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Sure, here is the translated text:
```
II. (40 points) Given that $a, b, c, d, e$ are integers, the equation
$$
a x^{5}-2 b x^{4}+3 c x^{3}-5 d x^{2}+7 e x-2.024=0
$$
has a positive integer solution. Prove: There exist infinitely many positive integers $k$ such that
$$
61 \mid \left(11 k^{5}-7 e k^{4}+5 d k^{3}-3 c k... | Let the positive integer solution of equation (1) be $x=m$.
First, we explain: $(m, 61)=1$.
Otherwise,
$$
\begin{array}{l}
61|m \Rightarrow 61|\left(a m^{5}-2 b m^{4}+3 m^{3}-5 d m^{2}+7 e m\right) \\
\Rightarrow 61 \mid 2024 .
\end{array}
$$
Contradiction.
$$
\begin{array}{l}
\text { By } a m^{5}-2 b m^{4}+3 \mathrm{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,628 |
Three, (50 points) If $a, b, c \in \mathbf{R}_{+}$, and satisfy $a+b+c = 2$, prove: $\sum \sqrt{\frac{a+b}{2}-a b} \geqslant \sqrt{2}$, where “$\sum$” denotes the cyclic sum. | $$
\begin{array}{l}
\text{Notice that, } a+b+c=2. \text{ Hence the original inequality} \\
\Leftrightarrow a+b+c-(a b+b c+c a)+ \\
2 \sum \sqrt{\left(\frac{a+b}{2}-a b\right)\left(\frac{b+c}{2}-b c\right)} \geqslant 2 \\
\Leftrightarrow 2 \sum \sqrt{\left(\frac{a+b}{2}-a b\right)\left(\frac{b+c}{2}-b c\right)} \\
\geqs... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,629 |
Four, (50 points) Find the maximum value of $m$ such that from the subsets of an $n$-element set, $m$ different subsets $A_{1}, A_{2}, \cdots, A_{m}$ can be selected, satisfying $\left(A_{i} \cap A_{k}\right) \subseteq A_{j}$ for all $1 \leqslant i<j<k \leqslant m$.
| Let's assume this $n$-element set is $M=\{1,2, \cdots, n\}$.
For any $i \in M$, let the subsets of $M$ that contain $i$ be $A_{i_{1}}, A_{i_{i}}, \cdots, A_{i_{i}}\left(i_{1}<i_{2}<\cdots<i_{t}\right)$.
Given that $\left(A_{i} \cap A_{k}\right) \subseteq A_{j}$ for all $1 \leqslant i<j<k \leqslant m$, it follows that ... | 2n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,630 |
Example 1 If $1$ and $\sqrt{2}$ are in the arithmetic sequence $\left\{x_{n}\right\}$, then it is impossible for this sequence to have three terms that form a geometric sequence.
(2002, Romanian Mathematics Competition) | 【Analysis】Let $x_{n}=x_{i}+(n-i) d$, and $x_{l}=1, x_{k}=\sqrt{2}$. Then
$$
d=\frac{\sqrt{2}-1}{k-l} \text {. }
$$
Assume in the sequence $\left\{x_{n}\right\}$, $x_{m} 、 x_{_{n}} 、 x_{p}$ are three consecutive terms forming a geometric sequence. Then
$$
\begin{array}{c}
x_{m}=x_{l}+(m-l) d \\
=1+\frac{m-l}{k-l}(\sqrt... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,631 |
Example 2 Let $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n+1}$ be positive numbers, and
$$
a_{2}-a_{1}=a_{3}-a_{2}=\cdots=a_{n+1}-a_{n} \geqslant 0 \text {. }
$$
Prove:
$$
\frac{1}{a_{2}^{2}}+\frac{1}{a_{3}^{2}}+\cdots+\frac{1}{a_{n}^{2}} \leqslant \frac{n-1}{2} \cdot \frac{a_{1} a_{n}+a_{2} a_{n+1}}{a_{1} a_{2} a_{n} a... | 【Analysis】Let's assume the common difference of the sequence $\left\{a_{n}\right\}$ is $d$. Clearly, when $d=0$, the equality in (1) holds.
Hereafter, let's assume $d \neq 0$.
Notice that, on the right side of inequality (1),
$$
\begin{array}{l}
\frac{a_{1} a_{n}+a_{2} a_{n+1}}{a_{1} a_{2} a_{n} a_{n+1}}=\frac{1}{a_{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,632 |
Example 3 A $12 \times 12$ grid is colored alternately in black and white (like a chessboard), and a "knight" can jump from one corner of a $3 \times 4$ (or $4 \times 3$) grid to the opposite corner. Prove: a knight cannot start from one square, pass through each square of the $12 \times 12$ grid exactly once, and fina... | Proof Assume there exists a "Knight's" path $C$ that meets the requirements of the problem.
First, consider the colors of the squares the knight lands on after an odd number of moves and after an even number of moves.
Notice that the diagonal squares of a $3 \times 4$ grid have different colors. Therefore, let
$A_{1}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,633 |
Five. (20 points) Let
$$
f(x)=\ln x-\frac{1}{2} a x^{2}-2 x \quad (a<0),
$$
and the function $f(x)$ has a monotonically decreasing interval.
(1) Find the range of values for $a$;
(2) If for any value of $a$ that satisfies the condition, $f(x)<b$ holds for all $x \in (0,1]$, find the range of values for the real number... | (1) From the given, we have
$$
f^{\prime}(x)=\frac{1}{x}-a x-2=-\frac{a x^{2}+2 x-1}{x}(x>0) \text {. }
$$
Since the function $f(x)$ has a monotonically decreasing interval, the inequality $f^{\prime}(x) \leqslant 0$, i.e., $a x^{2}+2 x-1 \geqslant 0$ has solutions on $(0,+\infty)$.
Noting that, $a0$.
Thus, $\Delta=4... | b \in\left(-\frac{3}{2},+\infty\right) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,635 |
Six. (20 points) Let $a_{1}, a_{2}, \cdots, a_{n}\left(n \in \mathbf{Z}_{+}\right)$ all be positive real numbers. Prove:
$$
\sum_{k=1}^{n} \frac{k^{2}}{a_{1}+a_{2}+\cdots+a_{k}}<4 \sum_{k=1}^{n} \frac{k}{a_{k}} .
$$ | Six, because $a_{i}>0(i=1,2, \cdots, n)$, so, by the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\sum_{i=1}^{k} a_{i}\right)\left(\sum_{i=1}^{k} \frac{i^{2}}{a_{i}}\right) \geqslant\left(\sum_{i=1}^{k} i\right)^{2}=\frac{k^{2}(k+1)^{2}}{4} . \\
\text { Hence } \sum_{k=1}^{n} \frac{k^{2}}{a_{1}+a_{2}+\c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,636 |
1. Let $a, b$ be rational numbers, and $|a|>0$, the equation $||x-a|-b|=5$ has three distinct roots. Then $b=(\quad)$.
(A) 1
(B) 3
(C) 5
(D) 6 | -,1. C.
From the given, we have
$$
|x \sim a|=b+5 \text { or } b-5 \text {. }
$$
And each equation has two, one, or zero solutions.
Since the original equation has three distinct roots, one equation has two solutions, and the other has one solution.
Therefore, one of the non-negative numbers $b+5, b-5$ must be 0.
Thus... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,637 |
2. Given that $a$, $b$, and $c$ are integers, and satisfy
$$
|a-b|^{2013}+|c-a|^{2013}=1 \text {. }
$$
Then $|a-b|+|b-c|+|c-a|=(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4 | 2. B.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,638 |
3. Observe the following equations:
$$
\begin{array}{l}
3^{1}=3,3^{2}=9,3^{3}=27,3^{4}=81, \\
3^{5}=243,3^{6}=729,3^{7}=2187, \cdots
\end{array}
$$
Then the last digit of $3+3^{2}+3^{3}+3^{4}+\cdots+3^{2014}$ is ( ).
(A) 2
(B) 3
(C) 7
(D) 9 | 3. A.
From the given equation, it is easy to know that the last digit repeats every four numbers. Since $2014=4 \times 503+2$, we have
$$
3+3^{2}+3^{3}+3^{4}+\cdots+3^{2014}
$$
the last digit is equivalent to that of
$$
3+9+7+1+\cdots+9
$$
which is 2. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,639 |
4. If any $n$ integers are taken, it is always possible to select three numbers whose sum is divisible by 3. Then the minimum value of $n$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 4. D.
On one hand, the sum of any three numbers chosen from 0, 1, 3, 4 is not divisible by 3.
On the other hand, when an integer is divided by 3, the remainder falls into one of three categories: 0, 1, or 2. For any five integers, if there are three with the same remainder when divided by 3, their sum can be divisibl... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,640 |
5. Given that $a$ and $b$ are integers, the linear equation in one variable $x$
$$
2 x-[2-(2 b-1) x]=a-2
$$
and $(2 b-1) x+3=7-[(2-b) x+3]$
have the same solution. Then $a b=(\quad)$.
(A) 0
(B) -6
(C) 0 or -6
(D) No solution | 5. C.
From the first equation, we get $x=\frac{a}{2 b+1}$.
From the second equation, we get $x=\frac{1}{b+1}$.
Since the two equations have the same solution, we have,
$$
\begin{array}{l}
\frac{a}{2 b+1}=\frac{1}{b+1} \\
\Rightarrow a=\frac{2 b+1}{b+1}=2-\frac{1}{b+1} .
\end{array}
$$
Given that $a$ and $b$ are integ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,641 |
6. As shown in Figure 1, in the Cartesian coordinate system, the vertex $A$ of the right triangle $\triangle O A B$ is on the positive $x$-axis, $B(3, \sqrt{3}), C\left(\frac{1}{2}, 0\right)$, and $P$ is a moving point on the hypotenuse $O B$. Then the minimum value of $P A+P C$ is
(A) $\frac{\sqrt{13}}{2}$
(B) $\frac{... | 6. B.
As shown in Figure 5, construct the symmetric point $D$ of $A$ with respect to $OB$, connect $CD$, intersecting $OB$ at point $P$, connect $AP$, and draw $DN \perp OA$ at point $N$. At this point, the value of $PA + PC$ is minimized.
Notice that $DP = PA$, then
$$
PA + PC = PD + PC = CD.
$$
It is easy to see t... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,642 |
Question 1 Given $a, b, c > 0$. Prove:
$$
\begin{array}{l}
\left(a^{3}+\frac{1}{b^{3}}-1\right)\left(b^{3}+\frac{1}{c^{3}}-1\right)\left(c^{3}+\frac{1}{a^{3}}-1\right) \\
\leqslant\left(a b c+\frac{1}{a b c}-1\right)^{3} .
\end{array}
$$
(3rd Chern Shiing-Shen Cup National High School Mathematics Olympiad) | Proof: Since the sum of any two of the product terms on the left side of the original inequality is positive, at most one of the three product terms is non-positive. Therefore, without loss of generality, we assume that all three product terms are positive.
Notice,
$$
\begin{array}{l}
\sum\left(a^{3}+\frac{1}{b^{3}}-1\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,644 |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$ (for example, $[2.1]=2,[1]=1)$. If the real number $a$ satisfies
$$
a-\frac{5}{a}+\frac{4 \sqrt{a(a-3)}}{a}=3 \text {, }
$$
then $[a]=$ $\qquad$ . | 3. -1 or 3.
The original equation is equivalent to
$a^{2}-5+4 \sqrt{a(a-3)}=3 a (a<0$ or $a \geqslant 3)$.
Let $x=\sqrt{a(a-3)} \geqslant 0$. Then
$x^{2}+4 x-5=0 \Rightarrow x_{1}=1, x_{2}=-5$ (discard).
Thus $\sqrt{a(a-3)}=1 \Rightarrow a^{2}-3 a-1=0$
$\Rightarrow a_{1}=\frac{3+\sqrt{13}}{2}, a_{2}=\frac{3-\sqrt{13}}... | -1 \text{ or } 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,646 |
4. As shown in Figure 3, in trapezoid $ABCD$, $DC \parallel AB$, $\frac{DC}{AB}=\frac{1}{3}$, $MN$ is the midline, $EF \parallel AB$ and passes through the intersection of $AC$ and $BD$, points $E$ and $F$ are on $AD$ and $BC$ respectively. Then the ratio of the areas of trapezoids $CDEF$, $FEMN$, and $NMAB$ is equal t... | 4.5:7:20.
It is easy to prove that trapezoid $CDEF$ is similar to trapezoid $NMAB$,
and trapezoid $CDMN$ is similar to trapezoid $FEAB$.
Let $DC=1$. Then $AB=3, MN=2, EF=\frac{3}{2}$.
Let the area of trapezoid $CDEF$ be 1. Then the area of trapezoid $NMAB$ is 4. Let the area of trapezoid $FEMN$ be $x$.
Notice that, $\... | 5:7:20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,647 |
One, (20 points) As shown in Figure 4, given that $PC$ is tangent to $\odot O$ at point $C$, $AC$ is the diameter of $\odot O$, $PEF$ is a secant of $\odot O$, and $AE, AF$ intersect line $PO$ at points $B, D$ respectively. Prove:
$$
AB=DC, BC=AD \text{. }
$$ | As shown in Figure 7, draw $C Q \perp P D$ at point $Q$, and connect $E O$, $E Q$, $E C$, $O F$, $Q F$, $C F$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,648 |
II. (25 points) Given the quadratic function $y=a x^{2}+b x+c$ $(a>0)$, the graph of which intersects with the graph of the linear function $y=x$ at two points with x-coordinates $x_{1}$ and $x_{2}$, and $0<x_{1}<x_{2}<\frac{1}{a}$.
(1) Try to express $b$ and $c$ in terms of $a$, $x_{1}$, and $x_{2}$;
(2) If $0<t<x_{1}... | (1) From the given information, we have
$$
\begin{array}{l}
a x^{2}+b x+c=x \\
\Rightarrow a x^{2}+(b-1) x+c=0 .
\end{array}
$$
Then, $x_{1}+x_{2}=-\frac{b-1}{a}, x_{1} x_{2}=\frac{c}{a}$.
Thus, $b=-a\left(x_{1}+x_{2}\right)+1, c=a x_{1} x_{2}$.
(2) When $0<t<x_{1}<x_{2}<\frac{1}{a}$, we get
$$
a\left(t-x_{1}\right)\l... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,649 |
Three, (25 points) Several people gather, some of whom know each other, and the following conditions are met:
(1) If two people have an equal number of acquaintances, then they do not have any common acquaintances;
(2) There is one person who has at least 56 acquaintances.
Prove: It is possible to find a participant wh... | Three, let the person with the most acquaintances among the party-goers (if there is more than one, choose any one of them) be denoted as $A$.
Suppose $A$ knows $n$ people $B_{1}, B_{2}, \cdots, B_{n}$.
Since any two people $B_{i}, B_{j}$ have $A$ as a common acquaintance, by condition (1), the number of acquaintances ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,650 |
1. A bag contains 8 white balls and 2 red balls. Each time a ball is randomly drawn from the bag, and then a white ball is put back. What is the probability that all red balls are exactly drawn by the fourth draw? | $$
-1.0 .0434
$$
The probability of exactly taking all the red balls on the fourth draw is
$$
\begin{array}{l}
\frac{2}{10} \times\left(\frac{9}{10}\right)^{2} \times \frac{1}{10}+\frac{8}{10} \times \frac{2}{10} \times \frac{9}{10} \times \frac{1}{10}+\left(\frac{8}{10}\right)^{2} \times \frac{2}{10} \times \frac{1}{... | 0.0434 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,651 |
2. Given the function $f(x)=\sqrt{x+2}+k$, and there exist $a, b$ $(a<b)$ such that the range of $f(x)$ on $[a, b]$ is $[a, b]$. Then the range of the real number $k$ is $\qquad$ | 2. $k \in\left(-\frac{9}{4},-2\right]$.
Notice that the domain of the function $f(x)$ is $\{x \mid x \geqslant-2\}$.
Since $f(x)$ is monotonically increasing, the range of $f(x)$ on $[a, b]$ being $[a, b]$ is equivalent to the equation $f(x)=\sqrt{x+2}+k=x$ having two distinct real roots, i.e.,
$$
x^{2}-(2 k+1) x+k^{2... | k \in\left(-\frac{9}{4},-2\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,652 |
3. In a regular quadrilateral pyramid $P-ABCD$, it is known that $AB=3$, and the dihedral angle formed by the side faces $PAD$ and $CPD$ is $\frac{3 \pi}{2}$. Then the volume of the circumscribed sphere of the quadrilateral pyramid is | 3. $\frac{243 \pi}{16}$.
Draw perpendiculars from points $C$ and $A$ to $PD$, then the perpendiculars must intersect $PD$ at a point $Q$, and
$$
\angle AQC = \frac{2 \pi}{3}, \quad AQ = CQ.
$$
Since $AC = \sqrt{2} AB = 3 \sqrt{2}$, we have
$$
\begin{array}{l}
AQ = CQ = \frac{AC}{\sin 120^{\circ}} \cdot \sin \frac{180... | \frac{243 \pi}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,653 |
4. The minimum value of the function $f(x)=\sqrt{x^{2}+4}+\sqrt{x^{2}-4 x+5}$ is $\qquad$ . | 4. $\sqrt{13}$.
Notice,
$$
\begin{array}{l}
\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}} \\
\geqslant \sqrt{(a+c)^{2}+(b+d)^{2}} .
\end{array}
$$
Then $f(x)=\sqrt{x^{2}+4}+\sqrt{x^{2}-4 x+5}$
$$
\begin{array}{l}
=\sqrt{x^{2}+2^{2}}+\sqrt{(2-x)^{2}+1^{2}} \\
\geqslant \sqrt{2^{2}+3^{2}}=\sqrt{13} .
\end{array}
$$ | \sqrt{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,654 |
Question 2 Given $a, b, c \geqslant 1$, and $a+b+c=9$. Prove:
$$
\sqrt{a}+\sqrt{b}+\sqrt{c} \geqslant \sqrt{b c+c a+a b} \text {, }
$$
with equality if and only if $a=b=c=3$. ${ }^{[1]}$
(The 3rd Chern Shiing-Shen Cup National High School Mathematics Olympiad)
This article provides another proof and relaxes the origina... | Assume without loss of generality that $c$ is the smallest. Then $\frac{1}{8} \leqslant c \leqslant 3$.
Thus, the original inequality is equivalent to
$$
\begin{array}{l}
(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2} \geqslant b c+c a+a b \\
\Leftrightarrow 2(\sqrt{b c}+\sqrt{c a}+\sqrt{a b})-a b+\left(c^{2}-9 c+9\right) \geqslant ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,655 |
5. The solution set of the equation $4 \sin x(1+\cos x)=3 \sqrt{3}$ is | 5. $x=2 k \pi+\frac{\pi}{3}(k \in \mathbf{Z})$.
Squaring both sides of the original equation, we get
$$
\begin{array}{l}
27=16 \sin ^{2} x(1+\cos x)^{2} \\
=16\left(1-\cos ^{2} x\right)\left(1+2 \cos x+\cos ^{2} x\right) \\
\Rightarrow 16 \cos ^{4} x+32 \cos ^{3} x-32 \cos x+11=0 \\
\Rightarrow(2 \cos x-1)^{2}\left(4 ... | x=2 k \pi+\frac{\pi}{3}(k \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,656 |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
a_{k}=\left[\frac{2014}{k}\right](k=1,2, \cdots, 100) \text {. }
$$
Then, among these 100 integers, the number of distinct integers is | 6.69.
When $01$, $[a+b]>[a]$.
$$
\begin{array}{l}
\text { Also } \frac{2014}{k}-\frac{2014}{k+1}=\frac{2014}{k(k+1)} \\
\Rightarrow \frac{2014}{k}=\frac{2014}{k+1}+\frac{2014}{k(k+1)} .
\end{array}
$$
When $k \leqslant 44$, $\frac{2014}{k(k+1)}>1$;
When $k \geqslant 45$, $\frac{2014}{k(k+1)}<1$.
Therefore, when $k \l... | 69 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,657 |
7. Let $A$ and $B$ be two different subsets of the set $\{a, b, c, d, e\}$, such that set $A$ is not a subset of set $B$, and $B$ is not a subset of set $A$. Then the number of different ordered pairs $(A, B)$ is | 7.570.
Notice that, the set $\{a, b, c, d, e\}$ has $2^{5}$ subsets, and the number of different ordered pairs $(A, B)$ is $2^{5}\left(2^{5}-1\right)$.
If $A \subset B$, and suppose the set $B$ contains $k(1 \leqslant k \leqslant 5)$ elements, then the number of ordered pairs $(A, B)$ satisfying $A \subset B$ is
$$
\... | 570 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,658 |
8. Function
$$
y=\frac{m x^{6}+n x^{5}+4 x^{4}-\left(m x^{2}+n x\right)\left(x^{2}-1\right)-4 x^{2}+4}{x^{6}+1}
$$
has a minimum value of 1 and a maximum value of 6. Then $m+n=$ | 8. $3+2 \sqrt{6}$.
Transform the given function to get
$$
\begin{array}{l}
y=\frac{m x^{2}+n x+4}{x^{2}+1} \\
\Rightarrow(m-y) x^{2}+n x+4-y=0 .
\end{array}
$$
Consider it as a quadratic equation in $x$, by $\Delta \geqslant 0$ we get
$$
\begin{array}{l}
n^{2}-4(m-y)(4-y) \geqslant 0 \\
\Rightarrow 4 y^{2}-(4 m+16) y... | 3+2 \sqrt{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,659 |
9. (16 points) Given the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{2 x^{3}}{x+1}+1, & x \in\left(\frac{1}{2}, 1\right] ; \\
-\frac{13}{3} x+\frac{7}{6}, & x \in\left[0, \frac{1}{2}\right] .
\end{array}\right.
$$
Let the range of the function $f(x)$ be $A$, and let real numbers $a, b, c \in A$. Prove: $a b c + 4 ... | 2,9. Notice that when $x \in\left(\frac{1}{2}, 1\right]$, we have
$$
f^{\prime}(x)=\frac{6 x^{2}(x+1)-2 x^{3}}{(x+1)^{2}}=\frac{4 x^{3}+6 x^{2}}{(x+1)^{2}}>0 \text {. }
$$
Thus, $f(x)$ is monotonically increasing on $x \in\left(\frac{1}{2}, 1\right]$.
Therefore, when $x \in\left(\frac{1}{2}, 1\right]$, $f(x) \in\left(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,660 |
10. (20 points) Given points $M(-1,0), N(1,0)$, the perimeter of $\triangle M N Q$ is 6, and the trajectory of the moving point $Q$ is the curve $C$. $P$ is any point on the circle $x^{2}+y^{2}=4$ (not on the x-axis), and $P A, P B$ are tangent to the curve $C$ at points $A, B$ respectively. Find the maximum value of t... | 10. According to the problem, the trajectory of the moving point $Q$ is an ellipse with foci at $M$ and $N$, and a major axis length of $2a = 4$ (excluding the intersection points with the $x$-axis).
Thus, the equation of the curve $C$ is
$$
\frac{x^{2}}{4}+\frac{y^{2}}{3}=1(y \neq 0).
$$
Let the points be $P\left(x_{... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,661 |
11. (20 points) Let $r$ be a positive integer, and the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=\frac{n a_{n}+2(n+1)^{2 r}}{n+2}(n=1,2, \cdots) \text {. }
$$
Prove: For any positive integer $n, a_{n}$ is always a positive integer. | 11. From the given, we have
$$
\begin{array}{l}
(n+2) a_{n+1}=n a_{n}+2(n+1)^{2 r} \\
\Rightarrow(n+1)(n+2) a_{n+1} \\
\quad=n(n+1) a_{n}+2(n+1)^{2 r+1} .
\end{array}
$$
Let $b_{n}=n(n+1) a_{n}$. Then
$$
b_{1}=2, b_{n+1}=b_{n}+2(n+1)^{2 r+1} \text {. }
$$
Thus, $b_{n}=b_{1}+\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,662 |
One. (40 points) As shown in Figure 1, in the isosceles right triangle $\triangle ABC$, $\angle C=90^{\circ}$, $M$ is the midpoint of the hypotenuse $AB$, $N$ is the midpoint of the leg $AC$, connect $BN$, and draw $CH \perp BN$, with the foot of the perpendicular being $H$. Prove: $CH^2 = AH \cdot MH$.
保留源文本的换行和格式，直接... | $$
\begin{array}{l}
\text { As shown in Figure 2, connect } M C. \text { Then } C M \perp M B. \\
\text { Since } C H \perp B N, \text { therefore, } \\
C, H, M, B \text { are concyclic. } \\
\text { Hence } \angle M H B = \angle M C B \\
= 45^{\circ}.
\end{array}
$$
$$
\begin{array}{l}
\text { Given } A N^{2} = C N^{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,663 |
II. (40 points) For a pile of 100 stones, perform the following operations: each time, arbitrarily divide any pile with more than 1 stone into two non-empty piles, until each pile has 1 stone (100 piles) in total. Prove:
(1) No matter how the operations are performed, there must be a moment when there are 20 piles, the... | (1) It is easy to see that after 19 steps, the stones are divided into 20 piles, and we denote the sum of the largest 20 piles after the $i$-th step as $S_{i}$.
Obviously, $S_{i}$ is monotonically non-increasing.
If there does not exist $S_{i}=60$, then $S_{i} \geqslant 61$, and $S_{i+1} \leqslant 59$.
Let the number o... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,664 |
Three. (50 points) Let $[x]$ denote the greatest integer not exceeding the real number $x$. Prove:
(1) The solution to the equation $[x]^{3}+x^{2}=x^{3}+[x]^{2}$ is an integer;
(2) The equation $\left[x^{3}\right]+x^{2}=x^{3}+\left[x^{2}\right]$ has a non-integer solution. | (1) Proof by contradiction.
If the equation has a non-integer solution $x$, let $[x]=A, \{x\}=B$. Then $A \in \mathbf{Z}, B \in (0,1)$.
Thus, the original equation becomes
$$
B^{2}+(3 A-1) B+3 A^{2}-2 A=0 \text{. }
$$
Then $\Delta=-3 A^{2}+2 A+1 \geqslant 0$.
Solving this, we get $-\frac{1}{3} \leqslant A \leqslant 1$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,665 |
Example 1 Let $0 \leqslant x, y, z \leqslant 1$. Prove:
$$
x y z+(1-x)(1-y)(1-z) \leqslant 1 \text {. }
$$
(2010, Slovenia Mathematical Olympiad) | Prove that let $x+y+z=3u$,
$$
\begin{array}{l}
xy+yz+zx=3v^2, \ xyz=w^3. \\
\text{Given } 0 \leqslant x, y, z \leqslant 1, \text{ we know } 0 \leqslant u \leqslant 1. \\
\text{Thus, } 0 \leqslant w \leqslant v \leqslant u \leqslant 1. \\
\text{Also, } (1-x)(1-y)(1-z) \\
=1-(x+y+z)+(xy+yz+zx)-xyz \\
=1-3u+3v^2-w^3,
\end... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,666 |
Four, (50 points) Find all positive integers $a, b$ such that the polynomial
$$
f(x)=\frac{x^{5}+a}{b}
$$ | Four, $f(n) \in \mathbf{Z}$ if and only if $b \mid\left(n^{5}+a\right)$.
Obviously, $b=1$ (for any $a$) satisfies the condition.
Assume $b>1$.
Suppose $f(n-1) 、 f(n) 、 f(n+1) \in \mathbf{Z}$.
Then $(n-1)^{5}+a 、 n^{5}+a 、(n+1)^{5}+a$ are all divisible by $b$.
These three numbers alternate in parity, hence $b$ is odd.
A... | b=1, a \text{ being any integer, or } b=11, a \text{ being any positive integer that is congruent to } \pm 1 \text{ modulo } b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,667 |
In acute $\triangle ABC$, it is known that $H$ is the orthocenter. Connecting $AH$, $BH$, $CH$, the circumcircle of $\triangle BHC$ is $\odot O$. The circle with $AH$ as diameter and the circle with $AB$ as diameter intersect $\odot O$ at points $D$ and $E$, respectively. Extending $AE$ intersects $HC$ at point $F$, an... | Prove: As shown in Figure 2, connect $B E$, extend $B O$, intersecting $\odot O$ at point $K$, and connect $K H, K C$.
Then $\angle B E K=90^{\circ}$.
Since $\angle A E B=90^{\circ}$, points $A, E, K$ are collinear.
Extend $A H$, intersecting $B C$ at point $Q$.
By $\angle B C K=\angle A Q C=90^{\circ} \Rightarrow A H ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,668 |
Find the smallest constant $\lambda$ such that
$$
a_{1} a_{2} \cdots a_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-n+\lambda\right) \leqslant \lambda \text {, }
$$
for any integer $n \geq 2$ and any sequence $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ satisfying
$$
\begin{array}{l}
a_{1}, a_{2}, \cd... | Solve for the required $\lambda_{\text {min }}=e$.
Define
$$
\begin{aligned}
E_{n}\left(a_{1}, a_{2}, \cdots, a_{n}\right) \\
=\frac{\lambda}{a_{1} a_{2} \cdots a_{n}}-\lambda+\left(n-\frac{1}{a_{1}}-\frac{1}{a_{2}}-\cdots-\frac{1}{a_{n}}\right) .
\end{aligned}
$$
First, prove: $\lambda \geqslant e$.
Notice that the o... | e | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,669 |
385 A certain school's 2014 graduates numbered 2014 students. The school's six leaders must sign each student's graduation album. It is known that each leader must and can only use one of the three designated colored pens, and the color choices for the six leaders signing 2014 albums can be represented as a $6 \times 2... | Let 1, 2, 3 represent three colors, and the colors used by six school leaders for signing in each memorial album be represented by a six-element ordered array
$$
\left(x_{1}, x_{2}, \cdots, x_{6}\right)\left(x_{i} \in\{1,2,3\}\right)
$$
This array is referred to as the sequence group $\left(x_{1}, x_{2}, \cdots, x_{6}... | 64 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,670 |
Given a sequence of positive numbers $\left\{a_{n}\right\}_{n \geqslant 1}$ that satisfies
$$
a_{1}=a, a_{2}=b, a_{n}=\frac{a_{n-2}}{a_{n-1}}(n=3,4, \cdots) \text {. }
$$
Find the general term formula for $a_{n}$. | Let the sequence $\left\{f_{n}\right\}_{n \geqslant 1}$ satisfy
$$
\begin{array}{l}
f_{1}=1, f_{2}=1, \\
f_{n}=f_{n-1}+f_{n-2}(n=3,4, \cdots) .
\end{array}
$$
Then the general term formula for the Fibonacci sequence $\left\{f_{n}\right\}$ is
$$
f_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,671 |
Example 2 Given that $a, b, c$ are positive real numbers, and $abc=1$. Prove:
$$
a+b+c \leqslant a^{2}+b^{2}+c^{2} \text {. }
$$ | Prove that let $a+b+c=3u$, $ab+bc+ca=3v^2$, $abc=w^3$, where $u \geqslant v \geqslant w=1$.
From $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$ $=9u^2-6v^2$,
thus, equation (1) $\Leftrightarrow 3u \leqslant 9u^2-6v^2 \Leftrightarrow u+2v^2 \leqslant 3u^2$.
By property 1, we get $u \geqslant v$. Then $u+2v^2 \leqslant u+2u^2$.
We ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,672 |
Example 3 Given that $a, b, c$ are positive real numbers. Prove:
$$
\begin{array}{l}
\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)} \\
\geqslant 1+\sqrt{1+\sqrt{\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)}} .
\end{array}
$$ | Prove that for $a+b+c=3 u, a b+b c+c a=3 v^{2}, a b c=w^{3}$, where $u \geqslant v \geqslant w>0$:
$$
\begin{array}{l}
\text { Since } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{b c+c a+a b}{a b c}=\frac{3 v^{2}}{w^{3}}, \\
\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \\
=\frac{(a b+b c+c a)^{2}-2 a b c(a+b+c)}{(a b ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,673 |
Example 4 Given that $a, b, c$ are positive real numbers, and $a+b+c=3$. Prove:
$$
a b c+\frac{12}{a b+b c+c a} \geqslant 5 .
$$ | Prove that if $a+b+c=3 u$, $ab+bc+ca=3 v^{2}, abc=w^{3}$, then $0<w \leqslant v \leqslant u=1$.
Thus, equation (1) $\Leftrightarrow w^{3}+\frac{4}{v^{2}} \geqslant 5 \Leftrightarrow v^{2} w^{3} \geqslant 5 v^{2}-4$.
By property 4, we know $w^{3} \geqslant 4 u v^{2}-3 u^{3}$. Therefore, it suffices to prove $\left(4 u v... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,674 |
Example 5 Given $a, b, c \geqslant 0$, and $a+b+c=1$. Prove:
$$
0 \leqslant a b+b c+c a-2 a b c \leqslant \frac{7}{27} \text {. }
$$ | Prove that if $a+b+c=3 u$, $a b+b c+c a=3 v^{2}, a b c=w^{3}$, then $0 \leqslant w \leqslant v \leqslant u=\frac{1}{3}$.
Thus, equation (1) $\Leftrightarrow 0 \leqslant 3 v^{2}-2 w^{3} \leqslant \frac{7}{27}$.
Since $3 u=1$, then
equation (2) $\Leftrightarrow 0 \leqslant 9 u v^{2}-2 w^{3} \leqslant \frac{7}{27}$.
By pr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,675 |
Example 1 Let positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1} x_{2} \cdots x_{n}=1$. Prove:
$$
\left(\sqrt{2}+x_{1}\right)\left(\sqrt{2}+x_{2}\right) \cdots\left(\sqrt{2}+x_{n}\right) \geqslant(\sqrt{2}+1)^{n} \text {. }
$$
(2014, Peking University and other universities' joint independent admission ... | Proof 1 Note that,
$$
\begin{array}{l}
\left(\sqrt{2}+x_{1}\right)\left(\sqrt{2}+x_{2}\right) \cdots\left(\sqrt{2}+x_{n}\right) \\
=(\sqrt{2})^{n}+(\sqrt{2})^{n-1} \sum_{i=1}^{n} x_{i}+(\sqrt{2})^{n-2} \sum_{10 \text {. }
\end{array}
$$
Therefore, $f(x)$ is a convex function.
By Jensen's inequality, we have
$$
\begin{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,676 |
Example 2 Given $n \in \mathbf{Z}_{+}, x \leqslant n$. Prove:
$$
n-n\left(1-\frac{x}{n}\right)^{n} \mathrm{e}^{x} \leqslant x^{2} \text {. }
$$
(2014, Tsinghua University and other schools' independent recruitment examination) | Prove that for $f(x)=\mathrm{e}^{x}-x-1$, then $f^{\prime}(x)=\mathrm{e}^{x}-1$.
When $x \geqslant 0$, $f^{\prime}(x) \geqslant 0$;
When $x-1)$, we get
$$
\begin{array}{l}
n\left[\left(1-\frac{x}{n}\right) \mathrm{e}^{\frac{x}{n}}\right]^{n} \\
\geqslant n\left[\left(1-\frac{x}{n}\right)\left(1+\frac{x}{n}\right)\right... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,677 |
Example 3 Real-coefficient polynomials $f_{i}(x)=a_{i} x^{2}+b_{i} x+c_{i}$ $\left(a_{i}>0, i=1,2, \cdots, 2011\right)$, and $\left\{a_{i}\right\} 、\left\{b_{i}\right\} 、\left\{c_{i}\right\}$ are all arithmetic sequences. If $F(x)=\sum_{i=1}^{204} f_{i}(x)$ has real roots, then at most how many polynomials in $\left\{f... | 【Analysis】From $\left\{a_{i}\right\}$ being an arithmetic sequence, we know
$$
\sum_{i=1}^{2011} a_{i}=(2 \times 1005+1) a_{1006}=2011 a_{1006} \text {. }
$$
Similarly, $\sum_{i=1}^{2011} b_{i}=2011 b_{1006}$,
$$
\sum_{i=1}^{2011} c_{i}=2011 c_{1006} \text {. }
$$
Therefore, $F(x)=\sum_{i=1}^{200} f_{i}(x)=2011 f_{10... | 1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,678 |
Example 3 Let $n$ be a positive integer no less than 2, and $a_{1}, a_{2}, \cdots, a_{n} \in (0,1]$. Prove:
$$
\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{a_{i}+1}\right) \geqslant n\left(\prod_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}+1}\right) .
$$
(2014, Nankai University Mathe... | Proof 1:
$$
\begin{array}{l}
a_{1}, a_{2}, \cdots, a_{n} \in(0,1] \\
\Rightarrow 00 \text {. } \\
\text { Hence }\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{a_{i}+1}-n\right)\left(\prod_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}+1}\right) \\
=\left(\sum_{i=1}^{n} a_{i}\right)\left(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,679 |
Example 4 Given $x, y \in\left(0, \frac{\pi}{2}\right)$, and $\tan x \cdot \tan y \geqslant 3$. Prove:
$$
\frac{\cos x+\cos y}{2} \geqslant \sqrt{\frac{\cos x \cdot \cos y}{\cos (x-y)}} .
$$
(2014, Nankai University Mathematics Experimental Class Admission Exam) | Let $\tan x=a>0, \tan y=b>0$. Then
$$
\begin{array}{l}
a b \geqslant 3, \cos x=\frac{1}{\sqrt{1+a^{2}}}, \cos y=\frac{1}{\sqrt{1+b^{2}}} . \\
\text { Also } \sqrt{\frac{\cos x \cdot \cos y}{\cos (x-y)}}=\sqrt{\frac{\cos x \cdot \cos y}{\cos x \cdot \cos y+\sin x \cdot \sin y}} \\
=\sqrt{\frac{1}{1+\tan x \cdot \tan y}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,680 |
Example 5 Let $f(x)$ be differentiable on $\mathbf{R}$, and for any $x_{0} \in \mathbf{R}$, it satisfies
$00)$.
Prove: (1) $f^{\prime}\left(x_{0}\right)<\frac{f\left(x+x_{0}\right)-f\left(x_{0}\right)}{x}$;
(2) If $|f(x)| \leqslant 1$, then $\left|f^{\prime}(x)\right| \leqslant 4$.
(2014, Excellence Alliance Independen... | Prove (1) Since $x>0$, for any $x_{0} \in \mathbf{R}$, we have $x+x_{0}>x_{0}$.
From $f^{\prime}\left(x+x_{0}\right)>f^{\prime}\left(x_{0}\right)$, we know that $f^{\prime}(x)$ is an increasing function on $\mathbf{R}$.
By the differentiability of $f(x)$ on $\mathbf{R}$ and the Lagrange Mean Value Theorem, we know th... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 727,681 |
In $\triangle ABC$, points $P, Q, R$ are on sides $BC, CA, AB$ respectively. Circles $\Gamma_{A}, \Gamma_{B}, \Gamma_{C}$ are the circumcircles of $\triangle AQR$, $\triangle BRP$, $\triangle CPQ$ respectively. Segment $AP$ intersects circles $\Gamma_{A}, \Gamma_{B}, \Gamma_{C}$ at points $X, Y, Z$ respectively. Prove ... | Prove as shown in Figure 1, draw $MN \parallel BC$, intersecting $AB$ and $AC$ at points $M$ and $N$ respectively.
Connect $XR$, $XQ$, $YM$, and $ZN$.
Since $\angle AMX = \angle ABP = \angle AYR$, points $X$, $M$, $R$, and $Y$ are concyclic.
Similarly, points $Z$, $Q$, $N$, and $X$ are concyclic.
Thus, $\angle XQN = \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,682 |
In $\triangle A B C$, it is known that $\angle B A C=40^{\circ}$, $\angle A B C=60^{\circ}$, and points $D$ and $E$ are on sides $A C$ and $A B$ respectively, such that $\angle C B D=40^{\circ}$, $\angle B C E=70^{\circ}$. Point $F$ is the intersection of $B D$ and $C E$, and $A F$ is connected. Prove: $A F \perp B C$.... | Proof As shown in Figure 1, let
$$
\angle B A F=\alpha \text {. }
$$
Then $\angle C A F=40^{\circ}-\alpha$.
From the given conditions, we have
$$
\begin{array}{l}
\angle A B D=20^{\circ}, \\
\angle A C E=10^{\circ}, \\
\angle C B D=40^{\circ}, \\
\angle B C E=70^{\circ} .
\end{array}
$$
By the trigonometric form of C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,683 |
Given positive numbers $a, b, c, d$ satisfying $a+b+c+d=4$.
Prove: $\frac{a}{a^{3}+8}+\frac{b}{b^{3}+8}+\frac{c}{c^{3}+8}+\frac{d}{d^{3}+8} \leqslant \frac{4}{9}$. $(2011$, Baltic Way Mathematical Contest) | Given $n(n \in \mathbf{Z}, n \geqslant 3)$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $a_{1}+a_{2}+\cdots+a_{n}=p$. Then
$$
\sum_{i=1}^{n} \frac{a_{i}}{a_{i}^{n-1}+n(n-2)} \leqslant \frac{n p}{(n-1)[p+n(n-2)]}.
$$
Proof: By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,684 |
1. The function
$$
y=4 \cos x+\cos 2 x(x \in \mathbf{R})
$$
has the range $\qquad$ | $-1 .[-3,5]$.
Let $t=\cos x \in[-1,1]$. Then $y=2 t^{2}+4 t-1=2(t+1)^{2}-3$
Thus, the range of the function is $[-3,5]$. | [-3,5] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,685 |
2. Let the complex number $z$ satisfy $|z|=1$. Then the maximum value of $\left|z^{2}-z+1\right|$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.3.
Solution 1 By the triangle inequality,
$$
\begin{array}{l}
\left|z^{2}-z+1\right|=\left|\left(z-\frac{1}{2}\right)^{2}+\frac{3}{4}\right| \\
\leqslant\left|z-\frac{1}{2}\right|^{2}+\frac{3}{4} \leqslant 3 .
\end{array}
$$
When $z=-1$, the above expression achieves its maximum value of 3.
Solution 2 Let $z=\cos \... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 727,686 |
3. As shown in Figure 1, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$ respectively. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
t... | 3. 2 .
Solution 1 Notice that,
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} .
$$
Since points $M, O, N$ are collinear, we have,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. }
$$
Solution 2 Since points $M, ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,687 |
4. If the inequality about $x$
$$
|x+a|<|x|+|x+1|
$$
has the solution set $\mathbf{R}$, then the range of values for $a$ is | 4. $00$.
It is easy to verify that when $0<a<1$, we always have
$|x+a|<\max \{|x|,|x+1|\}$.
In conclusion, the range of values for $a$ is $0<a<1$. | 0<a<1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,688 |
Example 4 In an arithmetic sequence $a_{1}, a_{2}, \cdots$ composed of positive integers, for any $n \in \mathbf{Z}_{+}$, we have $2005 \mid a_{n} a_{n+31}$. Prove: $2005 \mid a_{n}$. ${ }^{\text {[2] }}$
(31st Russian Mathematical Olympiad (4th Round)) | 【Analysis】Let the arithmetic sequence $\left\{a_{n}\right\} \subseteq \mathbf{Z}_{+}$ have a common difference of $d$.
Since $2005=5 \times 401$, the condition “for any $n \in \mathbf{Z}_{+}$, $2005 \mid a_{n} a_{n+31}$” is equivalent to “$5 \mid a_{n}$ or $5 \mid a_{n+31}$, and $401 \mid a_{n}$ or $401 \mid a_{n+31}$”... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,689 |
5. Let positive numbers $a$, $b$, $c$ satisfy $4a + b = abc$. Then the minimum value of $a + b + c$ is $\qquad$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5.6.
From the problem, we know that $c=\frac{1}{a}+\frac{4}{b}$.
Then $a+b+c=a+b+\frac{1}{a}+\frac{4}{b} \geqslant 6$.
When and only when $a=1, b=2$, $a+b+c$ takes the minimum value of 6. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,690 |
6. For any real number $x$, it always holds that
$$
\log _{a}(\sin x+\cos x)^{2} \geqslant-2 \text {. }
$$
Then the range of values for $a$ is $\qquad$ | $6.01$ when,
$$
\begin{array}{l}
{\left[\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\right]^{2} \geqslant \frac{1}{a^{2}}} \\
\Rightarrow a^{2} \geqslant \frac{1}{2 \sin ^{2}\left(x+\frac{\pi}{4}\right)} .
\end{array}
$$
Since $\frac{1}{2 \sin ^{2}\left(x+\frac{\pi}{4}\right)}$ has no upper bound, therefore, $a<1$. At ... | 0<a \leqslant \frac{\sqrt{2}}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,691 |
7. Given
$$
\begin{array}{l}
A \cup B \cup C=\{a, b, c, d, e\}, A \cap B=\{a, b, c\}, \\
c \in A \cap B \cap C .
\end{array}
$$
Then the number of sets $\{A, B, C\}$ that satisfy the above conditions is. $\qquad$ | 7. 100 .
As shown in Figure 3, the set $A \cup B \cup C$ can be divided into seven mutually exclusive regions, denoted as $1, 2, \cdots, 7$.
It is known that element $c$ is in region 7, elements $a, b$ can appear in each of the regions 4, 7, with 4 possibilities; elements $d, e$ can appear in each of the regions $1, ... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,692 |
8. Let the function $f(x)$ be defined on $\mathbf{R}$, for any $x \in \mathbf{R}$, $f(x+1006)=\frac{1}{2}+\sqrt{f(x)-f^{2}(x)}$,
and $f(-1005)=\frac{3}{4}$.
Then $f(2013)=$ | 8. $\frac{1}{2}+\frac{\sqrt{3}}{4}$.
From the problem, we know
$$
\begin{array}{l}
f(1)=\frac{1}{2}+\sqrt{f(-1005)-f^{2}(-1005)} \\
=\frac{1}{2}+\frac{\sqrt{3}}{4}, \\
f(1007)=\frac{1}{2}+\sqrt{f(1)-f^{2}(1)}=\frac{3}{4}, \\
f(2013)=f(1007+1006) \\
=\frac{1}{2}+\sqrt{\frac{3}{4}-\frac{9}{16}}=\frac{1}{2}+\frac{\sqrt{3... | \frac{1}{2}+\frac{\sqrt{3}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,693 |
9. Use five different colors to color the six vertices of the triangular prism $A B C-D E F$, requiring each point to be colored with one color, and the two endpoints of each edge to be colored with different colors. Then the number of different coloring methods is $\qquad$.
| 9. 1920.
Transform the adjacency relationship of the vertices of a triangular prism into the regional relationship as shown in Figure 4, then the coloring methods for vertices $A$, $B$, and $C$ are $\mathrm{A}_{5}^{3}$ kinds.
For each determined coloring scheme of points $A$, $B$, and $C$, if we temporarily do not co... | 1920 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,694 |
10. If real numbers $b, c$ satisfy $b^{2}+c^{2}=1$, and
$$
f(x)=a x+b \sin x+c \cos x
$$
has two perpendicular tangent lines on its graph, then the range of values for $a$ is $\qquad$. | 10. $\{0\}$.
From the problem, we have
$$
\begin{aligned}
f(x) & =a x+\sqrt{b^{2}+c^{2}} \sin (x+\varphi) \\
\therefore \quad & =a x+\sin (x+\varphi) \\
\Rightarrow & f^{\prime}(x)=a+\cos (x+\varphi)
\end{aligned}
$$
where, $\sin \varphi=\frac{c}{\sqrt{b^{2}+c^{2}}}, \cos \varphi=\frac{b}{\sqrt{b^{2}+c^{2}}}$.
If the... | \{0\} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,695 |
11. (15 points) As shown in Figure 2, in the rectangular prism $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, it is known that $A D=1, A B=2, A A_{1}=c$. If there exists a point $P$ on the diagonal $B D_{1}$ such that $P B_{1} \perp P C_{1}$, find the range of values for $c$. | II. 11. Taking $D$ as the origin, and $\overline{D A}, \overline{D C}, \overline{D D_{1}}$ as the positive directions of the $x, y, z$ axes, respectively, we establish a spatial rectangular coordinate system. Thus,
$$
B(1,2,0), B_{1}(1,2, c), C_{1}(0,2, c), D_{1}(0,0, c) \text {. }
$$
Let $\overrightarrow{D_{1} P}=\la... | 0<c \leqslant \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,696 |
12. (15 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram. | 12. Given that $\left|F_{1} F_{2}\right|=2$.
As shown in Figure 5, the inscribed quadrilateral $\square A B C D$ of the ellipse has a pair of opposite sides $B C$ and $A D$ passing through the foci $F_{1}$ and $F_{2}$, respectively.
Obviously, the center of symmetry of $\square A B C D$ is the origin.
Figure 5
Let th... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,697 |
13. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
S_{n}=1-a_{n}\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $c_{n}=\frac{1}{... | 13. (1) From the problem, we have
$$
S_{n+1}=1-a_{n+1} \text {. }
$$
Subtracting the given equation from the above equation, we get
$$
a_{n+1}=-a_{n+1}+a_{n} \Rightarrow a_{n+1}=\frac{1}{2} a_{n}\left(n \in \mathbf{Z}_{+}\right)
$$
Since $a_{1}=\frac{1}{2}$, we have
$$
a_{n}=\frac{1}{2} \times\left(\frac{1}{2}\right)... | P_{n}>2 n-\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,698 |
Example 5 Let the set $X=\{1,2, \cdots, 100\}, A$ be any 10-element subset of $X$. Prove: $X \backslash A$ must contain a 10-element arithmetic sequence. ${ }^{[3]}$
(24th Lithuanian National Team Selection Exam) | 【Analysis】Arrange the elements of set $X$ into the following $10 \times 10$ array:
\begin{tabular}{ccccc}
1 & 2 & 3 & $\cdots$ & 10 \\
11 & 12 & 13 & $\cdots$ & 20 \\
21 & 22 & 23 & $\cdots$ & 30 \\
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
91 & 92 & 93 & $\cdots$ & 100
\end{tabular}
Thus, the ten number... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,700 |
1. Lya has 1 cent and 5 cent coins totaling 13. If she had one more 5 cent coin, then the number of 5 cent and 1 cent coins would be the same. Therefore, Lya originally had ( ) cents in total.
(A) 33
(B) 35
(C) 37
(D) 39
(E) 41 | 1. C.
$$
\frac{13+1}{2}(5+1)-5=37
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,701 |
2. Simplify: $\frac{2^{3}+2^{3}}{2^{-3}+2^{-3}}=(\quad)$.
(A) 16
(B) 24
(C) 32
(D) 48
(E) 64 | 2. E.
$$
\frac{2^{3}+2^{3}}{2^{-3}+2^{-3}}=\frac{2^{3}}{2^{-3}}=2^{6}=64
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,702 |
3. Randy drove on a trip, knowing that the first $\frac{1}{3}$ of the journey was on a gravel road, the next 20 miles were on a sidewalk, and finally, the remaining $\frac{1}{5}$ of the journey was on a dirt road. Then Randy's total distance traveled on this trip was ( ) miles.
(A) 30
(B) $\frac{400}{11}$
(C) $\frac{75... | 3. E.
$$
20 \div\left(1-\frac{1}{3}-\frac{1}{5}\right)=\frac{300}{7}
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,703 |
4. Susie bought 4 muffins and 3 bananas, Calvin spent twice the amount to buy 2 muffins and 16 bananas. Then the unit price of a muffin is ( ) times the unit price of a banana.
(A) $\frac{3}{2}$
(B) $\frac{5}{3}$
(C) $\frac{7}{4}$
(D) 2
(E) $\frac{13}{4}$ | 4. B.
Let the price of a pancake and a banana be $x$ yuan per piece and $y$ yuan per one, respectively. Then $2(4 x+3 y)=2 x+16 y \Rightarrow x=\frac{5}{3} y$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,704 |
5. As shown in Figure 1, Doug built a square window, inlaid with eight rectangular glass panes of equal size, with the length to width ratio of each pane being $5: 2$. It is known that each glass pane has a 2-inch border around it. Then the side length of the square window is ( ).
(A) 26
(B) 28
(C) 30
(D) 32
(E) 34 | 5. A.
Let the length of the glass be $5 x$, and the width be $2 x$. Then
$$
\begin{array}{l}
5 x+5 x+2 \times 3=2 x+2 x+2 x+2 x+2 \times 5 \\
\Rightarrow x=2 .
\end{array}
$$
Therefore, the side length of the square window is $5 \times 2 \times 2+2 \times 3=26$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,705 |
6. Aorven brought enough money to buy 30 balloons to the store. Upon arriving, he found that the balloons were on sale: if the first balloon is bought at the original price, then the second balloon can be bought at a $\frac{1}{3}$ discount off the original price. How many balloons can Aorven buy at most?
(A) 33
(B) 34
... | 6. C.
Assume the original price of a balloon is $x$ dollars. Alvin spends $x+\left(1-\frac{1}{3}\right) x=\frac{5}{3} x$ dollars on two balloons. Therefore, he can buy at most $2 \times \frac{30 x}{\frac{5}{3} x}=36$ balloons. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,706 |
7. Let $A>B>0$, and $A$ is $x \%$ larger than $B$. Then $x=(\quad)$.
(A) $100\left(\frac{A-B}{B}\right)$
(B) $100\left(\frac{A+B}{B}\right)$
(C) $100\left(\frac{A+B}{A}\right)$
(D) $100\left(\frac{A-B}{A}\right)$
(E) $100 \frac{A}{B}$ | 7. A. From $A=B(1+x \%) \Rightarrow x=100\left(\frac{A}{B}-1\right)$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,707 |
8. Given that a truck travels $\frac{b}{6}$ feet in $t$ seconds, and 1 yard equals 3 feet. Then the truck can travel ( ) yards in 3 minutes.
(A) $\frac{b}{1080 t}$
(B) $\frac{30 t}{b}$
(C) $\frac{30 b}{t}$
(D) $\frac{10 t}{b}$
(E) $\frac{10 b}{t}$ | 8. E.
$$
\frac{\frac{b}{6 t} \times 3 \times 60}{3}=\frac{10 b}{t} .
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,708 |
9. Given $w, z \in \mathbf{R}$, and
$$
\frac{\frac{1}{w}+\frac{1}{z}}{\frac{1}{w}-\frac{1}{z}}=2014 \text {. }
$$
Then $\frac{w+z}{w-z}=(\quad)$.
(A) -2014
(B) $-\frac{1}{2014}$
(C) $\frac{1}{2014}$
(D) 1
(E) 2014 | 9. A.
$$
\frac{\frac{1}{w}+\frac{1}{z}}{\frac{1}{w}-\frac{1}{z}}=\frac{z+w}{z-w}=2014 \Rightarrow \frac{w+z}{w-z}=-2014 .
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,709 |
10. Let $A$, $B$, $C$, $D$ be four different natural numbers less than 10, and satisfy:
$$
\begin{array}{rrrr}
A B B C B \\
+B C A D A \\
\hline D & B & D & D
\end{array}
$$
Then $D$ has ( ) different values.
(A) 2
(B) 4
(C) 7
(D) 8
(E) 9 | 10. C.
From the problem, we know that $C=0, A+B=D$.
Since $A, B, D$ are different digits, and $A, B$ are the leading digits, therefore, $A, B$ are not 0.
Thus, $3=1+2 \leqslant A+B=D \leqslant 9$.
Therefore, $D=3,4,5,6,7,8,9$. There are seven different values in total. | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 727,710 |
Example 6 Let $n>6$, all positive integers not greater than $n$ and coprime with $n$
$$
1=a_{1}<a_{2}<\cdots<a_{\varphi(n)}=n-1
$$
form an arithmetic sequence exactly. Find $n$. | 【Analysis】When $n>6$, $\varphi(n) \geqslant 3$.
When the common difference $d=1$, all positive integers not greater than $n$ are coprime with $n$, i.e., $n$ is a prime number greater than 6.
When $d=2$, all positive odd numbers not greater than $n$, $1,3,5, \cdots$, $n-1$ are coprime with $n$.
Thus, $n=2^{\alpha}(\alp... | n \text{ is a prime number greater than 6 or an integer power of 2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,711 |
11. For consumers, a simple $n \%$ discount is more suitable than the following plans:
【Plan 1】Two consecutive $15 \%$ discounts;
【Plan 2】Three consecutive $10 \%$ discounts;
【Plan 3】A $25 \%$ discount followed by an additional $5 \%$ discount.
Then the smallest possible positive integer $n$ is ( ).
(A) 27
(B) 28
(C) 2... | 11. C.
Let the original price of the product be $x$ yuan. Then, the unit price after discount according to Plan 1 is
$$
(1-15 \%)^{2} x=0.7225 x \text {; }
$$
The unit price after discount according to Plan 2 is
$$
(1-10 \%)^{3} x=0.729 x \text {; }
$$
The unit price after discount according to Plan 3 is
$$
(1-25 \%... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,712 |
12. The largest factor of 2014000000 is the number itself, the fifth largest factor of this number is ( ).
(A) 125875000
(B) 201400000
(C) 251750000
(D) 402800000
(E) 503500000 | 12. C.
Studying the problem from the opposite side.
Since the factors of 2014000000 from smallest to largest are 1, 2, $4, 5, 8, \cdots$, the fifth factor is 8. Therefore, the fifth factor from largest to smallest is
$$
2014000000 \div 8 = 251750000.
$$ | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,713 |
13. As shown in Figure 2, given that a regular hexagon with a side length of 1 is surrounded by six congruent regular hexagons. Then the area of $\triangle A B C$ is ( ).
(A) $2 \sqrt{3}$
(B) $3 \sqrt{3}$
(C) $1+3 \sqrt{2}$
(D) $2+2 \sqrt{3}$
(E) $3+2 \sqrt{3}$ | 13. B.
By symmetry, we know that $\triangle ABC$ is an equilateral triangle. Therefore,
$$
S_{\triangle ABC}=\frac{\sqrt{3}}{4} AB^{2}=\frac{\sqrt{3}}{4}\left(4 \sin 60^{\circ}\right)^{2}=3 \sqrt{3} .
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,714 |
14. Danika drove her new car for an integer number of hours, with the car averaging 55 miles per hour. At the start, the odometer read $\overline{a b c}$, where $\overline{a b c}(a \geqslant 1, a+b+c \leqslant 7)$ is a three-digit number, and when she stopped, the odometer read $\overline{c b a}$. Then $a^{2}+b^{2}+c^{... | 14. D.
Let Danyika drove for $n\left(n \in \mathbf{Z}_{+}\right)$ hours.
From the problem, we have
$$
\begin{array}{l}
100 a+10 b+c+55 n=100 c+10 b+a \\
\Rightarrow 9(c-a)=5 n .
\end{array}
$$
Since $(5,9)=1$, it follows that, 5 divides $(c-a)$.
Thus, $(a, c ; n)$
$$
=(1,6,9),(2,7,9),(3,8,9),(4,9,9) \text {. }
$$
Al... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,715 |
15. As shown in Figure 3, in rectangle $A B C D$, it is known that $C D=2 B C$, and $E, F$ are on side $A B$, such that $D E, D F$ trisect $\angle A D C$. Then $\frac{S_{\triangle D E F}}{S_{\text {rectangle } A B C D}}=(\quad)$.
(A) $\frac{\sqrt{3}}{6}$
(B) $\frac{\sqrt{6}}{8}$
(C) $\frac{3 \sqrt{3}}{16}$
(D) $\frac{1... | 15. A.
From the problem, we know that $\angle A D C=90^{\circ}$, $\angle A D E=\angle E D F=\angle F D C=30^{\circ}$.
Thus, $A E=\frac{\sqrt{3}}{3} A D, A F=\sqrt{3} A D$.
Therefore, $E F=\frac{2 \sqrt{3}}{3} A D$.
Since $C D=2 A D$, we have
$$
\frac{S_{\triangle D E F}}{S_{\text {trapezoid } A B C D}}=\frac{\frac{1}{... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,716 |
16. Four fair cubic dice are rolled once. The probability that at least three of the four dice show the same number is ( ).
(A) $\frac{1}{36}$
(B) $\frac{7}{72}$
(C) $\frac{1}{9}$
(D) $\frac{5}{36}$
(E) $\frac{1}{6}$ | 16. B.
In two scenarios:
(1) If the numbers on the top faces of the four dice are all the same, then the probability is
$$
1 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{6^{3}} ;
$$
(2) If three of the four dice have the same number on the top face, then the probability is
$$
\mathrm{C}_{4}^{1} \t... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,717 |
17. If $2^{n} \|\left(10^{1002}-4^{501}\right)$, then $n=(\quad)$.
(A) 1002
(B) 1003
(C) 1004
(D) 1005
(E) 1006 | 17. B.
Notice,
$$
\begin{array}{l}
10^{1002}-4^{501}=2^{1002} \times 5^{1002}-2^{1002} \\
= 2^{1002} \times (5^{1002}-1) \\
= 2^{1002} \times (5^{501}-1)(5^{501}+1) \\
= 2^{1002} \times (5-1)(5^{500}+5^{499}+\cdots+1) \times \\
(5+1)(5^{500}-5^{499}+5^{498}-5^{497}+\cdots+1) \\
= 2^{1005} \times (5^{500}+5^{499}+\cdo... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,718 |
18. 11 positive integers are arranged in a sequence, with their mean, median, and mode being $10$, $9$, and $8$ respectively. The maximum possible value of the largest number in this sequence is ( ).
(A) 24
(B) 30
(C) 31
(D) 33
(E) 35 | 18. E.
Since 9 is the median, let the 11 positive integers in ascending order be
$a, b, c, d, e, 9, f, g, h, i, j$.
Since 8 is the mode, 8 must appear at least twice.
We will consider different cases.
(1) When 8 appears twice, let the 11 numbers be
$a, b, c, 8, 8, 9, f, g, h, i, j$.
To make the maximum number $j$ as l... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,719 |
19. Let circles $\Gamma_{1}$ and $\Gamma_{2}$ be two concentric circles with center $O$, and radii $2$ and $1$, respectively. Then the probability that a randomly chosen chord of the larger circle intersects the smaller circle is ( ).
(A) $\frac{1}{6}$
(B) $\frac{1}{4}$
(C) $\frac{2-\sqrt{2}}{2}$
(D) $\frac{1}{3}$
(E) ... | 19. D.
As shown in Figure 6, take any point $A$ on the circumference of the larger circle $\Gamma_{1}$. Draw two tangents $AB$ and $AC$ from $A$ to the smaller circle $\Gamma_{2}$, intersecting the circle $\Gamma_{1}$ at points $B$ and $C$, respectively, and connect $BC$. The chord $AA'$, which intersects the circle $... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,720 |
20. The number of integers that satisfy the inequality $x^{4}-51 x^{2}+50<0$ is ( ).
(A) 8
(B) 10
(C) 12
(D) 14
(E) 16 | 20. C.
Notice,
$$
\begin{array}{l}
x^{4}-51 x^{2}+50=\left(x^{2}-1\right)\left(x^{2}-50\right) \\
=(x-1)(x+1)(x-\sqrt{50})(x+\sqrt{50})<0 .
\end{array}
$$
Solving, we get
$$
\begin{array}{l}
x \in\{x \in \mathbf{Z} \mid x \in(-\sqrt{50},-1) \cup(1, \sqrt{50})\} \\
=\{ \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 7\},
\end{... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 727,721 |
Example 7 Determine the least possible value of the largest term in an arithmetic sequence composed of seven distinct primes. ${ }^{[4]}$
(2005, British Mathematical Olympiad) | 【Analysis】Let these seven different prime numbers be $p_{i}(i=1,2$, $\cdots, 7)$, and
$$
\begin{array}{l}
p_{2}=p_{1}+d, p_{3}=p_{1}+2 d, p_{4}=p_{1}+3 d, \\
p_{5}=p_{1}+4 d, p_{6}=p_{1}+5 d, p_{7}=p_{1}+6 d,
\end{array}
$$
where the common difference $d \in \mathbf{Z}_{+}$.
Since $p_{2}=p_{1}+d$ and $p_{3}=p_{1}+2 d$... | 907 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,722 |
21. In trapezoid $ABCD$, it is known that $AB \parallel CD, AB=33$, $CD=21$, the lengths of the two non-parallel sides are 10 and 14, and $\angle A, \angle B$ are both acute angles. Then the length of the shorter diagonal of the trapezoid is ( ).
(A) $10 \sqrt{6}$
(B) 25
(C) $8 \sqrt{10}$
(D) $18 \sqrt{2}$
(E) 26 | 21. B.
As shown in Figure 7, draw $DE \parallel BC$, intersecting $AB$ at point $E$.
$$
\begin{array}{l}
\text{Then } DE=10, \\
AE=12, \\
AD=14 .
\end{array}
$$
By the cosine rule,
$$
\cos A=\frac{14^{2}+12^{2}-10^{2}}{2 \times 14 \times 12}=\frac{5}{7} \text{. }
$$
In $\triangle ABD$,
$$
BD^{2}=AB^{2}+AD^{2}-2 AB \... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,723 |
23. As shown in Figure 5, a sphere is tangent to both the base and the side of a frustum, and the volume of the frustum is twice that of the sphere. Then the ratio of the radius of the lower base to the radius of the upper base of the frustum is ( ).
(A) $\frac{3}{2}$
(B) $\frac{1+\sqrt{5}}{2}$
(C) $\frac{3+\sqrt{5}}{2... | 23. C.
Consider the axial section as shown in Figure 9.
Let the radii of the lower and upper bases of the frustum be $R_{1}$ and $R_{2}$ respectively, with $R_{1} > R_{2}$, and the radius of the inscribed sphere be $r$.
Since the volume of the frustum is twice the volume of the sphere, we have:
$$
\begin{array}{l}
2 \... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,725 |
24. Arrange the numbers $1,2,3,4,5$ in a circle. If there exists an integer $n(1 \leqslant n \leqslant 15)$, such that the sum of any consecutive numbers in the arrangement is not equal to $n$, then the arrangement is called a "bad arrangement". For circular arrangements obtained by rotation or reflection, they are con... | 24. B.
The number of circular permutations of the numbers $1, 2, 3, 4, 5$ arranged on a circle is $4!=24$. Next, verify which of these are bad permutations.
If $n \in \{1,2,3,4,5\}$, then in any circular permutation, there is a number that equals $n$;
If $n=15$, then the sum of all numbers in any circular permutatio... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,726 |
25. On the surface of a small pond, there are 11 lotus leaves, marked as $0,1, \cdots, 10$. A frog is on the 1st lotus leaf, and the probability of it jumping to the $N(0<N<10)$-marked lotus leaf is equal to the sum of the probability of it being on the $N-1$-marked lotus leaf multiplied by $\frac{N}{10}$ and the proba... | 25. C.
Let the probability that the frog can escape when jumping from the 1st lily pad to the $k$th lily pad be $N_{k}(k=0,1, \cdots, 10)$.
By symmetry, when the frog jumps to the 5th lily pad, the probability of being eaten and the probability of escaping are both $\frac{1}{2}$.
From the problem, we have
$$
\left\{\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,727 |
1. Given sets $P=\{1,|a|\}, Q=\left\{2, b^{2}\right\}$ are subsets of the universal set $U=\left\{1,2,3, a^{2}+b^{2}+a+b\right\}$, and $\complement_{U}(P \cup Q)=\{6\}$. Which of the following conclusions is correct? ( ).
(A) $a=3, b=1$
(B) $a=3, b=-1$
(C) $a=-3, b=1$
(D) $a=-3, b=-1$ | $$
\begin{array}{l}
\text { I. 1. D. } \\
\text { From } \complement_{U}(P \cup Q)=\{6\} \Rightarrow a^{2}+b^{2}+a+b=6 .
\end{array}
$$
Obviously, only option D fits. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,728 |
2. Given complex numbers $z_{1}$ and $z_{2}$, and
$$
\left|z_{1}\right|=2\left|z_{2}\right|=2,\left|z_{1}+z_{2}\right|=\sqrt{7} \text {. }
$$
Then $\left|z_{1}-z_{2}\right|=(\quad)$.
(A) $\sqrt{5}$
(B) $\sqrt{7}$
(C) 3
(D) $\sqrt{3}$ | 2. D.
It is known that,
$$
\left|z_{1}-z_{2}\right|=\sqrt{2\left|z_{1}\right|^{2}+2\left|z_{2}\right|^{2}-\left|z_{1}+z_{1}\right|^{2}}=\sqrt{3} \text {. }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,729 |
3. Let $\angle A, \angle B, \angle C$ be the three interior angles of $\triangle ABC$. Proposition $P: \angle A=\angle B$;
Proposition $Q: \sin \angle A=\sin \angle B$.
Then $\neg P$ is $\neg Q$'s ( ) condition.
(A) sufficient but not necessary
(B) necessary but not sufficient
(C) sufficient and necessary
(D) neither s... | 3. C.
In $\triangle A B C$,
$$
\angle A \neq \angle B \Leftrightarrow \sin A \neq \sin B .
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,730 |
4. Given the geometric sequence $\left\{a_{n}\right\}: a_{1}=5, a_{4}=625$. Then $\sum_{k=1}^{2014} \frac{1}{\log _{5} a_{k} \cdot \log _{5} a_{k+1}}=(\quad)$.
(A) $\frac{2014}{2015}$
(B) $\frac{2013}{2014}$
(C) $\frac{2012}{4028}$
(D) $\frac{2013}{4030}$ | 4. A.
Let the common ratio of the geometric sequence be $q$. Then, according to the problem,
$$
\begin{array}{l}
625=5 q^{3} \Rightarrow q=5 \\
\Rightarrow \sum_{k=1}^{2014} \frac{1}{\log _{5} a_{k} \cdot \log _{5} a_{k+1}}=\sum_{k=1}^{2014} \frac{1}{k(k+1)}=\frac{2014}{2015} .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,731 |
5. Given $\odot O_{1}:(x+2)^{2}+(y-1)^{2}=1$ and $\odot O_{2}$ : $x^{2}+(y+1)^{2}=1$ are symmetric with respect to the line $l$. Then the equation of $l$ is ( ).
(A) $x+y+1=0$
(B) $x-y+1=0$
(C) $x-y-1=0$
(D) $x+y-1=0$ | 5. B.
Notice that, $o_{1}(-2,1), o_{2}(0,-1)$.
From the problem, we know that the line $l$ is the perpendicular bisector of the segment $O_{1} O_{2}$.
Therefore, the equation of the line $l$ is $x-y+1=0$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,732 |
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