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Example 10 When $a$ takes all real values from 0 to 5, the number of integer $b$ that satisfies $3 b=a(3 a-8)$ is $\qquad$
(1997, National Junior High School Competition) | Analysis: From $3 b=a(3 a-8)$, we have $b=a^{2}-$ $\frac{8}{3} a$. This is a quadratic function, and its graph is a parabola. When $a$ takes all real numbers from 0 to 5, finding the number of integer $b$ is equivalent to finding the number of integers between the maximum and minimum values of $b$.
Solution: First, dr... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. For a real number $a>1$, written as a reduced fraction $a=\frac{q}{p}$, $(p, q)=1$. The number of values of $a$ that satisfy $p q=30!$ is $\qquad$.
| 5.512.
30! has 10 prime factors, $\left(30!=2^{26} \times 3^{14} \times 5^{7} \times 7^{7} \times 11^{2} \times 13^{2} \times 17 \times 19 \times 23 \times 29\right),(p, q)=$ 1. Each prime factor $p_{i}^{\circ}$ is either all in the numerator or all in the denominator, which gives $2^{10}$ cases in total, with those gr... | 512 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. In $\triangle A B C$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given $a^{2}+b^{2}-7 c^{2}=0$, then $\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=$ | 6.3 .
$$
\begin{array}{l}
\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}=\frac{\cos C \sin A \sin B}{\sin (A+B) \sin C} \\
=\cos C \cdot \frac{\sin A \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. On the front and back of four cards, 0 and 1, 0 and 2, 3 and 4, 5 and 6 are written respectively. By placing any three of them side by side to form a three-digit number, a total of $\qquad$ different three-digit numbers can be obtained. | 6.124.
When the hundreds digit is 1 (or 2): $P_{3}^{2} \times 2 \times 2 \times 2=48$.
When the hundreds digit is 3 (or $4,5,6$), and the tens or units digit uses 0 pieces, there are $(05,06,50,60,00)$ 5 repetitions, so
$$
\left(P_{3}^{2} \times 2 \times 2-5\right) \times 4=76 .
$$
Therefore, the total number of diff... | 124 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{c}
\text { II. (50 points) }(1) \text { For } 0 \leqslant x \leqslant 1, \text { find the range of the function } \\
h(x)=(\sqrt{1+x}+\sqrt{1-x}+2) . \\
\left(\sqrt{1-x^{2}}+1\right)
\end{array}
$$
(2) Prove: For $0 \leqslant x \leqslant 1$, there exists a positive number $\beta$ such that the inequal... | $$
\begin{array}{l}
=(1) 00$, the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{a}}{\beta}(x \in[0,1])$ does not hold. Conversely, i.e., $-\frac{2 x^{2}}{h(x)} \leqslant-\frac{x^{0}}{\beta}$, which means $x^{2 \cdots} \geqslant \frac{h(x)}{2 \beta}$ holds.
Since $2-\alpha>0$, let $x \rightarrow 0$, we get
$$
0... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
For a constant $p \in N$, if the indeterminate equation $x^{2}+y^{2}=$ $p(x y-1)$ has positive integer solutions, prove that $p=5$ must hold. | Proof: Let $x_{0}, y_{0}$ be positive integer solutions of the equation. If $x_{0}=y_{0}$, substituting gives $p=(p-2) x_{0}^{2}$.
$$
\therefore x_{0}^{2}=\frac{p}{p-2}=1+\frac{2}{p-2} \in \mathbb{N} \text { : }
$$
Then $x_{0}^{2}=2$ or 3, which contradicts $x_{0} \in \mathbb{N}$.
Assume $x_{0}>y_{0} \geqslant 2$. Con... | 5 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 15 Find the minimum value of $\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Construct right triangles $\triangle P A C$ and $\triangle P B I$ as shown in Figure 1, such that
$$
\begin{array}{l}
A C=1, B D=2, P C \\
=x, C I=4, \text { and }
\end{array}
$$
$P C$ and $P D$ lie on line $l$. Then the problem of finding the minimum value is converted to "finding a point $I$ on line $l$ suc... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 16 The greatest integer not exceeding $(\sqrt{7}+\sqrt{5})^{6}$ is $\qquad$ . | Solution: Let $a=\sqrt{7}+\sqrt{5}, b=\sqrt{7}-\sqrt{5}$. Then
$$
\begin{array}{l}
a+b=2 \sqrt{7}, a b=2 . \\
a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b) \\
=44 \sqrt{7}, \\
a^{6}+b^{6}=\left(a^{3}+b^{3}\right)^{2}-2(a b)^{3}=13536 . \\
\because 0<b<1, \\
\therefore 0<b^{6}<1 .
\end{array}
$$
Therefore, the integer part of $a^{6... | 13535 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Define a positive integer $n$ to be a "tail" of a factorial if there exists a positive integer $m$ such that the decimal representation of $m$! ends with exactly $n$ zeros. How many positive integers less than 1992 are not tails of a factorial? | Solution: Clearly, the number of trailing zeros in $m!$ = the number of times 10 is a factor in $m!$ = the number of times 5 is a factor in $m!$.
Thus, we have $f(m)=\sum_{k=1}^{+\infty}\left[\frac{m}{5^{k}}\right]$.
From 1 CुS $1=\sum_{k=1}^{\infty}\left[\frac{m}{5^{k}}\right]7964$.
It is also easy to see that when $5... | 396 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 If the fractional parts of $9+\sqrt{13}$ and $9-\sqrt{13}$ are $a$ and $b$ respectively, then $a b-4 a+3 b-2=$ $\qquad$ | Solution: $\because 3<\sqrt{13}<4$,
$\therefore 9+\sqrt{13}$ has an integer part of 12, and a decimal part
$$
\begin{array}{l}
a=\sqrt{13}-3 . \\
\because-4<-\sqrt{13}<-3,
\end{array}
$$
i.e., $0<4-\sqrt{13}<1$,
$\therefore 9-\sqrt{13}$ has an integer part of 5, and a decimal part $b$ $=4-\sqrt{13}$.
Thus, $a b-4 a+3... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 We notice that $6!=8 \cdot 9 \cdot 10$. Try to find the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers. | Solution: According to the requirements of the problem, write $n!$ as
$$
n!=n(n-1) \cdot \cdots \cdot 5 \cdot 24 \text {. }
$$
Therefore, $n+1 \leqslant 24, n \leqslant 23$. Hence, the maximum value of $n$ is 23. At this point, we have
$$
23!=24 \times 23 \times \cdots \times 5 .
$$
The right-hand side of the above e... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Write a given rational number as a reduced fraction, and calculate the product of the resulting numerator and denominator. How many rational numbers between 0 and 1, when processed this way, yield a product of 20!? | 5. Solution: There are 8 prime numbers within 20: $2,3,5,7,11$. $13,17,19$. Each prime number either appears in the numerator or the denominator, but not in both. Therefore, we can form $2^{\times}$ irreducible fractions $\frac{p}{q}$ such that $pq=20!$. Since in $\frac{p}{q}$ and $\frac{q}{p}$, only one lies in the in... | 128 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Simplify $\frac{2 b-a-c}{(a-b)(b-c)}$
$$
+\frac{2 c-a-b}{(b-c)(c-a)}+\frac{2 a-b-c}{(c-a)(a-b)} .
$$ | $\begin{array}{l}\text { Solution: Original expression }=\frac{(b-c)-(a-b)}{(a-b)(b-c)} \\ \quad+\frac{(c-a)-(b-c)}{(b-c)(c-a)}+\frac{(a-b)-(c-a)}{(c-a)(a-b)} \\ =\frac{1}{a-b}-\frac{1}{b-c}+\frac{1}{b-c}-\frac{1}{c-a} \\ \quad+\frac{1}{c-a}-\frac{1}{a-b}=0 .\end{array}$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given that $\alpha$ is a root of the equation $x^{2}-x-1=0$. Try to find the value of $\alpha^{18}+323 \alpha^{-6}$.
Translating the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given: $\because \alpha$ is a root of $x^{2}-x-1=0$,
$$
\therefore \alpha^{2}-\alpha-1=0 \text {. }
$$
Thus, since $\alpha \neq 0$, we have $\alpha-\alpha^{-1}=1$.
Therefore, $\alpha^{2}+\alpha^{-2}=\left(\alpha-\alpha^{-1}\right)^{2}+2=3$,
$$
\begin{array}{l}
\alpha^{4}+\alpha^{-4}=7, \alpha^{6}+\alpha^{-6}=18, \\
\a... | 5796 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 If $x=\sqrt{19-8 \sqrt{3}}$. Then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$
Preserve the original text's line breaks and format, output the translation result directly. | Given: From the above, $x=\sqrt{(4-\sqrt{3})^{2}}=4-\sqrt{3}$, thus $x-4=-\sqrt{3}$.
Squaring and simplifying the above equation yields
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore, the denominator $=\left(x^{2}-8 x+13\right)+2=2$.
By long division, we get
the numerator $=\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { 2. Compare the sizes: } \frac{1}{2}(\sqrt{1998}-\sqrt{2000}) \\
\sqrt{1998}-\sqrt{1999} .
\end{array}
$$
3. Given that $\alpha$ is a root of the equation $x^{2}-3 x-1=0$. Then $\alpha^{6}+120 \alpha^{-2}=$ $\qquad$ . | Answer: >Answer:1309 | 1309 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. $n(\geqslant 5)$ football teams participate in a round-robin tournament. Each pair of teams plays one match, with the winning team getting 3 points, the losing team getting 0 points, and each team getting 1 point in the case of a draw. The team that finishes in 3rd to last place scores fewer points than all the tea... | Solution: Let the $n$ teams be $A_{1}, A_{2}, \cdots, A_{n-3}, B, C_{1}$, $C_{2}$.
Since the $A$ group teams score more but have fewer wins, and the $C$ group teams score less but have more wins, each team in the $A$ group must have at least 8 draws, team $B$ must have at least 1 win, 3 losses, and 4 draws, and each t... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a, b, c, d$ be integers, and $a<2b, b<3c, c<4d$. Given $d<100$, then the maximum possible value of $a$ is ( ).
(A) 2367
(B) 2375
(C) 2391
(D) 2399 | $-.1 .(\mathrm{A})$.
Then, if $d=99$, we have $c<4 \times 99=395$. Taking $c=395$, we get $b<3 \times 395=1185$. Taking $b=1184$, we get $a<2368$. Therefore, the greatest integer for $a$ is 2367. | 2367 | Inequalities | MCQ | Yes | Yes | cn_contest | false |
4. For a $4 n+2$-sided polygon $A_{1} A_{2} \cdots A_{4 n-2}$ (where $n$ is a natural number), each interior angle is an integer multiple of $30^{\circ}$, and $\angle A_{1}=\angle A_{2}=\angle A_{3}=90^{\circ}$, then the possible values of $n$ are $\qquad$ | 4. 1 .
From the problem, we get
$$
\begin{array}{l}
90^{\circ} \times 3+(4 n+2-3) \times 30^{\circ} \mathrm{k} \\
=(4 n+2-2) \times 180^{\circ},
\end{array}
$$
where $k \geqslant 4 n-1$.
Simplifying, we get $(4 n-1) k=24 n-9$,
which means $24 n-9 \geqslant(4 n-1)^{2}$.
Thus, $8 n^{2}-16 n+5 \leqslant 0$.
Clearly, $0... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. $a_{1}, a_{2}, \cdots, a_{6}$ are six distinct natural numbers whose sum is 23. Then, $a_{1} a_{2}+a_{2} a_{3}+\cdots+$ $a_{5} a_{6}+a_{6} a_{1}$ has the minimum value of ( ).
(A) 62
(B) 64
(C) 65
(D) 67 | 5. (B).
For 6 numbers whose sum is 21, arranged as shown in Figure 2, i.e., $a_{1}=1, a_{2}=6$, $\cdots, a_{6}=5$, the required "sum value" $a_{1} a_{2} +\cdots+a_{6} a_{1}$ is the smallest. At this time, the largest number 6 is surrounded by 1,2; the second largest number 5 is surrounded by 1,3; $\cdots$. Any adjustm... | 64 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. The sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers is 117. For all such $m$ and $n$, the maximum value of $3m + 2n$ is $\qquad$ . | 3.37.
Let the $m$ positive even numbers be $a_{i}$, then
$$
\sum_{i=1}^{m} a_{i} \geqslant 2+4+\cdots+2 m=m(m+1) \text {. }
$$
Let the $n$ positive odd numbers be $b_{j}$, then
$$
\sum_{j=1}^{n} b_{j} \geqslant 1+3+\cdots+(2 n-1)=n^{2} .
$$
Thus, the problem reduces to finding the maximum value of $y=3 m+2 n$ under ... | 37 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. The solution set of the inequality $\frac{1}{x-1}+\frac{2}{x-2} \geqslant \frac{3}{2}$, is the union of some non-overlapping intervals with a total length of $\qquad$. | 5.2.
For $g(x)=(x-1)(x-2)>($ or $\text { (or } \leqslant \text { ) } 0 \text {. }$
The graph of $y=f(x)$ is a parabola opening upwards, and the graph of $y=$ $g(x)$ is a parabola opening upwards.
For $f(x):$ when $x=1$, $y0$.
Therefore, the graph intersects the
$x$-axis at two points
$$
\begin{array}{l}
\left(x_{1}, 0... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}\right.$, $\left.B_{2}, \cdots, B_{k}\right\}$ is a family of non-empty subsets of $A$, when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Find the maximum value of $k$. | Solution: There are 175 subsets of one, two, and three elements, which clearly meet the requirements, so $k \geqslant 175$.
Assume $k>175$, then among $\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$, there is at least one subset with more than 3 elements. Without loss of generality, let $B_{i}$ have more than 3 elements,... | 175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $1998^{n} \mid 1999$!, then the maximum value of the positive integer $n$ is | $=1.55$.
$$
\because 1998=2 \times 3^{3} \times 37,3^{3}<37 .
$$
$\therefore 1999$ ! contains the factor of the form $37^{m}$, where the maximum value of $m$ is the maximum value of $n$.
$$
\therefore n \leqslant \sum_{i=1}^{\infty}\left[\frac{1999}{37^{i}}\right]=55 \text {. }
$$
Therefore, the maximum value of $n$ i... | 55 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 The first 100 natural numbers are arranged in a certain order, then the sum of every three consecutive numbers is calculated, resulting in 98 sums, of which the maximum number of sums that can be odd is how many?
(21st Russian Mathematical Olympiad) | Solution: First, it is impossible for all 98 sums to be odd; otherwise, the arrangement of these 100 natural numbers could only be one of the following:
(1) odd odd odd odd....$;$
(2) odd even even odd even even
(3) even odd even even odd even
(4) even even odd even even odd
These four cases contradict the sequence of ... | 97 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the equation $x+8 y+8 z=n(n \in \mathbf{N})$ has 666 sets of positive integer solutions $(x, y, z)$. Then, the maximum value of $n$ is . $\qquad$ | 4.304.
When $m>1, m \in \mathbf{N}$,
$y+z=m$ has $m-1$ sets of positive integer solutions.
The original equation also has $m-1$ sets of positive integer solutions.
$$
\begin{array}{l}
\because 1+2+\cdots+(m-1)=666, \\
\therefore \frac{1}{2} m(m-1)=666 .
\end{array}
$$
Solving gives $m=37$ or -36 (discard).
$$
\theref... | 304 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that for every real number $x$ and $y$, the function $f(x)$ satisfies $f(x)+f(y)=f(x+y)+x y$. If $f(1)=m$, then the number of positive integer pairs $(m, n)$ that satisfy $f(n)=1998$ is $\qquad$. | 6.16.
Let $y=1$, we get
$$
\begin{array}{l}
f(x)+f(1)=f(x+1)+x, \\
\therefore f(x+1)-f(x)=f(1)-x, \\
\therefore f(x)-f(x-1)=f(1)-(x-1), \\
\quad f(x-1)-f(x-2)=f(1)-(x-2), \\
\quad \cdots \cdots . \\
\quad f(2)-f(1)=f(1)-1 .
\end{array}
$$
Adding the above $x-1$ equations, we get
$$
\begin{array}{l}
f(x)-f(1)=(x-1) f(... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, for what real number $x$ does $y=x^{2}-x+1+$ $\sqrt{2(x+3)^{2}+2\left(x^{2}-5\right)^{2}}$ have a minimum value? What is the minimum value? | As shown in Figure 5, let \( P\left(x, x^{2}\right) \) be a point on the parabola \( y=x^{2} \). The distance from \( P \) to the line \( y=x-1 \) is
\[
\begin{aligned}
|P Q| & =\frac{\left|x^{2}-x+1\right|}{\sqrt{(-1)^{2}+1^{2}}} \\
& =\frac{\sqrt{2}}{2} \left| x^{2}-x+1 \right| \\
& =\frac{\sqrt{2}}{2}\left(x^{2}-x+1... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. Find the last four digits of $2^{1999}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\varphi(625)=\varphi\left(5^{4}\right)=5^{4} \times\left(1-\frac{1}{5}\right)=500 \text {. }
$$
Since $(2,625)=1$, by Euler's theorem, we get $2^{500} \equiv 1(\bmod 625)$.
$$
\begin{array}{l}
\therefore 2^{2000} \equiv 1(\bmod 625) . \\
\therefore \text { Let } 2^{2000}=625 m+1(m \in \mathbf{N}) .
\end{array}
$$
... | 4688 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
74. Let the set $A=\{1,2,3, \cdots, 1997\}$, for any 999-element subset $X$ of $A$, if there exist $x, y \in X$, such that $x<y$ and $x \mid y$, then $X$ is called a good set. Find the largest natural number $a(a \in A)$, such that any 999-element subset containing $a$ is a good set. | Solution: We prove that $\max a=665$.
First, we prove that $a \leqslant 665$. Clearly, the 999-element subset $X_{0}$ $=\{999,1000,1001, \cdots, 1997\}$ does not contain any $x, y \in$ $X_{0}$ such that $x1997$, i.e., larger than the maximum element of $X_{0}$, which holds. Thus, $a$ cannot be any of the numbers $999,1... | 665 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most could have bought the book that was purchased by the fewest people?
(1993, China Mathematical Olympiad) | Solution: Let's assume that person A bought three books, and since A has at least one book in common with each of the other 9 people, among A's three books, the book purchased by the most people is bought by no less than 4 people.
If the book purchased by the most people is bought by 4 people, then all three books bou... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $S=\{1,2,3,4\}$. An $n$-term sequence: $q_{1}, q_{2}, \cdots, q_{n}$ has the following property: For any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted by $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997, Shan... | Solution: First, each number in $S$ appears at least twice in the sequence $q_{1}, q_{2}$, $\cdots, q_{n}$, otherwise, since there are 3 binary subsets containing a certain number, but in the sequence, the number of adjacent pairs containing this number is at most 2, therefore, $n \geqslant 8$.
Moreover, the 8-term se... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number, the number in the $i$-th row and $j$-th column is denoted as $x_{ij}$, as shown in Table 1. Then, the numbers in each column of Table 1 are rearranged in non-increasing order as $x_{1j}^{\prime} \geqslant x_{2j}^{\pri... | Solution: First, consider a row $x_{r 1}, x_{r 2}$, $\cdots, x_{r 25}$ in Table 1, which must appear in the first few items of Table 2, because $100 \times 24 = 2400$, while $97 \times 25 = 2425$. Therefore, a row in Table 1 must appear in the first 97 items of Table 2.
Thus, when $i \geqslant 97$, $x_{i j}^{\prime} \... | 97 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let $S=\{1,2,3, \cdots, 280\}$. Find the smallest natural number $n$ such that every subset of $S$ with $n$ elements contains 5 pairwise coprime numbers.
(32nd IMO) | Solution: Let $A_{1}=\{S$ where the numbers are divisible by 2 $\}, A_{2}=$ $\{S$ where the numbers are divisible by 3 $\}, A_{3}=\{S$ where the numbers are divisible by 5 $\}, A_{4}=\{S$ where the numbers are divisible by 7 $\}$, and denote $A=$ $A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$. It is easy to see that $|A|=216... | 217 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For $n \in \mathbf{N}$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=17$. If there exists a unique $n$ such that $S_{n}$ is also an integer, find the value of $n$. | Solution: Based on the structure of $\sqrt{(2 k-1)^{2}+a_{k}^{2}}$, we construct the complex number $z_{k}=(2 k-1)+\sigma_{k} i, k=1,2, \cdots, n$.
Then $\sqrt{(2 k-1)^{2}+a_{k}^{2}}=\left|(2 k-i)+a_{k} i\right|$.
Thus $\sum_{k=1}^{n} \vee \sqrt{(2 k-1)^{2}+a_{k}^{2}}$
$=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} i\right|$
$=\... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let the complex numbers $z_{1}$ and $z_{2}$ satisfy
$$
\left|z_{1}\right|=\left|z_{1}+z_{2}\right|=3,\left|z_{1}-z_{2}\right|=3 \sqrt{3} \text {. }
$$
Find the value of $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|$.
(1991, National High School Mathematics... | Solution: From the given, we have
$$
\begin{aligned}
9 & =\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2} \\
& =\left(z_{1}+z_{2}\right)\left(\overline{z_{1}+z_{2}}\right) \\
& =\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right), \\
27 & =\left|z_{1}-z_{2}\right|^{2}=\lef... | 4000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Six, let a large cube of $4 \times 4 \times 4$ be composed of 64 unit cubes. Select 16 of these unit cubes to be painted red, such that in the large cube, each $1 \times 1 \times 4$ small rectangular prism composed of 4 unit cubes contains exactly 1 red cube. How many different ways are there to select the 16 red cubes... | Solution: Choose a $4 \times 4$ side of the $4 \times 4 \times 4$ large cube as the base plane. Divide the large cube into four layers parallel to the base plane (each layer being $4 \times 4 \times 1$ in size), and number these layers as $1, 2, 3, 4$. Project each red unit cube onto the base plane and write the layer ... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Real numbers $a, b$ satisfy $\sqrt{a^{2}}-2 a+1+$ $\sqrt{36-12 a+a^{2}}=10-|b+3|-|b-2|$. Then the maximum value of $a^{2}+b^{2}$ is $\qquad$ | 3. 45 | 45 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (Full marks 15 points) For the rectangle $\triangle B C(1)$, $A B=20$ cm, $B C=10$ cm. If points $M, N$ are taken on $A C$ and $A B$ (as shown in Figure 2), to make the value of $B M+M N$ the smallest, find this minimum value. | Three, as shown in Figure 1, construct the symmetric point $B'$ of $B$ with respect to $AC$, and connect $AB'$. Then the symmetric point of $N$ with respect to $AC$ is the point $N'$ on $AB$. At this time, the minimum value of $B$ and $M$ to $V$ is equal to the minimum value of $B \rightarrow M \rightarrow N'$, which i... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 The elements of set $A$ are all integers, the smallest of which is 1, and the largest is 100. Except for 1, each element is equal to the sum of two numbers (which can be the same) in set $A$. Find the minimum number of elements in set $A$.
| Solution: Construct a set with as many elements as possible to satisfy the conditions, such as $\{1,2,3,5,10,20,25,50, 9\}$.
Extend $\left\{1,2, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, 100\right\}$ to also satisfy the conditions, then $x_{1} \leqslant 4, x_{2} \leqslant 8, x_{3} \leqslant 16, x_{4} \leqslant 32, x_{5}$ $\l... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
II. (Full marks 15 points) $f(n)$ is a strictly increasing function defined on $\mathbf{N}$ and taking integer values (if for any $x_{1}, x_{2} \in A$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right)<f\left(x_{2}\right)$, then $f(x)$ is called a strictly increasing function on $A$). When $m, n$ are coprime, $f(m n)=f(... | II. H problem, $j(n)$ is defined on $\mathbf{N}$ with integer values and is a strictly increasing function.
$\because f(19)=f(1 \times 19)=f(1) \cdot f(19)$, but $f(19)=$ $19 \neq 0$,
$$
\therefore f(1)=1 \text {. }
$$
At this point, by $f(1)<f(2)<\cdots<f(19), f(1)=1$, $f(19)=19$, and $f(2), f(3), \cdots, f(18)$ are ... | 1862 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 15 points) There are 1998 athletes with numbers from 1 to 1998. Select some of these athletes to participate in the honor guard, but ensure that among the remaining athletes, no one's number is equal to the product of the numbers of any two other athletes. How many athletes can be selected for the hon... | Five, the selectable numbers are $2,3,4, \cdots, 43,44$, these 43 athletes can be chosen as flag bearers to meet the requirements of the problem. The reason is as follows:
After selecting these 43 athletes as flag bearers, the product of any two numbers among the remaining athletes' numbers (excluding number 1) will b... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 For a finite set $A$, there exists a function $f: N \rightarrow$ $A$, with the following property: if $i, j \in N$, and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements must the set $A$ have at least? | Solution: Since the absolute value of the difference between any two numbers among $1,3,6,8$ is a prime number, according to the problem, $f(1)$, $f(3)$, $f(6)$, and $f(8)$ are four distinct elements in $A$, thus $|A| \geqslant 4$.
On the other hand, if we let $A=\{0,1,2,3\}$, and the mapping $f: N \rightarrow A$ is d... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 1, in trapezoid $ABCD$, $DC \parallel AB$, $DC : AB = 1 : 2$, $MN \parallel BD$ and bisects $AC$. If the area of trapezoid $ABCD$ is $\overline{ab}$, and $S_{\triangle AMN} = \overline{ba}$, then $\bar{a} + \bar{b} + \overline{ba} =$ | ニ、1.99.
$\triangle A M N \backsim \triangle A B D$, similarity ratio $=\frac{\frac{1}{2} A C}{\frac{2}{3} A C}=\frac{3}{4}$,
$$
\begin{array}{l}
S_{\triangle A M N}=\frac{9}{16} S_{\triangle A B D}=\frac{9}{16} \times \frac{2}{3} S_{A B C D}=\frac{3}{8} S_{A B C D} \Rightarrow(10 b \\
+a)=\frac{3}{8}(10 a+b) \Rightarro... | 99 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. The solution set of the equation $\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$ is $\qquad$ . | Let $y=\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$, then we have
$$
\left\{\begin{array}{l}
5^{y}=3^{x}+4^{x}, \\
4^{y}=5^{x}-3^{x}
\end{array}\right.
$$
(1) + (2) gives $5^{y}+4^{y}=5^{x}+4^{x}$.
However, the function $f(x)=5^{x}+4^{x}$ is an increasing function, so from (3) we have
$$
f(y)=f(... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that the pure imaginary numbers $x_{1}, x_{2}, \cdots, x_{1999}$ have a modulus of 1. Then the remainder when $x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{1998} x_{1999}+x_{1999} x_{1}$ is divided by 4 is $\qquad$ | 4.1.
Obviously, if $x_{k}$ takes $i$ or $-i$, we have
$$
\begin{array}{l}
\left(i+x_{k}\right)\left(i-x_{k+1}\right) \\
=\left\{\begin{array}{l}
0, \text{ when } x_{k}=-i \text{ or } x_{k+1}=i; \\
-4, \text{ when } x_{k}=i \text{ and } x_{k+1}=-i
\end{array}\right.
\end{array}
$$
Both are multiples of 4, so the sum i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. There are 10 people with distinct heights standing in a row, and each person is either taller or shorter than the person in front of them. The number of arrangements that meet this condition is $\qquad$ (answer with a number).
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a repetition of the instr... | 5.512.
According to the problem, the person standing in the last position must be the tallest or the shortest. Otherwise, there would be people both taller and shorter than the 10th person, which contradicts the given conditions. Therefore, there are two ways to arrange the 10th position.
Let the number of arrangemen... | 512 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Using $1,2, \cdots, n$ to form an $n$-digit number without repeating digits, where 2 cannot be adjacent to 1 or 3, a total of 2400 different $n$-digit numbers are obtained. Then $n=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation ... | 6.7 .
Obviously, $n \geqslant 4$, otherwise 2 must be adjacent to 1 or 3.
When 2 is adjacent to 1, there are $2 \cdot(n-1)$! arrangements, and when 2 is adjacent to 3, there are also $2 \cdot(n-1)$! arrangements. When 2 is adjacent to both 1 and 3, there are $2 \cdot(n-2)$! arrangements. Therefore, the number of arran... | 7 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Four. (Full marks 20 points) As shown in Figure 8, point $O$ represents the sun, $\triangle A B C$ represents a triangular sunshade, and points $A$ and $B$ are two fixed points on the ground in the north-south direction. The sunlight $O C D$ from the due west direction casts the shadow of the sunshade onto the ground, ... | (1) As shown in Figure 17, draw the perpendicular line $O H$ from $O$ to the ground, connect $H D$ to intersect $A B$ at $E$, and connect $C E$. Then $H D$ is the projection of the oblique line $O D$ on the ground, and $\angle C D E = \theta$. Given that $A B$ is in the north-south direction and $C D$ is in the east-we... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 The sum of two positive integers is 1997 less than their product, and one of them is a perfect square. Then the larger number and the smaller (1997, Tianjin Junior High School Mathematics Competition) | Solution: Let the two positive integers be $\alpha, \beta$, and $\alpha>\beta$. Then
$$
\begin{array}{l}
\alpha+\beta+1997=\alpha \beta . \\
\therefore \alpha \beta-\alpha-\beta+1=1998,
\end{array}
$$
which means
$$
\begin{array}{l}
(\alpha-1)(\beta-1) \\
=1998=999 \times 2=333 \times 3 \times 2 .
\end{array}
$$
Sinc... | 663 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given $u+v=96$, and the quadratic equation $x^{2} + u x + v = 0$ has integer roots, then its largest root is $\qquad$
(1996, Anhui Province Partial Areas Junior High School Mathematics League) | Solution: Let the two integer roots of the equation be $x_{1}$ and $x_{2}$, then we have
$$
\begin{array}{l}
x_{1}+x_{2}=-u, x x_{2}=v . \\
\because u+v=96, \\
\therefore x_{1} x_{2}-\left(x_{1}+x_{2}\right)=v+u=96 .
\end{array}
$$
Thus, $x_{1} x_{2}-x_{1}-x_{2}+1=97$,
which means $\left(x_{1}-1\right)\left(x_{2}-1\ri... | 98 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $S$ be a set with 6 elements. In how many ways can two (not necessarily distinct) subsets of $S$ be chosen so that the union of the two subsets is $S$? The order of selection does not matter. For example, the pair of subsets $\{a, c\}, \{b, c, d, e, f\}$ and the pair of subsets $\{b, c, d, e, f\}, \{a, c\... | Solution: Let $S=A \cup B$, and without loss of generality, assume $|A| \leqslant|B|$ (denote the number of elements in set $S$ as $|S|$).
If $|A|=0$, then $A=\varnothing, B=S$, there is only 1 way to choose;
If $|A|=1$, then $|B|=6,5$. At this time, there are $\mathrm{C}_{6}^{1} \mathrm{C}_{6}^{6} +\mathrm{C}_{6}^{1... | 365 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $M=\{1,2,3, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: if $x \in A, 15 x \notin A$, then the maximum number of elements in $A$ is $\qquad$
(1995, National High School Mathematics Competition) | Solution: Construct subset $A$ as follows:
$1995 \div 15=133$, let $A_{1}=\{134,135, \cdots$, 1995\}, then $\left|A_{1}\right|=1862$ (elements);
$133 \div 15=8$ remainder 13, let $A_{2}=\{9,10, \cdots$, 133\}, then $\left|A_{2}\right|=125$ (elements);
Let $A_{3}=\{1,2,3,4,5,6,7,8\}$, then $\left|A_{3}\right|=$ 8 (elem... | 1870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let the set $M=\{1,2,3, \cdots, 1000\}$. For any non-empty subset $X$ of $M$, let $\alpha_{X}$ denote the sum of the largest and smallest numbers in $X$. Then the arithmetic mean of all such $\alpha_{X}$ is $\qquad$
(1991, National High School Mathematics Competition) | Solution: Construct the subset $X^{\prime}=\{1001-x \mid x \in X\}$, then all non-empty subsets can be divided into two categories: $X^{\prime}=X$ and $X^{\prime} \neq X$.
When $X^{\prime}=X$, it must be that $X^{\prime}=X=M$, thus, $\alpha_{X}=1001$.
When $X^{\prime} \neq X$, let $x$ and $y$ be the maximum and minim... | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 In hexagon $A B C D E F$, $\angle A=\angle B$ $=\angle C=\angle D=\angle E=\angle F$, and $A B+B C=$ $11, F A-C D=3$. Then $B C+D E=$ ?
(1994, Beijing Junior High School Mathematics Competition) | Solution: As shown in Figure 7, each angle of the hexagon is equal, all being $120^{\circ}$, so each exterior angle is $60^{\circ}$. Let the lines $A B, C D$, and $E F$ intersect at $P$, $Q$, and $R$, respectively, forming four equilateral triangles, i.e., $\triangle R A F$, $\triangle R P Q$, $\triangle B P C$, and $\... | 14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let the plane region $D$ be represented by $N(D)$, which denotes the number of all integer points (i.e., points on the xoy plane where both coordinates $x$ and $y$ are integers) belonging to $D$. If $A$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=10$, $y=1$ (incl... | Solution: Draw the figure in the Cartesian coordinate system, and it is easy to calculate from the figure
$$
\begin{array}{l}
N(A)=1^{2}+2^{2}+\cdots+10^{2} \\
=\frac{1}{6} \times 10(10+1)(2 \times 10+1)=385, \\
N(B)=\left(101-1^{2}\right)+\left(101-2^{2}\right)+\cdots \\
+\left(101-10^{2}\right) \\
=101 \times 10-\lef... | 1010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. The perimeter of $\triangle A B C$ is $24, M$ is the midpoint of $A B$. $M C=M A=5$. Then the area of $\triangle A B C$ is ( ).
(A) 12
(E) 16
(C) 24
(D) 30 | 2. (C)
$$
\begin{array}{l}
\because M A=M B=M C=5, \\
\therefore \angle A C B=90^{\circ} .
\end{array}
$$
Given the perimeter is 24, then $A C+B C=14, A C^{2}+B C^{2}=$ $10^{2}$
$$
\begin{array}{l}
\begin{aligned}
\therefore 2 A C \cdot B C & =(A C+B C)^{2}-\left(A C^{2}+B C^{2}\right) \\
& =14^{2}-10^{2}=4 \times 24 ... | 24 | Geometry | MCQ | Yes | Yes | cn_contest | false |
1. Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ | $$
\begin{array}{l}
\because(b-c)^{2}=4(a-b)(c-a), \\
b^{2}-2 b c+c^{2}=4 a c-4 b c+4 a b-4 a^{2}, \\
\therefore(b+c)^{2}-4 a(b+c)+4 a^{2}=0 .
\end{array}
$$
Therefore, $[2 a-(b+c)]^{2}=0$, which means $2 a=b+c$.
$$
\therefore \frac{b+c}{a}=2 \text {. }
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a$ and $b$ are integers, and satisfy
$$
\left(\frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}-\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}\right)\left(\frac{1}{a}-\frac{1}{b}\right) \frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}}=\frac{2}{3} \text {. }
$$
Then $a+b=$ $\qquad$ . | 3.3.
$$
\begin{array}{l}
\text { Left side }=\frac{a b}{a-b}=\frac{2}{3}, \\
\therefore(3 b-2)(3 a-2)=4 .
\end{array}
$$
Given that $a \neq b$ and they are integers, hence $3 b-2, 3 a-2$ can only take the values 1, 4 or -1, -4.
(1) Suppose $3 b-2=1, 3 a-2=4$.
Solving gives $b=1, a=2$. Therefore, $a+b=3$.
(2) Suppose $... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) A class participated in an intelligence competition, with three questions: $a$, $b$, and $c$. Each question either scores full marks or 0 points, where question $a$ is worth 20 points, and questions $b$ and $c$ are each worth 25 points. The competition results show that each student got at least one qu... | Let $x_{a}, x_{b}, x_{c}$ respectively represent the number of people who answered questions $a$, $b$, and $c$ correctly, then we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{a}+x_{b}=29, \\
x_{a}+x_{c}=25, \\
x_{b}+x_{c}=20 .
\end{array}\right. \\
\therefore x_{a}+x_{b}+x_{c}=37 .
\end{array}
$$
Solving, we ge... | 42 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. If the 5th term of the expansion of $\left(x \sqrt{x}-\frac{1}{x}\right)^{6}$ is $\frac{15}{2}$, then $\lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-a}\right)=$ | $\begin{array}{l}\text { II, 1.1. } \\ \because T_{5}=C_{6}^{4}(x \sqrt{x})^{2} \cdot\left(-\frac{1}{x}\right)^{4}=\frac{15}{x}, \\ \text { from } \frac{15}{x}=\frac{15}{2} \text {, we get } x^{-1}=\frac{1}{2} . \\ \therefore \lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-n}\right) \\ =\frac{\frac{1}{2}}{1-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. If the functions $f(x)$ and $g(x)$ are defined on $\mathbf{R}$, and
$$
\begin{array}{l}
f(x-y)=f(x) g(y)-g(x) f(y), f(-2) \\
=f(1) \neq 0, \text { then } g(1)+g(-1)=\ldots
\end{array}
$$
(Answer with a number). | 4. -1 .
$$
\begin{array}{l}
\because f(x-y)=f(x) g(y)-g(x) f(y), \\
\therefore f(y-x)=f(y) g(x)-g(y) f(x) \\
\quad=-[f(x) g(y)-g(x) f(y)]
\end{array}
$$
There is $f(x-y)=-f(y-x)$
$$
=-f[-(x-y)] \text {. }
$$
Then $f(-x)=-f(x)$, i.e., $f(x)$ is an odd function.
$$
\text { Hence } \begin{aligned}
f(1) & =f(-2)=f(-1-1) ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Seven, (Full marks 12 points) Given a sequence where each term is 1 or 2, the first term is 1, and there are $2^{k-1}$ twos between the $k$-th 1 and the $(k+1)$-th 1, i.e., $1,2,1,2,2,1,2,2,2,2,1,2,2,2,2,2,2,2,2,1, \cdots$
(1) Find the sum of the first 1998 terms of the sequence, $S_{1998}$;
(2) Does there exist a posi... | For the $k$-th segment, we take the $k$-th 1 and then follow it with $2^{k-1}$ 2's, making the $2^{k-1}+i$-th term the $k$-th segment of the sequence.
Suppose the 1958th term is in the $k$-th segment. Then $k$ is the smallest positive integer satisfying
$$
\dot{\kappa}+\left(1+2+2^{2}+\cdots+2^{k-1}\right)=2^{k}+k-1 \g... | 3985 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{4}+a_{6}+a_{8}+a_{10}+$ $a_{12}=120$, then the value of $2 a_{9}-a_{10}$ is ( ).
(A) 20
(B) 22
(C) 24
(D) 28 | \begin{array}{l}-1 .(\mathrm{C}) \\ \because a_{4}+a_{6}+a_{8}+a_{10}+a_{12} \\ =5 a_{1}+(3+5+7+9+11) d \\ =5\left(a_{1}+7 d\right)=120 \Rightarrow a_{1}+7 d=24, \\ \therefore 2 a_{9}-a_{10}=2\left(a_{1}+8 d\right)-\left(a_{1}+9 d\right) \\ =a_{1}+7 d=24 .\end{array} | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false |
5. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $c=10, \frac{\cos A}{\cos B}=\frac{b}{a}=\frac{4}{3}, P$ is a moving point on the incircle of $\triangle A B C$, and $d$ is the sum of the squares of the distances from $P$ to the vertices $A, B, C$, then $d_{\t... | 5. 160.
From the given information, we have $\angle C=90^{\circ}, a=$ $6, b=8, c=10$, and the inradius $r$ $=2$. Establish a rectangular coordinate system as shown, and let $P(x, y)$. Clearly, $(x-2)^{2}+(y-$ $2)^{2}=4$. Therefore, we have
$$
\begin{aligned}
d^{2} & =x^{2}+(y-8)^{2}+(x \\
& -6)^{2}+y^{2}+x^{2}+y^{2} \... | 160 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2: In a certain year, the total coal production of a coal mine, apart from a certain amount of coal used for civilian, export, and other non-industrial purposes each year, the rest is reserved for industrial use. According to the standard of industrial coal consumption of a certain industrial city in that year,... | Solution: Let the total coal production of the mine for the year be $x$, the annual non-industrial coal quota be $y$, and the industrial coal consumption of each industrial city for the year be $z$. Let $p$ be the number of years the coal can supply only one city. According to the problem, we have the system of equatio... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (Full marks 12 points) At the foot of the mountain is a pond, the scene: a steady flow (i.e., the same amount of water flows into the pond from the river per unit time) continuously flows into the pond. The pond contains a certain depth of water. If one Type A water pump is used, it will take exactly 1 hour to p... | Three, let the amount of water flowing into the pond from the spring every minute be $x$ m$^{3}$, each water pump extracts $y$ m$^{3}$ of water per minute, and the pond originally contains $z$ m$^{3}$ of water. It takes $t$ minutes for three water pumps to drain the pond. According to the problem, we have
$$
\left\{\be... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (Full marks 15 points) The unit digit of an $n$-digit number is 6. Moving the 6 to the front of the number while keeping the other digits in place results in a new $n$-digit number, which is 4 times the original $n$-digit number. Find the smallest positive number that satisfies the above conditions.
---
untran... | From the problem, we know that the first digit must be 1.
Expression:
$$
\begin{array}{r}
1 \cdots E D C B A 6 \\
\hline 6 \cdots E D C B A
\end{array}
$$
We get $A=4, B=8, C=3, D=5, F=1$. $\therefore$ The smallest number is 15384. | 15384 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 15 points) Expand new numbers according to the following rules:
Given two numbers $a$ and $b$, a new number can be expanded according to the rule $c = ab + a + b$. Any two numbers among $a$, $b$, and $c$ can then be used to expand another new number according to the rule, and so on. Each expansion of ... | (1) The first time, you can only get $1 \times 4+4+1=9$.
Since the goal is to get the maximum new number, the $\cdots$ time, take 4 and 9, to get $4 \times 9+4+9=49$.
Similarly, the second time, take 9 and 49, to get
$9 \times 49+9+49=441$.
(2) $\because a=a \cdot a+b=(a+1)(b+1)-1$,
$$
\therefore a+1=(a+1)(b+1).
$$
Ta... | 441 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$2.1999^{2000}$ divided by $10^{10}$, the remainder is $\qquad$ | $$
\begin{array}{l}
2.5996000001 . \\
1999^{2000}=(1-2000)^{2000} \\
=1-2000 \times 2000+\frac{1}{2} \times 2000 \times 1999 \times 2000^{2} \\
\quad-\frac{1}{16} \times 2000 \times 1999 \times 1998 \times 2000^{3}+\cdots \\
\quad+2000^{2000} .
\end{array}
$$
In the expression on the right side of the equation above, ... | 5996000001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given two points $A(0,1), B(6,9)$. If there is an integer point $C$ (Note: A point with both coordinates as integers is called an integer point), such that the area of $\triangle A B C$ is minimized. Then the minimum value of the area of $\triangle A B C$ is $\qquad$ | 5.1.
Since the slope of line $AB$ is $k=\frac{3}{4}$, the line passing through point $C$ and parallel to $AB$ can be expressed as $3 y-4 x=m$. Let the coordinates of point $C$ be $\left(x_{0}, y_{0}\right)$, where $x_{0}, y_{0}$ are integers. Therefore, we have
$$
3 y_{0}-4 x_{0}=m \text {. }
$$
Thus, $m$ is an integ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four, (Full marks 20 points) The sum of \( m \) different positive even numbers divisible by 5 and \( n \) different positive odd numbers divisible by 3 is \( M \). For all such \( m \) and \( n \), the maximum value of \( 5m + 3n \) is 123. What is the maximum value of \( M \)? Please prove your conclusion.
---
The ... | From the problem, we know
$$
\begin{aligned}
M \geqslant & 5(2+4+\cdots+2 m) \\
& +3[1+3+5+\cdots+(2 n-1)] \\
= & 5 m(m+1)+3 n^{2} .
\end{aligned}
$$
That is, \( 5\left(m^{2}+m\right)+3 n^{2} \leqslant M \),
$$
5\left(m+\frac{1}{2}\right)^{2}+3 n^{2} \leqslant M+\frac{5}{4} \text {. }
$$
By the Cauchy-Schwarz inequal... | 1998 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points) Given $0<\alpha_{i}<\frac{\pi}{4}(i=1$, $2,3,4)$, and $\sum_{i=1}^{4} \sin ^{2} \alpha_{i}=1$. Prove: $\sum_{i=1}^{4} \frac{\sin ^{2} \alpha_{i}}{\cos 2 \alpha_{i}} \geqslant$ 2. | Let $a_{i}=\sin ^{2} \alpha_{i}$, then $0<a_{i}<\frac{1}{2}$,
$$
\begin{array}{l}
\cos 2 \alpha_{i}=1-2 \sin ^{2} \alpha_{i}=1-2 a_{i} . \\
\frac{\sin ^{2} \alpha_{i}}{\cos 2 a_{i}}=\frac{a_{i}}{1-2 a_{i}}=2\left(\frac{a_{i}^{2}}{1-2 a_{i}}+\frac{a_{i}}{2}\right) \\
=\frac{2}{1-2 a_{i}}\left(a_{i}-\frac{1-2 a_{i}}{2}\r... | 2 | Inequalities | proof | Yes | Yes | cn_contest | false |
Three, (Full marks 50 points) If the hundreds digit of an $n$-digit natural number $N$ is 9, and the sum of its digits is $M$, where $n>3$, when the value of $\frac{N}{M}$ is the smallest, what is $N$?
---
Please note that the translation retains the original formatting and structure of the text. | $$
\begin{array}{l}
\text { Three, (1) When } n=4 \text {, let } N=\overline{a 9 b c} \text {. Then } \\
\frac{N}{M}=\frac{1000 a+900+10 b+c}{a+9+b+c} \\
=1+\frac{999 a+891+9 b}{a+9+b+c} \\
\geqslant 1+\frac{999 a+891+9 b}{a+9+b+9} \text { (equality holds when } c=9 \text {) } \\
=1+\frac{9(a+b+18)+990 a+729}{a+b+18} \... | 1999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 A store sells goods that cost 10 yuan each at 18 yuan each, and can sell 60 of them per day. After conducting a market survey, the store manager found that if the selling price of the goods (based on 18 yuan each) is increased by 1 yuan, the daily sales volume will decrease by 5; if the selling price of the g... | Solution: Let the selling price of each item be $x$ yuan, and the daily profit be $s$ yuan.
$$
\begin{array}{l}
\text { When } x \geqslant 18 \text {, we have } \\
s=[60-5(x-18)](x-10) \\
=-5(x-20)^{2}+500,
\end{array}
$$
That is, when the selling price of the item is increased to $x=20$ yuan, the daily profit $s$ is ... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $x=\frac{1}{\sqrt{3}+\sqrt{2}}, y=\frac{1}{\sqrt{3}-\sqrt{2}}$. Then, $x^{2}+y^{2}$ is
untranslated part: 轩隹
Note: The term "轩隹" does not have a clear meaning in this context and has been left untranslated. If you can provide more context or clarify the term, I can attempt to translate it accurately. | $=, 7.10$.
Let $x=\sqrt{3}-\sqrt{2}, y=\sqrt{3}+\sqrt{2}$, so,
$$
x^{2}+y^{2}=(\sqrt{3}-\sqrt{2})^{2}+(\sqrt{3}+\sqrt{2})^{2}=10 .
$$ | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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
$$
\left\{\begin{array}{l... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (Full marks 10 points) As shown in the figure, there is a natural number on each face of the cube. It is known that the sum of the two numbers on opposite faces is equal. Given that the numbers opposite to $12$, $9$, and $3$ are $a$, $b$, and $c$ respectively, find the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$. | Final 1. From the problem, we have $13+a=9+b=3+c$, which gives $a-b=-4, b-c=-6, c-a=10$. Therefore,
$$
\begin{array}{l}
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] \\
=\frac{1}{2}(16+36+100)=76 .
\end{array}
$$ | 76 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (Full marks 10 points) The numbers A, B, and C are 312, $270$, and 211, respectively. When these three numbers are divided by a natural number $A$, the remainder of dividing A is twice the remainder of dividing B, and the remainder of dividing B is twice the remainder of dividing C. Find this natural number $A$.... | Three, from the problem, we have
$$
270 \times 2-312=228, \quad 211 \times 2-270=152 \text {. }
$$
Then we have
\begin{tabular}{|c|c|c|}
\hline \multicolumn{2}{|c|}{$2 \mid \quad 228$} & 152 \\
\hline $2 T$ & 114 & 76 \\
\hline 19 & 57 & 38 \\
\hline & 3 & 2 \\
\hline
\end{tabular}
Therefore, 4 could be $2, 4, 19, 38... | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Based on market research analysis, a home appliance manufacturing company has decided to adjust its production plan and is preparing to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours). Moreover, the production of refrigerators should be at least... | 5. Weekly
Producing 30 air conditioners, 270 color TVs, and 60 refrigerators will maximize the output value, with the highest output value being 1050 thousand yuan. | 1050 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, let $x, y$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}$, $\cdots, a_{n}$, and satisfy the condition: when $x>y$, $x-y \geqslant \frac{x}{15}$. Find the maximum number of these natural numbers $n$. | $$
\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 |
Example 5 Let $A B C D E F$ be a hexagon, and a frog starts at vertex $A$. It can jump to one of the two adjacent vertices at each step. If it reaches point $D$ within 5 jumps, it stops jumping; if it does not reach point $D$ within 5 jumps, it stops after 5 jumps. How many different jumping sequences are possible from... | Analysis: Grasping the possible scenarios of the frog's jumping is the key to solving the problem.
Solution: According to the conditions, the frog's jumping methods can only result in two situations:
(1) Jump 3 times to reach point $D$, with 2 ways of jumping.
(2) Stop after 5 jumps (the first 3 jumps do not reach poi... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 There are 20 teams participating in the national football championship finals. To ensure that in any group of 3 teams, at least two teams have played against each other, how many matches must be played at minimum?
(1969, All-Soviet Union Mathematical Olympiad) | Analysis: The only possible competition schemes that meet the conditions are:
(1) No grouping, any two teams play a match.
(2) Divide into two groups, Group A and Group B, containing 1 team and 19 teams respectively, any two teams in Group B play a match.
(3) Divide into two groups, containing $k(2 \leqslant k \leqslan... | 90 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 In a sequence of coin tosses, the number of times a tail is followed by a head (denoted as "tail-head"), a head is followed by a tail (denoted as "head-tail"), a head is followed by a head (denoted as "head-head"), and a tail is followed by a tail (denoted as "tail-tail") can be counted. How many different se... | Analysis: From "反" being 1 more than "正", we know that the sequence starts and ends with "反" and "正" respectively. According to the conditions, it is easy to know that the count of "正" equals $2 \times 2 + 3 + 4 = 11$, and the count of "反" is $5 \times 2 + 3 + 4 = 17$. Since the "反" and "正" at the beginning and end are... | 560 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7 n+55$. Try to find the value of $n$.
(1994-1995, Chongqing and Four Other Cities Mathematics Competition) | Solution: Let $\frac{n^{2}-71}{7 n+55}=k$ (where $k$ is an integer),
then $n^{2}-7 k n-(55 k+71)=0$.
$$
\begin{aligned}
\Delta & =49 k^{2}+4(55 k+71) \\
& =49 k^{2}+220 k+284 .
\end{aligned}
$$
When $\Delta$ is a perfect square, the $n$ in equation (1) is an integer. And
$$
\begin{array}{l}
(7 k+15)^{2}<\Delta=49 k^{2... | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that the real number $x$ satisfies $\sqrt{x^{2}-x^{3}}=x$. $\sqrt{1-x}$. Then the range of values for $x$ is | 2.15.
From $2 x^{2}-6 x+y^{2}=0$ we get
$$
y^{2}=-2 x^{2}+6 x \geqslant 0 \text {. }
$$
Solving this, we get $0 \leqslant x \leqslant 3$.
Let $z=x^{2}+y^{2}+2 x$. Then
$$
\begin{array}{l}
z=x^{2}-2 x^{2}+6 x+2 x \\
=-x^{2}+8 x \\
=-(x-4)^{2}+16 . \\
\because 0 \leqslant x \leqslant 3,
\end{array}
$$
$\therefore$ whe... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in the figure, circle O is inscribed in $\triangle ABC$ touching sides $AC$, $BC$, and $AB$ at points $D$, $E$, and $F$ respectively, with $\angle C=90^{\circ}$. The area of $\triangle ABC$ is 6. Then $AF \cdot BF=$ $\qquad$ | 4.6 .
Let $A C=b, B C=a, A B=c, \odot O$ have a radius of $r$. Then
$$
\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 |
Three. (Full marks 25 points) A natural number that can be expressed as the difference of the squares of two natural numbers is called a "smart number." For example, $16=5^{2}-3^{2}$ is a "smart number." Try to answer the following questions:
(1) Is 1998 a "smart number"? Explain your reasoning.
(2) Arranged in ascendi... | Three, (1) 1998 is not a "smart number". If 1998 were a "smart number", we could set \(1998 = m^2 - n^2\) (where \(m\) and \(n\) are natural numbers), then \((m+n)(m-n) = 2 \times 999\).
Since \(m+n\) and \(m-n\) have the same parity, \((m+n)(m-n)\) is either odd or divisible by 4, hence it cannot be 1998. Therefore, 1... | 2665 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $f(x)=x^{2}+(\lg a+2) x+\lg b$, and $f(-1)=-2$, also $f(x) \geqslant 2 x$ for all $x \in R$. Then $a+b=(\quad)$.
(A) 100
(B) 110
(C) 120
(D) 130 | $-, 1 .(B)$
Since $f(-1)=-2$, then
$$
\begin{array}{l}
1-(\lg a+2)+\lg b=-2 \\
\Rightarrow \lg a=\lg b+1 \Rightarrow a=10 b .
\end{array}
$$
Since for all $x \in \mathbf{R}$, $f(x) \geqslant 2 x$,
$$
x^{2}+(\lg a) x+\lg b \geqslant 0 .
$$
Since the coefficient of $x^{2}$ is $1>0$, then
$$
... | 110 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Let $x$ and $y$ be real numbers, and satisfy
$$
\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 |
79. (1) Prove that 1998 cannot be expressed as the sum of any number of consecutive odd numbers;
(2) If the numbers from 1 to 1997 are connected using “+” and “-” signs, then no matter how many “-” signs are used, the result cannot be 1998;
(3) If the numbers from 1 to 1999 are connected using “+” and “-” signs, then w... | Solution: (1) If it can be expressed, let $n_{0}$ be the first odd number, then
$$
\begin{array}{l}
1998= n_{0}+\left(n_{0}+2\right)+\cdots+\left(n_{0}+2 k\right) \\
=\left(n_{0}+2 k\right)+\left(n_{0}+2 k-2\right)+\cdots \\
+\left(n_{0}+2\right)+n_{0} . \\
\therefore 2 \times 1998=(k+1)\left(2 n_{0}+2 k\right) .
\end... | 586 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 1 Let $x_{1}, x_{2}, \cdots, x_{9}$ all be positive integers, and $x_{1}<x_{2}<\cdots<x_{9}, x_{1}+x_{2}+\cdots+x_{9}=220$. Then, when the value of $x_{1}+x_{2}+\cdots+x_{5}$ is maximized, the minimum value of $x_{9}-x_{1}$ is $\qquad$
$(1992$, National Junior High School Mathematics League) | Solution: From $x_{1}+x_{2}+\cdots+x_{9}=220$,
we know $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}>110$,
or $x_{1}+x_{2}+\cdots+x_{5} \leqslant 110$.
From (1), then $x_{5} \geqslant 25$. Thus, $x_{6} \geqslant 26, x_{7} \geqslant$ $27, x_{8} \geqslant 28, x_{9} \geqslant 29$. We get
$$
\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 5 In hexagon $A B C D E F$, $\angle A=\angle B$ $=\angle C=\angle D=\angle E=\angle F$, and $A B+B C=$ $11, F A-C D=3$. Find $B C+D E$.
(1994, Beijing Junior High School Mathematics Competition) | Solution: As shown in Figure 5, let $F A$ and $C B$ intersect at $P$, and $F E$ and $C D$ intersect at $Q$. It is easy to see that quadrilateral $P C Q F$ is a parallelogram, and $\triangle A P B$ and $\triangle Q E D$ are both equilateral triangles. Therefore,
$$
\begin{aligned}
& F A + A B \\
= & F P = Q C \\
= & D E... | 14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Find the minimum value of the function with real variables $x$ and $y$
$$
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 |
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