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15. In the sequence $\left\{a_{n}\right\}$, it is known that $a_{1}=2$, and for all positive integers $n$, $a_{n+1}=a_{1} a_{2} \cdots a_{n}+1$. Prove:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \geqslant \frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n}}
$$
for all positive integers $n$. | 15. Since $a_{1}=2$, and for all positive integers $n$ we have $a_{n+1}=a_{1} a_{2} \cdots a_{n}+1$, therefore, for all positive integers $n$ we have $a_{n} \geqslant 2, a_{1} a_{2} \cdots a_{n} \geqslant 2^{n}$.
. We will prove by mathematical induction that for all positive integers $n$,
$$
1-\left(\frac{1}{a_{1}}+\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,038 |
1. Simplify the trigonometric rational expression
$$
\frac{\cos ^{4} x+\sin ^{4} x+\sin ^{2} x \cdot \cos ^{2} x}{\sin ^{6} x+\cos ^{6} x+2 \sin ^{2} x \cdot \cos ^{2} x}
$$
The value is ( ).
(A) 1
(B) $\sin x+\cos x$
(C) $\sin x \cdot \cos x$
(D) $1+\sin x \cdot \cos x$ | 1. A.
$$
\begin{array}{l}
\sin ^{6} x+\cos ^{6} x+2 \sin ^{2} x \cdot \cos ^{2} x \\
=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x-\right. \\
\left.\sin ^{2} x \cdot \cos ^{2} x\right)+2 \sin ^{2} x \cdot \cos ^{2} x \\
=\sin ^{4} x+\cos ^{4} x+\sin ^{2} x \cdot \cos ^{2} x
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,039 |
2. Let $p:\left(x^{2}+x+1\right) \sqrt{x+3} \geqslant 0$ , $q: x \geqslant-2$.
Then $p$ is a () condition for $q$.
(A) sufficient but not necessary
(B) necessary but not sufficient
(C) both sufficient and necessary
(D) neither sufficient nor necessary | 2. B.
Notice that
$$
\left(x^{2}+x+1\right) \sqrt{x+3} \geqslant 0 \Leftrightarrow x \geqslant-3 \text {. }
$$
Therefore, if $p$ holds, it does not necessarily imply that $q$ holds. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 724,040 |
4. Let $a, b$ be two perpendicular unit vectors. Given $\overrightarrow{O P}=a, \overrightarrow{O Q}=b, \overrightarrow{O R}=r a+k b$. If $\triangle P Q R$ is an equilateral triangle, then the values of $k, r$ are ( ).
(A) $k=r=\frac{-1 \pm \sqrt{3}}{2}$
(B) $k=\frac{-1 \pm \sqrt{3}}{2}, r=\frac{1 \pm \sqrt{3}}{2}$
(C)... | 4. C.
Notice
$$
\begin{array}{l}
|P Q|=|Q R|=|P R| \\
\Rightarrow \sqrt{r^{2}+(k-1)^{2}}=\sqrt{(r-1)^{2}+k^{2}}=\sqrt{2} \\
\Rightarrow r=k=\frac{1 \pm \sqrt{3}}{2} .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,042 |
5. In a regular triangular prism $A B C=A_{1} B_{1} C_{1}$, if $A B$ $=\sqrt{2} B B_{1}$, then the angle formed by $C A_{1}$ and $C_{1} B$ is ( ).
(A) $60^{\circ}$
(B) $75^{\circ}$
(C) $90^{\circ}$
(D) $105^{\circ}$ | 5. C.
Let the line containing $A_{1} B_{1}$ be the $x$-axis, the line perpendicular to $A_{1} B_{1}$ in the plane $A_{1} B_{1} C_{1}$ be the $y$-axis, and the line containing $B B_{1}$ be the $z$-axis to establish a spatial rectangular coordinate system. Then
$$
\begin{array}{l}
A_{1}(\sqrt{2}, 0,0), C_{1}\left(\frac{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,043 |
6. Let $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be an arithmetic sequence and a geometric sequence, respectively, and $a_{1}=b_{1}=4, a_{4}=b_{4}=1$. Then which of the following conclusions is correct? ( ).
(A) $a_{2}>b_{2}$
(B) $a_{3}b_{5}$
(D) $a_{6}>b_{6}$ | 6. A.
Let the common difference of the arithmetic sequence be $d$, and the common ratio of the geometric sequence be $q$.
Given $a_{1}=b_{1}=4, a_{4}=b_{4}=1$, we get
$d=-1, q=\frac{\sqrt[3]{2}}{2}$.
Thus, $a_{2}=3, b_{2}=2 \sqrt[3]{2} ; a_{3}=2, b_{3}=\sqrt[3]{4}$;
$a_{5}=0, b_{5}=\frac{\sqrt[3]{2}}{2} ; a_{6}=-1, b_... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,044 |
7. If $x \in \mathbf{R}_{+}$, then the term with the largest coefficient in the binomial expansion of $(1+2 x)^{15}$ is ( .).
(A) The 8th term
(B) The 9th term
(C) The 8th term and the 9th term
(D) The 11th term | 7. D.
$$
\begin{array}{l}
T_{r+1}=2^{\prime} \mathrm{C}_{15}^{r} . \\
\text { By } T_{r} \leqslant T_{r+1}, T_{r+2} \leqslant T_{r+1} \\
\Rightarrow \frac{29}{3} \leqslant r \leqslant \frac{32}{3} \Rightarrow r=10 .
\end{array}
$$
Therefore, the 11th term is the largest. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,045 |
8. Let $f(x)=\cos \frac{x}{5}, a=f\left(\log _{0} \frac{1}{\pi}\right)$,
$$
b=f\left(\log _{\pi} \frac{1}{\mathrm{e}}\right), c=f\left(\log _{\frac{1}{4}} \frac{1}{\pi^{2}}\right) .
$$
Then the correct relationship among the following is ( ).
(A) $a>b>c$
(B) $b>c>a$
(C) $c>a>b$
(D) $b>a>c$ | 8. D.
The function $f(x)=\cos \frac{x}{5}$ is an even function, and $g(x)=\cos x$ is decreasing on $\left(0, \frac{\pi}{2}\right)$.
Moreover, $\log _{e} \frac{1}{\pi}=-\log _{a} \pi, \log _{x} \frac{1}{e}=-\frac{1}{\log _{e} \pi}$, $\log _{\frac{1}{4}} \frac{1}{\pi^{2}}=2 \log _{e} \pi$,
Therefore, $0a>c$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,046 |
2. There are three numbers arranged in sequence: $3, 9, 8$. For any two adjacent numbers, the difference between the right number and the left number is written between these two numbers, resulting in a new sequence $3, 6, 9, -1, 8$, which is called the first operation; after the second similar operation, a new sequenc... | For convenience, let the sequence of $n$ numbers be $a_{1}, a_{2}, \cdots, a_{n}$. According to the problem, the newly added numbers are $a_{2}-a_{1}, a_{3}-a_{2}, \cdots, a_{n}-a_{n-1}$. Therefore, the sum of the newly added numbers is
$$
\begin{array}{l}
\left(a_{2}-a_{1}\right) + \left(a_{3}-a_{2}\right) + \cdots + ... | 520 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,047 |
11. Satisfy the equation
$$
\begin{array}{l}
\sqrt{x-2009-2 \sqrt{x-2010}}+ \\
\sqrt{x-2009+2 \sqrt{x-2010}}=2 .
\end{array}
$$
All real solutions are | $$
\text { II, 11.2 } 010 \leqslant x \leqslant 2011 \text {. }
$$
Transform the original equation to get
$$
\begin{array}{l}
\sqrt{(\sqrt{x-2010}-1)^{2}}+\sqrt{(\sqrt{x-2010}+1)^{2}}=2 \\
\Rightarrow 0 \leqslant \sqrt{x-2010} \leqslant 1 \\
\Rightarrow 2010 \leqslant x \leqslant 2011 .
\end{array}
$$ | 2010 \leqslant x \leqslant 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,050 |
12. If $x \in \mathbf{R}$, then the function
$$
f(x)=2 \sin \frac{x}{2}+3 \cos \frac{x}{3}
$$
has the smallest positive period of $\qquad$ . | 12. $12 \pi$.
Since the period of $2 \sin \frac{x}{2}$ is $4 \pi$, and the period of $3 \cos \frac{x}{3}$ is $6 \pi$, the period of the function $f(x)$ is $12 \pi$. | 12 \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,051 |
13. Let $P$ be a moving point on the circle $x^{2}+y^{2}=36$, and point $A(20,0)$. When $P$ moves on the circle, the equation of the trajectory of the midpoint $M$ of line segment $P A$ is $\qquad$. | 13. $(x-10)^{2}+y^{2}=9$.
Let $M(x, y) 、 P\left(x_{0}, y_{0}\right)$. Then
$$
\begin{array}{l}
x=\frac{x_{0}+20}{2}, y=\frac{y_{0}}{2} \\
\Rightarrow x_{0}=2 x-20, y_{0}=2 y .
\end{array}
$$
Since point $P$ is on the circle, we have
$$
(2 x-20)^{2}+(2 y)^{2}=36 \text {. }
$$
Therefore, the trajectory of point $M$ is... | (x-10)^{2}+y^{2}=9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,052 |
15. Let $z$ be an imaginary number, $w=z+\frac{1}{z}$, and $-1<w<2$. Then the range of the real part of $z$ is $\qquad$ . | 15. $\left(-\frac{1}{2}, 1\right)$.
Let $z=a+b$ i. Then
$$
\begin{array}{l}
-1<a+b \mathrm{i}+\frac{a-b \mathrm{i}}{a^{2}+b^{2}}<2 \\
\Rightarrow b-\frac{b}{a^{2}+b^{2}}=0 \\
\Rightarrow b=0 \text { or } a^{2}+b^{2}=1 .
\end{array}
$$
When $b=0$; no solution;
When $a^{2}+b^{2}=1 \Rightarrow-\frac{1}{2}<a<1$, which me... | \left(-\frac{1}{2}, 1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,054 |
16. Let $f(x)=k\left(x^{2}-x+1\right)-x^{4}(1-x)^{4}$. If for any $x \in[0,1]$, we have $f(x) \geqslant 0$, then the minimum value of $k$ is $\qquad$ . | 16. $\frac{1}{192}$.
It is easy to see that $k \geqslant \frac{x^{4}(1-x)^{4}}{x^{2}-x+1}$.
Notice that
$$
x^{2}-x+1=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4} \text {. }
$$
Therefore, when $x=\frac{1}{2}$, $x^{2}-x+1$ reaches its minimum value $\frac{3}{4}$. At this point, $x^{4}(1-x)^{4}$ reac... | \frac{1}{192} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,055 |
17. Let $p, q \in \mathbf{R}, f(x)=x^{2}+p|x|+q$. When the function $f(x)$ has more than one zero, the maximum value of $f(x)$ on the closed interval with its smallest and largest zeros as endpoints is $\qquad$ . | 17.0 or $q$.
Note that the function $f(x)=x^{2}+p|x|+q$ is an even function. By symmetry and the graph, we know that the maximum value of $f(x)$ on the closed interval with its smallest and largest zeros as endpoints is 0 or $q$. | q | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,056 |
18. Let the sequence be
$$
\frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{1}{3}, \frac{2}{2}, \frac{3}{1}, \cdots, \frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}, \cdots
$$
Question: (1) What is the value of the 2010th term in this sequence?
(2) In this sequence, what is the index of the 2010th term that has a value of... | Three, 18. (1) Group the sequence:
$$
\left(\frac{1}{1}\right),\left(\frac{1}{2}, \frac{2}{1}\right), \cdots,\left(\frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}\right), \cdots
$$
Notice that
$$
\begin{array}{l}
1+2+\cdots+62=1953, \\
1+2+\cdots+63=2016 .
\end{array}
$$
Thus, the 2010th term of the sequence belongs ... | \frac{57}{7} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,057 |
3. A chess piece moves on an $n \times n$ grid, and at each step, it can move to the adjacent cell above, to the adjacent cell on the right, or to the opposite corner cell in the left-down direction (as shown in Figure 1). Can it visit each cell exactly once and end at the cell to the right of its starting position? | Let the sum of the row number and column number of the square where the piece is located be denoted as $S$. In each step, $S$ either increases by 1 or decreases by 2, which means that after each move, the remainder of $S$ modulo 3 definitely increases by 1. Since there are a total of $n^{2}-1$ steps, and the $S$ at the... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,058 |
19. There are 10 red, 10 black, and 10 white balls. Now, all of them are to be placed into two bags, A and B, with the requirement that each bag must contain balls of all three colors, and the product of the number of balls of each color in bags A and B must be equal. How many ways are there to do this? | 19. Let the number of red, black, and white balls in bag A be $x$, $y$, and $z$ respectively. Then $1 \leqslant x, y, z \leqslant 9$, and
$$
x y z=(10-x)(10-y)(10-z) \text {, }
$$
i.e., $x y z=500-50(x+y+z)+5(x y+y z+z x)$.
Thus, $5 \mid x y z$.
Therefore, one of $x, y, z$ must be 5.
Assume $x=5$, substituting into eq... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,059 |
20. Let the ellipse be $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1), \operatorname{Rt} \triangle A B C$ with $A(0,1)$ as the right-angle vertex, and sides $A B, B C$ intersecting the ellipse at points $B, C$. If the maximum area of $\triangle A B C$ is $\frac{27}{8}$, find the value of $a$. | 20. Let $l_{A B}: y=k x+1(k>0)$. Then $l_{A C}: y=-\frac{1}{k} x+1$.
From $\left\{\begin{array}{l}y=k x+1, \\ \frac{x^{2}}{a^{2}}+y^{2}=1,\end{array}\right.$,
$\left(1+a^{2} k^{2}\right) x^{2}+2 a^{2} k x=0$
$\Rightarrow x_{B}=\frac{-2 a^{2} k}{1+a^{2} k^{2}}$.
Thus, $|A B|=\sqrt{1+k^{2}} \cdot \frac{2 a^{2} k}{1+a^{2}... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,060 |
21. Let $D$, $E$, $F$ be points on the sides $BC$, $CA$, $AB$ of $\triangle ABC$, respectively. Denote
$$
\alpha=\frac{BD}{BC}, \beta=\frac{CE}{CA}, \gamma=\frac{AF}{AB}.
$$
Prove: $S_{\triangle DEF} \geqslant \alpha \beta \gamma S_{\triangle ABC}$. | Notice that
$$
\frac{S_{\triangle B F D}}{S_{\triangle A B C}}=\frac{B D \cdot B F \sin B}{B C \cdot B A \sin B}=\alpha(1-\gamma) \text {. }
$$
Similarly, $\frac{S_{\triangle D B C}}{S_{\triangle A B C}}=\beta(1-\alpha)$,
$$
\frac{S_{\triangle E F}}{S_{\triangle A B C}}=\gamma(1-\beta) \text {. }
$$
Then $\frac{S_{\t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,061 |
22. (1) Let $a>0$. If the coordinates of a point in the plane are all integers, then it is called a "lattice point". Given that the curve $y=a x^{3}$ passes through the lattice point $(n, m)$, and the number of lattice points on the curve segment corresponding to $1 \leqslant x \leqslant n$ is denoted by $N$. Prove:
$$... | 22. (1) Consider the region $0<x \leqslant n, 0<y \leqslant m$, and the number of lattice points in this region is $n m$. This region is composed of region $E: 0<x \leqslant n, 0<y \leqslant a x^{3}$, and region $F$:
$$
0<y \leqslant m, 0<x \leqslant \sqrt[3]{\frac{y}{a}}
$$
In region $E$, the number of lattice points... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,062 |
1. If the arithmetic mean of two positive numbers is $2 \sqrt{3}$, and the geometric mean is $\sqrt{3}$, what is the difference between these two numbers? | 1. Let $x, y$ represent two numbers. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y=4 \sqrt{3}, \\
x y=3
\end{array}\right. \\
\Leftrightarrow|x-y|=\sqrt{(x-y)^{2}} \\
=\sqrt{(x+y)^{2}-4 x y}=6,
\end{array}
$$
which means the difference between the two numbers is 6. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,063 |
2. It is known that fresh shiitake mushrooms contain $90 \% \sim 99 \%$ water, while dried shiitake mushrooms contain $30 \% \sim 45 \%$ water. Then, under the influence of drying, by what maximum factor can the weight of fresh shiitake mushrooms be reduced? | 2. Let the weights of fresh mushrooms, baked mushrooms be $m_{1} \mathrm{~g}, m_{2} \mathrm{~g}$, and the weight of dried mushrooms be $x \mathrm{~g}$ (unknown). Then the range of the proportion of dried mushrooms in fresh mushrooms and baked mushrooms is
$$
\begin{array}{l}
0.01=1-0.99 \\
\leqslant \frac{x}{m_{1}} \le... | 70 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,064 |
3. For each value of $a$, find all values of $x$ that satisfy the equation
$$
\begin{array}{l}
\log _{5}\left[\frac{(x+1)^{2}}{x}-a\right] \\
=\log _{5} \frac{(x+1)^{2}}{x}-\log _{5} a
\end{array}
$$ | 3. Notice that
$$
\begin{array}{l}
\log _{5}\left[\frac{(x+1)^{2}}{x}-a\right] \\
=\log _{5} \frac{(x+1)^{2}}{x}-\log _{5} a \\
\Leftrightarrow\left\{\begin{array}{l}
\log _{5}\left[\frac{(x+1)^{2}}{x}-a\right]=\log _{5} \frac{(x+1)^{2}}{a x}, \\
a>0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
\frac{(x... | x=a-1 \text{ or } \frac{1}{a-1} \text{ when } a>1; \text{ no solution when } a \leqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,065 |
5. Given two circles $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent at point $A$, circle $\Gamma$ is externally tangent to $\Gamma_{1}$ and $\Gamma_{2}$ at points $B$ and $C$ respectively. Extend the chord $B A$ of circle $\Gamma_{1}$ to intersect circle $\Gamma_{2}$ at point $D$, extend the chord $C A$ of circle... | 5. First, prove that quadrilateral $B C G F$ is a rectangle:
In fact, let $K L, B M, C N$ be the common tangents of circles $\Gamma_{1}$ and $\Gamma_{2}$, $\Gamma_{1}$ and $\Gamma$, $\Gamma_{2}$ and $\Gamma$ (Figure 2). According to the inscribed angle and the angle between a tangent and a chord, we have
$$
\angle A B... | 13 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,067 |
6. Find all pairs $(x, y)$ such that for each pair of values, the numbers
$u=\sqrt{4+x^{3}-9 x}-x-3^{y}$ and $v=2-x-3^{y}$ satisfy the following three conditions:
(1) If $|u|>|v|$, then $u>0$;
(2) If $|u|v$;
(3) If $|u|=|v|$, then $u>0>v$.
保留了源文本的换行和格式。 | 6. Three conclusions hold simultaneously
$$
\begin{array}{l}
\Leftrightarrow u>v \\
\Leftrightarrow \sqrt{4+x^{3}-9 x}>2 \\
\Leftrightarrow 4+x^{3}-9 x>4 \\
\Leftrightarrow x(x-3)(x+3)>0 \\
\Leftrightarrow-33 .
\end{array}
$$
Therefore, $-33$, and $y$ can be any number. | -3 < x < 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,068 |
Example 1 Given real numbers $x, y$ satisfy
$$
\begin{array}{l}
\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right) \\
=2008 .
\end{array}
$$
Then the value of $3 x^{2}-2 y^{2}+3 x-3 y-2007$ is ). [1]
(A) -2008
(B) 2008
(C) -1
(D) 1
(2008, National Junior High School Mathematics Competition) | Solve: From the given, we have
$$
\begin{array}{l}
x-\sqrt{x^{2}-2008}=\frac{2008}{y-\sqrt{y^{2}-2008}} \\
=y+\sqrt{y^{2}-2008}, \\
y-\sqrt{y^{2}-2008}=\frac{2008}{x-\sqrt{x^{2}-2008}} \\
=x+\sqrt{x^{2}-2008} .
\end{array}
$$
From equations (1) and (2), we know
$$
x=y \text {. }
$$
Substituting equation (3) into equa... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,069 |
Example 2 Given non-zero numbers $a, b$ satisfy
$$
|2 a-4|+|b+2|+\sqrt{(a-3) b^{2}}+4=2 a \text {. }
$$
Then the value of $a+b$ is ( ).
(2]
(A) -1
(B) 0
(C) 1
(D) 2
（2009, "Mathematics Weekly" Cup National Junior High School Mathematics Competition） | From the non-negativity of the radicand in the square root, we have
$$
\begin{array}{l}
(a-3) b^{2} \geqslant 0 \\
\Rightarrow a \geqslant 3 \\
\Rightarrow|2 a-4|=2 a-4 .
\end{array}
$$
Thus, the original expression becomes
$$
\begin{array}{l}
|b+2|+\sqrt{(a-3) b^{2}}=0 \\
\Rightarrow b+2=0,(a-3) b^{2}=0 \\
\Rightarro... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,070 |
Example 3 Let $n \in \mathbf{N}_{+}$. Prove: $\{n \sqrt{7}\}>\frac{11}{20 n}$. | 【Analysis】Let $A_{n}=[n \sqrt{7}], \alpha_{n}=\{n \sqrt{7}\}$. Then $n \sqrt{7}=A_{n}+\alpha_{n}>A_{n} \Rightarrow 7 n^{2}>A_{n}^{2}$,
which means $7 n^{2}-A_{n}^{2} \geqslant 1$.
Also, $x^{2}=0,1,2,4(\bmod 7)$, so $7 n^{2}-A_{n}^{2} \neq 1,2$.
Therefore, $7 n^{2}-A_{n}^{2} \geqslant 3$
$$
\begin{array}{l}
\Rightarrow\... | \alpha_{n}>\frac{11}{20 n} | Number Theory | proof | Yes | Yes | cn_contest | false | 724,071 |
10. (20 points) Let $a_{1}, a_{2}, a_{3}$ be in arithmetic progression, $a_{1}+a_{2}+a_{3}=18$; $b_{1}, b_{2}, b_{3}$ be in geometric progression, $b_{1} b_{2} b_{3}=64$. If $a_{1}+b_{1}, a_{2}+b_{2}, a_{3}+b_{3}$ are positive integers and form a geometric progression, find the maximum value of $a_{3}$. | 10. Let \(a_{1}=6-d, a_{2}=6, a_{3}=6+d\);
\[
b_{1}=\frac{4}{q}, b_{2}=4, b_{3}=4 q \text{. }
\]
Then, by the conditions, \(6-d+\frac{4}{q}\) and \(6+d+4 q\) are both positive integers; and
\[
\begin{array}{l}
\left(6-d+\frac{4}{q}\right)(6+d+4 q)=\left(a_{2}+b_{2}\right)^{2}=100 . \\
\text{ Hence }\left\{\begin{array... | \frac{111+9 \sqrt{97}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,072 |
11. (20 points) Given an acute triangle $\triangle ABC$, let
$$
S=\cos A \cdot \cos B \cdot \cos C \text {. }
$$
Prove: $\sum \frac{S+\sin ^{2} A}{(2 S+\sin 2 A)^{2}} \geqslant \frac{1}{4 S}$,
where, " $\sum$ " denotes the cyclic sum. | 11. Equation (1)
$$
\begin{array}{l}
\Leftrightarrow \sum \frac{\cos A \cdot \cos B \cdot \cos C+\sin ^{2} A}{(\cos A \cdot \cos B \cdot \cos C+\sin A \cdot \cos A)^{2}} \\
\geqslant \frac{1}{\cos A \cdot \cos B \cdot \cos C} \\
\Leftrightarrow \sum \frac{1+\frac{\sin ^{2} A}{\cos A \cdot \cos B \cdot \cos C}}{\left(1+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,073 |
One, (40 points) Given an acute triangle $\triangle ABC$ with its circumcircle $\odot O_{1}$, and a circle $\odot O_{2}$ passing through points $A$ and $O_{1}$ intersects lines $AB$ and $AC$ at points $P$ and $Q$ different from $A$. Prove that $BC = PQ$ if and only if the angle between $BC$ and $PQ$ is $60^{\circ}$. | Let the angle between lines $k$ and $l$ be denoted as $\theta(k, l)$.
Extend $A O_{1}$ to intersect $\odot O_{1}$ at point $O^{\prime}$.
(1) If $P$ and $Q$ are on the extensions of $B A$ and $C A$ respectively, then point $A$ is inside $\triangle O_{1} P Q$, which contradicts the fact that $A, O_{1}, P, Q$ are concycli... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,074 |
II. (40 points) In an $n \times 2n$ (where $n \in \mathbf{N}, n > 1$) grid, each cell is filled with a “+” or “-” sign. If an ordered integer sequence $\left(x_{1}, x_{2}, \cdots, x_{2n}\right)$ has the following properties:
(i) $\sum_{i=1}^{2n}\left|x_{i}\right| \neq 0$;
(ii) $\left|x_{i}\right| \leqslant n$ (for $i=1... | Certainly! Here is the translation of the provided text into English, preserving the original formatting and line breaks:
```
Consider the ordered integer sequence $\left(x_{1}, x_{2}, \cdots, x_{2 n}\right)$ $\left(\left|x_{i}\right| \leqslant n, i=1,2, \cdots, 2 n\right)$, there are $(2 n+1)^{2 n}$ of them.
For eac... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,075 |
Four, (50 points) Prove: the sequence
$$
a_{n}=\frac{1}{4}\left[(1+\sqrt{2})^{2 n+1}+(1-\sqrt{2})^{2 n+1}+2\right](n>1)
$$
contains no perfect squares. | $$
\begin{array}{l}
\text { Assume } \frac{1}{4}\left[(1+\sqrt{2})^{2 n+1}+(1-\sqrt{2})^{2 n+1}+2\right] \\
=M^{2}\left(M \in \mathbf{N}_{+}, n>1\right) .
\end{array}
$$
Then $2 M^{2}-1=\frac{1}{2}\left[(1+\sqrt{2})^{2 n+1}+(1-\sqrt{2})^{2 n+1}\right]$
$\Rightarrow\left(2 M^{2}-1\right)^{2}+1$
$=\frac{1}{4}\left[(1+\s... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,077 |
Given a square $ABCD$, points $E$ and $F$ are on the extensions of $AB$ and $BC$ respectively, and $AE = EF + FC$. Prove: $\angle EDF = 45^{\circ}$. | Proof As shown in Figure 2,
rotate Rt $\triangle C D F$ 90° clockwise around point $D$, to get Rt $\triangle A D F^{\prime}$. Connect $F F^{\prime}$.
$$
\begin{array}{l}
\text { From } A E=E F+F C \\
=E F+A F^{\prime} \\
=E F^{\prime}+A F^{\prime} \\
\Rightarrow E F=E F^{\prime} .
\end{array}
$$
Since $D F=D F^{\prim... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,078 |
Given positive integers $a_{1}, a_{2}, \cdots, a_{n}\left(a_{1}<\right.$ $\left.a_{2}<\cdots<a_{n}\right), k \in \mathbf{R}_{+}$, and $k \geqslant 1$. Prove:
$$
\sum_{i=1}^{n} a_{i}^{2 k+1} \geqslant\left(\sum_{i=1}^{n} a_{i}^{k}\right)^{2} .
$$ | First, we prove using mathematical induction:
$$
2 \sum_{i=1}^{n} a_{i}^{k} \leqslant\left(a_{n}+1\right)^{k} a_{n} \text {. }
$$
When \( n=1 \), it is easy to see that the inequality (2) holds.
Assume that when \( n=m \), the inequality (2) holds, i.e.,
$$
2 \sum_{i=1}^{m} a_{i}^{k} \leqslant\left(a_{m}+1\right)^{k} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,079 |
Let $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ be a function, and for any positive integers $m, n$, we have
$$
f(f(m)+f(n))=m+n .
$$
Find the value of $f(2011)$. | Since $f(x)$ is a function from the set of positive integers to the set of positive integers, let
$$
f(1)=p\left(p \in \mathbf{N}_{+}\right) \text {. }
$$
If $p>1$, then $p \geqslant 2$.
$$
\text { Let } p=1+b\left(b \in \mathbf{N}_{+}\right), f(b)=c\left(c \in \mathbf{N}_{+}\right) \text {. }
$$
On one hand,
$$
\beg... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,080 |
Example 4 Find $n \in \mathbf{N}_{+}$, such that
$$
n-[n\{\sqrt{n}\}]=2 \text{. }
$$
(2002, Bulgarian Mathematical Olympiad) | 【Analysis】Let
$$
\sqrt{n}=A+\alpha(A=[\sqrt{n}], \alpha=\{\sqrt{n}\}) \text {. }
$$
Then $n=(A+\alpha)^{2}$.
When $\alpha=0$, $n=A^{2}=2$. At this time, $A$ cannot be an integer.
When $\alpha \in(0,1)$,
$$
\begin{aligned}
2 & =n-[n\{\sqrt{n}\}]=n-[n \sqrt{n}-A n] \\
& =n+A n-[n \sqrt{n}],
\end{aligned}
$$
which means... | n=2,8,15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,081 |
Example 5 Let $n \in \mathbf{N}_{+}$, and $p$ be a prime number such that $p<n$. Prove:
$$
p \left\lvert\,\left(\mathrm{C}_{n}^{p}-\left[\frac{n}{p}\right]\right)\right. \text {. }
$$
(2003, Croatian Mathematical Olympiad) | 【Analysis】The key to this problem is how to properly represent $\left[\frac{n}{p}\right]$. In the sequence of $p$ consecutive integers $n, n-1, \cdots, n-p+1$, there exists a unique integer (denoted as $m$) that is divisible by $p$. Clearly, $\left[\frac{n}{p}\right]=\frac{m}{p}$.
$$
\begin{array}{l}
\text { Then } k=\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,082 |
Example 6 Let $k, l$ be two given positive integers. Prove: there are infinitely many positive integers $n>k$, such that $\mathrm{C}_{n}^{k}$ is coprime with $l$. ${ }^{[3]}$
(2009, National High School Mathematics Joint Competition) | Prove a common formula related to the Gauss function: the exponent of a prime $p$ in the arithmetic factorization of $n!$ is
$$
V_{p}(n!)=\sum_{\alpha=1}^{+\infty}\left[\frac{n}{p^{\alpha}}\right] .
$$
Let the arithmetic factorization of a positive integer $l$ be
$$
p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{j}^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,083 |
Example 7 Let $n \in \mathbf{N}_{+}$. Prove:
$$
\sum_{k=0}^{+\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right]=n
$$ | Proof 1: Mathematical Induction.
When $n=1$, the equation obviously holds.
Assume when $n=m$, the equation holds.
Then when $n=m+1$, let
$m+1=2^{\alpha}(2 \beta+1)$.
Thus, $\left[\frac{m+1+2^{\alpha}}{2^{\alpha+1}}\right]=\frac{2^{\alpha}(2 \beta+2)}{2^{\alpha+1}}$
$=\beta+1=\left[\frac{m+2^{\alpha}}{2^{\alpha+1}}\righ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,084 |
Example 8 Let $k \in\{1,2,3,4,5\}, m \in \mathbf{N}_{1}$, and denote $f(m, k)=\sum_{i=1}^{5}\left[m \sqrt{\frac{k+1}{i+1}}\right]$.
Then for any positive integer $n$, there exists a unique pair $(m, k)$ such that
$$
f(m, k)=n .{ }^{[4]}
$$
(2007, National High School Mathematics Competition) | 【Analysis】Let $a_{1}=\left[\sqrt{\frac{k+1}{2}} m\right]$. Then
$$
a_{1} \leqslant \sqrt{\frac{k+1}{2}} m < a_{1}+1,
$$
which implies $\sqrt{2} a_{1} \leqslant \sqrt{k+1} m < \sqrt{2}\left(a_{1}+1\right)$.
Let the set $A_{1}=\left\{\sqrt{2} n \mid n \in \mathbf{N}_{+}\right\}$,
$$
B_{1}=A_{1} \cap[0, \sqrt{k+1} m].
$$... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,085 |
Example 9 Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1} \in(1,2) \text {, and } a_{n+1}=a_{n}+\frac{n}{a_{n}} \text {. }
$$
Prove: There exists at most one pair of positive integers $(i, j)(i<j)$ such that $a_{i}+a_{j} \in \mathbf{Z}$. ${ }^{[s]}$
(35th Russian Mathematical Olympiad (11th grade)) | 【Analysis】The recursive relationship of this sequence is not a common linear recursive relationship, and the initial value is uncertain. The trial method can be used to observe the characteristics of the sequence $\left\{a_{n}\right\}$.
Since $a_{1} \in(1,2)$, the most intuitive initial value is $a_{1}=\frac{3}{2}=1 \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,086 |
Example 10. Define the sequence $\left\{a_{n}\right\}$:
$$
a_{1}=2008, a_{n+1}=n\left[\frac{a_{n}}{n}\right]+n \text {. }
$$
Prove: The sequence $\left\{a_{n}\right\}$ has an infinite arithmetic subsequence $\left\{a_{n_{k}}\right\}$ and an infinite geometric subsequence $\left\{a_{n}\right\}$. | 【Analysis】The initial value $a_{1}=2008$ of the sequence $\left\{a_{n}\right\}$ is too large to facilitate mathematical experiments. This problem may be unrelated to 2008, so we choose a suitable initial value for the experiment, for example, $a_{1}=9$. It is easy to calculate:
$$
\begin{array}{l}
a_{2}=10, a_{3}=12, a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,087 |
Example 4 Let $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ be complex numbers, and
$$
f(z)=\prod_{j=1}^{n}\left(z-\alpha_{j}\right) .
$$ | Proof: There exists a complex number $z_{0}$ satisfying $\left|z_{0}\right|=1$, such that
$$
\left|f\left(z_{0}\right)\right| \geqslant \frac{\prod_{i=1}^{n}\left(1+\left|\alpha_{j}\right|\right)}{2^{n-1}}.
$$
【Analysis and Proof】First, take a step back and consider a special case.
Assume $\alpha_{j}$ lies on the unit... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,088 |
Example 1 Let $\angle BAC$ be the smallest interior angle of $\triangle ABC$. Points $B$ and $C$ divide the circumcircle of this triangle into two arcs. Let $U$ be a point on the arc that does not contain point $A$ and is different from $B$ and $C$. The perpendicular bisectors of segments $AB$ and $AC$ intersect segmen... | Prove as shown in Figure 1, extend $B T$ and $C W$ to intersect the circumcircle at points $S$ and $H$.
Since the perpendicular bisectors of segments $A B$ and $A C$ intersect the circumcircle at the center $O$, any line through the center is an axis of symmetry of the circle.
With $O V$ as the axis of symmetry, we ha... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,089 |
Example 3 Calculate the value of
$(\sqrt{30}+\sqrt{21}-\sqrt{3})(\sqrt{3}+\sqrt{10}-\sqrt{7})$ is ( ). ${ }^{[3]}$
(A) $6 \sqrt{7}$
(B) $-6 \sqrt{7}$
(C) $20 \sqrt{3}+6 \sqrt{7}$
(D) $20 \sqrt{3}-6 \sqrt{7}$
$(2007$, National Junior High School Mathematics Competition, Tianjin Preliminary Contest) | $$
\begin{array}{l}
=\sqrt{3}[\sqrt{10}+(\sqrt{7}-\sqrt{3})][\sqrt{10}-(\sqrt{7}-\sqrt{3})] \\
=\sqrt{3}(10-7+2 \sqrt{21}-3)=6 \sqrt{7} .
\end{array}
$$
Therefore, the correct answer is (A). | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,090 |
Example 2 In the plane, $\odot O_{1}$ and $\odot O_{2}$ intersect, and $A$ is one of the intersection points. Two moving points $M_{1}$ and $M_{2}$ start from point $A$ simultaneously, moving with the same angular velocity along $\odot O_{1}$ and $\odot O_{2}$ in the same direction, and return to point $A$ after each c... | 【Analysis】(1) Investigate the movement rules of two moving points $M_{1} 、 M_{2}$, and find the position of the fixed point $P$.
As shown in Figure 2, according to the problem, $\overparen{A M}_{1}^{\circ}=\overparen{A M}_{2}$.
Then $\angle A B M_{1}+\angle A B M_{2}=180^{\circ}$, which means the two moving points $M_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,091 |
Example 3 In $\triangle ABC$, it is known that $X$ is a moving point on line $BC$, and point $C$ is between points $B$ and $X$. The incircles of $\triangle ABX$ and $\triangle ACX$, $\odot O_{1}$ and $\odot O_{2}$, have two distinct intersection points $P$ and $Q$. Prove that $PQ$ passes through a fixed point independe... | 【Analysis】(1) As shown in Figure 4, place the known $\triangle ABC$ in a Cartesian coordinate system with $B$ as the origin and $BX$ coinciding with the $x$-axis. Let $\odot O_{1}$ and $\odot O_{2}$ be the incircles of $\triangle ABX$ and $\triangle ACX$, respectively, and let the points of tangency with the sides be $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,092 |
2. Given real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
a+b+c+d=6, \\
a^{2}+b^{2}+c^{2}+d^{2}=12 .
\end{array}
$$
Prove:
$$
\begin{array}{l}
36 \leqslant 4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)-\left(a^{4}+b^{4}+c^{4}+d^{4}\right) \\
\leqslant 48 .
\end{array}
$$ | 2. Let $x=a-1, y=b-1, z=c-1, t=d-1$.
$$
\begin{array}{l}
\text { Then } 4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)-\left(a^{4}+b^{4}+c^{4}+d^{4}\right) \\
=-\left[(a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\right]+ \\
6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-4(a+b+c+d)+4 \\
=-\left[(a-1)^{4}+(b-1)^{4}+(c-1)^{4}+\right. \\
\left.(d... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,093 |
3. Given $x_{1}, x_{2}, \cdots, x_{100}$ are non-negative real numbers, and for $i=1,2, \cdots, 100$, we have $x_{i}+x_{i+1}+x_{i+2} \leqslant 1$, where $x_{101}=x_{1}, x_{102}=x_{2}$. Find the maximum value of the sum $S=\sum_{i=1}^{100} x_{i} x_{i+2}$. | 3. The maximum value of $S$ is $\frac{25}{2}$.
For $i=1,2, \cdots, 50$, set $x_{2 i}=0, x_{2 i-1}=\frac{1}{2}$. Then $S=50 \times\left(\frac{1}{2}\right)^{2}=\frac{25}{2}$.
Below is the proof: For all $x_{1}, x_{2}, \cdots, x_{100}$ that satisfy the conditions, we have
$$
S \leqslant \frac{25}{2} \text {. }
$$
Since... | \frac{25}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,094 |
4. The sequence $x_{1}, x_{2}, \cdots$ is defined as follows:
$$
x_{1}=1, x_{2 k}=-x_{k}, x_{2 k-1}=(-1)^{k+1} x_{k},
$$
where $k \geqslant 1$. Prove: For all $n \geqslant 1$,
$$
x_{1}+x_{2}+\cdots+x_{n} \geqslant 0
$$ | 4. From the definition of $x_{i}$, for each positive integer $k$ we have
$$
\begin{array}{l}
x_{4 k-3}=x_{2 k-1}=-x_{4 k-2}, \\
x_{4 k-1}=x_{4 k}=-x_{2 k}=x_{k} .
\end{array}
$$
Let $S_{n}=\sum_{i=1}^{n} x_{i}$. Then
$$
\begin{array}{l}
S_{4 k}=\sum_{i=1}^{k}\left[\left(x_{4 i-3}+x_{4 i-2}\right)+\left(x_{4 i-1}+x_{4 ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,095 |
5. Let $Q_{+}$ be the set of all positive rational numbers. Find all functions $f: \mathbf{Q}_{+} \rightarrow \mathbf{Q}_{+}$ such that for all $x, y \in \mathbf{Q}_{+}$, we have
$$
f\left(f^{2}(x) y\right)=x^{3} f(x y) .
$$ | 5. The function that satisfies the condition is $f(x)=\frac{1}{x}$.
In equation (1), take $y=1$. Then
$$
f\left(f^{2}(x)\right)=x^{3} f(x) \text {. }
$$
If $f(x)=f(y)$, then
$$
\begin{array}{l}
x^{3}=\frac{f\left(f^{2}(x)\right)}{f(x)}=\frac{f\left(f^{2}(y)\right)}{f(y)}=y^{3} \\
\Rightarrow x=y .
\end{array}
$$
The... | f(x)=\frac{1}{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,096 |
6. Given two functions $f, g: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$, satisfying for all positive integers $n$,
$$
\begin{array}{l}
f(g(n))=f(n)+1, \\
g(f(n))=g(n)+1 .
\end{array}
$$
Prove: For all positive integers $n$,
$$
f(n)=g(n) .
$$ | $$
\begin{array}{l}
\text { 6. For any positive integer } k \text { and any function } h \text { : } \\
\mathbf{N}_{+} \rightarrow \mathbf{N}_{+} \text {, define } \\
h^{k}(x)=\underbrace{h(h(\cdots, h(x) \cdots))}_{k \uparrow},
\end{array}
$$
In particular, $h^{0}(x)=x$.
For any positive integer $k$, we have
$f\left... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,097 |
1. If $a+b=2, \frac{(1-a)^{2}}{b}+\frac{(1-b)^{2}}{a}=-4$, then $a b=(\quad)$.
(A) 1
(B) -1
(C) $-\frac{1}{2}$
(D) $\frac{1}{2}$ | $\begin{array}{l}\text { I. 1. B. } \\ \text { From } \frac{(1-a)^{2}}{b}+\frac{(1-b)^{2}}{a}=-4 \\ \Rightarrow a(1-a)^{2}+b(1-b)^{2}=-4 a b \\ \Rightarrow(a+b)-2\left(a^{2}+b^{2}\right)+a^{3}+b^{3}+4 a b=0 \\ \Rightarrow 2-2\left(a^{2}+b^{2}\right)+2\left(a^{2}-a b+b^{2}\right)+4 a b=0 \\ \Rightarrow a b=-1 .\end{arra... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,099 |
2. Given that the lengths of two altitudes of $\triangle A B C$ are 5 and 20. If the length of the third altitude is also an integer, then the maximum length of the third altitude is ( ).
(A) 5
(B) 6
(C) 7
(D) 8 | 2. B.
Let the area of $\triangle A B C$ be $S$, and the length of the third altitude be $h$. Then the lengths of the three sides are $\frac{2 S}{5}, \frac{2 S}{20}, \frac{2 S}{h}$.
Thus, $\frac{2 S}{5}-\frac{2 S}{20}<\frac{2 S}{h}<\frac{2 S}{5}+\frac{2 S}{20}$.
Solving this, we get $4<h<\frac{20}{3}$.
Therefore, the m... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,100 |
Example 4 Simplify $\frac{\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}$. The result is ( ). [4]
(A) $\sqrt{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) 2
(D) $\frac{1}{2}$
(2010, National Junior High School Mathematics League, Jiangxi Province Preliminary) | Notice
$$
\begin{array}{l}
13+\sqrt{48}=(2 \sqrt{3})^{2}+1+4 \sqrt{3}=(1+2 \sqrt{3})^{2}, \\
5-(1+2 \sqrt{3})=4-2 \sqrt{3}=(\sqrt{3}-1)^{2}, \\
3+(\sqrt{3}-1)=2+\sqrt{3}=\frac{4+2 \sqrt{3}}{2} \\
=\frac{(\sqrt{3}+1)^{2}}{2}=\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)^{2} .
\end{array}
$$
Therefore, the original expressio... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,101 |
3. The number of solutions to the equation $\left|x^{2}-1\right|=(4-2 \sqrt{3})(x+2)$ is ().
(A) 1
(B) 2
(C) 3
(D) 4 | 3. C.
When $|x| \geqslant 1$, the equation is
$$
x^{2}-1=(4-2 \sqrt{3})(x+2) \text {. }
$$
Solving it, we get $x_{1}=\sqrt{3}, x_{2}=4-3 \sqrt{3}$, both satisfying
$$
|x| \geqslant 1 \text {. }
$$
When $|x|<1$, the equation is
$$
1-x^{2}=(4-2 \sqrt{3})(x+2) \text {. }
$$
Solving it, we get $x_{3}=\sqrt{3}-2$, satis... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,102 |
4. There are segments of lengths $1, 2, \cdots, 9$, one of each. Now, select several segments to form a "segment group", and with this group of segments, a square can be exactly formed. Then the number of such segment groups is ( ).
(A) 5
(B) 7
(C) 9
(D) 11 | 4. C.
Obviously, using these segments to form a square requires at least 7 segments, and the side length of the square is at least 7.
Since $1+2+\cdots+9=45$, the side length of the square is no more than $\left[\frac{45}{4}\right]=11$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Notic... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,103 |
5. As shown in Figure 1, in rhombus $A B C D$, it is known that $A B=3, D F$ $=1, \angle D A B=$ $60^{\circ}, \angle E F G=$ $15^{\circ}, F G \perp B C$. Then $A E=(\quad$.
(A) $1+\sqrt{2}$
(B) $\sqrt{6}$
(C) $2 \sqrt{3}-1$
(D) $1+\sqrt{3}$ | 5. D.
Draw a perpendicular from point $F$ to $AB$, with the foot of the perpendicular being $H$.
Given $\angle DAB = 60^{\circ}$ and $AF = AD - FD = 2$, we have
$$
\begin{array}{l}
\angle AFH = 30^{\circ}, AH = 1, FH = \sqrt{3}. \\
\text{Also, } \angle EFH = \angle AFG - \angle AFH - \angle EFG \\
= 45^{\circ}.
\end{a... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,104 |
6. Given $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$,
$$
\begin{array}{l}
\frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}, \\
\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} . \\
\text { Then } \frac{2}{x}+\frac{3}{y}+\frac{4}{z}=(\quad \text { ). }
\end{array}
$$
(A) 1
(B) $\frac{3}{2}$
(C) 2
(D) $\frac{5}{2}$ | 6. C.
From the given equations, we have
$$
\begin{array}{l}
x+y+z=\frac{z x+x y}{2}=\frac{x y+y z}{3}=\frac{y z+z x}{4} \\
\Rightarrow \frac{x y}{1}=\frac{y z}{5}=\frac{z x}{3} \Rightarrow z=3 y=5 x .
\end{array}
$$
Substituting into $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$, we get
$$
\frac{1}{x}+\frac{1}{\frac{5}{3} ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,105 |
1. In $\triangle A B C$, it is known that $\angle B=2 \angle A, B C=$ $2, A B=2+2 \sqrt{3}$. Then $\angle A=$ $\qquad$ . | $=1.15^{\circ}$.
As shown in Figure 5, draw $CE \perp AB$ at point $E$, extend $AB$ to $D$ such that $BD = BC$, and connect $CD$.
Then $\angle D = \angle BCD = \frac{1}{2} \angle ABC = \angle A$.
Therefore, $AE = DE = \frac{1}{2} AD = \frac{1}{2}(AB + BD) = 2 + \sqrt{3}$,
$BE = AB - AE = (2 + 2\sqrt{3}) - (2 + \sqrt{3}... | 15^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,106 |
2. For the quadratic function $y=x^{2}+b x+c$, the vertex of its graph is $D$, and it intersects the positive x-axis at points $A$ and $B$ from left to right, and the positive y-axis at point $C$. If $\triangle A B D$ and $\triangle O B C$ are both isosceles right triangles (where $O$ is the origin), then $b+2 c=$ | 2. 2 .
From the given information, we have
$$
\begin{array}{l}
C(0, c) 、 A\left(\frac{-\dot{b}-\sqrt{b^{2}-4 c}}{2}, 0\right), \\
B\left(\frac{-b+\sqrt{b^{2}-4 c}}{2}, 0\right), D\left(-\frac{b}{2},-\frac{b^{2}-4 c}{4}\right) .
\end{array}
$$
Draw $D E \perp A B$ at point $E$. Then $2 D E=A B$,
which means $2 \times ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,107 |
3. The value of the positive integer $n$ that makes $2^{n}+256$ a perfect square is $\qquad$ | 3. 11.
When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$, if it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a natural number). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, then $k=1$, thus, $n=11$. | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,108 |
4. As shown in Figure 2, given that $A B$ is the diameter of $\odot O$, chord $C D$ intersects $A B$ at point $E$, a tangent line through $A$ intersects the extension of $C D$ at point $F$, and $D$ is the midpoint of $E F$. If $D E=\frac{3}{4} C E, A C=$ $8 \sqrt{5}$, then $A B=$ . $\qquad$ | 4.24.
Let $C E=4 x, A E=y$. Then
$$
D F=D E=3 x, E F=6 x \text {. }
$$
Connect $A D, B C$.
Since $A B$ is the diameter of $\odot O$ and $A F$ is the tangent of $\odot O$, we have
$$
\angle E A F=90^{\circ}, \angle A C D=\angle D A F \text {. }
$$
Since $D$ is the midpoint of the hypotenuse $E F$ of the right triangl... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,109 |
One. (20 points) Given three different real numbers $a, b, c$ satisfying $a-b+c=3$, the equations
$$
x^{2}+a x+1=0 \text { and } x^{2}+b x+c=0
$$
have one common real root, and the equations
$$
x^{2}+x+a=0 \text { and } x^{2}+c x+b=0
$$
also have one common real root. Find the values of $a, b, c$. | I. Sequentially number the four equations given in the problem as (1), (2), (3), and (4).
Let $x_{1}$ be a common real root of equations (1) and (2).
(1) - (2) gives $x_{1}=\frac{c-1}{a-b}$.
Let $x_{2}$ be a common real root of equations (3) and (4).
(3) - (4) gives $x_{2}=\frac{a-b}{c-1}$.
Thus, $x_{1} x_{2}=1$.
Since... | a=-2, b=-3, c=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,110 |
II. (25 points) As shown in Figure 3, in quadrilateral $ABCD$, it is known that $\angle BAD=60^{\circ}$, $\angle ABC=90^{\circ}$, $\angle BCD=120^{\circ}$, diagonal $AC$ intersects with $BD$ at point $S$, and $DS=2SB$, $P$ is the midpoint of $AC$. Prove:
(1) $\angle PBD=30^{\circ}$;
(2) $AD=DC$. | (1) From the given, we have $\angle A D C=90^{\circ}$.
Thus, points $A, B, C, D$ are concyclic, with $A C$ as the diameter, and $P$ as the center of the circle.
As shown in Figure 6, draw
$P M \perp B D$ at point
$M$, and connect $D P$. It is
evident that $M$ is the midpoint of $B D$.
Then $\angle B P M$
$$
=\frac{1}{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,111 |
Example 5 When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}
$$
The value equals $\qquad$ [5]
(2009, Beijing Middle School Mathematics Competition (Grade 8)) | Solution 1
$$
p=\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \text {. }
$$
Then $p^{2}=2 x+2 \sqrt{x^{2}-(2 \sqrt{x-1})^{2}}$
$$
=2 x+2 \sqrt{(x-2)^{2}} \text {. }
$$
Since $1 \leqslant x \leqslant 2$, therefore,
$$
\begin{array}{l}
p^{2}=2 x+2 \sqrt{(x-2)^{2}} \\
=2 x+2(2-x)=4 .
\end{array}
$$
And $p \geqslant 0$, s... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,112 |
Three. (25 points) Given $m, n, p (m<n)$ are positive integers. Let $A(-m, 0), B(n, 0), C(0, p)$, and $O$ be the origin. If $\angle ACB=90^{\circ}$, and
$$
OA^2 + OB^2 + OC^2 = 3(OA + OB + OC),
$$
(1) Prove: $m+n=p+3$;
(2) Find the analytical expression of the quadratic function whose graph passes through points $A, B,... | (1) From $\angle A C B=90^{\circ}, O C \perp A B$, we get
$$
O A \cdot O B=O C^{2} \text {, }
$$
which means $m n=p^{2}$.
$$
\begin{array}{l}
\text { From } O A^{2}+O B^{2}+O C^{2} \\
=3(O A+O B+O C),
\end{array}
$$
we get
$$
\begin{array}{l}
m^{2}+n^{2}+p^{2}=3(m+n+p) . \\
\text { Also, } m^{2}+n^{2}+p^{2} \\
=(m+n+... | y=-\frac{1}{2} x^{2}+\frac{3}{2} x+2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,113 |
II. (25 points) As shown in Figure 4, let $P$ be a point inside the acute triangle $\triangle ABC$. Draw perpendiculars from $P$ to $BC$, $AC$, and $AB$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively. Let $BM$ be the angle bisector of $\angle ABC$, and the extension of $MP$ intersects $AB$ at... | II. As shown in Figure 7, draw $M M_{1} \perp B C$ at point $M_{1}$, $M M_{2} \perp A B$ at point $M_{2}$, $N N_{1} \perp B C$ at point $N_{1}$, and $N N_{2} \perp A C$ at point $N_{2}$.
Let $N P=\lambda N M$.
Since $N N_{1} / / P D / / M M_{1}$, we have
$$
N_{1} D=\lambda N_{1} M_{1} \text {. }
$$
If $N N_{1}M M_{1}$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,114 |
1. Given sets
$$
\begin{array}{l}
A=\left\{(x, y) \mid y^{2}=8 x, x \in \mathbf{R}\right\}, \\
B=\left\{(x, y) \mid y=x^{2}, x \in \mathbf{R}\right\} .
\end{array}
$$
Then $A \cap B=(\quad)$.
(A) $[0,+\infty)$
(B) $\{(0,0),(2,4)\}$
(C) $\{0,2\}$
(D) $\{(0,0),(1,2 \sqrt{2})\}$ | - 1. B.
$A \cap B$ represents the set of points formed by the intersection of the two parabolas. Solving the system of equations, we get
$$
(x, y)=(0,0),(2,4) \text {. }
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,115 |
2. Given the complex numbers $z_{1}=m+2 \mathrm{i}, z_{2}=3-4 \mathrm{i}$. If $\frac{z_{1}}{z_{2}}$ is a real number, then the value of the real number $m$ is ( ).
(A) $\frac{8}{3}$
(B) $\frac{3}{2}$
(C) $-\frac{8}{3}$
(D) $-\frac{3}{2}$ | 2. D.
Notice that
$$
\frac{z_{1}}{z_{2}}=\frac{m+2 \mathrm{i}}{3-4 \mathrm{i}}=\frac{(3 m-8)+(6+4 m) \mathrm{i}}{25} \text {. }
$$
Thus, $6+4 m=0 \Rightarrow m=-\frac{3}{2}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,116 |
3. The number of lines that are tangent to the circle $(x-2)^{2}+y^{2}=1$ and have equal intercepts on both coordinate axes is ( ).
(A) 2
(B) 3
(C) 4
(D) 6 | 3. C.
From the combination of numerical and graphical analysis, we know that the lines tangent to the circle and having equal intercepts on the two coordinate axes are divided into two categories:
(1) There are 2 lines passing through the origin that satisfy the condition;
(2) There are also 2 lines with a slope of -1... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,117 |
7. Given the parabola $C: y^{2}=8 x$ with focus $F$, the directrix intersects the $x$-axis at point $K$. Point $A$ is on the parabola $C$, and $|A K|=\sqrt{2}|A F|$. Then the area of $\triangle A F K$ is ( ).
(A) 4
(B) 8
(C) 16
(D) 32 | 7. B.
Since the focus of the parabola $C: y^{2}=8 x$ is $F(2,0)$, and the directrix is $x=-2$, we have $K(-2,0)$.
Let $A\left(x_{0}, y_{0}\right)$. Draw a perpendicular line $A B$ from point $A$ to the directrix, then $B\left(-2, y_{0}\right)$.
Given that $|A K|=\sqrt{2}|A F|$, and
$$
A F=A B=x_{0}-(-2)=x_{0}+2 \text... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,119 |
8. Let $f(x)$ be a continuous even function, and when $x>0$, $f(x)$ is a strictly monotonic function. Then the sum of all $x$ that satisfy $f(x) = f\left(\frac{x+3}{x+4}\right)$ is ( ).
(A) -3
(B) -8
(C) 3
(D) 8 | 8. B.
When $f(x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $x=\frac{x+3}{x+4}$, we get $x^{2}+3 x-3=0$. At this point,
$$
x_{1}+x_{2}=-3 \text {. }
$$
Since $f(x)$ is a continuous even function, another scenario is $f(-x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $-x=\frac{x+3}{x+4}$, which gives
$$
x^{2}+5 x+3=0 \text {. }
... | -8 | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,120 |
9. If the graph of the function $y=\sqrt{3} \cos x-\sin x$ is shifted to the right by $m(m>0)$ units, and the resulting graph is symmetric about the $y$-axis, then the minimum value of $m$ is ( ).
(A) $\frac{\pi}{3}$
(B) $\frac{2 \pi}{3}$
(C) $\frac{\pi}{6}$
(D) $\frac{5 \pi}{6}$ | 9. C.
Notice that
$$
y=\sqrt{3} \cos x-\sin x=2 \cos \left(x+\frac{\pi}{6}\right),
$$
the equation of the axis of symmetry is $x=k \pi-\frac{\pi}{6}(k \in \mathbf{Z})$. Therefore, $m_{\min }=\frac{\pi}{6}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,121 |
10. A die is thrown twice, and the numbers facing up are $m$ and $n$ respectively. Then the probability that the function $y=\frac{2}{3} m x^{3}-n x+1$ is an increasing function on $[1,+\infty)$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) $\frac{5}{6}$ | 10. D.
Notice that $f^{\prime}(x)=2 m x^{2}-n \geqslant 0$ always holds on $[1,+\infty]$, which means $2 x^{2} \geqslant \frac{n}{m}$ always holds, or equivalently, $2 m \geqslant n$. However, the pairs that satisfy $2 m < n$ are $(1,3)$, $(1,4)$, $(1,5)$, $(1,6)$, $(2,5)$, and $(2,6)$, a total of six. Therefore, $1-\... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,122 |
Example 6 Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ [6] $(2008$, National Junior High School Mathematics Competition) | Solve: Given $a=\frac{\sqrt{5}-1}{2} \Rightarrow 2 a+1=\sqrt{5}$.
Square both sides and simplify to get
$$
a^{2}+a=1 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=\frac{a^{3}\left(a^{2}+a\right)-2 a^{3}-\left(a^{2}+a\right)+2}{a(a+1)(a-1)} \\
=\frac{1-a^{3}}{\left(a^{2}+a\right)(a-1)} \\
=-\... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,123 |
11. Given a point $P$ in space, the distances from $P$ to three mutually perpendicular rays $O A, O B, O C$ are $\sqrt{3}, 2, \sqrt{5}$, and the feet of the perpendiculars are $A^{\prime}, B^{\prime}, C^{\prime}$, respectively. Then the volume of the tetrahedron $P-A^{\prime} B^{\prime} C^{\prime}$ is ( ).
(A) $\sqrt{5... | 11. D.
Let the lengths of $O A^{\prime} 、 O B^{\prime} 、 O C^{\prime}$ be $a 、 b 、 c$. Then
$$
\begin{array}{l}
a^{2}+b^{2}=5, b^{2}+c^{2}=3, c^{2}+a^{2}=4 \\
\Rightarrow a^{2}=3, b^{2}=2, c^{2}=1 .
\end{array}
$$
By volume segmentation, we get $V_{\text {E }}=a b c$, then
$$
V_{P-A^{\prime} B^{\prime} C}=a b c-\frac... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,124 |
12. Given sets
$$
M=\{1,2,3\}, N=\{1,2,3,4\} \text {. }
$$
Define the function $f: M \rightarrow N$, and points $A(1, f(1))$, $B(2, f(2))$, $C(3, f(3))$. If the incenter of $\triangle A B C$ is $D$, and $\overrightarrow{D A}+\overrightarrow{D C}=\lambda \overrightarrow{D B}(\lambda \in \mathbf{R})$, then the number of... | 12. A.
Let $K$ be the midpoint of $A C$.
From $\overrightarrow{D A}+\overrightarrow{D C}=\lambda \overrightarrow{D B}(\lambda \in \mathbf{R})$, we know that $D$, $B$, and $K$ are collinear. This implies $A B=B C$.
Let $f(1)=x, f(2)=y, f(3)=z$.
By combining geometric and numerical analysis, we know $x=z$, and $z \neq y... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,125 |
13. If the constant term in the expansion of $\left(a \sqrt{x}-\frac{1}{\sqrt{x}}\right)^{6}$ is -160, then $\int\left(3 x^{2}-1\right) \mathrm{d} x=$. $\qquad$ | $$
\begin{array}{l}
T_{r+1}=\mathrm{C}_{6}^{r}(a \sqrt{x})^{6-r}\left(-\frac{1}{\sqrt{x}}\right)^{r} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{\frac{6-r}{2}-\frac{r}{2}} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{3-r} .
\end{array}
$$
Let $3-r=0$. Then $r=3$.
The constant term is
$$
\begin{array}{l}
-\mathrm{C}_{6}^{3}... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,126 |
14. Let the function
$$
f(x)=\left\{\begin{array}{ll}
(3-a) x-3, & x \leqslant 7 ; \\
a^{x-6} & x>7,
\end{array}\right.
$$
The sequence $\left\{a_{n}\right\}$ satisfies $a_{n}=f(n)\left(n \in \mathbf{N}_{+}\right)$, and the sequence $\left\{a_{n}\right\}$ is an increasing sequence. Then the range of the real number $a... | 14. $(2,3)$.
Since the sequence $\left\{a_{n}\right\}$ is an increasing sequence, we have,
$$
\begin{array}{l}
\left\{\begin{array}{l}
a-3>0, \\
a>1
\end{array} \Rightarrow 12 .
\end{array}
$$
Therefore, the range of the real number $a$ is $(2,3)$. | (2,3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,127 |
15. Let the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ have its right focus at $F$. Draw a perpendicular from $F$ to one of the asymptotes of the hyperbola $C$, intersecting the hyperbola at point $M$ and the asymptote at point $N$. If $|F N|=a$, and $\overrightarrow{F M}=\lambda \overrightarrow... | 15.1.
Since $|F N|=a=b$, therefore, $c=\sqrt{2} a$.
Let $l_{F N}: y=-x+c$.
Then, solving it together with the hyperbola equation yields $M\left(\frac{3 c}{4}, \frac{c}{4}\right)$. And $N\left(\frac{c}{2}, \frac{c}{2}\right)$, so $\overrightarrow{F M}=\overrightarrow{M N}$.
Therefore, $\lambda=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,128 |
16. Given that a sphere is tangent to the three sides and two bases of a regular triangular prism. If the volume of the sphere is $\frac{32 \pi}{3}$, then the volume of the triangular prism is $\qquad$ | $16.48 \sqrt{3}$.
From the volume, the radius of the sphere $R=2$, the height of the triangular prism is 4, and the side length of the base is $4 \sqrt{3}$.
Thus, $V=\frac{\sqrt{3}}{4}(4 \sqrt{3})^{2} \times 4=48 \sqrt{3}$. | 48 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,129 |
17. (10 points) Let $\triangle A B C$ have $\angle A$, $\angle B$, and $\angle C$ opposite sides $a$, $b$, and $c$ respectively, and $a \cos C+\frac{1}{2} c=b$.
(1) Find the size of $\angle A$;
(2) If $a=1$, find the maximum value of the inradius $R$ of $\triangle A B C$. | (1) By the sine rule, we have
$$
\sin A \cdot \cos C+\frac{1}{2} \sin C=\sin B \text {. }
$$
Since $\sin B=\sin (A+C)$, the original equation becomes
$$
\frac{1}{2} \sin C=\cos A \cdot \sin C \Rightarrow \cos A=\frac{1}{2} \text {. }
$$
Since $\angle A$ is an interior angle of the triangle, $\angle A=\frac{\pi}{3}$.
... | \frac{\sqrt{3}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,130 |
18. (12 points) To investigate the mathematics exam scores of students citywide, 10 students each from Class A and Class B of a certain middle school were randomly selected, and the following scores (in points) were obtained.
$$
\begin{aligned}
\text { Class A: } & 132,108,112,121,113,121,118, \\
& 128,118,129; \\
\tex... | 18. (1) Stem-and-leaf plot omitted, Class B has a higher average level.
(2) The number of excellent students in Class A is 5, and the number of excellent students in Class B is 6.
Let the event that none of the three randomly selected students are excellent be $A_{1}$; the event that one of the three randomly selected... | \frac{33}{20} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,131 |
19. (12 points) Given the sequence $a_{n}$ with the first term $a_{1}=4$, the sum of the first $n$ terms is $S_{n}$, and
$$
S_{n+1}-3 S_{n}-2 n-4=0\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let the function
$$
f(x)=a_{n} x+a_{n-1} x^{2}+... | 19. (1) From $S_{n+1}-3 S_{n}-2 n-4=0$, we get
$$
S_{n}-3 S_{n-1}-2 n+2-4=0 \text {. }
$$
Subtracting the two equations, we get
$$
\begin{array}{l}
a_{n+1}-3 a_{n}-2=0 \\
\Rightarrow a_{n+1}+1=3\left(a_{n}+1\right)(n \geqslant 2) .
\end{array}
$$
Also, from the given, $a_{2}=14$, thus,
$$
a_{2}+1=3\left(a_{1}+1\right... | b_{n}=\frac{5 \times 3^{n+1}-15}{4}-\frac{n(n+6)}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,132 |
20. (12 points) As shown in Figure 2, $\triangle ABC$ is inscribed in $\odot O$, $AB = AC$, line $MN$ is tangent to $\odot O$ at point $C$, chord $BD \parallel MN$, and $AC$ intersects $BD$ at point $E$.
(1) Prove:
$\triangle ABE \cong \triangle ACD$;
(2) If $AB = 6, BC = 4$, find $AE$. | 20. (1) In $\triangle A B E$ and $\triangle A C D$, since
$A B=A C, \angle A B E=\angle A C D$,
$\angle B A E=\angle E D C, B D / / M N$,
therefore, $\angle E D C=\angle D C N$.
From the fact that line $M N$ is a tangent to the circle,
$\angle D C N=\angle C A D$.
Thus, $\angle B A E=\angle C A D$.
Therefore, $\triangl... | \frac{10}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 724,133 |
Example 7 Given $a=\sqrt{5}-1$. Then the value of $2 a^{3}+7 a^{2}-2 a$ -12 is $\qquad$ [7]
(2010) "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | Given $a=\sqrt{5}-1 \Rightarrow a+1=\sqrt{5}$. Squaring both sides, we get
$$
a^{2}+2 a=4 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=2 a\left(a^{2}+2 a\right)+3 a^{2}-2 a-12 \\
=3 a^{2}+6 a-12 \\
=3\left(a^{2}+2 a\right)-12=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,134 |
21. (12 points) Given a fixed point $A(1,0)$ and two moving points $E, F$ on the fixed line $x=-1$, satisfying $\overrightarrow{A E} \perp \overrightarrow{A F}$, a moving point $P$ satisfies $\overrightarrow{E P} / / \overrightarrow{O A}, \overrightarrow{F O} / / \overrightarrow{O P}$ (where $O$ is the origin).
(1) Fin... | 21. (1) Let $P(x, y)$, $E\left(-1, y_{1}\right)$, $F\left(-1, y_{2}\right)$ ($y_{1}$, $y_{2}$ are not 0).
From $\overrightarrow{E P} / / \overrightarrow{O A}$, we get $y_{1}=y$, i.e., $E(-1, y)$.
From $\overrightarrow{F O} / / \overrightarrow{O P}$, we get $y_{2}=-\frac{y}{x}$, i.e.,
$$
F\left(-1,-\frac{y}{x}\right).
$... | -12<k<0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,135 |
22. (12 points) Given the function $f(x)=\frac{\ln x}{x}-1$.
(1) Determine the monotonicity of the function $f(x)$;
(2) Let $m>0$, find the maximum value of $f(x)$ on $[m, 2m]$; 7
(3) Prove that for any $n \in \mathbf{N}_{+}$, the inequality
$$
\ln \left(\frac{1+n}{n}\right)^{c}<\frac{1+n}{n}
$$
holds. | 22. (1) The domain of the function $f(x)$ is $(0,+\infty)$.
From $f^{\prime}(x)=\frac{1-\ln x}{x^{2}}$, let $f^{\prime}(x)=0$, we get $1-\ln x=0 \Rightarrow x=\mathrm{e}$.
When $0<x<\mathrm{e}$, $f^{\prime}(x)=\frac{1-\ln x}{x^{2}}>0$;
When $x>\mathrm{e}$, $f^{\prime}(x)=\frac{1-\ln x}{x^{2}}<0$, so,
$$
\begin{array}{... | proof | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 724,136 |
1. Let the set
$$
M=\{x \mid 0<x \leqslant 3\}, N=\{0<x \leqslant 2\} \text {. }
$$
Then “ $a \in M$ ” is “ $a \in N$ ” ( ).
(A) a sufficient but not necessary condition
(B) a necessary but not sufficient condition
(C) a necessary and sufficient condition
(D) neither a sufficient nor a necessary condition | 1. B.
Since $N \subseteq M$, " $a \in M$ " is a necessary but not sufficient condition for " $a \in N$ ". | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,137 |
2. Let $S_{n}$ be the sum of the first $n$ terms of the arithmetic sequence $\left\{a_{n}\right\}$. If $S_{5}=S_{9}$, then $a_{3}: a_{5}=(\quad)$
(A) $9: 5$
(B) $5: 9$
(C) $3: 5$
(D) $5: 3$ | 2. A.
$$
S_{5}=5 a_{3}, S_{9}=9 a_{5} \text {. }
$$
From $5 a_{3}=9 a_{5}$, we get $a_{3}: a_{5}=9: 5$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,138 |
3. Let $f(x)=\frac{x-1}{x+1}$, and denote $f_{1}(x)=f(x)$. If $f_{n+1}(x)=f\left(f_{n}(x)\right)$, then $f_{2010}(x)=(\quad)$.
(A) $x$
(B) $-\frac{1}{x}$
(C) $\frac{x-1}{x+1}$
(D) $\frac{1+x}{1-x}$ | 3. B.
From the problem, we can predict that $f_{n}(x)$ is periodic. To find the period, using the substitution method makes the calculation straightforward.
Let $a_{1}=f_{1}(2), a_{n}=f_{n}(2)$. Then
$$
\begin{array}{l}
a_{1}=\frac{1}{3}, a_{2}=\frac{\frac{1}{3}-1}{\frac{1}{3}+1}=-\frac{1}{2}, \\
a_{3}=\frac{-\frac{1}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,139 |
4. Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $x^{2}-y^{2}=4$, and $P$ be any point on the hyperbola. A perpendicular line is drawn from $F_{1}$ to the angle bisector of $\angle F_{1} P F_{2}$, with the foot of the perpendicular being $M$. Then the equation of the locus of point $M$ is ( ).
(A) $x^{2}+y^{... | 4. C.
Extend $F_{1} M$ to intersect with $P F_{2}$ (or the extension of $P F_{2}$) at point $N$. Then $\triangle P F_{1} N$ is an isosceles triangle, with $M$ being the midpoint of the base $F_{1} N$. Connect $O M$. Then
$$
O M=\frac{1}{2} F_{2} N=\frac{1}{2}\left|P F_{1}-P F_{2}\right|=2 .
$$
Therefore, the answer i... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,140 |
5. Assign 5 students to three dormitories $A$, $B$, and $C$, with at least 1 student in each dormitory. Among them, student Jia cannot be assigned to dormitory $A$. Then the number of different allocation schemes is ) kinds.
(A) 76
(B) 100
(C) 132
(D) 150 | 5. B.
First, consider the allocation schemes for three dormitory slots. Since $5=3+1+1=2+2+1$, the problem can be divided into the following categories based on the allocation of slots.
(1) $5=3+1+1$.
$A$ dormitory 3 people, $B$ dormitory 1 person, $C$ dormitory 1 person, there are $\mathrm{C}_{4}^{3} \mathrm{C}_{2}^{... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,141 |
6. Let the internal angles $\angle A, \angle B, \angle C$ of $\triangle ABC$ correspond to the sides $a, b, c$ which form a geometric sequence. Then
$$
\frac{\sin A + \cos A \cdot \tan C}{\sin B + \cos B \cdot \tan C}
$$
the range of values is ( ).
(A) $(0,+\infty)$
(B) $\left(\frac{\sqrt{5}-1}{2},+\infty\right)$
(C) ... | 6. D.
Let the common ratio of $a$, $b$, and $c$ be $q$. Then $b=a q$, $c=a q^{2}$.
$$
\begin{array}{l}
\text { and } \frac{\sin A+\cos A \cdot \tan C}{\sin B+\cos B \cdot \tan C} \\
=\frac{\sin A \cdot \cos C+\cos A \cdot \sin C}{\sin B \cdot \cos C+\cos B \cdot \sin C} \\
=\frac{\sin (A+C)}{\sin (B+C)}=\frac{\sin (\p... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,142 |
7. Given that the three interior angles $\angle A, \angle B$, and $\angle C$ of $\triangle A B C$ form an arithmetic sequence, and the sides $A B=1, B C=4$. Then the length of the median $A D$ on side $B C$ is $\qquad$ . | 7. $\sqrt{3}$.
It is known that $\angle B=60^{\circ}$. By the cosine rule, $A D=\sqrt{3}$. | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,143 |
$\begin{aligned} \text { 8. } 1+\frac{1}{1+2}+\cdots+\frac{1}{1+2+\cdots+2010} \\ =\end{aligned}$ | 8. $\frac{4020}{2011}$.
$$
\begin{array}{l}
\text { Let } a_{n}=\frac{1}{1+2+\cdots+n}=\frac{2}{n(n+1)} \\
=2\left(\frac{1}{n}-\frac{1}{n+1}\right) .
\end{array}
$$
Then $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$
$$
\begin{array}{l}
=2\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}... | \frac{4020}{2011} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,144 |
Example 8 Given $a>0, b>0$, and $\sqrt{a}(\sqrt{a}+4 \sqrt{b})=3 \sqrt{b}(\sqrt{a}+2 \sqrt{b})$. Then the value of $\frac{a+6 \sqrt{a b}-8 b}{2 a-3 \sqrt{a b}+2 b}$ is ( ).
(A) 1
(B) 2
(C) $\frac{19}{11}$
(D) $\sqrt{2}$
(2007, Shandong Province Junior High School Mathematics Competition) | From the given, we easily obtain
$$
a+\sqrt{a b}-6 b=0 \text {. }
$$
Thus, $(\sqrt{a}-2 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0$.
Since $a>0, b>0$, then $\sqrt{a}+3 \sqrt{b} \neq 0$.
Therefore, $\sqrt{a}-2 \sqrt{b}=0 \Rightarrow a=4 b$. Hence, $\frac{a+6 \sqrt{a b}-8 b}{2 a-3 \sqrt{a b}+2 b}=2$.
So, the answer is (B). | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,145 |
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