problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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4. As shown in the figure, $AB$ is the diameter of the semicircle $\odot O$, $CD \perp AB$. Let $\angle COD = \theta$, then $\frac{AD}{BD}$ - $\tan^2 \frac{\theta}{2}=$ $\qquad$ | 4. 1.
Connect $A C$, then $\angle C A D=\frac{1}{2} \angle \theta$, and $\operatorname{tg} \frac{\theta}{2}=\operatorname{tg} \angle C A D=\frac{C D}{A D}$.
According to the projection theorem, we know
$$
\begin{array}{l}
C D^{2}=A D \cdot B D . \\
\therefore \frac{A D}{B D} \operatorname{tg}^{2} \frac{\theta}{2}=\fra... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. Let $s_{1}, x_{2}, c_{3}, x_{4}$ all be positive real numbers, and $x_{1} + x_{2} + x_{3} - x_{4} = \pi$. Find the minimum value of the expression
\[
\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)
- \left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x... | Let $x_{1}+x_{2}$ be a constant. Since
$$
\begin{array}{l}
\sin x_{1} \sin x_{2} \\
=\frac{1}{2}\left[\cos \left(x_{1}-x_{2}\right)-\cos \left(x_{1}+x_{2}\right)\right],
\end{array}
$$
we know that the value of $\sin x_{1} \sin x_{2}$ increases as $\left|x_{1}-x_{2}\right|$ decreases. Let the expression in question be... | 81 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. On a plane, there are $n$ lines, among which $k$ lines are pairwise parallel, and $t$ lines pass through the same point $(k+t<n)$, in addition, any two intersect, and any 3 do not share a point. How many regions do they divide the plane into: | The number of intersection points is given by $p_{n}=C_{n}^{2}-C_{k}^{2}-C_{t}^{2}+(t-1)$. According to Theorem 1, the number of regions they divide the plane into is
$$
S_{n}=C_{n}^{2}-C_{1}^{2}-C_{t}^{2}+n+t.
$$
The figure below gives an example F for $n=8, k=4, t=3$. In this case, $S_{8}=C_{8}^{2}-C_{4}^{2}-C_{3}^{... | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. In $\triangle ABC$, $G$ is the centroid, and $P$ is a point inside the triangle. The line $PG$ intersects the lines $BC$, $CA$, and $AB$ at $A'$, $B'$, and $C'$, respectively. Prove that: $\frac{A'P}{A'G} + \frac{B'P}{B'G} + \frac{C'P}{C'G} = 3$
(1991, Huanggang Region Junior High School Competition) | Prove that by connecting $B G, G C, P B, P C$, and drawing $G G^{\prime} \perp B C$ at $G^{\prime}$, $P P^{\prime} \perp B C$ at $P^{\prime}$, then
$$
\begin{array}{c}
P P^{\prime} / / C G^{\prime}, \frac{P P^{\prime}}{G G^{\prime}}=\frac{A^{\prime} P^{2}}{A^{\prime} G} . \\
\text { Also, } \frac{S_{\triangle P B C}}{S... | 3 | Geometry | proof | Yes | Yes | cn_contest | false |
2. The two roots of the equation $x^{2}-3|x|-4=0$ are $x_{1}$, $x_{2}$, then $x_{1}+x_{2}=$ $\qquad$ | 2. 0
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
B-4. $p(x)$ is a non-zero polynomial of degree less than 1992, and $p(x)$ has no common factor with $x^{3}-x$. Let
$$
\frac{d^{1992}}{d x^{1992}}\left(\frac{p(x)}{x^{3}-x}\right)=\frac{f(x)}{g(x)},
$$
where $f(x), g(x)$ are polynomials. Find the smallest possible degree of $f(x)$. | The minimum possible degree of $f(x)$ is 3984.
By the division algorithm, we let $p(x) = (x^3 - x)q(x) + r(x)$, where $q(x)$ and $r(x)$ are polynomials, and the degree of $r(x)$ is less than 3, and the degree of $q(x)$ is less than 1989. Then,
$$
\cdot \frac{d^{1992}}{d x^{1992}}\left(\frac{p(x)}{x^3 - x}\right) = \fra... | 3984 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. (Question 2 of the 5th Test, *The American Mathematical Monthly*, pages 119 to 121, 1953)
Given a positive integer $n \geqslant 3$, for $n$ complex numbers $z_{1}, z_{2}, \cdots, z_{n}$ with modulus 1, find
$$
\min _{x_{1}, z_{2} \cdots, s_{n}}\left[\max _{\omega \in C,|\omega|=1} \prod_{j=1}^{n}\left|\omega-z_{j}\... | $$
\begin{array}{l}
f(u)=\prod_{j=1}^{n}\left(u-x_{j}\right) \\
=u^{n}+C_{n-1} u^{n-1}+\cdots+C_{1} u+C_{0} .
\end{array}
$$
Here, $\left|C_{0}\right|=\left|z_{1} z_{2} \cdots z_{n}\right|=1$. Let $C_{0}=e^{i\theta}, 0 \leqslant \theta<2 \pi$. Take $n$ complex numbers of modulus 1, $\varepsilon_{k}=e^{i \frac{2 x+\the... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (35 points) A group of students took an exam with 3 multiple-choice questions, each with four options. It is known that any two students in this group have at most one answer in common, and if one more student is added, regardless of their answers, the above property no longer holds. How many students are there ... | Three, this group of students must have at least 8 people.
First, consider if this group of students has 8 people, then it can meet the conditions of the problem.
Let the answers of these 8 people be $(1,1,1),(1,2,2),(2,2,1)$, $(2,1,2),(3,4,3),(3,3,4),(4,3,3),(4,4,4)$, which will suffice.
On the other hand, assume thi... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) For positive numbers $a, b$, define the operation * as follows: when $a \leqslant b$, $a * b=a b$; when $b<a<2 b$, $a * b=b^{2}+a-b$; when $a \geqslant 2 b$, $a * b=a^{2}+b-a$. If $3 * x=8$, find the value of $\left[x^{3}-4 x\right]$ (where $[x]$ represents the greatest integer not exceeding $x$). | If $3 \leqslant x$, then $3 * x=3 x \geqslant 3 \cdot 3>8$. This does not satisfy $3 * x=8$.
If $x<3<2 x$, then $3 * x=x^{2}+3-x=8$, which means $x^{2}-x-5=0$.
Thus, $x=\frac{1+\sqrt{21}}{2}$ (taking the positive root) (since $\sqrt{21}<5$ satisfies $x<3<2 x$).
If $3 \geqslant 2 x$, then $3 * x=3^{2}+x-3=6+x \leqslan... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. Let the sequences $\left\{x_{n}\right\},\left\{y_{n}\right\}$ satisfy $x_{n}+y_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n}$. Find the sum of the first 1994 terms of the sequence $\left\{x_{n}\right\}$, $S_{1994}$. | $\begin{array}{l}\text { Let } \begin{array}{l} \text { } F:=x_{n}+y_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n}, F^{\prime} \\ =x_{h}-j_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n} . \text { Then } F+F^{\prime}=2 x_{n} \\ =\omega^{n}+(\bar{\omega})^{n}\left(\text { where } \omega=\frac{-1+\sqrt{3} i}{2}\right) . \t... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. Let $f(x)=\frac{4^{x}}{4^{x}+2}$. Try to find the value of $f\left(\frac{1}{1001}\right)$ $+f\left(\frac{2}{1001}\right)+\cdots+f\left(\frac{1000}{1001}\right)$. (1986, National High School Competition) | Let $F=f(x)=\frac{4^{x}}{4^{x}+2}, F^{\prime}=f(y)=$ $\frac{4^{y}}{4^{y}+2}$. If $x+y=1$, then $f(x)+f(y)=1$. In fact, $f(x)+f(y)=f(x)+f(1-x)=\frac{4^{x}}{4^{x}+2}$ $+\frac{4^{1-x}}{4^{1-x}+2}=1$. Therefore, $f\left(\frac{1}{1001}\right)+f\left(\frac{2}{1001}\right)$ $+\cdots+f\left(\frac{999}{1001}\right)+f\left(\frac... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9. For the set $\{1,2,3, \cdots, n\}$ and any of its non-empty subsets, define the "alternating sum" as follows: rearrange the elements in decreasing order, then alternately add and subtract the subsequent numbers starting from the largest. For example, for $\{1,2,4,6,9\}$, rearranged as $(9,6,4,2,1)$, the alte... | Solution: Let's define the alternating sum of the elements of the empty set to be 0. Divide the $2^{n}$ subsets of $\{1,2, \cdots, n\}$ into two categories: one category contains $n$ and has $2^{n-1}$ subsets, and the other category does not contain $n$, also having $2^{n-1}$ subsets. Pair the subset $F=\left\{n, a_{1}... | 448 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $E=\{1,2, \cdots, 200\}, G=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$ $\subset E$, and $G$ has the following two properties:
(i) For any $1 \leqslant i \leqslant j \leqslant 100$, it always holds that $a_{i}+a_{j} \neq 201$;
(ii) $\sum_{i=1}^{130} a_{i}=10080$.
Prove: The number of odd numbers in $G$ is a mul... | 6. Divide the elements of $E$ into 100 subsets such that the sum of the two numbers in each subset is 201: $E_{1}=\{1,200\}, E_{2}=\{2,199\}, \cdots, E_{100}=\{100, 101\}$. According to property (i) of $G$, for any $1 \leqslant i \leqslant 100$, the two elements in $E_{i}$ cannot both be in $G$. Since the number of ele... | 1349380 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example. Given $f(x)=[3 x]+[4 x]+[12 x]+[16 x]$.
(1) Find the number of positive integer values $M$ that $f(x)$ takes when $0<x \leqslant 1$.
(2) Is 97 a value of $f(x)$? | $$
\begin{array}{l}
\text { (1) } \sum n_{i}=3+4+12+16=35 . \\
\sum_{1<i<j<1}\left(n_{i}, n_{j}\right)=(3,4)+(3,12)+(3,16)+(4,12)+(4,16)+(12,16)=17. \\
\sum_{1<i<j<k<t}\left(n_{i}, n_{j}, n_{k}\right)=(3,4,12)+(3,4,16)+(3,12,16)+(4,12,16)=7. \\
\left(n_{i}, n_{2}, n_{3}, n_{!}\right)=(3,4,12,16)=1, \\
\end{array}
$$
B... | 97 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\sin y \cos y+a=0$. Then $\cos (x+2 y)=$ $\qquad$ . | 2.1.
Given that $x^{3}+\sin x=2 a=(-2 y)^{3}+\sin (-2 y)$. Let $f(t)=t^{3}+\sin t$, then $f(x)=f(-2 y)$. The function $f(t)=t^{3}+\sin t$ is strictly increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, $x=-2 y, x + 2 y=0$. Hence, $\cos (x+2 y)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the point sets
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(x-3)^{2}+(y-4)^{2} \leqslant\left(\frac{5}{2}\right)^{2}\right.\right\}, \\
B=\left\{(x, y) \left\lvert\,(x-4)^{2}+(y-5)^{2}>\left(\frac{5}{2}\right)^{2}\right.\right\} .
\end{array}
$$
Then the number of integer points (i.e., points with both c... | 3. 7.
As shown in the figure, circles $E$ and $F$ intersect at points $M$ and $N$. The entire figure is symmetric about the line connecting the centers $E F$. Among them, $A \cap B$ is a crescent shape $S$ near the origin $O$ on the lower left. The x-coordinate of the points in $S$ can be 1, 2, 3, 4, and the y-coordin... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given 95 numbers $a_{1}, a_{2}, \cdots, a_{95}$, each of which can only take one of the two values +1 or -1. Then, the minimum positive value of the sum of their pairwise products $a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{81} a_{23}$ is $\qquad$ . | 6.13.
Let $N=a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{98} a_{95}$.
Suppose $a_{1}, a_{2}, \cdots, a_{95}$ contain $m \uparrow+1, n \uparrow-1$, then $m+n=95$.
Multiplying (1) by 2 and adding $a_{1}^{2}+a_{2}^{2}+\cdots+a_{95}^{2}=95$, we get
$$
\left(a_{1}+a_{2}+\cdots+a_{95}\right)^{2}=2 N+95 \text{. }
$$
Also, $a_{1}+a_{2... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
II. (This question is worth 25 points) Arrange all positive integers that are coprime with 105 in ascending order. Find the 1000th term of this sequence. | $$
\begin{array}{r}
\text { 2. Let } S=\{1,2, \cdots, 105\}, A_{1}=\{x \mid x \in S, \text { and } 3 \mid \\
x\}, A_{2}=\{x \mid x \in S, 45 \mid x\}, A_{3}=\{x \mid x \in S, \text { and } 7 \mid
\end{array}
$$
$$
\begin{array}{l}
\left|\overline{A_{1}} \cap \overline{A_{2}} \cap \overline{A_{3}}\right|=| S \mid-\left(... | 2186 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (This question is worth 35 points) Given a set of points on a plane $P=\left\{P_{1}\right.$, $\left.P_{2}, \cdots, P_{1004}\right\}$, where no three points in $P$ are collinear. Divide all the points in $P$ arbitrarily into 83 groups, such that each group has at least 3 points, and each point belongs to exactly o... | (1) Let $m(G)=m_{0}, G$ be obtained by the partition $X_{1}, X_{2}, \cdots, X_{\text {в3 }}$, where $X_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots$, 83.
$$
\begin{array}{l}
\text { Let }\left|X_{i}\right|=x_{i}(i=1,2, \cdots, 83) \text {, then } x_{1}+x_{2}+\cdots \\
+x_{83}=1994, \text { and } \\
m_... | 168544 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Through a non-central point inside a regular dodecagon, what is the maximum number of different diagonals that can be drawn?
保留源文本的换行和格式，这里的翻译结果应该是：
Through a non-central point inside a regular dodecagon, what is the maximum number of different diagonals that can be drawn?
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。
Throug... | Let the regular dodecagon be $A_{1} A_{2} \cdots A_{12}$.
First, prove that the 4 diagonals $A_{1} A_{5}, A_{2} A_{6}, A_{3} A_{8}, A_{4} A_{11}$ intersect at a non-central point inside the regular dodecagon.
Since $\overparen{A_{5} A_{8}}=\overparen{A_{8} A_{11}}$, $\overparen{A_{11} A_{1}}=\overparen{A_{1} A_{3}}$,... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Given 9 points in space, where any 4 points are not coplanar. Connect several line segments among these 9 points so that there is no tetrahedron in the graph. How many triangles can there be at most in the graph?
Translate the above text into English, please keep the original text's line breaks and format, and output ... | First, prove the following conclusion:
In a space graph with $n$ points, if there are no triangles, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$.
Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, where the number of edges from $A_{1}$ is the maximum, say $k$ edges $A_{1} A_{n}, A_{1} A_... | 27 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. The equation $x^{2}+m x+1=0$ and the equation $x^{2}-x$ $-m=0$ have one common root, then $m=$ $\qquad$ | Two equations have a common root $x=-1$. Substituting this root into either equation, we get $m=2$.
| 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The solution set of the equation $x^{2}+x-1=x e^{x^{2}-1}+\left(x^{2}-1\right) e^{x}$ is $A$ (where, $e$ is an irrational number, $e=2.71828$ $\cdots$). Then the sum of the squares of all elements in $A$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 5. (C)
Let $y=x^{2}-1$. The original equation becomes
$$
x+y=x e^{y}+y e^{x} \text {, }
$$
which is $x\left(e^{y}-1\right)+y\left(e^{x}-1\right)=0$.
Since $x>0$ implies $e^{x}-1>0$, i.e., $x$ and $e^{x}-1$ always have the same sign, $y$ and $e^{y}-1$ also always have the same sign.
Therefore, $x\left(e^{y}-1\right) ... | 2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. The numbers $2 x, 1, y-1$ form an arithmetic sequence in the given order, and $y+3,|x+1|+|x-1|$, $\cos (\arccos x)$ form a geometric sequence in the given order. Then $x+y+x y=$ | 4. 3.
Since $2 x, 1, y-1$ form an arithmetic sequence, we have
$$
2 x+y-1=2 \Rightarrow y=3-2 x \text {. }
$$
And $\cos (\arccos x)=x,-1 \leqslant x \leqslant 1$, it follows that,
$$
|x+1|+|x-1|=2 \text {. }
$$
From the second condition, we get $\quad(y+3) x=4$.
Substituting (1) into (2) yields
$$
2 x^{2}-6 x+4=0 \t... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. From 30 people with distinct ages, select two groups, the first with 12 people and the second with 15 people, such that the oldest person in the first group is younger than the youngest person in the second group. How many ways are there to select these groups? | 5. 4060.
Assume the ages of these 30 people, from smallest to largest, are
$$
a_{1}<a_{2}<a_{3}<\cdots<a_{29}<a_{30} .
$$
Let the first group of 12 people be $a_{i_{1}}<a_{i_{2}}<\cdots<a_{i_{12}}$,
and the second group of 15 people be $a_{j_{1}}<a_{j_{2}}<\cdots<a_{j_{15}}$,
which are the ages of the two selected gr... | 4060 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (Full marks 30 points) Let $f(x)=x^{2}-(4 a-2) - x-6 a^{2}$ have the minimum value $m$ on the interval $[0,1]$. Try to write the expression for $m$ in terms of $a$, $m=F(a)$. And answer: For what value of $a$ does $m$ achieve its maximum value? What is this maximum value?
---
The translation maintains the origin... | $$
-, f(x)=[x-(2 a-1)]^{2}-10 a^{2}+4 a-1 .
$$
(i) When $2 a-1 \leqslant 0$, i.e., $a \leqslant \frac{1}{2}$, $f(x)$ is \. on $[0,1]$. At this time, $m=f(0)=-6 a^{2}$.
(ii) When $2 a-1 \geqslant 1$, i.e., $a \geqslant 1$, $f(x)$ is L. on $[0,1]$. At this time, $m=f(1)=-6 a^{2}-4 a+3$.
(iii) When $0<2 a-1<1$, i.e., $\fr... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9.6 On 1000 cards, write down the natural numbers 1 to 1000 (one number per card), and then use these cards to cover 1000 squares in a $1 \times 1994$ rectangle (the rectangle consists of 1994 $1 \times 1$ squares. The size of the cards is also $1 \times 1$). If the square to the right of the square containing the card... | 9.6 Notice that the card with 1 written on it will not be moved, so the card with 2 written on it can be moved at most once, the card with 3 written on it can be moved at most twice, and so on. Thus, for any $n \leqslant 1000$, the card with $n$ written on it can be moved at most $n-1$ times (because we can only move i... | 500000 | Combinatorics | proof | Yes | Yes | cn_contest | false |
10. 1 Given? three quadratic trinomials; $P_{1}(x)=x^{2}+p_{1} x$ $+q_{1}, P_{2}(x)=x^{2}-p_{2} x+q_{2}$ and $P_{3}(x)=x^{2}+p_{3} x+q_{3}$. Prove: The equation $\left|P_{1}(x)\right|+\left|P_{2}(x)\right|=\left|P_{3}(x)\right|$ has at most 8 real roots. | 10.1 Each root of the original equation should be a root of a quadratic trinomial of the form $\pm p_{1}: p_{2} \pm p_{3}$, and with different choices of signs, there are 8 such quadratic trinomials. Since the coefficient of $x^{2}$ has the form $\pm 1 \pm 1 \pm 1$, the coefficient of $x^{2}$ will never be 0 regardless... | 8 | Algebra | proof | Yes | Yes | cn_contest | false |
10. 8 Class
30 students in total, each
student in the class
has the same number of friends.
How many students at most can there be who are
better than the majority of their friends' performance? (Assuming for any two students in the class, their performance can be compared to determine who is better or worse.)
保留源文本... | 10.8 䇾案, 25 个.
We refer to a student who performs better than the majority of their friends as a "good student". Let there be $x$ "good students", and each student has $k$ friends.
The best student is better in all $k$ "friend pairs", while each of the other "good students" is better in at least $\left[\frac{k}{2}\r... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a$ be the decimal part of $\sqrt{3}$, $b$ be the decimal part of $\sqrt{2}$: $\frac{a}{(a-b) b}$ has an integer part of $\qquad$ | $$
\begin{array}{l}
\text { 3. } \frac{a}{(a-b) b}=\frac{\sqrt{3}-1}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{(\sqrt{3}-\sqrt{2})+(\sqrt{2}-1)}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}-\sqrt{2}} \\
=\sqrt{3}+2 \sqrt{2}+1 \\
\approx 5.56 .
\end{array}
$$
The integer part is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $p, q, \frac{2 p-1}{q}, \frac{2 q-1}{p}$ all be integers, and $p>1, q>$
1. Then $p+q=$ $\qquad$ . | 6. $p+q=8$.
From $\frac{2 q-1}{p}, \frac{2 p-1}{q}$ both being positive integers, we know that one of them must be less than 2. Otherwise, $\frac{2 q-1}{p}>2, \frac{2 p-1}{q}>2$, leading to $2 p-1+2 q-1>2 p+2 q$, which is a contradiction. Assume $0<\frac{2 q-1}{p}<2$, then it must be that $2 q-1=p$. It is easy to see ... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, let the three sides of $\triangle A B C$ be $a, b, c$, and all are numbers. If it satisfies $\angle A=3 \angle B$, try to find the minimum value of the perimeter and provide a proof. | From the Law of Sines, we have \(\frac{a}{b} = \frac{\sin 3B}{\sin B} = (2 \cos B)^2 - 1\), \(\frac{c}{b} = (2 \cos B)^3 - 4 \cos B\). Since \(2 \cos B = \frac{a^2 + c^2 - b^2}{ac} \in \mathbb{Q}\), there must exist positive integers \(p, q\) with \((p, q) = 1\) such that \(2 \cos B = \frac{p}{q}\). Therefore, \(\frac{... | 21 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
There are 1994 points on a circle, which are painted in several different colors, and the number of points of each color is different. Now, take one point from each color set to form a polygon with vertices of different colors inside the circle. To maximize the number of such polygons, how many different colors should ... | Let 1994 points be colored with $n$ colors, and the number of points of each color, in ascending order, is $m_{1}1$. In fact, if $m_{1}=1$, because $m_{1} m_{2} \cdots m_{n}m_{k+1} m_{k}$.
Therefore, when $m^{\prime}{ }_{n+1}, m^{\prime} \star$ replace $m_{\star+1}, m_{\star}$, the value of $M$ increases, leading to a ... | 61 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. As shown in the figure, in pentagon $A B C D E$, $\angle B=\angle A E D=$ $90^{\circ}, A B=C D=A E=B C+D E=1$. Find the area of this pentagon. (1992, Beijing Junior High School Grade 2 Mathematics Competition) | Analysis We immediately think of connecting $A C, A D$, because there are two right-angled triangles. But we also find that it is not easy to directly calculate the area of each triangle. Given the condition $B C+D E=1$, we wonder if we can join $B C$ to one end of $D E$, making $E F=B C$. Connect $A F$. Then we find $... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. As shown in the figure, let $D$ and $E$ be on the sides $AC$ and $AB$ of $\triangle ABC$, respectively. $BD$ and $CE$ intersect at $F$, $AE = EB$, $\frac{AD}{DC} = \frac{2}{3}$, and $S_{\triangle ABC} = 40$. Find $S_{\text{quadrilateral AEFD}}$. (6th National Junior High School Mathematics Correspondence Com... | Analyzing, first connect $A F$ and then set:
$$
S_{\triangle A E P}=x, S_{\triangle A P D}=y, S_{\triangle C D F}=z, S_{\triangle B E P}=t\left(S_{\triangle B C F}\right.
$$
$=u$ ), to find $S_{\text {quadrilateral } A E F D}$, we only need to find the values of $x$ and $y$.
We easily get:
$$
\begin{array}{l}
x+y+z=20,... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $S=\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$ that satisfy: (1) $\left|A_{i}\right|=5, i=1,2, \cdots, k_{i}$ (2) $\mid A_{i} \cap$ $A_{j} \mid \leqslant 2, i, j=1,2, \cdots, k, i \neq j$. Find the maximum value of $k$. (6th test, 1st question, provided by Wu Chang) | First, prove that each element in $S$ belongs to at most 3 of the sets $A_{1}, A_{2}, \cdots, A_{k}$.
By contradiction, suppose there is an element in $S$ that belongs to at least 4 subsets. Without loss of generality, assume $1 \in A_{1}, A_{2}, A_{3}, A_{4}$.
(1) If each of the other elements in $S$ belongs to at mo... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. It is known that a total of 12 theater groups are participating in a 7-day drama festival. Each day, some of the groups perform on stage, while the groups not performing watch from the audience. It is required that each group must see the performances of all other groups. What is the minimum number of performances n... | Certainly, here is the translation of the provided text into English, preserving the original formatting:
```
It is impossible for 3 troupes to perform only for 2 days and meet the conditions of the problem. By the pigeonhole principle, on one day, at least two troupes must perform, and on that day, these two troupes ... | 22 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
(1) (This question is worth 35 points) Let the set $A=\{1,2,3, \cdots, 366\}$. If a binary subset $B=\{a, b\}$ of $A$ satisfies 17|(a $+b$), then $B$ is said to have property $P$.
(1) Find the number of all binary subsets of $A$ that have property $P$;
(2) Find the number of all pairwise disjoint binary subsets of $A$ ... | (1) $a+b$ is divisible by 17 if and only if $a+b \equiv 0 \pmod{17}$, i.e., $\square$
$$
\begin{array}{l}
a \equiv k \pmod{17}, b \equiv 17-k \pmod{17}, \\
k=0,1,2, \cdots, 16 .
\end{array}
$$
Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: [0], [1], ..., [16]. Since $366=17 \time... | 3928 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. From the numbers $1,2, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, and $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways of selection are there that meet the conditions? (1989, National High School Competition) | Destruct the following model:
Take 10 identical white balls and arrange them in a row, take 5 identical black balls and divide them into 3 groups in the order of $2,2,1$, then insert them into the 10 gaps between the white balls, from left to right, excluding the left end but including the right end. There are $C_{10}^... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Among the 95 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 95^{2}$, the numbers with an odd digit in the tens place total $\qquad$.
untranslated part: $\qquad$ | In $1^{2}, 2^{2}, \cdots, 10^{2}$, it is found through calculation that the tens digit is odd only for $4^{2}=16, 6^{2}=36$. A two-digit square number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2},
$$
Therefore, $b$ can only be 4 or 6. That is, in every 10 consecutive numbers, there are two numbers whose ... | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $\alpha$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$, then the value of $\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}$ is. $\qquad$ . | 2. 20 .
Factorization yields
$$
\begin{array}{l}
\alpha^{3}-1=(\alpha-1)\left(\alpha^{2}+\alpha+1\right), \\
\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}=\alpha^{2}(\alpha-1)(\alpha+1)^{2} .
\end{array}
$$
$\alpha$ satisfies $\alpha^{2}+\alpha-\frac{1}{4}=0$, hence $\alpha \neq 1$.
$$
\frac{\alpha^{3}-1}{\alpha^{5}+\al... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ . | 3. 1.
The formula yields
$$
\begin{aligned}
y & =(x-1)^{2}+x+\frac{1}{x}-1 \\
& =(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1 .
\end{aligned}
$$
When $x=1$, both $(x-1)^{2}$ and $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}$ simultaneously take the minimum value of 0, so the minimum value of $y=x^{2}-x+\f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The "Monkey Cycling" performance in the circus is carried out by 5 monkeys using 5 bicycles, with each monkey riding at least once, but no monkey can ride the same bicycle more than once. After the performance, the 5 monkeys rode the bicycles $2, 2, 3, 5, x$ times respectively, and the 5 bicycles were ridden $1, 1, ... | ,- 1 (B).
Every time the monkey rides a bike, the monkey and the bike each get one count, so the number of times the monkey rides a bike = the number of times the bike is ridden, i.e., $2+2+3+5+x=1+1+2+4+y$. Simplifying, we get $x+4=y$. From $x \geqslant 1$, we get $y \geqslant 5$. Also, from the problem, $y \leqslant ... | 6 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
For each natural number $n$, the parabola $y=\left(n^{2}+n\right) x^{2}$
$-(2 n+1) x+1$ intersects the $x$-axis at points $A_{n}, B_{n}$. Let $\left|A_{n} B_{n}\right|$ denote the distance between these two points. Then $\left|A_{1} B_{1}\right|+\left|A_{2} B_{2}\right|+\cdots$ $+\left|A_{1995} B_{1995}\right|$ is $\qq... | $$
\begin{aligned}
& \text { II.1. } 1995 \\
& \because y=(n x-1)[(n+1) x-1], \\
x_{1}= & \frac{1}{n}, x_{2}=\frac{1}{n+1}, \\
& \therefore\left|A_{n} B_{n}\right|=\frac{1}{n}-\frac{1}{n+1} .
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
& \left|A_{1} B_{1}\right|+\left|A_{2} B_{2}\right|+\cdots+\left|A_{1005} B_{100... | 1995 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. The ice numbers $a_{1}, u_{2}, \cdots, a_{9}$ can only take two different values, +1 or -1. Then, the maximum value of the expression $a_{1} a_{5} a_{9}-a_{1} a_{6} a_{8}+a_{2} a_{8} a_{7}-$ $a_{2} a_{4} a_{9}+a_{3} a_{4} a_{8}-a_{3} a_{5} a_{7}$ is $\qquad$. | 4. 4 .
Since each term in the sum can only be 11 or -1, the sum is even (the parity of $a-b$ and $a+b$ is the same), so the maximum value of the sum is at most 6. It is easy to see that the sum cannot be 6, because in this case $a_{1} a_{5} a_{9}, a_{2} a_{6} a_{7}, a_{3} a_{6} a_{8}$ should all be +1, making their pr... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the function $f(x)$ have a domain and range both equal to $R$, and for any $a, b \in R$ there is $f[a f(b)]=a b$. Then the value of $|f(1995)|$ is $\qquad$ | Ni, i. 1995.
For the known functional equation, substituting $a=1, a=b, a=f(b)$, we have $f(f(b))=b, f(b f(b))=b^{2}, f\left[f^{2}(b)\right]=b f(b)$. Applying the function $f$ to the third equation, $f\left\{f\left[f^{2}(b)\right]\right\}=f(b f(b))=b^{2}$. Thus, from $f(J(b))=b$, we have $f^{2}(b)=b^{2}$, i.e., $|f(b)|... | 1995 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $f(x)=a x^{2}+b x+c(a, b, c \in R, a \neq 0)$. If for $|x| \leqslant 1$, $|f(x)| \leqslant 1$, then for $|x| \leqslant 1$, the maximum value of $|2 a x+b|$ is $\qquad$ | 4. 4 .
Given $f(0)=c, f(1)=a+b+c, f(-1)=a-b+c$,
we solve to get $a=\frac{f(1)+f(-1)-2 f(0)}{2}, b=\frac{f(1)-f(-1)}{2}$.
For $|x| \leqslant 1$, $|f(x)| \leqslant 1$, we have
$$
\begin{array}{l}
|2 a x+b| \\
=\left|[f(1)+f(-1)-2 f(0)] x\right. \\
\left.+\frac{f(1)-f(-1)}{2} \right| \\
=\left|\left(x+\frac{1}{2}\right)... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
(4) (Total 20 points) In the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $M$ is a point on $AB$, and $AM^2 + BM^2 + CM^2 = 2AM + 2BM + 2CM - 3$. If $P$ is a moving point on the segment $AC$, and $\odot O$ is the circle passing through points $P, M, C$, and a line through $P$ parallel to $AB$ intersects $\o... | (1) From the given, we have
$$
(A M-1)^{2}+(B M-1)^{2}+(C M-1)^{2}=0,
$$
which implies $A M=B M=C M=1$, meaning $M$ is the midpoint of $A B$.
(2) From $\left\{\begin{array}{l}M A=M C \Rightarrow \angle A=\angle M C A, \\ P D / / A B: \angle A=\angle C P D\end{array}\right.$
$\Rightarrow \angle M C A=\angle C P D$, thu... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five, (20 points) A mall installs an escalator from the first floor to the second floor, moving at a uniform speed that is twice the speed of a child. It is known that a boy passed 27 steps to reach the top of the escalator, while a girl walked 18 steps to reach the top. (Assume the boy and girl each step on one step a... | (1) Let the girl's speed be $x$ levels/min, the escalator's speed be $y$ levels/min, the boy's speed be $2x$ levels/min, and the stairs have $s$ levels, then
$$
\left.\begin{array}{l}
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y}
\end{array}\right. \\
\Rightarrow \frac{13.5}{18}=... | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. In the table below, the average of any three adjacent small squares in the upper row is $1(x \neq 0)$, and the square of any four adjacent small squares in the lower row is also 1. Then the value of $\frac{x^{2}+y^{2}+z^{2}-32}{y z+16}$ is $ـ$. $\qquad$ | ii. 3 .
From the problem, we know that in the above row, the numbers in every two small squares separated by one are equal, thus we can deduce
$$
\frac{x+7-y}{3}=1 \text {. }
$$
Similarly, for the row below, we get
$$
\begin{array}{l}
\frac{z-x+9.5-1.5}{4}=1 . \\
\begin{array}{l}
\therefore y=x+4, z=x-4, \text { then... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12. Is there a north-facing 100 petals arrangement, such that they exactly have 1985 intersections. | Solution 1: Since $x$ parallel lines and another set of $100 - x$ parallel lines can yield $x(100-x)$ intersection points, and
$$
x(100-x)=1985
$$
has no integer solutions, we can consider a grid formed by 99 lines. Since
$$
x(99-x)<1955
$$
has solutions $x \leqslant 26$ or $x \geqslant 73, x \in \mathbb{N}$, and
$$
... | 1985 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. In a tennis tournament, $n$ women and $2 n$ men participate, and each player plays against all other players exactly once. If there are no ties, the ratio of the number of games won by women to the number of games won by men is 7:5. Then $n=$
| 6. $n=3$.
Let $k$ represent the number of times the woman wins over Luo Ziyue.
Here, we set $\frac{k+\mathrm{C}_{0}^{2}}{\mathrm{C}_{\mathrm{jn}}^{2}}=\frac{7}{12}, k=\frac{7}{12} \mathrm{C}_{3 n}-\mathrm{C}_{n}^{2}$. From $k \leq 2 n^{2}$, we can solve to get $n \leq 3$.
When $n=1$, we get $k=\frac{7}{4}$; when $n=... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at $D$, $P$ is the midpoint of $AD$, $BP$ intersects $AC$ at $E$, $EF \perp BC$ at $F$. Given $AE=3$, $EC=$ 12. Find the length of $EF$. | Analysis: Consider $\triangle A B C$ as its machine triangle, given that $E$, $P$ are the fixed ratio points of $A C, A D$ respectively, hence from (*) we get $\left(1+\frac{C D}{D B}\right) \cdot \frac{3}{12}=1$, i.e., $\frac{C D}{D B}=3$.
From the projection theorem, we can find $B D=\frac{5}{2} \sqrt{3}, D C$ $=\fr... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Among the 35 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 35^{2}$, the numbers with an odd digit in the tens place are $\qquad$ in total. | 3. 7. Among $1^{2}, 2^{2}, \cdots, 9^{2}$, those with an odd tens digit are only $4^{2}=16, 6^{2}=36$.
The square of a two-digit number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2} \text {. }
$$
It is evident that, when the unit digit is $A$ and $\hat{C}$, the tens digit of the square is odd. That is, on... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\frac{a}{a^{3}+a^{2} b+6 b^{2}+b^{3}}+\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}} \\
+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$ | \[
\begin{aligned}
=、 \text { Original expression } & =\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\
& +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\
= & \frac{a^{2}+b^{2} .}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\
&... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Six, a boat sails between
$A, B$ two docks. When sailing downstream, it takes 40 minutes to be 4 kilometers away from the destination, and when sailing upstream, it needs $1 \frac{1}{3}$ hours to arrive. It is known that the upstream speed is 12 kilometers per hour. Find the speed of the boat in still water. | Six, let the speed of the current be $x$ kilometers/hour. Then
$$
\frac{2}{3} x+4=1 \frac{1}{3} \times 12 \text {. }
$$
Solving for $x$ gives $x=18$.
Therefore, the speed of the boat in the stream is
$$
(18+12) \div 2=15 \text { (kilometers/hour). }
$$
Answer: The speed of the boat in still water is 15 kilometers/hou... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the equation $2 x^{2}+2 k x-13 k+1=0$ has two real roots whose squares sum to 13. Then $k=$ $\qquad$ . | 1.1. By the relationship between roots and coefficients, we solve to get $k=1$ or $k=$ -14. Substituting these values back into the original equation, we find that $k=-14$ does not satisfy the conditions, so it is discarded; $k=1$ satisfies the conditions. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. A four-digit number has the following property: dividing this four-digit number by its last two digits yields a perfect square (if the tens digit is zero, then divide by the units digit), and this perfect square is exactly the square of the first two digits plus 1. For example, $4802 \div 2$ $=2401=49^{2}=(48+1)^{2}... | 5. 1085 Let this four-digit number be $100 c_{1}+c_{2}$, where $10 \leqslant c_{1} \leqslant 99, 1 \leqslant c_{2} \leqslant 99$. According to the problem, we have
$$
\begin{array}{l}
100 c_{1}+c_{2}=\left(c_{1}+1\right)^{2} c_{2}=c_{1}^{2} c_{2}+2 c_{1} c_{2}+c_{2}, \\
\therefore 100 c_{1}=c_{1} c_{2}\left(c_{2}+2\ri... | 1085 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
II. (30 points) Given positive numbers $m, n$ are the roots of the quadratic equation $x^{2}+$ $p x+q=0$, and $m^{2}+n^{2}=3, m n=1$. Find the value of the polynomial $x^{3}-(\sqrt{5}-1) x^{2}-(\sqrt{5}-1) x+1994$. | $$
=, \because m, n>0, \therefore m+n=5 .
$$
Thus, $p=-(m+n)=-\sqrt{5}, q=mn=1$,
$$
\therefore x^{2}-\sqrt{5} x+1=0 \text {. }
$$
$$
\text { Also } \begin{array}{l}
x^{3}-(\sqrt{5}-1) x^{2}-(\sqrt{5}-1) x+1994 \\
=\left(x^{3}+1\right)-(\sqrt{5}-1) x(x+1)+1993 \\
=(x+1)\left[\left(x^{2}-x+1\right)-(\sqrt{5}-1) x\right... | 1993 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that the area of $\triangle A B C$ is $1, D$ is the midpoint of $B C, E$, $F$ are on $A C, A B$ respectively, and $S_{\triangle B D P}=\frac{1}{5}, S_{\triangle C D E}=\frac{1}{3}$. Then $S_{\triangle D S P}=$ $\qquad$ | (7)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. You asked me to translate a text, but the text you provided is just a number in parentheses. If you need a ... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. The square of the first and last digits of a four-digit number adds up to 13, and the square of the middle two digits adds up to 85. If 1089 is subtracted from this number, the result is a four-digit number written with the same digits but in reverse order. The original four-digit number is $\qquad$ | 8. 3762 | 3762 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, (18 points) In a round-robin football tournament with $n$ teams (i.e., each team must play a match against every other team), each match awards 2 points to the winning team, 1 point to each team in the case of a draw, and 0 points to the losing team. The result is that one team has more points than any other team... | Let the team with the highest score be Team $A$, and assume Team $A$ wins $k$ games, draws $m$ games, and thus Team $A$'s total points are $2k + m$.
From the given conditions, every other team must win at least $k+1$ games, and their points must be no less than $2(k+1)$. Therefore, we have $2k + m > 2(k+1)$, which sim... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. If $m$ satisfies the equation
$$
\begin{array}{l}
\sqrt{3 x+5 y-2-m}+\sqrt{2 x+3 y-m} \\
=\sqrt{x-199+y} \cdot \sqrt{199-x-y},
\end{array}
$$
try to determine the value of $m$. | Analyzing this problem, we have one equation with three unknowns, so we can only solve it cleverly. Observing the characteristics of each algebraic expression in the original equation, we know that
$$
x-199+y \geqslant 0,
$$
and $199-x-y \geqslant 0$.
Thus, $x+y \geqslant 199$,
and $x+y \leqslant 199$.
Therefore, $x+y... | 201 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) There are both coins and banknotes for 1 fen, 2 fen, and 5 fen. To exchange one jiao into several 1 fen, 2 fen, and 5 fen coins or banknotes, how many different methods are there? | Three, suppose one dime can be exchanged for $x, y, z$ coins of 1 cent, 2 cents, and 5 cents respectively. Then we have
$$
x+2 y+5 z=10 \text {. }
$$
Solving this, we get $(x, y, z)=(0,0,2),(0,5,0),(1,2$,
$$
\begin{array}{l}
1),(2,4,0),(3,1,1)(4,3,0),(5,0,1),(6, \\
2,0),(8,1,0),(10,0,0) .
\end{array}
$$
Suppose $m$ c... | 134 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, here $a, b, c, d$ are real numbers. It is known that $f(1)=5, f(2)=10$, $f(3)=15$, then $f(8)+f(-4)$ is $(\quad)$.
(A) 2500
(B) uncertain
(C) 2540
(D) 860 | 4. (C).
$$
\begin{array}{c}
f(x)-5 x \\
=(x-1)(x-2)(x-3)(x+h), \\
f(8)-40=210(8+h), \\
f(-4)+20=-210(h-4), \\
\text { then, } f(8)+f(-4)=2540 .
\end{array}
$$ | 2540 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 12. Let $M, x, y$ be positive integers, and $\sqrt{M-\sqrt{28}}=\sqrt{x}-\sqrt{y}$. Then the value of $x+y+M$ is ( ).
(1994, Hope Cup Mathematics Competition) | Given $M-\sqrt{28}=(x+y)$ $-\sqrt{4 x y}$. Therefore, $x+y=M$, and $4 x y=28$. Since $x, y$ are positive integers, and it is specified that $x \geqslant y$, from $x y=7$, we get $x=7, y=1, M=8, x+y+M=16$. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Find the smallest positive integer $n$, such that in any 9-vertex and $n$-edge graph with edges colored in two colors, there must exist a monochromatic triangle.
(33rd IMO, Problem 3) | Since $K_{2}$ can have one edge removed and be colored with 0 colors to ensure it contains no monochromatic triangles, and $K_{5}$ can be colored with 2 colors to ensure it contains no monochromatic triangles. $K$ can be divided into 4 $K_{2}$'s and 1 $K_{1}$ with no common vertices, so in $K$, 4 edges can be removed a... | 33 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
The last digit is not 0, after removing one digit (but not the first digit) from it, it becomes a smaller number that is an integer multiple of the resulting number (i.e., the original number is an integer multiple of the resulting number -- translator's note).
Translate the above text into English, please retain ... | 57. 19. Let the original number be $b$ times the later number, i.e.,
$$
\begin{array}{l}
\overline{a_{k} \cdots a_{l+1}} \cdot 10^{t}+a_{t} \cdot 10^{t-1}+\overline{a_{i-1} \cdots a_{1}} \\
=b\left(\overline{a_{k} \cdots a_{l+1}} \cdot 10^{t-1}+\overline{a_{t-1} \cdots a_{1}}\right) .
\end{array}
$$
It is easy to see ... | 180625 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. During the New Year festival, the school bought a chocolate bar for each student. It is known that if the chocolate bars are packed 20 to a box, then 5 more boxes are used compared to when they are packed 24 to a box. How many students are there? | 3. If each tray contains 20 pieces and requires $x$ boxes per week, then, $20 x=24(x-5)$. Solving for $x$ gives $x=30$. Therefore, the number of students is $20 \times 30=600$. | 600 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. On the diagonal $BD$ of square $ABCD$, take two points $E$ and $F$, such that the extension of $AE$ intersects side $BC$ at point $M$, and the extension of $AF$ intersects side $CD$ at point $N$, with $CM = CN$. If $BE = 3$, $EF = 4$, what is the length of the diagonal of this square? | 3. From the condition, we get
$\triangle A^{\prime} B M I \triangle \triangle A C M$.
$\therefore \angle B E=\angle D A E$.
Also, $A B=A D$,
$\angle A B E=\angle A D F$,
$\therefore \triangle A B E \simeq \triangle A D F$.
Therefore, $D F=B E=3$.
$\therefore A C=B D=10$. | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8. Calculate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}$
$$
\begin{array}{l}
+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7}) \\
\cdot(-\sqrt{5}+\sqrt{6}+\sqrt{7}) .
\end{array}
$$ | Analyze the repeated application of the difference of squares formula to get the result as 104. | 104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. For positive $x$, we have $x^{2}+\frac{1}{x^{2}}=7$. Prove that $x^{5}+\frac{1}{x^{5}}$ is a number, and calculate the result. | 4. From the equation $\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2$ and $x+\frac{1}{x}=3$.
We get
$$
\begin{array}{l}
27:=\left(x+\frac{1}{x}\right)^{3} \\
=x^{3}+3 x^{2} \cdot \frac{1}{x}+3 x \cdot \frac{1}{x^{2}}+\frac{1}{x^{3}} \\
=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right) .
\end{array}
$$
Thus, ... | 123 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. Calculate $(\sqrt{1300}]+[\sqrt{1901}]+\cdots$ $+[\sqrt{1993}]=(\quad)$.
(1993, Hope Cup Mathematics Competition) | $$
\begin{array}{l}
\text { Analysis } \because 43<\sqrt{1900}, \sqrt{1901}, \cdots, \\
\sqrt{1935}<44,44 \leqslant \sqrt{1936}, \sqrt{1937}, \cdots, \\
\sqrt{1993}<45, \\
\therefore[\sqrt{1900}],[\sqrt{1901}], \cdots,[\sqrt{1935}] \text { are all } 43, \text { a total of 36, } [\sqrt{1936}],[\sqrt{1937}], \cdots, [\sq... | 4100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 3 .
From $[\lg x] \leqslant \lg x$, we get $\lg ^{2} x-\lg x-2 \leqslant 0$, which means $-1 \leqslant \lg x \leqslant 2$.
When $-1 \leqslant \lg x<0$, we have $[\lg x]=-1$. Substituting into the original equation, we get $\lg x= \pm 1$, but $\lg x=1$ is not valid, so $\lg x=-1, x_{1}=\frac{1}{10}$.
When $0 \leqs... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. In the Cartesian coordinate plane, the number of integer points (i.e., points with both coordinates as integers) that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y \leqslant 3 x, \\
y \geqslant \frac{1}{3} x, \\
x+y \leqslant 100
\end{array}\right.
$$
is | $$
1+2+3+\cdots+101=\frac{102 \times 101}{2}=5151 \uparrow .
$$
The region enclosed by $y=\frac{1}{3} x, x+y=100$, and the $x$-axis (excluding the boundary $y=\frac{1}{3} x$) contains
$$
\begin{aligned}
& (1+1+1+1)+(2+2+2+2)+\cdots \\
& +(25+25+25+25) \\
= & 4 \times(1+2+\cdots+25)=1300 \text { points. }
\end{aligned}... | 2551 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. Color each vertex of a square pyramid with one color, and make the endpoints of the same edge have different colors. If only 5 colors are available, the total number of different coloring methods is $\qquad$ | $5,4,20$.
Suppose, for the pyramid $S-ABCD$, the vertices $S, A, B$ are colored with different colors, and they have a total of $5 \times 4 \times 3=60$ coloring methods.
When $S, A, B$ are already colored, assume their colors are $1, 2, 3$ respectively; if $C$ is colored with color 2, then $D$ can be colored with one... | 420 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $M=\{1,2,3, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: if $x \in A$, then $15 x \notin A$. Then the maximum number of elements in $A$ is $\qquad$ . | 6. 1870.
Let $n(A)$ denote the number of elements in the set $A$.
By the problem's setup, at least one of the numbers $k$ and $15k$ (where $k=9,10, \cdots, 133$) does not belong to $A$, so at least $125(=133-9+1)$ numbers do not belong to $A$, i.e., $n(A) \leqslant 1995-125=1870$. On the other hand, we can take $A=\{1... | 1870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
II. (This question is worth 25 points) Find all real numbers $p$ such that the cubic equation $5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p$ has three roots that are all natural numbers. | From observation, it is easy to see that $x=1$ is a root of the original cubic equation. By synthetic division, the original cubic equation can be reduced to a quadratic equation:
$$
5 x^{2}-5 p x+66 p-1=0 \text {. }
$$
The three roots of the original cubic equation are natural numbers, and the two roots of the quadra... | 76 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Find the smallest prime $p$ that cannot be expressed as $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers. | It has been verified that $2,3,5,7,11,13,17,19,23,29,31$, and 37 can all be expressed in the form $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers:
$$
\begin{array}{l}
2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, \\
7=2^{3}-3^{0}, 11=3^{3}-2^{4}, 13=2^{4}-3^{1}, \\
17=3^{4}-2^{6}, 19=3^{3}-2^{3},... | 41 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. 21 people participate in an exam, with the test paper containing 15 true/false questions. It is known that for any two people, there is at least one question that both of them answered correctly. What is the minimum number of people who answered the most questions correctly? Please explain your reasoning. | For the $i$-th problem, $a_{i}$ people answered correctly, thus there are exactly
$$
b_{i}=C_{s_{i}}^{z}
$$
pairs of people who answered the $i$-th problem correctly $(i=1,2, \cdots, 15)$.
Below, we focus on the sum $\sum_{i=1}^{15} b_{i}$.
Let $a=\max \left\{a_{1}, a_{2}, \cdots, a_{15}\right\}$, then we have
$$
15 C... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $S=\left\{A=\left(a_{1}, \cdots, a_{8}\right) \mid a_{i}=0\right.$ or $1, i=1, \cdots$, 8\}. For two elements $A=\left(a_{1}, \cdots, a_{8}\right)$ and $B=\left(b_{1}\right.$, $\cdots, b_{8}$ ) in $S$, denote
$$
d(A, B)=\sum_{i=1}^{\delta}\left|a_{i}-b_{i}\right|,
$$
and call it the distance between $A$ and $B$... | Solution One
(I) First, we point out that the sum of the weights of any two codewords in $\mathscr{D}$ does not exceed 11; otherwise, if
$$
\omega(X)+\omega(Y) \geqslant 12,
$$
then because $12-8=4$, these two codewords must share at least four positions with 1s, and the distance between them
$$
a^{\prime}\left(X, y^{... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}(n>4)$ where all interior angles are integer multiples of $15^{\circ}$, and $\angle A_{1}+\angle A_{2}+\angle A_{3}=$ $285^{\circ}$. Then, $n=$ | 2. 10 (Hint: The sum of the $n-3$ interior angles other than the male one is $(n-2) \cdot$ $180^{\circ}-285^{\circ}$, it can be divided by $n-3$, and its quotient should also be an integer multiple of $15^{\circ}$.） | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. A football made of leather in black and white colors is shown as follows. It is known that this football has 12 black leather pieces. Therefore, this football has white leather pieces.
保留源文本的换行和格式，直接输出翻译结果。 | 4. 20 (Hint: Each black piece is a pentagon, and the total number of edges for 12 pieces is: $5 \times 12=60$ (edges). Each white piece is a hexagon, and three of its edges are adjacent to the edges of black pieces. That is, three edges of black pieces determine one white piece. Therefore, the number of white pieces is... | 20 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. For real numbers $x, y$, define a new operation: $x * y=a x+b y+$ $c$, where $a, b, c$ are constants, and the right side of the equation is the usual addition and multiplication operations. It is known that $3 * 5=15,4 * 7=28$. Then, 1*1= $\qquad$ | 7. -11
Brief solution: According to the definition, we know that $3 * 5=3a+5b+c=15, 4 * 7$ $=4a+7b+c=28$, and $1 * 1=a+b+c=3(3a+5b+$c) $-2(4a+7b+c)=3 \times 15-2 \times 28=-11$. | -11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (12 points) Given points $P_{1}\left(x_{1}, 1994\right), P_{2}\left(x_{2}, 1994\right)$ are two points on the graph of the quadratic function $y=a x^{2}+b x+7(a \neq 0)$. Try to find the value of the quadratic function $y=x_{1}+x_{2}$. | Three, since $P_{1}, P_{2}$ are two points on the graph of a quadratic function, we have
$$
\begin{array}{l}
a x_{1}^{2}+b x_{1}+7=1994, \\
a x_{2}^{2}+b x_{2}+7=1994 .
\end{array}
$$
Subtracting the two equations and simplifying, we get
$$
\left(x_{1}-x_{2}\right)\left[a\left(x_{1}+x_{2}\right)+b\right]=0 .
$$
Since... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. (India) The sequence of positive integers $\left\{f_{n}\right\}_{m=1}^{m}$ is defined as follows:
$f(1)=1$, and for $n \geqslant 2$,
$$
f(n)=\left\{\begin{array}{l}
f(n-1)-n, \text { when } f(n-1)>n ; \\
f(n-1)+n, \text { when } f(n-1) \leqslant n .
\end{array}\right.
$$
Let $S=\{n \in \mathbb{N} \mid f(n)=1993\}$.... | (i) We point out that if $f(n)=1$, then $f(3n+3)=1$. From $f(n)=1$, we can sequentially obtain according to the definition:
$$
\begin{array}{l}
f(n+1)=n+2, f(n+2)=2n+4, \\
f(n+3)=n+1, f(n+4)=2n+5, \\
f(n+5)=n, \quad f(n+6)=2n+6 \\
f(n+7)=n-1, \cdots
\end{array}
$$
It is evident that the sequence $f(n+1), f(n+3), f(n+5... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Nephew 8. Find the number of lattice points inside and on the boundary of the triangle formed by the line $y=\frac{2}{3} x-\frac{1}{2}, x=10$ and the $x$-axis. | Given the function $y=\frac{2}{3} x-\frac{1}{2}$, when $x$ takes the values $1,2, \cdots$, 10 (note: when $x=\frac{3}{4}$, $y=0$), the corresponding $y$ values are $\frac{1}{6}, \frac{5}{6}, \frac{3}{2}, \frac{13}{6}, \frac{17}{6}, \frac{7}{2}, \frac{25}{6}, \frac{29}{6}, \frac{11}{2}, \frac{37}{6}$. The number of inte... | 37 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. In $\triangle A B C$, $\angle C=3 \angle A, a=27, c=48$. Then $b$ $=$ ـ $\qquad$ | 4. 35 .
As shown. Draw the trisectors of $\angle C$ as $C D$ and $C E$.
$$
\begin{array}{l}
\text { Let } C D=1, A N=x, \\
A E=E C=y .
\end{array}
$$
From $\triangle C D B \sim \triangle A C B$, we have
$$
\begin{array}{l}
\frac{a}{c}=\frac{t}{b}=\frac{c-x}{a}, \\
t=\frac{a b}{c}, x=\frac{c^{2}-a^{2}}{c} .
\end{array... | 35 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, from the 1995 natural numbers $1,2,3, \cdots$, 1995, remove some numbers so that in the remaining numbers, no number is equal to the product of any two other numbers. How many numbers must be removed at a minimum to achieve this? | (1) First, prove: When any 42 numbers are removed, there will be three numbers among the remaining numbers, such that the product of two of them equals the third.
We consider the array $(2,87,2 \times 87),(3,86,3 \times 86)$, $\cdots,(44,5,44 \times 45)$,
i.e., $(x, 89-x, x(89-x))(2 \leqslant x \leqslant 44)$.
$$
\beg... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $[x]$ denotes the greatest integer not exceeding $x$, $a, b, c \in$ $R^{+}, a+b+c=1$, let $M=\sqrt{3 a+1}+\sqrt{3 b+1}+$ $\sqrt{3 c+1}$. Then the value of $[M]$ is $($. | 5. B.
Given $a \in(0,1), b \in(0,1), c \in(0,1) \Rightarrow a^{2} & \sqrt{a^{2}+2 a+1}+\sqrt{b^{2}+2 b+1} \\
& +\sqrt{c^{2}+2 c+1} \\
= & (a+b+c)+3=4 .
\end{aligned}
$$
And $M=\sqrt{(3 a+1) \cdot 1}+\sqrt{(3 b+1) \cdot 1}$
$$
\begin{aligned}
& +\sqrt{(3 c+1) \cdot 1} \\
& \frac{(3 a+1)+1}{2}+\frac{(3 b+1)+1}{2}+\frac... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Given that there are exactly 600 triangles with integer sides, all of different lengths, and the longest side is exactly $n$. Find $n$.
The longest side of the triangle is $n$, and the other two sides are $a$ and $b$ with $a < b < n$. The triangle inequality theorem states that:
1. $a + b > n$
2. $a + n > b$
3. $b... | Analysis: Since one side of the triangle is fixed at length $n$, and the three side lengths $x, y, n$ are all integers, the number of such triangular shapes corresponds one-to-one with the lattice points $(x, y)$ in the coordinate plane. Therefore, the lattice point method can be used to solve this problem.
Let the le... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the functions $y=2 \cos \pi x(0 \leqslant x \leqslant 2)$ and $y=2(x \in$ $R$ ) whose graphs enclose a closed plane figure. Then the area of this figure is $\qquad$ . | 2. As shown in the figure, by symmetry, the area of $CDE$ = the area of $AOD$ + the area of $BCF$, which means the shaded area is equal to the area of square $OABF$, so the answer is 4. | 4 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
6. Now we have 1 one-tenth yuan note, 1 one-tenth yuan note, 1 six-tenth yuan note, 5 one yuan notes, and 2 two yuan notes. The number of different amounts (excluding the case of not paying) that can be made is $\qquad$ | 6. Using: Jiao Shi Ji to match 0 Jiao, 1 Jiao, 2 Zhong different denominations. That is, $(1+i)$ kinds.
Using 5 Jiao can match 0 Jiao, 5 Jiao, which is $(1+1)$ kinds; Yuan. That is, $(5+1)$ kinds;
Using 5 Yuan can match 0 Yuan, 5 Yuan, 10 Yuan, which is $(2+1)$ kinds.
Considering the amounts to be paid are 5.5, 1.5, 2... | 127 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. If the natural numbers $a, x, y$ satisfy $\sqrt{a-2 \sqrt{6}}=\sqrt{x} -\sqrt{y}$, then the maximum value of $a$ is $\qquad$. | $$
=, 1, a_{\max }=7 \text {. }
$$
Squaring both sides of the known equation yields $a-2 \sqrt{6}=x+y-$ number, it must be that $x+y=\alpha$ and $x y=6$. From the known equation, we also know that $x>y$, so it can only be $x=6, y=1$ or $x=3, y=2$. Therefore, $a=7$ or 2, the maximum value is 7. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For the quadratic function $y=x^{2}-4 a x+5 a^{2}$ $-3 a$ in terms of the independent variable $x$, the minimum value $m$ is a function of $a$, and $a$ satisfies the inequality $0 \leqslant$ $a^{2}-4 a-2 \leqslant 10$. Then the maximum value of $m$ is $\qquad$ $ـ$ | 2. 18 .
Solving $0 \leqslant a^{2}-4 a-2 \leqslant 10$ yields $-2 \leqslant a \leqslant 2-\sqrt{6}$ or $2+\sqrt{6} \leqslant a \leqslant 6$. Also, $m=a^{2}-3 a$, considering the quadratic function, $m=a^{2}-3 a$ when $-2 \leqslant a \leqslant 2-\sqrt{6}$ and $2+\sqrt{6} \leqslant a \leqslant 6$, we can see that the ma... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Person A and Person B start from the same point $A$ on a circular track at the same time and run in opposite directions. Person A's speed is $5 \mathrm{~m}$ per second, and Person B's speed is $7 \mathrm{~m}$ per second. They stop running when they meet again at point $A$ for the first time. During this period, they... | 3. 12 .
Two $\lambda \mu A$ start from $A$ and meet once (not at point $A$). At the moment of their first meeting, the two have run a certain distance. From the ratio of their speeds, it can be deduced that person A has run $\frac{5}{12}$ of a lap, and person B has run $\frac{7}{12}$ of a lap. Therefore, after the $n$... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
ニ、(Full score 25 points) Write 1995 equations of the form $A_{i} x^{2}+B_{i} x+C_{i}=0(i=1,2, \cdots, 1995)$ on the blackboard. Two people, A and B, take turns to play a game. Each time, only one non-zero real number is allowed to replace one of the $A_{i}, B$ or $C$ in the $n$ equations $(i=1,2, \cdots, 1995)$. Once a... | Second, the answer is that Jia can get at most 998 rootless equations.
If Jia first fills in the coefficient of the linear term $B_{1}$ with 1, and tries to fill in the linear term coefficient of each equation with 1 as much as possible, at this time, Jia has at least $\frac{1995+1}{2}=998$ chances.
If Yi fills in the... | 998 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The Chinese Football League A has 12 football clubs participating, with a home and away double round-robin system, meaning any two teams play each other twice, once at home and once away. A win earns 3 points, a draw earns 1 point each, and a loss earns 0 points. At the end of the league, teams are ranked according ... | $=1.46$ points.
Suppose the score difference between the $k$-th and the $(k+1)$-th is the largest. Since the total score of each match between any two teams is at most 3 points and at least 2 points, the maximum total score of the first $k$ teams playing against each other is $3 \cdot 2 C_{k}^{2}=6 C_{k}^{2}$. Addition... | 46 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In the plane, a parabola divides the plane into two parts, two parabolas can divide the plane into at most seven parts. Then 10 parabolas can divide the plane into $\qquad$ parts. | 3. 191 .
Let $n$ parabolas divide the plane into $a_{n}$ regions. When adding another parabola to maximize the number of regions, this new parabola intersects each of the $n$ given parabolas at 4 points, thus the $n+1$ parabolas are divided into $4n+1$ segments, each of which splits an original region into two. This l... | 191 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Consider the hyperbola \((x-2)^{2}-\frac{y^{2}}{2}=1\). A line \(l\) is drawn through the right focus of the hyperbola, intersecting the hyperbola at points \(A\) and \(B\). If \(|A B|=4\), then the number of such lines is \(\qquad\). | 4. 3 lines.
The coordinates of the right focus are $(2+\sqrt{3}, 0)$. The length of the chord perpendicular to the real axis through the right focus is exactly 4. The length of the chord $d$ formed by passing through the right focus and intersecting both branches of the hyperbola has the range $d>2$, so there are two ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
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