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8. Let the unit disk be $D=\left\{(x, y) \in \mathbf{R}^{2} \mid x^{2}+y^{2} \leqslant\right.$ 1\}, and define the width of a strip formed by two parallel lines as the distance between these lines. If the unit disk can be covered by some strips, prove: the sum of the widths of these strips is at least 2.
(19th Iranian ... | Solution: Consider the unit sphere that contains this unit circle. Let the unit circle belong to plane II. For each strip, replace the two parallel lines that form this strip with two planes that pass through these lines and are perpendicular to plane $I$, then for each strip, we can obtain two parallel planes, which w... | 2 | Geometry | proof | Yes | Yes | cn_contest | false |
8. As shown in Figure $3, A C=$ $B C, A C \perp B C$ at point $C, A B=A D=B D$, $C D=C E=D E$. If $A B=\sqrt{2}$, then $B E=$ | 8. 1 .
In $\triangle A D C$ and $\triangle B D C$, given $A D=D B, D C=D C$, $A C=B C$, we can conclude that
$\triangle A D C \cong \triangle B D C, \angle A D C=\angle B D C$.
Since $\angle A D B=60^{\circ}$, it follows that,
$\angle A D C=\angle B D C=\angle E D B=30^{\circ}$.
Therefore, $D B \perp C E, B C=B E$.
Mo... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. As shown in Figure 3, $A_{0} A_{1}$ is the diameter of a semicircle, $A_{0} A_{1}=2$, $A_{2}, A_{3}, \cdots, A_{k}$, $A_{k+1}, \cdots$, are points on the semicircle, $\angle A_{0} A_{1} A_{2}=1^{\circ}, \angle A_{1} A_{2} A_{3}=2^{\circ}$, $\angle A_{2} A_{3} A_{4}=3^{\circ}, \cdots, \angle A_{k-1} A_{k} A_{k+1}=k^{... | 2. 11 .
As shown in Figure 6, connect
$$
A_{k} O, A_{k+1} O, A_{k} A_{1} \text {, }
$$
then
$$
\begin{array}{l}
\angle O A_{k} A_{k+1}= \\
\angle A_{1} A_{k} A_{k+1}+\angle A_{1} A_{k} O \\
=\angle A_{1} A_{k} A_{k+1}+\angle A_{k} A_{1} O=\left(\frac{k(k+1)}{2}\right)^{\circ} .
\end{array}
$$
Since $A_{k} A_{k+1}60^... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given that $a, b, c$ are real numbers. The functions are $y_{1} = a x^{2} + b x + c, y_{2} = a x + b (a > 0)$. When $-1 \leqslant x \leqslant 1$, it is given that $-1 \leqslant y_{1} \leqslant 1$ and $y_{2}$ has a maximum value of 2. Try to find the area of the figure formed by connecting in sequence... | Three, from $a>0$, we know that $y_{2}$ increases as $x$ increases.
Therefore, $a+b=2$.
When $x=0,1$, $-1 \leqslant c \leqslant 1, -1 \leqslant a+b+c \leqslant 1$, so $-1 \leqslant c=(a+b+c)-2 \leqslant 1-2 \leqslant-1$.
Thus, $c=-1$. Therefore, when $x=0$, $y_{1}=-1$ is the minimum value of $y_{1}=$ $a x^{2}+b x+c$ in... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $\lg a<0, \lg b<0, \lg c<0$, and $\lg (a+b+c)=0$. Then the maximum value of $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right)$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 0 .
Given that $a, b, c$ are positive numbers less than 1, and $a+b+c=1$. We have
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
$\geqslant 9$ (Cauchy-Schwarz inequality).
The equality holds when $a=b=c=\frac{1}{3}$.
Transforming the inequality (1), we get
$$
\begi... | 0 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
6. The sum of $n$ consecutive natural numbers starting from the positive integer $m$ is 2004, and $(m, n)>1$ (not coprime). Then the greatest common divisor $(m, n, 2004)=$
| 6. 12 .
According to the problem, $(m, n)>1$ and $m+(m+1)+\cdots+(m+n-1)=2004$, that is,
$$
\frac{(2 m+n-1) n}{2}=2004 \text {. }
$$
And $(2 m+n-1) n$
$$
=1 \times 4008=3 \times 1336=167 \times 24=501 \times 8 \text {, }
$$
where $2 m+n-1$ and $n$ are one odd and one even, and $2 m+n-1$ is greater than $n$. Therefor... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, let $M$ be a set of $n$ points in the plane, satisfying:
(1) There exist 7 points in $M$ that are the 7 vertices of a convex heptagon;
(2) For any 5 points in $M$, if these 5 points are the 5 vertices of a convex pentagon, then this convex pentagon contains at least one point from $M$ inside it.
Find the minimum... | Three, prove $n \geqslant 11$.
Consider a convex heptagon $A_{1} A_{2} \cdots A_{7}$ with vertices in $M$, and connect $A_{1} A_{5}$. By condition (2), there is at least one point in $M$ within the convex pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$, denoted as $P_{1}$. Connect $P_{1} A_{1}$ and $P_{1} A_{5}$. Then, in the... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $x, y \in \mathbf{R}$, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{2003}+2002(x-1)=-1, \\
(y-2)^{2008}+2002(y-2)=1 .
\end{array}\right.
$$
Then $x+y=$ | 7. 3
Construct the function $f(t)=t^{2008}+2002 t$. It is easy to see that $f(t)$ is an odd function on $\mathbf{R}$, and it is also a monotonically increasing function. From this, we can get $f(x-1)=-f(y-2)$, which means $f(x-1)=f(2-y)$. Therefore, $x-1=2-y, x+y=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Let $x, y, z$ be positive real numbers, and $x+y+z=1$. Find the minimum value of the function
$$
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}
$$
and provide a proof. | 14. Consider the function $g(t)=\frac{t}{1+t^{2}}$, we know that $g(t)$ is an odd function. Since when $t>0$, $\frac{1}{t}+t$ is decreasing in $(0,1)$, it is easy to see that $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in $(0,1)$. For $t_{1} \backslash t_{2} \in(0,1)$ and $t_{1}<t_{2}$, we have
$$
\left(t_{1}-t_{2}\ri... | 0 | Algebra | proof | Yes | Yes | cn_contest | false |
8. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $a, b, c$ form an arithmetic sequence, and $c=10, a \cos A=$ $b \cos B, A \neq B$, then the inradius of $\triangle A B C$ is $\qquad$ | 8. 2 .
Let the inradius of $\triangle ABC$ be $r$. Since $a \cos A = b \cos B$, by the Law of Sines, we get $b \sin A = a \sin B$. Therefore, $\sin 2A = \sin 2B$.
Since $A \neq B$, then $A + B = 90^{\circ}$. Thus, $\triangle ABC$ is a right triangle, $\angle C = 90^{\circ}, a^2 + b^2 = c^2$.
Also, since $c = 10$, and ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Real numbers $x, y, z$ satisfy $x=y+\sqrt{2}, 2 x y+$ $2 \sqrt{2} z^{2}+1=0$. Then the value of $x+y+z$ is $\qquad$ | $=1.0$.
Since $x=y+\sqrt{2}$, we have,
$$
\begin{array}{l}
(x-y)^{2}=2, (x+y)^{2}-4 x y=2, \\
2 x y=\frac{1}{2}(x+y)^{2}-1 .
\end{array}
$$
Substituting this into $2 x y+2 \sqrt{2} z^{2}+1=0$ yields
$$
\frac{1}{2}(x+y)^{2}+2 \sqrt{2} z^{2}=0 \text {. }
$$ | 1.0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Points $A(-4,0)$ and $B(2,0)$ are two fixed points on the $x O y$ plane, and $C$ is a moving point on the graph of $y=-\frac{1}{2} x+2$. How many right triangles $\triangle A B C$ can be drawn that satisfy the above conditions? | 3. 4 .
As shown in Figure 6, draw perpendiculars from $A$ and $B$ to the $x$-axis, intersecting the line $y=-\frac{1}{2} x + 2$ at points $C_{1}$ and $C_{2}$, respectively; with $AB$ as the diameter, draw a semicircle intersecting the line
$$
y=-\frac{1}{2} x + 2 \text { at }
$$
points $C_{3}$ and $C_{4}$. Then,
$\t... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $p>0, q>0$, and satisfy $2 p+\sqrt{p q}-$ $q+\sqrt{p}+\sqrt{q}=0$. Then $(2 \sqrt{p}-\sqrt{q}+2)^{3}=$ $\qquad$ | 2. 1 .
The original expression can be transformed into
$$
2(\sqrt{p})^{2}+\sqrt{p q}-(\sqrt{q})^{2}+(\sqrt{p}+\sqrt{q})=0 \text {, }
$$
which is $(\sqrt{p}+\sqrt{q})(2 \sqrt{p}-\sqrt{q}+1)=0$.
Since $p>0, q>0$, then $\sqrt{p}+\sqrt{q}>0$. Therefore, $2 \sqrt{p}-\sqrt{q}+1=0$, which means $2 \sqrt{p}-\sqrt{q}+2=1$.
Th... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) A research study group from Class 1, Grade 3 of Yuhong Middle School conducted a survey on students' lunch time at the school canteen. It was found that within a unit of time, the number of people buying lunch at each window and the number of people choosing to eat outside due to unwillingness to wait ... | Let each window sell lunch to $x$ people per minute, and $y$ people go out to eat per minute, with the total number of students being $z$ people, and assume that at least $n$ windows need to be open simultaneously. According to the problem, we have
$$
\left\{\begin{array}{l}
45 x=z-45 y, \\
2 \times 30 x=z-30 y, \\
20 ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In a tetrahedron $ABCD$ with volume 12, points $E$, $F$, and $G$ are on edges $AB$, $BC$, and $AD$ respectively, such that $AE = 2EB$, $BF = FC$, and $AG = 2GD$. A plane through points $E$, $F$, and $G$ intersects the tetrahedron in a section $EFGH$, and the distance from point $C$ to this section is 1. The area of ... | 2. 7 .
As shown in Figure 3, it is easy to know that $G E / /$ $D B$. Since $D B / /$ plane $E F H G$, then $H F$ // BD. Therefore, $H$ is also the midpoint of $D C$.
Let the distances from points $A, B, C, D$ to the section $E F H G$ be $h_{a}, h_{b}, h_{c}, h_{d}$, respectively. We have
$$
\begin{array}{l}
h_{b}=h_... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $f(x)=a x^{2}+b x+c$ be a quadratic trinomial with integer coefficients. If integers $m, n$ satisfy $f(m)-f(n)=1$. Then $|m-n|=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. 1 .
$$
\begin{array}{l}
f(m)-f(n)=a\left(m^{2}-n^{2}\right)+b(m-n) \\
=(m-n)[a(m+n)+b],
\end{array}
$$
then $(m-n)(a m+a n+b)=1$.
Since $m-n$ and $a m+a n+b$ are both integers, it must be that
$$
|m-n|=1 \text{. }
$$ | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
8. Given real numbers $a, b, x, y$ satisfy $a+b=x+y=2$, $a x+b y=5$. Then the value of $\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right)$ is $\qquad$ . | 8. -5 .
Given $a+b=x+y=2$, we have
$$
(a+b)(x+y)=a x+b y+a y+b x=4 \text{. }
$$
Since $a x+b y=5$, then,
$$
\begin{array}{l}
a y+b x=-1 . \\
\text{ Thus, }\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right) \\
=(a y+b x)(a x+b y)=-5 .
\end{array}
$$ | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. $n$ is a positive integer, $f(n)=\sin \frac{n \pi}{2}$. Then
$$
f(1991)+f(1992)+\cdots+f(2003)=
$$
$\qquad$ | 7. -1 .
It is easy to know that $f(1991)=-1, f(1992)=0, f(1993)=1$, $f(1994)=0, \cdots, f(2003)=-1$. Therefore, $f(1991)+f(1992)+\cdots+f(2003)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The function defined on the set of positive integers is
$$
f(x)=\left\{\begin{array}{ll}
3 x-1, & x \text { is odd, } \\
\frac{x}{2}, & x \text { is even. }
\end{array}\right.
$$
Let $x_{1}=12, x_{n+1}=f\left(x_{n}\right), n \in \mathbf{N}$, then the number of elements in the set $\left\{x \mid x=x_{n}\right.$, $n ... | 3. 7 .
From $x_{1}=12$ we get $x_{2}=6$, then $x_{3}=3, x_{4}=8, x_{5}=4$, $x_{6}=2, x_{7}=1, x_{8}=2, x_{9}=1, \cdots$, and thereafter it is a cycle of $2,1,2,1$, $\cdots$. Therefore, $x_{n}$ takes a total of 7 different values, i.e., the set $|x| x$ $\left.=x_{n}, n \in \mathbf{N}\right\}$ contains 7 elements. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{49}-\frac{1}{50}\right) \div\left(\frac{1}{20}+\frac{1}{77}+\cdots+\frac{1}{50}\right) \\
= \\
\end{array}
$$ | $=γ 1.1$
$$
\begin{array}{l}
\text { Original expression }=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{49}+\frac{1}{50}\right)-\right. \\
\left.\quad\left(1+\frac{1}{2}+\cdots+\frac{1}{25}\right)\right] \div\left(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\right) \\
=\left(\frac{1}{26}+\frac{1}{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of real roots of the equation $x^{2}|x|-5 x|x|+2 x=0$ is $\qquad$.
The equation $x^{2}|x|-5 x|x|+2 x=0$ has $\qquad$ real roots. | 3.4 .
From the given, we have $x(x|x|-5|x|+2)=0$.
Thus, $x=0$ or $x|x|-5|x|+2=0$.
(1) When $x>0$, $x^{2}-5 x+2=0$, which has two distinct positive real roots;
(2) When $x<0$, $-x^{2}+5 x+2=0$, which has one negative real root.
Therefore, the original equation has 4 real roots. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) The integers $x_{0}, x_{1}, \cdots, x_{2000}$ satisfy the conditions: $x_{0}=0,\left|x_{1}\right|=\left|x_{0}+1\right|,\left|x_{2}\right|=\left|x_{1}+1\right|$, $\cdots,\left|x_{2004}\right|=\left|x_{2003}+1\right|$. Find the minimum value of $\mid x_{1}+x_{2}+\cdots+$ $x_{2004} \mid$. | Three, from the known we can get
$$
\left\{\begin{array}{l}
x_{1}^{2}=x_{0}^{2}+2 x_{0}+1, \\
x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \\
\cdots \cdots . \\
x_{2004}^{2}=x_{2003}^{2}+2 x_{2003}+1 .
\end{array}\right.
$$
Thus, $x_{2004}^{2}=x_{0}^{2}+2\left(x_{0}+x_{1}+\cdots+x_{2003}\right)+2004$.
Given $x_{0}=0$, then
$$
\begi... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The four sides of the quadrilateral pyramid are isosceles triangles with a leg length of $\sqrt{7}$ and a base length of 2. Then the maximum possible volume of this quadrilateral pyramid is | 5. 3 .
There are three scenarios for the quadrilateral pyramid that meet the requirements.
(1) As shown in Figure 4, all four lateral edges are $\sqrt{7}$, and the volume is calculated as $V_{1}=\frac{4}{3} \sqrt{5}$.
(2) As shown in Figure 5, two lateral edges are $\sqrt{7}$.
Construct $O E \perp$ plane $A B C D$, a... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 3: In front of each number in $1, 2, 3, \cdots, 1989$, add a β+β or β-β sign to make their algebraic sum the smallest non-negative number, and write out the equation.
(1989, All-Russian Mathematical Olympiad) | Proof: First, we prove that the algebraic sum is an odd number.
Consider the simplest case:
If all are filled with β+β, then at this moment
$$
1+2+\cdots+1989=995 \times 1989
$$
is an odd number.
For the general case, it only requires adjusting some β+β to β-β.
Since $a+b$ and $a-b$ have the same parity, the parity o... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given that $\alpha, \beta$ are the roots of the equation $x^{2}-x-1=0$. Then the value of $\alpha^{4}+3 \beta$ is $\qquad$
(2003, National Junior High School Mathematics Competition, Tianjin Preliminary Contest) | Explanation: $\alpha^{4}+3 \beta$ is not a symmetric expression of the two roots of the equation, and it is obviously impossible to directly substitute it using Vieta's formulas. We can construct the dual expression $\beta^{4}+3 \alpha$ of $\alpha^{4}+3 \beta$, and calculate the sum and difference of the two expression... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $a b c \neq 0$, and $a+b+c=0$. Then the value of the algebraic expression $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0 | $\begin{array}{l}\text { I. 1.A. } \\ \text { Original expression }=\frac{-(b+c) a}{b c}+\frac{-(a+c) b}{a c}+\frac{-(a+b) c}{a b} \\ =-\left(\frac{a}{b}+\frac{a}{c}\right)-\left(\frac{b}{a}+\frac{b}{c}\right)-\left(\frac{c}{a}+\frac{c}{b}\right) \\ =\frac{a}{a}+\frac{b}{b}+\frac{c}{c}=3 .\end{array}$ | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. The minimum value of the function $f(x)=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$ is $\qquad$ . | 2. 5 .
Obviously, if $xf(-x)$.
Therefore, when $f(x)$ takes its minimum value, there must be $x \geqslant 0$.
As shown in Figure 6, draw a line segment $A B=4$, $A C \perp A B, D B \perp A B$, and $A C=$ $1, B D=2$. For any point $O$ on $A B$, let $O A=x$, then
$$
\begin{array}{l}
O C=\sqrt{x^{2}+1}, \\
O D=\sqrt{(4-x... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$2 . a γ b γ c$ are non-zero real numbers, and $a+b+c \neq 0$. If $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$, then $\frac{(a+b)(b+c)(c+a)}{a b c}$ equals ( ).
(A) 8
(B) 4
(C) 2
(D) 1 | 2.A.
Let $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=k$, then $a+b+c=k(a+b+c)$.
Since $a+b+c \neq 0$, we have $k=1$.
Therefore, the original expression $=\frac{2 c \cdot 2 a \cdot 2 b}{a b c}=8$. | 8 | Algebra | MCQ | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 1. If } n \text { satisfies }(n-2003)^{2}+(2004-n)^{2} \\ =1 \text {, then }(2004-n)(n-2003)=\end{array}$ | $\begin{array}{l} \text { II.1.0. } \\ \text { From }(n-2003)^{2}+2(n-2003)(2004-n)+ \\ (2004-n)^{2}-2(n-2003)(2004-n) \\ =(n-2003+2004-n)^{2}-2(n-2003)(2004-n) \\ =1-2(n-2003)(2004-n)=1, \\ \text { we get }(n-2003)(2004-n)=0 .\end{array}$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If real numbers $x, y$ satisfy $x \geqslant 0$, and
$$
\max \{1-x, x-1\} \leqslant y \leqslant x+2 \text {, }
$$
then the minimum value of the bivariate function $u(x, y)=2 x+y$ is
$\qquad$ . | 6.1.
From the given information, we have
$$
\begin{array}{l}
u(x, y)=2 x+y \geqslant 2 x+\max \{1-x, x-1\} \\
=\max \{2 x+(1-x), 2 x+(x-1)\} \\
=\max \{x+1,3 x-1\} \\
\geqslant \max \{1,-1\}=1,
\end{array}
$$
i.e., $u(x, y) \geqslant 1$, with equality holding if and only if $x=0, y=1$. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Given $a>0$, and
$$
\sqrt{b^{2}-4 a c}=b-2 a c \text {. }
$$
Find the minimum value of $b^{2}-4 a c$.
(2004, "TRULY ${ }^{\circledR}$ Xinli Cup" National Junior High School Mathematics Competition) | Let $y=a x^{2}+b x+c$.
From $a<0$, we know $\Delta=b^{2}-4 a c>0$.
Therefore, the graph of this quadratic function is a parabola opening downwards, and it intersects the $x$-axis at two distinct points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$.
Since $x_{1} x_{2}=\frac{c}{a}<0$, let's assume $x_{1}<x_{2}$.
The... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $S=\{1,2,3,4\}$, and the sequence $a_{1}, a_{2}, \cdots, a_{n}$ has the following property: for any non-empty subset $B$ of $S$, there are consecutive $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997, Shanghai High School Mathematics Competition) | (Since a binary subset containing a fixed element of $S$ has 3 elements, any element of $S$ appears at least twice in the sequence. Therefore, the minimum value of $n$ is estimated to be 8. On the other hand, an 8-term sequence: $3,1,2,3,4,1,2,4$ satisfies the condition, so the minimum value of $n$ is 8.) | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 13 Unload several boxes from a cargo ship, with a total weight of $10 \mathrm{t}$, and the weight of each box does not exceed $1 \mathrm{t}$. To ensure that these boxes can be transported away in one go, how many trucks with a carrying capacity of $3 \mathrm{t}$ are needed at least?
(1990, Jiangsu Province Juni... | First, note that the weight of each box does not exceed $1 \mathrm{t}$, so the weight of the boxes that each vehicle can carry at one time will not be less than $2 \mathrm{t}$; otherwise, another box can be added.
Let $n$ be the number of trucks needed, and the weights of the boxes they carry be $a_{1}, a_{2}, \cdots,... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Three, (16 points) Given real numbers $a, b, c$ satisfy $a+b+c=2$, $abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$. | (1) Let's assume $a=\max \{a, b, c\}$. From the given conditions, we have
$$
a>0, b+c=2-a, bc=\frac{4}{a} \text{. }
$$
Therefore, $b$ and $c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$. Thus,
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \times \frac{4}{a} \geqslant 0 \\
\Leftrightarrow a^{... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $k$ is an integer. If the quadratic equation $k x^{2}+(2 k+3) x+1$ $=0$ has rational roots, then the value of $k$ is $\qquad$ | 2. -2 .
From the given, $\Delta_{1}=(2 k+3)^{2}-4 k$ is a perfect square. Let $(2 k+3)^{2}-4 k=m^{2}(m$ be a positive integer), that is,
$$
4 k^{2}+8 k+9-m^{2}=0 \text {. }
$$
Considering equation (1) as a quadratic equation in $k$, by the problem's condition, it has integer roots, so the discriminant of equation (1)... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ .
(2003, National Junior High School Mathematics League) | (Solution: $m=4$ ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5,5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$ such that for all integers $n$, $a_{n} = b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$.... | Explanation: Divide the sequence $\{1,3,3,3,5,5,5,5,5, \cdots\}$ into groups:
(1), $(3,3,3),(5,5,5,5,5), \cdots,(2 k-1, 2 k-1, \cdots, 2 k-1), \cdots$,
where the $k$-th group consists of $2 k-1$ occurrences of $2 k-1$. If $a_{n}$ is in the $k$-th group, then $a_{n}=2 k-1$. Therefore,
$$
\begin{array}{l}
1+3+\cdots+(2 k... | 2 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Example 5 Find all positive integers $k$ such that for any positive numbers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a, b, c$.
(First China Girls Mathematical Olympiad)
Analysis: To find $k$, we can first determine the upper... | Solution: It is easy to know that $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
Therefore, the required $k>5$.
Take $a, b, c$ that do not form a triangle ($a=1, b=1, c=2$), then we have
$$
k(1 \times 1+1 \times 2+2 \times 1) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right).
$$
This gives $k \leqslant 6$.
Therefore, if $55\left(... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
(1999, National Junior High School Mathematics League) This problem has a resemblance to a 1979 college entrance exam question. | Example $\mathbf{3}^{\prime}$ If $(z-x)^{2}-4(x-y)(y-z)=0$, prove that $x, y, z$ form an arithmetic sequence.
In Example 3, taking $a=b=c$ can guess the answer to be 2. But this poses a risk of "root reduction," because the condition is a quadratic expression, while the conclusion
$$
\frac{b+c}{a}=2 \Leftrightarrow b+... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. In the Cartesian coordinate system $x 0 y$, given two points $M(-1,2)$ and $N(1,4)$, point $P$ moves on the $x$-axis. When $\angle M P N$ takes the maximum value, the x-coordinate of point $P$ is $\qquad$ $-$.
| 12.1.
The center of the circle passing through points $M$ and $N$ lies on the perpendicular bisector of line segment $MN$, which is $y=3-x$. Let the center of the circle be $S(a, 3-a)$, then the equation of circle $S$ is $(x-a)^{2}+(y-3+a)^{2}=2\left(1+a^{2}\right)$.
For a fixed-length chord, the inscribed angle subt... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $a, b, c \in \mathbf{N}_{+}$, and the parabola $f(x) = ax^{2} + bx + c$ intersects the $x$-axis at two different points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a + b + c$.
(1996, National Junior High School Mathematics Competition) | Let $f(x)=a x^{2}+b x+c, A\left(x_{1}, 0\right)$, and $B\left(x_{2}, 0\right)$. According to the problem, we have
$$
\left\{\begin{array}{l}
\Delta=b^{2}-4 a c>0, \\
f(1)=a+b+c>0, \\
f(-1)=a-b+c>0 .
\end{array}\right.
$$
From (1), we get $b>2 \sqrt{a c}$; from (3), we get $a+c>b$.
$$
\begin{array}{l}
a+c>b \Rightarrow... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a, b$ are the real roots of the equation $x^{2}+2 p x+3=0$, $c, d$ are the real roots of the equation $x^{2}+2 q x+3=0$, and $e, v$ are the real roots of the equation $x^{2}+2\left(p^{2}-q^{2}\right) x+3=0$, where $p, q$ are real numbers. Then, the value of $\frac{(a-c)(b-c)(a+d)(b+d)}{e+v}$ is $\qquad$ ... | 3.6 .
From the relationship between roots and coefficients, we get
$$
\begin{array}{l}
a+b=-2 p, a b=3 ; c+d=-2 q, c d=3 ; \\
e+f=-2\left(p^{2}-q^{2}\right), e f=3 . \\
\text { Therefore } \frac{(a-c)(b-c)(a+d)(b+d)}{e+f} \\
=\frac{\left[a b-(a+b) c+c^{2}\right]\left[a b+(a+b) d+d^{2}\right]}{e+f} \\
=\frac{\left(c d+... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) In the quadratic function $f(x)=a x^{2}+b x$ $+c$, $a$ is a positive integer, $a+b+c \geqslant 1, c \geqslant 1$, and the equation $a x^{2}+b x+c=0$ has two distinct positive roots less than 1. Find the minimum value of $a$.
δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζε¦δΈοΌ
```
One, (20 points) In the quadratic function $f(x)... | Given the equation $a x^{2}+b x+c=0$ has two roots $x_{1} γ x_{2}(0<x_{1}<x_{2})$, and $x_{1} x_{2}=\frac{c}{a}=\frac{3}{5}$, we have $a x_{1} x_{2}=3$.
Since $x_{1}+x_{2}=\frac{11}{10}$, we have $a(x_{1}+x_{2})=\frac{11}{10}a>4.4$.
Given $a>4$, and $a$ is a positive integer, so, $a \geqslant 5$.
Taking $f(x)=5\left... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Try to determine, for any $n$ positive integers, the smallest positive integer $n$ such that at least 2 of these numbers have a sum or difference that is divisible by 21.
| Three, let $A_{i}$ represent all numbers that leave a remainder of $i$ when divided by 21; $i=0,1$, $\cdots, 20$; let $B_{j}=A_{j} \cup A_{21-j}, j=1,2, \cdots, 10, B_{11}=A_{0}$.
By taking 12 numbers, there will be at least 2 numbers belonging to the same category among the 11 categories $B_{1}, B_{2}$, $\cdots, B_{1... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Consider the binomial coefficients $\mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ as a sequence. When $n \leqslant 2004\left(n \in \mathbf{N}_{+}\right)$, the number of sequences in which all terms are odd is $\qquad$ groups. | 2.10.
From $\mathrm{C}_{n}^{r}=\mathrm{C}_{n-1}^{r-1}+C_{n-1}^{r}$, we know that the 2004 numbers $n=1,2, \cdots, 2004$ can be arranged in the shape of Pascal's Triangle, and the even and odd numbers can be represented by 0 and 1, respectively, as shown in Figure 3.
The pattern of the changes in the parity of each ter... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=a_{n}^{2}+2 a_{n}+3, n \in \mathbf{N}_{+}$, try to find
$$
M=a_{m}^{2}+b_{n}^{2}+m^{2}+n^{2}-2\left(a_{m} b_{n}+m n\right)
$$
$\left(m γ n \in \mathbf{N}_{+}\right)$ the minimum value. | (1) It is easy to know that when $n \geqslant 2, n \in \mathbf{N}_{+}$, we have
$4\left(S_{n}-S_{n-1}\right)=\left(a_{n}+1\right)^{2}-\left(a_{n-1}+1\right)^{2}$.
Thus, $\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}-2\right)=0$.
Since $a_{n}+a_{n-1}>0$, it follows that,
$$
a_{n}-a_{n-1}=2 \text {. }
$$
It is easy to s... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given $a>0$, and
$$
\sqrt{b^{2}-4 a c}=b-2 a c \text{. }
$$
Find the minimum value of $b^{2}-4 a c$. | Solution 1: The given condition is
$$
\frac{-b+\sqrt{b^{2}-4 a c}}{2 a c}=-1(a c \neq 0),
$$
This indicates that the quadratic equation
$$
a c x^{2}+b x+1=0
$$
has a real root $x=-1$. Substituting into equation (1) yields
$$
a c=b-1 \text {. }
$$
Transform equation (3) into the overall structure of the discriminant
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If the decimal parts of $7+\sqrt{7}$ and $7-\sqrt{7}$ are $a$ and $b$ respectively, then $a b-3 a+2 b+1=$ $\qquad$ . | 2.0.
Since $2<\sqrt{7}<3$, therefore, the decimal part of $7+\sqrt{7}$ is $a=\sqrt{7}-2$. Also, because $-3<-\sqrt{7}<-2$, then $0<3-\sqrt{7}<1$. Therefore, the decimal part of $7-\sqrt{7}$ is $b=3-\sqrt{7}$. Hence
$$
\begin{array}{l}
a b-3 a+2 b+1 \\
=(a+2)(b-3)+7=\sqrt{7} \times(-\sqrt{7})+7=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. A bookstore is holding a summer book fair, where book buyers can enjoy the following discounts:
(1) For a single purchase not exceeding 50 yuan, no discount is given;
(2) For a single purchase exceeding 50 yuan but not exceeding 200 yuan, a 10% discount is given on the marked price;
(3) For a single purchase exceedi... | 4.3.
Obviously, 81 and 126 are both the prices after the discount. If not discounted, the combined price should be
$$
(81+126) \div 0.9=230 \text { (yuan). }
$$
Since 230 yuan exceeds 200 yuan, according to (3), it should be
$$
200 \times 0.9+30 \times 0.8=204 \text { (yuan) } \text {. }
$$
Thus, we have $81+126-204... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $\{x\}$ denote the fractional part of the real number $x$. If $a=$ $(5 \sqrt{13}+18)^{2005}$, then $a \cdot\{a\}=$ $\qquad$ | 5.1.
Let $b=(5 \sqrt{13}-18)^{2005}$.
Since $0<5 \sqrt{13}-18<1$, we have $0<b<1$.
By the binomial theorem, we know
$$
\begin{aligned}
a= & (5 \sqrt{13})^{2005}+\mathrm{C}_{2005}^{1}(5 \sqrt{13})^{2004} \times 18^{1}+\cdots+ \\
& \mathrm{C}_{2005}^{r}(5 \sqrt{13})^{2005-r} \times 18^{r}+\cdots+18^{2005}, \\
b= & (5 \s... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $a \neq b$, and
$$
a^{2}-4 a+1=0, b^{2}-4 b+1=0 \text {. }
$$
Find the value of $\frac{1}{a+1}+\frac{1}{b+1}$. | Solution 1: Since
$$
\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a b+(a+b)+1},
$$
we only need to find the values of $a b$ and $a+b$.
From the given conditions, $a$ and $b$ are the roots of the quadratic equation $x^{2}-4 x+1=0$, so
$$
a+b=4, a b=1 \text {. }
$$
Therefore, $\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 As shown in Figure 14, villages $A$, $B$, and $C$ are located along an east-west oriented highway, with $AB=2 \text{ km}$, $BC=3 \text{ km}$. To the due north of village $B$ is village $D$, and it is measured that $\angle ADC=45^{\circ}$. Now the $\triangle ADC$ area is planned to be a development zone, exce... | Solution: As shown in Figure 15, construct the axisymmetric figure of Rt $\triangle A D B$ with respect to the line containing $D A$, which is Rt $\triangle A D B_{1}$. It is easy to know that
Rt $\triangle A D B \cong$
Rt $\triangle A D B_{1}$.
Construct the axisymmetric figure of Rt $\triangle C D E$ with respect to ... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7.2. There are 12 people in the room,
some of whom always lie, while the rest always tell the truth. One of them says: βThere are no honest people hereβ; another says: βThere is at most 1 honest person hereβ; the third person says: βThere are at most 2 honest people hereβ; and so on, until the 12th person says: βThere... | 7.2. There are 6 honest people in the room.
Obviously, the first few people are not honest. If the 6th person is telling the truth, then there are no more than 5 honest people in the room. On the other hand, from him onwards, the next 7 people are telling the truth, and thus, they should all be honest, which is imposs... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
7.2. Does there exist a number of the form $1000 \cdots 001$ that can be divided by a number of the form $111 \cdots 11$? | 7.2. Only for 11
In fact, $1000 \cdots 001=99 \cdots 9+2$, and the remainder of $99 \cdots 9$ divided by $111 \cdots 11$ is a number of the form $99 \cdots 9$ with fewer digits than the divisor. Therefore, if $99 \cdots 9+2$ can be divided by $111 \cdots 11$, then $99 \cdots 9+2$ must equal $111 \cdots 11$. This is on... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $k$ be a real number, and the quadratic equation $x^{2}+k x+k+1=0$ has two real roots $x_{1}$ and $x_{2}$. If $x_{1}+2 x_{2}^{2}=k$, then $k$ equals $\qquad$ . | 2.5.
From $\Delta \geqslant 0$ we get
$$
k^{2}-4 k-4 \geqslant 0 \text {. }
$$
Since $x_{2}^{2}=-k x_{2}-k-1$, then,
$$
\begin{array}{l}
x_{1}+2 x_{2}^{2}=k \Rightarrow x_{1}-2 k x_{2}=3 k+2 \\
\Rightarrow\left(x_{1}+x_{2}\right)-(2 k+1) x_{2}=3 k+2 \\
\Rightarrow-k-(2 k+1) x_{2}=3 k+2 \\
\Rightarrow-(2 k+1) x_{2}=2(... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of integers $x$ that make $x^{2}+x-2$ a perfect square is $\qquad$.
Make the above text in English, please keep the original text's line breaks and format, and output the translation result directly. | 3.4 .
Let $x^{2}+x-2=y^{2}, y \in \mathbf{N}$, then $4 x^{2}+4 x-8=(2 y)^{2}$,
i.e., $(2 x-2 y+1)(2 x+2 y+1)=3^{2}$.
Therefore, we have $\left\{\begin{array}{l}2 x-2 y+1=1, \\ 2 x+2 y+1=9 ;\end{array}\left\{\begin{array}{l}2 x-2 y+1=3, \\ 2 x+2 y+1=3 ;\end{array}\right.\right.$ $\left\{\begin{array}{l}2 x-2 y+1=-3, \\... | 4 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Five, (20 points) Find the smallest natural number $k$, such that for any $x \in [0,1]$ and $n \in \mathbf{N}_{+}$, the inequality
$$
x^{k}(1-x)^{n}<\frac{1}{(1+n)^{3}}
$$
always holds. | Given $x \in [0,1]$, by the AM-GM inequality, we have
$$
\begin{array}{l}
x^{k}(1-x)^{n} \\
=\underbrace{x \cdots \cdots x}_{k \uparrow} x \cdot \underbrace{(1-x) \cdot(1-x) \cdots \cdots(1-x)}_{n \uparrow} \\
=\left(\frac{n}{k}\right)^{n} \underbrace{x \cdots \cdots x}_{k \uparrow} \cdot \underbrace{\frac{k(1-x)}{n} \... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}$ $=p$. Then $p^{3}+p^{2}+p=$ $\qquad$ | 4.1 .
From the given, we have
$$
\begin{array}{l}
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y}=\frac{x+y-z}{x+y+z}, \\
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z} .
\end{array}
$$
Then \( p^{3}+p^{2}+p=1 \). | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the minimum value of the function with real variables $x$ and $y$
$$
u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}
$$
(2nd "Hope Cup" National Mathematics Invitational Competition) | Explanation: The original formula is transformed to
$$
u(x, y)=\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}+(x-y)^{2}-2 \text {. }
$$
Consider the points $P_{1}\left(x, \frac{9}{x}\right), P_{2}\left(y,-\sqrt{2-y^{2}}\right)$. When $x \in \mathbf{R}(x \neq 0)$, point $P_{1}$ lies on a hyperbola with the coordinate axes... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7.5. If $a, b, c, d, e, f, g, h, k$ are all 1 or -1, try to find the maximum possible value of
$$
a e k - a f h + b f g - b d k + c d h - c e g
$$ | 7.5. Since each term in the expression $a e k - a f h + b f g - b d k + c d h - c e g$ is either 1 or -1, the value of the expression is even. However, the expression cannot equal 6. If it did, then $a e k$, $b f g$, and $c d h$ would all have to be 1, making their product 1, while $a f h$, $b d k$, and $c e g$ would a... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $x, y$ satisfy $x^{3}+y^{3}=2$. Then the maximum value of $x+y$ is $\qquad$ . | Ni.1.2.
Let $x+y=k$, it is easy to know that $k>0$.
From $x^{3}+y^{3}=2$, we get
$$
(x+y)\left(x^{2}-x y+y^{2}\right)=2 \text {. }
$$
Thus, $x y=\frac{1}{3}\left(k^{2}-\frac{2}{k}\right)$.
From this, we know that $x, y$ are the two real roots of the equation about $t$
$$
t^{2}-k t+\frac{1}{3}\left(k^{2}-\frac{2}{k}\ri... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three piles of stones have the numbers 21, 10, and 11, respectively. Now, the following operation is performed: each time, 1 stone is taken from any two piles, and then these 2 stones are added to the other pile. The question is:
(1) Can the number of stones in the three piles be 4, 14, and 24 after several such opera... | (1) It can be achieved.
The minimum number of operations required is 6, for example:
$$
\begin{array}{l}
(21,10,11) \rightarrow(23,9,10) \rightarrow(22,8,12) \\
\rightarrow(24,7,11) \rightarrow(23,6,13) \rightarrow(25,5,12) \\
\rightarrow(24,4,14) .
\end{array}
$$
Since the minimum number of stones in one pile decreas... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the sequence
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}+\frac{2}{3}, \cdots, \\
a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1} .
\end{array}
$$
Then $\lim _{n \rightarrow \infty}\left(\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}\right)=$ ( ).
(A) 2
(B) 3
(C)... | 6.C.
$$
\begin{array}{l}
\text { Because } a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1} \\
=\frac{n(n+1)}{2(n+1)}=\frac{n}{2}, \\
\text { so } \frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}} \\
=\frac{4}{1 \times 2}+\frac{4}{2 \times 3}+\cdots+\frac{4}{n(n+1)} \\
=4\left[\left(1-\fr... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
II. (50 points) An old man divides his savings of $m$ gold coins among his $n$ children ($m, n$ are positive integers greater than 1). First, he gives the eldest child 1 gold coin and $\frac{1}{7}$ of the remainder; then, from the remaining coins, he gives the second child 2 gold coins and $\frac{1}{7}$ of the remainde... | Let the number of gold coins left after giving to the $k$-th child be $a_{k}$, then $a_{0}=m, a_{n-1}=n$,
$$
a_{k}=a_{k-1}-\left(k+\frac{a_{k-1}-k}{7}\right)=\frac{6}{7}\left(a_{k-1}-k\right) \text {. }
$$
Thus, $a_{k}+6 k-36=\frac{6}{7}\left[a_{k-1}+6(k-1)-36\right]$.
This indicates that the sequence $b_{k}=a_{k}+6 k... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$1 . m$ is what integer when the equation
$$
\left(m^{2}-1\right) x^{2}-6(3 m-1) x+72=0
$$
has two distinct positive integer roots? | (Tip: $m^{2}-1 \neq 0, m \neq \pm 1$. Since $\Delta=36(m-3)^{2}>$ 0, hence $m \neq 3$. Using the quadratic formula, we get $x_{1}=\frac{6}{m-1}, x_{2}=$ $\frac{12}{m+1}$. Therefore, $(m-1)|6,(m+1)| 12$. So $m=2$, at this point, $x_{1}=6, x_{2}=4$.) | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. From 3 boys and $n$ girls, any 3 people are selected to participate in a competition, given that the probability of having at least 1 girl among the 3 people is $\frac{34}{35}$. Then $n=$ $\qquad$ . | 13.4. From the condition, $1-\frac{\mathrm{C}_{3}^{3}}{\mathrm{C}_{n+3}^{3}}=\frac{34}{35}$, solving for $n$ yields $n=4$. | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
14. There are 10 table tennis players participating in a round-robin tournament. The match results show that there are no draws, and among any 5 players, there is 1 player who wins against the other 4, and 1 player who loses to the other 4. Then the number of players who won exactly two matches is $\qquad$. | 14.1.
It can be proven that under the given conditions, no two players have the same number of wins. Therefore, the number of wins for 10 players are 10 different numbers: $0,1, \cdots, 9$. Hence, the number of players who win exactly two games is 1.
If not, suppose there exist $A$ and $B$ with the same number of win... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $m$ be a non-zero integer, and the quadratic equation in $x$, $m x^{2}-(m-1) x+1=0$, has rational roots. Find the value of $m$.
| (Given: Let $\Delta=(m-1)^{2}-4 m=n^{2}, n$ be a non-negative integer, then $(m-3)^{2}-n^{2}=8$, i.e., $(m-3-n)(m-3+n)=$ 8. Following Example 2, we get $\left\{\begin{array}{l}m=6, \\ n=1\end{array}\right.$ or $\left\{\begin{array}{l}m=0, \\ n=1\end{array}\right.$ (discard). Therefore, $m=6$, and the two roots of the e... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6.4. There is a 36-digit number, in which the digits $1,2, \cdots, 9$ each appear 4 times, and except for 9, all other digits are less than the digit that follows them. It is known that the first digit of the number is 9. What is the last digit of the number? Please provide all possible answers and prove that there are... | 6.4. The last digit of this number is 8.
According to the problem, only 9 can follow 8. If all four 8s are located within the number, then each of them is followed by one 9, and adding the one 9 at the beginning, there are a total of 5 nines. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7.2. Person A and Person B perform division with remainder on the same number. A divides it by 8, and B divides it by 9. It is known that the sum of the quotient obtained by A and the remainder obtained by B equals 13. Try to find the remainder obtained by A.
| 7.2. Let $a$ and $b$ represent the quotient and remainder obtained by Jia, and let $c$ and $d$ represent the quotient and remainder obtained by Yi. Thus, we have
$$
8 a+b=9 c+d, a+d=13 \text{. }
$$
Substituting $d=13-a$ into the first equation, we get $9(a-c)=$ $13-b$, so $13-b$ is divisible by 9. Since $b$ can only b... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6.3. Several boys and 5 girls are sitting around a round table, and there are 30 pieces of bread on the plate on the table. Each girl takes 1 piece of bread from the plate for each boy she knows, and then each boy takes 1 piece of bread from the plate for each girl he does not know, at which point the bread on the plat... | 6.3. There are $n$ boys, so there are $5 n$ different βboy-girlβ pairs. In each such βpairβ, 1 piece of bread was taken, because if the 2 people in the pair know each other, the girl took 1 piece of bread for the boy; and if they do not know each other, the boy took 1 piece of bread for the girl. Therefore, $5 n=$ $30,... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
13. Given that $f(x)$ is defined on $(-1,1)$, $f\left(\frac{1}{2}\right)=-1$, and satisfies $x, y \in (-1,1)$, we have $f(x) + f(y) = f\left(\frac{x+y}{1+xy}\right)$.
(1) The sequence $\{x_n\}$ satisfies
$$
x_1 = \frac{1}{2}, \quad x_{n+1} = \frac{2x_n}{1 + x_n^2}.
$$
Let $a_n = f(x_n)$. Find the general term formula ... | $$
\begin{array}{l}
\text { Therefore, } 1+f\left(\frac{1}{b_{1}}\right)+f\left(\frac{1}{b_{2}}\right)+\cdots+f\left(\frac{1}{b_{2000}}\right)+f\left(\frac{1}{2004}\right) \\
=1+\left[f\left(\frac{1}{2}\right)-f\left(\frac{1}{3}\right)\right]+\left[f\left(\frac{1}{3}\right)-f\left(\frac{1}{4}\right)\right]+ \\
\quad \c... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $a, b, c$ are real numbers, and $2a + b + c = 5, b - c = 1$. Then the maximum value of $ab + bc + ca$ is | From $2 a+b+c=5, b-c=1$, we can obtain
$$
b=3-a, c=2-a \text {. }
$$
Therefore, $a b+b c+c a$
$$
\begin{array}{l}
=a(3-a)+(3-a)(2-a)+a(2-a) \\
=-a^{2}+6 \leqslant 6 .
\end{array}
$$ | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 2, quadrilateral $ABCD$ is inscribed in $\odot O$, with $BD$ being the diameter of $\odot O$, and $\overparen{AB}=\overparen{AD}$. If $BC + CD = 4$, then the area of quadrilateral $ABCD$ is $\qquad$ . | 3.4 .
From $\overparen{A B}=\overparen{A D}$, we know $A B=A D$.
Also, $B D$ is the diameter of $\odot 0$, so $\angle A=\angle C=90^{\circ}$.
Let $A B=A D=x, B C=y, C D=z$.
By the Pythagorean theorem, we have
$$
x^{2}+x^{2}=y^{2}+z^{2} \text {. }
$$
Thus, $2 x^{2}=y^{2}+z^{2}=(y+z)^{2}-2 y z$.
Also, $S_{\text {quadri... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given that $a$ is a positive integer not greater than 2005, and $b, c$ are integers, the parabola $y=a x^{2}+b x+c$ is above the $x$-axis and passes through points $A(-1,4 a+7)$ and $B(3,4 a-1)$. Find the minimum value of $a-b+c$.
The parabola $y=a x^{2}+b x+c$ passes through points $A(-1,4 a+7)$ an... | Three, since $a$ is a positive integer, the parabola opens upwards.
The parabola $y=a x^{2}+b x+c$ is above the $x$-axis, so,
$$
b^{2}-4 a c<0.
$$
Given $b=-2-2a$ and $c=a+5$, we have
$$
b^{2}-4 a c=(-2-2a)^{2}-4a(a+5)=4+8a+4a^{2}-4a^{2}-20a=4-12a<0.
$$
This implies
$$
a>\frac{1}{3}.
$$
Since $a$ is a positive integer ... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. A cube with an edge length of a certain integer is cut into 99 smaller cubes, 98 of which are unit cubes with an edge length of 1, and the other cube also has an integer edge length. Then its edge length is $\qquad$ | 4.3.
Let the edge length of the original cube be $a$, and the edge length of the other cube after cutting be $b\left(a, b \in \mathbf{Z}_{+}, a>b\right)$. Then
$$
a^{3}-98=b^{3} \text {, }
$$
i.e., $(a-b)\left(a^{2}+a b+b^{2}\right)=98$.
Thus, $\left\{\begin{array}{l}a-b=1, \\ a^{2}+a b+b^{2}=98 ;\end{array}\right.$
... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given a non-constant sequence $\left\{a_{i}\right\}$ satisfies
$$
a_{i}^{2}-a_{i-1} a_{i}+a_{i-1}^{2}=0 \text {, }
$$
and $a_{i+1} \neq a_{i-1}, i=1,2, \cdots, n$.
For a given positive integer $n, a_{1}=a_{n+1}$. Then
$$
\sum_{i=0}^{n-1} a_{i}=
$$ | 6.0 .
Given $a_{1}^{2}-a_{1-1} a_{i}+a_{1-1}^{2}=0$, so,
$$
a_{i+1}^{2}-a_{i} a_{i+1}+a_{i}^{2}=0 \text {. }
$$
Subtracting the two equations, we get
$$
\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}\right)-a_{i}\left(a_{i+1}-a_{i-1}\right)=0 .
$$
That is, $\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}-a_{i}\... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4: On a circular road in a certain town, there are five primary schools in sequence: No.1 Primary School, No.2 Primary School, No.3 Primary School, No.4 Primary School, and No.5 Primary School. They have 15, 7, 11, 3, and 14 computers respectively. To make the number of computers in each school the same, how ma... | Solution: As shown in Figure 4, let $A, B, C, D, E$ represent the first to fifth schools, respectively, and let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the number of computers transferred to the neighboring schools. Let the total number of transfers be $y$. According to the problem, we have
$$
\begin{array}{l}
7+x_{1}-x... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
$13.2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5$ can achieve the minimum value of $\qquad$ . | 13. -10 .
Original expression $=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10$. When $x=2, y=-1$, it has the minimum value -10. | -10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $a=1+2+\cdots+2004$. Then the remainder when $a$ is divided by 17 is | 8.1.
$$
\begin{array}{l}
a=1+2+\cdots+2004 \\
=\frac{2004 \times 2005}{2}=2009010 .
\end{array}
$$
Then 2009010 divided by 17 gives a quotient of 118177, with a remainder of 1. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given $f(x)=x^{2}+x-1$. If $a b^{2} \neq 1$, and $f\left(a^{-1}\right)=f\left(b^{2}\right)=0$, then $\frac{a}{1+a b^{2}}=$ $\qquad$ . | 9. -1 .
Given $f(x)=x^{2}+x-1, f\left(a^{-1}\right)=f\left(b^{2}\right)=0, a b^{2} \neq 1$, we know that $a^{-1}$ and $b^{2}$ are the two real roots of $f(x)=x^{2}+x-1$.
By Vieta's formulas, we get $\frac{1}{a}+b^{2}=-1, \frac{b^{2}}{a}=-1$. Thus, $\frac{1}{a}+b^{2}=\frac{b^{2}}{a}=-1$.
Therefore, we have $1+a b^{2}=-... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. A test paper has 4 multiple-choice questions, each with three options (A), (B), (C). Several students take the exam, and after grading, it is found that: any 3 students have 1 question where their answers are all different. How many students can take the exam at most? | (ζη€Ί: If 10 people take the exam, then for the 1st question, at least 7 people choose two options: for the 2nd question, at least 5 people among these 7 choose two options; for the 3rd question, at least 4 people among these 5 choose two options; for the 4th question, at least 3 people among these 4 choose two options. ... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 1, there is a rectangular piece of paper $A B C D, A B=8$, $A D=6$. The paper is folded so that the edge $A D$ lies on the edge $A B$, with the fold line being $A E$. Then, $\triangle A E D$ is folded along $D E$ to the right, and the intersection of $A E$ and $B C$ is point $F$. The area of $\tri... | - .1.A.
From the folding process, we know $D E=A D=6, \angle D A E=\angle C E F=$ $45^{\circ}$. Therefore, $\triangle C E F$ is an isosceles right triangle, and $E C=8-6$ $=2$. Hence, $S_{\triangle C E F}=2$. | 2 | Geometry | MCQ | Yes | Yes | cn_contest | false |
7. In the Cartesian coordinate system, the parabola
$$
y=x^{2}+m x-\frac{3}{4} m^{2}(m>0)
$$
intersects the $x$-axis at points $A$ and $B$. If the distances from points $A$ and $B$ to the origin are $O A$ and $O B$, respectively, and satisfy $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, then the value of $m$ is $\qquad$. | 7. 2 .
Let the roots of the equation $x^{2}+m x-\frac{3}{4} m^{2}=0$ be $x_{1}$ and $x_{2}$, and $x_{1}<0<x_{2}$.
From $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, we know $O A>O B$.
Also, since $m>0$, the axis of symmetry of the parabola is to the left of the $y$-axis, thus,
$$
O A=\left|x_{1}\right|=-x_{1}, O B=x_{2} ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (20 points) Given $a, b, c \in \mathbf{R}_{+}$, and
$$
\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1 \text{. }
$$
Prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant 12$. | Given, we have
$$
\frac{1}{1+\frac{1}{a}}+\frac{1}{1+\frac{1}{b}}+\frac{1}{1+\frac{1}{c}}=1 \text {. }
$$
Let $x=\frac{1}{1+\frac{1}{a}}, y=\frac{1}{1+\frac{1}{b}}, z=\frac{1}{1+\frac{1}{c}}$. Then
$$
x+y+z=1 \text {. }
$$
From $\frac{1}{a}=\frac{1}{x}-1, \frac{1}{b}=\frac{1}{y}-1, \frac{1}{c}=\frac{1}{z}-1$, we get
... | 12 | Inequalities | proof | Yes | Yes | cn_contest | false |
4. (24 points)(1) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals 2005, find the minimum value of $n$ and explain the reason;
(2) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals $2... | 4. (1) Since $2005=1728+125+125+27=12^{3}+$ $5^{3}+5^{3}+3^{3}$, therefore, there exists $n=4$, such that $n_{\min } \leqslant 4$.
Also, $10^{3}=1000,11^{3}=1331,12^{3}=1728,13^{3}=$ 2197, and $12^{3}3 \times 8^{3}$, so the edge length of the largest cube can only be $9,10, 11$ or 12. And
$$
\begin{array}{l}
20050,
\e... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If $\frac{1+a}{1-a}=\frac{1-b}{1+b}$, then, $(2+a)(2+b)+b^{2}$
is equal to ( ).
(A) 4
(B) -4
(C) 2
(D) -2 | 1. A.
From the given, we have $(1+a)(1+b)=(1-a)(1-b)$, simplifying to $a+b=0$, which means $a=-b$.
Therefore, $(2+a)(2+b)+b^{2}=(2-b)(2+b)+b^{2}=4$. | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. Place the natural numbers $1,2, \cdots, 2 n$ randomly on a circle. It is found that among all sets of three consecutive numbers, there are $a$ sets where all three numbers are odd, $b$ sets where exactly two numbers are odd, $c$ sets where only one number is odd, and $d$ sets where all three numbers are even. If $a ... | 4. -3 .
If each number is counted 3 times, a total of $6 n$ numbers are counted ($3 n$ odd numbers, $3 n$ even numbers), then the system of equations can be set up as
$$
\left\{\begin{array}{l}
3 a+2 b+c=3 n, \\
2 c+b+3 d=3 n .
\end{array}\right.
$$
By eliminating $n$ from equations (1) and (2), we get
$$
\frac{b-c}{... | -3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given a triangle $\triangle A B C$ with side lengths $4,5,6$ respectively, the circumcircle of $\triangle A B C$ is a great circle of sphere $O$, and $P$ is a point on the sphere. If the distances from point $P$ to the three vertices of $\triangle A B C$ are all equal, then the volume of the tetrahedron $P-A B C$ is... | 1.10 .
Since $P A=P B=P C$, the projection of point $P$ on the plane $A B C$ is the circumcenter $O$ of $\triangle A B C$, i.e., $P O \perp$ plane $A B C$, and $P O$ equals the radius $R$ of sphere $O$. Therefore,
$$
V_{P-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot P O=\frac{1}{3} \cdot \frac{a b c}{4 R} \cdot R=10 \... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Three positive integers $a$, $b$, and $c$ satisfy the conditions:
(1) $a<b<c<30$;
(2) For some positive integer base, the logarithms of $a(2 b-a)$ and $c^{2}+$ $60 b-11 a$ are 9 and 11, respectively.
Then the value of $a-2 b+c$ is $\qquad$. | 3. -4 .
Let the base be a positive integer $n(n \geqslant 2)$, then from condition (2) we get
$$
\left\{\begin{array}{l}
a(2 b-a)=n^{9} . \\
c^{2}+60 b-11 a=n^{11} .
\end{array}\right.
$$
When $n \geqslant 3$, from equation (1) we get
$$
3^{9} \leqslant n^{9}=a(2 b-a)2^{9}$, i.e., $b \geqslant 23$.
$$
Thus, we have
$$... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $\alpha, \beta, \gamma \in \mathbf{R}$,
$$
\begin{aligned}
u= & \sin (\alpha-\beta)+\sin (\beta-\gamma)+ \\
& \sin (\gamma-\alpha) .
\end{aligned}
$$
Then $u_{\text {max }}+u_{\text {min }}=$ | 6.0.
$$
\begin{aligned}
u & =\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \gamma+\sin \gamma \cdot \cos \alpha- \\
& =\left|\begin{array}{lll}
\cos \alpha \cdot \sin \beta-\cos \beta \cdot \sin \gamma-\cos \gamma \cdot \sin \alpha \\
\sin \alpha & \cos \alpha & 1 \\
\sin \beta & \cos \beta & 1 \\
\sin \gamma & \c... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 (1) In a $4 \times 4$ grid paper, some small squares are colored red, and then two rows and two columns are crossed out. If no matter how they are crossed out, there is at least one red small square that is not crossed out, how many small squares must be colored at least?
(2) If the β$4 \times 4$β grid paper ... | Solution: (1) If the number of colored small squares is less than or equal to 4, then we can appropriately strike out two rows and two columns to remove all the colored small squares.
If the number of colored small squares is 5, then by the pigeonhole principle, there must be at least one row with 2 or more colored sm... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $P(x)$ be the product of the digits in the decimal representation of $x$. Try to find all positive integers $x$ such that
$$
P(x)=x^{2}-10 x-22
$$
holds. | Solution: Let $x=\overline{a_{n} a_{n-1} \cdots a_{1}}$.
When $n \geqslant 2$, according to (1) and (2), we have the estimate
$$
x \geqslant 10^{n-1} \cdot a_{n}>9^{n-1} \cdot a_{n} \geqslant P(x) \text {, }
$$
thus $x^{2}-10 x-22<x$.
Solving this, we get $0<x<13$.
Upon verification, $x=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 On a plane, there are 7 points, and some line segments can be connected between them, so that any three points among the 7 points must have two points connected by a line segment. How many line segments are needed at least? Prove your conclusion. | Proof: First, to meet the requirements of the problem, the number of line segments to be connected must be greater than or equal to 9.
Below, we discuss in 4 cases.
(1) If among the 7 points, there exists 1 point not connected to any other points, by the problem's condition, the remaining 6 points must each be connecte... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false |
3. (25 points) $a$, $b$, $c$ are positive integers, and $a^{2}+b^{3}=$ $c^{4}$. Find the minimum value of $c$.
δΏηδΊζΊζζ¬ηζ’θ‘εζ ΌεΌγ | 3. Clearly, $c>1$. From $b^{3}=\left(c^{2}-a\right)\left(c^{2}+a\right)$, if we take $c^{2}-a=b, c^{2}+a=b^{2}$, then at this time
$$
c^{2}=\frac{b(b+1)}{2} \text {. }
$$
Examining $b$ from small to large to make the right side a perfect square, we find that when $b=8$, $c=6$, and thus, $a=28$.
Next, we show that ther... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. $f(x)$ is a function defined on $\mathbf{R}$ that is odd, and its smallest positive period is 2. Then the value of $f(-1)$ is $\qquad$ . | 1.0.
Since $f(x+2)=f(x)$, let $x=-1$, then we have $f(-1+2)=f(-1)$,
which means $f(1)=f(-1)=-f(1)$.
Therefore, $f(-1)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find $\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1)\left(\mathrm{C}_{2 n}^{k}\right)^{-1}$.
Analysis: Considering the use of identity (III), we can obtain
$$
\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}} \text {. }... | Let the original expression be $y_{n}$, from the analysis we have
$$
y_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{C_{2 n}^{2 n-k-1}} \text {. }
$$
By making a change of the summation index, set $l=2 n-$ $k-1$, when $k$ varies from 0 to $2 n-1$, $l$ varies from $2 n-1$ to 0. Thus,
$$
\begin{array}{l}
y_{n}=\sum_{l=0... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 1, in rectangle $A B C D$, $A B=2, B C=$ $\sqrt{3}, E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. Line segments $D E$, $B F$, and $E F$ intersect diagonal $A C$ at points $M$, $N$, and $P$ respectively. The number of right triangles formed by the line segments in Figure 1 is ( ).
(... | 4.B.
Let the notation $[A]$ represent the number of right-angled triangles with $A$ as the right-angle vertex, then $[A]=2$,
Similarly, $[B]=1,[C]=2,[D]=1,[E]=3$,
$$
[F]=3,[M]=[N]=[P]=0 \text {. }
$$
Therefore, the number of right-angled triangles is
$$
2+1+2+1+3+3=12 \text {. }
$$ | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
2. Fill in $n$ distinct numbers on a circle so that for every three consecutive numbers, the middle number is equal to the product of the two numbers on its sides. Then $n=$ $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directl... | 2.6.
According to the problem, $n \geqslant 3$. Take any two adjacent numbers $a, b(a \neq b)$, and by the problem's condition, the numbers on the circle can be written in sequence as
$$
a, b, \frac{b}{a}, \frac{1}{a}, \frac{1}{b}, \frac{a}{b}, a, b, \cdots
$$
Starting from the 7th number, the sequence of numbers has... | 6 | Number Theory | proof | Yes | Yes | cn_contest | false |
10.1. Try to find the smallest positive integer that cannot be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are all positive integers. | 10.1.11.
From the problem, we have
$$
\begin{array}{l}
1=\frac{4-2}{4-2}, 3=\frac{8-2}{4-2}, 5=\frac{16-1}{4-1}=\frac{2^{5}-2}{2^{3}-2}, \\
7=\frac{16-2}{4-2}, 9=2^{3}+1=\frac{2^{6}-1}{2^{3}-1}=\frac{2^{7}-2}{2^{4}-2}, \\
2=2 \times 1=\frac{2^{3}-2^{2}}{2^{2}-2}, \cdots, 10=2 \times 5=\frac{2^{6}-2^{2}}{2^{3}-2} .
\en... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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