $C$ on side $BC$ be $V, D, U$; and the three equal division points from point $C$ to $A$ on side $CA$ be $N, E, M$. Denote the intersection p... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 What is the maximum number of rational points (points with both coordinates being rational numbers) that can lie on a circle in the plane, given that the center of the circle is not a rational point. | γAnalysisγIf $A, B, C$ are three rational points on a circle,
then the midpoint $D$ of $AB$ is a rational point, the slope of $AB$ is a rational number or infinite, so the equation of the perpendicular bisector of $AB$ is a linear equation with rational coefficients. Similarly, the equation of the perpendicular bisec... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $A, B, C$ be three non-collinear lattice points on a plane, and the side lengths of $\triangle ABC$ are all positive integers. Find the minimum value of $AB$ and the minimum perimeter. | γAnalysisγIf $A B=1$, we might as well set $A(0,0), B(1,0)$. Then $|A C-B C|<A B=1$, which can only be $A C=B C$.
Thus, point $C$ lies on the perpendicular bisector of $A B$.
Therefore, the x-coordinate of point $C$ is $\frac{1}{2}$, meaning $C$ cannot be a lattice point.
Hence, $A B$ cannot be 1.
If $A B=2$, we might ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 On a plane, there exist $n$ points, no three of which are collinear, and when these $n$ points are arbitrarily labeled as $A_{1}, A_{2}, \cdots, A_{n}$, the broken line $A_{1} A_{2} \cdots A_{n}$ does not intersect itself. Find the maximum value of $n$.
| γAnalysisγWhen $n=2,3$, it is obviously true.
When $n=4$, if the convex hull of the four points is a triangle, then it satisfies the condition.
Next, we show that when $n \geqslant 5$, it is impossible to satisfy the condition.
Consider only five points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ among these $n$ points.
If th... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Find the smallest $n \in \mathbf{N}_{+}$, such that for any finite point set $M$ in the plane, if any $n$ points in $M$ can be covered by two lines, then there must exist two lines that can cover the point set $M$. | The required minimum value is 6.
When $|M| \leqslant 6$, the proposition is obviously true.
If $|M|>6$, for any six points $A_{1}, A_{2}, \cdots, A_{6}$ in $M$, they can be covered by two lines $l_{1}$ and $l_{2}$. Without loss of generality, assume that $l_{1}$ contains at least three points $A_{1}, A_{2}, A_{3}$. Let... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. (50 points) Given that $a$, $b$, and $c$ are three distinct real numbers. If any two of the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$
have exactly one common root, find the value of $a^{2}+b^{2}+$ $c^{2}$. | 2. From equations (1) and (2), we know their common root is $p=\frac{b-c}{b-a}$.
Similarly, the common roots of equations (2) and (3), and equations (1) and (3) are $q=\frac{c-a}{c-b}$ and $r=\frac{a-b}{a-c}$, respectively.
Thus, $p q r=-1$.
If any two of $p, q, r$ are equal, assume $p=q$, then the three equations hav... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4.12 The acrobats are numbered $1, 2, \cdots$, 12. They are to be arranged in two circles, $A$ and $B$, each with six people. In circle $B$, each acrobat stands on the shoulders of two adjacent acrobats in circle $A$. If the number of each acrobat in circle $B$ is equal to the sum of the numbers of the two acrobats bel... | 4. Let the sums of the elements in circles $A$ and $B$ be $x$ and $y$ respectively. Then $y=2x$. Therefore,
$$
3x = x + y = 1 + 2 + \cdots + 12 = 78.
$$
Solving for $x$ gives $x = 26$.
Clearly, $1, 2 \in A$ and $11, 12 \in B$.
Let $A = \{1, 2, a, b, c, d\} (a < b < c < d)$. Then $a + b + c + d = 23$, and $a \geq 3, 8 ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Remove any $2 \times 2$ small square from the corner of a $3 \times 3$ grid, and the remaining shape is called a "corner shape" (Figure 3 is an example of a corner shape). Now, place some non-overlapping corner shapes in a $10 \times 10$ grid (Figure 4). The boundaries of the corner shapes must coincide with the bou... | 7. First, $k_{\max }$
$<8$. This is because, if eight corner shapes are placed in the manner shown in Figure 9, it is impossible to place another corner shape in the grid.
Next, we prove that after placing seven corner shapes arbitrarily, it is still possible to place another complete corner shape.
Cover the 5th and ... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 4, in the Cartesian coordinate system, $O$ is the origin, the diagonals of $\square A B O C$ intersect at point $M$, and the hyperbola $y=\frac{k}{x}(x<0)$ passes through points $B$ and $M$. If the area of $\square A B O C$ is 24, then $k=$ . $\qquad$ | 2, 1. -8.
Let $M\left(\frac{k}{y}, y\right)$. Then $B\left(\frac{k}{2 y}, 2 y\right), C\left(\frac{3 k}{2 y}, 0\right)$.
From $S_{\text {OABOC }}=4 S_{\triangle O C M}=2\left|x_{c} y_{M}\right|$
$\Rightarrow 24=2\left|\frac{3 k}{2}\right| \Rightarrow|k|=8 \Rightarrow k=-8$. | -8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find all odd prime numbers $p$ such that $p \mid \sum_{k=1}^{103} k^{p-1}$.
untranslated text remains the same as requested. | If $p>103$, then for $1 \leqslant k \leqslant 103$, we have
$$
\begin{array}{l}
k^{p-1} \equiv 1(\bmod p), \\
\sum_{k=1}^{103} k^{p-1} \equiv 103(\bmod p) .
\end{array}
$$
Therefore, $p \leqslant 103$.
Let $103=p q+r(0 \leqslant r < q)$, then $r=q, 103=p q+r=(p+1) r$. Since 103 is a prime number, we get $p=102, r=1$, ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Find all positive integers that are coprime with all terms of the sequence $\left\{a_{n}\right\}$ satisfying
$$
a_{n}=2^{n}+3^{n}+6^{n}-1\left(n \in Z_{+}\right)
$$ | Solution: Clearly, $\left(1, a_{n}\right)=1$.
Let $m(m>1)$ be a positive integer that is coprime with all terms in $\left\{a_{n}\right\}$, and let $p$ be a prime factor of $m$.
If $p>3$, then by Fermat's Little Theorem,
$$
\begin{array}{l}
2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), \\
6^{p-1} \equiv 1(\bmod ... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Prove: For any positive integer $n, 3^{n}+2 \times 17^{n}$ is not a multiple of 5, and find the smallest positive integer $n$, such that
$$
11 \mid\left(3^{n}+2 \times 17^{n}\right) .
$$ | $$
\begin{array}{l}
3^{2 k}+2 \times 17^{2 k} \equiv(-1)^{k}+2 \times 2^{2 k} \\
\equiv 3(-1)^{k}(\bmod 5) \\
3^{2 k+1}+2 \times 17^{2 k+1} \equiv 3(-1)^{k}+4(-1)^{k} \\
\equiv 2(-1)^{k}(\bmod 5)
\end{array}
$$
Therefore, $3^{n}+2 \times 17^{n}$ is not a multiple of 5.
$$
\begin{array}{l}
\text { Also, } 3^{n}+2 \time... | 4 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 5 What is the minimum degree of the highest term of a polynomial with rational coefficients that has $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots?
(2013, Joint Autonomous Admission Examination of Peking University and Other Universities) | Notice that the polynomial
$$
f(x)=\left(x^{2}-2\right)\left[(x-1)^{3}-2\right]
$$
has roots $\sqrt{2}$ and $1-\sqrt[3]{2}$, and its degree is 5. Therefore, the degree of the highest term of a rational-coefficient polynomial with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots cannot be less than 5.
If there exists a rationa... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $x$ is an integer, and satisfies the inequality system
$$
\left\{\begin{array}{l}
x-1>0, \\
2 x-1<4,
\end{array}\right.
$$
then $x=$ $\qquad$ | $$
=, 1.2 \text {. }
$$
From the given, we know that $1<x<\frac{5}{2}$. Therefore, the integer $x=2$. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Let real numbers $x, y, z$ satisfy
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=27 \text {. }
$$
Then the maximum value of $|y-z|$ is $\qquad$ | 3. 6 .
The original equation is equivalent to a quadratic equation in $x$
$$
\begin{array}{l}
x^{2}-(y+z) x+y^{2}+z^{2}-y z-27=0 . \\
\text { And } \Delta=(y+z)^{2}-4\left(y^{2}+z^{2}-y z-27\right) \geqslant 0 \\
\Rightarrow(y-z)^{2} \leqslant 36 \Rightarrow|y-z| \leqslant 6 .
\end{array}
$$
When $|y-z|=6$, and $x=\f... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $x_{1}, x_{2}, \cdots, x_{15}$ take values of 1 or -1. Let
$$
S=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{15} x_{1} \text {. }
$$
Then the smallest positive integer that $S$ can take is $\qquad$ | 4.3.
Let $y_{i}=x_{i} x_{i+1}(i=1,2, \cdots, 15)$, with the convention that $x_{16}=$ $x_{1}$. Then $y_{i}=1$ or -1.
In $y_{1}, y_{2}, \cdots, y_{15}$, let there be $a$ values that are 1 and $b$ values that are -1. Clearly, $a+b=15$.
Also, $1^{a}(-1)^{b}=y_{1} y_{2} \cdots y_{15}=\left(x_{1} x_{2} \cdots x_{15}\righ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. (25 points) As shown in Figure 1, given that $E$ is a point on side $A B$ of square $A B C D$, and the symmetric point of $A$ with respect to $D E$ is $F, \angle B F C = 90^{\circ}$. Find the value of $\frac{A B}{A E}$. | 2. As shown in Figure 4, extend $E F$ to intersect $B C$ at point $M$, connect $D M$, and let it intersect $C F$ at point $G$.
Then, Rt $\triangle D F M \cong$ Rt $\triangle D C M$. Therefore, $\angle F D M = \angle M D C$, and $F M = C M$.
Thus, $M$ is the midpoint of $B C$.
$$
\begin{array}{l}
\text { Also, } \angle ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. (25 points) On a circle, there are $n$ different positive integers $a_{1}$, $a_{2}, \cdots, a_{n}$ placed in a clockwise direction. If for any number $b$ among the ten positive integers $1, 2, \cdots, 10$, there exists a positive integer $i$ such that $a_{i}=b$ or $a_{i}+a_{i+1}=b$, with the convention that $a_{n+1}... | 3. From the conditions, we know that the $2n$ numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n}+a_{1}$ should include the ten positive integers $1,2, \cdots, 10$. Therefore, $2 n \geqslant 10 \Rightarrow n \geqslant 5$.
When $n=5$, the ten numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Find the largest positive integer $n(n \geqslant 3)$, such that there exists a convex $n$-gon, where the tangent values of all its interior angles are integers.
(Proposed by the Problem Committee) | On the one hand, since each interior angle of a regular octagon is $135^{\circ}$, and its tangent value is -1, $n=8$ satisfies the condition.
On the other hand, if $n \geqslant 9$, let the exterior angles of the $n$-sided polygon be $\angle A_{1}, \angle A_{2}, \cdots, \angle A_{n}\left(0<\angle A_{1} \leqslant \angle... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the remainder when $10^{10}(100$ ones$)$ is divided by 7. | \begin{array}{l}\text { Sol } 10^{10} \equiv(7+3)^{10^{10}} \equiv 3^{10} \\ \equiv(7+2)^{50 \cdots 0} \equiv 2^{5 \times 10^{9}} \equiv 2^{3 \times 106 \cdots 6+2} \\ \equiv 4(7+1)^{166 \cdots 6} \equiv 4(\bmod 7) .\end{array} | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Seven, let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1}=\left(1+\frac{k}{n}\right) a_{n}+1(n=1,2, \cdots) .
\end{array}
$$
Find all positive integers $k$ such that every term in the sequence $\left\{a_{n}\right\}$ is an integer.
(Zhang Lei) | When $k=1$, $a_{2}=3, a_{3}=\frac{11}{2}$, which does not satisfy the condition.
When $k=2$, by the given condition we have
$$
\frac{a_{n+1}}{(n+1)(n+2)}=\frac{a_{n}}{n(n+1)}+\frac{1}{(n+1)(n+2)} \text {. }
$$
Thus, $\frac{a_{n}}{n(n+1)}=\frac{a_{1}}{1 \times 2}+\sum_{i=2}^{n} \frac{1}{i(i+1)}=1-\frac{1}{n+1}$.
Theref... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $S$ be a set of $n(n \geqslant 5)$ points in the plane. If any four points chosen from $S$ have at least one point connected to the other three, then which of the following conclusions is correct? $\qquad$
(1) There is no point in $S$ that is connected to all other points;
(2) There is at least one point in $S$ ... | -γ1. (2).
In the point set $S$, all points are connected to each other, which clearly satisfies the problem. Therefore, conclusions (1) and (4) are incorrect.
Suppose $A$, $B$, and $C$ are three points in the point set $S$ that are not connected to each other, but the remaining $n-3$ points are all connected to each o... | 2 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
6. Given that the three vertices of $\triangle A B C$ are all on the parabola $y^{2}=2 p x(p>0)$, and the centroid of $\triangle A B C$ is exactly the focus of the parabola. If the equation of the line on which side $B C$ lies is $4 x+y$ $-20=0$, then $p=$ $\qquad$ . | 6. $p=8$.
Let $A\left(\frac{y_{1}^{2}}{2 p}, y_{1}\right), B\left(\frac{y_{2}^{2}}{2 p}, y_{2}\right), C\left(\frac{y_{3}^{2}}{2 p}, y_{3}\right)$.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
y^{2}=2 p x, \\
4 x+y-20=0
\end{array}\right. \\
\Rightarrow 2 y^{2}+p y-20 p=0 .
\end{array}
$$
From the given ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $\tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$, $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}$, $\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha=-\frac{17}{5}$.
Then $\tan (\alpha+\beta+\gamma)=$ $\qquad$ [5]
(2012, Xin Zhi Cup Shanghai High School Mathemati... | Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then $x+y+z=\frac{17}{6}$,
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}$,
$\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5}$.
(1) $\div$ (3) gives
$x y z=-\frac{5}{6}$.
(2) $\times$ (4) gives $x y+y z+z x=\frac{2}{3}$.
Therefore, $\tan (\alpha+\beta+\gamma)... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.
[2] | Notice that,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1006} \text {. }
$$
Clearly, $0<(5-2 \sqrt{6})^{1006}<1$,
$$
\begin{array}{l}
(5+2 \sqrt{6})^{1006}+(5-2 \sqrt{6})^{1006} \\
=2\left(C_{1006}^{0} 5^{1006}+C_{1006}^{2} 5^{1004} \times 24+\cdots+\right. \\
\left.\quad C_{1006}^{1006} 24^{503}\right) \in \mathbf... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $x=(15+\sqrt{220})^{19}+(15+\sqrt{200})^{22}$. Find the unit digit of the number $x$.
| Solve the conjugate expression
$$
y=(15-\sqrt{220})^{19}+(15-\sqrt{220})^{82} \text {. }
$$
Then $x+y$
$$
\begin{aligned}
= & (15+\sqrt{220})^{19}+(15-\sqrt{220})^{19}+ \\
& (15+\sqrt{220})^{82}+(15-\sqrt{220})^{82} .
\end{aligned}
$$
By the binomial theorem, for any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
A=\{2,0,1,3\}, \\
B=\left\{x \mid -x \in A, 2-x^{2} \notin A\right\} .
\end{array}
$$
Then the sum of all elements in set $B$ is | ,$- 1 .-5$.
It is easy to know that $B \subseteq\{-2,0,-1,-3\}$.
When $x=-2,-3$, $2-x^{2}=-2,-7 \notin A$; when $x=0,-1$, $2-x^{2}=2,1 \in A$.
Therefore, the set $B=\{-2,-3\}$.
Thus, the sum of all elements in set $B$ is -5. | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In the Cartesian coordinate system $x O y$, it is known that points $A$ and $B$ lie on the parabola $y^{2}=4 x$, and satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B}=-4, F$ is the focus of the parabola. Then $S_{\triangle O F A} \cdot S_{\triangle O F B}=$ $\qquad$ . | 2. 2 .
From the problem, we know the point $F(1,0)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Thus, $x_{1}=\frac{y_{1}^{2}}{4}, x_{2}=\frac{y_{2}^{2}}{4}$.
Then $-4=\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2}$
$$
\begin{array}{l}
=\frac{1}{16}\left(y_{1} y_{2}\right)^{2... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, it is known that
$$
\sin A=10 \sin B \cdot \sin C, \cos A=10 \cos B \cdot \cos C \text {. }
$$
Then $\tan A=$ $\qquad$ | 3. 11 .
From $\sin A-\cos A$
$$
\begin{array}{l}
=10(\sin B \cdot \sin C-\cos B \cdot \cos C) \\
=-10 \cos (B+C)=10 \cos A \\
\Rightarrow \sin A=11 \cos A \\
\Rightarrow \tan A=11 .
\end{array}
$$ | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$
$(2009$, National High School Mathematics League Jilin Province Preliminary) | Hint: The fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is $1-(\sqrt{2}-\sqrt{3})^{2010}$.
Since $0.9<1-(\sqrt{2}-\sqrt{3})^{2010}<1$, the first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is 9. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The number of positive integers $n$ such that $n+1$ divides $n^{2012}+2012$ is $\qquad$ . | From the problem, we know
$$
\begin{array}{l}
n^{2012}+2012 \equiv(-1)^{2012}+2012 \\
=2013 \equiv 0(\bmod n+1) .
\end{array}
$$
Since $2013=3 \times 11 \times 61$, $n$ has $2^{3}-1=7$ solutions. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) As shown in Figure 5, $P$ is a point outside circle $\odot O$, $PA$ is tangent to $\odot O$ at point $A$, and $PBC$ is a secant of $\odot O$. $AD \perp PO$ at point $D$. If $PB=4$, $CD=5$, $BC=6$, find the length of $BD$. | Connect $O A, O B, O C$. Then $O B=O C$. It is easy to see that
$$
\begin{array}{l}
P A \perp O A, \\
P A^{2}=P B \cdot P C=P D \cdot P O \\
\Rightarrow \frac{P O}{P C}=\frac{P B}{P D} .
\end{array}
$$
Since $\angle B P D$ is a common angle, we have
$$
\begin{array}{l}
\triangle P O B \sim \triangle P C D . \\
\text {... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. For each positive integer $n$, let the tangent line to the curve $y=x^{n+1}$ at the point $(1,1)$ intersect the $x$-axis at a point with abscissa $x_{n}$. Let $a_{n}=\lg x_{n}$. Then
$$
a_{1}+a_{2}+\cdots+a_{99}=
$$
$\qquad$ | 3. -2 .
Given $y=x^{n+1}$, we know $y^{\prime}=(n+1) x^{n}$.
Then according to the condition, we have
$$
\begin{array}{l}
\frac{0-1}{x_{n}-1}=y^{\prime}(1)=n+1 \Rightarrow x_{n}=\frac{n}{n+1} . \\
\text { Therefore, } a_{1}+a_{2}+\cdots+a_{99}=\lg \left(x_{1} x_{2} \cdots x_{99}\right) \\
=\lg \left(\frac{1}{2} \times... | -2 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the ellipse $C$ be:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{20}=1(a>2 \sqrt{5})
$$
with its left focus at $F$, and point $P(1,1)$. It is known that there exists a line $l$ passing through point $P$ intersecting the ellipse at points $A$ and $B$, with $M$ being the midpoint of $A B$, such that $|F M|$ i... | 11. By the median length formula, we have
$|F M|^{2}=\frac{1}{2}\left(|F A|^{2}+|F B|^{2}\right)-\frac{1}{4}|A B|^{2}$.
Also, $|F M|^{2}=|F A||F B|$, so
$|A B|^{2}=2(|F A|-|F B|)^{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and
$c=\sqrt{a^{2}-20}$.
Then $|F A|-|F B|=\frac{c}{a} x_{1}-\frac{c}{a} ... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Remove a $2 \times 2$ small square from any corner of a $3 \times 3$ grid, and the remaining shape is called a "corner shape". Now, place some non-overlapping corner shapes in a $9 \times 9$ grid, with the requirement that the boundaries of the corner shapes coincide with the boundaries or grid lines... | Three, first, $k_{\max }<6$. This is because, if 6 corner shapes are placed in the manner shown in Figure 4, it is impossible to place another complete corner shape on this grid.
Now, place 5 corner shapes in any manner.
Next, we will prove that it is still possible to place another complete corner shape.
Consider th... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leqslant 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at this time.
(Ninth China Southeast Mathematical Olympiad) | Let
$$
\begin{array}{l}
f(x)=a \cos x+b \cos 2 x+c \cos 3 x+d \cos 4 x . \\
\text { By } f(0)=a+b+c+d, \\
f(\pi)=-a+b-c+d, \\
f\left(\frac{\pi}{3}\right)=\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2},
\end{array}
$$
then \(a+b-c+d\)
$$
=f(0)+\frac{2}{3} f(\pi)+\frac{4}{3} f\left(\frac{\pi}{3}\right) \leqslant 3 .
$$
Equality... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
17. A moving point moves on the integer points in the first quadrant of the Cartesian coordinate system (including the integer points on the $x$-axis and $y$-axis of the first quadrant), with the movement rules being $(m, n) \rightarrow(m+1, n+1)$ or $(m, n) \rightarrow$ $(m+1, n-1)$. If the moving point starts from th... | $\begin{array}{l}\text { 17. } 9 \\ C_{6}^{2}-C_{6}^{1}=9\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 17. } 9 \\ C_{6}^{2}-C_{6}^{1}=9\end{array}$
Note: The original text is already in a mathematical format, which is universal and does not require translation. However, if you intended to have... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure $4, \triangle A B C$ has an incircle $\odot O_{1}$ that touches side $B C$ at point $D, \odot O_{2}$ is the excircle of $\triangle A B C$ inside $\angle A$. If $O_{1} B=6, O_{1} C=3, O_{1} D=2$, then $O_{1} O_{2}=$ $\qquad$ | $$
\begin{aligned}
& \angle O_{1} B O_{2}=\angle O_{1} C O_{2}=90^{\circ} \\
\Rightarrow & O_{1} γ C γ O_{2} γ B \text { are concyclic } \\
\Rightarrow & \angle O_{1} C D=\angle O_{1} O_{2} B \\
\Rightarrow & \triangle O_{1} D C \backsim \triangle O_{1} B O_{2} \\
\Rightarrow & O_{1} O_{2}=\frac{O_{1} B \cdot O_{1} C}{... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Among $m$ students, it is known that in any group of three, two of them know each other, and in any group of four, two of them do not know each other. Then the maximum value of $m$ is $\qquad$ | 3. 8 .
When $m=8$, the requirement is satisfied.
It only needs to prove: $m \leqslant 8$.
First, prove that the following two scenarios are impossible.
(1) If a student $A$ knows at least 6 people, by Ramsey's theorem, among these 6 people, there exist 3 people who either all know each other or all do not know each ot... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. If $n$ is a positive integer greater than 1, then
$$
\begin{array}{l}
\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cos \frac{6 \pi}{n}+\cdots+\cos \frac{2 n \pi}{n} \\
=
\end{array}
$$ | 5. 0 .
$$
\sum_{k=1}^{n} \cos \frac{2 k \pi}{n}=\operatorname{Re} \sum_{k=1}^{n} \mathrm{e}^{\frac{2 k \pi i}{n}}=0 .
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 A philanthropist recruits members for his club in the following way: each member can introduce two others to join, where these two are not introduced by anyone else; and each new member can also introduce two others to join. For a member $A$, the members introduced by $A$ and the members introduced by those i... | Solve using reverse thinking.
For this, let $200=k$, and denote the minimum number of members when exactly $r$ people receive coupons as $f(r)$. Members who join without being introduced (or without a superior) are called "bosses", and the members they introduce are called "followers". Each boss along with all their fo... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If the positive integer $m$ makes it true that for any set of positive numbers $a_{1}, a_{2}, a_{3}, a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$
then the minimum value of the pos... | Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$
Verification shows that $m=1, m=2$ do not meet the... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { Example } 6 \text { Let } u=1+\frac{x^{3}}{3!}+\frac{x^{6}}{6!}+\cdots, \\
v=\frac{x}{1!}+\frac{x^{4}}{4!}+\frac{x^{7}}{7!}+\cdots, w=\frac{x^{2}}{2!}+\frac{x^{5}}{5!}+\frac{x^{8}}{8!}+\cdots
\end{array}
$$
Prove: $u^{3}+v^{3}+w^{3}-3 u v w=1$. | Let $\lambda=\mathrm{e}^{\frac{2 \pi}{3}}$ be a unit cube root.
Then $1+\lambda+\lambda^{2}=0$.
Thus $u^{3}+v^{3}+w^{3}-3 u v w$
$$
=(u+v+w)\left(u+\lambda v+\lambda^{2} w\right)\left(u+\lambda^{2} v+\lambda w\right)
$$
(By Corollary 3)
$$
=\mathrm{e}^{x} \mathrm{e}^{\lambda x} \mathrm{e}^{\lambda^{2} x}=\mathrm{e}^{0}... | 1 | Algebra | proof | Yes | Yes | cn_contest | false |
2. Given $x, y, z \in \mathbf{R}_{+}$, satisfying $x^{2}+y^{2}+z^{2}=1$. Then $\min \left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)=$ $\qquad$ . | Notice,
$$
\begin{array}{l}
\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)\left(x^{2}+y^{2}+z^{2}\right) \\
\geqslant 3 \sqrt[3]{\frac{1}{x^{2}} \cdot \frac{1}{y^{2}} \cdot \frac{1}{z^{2}}} \times 3 \sqrt[3]{x^{2} y^{2} z^{2}}=9,
\end{array}
$$
when and onl... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. Calculate:
$$
\frac{2013^{2}+2011}{2011^{2}-2013} \times \frac{4020^{2}-8040}{2011 \times 2014-4}=
$$
$\qquad$ | $=1.4$.
Let $a=2$ 011. Then
$$
\begin{array}{l}
\text { Original expression }=\frac{(a+2)^{2}+a}{a^{2}-a-2} \cdot \frac{(2 a-2)^{2}-4(a-1)}{a(a+3)-4} \\
=\frac{(a+1)(a+4)}{(a-2)(a+1)} \cdot \frac{4(a-1)(a-2)}{(a+4)(a-1)}=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. A sequence of numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is constructed according to the following rule: $a_{1}=7, a_{k}=a_{k-1}^{2}$'s digit sum $+1(k=2,3, \cdots)$. For example, $a_{2}=14, a_{3}=17$, and so on.
Then $a_{2013}=$ $\qquad$ . | 2.8.
From the problem, we know
$$
\begin{array}{l}
a_{1}=7, a_{2}=14, a_{3}=17, a_{4}=0, \\
a_{5}=5, a_{6}=8, a_{7}=11, a_{8}=5 .
\end{array}
$$
Thus, $a_{8}=a_{5}$, meaning from $a_{5}$ onwards, the sequence repeats with a period of 3.
Therefore, $a_{2013}=a_{6+3 \times 669}=8$. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1$, and $a_{2 n}=a_{n}, a_{2 n+1}=a_{n}+1\left(n \in \mathbf{Z}_{+}\right)$.
Then $a_{2013}=$ . $\qquad$ | 2.9.
From the problem, we know
$$
\begin{array}{l}
a_{2013}=a_{1006}+1=a_{503}+1=a_{251}+2 \\
=a_{125}+3=a_{62}+4=a_{31}+4=a_{15}+5 \\
=a_{7}+6=a_{3}+7=a_{1}+8=9 .
\end{array}
$$ | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $M$ be a moving point on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. Given points $F(1,0)$ and $P(3,1)$. Then the maximum value of $2|M F|-|M P|$ is $\qquad$. | 2.1.
Notice that $F$ is the right focus of the ellipse, and the right directrix of the ellipse is $l: x=4$. Then $2|M F|$ is the distance from point $M$ to $l$.
Draw a perpendicular line $M A$ from point $M$ to $l$, and draw a perpendicular line $P B$ from point $P$ to $M A$, where $A$ and $B$ are the feet of the per... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x, y \in \mathbf{R}$, and $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $x+y-x y$ is $\qquad$ . | 3. 1 .
Notice that, $x+y-x y=x(1-y)+y$.
When $y$ is fixed, the expression can only achieve its maximum value when $x$ is as large as possible; similarly, when $x$ is fixed, $y$ should also be as large as possible. Therefore, we might as well assume that $x$ and $y$ are non-negative, and $x^{2}+y^{2}=1$.
Let $x+y=t(t \... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the function
$$
f(x)=A \cos \left(\omega x+\frac{\pi}{4} \omega\right)(A>0)
$$
is decreasing on $\left(0, \frac{\pi}{8}\right)$. Then the maximum value of $\omega$ is | Ni,6.8.
Assume $\omega>0$.
To make $f(x)$ a decreasing function in $\left(0, \frac{\pi}{8}\right)$, combining the image of the cosine-type function, we must have
$$
\begin{array}{l}
\frac{T}{2} \geqslant \frac{\pi}{8} \Rightarrow \frac{\pi}{\omega} \geqslant \frac{\pi}{8} \Rightarrow \omega \leqslant 8 . \\
\text { Whe... | 8 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
1. For the real number $x$, the functions are
$$
f(x)=\sqrt{3 x^{2}+7}, g(x)=x^{2}+\frac{16}{x^{2}+1}-1,
$$
then the minimum value of the function $g(f(x))$ is . $\qquad$ | $-, 1.8$.
From the problem, we have
$$
g(f(x))=3 x^{2}+7+\frac{16}{3 x^{2}+8}-1.
$$
Let $t=3 x^{2}+8(t \geqslant 8)$. Then
$$
h(t)=g(f(x))=t+\frac{16}{t}-2 \text{. }
$$
It is easy to see that $h(t)$ is a monotonically increasing function on the interval $[8,+\infty)$. Therefore, $h(t) \geqslant h(8)=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $a_{1}, a_{2}, \cdots, a_{10}$ and $b_{1}, b_{2}, \cdots, b_{10}$ are 20 distinct real numbers. If the equation
$$
\begin{array}{l}
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right| \\
=\left|x-b_{1}\right|+\left|x-b_{2}\right|+\cdots+\left|x-b_{10}\right|
\end{array}
$$
has a finite numb... | 7.9.
$$
\text { Let } \begin{aligned}
f(x)= & \left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right|- \\
& \left|x-b_{1}\right|-\left|x-b_{2}\right|-\cdots-\left|x-b_{10}\right| .
\end{aligned}
$$
Thus, by the problem statement, $f(x)=0$.
Let $c_{1}<c_{2}<\cdots<c_{20}$ be the elements of the set
$$
\l... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (16 points) Given the sequence $\left\{F_{n}\right\}$ satisfies
$$
\begin{array}{l}
F_{1}=F_{2}=1, \\
F_{n+2}=F_{n+1}+F_{n}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
If $F_{a} γ F_{b} γ F_{c} γ F_{d}(a<b<c<d)$ are the side lengths of a convex quadrilateral, find the value of $d-b$. | From the given, we know that $F_{a}+F_{b}+F_{c}>F_{d}$.
If $c \leqslant d-2$, then
$$
F_{a}+\left(F_{b}+F_{c}\right) \leqslant F_{a}+F_{d-1} \leqslant F_{d},
$$
which is a contradiction.
Therefore, $c=d-1$.
Thus, the side lengths of the quadrilateral are $F_{a} γ F_{b} γ F_{d-1} γ F_{d}$.
If $b \leqslant d-3$, then
$$... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. Let the equation $x^{2}-m x-1=0$ have two real roots $\alpha, \beta (\alpha<\beta)$, and the function $f(x)=\frac{2 x-m}{x^{2}+1}$.
(1) Find the value of $\alpha f(\alpha)+\beta f(\beta)$;
(2) Determine the monotonicity of $f(x)$ in the interval $(\alpha, \beta)$, and provide a proof;
(3) If $\lambda, \mu$ are both... | Three, 13. (1) Given that $\alpha, \beta$ are the roots of the equation $x^{2}-m x-1=0$, we know
$$
\begin{array}{l}
\alpha+\beta=m, \alpha \beta=-1 . \\
\text { Then } f(\alpha)=\frac{2 \alpha-m}{\alpha^{2}+1}=\frac{2 \alpha-(\alpha+\beta)}{\alpha^{2}-\alpha \beta} \\
=\frac{\alpha-\beta}{\alpha(\alpha-\beta)}=\frac{1... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
4. For the convex pentagon $A B C D E$, the side lengths are sequentially $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. It is known that a quadratic trinomial in $x$ satisfies:
When $x=a_{1}$ and $x=a_{2}+a_{3}+a_{4}+a_{5}$, the value of the quadratic trinomial is 5;
When $x=a_{1}+a_{2}$, the value of the quadratic trinomial i... | 4. 0 .
Let the quadratic trinomial be
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {, }
$$
and let the axis of symmetry of its graph be $x=x_{0}$.
By the problem, we know
$$
f\left(a_{1}\right)=f\left(a_{2}+a_{3}+a_{4}+a_{5}\right)=5,
$$
and $\square$
$$
\begin{array}{l}
a_{1} \neq a_{2}+a_{3}+a_{4}+a_{5} . \\
\text { Henc... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Find the smallest integer \( n (n > 1) \), such that there exist \( n \) integers \( a_{1}, a_{2}, \cdots, a_{n} \) (allowing repetition) satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2013 .
$$
12. (20 points) Let positive integers \( a, b, c, d \) satisfy
$$
a^{2}=c(d+13), b^{2}=c(d-1... | 11. Since $a_{1} a_{2} \cdots a_{n}=2013$, it follows that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. From $a_{1}+a_{2}+\cdots+a_{n}=2013$ being odd, we know that $n$ is odd (otherwise, the sum of an even number of odd numbers should be even, which is a contradiction).
If $n=3$, then $a_{1}+a_{2}+a_{3}=a_{1} a_... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Write all positive integers in ascending order in a row. Then the 2013th digit from left to right is | $-1.7$
All single-digit numbers occupy 9 positions, all two-digit numbers occupy $2 \times 90=180$ positions, and next come the three-digit numbers in sequence. Since $2013-9-180=1824$, and $\frac{1824}{3}=608$, because $608+99=707$, the 2013th digit is the last digit of the three-digit number 707, which is 7. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Equation
$$
\sin \pi x=\left[\frac{x}{2}-\left[\frac{x}{2}\right]+\frac{1}{2}\right]
$$
The sum of all real roots of the equation in the interval $[0,2 \pi]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.12. Let $\left\{\frac{x}{2}\right\}=\frac{x}{2}-\left[\frac{x}{2}\right]$. Then for any real number $x$, we have $0 \leqslant\left\{\frac{x}{2}\right\}<1$.
Thus, the original equation becomes
$$
\sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right] \text {. }
$$
(1) If $0 \leqslant\left\{\frac{x}{2}\right\}<... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $f(x)$ is an increasing function on $\mathbf{R}$, and for any $x \in$ $\mathbf{R}$, we have
$$
f\left(f(x)-3^{x}\right)=4 .
$$
Then $f(2)=$ | 8. 10 .
From the problem, we know that $f(x)-3^{x}$ is a constant. Let's assume $f(x)-3^{x}=m$.
Then $f(m)=4, f(x)=3^{x}+m$.
Therefore, $3^{m}+m=4 \Rightarrow 3^{m}+m-4=0$.
It is easy to see that the equation $3^{m}+m-4=0$ has a unique solution $m=1$.
Thus, $f(x)=3^{x}+1$,
and hence, $f(2)=3^{2}+1=10$. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that the elements of set $A$ are all integers, the smallest is 1, and the largest is 200, and except for 1, every number in $A$ is equal to the sum of two numbers (which may be the same) in $A$. Then the minimum value of $|A|$ is $\qquad$ ( $|A|$ represents the number of elements in set $A$). | 9. 10 .
It is easy to know that the set
$$
A=\{1,2,3,5,10,20,40,80,160,200\}
$$
meets the requirements, at this time, $|A|=10$.
Next, we will show that $|A|=9$ does not meet the requirements.
Assume the set
$$
A=\left\{1, x_{1}, x_{2}, \cdots, x_{7}, 200\right\},
$$
where $x_{1}<x_{2}<\cdots<x_{7}$ meets the require... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. Given the set
$$
P=\left\{x \mid x=7^{3}+a \times 7^{2}+b \times 7+c, a γ b γ c\right. \text { are positive integers not }
$$
exceeding 6 $\}$.
If $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ elements in set $P$ that form an arithmetic sequence, find the maximum value of $n$. | 15. (1) Clearly,
$$
\begin{array}{l}
7^{3}+7^{2}+7+1, 7^{3}+7^{2}+7+2, \\
7^{3}+7^{2}+7+3, 7^{3}+7^{2}+7+4, \\
7^{3}+7^{2}+7+5, 7^{3}+7^{2}+7+6
\end{array}
$$
These six numbers are in the set $P$ and form an arithmetic sequence.
(2) Prove by contradiction: Any seven different numbers in the set $P$ cannot form an arit... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $A=\{2,4, \cdots, 2014\}, B$ be any non-empty subset of $A$, and $a_{i} γ a_{j}$ be any two elements in set $B$. There is exactly one isosceles triangle with $a_{i} γ a_{j}$ as side lengths. Then the maximum number of elements in set $B$ is $\qquad$ | 7. 10 .
By symmetry, without loss of generality, assume $a_{i}<a_{j}$. Then there must exist an isosceles triangle with $a_{j}$ as the waist and $a_{i}$ as the base, and there is only one isosceles triangle with $a_{i}$ and $a_{j}$ as side lengths.
Thus, $a_{i}+a_{i} \leqslant a_{j} \Rightarrow a_{j} \geqslant 2 a_{i}... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 2, given that $D$ is the intersection of the tangents to the circumcircle $\odot O$ of $\triangle A B C$ at points $A$ and $B$, the circumcircle of $\triangle A B D$ intersects line $A C$ and segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. I... | Let $C D$ intersect $A B$ at point $M$.
If the center $O$ is not on the line segment $B C$, from $\frac{B C}{B F}=2$, we know that $F$ is the midpoint of $B C$.
Then, by the perpendicular diameter theorem,
$$
F O \perp B C \Rightarrow \angle O F B=90^{\circ} \text {. }
$$
Since $D A$ and $D B$ are tangent to $\odot O$... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. In $\triangle A B C$, it is known that $a, b, c$ are the sides opposite to $\angle A$, $\angle B$, and $\angle C$ respectively, and $a c+c^{2}=b^{2}-a^{2}$. If the longest side of $\triangle A B C$ is $\sqrt{7}$, and $\sin C=2 \sin A$, then the length of the shortest side of $\triangle A B C$ is $\qquad$. | 6. 1 .
From $a c+c^{2}=b^{2}-a^{2}$
$\Rightarrow \cos B=-\frac{1}{2} \Rightarrow \angle B=\frac{2 \pi}{3}$.
Thus, the longest side is $b$.
Also, $\sin C=2 \sin A \Rightarrow c=2 a$.
Therefore, $a$ is the shortest side.
By the cosine rule,
$$
(\sqrt{7})^{2}=a^{2}+4 a^{2}-2 a \times 2 a \times\left(-\frac{1}{2}\right) \... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has one focus at $F_{1}(-\sqrt{3}, 0)$, and passes through the point $H\left(\sqrt{3}, \frac{1}{2}\right)$. Let the upper and lower vertices of the ellipse $E$ be $A_{1}$ and $A_{2}$, respectively, and let $P$ be any point on the ellipse differ... | 12. From the problem, we have
$$
a^{2}-b^{2}=3, \frac{3}{a^{2}}+\frac{1}{4 b^{2}}=1 \text {. }
$$
Solving, we get $a^{2}=4, b^{2}=1$.
Thus, the equation of the ellipse $E$ is $\frac{x^{2}}{4}+y^{2}=1$.
From this, we know the points $A_{1}(0,1), A_{2}(0,-1)$. Let point $P\left(x_{0}, y_{0}\right)$.
Then $l_{P A_{1}}: y... | 2 | Geometry | proof | Yes | Yes | cn_contest | false |
15. Given that $a, b, c$ are positive real numbers. Prove:
$$
\frac{\sqrt{a^{2}+3 b c}}{a}+\frac{\sqrt{b^{2}+3 a c}}{b}+\frac{\sqrt{c^{2}+3 a b}}{c} \geqslant 6 \text {. }
$$ | 15. Notice,
$$
\begin{array}{l}
\frac{\sqrt{a^{2}+3 b c}}{a}=\frac{\sqrt{a^{2}+b c+b c+b c}}{a} \\
\geqslant \frac{\sqrt{4 \sqrt[4]{a^{2} b^{3} c^{3}}}}{a}=\frac{2 \sqrt[8]{a^{2} b^{3} c^{3}}}{a} .
\end{array}
$$
Similarly, $\frac{\sqrt{b^{2}+3 a c}}{b} \geqslant \frac{2 \sqrt[8]{a^{3} b^{2} c^{3}}}{b}$,
$$
\begin{arr... | 6 | Inequalities | proof | Yes | Yes | cn_contest | false |
6. Given $a, b, c \in \mathbf{R}_{+}$, and
$$
a+b+c=12, a b+b c+c a=45 \text{. }
$$
Then $\min \max \{a, b, c\}=$ $\qquad$ | 6. 5 .
Let $a=\max \{a, b, c\}$.
From $a+b+c=12$, we get $a \geqslant 4$.
$$
\begin{array}{l}
\text { By }(a-b)(a-c) \geqslant 0 \\
\Rightarrow a^{2}-a(12-a)+b c \geqslant 0 \\
\Rightarrow b c \geqslant 12 a-2 a^{2} . \\
\text { Also } 45=a b+b c+c a=b c+a(12-a) \\
\geqslant 12 a-2 a^{2}+a(12-a),
\end{array}
$$
Then ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. The unit digit of $\left[\frac{10^{10000}}{10^{100}+9}\right]$ is | 8. 1 .
Notice that,
$$
\frac{10^{10000}}{10^{100}+9}=\frac{\left(10^{100}\right)^{100}-3^{200}}{10^{100}+9}+\frac{3^{200}}{10^{100}+9} \text {. }
$$
And $\left(10^{100}\right)^{100}-3^{200}=\left[\left(10^{100}\right)^{2}\right]^{50}-\left(9^{2}\right)^{50}$, so $\left[\left(10^{100}\right)^{2}-9^{2}\right] \mid\left... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Given $\odot O: x^{2}+y^{2}=4$, circle $M$ :
$$
(x-5 \cos \theta)^{2}+(y-5 \sin \theta)^{2}=1(\theta \in \mathbf{R}) \text {, }
$$
Through any point $P$ on circle $M$, draw two tangents $P E$ and $P F$ to $\odot O$, with the points of tangency being $E$ and $F$. Try to find the minimum value of $\overri... | 9. The center of circle $M$ is on the circle $x^{2}+y^{2}=25$.
Let $|P E|=|P F|=d$.
In the right triangle $\triangle P E O$, it is easy to see that
$4 \leqslant|P O| \leqslant 6,|O E|=2$.
Thus, $2 \sqrt{3} \leqslant d \leqslant 4 \sqrt{2}$.
Also, $\overrightarrow{P E} \cdot \overrightarrow{P F}=|\overrightarrow{P E}||\... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let
$$
\begin{array}{l}
f(x)=a_{1} x^{2013}+a_{2} x^{2012}+\cdots+a_{2013} x+a_{2014} \\
=x^{13}\left(x^{10}+x^{2}+x\right)^{2000}, \\
b_{0}=1, b_{1}=2, b_{n+2}+b_{n}=b_{n+1}(n \in \mathbf{N}) .
\end{array}
$$
Find the value of $\sum_{i=1}^{2013} a_{i} b_{i}$. | 11. From the recurrence relation of $\left\{b_{n}\right\}$, we have
$$
\begin{array}{l}
b_{2}=1, b_{3}=-1, b_{4}=-2, b_{5}=-1, \\
b_{6}=1=b_{0}, b_{7}=2=b_{1} .
\end{array}
$$
Thus, $b_{6 k}=1, b_{6 k+1}=2, b_{6 k+2}=1, b_{6 k+3}=-1$, $b_{6 k+4}=-2, b_{6 k+5}=-1(k \in \mathbf{N})$.
Let $\lambda=\frac{1}{2}+\frac{\sqrt... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. If the complex number $x$ satisfies $x+\frac{1}{x}=-1$, then $x^{2013}+\frac{1}{x^{2013}}=$ $\qquad$ . | 8. 2 .
Given that $x^{2}+x+1=0$.
Since the discriminant $\Delta=-3<0$, $x$ is a complex number.
Also, $x^{3}-1=(x-1)\left(x^{2}+x+1\right)=0$
$\Rightarrow x^{3}=1$.
Therefore, $x^{2013}+\frac{1}{x^{2013}}=\left(x^{3}\right)^{671}+\frac{1}{\left(x^{3}\right)^{671}}=1+1=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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