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Example 2 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
1,1,2,2,2,2,2, \cdots, \underbrace{n, n, \cdots, n}_{(3 n-1) \uparrow}, \cdots \text {. }
$$
Find $a_{n}$ and $S_{n}$. | Solution: By the formula for the sum of an arithmetic sequence, the sum of the first $(m-1)$ segments of the sequence $a_{n}$ is
$$
\begin{array}{l}
2+5+8+\cdots+(3 m-4) \\
=\frac{1}{2}(m-1)(3 m-2) .
\end{array}
$$
It is evident that, if and only if
$$
\frac{1}{2}(m-1)(3 m-2)+1 \leqslant n \leqslant \frac{1}{2} m(3 m+... | a_{n}=\left[\frac{5+\sqrt{24 n-23}}{6}\right], \quad S_{n}=m n-\frac{1}{2} m^{2}(m-1), \text{ where } m=\left[\frac{5+\sqrt{24 n-23}}{6}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,810 |
Example 3 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
2,5,5,5, \cdots, \underbrace{(3 n-1), \cdots,(3 n-1)}_{(2 n-1) \uparrow}, \cdots \text {. }
$$
Find $a_{n}$ and $S_{n}$. | Solution: By the formula for the sum of an arithmetic sequence, the sum of the first $(m-1)$ segments of the sequence $\left\{a_{n}\right\}$ is
$$
1+3+5+\cdots+(2 m-3)=(m-1)^{2} \text {. }
$$
It is evident that, if and only if
$$
(m-1)^{2}+1 \leqslant n \leqslant m^{2}
$$
then, $a_{n}=3 m-1$. Solving the inequality (... | 3 m n-n-\frac{1}{2} m(m-1)(2 m-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,811 |
Example 4 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
3,3,7,7, \cdots,(4 n-1),(4 n-1), \cdots \text {. }
$$
Find $a_{n}$ and $S_{n}$. | Solution: This problem involves an equal-segment increment sequence, where the number of items in each segment (repeated items) is equal. Using the formula for the sum of an arithmetic sequence, the sum of the first $(m-1)$ segments of the sequence $\left\{a_{n}\right\}$ is
$$
\underbrace{2+2+\cdots+2}_{(m-1) \uparrow}... | a_{n}=4\left[\frac{n+1}{2}\right]-1, \quad S_{n}=4 m n-n-4 m(m-1), \text{ where } m=\left[\frac{n+1}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,812 |
Example 5 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
\underbrace{1,1, \cdots, 1}_{n \uparrow}, \underbrace{2,2, \cdots, 2}_{n \uparrow}, \cdots, \underbrace{n, n, \cdots, n}_{n \uparrow}, \cdots \text {. }
$$
Find $a_{n}$ and $S_{n}$. | Solution: This problem still involves an equal-segment incrementing sequence, where the sum of the first $(m-1)$ segments is
$$
\underbrace{k+k+\cdots+k}_{(m-1) \uparrow}=k(m-1) .
$$
It is clear that when and only when
$$
k(m-1)+1 \leqslant n \leqslant k m
$$
then, $a_{n}=m$. Solving the inequality (5) for $m$, we ge... | S_{n} = m n - \frac{1}{2} k m (m-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,813 |
Example 6 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
1,2.3,4,2,3,4,5,3.4,5,6, \cdots \text {. }
$$
Find $a_{n}$ and $S_{n}$, and calculate the values of $a_{80}$ and $S_{1921}$. | Given the sequence $1,2,3,4,5,6,7,8,9,10,11,12$, $\cdots$ has a general term of $n$, and the sequence $0,0,0,0,3,3,3,3,6,6$. $6,6, \cdots$ has a general term of $3\left[\frac{n-1}{4}\right]$. Subtracting the corresponding terms, we get the sequence $1,2,3,4,2,3,4,5,3,4,5,6 \cdots$ with the general term:
$$
\begin{array... | a_{80} = 23, S_{1921} = 465121 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,815 |
For example, $5 \alpha$ is a real number greater than zero. It is known that there exists a unique real number $k$, such that the equation about $x$
$$
x^{2}+\left(k^{2}+\alpha k\right) x+1999+k^{2}+\alpha k=0
$$
has two roots that are prime numbers. Find the value of $\alpha$. | Analysis: Since $\alpha, k$ are real numbers, the discriminant method cannot solve the problem.
Let the two roots of the equation be $x_{1}, x_{2}$, and $x_{1} \leqslant x_{2}, x_{1}, x_{2}$ are both prime numbers, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-k^{2}-\alpha k, \\
x_{1} x_{2}=1999+k^{2}+\alpha k .
\end{a... | 2 \sqrt{502} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,816 |
Example 7 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
0, \underbrace{1,1, \cdots, 1}_{k i}, \underbrace{2,2, \cdots,}_{k \uparrow}, 2 \underbrace{3,3, \cdots, 3}_{k \uparrow}, \cdots \text {. }
$$
Find $a_{n}$ and $S_{n}$. When $k=3$, calculate the values of $a_{2002}$ and $S_{2002}$. | Solution: This problem can be regarded as an equal-segment increment sequence, the sum of the first $m$ segments is
$$
1+\underbrace{k+k+\cdots+k}_{(m-1)+}=k(m-1)+1 .
$$
It is evident that, if and only if
$$
k(m-1)+2 \leqslant n \leqslant k m+1
$$
then, $a_{n}=m$. Solving the inequality (6) for $m$, we get
$\frac{n-1... | 668334 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,817 |
Example 8 Given the sequence $\left\{a_{n}\right\}$ with terms
$$
1,1,1,2,3,3,3,4,5,5,5,6,7,7,7,8, \cdots \text {. }
$$
Find $a_{n}$ and $S_{n}$. | Solution: The general term of the sequence $1,1,1,1,2,2,2,2,3,3,3,3,4,4$,
$4,4, \cdots$ is $\left[\frac{n+3}{4}\right]$.
The general term of the sequence $0,0,0,1,1,1,1,2,2,2,2,3,3,3$, $3,4, \cdots$ is $\left[\frac{n}{4}\right]$.
Adding the corresponding terms, we get the sequence $1,1,1,2,3,3,3,4$, $5,5,5,6,7,7,7,8, ... | \frac{n^{2}+n+2}{4}-\frac{i^{2}-3 i-4 k+2}{4} | Other | math-word-problem | Yes | Yes | cn_contest | false | 712,818 |
In 2001, the sixth question of the China Mathematical Olympiad National Training Team Selection Examination was as follows:
$$
\text { Let } F=\max _{1 \leq x \leq 3}\left|x^{3}-a x^{2}-b x-c\right| \text {. Find the minimum value of } F \text { when } a 、 b \text { 、 }
$$
$c$ take all real numbers. | Solve: First, find $F^{\prime}=\max _{-1<x \leqslant 1}\left|x^{3}-a_{1} x^{2}-b_{1} x-c_{1}\right|$ $\left(a_{1}, b_{1}, c_{1} \in \mathbf{R}\right)$'s minimum value.
$$
\begin{array}{l}
\text { Let } f(x)=x^{3}-a_{1} x^{2}-b_{1} x-c_{1} \\
(-1 \leqslant x \leqslant 1),
\end{array}
$$
It is easy to know that $4 f(1)-... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,819 |
In the 30th IMO training questions, there is a problem:
For positive numbers $x 、 y 、 z$ satisfying $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$. | Given $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n, n \geqslant 3)$, $\sum_{i=1}^{n} a_{i}^{n-1}=1$. Then $\sum_{i=1}^{n} \frac{a_{i}^{n-2}}{1-a_{i}^{n-1}} \geqslant \frac{n}{n-1} \sqrt[n-1]{n}$.
Inspired by this, the author found that it can be further generalized to:
Given $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n, ... | \frac{n}{k}\left(\frac{c}{k+1}\right)^{\frac{a_{1}-k a_{2}}{k a_{2}}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,820 |
Text [1] includes an inequality proposed and proved by D.M. Milosevic in 1987:
Let the sides of $\triangle A B C$ be $a, b, c$, and the corresponding altitudes be $h_{a}, h_{b}, h_{c}$, with the circumradius and inradius being $R, r$ respectively. Then
$$
\begin{array}{l}
\frac{a}{h_{b}+h_{c}}+\frac{b}{h_{c}+h_{a}}+\f... | Proof: Let the volume of tetrahedron \( A_{1} A_{2} A_{3} A_{4} \) be \( V \).
\[
\begin{aligned}
\because V & =\frac{1}{3} S_{1} h_{1}=\frac{1}{3} S_{2} h_{2}=\frac{1}{3} S_{3} h_{3} \\
& =\frac{1}{3} S_{4} h_{4},
\end{aligned}
\]
\(\therefore\) the left side of equation (2)
\[
\begin{array}{l}
=\frac{S_{1}}{\frac{3 V... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,821 |
The following inequality can be found in many relevant journals:
If $x, y \in \mathbf{R}^{+}$,
then $\left(x^{2}+y^{2}\right)^{\frac{1}{2}}>\left(x^{3}+y^{3}\right)^{\frac{1}{3}}$.
Below, the author provides two generalizations of formula (1):
Generalization 1: If $x, y \in \mathbf{R}^{+}, m, n \in \mathbf{N}$ and $n>m... | To prove: The inequality to be proved is equivalent to the following inequality
i.e. $\square$
$$
\begin{array}{l}
\frac{\left(\sum_{i=1}^{n} a_{i}^{t}\right)^{\frac{1}{t}}}{\left(\sum_{i=1}^{n} a_{i}^{s}\right)^{\frac{1}{4}}}>1, \\
{\left[\left(\frac{a_{1}^{s}}{\sum a_{i}^{s}}\right)^{\frac{t}{3}}+\cdots+\left(\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,822 |
Text [1] gives and proves the following proposition:
Proposition If $A_{1} \cup A_{2} \cup \cdots \cup A_{m}=\left\{a_{1}, a_{2}\right.$, $\left.\cdots, a_{\mathrm{n}}\right\}$, then the number of groups of sets $A_{1}, A_{2}, \cdots, A_{m}$ is $\left(2^{m}-1\right)^{n}$ groups.
Obviously, among the $\left(2^{m}-1\rig... | Prove: Let $A_{1} \cup A_{2} \cup \cdots \cup A_{m}=\left\{a_{1}, a_{2}, \cdots\right.$, $a_{n}$ |, the number of sets $A_{1}, A_{2}, \cdots, A_{m}$ is $f(m, n)$. From the above proposition, we know that
$$
f(m, n)=\left(2^{m}-1\right)^{n} .
$$
First, use $f(m, n)$ to establish a recursive relation for $g(m, n)$.
Sup... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,823 |
1. Let $a<b<0, a^{2}+b^{2}=4 a b$. Then the value of $\frac{a+b}{a-b}$ is ( ).
(A) $\sqrt{3}$
(B) $\sqrt{6}$
(C) 2
(D) 3 | -.1.(A).
Since $(a+b)^{2}=6 a b,(a-b)^{2}=2 a b$, and given $a<0$, we get $a+b=-\sqrt{6 a b}, a-b=-\sqrt{2 a b}$. Therefore, $\frac{a+b}{a-b}=\sqrt{3}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,824 |
2. Given $a=1999 x+2000, b=1999 x+2001, c$ $=1999 x+2002$. Then the value of the polynomial $a^{2}+b^{2}+c^{2}-a b-b c-$ $c a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 2. (D).
$$
\begin{array}{l}
\because a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] . \\
\text { Also } a-b=-1, b-c=-1, c-a=2, \\
\therefore \text { the original expression }=\frac{1}{2}\left[(-1)^{2}+(-1)^{2}+2^{2}\right]=3 .
\end{array}
$$ | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,825 |
Example 6 Let the quadratic equation $\left(k^{2}-6 k+8\right) \cdot x^{2}+\left(2 k^{2}-6 k-4\right) x+k^{2}=4$ have two integer roots. Find all real values of $k$ that satisfy the condition. | Analysis: The expression of the equation is relatively complex, and the discriminant method and Vieta's formulas are not applicable. Transform the original equation to get
$$
\begin{array}{l}
(k-2)(k-4) x^{2}+\left(2 k^{2}-6 k-4\right) x \\
+(k-2)(k+2)=0 .
\end{array}
$$
Factorizing, we get
$$
[(k-2) x+k+2][(k-4) x+k-... | 6,3, \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,827 |
4. Let $a, b, c$ be real numbers, $x=a^{2}-2 b+\frac{\pi}{3}, y=b^{2}-$ $2 c+\frac{\pi}{6}, z=c^{2}-2 a+\frac{\pi}{2}$. Then among $x, y, z$, at least -- values ( ).
(A) are greater than 0
(B) are equal to 0
(C) are not greater than 0
(D) are less than 0 | 4. (A).
Since $x+y+z=(a-1)^{2}+(b-1)^{2}+(c-1)^{2}+\pi$ $-3>0$, then at least one of $x, y, z$ is greater than 0. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,828 |
5. Let the equation $a x^{2}+(a+2) x+9 a=0$ with respect to $x$ have two distinct real roots $x_{1} 、 x_{2}$, and $x_{1}<\frac{2}{5}<x_{2}$
(A) $a>0$
(B) $-\frac{2}{7}<a<0$
(C) $a<-\frac{2}{7}$
(D) $-\frac{2}{11}<a<0$ | 5. (D).
It is clear that $a \neq 0$. The original equation can be transformed into
$$
x^{2}+\left(1+\frac{2}{a}\right) x+9=0 \text {. }
$$
Let $y=x^{2}+\left(1+\frac{2}{a}\right) x+9$, then this parabola opens upwards. Since $x_{1}<1<x_{2}$, when $x=1$, $y<0$. That is, $1+\left(1+\frac{2}{a}\right)+9<0$. Solving this... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,829 |
6. Given that $A_{1} A_{2} A_{3} \cdots A_{9}$ is a regular nonagon, $A_{1} A_{2}=$ $a, A_{1} A_{3}=b$. Then $A_{1} A_{5}$ equals ( ).
(A) $\sqrt{a^{2}+b^{2}}$
(B) $\sqrt{a^{2}+a b+b^{2}}$
(C) $\frac{1}{2}(a+b)$
(D) $a+b$ | 6. (D).
As shown in Figure 5, we have
$$
\begin{array}{l}
\angle A_{2} A_{1} A_{5}=\angle A_{4} A_{5} A_{1} \\
=\frac{3 \times 180^{\circ}-\frac{7 \times 180^{\circ}}{9} \times 3}{2} \\
=60^{\circ} .
\end{array}
$$
Thus, $\angle A_{1} A_{5} B$ and $\triangle A_{2} A_{4} B$ are both equilateral triangles.
Therefore, $... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,830 |
7. Let $x_{1}, x_{2}$ be the two real roots of the quadratic equation $x^{2} + a x + a = 2$. Then the maximum value of $\left(x_{1}-2 x_{2}\right)\left(x_{2}-2 x_{1}\right)$ is $\qquad$ . | $$
\begin{array}{l}
\text { II.7. }-\frac{63}{8} . \\
\because \Delta=a^{2}-4(a-2)=(a-2)^{2}+4>0,
\end{array}
$$
Then $x_{1}+x_{2}=-a, x_{1} x_{2}=a-2$.
$$
\begin{array}{l}
\therefore\left(x_{1}-2 x_{2}\right)\left(x_{2}-2 x_{1}\right)=-2 x_{1}^{2}-2 x_{2}^{2}+5 x_{1} x_{2} \\
=-2\left(x_{1}+x_{2}\right)^{2}+9 x_{1} x... | -\frac{63}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,831 |
8. Given that $a$ and $b$ are the x-coordinates of the points where the parabola $y=(x-c)(x-c-d)-2$ intersects the x-axis, and $a<b$. Then the value of $|a-c|+|c-b|$ is $\qquad$ | $8 . b-a$.
When $x=c$, $y=-2$, which means the point $(c,-2)$ is on the parabola and below the $x$-axis. Since the parabola $y=(x-c)(x-c-d)-2$ opens upwards, from the graph we can see that the value of $c$ lies between $a$ and $b$, i.e., $a<c<b$. Simplifying, we get $b-a$. | b-a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,832 |
9. As shown in Figure 1, in $\triangle A B C$, $\angle A B C=60^{\circ}$, point $P$ is a point inside $\triangle A B C$ such that $\angle A P B=\angle B P C=\angle C P A$, and $P A=8, P C=6$. Then $P B=$ $\qquad$ | $9.4 \sqrt{3}$.
It is known that $\angle A P B=\angle B P C=\angle C P A=120^{\circ}$.
$$
\begin{array}{l}
\because \angle P A B=180^{\circ}-120^{\circ}-\angle P B A \\
=60^{\circ}-\angle P B A=\angle P B C \text {. } \\
\therefore \triangle P A B \backsim \triangle P B C, \frac{P A}{P B}=\frac{P B}{P C} \text {. } \\
... | 4 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,833 |
10. As shown in Figure 2, the diameter $A B=a \mathrm{~cm}$ of the large circle $\odot O$, circles $\odot O_{1}$ and $\odot O_{2}$ are drawn with $O A$ and $O B$ as diameters, respectively, and two equal circles $\odot O_{3}$ and $\odot O_{4}$ are drawn in the space between $\odot O$ and $\odot O_{1}$ and $\odot O_{2}$... | 10. $\frac{1}{6} a^{2}$.
Connecting $\mathrm{O}_{3} \mathrm{O}_{4}$ must pass through point $\mathrm{O}$, then $\mathrm{O}_{3} \mathrm{O}_{4} \perp \mathrm{AB}$. Let the radii of $\odot \mathrm{O}_{3}$ and $\odot \mathrm{O}_{4}$ be $x \mathrm{~cm}$. Since the radii of $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ ... | \frac{1}{6} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,834 |
11. The number of integers $n$ that satisfy $\left(n^{2}-n-1\right)^{n+2}=1$ is.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task ... | 11.4 .
From $n+2=0, n^{2}-n-1 \neq 0$, we get $n=-2$;
From $n^{2}-n-1=1$, we get $n=-1, n=2$;
From $n^{2}-n-1=-1$ and $n+2$ is even, we get $n=0$.
Therefore, $n=-1,-2,0,2$ for a total of 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,835 |
12. The marked price of a product is $p$ \% higher than the cost. When the product is sold at a discount, to avoid a loss, the discount on the selling price (i.e., the percentage reduction) must not exceed $d$ \%. Then $d$ can be expressed in terms of $p$ as $\qquad$ | 12. $\frac{100 p}{100+p}$.
Let the cost of the item be $a$, then we have
" $(1+p \%) (1-d \%) = a$, solving for $d$ gives $d = \frac{100 p}{100+p}$. | \frac{100 p}{100+p} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,836 |
13. For a certain project, if contracted to Team A and Team B, it will be completed in $2 \frac{2}{5}$ days, costing 180,000 yuan; if contracted to Team B and Team C, it will be completed in $3 \frac{3}{4}$ days, costing 150,000 yuan; if contracted to Team A and Team C, it will be completed in $2 \frac{6}{7}$ days, cos... | Three, 13. Let the number of days required for A, B, and C to complete the task individually be \( x, y, z \) respectively. Then,
$$
\left\{\begin{array} { l }
{ \frac { 1 } { x } + \frac { 1 } { y } = \frac { 5 } { 1 2 } . } \\
{ \frac { 1 } { y } + \frac { 1 } { z } = \frac { 4 } { 1 5 } , } \\
{ \frac { 1 } { z } +... | 177000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,837 |
Example 7 Let $\alpha$ be an integer. If there exist integers $b$ and $c$ such that $(x+\alpha)(x-15)-25=(x+b)(x+c)$ holds, find the possible values of $\alpha$.
| Analysis: This problem can be transformed into: For what value of $\alpha$ does the equation $(x+\alpha)(x-15)-25=0$ have integer roots.
The equation can be rewritten as
$x^{2}-(15-\alpha) x-15 \alpha-25=0$.
Considering it as a linear equation in $\alpha$, we rearrange to get
$$
\alpha(x-15)=-x^{2}+15 x+25 \text {. }
$... | 9, -15, -39 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,838 |
14. As shown in Figure 3, the cyclic hexagon $A B C D E F$ satisfies $A B=C D=E F$, and the diagonals $A D, B E, C F$ intersect at a point $Q$. Let the intersection of $A D$ and $C E$ be $P$. Prove:
(1) $\frac{Q D}{E D}=\frac{A C}{E C}$;
(2) $\frac{C P}{P E}=\frac{A C^{2}}{C E^{2}}$. | 14. (1) Connect $A E$. From $A B=C D=E F$ we know $\overparen{A B}=\overparen{C D}=\overparen{E F}$.
In $\angle Q D E$ and $\angle A C E$,
$$
\begin{array}{l}
\angle Q E D=\angle B E C+\angle C E D \\
=\angle B E C+\angle A E B=\angle A E C . \\
\angle Q D E=\angle A C E .
\end{array}
$$
Therefore, $\angle Q D E \subs... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,839 |
15. If for all integer values of $x$, the quadratic trinomial $a x^{2} + b x + c$ is a perfect square (i.e., the square of an integer), prove:
(1) $2a$, $2b$ are integers;
(2) $a$, $b$, $c$ are integers, and $c$ is a perfect square.
Conversely, if (2) holds, is $a x^{2} + b x + c$ a perfect square for all integer value... | 15. (1) Let $x=0$, we get $c=$ a square number $l^{2}$;
Let $x= \pm 1$, we get $a+b+c=m^{2}, a-b+c=n^{2}$, where $m, n$ are integers. Therefore,
$$
2 a=m^{2}+n^{2}-2 c, \quad 2 b=m^{2}-n^{2}
$$
are both integers.
(2) If $2 b$ is an odd number $2 k+1$ (where $k$ is an integer), let $x=4$ we get $16 a+4 b+l^{2}=h^{2}$,... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,840 |
$-、 \triangle A B C$ has side lengths $a, b, c$ with $b < c$, and $A D$ is the angle bisector of $\angle A$, with point $D$ on $B C$.
(1) Find the necessary and sufficient condition for the existence of points $E, F$ (not endpoints) on segments $A B, A C$ respectively, such that $B E = C F$ and $\angle B D E = \angle C... | (1) As shown in Figure 1, if there exist points $E$ and $F$ on segments $AB$ and $AC$ respectively, such that $BE = CF$ and $\angle BDE = \angle CDF$, then because the distances from point $D$ to $AB$ and $AC$ are equal, we have
$S_{\triangle BDE} = S_{\triangle CDF}$.
Thus, we have
$BD \cdot DE = DC \cdot DF$.
By the ... | BE = \frac{a^2}{b + c} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,841 |
Sure, here is the translation:
```
II. Let the polynomial sequence $\left\{P_{n}(x)\right\}$ satisfy:
$$
P_{1}(x)=x^{2}-1, P_{2}(x)=2 x\left(x^{2}-1\right) .
$$
and $P_{n+1}(x) P_{n-1}(x)=\left(P_{n}(x)\right)^{2}-\left(x^{2}-1\right)^{2}$, $n=2,3, \cdots$.
Let $s_{n}$ be the sum of the absolute values of the coeffi... | From equation (1), we have
\[ P_{n+1}(x) P_{n-1}(x) - P_{n}^2(x) \]
\[ = P_{n}(x) P_{n-2}(x) - P_{n-1}^2(x), \quad n=3,4, \cdots, \]
\[ P_{n-1}(x) \left[ P_{n+1}(x) + P_{n-1}(x) \right] \]
\[ = P_{n}(x) \left[ P_{n}(x) + P_{n-2}(x) \right]. \]
By induction, we get
\[ \frac{P_{n+1}(x) + P_{n-1}(x)}{P_{n}(x)} = \frac{P_{... | k_{2m} = m + 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,842 |
Three, 18 football teams are participating in a single round-robin tournament, meaning each round the 18 teams are divided into 9 groups, with each group's two teams playing one match. In the next round, the teams are regrouped to play, for a total of 17 rounds, ensuring that each team plays one match against each of t... | ```
3. Consider the following competition program:
1.(1.2)(3.4)(5.6)(7.8)(9,18)
(10,11)(12,13)(14,15)(16,17)
2.(1,3)(2.4)(5,7)(6.9)(8,17)
(10,12)(11,13)(14,16)(15.18)
3.(1.4)(2.5)(3.6)(8.9)(7.16)
(10,13)(11,14)(12,15)(17,18)
4.(1,5)(2.7)(3.8)(4,9)(6.15)
(10.14)(11,16)(12.17)(13.18)
5.(1,6)(2,8)(3,9)(4,7... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,843 |
For any 4 distinct points $P_{1}, P_{2}, P_{3}, P_{4}$ in the plane, find the minimum value of the ratio
$$
\frac{\sum_{1 \leqslant i<j \leqslant 4} P_{4} P_{j}}{\min _{1 \leqslant i<4} P_{4} P_{j}}
$$
(Wang Jianwei) | Four, first prove the following lemma.
Lemma In $\triangle A B C$, if $A B \geqslant m, A C \geqslant m, \angle B A C = \alpha$. Then $B C \geqslant 2 m \sin \frac{\alpha}{2}$.
Proof of the lemma: Draw the angle bisector $A D$ of $\angle A$. By the Law of Sines, we have
$$
\begin{aligned}
B C= & B D+D C \\
= & A B \fra... | 5+\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,844 |
Five, points in the plane whose both coordinates are rational numbers are called rational points. Prove: the set of all rational points in the plane can be divided into 3 pairwise disjoint sets, satisfying the conditions:
(i) In any circle centered at each rational point, there must be 3 points belonging to these 3 set... | Five, any rational point can be uniquely written in the form $\left(\frac{u}{u}, \frac{r}{w}\right)$. Here, $u, r, w$ are all integers, $w>0$, and $(u, r, u)=1$.
$$
\begin{array}{l}
\text { Let } A=\left\{\left.\left(\frac{u}{u}, \frac{r}{u}\right) \right\rvert\, 2 \nmid u\right\}, \\
B=\left\{\left(\frac{u}{u} \cdot \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,845 |
2. Let $f(x)=x^{3}, g(x)=A x+B$. Then the values of $A$ and $B$ that satisfy $f(1)=$ $g(1)$ are ().
(A) $A=0, B=1$
(B) $A=1, B=0$
(C) $A=3, B=-2$
(D) $A=-2, B=3$
(E) All of the above | 2. (E)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,848 |
Example 1 For any $n \geqslant 2$, prove: there exist $n$ consecutive positive integers, none of which is of the form $p^{q}$. Here $p$ is a prime, $q$ is a positive integer greater than 1. | Prove: It is only necessary to find $n$ consecutive positive integers, each of which has at least two different prime factors. For example, take $n$ consecutive positive integers
$a_{k}=[(n+1)!]^{2}+k(k=2,3, \cdots, n+1)$.
Clearly, $k \mid a_{k}$. Since $a_{k}=k\left[\frac{[(n+1)!]^{2}}{k}+1\right]$, and $k \left\lvert... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,849 |
3. The inverse function of $f(x)=x^{3}-1$ is $(\quad)$.
(A) $f^{-1}(x)=(x+1)^{\frac{1}{3}}$
(B) $f^{-1}(x)=\left(x^{3}-1\right)^{-1}$
(C) $f^{-1}(x)=(x+1)^{-\frac{1}{3}}$
(D) The inverse function does not exist
$(\mathrm{E}) *$ | 3. (A)
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,850 |
$5 .\left(2^{n}\right)^{2}-1$ The largest prime factor is ( ). $\begin{array}{lllll}\text { (A) } 3 & \text { (B) } 7 & \text { (C)31 } & \text { (D) } 5 & \text { (E) } 13\end{array}$ | 5. (E)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,852 |
6. Simplify $\frac{\left(x^{2}+1\right)\left(x^{2}-1\right)}{x^{3}+x^{2}+x+1}$, we get $(\quad)$.
(A) $x-1$
(B) $x+1$
(C) $1-x$
(D) $-1-x$
(E) * | 6. (A)
Translate the text above into English, preserving the original text's line breaks and formatting, and output the translation result directly. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,853 |
7. The value of $x$ that satisfies $\log _{t} 10=\log _{4} 100$ is $(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{1}{10}$
(C) $\ln 2$
(D) 2
(E) $\log _{2} 2$ | 7.
(D)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,854 |
8. $\ln \left(\frac{3}{\sqrt{3}}\right)-\ln (2+\sqrt{3})=(\quad)$.
(A) $\ln 3-\ln (2-\sqrt{3})$
(B) $\ln \sqrt{3}+\ln (2-\sqrt{3})$
(C) $\ln \sqrt{3}-\ln (2-\sqrt{3})$
(D) $\ln 3+\ln (2-\sqrt{3})$
(E) * | 8. (B).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,855 |
10. If $r_{1}, r_{2}$ are the roots of the equation $x^{2}-x+1=0$, then, $\frac{1}{r_{1}}+\frac{1}{r_{2}}=(\quad)$.
(A) 0
(B) 1
(C) -1
(D) $i \sqrt{3}$
(k) $-i \sqrt{3}$ | 10. (B).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,857 |
12. If $(x, y, z)$ is
$$
\left\{\begin{array}{l}
5 x-3 y+2 z=3, \\
2 x+4 y-z=7, \\
x-11 y+4 z=3
\end{array}\right.
$$
the value of $z$ is ( ).
(A) 0
(B) -1
(C) 1
(D) $\frac{4}{3}$
(E) $*$ | 12. (E).
(1) $-(3)$ yields
$4 x+8 y-2 z=0$. That is, $2 x+4 y-z=0$, which contradicts (2). Therefore, the system of equations has no solution. | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,859 |
13. If the function $f$ satisfies $f\left(n^{2}\right)=f(n)+2, n \geqslant 2$ and $f(2)=1$, then, $f(256)=(\quad)$.
(A) 3
(B) 5
(C) 7
(D) 9
$(\mathrm{E}) *$ | 13. (C).
From $f\left(n^{2}\right)=f(n)+2$, we have $f\left(n^{2^{4}}\right)=f\left(n^{2^{t-1}}\right)+2$, for $\forall k \in \mathbf{N}$. Then $f\left(n^{2^{2}}\right)=f(n)+2 k$. Therefore, $f(256)=f\left(2^{2^{3}}\right)=f(2)+2 \times 3=7$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,861 |
14. The set of real numbers $x$ that satisfy $\frac{1}{x+1}>\frac{1}{x-2}$ is ( ).
(A) $x>2$
(B) $x<-1$
(C) $x<2$ or $x<-1$
(D) $-1<x<2$
(E) $*$ | 14. (D).
When $x0>x-2$, the inequality holds: When $x>2$, we have $x+1>x-2>0$, thus, $\frac{1}{x-2}>$ $\frac{1}{x+1}$. Then $\frac{1}{x+1}>\frac{1}{x-2} \Leftrightarrow-1<x<2$. | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 712,862 |
16. Given positive integers $x, y$, then $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5}$ has ( ) solutions of the form $(x, 1)$.
(A) 0
(B) 1
(C) 2
(D) More than 2, but finite
(E) Infinite
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result direc... | 16. (C).
$$
\text { Given } \frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5} \Rightarrow 50 y=x^{2}(y+5) \Rightarrow x^{2}=\frac{50 y}{y+5} \text {. }
$$
If $(y, 5)=1$, then $(y, y+5)=1$, so $(y+5) \mid 50=5^{2} \times 2$. Since $y+5>2$, then $(y+5) \mid 5^{2}$, which contradicts $(y+5,5)=1$. Therefore, $5 \mid y$.
Let $y=5 y... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,864 |
18. The value of $x$ that satisfies $\sqrt{10}=\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-1}$ is $(\quad)$.
(A) $\frac{1}{2} \cdot \ln (4-\sqrt{15})$
(B) $\frac{1}{2} \ln (4+\sqrt{15})$
(C) $\frac{1}{2} \ln (4+\sqrt{17})$
(1)) $\frac{1}{2} \ln (4-\sqrt{17})$
(E) $\ln \left(\frac{\sqrt{10}+\sqrt{14}}{2}\right)$ | 18. (B).
Let $e^{2}=t$. Then $\mathrm{e}^{-1}=\frac{1}{t}$.
We have $t+\frac{1}{t}=\sqrt{10}$. That is, $t^{2}-\sqrt{10} t+1=0$.
Solving, we get $t=\frac{\sqrt{10} \pm \sqrt{6}}{2}, x=\ln t=\ln \left(\frac{\sqrt{10} \pm \sqrt{6}}{2}\right)$.
$$
\begin{array}{l}
\because \frac{\sqrt{10}+\sqrt{6}}{2}>1, \frac{\sqrt{10}-... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,866 |
19. The number of rational values that make $f(x)=-2 x^{5}+4 x^{3}-2 x^{2}+5$ equal to 0 is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(E) $\neq$ | 19. (E).
Let $f(x)=0$ have a rational root $x_{1}=\frac{p}{q}$. Where $p, q \in \mathbb{Z}$. $(p, q)=1, q \neq 0$. Then
$$
f\left(x_{1}\right)=-2\left(\frac{p}{q}\right)^{5}+4\left(\frac{p}{q}\right)^{3}-2\left(\frac{p}{q}\right)^{2}+5=0 .
$$
That is, $2 p^{5}=\left(4 p^{3}-2 p^{2} q+5 q^{3}\right) q^{2}$.
Thus, $q^{... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,867 |
20. If $b$ is a real number, then the number of negative real roots of $f(x)=x^{4}+(6-b)x^{3}+(11-6b)x^{2}+(6-11b)x-6b$ could be ( ).
(A) dependent on $b$
(B) 1 or 3
(C) 1
(D) 0
(E) * | 20. (B).
$$
\begin{array}{l}
x^{4}+(6-b) x^{3}+(11-6 b) x^{2}+(6-11 b) x-6 b \\
=\left(x^{3}+6 x^{2}+11 x+6\right)(x-b) \\
=(x+1)(x+2)(x+3)(x-b) . \\
\because b>0, \\
\therefore f(x)=0 \text { has 3 negative real roots: }-1,-2,-3 .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,868 |
22. Let $s$ be a square with side length $x$. If the change in side length from $x_{0}$ to $x_{0}+h$ is approximated by $\frac{\mathrm{d} A}{\mathrm{~d} x}\left(x_{0}\right) \cdot h$ for the change in area of $S$, then the absolute error of this approximation is ( ).
(A) $h^{2}$
(B) $2 h x_{0}$
(C) $x_{0}^{2}$
(D) $h$
... | 22. (A).
Since the area of $S$ is $A(x)=x^{2}$. Then when $x$ changes from $x_{0}$ to $x_{0}+h$, the absolute value of the change in $A(x)$ is
$$
A\left(x_{0}+h\right)-A\left(x_{0}\right)=\left(x_{0}+h\right)^{2}-x_{0}^{2}=2 h x_{0}+h^{2} \text {. }
$$
Also, $\frac{\mathrm{d} A}{\mathrm{~d} x}\left(x_{0}\right) \cdot... | A | Calculus | MCQ | Yes | Yes | cn_contest | false | 712,870 |
Example 3 Proof: The set of positive integers $\mathbf{N}^{+}$ can be partitioned into two subsets $A$ and $B$, such that no three numbers in $A$ form an arithmetic progression, and there does not exist an infinite arithmetic progression formed by numbers in $B$. | Proof: Let $A=\left\{n!+n \mid n \in \mathbf{N}^{+}\right\}, B=\mathbf{N}-A$. If $B$ contains an infinite arithmetic sequence with the first term $a_{1}$ and common difference $d$, then one term of this sequence is
$$
\begin{array}{l}
a_{1}+\left[\frac{\left(a_{1}+d\right)!}{d}+1\right] d \\
=\left(a_{1}+d\right)!+\lef... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,871 |
23. $f(x)=\left[\cos ^{2}\left(x^{2}+\pi\right)+\sin ^{2} x^{2}\right] \cos \left(x^{3}+2 x\right)$ The derivative is ( ).
(A) $\left[8 x \cos x^{2} \cdot \sin x^{2}\right] \cos \left(x^{3}+2 x\right)-\left[\cos ^{2}\left(x^{2}+\pi\right)\right.$ $\left.+\sin ^{2} x^{2}\right] \sin \left(x^{3}+2 x\right) \cdot\left(3 x... | 23. (1).
\[
\begin{array}{l}
\because f(x)=\left[\cos ^{2}\left(x^{2}+\pi\right)+\sin ^{2} x^{2}\right] \cos \left(x^{3}+2 x\right) \\
=\left[\cos ^{2} x^{2}+\sin ^{2} x^{2}\right] \cos \left(x^{3}+2 x\right) \\
=\cos \left(x^{3}+2 x\right) \text {. } \\
\therefore f^{\prime}(x)=-\sin \left(x^{3}+2 x\right) \cdot\left(... | D | Calculus | MCQ | Yes | Yes | cn_contest | false | 712,872 |
25. The range of the function $f(x)=\frac{\sqrt{x}+4-3}{x-5}$ is ( ).
(A) $\left(0, \frac{1}{3}\right]$
(B) $(-\infty, 0) \cup(0, \infty)$
(C) $\left(0, \frac{1}{6}\right) \cup\left(\frac{1}{6}, \frac{1}{3}\right]$
(D) $(0, \infty)$
(E) * | 25. (C)
Let $z=x-5$. Then
$$
\begin{array}{l}
f(x)=\frac{\sqrt{x+4}-3}{x-5}=\frac{\sqrt{z+9}-3}{z}=\frac{1}{\sqrt{z+9}+3} . \\
\because z+9 \geqslant 0, \text { and } z \neq 0 . \\
\therefore \sqrt{z+9}+3 \geqslant 3 . \text { and } \sqrt{z+9}+3 \neq 6 . \\
\therefore 0 \leqslant \frac{1}{\sqrt{z+9}+3} \leqslant \frac... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,874 |
26. Let $f(x)=\left\{\begin{array}{l}1, x \text { is rational } \\ 0, x \text { is irrational }\end{array}\right.$ For all $x$, the function $g(x)$ that satisfies $x f(x) \leqslant g(x)$ is $(\quad$.
(A) $g(x)=\sin x$
(B) $g(x)=x$
(C) $g(x)=x^{2}$
(D) $g(x)=|x|$
(E) * | 26. (1).
$$
\begin{array}{l}
\because \sin 2<2=2 f(2),(-\sqrt{2})<0=(-\sqrt{2}) \cdot f(-\sqrt{2}), \\
\left(\frac{1}{2}\right)^{2}<\frac{1}{2}=\frac{1}{2} f\left(\frac{1}{2}\right) .
\end{array}
$$
$\therefore(A) 、(B) 、(C)$ do not meet the requirements of the problem.
Also, $x f(x)=\left\{\begin{array}{l}x, x \text { ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,875 |
27 . The center of the circle passing through the points $(4,5),(6,3)$ and $(2,1)$ is ).
(A) $(8,7)$
(B) $(4.3)$
(C) $\left(\frac{13}{5}, \frac{8}{5}\right)$
(D) $\left(\frac{11}{3}, \frac{8}{3}\right)$
(E) * | 27. (D).
Let $A(4,5), B(6,3), C(2,1)$. It is easy to see that the midpoint of $B C$ is $(4,2)$, and the slope is $\frac{1}{2}$. Therefore, the perpendicular bisector of $B C$ is
$$
y-2=-2(x-4) .
$$
We also know that the midpoint $D$ of $A B$ is $(5,4)$, and $A C=B C$, so the perpendicular bisector of $A B$ is
$$
\fra... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,876 |
28. There is a point $P(x, y)$ on the plane. If the distance from point $P$ to $(3,0)$ and the distance from point $P$ to $(0,3)$ sum up to 6, then point $P$ must satisfy ( ).
(A) $3(x-2)^{2}+3(y-2)^{2}+3 x y=24$
(B) $3(x-2)^{2}+3(y-2)^{2}+2 x=24$
(C) $3(x-2)^{2}+3(y-2)^{2}+3 x y=18$
(D) $3(x-2)^{2}+3(y-2)^{2}+2 x y=18... | 28. (B).
From the given, we have
$$
\begin{array}{l}
\sqrt{(x-3)^{2}+y^{2}}-\sqrt{x^{2}+(y-3)^{2}}=6 \\
\Rightarrow 2 \sqrt{x^{2}+(y-3)^{2}}=6+(x-y) \\
\Rightarrow 3\left(x^{2}+y^{2}\right)-12(x+y)+2 x y=0 \\
\Rightarrow 3(x-2)^{2}+3(y-2)^{2}+2 x y=24 .
\end{array}
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,877 |
29. Tossing a fair coin 10 times, the probability of getting exactly three tails is ( ).
(A) $\frac{15}{128}$
(B) $\frac{3}{10}$
(C) $\frac{21}{100}$
(J) $\frac{1}{8}$
$(\mathrm{E}) *$ | 29. (A).
Due to tossing a coin 10 times, there are $2^{10}$ possible outcomes, among which there are $\mathrm{C}_{10}^{3}$ outcomes with exactly 3 tails. Thus, the probability is
$$
\frac{C_{10}^{3}}{2^{10}}=\frac{10 \cdot 9 \cdot 8}{6 \cdot 2^{10}}=\frac{15}{128} .
$$ | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 712,878 |
30. As shown in Figure 2, $C$ is a point on ray $\overrightarrow{A E}$, and $B D$ bisects $\angle A B C$. If $\angle B A E = 30^{\circ}$, then for all possible positions of $B, C$, the range of $\angle C B D$ is ( ).
(A) $0^{\circ}-30^{\circ}$
(B) $0^{\circ}-60^{\circ}$
(C) $0^{\circ} \sim 75^{\circ}$
(D) $0^{\circ} \s... | 30. (C).
$$
\begin{array}{l}
\because B D \text { bisects } \angle A B C . \\
\begin{aligned}
\therefore \angle C B D & =\frac{1}{2}\left(180^{\circ}-\angle A-\angle B C A\right) \\
& =\frac{1}{2}\left(150^{\circ}-\angle B C A\right) \leqslant 75^{\circ} .
\end{aligned}
\end{array}
$$
When $B \rightarrow A$, $\angle B... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,879 |
32. On a twelve-sided die, the numbers $1,2,3,4$ are each marked on two faces, and the numbers 5,6,7,8 are each marked on one face. Observing, it is found that the probability of each of the 12 faces appearing is the same. The probability that the sum of the results of two rolls of this twelve-sided die is 6 is ().
(A)... | 32. (A).
$$
6=5+1=4+2=3+3=2+4=1+5 \text {. }
$$
Due to the probability of the first roll showing $5,4,3,2,1$ being $\frac{1}{12}, \frac{1}{6}, \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}$; the probability of the second roll showing $1,2,3,4,5$ being $\frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{12}... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 712,881 |
1 Elimination Method
Example 1 Given that $x, y, z$ are real numbers, and $x+2y-z=6, x-y+2z=3$. Then, the minimum value of $x^{2}+y^{2}+z^{2}$ is $\qquad$
(2001, Hope Cup Junior High School Mathematics Competition) | Solution: From the given, we can solve for $y=5-x, z=4-x$. Then
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=x^{2}+(5-x)^{2}+(4-x)^{2} \\
=3(x-3)^{2}+14 . \\
\because 3(x-3)^{2} \geqslant 0, \quad \therefore x^{2}+y^{2}+z^{2} \geqslant 14 .
\end{array}
$$
Therefore, the minimum value of $x^{2}+y^{2}+z^{2}$ is 14. | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,882 |
2 Factorization Method
Example 2 Let $a, b, c$ be distinct natural numbers, and $a b^{2} c^{3}=1350$. Then the maximum value of $a+b+c$ is $\qquad$
(1990, Wu Yang Cup Junior High School Mathematics Competition) | Solution: $\because 1350=2 \times 5^{2} \times 3^{3}=150 \times 3^{2} \times 1^{3}$, $\therefore$ when $a=150, b=3, c=1$, the maximum value of $a+b+c$ is 154. | 154 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,883 |
Example 11 Let $x_{1}, x_{2}, \cdots, x_{7}$ be natural numbers. And $x_{1}$
$$
<x_{2}<\cdots<x_{6}<x_{7} \text {, and } x_{1}+x_{2}+\cdots+x_{7}=
$$
159. Find the maximum value of $x_{1}+x_{2}+x_{3}$.
(1997, Anhui Province Junior High School Mathematics Competition) | Solution: $\because 159=x_{1}+x_{2}+\cdots+x_{7}$
$$
\begin{array}{l}
\geqslant x_{1}+\left(x_{1}+1\right)+\left(x_{1}+2\right)+\cdots+\left(x_{1}+6\right) \\
=7 x_{1}+21, \\
\therefore x_{1} \leqslant 19 \frac{5}{7} .
\end{array}
$$
Therefore, the maximum value of $x_{1}$ is 19.
$$
\begin{array}{l}
\text { Also, } \b... | 61 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,884 |
4. As shown in Figure $2, D, E$ are points on the sides $AC, AB$ of $\triangle ABC$, respectively, and $BD, CE$ intersect at point $O$. If $S_{\triangle OCD}=2, S_{\triangle OBE}=3, S_{\triangle OBC}=4$, then, $S_{\text {quadrilateral } A O D E}=$ $\qquad$ | 4. $\frac{39}{5}$.
As shown in Figure 5, let $S_{\triangle O O D}=x$, $S_{\triangle \text { SOE }}=y$. Then we have
$$
\frac{x}{3+y}=\frac{2}{4}, \frac{y}{2+x}=\frac{3}{4} .
$$ | \frac{39}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,889 |
5. As shown in Figure 3, each face of the cube is written with a natural number, and the sum of the two numbers on opposite faces is equal. If the number opposite to 10 is a prime number $a$, the number opposite to 12 is a prime number $b$, and the number opposite to 15 is a prime number $c$, then $a^{2}+b^{2}+c^{2}-a ... | 5.19.
Given that $10+a=12+b=15+c$. Therefore, $c=2$. Then $a=7, b=5$. | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,890 |
Three. (20 points) A company produces computers, and in 1997, the average production cost per unit was 5000 yuan, with a pure profit margin of 20% used to set the factory price. Starting from 1998, the company strengthened management and technological transformation, leading to a gradual reduction in production costs. ... | Three, the ex-factory price in 1997 is
$$
5000(1+20 \%)=6000 \text { (yuan). }
$$
Let the production cost per computer in 2001 be $x$ yuan. Then $x(1+50 \%)=6000 \times 80 \%$. Solving for $x$ gives $x=3200$ (yuan).
Let the annual reduction in production cost be $y$. Then
$$
(1-y)^{4} \times 5000=3200 \text {. }
$$
S... | 11\% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,892 |
Four, (20 points) As shown in Figure 4, $P$ is a point outside $\odot O$, $PA$ is tangent to $\odot O$ at $A$, and $PBC$ is a secant of $\odot O$. $AD \perp PO$ at $D$. Prove:
$$
PB: BD = PC: CD.
$$ | $$
\begin{array}{l}
PA^2 = PD \cdot PO \\
= PB \cdot PC.
\end{array}
$$
Therefore, points $B$, $C$, $O$, and $D$ are concyclic. We have
$$
\begin{array}{l}
\triangle PCD \sim \triangle POB. \\
\text{Then } \frac{PC}{CD} = \frac{PO}{OB} = \frac{PO}{OC}. \\
\text{Also, } \because \triangle POC \sim \triangle PBD, \\
\th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,893 |
Example 12 If $m$ and $n$ are positive integers, and $m \leqslant$ $1996, r=2-\frac{m}{n}>0$, then the minimum value of $r$ is $\qquad$
(1996, Shanghai Junior High School Mathematics Competition) | Solution: From $r=2-\frac{m}{n}>0$, we get $\frac{m}{n}<2$.
If we take $m=1996, n=999$,
then $\frac{m}{n}=\frac{1996}{999}$;
If we take $m=1995, n=998$,
then $\frac{m}{n}=\frac{1995}{998}$;
Clearly, $\frac{1995}{998}>\frac{1996}{999}$.
If we take $m=1994, n=998$, then $\frac{m}{n}=\frac{1994}{998}$;
If we take $m=1993,... | \frac{1}{998} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,895 |
2. Given positive numbers $a$ and $b$, the following propositions are given:
(1) If $a=1, b=1$, then $\sqrt{a b} \leqslant 1$;
(2) If $a=\frac{1}{2}, b=\frac{5}{2}$, then $\sqrt{a b} \leqslant \frac{3}{2}$;
(3) If $a=2, b=3$, then $\sqrt{a b} \leqslant \frac{5}{2}$;
(4) If $a=1, b=5$, then $\sqrt{a b} \leqslant 3$.
Ba... | 2. $\left(\frac{13}{2}\right)^{2}$ | \left(\frac{13}{2}\right)^{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,897 |
6. Given in $\triangle A B C$, $\angle A, \angle B$ are acute angles, and $\sin A$ $=\frac{5}{13}, \tan B=2, A B=29 \mathrm{~cm}$. Then the area of $\triangle A B C$ is $\qquad$ $\mathrm{cm}^{2}$ | 6.145 .
Draw a perpendicular from point $C$ to $AB$. Let the foot of the perpendicular be $D$.
$\because \sin A=\frac{5}{13}=\frac{CD}{AC}$, let $m>0$,
$\therefore CD=5m, AC=13m$.
$\because \tan B=\frac{CD}{BD}=2$, we can set $n>0, CD=2n, BD=n$,
$\therefore BD=n=\frac{CD}{2}=\frac{5}{2}m$.
$\therefore AD=\sqrt{(13m)^2... | 145 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,901 |
7. (10 points) Observe: $1 \times 2 \times 3 \times 4+1=5^{2}$.
$$
\begin{array}{l}
2 \times 3 \times 4 \times 5+1=11^{2}, \\
3 \times 4 \times 5 \times 6+1=19^{2},
\end{array}
$$
(1) Please write a general conclusion and provide a proof;
(2) According to (1), calculate $2000 \times 2001 \times 2002 \times 2003+1$ (exp... | For the natural number $n$, we have
$$
\begin{array}{l}
n(n+1)(n+2)(n+3)+1 \\
=\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1 \\
=\left(n^{2}+3 n\right)^{2}+2\left(n^{2}+3 n\right)+1=\left(n^{2}+3 n+1\right)^{2} .
\end{array}
$$
(2) From (1), we get
$$
2000 \times 2001 \times 2002 \times 2003+1=4006001^{2} .
$$ | 4006001^2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,902 |
9. (10 points) In a math competition, the organizing committee decided to use the funds sponsored by NS Company to purchase a batch of prizes. If one NS calculator and 3 copies of the "Math Competition Lecture" book are considered as one set of prizes, then 100 sets of prizes can be bought; if one NS calculator and 5 c... | 9. Let each calculator cost $x$ yuan, each book "Lecture on Mathematical Competition" cost $y$ yuan, and the total amount of money be $s$ yuan. Then we have
$$
100(x+3 y)=s=80(x+5 y) \text {. }
$$
Simplifying, we get $x=5 y$. Solving, we get $s=800 y$.
Thus, this amount of money can buy 800 copies of "Lecture on Mathe... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,904 |
10. (15 points) As
shown in Figure 3, $O B$ is a chord of the circle $\odot O_{1}$ with center $(0, a)$ and radius $a$. A tangent to $\odot O_{1}$ is drawn through point $B$, and $P$ is any point on the minor arc $\overparen{O B}$. Perpendiculars are drawn from $P$ to $O B$, $A B$, and $O A$, with the feet of the perp... | 10. (1) Hint: Connect $E D, D F$, and prove $\triangle F D P \sim \triangle D E P$;
$$
\begin{array}{l}
\text { (2) } D\left(-\frac{\sqrt{3}}{4} a, \frac{3}{4} a\right), E\left(-\frac{3 \sqrt{3}}{4} a, \frac{3}{4} a\right), \\
F\left(-\frac{\sqrt{3}}{2} a, 0\right), P\left(-\frac{\sqrt{3}}{2} a, \frac{a}{2}\right), \\
... | S_{\triangle D E F}=\frac{3 \sqrt{3}}{16} a^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 712,905 |
Example 13 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
(1996, National Junior High School Mathematics League... | Solution: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $x_{1} < 0 < x_{2}$,
then $x_{1} < 0, \\
\therefore b > 2 \sqrt{a c} . \\
\text{Also, } \because |O A| = |x_{1}| > 1$. Therefore, the parabola opens upwards, and when $x = -1$, $y > 0$, so $a(-1)^{2} + b(-1) + c > 0$, which means $b < a + c + ... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,906 |
11. (10 points) If $a, b, c, d > 0$, prove: In the equations
$$
\begin{array}{l}
\frac{1}{2} x^{2}+\sqrt{2 a+b} x+\sqrt{c d}=0 ; \\
-\frac{1}{2} x^{2}+\sqrt{2 b+c} x+\sqrt{a d}=0 \\
\frac{1}{2} x^{2}+\sqrt{2 c+d} x+\sqrt{a b}=0 ; \\
\frac{1}{2} x^{2}+\sqrt{2 d+a} x+\sqrt{b c}=0
\end{array}
$$
at least two of the equa... | 11. Write down the discriminants $\Delta_{1}, \Delta_{2}, \Delta_{3}, \Delta_{4}$ of these four equations.
Notice that $\Delta_{1}+\Delta_{3}>0 . \Delta_{2}+\Delta_{4}>0$,
so at least two of $\Delta_{1}, \Delta_{2}, \Delta_{3}, \Delta_{4}$ are greater than zero, meaning that at least two of the four equations have une... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,907 |
3. The parabola obtained by rotating the parabola $y=x^{2}-2 b x$ (where $b$ is a real number) $180^{\circ}$ about its vertex is ( ).
(A) $y=-x^{2}-2 b x$
(B) $y=-x^{2}-2 b x+2 b^{2}$
(C) $y=-x^{2}+2 b x$
(D) $y=-x^{2}+2 b x-2 b^{2}$ | 3.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,911 |
Example 14 Let $a, b, c, a+b-c, a+c-b, b+c-a, a+b+c$ be seven distinct prime numbers, and the sum of two of $a, b, c$ is 800. Let $d$ be the difference between the largest and smallest of these seven prime numbers. Find the maximum possible value of $d$.
(2001, China Mathematical Olympiad) | $$
\begin{array}{l}
\text { Let } a<b<c<d \text { be prime numbers, and } a+b, a+c, b+c \text { are also prime numbers. } \\
\text { Without loss of generality, let } a< b, \\
\therefore c<a+b<a+c<b+c .
\end{array}
$$
Also, since one of $a+b$, $a+c$, $b+c$ is
800,
$$
\therefore c<800 \text {. }
$$
Since $799=17 \time... | 1594 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,917 |
4. Given in $\triangle A B C$, $\angle C=90^{\circ}, A C=4, B C=3$. $C D$ is the altitude on $A B$, $O_{1} 、 O_{2}$ are the incenters of $\triangle A C D 、 \triangle B C D$ respectively. Then $\mathrm{O}_{1} \mathrm{O}_{2}=$ $\qquad$ . | $4 \cdot \sqrt{2}$ | 4 \cdot \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,918 |
Three. (20 points) As shown in Figure 2, $\triangle A O B$ is an equilateral triangle, and the coordinates of point $B$ are $(2,0)$. A line $l$ is drawn through point $C$ $(-2,0)$, intersecting $A O$ at $D$ and $A B$ at $E$, such that the areas of $\triangle A D E$ and $\triangle D C O$ are equal. Find the function exp... | Three, $\because \triangle A D E$ and $\triangle D C O$ have equal areas,
$\therefore \angle A O B$ and $\triangle C B E$ have equal areas.
The area of $\triangle A O B$ is $\sqrt{3}$. Let the coordinates of point $E$ be $\left(x_{0}, y_{0}\right)$, then the area of $\triangle C B E$ is $2 y_{0}$. From $2 y_{0}=\sqrt{3... | y=\frac{\sqrt{3}}{7}(x+2) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,920 |
Four, (20 points) Given the quadratic function $y=x^{2}-x-2$ and the real number $a>-2$. Find:
(1) The minimum value of the function in $-2<x \leqslant a$;
(2) The minimum value of the function in $a \leqslant x \leqslant a+2$. | For the function $y=x^{2}-x-2$, the graph is shown in Figure 4.
(1) When $-2<a<\frac{1}{2}$,
$y_{\text {min }}=\left.y\right|_{x=a}=a^{2}-a-2$;
When $a \geqslant \frac{1}{2}$, $y_{\text {min }}=$
$$
\left.y\right|_{x=\frac{1}{2}}=-\frac{9}{4} \text {. }
$$
(2) When $-2<a$ and $a+2<\frac{1}{2}$, i.e., $-2<a<-\frac{3}{2... | y_{\text {min }} = \begin{cases}
a^{2} - a - 2 & \text{if } -2 < a < \frac{1}{2} \\
-\frac{9}{4} & \text{if } a \geq \frac{1}{2} \\
a^{2} + 3a & \text{if } -2 < a < -\frac{3}{2} \\
-\frac | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,921 |
Five. (20 points) As shown in Figure 3, given that $AB$ is the diameter of $\odot O$, $C$ is a point on $\odot O$, extend $BC$ to $D$ such that $CD = BC$, $CE \perp AD$, the foot of the perpendicular is $E$, $BE$ intersects $\odot O$ at $F$, $AF$ intersects $CE$ at $P$. Prove:
$$
PE = PC.
$$ | Extend $DA$ to intersect $\odot O$ at $K$, and connect $BK, OC$.
$$
\begin{array}{l}
\because AB \text{ is the diameter of } \odot O, \therefore BK \perp DA. \\
\text{Also, } \because CE \perp AD, \\
\therefore CE \parallel BK. \\
\text{Thus, } \angle 1 = \angle 2. \\
\text{Also, } \because A, K, B, F
\end{array}
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,922 |
Six. (20 points) A company plans
to donate $N$ million yuan (where $130<N<150$) to $n$ Hope Primary Schools. The allocation method is as follows: give the first school 4 million yuan and $\frac{1}{m}$ of the remaining amount; give the second school 8 million yuan and $\frac{1}{m}$ of the new remaining amount; give the ... | Six, let each school receive $x$ ten thousand yuan.
From the problem, we have $x=\frac{N}{n}$.
Also, since the $n$-th school receives $x=4n$, with the remainder being zero,
$$
\begin{array}{l}
\therefore N=(4n) \times n=4n^2. \\
\text{Also, since } 130<N<150,
\end{array}
$$
i.e., $130<4n^2<150$, and $n$ is an integer,... | N=144, n=6, x=24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,923 |
1. Given that $a_{1}, a_{2}, \cdots, a_{2002}$ are all positive numbers, and satisfy
$$
\begin{array}{l}
M=\left(a_{1}+a_{2}+\cdots+a_{2001}\right)\left(a_{2}+a_{3}+\cdots+a_{2001}\right), \\
N=\left(a_{1}+a_{2}+\cdots+a_{2002}\right)\left(a_{2}+a_{3}+\cdots+a_{2001}\right) .
\end{array}
$$
Then the relationship betwe... | $-1 .(\mathrm{A})$.
Let $T=a_{2}+a_{3}+\cdots+a_{2001}$. Then we have
$$
\begin{aligned}
M & =\left(a_{1}+T\right)\left(T+a_{2002}\right) \\
& =T^{2}+a_{1} T+a_{2002} T+a_{1} a_{2002}, \\
N & =\left(a_{1}+T+a_{2002}\right) T=T^{2}+a_{1} T+a_{2002} T . \\
\because & M-N=a_{1} a_{2002}>0, \therefore M>N .
\end{aligned}
$... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,924 |
2. The area of trapezoid $ABCD$ is $S, AB \parallel CD$, and $AB=b$, $CD=a(a<b)$, the diagonals $AC$ and $BD$ intersect at point $O$, and the area of $\triangle BOC$ is $\frac{1}{7} S$. Then $\frac{a}{b}=(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{4-\sqrt{15}}{2}$
(C) $\frac{5-\sqrt{21}}{2}$
(D) $\frac{6-\sqrt{26}}{2}$ | 2. (C).
As shown in Figure 4, we know
$$
\begin{array}{l}
S_{1}+S_{2}=\frac{5}{7} S . \\
\text { Also, } \frac{S_{2}}{S_{\triangle B A C}}=\frac{S_{2}}{\frac{1}{7} S}=\frac{O D}{O B}, \\
\frac{S_{\triangle B A C}}{S_{1}}=\frac{\frac{1}{7} S}{S_{1}}=\frac{O C}{O A} .
\end{array}
$$
Given $A B \parallel C D$, we have $... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,925 |
3. A store sells three models of computers: $A$, $B$, and $C$. It is known that in July, the sales revenue of model $B$ accounts for $55\%$ of the total sales revenue of these three models. In August, the sales revenue of models $A$ and $C$ decreased by $12\%$ compared to July, but the store's total sales revenue incre... | 3. (D).
Let the sales revenue of model $B$ computers in July be $x$ yuan, and the sales revenue of models $A$ and $C$ computers be $y$ yuan. The sales revenue of model $B$ computers in August increased by $z \%$ compared to July. Then we have
$$
\left\{\begin{array}{l}
x=(x+y) 55 \%, \\
x(1+z \%) + y(1-12 \%) = (x+y)(... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,926 |
4. As shown in Figure $1, A B$ is the diameter of $\odot O$. The radius is $\frac{5}{2}, B C=3$, points $E$ and $F$ are taken on chords $A B$ and $A C$ respectively, such that line segment $E F$ divides $\triangle A B C$ into two parts of equal area. Then the minimum value of line segment $E F$ is ( ).
(A)2
(B) $2 \sqr... | 4. (A).
$\because AB$ is the diameter of $\odot O$, $\therefore \angle C=90^{\circ}$.
Also, $\because$ the radius is $\frac{5}{2}$, $\therefore AB=5, BC=3$. Thus, $AC=4$. Therefore, $S_{\triangle ABC}=\frac{1}{2} \times 3 \times 4=6$. Let $AE=x, AF=y$.
Since $S_{\triangle AEF}=\frac{1}{2} x y \sin A=\frac{1}{2} S_{\tr... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,927 |
Example 1 In the right triangle $\triangle ABC$, the hypotenuse $AC=2$, $O$ is the midpoint of $AC$, and $I$ is the incenter of $\triangle ABC$. Find the minimum value of $OI$. | Solution: As shown in Figure 1, with $O$ as the center and $OA$ as the radius, draw a circle. Extend $BI$ to intersect $\odot O$ at $M$. Then
$$
\begin{array}{l}
\angle IAM \\
=\angle OAM + \angle OAI \\
=\angle CAM + \angle OAI \\
=\angle CBM + \angle OAI = \frac{\pi}{4} + \frac{1}{2} \angle A \\
=\angle ABI + \angle ... | \sqrt{2} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,928 |
5. As shown in Figure 2, in parallelogram $A B C D$, $A B=15$, a circle is drawn through point $D$ that is tangent to $A B$ and $B C$, and intersects sides $A D$ and $C D$ at points $E$ and $F$, respectively, with $5 A E=4 D E$ and $8 C F=D F$. Then $B H$ equals ( ).
(A) 5
(B) 6
(C) 7
(D) 8 | 5. (C).
$\because 8 C F=D F, \therefore C F=15 \times \frac{1}{9}=\frac{5}{3}$.
By the secant-tangent theorem, we have $C H^{2}=C F \cdot C D$,
thus $C H=5$. Let $B C=x>0$.
$\because B H=x-5=B G, \therefore A G=20-x$.
Also, $\because 5 A E=4 D E, \therefore D E=\frac{5}{9} x, A E=\frac{4}{9} x$.
By the secant-tangent t... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,929 |
6. A city's waterworks, the reservoir originally has 500 tons of water. In one day, while water is being added to the reservoir, water is also being supplied to users from the reservoir. $x(0 \leqslant x \leqslant 24)$ hours of water supplied to users is $80 \sqrt{24 x}$ tons. When 96 tons of water are added to the res... | 6. (B).
From the given, we have
$$
y=500+96 x-80 \sqrt{24 x}(0 \leqslant x \leqslant 24) \text {. }
$$
Let $\sqrt{24 x}=t(0 \leqslant t \leqslant 24)$. Then
$$
y=4 t^{2}-80 t+500=4(t-10)^{2}+100 \text {. }
$$
Therefore, when $t=10$, the minimum value of $y$ is 100. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,930 |
1. Let $[A]$ denote the greatest integer less than or equal to $A$, and set $A=38+$ $17 \sqrt{5}$. Then $A^{2}-A[A]=$ $\qquad$ . | $\begin{array}{l}\text { II.1.1. } \\ \because A=38+17 \sqrt{5}=(\sqrt{5}+2)^{3} \text {, let } B=(\sqrt{5}-2)^{3} \text {, } \\ \therefore A-B=76 . \\ \text { Also } \because 0<(\sqrt{5}-2)^{3}<1 \text {, } \\ \therefore[A]=76 \text {, then } A-[A]=B . \\ \text { Therefore } A^{2}-A[A]=A(A-[A])=A B=1 .\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,931 |
2. Let $a$ be a real root of the equation $x^{2}-2002 x+1=0$. Then $a^{2}-2001 a+\frac{2002}{a^{2}+1}=$ $\qquad$ . | 2.2001 .
$\because \alpha$ is a real root of the equation $x^{2}-2002 x+1=0$, then
$$
\begin{array}{l}
\alpha^{2}-2002 \alpha+1=0 . \\
\therefore \alpha+\frac{1}{\alpha}=2002 .
\end{array}
$$
Therefore, $\alpha^{2}-2001 \alpha+\frac{2002}{\alpha^{2}+1}=\alpha-1+\frac{1}{\alpha}=2001$. | 2001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,932 |
3. As shown in Figure 3, given two perpendicular lines intersecting at point $O$, particle 甲 moves from point $A$ to point $C$ from west to east at a speed of 1.5 cm/s, and particle 乙 moves from point $B$ to point $D$ from south to north at a speed of 2.5 cm/s. If $A O=3$ cm, $B O=4$ cm, then after $\qquad$ seconds, th... | 3. $\frac{14}{11}$ or $\frac{50}{29}$ or $\frac{16}{9}$.
As shown in Figure 5, let the particle 甲 reach position $C_{1}$ and particle 乙 reach position $D_{1}$ after $t$ seconds. At this time, $\triangle C_{1} O D_{1} \backsim \triangle B O A$, so $\frac{C_{1} O}{B O}=\frac{D_{1} O}{A O}$, which gives $\frac{3-1.5 t}{4... | \frac{14}{11} \text{ or } \frac{50}{29} \text{ or } \frac{16}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,933 |
4. In Rt $\triangle A B C$, $\angle A=90^{\circ}, A D \perp B C$ at $D$, the angle bisector of $\angle B$ intersects $A D$ at $M$, and intersects $A C$ at $N$, and $A B-A M=\sqrt{3}-1, B M \cdot B N=2$. Then $\frac{B C-A B}{C N}=$ $\qquad$ | 4. $\frac{\sqrt{3}}{2}$.
As shown in Figure 6, with point $A$ as the center and $AN$ as the radius, a circle is drawn intersecting $AB$ at point $F$.
$$
\begin{array}{c}
\because \angle 2+\angle 3=\angle 4+\angle 5 \\
=90^{\circ}, \text { and } \angle 3=\angle 4, \\
\therefore \angle 5=\angle 2 . \\
\text { Also, } \b... | \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,934 |
一,(20 points) Given that $a, b, c, d$ are all non-zero real numbers. Substituting $x=a$ and $b$ into $y=x^{2}+c x$, the value of $y$ is 1 in both cases; substituting $x=c$ and $d$ into $y=x^{2}+a x$, the value of $y$ is 3 in both cases. Try to find the value of $6 a+2 b+3 c+2 d$. | Given the problem, we have
$$
\left\{\begin{array}{l}
a^{2}+a c=1, \\
b^{2}+b c=1, \\
c^{2}+a c=3, \\
d^{2}+a d=3 .
\end{array}\right.
$$
(1) + (3) gives $(a+c)^{2}=4 \Rightarrow a+c= \pm 2$.
Thus, we know $a+c \neq 0$.
(3) $\div$ (1) gives $c=3 a$,
so $\left\{\begin{array}{l}a=\frac{1}{2}, \\ c=\frac{3}{2}\end{array... | \pm \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,935 |
In $\triangle A B C$, $A D$ is the bisector of $\angle B A C$ and intersects $B C$ at $D$. Circles $\odot B, \odot C$ are constructed with $B D, C D$ as radii, respectively. These circles intersect line $A D$ at $E$ and $F$. Prove:
$$
A E \cdot A F + B D \cdot C D = A B \cdot A C .
$$ | As shown in Figure 7, let $\odot O$ intersect $AD$ at point $M$, and connect $CF$, $BE$, and $BM$.
$$
\because \angle ADB + \angle FDC = 180^{\circ},
$$
$$
\begin{array}{l}
\angle AFC + \angle DFC \\
= 180^{\circ}, \angle FDC = \\
\angle DFC,
\end{array}
$$
$$
\therefore \angle ADB = \angle AFC.
$$
$$
\text{Also, } \be... | AE \cdot AF + BD \cdot CD = AB \cdot AC | Geometry | proof | Yes | Yes | cn_contest | false | 712,936 |
Three. (25 points) There are 20 weights, all of which are integers, such that any integer weight $m(1 \leqslant m \leqslant 2002)$ can be balanced by placing it on one pan of a scale and some of the weights on the other pan. What is the smallest possible value of the heaviest weight among these 20 weights? | Let's assume the weights of these 20 weights are \(a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{20}\), where \(a_{i} (1 \leqslant i \leqslant 20)\) are positive integers. It is easy to see that
\[
a_{1}=1, a_{k+1} \leqslant a_{1}+a_{2}+\cdots+a_{k}+1 (1 \leqslant k \leqslant 19).
\]
\[
\text{Then } a_{2} \leqsla... | 146 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,937 |
1. If $f(x)$ is a decreasing function on $\mathbf{R}$, and the graph of $f(x)$ passes through the points $A(0,3)$ and $B(3,-1)$, then the solution set of the inequality $|f(x+1)-1|<2$ is ( ).
(A) $(-\infty, 3)$
(B) $(-\infty, 2)$
(C) $(0,3)$
(D) $(-1,2)$ | -、1.(D).
It is easy to get $-1<f(x+1)<3$.
$\because f(0)=3, f(3)=-1$, and $f(x)$ is a decreasing function,
$\therefore$ the solution to the inequality $-1<f(x)<3$ is $0<x<3$.
Therefore, when $-1<f(x+1)<3$, we have $0<x+1<3$. | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 712,938 |
Example 2 Given a square $A B C D$ in the plane. Find the minimum value of the ratio $\frac{O A+O C}{O B+O D}$, where $O$ is any point in the plane.
(1993, St. Petersburg City Mathematical Selection Exam) | Solution: As shown in Figure 2, it is easy to see that when $O$ is at point $A$, we have
$$
\begin{array}{l}
\frac{O A+O C}{O B+O D} \\
=\frac{A C}{A B+A D}=\frac{\sqrt{2}}{2} .
\end{array}
$$
When $O$ is inside the square, such as at the center of the square, the ratio is 1;
When $O$ is outside the square, such as o... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,939 |
2. If the graph of the function $f(x)=a^{2} \sin 2 x+(a-2) \cos 2 x$ is symmetric about the line $x=-\frac{\pi}{8}$, then the value of $a$ is ( ).
(A) $\sqrt{2}$ or $-\sqrt{2}$
(B) 1 or -1
(C) 1 or -2
(D) -1 or 2 | 2. (C).
$\because x=-\frac{\pi}{8}$ is the axis of symmetry of $f(x)$,
$$
\therefore f(0)=f\left(-\frac{\pi}{4}\right) \text {. }
$$
Then $a-2=-a^{2} \Rightarrow a^{2}+a-2=0 \Leftrightarrow a=1$ or $a=-2$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,940 |
3. Let the equation of the ellipse be $\frac{x^{2}}{3}+y^{2}=1$, and $A(0,-1)$ be one endpoint of the minor axis, $M$ and $N$ be two distinct points on the ellipse. If there always exists an isosceles $\triangle A M N$ with $M N$ as the base, then the range of the slope $k$ of the line $M N$ is ( ).
(A) $(-1,0]$
(B) $[... | 3. (C).
Let $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)$, it is easy to know that $x_{1} \neq x_{2}$. Suppose the equation of $MN$ is $y=k x+b$. Substituting into the ellipse equation $x^{2}+3 y^{2}=3$ yields
$$
\left(1+3 k^{2}\right) x^{2}+6 k b x+3 b^{2}-3=0 \text {. }
$$
Its two roots are $x_{1}$ ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,941 |
4. $f(x)$ is a function defined on $\mathbf{R}$, and for any $x$ it satisfies $f(x+1)=-f(x)$. It is known that when $x \in(2,3]$, $f(x)=x$. Then, when $x \in(-2,0]$, the expression for $f(x)$ is ( ).
(A) $f(x)=x+4$
(B) $f(x)=\left\{\begin{array}{ll}x+4, & x \in(-2,-1] \\ -x+2, & x \in(-1,0]\end{array}\right.$
(C) $f(x)... | 4. (C).
$$
\begin{array}{l}
\because f(x+1)=-f(x), \\
\quad \therefore f(x+2)=-f(x+1)=f(x) .
\end{array}
$$
Thus, $f(x)$ is a periodic function with a period of 2. When $x \in (2,3]$, we have $f(x)=x$. When $x \in (3,4]$, $x-1 \in (2,3]$, so $f(x-1)=x-1$. Therefore, $f(x)=-f(x-1)=1-x$, hence $f(x)=\left\{\begin{array}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,942 |
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