problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
9.1. A simplest fraction is equal to the sum of two simplest fractions with denominators 600 and 700, respectively. Find the smallest possible value of the denominator of such a simplest fraction. | 9. 1. $2^{3} \times 3 \times 7=168$.
Let the two simplest fractions be $\frac{a}{600}$ and $\frac{b}{700}$. Then $(a, 6)=(b, 7)=1$.
Thus, the sum $\frac{7 a+6 b}{4200}$ has a numerator that is coprime with 6 and 7.
Since $4200=2^{3} \times 3 \times 7 \times 5^{2}$, after canceling out the common factors, the denominat... | 168 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,752 |
9. 2. Given that $B D$ is the angle bisector of $\triangle A B C$ (point $D$ lies on segment $A C$), and intersects the circumcircle $\Gamma$ of $\triangle A B C$ at points $B$ and $E$. The circle $\omega$ with diameter $D E$ intersects circle $\Gamma$ at points $E$ and $F$. Prove: The line symmetric to $B F$ with resp... | 9. 2. Suppose $A B<B C$. Let $M$ be the midpoint of $A C$, and the second intersection points of lines $B M$ and $F M$ with circle $\Gamma$ be $F^{\prime}$ and $B^{\prime}$, respectively.
Since $\angle A B E=\angle C B E$, and $E$ is the midpoint of arc $\overparen{A C}$, points $E$ and $M$ both lie on the perpendicul... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,753 |
Example 7 As shown in Figure 12, in trapezoid $A B C D$, $A D / /$ $B C$, squares
$A B G E$ and square
$D C H F$ are constructed
on the two legs
$A B$ and $C D$ respectively,
$A Q=Q D$,
$B R=R C$. Prove:
$E F \perp Q R$. | Analysis: The idea in Text [3] is quite ingenious, essentially still transforming the figure into a basic figure of a bundle of parallel lines, then stretching and compressing the upper and lower bases of the trapezoid to deform it into a triangle, thus proving the theorem.
Proof: Note that
$$
A Q=Q D, B R=R C, A D / /... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,754 |
9. 3. Given a positive integer $n(n>1)$. An integer $a>n^{2}$ satisfies that for every $i(i=1,2, \cdots, n)$, the set $\{a+1, a+2, \cdots, a+n\}$ contains a multiple of $n^{2}+i$. Prove: $a>n^{4}-n^{3}$. | 9.3. First note that the distance between any two of $a+i(i=1,2, \cdots, n)$ does not exceed $n-1$.
For each $i(1 \leqslant i \leqslant n)$, by the condition, there exists a positive integer $a_{i}$ such that
$$
a_{i}\left(n^{2}+i\right) \in\{a+1, a+2, \cdots, a+n\} .
$$
Obviously, $a_{1}>1$.
If $a_{1} \leqslant a_{2... | a>n^{4}-n^{3} | Number Theory | proof | Yes | Yes | cn_contest | false | 721,755 |
9.4. In a regular 100-gon, each vertex is covered by a cloth. It is known that exactly one of the cloths hides a coin. The following action is called an operation: choose any four cloths to check if there is a coin underneath, and after each operation, the cloths are returned to their original positions, while the coin... | 9.4.33 times.
Assume that a regular 100-gon is placed on a rotatable horizontal circular table, and the vector from the center to one vertex points due north. The initial positions of the vertices of the polygon are defined from due north in a counterclockwise direction as $0, 1, \cdots, 99$. Each operation and coin t... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 721,756 |
9. 5. Real numbers $a, b, c$ satisfy
$$
\sum(a+b)=a b c, \sum\left(a^{3}+b^{3}\right)=a^{3} b^{3} c^{3} \text {. }
$$
Prove: $a b c=0$. | 9. 5. First note that for any real numbers $x, y$, we have $x^{2}-x y+y^{2} \geqslant|x y|$.
The equality holds if and only if $x=y$.
If $a b c \neq 0$, then dividing the two expressions yields
$$
\begin{array}{l}
\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right) \\
=a^{2} b^{2} c^{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,757 |
9.6. Can all positive integers be colored using 2009 colors such that:
(1) Each color is used for infinitely many numbers;
(2) There do not exist three pairwise distinct positive integers \(a\), \(b\), and \(c\) of different colors such that \(a = bc\)? | 9.6. Can.
Take 2008 prime numbers $p_{1}, p_{2}, \cdots, p_{2008}$, and
$$
p_{1}<p_{2}<\cdots<p_{2008} \text {. }
$$
Construct subsets $A_{1}, A_{2}, \cdots, A_{2009}$ of the set of positive integers as follows: $A_{1}$ consists of all multiples of $p_{1}$; $A_{2}$ consists of all multiples of $p_{2}$ that are not mu... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,758 |
9.7. A diagonal of eight squares on a chessboard is called a "fence". A rook starts from a square outside the fence on the chessboard and moves, satisfying the following conditions:
(1) It stays on any square of the chessboard at most once;
(2) It never stays on a square of the fence.
Find the maximum number of times ... | 9.7.47 times.
Let the square at the $i$-th row and $j$-th column be denoted as $(i, j)$. Suppose the eight squares occupied by the fence are $(i, i) (i=1,2, \cdots, 8)$. The non-fence squares are divided into four categories $A, B, C, D$:
$$
\begin{aligned}
A= & \{(i, j) \mid 2 \leqslant j+1 \leqslant i \leqslant 4\} ... | 47 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,759 |
9.8. Given two triangles $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ with equal areas, determine whether it is always possible to construct, using a compass and straightedge, a triangle $\triangle A_{2} B_{2} C_{2}$ congruent to $\triangle A_{1} B_{1} C_{1}$, such that $A A_{2} / / B B_{2} / / C C_{2}$ | 9.8. Always possible.
If $\triangle A B C \cong \triangle A_{1} B_{1} C_{1}$, then translating $\triangle A B C$ is sufficient. Now assume they are not congruent, and without loss of generality, let $A B < A_{1} B_{1}$.
Construct $\triangle A^{\prime} B^{\prime} C$ such that $A^{\prime} B^{\prime}$ does not intersect ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,760 |
1. Given
$$
(\sqrt{3} x+\sqrt{2})^{5}=a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0} \text {. }
$$
then $\left(a_{5}+a_{3}+a_{1}\right)^{2}-\left(a_{4}+a_{2}+a_{0}\right)^{2}$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1 | 1. D.
Let $x=1$, we get
$$
a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=(\sqrt{3}+\sqrt{2})^{5} \text {; }
$$
Let $x=-1$, we get
$$
\begin{array}{l}
-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0}=(-\sqrt{3}+\sqrt{2})^{5}, \\
a_{5}-a_{4}+a_{3}-a_{2}+a_{1}-a_{0}=(\sqrt{3}-\sqrt{2})^{5} . \text { (2) }
\end{array}
$$
From equations (1) a... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,761 |
2. As shown in Figure 1, given $A(-3,0)$, $B(0,-4)$, and $P$ is any point on the hyperbola $y=\frac{12}{x} (x > 0)$. A perpendicular line from point $P$ to the x-axis meets at point $C$, and a perpendicular line from point $P$ to the y-axis meets at point $D$. Then the minimum value of the area $S$ of quadrilateral $A ... | 2. C.
Let $P\left(x, \frac{12}{x}\right)$. Then $C(x, 0)$ and $D\left(0, \frac{12}{x}\right)$.
It is easy to see that $C A=x+3, D B=\frac{12}{x}+4$, so
$$
S=\frac{1}{2} C A \cdot D B=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \text {. }
$$
Simplifying, we get $S=2\left(x+\frac{9}{x}\right)+12$.
Since $x>0, \frac{9}{... | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,762 |
3. Given that $b$ is a positive number, $a$ is the fractional part of $b$, and $a^{2}+b^{2}=27$. Then the value of $a b$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3. A.
Let $b=a+m\left(m \in \mathbf{N}_{+}, 0m^{2} \Rightarrow m4 .
\end{array}
$
Therefore, $4<m<3 \sqrt{3} \Rightarrow m=5$.
Thus, $5^{2}+10 a+2 a^{2}=27$.
Solving for $a$, we get $a=\frac{-5+\sqrt{29}}{2}$ (negative value discarded).
Hence, $a+b=a+a+5=2 a+5=\sqrt{29}$.
Therefore, $a b=\frac{1}{2}\left[(a+b)^{2}-\l... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,763 |
4. Given $a^{3}+b^{3}+3 a b=1$. Then the value of $a+b$ is ( ).
(A) 1
(B) -2
(C) 1 or -2
(D) None of the above | 4. C .
Let $a+b=x$. Then
$$
\begin{array}{l}
(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b) \\
\Rightarrow x^{3}-3 a b x+3 a b=1 \\
\Rightarrow x^{3}-1-3 a b(x-1)=0 \\
\Rightarrow(x-1)\left(x^{2}+x+1-3 a b\right)=0
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow x=1 \text { or } x^{2}+x+1=3 a b \\
\Rightarrow a+b=1 \text { or } x^{2... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,764 |
1. Given circles $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, with radii $5$ and $7$ respectively, and $O_{1} O_{2}=6$. A line through point $A$ intersects $\odot O_{1}$ and $\odot O_{2}$ at points $C$ and $D$, respectively. Points $P$ and $O$ are the midpoints of segments $CD$ and $O_{1} O_{2}$, re... | (Draw diameter $A E$ and $A F$ of $\odot O_{1}$ and $\odot O_{2}$ through point $A$, and extend $A O$ to intersect $E F$ at point $G$. Then $E, B, F$ are collinear, and $E F \parallel O_{1} O_{2}$. It is easy to see that $P G$ is the midline of the right trapezoid $C E F D$. $O P=\frac{1}{2} A G=A O=2 \sqrt{7}$. ) | 2\sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,765 |
5. Let the three sides of $\triangle A B C$ be $B C=a, C A=b$, $A B=c$. If $\tan \frac{A}{2}=\frac{a}{b+c}, \tan \frac{B}{2}=\frac{b}{b+c}$, then $\triangle A B C$ is ( ).
(A) Isosceles triangle
(B) Right triangle
(C) Isosceles triangle or right triangle
(D) Isosceles right triangle | 5. C.
Let $p=\frac{1}{2}(a+b+c)$, and the inradius of $\triangle ABC$ be $r$. Then
$$
\begin{array}{l}
\tan \frac{A}{2}=\frac{r}{p-a}=\frac{a}{b+c}, \\
\tan \frac{B}{2}=\frac{r}{p-b}=\frac{b}{b+c} .
\end{array}
$$
Dividing (1) by (2) gives
$$
\begin{array}{l}
\frac{p-b}{p-a}=\frac{a}{b+c} \cdot \frac{a+c}{b} \\
\Right... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,766 |
6. For three numbers $a, b, c$, use $\min \{a, b, c\}$ to represent the smallest of these three numbers, for example,
$$
\begin{array}{l}
\min \{-1,2,3\}=-1, \\
\min \{-1,2, a\}=\left\{\begin{array}{cc}
a, & a \leqslant-1 ; \\
-1, & a>-1 .
\end{array}\right.
\end{array}
$$
Then, the maximum value of $\min \left\{x+1,(... | 6. D.
Draw the graphs of $y=x+1$, $y=(x-1)^{2}$, and $y=2-x$ as shown in Figure 4.
From the graphs, we know that the maximum value of $\min \left\{x+1,(x-1)^{2}, 2-x\right\}$ is 1. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,767 |
1. Let $a<b, f(x)=|x-a|-|x-b|$. Then the range of $f(x)$ is $\qquad$ | 1. $a-b \leqslant f(x) \leqslant b-a$.
If $x \leqslant a$, then
$$
f(x)=a-x-(b-x)=a-b \text{; }
$$
If $a<x<b$, then
$$
f(x)=x-a-(b-x)=-a-b+2 x \text{, }
$$
thus $a-b<f(x)<b-a$;
If $x \geqslant b$, then
$$
f(x)=x-a-(x-b)=b-a \text{. }
$$
In summary, $a-b \leqslant f(x) \leqslant b-a$. | a-b \leqslant f(x) \leqslant b-a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,768 |
2. As shown in Figure 2, the first polygon is "expanded" from an equilateral triangle, with the number of sides denoted as $a_{3}$, the second polygon is "expanded" from a square, with the number of sides denoted as $a_{4}, \cdots \cdots$ and so on. The polygon "expanded" from a regular $n(n \geqslant 3)$-sided polygon... | 2. 2009 .
From the extended definition, we know
$$
\begin{array}{l}
a_{3}=12=3 \times 4, \\
a_{4}=20=4 \times 5, \\
\cdots \cdots \\
a_{n}=n(n+1) . \\
\text { Also, } \frac{1}{a_{n}}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \\
\Rightarrow \frac{1}{a_{3}}+\frac{1}{a_{4}}+\cdots+\frac{1}{a_{n}} \\
=\left(\frac{1}{3}-\... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,769 |
4. Given $41^{x}=2009,7^{y}=2009$. Then the value of $\frac{1}{x}+\frac{2}{y}$ is . $\qquad$ | 4. 1 .
From $41^{x}=2009$, we get $41^{x y}=2009^{y}$.
$$
\begin{array}{l}
\text { Also } 7^{y}=2009 \Rightarrow 7^{2 y}=2009^{2} \\
\Rightarrow 49^{y}=2009^{2} \Rightarrow 49^{x y}=2009^{2 x} .
\end{array}
$$
$$
\text { Also } 41 \times 49=2009 \Rightarrow 41^{x y} \times 49^{x y}=2009^{x y} \text {. }
$$
Therefore ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,771 |
$$
\begin{array}{l}
\text { I. (20 points) If } \\
\frac{\left(1-b^{2}\right)\left(1-c^{2}\right)}{b c}+\frac{\left(1-c^{2}\right)\left(1-a^{2}\right)}{c a}+ \\
\frac{\left(1-a^{2}\right)\left(1-b^{2}\right)}{a b}=4,
\end{array}
$$
Prove: $a+b+c=a b c$ or $a b+b c+c a=0$. | $$
\begin{array}{l}
\frac{1-b^{2}-c^{2}+b^{2} c^{2}}{b c}+\frac{1-c^{2}-a^{2}+c^{2} a^{2}}{c a}+ \\
\frac{1-a^{2}-b^{2}+a^{2} b^{2}}{a b} \\
=\frac{1}{b c}-\frac{b}{c}-\frac{c}{b}+b c+\frac{1}{a c}-\frac{c}{a}-\frac{a}{c}+a c+ \\
\frac{1}{a b}-\frac{a}{b}-\frac{b}{a}+a b \\
=\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,772 |
II. (25 points) Given that $\odot O$ is the incircle of $\triangle A B C$, touching the sides $A B$, $B C$, and $C A$ at points $E$, $F$, and $D$ respectively. Prove:
$\triangle A B C$ is a right triangle $\Leftrightarrow S_{\triangle A B C}=A D \cdot B D$. | II. Sufficiency.
Let $A D=A F=x, B D=B E=y, C E=C F=$ $z$, and let $p=\frac{1}{2}(a+b+c)=x+y+z$. Then
$$
\begin{array}{l}
a=y+z, b=z+x, c=x+y, \\
p-a=x, p-b=y, p-c=z .
\end{array}
$$
By Heron's formula, we have
$$
\begin{array}{l}
S_{\triangle A B C}=\sqrt{p(p-a)(p-b)(p-c)} \\
=\sqrt{x y z(x+y+z)} .
\end{array}
$$
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,773 |
Three. (25 points) Does there exist a right triangle with a hypotenuse of $\sqrt{2009}$ and two legs of integer lengths? If it exists, find the lengths of the two legs; if not, explain why.
---
Translated as requested, maintaining the original format and line breaks. | Three, assuming there exists such a right-angled triangle. Let the lengths of the two legs be \( x \) and \( y \). Then \( x^2 + y^2 = 2009 \).
Since 2009 is an odd number, \( x \) and \( y \) must be one odd and one even. Without loss of generality, let \( x = 2k \) and \( y = 2l + 1 \). Then
\[ 4k^2 + 4l(l+1) = 2008... | 28, 35 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,774 |
1. The solution set of the inequality
$$
(x+1)^{3}\left(x^{3}+5 x\right)<10(x+1)^{2}+8
$$
is $\qquad$ . | -1. $(-2,1)$.
Let $f(x)=x^{3}+5 x$. It is easy to know that $f(x)$ is an increasing function on $\mathbf{R}$.
When $x>-1$, the original inequality is
$$
\begin{array}{l}
\left(\frac{2}{x+1}\right)^{3}+\frac{10}{x+1}>x^{3}+5 x \\
\Rightarrow f\left(\frac{2}{x+1}\right)>f(x) \Rightarrow \frac{2}{x+1}>x \\
\Rightarrow-1<x... | (-2,1) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,775 |
2. Given a line segment $AB$ as the diameter, construct a semicircle on one side, and on the other side, construct a rectangle $ABCD$ such that $AB = \sqrt{2} AD$. Let $P$ be any point on the semicircle, and let $PC$ and $PD$ intersect $AB$ at points $E$ and $F$ respectively. Prove that:
$$
AE^2 + BF^2 = AB^2.
$$ | ( Hint: Connect $P A$ and $P B$ intersecting line $C D$ at points $M$ and $N$. Then $M C^{2}+N D^{2}=M N^{2}$, and $A B / / C D$. Therefore, $A E^{2}+B F^{2}=A B^{2}$. ) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,776 |
2. Given a positive integer $k(1 \leqslant k \leqslant 9)$. Let $\underset{n \uparrow \cdots k}{k k \cdots}$ denote the $n$-digit positive integer in decimal notation where each digit is $k$. If for any positive integer $n$, the function $f(x)$ satisfies
$$
f(\underbrace{\overline{k k \cdots k}}_{n \uparrow})=\underbra... | 2. $f(x)=\frac{9}{k} x^{2}+2 x$.
Notice that
$$
\begin{array}{l}
\underbrace{k k \cdots k}_{n \uparrow}=k\left(10^{n-1}+10^{n-2}+\cdots+10+1\right) \\
=\frac{k}{9}\left(10^{n}-1\right)
\end{array}
$$
Let $x=\frac{k}{9}\left(10^{n}-1\right)$, i.e., $10^{n}=\frac{9}{k} x+1$.
$$
\begin{array}{l}
\text { Hence } \frac{\o... | f(x)=\frac{9}{k} x^{2}+2 x | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,777 |
3. Among the natural numbers from 1 to 144, the number of ways to pick three numbers that form an increasing geometric progression with an integer common ratio is $\qquad$ . | 3. 78.
Let the three numbers $a$, $a q$, and $a q^{2}$ form an increasing geometric sequence. Then
$$
1 \leqslant a < a q < a q^{2} \leqslant 144 \text{.}
$$
From this, we have $2 \leqslant q \leqslant 12$.
When $q$ is fixed, the number of integers $a$ such that the three numbers $a$, $a q$, and $a q^{2}$ are integer... | 78 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,778 |
4. If 4 dice are thrown at random, the probability that the sum of the numbers on two of them is 9 is $\qquad$ . | 4. $\frac{145}{324}$.
Let $A_{i}=\{4$ dice do not show $i\}(i=1,2, \cdots, 6)$.
Then $\left|\overline{A_{3}} \cap \overline{A_{6}}\right|=\left|\overline{A_{3} \cup A_{6}}\right|=6^{4}-\left|A_{3} \cup A_{6}\right|$
$$
\begin{array}{l}
=6^{4}-\left(\left|A_{3}\right|+\left|A_{6}\right|-\left|A_{3} \cap A_{6}\right|\ri... | \frac{145}{324} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,779 |
5. Let $A$ and $B$ be moving points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $\overrightarrow{O A} \cdot \overrightarrow{O B}=0$, where $O$ is the origin. Then the distance from $O$ to the line $A B$ is $\qquad$. | 5. $\frac{a b}{\sqrt{a^{2}+b^{2}}}$.
Let $A\left(x_{1}, y_{1}\right) 、 B\left(-k y_{1}, k x_{1}\right)$, and the distance from point $O$ to $A B$ be $d$. Then
$$
|O A|^{2}=x_{1}^{2}+y_{1}^{2},|O B|^{2}=k^{2}|O A|^{2} .
$$
From $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1, \frac{y_{1}^{2}}{a^{2}}+\frac{x_{1}^{2}... | \frac{a b}{\sqrt{a^{2}+b^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,780 |
6. Let $z \in \mathbf{C}, f(z)=a z^{2}+b z+c$ be a polynomial with complex coefficients, and suppose that when $|z| \leqslant 1$, $|f(z)| \leqslant 1$. Then, when $|z| \leqslant 1$, the range of $|a z+b|$ is $\qquad$ | $$
\text { 6. } 0 \leqslant|a z+b| \leqslant 2 \text {. }
$$
When $|z| \leqslant 1$, because
$$
\left|\frac{z+1}{2}\right| \leqslant 1, \left|\frac{z-1}{2}\right| \leqslant 1 \text {, }
$$
so, $\left|f\left(\frac{z+1}{2}\right)\right| \leqslant 1, \left|f\left(\frac{z-1}{2}\right)\right| \leqslant 1$.
Therefore, $|a ... | 0 \leqslant|a z+b| \leqslant 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,781 |
8. If for any positive numbers $x, y$, it can be made that $a=x+y$, $b=\sqrt{x^{2}+x y+y^{2}}$, $c=m \sqrt{x y}$ are the three sides of a triangle, then the range of the positive number $m$ is $\qquad$ | 8. $2-\sqrt{3}b$.
$a, b, c$ are the three sides of a triangle
$$
\begin{aligned}
\Leftrightarrow & a-b<c<a+b \\
\Leftrightarrow & x+y-\sqrt{x^{2}+x y+y^{2}}<m \sqrt{x y} \\
& <x+y+\sqrt{x^{2}+x y+y^{2}}
\end{aligned}
$$
$$
\Leftrightarrow \frac{x+y}{\sqrt{x y}}-\frac{\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}}<m<\frac{x+y}... | 2-\sqrt{3} < m < 2+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,783 |
9. (14 points) Let positive real numbers $x, y, z$ satisfy $xyz=1$. Try to find the maximum value of
$$
f(x, y, z)=(1-yz+z)(1-xz+x)(1-xy+y)
$$
and the values of $x, y, z$ at that time. | $$
\begin{array}{l}
\left\{\begin{array} { l }
{ 1 - y z + z < 0 , } \\
{ 1 - x z + x < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+x z<1, \\
y+x y<1
\end{array}\right.\right. \\
\Rightarrow(x+x z)(y+x y)<1 \\
\Leftrightarrow x+x y+x^{2} y<0,
\end{array}
$$
Contradiction. Therefore, among $1-y z+z$, $1-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,784 |
10. (15 points) Given that $\left\{a_{n}\right\}$ is a sequence of non-negative integers, satisfying
$$
a_{1}=0, a_{2}=3 \text {, }
$$
and $a_{n+1} a_{n}=\left(a_{n-1}+2\right)\left(a_{n-2}+2\right)(n \geqslant 3)$.
If $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, prove that: $S_{n}... | 10. When $n=3$, since $a_{4} a_{3}=10$, thus, $a_{3}=1,2,5,10$;
When $n=4$, $a_{5} a_{4}=5\left(a_{3}+2\right)$, if $a_{3}=1, 5$, then $a_{4}=10,2$, but $a_{5} \notin \mathbf{N}$;
When $n=5$, $a_{6} a_{5}=\left(a_{4}+2\right)\left(a_{3}+2\right)$, if $a_{3}=10$, then $a_{4}=1, a_{5}=60$, but $a_{6} \notin \mathbf{N}$.... | S_{n} \leqslant \frac{n(n+1)}{2}, \text{ equality holds when } n \text{ is even} | Algebra | proof | Yes | Yes | cn_contest | false | 721,785 |
11. (15 points) Given that skew lines $a$ and $b$ form a $60^{\circ}$ angle, their common perpendicular segment is $E F$, and $|E F|=2$. The length of segment $A B$ is 4, with endpoints $A$ and $B$ moving on $a$ and $b$ respectively. Find the equation of the trajectory of the midpoint $P$ of segment $A B$. | 11. Obviously, point $P$ lies in the plane $\alpha$ passing through the midpoint $O$ of $E F$ and parallel to $a$ and $b$.
As shown in Figure 2, let the projections of $a$ and $b$ in $\alpha$ be $a^{\prime}$ and $b^{\prime}$, and the projections of $A$ and $B$ in $\alpha$ be $A^{\prime}$ and $B^{\prime}$, respectively.... | \frac{x^{2}}{9}+y^{2}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,786 |
Example 1 As shown in Figure 1, in $\square A B C D$, $P$ is the midpoint of side $B C$. A line parallel to $B D$ is drawn through $P$ and intersects $C D$ at point $Q$. Connect $P A$, $P D$, $Q A$, and $Q B$. In Figure 1, apart from $\triangle A B P$, there are ( ) triangles with the same area as $\triangle A B P$.
(A... | Explanation: As shown in Figure 1, from the given conditions, we know that $Q$ is the midpoint of $C D$. Therefore, in Figure 1, the triangles that have the same area as $\triangle A B P$ are:
(1) Two triangles that are equal in base and height to $\triangle A B P$ $(\triangle B P D, \triangle P C D)$;
(2) Two triangle... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,787 |
Example 2 As shown in Figure 2, given the side lengths of $\triangle D E F$ are $1, \sqrt{3}, 2$, the regular hexagonal grid is composed of 24 equilateral triangles with side length 2. Using the vertices of these equilateral triangles to draw $\triangle A B C$, such that $\triangle A B C \backsim \triangle D E F$. If t... | Explanation: By drawing, we can create a triangle similar to $\triangle D E F$. Triangles with different similarity ratios are shown in Figure 3 as three types.
Therefore, there are 3 different values of $k$. Hence, the answer is (C). | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,788 |
2. As shown in Figure 7, the shaded part is composed of three small squares on the grid paper, and such a pattern is called an "L-shape". Therefore, on a grid paper composed of $4 \times 5$ small squares, the number of different L-shape patterns that can be drawn is ( ) .
(A) 16
(B) 32
(C) 48
(D) 64
(2006, National Jun... | (提示: each $2 \times 2$ small square L-shaped pattern has four different ways to be drawn, and there are 12 different positions for the $2 \times 2$ small square patterns, so the number of L-shaped patterns in different positions is $12 \times 4=48$.) | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 721,789 |
$$
\begin{array}{l}
A=\left\{x^{2}, 1+2 z^{2}, x y+y z+z x\right\}, \\
B=\left\{1, y^{2}, 2+3 z^{2}\right\},
\end{array}
$$
satisfy $A=B$, and $x, y, z$ are real numbers. Find the values of $x, y, z$. | Solution: Discuss the problem by cases.
(1) When $1+2 z^{2}=1$, $z=0$, at this time, $A=\left\{x^{2}, 1, x y\right\}, B=\left\{1, y^{2}, 2\right\}$.
It is easy to solve that $x= \pm \sqrt{2}, y=0$ satisfies the requirements of the problem.
(2) When $x^{2}=1$, i.e., $x= \pm 1$,
$$
\begin{array}{l}
A=\left\{1,1+2 z^{2}, ... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,790 |
3. The line $l: y=p x$ (where $p$ is an integer not equal to 0)
intersects the line $y=x+10$ at exactly an integer point. Then, the number of lines $l$ that satisfy this condition is ( ).
(A) 6
(B) 7
(C) 8
(D) infinitely many
(2007, National Junior High School Mathematics Competition, Zhejiang Province Preliminary) | (Solve the system of equations $y=p x$ and $y=x+10$ to get $x=\frac{10}{p-1}$. For $x$ to be an integer, $p-1= \pm 1, \pm 2$, $\pm 5, \pm 10$. Therefore, $p=2,3,-1,6,-4,11,-9$. Hence, there are 7 such lines.) | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,791 |
4. As shown in Figure 8, point $A$ is a grid point (vertex of a small square) in a $5 \times 5$ grid. Each small square in the grid has a side length of 1. How many grid-point isosceles right triangles (with all three vertices being grid points) have an area of $\frac{5}{2}$ and one vertex at point $A$?
(A) 10
(B) 12
(... | (提示: The area of an isosceles right triangle is $\frac{5}{2}$, then the two legs are $\sqrt{5}$.
$$
\text { Point } A \text { is }
$$
the right-angle vertex
(as shown in Figure 9
(a), or
point $A$ is the base
vertex (as shown in
Figure 9 (b)), there are 8 right triangles in each case,
a total of 16.) | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,792 |
5. The number of integer points within the square (including the four sides) formed by the four lines $y=x+10, y=-x+10, y=$ $x-10, y=-x-10$ in the Cartesian coordinate system is $\qquad$ . | (Tip: By plotting, it is known that the number of integer points in each quadrant is 45. The number of integer points on the coordinate axes is 41, so, the number of integer points is $(4 \times 45+41=) 221$. | 221 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,793 |
Example 1 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \leqslant \frac{3 \sqrt{2}}{2} .
$$ | Explanation: Directly using the condition $a b c=1$ to reduce it to a one-variable problem is difficult. However, we can first treat $a b$ as a whole to reduce the three-variable problem to a two-variable one, and then use inequalities to further reduce it to a one-variable problem.
Assume $a \leqslant b \leqslant c$,... | \frac{3 \sqrt{2}}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,794 |
Example 2 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\frac{ab}{a^{5}+b^{5}+ab}+\frac{bc}{b^{5}+c^{5}+bc}+\frac{ca}{c^{5}+a^{5}+ca} \leqslant 1 .
$$ | Notice that
$$
\begin{array}{l}
a^{5}+b^{5}-a^{2} b^{2}(a+b) \\
=\left(a^{2}-b^{2}\right)\left(a^{3}-b^{3}\right) \geqslant 0 .
\end{array}
$$
Then $\frac{a b}{a^{5}+b^{5}+a b}=\frac{a b \cdot a b c}{a^{5}+b^{5}+a b \cdot a b c}$
$$
\begin{array}{l}
=\frac{a^{2} b^{2} c}{a^{5}+b^{5}+a^{2} b^{2} c} \leqslant \frac{a^{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,795 |
Example $\mathbf{3}$ Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \text { [2]. }
$$ | Explanation: The left side of the inequality to be proven is somewhat "top-light and bottom-heavy" compared to the right side. We can use the reciprocal substitution $a=\frac{1}{x}, b=\frac{1}{y}$, $c=\frac{1}{z}$ to make its form more "harmonious".
Let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$. Then $x y z=1$.
It ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,796 |
Example 4 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\frac{1}{1+2a}+\frac{1}{1+2b}+\frac{1}{1+2c} \geqslant 1^{[2]} \text{. }
$$ | Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}_{+}\right)$.
$$
\begin{array}{l}
\text { Then } \frac{1}{1+2 a}+\frac{1}{1+2 b}+\frac{1}{1+2 c} \\
=\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z} \text {. }
\end{array}
$$
By Nesbitt's inequality, we have
$$
\begin{array}{l}
{[y(y+2 x)+z(z+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,797 |
Example 5 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove that:
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c} \geqslant 4 .
$$ | Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}_{+}\right)$.
Assume without loss of generality that $x \geqslant y \geqslant z$. Then
$$
\begin{array}{l}
y^{2} z+z^{2} x+x^{2} y-\left(x^{2} z+y^{2} x+z^{2} y\right) \\
=(x-y)(y-z)(x-z) \geqslant 0 . \\
\text { Hence } \frac{1}{a}+\frac{1}{b}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,798 |
Example 6 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove that:
$$
\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leqslant 1 .
$$ | Let $a=x^{3}, b=y^{3}, c=z^{3}(x, y, z \in \mathbf{R}_{+})$. Then $xyz=1$.
Notice that $x^{3}+y^{3} \geqslant x^{2} y+y^{2} x$. Thus,
$$
\begin{array}{l}
\frac{1}{1+a+b}=\frac{1}{1+x^{3}+y^{3}} \\
\leqslant \frac{1}{1+x^{2} y+y^{2} x}=\frac{1}{xyz+x^{2} y+y^{2} x} \\
=\frac{1}{xy(x+y+z)}=\frac{z}{x+y+z} .
\end{array}
$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,799 |
Example 3 If 20 points divide a circle into 20 equal parts, then the number of regular polygons that can be formed with vertices chosen only from these 20 points is ( ).
(A) 4
(B) 8
(C) 12
(D) 24
(1996, National Junior High School Mathematics League) | Explanation: If you draw the figure directly, it is hard to get the correct answer. If you consider the characteristics of a regular polygon, list a simple equation for the distribution characteristics of points other than the vertices of the regular polygon, and then conduct a case analysis, it is easy to get the answ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 721,800 |
Example 7 Let real numbers $x, y, z$ all be not equal to 1, and satisfy $xyz=1$. Prove:
$$
\frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geqslant 1^{[3]} .
$$ | Let $\frac{x}{x-1}=a, \frac{y}{y-1}=b, \frac{z}{z-1}=c$. Then $x=\frac{a}{a-1}, y=\frac{b}{b-1}, z=\frac{c}{c-1}$.
From $x y z=1$, we get
$$
a b c=(a-1)(b-1)(c-1),
$$
which simplifies to
$$
a+b+c-1=a b+b c+c a \text {. }
$$
Therefore, $a^{2}+b^{2}+c^{2}$
$$
\begin{array}{l}
=(a+b+c)^{2}-2(a b+b c+c a) \\
=(a+b+c)^{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,801 |
2. If $x>0, y>0, z>0$, and $x y z=1$, prove that:
$$
1<\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}<2 \text {. }
$$
$(2008$, National High School Mathematics League Shandong Province Preliminary) | Let $x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c}$. Then
$$
\begin{aligned}
1 & =\sum \frac{a}{a+b+c}<\sum \frac{a}{a+b}=\sum \frac{1}{1+x} \\
& \left.<\sum \frac{a+c}{a+b+c}=2 .\right)
\end{aligned}
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,802 |
3. Let $a, b, c \in \mathbf{R}_{+}$, and satisfy $a b c=1$. Prove: $\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1$.
(41st IMO) | Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}_{+}\right)$. Then the original inequality
$$
\Leftrightarrow(x-y+z)(y-z+x)(z-x+y) \leqslant x y z \text {. }
$$
Also, the sum of any two of $x-y+z$, $y-z+x$, $z-x+y$ is greater than 0, so at most only one of them is less than or equal to 0.
(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,803 |
4. Let $p$, $q$, $r$ be positive numbers, and $pqr=1$. Prove: for all $n \in \mathbf{N}$, we have
$$
\frac{1}{p^{n}+q^{n}+1}+\frac{1}{q^{n}+r^{n}+1}+\frac{1}{r^{n}+p^{n}+1} \leqslant 1 .
$$
(2004, Baltic Way Mathematical Contest) | (提示: Let $p^{n}=a, q^{n}=b, r^{n}=c$. Then $a b c=1$. Hence the original inequality
$$
\Leftrightarrow \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \leqslant 1 .
$$
Since $a+b \geqslant\left(a^{\frac{1}{3}} b^{\frac{1}{3}}\right)\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)$, we have
$$
\begin{array}{l}
\sum \frac{1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,804 |
5. Given $a b c=1, a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+a+c} \leqslant 1 .
$$ | (Introduction of parameter $t$, let
$$
\begin{array}{l}
\frac{1}{1+a+b} \leqslant \frac{c^{t}}{a^{t}+b^{t}+c^{t}} \\
\Leftrightarrow a c^{t}+b c^{t} \geqslant a^{t}+b^{t} \\
\Leftrightarrow a+b \geqslant a^{2 t} b^{t}+b^{2 t} a^{t} .
\end{array}
$$
When $t>0$, by the rearrangement inequality, it is easy to get
$$
a^{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,805 |
Question 1 As shown in Figure 1, $M$ and $N$ are the midpoints of arcs $\overparen{B C}$ and $\overparen{A C}$, respectively, on the circumcircle $\Gamma$ of acute triangle $\triangle A B C (\angle A<\angle B)$. A line through point $C$ parallel to $M N$ intersects circle $\Gamma$ at point $P$, and $I$ is the incenter ... | Proof 1: To prove this problem, we first provide a lemma.
Lemma Let $I$ be a point inside $\triangle ABC$, and the line $AI$ intersects the circumcircle of $\triangle ABC$ at point $D$. Then $I$ is the incenter of $\triangle ABC$ if and only if $BD = DI = DC$.
Proof omitted.
Back to the original problem.
(1) As shown i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,806 |
The second question: Prove the inequality
$$
-1<\sum_{k=1}^{n} \frac{k}{k^{2}+1}-\ln n \leqslant \frac{1}{2}(n=1,2, \cdots) \text {. }
$$ | $$
\begin{array}{l}
\text { Proof 1: First, we prove: } \\
\sum_{k=1}^{n-1} \frac{1}{k+1}1) \text {, } \\
\text { i.e., } \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\sum_{k=1}^{n} \frac{k}{k^{2}+k}=\sum_{k=1}^{n} \frac{1}{k+1} \\
=\sum_{k=0}^{n} \frac{1}{k+1}-1>\ln (n+2)-1>\ln n-1 \text {. } \\
\text { i.e., } \sum_{k=1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,807 |
Let $p$ be a prime number greater than 3. Prove that there exist several integers $a_{1}, a_{2}, \cdots, a_{t}$ satisfying the condition
$$
-\frac{p}{2}<a_{1}<a_{2}<\cdots<a_{t}<\frac{p}{2},
$$
such that the product $\frac{p-a_{1}}{\left|a_{1}\right|} \cdot \frac{p-a_{2}}{\left|a_{2}\right|} \cdots \cdots \cdot \frac{... | Prove: Take an integer $a_{0}$ satisfying $0<\left|a_{0}\right|<\frac{p}{2}$. Then there exists a unique positive integer $l_{0}$, such that
$$
\frac{p}{2 \times 3^{l_{0}}}<\left|a_{0}\right|<\frac{p}{2 \times 3^{l_{0}-1}}.
$$
Let $a_{1}=p-3^{l_{0}}\left|a_{0}\right|$. Then $0<\left|a_{1}\right|<\frac{p}{2}$.
Therefor... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,809 |
Question Arrange all positive integers $m$ whose digits are no greater than 3 in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2007}=$ $\qquad$
$(2007$, National High School Mathematics League Jiangxi Province Preliminary Contest) | Another solution: Considering that $a_{n}$ is not continuous in decimal, and the digits of $a_{n}$ are all no greater than 3 (i.e., $0, 1, 2, 3$), then the form of $a_{n}$ is a sequence of consecutive integers in quaternary, thus simplifying the problem.
Let the set $\left\{\left(a_{n}\right)_{4}\right\}=\{$ consecuti... | 133113 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,810 |
Example 4 As shown in Figure 4, connect the midpoints of each side of a square with side length 1, and connect the diagonals of the square. Then the total number of isosceles right triangles in the figure is ( ).
(A) 16
(B) 32
(C) 22
(D) 44
(2009, National Junior High School Mathematics League, Nanchang City Competitio... | Explanation: Obviously, the number is large, so the exhaustive method cannot be used. If we classify and count the shapes of isosceles right triangles based on the "hypotenuse length" or "leg length," it is easy to obtain the correct result.
Notice that, there are 4 isosceles right triangles with a hypotenuse of $\sqr... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,811 |
For example, the five-digit numbers formed by the digits $1, 2, 3, 4, 5$ with repetition, arranged in ascending order. Ask:
(1) What are the positions of 22435 and 43512?
(2) What is the 200th number? ${ }^{[1]}$ | Analysis: Let the $n$-th number after sorting be denoted as $a_{n}$. Since the numbers do not contain 0, we cannot directly use the quinary system. Therefore, subtract 11111 from all $a_{n}$, and denote the result as $b_{n}$. Thus, we can draw the following conclusions:
(i) Each number in $b_{n}$ consists of some of th... | 12355 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,812 |
Promotion 1 Select $n$ $(1 \leqslant n \leqslant 10)$ consecutive digits $a_{1}, a_{2}, \cdots, a_{n}$ from the ten digits $0,1, \cdots, 9$, to form an $n$-digit number $P$, where, without loss of generality, let
$$
a_{1}=a_{2}-1=a_{3}-2=\cdots=a_{n}-n+1 \text {. }
$$
Arrange all $P$ in ascending order, find what $P_{... | Solution: First, we make the following stipulations:
For $x, y, k \in \mathbf{N} (k$ being the base), $k \geqslant 2$, if $(y)_{k}=x$, then $y=\varphi_{k}(x)$.
For any $P_{i}$, let $Q_{i}=P_{i}-\underbrace{a_{1} a_{1} \cdots a_{1}}_{n \uparrow}$. Thus, $\left\{Q_{i}\right\}$ satisfies the requirement of being a sequen... | P_{m}=\varphi_{n}(m-1)+\underbrace{a_{1} a_{1} \cdots a_{1}}_{n \uparrow} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,813 |
Promotion 2 Select $n$ $(1 \leqslant n \leqslant 10)$ digits $a_{1}, a_{2}, \cdots, a_{n}$ from the ten digits $0,1, \cdots, 9$, to form an $n$-digit number $P$. Sort all $P$ in ascending order, find what $P_{m}$ equals to. | Solution: As defined above.
Let $b_{1}, b_{2}, \cdots, b_{n}$ be a permutation of $a_{1}, a_{2}, \cdots, a_{n}$ in ascending order. Now, we analyze the numbers formed by $b_{i}(i=1,2, \cdots, n)$.
Construct a mapping $f$: replace each $b_{i}$ in every number $P$ with $i-1$, forming a new number $Q$ (for example, $\lef... | P_{m}=f^{-1}\left(\varphi_{n}(m-1)\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,814 |
1. Given $x+\frac{1}{x}=7(0<x<1)$. Then the value of $\sqrt{x}-\frac{1}{\sqrt{x}}$ is ( ).
(A) $-\sqrt{7}$
(B) $-\sqrt{5}$
(C) $\sqrt{7}$
(D) $\sqrt{5}$ | - 1. B.
$$
\text { Since }\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}=x+\frac{1}{x}-2=5(0<x<1) \text {, }
$$
Therefore, $\sqrt{x}-\frac{1}{\sqrt{x}}=-\sqrt{5}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,815 |
2. If one root of the equation $2 x^{2}-3 x+m=0$ is greater than -2 and less than -1, and the other root is greater than 2 and less than 3, then the range of values for $m$ is ( ).
(A) $m<\frac{9}{8}$
(B) $-14<m<\frac{9}{8}$
(C) $-9<m<-5$
(D) $-14<m<-2$ | 2. C.
Let $f(x)=2 x^{2}-3 x+m$.
From the given information, we can roughly sketch the graph of $y=f(x)$, knowing that $f(-2)>0, f(-1)>0$. Solving these inequalities simultaneously yields $-9<m<-5$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,816 |
3. A certain section of a road consists of three equal-length segments: an uphill, a flat, and a downhill section. It is known that a car travels at average speeds of $v_{1}, v_{2}, v_{3}$ on these three segments, respectively. Then, the average speed of the car over this section of the road is ( ).
(A) $\frac{v_{1}+v_... | 3. D.
Let the length of this section of the road be $3 s$. Then the time taken by the car to travel on each section is $t_{i}=\frac{s}{v_{i}}(i=1,2,3)$. Therefore, the average speed of the car on this section of the road is
$$
\bar{v}=\frac{3 s}{t_{1}+t_{2}+t_{3}}=\frac{3 s}{\frac{s}{v_{1}}+\frac{s}{v_{2}}+\frac{s}{v_... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,817 |
4. Given in rectangle $A B C D$, $A B=72, A D=$ 56. If $A B$ is divided into 72 equal parts, and parallel lines to $A D$ are drawn through each division point; and if $A D$ is divided into 56 equal parts, and parallel lines to $A B$ are drawn through each division point, then these parallel lines divide the entire rect... | 4. D.
According to the problem, establish a Cartesian coordinate system such that $A(0,0)$, $B(72,0)$, and $D(0,56)$. Then, $C(72,56)$.
Since $AC$ intersects with every horizontal line (including $AB$ and $DC$) and every vertical line (including $AD$ and $BC$), there are $57 + 73 = 130$ intersection points (including... | 120 | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,818 |
5. Given a square $A B C D$ with side length 1, $E$ is the midpoint of side $C D$, and a moving point $P$ travels along the perimeter of the square $A B C D$ in the direction $A$ $\rightarrow B \rightarrow C \rightarrow E$. Let the distance traveled by point $P$ be $x$, and the area of $\triangle A P E$ be $y$. Then th... | 5. A.
As shown in Figure 5 (a), when $0 < x \leqslant 1$, $A P = x$, hence $y = \frac{1}{2} x$;
As shown in Figure 5 (b), when $1 < x \leqslant 2$, $B P = x - 1$,
$$
\begin{array}{l}
C P = 2 - x \text{. Therefore,} \\
y = S_{\text{square } A B C D} - S_{\triangle A B P} - S_{\triangle E C P} - S_{\triangle A D E} \\
=... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,819 |
6. A die is rolled twice. If the number of dots that appear on the first roll is $x$, and the number of dots that appear on the second roll is $y$, then the probability that the point $M(x, y)$ determined by $x$ and $y$ lies on the hyperbola $y=\frac{6}{x}$ is . $\qquad$ | ii. $6 \cdot \frac{1}{9}$.
Point $M(x, y)(x, y \in\{1,2, \cdots, 6\})$ totals 36, among which, points $(1,6) 、(6,1) 、(2,3) 、(3,2)$ lie on the hyperbola $y=\frac{6}{x}$. Therefore, the required probability equals $\frac{1}{9}$. | \frac{1}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,820 |
7. Calculate $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}}(n \geqslant 2$, $n \in \mathbf{N})$ 的值为
The value of $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}... | 7. 100 .
$$
\begin{array}{l}
\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}} \\
=\sqrt[n]{\left(10^{n}-1\right)^{2}+2 \times 10^{n}-1} \\
=\sqrt[n]{10^{2 n}}=100 .
\end{array}
$$ | 100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,821 |
Example 5 Using weights labeled $1 \mathrm{~g}, 2 \mathrm{~g}, 6 \mathrm{~g}, 26 \mathrm{~g}$, each one in quantity, to weigh objects on a balance scale without graduations. If weights can be placed on both ends of the balance, the number of different gram weights (positive integer weights) that can be measured is ( ).... | Explanation: If we consider each case individually, there will inevitably be omissions; if we categorize by the number of weights on each side of the balance, the approach is clear and less prone to omissions.
(1) When one weight is placed on one side of the balance and no weight is placed on the other side, the grams ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 721,822 |
8. If $p$ is a prime number, and $p+3$ divides $5p$, then the last digit of $p^{2009}$ is $\qquad$ . | 8. 2 .
Since the positive divisors of $5p$ are $1, 5, p, 5p$, and $(p+3)$ divides $5p$, therefore, $p=2$.
Thus, $p^{2009}=2^{2009}=\left(2^{4}\right)^{502} \times 2=2 \times 16^{502}$.
Given that the last digit of $16^{502}$ is 6, it follows that the last digit of $p^{2009}$ is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,823 |
9. As shown in Figure 2, in quadrilateral $A B C D$, $\angle A C B=$ $\angle B A D=105^{\circ}, \angle A B C=\angle A D C=45^{\circ}$. If $A B$ $=2$, then the length of $C D$ is $\qquad$. | 9. 2 .
As shown in Figure 6, draw $EA \perp AB$ intersecting the extension of $BC$ at point $E$.
$$
\begin{array}{l}
\text { Then } \angle AEB=45^{\circ} \\
\quad=\angle ADC, \\
AE=AB=2 . \\
\text { Also } \angle DAC=\angle DAB-\angle CAB \\
=\angle DAB-\left(180^{\circ}-\angle ABC-\angle ACB\right) \\
=75^{\circ}=180... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,824 |
10. As shown in Figure 3, fill in the 10 spaces in the annulus with the numbers $1,2, \cdots, 10$. Add up the absolute values of the differences between all adjacent cells (cells that share a common edge). If this sum is to be maximized, then the maximum value is $\qquad$ | 10.50.
Let the two numbers in adjacent cells be $a$ and $b$ ($a > b$). Then $|a-b| = a-b$, and there are 10 differences. To maximize the sum of these 10 differences, the 10 minuends $a$ should be as large as possible, and the 10 subtrahends $b$ should be as small as possible. Since each number is adjacent to two other... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,825 |
11. Given $\frac{x y}{x+y}=2, \frac{x z}{x+z}=3, \frac{y z}{y+z}=4$. Find the value of $7 x+5 y-2 z$. | Three, 11. Given
$$
\begin{array}{l}
\frac{1}{2}=\frac{1}{x}+\frac{1}{y}, \frac{1}{3}=\frac{1}{x}+\frac{1}{z}, \\
\frac{1}{4}=\frac{1}{y}+\frac{1}{z} .
\end{array}
$$
Solving simultaneously, we get $x=\frac{24}{7}, y=\frac{24}{5}, z=24$.
Therefore, $7 x+5 y-2 z=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,826 |
12. Starting from the equilateral triangle shown in Figure 4 (a), divide each side into three equal segments, and draw a small equilateral triangle outward on the middle segment of each side, forming the six-pointed star shape shown in Figure 4 (b); then, on each side of the six-pointed star, use the same method to dra... | 12. As shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline & \begin{tabular}{l}
First \\
Growth
\end{tabular} & \begin{tabular}{l}
Second \\
Growth
\end{tabular} & \begin{tabular}{l}
Third \\
Growth
\end{tabular} & $\cdots$ & \begin{tabular}{l}
$n$-th \\
Growth
\end{tabular} \\
\hline \begin{tabular... | \frac{4^{n}}{3^{n-1}} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,827 |
13. Given that $m$ and $n$ are positive integers, the equation $x^{2}-m n x+(m+n)=0$ has positive integer solutions. Find the values of $m$ and $n$. | 13. Let the equation $x^{2}-m n x+(m+n)=0$ have two roots $\alpha, \beta$. Then $\left\{\begin{array}{l}\alpha+\beta=m n, \\ \alpha \beta=m+n .\end{array}\right.$
Given that $m, n, \alpha, \beta$ are all positive integers, without loss of generality, assume
$$
\alpha \geqslant \beta \geqslant 1, m \geqslant n \geqslant... | (5,1),(2,2),(3,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,828 |
14. Given that point $P$ is a point inside the acute triangle $\triangle A B C$ and minimizes $P A+P B+P C$. Determine the position of point $P$ and prove your conclusion. | 14. As shown in Figure 8,
Construct equilateral triangles $\triangle A C B^{\prime}$ and $\triangle B C A^{\prime}$ outwardly on sides $A C$ and $B C$ respectively, and connect $B B^{\prime}$ and $A A^{\prime}$, intersecting at point $P$. Then point $P$ is the desired point.
In fact, it is easy to prove that
$\triangl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,829 |
2. Given the line $l: x+y-9=0$ and the circle $\odot M: 2 x^{2}+$ $2 y^{2}-8 x-8 y-1=0$, point $A$ is on $l$, and $B, C$ are two points on $\odot M$. In $\triangle A B C$, $\angle B A C=45^{\circ}, A B$ passes through the center $M$. Then the range of the x-coordinate of point $A$ is $\qquad$ | 2. $[3,6]$.
Let $A(a, 9-a)$. Then the distance from the center $M$ to the line $A C$ is $d=|A M| \sin 45^{\circ}$.
From the condition that the line $A C$ intersects with $\odot M$, we get $d \leqslant \frac{\sqrt{34}}{2}$.
Solving this, we get $3 \leqslant a \leqslant 6$. | [3,6] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,831 |
3. In the coordinate plane, there are two regions $M$ and $N$. $M$ is defined by
\[
\left\{\begin{array}{l}
y \geqslant 0, \\
y \leqslant x, \\
y \leqslant 2-x,
\end{array} \quad N\right.
\]
is a region that varies with $t$, determined by the inequality $t \leqslant x \leqslant t+1$, where the range of $t$ is $0 \leqs... | 3. $-t^{2}+t+\frac{1}{2}$.
As shown in Figure 2, according to the problem, we have
$$
\begin{array}{l}
=1-\frac{1}{2} t^{2}- \\
\frac{1}{2}(1-t)^{2} \\
=-t^{2}+t+\frac{1}{2} \text {. } \\
\end{array}
$$ | -t^{2}+t+\frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,832 |
Example 6 Xiao Ming divides a cube with an edge length of 4 into 29 smaller cubes with integer edge lengths. Then the number of smaller cubes with an edge length of 1 is ( ).
(A) 22
(B) 23
(C) 24
(D) 25 | Explanation 1: If a cube with an edge length of 3 is cut out, then there can only be 1 cube with an edge length of 3. The remaining ones are all cubes with an edge length of 1 (37 in total), which does not meet the requirements.
Let the number of cubes with an edge length of 2 be $x$, and the number of cubes with an e... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,833 |
4. The smallest positive integer $a$ that makes the inequality
$$
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}<a-2007 \frac{1}{3}
$$
hold for all positive integers $n$ is $\qquad$ | 4. 2009 .
Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$.
Obviously, $f(n)$ is monotonically decreasing. From the maximum value of $f(n)$, $f(1)<a-2007 \frac{1}{3}$, we get $a=2009$. | 2009 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,834 |
5. For any two points $P, Q$ on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, if $O P \perp O Q$, then the minimum value of $|O P| \cdot|O Q|$ is $\qquad$. | 5. $\frac{2 a^{2} b^{2}}{a^{2}+b^{2}}$.
Let $P(|O P| \cos \theta,|O P| \sin \theta)$,
$$
Q\left(|O Q| \cos \left(\theta \pm \frac{\pi}{2}\right),|O Q| \sin \left(\theta \pm \frac{\pi}{2}\right)\right) \text {. }
$$
Since points $P$ and $Q$ are on the ellipse, we have
$$
\begin{array}{l}
\frac{1}{|O P|^{2}}=\frac{\cos... | \frac{2 a^{2} b^{2}}{a^{2}+b^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,835 |
6. If the equation $\lg k x=2 \lg (x+1)$ has only one real root, then the range of values for $k$ is . $\qquad$ | $$
\left\{\begin{array}{l}
k<0, \\
x+1>0, \\
x^{2}+(2-k) x+1=0 .
\end{array}\right.
$$
From equation (3) we get
$\Delta=k^{2}-4 k \geqslant 0 \Rightarrow k \leqslant 0$ or $k \geqslant 4$.
Let the two roots of equation (3) be $x_{1}$ and $x_{2}\left(x_{1} \leqslant x_{2}\right)$.
(1) When $k<0$.
$$
\left\{\begin{array... | k<0 \text{ or } k=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,836 |
7. A number table composed of several rows of numbers, starting from the second row, each number in the row is equal to the sum of the two numbers above it, the last row contains only one number, the first row is the first 100 positive integers arranged in ascending order. Then the number in the last row is (exponentia... | 7. $101 \times 2^{98}$.
It is easy to know:
(1) The table has a total of 100 rows;
(2) Each row forms an arithmetic sequence, and the common differences are
$$
d_{1}=1, d_{2}=2, d_{3}=2^{2}, \cdots, d_{99}=2^{98} ;
$$
(3) Let the first number of the $n(n \geqslant 2)$-th row be $a_{n}$. Then $a_{100}$ is the desired v... | 101 \times 2^{98} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,837 |
8. A station has exactly one bus arriving between $8:00 \sim 9:00$ and $9:00 \sim 10:00$ every morning, but the arrival times are random, and the two arrival times are independent of each other, as shown in Table 1. A passenger arrives at the station at $8:20$. What is the expected waiting time for the passenger (round... | 8. 27 .
The distribution of passenger waiting times is shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline \begin{tabular}{c}
Waiting Time \\
(min)
\end{tabular} & 10 & 30 & 50 & 70 & 90 \\
\hline Probability & $\frac{1}{2}$ & $\frac{1}{3}$ & $\frac{1}{6} \times \frac{1}{6}$ & $\frac{1}{2} \times \frac{1}... | 27 | Other | math-word-problem | Yes | Yes | cn_contest | false | 721,838 |
1. (14 points) Let the line $l: y=k x+m(k, m \in$ Z) intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ at two distinct points $A, B$, and intersect the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ at two distinct points $C, D$. Question: Does there exist a line $l$ such that the vector $\overrightarrow{A C}... | Given the system of equations:
$$
\left\{\begin{array}{l}
y=k x+m, \\
\frac{x^{2}}{16}+\frac{y^{2}}{12}=1,
\end{array}\right.
$$
eliminating \( y \) and simplifying, we get:
$$
\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0.
$$
Let \( A\left(x_{1}, y_{1}\right) \) and \( B\left(x_{2}, y_{2}\right) \). Then:
$$
\beg... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,839 |
2. (15 points) Given $p, q(q \neq 0)$ are real numbers, the equation $x^{2}-p x+q=0$ has two real roots $\alpha, \beta$, and the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=p, a_{2}=p^{2}-q, a_{n}=p a_{n-1}-q a_{n-2}(n=$ $3,4, \cdots)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$ (expr... | 2. Solution 1: (1) By Vieta's formulas, we know $\alpha \beta=q \neq 0$. Also, since $\alpha+\beta=p$, we have
$$
a_{n}=p a_{n-1}-q a_{n-2}
$$
$$
=(\alpha+\beta) a_{n-1}-\alpha \beta a_{n-2}(n=3,4, \cdots) \text {. }
$$
Rearranging gives $a_{n}-\beta a_{n-1}=\alpha\left(a_{n-1}-\beta a_{n-2}\right)$.
Let $b_{n}=a_{n+1... | S_{n}=3-\frac{n+3}{2^{n}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,840 |
3. (15 points) Find the maximum and minimum values of the function
$$
y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}
$$ | 3. The domain of the function is $[0,13]$.
$$
\begin{array}{l}
\text { Given } y=\sqrt{x}+\sqrt{x+27}+\sqrt{13-x} \\
=\sqrt{x+27}+\sqrt{13+2 \sqrt{x(13-x)}} \\
\geqslant \sqrt{27}+\sqrt{13}=3 \sqrt{3}+\sqrt{13},
\end{array}
$$
we know that the equality holds when $x=0$.
Thus, the minimum value of $y$ is $3 \sqrt{3}+\s... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,841 |
One. (50 points) As shown in Figure 1, $M$ and $N$ are the midpoints of the arcs $\overparen{BC}$ and $\overparen{AC}$, respectively, on the circumcircle $\Gamma$ of the acute triangle $\triangle ABC (\angle A < \angle B)$. A line through point $C$ parallel to $MN$ intersects the circle $\Gamma$ at point $P$, and $I$ i... | (1) As shown in Figure 3, connect $N I$ and $M I$.
Since $P C \parallel M N$ and points $P, C, M, N$ are concyclic, quadrilateral $P C M N$ is an isosceles trapezoid. Therefore,
$$
N P = M C, \quad P M = N C.
$$
Connect $A M$ and $C I$. Then $A M$ and $C I$ intersect at point $I$.
Since $\angle M I C$
$$
\begin{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,842 |
Example 7 satisfies that there are right-angled triangles with two legs of integer lengths, and the perimeter is exactly an integer multiple of the area. There are ( ) such triangles.
(A) 1
(B) 2
(C) 3
(D) infinitely many
(2007, National Junior High School Mathematics Competition, Zhejiang Province) | Explanation: Let the lengths of the two legs of a right triangle be $a$ and $b$ ($a \leqslant b$). Then,
$$
a+b+\sqrt{a^{2}+b^{2}}=k \cdot \frac{1}{2} a b\left(a, b, k \in \mathbf{N}_{+}\right) \text {. }
$$
Simplifying, we get $(k a-4)(k b-4)=8$.
Therefore, $\left\{\begin{array}{l}k a-4=1, \\ k b-4=8\end{array}\right... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,844 |
Three, (50 points) Let $k, l$ be two given positive integers. Prove: there are infinitely many positive integers $m(m \geqslant k)$, such that $\mathrm{C}_{m}^{k}$ is coprime with $l$. | For any positive integer $t$, let
$$
m=k+t l(k!) \text {. }
$$
We will prove: $\left(\mathrm{C}_{m}^{k}, l\right)=1$.
Let $p$ be any prime factor of $l$, we only need to prove:
$p \nmid \mathrm{C}_{m}^{k}$.
If $p \nmid k!$, then by
$$
\begin{array}{l}
k!\mathrm{C}_{m}^{k}=\prod_{i=1}^{k}(m-k+i) \\
\equiv \prod_{i=1}^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,845 |
Four. (50 points) In the $3 \times 9$ table of non-negative numbers
$$
P=\left(\begin{array}{llll}
x_{11} & x_{12} & \cdots & x_{19} \\
x_{21} & x_{22} & \cdots & x_{29} \\
x_{31} & x_{32} & \cdots & x_{39}
\end{array}\right)
$$
each row contains distinct numbers, the sum of the three numbers in each of the first six ... | (1) Suppose the minimum value $u_{i}=\min \left\{x_{i 1}, x_{i 2}, x_{i 3}\right\}$ $(i=1,2,3)$ is not taken from different columns of the number table $S$. Then there exists a column that does not contain any $u_{i}$. Without loss of generality, assume $u_{i} \neq x_{i 2}(i=1,2,3)$.
Since any two elements in the same... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 721,846 |
10.1. Find all positive integers $n$, such that there exist non-zero real numbers $a, b, c, d$, satisfying the polynomial
$$
(a x+b)^{1000}-(c x+d)^{1000}
$$
after expansion and combining like terms, exactly $n$ non-zero terms remain. | 10.1. $n=500,1000,1001$.
Obviously, there exist polynomials with 1001 and 1000 non-zero terms:
$$
\begin{array}{l}
(2 x+2)^{1000}-(x+1)^{1000}, \\
(2 x+1)^{1000}-(x+1)^{1000} .
\end{array}
$$
Assume the coefficients of the $x^{i}$ and $x^{j}$ terms $(i>j)$ are 0. Then
$a^{i} b^{1000-i}=c^{i} d^{1000-i}$,
$a^{j} b^{10... | n=500,1000,1001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,847 |
10.3. The function $f(x)=\prod_{i=1}^{2000} \cos \frac{x}{i}$ changes sign how many times in the interval $\left[0, \frac{2009 \pi}{2}\right]$? | 10.3. 75 times.
Let $n=2009$. Consider the function $\cos \frac{x}{k}$. It changes sign at $x=\frac{k(2 m+1) \pi}{2}$. This indicates that the zeros of $f(x)$ are $x_{i}=\frac{i \pi}{2}(1 \leqslant i \leqslant n)$.
We only need to consider the sign change of $f(x)$ at $x_{i}(i=1,2, \cdots, n-1)$.
$\cos \frac{x}{k}$ c... | 75 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 721,848 |
10. 4. In a regular 2009-gon, a non-negative integer not exceeding 100 is placed at each vertex. Adding 1 to the numbers at two adjacent vertices is called an operation on these two adjacent vertices. For any given two adjacent vertices, the operation can be performed at most $k$ times. Find the minimum value of $k$ su... | 10. 4. $k_{\min }=100400$.
Let the numbers at each vertex be $a_{1}, a_{2}, \cdots, a_{2009}$. Let $N=100400$.
$$
\begin{array}{l}
\text { (1) Let } a_{2}=a_{4}=\cdots=a_{2008}=100, \\
a_{1}=a_{3}=\cdots=a_{2008}=0, \\
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right) .
\end{arr... | 100400 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,849 |
10.5. Let $k$ be a positive integer. An infinite strictly increasing sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integer $n$, we have $a_{n+1}-a_{n} \leqslant k$, and $a_{n}$ is a multiple of 1005 or 1006, but not a multiple of 97. Find the smallest possible value of $k$. | 10.5. $k_{\text {min }}=2010$.
Given a positive integer $N$, such that
$$
a_{1}<1005 \times 1006 \times 97 \times N=D \text {. }
$$
Since $D$ is a multiple of 97, it is not any term in the sequence $a_{n}$, hence there exists a positive integer $n$, such that $a_{n}<D<a_{n+1}$.
The largest multiple of 1005 or 1006 l... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,850 |
10.6. There is a kingdom consisting of $n$ cities, some of which are connected by roads, satisfying:
(1) All roads do not intersect (if there is a road between two cities, they are called adjacent);
(2) For any two cities, one can travel from one city to another along the roads (possibly passing through other cities);
... | 10.6. Prove the proposition by induction on $n$.
When $n=1,2$, the proposition is obviously true.
Below, assume that when $n \geqslant 3$ and the proposition holds for the number of cities less than $n$.
Consider each city as a point, and the roads between cities as edges connecting them. This results in a graph. By ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,851 |
10.7. Given the incenter of a non-isosceles $\triangle ABC$ is $I$, and the incircle touches sides $AB$, $BC$, and $CA$ at points $C_{1}$, $A_{1}$, and $B_{1}$, respectively. Let the incircles of quadrilaterals $B A_{1} I C_{1}$ and $C A_{1} I B_{1}$ be circles $\omega_{B}$ and $\omega_{C}$, respectively. Prove: The ot... | 10.7. Let circle $\omega_{B}$ be tangent to $B A_{1}$, $I A_{1}$, and $B C_{1}$ at points $K_{B}$, $L_{B}$, and $M_{B}$, respectively. Let circle $\omega_{C}$ be tangent to $C A_{1}$, $I A_{1}$, and $C B_{1}$ at points $K_{C}$, $L_{C}$, and $M_{C}$, respectively. Let $o_{B}$ and $o_{C}$ be the centers of circles $\omeg... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,852 |
10.8. Let $x, y$ be integers in the interval $[2,100]$. Prove: There exists a positive integer $n$, such that $x^{2^{n}}+y^{2^{n}}$ is composite. | 10. 8. If $x=y$, let $n=1$. Then $x^{2}+y^{2}$ is an even number greater than 2, hence it is a composite number.
Now assume $x \neq y$.
Next, we prove: there exists a positive integer $n$, such that $x^{2^{n}}+$ $y^{2^{n}}$ is a multiple of 257, and not equal to 257.
Suppose $x^{2^{n}}+y^{2^{n}}=257$. Let $a=x^{2^{n-1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,853 |
1. If $m=\frac{(|2 a+1|+|2 b-1|)^{2}-4(a+b)^{2}}{(2 a+1)(2 b-1)}$, then the value of $m(m+4)$ is ( ).
(A) $-\frac{1}{2}$
(B) 0
(C) $\frac{1}{4}$
(D) $\frac{1}{2}$ | - 1. B.
Let $2a+1=x, 2b-1=y$. Then $m=\frac{(|x|+|y|)^{2}-(x+y)^{2}}{xy}=\frac{2|xy|}{xy}-2$. Therefore, $m(m+4)=(m+2)^{2}-4=\frac{4x^{2}y^{2}}{x^{2}y^{2}}-4=0$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,854 |
Example 8 In the Cartesian coordinate system, a point whose both the x-coordinate and y-coordinate are integers is called an integer point. The image of the quadratic function $y=-x^{2}+6 x-\frac{27}{4}$ and the x-axis enclose a closed figure which is colored red. The number of integer points inside this red region and... | Explanation: As shown in Figure 5, the graph of the quadratic function $y=-x^{2}+6 x-\frac{27}{4}$ intersects the $x$-axis at two points $\left(\frac{3}{2}, 0\right)$ and $\left(\frac{9}{2}, 0\right)$.
Between $x=\frac{3}{2}$ and $x=\frac{9}{2}$, there are three integers: $2, 3, 4$.
When $x=2, 4$, $y=\frac{5}{4}$, th... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,855 |
2. In $\triangle A B C$, it is known that $D$ is the midpoint of $A B$, point $E$ is on $A C$, and $\angle A E D=90^{\circ}+\frac{1}{2} \angle C$. Then $B C+2 A E-A C$ is ( ).
(A) Positive
(B) 0
(C) Negative
(D) Uncertain | 2. B.
As shown in Figure 1, draw
$D F / / B C$ intersecting $A C$ at
point $F$. Then $\angle D F E=$
$$
\begin{array}{l}
\angle C, A C=2 A F, \\
B C=2 D F .
\end{array}
$$
In $\triangle D E F$, draw $F G \perp D E$ intersecting $D E$ at point $G$.
$$
\text { Given } \angle A E D=90^{\circ}+\frac{1}{2} \angle C=90^{\c... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,856 |
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