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Example 5 A public bus company, after two technological innovations, adjusted the seating from New Year's Day 1957, allowing each bus to carry 6 more people, so that the number of people each 5 trips could carry exceeded 270. From New Year's Day 1958, a trailer was added, allowing each bus to carry 98 more people than ... | Solution: Let the number of people each car can carry before New Year's Day 1957 be $x$, then on New Year's Day 1957 and 1958, each car can carry $x+6$ and $x+98$ people, respectively. According to the problem, we have $\left\{\begin{array}{l}5(x+6)>270, \\ 3(x+98)>8(x+6) .\end{array}\right.$
Solving, we get $\left\{\b... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. A pipe burst occurred in a low-lying area by the riverbank, and river water is continuously gushing out, assuming the water gushing out per minute is constant. If two water pumps are used to pump out the water, it takes 40 minutes to finish; if four water pumps are used, it takes 16 minutes to finish. If the water ... | 12.6 .
Let the amount of water that has already gushed out before the pumping starts be $u$ cubic meters, the rate of water gushing out per minute be $b$ cubic meters, and the amount of water each pump can extract per minute be $c$ cubic meters $(c \neq 0)$. From the given conditions, we have
$$
\left\{\begin{array}{l... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, let $x, y$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}$, $\cdots, a_{n}$, and satisfy the condition: when $x>y$, $x-y \geqslant \frac{x}{15}$. Find the maximum number of these natural numbers $n$. | $$
\begin{array}{l}
\text { Let's assume } a_{1}=1, d_{4} \geqslant 2 ; \\
a_{5}=a_{4}+d_{4} \geqslant 6, d_{5} \geqslant \frac{6^{2}}{19-6}>2, d_{5} \geqslant 3 ; \\
a_{6} \geqslant 9, d_{6} \geqslant \frac{9^{2}}{19-9}>8, d_{6} \geqslant 9 ; \\
a_{7} \geqslant 18, d_{7} \geqslant \frac{18^{2}}{19-18}=324 ; \\
a_{8} \... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Suppose $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=$ 0 , and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is (1990, Hefei Junior High School Mathematics Competition) | Given $a^{2}+2 a-1=0$, we know $a \neq 0$.
$$
\therefore\left(\frac{1}{a}\right)^{2}-2\left(\frac{1}{a}\right)-1=0 \text {. }
$$
From $b^{4}-2 b^{2}-1=0$, we get
$$
\left(b^{2}\right)^{2}-2\left(b^{2}\right)-1=0 \text {. }
$$
From $1-a b^{2} \neq 0$, we know $\frac{1}{a} \neq b^{2}$.
From (1) and (2), we know $\frac{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in the figure, circle O is inscribed in $\triangle ABC$ touching sides $AC$, $BC$, and $AB$ at points $D$, $E$, and $F$ respectively, with $\angle C=90^{\circ}$. The area of $\triangle ABC$ is 6. Then $AF \cdot BF=$ $\qquad$ | 4.6 .
Let $A C=b, B C=a, A B=c, \odot O$ have a radius of $r$. Then
$$
\begin{array}{l}
r=\frac{a+b-c}{2} \cdot a^{2}+i^{2}=c^{2} . \\
\therefore A F \cdot B F=A C \cdot B E \\
=(b-r)(a-r) \\
=\frac{b+c-a}{2} \cdot \frac{a+c-b}{2} \\
=\frac{1}{4}\left[c^{2}-(a-b)^{2}\right] \\
=\frac{1}{2} a b=6 .
\end{array}
$$ | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x$ and $y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{5}+1999(x-1)=2519, \\
(y-1)^{5}+1999(y-1)=-2519 .
\end{array}\right.
$$
Then the value of $x+y$ is $\qquad$ | 1.2.
Let $f(t)=t^{5}+1999 t$. Then $f(t)$ is increasing on $R$, and $f(x-1)=-f(y-1)$. Since $f(y-1)=-f(1-y)$, then
$$
\begin{array}{l}
f(x-1)=f(1-y) . \\
\therefore x-1=1-y \Rightarrow x+y=2 .
\end{array}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant\right.$ $\left.180^{\circ}\right)$, and the complex numbers $z, (1+i)z, 2\bar{z}$ correspond to three points $P, Q, R$ on the complex plane. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by... | 5.3.
Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
\begin{aligned}
w+z & =2 \bar{z}+(1+\mathrm{i}) z, \text { i.e., } w=2 \bar{z}+\mathrm{i} z . \\
\therefore|w|^{2} & =(2 \bar{z}+\mathrm{i} z)(2 z-\mathrm{i} \bar{z}) \\
& =4+1+2 \mathrm{i}\left(z^{2}-\bar{z}^{2}\ri... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points) There is a quantity $W$, after "modeling" the relationship is given by
$$
W=\frac{1}{c}\left(\frac{3 a}{\sqrt{1-u^{2}}}+\frac{b}{\sqrt{1-t^{2}}}\right),
$$
where $a, b, c, u, t$ are all positive, $u<1, t<1$, and satisfy $a t+b u=c, a^{2}+2 b c u=b^{2}+c^{2}$. Please design a method to find... | Given $a^{2}=b^{2}+c^{2}-2 b c u$. All $a, b, c, u$ are positive, and $ub^{2}+c^{2}-2 b c=(b-c)^{2}$.
From this, we get $|b-c|<a<b+c$.
Therefore, the positive numbers $a, b, c$ can be the lengths of the three sides of a triangle. Let the vertices opposite these sides be $A, B, C$, respectively. By transforming the con... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $x_{1}, x_{2}, \cdots, x_{9}$ all be positive integers, and $x_{1}<x_{2}<\cdots<x_{9}, x_{1}+x_{2}+\cdots+x_{9}=220$. Then, when the value of $x_{1}+x_{2}+\cdots+x_{5}$ is maximized, the minimum value of $x_{9}-x_{1}$ is $\qquad$
$(1992$, National Junior High School Mathematics League) | Solution: From $x_{1}+x_{2}+\cdots+x_{9}=220$,
we know $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}>110$,
or $x_{1}+x_{2}+\cdots+x_{5} \leqslant 110$.
From (1), then $x_{5} \geqslant 25$. Thus, $x_{6} \geqslant 26, x_{7} \geqslant$ $27, x_{8} \geqslant 28, x_{9} \geqslant 29$. We get
$$
\begin{aligned}
\left(x_{1}+x_{2}+x_{3}+x_{4}+... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given $p^{3}+q^{3}=2$, where $p, q$ are real numbers. Then the maximum value of $p+q$ is $\qquad$
(1987, Jiangsu Province Junior High School Mathematics Competition) | Solution: Let $s=p+q$.
From $p^{3}+q^{3}=2$ we get
$(p+q)\left(p^{2}+q^{2}-p q\right)=2$,
$(p+q)^{2}-3 p q=\frac{2}{s}$,
Thus, $p q=\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)$.
From (1) and (2), $p$ and $q$ are the two real roots of the equation
$$
x^{2}-s x+\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)=0
$$
We know $\Delt... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, $5 x, y, z$ are real numbers, and satisfy $x+y+z=0, xyz=2$. Find the minimum value of $|x|+|y|+|z|$.
(1990, Beijing Junior High School Mathematics Competition). | Solution: From the problem, we know that among $x, y, z$, there are 2 negative numbers and 1 positive number. Without loss of generality, let $x>0, y<0, z<0$. When $x>0, -y>0, -z>0$, we have
$(-y)(-z) \leqslant \left[\frac{(-y)+(-z)}{2}\right]^{2}=\frac{x^{2}}{4}$,
which means $2 x=\frac{4}{(-y)(-z)} \geqslant \frac{4}... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 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 for High School Grade 2) | Solution: Completing the square, we get
$$
u=(x-y)^{2}+\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}-2 \text {. }
$$
Consider points $P_{1}\left(x, \frac{9}{x}\right)$, $P_{2}\left(y,-\sqrt{2-y^{2}}\right)$ on the plane. When $x \in \mathbf{R}, x \neq 0$, the trajectory of $P_{1}$ is a hyperbola with the two coordinate ... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 If $x, y$ are real numbers, find the minimum value of $S=5 x^{2}-4 x y$ $+y^{2}-10 x+6 y+5$. | Solution: From the given, we have
$$
\begin{array}{l}
5 x^{2}-(10+4 y) x+y^{2}+6 y+5-S=0 . \\
\Delta_{x}=-4\left(y^{2}+10 y-5 S\right) \geqslant 0 . \\
\text { Hence, } 5 S \geqslant y^{2}+10 y=(y+5)^{2}-25 \\
\quad \geqslant-25 .
\end{array}
$$
Thus, $5 S \geqslant y^{2}+10 y=(y+5)^{2}-25$
$\therefore S \geqslant-5$,... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 Let $x, y$ be positive numbers, and $x+y=1$. Find the minimum value of the function $W=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$. (Adapted from the 3rd Canadian Mathematical Competition) | Solution: Without loss of generality, let $x \geqslant y$, and set $x=\frac{1}{2}+t$, then $y=$
$$
\begin{array}{l}
\frac{1}{2}-t, 0 \leqslant t<\frac{1}{2} . \\
\therefore W=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \\
=\frac{3+2 t}{1+2 t} \cdot \frac{3-2 t}{1-2 t}=\frac{9-4 t^{2}}{1-4 t^{2}} \\
\geqslant ... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
16. Find the smallest integer \( n (n \geq 4) \), such that from any \( n \) integers, four different numbers \( a, b, c, d \) can be selected, making \( a + b - c - d \) divisible by 20. | Solution: First, we consider the different residue classes modulo 20. For a set with $k$ elements, there are $\frac{1}{2} k(k - 1)$ integer pairs. If $\frac{1}{2} k(k - 1) > 20$, i.e., $k \geq 7$, then there exist two pairs $(a, b)$ and $(c, d)$ such that $a + b \equiv c + d \pmod{20}$, and $a, b, c, d$ are all distinc... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $x$ be a real number, and $f(x)=|x+1|+|x+2|+|x+3|+|x+4|+|x+5|$. Find the minimum value of $f(x)$. | Analysis: According to the geometric meaning of absolute value, draw the points $A, B, C, D, E$ corresponding to the real numbers $-1, -2, -3, -4, -5$ on the number line, as shown in Figure 1. Let $x$ correspond to the moving point $P$, then $f(x)=|PA|+|PB|+|PC|+|PD|+|PE| \geqslant |CB|+|CD|+|CA|+|CE|=2+4=6$, that is, ... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $f(x)=|1-2 x|, x \in[0$, 1]. How many real solutions does the equation $f(f(f(x)))=\frac{x}{2}$ have? | Solution: To find the number of real solutions using the graphical method. First, draw the graph of $f(x) = |2x - 1|$, and then double the y-coordinates of all points while keeping the x-coordinates unchanged. Next, shift the obtained graph down by 1 unit (as shown in Figure 2), and then reflect the part of the graph b... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let the function $f_{3}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed part of the figure enclosed by the graph of the function $y=f_{2}(x)$ and the $x$-axis is $\qquad$ | Analysis: First, draw the graph of $f_{0}(x)=|x|$ and translate it downward by 1 unit to get the graph of $y=f_{0}(x)-1$, from which we obtain the graph of $f_{1}(x)=\left|f_{0}(x)-1\right|$ (as shown in Figure 7).
Next, translate the graph of $f_{1}(x)$ downward by 2 units to get the graph of $y=f_{1}(x)-2$, and refl... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7: A town has five primary schools along a circular road, sequentially named First, Second, Third, Fourth, and Fifth Primary Schools. They have 15, 7, 11, 3, and 14 computers, respectively. To make the number of computers equal in each school, some computers need to be transferred to neighboring schools: First ... | Analysis: Let $A, B, C,$
$D, E$ represent the five primary schools in a clockwise order, and let them sequentially transfer $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ computers to their neighboring schools, as shown in Figure 9. Then,
\[
\begin{array}{c}
7 + x_{1} - x_{2} = 11 + x_{2} \\
- x_{3} = 3 + x_{3} - x_{4} = 14 + x_{... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 On a circular road, there are four middle schools arranged in sequence: $A_{1}, A_{2}, A_{3}, A_{4}$. They have 15, 8, 5, and 12 color TVs, respectively. To make the number of color TVs in each school the same, some schools are allowed to transfer color TVs to adjacent schools. How should the TVs be transferr... | Solution: Let $A_{1}$ high school transfer $x_{1}$ color TVs to $A_{2}$ high school (if $x_{1}$ is negative, it means $A_{2}$ high school transfers $r_{1}$ color TVs to $A_{1}$ high school. The same applies below), $A_{2}$ high school transfers $x_{2}$ color TVs to $A_{3}$ high school, $A_{3}$ high school transfers $x_... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 (Fill in the blank Question 1) Given $\frac{1}{4}(b-c)^{2}=$ $(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ . | Solution 1: From the given, we have
$$
\begin{aligned}
0 & =\frac{1}{4}[-(a-b)-(c-a)]^{2}-(a-b)(c-a) \\
& =\frac{1}{4}\left[(a-b)^{2}+(c-a)^{2}-2(a-b)(c-a)\right] \\
& =\frac{1}{4}[(a-b)-(c-a)]^{2} \\
& =\frac{1}{4}(2 a-b-c)^{2} .
\end{aligned}
$$
Thus, $\frac{b+c}{a}=2$.
Solution 2: The given condition indicates that... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $[x]$ represents the greatest integer not exceeding $x$. Then the number of solutions to the equation
$$
3^{2 x}-\left[10 \times 3^{x+1}\right]+\sqrt{3^{2 x}-10 \times 3^{x+1}+82}=-80
$$
is | 5.2.
The original equation can be transformed into
$$
\begin{array}{l}
3^{2 x}-\left[10 \times 3^{x+1}\right]+82 \\
+\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-2=0 . \\
\therefore\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}+2\right) \\
\quad \cdot\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-1... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{r}
\text { 5. Given a sequence } z_{0}, z_{1}, \cdots, z_{n}, \cdots \text { satisfying } z_{0}=0, z_{1} \\
=1, z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right), \alpha=1+\sqrt{3} \mathrm{i}, n=1,2,
\end{array}
$$
5. Given a sequence of complex numbers $z_{0}, z_{1}, \cdots, z_{n}, \cdots$ satisfying $z_... | 5.5.
Since $z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right)=\alpha^{2}\left(z_{n-1}-z_{n-2}\right)$ $=\cdots=\alpha^{n}\left(z_{1}-z_{0}\right)=\alpha^{n}$, therefore,
$$
z_{n}-z_{n-1}=\alpha^{n-1}, \cdots, z_{1}-z_{0}=\alpha^{0}=1 \text {. }
$$
Adding the above $n$ equations, we get
$$
z_{n}=\alpha^{n-1}+\cdots+\alph... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 20 points) For every pair of real numbers $x, y$, the function $f(t)$ satisfies $f(x+y)=f(x)+f(y)+xy+1$. If $f(-2)=-2$, find the number of integer solutions $a$ that satisfy $f(a)=a$.
Translate the above text into English, please retain the original text's line breaks and format, and output the tran... | Let $x=y=0$, we get $f(0)=-1$; let $x=y=-1$, from $f(-2)=-2$ we get $f(-1)=-2$. Also, let $x=1$, $y=-1$ to get $f(1)=1$. Then let $x=1$, we have
$$
f(y+1)=f(y)+y+2 \text {. }
$$
Therefore, $f(y+1)-f(y)=y+2$, which means when $y$ is a positive integer, $f(y+1)-f(y)>0$.
From $f(1)=1$, we know that for all positive inte... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points) A county is located in a desert area, and people have been engaged in a tenacious struggle against nature for a long time. By the end of 1998, the county's greening rate had reached $30 \%$. Starting from 1999, the following situation will occur every year: $16 \%$ of the original desert ar... | (1) Let the current desert area be $b_{1}$, and after $n$ years, the desert area be $b_{n+1}$. Thus, $a_{1}+b_{1}=1, a_{n}+b_{n}=1$.
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} a_{n}$, which is $\frac{96}{1... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given three real
numbers $x_{1}, x_{2}, x_{3}$, any one of them plus five times the product of the other two equals 6. The number of such triples $\left(x_{1}, x_{2}\right.$, $x_{3}$ ) is. | 4.5
From the problem, we have
$$
\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{1} x_{3}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}
$$
(1) - (2) gives $\left(x_{1}-x_{2}\right)\left(5 x_{3}-1\right)=0$.
Thus, $x_{1}=x_{2}$ or $x_{3}=\frac{1}{5}$.
Similarly, $x_{2}=x_{3}$ or $x_{1}=\frac{1}{5}$, $x_{3}=x_{1}$ or... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 If $f(x)=\left(2 x^{5}+2 x^{4}-53 x^{3}-\right.$ $57 x+54)^{1998}$, then $f\left(\frac{\sqrt{1}-\frac{1}{2}-i}{2}\right)=$ $\qquad$ . | Let $\frac{\sqrt{111}-1}{2}=x$, then we have
$$
\begin{array}{l}
2 x^{2}+2 x-55=0 \\
\because 2 x^{5}+2 x^{4}-53 x^{3}-57 x+54 \\
=x^{3}\left(2 x^{2}+2 x-55\right)+x\left(2 x^{2}+2 x\right. \\
\quad-55)-\left(2 x^{2}+2 x-55\right)-1 \\
\therefore f\left(\frac{\sqrt{111}-1}{2}\right)=(-1)^{1998}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, after expansion and combining like terms, the sum of the coefficients of the odd powers of $x$ is
保留源文本的换行和格式，直接输出翻译结果如下：
```
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, ... | Solution: Let $f(x)=\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}\right.$
$$
\begin{aligned}
& -3 x-3)^{2001} \\
= & a_{0}+a_{1} x+a_{2} x^{2}+\cdots \\
& +a_{4001} x^{4001}+a_{4002} x^{4002} .
\end{aligned}
$$
Let $x=1$ and $x=-1$, we get
$$
\begin{array}{l}
a_{0}+a_{1}+a_{2}+\cdots+a_{4001}+a_{4002} \\
=f(1)=0, \\
a_{... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\cdots+1994^{2}$. | Solution: Since this problem only requires the unit digit of the sum, we only need to consider the unit digit of each number. Thus, the original problem simplifies to finding the unit digit of
$$
\underbrace{i^{2}+2^{2}+3^{2}+4^{2}+\cdots+9^{2}}_{\text {199 groups }}+1^{2}+2^{2}+3^{2}+4^{2}
$$
The unit digits follow a... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Let $1995 x^{3}=1996 y^{3}=1997 z^{3}$,
$$
\begin{array}{l}
x y z>0, \text { and } \sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}} \\
=\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997} . \\
\quad \text { Then } \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=
\end{array}
$$
Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$ | Solution: Let $1995 x^{3}=1996 y^{3}=1997 z^{3}=k$, obviously $k \neq 0$. Then we have
$$
1995=\frac{k}{x^{3}}, 1996=\frac{k}{y^{3}}, 1997=\frac{k}{z^{3}} .
$$
From the given, we get
$$
\sqrt{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}=\sqrt[3]{\frac{k}{x^{3}}}+\sqrt[3]{\frac{k}{y^{3}}}+\sqrt[3]{\frac{k}{z^{3}}}>0 .
$$
Thus... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given real numbers $a, b$ satisfy $3 a^{4}+2 a^{2}-4=0$ and $b^{4}+b^{2}-3=0$. Then $4 a^{-4}+b^{4}=$ ( ).
(A) 7
(B) 8
(C) 9
(D) 10 | Solution: Transform the equation $3 a^{4}+2 a^{2}-4=0$ into $\frac{4}{a^{4}}-$ $\frac{2}{a^{2}}-3=0$, and it is known that $-\frac{2}{a^{2}}$ and $b^{2}$ are the two distinct real roots of the equation $x^{2}+x-3=0$.
Let $-\frac{2}{a^{2}}=x_{1}, b^{2}=x_{2}$. By Vieta's formulas, we have $x_{1}+x_{2}=-1, x_{1} x_{2}=-3... | 7 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 13 Given $a+b+c=0, a^{3}+b^{3}+c^{3}$ $=0$. Find the value of $a^{15}+b^{15}+c^{15}$. | Solution: From $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)$ - $\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)$, we have $a b c=0$. Therefore, at least one of $a, b, c$ is 0.
Assume $c=0$, then $a, b$ are opposites,
$$
\therefore a^{15}+b^{15}+c^{15}=0 \text {. }
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example: 15 Given positive integers $x, y, z$ satisfy $x^{3}-y^{3}-z^{3}=3 x y z, x^{2}=2(y+z)$. Find the value of $x y+y z+z x$.
---
The translation is provided as requested, maintaining the original text's line breaks and format. | $$
\begin{array}{l}
\text { Solution: } x^{3}-y^{3}-z^{3}-3 x y z \\
=x^{3}-(y+z)^{3}-3 x y z+3 y^{2} z+3 y z^{2} \\
=(x-y-z)\left(x^{2}+x y+x z+y^{2}+2 y z\right. \\
\left.+z^{2}\right)-3 y z(x-y-z) \\
=(x-y-z)\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right) \\
=\frac{1}{2}(x-y-z)\left[(x+y)^{2}+(y-z)^{2}\right. \\
\left.+(... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, let $a, b, c, d$ be four distinct real numbers such that
$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$, and $a c=b d$.
Find the maximum value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$. | Three, let $x=\frac{a}{b}, y=\frac{b}{c}$, then from $a c=b d$ we know $\frac{c}{d}=\frac{b}{a}=\frac{1}{x}, \frac{d}{a}=\frac{c}{b}=\frac{1}{y}$. Thus, the problem becomes finding the maximum value of $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}$ under the constraint $x \neq 1, y \neq 1, x+y+\frac{1}{x}+\frac{1}{y}=4$.
... | -12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Simplify $\frac{a+1}{a+1-\sqrt{1-a^{2}}}+\frac{a-1}{\sqrt{1-a^{2}}+a-1}$ $(0<|a|<1)$ The result is $\qquad$ | $\begin{array}{l}\text { Original expression }=\frac{(\sqrt{a+1})^{2}}{\sqrt{a+1}(\sqrt{a+1}-\sqrt{1-a})} \\ \quad+\frac{-(\sqrt{1-a})^{2}}{\sqrt{1-a}(\sqrt{1+a}-\sqrt{1-a})}=1 .\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $(x-1)^{2}$ divides the polynomial $x^{4}+a x^{3}-$ $3 x^{2}+b x+3$ with a remainder of $x+1$. Then $a b=$ $\qquad$ . | II, 1.0.
From the given, we have
$$
\begin{array}{l}
x^{4}+a x^{3}-3 x^{2}+b x+3 \\
=(x-1)^{2}\left(x^{2}+\alpha x+\beta\right)+x+1 \\
=x^{4}+(\alpha-2) x^{3}+(\beta+1-2 \alpha) x^{2}+(1+\alpha- \\
2 \beta) x+1+\beta .
\end{array}
$$
Then $a=\alpha-2$,
$$
\begin{array}{l}
-3=\beta+1-2 \alpha, \\
b=1+\alpha-2 \beta, \\... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3: A certain area currently has 10,000 hectares of arable land. It is planned that in 10 years, the grain yield per unit area will increase by $22\%$, and the per capita grain possession will increase by $10\%$. If the annual population growth rate is $1\%$, try to find the maximum number of hectares by which t... | Solution: Let the average annual reduction of arable land be $x$ hectares, and let the current population of the region be $p$ people, with a grain yield of $M$ tons/hectare. Then
$$
\begin{array}{l}
\frac{M \times(1+22 \%) \times\left(10^{4}-10 x\right)}{p \times(1+1 \%)^{10}} \\
\geqslant \frac{M \times 10^{4}}{p} \t... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $D$ be a point on the side $AB$ of $\triangle ABC$, point $D$ moves along a direction parallel to $BC$ to point $E$ on side $AC$; then from point $E$ along a direction parallel to $AB$ to point $F$ on side $BC$; then from point $F$ along a direction parallel to $CA$ to point $G$ on side $AB$, $\cdots \cdo... | Analysis: We can estimate through graphical experiments that point $D$ can return to the original starting point after 6 times.
In fact, as shown in
Figure 1, by the intercept theorem of parallel lines,
we get
$$
\begin{array}{l}
\frac{A D}{B D}=\frac{A E}{E C} . \\
=\frac{B F}{F C}=\frac{B G}{A G} \\
=\frac{C H}{A H}... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given three real numbers $x_{1}, x_{2}, x_{3}$, any one of these numbers plus five times the product of the other two always equals 6. The number of such triples $\left(x_{1}, x_{2}, x_{3}\right)$ is $\qquad$.
$(1995$, Dongfang Airlines Cup - Shanghai Junior High School Mathematics Competition) | Solving the system of equations given by the problem, we have:
$$
\left\{\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{3} x_{1}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}\right.
$$
(1) - (2) gives
$$
\left(x_{1}-x_{2}\right)\left(1-5 x_{3}\right)=0 \text {. }
$$
(2) - (3) gives
$$
\left(x_{2}-x_{3}\right)\left(1... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Real numbers $x, y$ satisfy $x \geqslant y \geqslant 1$ and $2 x^{2}-x y$ $-5 x+y+4=0$. Then $x+y=$ $\qquad$ | 3.4 .
From the given equation, we know that $2 x^{2}-5 x+4=y(x-1) \leqslant x(x-1)$, which leads to $x^{2}-4 x+4 \leqslant 0$, or $(x-2)^{2} \leqslant 0$. Therefore, $x=2$.
Substituting $x=2$ into the given equation yields $y=2$, hence $x+y=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $n$ be a natural number, and $a_{n}=\sqrt[3]{n^{2}+2 n+1}$ $+\sqrt[3]{n^{2}-1}+\sqrt[3]{n^{2}-2 n+1}$. Then the value of $\frac{1}{a_{1}}+\frac{1}{a_{3}}+\frac{1}{a_{5}}$ $+\cdots+\frac{1}{a_{997}}+\frac{1}{a_{999}}$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 4. (C).
$$
\begin{array}{l}
a_{n}=(\sqrt[3]{n+1})^{2}+\sqrt[3]{n+1} \cdot \sqrt[3]{n-1}+(\sqrt[3]{n-1})^{2}, \\
\frac{1}{a_{n}}=\frac{1}{2}(\sqrt[3]{n+1}-\sqrt[3]{n-1}) . \\
\therefore \sum_{i=1}^{300} \frac{1}{a_{2 i}-1}=\frac{1}{2}[(\sqrt[3]{2}-0)+(\sqrt[3]{4}-\sqrt[3]{2}) \\
\quad+\cdots+(\sqrt[3]{1000}-\sqrt[3]{998... | 5 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. Given that $a, b, c, d$ are all real numbers, and $a+b+c+d=4, a^{2}+b^{2}+c^{2}$ $+d^{2}=\frac{16}{3}$. Then the maximum value of $a$ is $\qquad$ . | 4. 2 .
Construct the function
$$
y=3 x^{2}-2(b+c+d) x+\left(b^{2}+c^{2}+d^{2}\right) \text {. }
$$
Since $y(x-b)^{2}+(x-c)^{2}+(x-d)^{2} \geqslant 0$, and the graph
is a parabola opening upwards, we have
$$
\Delta=4(b+c+d)^{2}-12\left(b^{2}+c^{2}+d^{2}\right) \leqslant 0,
$$
which simplifies to $(4-a)^{2}-3\left(\fr... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$. Then $m^{5}+n^{5}=$ $\qquad$ .
(From the riverbed Jiangsu Province Junior High School Mathematics Competition) | Solution: Construct a quadratic equation $x^{2}-x-1=0$. From the given conditions, we know that $m$ and $n$ are the two roots of the equation $x^{2}-x-1=0$, so $m+n=1, m n=-1$. Then
$$
\begin{array}{l}
m^{2}+n^{2}=(m+n)^{2}-2 m n=3 . \\
m^{4}+n^{4}=\left(m^{2}+n^{2}\right)^{2}-2 m^{2} n^{2}=7 \text {. } \\
\therefore m... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a, b, c$ are non-zero real numbers, and $a+b+c$ $=0$. Then the value of $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ is . $\qquad$ | 2. -3 | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 30 points) There are several weights of 9 grams and 13 grams each. To weigh an object of 3 grams on a balance, what is the minimum number of such weights needed? Prove your conclusion.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the... | On the three pans of a balance, assuming that when the balance is level, 9 grams are used, and the number is an integer. Similarly, assuming that 13 grams of weights are used $|y|$ times. Therefore, when the balance is level and measures a 3-gram object, there should be
$$
9 x+13 y=3 \text {. }
$$
The problem becomes ... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. The integer part of $\frac{1}{3-\sqrt{7}}$ is $a$, and the fractional part is $b$. Then $a^{2}+(1+\sqrt{7}) a b=$ $\qquad$ . | 3.10 . From $\frac{1}{3-\sqrt{7}}=\frac{3+\sqrt{7}}{2}$, we know that $2<\frac{3+\sqrt{7}}{2}<3$, thus $a=2, b=\frac{1}{3-\sqrt{7}}-2=\frac{\sqrt{7}-1}{2}$. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In flood control and rescue operations, a depression near the river dike has experienced a pipe burst, with $x$ cubic meters of river water already rushing in, and water continues to flow in at a rate of $y$ cubic meters per minute. Now, a water extraction and plugging project needs to be carried out. If 1 water pum... | 5.4
Let at least $n$ water pumps be required. According to the problem, we have
$$
\left\{\begin{array}{l}
x+30 y=30 z, \\
x+10 y=20 z, \\
x+5 y=n \cdot 5 z .
\end{array}\right.
$$
From equations (1) and (2), we can solve for $x=15 z, y=0.5 z$.
Substituting into equation (3), we get
$15 z+0.5 z \leqslant 5 n z$, henc... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (Total 20 points) On the 6 lines of the three edges of the tetrahedron $V-A B C$, what is the maximum number of pairs of lines that are perpendicular to each other? (Each pair consists of two lines)
保留源文本的换行和格式，翻译结果如下：
Four. (Total 20 points) On the 6 lines of the three edges of the tetrahedron $V-A B C$, what ... | Four, prove in three steps that there are at most 6 pairs of mutually perpendicular lines.
(1) 6 pairs can be achieved.
When $V A \perp$ plane $A B C$ and $A B \perp A C$, by the property of line perpendicular to a plane, we have $V A \perp A B, V A \perp B C, V A \perp C A$.
By the theorem of three perpendiculars, we ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. Let $m$ be a real number not less than -1, such that the equation $x^{2}+2(m-2) x+m^{2}-3 m+3=0$ has two distinct real roots $x_{1}$ and $x_{2}$.
(1) If $x_{1}^{2}+x_{2}^{2}=6$. Find the value of $m$;
(2) Find the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$. | $$
\begin{aligned}
\Delta & =4(m-2)^{2}-4\left(m^{2}-3 m+3\right) \\
& =-4 m+4>0 .
\end{aligned}
$$
Then $m<1$.
Combining the given conditions, we have $-1 \leqslant m<1$.
$$
\begin{array}{l}
\text { (1) } \because x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
=4(m-2)^{2}-2\left(m^{2}-3 m+3\right) ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (16 points) An engineering
team contracted two projects, Project A and Project B, with the workload of Project A being twice that of Project B. In the first half of the month, all workers worked on Project A, and in the second half of the month, the workers were divided into two equal groups, one group continued... | Three, suppose this construction team has $x$ people, and the monthly work volume per person is 1. Then the work volume for site A is $\frac{\bar{x}}{2}+\frac{1}{2}$ $\frac{x}{2}$, and the work volume for site B is $\frac{x}{2} \times \frac{1}{2}+1$. According to the problem, we get $\frac{x}{2}+\frac{x}{4}=2\left(\fra... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the sum of all solutions to the equation $[3 x+1]=2 x-\frac{1}{2}$.
(1987, All-China Junior High School Mathematics Competition) | Solution: We can consider $[3 x+1]$ as a whole, for example, an integer $t$, thus transforming the given equation into an inequality related to $t$.
Let $[3 x+1]=t$, then $t$ is an integer, and
$$
0 \leqslant(3 x+1)-t<1 \text {. }
$$
Thus, the original equation becomes
$$
t=2 x-\frac{1}{2} \text {, }
$$
which means $... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Find the smallest positive integer $n$ such that the equation $\left[\frac{10^{n}}{x}\right]=1989$ has an integer solution $x$.
(1989, Soviet Union Mathematical Olympiad) | (Hint: Using the inequality $\frac{10^{n}}{x}-1<\left[\frac{10^{n}}{x}\right] \leqslant \frac{10^{n}}{x}$, solve for $x=5026$, at this point $n=7$.) | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
In 1935, the famous mathematicians P. Erdos and G. Szekeres proved a well-known proposition:
For any positive integer $n \geqslant 3$, there exists a number $f(n)$, such that if and only if $m \geqslant f(n)$, any set of $m$ points in the plane, with no three points collinear, contains $n$ points that form a convex $n... | Lemma 1: Let the vertex set of an arbitrary convex hull $K$ be denoted as $S$. If there are $m$ points inside the convex hull forming a convex $m$-gon, and this convex $m$-gon has a side $A_{2} A_{3}$, with its adjacent sides $A_{1} A_{2}$ and $A_{3} A_{4}$ (where $A_{1}, A_{2}, A_{3}, A_{4}$ are all vertices), and if ... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false |
4. A magician has one hundred cards, each with a number from 1 to 100. He places these one hundred cards into three boxes, one red, one white, and one blue. Each box must contain at least one card.
A participant selects two of the three boxes and then picks one card from each of the selected boxes, announcing the sum ... | Solution: There are 12 different methods. Consider the integers from 1 to 100. For simplicity, define the color of the box in which integer $i$ is placed as the color of that integer. Use $\mathrm{r}$ to represent red, $\mathrm{w}$ to represent white, and $\mathrm{b}$ to represent blue.
(1) There exists some $i$ such t... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 䒴 $\frac{4 x}{x^{2}-4}=\frac{a}{x+2}+\frac{b}{x-2}$, then the value of $a^{2}+$ $b^{2}$ is $\qquad$
(1996, Hope Cup National Mathematics Invitational Competition) | Solution: From the problem, we know that the given equation is not an equation in $x$, but an identity in $x$. From
$$
\frac{a}{x+2}+\frac{b}{x-2}=\frac{(a+b) x-2(a-b)}{x^{2}-4}
$$
we get the identity
$$
\begin{array}{l}
\frac{4 x}{x^{2}-4}=\frac{(a+b) x-2(a-b)}{x^{2}-4}, \\
4 x=(a+b) x-2(a-b) .
\end{array}
$$
Accord... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Suppose $1995 x^{3}=1996 y^{3}=1997 z^{3}$, $x y z>0$, and $\sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}}=$ $\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997}$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$
(1996, National Junior High School Mathematics League) | Given that there are three independent equations (equations) and exactly three letters, it should be possible to solve for $x, y, z$, but in practice, it is quite difficult. Inspired by the "chain equality" type problem from the previous question, we can introduce the letter $k$ - try.
Assume $1995 x^{3}=1996 y^{3}=19... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 Given that $x, y, z$ are 3 non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $s=2x+y-z$, then what is the sum of the maximum and minimum values of $s$?
(1996, Tianjin Junior High School Mathematics Competition) | The given equations and the expression for $s$ can be viewed as a system of equations in $x$, $y$, and $z$:
$$
\left\{\begin{array}{l}
3 x+2 y+z=5, \\
x+y-z=2, \\
2 x+y-z=s .
\end{array}\right.
$$
Solving this system, we get $\left\{\begin{array}{l}x=s-2, \\ y=5-\frac{4 s}{3}, \\ z=-\frac{s}{3}+1 .\end{array}\right.$
... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 There are two rivers, A and B, each with a flow rate of $300 \mathrm{~m}^{3} / \mathrm{s}$, which converge at a certain point and continuously mix. Their sediment contents are $2 \mathrm{~kg} / \mathrm{m}^{3}$ and $0.2 \mathrm{~kg} / \mathrm{m}^{3}$, respectively. Assume that from the convergence point, there... | Solution: Let the sand content of two water flows be $a \, \text{kg} / \text{m}^{3}$ and $b \, \text{kg} / \text{m}^{3}$, and the water volumes flowing through in a unit time be $p \, \text{m}^{3}$ and $q \, \text{m}^{3}$, respectively. Then the sand content after mixing is
$$
\frac{a p + b q}{p + q} \, \text{kg} / \te... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given 19 drawers arranged in a row, place $n$ identical balls into them such that the number of balls in each drawer does not exceed the number of balls in the drawer to its left. Let the number of such arrangements be denoted by $F_{m, n}$.
(1) Find $F_{1, n}$;
(2) If $F_{m, 0}=1$, prove:
$$
F_{m, n}=\left\{\begin{arr... | Solution: (1) When $m=1$, there is obviously only one way, i.e., $F_{1, n}=1$.
(2) When $m>n \geqslant 1$, for any method that meets the conditions, there will be no balls in all the drawers to the right of the $n$-th drawer, so $F_{m, n}=F_{n, n}$.
When $1$ 0, if we reduce the number of balls in each drawer by 1, we ... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-$ $a)$, and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
(1999, National Junior High School Mathematics Competition) | Solution: According to the characteristics of the required expression, the given equation can be regarded as a quadratic equation in $a$ or $b+c$. Therefore, there are two methods of solution.
Solution One: Transform the given equation into a quadratic equation in $a$
$$
a^{2}-(b+c) a+\frac{1}{4}(b+c)^{2}=0 .
$$
Its ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given positive integers $k, m, n$, satisfying $1 \leqslant k \leqslant m \leqslant n$. Try to find
$$
\sum_{i=0}^{n}(-1)^{i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}
$$
and write down the derivation process.
(Xu Yichao, provided) | II. The answer to this question is 0. Below, we will prove this combinatorial identity by constructing polynomials and using interpolation and difference methods.
Method 1: Construct the polynomial
$$
\begin{aligned}
f(x) & =\sum_{i=0}^{n} a_{i} x(x+1) \cdots(x+i-1)(x+i+1) \\
& \cdots(x+n)-(x-m-1) \cdots(x-m-n) .
\end{... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Given real numbers $x, y, z$ satisfy $x+y=5$, $z^{2}=x y+y-9$. Then, $x+2 y+3 z=$ $\qquad$
(1995, Zu Chongzhi Cup Junior High School Mathematics Invitational Competition) | Solution: The problem provides two equations with three unknowns. Generally speaking, when the number of unknowns exceeds the number of equations, it is impossible to solve for each unknown individually. However, under the condition that $x$, $y$, and $z$ are all real numbers, a solution may still be possible. In fact,... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a<b<c<d$. If variables $x, y, z, t$ are a permutation of $a, b, c, d$, then the expression
$$
\begin{array}{l}
n(x, y, z, t)=(x-y)^{2}+(y-z)^{2} \\
\quad+(z-t)^{2}+(t-x)^{2}
\end{array}
$$
can take different values. | $=、 1.3$.
If we add two more terms $(x-z)^{2}$ and $(y-t)^{2}$ to $n(x, y, z, t)$, then $n(x, y, z, t)+(x-z)^{2}+(y-t)^{2}$ becomes a fully symmetric expression in terms of $x, y, z, t$. Therefore, the different values of $n(x, y, z, t)$ depend only on the different values of $(x-z)^{2}+(y-t)^{2}=\left(x^{2}+y^{2}+z^{2... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the positive direction of the $y$-axis at point $B$. Find the number of isosceles right triangles inscribed in the ellipse with point $B$ as the right-angle vertex.
---
Please note that the translation pres... | Let the right-angled isosceles triangle inscribed in the ellipse be $\triangle A B C$, and suppose the equation of $A B$ is
\[
\left\{\begin{array}{l}
x=t \cos \alpha, \\
y=b+t \sin \alpha .
\end{array}\right.
\]
(t is a parameter)
Substituting (1) into $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, we get
\[
\left(b^{2} \cos ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 50 points) On a straight ruler of length $36 \mathrm{~cm}$, mark $n$ graduations so that the ruler can measure any integer $\mathrm{cm}$ length in the range $[1,36]$ in one go. Find the minimum value of $n$.
| Three, if the ruler is marked with 7 (or fewer than 7) graduations, we can prove that it is impossible to measure any integer length in the range [1, 6] cm in a single measurement.
In fact, 7 graduations, including the two end lines of the ruler, total 9 lines, which have 36 different combinations. Therefore, 7 gradua... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given that the bases of two congruent regular triangular pyramids are glued together, exactly forming a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$
(1996, National High School League) | Analysis: There are two difficulties in this problem: one is to determine which of $AC$ and $CD$ is 2 in length; the other is to determine which of the line segments $AC$, $CD$, and $AB$ represents the distance between the farthest two vertices.
Solution: As shown in Figure 6, construct $CE \perp AD$, and connect $EF$... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Select $k$ edges and face diagonals from a cube such that any two line segments are skew lines. What is the maximum value of $k$? | (提示:考察线段 $A C 、 B C_{1} 、 D_{1} B_{1} 、 A_{1} D$, 它们所在的直线两两都是异面直线. 若存在 5 条或 5 条以上满足条件的线段, 则它们的端点相异, 且不少于 10 个, 这与正方体只有 8 个端点矛盾, 故 $k$ 的最大值是 4.)
(Translation: Consider the line segments $A C, B C_{1}, D_{1} B_{1}, A_{1} D$, the lines on which they lie are pairwise skew lines. If there are 5 or more line segments that s... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.5. Let $M$ be a finite set of numbers. It is known that from any 3 elements of it, two numbers can be found whose sum belongs to $M$. How many elements can $M$ have at most? | 10.5.7.
An example of a set of numbers consisting of 7 elements is: $\{-3,-2, -1,0,1,2,3\}$.
We will prove that for $m \geqslant 8$, any set of numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ does not have the required property. Without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>\cdots>a_{m}$ and $a_... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the two real roots of $x^{2}-p x+q=0$ be $\alpha, \beta$, and the quadratic equation with $\alpha^{2}, \beta^{2}$ as roots is still $x^{2}-p x+q=0$. Then the number of pairs $(p, q)$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 0 | 5. (B).
From the problem,
$\left\{\begin{array}{l}\alpha+\beta=p, \\ \alpha \beta=q\end{array}\right.$ and $\left\{\begin{array}{l}\alpha^{2}+\beta^{2}=p, \\ \alpha^{2} \beta^{2}=q .\end{array}\right.$
Thus, $q^{2}=q$. Therefore, $q=0$ or $q=1$.
And we have $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$, whi... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Real numbers $a, b, c$ are all non-zero, and $a+b+c=$
0. Then
$$
=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)
$$ | \begin{aligned} \text { II.1. } & -3 . \\ \text { Original expression }= & a\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)+b\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right) \\ & +c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3 \\ = & \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(a+b+c)-3=-3 .\end{aligned} | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 2,
Square $A B C D$ has a side length of $1, E$ is a point on the extension of $C B$, connect $E D$ intersecting $A B$ at $P$, and $P E$ $=\sqrt{3}$. Then the value of $B E-P B$ is $\qquad$ | 4.1.
Let $B E=x, P B=y$, then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=(\sqrt{3})^{2}, \\
\frac{y}{1}=\frac{x}{x+1} .
\end{array}\right.
$$
From (2), we have $x-y=x y$.
From (1) and (3), we have
$$
\begin{array}{l}
(x-y)^{2}+2(x-y)-3=0, \\
(x-y+3)(x-y-1)=0 .
\end{array}
$$
Clearly, $x>y, x-y+3>0$.
Therefore, $x-y-1=0,... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $a_{n}=\log _{n}(n+1)$, let $\sum_{n=2}^{1023} \frac{1}{\log _{100} a_{n}}$ $=\frac{q}{p}$, where $p, q$ are integers, and $(p, q)=1$. Then $p+q=(\quad)$.
(A) 3
(B) 1023
(C) 2000
(D) 2001 | -1 (A).
$$
\begin{array}{l}
\sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\sum_{n=2}^{1023} \log _{100} a_{n} \\
=\log _{100}\left(a_{2} a_{3} \cdots a_{1023}\right) . \\
\text{Since } a_{n}=\log _{n}(n+1)=\frac{\lg (n+1)}{\lg n}, \\
\text{therefore } a_{2} a_{3} \cdots a_{1023}=\frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 1 Given that the polynomial $3 x^{3}+a x^{2}+b x+1$ can be divided by $x^{2}+1$, and the quotient is $3 x+1$. Then, the value of $(-a)^{i}$ is $\qquad$
（Dinghe Hesheng Junior High School Math Competition） | Solution: According to the polynomial identity, we have
$$
3 x^{3}+a x^{2}+b x+1=\left(x^{2}+1\right)(3 x+1) \text {. }
$$
Taking $x=1$ gives $a+b+4=8$.
Taking $x=-1$ gives $a-b-2=-4$.
Solving these, we get $a=1, b=3$.
$$
\therefore(-a)^{b}=(-1)^{3}=-1 \text {. }
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Third question: There are $n$ people, and it is known that any two of them make at most one phone call. Any $n-2$ of them have the same total number of phone calls, which is $3^{k}$ times, where $k$ is a natural number. Find all possible values of $n$.
---
The translation maintains the original text's format and line... | Solution 1: Clearly, $n \geqslant 5$. Let the $n$ people be $n$ points $A_{1}, A_{2}, \cdots, A_{n}$. If $A_{i}$ and $A_{j}$ make a phone call, then connect $A_{i} A_{j}$. Therefore, there must be line segments among these $n$ points. Without loss of generality, assume there is a line segment between $A_{1} A_{2}$:
If... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy $a+b=2$. Then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$
(1990, National High School Mathematics Competition) | Solution: $\because a, b>0$,
$$
\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1 \text {. }
$$
Thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}}$ $\geqslant 1$.
When $a=b=1$, the above expression $=1$, hence the minimum value is 1. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 When $s$ and $t$ take all real numbers, then the minimum value that $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ can achieve is $\qquad$
(1989, National High School Mathematics Competition) | Solution: As shown in Figure 1, the distance squared between any point on the line
$$
\left\{\begin{array}{l}
x=s+5, \\
y=s
\end{array}\right.
$$
and any point on the elliptical arc
$$
\left\{\begin{array}{l}
x=3|\cos t|, \\
y=2|\sin t|
\end{array}\right.
$$
is what we are looking for.
Indeed, the shortest distance ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The ground floor of a hotel has 5 fewer rooms than the second floor. A tour group has 48 people. If all are arranged to stay on the ground floor, each room can accommodate 4 people, but there are not enough rooms; if each room accommodates 5 people, some rooms are not fully occupied. If all are arranged to stay on t... | 5. (B).
Let the number of guest rooms on the ground floor be $x$, then the number of rooms on the second floor is $x+5$. According to the problem, we have the system of inequalities
$$
\left\{\begin{array}{l}
\frac{48}{5}<x<\frac{48}{4}, \\
\frac{48}{4}<x+5<\frac{48}{3} .
\end{array}\right.
$$
which simplifies to $\l... | 10 | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false |
1. Cut a wire of length $143 \mathrm{~cm}$ into $n$ small segments $(n \geqslant 3)$, with each segment no less than $1 \mathrm{~cm}$. If no three segments can form a triangle, the maximum value of $n$ is | 2.1.10.
Each segment should be as small as possible, and the sum of any two segments
should not exceed the third segment. A $143 \mathrm{~cm}$ wire should be divided into 10
such segments:
$1 \mathrm{~cm}, 1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}, 8 \mathrm{~cm}$,
$13 \mathrm{~cm}, 21 \mathrm{~cm}... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, and $x_{n+1}=$ $\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$, then $x_{1999}-x_{601}=$ $\qquad$ . | 3.0 .
From $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$ we get $x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$.
Let $x_{n}=\tan \alpha_{n}$, then
$$
x_{n+1}=\tan \alpha_{n+1}=\tan \left(\alpha_{n}+\frac{\pi}{6}\right) \text {. }
$$
Therefore, $x_{n+6}=x_{n}$,
which means the sequence $\left... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. If the surface area and volume of a circular cone are divided into upper and lower parts by a plane parallel to the base in the ratio $k$, then the minimum value of $c$ that makes $k c>1$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the tran... | 4.7.
As shown in Figure 3, let $C O_{1}=r_{1}, A O=r_{2}$, the surface area of the smaller cone is $S_{1}$. The lateral surface area of the frustum is $S_{2}$, then
$$
\frac{S_{2}+S_{\text {base }}}{S_{1}}=\frac{1}{k},
$$
which means
$$
\frac{S_{1}}{S_{\text {lateral }}+S_{\text {base }}}=\frac{k}{k+1} \text {. }
$$
... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
6.12 friends have a weekly dinner together, each week they are divided into three groups, each group 4 people, and different groups sit at different tables. If it is required that any two of these friends sit at the same table at least once, then at least how many weeks are needed. | 6.5.
First, for any individual, sitting with 3 different people each week, it would take at least 4 weeks.
Second, there are $C_{12}^{2}=66$ pairs among 12 people. Each table has $\mathrm{C}_{4}^{2}=6$ pairs, so in the first week, $3 \times 6=18$ pairs get to know each other.
Since 4 people sit at 3 tables, after th... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. Does there exist a prime number that remains prime when 16 and 20 are added to it? If so, can the number of such primes be determined? | Solution: Testing with prime numbers $2, 3, 5, 7, 11, 13 \cdots$, we find that 3 meets the requirement. Below, we use the elimination method to prove that apart from 3, no other prime number satisfies the requirement.
Divide the natural numbers into three categories: $3n, 3n+1, 3n+2$ $(n \in \mathbb{N})$,
$\because 2n... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $a_{1}=1, a_{2}=\frac{5}{2}, a_{n+1}=$ $\frac{a_{n}+b_{n}}{2}, b_{n+1}=\frac{2 a_{n} b_{n}}{a_{n}+b_{n}}$. Prove:
$$
\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .
$$ | Proof: From the characteristic of the simultaneous recurrence relations, we know that $\left\{a_{n}\right\}$ is an arithmetic mean sequence, and $\left\{b_{n}\right\}$ is a harmonic mean sequence. By considering the inequality of means, we can insert the geometric mean between $\left\{a_{n}\right\}$ and $\left\{b_{n}\r... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
As shown in Figure $3, D$ and $E$ are points on side $BC$ of $\triangle ABC$, and $F$ is a point on the extension of $BC$. $\angle DAE = \angle CAF$.
(1) Determine the positional relationship between the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$, and prove your conclusion.
(2) If the radiu... | Three, (1) Two circles are externally tangent.
Draw the tangent line $l$ of $\odot A B D$, then $\angle 1=\angle B$.
$$
\begin{array}{l}
\because \angle 3=\angle B+\angle C, \\
\therefore \angle 3=\angle 1+\angle C . \\
\because \angle 1+\angle 2=\angle 3=\angle 1+\angle C, \\
\therefore \angle 2=\angle C .
\end{array}... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $m$ and $n$ are integers, the equation
$$
x^{2}+(n-2) \sqrt{n-1} x+m+18=0
$$
has two distinct real roots, and the equation
$$
x^{2}-(n-6) \sqrt{n-1} x+m-37=0
$$
has two equal real roots. Find the minimum value of $n$, and explain the reasoning. | 1. $\left\{\begin{array}{l}n \geqslant 1, \\ (n-2)^{2}(n-1)-4(m+18)>0, \\ (n-6)^{2}(n-1)-4(m-37)=0 .\end{array}\right.$
(2) - (3), and rearrange to get
$$
(n-4)(n-1)>27.5 \text {, }
$$
then $n \geqslant 8$.
When $n=8$, $m=44$, which satisfies the given conditions. Therefore, the minimum value of $x$ is 8. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $f(x)=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x$ is an even function, $g(x)=2^{x}+\frac{a+b}{2^{x}}$ is an odd function, where $a 、 b \in \mathbf{C}$. Then $a+a^{2}+a^{3}+\cdots+a^{2000}+b+b^{2}$ $+b^{3}+\cdots+b^{2000}=$ | 3. -2 .
From the known $f(-x)=f(x), g(-x)=-g(x)$ we get
$$
\begin{array}{l}
\log _{\frac{1}{3}}\left(3^{-x}+1\right)-\frac{1}{2} a b x \\
=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x . \\
2^{-x}+\frac{a+b}{2^{-x}}=-\left(2^{x}+\frac{a+b}{2^{x}}\right) .
\end{array}
$$
Simplifying, from equation (1) we g... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in Figure 1, if
the three sides of $\triangle A B C$ are
$n+x, n+2 x, n$
$+3 x$, and the height $A D$ from $B C$ is $n$, where $n$
is a positive integer, and $0<x \leqslant$
1. Then the number of triangles that satisfy the above conditions is $\qquad$. | 6.12 .
Let $a=n+x, b=n+2 x, c=n+3 x, s=\frac{1}{2}(a+b+c)$. By Heron's formula and the general area formula, we have
$$
\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2} n(n+2 x) .
$$
Simplifying, we get $12 x=n$.
And $0<x=\frac{n}{12} \leqslant 1$,
so $0<n \leqslant 12$,
which means $n$ has exactly 12 possibilities (correspondin... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $x, y \in \mathbf{R}$. If $2 x, 1, y-1$ form an arithmetic sequence, and $y+3,|x+1|+|x-1|, \cos (\arccos x)$ form a geometric sequence, then the value of $(x+1)(y+1)$ is $\qquad$ . | 2.4.
Given that $2 x, 1, y-1$ form an arithmetic sequence, we have
$$
y=3-2 x \text{. }
$$
From $\cos (\arccos x)=x$ and $-1 \leqslant x \leqslant 1$, it follows that
$$
|x+1|+|x-1|=2 \text{. }
$$
Thus, $y+3, 2, x$ form a geometric sequence, which gives
$$
x(y+3)=4 \text{. }
$$
Substituting (1) into (2), we get $2 ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a, b$ be two positive numbers, and $a>b$. Points $P, Q$ are on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the line connecting point $A(-a, 0)$ and $Q$ is parallel to the line $O P$, and intersects the $y$-axis at point $R$, where $O$ is the origin, then $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=$ $\q... | 5.2.
Let $A Q:\left\{\begin{array}{l}x=-a+t \cos \theta \\ y=t \sin \theta\end{array}\right.$ ( $t$ is a parameter ).
Substitute (1) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$t=\frac{2 a b^{2} \cos \theta}{b^{2} \cos ^{2} \theta + a^{2} \sin ^{2} \theta}$.
Then $|A Q|=\frac{2 a b^{2}|\cos \theta|}{b^{2... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. Given the equation about $x$
Figure 3 $\left(a^{2}-1\right)\left(\frac{x}{x-1}\right)^{2}-(2 a+7)\left(\frac{x}{x-1}\right)+1=0$ has real roots.
(1) Find the range of values for $a$;
(2) If the two real roots of the original equation are $x_{1}$ and $x_{2}$, and $\frac{x_{1}}{x_{1}-1} + \frac{x_{2}}{x_{2}-1}=\frac... | 15.1) Let $\frac{x}{x-1}=t$, then $t \neq 1$. The original equation can be transformed into $\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0$.
When $a^{2}-1=0$, i.e., $a= \pm 1$, the equation becomes
$-9 t+1=0$ or $-5 t+1=0$,
which means $\frac{x}{x-1}=\frac{1}{9}$ or $\frac{x}{x-1}=\frac{1}{5}$.
Thus, $x=-\frac{1}{8}$ or $x=... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (16 points) (1) In a $4 \times 4$ grid paper, some small squares are painted red, and then 2 rows and 2 columns are crossed out. If no matter how they are crossed out, at least one red small square remains uncrossed, how many small squares must be painted red at least? Prove your conclusion.
(2) If the “$4 \time... | Three, (1) At least 7 cells need to be colored.
If the number of colored cells $\leqslant 4$, then by appropriately crossing out 2 rows and 2 columns, all the colored cells can be crossed out.
If the number of colored cells is 5, then at least one row has 2 cells colored. By crossing out this row, the remaining colore... | 5 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 3 There is a type of sports competition with $M$ events, and athletes $A$, $B$, and $C$ participate. In each event, the first, second, and third places receive $p_{1}$, $p_{2}$, and $p_{3}$ points respectively, where $p_{1}$, $p_{2}$, $p_{3} \in \mathbf{Z}^{+}$, and $p_{1}>p_{2}>p_{3}$. In the end, $A$ scores 2... | Solution: Consider the total score of three people, we have
$$
\begin{array}{l}
M\left(p_{1}+p_{2}+p_{3}\right)=22+9+9=40 . \\
\because p_{1}, p_{2}, p_{3} \in \mathbf{Z}^{+}, \text { and } p_{1}>p_{2}>p_{3}, \\
\therefore p_{1}+p_{2}+p_{3} \geqslant 3+2+1=6 .
\end{array}
$$
Thus, $6 M \leqslant 40, M \leqslant 6$.
Si... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example $1\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}=$ $\qquad$
(1999, National Junior High School Mathematics League Wuhan Selection Competition) | Let $3^{999}=a, 5^{099}=b, 7^{999}=c$, then
$$
\begin{array}{l}
\text { Original expression }=\frac{c}{a} \sqrt{\frac{a^{2}+(a b)^{2}}{c^{2}+(b c)^{2}}} \\
=\frac{c}{a} \sqrt{\frac{a^{2}\left(1+b^{2}\right)}{c^{2}\left(1+b^{2}\right)}} . \\
=\frac{c}{a} \cdot \frac{a}{c}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given $a^{2}+a+1=0$. Then, $a^{1992}+$ $a^{322}+a^{2}=$ $\qquad$
(Harbin 15th Junior High School Mathematics Competition) | Given the condition $a \neq 1$, from $(a-1)(a^2 + a + 1) = 0$, we get $a^3 - 1 = 0$, hence $a^3 = 1$.
Substituting 1 for $a^3$ and 0 for $a^2 + a + 1$, we get
$$
\begin{aligned}
\text { Original expression } & =\left(a^{3}\right)^{664}+\left(a^{3}\right)^{107} a+a^{2} \\
& =1^{664}+1^{107} a+a^{2} \\
& =1+a+a^{2}=0 .
\... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. In a $3 \times 3$ square grid, fill in the numbers as shown in the table below. The operation on the table is as follows: each operation involves adding a number to two adjacent numbers in the grid (adjacent means two small squares that share a common edge).
\begin{tabular}{|l|l|l|}
\hline 0 & 3 & 2 \\
\hline 6 & 7 ... | (Tip: (1) After 5 operations, all cells can be 0.
(2) As shown in the table below, the invariant $S=a-$ $b+c-d+e-f+g-h+k$.
\begin{tabular}{|l|l|l|}
\hline$a$ & $b$ & $c$ \\
\hline$d$ & $e$ & $f$ \\
\hline$g$ & $h$ & $k$ \\
\hline
\end{tabular}
The initial state is
$$
S=0-3+2-6+7-0+4-9+5=0,
$$
The target state is
$$
S... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
3. The solution to the equation $\sqrt{13-\sqrt{13+x}}$ $=x$ is $\qquad$ | 3.3.
Let $\sqrt{13+x}=y$ then
$$
\left\{\begin{array}{l}
\sqrt{13-y}=x, \\
\sqrt{13+x}=y .
\end{array}\right.
$$
(2) $)^{2}-(1)^{2}$ gives $(x+y)(x-y+1)=0$.
From the original equation, we know $x>0$, and from (2), $y>0$, so $x+y \neq 0$. Thus, $x-y+1=0$, which means $y=x+1$. Therefore, it is easy to get $x=3$ (discar... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) A chemical plant, starting from January this year, if it does not improve its production environment and continues to produce as it is, will earn 700,000 yuan per month. At the same time, it will receive penalties from the environmental protection department, with the first month's penalty being 30,0... | Three, let the cumulative income for $n$ months without modifying the equipment, under the original conditions, be $a(n)$, and let the cumulative income for $n$ months after modifying the equipment be $b(n)$. From the given conditions, we have $a(n)=70 n$, and we can assume
$$
b(n)=a n^{2}+b n+c, \quad n \leqslant 5.
$... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x, y, z \in \mathbf{R}, x y+y z+z x=-1$. Then the minimum value of $x^{2}+5 y^{2}+8 z^{2}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.4 .
$$
\begin{array}{l}
\text { Given }(x+2 y+2 z)^{2}+(y-2 z)^{2} \geqslant 0 , \\
x^{2}+5 y^{2}+8 z^{2} \geqslant-4(x y+y z+z x)=4 ,
\end{array}
$$
we have $x=\frac{3}{2} , y=-\frac{1}{2} , z=-\frac{1}{4}$ ,
or $x=-\frac{3}{2}, y=\frac{1}{2}, z=\frac{1}{4}$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $S_{n}$ be the sum of the elements of all 3-element subsets of the set $A=\left\{1, \frac{1}{2}, \cdots, \frac{1}{2^{n-1}}\right\}$. Then $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=$ $\qquad$ . | 6.1 .
For any element in set $A$, it appears in a subset containing 3 elements $\mathrm{C}_{n-1}^{2}$ times, then
$$
\begin{array}{l}
S_{n}=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^{n-1}}\right) \mathrm{C}_{n-1}^{2} . \\
\text { Therefore, } \lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=\lim _{n \rightarrow \infty} \f... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (20 points) The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=$ $b_{0}=1, a_{n}=a_{n-1}+2 b_{n-1}, b_{n}=a_{n-1}+b_{n-1}$, $(n=1,2 \ldots)$. Find the value of $a_{2001}^{2}-2 b_{2001}^{2}$. | $$
\begin{array}{l}
a_{0}^{2}-2 b_{0}^{2}=-1, \\
a_{1}^{2}-2 b_{1}^{2}=1, \\
\ldots \ldots
\end{array}
$$
Conjecture $a_{n}^{2}-2 b_{n}^{2}=(-1)^{n+1}(n=0,1 \ldots)$.
We will prove this by mathematical induction.
Assume the proposition holds for $n=k$, that is,
$$
a_{k}^{2}-2 b_{k}^{2}=(-1)^{k+1} \text {. }
$$
Then f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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