problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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Question 3 Given $a b c=-1, a^{2} b+b^{2} c+c^{2} a=t$, $\frac{a^{2}}{c}+\frac{b}{c^{2}}=1$. Try to find the value of $a b^{5}+b c^{5}+c a^{5}$. | 【Analysis】Using $a b c=-1$ can make common substitutions
$$
a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x} \text {. }
$$
This problem involves three letters, and can also be solved by the method of elimination.
Solution 1 Let $a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x}$.
$$
\begin{array}{l}
\text { Then } \frac{a^... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,106 |
Question 4 In a plane, there are parallel lines spaced $d$ apart. Prove: the probability that an arbitrarily placed needle $l$ intersects a line is
$$
P=\frac{2 l}{\pi d} .
$$ | 【Analysis】This is the famous Buffon's Needle Problem. The key is how to convert it into a geometric probability model.
Proof As shown in Figure 1, let the midpoint of the needle be $M$, and use $x$ to denote the distance from point $M$ to the nearest parallel line, and use $\theta$ to denote the angle from the paralle... | \frac{2 l}{\pi d} | Geometry | proof | Yes | Yes | cn_contest | false | 726,107 |
Question 1 As shown in Figure 1, let $A D$ be the altitude of $\triangle A B C$, and the semicircle with $B C$ as its diameter and on the same side as point $A$ intersects $A B$, $A C$, and $A D$ at points $F$, $E$, and $X$, respectively. The circumcircle of $\triangle D E X$ intersects $B C$ at point $L$ (not coincidi... | Proof As shown in Figure 1, take the midpoint $O$ of $BC$. Clearly, $O$ is the center of the semicircle.
Connect $XN$, $EL$, and $CF$.
It is easy to see that points $A$, $F$, $D$, and $C$ are concyclic, and points $D$, $F$, $X$, and $N$ are concyclic.
$$
\begin{array}{l}
\text { Then } \angle FXN = \angle BDF = \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,109 |
Let $I_{a}$ be the excenter of $\triangle ABC$ opposite to $\angle A$, and let $P, Q$ be the points of tangency of the excircle $\odot I_{u}$ with the lines $AB, AC$, respectively. Let $PQ$ intersect $I_{u}B, I_{u}C$ at points $D, E$, and let $DC$ intersect $BE$ at point $A_{1}$. Similarly define $B_{1}, C_{1}$. Prove ... | Prove that, as shown in Figure 2, taking two excenters $I_{b}$ and $I_{c}$, it is easy to see that $I_{b}$, $A$, $I_{c}$, $I_{c}$, $B$, $I_{a}$, $I_{a}$, $C$, $I_{b}$ are collinear, respectively.
Let the internal angles of $\triangle ABC$ be $\angle A$, $\angle B$, $\angle C$. Connect $I_{a}P$ and $I_{u}Q$.
It is easy... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,110 |
Example 2 (Bachelor Problem) For a positive integer $n$, $[x]$ represents the greatest integer not exceeding the number $x$. Prove:
$$
\sum_{k=0}^{n} 2^{k} C_{n}^{k} C_{n-k}^{\left[\frac{n-k}{2}\right]}=C_{2 n+1}^{n} .
$$ | Proof using the combinatorial model construction method.
Assume a community has $n+1$ households, including $n$ married couples and one single person, totaling $2n+1$ people. Now, $n$ people are selected to participate in an activity, which can be done in $\mathrm{C}_{2 n+1}^{n}$ ways.
On the other hand, this event c... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,111 |
Example 1 Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than 1. Then
$\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=$ $\qquad$ [2]
(2012, Zhejiang Province High School Mathematics Competition) | When the common ratio $q=1$,
$$
\begin{array}{l}
a_{n}=a_{1}, \\
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=2011 .
\end{array}
$$
When the common ratio $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1}... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,112 |
Example 2 A function $f(x)$ defined on $(-1,1)$ satisfies
(1) For any $x, y \in(-1,1)$, we have
$$
f(x)+f(y)=f\left(\frac{x+y}{1+x y}\right) \text {; }
$$
(2) When $x \in(-1,0)$, $f(x)>0$. | Proof: Let $x=y=0$, we get $f(0)=0$.
Let $y=-x$, then
$$
f(x)+f(-x)=f(0)=0.
$$
Therefore, $f(x)$ is an odd function.
$$
\begin{array}{l}
\text { Also, } f\left(\frac{1}{n^{2}+5 n+5}\right) \\
=f\left(\frac{1}{(n+2)(n+3)-1}\right) \\
=f\left(\frac{\frac{1}{n+2}+\left(-\frac{1}{n+3}\right)}{1+\frac{1}{n+2}\left(-\frac{1... | f\left(\frac{1}{3}\right) | Algebra | proof | Yes | Yes | cn_contest | false | 726,113 |
Example 3 Given
$$
a_{k}=\frac{k+2}{k!+(k+1)!+(k+2)!} \text {. }
$$
Then the sum of the first 100 terms of the sequence $\left\{a_{n}\right\}$ is $\qquad$
(2006, Shanghai Jiao Tong University Independent Admission Examination) | Notice that,
$$
\begin{aligned}
a_{k} & =\frac{k+2}{k!+(k+1)!+(k+2)!} \\
& =\frac{k+2}{k![1+(k+1)+(k+2)(k+1)]} . \\
& =\frac{k+2}{k!(k+2)^{2}}=\frac{1}{k!(k+2)} \\
& =\frac{k+1}{(k+2)!}=\frac{1}{(k+1)!}-\frac{1}{(k+2)!} .
\end{aligned}
$$
Therefore, $\sum_{k=1}^{100} a_{k}=\frac{1}{2}-\frac{1}{102!}$.
[Note] Related t... | \frac{1}{2}-\frac{1}{102!} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,114 |
Example 4 Let $p, q$ be the roots of the quadratic equation
$$
x^{2}+2 a x-1=0(a>0)
$$
where $p>0$. Let
$$
y_{1}=p-q, y_{n+1}=y_{n}^{2}-2(n=1,2, \cdots) .
$$ | Prove: $\lim _{n \rightarrow \infty}\left(\frac{1}{y_{1}}+\frac{1}{y_{1} y_{2}}+\cdots+\frac{1}{y_{1} y_{2} \cdots y_{n}}\right)=p$.
(2010, Joint Autonomous Admissions Examination of Tsinghua University and Other Schools)
By Vieta's formulas, we know
$$
\begin{array}{l}
p q=-1, y_{1}=p-q=p+\frac{1}{p}, \\
y_{2}=y_{1}^{... | p | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,115 |
Example 6 Given $a>0, b>0$. Prove:
$$
\sum_{k=1}^{n} \frac{1}{a+k b}<\frac{n}{\sqrt{\left(a+\frac{1}{2} b\right)\left(a+\frac{n+1}{2} b\right)}} \text {. }
$$
(2007, Peking University Independent Admission Examination) | $$
\begin{array}{l}
\sum_{k=1}^{n} \frac{1}{a+k b} \leqslant \sqrt{n \sum_{k=1}^{n} \frac{1}{(a+k b)^{2}}} \\
<\sqrt{n \sum_{k=1}^{n} \frac{1}{\left[a+\left(k-\frac{1}{2}\right) b\right]\left[a+\left(k+\frac{1}{2}\right) b\right]}} \\
=\sqrt{\frac{n}{b} \sum_{k=1}^{n}\left[\frac{1}{a+\left(k-\frac{1}{2}\right) b}-\frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,116 |
Example 6 Let $n$ be a positive integer, and $a_{i}(i=1,2, \cdots, n)$ be non-negative real numbers. Prove:
$$
\begin{array}{l}
\frac{1}{1+a_{1}}+\frac{a_{1}}{\left(1+a_{1}\right)\left(1+a_{2}\right)}+\cdots+ \\
\frac{a_{1} a_{2} \cdots a_{n-1}}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)} \leqs... | Prove that,
$$
\begin{array}{l}
\frac{1}{1+a_{1}}=1-\frac{a_{1}}{1+a_{1}}, \\
\frac{\prod_{i=1}^{k} a_{k}}{\prod_{j=1}^{k+1}\left(1+a_{j}\right)}=\prod_{j=1}^{k} \frac{a_{j}}{1+a_{j}}-\prod_{j=1}^{k+1} \frac{a_{j}}{1+a_{j}} .
\end{array}
$$
By adding the above equations, we get
$$
\begin{array}{l}
\frac{1}{1+a_{1}}+\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,117 |
1. Given $a_{n}=\frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}$. Then $a_{1}+a_{2}+\cdots+a_{99}=$ $\qquad$
$(2008$, Shanghai Jiao Tong University Winter Camp) | $$
\begin{array}{l}
\text { Hint: } a_{n}=\frac{1}{\sqrt{n} \sqrt{n+1}(\sqrt{n}+\sqrt{n+1})} \\
=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n} \sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}} .
\end{array}
$$
Answer: $\frac{9}{10}$. | \frac{9}{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,118 |
$$
2.1 \times 1!+2 \times 2!+\cdots+n \times n!=
$$
$\qquad$ (2007, Shanghai Jiao Tong University Independent Admission Examination) | Hint: $n \times n!=(n+1)!-n!$ Answer: $(n+1)!-1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | (n+1)!-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,119 |
3. Prove: $1+\frac{1}{\sqrt{2^{3}}}+\frac{1}{\sqrt{3^{3}}}+\cdots+\frac{1}{\sqrt{n^{3}}}<3$.
(2004, Fudan University Admission Test for Exceptional Students; 2011, Hebei Province High School Mathematics Competition) | $$
\begin{array}{l}
\frac{1}{\sqrt{k^{3}}}<\frac{(\sqrt{k}+\sqrt{k-1})}{\sqrt{k}} \cdot \frac{(\sqrt{k}-\sqrt{k-1})}{\sqrt{k} \sqrt{k-1}} \\
<2 \times \frac{\sqrt{k}-\sqrt{k-1}}{\sqrt{k} \sqrt{k-1}}=2\left(\frac{1}{\sqrt{k-1}}-\frac{1}{\sqrt{k}}\right) .
\end{array}
$$
Note that, | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,120 |
4. If $a_{1}, a_{2}, \cdots, a_{n}$ are distinct positive integers, when $\alpha \geqslant 2$, prove:
$$
\left(\frac{1}{a_{1}}\right)^{\alpha}+\left(\frac{1}{a_{2}}\right)^{\alpha}+\cdots+\left(\frac{1}{a_{n}}\right)^{\alpha}<2 .
$$
(2003, Fudan University Admission Examination for Recommended Students) | Suppose $\mathrm{I} \leqslant a_{1}<a_{2}<\cdots<a_{n}$. Then $0<\frac{1}{a_{i}} \leqslant 1 \Rightarrow\left(\frac{1}{a_{i}}\right)^{\alpha} \leqslant\left(\frac{1}{a_{i}}\right)^{2}$.
Thus, the left side of equation (1)
$$
\begin{array}{l}
\leqslant\left(\frac{1}{a_{1}}\right)^{2}+\left(\frac{1}{a_{2}}\right)^{2}+\cd... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,121 |
Example 3 The Euler's totient function $\varphi(n)$ is a function defined on the set of positive integers. For each positive integer $n$, the value of $\varphi(n)$ is the number of integers in the set $\{1,2, \cdots, n\}$ that are coprime to $n$. Prove: If positive integers $m, n$ are coprime, then
$$
\varphi(m n)=\var... | Prove by using the counting model construction method. Construct a finite set $A$ and $R$, and establish a correspondence between them. For this, consider the set of $m n$ numbers:
$$
\begin{array}{l}
\Lambda=\left\{a_{i j} \mid a_{i j}=i n+j m, i=1,2, \cdots, m ; j=1,2, \cdots, n\right\}, \\
R=\left\{r_{i j} \mid i=1,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,122 |
1. As shown in Figure 1, the side length of the regular hexagon $A_{1} B_{1} C_{1} D_{1} E_{1} F_{1}$ is 1, and its six diagonals form another regular hexagon $A_{2} B_{2} C_{2} D_{2} E_{2} F_{2}$. This process continues. Then the sum of the areas of all these hexagons is $\qquad$ | -1. $\frac{9 \sqrt{3}}{4}$.
Let the area of the $n$-th regular hexagon be $a_{n}$. Then $a_{1}=6 \times \frac{\sqrt{3}}{4} \times 1^{2}=\frac{3 \sqrt{3}}{2}$.
It is easy to see that $A_{2} B_{2}=A_{2} B_{1}=\frac{\sqrt{3}}{3}, \frac{a_{2}}{a_{1}}=\frac{1}{3}$.
In general, $\frac{a_{n+1}}{a_{n}}=\frac{1}{3}$.
Therefore... | \frac{9 \sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,123 |
2. Given positive integers $a_{1}, a_{2}, \cdots, a_{10}$ satisfy
$$
\frac{a_{j}}{a_{i}}>\frac{2}{3}(1 \leqslant i \leqslant j \leqslant 10) \text {. }
$$
Then the minimum possible value of $a_{10}$ is $\qquad$ . | 2. 92.
From $a_{1} \geqslant 1, a_{2}>\frac{3}{2} a_{1} \geqslant \frac{3}{2}\left(a_{2} \in \mathbf{N}_{+}\right)$, we get $a_{2} \geqslant 2$.
Similarly, $a_{3}>\frac{3}{2} a_{2} \geqslant 3, a_{3} \geqslant 4$;
$$
\begin{array}{l}
a_{4}>\frac{3}{2} a_{3} \geqslant 6, a_{4} \geqslant 7 ; \\
a_{5}>\frac{3}{2} a_{4} \... | 92 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,124 |
3. Given $\tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$, $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}$, $\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha=-\frac{17}{5}$. Then $\tan (\alpha+\beta+\gamma)=$ $\qquad$ | 3. 11 .
Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then the given equations can be written as
$$
\begin{array}{l}
x+y+z=\frac{17}{6}, \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}, \\
\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5} . \\
\text { From (1) } \div \text { (3) we get } \\
x y z=-\frac{... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,125 |
4. Given the equation about $x$
$$
\lg k x=2 \lg (x+1)
$$
has only one real solution. Then the range of the real number $k$ is | 4. $(-\infty, 0) \cup\{4\}$.
Notice that,
$$
\lg k x=2 \lg (x+1) \Leftrightarrow\left\{\begin{array}{l}
x>-1, \\
k x=(x+1)^{2} .
\end{array}\right.
$$
From the graph, we know that when $k-1)$ there is only one common point; when $k>0$, the line $y=k x$ and the curve $y=(x+1)^{2}$ $(x>-1)$ have only one common point i... | (-\infty, 0) \cup\{4\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,126 |
5. As shown in Figure 2, $\triangle A E F$ is an inscribed triangle in the square $A B C D$ with side length $x$. If $\angle A E F=90^{\circ}$, $A E=a, E F=b(a>b)$, then $x=$ | 5. $\frac{a^{2}}{\sqrt{a^{2}+(a-b)^{2}}}$.
Let $\angle A E B=\theta$ ( $\theta$ is an acute angle). Then $a \cos \theta+b \sin \theta=x=a \sin \theta$.
Thus $(a-b) \sin \theta=a \cos \theta$
$\Rightarrow \tan \theta=\frac{a}{a-b}$
$\Rightarrow \sin \theta=\frac{a}{\sqrt{a^{2}+(a-b)^{2}}}$.
Therefore, $x=a \sin \theta=... | \frac{a^{2}}{\sqrt{a^{2}+(a-b)^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,127 |
6. Equation
$$
2^{m} \times 3^{n}-3^{n+1}+2^{m}=13
$$
The non-negative integer solution $(m, n)=$ $\qquad$ | 6. $(3,0),(2,2)$.
The original equation can be transformed into
$$
\begin{array}{l}
\left(2^{m}-3\right)\left(3^{n}+1\right)=10 \\
\Rightarrow\left\{\begin{array}{l}
3^{n}+1=2,5,10 ; \\
2^{m}-3=5,2,1 .
\end{array}\right. \\
\Rightarrow(m, n)=(3,0),(2,2) .
\end{array}
$$ | (3,0),(2,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,128 |
7. A pocket contains five equally sized small balls, of which two are red, two are white, and one is black. The probability that the colors of two adjacent balls are different when the five balls are drawn in sequence is $\qquad$ (answer with a number).
| 7. $\frac{2}{5}$.
Consider the five balls as distinct. There are $5!=120$ possible ways to draw the five balls one by one from the bag.
There are $2!\times 4!$ possible ways for the two white balls to be adjacent; there are $2!\times 4!$ possible ways for the two red balls to be adjacent; there are $2!\times 2!\times... | \frac{2}{5} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,129 |
8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, \\
a_{n+2}=\frac{2(n+1)}{n+2} a_{n+1}-\frac{n}{n+2} a_{n}(n=1,2, \cdots) .
\end{array}
$$
If $a_{m}>2+\frac{2011}{2012}$, then the smallest positive integer $m$ is . $\qquad$ | 8.4025.
$$
\begin{array}{l}
\text { Given } a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
\begin{array}{l}
\Rightarrow a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n-1}-a_{n-2}\right) \\
=\frac{n-2}{n} \cdot \frac{n-3}{n-1}\left(a_{n-2}-a_{n-3}\right)=\cdots \\
= \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \cdots \fr... | 4025 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,130 |
9. (14 points) In $\square A B C D$, it is known that $A B=x$, $B C=1$, the diagonals $A C$ and $B D$ intersect at point $O$, and $\angle B O C = 45^{\circ}$. Let the distance between the lines $A B$ and $C D$ be $h(x)$. Find the expression for $h(x)$ and state the range of $x$.
In $\square A B C D$, it is known that ... | ```
9. From the property that the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of its four sides, we have
$$
\begin{array}{l}
O B^{2}+O C^{2}=\frac{1}{2}\left(A B^{2}+B C^{2}\right) \\
=\frac{1}{2}\left(x^{2}+1\right) .
\end{array}
$$
In $\triangle O B C$, by the cosine rule, we... | h(x) = \frac{x^2 - 1}{2x}, \quad 1 < x \leq \sqrt{2} + 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,131 |
10. (14 points) Given a real number $a>1$. Find the minimum value of the function
$$
f(x)=\frac{(a+\sin x)(4+\sin x)}{1+\sin x}
$$ | 10. Note that,
$$
\begin{array}{l}
f(x)=\frac{(a+\sin x)(4+\sin x)}{1+\sin x} \\
=1+\sin x+\frac{3(a-1)}{1+\sin x}+a+2 .
\end{array}
$$
When $0 < t \leq 2$, the function
$$
y=t+\frac{3(a-1)}{t}
$$
is decreasing in $(0,2]$, thus,
$$
\begin{array}{l}
f(x)_{\text {min }}=f(1)=2+\frac{3(a-1)}{2}+a+2 \\
=\frac{5(a+1)}{2} ... | \frac{5(a+1)}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,132 |
11. (16 points) Given positive real numbers $x, y, z$ satisfy
$$
9xyz + xy + yz + zx = 4 \text{.}
$$
Prove: (1) $xy + yz + zx \geqslant \frac{4}{3}$;
$$
\text{(2) } x + y + z \geqslant 2 \text{.}
$$ | 11. (1) Let $t=\sqrt{\frac{x y+y z+z x}{3}}$.
By the AM-GM inequality, we have
$$
\begin{array}{l}
x y z=[\sqrt[3]{(x y)(y z)(z x)}]^{\frac{3}{2}} \\
\leqslant\left(\frac{x y+y z+z x}{3}\right)^{\frac{3}{2}} .
\end{array}
$$
Thus, $4=9 x y z+x y+y z+z x \leqslant 9 t^{3}+3 t^{2}$
$$
\Rightarrow(3 t-2)\left(3 t^{2}+3 ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,134 |
12. (16 points) Given an integer $n(n \geqslant 3)$, let $f(n)$ be the minimum number of elements in a subset $A$ of the set $\left\{1,2, \cdots, 2^{n}-1\right\}$ that satisfies the following two conditions:
(i) $1 \in A, 2^{n}-1 \in A$;
(ii) Each element in subset $A$ (except 1) is the sum of two (possibly the same) e... | 12. (1) Let set $A \subseteq\left\{1,2, \cdots, 2^{3}-1\right\}$, and $A$ satisfies (i) and (ii). Then $1 \in A, 7 \in A$.
Since $\{1, m, 7\}(m=2,3, \cdots, 6)$ does not satisfy (ii), hence $|A|>3$.
Also, $\{1,2,3,7\},\{1,2,4,7\},\{1,2,5,7\}$,
$\{1,2,6,7\},\{1,3,4,7\},\{1,3,5,7\}$,
$\{1,3,6,7\},\{1,4,5,7\},\{1,4,6,7\}... | 108 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,135 |
1. The area of the inscribed square in the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ is | $-1 . \frac{450}{17}$.
By the symmetry of the ellipse, we know that the sides of the inscribed square should be parallel to the coordinate axes, and the center is at the origin. Therefore, the equations of the diagonals are $y= \pm x$.
Substituting into the ellipse equation, we get
$$
x^{2}\left(\frac{1}{5^{2}}+\frac{1... | \frac{450}{17} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,136 |
2. Let two acute angles $\alpha, \beta$ satisfy
$$
\begin{array}{l}
(\sin \alpha+\cos \alpha)(\sin \beta+\cos \beta)=2 \text {. } \\
\text { Then }(\sin 2 \alpha+\cos 3 \beta)^{2}+(\sin 2 \beta+\cos 3 \alpha)^{2} \\
=
\end{array}
$$ | $2.3-2 \sqrt{2}$.
From the given, we know
$$
\begin{array}{l}
\sin (\alpha+\beta)+\cos (\alpha-\beta)=2 \\
\Rightarrow \sin (\alpha+\beta)=\cos (\alpha-\beta)=1 \\
\Rightarrow \alpha+\beta=90^{\circ}, \alpha=\beta \\
\Rightarrow \alpha=\beta=45^{\circ} .
\end{array}
$$
Then $(\sin 2 \alpha+\cos 3 \beta)^{2}+(\sin 2 \b... | 3-2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,137 |
3. In a regular tetrahedron $D-ABC$, it is known that the side lengths of the base $\triangle ABC$ are all 6, and the lengths of each lateral edge are all 5. $I$ is the incenter of the lateral face $\triangle DAB$. Then the volume of the tetrahedron $IABC$ is $\qquad$ | 3. $\frac{9 \sqrt{39}}{8}$.
Take the midpoint $M$ of $A B$.
Since $D A=D B$, then $D M \perp A B$, and point $I$ lies on side $D M$.
Thus $\frac{I M}{I D}=\frac{A M}{A D}=\frac{3}{5} \Rightarrow \frac{I M}{D M}=\frac{3}{8}$.
Let the projection of point $D$ on the base $\triangle A B C$ be $H$. Then
$$
\begin{array}{l}... | \frac{9 \sqrt{39}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,138 |
4. Given real numbers $a, b, c$ satisfy
$$
a+b+c=a^{2}+b^{2}+c^{2} \text {. }
$$
Then the maximum value of $a+b+c$ is $\qquad$ | 4.3.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
3(a+b+c) \\
=\left(1^{2}+1^{2}+1^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)
\end{array}
$$
$$
\begin{array}{l}
\geqslant(a+b+c)^{2} \\
\Rightarrow a+b+c \leqslant 3 .
\end{array}
$$
Equality holds if and only if \(a=b=c=1\). | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,139 |
5. The maximum value of the function $y=x\left(1+\sqrt{1-x^{2}}\right)$ is
保留了源文本的换行和格式。 | 5. $\frac{3 \sqrt{3}}{4}$.
From the problem, we know $|x| \leqslant 1$.
Since we are looking for the maximum value, we can assume $0 < x \leqslant 1$.
Let $x=\sin \alpha\left(\alpha \in\left(0, \frac{\pi}{2}\right]\right)$. Then
$$
\begin{aligned}
y= & \sin \alpha(1+\cos \alpha) \\
\Rightarrow & y^{\prime}=\cos \alpha... | \frac{3 \sqrt{3}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,140 |
6. Given three distinct positive integers $a, b, c$ form a geometric sequence, and their sum is 111. Then $\{a, b, c\}=$ $\qquad$ | 6. $\{1,10,100\},\{27,36,48\}$.
Let $a<b<c$ and the common ratio be $q$. Then $b=a q, c=a q^{2} \Rightarrow q=\frac{b}{a}$ is a rational number.
Express $q$ as a reduced fraction, denoted as $q=\frac{n}{m}$, where $m<n$, and $(m, n)=1$.
Since $c=a \cdot \frac{n^{2}}{m^{2}}$ is a positive integer $\Rightarrow m^{2} \mi... | \{1,10,100\},\{27,36,48\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,141 |
7. Arrange the positive integers whose sum of digits is 5 in ascending order to form a sequence. Then 2012 is the $\qquad$th term of this sequence. | 7.38.
To represent 5 as the sum of no more than four positive integers, there are six methods, that is
$$
\begin{array}{l}
5=1+4=2+3=1+1+3 \\
=1+2+2=1+1+1+2 .
\end{array}
$$
When filling them into a $1 \times 4$ grid, positions that are not filled are supplemented with 0. Then $\{5\}$ has 3 ways of filling; $\{1,4\}$... | 38 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,142 |
8. Given that $18^{2}=324, 24^{2}=576$, they are formed by the permutation of two consecutive digits $2,3,4$ and $5,6,7$ respectively; and $66^{2}=4356$ is formed by the permutation of four consecutive digits $3, 4, 5, 6$. Then the next such square number is $\qquad$
| 8.5476.
For any square number, its last digit can only be $0, 1, 4, 5, 6, 9$, and
$$
\begin{array}{l}
(10 a)^{2}=100 a^{2}, \\
(10 a+5)^{2}=100 a^{2}+100 a+25, \\
(10 a \pm 4)^{2}=100 a^{2} \pm 80 a+16 ; \\
(10 a \pm 1)^{2}=100 a^{2} \pm 20 a+1, \\
(10 a \pm 3)^{2}=100 a^{2} \pm 60 a+9 ; \\
(10 a \pm 2)^{2}=100 a^{2} ... | 5476 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,143 |
Example 1 As shown in Figure 1, given that the area of $\triangle ABC$ is 1, points $D$, $E$, and $F$ are on sides $BC$, $CA$, and $AB$ respectively, with $BD=2DC$, $CE=2EA$, and $AF=2FB$. $AD$, $BE$, and $CF$ intersect pairwise at points $P$, $Q$, and $R$. Find the area of $\triangle PQR$.
(2009, University of Science... | Solve: By applying Ceva's Theorem to $\triangle A D C$ cut by line $B P E$, we get
$$
\begin{array}{l}
\frac{A P}{P D} \cdot \frac{D B}{B C} \cdot \frac{C E}{E A}=1 \\
\Rightarrow \frac{A P}{P D} \cdot \frac{2}{3} \cdot \frac{2}{1}=1 \Rightarrow \frac{A P}{P D}=\frac{3}{4} \\
\Rightarrow S_{\triangle N A B}=\frac{3}{7}... | \frac{1}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,144 |
9. (18 points) As shown in Figure 1, given the parabola $y=x^{2}$ with vertex $O$, $AB$ is a chord of length 2 passing through the focus $F$, and $D$ is the intersection of the perpendicular bisector of $AB$ with the $y$-axis. Find the area of quadrilateral $AOBD$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果... | 9. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, with the focus $F\left(0, \frac{1}{4}\right)$, and the directrix equation as $y=-\frac{1}{4}$.
By the definition of a parabola, we have
$$
\begin{array}{l}
2=A F+B F=y_{1}+y_{2}+\frac{1}{2} \\
\Rightarrow y_{1}+y_{2}=\frac{3}{2} .
\end{array}
$$
Let the ... | \frac{5 \sqrt{2}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,145 |
10. (18 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=1$, for each $n \in \mathbf{N}, a_{4 n+1}, a_{4 n+2}, a_{4 n+3}$ form an arithmetic sequence with a common difference of 2, and $a_{4 n+3}, a_{4 n+4}, a_{4 n+5}$ form a geometric sequence with a common ratio of $\frac{1}{2}$. Prove that $... | 10. Clearly, all terms of the sequence are positive.
To explore the structure of the sequence, we can list the initial terms:
$$
1,3,5, \frac{5}{2}, \frac{5}{4}, \frac{13}{4}, \frac{21}{4}, \frac{21}{8}, \frac{21}{16}, \frac{53}{16}, \frac{85}{16}, \frac{85}{32},
$$
$\frac{85}{64}, \frac{213}{64}, \frac{341}{64}$.
For... | \frac{16}{3} | Algebra | proof | Yes | Yes | cn_contest | false | 726,146 |
11. (25 points) Prove: For each positive integer $n$, there exists a positive integer $p(n)$, such that the sequence of the first $p(n)$ positive integers $1,2, \cdots, p(n)$ can be divided into $n$ segments in such a way that the sum of the numbers in each segment is a perfect square. | 11. Notice that 1 is a square number, take
$$
p(1)=1 \text {. }
$$
Also, since $2+3+4=9=3^{2}$, the second segment can take three numbers, i.e., take $p(2)=1+3=4$;
Furthermore, since $5+6+\cdots+13=81=3^{4}$, the third segment can take nine numbers, i.e., take
$$
p(3)=1+3+3^{2}=13 \text {; }
$$
$\qquad$
In general, c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,147 |
12. (25 points) As shown in Figure 2, in the acute triangle $\triangle ABC$, it is known that $T$ is any point on the altitude $AD$, $BT$ intersects $AC$ at point $E$, $CT$ intersects $AB$ at point $F$, $EF$ intersects $AD$ at point $G$, and a line $l$ through $G$ intersects $AB$, $AC$, $BT$, and $CT$ at points $M$, $N... | 12. As shown in Figure 3.
By applying Ceva's Theorem to $\triangle ABE$ cut by $CF$, we get $\frac{AC}{CE} \cdot \frac{ET}{TB} \cdot \frac{BF}{FA}=1$.
By applying Ceva's Theorem to $\triangle ATE$ cut by $BC$, we get $\frac{AD}{DT} \cdot \frac{TB}{BE} \cdot \frac{EC}{CA}=1 \Rightarrow \frac{AD}{DT}=\frac{BE \cdot AC}{B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,148 |
1. Given $4^{a}-3 a^{b}=16, \log _{2} a=\frac{a+1}{b}$. Then $a^{b}=$ $\qquad$ . | - 1. 16.
From $\log _{2} a=\frac{a+1}{b} \Rightarrow a^{b}=2^{a+1}$.
Substituting into $4^{a}-3 a^{b}=16$, we get
$2^{2 a}-6 \times 2^{a}-16=0$.
Solving, we get $2^{a}=8$ or -2 (discard).
Therefore, $a^{b}=2^{a+1}=16$. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,149 |
3. Given that the base edge length of a regular tetrahedron is 6, and the side edge is 4. Then the radius of the circumscribed sphere of this regular tetrahedron is $\qquad$ | 3. 4 .
From the problem, we know that the distance from point $A$ to the base $B C D$ is 2, and the radius of its circumscribed sphere is $R$. Then
$$
R^{2}-12=(R-2)^{2} \Rightarrow R=4 \text {. }
$$ | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,151 |
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $S_{4} \leqslant 4, S_{5} \geqslant 15$, then the minimum value of $a_{4}$ is $\qquad$ . | 4. 7 .
Let the common difference be $d$. From the given conditions, we have
$$
\begin{array}{l}
2 a_{4} \leqslant 2+3 d, d \leqslant a_{4}-4 \\
\Rightarrow 2 a_{4} \leqslant 2+3 d \leqslant 2+3\left(a_{4}-3\right) \\
\Rightarrow a_{4} \geqslant 7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,152 |
5. Let $n \in \mathbf{N}_{+}$, and
$$
n^{4}+2 n^{3}+5 n^{2}+12 n+5
$$
be a perfect square. Then $n=$ $\qquad$ | 5.1 or 2 .
$$
\begin{array}{l}
\text { Given }\left(n^{2}+n+2\right)^{2} \\
<n^{4}+2 n^{3}+5 n^{2}+12 n+5 \\
<\left(n^{2}+n+4\right)^{2} \\
\Rightarrow n^{4}+2 n^{3}+5 n^{2}+12 n+5=\left(n^{2}+n+3\right)^{2} \\
\Rightarrow n=1 \text { or } 2 \text {. }
\end{array}
$$ | 1 \text { or } 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,153 |
6. If the inequality
$$
a \sin ^{2} x+\cos x \geqslant a^{2}-1
$$
holds for any $x \in \mathbf{R}$, then the range of real number $a$ is $\qquad$ | 6. $a=0$.
Let $\cos x=-1, \sin x=0$.
From the given condition, we have $a^{2} \leqslant 0 \Rightarrow a=0$.
When $a=0$, the original equation is obviously always true. | a=0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,154 |
Example 2 Given that $O$ is a point inside $\triangle ABC$ and satisfies $\angle BAO = \angle CAO = \angle CBO = \angle ACO$. Prove: The side lengths of $\triangle ABC$ form an arithmetic sequence.
(2011, Peking University Admission Examination for Recommended Students) | 【Analysis】From the conditions, we first consider applying the trigonometric form of Ceva's Theorem, and then combining it with analytical methods, we can find a way to solve the problem.
$$
\begin{array}{l}
\Leftrightarrow \sin ^{2} 2 \alpha=\sin (\alpha+\beta) \cdot \sin (\alpha+\gamma) \\
\Leftrightarrow 1-\cos 4 \al... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,155 |
7. In $\triangle A B C$, it is known that $A B=10$, and the height from $A B$ is 3. If $A C \cdot B C$ is minimized, then $A C+B C$ $=$ . $\qquad$ | 7. $4 \sqrt{10}$.
From the formula for the area of a triangle, we have
$$
\frac{1}{2} A C \cdot B C \sin C=\frac{1}{2} \times 3 \times 10=15 \text {. }
$$
Therefore, $A C \cdot B C \geqslant 30$.
The equality holds if and only if $\angle C=90^{\circ}$.
Then, $A C^{2}+B C^{2}=100$.
Thus, $(A C+B C)^{2}=A C^{2}+B C^{2}... | 4 \sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,156 |
8. In a given regular $n(n \geqslant 10)$-sided polygon, choose $k$ points from the $n$ vertices such that among these $k$ points, there exist four points that are the vertices of a quadrilateral, and this quadrilateral has: one side that is a side of the given regular $n$-sided polygon. Then the minimum value of $k$ i... | 8. $\left[\frac{3 n}{4}\right]+1$.
Label the $n$ vertices of this regular $n$-gon in sequence as $A_{1}, A_{2}, \cdots, A_{n}$. A necessary and sufficient condition for a quadrilateral to have three sides as the given regular $n$-gon is that its four vertices are four consecutive vertices of the regular $n$-gon.
Let ... | \left[\frac{3 n}{4}\right]+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,157 |
II. (16 points) Let the side lengths opposite to two interior angles of $\triangle A B C$ be $a, b, c$ respectively, and $a+b+c=16$. Find
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
the value. | $$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 4 R^{2}\left(\sin ^{2} B \cdot \cos ^{2} \frac{C}{2}+\sin ^{2} C \cdot \cos ^{2} \frac{B}{2}+\right. \\
\left.2 \sin B \cdot \sin C \cdot \cos \frac{B}{2} \cdot \cos \frac... | 64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,158 |
Three. (20 points) As shown in Figure 1, given the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
with its right focus at $F$, the right directrix $l$ intersects the $x$-axis at point
$N$, and a perpendicular line $P M \perp l$ is drawn from a point $P$ on the ellipse to point $M$.
If $P N$ bisects $\a... | $$
\begin{array}{l}
\text { Three, since } \angle N P F=\angle N P M=\angle P N F, \text { therefore, } \\
P F=N F .
\end{array}
$$
Also, since quadrilateral $O F M P$ is a parallelogram, then $P M=O F$.
Thus $e=\frac{P F}{P M}=\frac{N F}{O F}=\frac{O N-O F}{O F}$
$$
=\frac{\frac{a^{2}}{c}}{c}-1=\frac{1}{e^{2}}-1 \tex... | e>\frac{2}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 726,159 |
Four, (20 points) Solve the equation for $x$
$$
\left(\cos ^{2} \frac{\theta}{2}\right) x^{3}+\left(3 \cos ^{2} \frac{\theta}{2}-4\right) x+\sin \theta=0 .
$$ | If $\cos \frac{\theta}{2}=0$, then $\sin \theta=0$. Hence $x=0$.
If $\sin \frac{\theta}{2}=0$, then $\cos ^{2} \frac{\theta}{2}=1, \sin \theta=0$.
Thus, $x=0$ or $\pm 1$.
If $\cos \frac{\theta}{2} \neq 0$, and $\sin \frac{\theta}{2} \neq 0$, then $x \neq 0$, and the original equation becomes
$$
\begin{array}{l}
x^{3}-\... | x=0 \text{ or } x=2 \tan \frac{\theta}{2} \text{ or } x=-\tan \frac{\theta}{2} \pm \sec \frac{\theta}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,160 |
One. (40 points) Let $A$ and $B$ be the two intersection points of $\odot O_{1}$ and $\odot O_{2}$. Draw a line through $A$ that intersects $\odot O_{1}$ and $\odot O_{2}$ at points $C$ and $D$, respectively. Draw the tangents to $\odot O_{1}$ and $\odot O_{2}$ at $C$ and $D$, and draw perpendiculars from $B$ to these ... | As shown in Figure 3, let $C P$ and $D Q$ intersect at point $E$.
Then $\angle B D Q=180^{\circ}-\angle B A D$
$$
=\angle C A B=\angle B C P \text {. }
$$
Therefore, points $B, C, E, D$ are concyclic, meaning point $B$ lies on the circumcircle of $\triangle C D E$.
Draw $B H \perp C D$ at point $H$.
By the Simson line... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,161 |
II. (40 points) Let $\theta_{i}(i=1,2, \cdots, n)$ be real numbers, and $x_{i}=1+3 \sin ^{2} \theta_{i}$.
Prove:
$$
\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right) \leqslant\left(\frac{5 n}{4}\right)^{2} .
$$ | $$
1 \leqslant x_{i} \leqslant 4(i=1,2, \cdots, n) \text {. }
$$
By the AM-GM inequality, we have
$$
\begin{array}{l}
\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right) \\
=\left(\frac{x_{1}}{2}+\frac{x_{2}}{2}+\cdots+\frac{x_{n}}{2}\right)\left(\frac{2}{x_{1}}+\fr... | \left(\frac{5 n}{4}\right)^{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 726,162 |
Three. (50 points) The sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}=3, x_{n+1}=\left[\sqrt{2} x_{n}\right]\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Find all $n$ such that $x_{n} 、 x_{n+1} 、 x_{n+2}$ form an arithmetic sequence, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
保留... | Three, from $x_{n}, x_{n+1}, x_{n+2}$ forming an arithmetic sequence, we know
$$
2\left[\sqrt{2} x_{n}\right]=x_{n}+\left[\sqrt{2}\left[\sqrt{2} x_{n}\right]\right] \text {. }
$$
Also,
$$
x-1x_{n}+\sqrt{2}\left[\sqrt{2} x_{n}\right]-1 \\
>x_{n}+\sqrt{2}\left(\sqrt{2} x_{n}-1\right)-1 \\
=3 x_{n}-\sqrt{2}-1 .
\end{arr... | 1 \text{ or } 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,163 |
Four. (50 points) Let $p$ be a prime number, $n$ be a positive integer, and
$$
n=n_{0}+n_{1} p+\cdots+n_{t} p^{t} \text {, }
$$
where $n_{i} \in \mathbf{N}, 0 \leqslant n_{i} \leqslant p-1, i=0,1, \cdots, t$.
Let $S_{n}$ denote the set of ordered triples $(a, b, c)$ that satisfy the following conditions:
(1) $a, b, c ... | Let $p$ be a prime, $n \in \mathbf{N}_{+}$.
If $p^{\alpha} \operatorname{In}$, but $p^{\alpha+1} \chi_{n}$, then we denote $v_{p}(n)=\alpha$. Therefore,
$$
v_{p}(n!)=\sum_{k=1}^{+\infty}\left[\frac{n}{p^{k}}\right],
$$
where $[x]$ represents the greatest integer not exceeding the real number $x$.
If $n=n_{0}+n_{1} p+\... | \left|S_{n}\right|=\mathrm{C}_{n_{0}+2}^{2} \mathrm{C}_{n_{1}+2}^{2} \cdots \mathrm{C}_{n_{t}+2}^{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,164 |
1. $\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{2011}=(\quad)$.
(A) $\mathrm{i}$
(B) $-\mathrm{i}$
(C) $2^{2011}$
(D) $-2^{2011}$ | \begin{array}{l}-1 . B \\ \left(\frac{1+i}{1-i}\right)^{2011}=\left(\frac{2 i}{2}\right)^{2011}=i^{2011}=-i\end{array} | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,165 |
2. Let the sets be
$$
A=\{1,2, \cdots, 6\}, B=\{4,5,6,7\} \text {. }
$$
Then the number of sets $S$ that satisfy $S \subseteq A$ and $S \cap B \neq \varnothing$ is ( ).
(A) 57
(B) 56
(C) 49
(D) 8 | 2. B.
In fact, the set $\{1,2,3\}$ has 8 subsets, and the set $\{4,5,6\}$ has 7 proper subsets.
Therefore, there are 56 sets that satisfy the condition. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,167 |
3. "The three numbers $\lg x, \lg y, \lg z$ form an arithmetic sequence" is a () condition for "$y^{2}=x z$".
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and necessary
(D) Neither sufficient nor necessary | 3. A.
From the problem, we know
$$
\begin{array}{l}
2 \lg y=\lg x+\lg z \\
\Rightarrow \lg y^{2}=\lg x z \Rightarrow y^{2}=x z .
\end{array}
$$
However, the converse is not necessarily true. When $x, y, z=0$, it is meaningless. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,168 |
5. The graph of the function $y=\mathrm{e}^{2}$ and the lines $x=0, y=\mathrm{e}$ enclose an area which is ( ).
(A) 1
(B) e - 1
(C) e
(D) $2 e-1$ | 5. A.
$$
\int_{0}^{t}\left(e-e^{x}\right) d x=1
$$ | A | Calculus | MCQ | Yes | Yes | cn_contest | false | 726,170 |
6. Put ten identical balls into three boxes numbered $1, 2, 3$ (all ten balls must be placed each time), with the requirement that the number of balls in each box is no less than the number of the box. Then the number of such arrangements is ( ).
(A) 9
(B) 12
(C) 15
(D) 18 | 6. C.
By the partition method, we get $\mathrm{C}_{6}^{2}=15$. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,171 |
7. The maximum value of the function $y=\cos ^{3} x+\sin ^{2} x-\cos x$ is ( ).
(A) $\frac{32}{27}$
(B) $\frac{16}{27}$
(C) $\frac{8}{27}$
(D) $\frac{4}{27}$ | 7. A.
The original function can be transformed into
$$
y=\cos ^{3} x-\cos ^{2} x-\cos x+1 \text {. }
$$
Let $t=\cos x$. Then $t \in[-1,1]$.
By using the derivative method, the maximum point is found to be $t=-\frac{1}{3}$. Therefore, the maximum value of the function is $\frac{32}{27}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,172 |
9. Given the inequality $\frac{x-5}{x+1}<0$ with the solution set $M$. If $x_{0} \in M$, then the probability that $\log _{2}\left(x_{0}+1\right)<1$ is ( ).
(A) $\frac{1}{4}$
(B) $\frac{1}{3}$
(C) $\frac{1}{5}$
(D) $\frac{2}{5}$ | 9. B.
From the problem, we know that $M=\{x \mid-1<x<5\}$.
And the solution set of $\log _{2}\left(x_{0}+1\right)<1$ is
$$
\{x \mid-1<x<1\} \text {, }
$$
Therefore, the required probability is $\frac{1}{3}$. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 726,174 |
10. Given that the area of the section through the body diagonal $B D_{1}$ of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is $S$, and the maximum and minimum values of $S$ are denoted as $S_{\text {max }}$ and $S_{\text {min }}$ respectively. Then $\frac{S_{\text {max }}}{S_{\text {min }}}=(\quad)$.
(A) $\frac{\sqrt{3}}... | 10. C.
Notice that the section is divided into two identical triangles by $B D_{1}$, that is, the ratio of heights is the ratio of areas with $B D_{1}$ as the base. The maximum height is $\frac{\sqrt{6}}{3}$, and the minimum height is $\frac{\sqrt{2}}{2}$. Therefore, $\frac{S_{\max }}{S_{\min }}=\frac{2 \sqrt{3}}{3}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,175 |
11. Let $A$ and $B$ be two distinct points on the parabola $C: y^{2}=4 x$, and $F$ be the focus of the parabola $C$. If $\overrightarrow{F A}=-4 \overrightarrow{F B}$, then the slope of the line $A B$ is ().
(A) $\pm \frac{2}{3}$
(B) $\pm \frac{3}{2}$
(C) $\pm \frac{3}{4}$
(D) $\pm \frac{4}{3}$ | 11. D.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $\overrightarrow{F A}=-4 \overrightarrow{F B} \Rightarrow y_{1}=-4 y_{2}$.
Let $l_{A B}: y=k(x-1)$, and combine with the parabola to get
$$
\begin{array}{l}
k y^{2}-4 y-4 k=0 \\
\Rightarrow y_{1}+y_{2}=\frac{4}{k}, y_{1} y_{2}=-4 .
\end{array}
$... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,176 |
Example 3 As shown in Figure 3, in the acute triangle $\triangle ABC$, it is known that $BE \perp AC$ at point $E$, $CD \perp AB$ at point $D$, $BC=25$, $CE=7$, $BD=15$. If $BE$ and $CD$ intersect at point $H$, connect $DE$, and construct a circle with $DE$ as the diameter, which intersects $AC$ at another point $F$. F... | Solve for DF. From the given conditions, we have
$$
\begin{array}{l}
\cos B=\frac{3}{5}, \cos C=\frac{7}{25} \\
\Rightarrow \sin B=\frac{4}{5}, \sin C=\frac{24}{25} \\
\Rightarrow \sin A=\sin B \cdot \cos C+\sin C \cdot \cos B=\frac{4}{5}=\sin B \\
\Rightarrow \angle A=\angle B \Rightarrow AC=BC=25 .
\end{array}
$$
Fr... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,177 |
12. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{2}=\frac{1}{2} \text {, }
$$
and $a_{n}\left(a_{n-1}+a_{n+1}\right)=2 a_{n+1} a_{n-1}(n \geqslant 2)$.
Then the 2012th term of the sequence $\left\{a_{n}\right\}$ is ( ).
(A) $\frac{1}{2010}$
(B) $\frac{1}{2011}$
(C) $\frac{1}{2012}$
(D) $\frac{1}{... | 12. C.
$$
\begin{array}{l}
\text { Given } a_{n}\left(a_{n-1}+a_{n+1}\right)=2 a_{n+1} a_{n-1} \\
\Rightarrow \frac{1}{a_{n+1}}+\frac{1}{a_{n-1}}=2 \cdot \frac{1}{a_{n}}(n \geqslant 2) .
\end{array}
$$
Then $\left\{\frac{1}{a_{n}}\right\}$ is an arithmetic sequence, and $\frac{1}{a_{1}}=1, d=1$. Therefore, $a_{n}=\fra... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,178 |
13. Given the function $f(x)=2 \sin \omega x(\omega>0)$ is monotonically increasing on $\left[0, \frac{\pi}{4}\right]$, and the maximum value on this interval is $\sqrt{3}$. Then $\omega=$ $\qquad$ . | $=13 \cdot \frac{4}{3}$.
From the problem, we know
$$
\begin{array}{l}
\frac{\pi}{4} \omega \leqslant \frac{\pi}{2}, \text { and } 2 \sin \frac{\pi}{4} \omega=\sqrt{3} \\
\Rightarrow \omega=\frac{4}{3} .
\end{array}
$$ | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,179 |
14. Given real numbers $x, y$ satisfy $3 x+2 y-1 \geqslant 0$. Then the minimum value of $u=x^{2}+y^{2}+6 x-2 y$ is $\qquad$ . | 14. $-\frac{66}{13}$.
From $u=x^{2}+y^{2}+6 x-2 y$
$$
=(x+3)^{2}+(y-1)^{2}-10 \text {, }
$$
we know that the geometric meaning of the function $u$ is the square of the distance from any point in the feasible region to the point $(-3,1)$, minus 10. | -\frac{66}{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,180 |
16. Let $P$ be any point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ other than the endpoints of the major axis, $F_{1}$ and $F_{2}$ be the left and right foci respectively, and $O$ be the center. Then
$\left|P F_{1}\right|\left|P F_{2}\right|+|O P|^{2}=$ $\qquad$ . | 16. 25.
According to the definition of an ellipse and the cosine rule, the solution can be found. | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,181 |
17. (10 points) In $\triangle A B C$, the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively. Let
$$
\begin{array}{l}
\boldsymbol{m}=(2 \sin B, -\sqrt{3}), \\
\boldsymbol{n}=\left(\cos 2 B, 2 \cos ^{2} \frac{B}{2}-1\right), \text{ and } \boldsymbol{m} / / \boldsymbol{n} .
\end{a... | And $a^{2}+c^{2} \geqslant 2 a c$, substituting into the above equation we get $a c \leqslant 4$.
Equality holds if and only if $a=c=2$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} a c \sin B=\frac{\sqrt{3}}{4} a c \leqslant \sqrt{3}$, and equality holds in equation (1) if and only if $a=c=2$.
Three, 17. (1) From $m / ... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,182 |
18. (12 points) As shown in Figure 5, quadrilateral $P Q R S$ is a cyclic quadrilateral, $\angle P S R = 90^{\circ}$, a perpendicular line is drawn from point $Q$ to $P R$ and $P S$, with the feet of the perpendiculars being $H$ and $K$ respectively. Prove:
(1) $Q, H, K$, and $P$ are concyclic;
(2) $Q T = T S$.
| 18. (1) Since $\angle P H Q=\angle P K Q=90^{\circ}$, therefore, $Q, H, K, P$ are concyclic.
(2) From the fact that $Q, H, K, P$ are concyclic, we know
$\angle H K S=\angle H Q P$.
Also, $\angle P S R=90^{\circ}$, so $P R$ is the diameter of the circle.
Thus, $\angle P Q R=90^{\circ}, \angle Q R H=\angle H Q P$.
And $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,183 |
19. (12 points) A company decides whether to invest in a project through voting by three people: A, B, and C. Each of them has one "agree," one "neutral," and one "disagree" vote. When voting, each person must and can only cast one vote, and the probability of each person casting any of the three types of votes is $\fr... | 19. (1) The probability that the company decides to invest in this project is
$$
P=\mathrm{C}_{3}^{2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)+\mathrm{C}_{3}^{3}\left(\frac{1}{3}\right)^{3}=\frac{7}{27} .
$$
(2) The company decides to abandon the investment in this project and the voting results have at most... | \frac{7}{27}, \frac{13}{27} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,184 |
20. (12 points) Given the function $f(x)=\frac{2 x+3}{3 x}$, the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=f\left(\frac{1}{a_{n}}\right)\left(n \in \mathbf{N}_{+}\right) .
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $T_{n}=\sum_{i=1}^{2 n}(-1)^{i+1} a_{i} a_{... | $$
\begin{array}{l}
a_{n+1}=f\left(\frac{1}{a_{n}}\right)=\frac{2+3 a_{n}}{3}=a_{n}+\frac{2}{3} \\
\Rightarrow a_{n}=\frac{2}{3} n+\frac{1}{3} . \\
\text { (2) } T_{n}=\sum_{i=1}^{2 n}(-1)^{i+1} a_{i} a_{i+1} \\
=a_{2}\left(a_{1}-a_{3}\right)+a_{4}\left(a_{3}-a_{5}\right)+\cdots+ \\
\quad a_{2 n}\left(a_{2 n-1}-a_{2 n+... | -\frac{4}{9}(2n^2 + 3n) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,185 |
21. (12 points) Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$ pass through the point $(0,1)$, and have an eccentricity of $\frac{\sqrt{3}}{2}$.
(1) Find the equation of the ellipse $C$.
(2) Let the line $l: x=m y+1$ intersect the ellipse $C$ at points $A$ and $B$, and let the point $A$ be sym... | 21. (1) According to the problem, we have
$$
\left\{\begin{array}{l}
b=1, \\
\frac{c}{a}=\frac{\sqrt{3}}{2}, \\
a^{2}=b^{2}+c^{2}
\end{array} \Rightarrow a=2 .\right.
$$
Therefore, the equation of the ellipse $C$ is $\frac{x^{2}}{4}+y^{2}=1$.
(2) From $\left\{\begin{array}{l}\frac{x^{2}}{4}+y^{2}=1, \\ x=m y+1\end{arr... | (4,0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,186 |
22. (12 points) Let
$$
f(x)=\frac{a}{x}+x \ln x, \quad g(x)=x^{3}-x^{2}-3 \text {. }
$$
(1) When $a=2$, find the equation of the tangent line to the curve $y=f(x)$ at $x=1$;
(2) If there exist $x_{1}, x_{2} \in [0,2]$ such that
$$
g\left(x_{1}\right)-g\left(x_{2}\right) \geqslant M
$$
holds, find the maximum integer $... | 22. (1) When $a=2$,
$$
f(1)=2, f^{\prime}(1)=-1 \text {. }
$$
Therefore, the equation of the tangent line to the curve $y=f(x)$ at $x=1$ is
$$
y-2=-(x-1) \text {. }
$$
(2) From $g^{\prime}(x)=3 x^{2}-2 x=3 x\left(x-\frac{2}{3}\right)$, and
$$
g(0)=-3, g(2)=1, g\left(\frac{2}{3}\right)=-\frac{85}{27},
$$
we know
$$
\b... | a \geqslant 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,187 |
Example 4 As shown in Figure 4, given that $AB$ is the diameter of $\odot O$, $CE \perp AB$ at point $H$, intersects $\odot O$ at points $C$ and $D$, and $AB = 10, CD = 8, DE = 4, EF$ is tangent to $\odot O$ at point $F, BF$ intersects $HD$ at point $G$.
(1) Find $GH$;
(2) Connect $FD$, determine whether $FD$ is parall... | Solve (1) Connect $A F, O E, O F$. Then $A, F, G, H$ are concyclic.
Since $E F$ is a tangent, $O F \perp E F$.
Thus, $\angle F G E=\angle B A F=\angle E F G$.
Therefore, $E F=E G$.
Also, $O E^{2}=O H^{2}+H E^{2}=O F^{2}+E F^{2}$, so
$E F^{2}=O H^{2}+H E^{2}-O F^{2}=3^{2}+8^{2}-5^{2}=48$.
Hence, $E F=E G=4 \sqrt{3}$.
Th... | 8-4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,188 |
1. Given the function
$$
f(x)=\arcsin (\cos x) \text {. }
$$
then the smallest positive period of $f(f(f(x)))$ is $\qquad$ . | $-1 . \pi$.
From the fact that $\cos x$ is an even function, we know that $f(x)$ is an even function.
$$
\begin{array}{l}
\text { and } f(x+\pi)=\arcsin [\cos (x+\pi)] \\
=\arcsin (-\cos x)=-\arcsin (\cos x) \\
=-f(x),
\end{array}
$$
Therefore, $f(f(x+\pi))=f(-f(x))=f(f(x))$. Thus, the period of $f(f(x))$ is $\pi$.
He... | \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,189 |
2. Let real numbers $x, y$ satisfy
$$
x^{2}-8 x+y^{2}-6 y+24=0 \text {. }
$$
Then the maximum value of $x-2 y$ is $\qquad$ $\therefore$ | 2. $\sqrt{5}-2$.
From $x^{2}-8 x+y^{2}-6 y+24=0$, we get
$$
(x-4)^{2}+(y-3)^{2}=1 \text {. }
$$
When the line $x-2 y=k$ is tangent to the circle, it is the desired maximum value.
$$
\text { By } \frac{|4-6-k|}{\sqrt{1+2^{2}}}=1
$$
$\Rightarrow k=-2 \pm \sqrt{5}$ (negative value discarded). | \sqrt{5}-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,190 |
$\begin{array}{l}\text { 3. } \cos \frac{\pi}{11}-\cos \frac{2 \pi}{11}+\cos \frac{3 \pi}{11}-\cos \frac{4 \pi}{11}+\cos \frac{5 \pi}{11} \\ = \\ \text { (answer with a number). }\end{array}$ | 3. $\frac{1}{2}$.
Notice that,
$$
\begin{array}{l}
\sum_{k=1}^{5}(-1)^{k+1} \cos \frac{k \pi}{11} \\
=\frac{1}{2 \cos \frac{\pi}{22}} \sum_{k=1}^{5}(-1)^{k+1}\left[\cos \frac{(2 k+1) \pi}{22}+\cos \frac{(2 k-1) \pi}{22}\right] \\
=\frac{1}{2 \cos \frac{\pi}{22}}\left(\cos \frac{\pi}{22}+\cos \frac{11 \pi}{22}\right)=\... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,191 |
4. Let points $C$ and $D$ be on the semicircle with diameter $AB$, and line segments $AC$ and $BD$ intersect at point $E$. Given $AB=10$, $AC=8$, and $BD=5\sqrt{2}$. Then the area of $\triangle ABE$ is
Translate the above text into English, please keep the original text's line breaks and format, and output the transla... | 4. $\frac{150}{7}$.
As shown in Figure 2, connect $A D$ and $B C$. Draw $E M \perp A B$ at point $M$.
From the given information, we have
$$
B C=6 \text{. }
$$
Since $A B$ is the diameter of the semicircle, then
Rt $\triangle A C B \backsim \text{Rt} \triangle A M E$,
Rt $\triangle B M E \backsim \text{Rt} \triangle ... | \frac{150}{7} | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 726,192 |
5. Let two ellipses be
$$
\frac{x^{2}}{t^{2}+2 t-2}+\frac{y^{2}}{t^{2}+t+2}=1
$$
and $\frac{x^{2}}{2 t^{2}-3 t-5}+\frac{y^{2}}{t^{2}+t-7}=1$
have common foci. Then $t=$ $\qquad$ . | 5.3.
Given that the two ellipses have a common focus, we have
$$
\begin{array}{l}
t^{2}+2 t-2-\left(t^{2}+t+2\right) \\
=2 t^{2}-3 t-5-\left(t^{2}+t-7\right) \\
\Rightarrow t=3 \text { or } 2 \text { (rejected). }
\end{array}
$$
Therefore, $t=3$. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,193 |
6. As shown in Figure 1, given that the volume of the regular quadrilateral pyramid $P$ $A B C D$ is 1, $E$, $F$, $G$, $H$ are the midpoints of line segments $A B$, $C D$, $P B$, $P C$ respectively. Then the volume of the polyhedron $B E G-C F H$ is $\qquad$ | 6. $\frac{5}{16}$.
$$
x^{2}+1 \text {, }
$$
Transform the polyhedron. $g(x) \cdots)$ ).
$$
E B G-F C H^{\prime} \text {. }
$$
Then $V_{\text {sinik Bec }} g(f(x))$, and $f(x)$ is $2^{m}$ times
$$
=\frac{3}{4} V_{\text {tetrahedron } P A B D}-\frac{1}{8} V_{\text {tetrahedron } P B C D}=\frac{5}{16} \text {. }
$$ | \frac{5}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,194 |
7. The number of positive integers not exceeding 2012 and whose greatest common divisor with 210 is 1 is $\qquad$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7.460.
From $2012=210 \times 9+122$, and $\varphi(210)=$ 48, we know that there are 24 numbers between 1 and 105 that are coprime with 210, and 4 numbers between 106 and 122 that are coprime with 210.
Therefore, there are $48 \times 9+24+4=460$ (numbers) between 1 and 2012 that are coprime with 210. | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,195 |
9. (25 points) Given that the perimeter of $\triangle A B C$ is 1, and $\sin 2 A+\sin 2 B=4 \sin A \cdot \sin B$.
(1) Prove: $\triangle A B C$ is a right triangle;
(2) Find the maximum area of $\triangle A B C$. | (1) Notice,
\[
\begin{aligned}
0= & \sin 2 A+\sin 2 B-4 \sin A \cdot \sin B \\
= & 2 \sin (A+B) \cdot \cos (A-B)- \\
& 2[\cos (A-B)-\cos (A+B)] \\
= & 2 \sin C \cdot \cos (A-B)- \\
& 2[\cos (A-B)+\cos C] \\
= & -2 \cos (A-B)(1-\sin C)-2 \cos C \\
= & -2 \cos (A-B)\left(1-2 \sin \frac{C}{2} \cdot \cos \frac{C}{2}\right)... | \frac{3-2 \sqrt{2}}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 726,197 |
Example 1 (An Ancient Chinese Mathematical Problem) Emperor Taizong of Tang ordered the counting of soldiers: if 1,001 soldiers make up one battalion, then one person remains; if 1,002 soldiers make up one battalion, then four people remain. This time, the counting of soldiers has at least $\qquad$ people. | Let the first troop count be 1001 people per battalion, totaling $x$ battalions, then the total number of soldiers is $1001 x + 1$ people; let the second troop count be 1002 people per battalion, totaling $y$ battalions, then the total number of soldiers is $1002 y + 4$ people.
From the equality of the total number of ... | 1000000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,199 |
Example 2 The scoring rules for a certain football tournament are: 3 points for a win, 1 point for a draw, and 0 points for a loss. A team participated in 15 matches and accumulated 33 points. If the order of the matches is not considered, then the number of scenarios for the team's wins, draws, and losses is ( ).
(A) ... | Let the team win $x$ games, draw $y$ games, and lose $z$ games, where $x, y, z \in \mathbf{N}$, and $0 \leqslant x, y, z \leqslant 15$.
According to the problem, we set up the system of equations
$$
\left\{\begin{array}{l}
x+y+z=15, \\
3 x+y=33 .
\end{array}\right.
$$
From equation (2), we get
$$
x=11-\frac{y}{3} \tex... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,200 |
Example 1 Polynomial
$$
f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0},
$$
If there exist $n+1$ consecutive integers $m, m+1, \cdots, m+n$, such that $f(n+k) \in \mathbf{Z}(k=0,1, \cdots, n)$, then $f(x)$ is an integer-valued polynomial (when $x \in \mathbf{Z}$, $f(x) \in \mathbf{Z}$). | 【Analysis】Apply mathematical induction on $n$.
When $n=0$, $f(x)=c \in \mathbf{Z}$, the proposition is obviously true.
Assume the proposition holds for $n=k$.
When $n=k+1$, let
$f^{*}(x)=f(x+1)-f(x)$.
Then $f^{*}(x)=n a_{n} x^{n-1}+\cdots$.
Thus $\operatorname{deg} f^{*}(x)=n-1$, and $f^{*}(x)$ takes integer values at ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,201 |
5. Calculate:
$$
\sum_{k=0}^{2013}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2014}^{k}}=
$$
$\qquad$ | 5. 0 .
Let $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2 n}^{k}}$.
And $\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n}-k-1}$, so $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{\mathrm{C}_{2 n}^{2 n... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,202 |
$$
\begin{array}{l}
\text { 6. Let }(1+2 \sqrt{2}+3 \sqrt{3})^{n} \\
=a_{n}+b_{n} \sqrt{2}+c_{n} \sqrt{3}+d_{n} \sqrt{6} \text {, }
\end{array}
$$
where, $a_{n}, b_{n}, c_{n}, d_{n} \in \mathbf{N}$.
Then $\lim _{n \rightarrow+\infty} \frac{d_{n}}{a_{n}}=$ $\qquad$ | $$
\begin{array}{l}
\text { 6. } \frac{\sqrt{6}}{6} . \\
\text { Let } \lambda_{1}=1+2 \sqrt{2}+3 \sqrt{3}, \\
\lambda_{2}=1-2 \sqrt{2}+3 \sqrt{3}, \\
\lambda_{3}=1+2 \sqrt{2}-3 \sqrt{3}, \\
\lambda_{4}=1-2 \sqrt{2}-3 \sqrt{3} .
\end{array}
$$
Then $\left.\left|\lambda_{1}\right|>\max || \lambda_{2}|,| \lambda_{3}|,| ... | \frac{\sqrt{6}}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,203 |
7. If the distance from the center $O$ of $\odot O$ to the line $l$ in the plane where $\odot O$ lies is $d$, and the radius of the circle is $r(d>r>0)$, then the volume of the torus (tire) obtained by rotating $\odot O$ around the line $l$ once is $\qquad$ | 7. $2 \pi^{2} r^{2} d$
Construct a cylinder with a base radius of $r$ and a height of $4 \pi d$.
Take one quarter of the cylinder, flip it over, and then take half of the torus, and use a plane at a distance $h$ from the base to cut these two geometric bodies. The cross-sectional area of the torus is
$$
S=\pi\left(r_{... | 2 \pi^{2} r^{2} d | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,204 |
8. In $\triangle A B C$,
$$
\frac{\sin \frac{A}{2} \cdot \sin \frac{B}{2}+\sin \frac{B}{2} \cdot \sin \frac{C}{2}+\sin \frac{C}{2} \cdot \sin \frac{A}{2}}{\sin A+\sin B+\sin C}
$$
the maximum value is | 8. $\frac{\sqrt{3}}{6}$.
In $\triangle ABC$, there is the identity
$$
\sum \tan \frac{A}{2} \cdot \tan \frac{B}{2}=1,
$$
where “$\sum$” denotes the cyclic sum.
$$
\begin{array}{l}
\text { Then } \sum \sin \frac{A}{2} \cdot \sin \frac{B}{2} \\
=\sum \sqrt{\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \cos \frac{A}{2} ... | \frac{\sqrt{3}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,205 |
9. (16 points) Let $a, b, c > 0$, and $a+b+c=3$. Find the maximum value of $a^{2} b+b^{2} c+c^{2} a+a b c$.
| Let's assume $a \leqslant b \leqslant c$ or $c \leqslant b \leqslant a$.
Then $c(b-a)(b-c) \leqslant 0$
$$
\begin{aligned}
\Rightarrow & b^{2} c+c^{2} a \leqslant a b c+c^{2} b \\
\Rightarrow & a^{2} b+b^{2} c+c^{2} a+a b c \\
& \leqslant a^{2} b+c^{2} b+2 a b c \\
& =b(a+c)^{2}=b(3-b)^{2} \\
& \leqslant \frac{1}{2}\le... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,206 |
10. (20 points) Given the parabola $y^{2}=2 p x(p>0)$ with a chord $A B$ of length $l(l>0)$. Find the minimum distance from the midpoint of the chord $A B$ to the $y$-axis.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10. Let the endpoints of the chord be \( A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right) \), and the midpoint be \( M(x, y) \). Then
\[
\begin{array}{l}
y_{1}^{2}=2 p x_{1}, y_{2}^{2}=2 p x_{2}, \\
2 x=x_{1}+x_{2}, 2 y=y_{1}+y_{2} . \\
\text { And } l^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}... | x_{\min }=\frac{l-p}{2} \text{ when } l \geqslant 2 p; \, x_{\min }=\frac{l^{2}}{8 p} \text{ when } 0<l<2 p | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,207 |
11. (20 points) Given $x_{n}=\{\sqrt{2} n\}\left(n \in \mathbf{N}_{+}\right)$, where $\{x\}=x-[x],[x]$ represents the greatest integer not exceeding the real number $x$. If the sequence $\left\{x_{n}\right\}$ satisfies
$$
\left|(m-n)\left(x_{m}-x_{n}\right)\right| \geqslant \frac{1}{p}\left(p \in \mathbf{N}_{+}\right)
... | 11. First prove:
$$
|\sqrt{2} m-n|>\frac{1}{3 m}\left(m, n \in \mathbf{N}_{+} ; m \neq n\right) \text {. }
$$
Assume $|\sqrt{2} m-n| \leqslant \frac{1}{3 m}$.
(1) If $0 \leqslant \sqrt{2} m \div n \leqslant \frac{1}{3 m}$, then
$$
\begin{array}{l}
2 n \leqslant \sqrt{2} m+n \leqslant \frac{1}{3 m}+2 n \\
\Rightarrow 1... | p \geqslant 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,208 |
One, (40 points) Given that $O_{1}$ and $H$ are the circumcenter and orthocenter of $\triangle ABC$, respectively, $M$ is the midpoint of $BC$, and $O_{2}$ is the midpoint of $AM$; the circle $\odot O_{2}$ with diameter $AM$ intersects the circumcircle $\odot O_{1}$ of $\triangle ABC$ at another point $P$ different fro... | As shown in Figure 1, let $A H$ intersect $B C$ at point $D$, and $B H$ intersect $A C$ at point $E$. Let $\odot M$ be the circle with $B C$ as its diameter. Since $H$ is the orthocenter of $\triangle A B C$, we have
$$
A H \perp B C, B H \perp A C .
$$
Therefore, points $D$ and $E$ lie on $\odot O_{2}$ and $\odot M$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,209 |
II. (40 points) Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$. Find
$$
\frac{\left(1+\cos ^{2} \alpha\right)\left(1-\sin ^{4} \alpha \cdot \cos ^{4} \beta\right)\left(1-\sin ^{4} \alpha \cdot \sin ^{4} \beta\right)}{\sin ^{2} 2 \alpha \cdot \sin ^{2} 2 \beta}
$$
the minimum value. | $$
\begin{array}{l}
\text { II. Let } x=\sin ^{2} \alpha \cdot \cos ^{2} \beta, y=\sin ^{2} \alpha \cdot \sin ^{2} \beta, \\
z=\cos ^{2} \alpha\left(\alpha, \beta \in\left(0, \frac{\pi}{2}\right)\right) .
\end{array}
$$
Then $x+y+z=1$. Therefore,
$$
\begin{array}{l}
\frac{\left(1+\cos ^{2} \alpha\right)\left(1-\sin ^{... | \frac{32}{27} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,210 |
Three, (50) Find the maximum value of $n$ such that there are $n$ points in the plane, where among any three points, there must be two points whose distance is 1. | If there exist $n(n \geqslant 8)$ points satisfying the conditions of the problem, let $V=\left\{v, v_{1}, v_{2}, \cdots, v_{7}\right\}$ represent any eight of these points. When and only when the distance between two points is 1, connect an edge between these two points, forming a graph $G$.
If there exists a point (... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,211 |
Example 2 Let $a>3, f(x)$ be an $n$-degree polynomial. Then among the $n+2$ non-negative numbers
$$
\begin{array}{l}
\left|a^{0}-f(0)\right|,\left|a^{1}-f(1)\right|,\left|a^{2}-f(2)\right|, \\
\cdots,\left|a^{n+1}-f(n+1)\right|
\end{array}
$$
at least one is greater than 1. | 【Analysis】Apply mathematical induction on $n$.
When $n=0$, $f(x)=A$ (a constant polynomial).
It is easy to see that $|1-A|+|a-A| \geqslant|a-1|>2$, the proposition is obviously true.
Assume the proposition is true for $n=k$.
When $n=k+1$, consider the auxiliary function
$$
g(x)=\frac{f(x+1)-f(x)}{a-1} \text {. }
$$
Th... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,212 |
Four, (50) Let
$$
f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}
$$
be a polynomial with integer coefficients, where $n \in \mathbf{N}_{+}, n$ is odd, and $a_{n} \neq 0$. Prove: There exists a positive integer $m$, such that $f(m)$ is not a perfect square. | Four, Proof by Contradiction.
Assume that for any positive integer $m, f(m)$ is a perfect square.
Then $f(x) \geqslant 0\left(x \in \mathbf{N}_{+}\right)$.
Therefore, $a_{n}>0$.
Furthermore, as $x \rightarrow-\infty$, $f(x) \rightarrow-\infty$, so there exists a negative integer $r$, such that $f(r)-r\right)$,
$$
g(m)=... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,213 |
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