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Example 4 Let real numbers $a, b$ satisfy
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {. }
$$
Find the minimum value of $u=9 a^{2}+72 b+2$. | Notice,
$$
\begin{array}{l}
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \\
\Rightarrow(a-2 b)(3 a-4 b+5)=0 \\
\Rightarrow a-2 b=0 \text { or } 3 a-4 b+5=0 .
\end{array}
$$
(1) $a-2 b=0$.
Then $u=9 a^{2}+72 b+2=36 b^{2}+72 b+2$ $=36(b+1)^{2}-34$.
Thus, when $b=-1$, the minimum value of $u$ is -34.
$$
\text { (2) } 3 a-4 b+5=0 \t... | -34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,999 |
1. Given an arithmetic sequence $\left\{a_{n}\right\}(n=1,2, \cdots)$ with the first term $a$ and common difference $b$, and a geometric sequence $\left\{b_{n}\right\}$ with the first term $b$ and common ratio $a$, where $a$ and $b$ are both positive numbers, and $a_{1}<b_{1}<a_{2}<b_{2}<a_{3}$.
(1) Find the value of $... | (1) $a=2 ;(2) b=3$;
(3) $a_{1}+a_{2}+\cdots+a_{m}=2^{n-2}+3 \times 2^{2 n-3}$. | a=2, b=3, a_{1}+a_{2}+\cdots+a_{m}=2^{n-2}+3 \times 2^{2 n-3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,000 |
Example 1 Given
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac... | Solve: Consider the given system of equations as a fractional equation in terms of $t$
$$
\begin{array}{c}
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1 \\
\Rightarrow\left(t-1^{2}\right)\left(t-3^{2}\right)\left(t-5^{2}\right)\left(t-7^{2}\right)- \\
x^{2}\left(t-3^{2}\right... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,001 |
Example 2 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This line is not translated as it seems to be an instruction and not part of the text to be translated. If you need this line translated as well, please l... | Notice,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1000} \text {. }
$$
Consider the integer-coefficient quadratic equation $x^{2}-10 x+1=0$ with $5+2 \sqrt{6}$ as one of its roots, the other root being $5-2 \sqrt{6}$.
$$
\begin{array}{l}
\text { Let } a=5+2 \sqrt{6}, b=5-2 \sqrt{6}, \\
u_{n}=a^{n}+b^{n}(n=1,2, \cdo... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,002 |
Example 3 Let $a_{1}, a_{2}, \cdots, a_{2012}, b_{1}, b_{2}, \cdots, b_{2012}$ be distinct real numbers, and fill them into a $2012 \times 2012$ grid according to the following method: fill the number $a_{i}+b_{j}$ into the cell at the $i$-th row and $j$-th column. Let $P_{i}$ be the product of the numbers in the $i$-t... | Proof Let $P_{1}=P_{2}=\cdots=P_{2012}=d$, that is,
$$
\left(a_{i}+b_{1}\right)\left(a_{i}+b_{2}\right) \cdots\left(a_{i}+b_{2012}\right)=d,
$$
where $i=1,2, \cdots, 2012$.
Thus, $a_{1}, a_{2}, \cdots, a_{2012}$ are the 2012 distinct roots of the polynomial
$$
f(x)=\left(x+b_{1}\right)\left(x+b_{2}\right) \cdots\left(... | P_{i}+Q_{j}=0 | Algebra | proof | Yes | Yes | cn_contest | false | 726,003 |
Example 4 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{2}=7, a_{n+2}=\frac{a_{n+1}^{2}-1}{a_{n}} \text {. }
$$
Prove: For all $n \in \mathbf{N}_{+}, a_{n}$ are positive integers, and $9 a_{n} a_{n+1}+1$ is a perfect square. | Proof: Given $a_{3}=48$.
For all $n$, $a_{n+1}^{2}=a_{n} a_{n+2}+1$.
$$
\begin{array}{l}
\text { Also, } \frac{a_{n-1}+a_{n+1}}{a_{n}}=\frac{a_{n-1} a_{n+1}+a_{n+1}^{2}}{a_{n} a_{n+1}} \\
=\frac{\left(a_{n}^{2}-1\right)+\left(a_{n} a_{n+2}+1\right)}{a_{n} a_{n+1}}=\frac{a_{n}+a_{n+2}}{a_{n+1}} \\
\Rightarrow \frac{a_{n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,004 |
Example 5 In an acute $\triangle A B C$, prove:
$$
\sin A+\sin B+\sin C>\cos A+\cos B+\cos C \text {. }
$$ | To prove the problem by converting it to trigonometric functions, consider the circumcircle $\odot O$ of $\triangle ABC$.
As shown in Figure 1, assume $\odot O$ is a unit circle, and $D, E, F$ are the midpoints of sides $BC, CA, AB$ respectively.
Then, $\cos A = OD, \sin A = EF, \cos B = OE, \sin B = FD, \cos C = OF, ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,005 |
Example 6 Suppose there are $m+n$ positive integers $a_{1}, a_{2}, \cdots, a_{m}$ and $b_{1}, b_{2}, \cdots, b_{n}$, satisfying $\sum_{i=1}^{m} a_{i}=\sum_{j=1}^{n} b_{j}$. Prove: it is possible to select $m+n-1$ cells from an $m \times n$ grid, and fill each selected cell with a natural number, such that the sum of th... | Proof: Let $\sum_{i=1}^{m} a_{i}=\sum_{j=1}^{n} b_{j}=S\left(S \in \mathbf{N}_{+}\right)$.
In the first quadrant of the coordinate system, construct a square $O A C B$ with side length $S$, where the rays $O A$ and $O B$ are the positive directions of the $x$-axis and $y$-axis, respectively.
On $O A$, take $m-1$ poin... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,006 |
In $\triangle A B C$, let the angle bisectors of $\angle A$ and $\angle C$ intersect at point $I$, and intersect $C B$ and $A B$ at points $A_{1}$ and $C_{1}$, respectively. They also intersect the circumcircle of $\triangle A B C$ at points $A_{2}, C_{2}$. Let $K$ be the intersection of $A_{1} C_{2}$ and $A_{2} C_{1}$... | Proposition Let $P$ be a point in the plane of $\triangle ABC$, and $BP, CP$ intersect the opposite sides $CA, AB$ at points $E, F$, respectively, and intersect the circumcircle of $\triangle ABC$ at another point $M, N$. $EN$ and $FM$ intersect at point $Q$, and $QP$ intersects $BC$ at point $R$. Let $\angle PAB, \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,008 |
1. Two circles $\Gamma_{1}$ and $\Gamma_{2}$ with unequal radii intersect at points $A$ and $B$. Points $C$ and $D$ lie on circles $\Gamma_{1}$ and $\Gamma_{2}$, respectively, and segment $CD$ has $A$ as its midpoint. Extend $DB$ to intersect circle $\Gamma_{1}$ at point $E$, and extend $CB$ to intersect circle $\Gamma... | 1. (1) Since $C, A, B, E$ and $D, A, B, F$ are each four points on a circle, and $C A = A D$, by the power of a point theorem, we have
$$
C B \cdot C F = C A \cdot C D = D A \cdot D C = D B \cdot D E.
$$
Assume $l_{1}$ and $l_{2}$ do not intersect. Then
$$
C D \parallel E F \Rightarrow \frac{C F}{C B} = \frac{D E}{D B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,009 |
Example 5 Let $x$ be a real number. Then
$$
|x-1|+|x+1|+|x+5|
$$
the minimum value is $\qquad$ (s) | Let $y=|x-1|+|x+1|+|x+5|$. When $x<-5$,
$$
y=1-x-x-1-x-5=-3 x-5 \text{; }
$$
When $-5 \leqslant x<-1$,
$$
y=1-x-x-1+x+5=-x+5 \text{; }
$$
When $-1 \leqslant x<1$,
$$
y=1-x+x+1+x+5=x+7 \text{; }
$$
When $x \geqslant 1$,
$$
y=x-1+x+1+x+5=3 x+5 \text{. }
$$
By the properties of linear functions, we know that when $x=-... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,010 |
2. Determine all non-empty sets $S$ of integers that satisfy: if $m, n \in S$ (m and n can be the same), then $3 m-2 n \in S$.
| 2. If the set $S$ contains only one integer, then the set $S$ satisfies the condition.
Below, assume the set $S$ contains at least two different integers. Then, the difference between any two numbers in $S$ has a minimum positive value, denoted as $d$. Thus, there exists an integer $a$ such that
$$
\begin{array}{l}
a+... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,011 |
3. Find all positive real numbers $t$ such that there exists an infinite set $X$ of real numbers, for which, for any $x, y, z \in X$ (where $x, y, z$ can be the same), and any real number $a$ and positive real number $d$, we have
$$
\max \{|x-(a-d)|,|y-a|,|z-(a+d)|\}>t d
$$ | 3. (1) $01$.
Let $x_{i}=\lambda^{i}, X=\left\{x_{1}, x_{2}, \cdots\right\}$.
Assume there exist $a \in \mathbf{R}, d \in \mathbf{R}$. and $x_{i}, x_{j}, x_{k}$ satisfying
$$
\begin{array}{l}
\max \left\{\left|x_{i}-(a-d)\right|,\left|x_{j}-a\right|,\left|x_{k}-(a+d)\right|\right\} \leqslant t d, \\
\left\{\begin{array}... | 0<t<\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,012 |
4. Given an integer $n \geqslant 2$. Let $n$ non-empty finite sets $A_{1}, A_{2}, \cdots, A_{n}$ satisfy:
$$
\left|A_{i} \Delta A_{j}\right|=|i-j|(i, j \in\{1,2, \cdots, n\}),
$$
where
$$
X \Delta Y=\{a \mid a \in X, a \notin Y\} \cup\{a \mid a \in Y, a \notin X\} .
$$
Find the minimum value of $\left|A_{1}\right|+\l... | 4. For integer $n \geqslant 2$, prove that for any positive integer $k$,
$$
S_{2 k} \geqslant k^{2}+2, S_{2 k+1} \geqslant k(k+1)+2 \text {. }
$$
First, give two obvious conclusions:
(1) For finite sets $X$ and $Y$,
$$
|X|+|Y| \geqslant|X \Delta Y| \text {. }
$$
(2) If non-empty finite sets $X$ and $Y$ satisfy $|X \De... | \left[\frac{n^{2}}{4}\right]+2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,013 |
5. For a positive integer $n$ and an integer $i (0 \leqslant i \leqslant n)$, let $\mathrm{C}_{n}^{i}=c(n, i)(\bmod 2)(c(n, i) \in\{0,1\})$.
Let $f(n, q)=\sum_{i=0}^{n} c(n, i) q^{i}$.
Suppose $m, n, q \in \mathbf{N}_{+}$, and $q+1$ is not a power of 2. Prove: If $f(m, q) \mid f(n, q)$, then for any positive integer $r... | For any positive integer $u$, let the binary representation of $u$ be
$$
u=2^{d_{4}}+2^{d_{2}}+\cdots+2^{d_{k}},
$$
where $0 \leqslant d_{1}1$.
Proof of Proposition 2: In fact, the case $i=0$ is the given condition, from which we also know that $q>1$.
When $i>0$, note that $q^{2 i}+1 \equiv 1$ or $2(\bmod 4)$.
Since ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,014 |
6. Given positive integers $m, n$. Find the smallest integer $N (N \geqslant m)$: if an $N$-element integer set contains a complete residue system modulo $m$, then it has a non-empty subset whose elements sum to a multiple of $n$.
（Zhai Zhenhua, problem contributor） | 6. $\left.N=\max \left\{m, m+n-\frac{1}{2} m[m, n)+1\right]\right\}$
First, prove:
$N \geqslant \max \left\{m, m+n-\frac{1}{2} m[(m, n)+1]\right\}$. (1)
Let $d=(m, n), m=d m, n=d n$,
If $n>\frac{1}{2} m(d+1)$, consider a complete residue system modulo $m$: $x_{1}, x_{2}, \cdots, x_{n}$, whose remainders are $m, \uparr... | N=\max \left\{m, m+n-\frac{1}{2} m[(m, n)+1]\right\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,015 |
1. Given that the real number $x$ satisfies for any positive number $a$, $x^{2}<1+a$. Then the minimum value of $x$ is ( ).
(A) 0
(B) 1
(C) -1
(D) Does not exist: | -, 1. C.
From $a>0 \Rightarrow a+1>1$.
Then $x^{2}<1+a \Leftrightarrow x^{2} \leqslant 1$.
Therefore $-1 \leqslant x \leqslant 1$. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 726,016 |
2. Four functions
$$
\begin{array}{l}
y=\sin |x|, y=\cos |x|, \\
y=|\tan x|, y=-\ln |\sin x|
\end{array}
$$
with a period of $\pi$, which is monotonically decreasing on $\left(0, \frac{\pi}{2}\right)$ and is an even function is ( ).
(A) $y=\sin |x|$
(B) $y=\cos |x|$
(C) $y=|\tan x|$
(D) $y=-\ln |\sin x|$ | 2. D.
Since $y=\sin |x|$ is not a periodic function, option A is incorrect;
Since $y=\cos |x|$ is a periodic function with the smallest positive period of $2 \pi$, option B is incorrect;
Since $y=|\tan x|$ is monotonically increasing in $\left(0, \frac{\pi}{2}\right)$, option C is incorrect;
The function $y=-\ln |\s... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,017 |
3. Let the sequence $\left\{a_{n}\right\}$ have 11 terms,
$$
a_{1}=0, a_{11}=4 \text {, }
$$
and $\left|a_{k+1}-a_{k}\right|=1(k=1,2, \cdots, 10)$. Then the number of different sequences that satisfy the conditions is ( ).
(A) 100
(B) 120
(C) 140
(D) 160 | 3. B.
According to the problem, we have $a_{k+1}-a_{k}=1$ or $a_{k+1}-a_{k}=-1$. If there are $m$ ones, then there are $10-m$ negative ones. Therefore, $4=m-(10-m) \Rightarrow m=7$.
Thus, the number of sequences we are looking for is $\mathrm{C}_{10}^{7}=120$. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,018 |
4. Let $r>1$ be a real number. If the moving point $z$ on the complex plane satisfies $|z|=r$, then the trajectory of the moving point $\omega=z+\frac{1}{z}$ is an ellipse with a focal distance of ( ).
(A) 4
(B) $\frac{4}{r}$
(C) 2
(D) $\frac{2}{r}$ | 4. A.
Let $z=r(\cos \theta+\mathrm{i} \sin \theta)(\theta \in[0,2 \pi))$. Then $\omega=z+\frac{1}{z}=\left(r+\frac{1}{r}\right) \cos \theta+\mathrm{i}\left(r-\frac{1}{r}\right) \sin \theta$. Let $\omega=x+\mathrm{i} y$. Thus,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{x}{r+\frac{1}{r}}=\cos \theta, \\
\frac{y}{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,019 |
5. If the line $a x-b y+1=0$ bisects the circumference of $C$ : $x^{2}+y^{2}+2 x-4 y+1=0$, then the range of values for $a b$ is ( ).
(A) $\left(-\infty, \frac{1}{4}\right]$
(B) $\left(-\infty, \frac{1}{8}\right]$
(C) $\left(0, \frac{1}{4}\right]$
(D) $\left(0, \frac{1}{8}\right]$ | 5. B.
Since the circumference of the circle is bisected by the line, the center of the circle $(-1,2)$ lies on the given line, i.e.,
$$
\begin{array}{c}
-a-2 b+1=0 \Rightarrow a+2 b=1 \\
\Rightarrow a b=(1-2 b) b=-2 b^{2}+b \\
=-2\left(b-\frac{1}{4}\right)^{2}+\frac{1}{8} \leqslant \frac{1}{8} .
\end{array}
$$
Equali... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,020 |
Example 6 If $y=\sqrt{1-x}+\sqrt{x-\frac{1}{2}}$ has a maximum value of $a$ and a minimum value of $b$, then $a^{2}+b^{2}=$ $\qquad$ [6] | Solve: From $1-x \geqslant 0, x-\frac{1}{2} \geqslant 0 \Rightarrow \frac{1}{2} \leqslant x \leqslant 1$.
According to the problem, we have
$$
\begin{aligned}
y^{2} & =\left(\sqrt{1-x}+\sqrt{x-\frac{1}{2}}\right)^{2} \\
& =\frac{1}{2}+2 \sqrt{(1-x)\left(x-\frac{1}{2}\right)}: \\
& =\frac{1}{2}+2 \sqrt{-\left(x-\frac{3}... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,021 |
6. If a sphere of radius $R$ contains four identical smaller spheres of radius $r$, then the maximum possible value of $r$ is $(\quad)$.
(A) $\frac{\sqrt{3}}{2+\sqrt{3}} R$
(B) $\frac{\sqrt{6}}{3+\sqrt{6}} R$
(C) $\frac{1}{1+\sqrt{3}} R$
(D) $\frac{\sqrt{5}}{2+\sqrt{5}} R$ | 6. B.
As shown in Figure 2, connecting the centers of the four small balls $O_{1}$, $O_{2}$, $O_{3}$, and $O_{4}$ forms a tetrahedron. Let the projection of point $O_{1}$ on the base $O_{2}O_{3}O_{4}$ be $H$.
Then the center of the large ball $O$ lies on $O_{1}H$, and
$$
\begin{array}{l}
O_{2}H=\frac{\sqrt{3}}{3} \ti... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,022 |
7. Given the sets
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(x-1)^{2}+(y-2)^{2} \leqslant \frac{4}{5}\right.\right\}, \\
B=\{(x, y)|| x-1|+2| y-2 \mid \leqslant a\},
\end{array}
$$
and $A \subseteq B$. Then the range of the real number $a$ is | $$
\text { II.7. } a \geqslant 2 \text {. }
$$
Assume $x-1=x_{1}, y-2=y_{1}$. Then
$$
\begin{array}{l}
A=\left\{\left(x_{1}, y_{1}\right) \left\lvert\, x_{1}^{2}+y_{1}^{2} \leqslant \frac{4}{5}\right.\right\}, \\
B=\left\{\left(x_{1}, y_{1}\right)|| x_{1}|+2| y_{1} \mid \leqslant a\right\} . \\
\text { Let } x_{1}=t \... | a \geqslant 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,023 |
8. The solution to the equation $3 \times 16^{2}+2 \times 81^{x}=5 \times 36^{x}$ is | 8.0 or $\frac{1}{2}$.
The original equation can be transformed into
$$
3\left[\left(\frac{4}{9}\right)^{x}\right]^{2}-5\left(\frac{4}{9}\right)^{x}+2=0 \text {. }
$$
Let $\left(\frac{4}{9}\right)^{x}=t$. Then
$$
\begin{array}{l}
3 t^{2}-5 t+2=0 \Rightarrow t_{1}=1, t_{2}=\frac{2}{3} \\
\Rightarrow\left(\frac{4}{9}\ri... | x=0 \text{ or } \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,024 |
9. For any real number $x$, we have
$$
f(x)=\log _{a}\left(2+\mathrm{e}^{\mathrm{x}-1}\right) \leqslant-1 \text {, }
$$
then the range of real number $a$ is $\qquad$ | 9. $\left[\frac{1}{2}, 1\right)$.
When $0 < a < 1$, the condition is always satisfied.
When $a > 1$, from the condition we get
$$
2+\mathrm{e}^{x-1} \leqslant \frac{1}{a} \Rightarrow a \leqslant \frac{1}{2+\mathrm{e}^{x-1}}
$$
which is impossible to hold constantly.
Therefore, the range of real number $a$ is $\left[... | \left[\frac{1}{2}, 1\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,025 |
10. $\mathrm{U}$ knows
$$
a+3 b \perp 7 a-5 b \text {, and } a-4 b \perp 7 a-2 \text {. }
$$
Then the angle between $a$ and $b$ is $\qquad$ . | 10. $\frac{\pi}{3}$.
According to the problem, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
(a+3 b) \cdot(7 a-5 b)=0, \\
(a-4 b) \cdot(7 a-2 b)=0
\end{array}\right. \\
\Rightarrow|a|^{2}=|b|^{2}=2 a \cdot b \\
\Rightarrow \cos \langle a, b\rangle=\frac{a \cdot b}{|a||b|}=\frac{1}{2} \\
\Rightarrow\langle a, b\r... | \frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,026 |
11. In a square $A B C D$ with side length 4, if it is folded along $B D$ to form a $60^{\circ}$ dihedral angle, then the distance between the midpoint of side $B C$ and point $A$ is $\qquad$ | 11. $2 \sqrt{2}$.
Take the midpoints $O$ and $E$ of $B D$ and $B C$, respectively, and connect $A O$ and $C O$. Then, $\angle A O C=60^{\circ}$. Therefore,
$$
A C=A O=C O=2 \sqrt{2} \text {. }
$$
In $\triangle A B C$, by the median length formula, we have
$$
\begin{array}{l}
A E^{2}=\frac{1}{2}\left(A B^{2}+A C^{2}\r... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,027 |
13. (16 points) Given $p>0$, a line passing through the focus of the parabola $y^{2}=2 p x$ has a slope of $k$, and intersects the parabola at points $A$ and $B$. Let $f(k)=|A B|$.
(1) Try to find the analytical expression for $f(k)$.
(2) Does there exist a point $C$ on the parabola $1:$ such that $\triangle A B C$ is ... | Three, 13. (1) The focus of the parabola is $F\left(\frac{p}{2}, 0\right)$. Suppose the equation of the line is $y=k\left(x-\frac{p}{2}\right)$, substituting into $y^{2}=2 p x$, we get $k^{2} x^{2}-\left(p k^{2}+2 p\right) x+\frac{p^{2} k^{2}}{4}=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$x... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,029 |
14. (16 points) As shown in Figure 1, in a certain engineering project to measure the radius $R$ of an arc, two identical small balls are placed on the arc so that each contact point is tangent to the arc. The height difference between the balls is $h$, and the radius of the small balls is $r$. Try to express $R$ in te... | 14. Let the angle between the line connecting the centers of the two smaller circles and the line connecting the center of the larger circle be $\theta$. Then
$$
\cos \theta=\frac{(R-r)^{2}+(R-r)^{2}-(2 r)^{2}}{2(R-r)^{2}} .
$$
Also, $h=(R-r)-(R-r) \cos \theta=\frac{2 r^{2}}{R-r}$
$$
\Rightarrow R=r+\frac{2 r^{2}}{h} ... | 600 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,030 |
15. (20 points) Given the function $f(x)=\mathrm{e}^{x}-x$.
(1) If the function $g(x)=f(x)-a x^{2}-1$ has a derivative $g^{\prime}(x)$ that is increasing on $[0,+\infty)$, find the maximum value of the real number $a$;
(2) Prove:
$$
f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)+\cdots+f\left(\frac{1}{n}\right)>n\... | 15. (1) From
$$
g^{\prime}(x)=f^{\prime}(x)-2 a x=\left(\mathrm{e}^{x}-1\right)-2 a x
$$
being an increasing function on $[0,+\infty)$, we know that
$$
\left[g^{\prime}(x)\right]^{\prime}=\mathrm{e}^{x}-2 a \geqslant 0 \Rightarrow a \leqslant \frac{1}{2} \text {. }
$$
Thus, the maximum value of $a$ is $\frac{1}{2}$.
... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 726,031 |
Example 7 Given that $x_{1}, x_{2}, \cdots, x_{40}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{40}=58$. If the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ is $A$, and the minimum value is $B$, then $A+B=$ $\qquad$ | Solution: Since there are only a finite number of ways to write 58 as the sum of 40 positive integers, the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ exist.
Assume without loss of generality that $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{40}$.
If $x_{1}>1$, then
$$
x_{1}+x_{2}=\lef... | 494 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,032 |
16. (20 points) Let $\{a_n\}$ be a positive, increasing arithmetic sequence. Prove:
(1) For any $k, l \in \mathbf{N}_{+}$, when $l > k \geqslant 2$, the inequality
$$
\frac{a_{l+1}}{a_{k+1}} < \frac{a_{l}}{a_{k}} < \frac{a_{l-1}}{a_{k-1}}
$$
holds;
(2) For any $k \in \mathbf{N}_{+}$, when $k \geqslant 2$, the inequalit... | 16. (1) Given $a_{l}>a_{k}>0$, common difference $d>0$, we know
$$
\frac{a_{l+1}}{a_{k+1}}=\frac{a_{l}+d}{a_{k}+d}a_{2012 k+1}>0$, common difference $d>0$, we know
$$
\begin{array}{l}
\frac{a_{2012 k+2}}{a_{2012 k+1}}=\frac{a_{2012 k+1}+d}{a_{2012 k}+d}\prod_{n=1}^{2012} \prod_{m=1}^{k} \frac{a_{n k+m+1}}{a_{n k+m}}=\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,033 |
1. Four points $A, B, C, D$ in space are pairwise 1 unit apart, and points $P, Q$ move on line segments $AB, CD$ respectively. The minimum distance between point $P$ and $Q$ is | $-1 . \frac{\sqrt{2}}{2}$.
From the problem, we know that the tetrahedron $ABCD$ is a regular tetrahedron. Therefore, finding the minimum distance between points $P$ and $Q$ is equivalent to finding the distance between $AB$ and $CD$, as shown in Figure 1.
Take the midpoints $E$ and $F$ of $AB$ and $CD$, respectively... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,034 |
2. Let the vectors $\overrightarrow{O A}=(1,0), \overrightarrow{O B}=(1,1)$, and $O$ be the origin. A moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O A} \leqslant 1, \\
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O B} \leqslant 2
\end{array... | 2. 2 .
From the problem, the point $P(x, y)$ satisfies
$$
\begin{array}{l}
\left\{\begin{array}{l}
0 \leqslant x \leqslant 1, \\
0 \leqslant x+y \leqslant 2 .
\end{array}\right. \\
\text { Let }\left\{\begin{array}{l}
x+y=u, \\
y=v .
\end{array}\right.
\end{array}
$$
Then the point $Q(u, v)$ satisfies
$$
\left\{\begi... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,035 |
3. Let there be a non-empty set $A \subseteq\{1,2, \cdots, 7\}$, and when $a \in A$, it must also be that $8-a \in A$. Then the number of such sets $A$ is $\qquad$ . | 3. 15 .
Find the single element or binary element set that is congruent to $8-a$ in $A$:
$$
\begin{array}{l}
A_{1}=\{4\}, A_{2}=\{1,7\}, \\
A_{3}=\{2,6\}, A_{4}=\{3,5\} .
\end{array}
$$
The problem is equivalent to finding the number of non-empty subsets of $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$.
Thus, there are... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,036 |
4. Let $f(x)=\left\{\begin{array}{ll}x-[x], & x \leqslant 0 ; \\ f(x-1), & x>0,\end{array}\right.$ where $[x]$ denotes the greatest integer not exceeding the real number $x$. If $f(x)=k x+k(k>0)$
has: different real roots, then the range of the real number $k$ is . $\qquad$ | 4. $k \in\left[\frac{1}{4}, \frac{1}{3}\right)$. | \left[\frac{1}{4}, \frac{1}{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,037 |
5. Given the first seven digits of an 11-digit mobile phone number are 1390931. If the remaining four digits can only be 1, 3, 5 and each must appear at least once, then there are such mobile phone numbers.
untranslated: 个.
Note: The word "个" at the end of the sentence is not translated as it is a placeholder for t... | 5. 36 . | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,038 |
6. If $\tan x_{1} \cdot \tan x_{2} \cdots \cdots \tan x_{2012}=1$, then the maximum value of $\sin x_{1} \cdot \sin x_{2} \cdots \cdots \sin x_{2012}$ is | 6. $\frac{1}{2^{1006}}$.
From $\tan x_{1} \cdot \tan x_{2} \cdots \cdot \tan x_{2012}=1$, we get
$$
\begin{array}{l}
\sin x_{1} \cdot \sin x_{2} \cdots \cdots \sin x_{2012} \\
=\cos x_{1} \cdot \cos x_{2} \cdots \cdots \cos x_{2012} .
\end{array}
$$
Therefore, $2^{2012}\left(\sin x_{1} \cdot \sin x_{2} \cdots \cdot \... | \frac{1}{2^{1006}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,039 |
7. Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfy $f(0)=1$, and for any $x, y \in \mathbf{R}$ we have
$$
f(x y+1)=f(x) f(y)-f(y)-x+2 .
$$
Then $f(x)=$ $\qquad$ | $$
\text { 7. } x+1 \text {. }
$$
For any $x, y \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(x y+1)=f(x) f(y)-f(y)-x+2 \\
\Rightarrow f(x y+1)=f(y) f(x)-f(x)-y+2 . \\
\text { Therefore, } f(x) f(y)-f(y)-x+2 \\
=f(y) f(x)-\dot{f}(x)-y+2,
\end{array}
$$
which means $f(x)+y=f(y)+x$.
Let $y=0$, we get $f(x)=x+1$. | x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,040 |
8. Given real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$. Then the maximum value of $x y+y z$ is $\qquad$ . | 8. $\frac{\sqrt{2}}{2}$.
$$
\begin{array}{l}
\text { From }\left(x^{2}+\frac{1}{2} y^{2}\right)+\left(\frac{1}{2} y^{2}+z^{2}\right) \\
\geqslant 2 \sqrt{\frac{1}{2} x^{2} y^{2}}+2 \sqrt{\frac{1}{2} y^{2} z^{2}} \\
=\sqrt{2}(x y+y z),
\end{array}
$$
we get $x y+y z \leqslant \frac{\sqrt{2}}{2}$.
Therefore, the maximum... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,041 |
9. (14 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\frac{a_{n+1}+a_{n}-1}{a_{n+1}-a_{n}+1}=n\left(n \in \mathbf{N}_{+}\right) \text {, and } a_{2}=6 \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$.
(2) Let $b_{n}=\frac{a_{n}}{n+c}\left(n \in \mathbf{N}_{+}, c\ri... | (1) From the problem, we have
$$
\begin{array}{l}
\frac{a_{n+1}+a_{n}-1}{a_{n+1}-a_{n}+1}=n\left(n \in \mathbf{N}_{+}\right) \\
\Rightarrow(n-1) a_{n+1}-(n+1) a_{n}=-(n+1) .
\end{array}
$$
When $n \geqslant 2\left(n \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
\frac{a_{n+1}}{n+1}-\frac{a_{n}}{n-1}=-\frac{1}{n-1} \\... | 4-\frac{n+2}{2^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,042 |
1. Given real numbers $x, y, z$ satisfy
$$
x+y+z=5, \quad xy+yz+zx=3 \text{.}
$$
Then the maximum value of $z$ is $\qquad$. | Prompt: Example 2. Answer: $\frac{13}{3}$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,043 |
10. (14 points) Let $M$ be any point on the directrix of the parabola $y^{2}=2 p x(p>0)$. Draw tangents $l_{1}$ and $l_{2}$ from $M$ to the parabola, touching the parabola at points $A$ and $B$ (point $A$ is above the $x$-axis).
(1) Prove that the line $A B$ passes through a fixed point;
(2) Let $P$ be the midpoint of ... | Then $x_{1} x_{2}=\frac{m^{2}}{k^{2}}$.
$$
\text { Hence } \frac{m^{2}}{k^{2}}=\frac{p^{2}}{4} \Rightarrow \frac{m}{k}=-\frac{p}{2} \text {. }
$$
Thus, line $A B$ passes through the focus.
(2) From $k_{1} k_{2}=-\frac{2 p}{4 \sqrt{x_{1} x_{2}}}=-1$
$\Rightarrow l_{1} \perp l_{2} \Rightarrow M P=\frac{1}{2} A B$
$\Righ... | p | Geometry | proof | Yes | Yes | cn_contest | false | 726,044 |
11. (18 points) Let $a, b, c$ be positive real numbers, and $a+b+c=1$. Prove:
$$
\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right) \geqslant \frac{1}{2} .
$$ | 11. By the rearrangement inequality, we have
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a, \\
a^{2}+b^{2}+c^{2} \geqslant a c+b a+c b .
\end{array}
$$
Adding the two inequalities, we get
$$
\begin{array}{l}
2\left(a^{2}+b^{2}+c^{2}\right) \geqslant a(b+c)+b(c+a)+c(a+b) \\
\Rightarrow 2\left(a^{2}+b^{2}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,045 |
12. (18 points) A school's mathematics interest group consists of $m$ students, and the school has arranged $n$ instructors to guide them. During one of the group's activities, each pair of students asks each other one question, each student asks each instructor one question, and each instructor asks the entire group o... | 12. From the problem, we have
$$
\begin{array}{l}
m(m-1)+m n+n=51 \\
\Rightarrow m^{2}+(n-1) m+n-51=0 .
\end{array}
$$
Then $\Delta=(n-1)^{2}-4(n-51)$
$$
=n^{2}-6 n+205=(n-3)^{2}+196
$$
must be a perfect square.
Let $(n-3)^{2}+196=k^{2}\left(k \in \mathbf{N}_{+}\right)$. Then
$$
(n-3+k)(n-3-k)=-196 \text {, }
$$
whe... | (m, n)=(6,3) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,046 |
1. Given that circle $\Gamma_{1}$ and circle $\Gamma_{2}$ intersect at points $P$ and $Q$, segments $AC$ and $BD$ are chords of circles $\Gamma_{1}$ and $\Gamma_{2}$, respectively, such that $AB$ intersects ray $CD$ at point $P$, $AC$ intersects ray $BD$ at point $X$, and $Y$, $Z$ are points on circles $\Gamma_{1}$ and... | 1. As shown in Figure 1.
From the fact that $A, C, Q, P$ and $P, D, Q, B$ are each four points on a circle, we know
$$
\begin{array}{l}
\angle X A Q=\angle C A Q=\angle C P Q \\
=\angle D P Q=\angle D B Q=\angle X B Q .
\end{array}
$$
Therefore, $A, X, Q, B$ are four points on a circle.
Furthermore, from the fact that... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,047 |
2. Let real numbers $a_{i}, b_{i}\left(i=1,2, \cdots, n, n \in \mathbf{N}_{+}\right)$
satisfy
$$
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n},
$$
and
$$
\sum_{k=1}^{i} a_{k} \leqslant \sum_{k=1}^{i} b_{k}(i=1,2, \cdots, n-1),
$$
and $\sum_{k=1}^{n} a_... | 2. If for any real number $m$, the number of integer pairs $(i, j)$ satisfying $a_{i}-a_{j}=m$ is equal to the number of integer pairs $(k, l)$ satisfying $b_{k}-b_{t}=m$, then
$\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)=\sum_{1 \leqslant k<1 \leqslant n}\left(b_{k}-b_{l}\right)$.
Let $S_{n}=\sum_{k=1}^... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,048 |
3. Given that $\cos \theta\left(\theta \in\left(0, \frac{\pi}{2}\right)\right)$ is an irrational number, and there exists some positive integer $k$, such that $\cos k \theta$ and $\cos (k+1) \theta$ are both rational numbers. Prove: $\theta=\frac{\pi}{6}$. | 3. First, prove two lemmas.
Lemma 1 For any positive integer $n$, there exists an integer-coefficient polynomial $S_{n}(x)$ with the leading coefficient 1, such that
$S_{n}(2 \cos \alpha)=2 \cos n \alpha$.
Proof of Lemma 1 Use mathematical induction on $n$.
When $n=1,2$, we can take
$S_{1}(x)=x, S_{2}(x)=x^{2}-2$
res... | \theta=\frac{\pi}{6} | Number Theory | proof | Yes | Yes | cn_contest | false | 726,049 |
4. Do there exist positive integers $a, b$ such that for any positive integer $n, b^{n}-n$ is never divisible by $a$? Prove your conclusion. | 4. Does not exist.
First, we prove a lemma.
Lemma Given positive integers \(a, b\), for sufficiently large positive integer \(n\) we have
\[
b^{n+\phi(a)} = b^n \pmod{a}.
\]
Proof Let \(a = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}\), where \(p_1, p_2, \cdots, p_k\) are distinct primes, and \(\alpha_1, \alpha_2, \cdots, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,050 |
5. Given that $\triangle A B C$ is inscribed in circle $\Gamma$, the tangents to circle $\Gamma$ at points $B$ and $C$ intersect at point $T$, point $S$ lies on ray $B C$ such that $A S \perp A T$, points $B_{1}$ and $C_{1}$ lie on ray $S T$ ($C_{1}$ is between $B_{1}$ and $S$), and satisfy $B_{1} T=B T=C_{1} T$. Prove... | 5. As shown in Figure 2, let $M$ be the midpoint of segment $BC$.
First, prove: $\angle BAT = \angle CAM$.
Let $D$ be the other intersection point of line $AT$ with circle $\Gamma$.
It is easy to see that $\triangle TBD \sim \triangle TAB \Rightarrow \frac{BD}{AB} = \frac{TB}{TA}$.
Similarly, $\frac{CD}{AC} = \frac{TC}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,051 |
6. For the polynomial $P(x)$ with integer coefficients, $r(2i-1)(i=$ $1,2, \cdots, 512)$ represents the remainder when $P(2i-1)$ is divided by 1024. The array $(r(1), r(3), \cdots, r(1013))$ is called the "remainder sequence" of $P(x)$. If a remainder sequence is a permutation of $1,3, \cdots, 1023$, it is called "comp... | $$
\begin{array}{l}
\text { 6. Define the polynomial } \\
\left\{\begin{array}{l}
Q_{0}(x)=b_{0}, \\
Q_{k}(x)=b_{k} \prod_{i=1}^{k}(x+2 i-1),
\end{array}\right. \\
\text { where, } k=1,2, \cdots, 6, b_{0}=2^{10}, b_{1}=2^{9}, b_{2}=2^{7} \text {, } \\
b_{3}=2^{6}, b_{4}=2^{3}, b_{5}=2^{2}, b_{6}=2^{0} \text {. } \\
\en... | 2^{35} | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,052 |
1. Given two monomials about $x, y$: $8 x^{3-6} y^{2 b-3}$ and $-a x^{3-2 a} y^{b-a}$, their difference is still a monomial. Then the value of $a+b$ is ( ).
(A) 2
(B) 3
(C) $\pm 2$
(D) 2 or 3 | -1. D.
(1) When $a \neq 0$, the two monomials should be like terms, thus:
$$
\left\{\begin{array} { l }
{ 3 - b = 3 - 2 a , } \\
{ 2 b - 3 = b - a }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=1, \\
b=2
\end{array} \Rightarrow a+b=3\right.\right. \text {. }
$$
(2) When $a=0$, the difference of the two monomials ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,053 |
2. Let $(a, b)$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-a-2 b$ is . $\qquad$ | Hint: Use the method of completing the square. The original expression can be transformed into
$$
\left(a+\frac{1}{2} b-\frac{1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 \text {. }
$$
When $a+\frac{1}{2} b-\frac{1}{2}=0, b-1=0$, i.e., $a=0$, $b=1$, the original expression can achieve the minimum value of -1. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,054 |
2. As shown in Figure 1, given that the area of quadrilateral $A B C D$ inscribed in $\odot O$ is 1, and $A C \perp B D$. Then the area of quadrilateral $O A B C$ is ( ).
(A) 0.3
(B) 0.4
(C) 0.5
(D) 0.6 | 2. C.
Draw $O E \perp B D$ through point $O$ at point $E$.
Then $B E=D E=\frac{1}{2} B D$.
Thus $S_{\text {V抓边OABC }}=\frac{1}{2} A C \cdot B E=\frac{1}{4} A C \cdot B D$ $=\frac{1}{2} S_{\text {田迈㻆 } A B C D}=0.5$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,055 |
3. The number of all distinct integer solutions of the equation $2 x^{2}+5 x y+2 y^{2}=2012$ is ( ).
(A) 4
(B) 6
(C) 10
(D) 12 | 3. D.
The original equation can be transformed into
$$
(2 x+y)(x+2 y)=2^{2} \times 503 \text {. }
$$
Since the sum of $2 x+y$ and $x+2 y$ is divisible by 3, and $2^{2} \times 503$ can be factored into two integer factors (including negative factors), the sum of these two integer factors is also divisible by 3, so the... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,056 |
4. Given $a \geqslant b \geqslant c>0, a+b+c \leqslant 4$. Then the maximum value of $a^{2}+3 b^{2}+5 c^{2}$ is ( ).
(A) 18
(B) 16
(C) 14
(D) 12 | 4. B.
Notice,
$$
\begin{array}{l}
16=4^{2} \geqslant(a+b+c)^{2} \\
=a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 b c \\
\geqslant a^{2}+b^{2}+c^{2}+2 b^{2}+2 c^{2}+2 c^{2} \\
=a^{2}+3 b^{2}+5 c^{2},
\end{array}
$$
i.e., $a^{2}+3 b^{2}+5 c^{2} \leqslant 16$.
The equality holds if and only if $a=b=c=\frac{4}{3}$. Therefore, the max... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 726,057 |
5. In quadrilateral ABCD, it is known that $\angle A B C=60^{\circ}, B E$ bisects $\angle A B C$, intersecting $A D$ and $C D$ at points $E$ and $F$ respectively; $A E=3, D E=2, O$ is the circumcenter of $\triangle D E F$. The following conclusions are given:
(1) Quadrilateral $O E D F$ is a rhombus;
(2) $\triangle O ... | 5. D.
As shown in Figure 3, it is easy to see that
$$
\begin{array}{l}
D E=D F, \\
A E=A B=C D, \\
\angle A D C=60^{\circ} . \\
\text { Since } O E=O F,
\end{array}
$$
we know that $O D$ perpendicularly bisects $E F$, and $\triangle O D E$ and $\triangle O D F$ are both equilateral triangles.
Thus, quadrilateral $O E... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,058 |
$$
\left\{\begin{array}{l}
x-999 \geqslant 1000, \\
x+1 \leqslant a
\end{array}\right.
$$
has a finite number of real solutions. Then the value of $a$ is $\qquad$ . | Ni, 1.2000.
The solution set of the inequality is $1999 \leqslant x \leqslant a-1$.
From the problem, we know $a-1=1999 \Rightarrow a=2000$. | 2000 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,060 |
2. On the way from Xiaoming's house to the swimming pool, there are 200 trees. When going to the swimming pool and returning, Xiaoming ties red ribbons on some trees: 1: to make marks. When going to the swimming pool, he marks the 1st tree, the 6th tree, the 11th tree, …, each time skipping 4 trees without marking. On ... | 2. 140 .
According to the problem, the trees marked are the $5 x+1(x=0,1, \cdots, 39)$ and $8 y(y=1,2, \cdots, 25)$ ones, among which exactly 5 trees are marked twice.
Therefore, the number of trees that are not marked is
$$
200-(40+25-5)=140 \text {. }
$$ | 140 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,061 |
3. In a right-angled triangle table with one acute angle of $30^{\circ}$ and a hypotenuse of $4 \mathrm{~m}$, there are pockets at each corner (the size of the pockets can be ignored). A ball is placed in front of the pocket at the $60^{\circ}$ angle, and it is hit towards a point on the opposite side (not a vertex). A... | $3.4 \sqrt{13}$.
Construct rectangle $A B C D$ such that $A B=2 \sqrt{3} \text{ m}, B C=14 \text{ m}$, and divide rectangle $A B C D$ into several triangles congruent to the tabletop as shown in Figure 4.
Place the ball at point $B$, and strike it towards point $E$, which is point $D$. It is easy to see that the ball ... | 4 \sqrt{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,062 |
4. Given that the radius of $\odot O$ is 1. Then the area of the region formed by the orthocenters of all inscribed triangles in $\odot O$ is $\qquad$ . | 4. $9 \pi$.
If point $H$ is inside $\odot O$, and a chord $AD$ is drawn through $H$ such that $AH \leqslant HD$, a chord $BC$ is drawn perpendicular to and bisecting $HD$. Connect $BD$, $BH$, and extend $BH$ to intersect $AC$ at point $E$, as shown in Figure 5.
$$
\begin{array}{c}
\text { Then } \angle C=\angle D \\
=... | 9 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,063 |
One. (20 points) Given the parabola $y=a x^{2}+b x+c$ passes through the point $(b,-3)$, and $|a| c+b|c|=0$, and the inequality $a x^{2}+b x+c+3>0$ has no solution. Find all possible values of the triplet $(a, b, c)$.
---
Please note that the formatting and line breaks have been preserved as requested. | From the problem, we know that $(b, -3)$ is the vertex of the parabola, and $a < 0$. Therefore, the parabola is
$$
\begin{array}{l}
y=a(x-b)^{2}-3 \\
\Rightarrow b=-2 a b, c=a b^{2}-3<0 .
\end{array}
$$
From $|a| c+b|c|=0 \Rightarrow|a|=b$.
Thus, $(a, b, c)=\left(-\frac{1}{2}, \frac{1}{2},-\frac{25}{8}\right)$. | \left(-\frac{1}{2}, \frac{1}{2},-\frac{25}{8}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,064 |
3. Given that $x, y, z$ are real numbers, and satisfy
$$
x+2 y-5 z=3, x-2 y-z=-5 \text{. }
$$
Then the minimum value of $x^{2}+y^{2}+z^{2}$ is ( ).
(A) $\frac{1}{11}$
(B) 0
(C) 5
(D) $\frac{54}{11}$ | Hint: Use the method of elimination.
When $z=\frac{1}{11}$, the minimum value of $x^{2}+y^{2}+z^{2}$ is $\frac{54}{11}$. | \frac{54}{11} | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,065 |
II. (25 points) As shown in Figure 2, $M$ is a point inside the circle $\odot O$ with diameter $AB$, and the extensions of $AM$ and $BM$ intersect $\odot O$ at points $C$ and $D$, respectively. A perpendicular line $MN$ is drawn from point $M$ to $AB$ at point $N$. A tangent line is drawn from point $C$ to $\odot O$ an... | As shown in Figure 7, connect $O C$, $O D$, $O E$, $B C$, $O E$, and let $O E$ intersect $\odot O$ at point $F$. Clearly,
$$
\begin{array}{l}
\angle O C E=\angle O N E=90^{\circ} \\
\Rightarrow O, N, C, E \text { are concyclic } \\
\Rightarrow \angle B O C=\angle C E N .
\end{array}
$$
Thus, $\triangle O B C \backsim ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,066 |
Three, (25 points) Given that the positive integer $M$ when divided by the positive integer $N$ leaves a remainder of 2, and the sum of the reciprocals of all values of $N$ is $\frac{627}{670}$, with the number of all values of $N$ being less than 16. Find all possible values of $M$.
| Obviously, $N>2, N \mid (M-2)$.
Since $670=2 \times 5 \times 67$, $M-2$ has prime factors 2, 5, and 67. Therefore, the sum of the reciprocals of all positive divisors of $M-2$ is
$$
\frac{1}{1}+\frac{1}{2}+\frac{627}{670}=\frac{816}{335} .
$$
The number of positive divisors of $M-2$ is less than 18.
Since $(1+1)^{5}=32... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,067 |
1. If the function $f(x)=\ln \left(a \mathrm{e}^{x}-x-3\right)$ has a range of $\mathbf{R}$, then the range of the real number $a$ is $\qquad$ . | $-1 .\left(-\infty, \mathrm{e}^{2}\right]$.
Let $g(x)=a \mathrm{e}^{x}-x-3$.
Then the range of $f(x)$ is $\mathbf{R}$ if and only if $g(x)$ can take all values in $(0,+\infty)$.
When $a=0$, $g(x)=-x-3$. Clearly, $g(x)$ can take all values in $(0,+\infty)$.
Now consider the case where $a \neq 0$.
In this case, $g^{\prim... | \left(-\infty, \mathrm{e}^{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,068 |
2. Given a tetrahedron $P-ABC$ with the base $\triangle ABC$ being an equilateral triangle with side length 6, and $PA \perp$ plane $ABC$, $PA = 4$. If point $Q$ satisfies
$$
\overrightarrow{PQ}=\frac{1}{2}(\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}),
$$
then the volume of the tetrahedron $Q-ABC$ is $... | 2. $6 \sqrt{3}$.
Let the center of $\triangle ABC$ be $O$.
Then $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=0$
$$
\begin{aligned}
\Rightarrow & \overrightarrow{P Q}=\frac{1}{2}(\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}) \\
& =\frac{1}{2}[(\overrightarrow{P O}+\overrightarrow{O ... | 6 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,069 |
3. In $\triangle A B C$, it is known that $a, b, c$ are the sides opposite to $\angle A$, $\angle B$, $\angle C$ respectively. If
$$
\frac{a}{b}+\frac{b}{a}=4 \cos C, \cos (A-B)=\frac{1}{6},
$$
then $\cos C=$ | 3. $\frac{2}{3}$.
From the given and the cosine rule, we know
$$
\begin{array}{l}
\frac{a}{b}+\frac{b}{a}=4 \cdot \frac{a^{2}+b^{2}-c^{2}}{2 a b} \\
\begin{aligned}
\Rightarrow & a^{2}+b^{2}=2 c^{2} \\
\Rightarrow & 2 \sin ^{2} C=\sin ^{2} A+\sin ^{2} B \\
& =\frac{1-\cos 2 A}{2}+\frac{1-\cos 2 B}{2} \\
& =1-\cos (A+B... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,070 |
4. Let positive real numbers $a, b, c$ satisfy $\frac{2}{a}+\frac{1}{b}=\frac{\sqrt{3}}{c}$. Then the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is $\qquad$ . | 4. 9 .
From the given, we have $\frac{2 c}{a}+\frac{c}{b}=\sqrt{3}$.
By the Cauchy-Schwarz inequality and the AM-GM inequality, we get
$$
\begin{array}{l}
\frac{2 a^{2}+b^{2}}{c^{2}} \\
=\frac{1}{3}(2+1)\left[2\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}\right] \\
\geqslant \frac{1}{3}\left(\frac{2 a}{c}+... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,071 |
5. Let $a_{1}, a_{2}, \cdots, a_{6}$ be any permutation of $1,2, \cdots, 6$, and $f$ be a one-to-one mapping from $\{1,2, \cdots, 6\}$ to $\{1,2, \cdots, 6\}$, satisfying
$$
f(i) \neq i, f(f(i))=i(i=1,2, \cdots, 6) .
$$
Consider the number table
$$
A=\left[\begin{array}{cccccc}
a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & ... | 5. 10800 .
For a permutation $a_{1}, a_{2}, \cdots, a_{6}$, consider the one-to-one mapping satisfying
$$
f(i) \neq i, f(f(i))=i(i=1,2, \cdots, 6)
$$
For each such mapping $f$, the elements of set $A$ can be paired as $\{i, j\}$, such that
$$
f(i)=j, f(j)=i \text {. }
$$
Thus, for each permutation, the number of map... | 10800 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,072 |
6. Given the equation in $x$
$$
x^{3}-4 x^{2}+5 x+a=0(a \in \mathbf{R})
$$
has three real roots $x_{1}, x_{2}, x_{3}$. Then the maximum value of $\max \left\{x_{1}, x_{2}, x_{3}\right\}$ is $\qquad$ . | 6.2.
Assume $x_{3}=\max \left\{x_{1}, x_{2}, x_{3}\right\}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=4, \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=5
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x_{1}+x_{2}=4-x_{3}, \\
x_{1} x_{2}=5-x_{3}\left(x_{1}+x_{2}\right... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,073 |
7. Given that $F_{1}$ and $F_{2}$ are the left and right foci of a hyperbola $C$ centered at the origin with symmetric axes as the coordinate axes, a line $l$ passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. $I_{1}$ and $I_{2}$ are the incenter of $\triangle A F_{1} F_{2}$ and... | 7. $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$.
Let the equation of the hyperbola $C$ be
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0) \text {, }
$$
with the semi-focal distance $c=\sqrt{a^{2}+b^{2}}$.
Let the points where the incircle of $\triangle A F_{1} F_{2}$ touches the $x$-axis, $A F_{1}$, and $A F_{2}$ be $... | \frac{x^{2}}{4}-\frac{y^{2}}{12}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,074 |
8. Two Go teams, Team A and Team B, each have 5 players who compete in a broadcast match according to a predetermined order. The No. 1 players of both teams compete first, and the loser is eliminated; then the No. 2 player of the losing team competes with the winner, and the loser is eliminated again. This continues un... | 8. $\frac{1}{6}$.
Let the event that Team A wins be denoted as $A$. Then
$$
P(A)=P(\bar{A})=\frac{1}{2} \text {. }
$$
When Team B wins, all members of Team A have played, so the number of Team A members who did not play, $X=0$.
When Team A wins and $X=k(k=0,1, \cdots, 4)$, $4-k$ members of Team A are eliminated, and... | \frac{1}{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,075 |
Example 1 (Ticket Selling Problem) 2,012 tourists are queuing to buy admission tickets, each ticket costing 5 yuan, among them, 1006 people each hold a 5 yuan note, and the other 1006 people each hold a 10 yuan note. At the beginning, there is no change in the ticket vending machine. Determine the number of queuing met... | Solve: Replace 2012 with a general positive integer $2n$.
If a person holds a 5-yuan note, assign the value “+1” (i.e., the ticket machine can receive a 5-yuan note); if a person holds a 10-yuan note, assign the value “-1” (i.e., the ticket machine needs to give out a 5-yuan note as change).
Thus, a queue of $2n$ peop... | \frac{1}{n} \mathrm{C}_{2n}^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,076 |
Example 4 Let $[x]$ denote the greatest integer not exceeding the real number $x$. Prove: If $p=2 n+1$ is a prime number, then it must be a factor of $c=2^{n}+(-1)^{\left[\frac{n-1}{2}\right]}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation re... | $\sqrt{2}$, the line segment with a slope of -1 is denoted as $L^{-}$.
To prove that it is sufficient to construct an integer-coefficient equation that satisfies the following conditions:
(1) $p=2n+1$ is a root of the equation;
(2) The coefficient of the highest degree term is coprime with $p$;
(3) The constant term ha... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,077 |
Example 5 As shown in Figure 5, given $\triangle A B C$ with two altitudes $A D$ and $B E$ intersecting at point $H$, the circumcenter of the triangle is $O$. A line $O F \perp B C$ is drawn through $O$ and intersects $B C$ at point $F$, and $O H$ intersects $A F$ at point $G$. Then $S_{\triangle O F G}: S_{\triangle C... | Since $O$ is the circumcenter of $\triangle A B C$ and $O F \perp B C$, therefore, $F$ is the midpoint of side $B C$.
Thus, $A F$ is the median on side $B C$.
By Euler's theorem, the intersection point $G$ of $O H$ and $A F$ is the centroid of $\triangle A B C$.
Therefore, $F G: G A=1: 2$.
Also, $\triangle O F G \backs... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,078 |
12. (18 points) For any $n(n \geqslant 2)$ vectors $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n}$ in the plane, let $M$ denote the number of real pairs $(i, j)$ satisfying $\boldsymbol{\alpha}_{i} \cdot \boldsymbol{\alpha}_{j}<0$ $(i<j)$. Prove: $M \leqslant \frac{n^{2}}{3}$. | 12. Suppose in the vectors $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ there exists $\boldsymbol{\alpha}_{i} \neq \boldsymbol{\alpha}_{j}$, and $\boldsymbol{\alpha}_{i} \cdot \boldsymbol{\alpha}_{j} \geqslant 0$.
Let $N_{1}$ be the number of $k$ such that $\boldsymbol{\alpha}_{i} \cdot \boldsymbol{\alpha}_{k}<0$, and... | M \leqslant \frac{n^{2}}{3} | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,080 |
2. Given that the internal angles of $\triangle A B C$ are all greater than $30^{\circ}, \odot T$ intersects sides $B D, C A, A B$ at points $P, Q, K, L, M, N$ respectively, and these six points are arranged clockwise on $\odot T$. If $\triangle T Q L, \triangle T L M, \triangle T N P$ are equilateral triangles, prove:... | 2. (1) Let $\angle P T Q=2 x, \angle K T L=2 y$,
$\angle M T N=2 z$.
From $T P=T Q$
$$
\begin{array}{l}
\Rightarrow \angle T Q P=\angle T P Q=90^{\circ}-x \\
\Rightarrow \angle B P N=180^{\circ}-\angle T P N-\angle T P Q \\
=30^{\circ}+x .
\end{array}
$$
Similarly, $\angle B N P=30^{\circ}+z$.
Thus, $180^{\circ}-\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,082 |
3. In pentagon $A B C D E$, it is known that $B C / / A E, A B = B C + A E, \angle A B C = \angle C D E$. Let $M$ be the midpoint of $C E$, and $O$ be the circumcenter of $\triangle B C D$. If $\angle D M O = 90^{\circ}$, prove:
$$
2 \angle B D A = \angle C D E \text {. }
$$ | 3. Let the midpoint of $O D$ be $O^{\prime}, \odot O^{\prime}$ is the circle with $O D$ as its diameter. Then $O O^{\prime}$ passes through point $D$, and the two circles are internally tangent at point $D$.
Hence, $\odot O$ and $\odot O^{\prime}$ are homothetic with point $D$ as the homothetic center, and the homothe... | 2 \angle B D A = \angle C D E | Geometry | proof | Yes | Yes | cn_contest | false | 726,083 |
4. Let $\left\{a_{0}, a_{1}, \cdots\right\}$ and $\left\{b_{0}, b_{1}, \cdots\right\}$ be two infinite integer sequences, and
$$
\begin{array}{l}
\left(a_{n}-a_{n-1}\right)\left(a_{n}-a_{n-2}\right)+\left(b_{n}-b_{n-1}\right)\left(b_{n}-b_{n-2}\right) \\
=0(n \geqslant 2) .
\end{array}
$$
Prove: There exists a positiv... | 4. In the plane, consider the point $P_{j}=\left(a_{j}, b_{j}\right)$. Then
$k_{P_{n} P_{n-1}}=\frac{b_{n}-b_{n-1}}{a_{n}-a_{n-1}}$,
$k_{P_{n} P_{n-2}}=\frac{b_{n}-b_{n-2}}{a_{n}-a_{n-2}}$.
By the given condition, $k_{P_{n} P_{n-1}} k_{P_{n} P_{n-2}}=-1$.
Therefore, $P_{n} P_{n-1} \perp P_{n} P_{n-2}$, which means poin... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,084 |
3. Let $a>1$. Then when the graphs of $y=a^{x}$ and $y=\log _{a} x$ are tangent, $\ln \ln a=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. -1.
Since the two functions are inverse functions of each other and are symmetric about the line $y=x$, the point of tangency lies on $y=x$.
Let the point of tangency be $\left(x_{0}, y_{0}\right)$. Then
$$
\begin{array}{l}
x_{0}=a^{x_{0}}, \\
a^{x_{0}} \ln a=1 .
\end{array}
$$
Substituting equation (1) into equat... | -1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,086 |
4. In a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, points $E$, $F$, and $G$ are the midpoints of edges $A A_{1}$, $C_{1} D_{1}$, and $B C$ respectively. Then the volume of the tetrahedron $B_{1} E F G$ is $\qquad$ | 4. $\frac{3}{16}$.
As shown in Figure 2, draw a line through point $E$ parallel to $B_{1} G$, intersecting $A_{1} D_{1}$ at point $E^{\prime}$.
Then $V_{\text {prism } B_{1} E F C}=V_{\text {pyramid } B_{1} E^{\prime} F C}$
$$
\begin{array}{l}
=V_{\text {pyramid } G E^{\prime} F R_{1}} \\
=\frac{1}{3} S_{\triangle E^{... | \frac{3}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,087 |
5. Through the right focus $F$ of the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
draw $B_{1} B_{2} \perp x$ axis, intersecting the hyperbola at points $B_{1}$ and $B_{2}$. The line connecting $B_{2}$ and the left focus $F_{1}$ intersects the hyperbola at point $B$. Connecting $B_{1} B$ interse... | 5. $-\frac{a^{2}}{c}$.
From the given, we have point $B_{2}\left(c,-\frac{b^{2}}{a}\right)$.
By combining the equation of line $B_{2} F_{1}$ with the hyperbola equation, we get
$$
\left(1-\frac{b^{2}}{4 c^{2}}\right) x^{2}-\frac{b^{2}}{2 c} x-\frac{b^{2}}{4}-a^{2}=0 \text {. }
$$
By Vieta's formulas, we have
$$
c x_{... | -\frac{a^{2}}{c} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,088 |
Example 6 As shown in Figure 6, it is known that $\triangle A B C$ is inscribed in $\odot O$. A line $l$ is drawn through the midpoint $D$ of $B C$, parallel to $A C$. Line $l$ intersects $A B$ at point $E$, and intersects $\odot O$ at points $G$ and $F$, and intersects the tangent line of $\odot O$ at point $A$ at poi... | Since $A C / / P F$, we have,
$$
\begin{array}{l}
\angle B D E=\angle B C A=\angle P A E \\
\Rightarrow P, A, D, E \text{ are concyclic} \\
\Rightarrow B E \cdot A E=P E \cdot D E=6 \\
\Rightarrow B E=A E=\sqrt{6} .
\end{array}
$$
By the intersecting chords theorem,
$$
G E \cdot E F=B E^{2} \text{. }
$$
Thus, $G E=2,... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,089 |
6. Given $a_{1}, a_{2}, \cdots$ is a geometric sequence with the first term $a_{1}=a\left(a \in \mathbf{Z}_{+}\right)$ and common ratio $r\left(r \in \mathbf{Z}_{+}\right)$. Suppose
$$
\log _{4} a_{2}+\log _{4} a_{3}+\cdots+\log _{4} a_{12}=2013 .
$$
Then the number of ordered pairs $(a, r)$ that satisfy the condition... | 6. 62.
From the given, we have $a_{n}=a r r^{n-1}$, substituting into the given equation yields
$$
a^{11} r^{66}=2^{4026} \Rightarrow a r^{6}=2^{366} \text{. }
$$
Let $a=2^{x}, r=2^{y}(x, y \in \mathrm{N})$. Then $x+6y=366$.
Thus, the number of pairs satisfying the condition is 62.
_ 171
2. $\frac{1}{3}$.
By the uni... | 62 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,090 |
8. In the complex plane, the point corresponding to the complex number $z_{1}$ moves on the line segment connecting 1 and $\mathrm{i}$, and the point corresponding to the complex number $z_{2}$ moves on the circle centered at the origin with a radius of 1. Then the area of the region where the point corresponding to th... | 8. $2 \sqrt{2}+\pi$.
Let $z_{1}=t+\mathrm{i}(1-t)(0 \leqslant t \leqslant 1)$,
$z_{2}=\cos \theta+\mathrm{i} \sin \theta$.
Then $z_{1}+z_{2}=x+y \mathrm{i}$
$$
=t+\cos \theta+\mathrm{i}(1-t+\sin \theta) \text {. }
$$
Therefore, $(x-t)^{2}+[y-(1-t)]^{2}=1$ represents a set of circles with centers on the line $y=1-x$, ... | 2 \sqrt{2}+\pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,091 |
9. (16 points) Given $x, y \in \mathbf{R}$, and satisfying $(x+1)(y-1)=1$.
Find the minimum value of $\left(x^{2}+1\right)\left(y^{2}+1\right)$. | $$
\begin{array}{l}
(x+1)(y-1)=1 \\
\Rightarrow x y-x+y-1=1 \\
\Rightarrow(x y+1)+(y-x)=3 . \\
\text { And }\left(x^{2}+1\right)\left(y^{2}+1\right) \\
=(x y+1)^{2}+(y-x)^{2} \\
\geqslant \frac{1}{2}[(x y+1)+(y-x)]^{2}=\frac{9}{2} .
\end{array}
$$
When and only when $x y+1=y-x=\frac{3}{2}$, that is,
$$
x=\frac{\sqrt{1... | \frac{9}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,092 |
10. (20 points) (1) For any integer $n>1$, try to find two different positive integers $x, y$ such that
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{n} \text {. }
$$
(2) For any positive integer $n$, find the positive integer solutions $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ that satisfy the equation
$$
\frac{1}{x_{1}}+\frac... | 10. (1) It is easy to get $(x-n)(y-n)=n^{2}$.
$$
\begin{array}{l}
\text { Let } x-n=1, y-n=n^{2} \\
\Rightarrow x=n+1, y=n(n+1) \text {. }
\end{array}
$$
(2) When $n=1$, $x_{1}=2$.
From (1) we get
$$
\begin{array}{l}
\frac{1}{x_{1} x_{2} \cdots x_{n}} \\
=\frac{1}{x_{1} x_{2} \cdots x_{n}+1}+\frac{1}{x_{1} x_{2} \cdot... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,093 |
11. (20 points) Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, points $A(-1,0), B(a, 0)$, and a point $P\left(x_{0}, y_{0}\right)$ on the hyperbola $C$. Draw the tangent line $l$ through $P$, and drop perpendiculars from the foci $F_{1} 、 F_{2}$ to the tangent line $l$, with the feet of the perpend... | 11. (1) As shown in Figure 3, connect $P F_{1}$ and $P F_{2}$.
By the optical property of the hyperbola, we know that line $l$ bisects $\angle F_{1} P F_{2}$, i.e.,
$$
\angle F_{1} P H_{1} = \angle F_{2} P H_{2}.
$$
Extend $F_{1} H_{1}$ to intersect $P F_{2}$ at point $F_{1}^{\prime}$. Then $\triangle F_{1} P F_{1}^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,094 |
一、(40 Points) As shown in Figure 1, in $\triangle ABC$, it is known that $AD$ is the altitude on side $BC$, and the circle $\odot O$ with $AD$ as its diameter intersects $AC$ and $AB$ at points $E$ and $F$ respectively. Let $M$, $M_1$, and $M_2$ be the midpoints of $BC$, $BD$, and $DC$ respectively, $BE$ and $CF$ inter... | As shown in Figure 4, let $E M_{1}$ and $F M_{2}$ intersect $\odot O$ at points $K_{1}$ and $K_{2}$, respectively, and connect $B K_{1}$, $C K_{2}$, $D E$, and $D F$.
Let the circumcircles of $\triangle B E K_{1}$ and $\triangle C F K_{2}$ be $\odot O_{1}$ and $\odot O_{2}$, respectively.
By the projection theorem, we... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,095 |
II. (40 points) Prove: For any $x, y, z \in \mathbf{R}_{+}$, we have $8 x^{2} y^{2} z^{2} \geqslant \prod\left(x^{2}+x y+x z-y z\right)$, where “ ” denotes the cyclic symmetric product. | Since the original inequality is homogeneous, we may assume without loss of generality that
$$
x+y+z=1 \text {. }
$$
Thus, the original inequality is equivalent to
$$
8 x^{2} y^{2} z^{2} \geqslant(x-y z)(y-z x)(z-x y) \text {. }
$$
Notice that,
$$
\begin{array}{l}
(x-y z)+(y-z x)=(x+y)^{2}>0, \\
(y-z x)+(z-x y)=(y+z)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,096 |
Three. (50 points) Given the sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{0}=0, x_{n+2}=2 x_{n+1}+x_{n} \text {. }
$$
Given an odd prime $p$ and a positive integer $m$ such that
$$
\left(m, \frac{p^{2}-1}{2}\right)=1 \text {. }
$$
Prove: $p \mid x_{m}$ if and only if $p \mid x_{1}$. | Three, Sufficiency.
It is easy to prove by mathematical induction that for any $m \in \mathbf{N}_{+}$, we have $p \mid x_{m}$. Necessity.
(1) For any $k \in \mathbf{N}_{+}$, there exists $l \in \mathbf{N}_{+}$ such that $k \mid x_{i}$. Since the pair $(a, b)$ has at most $k^{2}$ combinations modulo $k$, the sequence
$$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,097 |
Four. (50 points) Consider a $6 \times 6$ grid, where each cell contains the number 0. Each operation involves first selecting three real numbers $a$, $b$, and $c$, then selecting a row and adding $a i^{2}+b i+c$ (where $i$ is the column number, $1 \leqslant i \leqslant 6$) to each cell in that row; or selecting a colu... | Four, cannot.
Proof by contradiction.
Assume that it is possible to transform the grid so that the four corners are all 1, while all other cells remain 0 after a finite number of operations. Consider the $4 \times 4$ subgrid in the top-left corner of the grid. Let the number in the $i$-th row and $j$-th column after a ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 726,098 |
In Rt $\triangle A B C$, it is known that $\angle A=$ $20^{\circ}, \angle B=90^{\circ}, A D$ is the bisector of $\angle B A C$, point $E$ is on side $A B$, and lines $C E$ and $D E$ are connected. If $\angle D C E=30^{\circ}$, find the degree measure of $\angle A D E$. | Solve As shown in Figure 2, construct $\angle A E F=20^{\circ}, E F$ intersects $A D$ at point $F$. Take point $G$ on side $A C$ such that $A G=A E$, and connect $F G$, $E G$, and $C F$.
Since $A D$ bisects $\angle E A G$, by its symmetry we know
$$
F E=F G,
$$
and
$$
\begin{array}{l}
\angle G F D=\angle E F D \\
=\an... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,099 |
Example 7 As shown in Figure 7, in the hexagon $A C_{1} B A_{1} C B_{1}$, it is known that $A C_{1}=A B_{1}$, $B C_{1}=B A_{1}$, $C A_{1}=C B_{1}$, and
$$
\begin{array}{l}
\angle A+\angle B+\angle C \\
=\angle A_{1}+\angle B_{1}+\angle C_{1} .
\end{array}
$$
Prove: The area of $\triangle A B C$ is half the area of the... | Prove that as shown in Figure 8, rotate $\triangle B C A_{1}$ to $\triangle B A_{1}^{\prime} C_{1}$.
Then $\triangle B C A_{1} \cong \triangle B A_{1}^{\prime} C_{1}$
$$
\begin{aligned}
\Rightarrow & \angle A B_{1} C=\angle A C_{1} A_{1}^{\prime}, \\
& C_{1} A_{1}^{\prime}=C A_{1}=B_{1} C, A C_{1}=A B_{1} \\
\Rightarro... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,100 |
As shown in Figure 3, $\odot O$ is the circumcircle of $\triangle ABC$, $\odot O_{1}, \odot O_{2}, \odot O_{3}$ are tangent to the sides $BC$, $CA$, $AB$ of $\triangle ABC$ at their midpoints $D$, $E$, $F$, respectively, and are tangent to $\odot O$ at points $G$, $H$, $I$, respectively. Let the areas of $\odot O$, $\o... | Let $B C=a, C A=b, A B=c$, and the radius of $\odot O$ be $R$. As shown in Figure 3, connect $O B, O C, O G$.
Obviously, points $D, O_{1}$ lie on $O G$.
It is easy to get $O D=R \cos A, D G=R(1-\cos A)$.
Thus, $S_{1}=\frac{1}{4} \pi R^{2}(1-\cos A)^{2}$.
Similarly, $S_{2}=\frac{1}{4} \pi R^{2}(1-\cos B)^{2}$,
$$
S_{3}=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,101 |
For $F_{k}=2^{2^{k}}+1(k \geqslant 0)$, let $p$ be a divisor of $F_{k}$. Find the smallest positive integer $r$ that satisfies $F_{k-1}^{t} \equiv 1(\bmod p)$, and prove that $p \equiv 1\left(\bmod 2^{k+2}\right)$. | First, we introduce a well-known conclusion.
Let $m \geqslant 1, (a, m)=1$, and the smallest positive integer $x$ that satisfies
$$
a^{x}=1(\bmod m)
$$
is called the order of $a$ modulo $m$, denoted by $\operatorname{ord}_{m}(a)$. Then for any positive integer $x$ we have
$$
\operatorname{ord}_{m}(a) \mid x \Leftrighta... | p \equiv 1\left(\bmod 2^{k+2}\right) | Number Theory | proof | Yes | Yes | cn_contest | false | 726,102 |
Given $x 、 y>0$, and $x \neq y, n(n>1)$ is a given positive integer. If $x^{n}-y^{n}=x^{n+1}-y^{n+1}$, prove:
$$
1<x+y<\frac{2 n}{n+1} \text {. }
$$ | Proof First, prove: $x+y>1$.
Notice,
$$
\begin{array}{l}
(x+y)\left(x^{n}-y^{n}\right) \\
=x^{n+1}-y^{n+1}+x y\left(x^{n-1}-y^{n-1}\right) .
\end{array}
$$
Then $x+y=\frac{x^{n+1}-y^{n+1}}{x^{n}-y^{n}}+x y \cdot \frac{x^{n-1}-y^{n-1}}{x^{n}-y^{n}}$
$$
\begin{aligned}
& >\frac{2\left(x^{n}+x^{n-1} y+x^{n-2} y^{2}+\cdot... | 1<x+y<\frac{2 n}{n+1} | Inequalities | proof | Yes | Yes | cn_contest | false | 726,103 |
Question 1 Let $[x]$ denote the greatest integer not exceeding the real number $x$. Prove:
$$
\sum_{i=0}^{\left[\frac{n}{3}\right]}\left[\frac{n-3 i}{2}\right]=\left[\frac{n^{2}+2 n+4}{12}\right]\left(n \in \mathbf{N}_{+}\right) .
$$ | 【Analysis】The expressions $\left[\frac{n}{3}\right]$ and $\left[\frac{n-3 i}{2}\right]$ appear in the problem. To calculate the left side of the expression, we can discuss $n$ modulo 6.
Proof: We discuss $n$ in six cases based on modulo 6.
When $n=6 t-3+r\left(t \in \mathbf{N}_{+}, r=0,1,2\right)$,
$$
\begin{array}{l}
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,104 |
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