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Five, (20 points) For the parabola $y^{2}=2 p x(p>0)$ with focus $F$, does there exist an inscribed isosceles right triangle such that one of its legs passes through $F$? If it exists, how many are there? If not, explain the reason.
---
Translate the above text into English, please retain the original text's line bre... | As shown in Figure 4, let \( A\left(x_{A}, y_{A}\right) \), \( B\left(x_{B}, y_{B}\right) \), \( C\left(x_{C}, y_{C}\right) \), and the inclination angle of \( AC \) be \( \theta \). Clearly, \( \theta \neq 0 \), \( \frac{\pi}{2} \). Suppose \( \theta \in \left(0, \frac{\pi}{2}\right) \). By the focal chord length form... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $a$ is a natural number, there exists a linear polynomial with integer coefficients and $a$ as the leading coefficient, which has two distinct positive roots less than 1. Then, the minimum value of $a$ is $\qquad$ . | 2.5.
Let $f(x)=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, $00$. Then from $f(0)$ and $f(1)$ being positive integers, we get $f(0) f(1) \geqslant 1$, that is, $a^{2} x_{1} x_{2}\left(1-x_{1}\right)\left(1-x_{2}\right) \geqslant 1$.
Also, $x(1-x) \leqslant \frac{1}{4}$, with equality when $x=\frac{1}{2}$.... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (20 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=3, a_{n}=$ $3^{a_{n-1}}(n \geqslant 2)$. Find the last digit of $a_{n}(n \geqslant 2)$. | Let's prove, when $n \geqslant 2$, we have
$$
a_{n}=4 m+3, m \in \mathbf{N} \text {. }
$$
Using mathematical induction:
(i) When $n=2$, $a_{2}=3^{3}=4 \times 6+3$, so equation (1) holds.
(ii) Assume when $n=k(k \geqslant 2)$, equation (1) holds, i.e.,
$$
a_{k}=4 m+3, m \in \mathbf{N} \text {. }
$$
Then $a_{k+1}=3^{a_... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given an integer $n > 3$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}$, $x_{n+1}, x_{n+2}$ satisfy the condition
$$
0 < x_{1} < x_{2} < \cdots < x_{n} < x_{n+1} < x_{n+2}.
$$
Find the minimum value of
$$
\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{\left... | Solution: (I) Let $t_{i}=\frac{x_{i+1}}{x_{i}}(>1), 1 \leqslant i \leqslant n+1$. The expression in the problem can be written as
$$
\frac{\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} i_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i=1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)}... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. If $16^{9 m}=a, 4^{37 n}=\frac{1}{a}$, then $(36 m+74 n-1)^{2000}$ | Ni.1.1.
Since $a=16^{9 m}=\left(2^{4}\right)^{9 m}=2^{36 m}$, and $\frac{1}{a}=4^{37 n}=2^{74 n}$, then $1=a \times \frac{1}{a}=2^{36 m} \cdot 2^{74 n}=2^{36 m+74 n}$,
we have $36 m+74 n=0$.
Therefore, $(36 m+74 n-1)^{2000}=(-1)^{20000}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 2, in
Rt $\triangle A B C$,
$\angle C=90^{\circ}$, point $M$
is the intersection of the three
medians of the triangle. Perpendiculars are drawn from $M$
to $A B$, $B C$, and $A C$,
with the feet of the perpendiculars being $D$, $E$, and $F$, respectively. If $A C=3$, $B C=12$, then the area of $\... | 4.4.
$$
\begin{array}{l}
AB = \sqrt{3^2 + 12^2} = \sqrt{153} = 3 \sqrt{17}, \\
S_{\triangle ABC} = \frac{1}{2} \times 3 \times 12 = 18.
\end{array}
$$
Draw a perpendicular from $C$ to $AB$, with the foot of the perpendicular at $H$, and let it intersect $EF$ at $G$. We have
$$
\begin{array}{l}
S_{\triangle ABC} = \fra... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. A piece of lead wire of length $2 n$ (where $n$ is a natural number and $n \geqslant 4$) is folded into a triangle with integer side lengths. Let $(a, b, c)$ represent a triangle with side lengths $a, b, c$ such that $a \leqslant b \leqslant c$.
(1) For the cases $n=4, 5, 6$, write down all the $(a, b, c)$ that sat... | 15. (1) When $n=4$, the length of the lead wire is 8. Then the only group of $(a, b, c)$ that satisfies the condition is $(2,3,3)$;
When $n=5$, the length of the lead wire is 10. Then the groups of $(a, b, c)$ that satisfy the condition are $(2,4,4),(3,3,4)$;
When $n=6$, the length of the lead wire is 12. Then the gr... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. The square number $y^{2}$ is
the sum of the squares of 11 consecutive integers. Then the smallest value of the natural number $y$ is
$\qquad$ | 3.11.
Let the middle number of these 11 consecutive integers be \( a \). Then
\[
\begin{aligned}
y^{2}= & (a-5)^{2}+(a-4)^{2}+(a-3)^{2}+(a-2)^{2} \\
& +(a-1)^{2}+a^{2}+(a+1)^{2}+(a+2)^{2} \\
& +(a+3)^{2}+(a+4)^{2}+(a+5)^{2} \\
= & 11 a^{2}+2\left(5^{2}+4^{2}+3^{2}+2^{2}+1^{2}\right) \\
= & 11\left(a^{2}+10\right) .
\e... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $a_{n}=6^{n}+8^{n}$. Then $a_{84} \equiv$ $\qquad$ $(\bmod 49)$ | 5.2.
$$
\begin{aligned}
a_{84}= & (7-1)^{84}+(7 \\
& +1)^{84} \\
= & 2\left(\mathrm{C}_{84}^{0} \cdot 7^{84}+\right. \\
& \mathrm{C}_{84}^{2} \cdot 7^{82}+\cdots \\
& \left.+\mathrm{C}_{84}^{82} \cdot 7^{2}+\mathrm{C}_{84}^{84}\right) \\
= & 45 \times M+2 . \\
a_{84}= & 2(\text { modulo } 49) .
\end{aligned}
$$ | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$=6$ integer solutions.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Let the two integer roots of the equation be $x_{1}$ and $x_{2}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-(10 a+b), \\
x_{1} x_{2}=10 b+a .
\end{array}\right.
$$
We have $10 x_{1}+10 x_{2}+x_{1} x_{2}=-99 a(1 \leqslant a \leqslant 9)$.
Thus $\left(x_{1}+10\right)\left(x_{2}+10\right)=100-99 a$.
By the p... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Through the vertex of isosceles $\triangle A B C$, draw a line intersecting the extension of the opposite side at $D$. If the resulting triangles are all isosceles, then $\triangle A B C$ has $\qquad$ such configurations. | 4.5.
Draw a straight line from the base angle to intersect with the extension of the opposite side, there are 5 $\triangle A B C$ that meet the conditions, with base angles of $45^{\circ} 、 72^{\circ} 、 36^{\circ} 、 \frac{180^{\circ}}{7}$ 、 $\frac{2 \times 180^{\circ}}{7}$.
Draw a straight line from the vertex angle t... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. The number of triangles with unequal integer sides and a perimeter less than 13 is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | (3)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 3 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 3 On a street $AB$, person A walks from $A$ to $B$, and person B rides a bike from $B$ to $A$. Person B's speed is 3 times that of person A. At this moment, a public bus departs from the starting station $A$ and heads towards $B$, and a bus is dispatched every $x$ minutes. After some time, person A notices that... | Solution: According to the problem, draw the graph as shown in Figure 4.
$A C, A_{1} C_{1}, A_{2} C_{2}, A_{3} C_{3}$ represent the motion graphs of buses departing every $x$ minutes, $A D$ and $B . V$ are the motion graphs of person A and person B, respectively. $E_{1}, E_{2}$ are the points where a bus catches up wit... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example $8 A$ and $B$ are two fixed points on a plane. Find a point $C$ on the plane such that $\triangle A B C$ forms an isosceles right triangle. There are $\qquad$ such points $C$. | Solution: As shown in Figure 5, the vertices of the two isosceles right triangles with $AB$ as the hypotenuse are $C_{1}$ and $C_{2}$; the vertices of the four isosceles right triangles with $AB$ as the legs are $C_{3}, C_{4}, C_{5}, C_{6}$.
Therefore, there are a total of 6 points $C$ that meet the conditions. | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
14. As shown in Figure 5, in $\square A B C D$, $P_{1}$, $P_{2}, \cdots$, $P_{n-1}$ are the $n$ equal division points on $B D$. Connect $A P_{2}$ and extend it to intersect $B C$ at point $E$, and connect $A P_{n-2}$ and extend it to intersect $C D$ at point $F$.
(1) Prove that $E F \parallel B D$;
(2) Let the area of ... | 14. (1) Since $A D / / B C, A B / / D C$. Therefore, $\left.\triangle P_{n-2} F I\right) \triangle \triangle P_{n-2} A B, \triangle P_{2} B E \subset \triangle P_{2} D . A$.
Thus, we know
$$
\frac{A P_{n-2}}{P_{n+2} F}=\frac{B P_{n}}{P_{n-2} D}=\frac{n-2}{2}, \frac{A P_{2}}{P_{2} E}=\frac{D P_{2}}{P_{2} B}=\frac{n-2}{2... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) Several containers are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each container does not exceed 1 ton. To ensure that these containers can be transported in one go, how many trucks with a carrying capacity of 3 tons are needed at least? | First, note that the weight of each container does not exceed 1 ton, so the weight of containers that each vehicle can carry at one time will not be less than 2 tons; otherwise, another container can be added.
Let $n$ be the number of vehicles, and the weights of the containers they carry be $a_{1}, a_{2}, \cdots, a_{... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 In Figure 1, there are 8 vertices, each with a real number. The real number at each vertex is exactly the average of the numbers at the 3 adjacent vertices (two vertices connected by a line segment are called adjacent vertices). Find
$$
a+b+c+d-(e+f+g+h)
$$ | The following solution has appeared in a journal:
Given
$$
\begin{array}{l}
a=\frac{b+e+d}{3}, \\
b=\frac{a+f+c}{3}, \\
c=\frac{b+g+d}{3}, \\
d=\frac{c+h+a}{3}.
\end{array}
$$
Adding the four equations yields
$$
\begin{array}{l}
a+b+c+d \\
=\frac{1}{3}(2 a+2 b+2 c+2 d+e+f+g+h),
\end{array}
$$
which simplifies to $a+b... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2: When A was B's current age, B was 10 years old; when B was A's current age, A was 25 years old. Who is older, A or B? How many years older? | Solution: Let the age difference between A and B be $k$ years, which is an undetermined constant. When $k>0$, A is older than B; when $k<0$, A is younger than B. Then, A's age $y$ and B's age $x$ have a linear relationship:
$$
y=x+k \text {. }
$$
After designing this dynamic process, the given conditions become 3 "mom... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If $\left(1+x+x^{2}+x^{3}\right)^{5}\left(1-x+x^{2}-\right.$ $\left.x^{3}\right)^{5}=a_{30}+a_{29} x+\cdots+a_{1} x^{29}+a_{0} x^{30}$, find $a_{15}$. | Let $f(x)=\left(1+x+x^{2}+x^{3}\right)^{5}$, then the original expression is
$$
F(x)=f(x) f(-x)
$$
which is an even function. Therefore, we have
$$
\begin{array}{l}
\boldsymbol{F}(x)=\frac{1}{2}[\boldsymbol{F}(x)+\boldsymbol{F}(-x)] \\
=a_{30}+a_{28} x^{2}+\cdots+a_{2} x^{28}+a_{0} x^{30},
\end{array}
$$
where all th... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 There are 5 medicine boxes, every 2 boxes contain one same medicine, each medicine appears in exactly 2 boxes, how many kinds of medicines are there? | Solution: Represent the medicine boxes as 5 points. When a medicine box contains the same medicine, draw a line segment between the corresponding points.
Since every 2 medicine boxes have one kind of the same medicine, a line should be drawn between every two points. Also, because each kind of medicine appears in exac... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a=1999 x+2000, b=1999 x+2001, c$ $=1999 x+2002$. Then the value of the polynomial $a^{2}+b^{2}+c^{2}-a b-b c-$ $c a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 2. (D).
$$
\begin{array}{l}
\because a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] . \\
\text { Also } a-b=-1, b-c=-1, c-a=2, \\
\therefore \text { the original expression }=\frac{1}{2}\left[(-1)^{2}+(-1)^{2}+2^{2}\right]=3 .
\end{array}
$$ | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
11. The number of integers $n$ that satisfy $\left(n^{2}-n-1\right)^{n+2}=1$ is.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task ... | 11.4 .
From $n+2=0, n^{2}-n-1 \neq 0$, we get $n=-2$;
From $n^{2}-n-1=1$, we get $n=-1, n=2$;
From $n^{2}-n-1=-1$ and $n+2$ is even, we get $n=0$.
Therefore, $n=-1,-2,0,2$ for a total of 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, 18 football teams are participating in a single round-robin tournament, meaning each round the 18 teams are divided into 9 groups, with each group's two teams playing one match. In the next round, the teams are regrouped to play, for a total of 17 rounds, ensuring that each team plays one match against each of t... | ```
3. Consider the following competition program:
1.(1.2)(3.4)(5.6)(7.8)(9,18)
(10,11)(12,13)(14,15)(16,17)
2.(1,3)(2.4)(5,7)(6.9)(8,17)
(10,12)(11,13)(14,16)(15.18)
3.(1.4)(2.5)(3.6)(8.9)(7.16)
(10,13)(11,14)(12,15)(17,18)
4.(1,5)(2.7)(3.8)(4,9)(6.15)
(10.14)(11,16)(12.17)(13.18)
5.(1,6)(2,8)(3,9)(4,7... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
16. Given positive integers $x, y$, then $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5}$ has ( ) solutions of the form $(x, 1)$.
(A) 0
(B) 1
(C) 2
(D) More than 2, but finite
(E) Infinite
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result direc... | 16. (C).
$$
\text { Given } \frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5} \Rightarrow 50 y=x^{2}(y+5) \Rightarrow x^{2}=\frac{50 y}{y+5} \text {. }
$$
If $(y, 5)=1$, then $(y, y+5)=1$, so $(y+5) \mid 50=5^{2} \times 2$. Since $y+5>2$, then $(y+5) \mid 5^{2}$, which contradicts $(y+5,5)=1$. Therefore, $5 \mid y$.
Let $y=5 y... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 13 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
(1996, National Junior High School Mathematics League... | Solution: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $x_{1} < 0 < x_{2}$,
then $x_{1} < 0, \\
\therefore b > 2 \sqrt{a c} . \\
\text{Also, } \because |O A| = |x_{1}| > 1$. Therefore, the parabola opens upwards, and when $x = -1$, $y > 0$, so $a(-1)^{2} + b(-1) + c > 0$, which means $b < a + c + ... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $[A]$ denote the greatest integer less than or equal to $A$, and set $A=38+$ $17 \sqrt{5}$. Then $A^{2}-A[A]=$ $\qquad$ . | $\begin{array}{l}\text { II.1.1. } \\ \because A=38+17 \sqrt{5}=(\sqrt{5}+2)^{3} \text {, let } B=(\sqrt{5}-2)^{3} \text {, } \\ \therefore A-B=76 . \\ \text { Also } \because 0<(\sqrt{5}-2)^{3}<1 \text {, } \\ \therefore[A]=76 \text {, then } A-[A]=B . \\ \text { Therefore } A^{2}-A[A]=A(A-[A])=A B=1 .\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the function $y=\frac{a-x}{x-a-1}$, the graph of its inverse function is symmetric about the point $(-1,4)$. Then the value of the real number $a$ is $\qquad$ . | 2.3.-
From the problem, we know that the graph of the function $y=\frac{a-x}{x-a-1}$ is centrally symmetric about the point $(4,-1)$.
$\because y=\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, we have $(y+1)[x-(a+1)]=-1$,
$\therefore$ the graph of the function is a hyperbola with its center at $(a+1,-1)$.
Also, $\because$ t... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Find the largest constant $k$, such that for all real numbers $a, b, c, d$ in $[0,1]$, the inequality
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant k\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
holds. | Solution: First, estimate the upper bound of $k$.
When $a=b=c=d=1$, we have $4 k \leqslant 4+4, k \leqslant 2$.
Next, we prove that for $a, b, c, d \in[0,1]$, it always holds that
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
First, we prove a... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the smallest positive integer $k$, such that for all $a$ satisfying $0 \leqslant a \leqslant 1$ and all positive integers $n$, we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$ | Solution: First, we aim to eliminate the parameter $a$, and then it will be easier to find the minimum value of $k$. Using the arithmetic-geometric mean inequality, we get
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{k+n}=\frac{k}{k+n} .
\... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If $xy=1$, then the minimum value of the algebraic expression $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is $\qquad$ .
(1996, Huanggang City, Hubei Province, Junior High School Mathematics Competition) | $$
\text { Sol: } \begin{aligned}
\because & \frac{1}{x^{4}}+\frac{1}{4 y^{4}}=\left(\frac{1}{x^{2}}\right)^{2}+\left(\frac{1}{2 y^{2}}\right)^{2} \\
& \geqslant 2 \cdot \frac{1}{x^{2}} \cdot \frac{1}{2 y^{2}}=1,
\end{aligned}
$$
$\therefore \frac{1}{x^{4}}+\frac{1}{4 y^{4}}$'s minimum value is 1.
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Person A and Person B go to a discount store to buy goods. It is known that both bought the same number of items, and the unit price of each item is only 8 yuan and 9 yuan. If the total amount spent by both on the goods is 172 yuan, then the number of items with a unit price of 9 yuan is $\qquad$ pieces.
Person A a... | 3.12.
Suppose each person bought $n$ items, among which $x$ items cost 8 yuan each, and $y$ items cost 9 yuan each. Then we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x + y = 2 n , } \\
{ 8 x + 9 y = 172 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=18 n-172, \\
y=172-16 n .
\end{array}\right.\right.... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five, (15 points) 1 and 0 alternate to form the following sequence of numbers:
$$
101,10101,1010101,101010101, \cdots
$$
Please answer, how many prime numbers are there in this sequence? And please prove your conclusion. | Obviously, 101 is a prime number.
Below is the proof that $N=\underbrace{101010 \cdots 01}_{k \uparrow 1}(k \geqslant 3)$ are all composite numbers (with $k-1$ zeros in between).
$$
\begin{aligned}
11 N= & 11 \times \underbrace{10101 \cdots 01}_{k \uparrow 1} \\
& =\underbrace{1111 \cdots 11}_{2 k \uparrow 1}=\underbra... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Under the conditions $x+2 y \leqslant 3, x \geqslant 0, y \geqslant 0$, the maximum value that $2 x+y$ can reach is $\qquad$
(2000, Hope Cup Junior High School Mathematics Competition Second Trial) | Solution: As shown in Figure 1, draw the line $x + 2y = 3$. The set of points satisfying the inequalities $x \geqslant 0, y \geqslant 0, x + 2y \leqslant 3$ is the region $\triangle ABO$ (including the boundaries) enclosed by the line and the $x$ and $y$ axes. To find the maximum value of $s = 2x + y$, we transform $s ... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given that a square has three vertices $A, B, C$ on the parabola $y=x^{2}$. Find the minimum value of the area of such a square.
(1998, Shanghai High School Mathematics Competition) | Solution: As shown in Figure 3, without loss of generality, assume that two of the three vertices are on the right side of the $y$-axis (including the $y$-axis). Let the coordinates of points $A$, $B$, and $C$ be $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively, an... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. When $s$ and $t$ take all real values, the minimum value that can be reached by $(s+5-3|\cos t|)^{2}$ $+(s-2|\sin t|)^{2}$ is $\qquad$
(1989, National High School Mathematics Competition) | (The original expression can be regarded as the square of the distance between any point on the line $\left\{\begin{array}{l}x=s+5, \\ y=s\end{array}\right.$ and any point on the ellipse arc $\left\{\begin{array}{l}x=3|\cos t| \\ y=2|\sin t|\end{array}\right.$. It is known that the square of the distance from the point... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given that $p$, $q$, $\frac{2p-1}{q}$, $\frac{2q-1}{p}$ are all integers, and $p>1$, $q>1$. Try to find the value of $p+q$. | Analysis: The conditions of this problem do not provide specific numbers. Yet, it requires finding the value of $p+q$, which poses a certain difficulty. If we analyze the given conditions one by one, we almost get no useful information. However, if we look at the conditions as a whole, i.e., consider $\frac{2 p-1}{q}$ ... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9.4 In Greek mythology, the "many-headed serpent" god is composed of some heads and necks, with each neck connecting two heads. With each strike of a sword, one can sever all the necks connected to a certain head $A$. However, head $A$ immediately grows new necks connecting to all the heads it was not previously connec... | 9.410 .
We will reformulate the problem using graph theory terminology, with heads as vertices, necks as edges, and a strike that cuts the necks connected to head $A$ as a "reversal" of vertex $A$. It is easy to see that if a vertex $X$ has a degree no greater than 10, then it is sufficient to perform a "reversal" on ... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of intersections of the function $y=x \cdot|x|-\left(4 \cos 30^{\circ}\right) x+2$ with the $x$-axis is $\qquad$ | 3.3 .
$$
y=\left\{\begin{array}{ll}
x^{2}-2 \sqrt{3} x+2, & x>0, \\
2, & x=0 . \\
-x^{2}-2 \sqrt{3} x+2, & x<0
\end{array}\right.
$$
When $x>0$, $y=x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 2 points;
When $x=0$, $y=2$ does not intersect the x-axis;
When $x<0$, $y=-x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 1 p... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always 3 numbers, where the sum of any two of them is still irrational.
| Three, take 4 irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, clearly they do not satisfy the condition, hence $n \geqslant 5$.
Consider 5 irrational numbers $a, b, c, d, e$. View them as 5 points. If the sum of two numbers is a rational number, then connect the corresponding two points with a red lin... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Determine the smallest natural number $k$, such that for any $a \in [0,1]$ and any $n \in \mathbf{N}$ we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$ | Solution: By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{n+k}=\frac{k}{n+k} .
\end{array}
$$
Thus, $a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}$.
Equality holds if and only i... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that the three altitudes of $\triangle A B C$ are $A D=3, B E=4, C F=5$, and the lengths of the three sides of this triangle are all integers. Then the minimum value of the length of the shortest side is ( ).
(A) 10
(B) 12
(C) 14
(D) 16 | 5. (B).
From $S_{\triangle A B C}=\frac{1}{2} B C \cdot A D=\frac{1}{2} C A \cdot B E=\frac{1}{2} A B \cdot C F$,
we get $3 B C=4 C A=5 A B$.
It is clear that $A B$ is the shortest side.
From $B C=\frac{5}{3} A B, C A=\frac{5}{4} A B$ and the lengths of $B C, C A, A B$ are all
integers, we know that $3 \mid A B$ and ... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
II. (25 points) Let $a$, $b$, and $c$ be three distinct real numbers, and $c \neq 1$. It is known that the equations $x^{2} + a x + 1 = 0$ and $x^{2} + b x + c = 0$ have a common root, and the equations $x^{2} + x + a = 0$ and $x^{2} + c x + b = 0$ also have a common root. Find the value of $a + b + c$. | Let the common root of the first two equations be $x_{1}$, then
$$
\begin{array}{l}
x_{1}^{2}+a x_{1}+1=0, \\
x_{1}^{2}+b x_{1}+c=0 .
\end{array}
$$
(2)
(1) - (2) gives $(a-b) x_{1}+(1-c)=0$.
$$
\because a \neq b, \quad \therefore x_{1}=\frac{c-1}{a-b} \text {. }
$$
Similarly, the common root of the last two equations... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. For any natural numbers $m, n$ satisfying $\frac{m}{n}<\sqrt{7}$, the inequality $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\lambda}{n^{2}}$ always holds. Find the maximum value of $\lambda$.
| (Let $G=|(m, n)| m<\sqrt{7} n, m, n \in \mathbf{N}$. $\lambda_{\text {max }}=\min _{(m, n \in 6} 17 n^{2}-m^{2}$, then perform $\bmod 7$ analysis on $7 n^{2}-m^{2}$, obtaining $\lambda_{\text {max }}=3$. ) | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. For the polynomial $\left(\sqrt{x}+\frac{1}{2 \sqrt[4]{x}}\right)^{n}$ expanded in descending powers of $x$, if the coefficients of the first three terms form an arithmetic sequence, then the number of terms in the expansion where the exponent of $x$ is an integer is $\qquad$ . | 8.3.
It is easy to find that the coefficients of the first three terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$. Since these three numbers form an arithmetic sequence, we have $2 \times \frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get $n=8$ and $n=1$ (the latter is discarded).
When $n=8$, $T_{r+1}=\mathrm{... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Given that $f(x)$ is a function defined on
$\mathbf{R}$, $f(1)=1$ and for any $x \in \mathbf{R}$ we have
$$
f(x+5) \geqslant f(x)+5, f(x+1) \leqslant f(x)+1 \text {. }
$$
If $g(x)=f(x)+1-x$, then $g(2002)=$ | 10.1.
$$
\begin{array}{l}
\text { From } g(x)=f(x)+1-x \text { we get } f(x)=g(x)+x-1 . \\
\text { Then } g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5 \text {, } \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 . \\
\text { Therefore, } g(x+5) \geqslant g(x), g(x+1) \leqslant g(x) . \\
\therefore g(x) \leqslant g(x+5) \leqslant g(x+4... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 In the school football championship, it is required that each team must play a match against all the other teams. Each winning team gets 2 points, a draw gives each team 1 point, and a losing team gets 0 points. It is known that one team scored the most points, but it played fewer matches than any other team.... | Explanation: We call the team $A$ with the highest score the winning team.
Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.
From the given conditions, every other team must play at least $n+1$ matches, meaning they score no less than $2(n+1)$ points. Therefo... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}$, $a \neq 0$ ) satisfy the following conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find... | 15. Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about $x=-1$. Therefore, $-\frac{b}{2a}=-1, b=2a$.
From (3), when $x=-1$, $y=0$, i.e., $a-b+c=0$.
From (1), $f(1) \geqslant 1$, and from (2), $f(1) \leqslant 1$,
thus $f(1)=1$, i.e., $a+b+c=1$.
Also, $a-b+c=0$, so $b=\frac{1}{2}, a=\frac{1}{4}, c=\frac{1... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $A B C D$ be a rectangle with an area of 2, and let $P$ be a point on side $C D$. Let $Q$ be the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B \geqslant... | Thus, $\frac{1}{2} P A \cdot P B \sin \angle A P B=1$,
which means $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$.
Equality holds only when $\angle A P B=90^{\circ}$. This indicates that point $P$ lies on the circle with $A B$ as its diameter, and this circle should intersect with $C D$,
Therefore, when $P A \... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
16. Given $a+b+c=0$, and $a, b, c$ are all non-zero. Then simplify $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ to | 16. -3 .
$$
\begin{aligned}
\text { Original expression }= & \left(\frac{a}{a}+\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{b}+\frac{c}{b}\right) \\
& +\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{c}\right)-\frac{a}{a}-\frac{b}{b}-\frac{c}{c} \\
= & \frac{0}{a}+\frac{0}{b}+\frac{0}{c}-3=-3 .
\end{aligned}
$$ | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Question: How many real roots does the equation $x^{2}|x|-5 x|x|+2 x=0$ have (where $|x|$ represents the absolute value of $x$)? | 1.4 .
The original equation simplifies to $x(x|x|-5|x|+2)=0$. After removing the absolute value signs and discussing, we get
$$
x_{1}=0, x_{2}=\frac{5+\sqrt{17}}{2}, x_{3}=\frac{5-\sqrt{17}}{2}, x_{4}=\frac{5-\sqrt{33}}{2}
$$
as the four real roots of the original equation. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $x_{1}=\sqrt[3]{3}, x_{2}=\left(x_{1}\right)^{\sqrt[3]{3}}$, for $n>1$ define $x_{n+1}$ $=\left(x_{n}\right)^{\sqrt[3]{3}}$. Find the smallest positive integer $n$ such that $x_{n}=27$. | 4.7.
It is known that $x_{n}=x_{1}^{x_{1}^{n-1}}(n>1)$, and $27=(\sqrt[3]{3})^{(\sqrt[3]{3})^{6}}$. Therefore, $n-1=6$. Hence, $n=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $\triangle A B C$ be an acute triangle, and construct isosceles triangles $\triangle D A C$, $\triangle E A B$, and $\triangle F B C$ outside $\triangle A B C$ such that $D A = D C$, $E A = E B$, and $F B = F C$, with $\angle A D C = 2 \angle B A C$, $\angle B E A = 2 \angle A B C$, and $\angle C F B = 2 \angle ... | Solution: Since $\angle ABC$ is an acute triangle, then $\angle ADC$, $\angle BEA$, $\angle CFB < \pi$. Therefore,
$$
\begin{array}{l}
\angle DAC = \frac{\pi}{2} - \frac{1}{2} \angle ADC = \frac{\pi}{2} - \angle BAC. \\
\angle BAE = \frac{\pi}{2} - \frac{1}{2} \angle BEA = \frac{\pi}{2} - \angle ABC. \\
\text{Thus, } \... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a_{1}=11^{11}, a_{2}=12^{12}, a_{3}=13^{13}$, and
$$
a_{n}=\left|a_{n-1}-a_{n-2}\right|+\left|a_{n-2}-a_{n-3}\right|, n \geqslant 4 \text {. }
$$
Find $a_{4^{4}}$. | For $n \geqslant 2$, define $s_{n}=\left|a_{n}-a_{n-1}\right|$. Then for $n \geqslant 5, a_{n}=s_{n-1}+s_{n-2}, a_{n-1}=s_{n-2}+s_{n-3}$. Thus, $s_{n}=$ $\left|s_{n-1}-s_{n-3}\right|$. Since $s_{n} \geqslant 0$, if $\max \left\{s_{n}, s_{n+1}, s_{n+2}\right\} \leqslant T$, then for all $m \geqslant n$, we have $s_{m} \... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 The three edge lengths of an isosceles tetrahedron are 3, $\sqrt{10}$, and $\sqrt{13}$. Then the radius of the circumscribed sphere of this tetrahedron is | Analysis: According to Basic Conclusion 10, the triangular pyramid can be expanded into a rectangular parallelepiped, making the known three edges the diagonals of the faces of the rectangular parallelepiped. At this point, the original triangular pyramid and the rectangular parallelepiped have the same circumscribed s... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, try to find all positive integers $k$, such that for any positive numbers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a, b, c$. | Three, due to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$,
thus $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
It can be known that $k>5$. Noting that $k$ is a positive integer, therefore, $k \geqslant 6$.
Since there does not exist a triangle with side lengths $1,1,2$, according to the problem, we have
$$
k(1 \times 1+1 \... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Lift Your Veil
A 101-digit natural number $A=\underbrace{88 \cdots 8}_{\text {S0 digits }} \square \underbrace{99 \cdots 9}_{\text {S0 digits }}$ is divisible by 7. What is the digit covered by $\square$? | Explanation: “ $\square$ ” is a veil, covering the number to be found. To lift the veil, whether approaching from the front or the back, there are 50 digits in between, which is too many. The key is to find a way to reduce the number of digits.
The difference between two multiples of 7 is still a multiple of 7.
We alr... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure $1, \angle A O B=30^{\circ}$, within $\angle A O B$ there is a fixed point $P$, and $O P$ $=10$, on $O A$ there is a point $Q$, and on $O B$ there is a fixed point $R$. To make the perimeter of $\_P Q R$ the smallest, the minimum perimeter is | 3.10 .
As shown in Figure 5, construct the symmetric points $P_{1}, P_{2}$ of $P$ with respect to the sides $O A, O B$ of $\angle A O B$. Connect $P_{1} P_{2}$, intersecting $O A, O B$ at $Q, R$. Connect $P Q, P R$. It is easy to see that
$$
P Q=P_{1} Q, P R=P_{2} R,
$$
which implies
$$
\begin{aligned}
l_{\triangle P... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given theorem: “For any prime number $n$ greater than 3, $b$ and $c$ satisfy the equation $2a + 5b = c$. Then $a + b + c$ is how many times the integer $n$? Prove your conclusion.”
Translate the above text into English, please retain the original text's line breaks and format, and output the trans... | Three, the maximum possible value of $n$ is 9.
First, we prove that $a+b+c$ is divisible by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$.
Thus, $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, with $r_{a} \neq 0$ and $r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, ... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 7 Multiplicative Magic Square
Figure 1 shows a partially filled magic square. Fill in the following nine numbers: $\frac{1}{4}, \frac{1}{2}, 1,2,4,8,16,32,64$ in the grid so that the product of the numbers in each row, column, and diagonal is the same. The number that should be filled in the “ $x$ ” cell is $\q... | Explanation: Label the unfilled cells with letters, as shown in Figure 2.
The product of the nine known numbers is
$$
\frac{1}{4} \times \frac{1}{2} \times 8 \times 1 \times 2 \times 32 \times 4 \times 16 \times 64=64^{3} \text {. }
$$
Therefore, the fixed product of the three numbers in each row, each column, and eac... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
1. $x, y$ are real numbers, and $\left(x+\sqrt{x^{2}+1}\right)(y+$ $\left.\sqrt{y^{2}+1}\right)=1$. Then $x+y=$ $\qquad$ . | 二、1.0.
Multiply both sides of the original equation by $\sqrt{x^{2}+1}-x$, we get $y+\sqrt{y^{2}+1}=\sqrt{x^{2}+1}-x$; multiply both sides of the original equation by $\sqrt{y^{2}+1}-y$, we get $x+\sqrt{x^{2}+1}=\sqrt{y^{2}+1}-y$. Adding the two equations immediately yields $x+y=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If $\left|\log \frac{a}{\pi}\right|<2$, then the number of $a$ for which the function $y=\sin (x+a)+\cos (x-a)(x \in \mathbf{R})$ is an even function is $\qquad$. | 2.10.
$$
\begin{aligned}
\because-2 & <\log \frac{\alpha}{\pi}<2, \therefore \frac{1}{\pi}<\alpha<\pi^{3} . \\
f(x)= & \sin (x+\alpha)+\cos (x+\alpha) \\
& =\sqrt{2} \sin \left(x+\alpha+\frac{\pi}{4}\right) .
\end{aligned}
$$
Since $f(x)$ is an even function,
$$
\therefore \sin \left(\frac{\pi}{4}+\alpha+x\right)=\sin ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Total 20 points) The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, n \in \mathbf{N}, y_{n}=\frac{1}{1+x_{n}}, \\
S_{n}=y_{1}+y_{2}+\cdots+y_{n}, P_{n}=y_{1} y_{2} \cdots y_{n} .
\end{array}
$$
Find $P_{n}+\frac{1}{2} S_{n}$. | $$
\begin{array}{l}
\text { Three, } \because x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, \\
\therefore x_{n+1}>x_{n}>0, x_{n+1}=x_{n}\left(1+x_{n}\right) . \\
\therefore y_{n}=\frac{1}{1+x_{n}}=\frac{x_{n}^{2}}{x_{n} x_{n+1}}=\frac{x_{n+1}-x_{n}}{x_{n} x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}} . \\
\therefore P_{n}=y... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. The number of sets $X$ that satisfy the condition $\{1,2,3\} \subseteq X \subseteq\{1,2,3,4,5,6\}$ is $\qquad$ . | ii. 7.8 items.
$X$ must include the 3 elements $1,2,3$, while the numbers 4,5,6 may or may not belong to $X$. Each number has 2 possibilities, so the total number of different $X$ is $2^{3}=8$. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. The function $y=f(x)$ defined on $\mathbf{R}$ has the following properties:
（1）For any $x \in \mathbf{R}$, $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1} 、 x_{2} \in \mathbf{R}, x_{1} \neq x_{2}$, $f\left(x_{1}\right)$ $\neq f\left(x_{2}\right)$.
Then the value of $f(0)+f(1)+f(-1)$ is $\qquad$ | 10.0 .
From $f(0)=f^{3}(0)$, we know $f(0)[1-f(0)][1+f(0)]=0$,
thus, $f(0)=0$ or $f(0)=1$, or $f(0)=-1$;
from $f(1)=f^{3}(1)$, similarly $f(1)=0$ or 1 or 1;
from $f(-1)=f^{3}(-1)$, similarly $f(-1)=0$ or 1 or -1.
However, $f(0)$, $f(1)$, and $f(-1)$ are pairwise distinct,
so $\{f(0), f(1), f(-1)\}=\{0,1,-1\}$.
Therefo... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (13 points) Let the quadratic equation in $x$, $2 x^{2}-t x-2=0$, have two roots $\alpha, \beta(\alpha<\beta)$.
(1) If $x_{1} 、 x_{2}$ are two different points in the interval $[\alpha, \beta]$, prove that $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4<0$;
(2) Let $f(x)=\frac{4 x-t}{x^{2}+1}$, and the maximum and minim... | $$
\begin{aligned}
14. (1) & \text{ From the conditions, we have } \alpha+\beta=\frac{t}{2}, \alpha \beta=-1. \\
& \text{ Without loss of generality, assume } \alpha \leqslant x_{1}4 x_{1} x_{2}-2(\alpha+\beta)\left(x_{1}+x_{2}\right)+4 \alpha \beta=4 x_{1} x_{2} \\
& -t\left(x_{1}+x_{2}\right)-4 .
\end{aligned}
$$
Th... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. (13 points) Given $a_{1}=1, a_{2}=3, a_{n+2}=(n+3) a_{n+1}$ $-(n+2) a_{n}$. If for $m \geqslant n$, the value of $a_{m}$ can always be divided by 9, find the minimum value of $n$.
| 15. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}$
$$
\begin{array}{l}
-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1) \cdot \\
\left(a_{n}-a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots+4 \cdot 3 \cdot\left(a_{2}-\right. \\
\left.a_{1}\right)=(n+2)!, \\
\text { Therefore, } a_{n}=a_{1}+\left(a_{2}-a_{1}\r... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. As shown in Figure 1, the area of square $A B C D$ is 256, point $F$ is on $A D$, and point $E$ is on the extension of $A B$. The area of right triangle $\triangle C E F$ is 200. Then the length of $B E$ is $(\quad)$.
(A) 10
(B) 11
(C) 12
(D) 15 | 2. (C).
It is easy to prove that $\mathrm{Rt} \triangle C D F \cong \mathrm{Rt} \triangle C B E$. Therefore, $C F=C E$.
Since the area of Rt $\triangle C E F$ is 200, i.e., $\frac{1}{2} \cdot C F \cdot C E=200$, hence $C E=20$.
And $S_{\text {square } B C D}=B C^{2}=256$, so $B C=16$.
By the Pythagorean theorem, $B E=... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
8. As shown in Figure $11, \angle A O B=$ $30^{\circ}, \angle A O B$ contains a fixed point $P$, and $O P=10, O A$ has a point $Q, O B$ has a fixed point $R$. If the perimeter of $\triangle P Q R$ is minimized, find its minimum value. | (Tip: Draw auxiliary lines with $O A$ and $O B$ as axes of symmetry. Answer:
10.) | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let the function $f_{0}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed figure formed by the graph of $y$ $=f_{2}(x)$ and the $x$-axis is $\qquad$
(1989, National High School Mathematics Competition) | Analysis: If it is not easy to directly draw the graph of $y=f_{2}(x)$, but we know the relationship between the graphs of $y=f(x)$ and $y=|f(x)|$. If we follow the sequence
$$
\begin{array}{l}
f_{0}(x)=|x| \rightarrow y=f_{0}(x)-1 \\
\rightarrow f_{1}(x)=\left|f_{0}(x)-1\right| \rightarrow y=f_{1}(x)-2 \\
\rightarrow ... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 The sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{2 n}, a_{2 n+1}$ forms an arithmetic sequence, and the sum of the terms with odd indices is 60, while the sum of the terms with even indices is 45. Then the number of terms $n=$ $\qquad$ | Analysis: From $\left\{\begin{array}{l}a_{1}+a_{3}+\cdots+a_{2 n+1}=60, \\ a_{2}+a_{4}+\cdots+a_{2 n}=45\end{array}\right.$ $\Rightarrow\left\{\begin{array}{l}(n+1) a_{n+1}=60, \\ n a_{n+1}=45\end{array} \Rightarrow \frac{n+1}{n}=\frac{4}{3}\right.$, solving gives $n=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 Given the three side lengths $a$, $b$, and $c$ of $\triangle ABC$ satisfy: (1) $a>b>c$, (2) $2b=a+c$, (3) $b$ is an integer, (4) $a^{2}+b^{2}+c^{2}=84$. Then the value of $b$ is $\qquad$ | Analysis: Starting from $2 b=a+c$, let $a=b+d$, $c=b-d(d>0)$, then
$$
(b+d)^{2}+b^{2}+(b-d)^{2}=84 .
$$
That is, $3 b^{2}+2 d^{2}=84$.
Obviously, $2 d^{2}$ is a multiple of 3. Also, by $b+c>a \Rightarrow$ $b>2 d$, substituting into (4) gives $2 d^{2}<12$. Therefore, $2 d^{2}=3,6,9$. Verification shows $b=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 19 Let $f(x)=|1-2 x|, x \in[0,1]$. Then, the number of solutions to the equation $f\{f[f(x)]\}=\frac{1}{2} x$ is $\qquad$ . | Analysis: Let $y=f(x), z=f(y), w=f(z)$, use “ $\rightarrow$ ” to indicate the change, we have
$$
\begin{array}{l}
x: 0 \rightarrow 1, y: 1 \rightarrow 0 \rightarrow 1, \\
z: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1, \\
w: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarr... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If $x \in \mathbf{R}$, find the maximum value of $F(x)=\min \{2 x+1$, $x+2,-x+6\}$.
(38th AHSME) | Solution: As shown in Figure 1, draw the graph of $F(x)$ (the solid part). From the graph, we can see that the maximum value of $F(x)$ is equal to the y-coordinate of the intersection point of $y=x+2$ and $y=-x+6$.
Therefore, the maximum value of $F(x)$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}$, $\cdots, b_{n}$ satisfying
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1 .
$$
Find the maximum value of $F=\min \left\{\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \cdots, \frac{a_{n}}{b_{n}}\right\}$.
(1979, G... | Solution: It is easy to see that when all the letters are equal, the value of $F$ is 1.
Below we prove that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$; $b_{1}, b_{2}, \cdots, b_{n}$, we have $F \leqslant 1$.
If not, then $F>1$.
$$
\text { Hence } \frac{a_{1}}{b_{1}}>1, \frac{a_{2}}{b_{2}}>1, \cdots, \frac{... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
7. If real numbers $x, y, z$ satisfy $x+\frac{1}{y}=4, y+\frac{1}{z}=1, z+$ $\frac{1}{x}=\frac{7}{3}$, then the value of $x y z$ is $\qquad$. | 7.1 .
Since $4=x+\frac{1}{y}=x+\frac{1}{1-\frac{1}{z}}=x+\frac{z}{z-1}$
$=x+\frac{\frac{7}{3}-\frac{1}{x}}{\frac{7}{3}-\frac{1}{x}-1}=x+\frac{7 x-3}{4 x-3}$, then
$4(4 x-3)=x(4 x-3)+7 x-3$,
which simplifies to $(2 x-3)^{2}=0$.
Thus, $x=\frac{3}{2}$.
Therefore, $z=\frac{7}{3}-\frac{1}{x}=\frac{5}{3}, y=1-\frac{1}{z}=\f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Given the quadratic function $y=a x^{2}+b x+c$ (where $a$ is a positive integer) whose graph passes through the points $A(-1,4)$ and $B(2,1)$, and intersects the $x$-axis at two distinct points. Then the maximum value of $b+c$ is $\qquad$ . | 10. -4 .
From the given, we have $\left\{\begin{array}{l}a-b+c=4, \\ 4 a+2 b+c=1,\end{array}\right.$ solving this yields $\left\{\begin{array}{l}b=-a-1, \\ c=3-2 a .\end{array}\right.$
Since the graph of the quadratic function intersects the $x$-axis at two distinct points, we have,
$$
\begin{array}{l}
\Delta=b^{2}-4 ... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$. | 14. (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$. From the problem, we know $a>0$, and $b+c=2-a, bc=\frac{4}{a}$. Therefore, $b, c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$, then
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \tim... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $0 \leqslant a-b \leqslant 1,1 \leqslant a+b \leqslant 4$. Then, when $a-2 b$ reaches its maximum value, the value of $8 a+2002 b$ is $\qquad$ . | 3.8.
$$
\begin{array}{l}
\text { Let } 0 \leqslant a-b \leqslant 1, \\
1 \leqslant a+b \leqslant 4, \\
\text { and } m(a-b)+n(a+b)=a-2 b .
\end{array}
$$
By comparing the coefficients of $a$ and $b$ on both sides, we get the system of equations and solve to find
$$
m=\frac{3}{2}, n=-\frac{1}{2} \text {. }
$$
Thus, $a-... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. A $9 \times 9$ grid of squares is colored in two colors, black and white, such that the number of black squares adjacent to each white square is greater than the number of white squares, and the number of white squares adjacent to each black square is greater than the number of black squares (squares sharing a commo... | Solution: A coloring scheme of a square matrix that satisfies the problem's conditions is called a good scheme. For a good coloring scheme, it is easy to prove the following conclusions:
(1) If two adjacent cells have the same color and are in different rows (or columns), then any two adjacent cells in these two rows (... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
17. Given a wire of length $150 \mathrm{~cm}$, it is to be cut into $n(n>2)$ smaller segments, each of which has an integer length of no less than $1(\mathrm{~cm})$. If no three segments can form a triangle, find the maximum value of $n$, and how many ways there are to cut the wire into $n$ segments that satisfy the co... | 17. Since the sum of $n$ segments is a fixed value of $150(\mathrm{~cm})$, to make $n$ as large as possible, the length of each segment must be as small as possible. Given that the length of each segment is no less than $1(\mathrm{~cm})$, and no three segments can form a triangle, the lengths of these segments can only... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
保留源文本的换行和格式，翻译结果如下：
```
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
``` | Three, let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a$ and $b$ remain unchanged, and only let $c$ vary, where $c$ can take any integer from 1 to 9.
Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that
when $c=1$, $P$ reaches its maximum value, so $c=1$.
Thus, $P=\frac{a+b+1}{a b}=\frac... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) In a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)(n$ $\geqslant 2)$ grid, each cell is filled with 1 or -1. If the number in any cell is equal to the product of the numbers in the cells that share an edge with it, then this filling method is called "successful". Find the total number of "successf... | Three, assuming there exists some successful filling method that contains -1. First, prove: if this successful filling method is symmetric about the middle column (row), then the middle column (row) is entirely 1.
Let $a_{0}=1, a_{2}^{n}=1$.
If $a_{1}=1$, by $a_{1}=a_{0} \times a_{2} \times 1$, we get $a_{2}=1$.
Simil... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c$ $=$ $\qquad$ . | -1 .
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$. Since $\triangle A B C$ is a right triangle, it follows that $x_{1} 、 x_{2}$ must have opposite signs. Thus, $x_{1} x_{2}=\frac{c}{a}<0$.
By the projection theorem, we know $|O C|^{2}=|A O| \cdot|B O|$, i.e., $c^{2}=\left|x_{1}\right| \cdot\left|x_{2}\right|=\... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ . | 2.4 .
From the problem, we have
$$
\left\{\begin{array}{l}
3 \times\left(-\frac{9}{5}\right)^{2}+m \times\left(-\frac{9}{5}\right)-2>0, \\
3 \times\left(\frac{3}{7}\right)^{2}+m \times\left(\frac{3}{7}\right)-2>0 .
\end{array}\right.
$$
Solving this, we get \(3 \frac{8}{21}<m<4 \frac{13}{45}\). Therefore, \(m=4\). | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (18 points) On a plane, there are 7 points, and some line segments can be connected between them, so that any 3 points among the 7 points must have 2 points connected by a line segment. How many line segments are needed at least? Prove your conclusion. | (1) If one of the 7 points is isolated (i.e., it is not connected to any other points), then the remaining 6 points must be connected in pairs, which requires at least $\frac{6 \times 5}{2}=15$ lines.
(2) If one of the 7 points is connected to only one other point, then the remaining 5 points must be connected in pairs... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false |
12. As shown in Figure 4, line $AB$ intersects $\odot O$ at points $A$ and $B$, point $O$ is on $AB$, and point $C$ is on $\odot O$, with $\angle AOC=40^{\circ}$. Point $E$ is a moving point on line $AB$ (not coinciding with point $O$), and line $EC$ intersects $\odot O$ at another point $D$. The number of points $E$ t... | 12.3.
Consider the position of point $E$: Point $E$ can be on the extension of line segment $O A$; Point $E$ can be on line segment $O A$ (excluding point $O$); Point $E$ can be on the extension of line segment $O B$; but point $E$ cannot be on line segment $O B$ (excluding point $O$). Therefore, point $E$ has a total... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
14. Given real numbers $a, b, c$, satisfying $a+b+c=0, a^{2}+b^{2}+c^{2}=6$. Then the maximum value of $a$ is | 14.2.
From the problem, we get $c=-(a+b)$, thus,
$$
a^{2}+b^{2}+[-(a+b)]^{2}=6,
$$
which simplifies to $b^{2}+a b+a^{2}-3=0$.
Since $b$ is a real number, the above equation, which is a quadratic equation in $b$, must have real roots, hence,
$$
\Delta=a^{2}-4\left(a^{2}-3\right) \geqslant 0, \quad -2 \leqslant a \leqs... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. A group of 17 middle school students went to several places for a summer social survey, with a budget for accommodation not exceeding $x$ yuan per person per day. One day, they arrived at a place with two hostels, $A$ and $B$. $A$ has 8 first-class beds and 11 second-class beds; $B$ has 10 first-class beds, 4 second... | 4. $x=10$.
If staying at location $A$, even choosing the most economical beds, the average accommodation cost per person will exceed 10 yuan (since $8 \times 11 + 14 \times 6 = 172$ (yuan), $172 \div 17 \approx 10.12$ (yuan)).
If staying at location $B$, with a reasonable choice of beds, the budget can be met, and th... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
2. If the polynomial $x^{2}-x+1$ can divide another polynomial $x^{3}+x^{2}+a x+b(a, b$ are constants). Then $a+b$ equals ( ).
(A) 0
(B) -1
(C) 1
(D) 2 | 2. (C).
By synthetic division, it is easy to obtain the remainder $r(x)=(a+1) x+b-2$. Since it can be divided exactly, we have $a+1=0$ and $b-2=0$, i.e., $a=-1, b=2$. Therefore, $a+b=1$. | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. $f(x)=\frac{x^{2}}{8}+x \cos x+\cos (2 x)(x \in \mathbf{R})$'s minimum value is $\qquad$ . | 4. -1 .
$$
\begin{array}{l}
f(x)=\frac{x^{2}}{8}+x \cos x+2 \cos ^{2} x-1 \\
=\frac{1}{8}(x+4 \cos x)^{2}-1 \geqslant-1 .
\end{array}
$$
Since the equation $\cos x=-\frac{x}{4}$ (as can be seen from the graph) has a solution, we have
$$
f(x)_{\min }=-1
$$ | -1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Try to find the smallest possible value of the positive integer $k$, such that the following proposition holds: For any $k$ integers $a_{1}, a_{2}, \cdots, a_{k}$ (equality is allowed), there must exist corresponding $k$ integers $x_{1}, x_{2}, \cdots, x_{k}$ (equality is also allowed), and $\left|x_... | Three, first prove that the proposition holds when $k=7$. For this, consider the sum
$$
S\left(y_{1}, y_{2}, \cdots, y_{7}\right)=y_{1} a_{1}+y_{2} a_{2}+\cdots+y_{7} a_{7} \text {, }
$$
where $y_{i} \in\{-1,0,1\}$.
There are $3^{7}=2187$ such sums, and since $2187>2003$, by the pigeonhole principle, there must be two... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $n$ be a given positive integer. Try to find non-negative integers $k, l$, satisfying $k+l \neq 0$, and $k+l \neq n$, such that
$$
s=\frac{k}{k+l}+\frac{n-k}{n-(k+l)}
$$
takes the maximum value. | If $l=0$, then $s=2$.
If $l>0$, let $x=k+l$, then $00 .
\end{array}
$
Therefore, $f(1)1$ when, $s$ reaches the maximum value 2;
(2) If $n=2$, then when $l=1, k=0$ or $l=0, 0 < k \neq 2$, $s$ reaches the maximum value 2;
(3) If $n \geqslant 3$, then when $l=n-1, k=0$, $s$ reaches the maximum value $n$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Find the smallest positive integer $n$, such that
$$
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2002}
$$
has integer solutions.
(Uzbekistan provided) | Solution: Since $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), 2002$ $=667 \times 3+1$, therefore,
$$
2002^{2002} \equiv 4^{2002} \equiv 4(\bmod 9) \text {. }
$$
Also, $x^{3} \equiv 0, \pm 1(\bmod 9)$, where $x$ is an integer, thus,
$$
x_{1}^{3}, x_{1}^{3}+x_{2}^{3}, x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \equiv 4(\bmod 9) ... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. For all non-negative integers $x, y$, find all functions $f: \mathbf{N} \rightarrow \mathbf{N}$, satisfying $f(3 x+2 y)=f(x) f(y)$, where $\mathbf{N}$ is the set of non-negative integers.
(53rd Romanian Mathematical Olympiad (Final)) | Solution: Let $x=y=0$, we get $f(0)=f(0)^{2}$. Therefore, $f(0)=0$ or $f(0)=1$.
If $f(0)=0$, for $x=0$ or $y=0$, we get $f(2 y)=$ $f(3 x)=0$ for all $x, y \in \mathrm{N}$. Let $f(1)=a$, then
$$
\begin{array}{l}
f(5)=f(3 \times 1+2 \times 1)=f(1) f(1)=a^{2}, \\
f(25)=f(3 \times 5+2 \times 5)=f(5) f(5)=a^{4} .
\end{array... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Consider a square on the complex plane, whose 4 vertices correspond to the 4 roots of a certain monic quartic equation with integer coefficients $x^{4}+p x^{3}+q x^{2}+r x+s=0$. Find the minimum value of the area of such a square.
| 3. According to the problem, the 4 roots of the equation can only be in two scenarios: 2 real roots and 1 pair of conjugate complex roots; 2 pairs of conjugate complex roots.
(1) If the 4 roots of the equation are 2 real roots and 1 pair of conjugate complex roots, then we can set these 4 roots as $a \pm b, a \pm b \ma... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find a point $P$ on the plane of equilateral $\triangle A B C$ such that $\triangle P A B$, $\triangle P B C$, and $\triangle P C A$ are all isosceles triangles. How many points $P$ with this property are there? | Solution: Let any side of the known triangle be denoted as $a$. When $a$ is the base of the sought isosceles triangle, point $P$ lies on the perpendicular bisector of $a$; when $a$ is one of the legs of the sought isosceles triangle, point $P$ lies on the circumference of the circle with the vertex of $\triangle ABC$ a... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given an $n \times n$ ($n$ is an odd number) chessboard where each unit square is colored in a checkerboard pattern, and the 4 corner unit squares are colored black. A figure formed by 3 connected unit squares in an L-shape is called a "domino". For what value of $n$ can all the black squares be covered by non-overl... | Solution: Let $n=2m+1$, consider the odd rows, then each row has $m+1$ black cells, with a total of $(m+1)^2$ black cells. Any two black cells cannot be covered by a single "domino", therefore, at least $(m+1)^2$ "dominoes" are needed to cover all the black cells on the chessboard. Since when $n=1,3,5$, we have $3(m+1)... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $T$ be a set of ordered triples $(x, y, z)$, where $x, y, z$ are integers, and $0 \leqslant x, y, z \leqslant 9$. Two players, A and B, play the following game: A selects a triple $(x, y, z)$ from $T$, and B has to guess A's chosen triple using several "moves". One "move" consists of: B giving A a triple $(a, b,... | Solution: Two "movements" are not enough. Because each answer is an even number between 0 and 54, i.e., there are 28 possible values for each answer. The maximum number of possible results from two "movements" is $28^{2}$, which is less than the 1000 possible values for $(x, y, z)$.
Below, we prove that 3 "movements" ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
4. The sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{k+1}=x_{k}^{2}+x_{k}$. Then the integer part of the sum $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{200 B}+1}$ is $\qquad$ | 4. 1 .
From the problem, we know that $\left\{x_{n}\right\}$ is an increasing sequence, and $x_{3}>1$. From $\frac{1}{x_{k+1}}=\frac{1}{x_{k}\left(x_{k}+1\right)}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$, we get $\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$.
Then $S=\sum_{k=1}^{2008} \frac{1}{x_{k}+1}=\sum_{k=1}^{20... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given an infinite sequence $\left\{a_{n}\right\}$ where all terms are positive integers, and the sum of any consecutive terms is not equal to 100. Find the minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$. | Solution: The minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$ is 3.
First, we prove: If all terms of the sequence $\left\{a_{n}\right\}$ are 1 or 2, then there must exist a continuous subsequence whose sum equals 100.
Consider the first 100 terms of the sequence $\left\{a_{n}\right\}$, and let
$$
S_{0}=0... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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