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7. If $A$ is the sum of the absolute values of all roots of the equation
$$
x=\sqrt{19}+\frac{91}{\sqrt{19}+\cdots \sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}
$$
Find $A^{2}$.
|
7. The fraction to the right of the equation can be simplified to $\frac{a x+b}{c x+d}$ $(a, b, c, d$ are real numbers), then the equation is a quadratic equation, with at most two roots. H $x=\sqrt{19}+\frac{91}{x}$ has roots that are also $x=\frac{\sqrt{19}-\sqrt{38}}{2}$, which are the only two roots of the original equation. Therefore, $A=\sqrt{383}$, i.e., $A^{2}=383$.
|
383
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. How many real numbers $a$ are there such that $x^{2}+a x+6 a$ $=0$ has only integer solutions?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note at the end is not part of the translation but is provided to clarify that the format and structure of the original text have been maintained in the translation.
|
8. Let $x^{2}+a x+6 a=$ O have integer solutions $m, n$ $(m \leqslant n)$. Then we have
$$
a=-(m+n), \quad 6 a=m n .
$$
Since $a$ must be an integer, we also have
$$
-6(m+n)=m n \text { . }
$$
H $(m+6)(n+6)=36$. Since $36=36 \times 1=18 \times 2=12 \times 3=6 \times 6$, the solutions satisfying $m \leqslant n$ are
$$
\begin{array}{l}
(-42,-7),(-24,-8),(-18,-9), \\
(-15,-10),(-12,-12),(-5,30), \\
(-4,12),(-3,6),(-2,3),(0,0) .
\end{array}
$$
The corresponding values of $a=-(m+n)$ are $49,32,27,25$,
$$
{ }^{2} 4,-25,-8,-3,-1,0 \text {. A total of 10. }
$$
|
10
|
Other
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. If $\sec x+\tan x=\frac{22}{7}, \csc x+\cot x$ $=\frac{m}{n}$, where $\frac{m}{n}$ is a reduced fraction, find $m+n$.
|
$$
\begin{array}{l}
\text { 9. } \because \sec ^{2} x-\tan ^{2} x=1 \text {, let } p=\frac{22}{7}, \\
\therefore \sec x-\tan x=\frac{1}{p}, \sec x+\tan x=p .
\end{array}
$$
Adding these two equations, we get $2 \sec x=p+\frac{1}{p}$. Subtracting them, we get
$$
\begin{aligned}
2 \tan x & =p-\frac{1}{p} . \\
& \therefore \cos x=\frac{2 p}{p^{2}+1}, \sin x=\frac{p^{2}-1}{p^{2}+1} . \\
& \therefore \csc x+\cot x=\frac{p+1}{p-1}=\frac{2}{15} \\
& \text { then } m+n=29+15=44 .
\end{aligned}
$$
|
44
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. For the string of three letters $\hat{\mathrm{i}}$ "aaa" and "bbb", they are transmitted through a circuit, with each string being sent one letter at a time. Due to issues with the line, each of the 6 letters has a $\frac{1}{3}$ chance of being transmitted incorrectly (an $a$ is received as a $b$, or a $b$ is received as an $a$), and the correctness of each letter's reception is independent of the others. Let $S_{\mathrm{a}}$ be the message received when "aaa" is transmitted, and $S_{\mathrm{b}}$ be the message received when "bbb" is transmitted. Let $P$ be the probability that $S_{\mathrm{a}}$ is lexicographically before $S_{\mathrm{b}}$. When $P$ is written as a reduced fraction, what is the numerator?
|
$x_{1}=a, y_{1}=b, i=1,2,3$.
Let $S_{1} \propto S_{2}$ represent that $S_{1}$ is lexicographically before $S_{2}$, and $P\left(S_{1} \propto S_{2}\right)$ represent the probability that $S_{1}$ is lexicographically before $S_{2}$. Due to the independence of the correct transmission of letters, we have
$$
\begin{array}{c}
P\left(S, \propto S_{\mathrm{b}}\right)=P\left(x_{1} x_{2} x_{3} \propto y_{1} y_{2} y_{\mathrm{s}}\right) \\
=P\left(x_{1} \propto y_{1}\right)+P\left(x_{1}=y_{1}, x_{2} \propto y_{2}\right) \\
+P\left(x_{1}=y_{1}, x_{2}=y_{2}, x_{3} \propto y_{3}\right) \\
=P\left(x_{1} \propto y_{1}\right)+P\left(x_{1}=y_{1}\right) P\left(x_{2} \propto y_{2}\right) \\
+P\left(x_{1}=y_{1}\right) P\left(x_{2} \propto y_{2}\right) P\left(x_{3} \propto y_{3}\right) .
\end{array}
$$
Since the necessary and sufficient condition for $x_{1} \propto y_{1}$ is $x_{1}=a, y_{1}=b$,
we have $P\left(x_{1} \propto y_{1}\right)=\frac{2}{3} \cdot \frac{2}{3}=\frac{4}{9}$.
Similarly, $P\left(x_{2} \propto y_{2}\right)=P\left(x_{3} \propto y_{3}\right)=\frac{4}{9}$.
Also, the necessary and sufficient condition for $x_{1}=y_{1}$ is that one transmission is correct and the other is incorrect.
Thus, $P\left(x_{1}=y_{1}\right)$
$$
\begin{array}{l}
=P\left(x_{1}=y_{1}=a\right)+P\left(x_{1}=y_{1}=b\right) \\
=\frac{2}{3} \cdot \frac{1}{3}+\frac{1}{3} \cdot \frac{2}{3}=\frac{4}{9}
\end{array}
$$
Similarly, $P\left(x_{2}=y_{2}\right)=\frac{4}{9}$.
Therefore, $P\left(S_{\mathrm{a}} \propto S_{\mathrm{b}}\right)$
$$
=\frac{4}{9}+\left(\frac{4}{9}\right)^{2}+\left(\frac{4}{9}\right)^{3}=\frac{532}{729} \text {. }
$$
Hence, the required number is 532.
|
532
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. As shown in the figure, 12 congruent disks are placed on the circumference of a circle $C$ with radius 1, such that they cover $C$. No two disks overlap, and adjacent disks are tangent to each other. The sum of the areas of these disks can be written as
$$
\pi(a-b \sqrt{c}) \text {, where }
$$
$a, b, c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$.
|
11. By the people's knowledge, 6-certainly passes through the tangency points of each circle, and each tangent line passes through the center $O$ of circle $C$ (as shown in the figure).
$P$ is the center of a circle,
$$
\begin{array}{l}
\angle P B O=90^{\circ}, \\
\angle B O A=\frac{1}{12} 2 \pi=\frac{\pi}{6}, \\
\angle P O B=\frac{\pi}{12} .
\end{array}
$$
Therefore, $P B=B O$
- $\operatorname{tg} \angle P O B=\operatorname{tg} \frac{\pi}{12}$
$$
\begin{array}{l}
=\operatorname{tg}\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=\frac{\ln \frac{\pi}{3} \cdots \operatorname{tg} \frac{\pi}{4}}{1+\operatorname{tg} \frac{\pi}{3} \operatorname{tg} \frac{\pi}{4}} \\
=\frac{\sqrt{3}-1}{1+\sqrt{3}}=2-\sqrt{3} .
\end{array}
$$
Thus, the area of each disk is $\pi(2-\sqrt{3})^{2}=$ $\pi(7-4 \sqrt{ } \overline{3})$. The sum of the areas of these disks is
$$
\begin{array}{l}
12 \pi(7-4 \sqrt{3})=\pi(84-48 \sqrt{3}) . \\
\text { Hence } a+b+c=84+48+j=135 .
\end{array}
$$
|
135
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Rhombus $P Q R S$ is inscribed in rectangle $A B C D$, such that $P, Q, R, S$ are interior points on $\overline{A B}, \overline{B C}, \overline{C D}, \overline{D A}$. Given that $P B=15, B Q=20, P R=30, Q S$ $=40$. If the reduced fraction $\frac{m}{n}$ is the perimeter of rectangle $A B C D$, find $m+n$.
|
12. As shown in the figure, let $x, y$ represent the lengths of $Q C, R C$ respectively. By symmetry, the lengths of $SA, PA$ are also $x, y$.
The diagonals of the rectangle intersect at the center H of the rectangle, and they are perpendicular to each other, thus forming eight right triangles. Six of these right triangles have side lengths of $15, 20, 25$, and the other two triangles ($\triangle A S P$, $\triangle C Q R$) have side lengths of $x, y, 25$.
The sum of the areas of these eight parts is
$$
6 \times(150)+2\left(\frac{1}{2} x y\right)=S_{\triangle B C D}=(20+x)(15+y) \text {, }
$$
Thus, $3 x+4 y=120$.
Also, $\because x^{2}+y^{2}=625$,
$\therefore 5 x^{2}-144 x+880=0$.
That is, $(5 x-44)(x-20)=0$.
When $x=20$, it leads to $B C=40$, which contradicts $P R=30$. Therefore, $x=\frac{44}{5}, y=\frac{117}{5}$.
Thus, the perimeter of $A B C D$ is
$$
2(15+20+x+y)=\frac{672}{5} \text {. }
$$
Therefore, $m+n=677$.
|
677
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. In a drawer, there are red and blue socks, no more than 1991 in total. If two socks are drawn without replacement, the probability that they are the same color is $\frac{1}{2}$. How many red socks can there be at most in this case?
|
13. Let the number of red and black balls be $x$ and $y$ respectively. Given that the probability of randomly picking two balls of different colors is $\frac{1}{2}$, we have $\frac{x y}{C_{x+y}^{2}}=\frac{1}{2}$.
Therefore, $(x+y)(x+y-1)=4 x y$,
Thus, $(x-y)^{2}=x+y$.
Hence, the total number of balls is a perfect square. Let $n=x-y$, then
$$
\begin{array}{l}
n^{2}=x+y, \quad \therefore x=\frac{n^{2}+n}{2}, \\
\because x+y \leqslant 1991, \quad \therefore|n| \leqslant \sqrt{1991}<45 .
\end{array}
$$
When $n=44$, $x$ reaches its maximum value of 990.
|
990
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. For a positive integer $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value. Where $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ are positive integers, and their sum is 17. There is a unique value of $n$ for which $S_{n}$ is an integer, find $n$.
|
15. Consider each $l_{i}=\sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the hypotenuse of a right-angled triangle, with the two legs being $2 k-1$ and $a_{b}$. When these right-angled triangles are placed together to form a ladder, let $A, B$ be the starting and ending points, respectively.
The distance from $A$ to $B$ is
$$
\begin{array}{l}
\left(\sum_{k=1}^{n} a_{b}\right)^{2}+\left(\sum_{k=1}^{n}(2 h-1)\right)^{2} \\
=\sqrt{17^{2}+n^{4}}
\end{array}
$$
Let $\sum_{k=1}^{a} t_{i}=\sqrt{17^{2}+n^{4}}$. And we can choose $a_{h}$ such that $S_{n}=\sqrt{17^{2}+n^{4}}$. When $S_{n}$ is an integer, then
$$
\begin{array}{l}
17^{2}=S_{u}^{2}-n^{4}=\left(S_{n}-n^{2}\right)\left(S_{u}+n^{2}\right) . \\
\text { By }\left\{\begin{array}{l}
S_{u}+n^{2}=17^{2} \\
S_{n}-n^{2}=1,
\end{array} \text { we solve to get } S_{n}=1+5, n=12\right. \text {. }
\end{array}
$$
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Let $x$ be a cube root of 1 different from 1, find the value of $x^{\text {D}}$ $+x^{2}$. $(n \in N)$
|
Solve: From $x^{3}=1$, i.e., $x^{3}-1=0$, which is also $(x-1)\left(x^{2}+x+1\right)=0$.
Since $x \neq 1$, it follows that $x^{2}+x=-1$.
When $n=3k$,
$$
x^{\mathrm{n}}+x^{2 n}=\left(x^{3}\right)^{k}+\left(x^{3}\right)^{2 k}=2 \text {; }
$$
When $n=3 k+1$,
$$
\begin{array}{l}
x^{\mathrm{n}}+x^{2 \mathrm{n}}=\left(x^{3}\right)^{k} \cdot x+\left(x^{3}\right)^{2 k} \cdot x^{2} \\
=x+x^{2}=-1 ;
\end{array}
$$
When $n=3 k+2$,
$$
\begin{array}{l}
x^{n}+x^{20} \\
=\left(x^{3}\right)^{1} \cdot x^{2}+\left(x^{3}\right)^{2 k} \cdot x^{3} \cdot x \\
=x^{2}+x=-1
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If two real-coefficient quadratic equations in $x$, $x^{2}+x+a=0$ and $x^{2}+a x+1=0$, have at least one common real root, then $a=$ $\qquad$
|
1. $-2 ;$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, there are 1990 points distributed on a straight line. We will mark the midpoints of all possible line segments with these points as endpoints. Try to find the minimum number of distinct midpoints that can be obtained.
untranslated text:
直线上外布着1990个点, 我们来标出以这些点为端点的一切可能的线段的 中点. 试求至少可以得出多少个互不重合的中点.
translated text:
There are 1990 points distributed on a straight line. We will mark the midpoints of all possible line segments with these points as endpoints. Try to find the minimum number of distinct midpoints that can be obtained.
|
Four, Solution: Let the two points that are farthest apart be denoted as $A, B$. Consider the 1988 line segments formed by $A$ and the other 1988 points excluding $B$. The midpoints of these 1988 line segments are all distinct, and their distances to point $A$ are less than $\frac{1}{2} A B$.
Similarly, for point $B$, we can obtain another 1988 distinct midpoints, which are all less than $\frac{1}{2} A B$ away from point $B$.
Including the midpoint of $A B$, we have a total of $1988 + 1988 + 1 = 3977$ distinct midpoints.
On the other hand, when the 1990 points are equally spaced, the number of distinct midpoints is exactly 3977.
In conclusion, the minimum number of distinct midpoints of all possible line segments is 3977.
|
3977
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. For $\triangle A B C=$ with sides $a, b, c$, construct squares outward on each side, with areas sequentially $S_{a}, S_{b}, S_{c}$. If $a+b+c=18$, find the minimum value of $S_{\mathrm{a}}+S_{\mathrm{b}}+S_{\mathrm{c}}$.
|
Given $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ $\geqslant 0$, which means $2\left(a^{2}+b^{2}+c^{2}\right)=2(a b+b c+c a)$. Therefore, $3\left(a^{2}+b^{2}+c^{2}\right) \geqslant a^{2}+b^{2}+c^{2}+2 a b$ $+2 a c+2 b c=(a+b+c)^{2}$. Thus, $S_{1}+S_{v}+S_{r} \geqslant \frac{18^{2}}{3}=108$.
When $a=b=c$, the equality holds, so the minimum value is 108.
This is the method of inequality.
|
108
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The teacher distributed a stack of books among A, B, C, L, E. $\frac{1}{4}$ of the books were given to $B$, then $\frac{1}{3}$ of the remaining books were given to $C$, and the rest were split equally between $D$ and $E$. If $E$ received 6 books, then the teacher originally had $\qquad$ books.
|
2. 48
|
48
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. (Zu Chongzhi Cup Mathematics Competition) The sum of all digits of the natural numbers $1, 2, 3, \ldots, 9999$ is $\qquad$
|
Solve $0+9999, 1+9998, \cdots, 4999+5000$
all sum to 9999 without any carry, so the sum of digits is more:
$$
4 \times 9 \times 5000 \div 2-5=89995 \text {. }
$$
This is called the method of integer value combination.
|
89995
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six, in all integers that start and end with 1 and alternate between 1 and 0 (i.e., $101$, $10101$, $1010101$, etc.), how many of them are prime numbers?
|
Six, Proof that there is only one prime number 101.
If $n \geqslant 2$, then $A=10^{2 n}+10^{2 n-2}+\cdots+10^{2}$
$$
+1=\frac{\left(10^{\mathrm{n}+1}-1\right)\left(10^{\mathrm{n}+1}+1\right)}{99} \text {. }
$$
When $n=2 m+1$, $\frac{10^{2 m+2}-1}{99}=10^{2 \mathrm{~m}}$ $+\cdots+10^{2}+1, A$ is a composite number;
When $n=2 m$, $9 \mid\left(10^{n+1}-1\right)$,
$11 \mid\left(10^{\mathrm{n}+1}+1\right), A$ is a composite number.
|
101
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $[x]$ denote the greatest integer not exceeding $x$. If
$$
\begin{array}{l}
f=[1 \mathrm{~g} 1]+[1 \mathrm{~g} 2]+[1 \mathrm{~g} 3]+\cdots+[1 \mathrm{~g} 1989] \\
+[1 \mathrm{~g} 1990], \text { then } f=
\end{array}
$$
|
3. $4863 ;$
|
4863
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. (Shanghai Mathematics, 1987)
$$
\log _{3}\left[(3+1)\left(3^{2}+1\right)\left(3^{4}+1\right) \cdot \ldots\right.
$$
- $\left.\left(3^{84}+1\right)+\frac{1}{2}\right]+\log _{3} 2$ The value is ( ).
(A) 32.
( B) 64.
(C) 128.
(D) None of the above.
|
$$
\begin{aligned}
\text { Original expression }= & \log _{3}(3-1)(3+1)\left(3^{2}+1\right) \\
& \cdot \cdots \cdot\left(3^{84}+1\right) \\
= & \log _{3} 3^{128}=128 .
\end{aligned}
$$
Expressing "3-1" as 2 in the second line is a kind of simplification. This can be referred to as the method of adding factors.
|
128
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Two cars start from the same location at the same time, driving in the same direction at the same speed. Each car can carry a maximum of 24 barrels of gasoline. They cannot use any other fuel during the journey. Each barrel of gasoline can make a car travel 60 kilometers. Both cars must return to the starting point, but they do not have to return at the same time. They can lend fuel to each other. To make one of the cars travel as far away from the starting point as possible, at what distance from the starting point should the other car turn back? How many kilometers in total has the car that travels the farthest traveled?
|
If so, A gives B $(24-2 x)$ barrels of gasoline. B continues to move forward, carrying $(24-2 x)+(24-x)=48-3 x$ barrels of gasoline. According to the problem, we should have
$$
48-3 x \leqslant 24 \text {. That is, } x \geqslant 8 \text {. }
$$
After A and B separate, the distance B continues to travel is
$$
\begin{array}{l}
S(x)=-\frac{(24-2 x)+(24-2 x)}{2} \cdot 60 \\
=30(48-4 x) \text {. } \\
\end{array}
$$
Therefore, when $x=8$, $S(x)$ is maximized $=S(8)=480$ (km).
Thus, the total distance B travels is $2[60 x+480]_{2}=8$ $=1920$ (km).
|
1920
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. There is a sequence of numbers which are $1, 5, 11, 19, 29$, $A, 55$, where $A=$ $\qquad$ .
|
3. 41;
|
41
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Find the positive integer solutions to the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$.
|
3. Solution Since $x, y, z$ are positive integers and
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{6} \leqslant 1 .
$$
Assume without loss of generality that $1<x \leqslant y \leqslant z$, then $\frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}$.
Thus, $\frac{1}{x}<\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{3}{x}$.
That is, $\frac{1}{x}<\frac{5}{6} \leqslant \frac{3}{x}$, solving this gives $\frac{5}{6}<x \leqslant \frac{18}{5}$, which means $x=2,3$.
When $x=2$, we can determine $y=4,5,6$.
When $x=3$, we can determine $y=3,4$.
In summary, we can determine $z=12,6,6,4$.
Therefore, when $1<x \leqslant y \leqslant z$, the solutions $(x, y, z)$ are
$$
\begin{array}{l}
(2,4,12),(2,6,6), \\
(3,3,6),(3,4,4)
\end{array}
$$
Four sets.
Since $x, y, z$ are symmetric in the equation, there are a total of 15 sets of solutions.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Among all possible four-digit numbers formed using the digits $1,9,9,0$, for each such four-digit number and a natural number $n$, their sum when divided by 7 does not leave a remainder of 1. List all such natural numbers $n$ in descending order.
$$
n_{1}<n_{2}<n_{3}<n_{4}<\cdots \cdots,
$$
Find: the value of $n_{1}$.
|
3. Solution
$1,0,9,0$ digits can form the following four-digit numbers: 1099, 1909, 1990, 9019, 9091, 9109, $9190,9901,9910$ for a total of nine. The remainders when these nine numbers are divided by 7 are $0,5,2,3,5,2,6$, 3 , 5 . Since $n$ and their sum cannot be divisible by 7 with a remainder of 1, $n$ cannot have a remainder of $1,3,6,5$ : $3,6,2,5,3$ when divided by 7. That is, $n$ cannot have a remainder of $1,2,3,5,6$ when divided by 7. Therefore, the remainder when $n$ is divided by 7 can only be 0 or 4. Since $n$ is a natural number, $n_{1}=4, n_{2}=7$, $n_{1} n_{2}=28$.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-c-a}{b}$ $=3$, and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$. Then $x-a-b-c=$
|
5. 0 ;
The above text has been translated into English, maintaining the original text's line breaks and format.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, A and B start from locations $A, B$ respectively, heading towards each other. They meet at point $C$ on the way, after which A takes 5 hours to reach $B$, and B takes $3 \frac{1}{5}$ hours to reach $A$; it is known that A walks 1 kilometer less per hour than B. Find the distance between $A$ and $B$.
|
Three, Hint: Let $A C=S_{1}, C B=S_{2}$, the time required for both to meet is /small hour, then we get $t=4$ hours. From the problem, we solve to get $S_{1}=16, S_{2}=20$. Therefore, $A B=35$ km.
|
35
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. In $\triangle A B C$, $A B=A C=2, B C$ side has 100 different points $P_{1}, P_{2}, \cdots, P_{1} 00$. Let $m_{1}=A P_{1}^{2}+B P_{1} \cdot P_{1} C(i=1,2, \cdots, 100)$, then $m_{1}+m_{2}+\cdots+m_{100}$ equals what? (1990 National Junior High School Mathematics League)
|
Solve $\because A B$ $=A C$, according to Corollary 1 we have
$$
\begin{array}{c}
A P_{1}^{2}=A B^{2} \\
-B P_{\mathrm{i}} \cdot P_{1} C, \\
\therefore A P_{1}^{2} \\
+B P_{1} \cdot P_{1} C \\
=A B^{2}, \\
\therefore m_{\mathrm{i}}=A P_{1}^{2}+B P_{1} \cdot P_{\mathrm{i}} C \\
=A B^{2}=2^{2}=4,
\end{array}
$$
that is
$$
\begin{array}{c}
m_{1}+m_{2}+\cdots+m_{100} \\
=4 \times 100=400 .
\end{array}
$$
|
400
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { Four, Prove: } 1991^{1992}+1993^{1994} \\
+1995^{1996}+1997^{1998}+1999^{2000}
\end{array}
$$
is divisible by 5.
|
Four, Hint: By discussing the regularity of the powers of the unit digits, we know that the unit digits of the terms in the sum are $1, 9, 5, 9, 1$, and their sum is 25. Therefore, the sum can be divisible by 5.
|
25
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. In $\triangle A B C$, $D$ is a point on side $B C$. It is known that $A B=13, A D=12, A C=15, B D=5$. What is $D C$? (3rd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition)
|
According to Stewart's Theorem, we have
$$
A D^{2}=A B^{2} \cdot \frac{C D}{B C}+A C^{2} \cdot \frac{B D}{B C}-B D \cdot D C \text {. }
$$
Let $D C=x$,
then $B C=5+x$.
Substituting the known data
into the above equation, we get
$$
\begin{array}{l}
12^{2}=13^{2} \\
\cdot \frac{x}{5+x}+15^{2} \cdot \frac{5}{5+x}-5 x,
\end{array}
$$
Solving this equation yields $x_{1}=9, x_{2}=-9$ (discard). Therefore, $D C=9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, given that $a$ is an integer, the equation $x^{2}+(2 a+1) x$ $+a^{2}=0$ has integer roots $x_{1}, x_{2}, x_{1}>x_{2}$. Try to find the value of $\sqrt[4]{x_{1}^{2}}-\sqrt[4]{x_{2}^{2}}$.
|
Three, Solution: From the problem, we know that the discriminant $4a + 1$ is a perfect square, so $a \geqslant 0$. Since $4a + 1$ is odd, we can set $(2k + 1)^2 = 4a + 1$, solving for $a = k(k + 1)$.
$$
x_{1,2} = \frac{1}{2} \left[ \left( -2k^2 - 2k - 1 \right) \pm \sqrt{(2k + 1)^2} \right].
$$
When $k \geqslant 0$, $x_1 = -k^2$, $x_2 = -(k + 1)^2$, then $\sqrt[4]{x_1^2} - \sqrt[4]{x_2^2} = -1$,
When $k < 0$, $x_1 = -(k + 1)^2$, $x_2 = -k^2$, then $\sqrt[4]{x_1^2} - \sqrt[4]{x_2^2} = -1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. As shown in the figure, points $A, B, C, D$ lie on the same circle, and $BC=DC=4, AE=6$. The lengths of segments $BE$ and $DE$ are both positive integers. What is the length of $BD$? (1988
National Junior High School Mathematics Competition)
|
Solve in $\triangle B C D$, $B C=D C$, H
Inference 1 gives
$$
C E^{2}=B C^{2}-B E \cdot D E,
$$
$\because A, B, C, D$ are concyclic,
$$
\begin{array}{l}
\therefore B E \cdot D E=A E \cdot C E, \\
\therefore C E^{2}=B C^{2}-A E \cdot C E=4^{2}-6 \times C E,
\end{array}
$$
which is $C E^{2}+6 \cdot C E-16=0$.
Solving gives $C E=2, C E=-8$ (discard),
$$
\begin{array}{l}
\therefore B E \cdot D E=A E \cdot C E=6 \times 2=12 . \\
\text { Also } B D<B C+C D=4+4=8, \\
\therefore\left\{\begin{array} { l }
{ B E = 4 , } \\
{ D E = 3 ; }
\end{array} \text { or } \left\{\begin{array}{l}
B E=3, \\
D E=4 .
\end{array}\right.\right. \\
\therefore B D=7 .
\end{array}
$$
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Factorize: $a^{3}+2 a^{2}-12 a+15$ $=$ $\qquad$ - If $a$ is a certain natural number, and the above expression represents a prime number, then this prime number is
|
3. $\left(a^{2}-3 a+3\right)(a+5), 7$;
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. A hostess is waiting for 7 or 11 children to arrive, and she has prepared 77 marbles as gifts. She puts these marbles into $n$ bags so that each child (whether 7 or 11) can receive several bags of marbles, and the 77 marbles are evenly distributed among these children. Find the minimum value of $n$.
|
10. Since these $n$ bags of sand can be divided into 7 portions, each containing 11 marbles, represented by 7 vertices $x_{1}, x_{2}, \cdots, x_{7}$. These $n$ bags of marbles can also be divided into 11 portions, each containing 7 marbles, represented by vertices $y_{1}, y_{2}, \cdots, y_{11}$. If a certain bag of marbles appears in both $x_{i}$ and $y_{i}$, then an edge is drawn between $x_{1}$ and $y$. This results in a bipartite graph $G=(X, Y, E)$, where $X=\left\{x_{1}, x_{2}, \cdots, x_{7}\right\}, Y=\left\{y_{1}, y_{2}, \cdots, y_{11}\right\}$.
This bipartite graph $G$ must be connected. If not, let $G^{\prime}=\left(X^{\prime}, Y^{\prime}, E^{\prime}\right)$ be a connected subgraph, with $\left|X^{\prime}\right|=a,\left|Y^{\prime}\right|=b$, where $a \leqslant 7, b \leqslant 11$, and the equalities cannot hold simultaneously. Since $G^{\prime}$ is connected, we have $11a = 7b$.
Since 7 and 11 are coprime, the above equation holds only when $a=7, b=11$. This is a contradiction.
Since $G$ is a connected graph, its number of edges is at least $7 + 11 - 1 = 17$, so $n \geqslant 17$.
When $n=17$, if the number of marbles in each bag is
$7,7,7,7,7,7,7,4,4,4,4$,
$3,3,3,1,1,1$,
then the requirements are met. Therefore, the minimum value of $n$ is 17.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Place several points on the unit sphere such that the distance between any two points is (1) at least $\sqrt{2}$; (2) greater than $\sqrt{2}$. Determine the maximum number of points and prove your conclusion.
|
13. (1) The maximum number of points is 6. If $A$ is one of these points, let's assume $A$ is at the North Pole, then the remaining points must all be in the Southern Hemisphere (including the equator). If there is only one point $B$ at the South Pole, then the rest of the points are all on the equator, in which case there are at most $2+4=6$ points.
If there is no point at the South Pole, it can be proven that the number of points does not exceed 5. Otherwise, there would be at least 5 points $A_{1}$, $A_{2}, \cdots, A_{5}$ in the Southern Hemisphere (including the equator). Let $A^{\prime}$ be the South Pole, and the arcs $A^{\prime} A_{1}(i=1,2, \cdots, 5)$ intersect the equator at $A_{1}^{\prime}$. Then, among $\angle A_{1}^{\prime} A^{\prime} A_{\mathrm{j}}^{\prime}(1 \leqslant i \neq j \leqslant 5)$, at least one angle is less than or equal to $72^{\circ}$. Suppose $\angle A_{1}^{\prime} A^{\prime} A_{2}^{\prime} \leqslant 72^{\circ}$, then in the spherical triangle $\triangle A_{1}^{\prime} A^{\prime} A_{2}^{\prime}$, the distance between any two points is less than $\sqrt{2}$, which is a contradiction. Therefore, the maximum number of points on the sphere is 6. (2) The maximum number of points is 4. The proof is similar to (1).
|
6
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. (China) Let $S=\{1,2,3, \ldots, 280\}$. Find the smallest natural number $n$ such that every subset of $S$ with $n$ elements contains 5 pairwise coprime numbers.
|
Let $A_{1}=\{S \mapsto$ all natural numbers divisible by $i\}, i=2,3,5,7$. Let $A=A_{2} \cup A_{3} \cup A \cup \cup A_{7}$. Using the principle of inclusion-exclusion, it is easy to calculate that the number of elements in $A$ is 216. Since any 5 numbers chosen from $A$ must have at least two numbers in the same $A_i$, they are not coprime, thus $n \geqslant 217$.
$$
\begin{array}{l}
\text { On the other hand, let } \\
B_{1}=\{1 \text { and all prime numbers in } S\}, \\
B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}\right\}, \\
B_{3}=\{2 \times 131,3 \times 89,5 \times 53, \\
7 \times 37,11 \times 23,13 \times 19\}, \\
B_{4}=\{2 \times 127,3 \times 83,5 \times 47, \\
7 \times 31,11 \times 19,13 \times 17\}, \\
B_{5}=\{2 \times 113,3 \times 79,5 \times 43, \\
7 \times 29,11 \times 17\} . \\
B_{8}=\{2 \times 109,3 \times 73,5 \times 41, \\
7 \times 23,11 \times 13\} .
\end{array}
$$
It is easy to see that the number of elements in $B_{1}$ is 60. Let $B=B_{1} \cup B_{2} \cup B_{3} \cup B_{4} \cup B_{5} \cup B_{8}$, then the number of elements in $B$ is 88. Thus, the number of elements in $S-B$ is 192. When 217 numbers are chosen from $S$, since $217-192=25$, there exists $1 \leqslant i \leqslant 6$ such that these 217 numbers include 5 numbers from $B$, and these 5 numbers are pairwise coprime. Therefore, $n \leqslant 217$. Hence, we have $n=217$.
|
217
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
21. Let the weight of the counterfeit coin be $a$, and the weight of the genuine coin be $b$ $(a \neq b)$. There are two piles of three coins each, and it is known that each pile contains exactly one counterfeit coin. How many times at least must a precise scale (not a balance) be used to find these two counterfeit coins?
(1) Assuming $a$ and $b$ are known, solve this problem.
(2) Assuming $a$ and $b$ are unknown, solve this problem.
|
21. (1) At least 3 weighings are required.
First, take one coin from each of the two piles, (1) if the weight is $2a$, then the real coins have been found; (2) if the weight is $2b$, then take one coin from the remaining two in each pile and weigh them separately, and the two fake coins can be found; (3) if the weight is $a+b$, take one out and weigh it again, then take another one from the remaining pile with the fake coin and weigh it, and the other fake coin can be found.
(2) At least 4 weighings are required.
First, take two coins from one pile and weigh them separately. (1) If the two coins weigh the same, then they are real coins. The remaining one is a fake coin, and $b$ is known. Take two more coins from the other pile and weigh them separately, and the fake coins can be found; (2) If the two coins weigh differently, one weighs $a$ and the other weighs $b$, at this point, take two coins from the other pile and weigh them together with the remaining one from this pile. If the weight is $3b$, then the two fake coins are found; if the weight is $a+2b$, take one of the two coins already taken from the other pile and weigh it again, and the two fake coins can also be found.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A cube with an edge length of 3 is composed of 27 unit cubes. How many lines pass through the centers of 3 unit cubes? How many lines pass through the centers of 2 unit cubes?
|
There are 49 lines passing through the centers of 3 unit cubes; there are
$$
C_{27}^{2}-3 \times 49=204 \text { (lines). }
$$
|
204
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. It is known that a parliament has 30 members, where any two are either friends or enemies, and each member has exactly 6 enemies. Any three members form a committee. Find the total number of such committees where the three members are either all friends or all enemies.
|
3. Let the set of all three-member committees that satisfy the requirements of the problem be denoted as $X$, and the number of elements in set $X$ be denoted as $x$. The number of other three-member committees is denoted as $y$, thus
$$
x+y=C_{30}^{s}=4060 \text {. }
$$
For any senator $a$, we denote the set of committees in which $a$ participates as $S_{\mathrm{a}}$. In $S_{\mathrm{a}}$, the number of committees where the other two members are both friends or both enemies of $a$ is $C_{B}^{2}+C_{2}^{2}=268$. For the 30 senators, there are a total of 8040 such committees. Clearly, in the above counting process, the committees in $X$ are counted three times, while the other committees are counted only once, hence we also have
$$
3 x+y=8040 \text { . }
$$
By solving the equations (1) and (2) simultaneously, we get $x=1990$.
|
1990
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Find the maximum value of the expression
$$
|| \cdots|| x_{1}-x_{2}\left|-x_{3}\right|-\cdots \mid-x_{1000}
$$
where $x_{1} ， x_{2}, \cdots, x_{1000}$ are different natural numbers from 1 to 1990.
|
7. Noting that when $x \geqslant 0, y \geqslant 0$, the following formulas hold:
$$
\begin{array}{l}
|x-y| \leqslant \max \{x, y\}, \\
\max \{\max \{x, y\}, z\}=\max \{x, y, z\},
\end{array}
$$
we can directly obtain
$$
\begin{array}{l}
\left.|| \cdots|| x_{1}-x_{2}\left|-x_{3}\right|-\cdots \mid-x_{\mathrm{n}}\right\} \\
\leqslant \max \left\{x_{1}, x_{2}, \cdots, x_{\mathrm{n}}\right\},
\end{array}
$$
where $n=1,2, \cdots, 1990$. Therefore, the value of the expression in question does not exceed $\max \left\{x_{1}, x_{2}, \cdots, x_{1990}\right\} = 1990$. However, since its value has the same parity as the sum
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{1990} \\
=1+2+\cdots+1990 \\
=995 \times 1991
\end{array}
$$
the value of the expression cannot be 1990, so the maximum possible value is 1989. The following specific example
$$
\begin{array}{l}
|\cdots||| 2-4|-5|-3|-\cdots|-(4 k+2) \\
|-(4 k+4)|-(4 k+5)|-(4 k+3)| \\
\cdots|-1986|-1988|-1989|-1987 \mid \\
\quad-1990|-1|=1989
\end{array}
$$
demonstrates that the maximum value sought is 1989.
|
1989
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. There are two fleas at the two endpoints of the line segment $[0, 1]$. Some points are marked within the line segment. Each flea can jump over the marked points such that the positions before and after the jump are symmetric about the marked point, and they must not jump out of the range of the segment $[0, 1]$. Each flea can independently jump once or stay in place, which counts as one step. How many steps are needed at minimum to ensure that the two fleas can always jump to the same small segment divided by the marked points on $[0, 1]$?
|
4. The segment [0, 1] is divided into smaller segments by certain fixed points, with lengths $\frac{17}{23}$ and $\frac{19}{23}$. Therefore, the two fleas cannot land on the same segment after each taking one step (see the following supplement). From this, we can conclude that the minimum number of steps required must be 1.
The following proof shows that, regardless of the number of fixed points and how they are placed on the segment $[0,1]$, it is always possible to make both fleas jump into the longest segment after two jumps (if there are multiple such segments, we can choose one of them). By symmetry, it is sufficient to prove this for the flea starting at point 0.
Let the length of the chosen longest segment be $s$, and $\alpha$ be one of its endpoints.
If $\alpha < s$ (see the following figure), the flea can jump over the fixed point $\alpha$ in one step and land in the selected segment $[\alpha, \alpha+s]$ (if $\alpha=0$, the flea is already in the selected segment and does not need to jump at all). If $\alpha \geqslant s$, consider the interval $\left[\frac{\alpha-s}{2}, \frac{\alpha+s}{2}\right]$, which has a length of $s$ (see the following figure). This interval must contain at least one fixed point $\beta$. Otherwise, the length of the segment containing this interval would be greater than $s$, which is impossible. By jumping over $\beta$, the flea will land at point $2 \beta \in [\alpha-s, \alpha+s]$. If $2 \beta \notin [\alpha, \alpha+s]$, then by jumping over point $\alpha$ again, the flea will definitely land in the selected segment $[\alpha, \alpha+s]$.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5 . Find the integer solution to the equation
$$
\left[\frac{x}{1!}\right]+\left[\begin{array}{c}
x \\
2!
\end{array}\right]+\cdots+\left[\frac{x}{10!}\right]=1001
$$
|
5. From the old equation, we know that $x$ is a positive integer not exceeding 1001, so $x<6$!. Therefore, the last five terms on the left side of the equation can be removed. Each positive integer $x<6$! can be uniquely expressed as
$$
x=a \cdot 5!+b \cdot 4!+c \cdot 3!+d \cdot 2!+e,
$$
where $a$, $b$, $c$, $d$, $e$ are non-negative integers, and $a \leqslant 5$, $b \leqslant 4$, $c \leqslant 3$, $d \leqslant 2$, $e \leqslant 1$. Substituting (1) into the known equation, we get
$$
\begin{array}{l}
206 a+41 b+10 c+3 d+e=1001 . \\
\text { Since } 41 b+10 c+3 d+e \leqslant 201,
\end{array}
$$
it follows that $800 \leqslant 206 a \leqslant 1001$. This implies $a=4$. This also means
$$
41 b+10 c+3 d+e=177 .
$$
By similar reasoning, we get $b=4$, $c=d=1$, $e=0$. Therefore, $x=4 \times 5!+4 \times 4!+3!+2!=584$.
|
584
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
```
11. (16th All-Russian Mathematics Competition) Replace the “*” in the following division problem with appropriate digits
$$
\begin{array}{l}
\text { * } 8 \text { * } \\
* * \sqrt{* * * * *} \\
-) * * * * * * \\
\frac{-) * *}{* *} \\
\frac{-) * * *}{0} \\
\end{array}
$$
```
|
From the fact that the divisor multiplied by 8 yields a two-digit number, we know the divisor is no greater than 12. Furthermore, since the divisor multiplied by the first or last digit of the quotient results in a three-digit number, we know the divisor is no less than 12.
Therefore, the divisor is 12.
And since 12 multiplied by the first or last digit of the quotient is a three-digit number, the first and last digits must both be 9, making the quotient 989.
The dividend is \(12 \times 989 = 11868\).
|
11868
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. (19th Austrian Mathematical Competition) Find $N=$ $19^{88}-1$ for all divisors $d=2^{2} \cdot 3^{b}$ (where $a, b$ are natural numbers) and determine the sum of these divisors $d$.
|
$$
\begin{aligned}
N & =(20-1)^{88}-1 \\
& =(1-4 \times 5)^{88}-1 \\
& =-C_{88}^{1} 4 \times 5+C_{88}^{2} 4^{2} \times 5^{2} \\
& -C_{88}^{8} 4^{8} \times 5^{3}+\cdots \\
& -C_{88}^{87} 4^{87} \times 5^{87}+C_{88}^{88} 4^{88} \times 5^{88} \\
& =-2^{5} \times 55+2^{8} M \\
& =2^{5}(-55+2 M),
\end{aligned}
$$
where $M$ is an integer.
Therefore, the highest power of 2 in the prime factors of $N$ is 5.
On the other hand,
$$
\begin{aligned}
N & =(1+2 \times 9)^{88}-1 \\
& =C_{88}^{1} 2 \times 9+C_{88}^{2} 2^{2} \times 9^{2}+\cdots \\
& +C_{88}^{88} 2^{88} \times 9^{88} \\
& =3^{2} \times 2 \times 88+3^{3} T \\
& =3^{2}(2 \times 88+3 T),
\end{aligned}
$$
where $T$ is an integer.
Therefore, the highest power of 3 in the prime factors of $N$ is 2.
Thus, $N=2^{5} \cdot 3^{2} \cdot L$, where $L$ does not contain 2 and 3 as factors.
Therefore, the sum of all divisors of $N$ of the form $d=2^{\mathrm{a}} \cdot 3^{\mathrm{b}}$ is
$$
\left(2+2^{2}+2^{3}+2^{4}+2^{5}\right)\left(3+3^{2}\right)=744 .
$$
|
744
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. (16th All-Russian Mathematics Competition) A desert is shaped like a half-plane, which is divided into many small squares of size $1 \times 1$. In the desert, 15 squares away from the boundary, there is a robot with energy $E=59$. The "energy consumption" of each small square is a natural number not greater than 5, and the "energy consumption" of any $5 \times 5$ square in the desert is 88. The robot can move into any adjacent small square (squares with a common side are considered adjacent), and each time it enters a square, its energy decreases by the energy consumption of that square.
Will the robot be able to leave the desert?
|
Solve As shown in the figure, use routes with arrows to represent five different stubborn movement routes of the robot, * indicates the position of the robot.
The total energy consumption of the small squares passed by these five routes does not exceed $3 \times 88 + 2 \times 10 + 2 \times 5 = 294$ (the above routes pass through 3 squares of size $5 \times 5$ once, pass over the two adjacent squares above and below the robot's original position twice, and pass over the two adjacent squares above and below the robot's original position once).
Therefore, among these five different routes, there is at least one route where the total energy consumption of the small squares passed is less than 59, so the robot can move along this route to get out of the desert.
|
59
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
18. (16th All-Russian Mathematics Competition) In the school football championship, 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 event of a draw, and 0 points to the losing team. It is known that one team has the highest score, but it has won fewer matches than any other team. How many teams must have participated at a minimum?
|
Let the team with the highest score, denoted as team $A$, be the champion. 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 win at least $n + 1$ matches, meaning their score is no less than $2(n + 1)$ points. Therefore,
$$
\begin{array}{l}
2n + m > 2(n + 1), \\
m \geq 3.
\end{array}
$$
Thus, there must be a team that draws with the champion team, and this team's score should be no less than $2(n + 1) + 1$ points. Therefore,
$$
\begin{array}{l}
2n + m > 2(n + 1) + 1, \\
m \geq 4.
\end{array}
$$
Let there be $s$ teams in total. The champion must win at least one match; otherwise, its score would not exceed $s - 1$ points. Any other team would score strictly less than $s - 1$ points, and the total score of all participating teams would be less than $s(s - 1)$ points, while the total score of $s$ teams is $s(s - 1)$ points, leading to a contradiction.
Thus, $m > 4, n \geq 1$, meaning the champion team $A$ must play at least 5 matches, implying there are at least 6 teams in the competition.
A score table for 6 teams can always be constructed to meet the given conditions, where the champion team $A$ wins the minimum number of matches:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & Score \\
\hline$A$ & & 1 & 1 & 1 & 1 & 2 & 6 \\
\hline$B$ & 1 & & 2 & 0 & 0 & 2 & 5 \\
\hline$C$ & 1 & 0 & & 0 & 2 & 2 & 5 \\
\hline$D$ & 1 & 2 & 2 & & 0 & 0 & 5 \\
\hline$E$ & 1 & 2 & 0 & 2 & & 0 & 5 \\
\hline$F$ & 0 & 0 & 0 & 2 & 2 & & 4 \\
\hline
\end{tabular}
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. The figure formed by the diagonals of squares with side length 1 is called a cross. Prove: In a circle with a radius of 100, the number of non-overlapping crosses that can be placed does not exceed 80000.
|
Prove that for each cross, consider a circle with the center of the line segment as the center and the half of the cross diagonal (diagonal length... half) $\frac{1}{2} \cdot \frac{\sqrt{2}}{2}=\frac{1}{2 \sqrt{2}}$ as the radius.
Then, in the figure, the intersection $O_{1} O_{2} \leqslant r_{1}+r_{2}=2 r$ $=\frac{1}{\sqrt{2}}$, and it is known that $\mathrm{O}_{2} A=\frac{1}{\sqrt{2}}$. Consider the rectangle $\mathrm{O}_{1} \mathrm{CO}_{2} \mathrm{D}$, because $\mathrm{O}_{2} \mathrm{~A} \geqslant \mathrm{O}_{2} \mathrm{O}_{1}$, so $\mathrm{BO}_{2} \mathrm{~A}$ is within the right angle $\mathrm{CO}_{2} \mathrm{D}$ of this rectangle. One of the four branches of the cross $\mathrm{O}_{2}$ must be inside $\angle \mathrm{CO}_{2} \mathrm{D}$. Therefore, it intersects with the first cross $O$.
Since $\frac{\pi \cdot 100^{2}}{\pi \cdot\left(\frac{1}{2 \sqrt{2}}\right)}=80000$, this indicates that in a circle with a radius of 100, no more than 80000 non-intersecting circles with a radius of $\frac{1}{2 \sqrt{2}}$ can be placed, and thus no more than 80000 non-intersecting crosses can be placed.
|
80000
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three, given any 5 points on a plane, where no three points are collinear and no four points are concyclic. If a circle passes through three of these points, and the other two points are respectively inside and outside the circle, then it is called a "good circle".
Let the number of good circles be $n$, find all possible values of $n$.
---
Note: The translation preserves the original text's line breaks and formatting.
|
Three, among: 5 points, take any two points $A, B$ and draw the line through $A$, $B$. If the other three points $C, D, E$ are on the same side of line $A B$, then consider $\angle A C B, \angle A D B, \angle A E B$. Without loss of generality, if $\angle A C B < 180^{\circ}$, then circle $A D B$ is the unique good circle; if $\angle A E B + \angle A C B = 180^{\circ}$, then circles $A C B, A D B, A E B$ are all good circles. This means that through two fixed points, there is either one or three good circles.
From 5 points, a total of 10 point pairs can be formed. Through each point pair, there is at least one good circle, so there are at least 10 good circles (including repeated counts). Each good circle passes through 3 point pairs, so there are at least 4 different good circles, and $n \geqslant 1$.
Connecting each pair of points among the 5 points with a line segment, each line segment is either a chord of a good circle or a common chord of 3 good circles. If there are at least 5 good circles, then they have at least 15 chords. Since there are only 10 line segments in total and each line segment contributes 1 or 3 to the count, the sum of 10 odd numbers is even, which cannot be 15, so there must be at least 6 different good circles. This means that at least 4 line segments are chords of good circles: 4 line segments have 8 endpoints, so there must be at least two line segments sharing a common endpoint, say $A B, A C$. Thus, $A B D$ and $A C D$ are both good circles, so there are at least three good circles through $A, D$.
Assume $A B, A C, A D, A E$ with the shortest one being $A B$, then $\angle A C B, \angle A D B, \angle A E B$ are all acute angles. If $C, D, E$ are on the same side of line $A B$, then there is only one good circle through $A, B$, so $C, D, E$ must be on different sides of line $A B$. In this case, since the sum of any two angles among $\angle A C B, \angle A D B, \angle A E B$ is less than $180^{\circ}$, there can only be one good circle through $A, B$, which is a contradiction.
In summary, the number of good circles $n$ must be 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, try to select 1999 integers from the set $\{1,2,3, \cdots, 3000\}$ to form a subset $A$, such that if $x \in A$, then $2 x \notin A$.
|
Five, the selection method is as follows:
1) Select 1500 integers from 1501 to 3000 to form $A_{1}$.
2) From 1 to 1500, remove numbers 751 to 1500 (these numbers multiplied by 2 are in $A_{1}$), and select 376 to 750, a total of 375 integers to form $A_{2}$.
3) From 1 to 375, remove 188 to 375 (these numbers multiplied by 2 are in $A_{2}$), and select 94 to 187 from 1 to 187, a total of 94 integers to form $A_{3}$.
4) From 1 to 93, remove 47 to 93 (these numbers multiplied by 2 are in $A_{3}$), and select 24 to 46 from 1 to 46, a total of 23 integers to form $A_{4}$.
5) From 1 to 23, remove 12 to 23 and select 6 to 11 and 1 (or 2), a total of 7 integers to form $A_{5}$.
Let $A=A_{1} \cup A_{2} \cup A_{3} \cup A_{4} \cup A_{5}$, then the number of elements in $A$ is: $1500+375+94+23+7$ $=1999$ (elements).
|
1999
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For $a \geqslant 1$, calculate the sum of the infinite series
$$
\begin{array}{l}
\frac{a}{a+1}+\frac{a^{2}}{(a+1)\left(a^{2}+1\right)} \\
+\frac{a^{4}}{\left.(a+1) a^{2}+1\right)\left(a^{4}+1\right)} \\
+\frac{a^{8}}{(a+1)\left(a^{2}+1\right)\left(a^{4}+1\right)\left(a^{8}+1\right)}+\cdots
\end{array}
$$
|
$\begin{array}{c}\text { I. When } a=1 \text {, } \Sigma=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \\ =\frac{\frac{1}{2}}{1-\frac{1}{2}}=1 \text {. When } a>1 \text {, we have } \Sigma_{1}=\frac{a^{2}-a}{a^{2}-1}, \\ \Sigma_{2}=\frac{a^{4}-a}{a^{4}-1}, \Sigma_{3}=\frac{a^{8}-a}{a^{8}-1}, \cdots, \Sigma_{n}=\frac{a^{2}-a}{a^{2^{n}}-1} \\ \text { Therefore, } \Sigma=\lim _{n \rightarrow \infty} \Sigma_{n}=\lim _{n \rightarrow \infty} \frac{a^{2^{n}}-a}{a^{2^{n}}-1}=1 .\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, let $A B C$ be an equilateral triangle, and $P$ a point on its incircle. Prove that $P A^{2}+P B^{2}+P C^{2}$ is a constant.
|
Four, as shown in the figure, establish
a rectangular coordinate system,
the coordinates of point $P$ are
$$
\begin{array}{l}
\left(\frac{\sqrt{3}}{3} \cos \theta,\right. \\
\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3} \\
\cdot \sin \theta) .
\end{array}
$$
Let $\Sigma=P A^{2}+P B^{2}+P C^{2}$, then
$$
\begin{array}{l}
\Sigma=\left[\left(\frac{\cos \theta}{\sqrt{3}}-1\right)^{2}+\left(\frac{1}{\sqrt{3}}+\frac{\sin \theta}{\sqrt{3}}\right)^{2}\right] \\
+\left[\left(\frac{\cos \theta}{\sqrt{3}}\right)^{2}+\left(\frac{\sin \theta}{\sqrt{3}}-\frac{2}{\sqrt{3}}\right)^{2}\right] \\
+\left[\left(\frac{\cos \theta}{\sqrt{3}}+1\right)^{2}+\left(\frac{1}{\sqrt{3}}+\frac{\sin \theta}{\sqrt{3}}\right)^{2}\right] \\
=\cos ^{2} \theta+\sin ^{2} \theta+4+\left(-\frac{2}{\sqrt{3}}\right. \\
\left.+\frac{2}{\sqrt{3}}\right) \cos \theta+\left(\frac{2}{3}-\frac{4}{3}+\frac{2}{3}\right) \\
\sin ^{2} \theta=5 \text{ (constant)}
\end{array}
$$
|
5
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Let $S=\{1,2,3,4\}, a_{1}, a_{2}, \cdots$, be any permutation ending with 1, i.e., for any permutation $\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ of the four numbers in $S$ that does not end with 1, $\left(b_{4} \neq 1\right)$, there exist $i_{1}, i_{2}, i_{3}, i_{4}$, such that $1 \leqslant i_{1}<i_{2}<i_{3}<i_{4} \leqslant k$, and $\left(a_{i_{1}}, a_{i_{2}}, a_{i_{3}}, a_{i_{4}}\right)=\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$. Find the minimum value of the number of terms $k$ in such a sequence.
|
II. Hint: First consider the minimum value of the number of terms $k$ in any sequence containing a permutation of $S$.
1. For $S=\{1,2,3\}$, it can be proven that a sequence with only 6 terms cannot contain any permutation of $\{1,2,3\}$.
2. For $S=\{1,2,3,4\}$, prove that a sequence with only 11 terms cannot contain any permutation of $\{1,2,3,4\}$.
Direct verification shows that the sequence 1, 2, 3, 4, 1, 2, 3, 1, 4, 2, 3, 1 contains any permutation of $S$, thus 12 is the minimum value.
Returning to the problem itself. If there is a sequence with fewer than 11 terms that contains any permutation of $S$ not ending in 1, then adding one more positive term to this sequence would result in a sequence containing all permutations of $S$, but with fewer than 12 terms, which is a contradiction. Additionally, 1, 2, 3, 4, 1, 2, 3, 1, 4, 2, 3 contains any permutation of $S$ not ending in 1. Therefore, the minimum value of $k$ is 11.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Find the minimum value of the function $f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \quad(x \in R)$.
|
$$
\begin{array}{l}
\text { Sol } \because\left(x^{2}+1\right)-\cos x \\
=x^{2}+(1-\cos x) \geqslant 0, \\
\left(x^{2}+1\right)-2 x=(x-1)^{2} \geqslant 0, \\
\therefore \quad x^{2}+1 \geqslant \cos x, x^{2}+1 \geqslant 2 x \\
\text { Therefore, } f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \\
=x^{2}+1 \geqslant 1,
\end{array}
$$
where the equality holds when $x=0$, hence the minimum value of $f(x)$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The average score of the participants in the Zu Chongzhi Cup Mathematics Invitational at a certain school is 75. Among them, the number of male participants is $80\%$ more than that of female participants, and the average score of female participants is $20\%$ higher than that of male participants. Therefore, the average score of female participants is — points.
|
1. 84
|
84
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. $A, B$ are two fixed points on a plane, find a point $C$ on the plane such that $\triangle A B C$ forms an isosceles triangle. There are $\qquad$ such points $C$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The blank space represented by $\qquad$ in the original text is kept as is in the translation.
|
4.6.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
6
|
Logic and Puzzles
|
other
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (1) Prove: $\sqrt{a^{2}+\frac{1}{b^{2}}+\frac{a^{2}}{(a b+1)^{2}}}$ $=\left|a+\frac{1}{b}-\frac{a}{a b+1}\right|$
(2) Using or not using (1), calculate
$$
\sqrt{1+1990^{2}+\frac{1990^{2}}{1991^{2}}}-\frac{1}{1991}
$$
|
Three, Hint: Square both sides of the equation, noting that
$$
\frac{a}{b}-\frac{a^{2}}{a b+1}-\frac{a}{b(a b+1)}=0,
$$
and then take the square root of both sides.
Note: In fact, the following equation also holds:
$$
\sqrt{a^{2}+b^{2}+\frac{a^{2} b^{2}}{(a+b)^{2}}}=\left|a+b-\frac{a b}{a+b}\right|
$$
Using equation (*) to solve part (2) will be more convenient. In fact, these two equations are special cases of the following conditional equation.
If $a+b+c=0$, then
$$
\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right| \text {. }
$$
This equation is an important conditional equation with a wide range of applications.
$$
\begin{array}{l}
\text { (2) } \sqrt{1+1990^{2}+\frac{1990^{2}}{1991^{2}}} \\
=\sqrt{1990^{2}+\frac{1}{1^{2}}+\frac{1990^{2}}{(1990 \cdot 1+1)^{2}}} \\
\end{array}
$$
|
1990
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Given the equation $x^{2}+(a-6) x+a=0$ ( $a$ $\neq 0$ ) with both roots being integers. Try to find the integer $a$. (1989, Sichuan Province Junior High School Mathematics Competition)
|
Let the two integer roots of the equation be \( y_{1} x_{1}, x_{2} \), and \( x_{1} \geqslant x_{2} \). By Vieta's formulas, we have
\[
\left\{\begin{array}{l}
x_{1}+x_{2}=6-a \\
x_{1} x_{2}=a
\end{array}\right.
\]
From (1) and (2), we get \( x_{1} x_{2} + x_{1} + x_{2} = 6 \), which simplifies to \( \left(x_{1}+1\right)\left(x_{2}+1\right)=7 \). Since \( a \neq 0 \), it follows that \( x_{1} \neq 0, x_{2} \neq 0 \). Thus, \( x_{1}=-2, x_{2}=-8 \). Therefore, \( a=16 \).
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Sure, here is the translation:
---
One, starting from the natural number 1, write down in sequence to form the following series of numbers: $12345678910111213 \cdots$. With each digit occupying one position, determine the digit at the 1992nd position.
|
In the following sequence of digits, there are 9 single-digit numbers, $2 \times 90$ two-digit numbers, and $3 \times 900$ three-digit numbers. From $(1992-9-2 \times 90) \div 3$ $=601$, we know that the 1992nd position is in the 601st three-digit number starting from 100, which is the unit digit of the natural number 700. Therefore, the digit at the required position is 0.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. The lengths of the three sides of $\triangle ABC$ are $BC=17$, $CA=18$, and $AB=19$. Through a point $P$ inside $\triangle ABC$, perpendiculars $PD$, $PE$, and $PF$ are drawn to the three sides of $\triangle ABC$ ($D$, $E$, and $F$ are the feet of the perpendiculars), and $BD + CE + AF = 27$. Find the length of $BD + BF$. (5th National Junior High School Mathematics Correspondence Competition)
|
Let $B D=x, B F=y$. Then $C D=17-x$, $A F=19-2$. Also, $B D+C E+A F=27$, so
$$
\begin{array}{l}
C E=27-B D-A F=8-x+y, \\
A B=18-C E=10+x-y .
\end{array}
$$
By Theorem 1 conclusion (2), we have
$$
\begin{array}{l}
x^{2}+(8-x+y)^{2}+(19-y)^{2} \\
=(17-x)^{2}+(10+x-y)^{2}+y^{2} .
\end{array}
$$
Simplifying, we get $2 x+2 y=8^{2}+19^{2}-17^{2}-10^{2}$. Therefore, $B D+B F=18$.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. As shown in Figure 3, through an internal point $P$ of $\triangle ABC$, three lines parallel to the three sides are drawn, resulting in three triangles $t_{1}, t_{2}$, and $t_{3}$ with areas 4, 9, and 49, respectively. Find the area of $\triangle ABC$.
(2nd American
Mathematical Invitational)
|
This example can directly apply the conclusion (4) of Theorem 2 to obtain
$$
S \triangle A B C=144
$$
|
144
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Given $-\frac{x-b}{c}+\frac{x-b-c}{a}$ $+\frac{x-c-a}{b}=3$,
and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$.
Then $x-a-b-c=$ $\qquad$ (8th Jincheng
Mathematics Competition)
|
From the known, we get $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(x-a-b-c)=0$, so we should fill in “0”. This is transforming the known to the unknown, the other is transforming the unknown to the known.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. If $x^{3}-x^{2}+x-2=0$. Then $x^{4}+2 x^{3}-2 x^{2}+x-1=$ $\qquad$ . (1991, Hubei Huanggang Region Mathematics Competition).
|
Left= $\begin{aligned} & \left(x^{4}-x^{3}+x^{2}-2 x\right) \\ & +\left(3 x^{3}-3 x^{2}+3 x-6\right)+5 \\ = & x \cdot 0+3 \cdot 0+5=5 .\end{aligned}$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Calculate: $\left(1^{2}+3^{2}+5^{2}+\cdots+99^{2}\right)-\left(2^{2}\right.$ $\left.+4^{2}+6^{2}+\cdots+100^{2}\right)$
|
$\begin{array}{l}\text { 2. Hint: Original expression }=\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right) \\ +\left(5^{2}-6^{2}\right)+\cdots+\left(99^{2}-100^{2}\right)=-5050 .\end{array}$
|
-5050
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. When a surveyor is measuring a piece of land, they first walk 100 meters from point $A$ at a bearing of $35^{\circ}$ north of east to point $B$, then from point $B$ at a bearing of $55^{\circ}$ north of west for 80 meters to point $C$, and from point $C$ at a bearing of $35^{\circ}$ south of west for 60 meters to point $D$, and finally return to $A$ by the shortest distance. Try to answer the following questions:
（1）Draw the diagram.
（2）What is the shape of this piece of land?
(3) Find the area of this piece of land.
|
3. Brief solution: ( 1 ) As shown in the figure.
( 2 ) This piece of land is a trapezoid.
$$
\begin{array}{c}
\text { ( } 3 \text { ) } S= \\
\frac{1}{2}(60+100) \times 80 . \\
=6400 \text { (square meters). }
\end{array}
$$
|
6400
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $a, b$ are integers, $a$ divided by 7 leaves a remainder of 3, and $b$ divided by 7 leaves a remainder of 5. When $a^{2}>4 b$, find the remainder when $a^{2}-4 b$ is divided by 7.
|
4. Slightly explained: Let $a=$
$$
7 m+3, b=7 n+5 \text {, }
$$
where $m, n$ are integers, then
$$
\begin{aligned}
a^{2}-4 b & =(7 m+3)^{2}-4(7 n+5) \\
& =7\left(7 m^{2}+6 m-4 n-2\right)+3 .
\end{aligned}
$$
Since $7 m^{2}+6 m-4 n-2$ is an integer,
thus $a^{2}-4 b$ leaves a remainder of 3 when divided by 7.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. There are two specifications of steel bars, A and B, and C. It is known that 2 bars of type A, 1 bar of type B, and 3 bars of type C are a total of 23 meters long; 1 bar of type A, 4 bars of type B, and 5 bars of type C are a total of 36 meters long. How long are 1 bar of type A, 2 bars of type B, and 3 bars of type C?
|
6. Slightly explained: Let the lengths of copper strips of specifications A, B, and C be $x, y, z$ meters, respectively, then
$$
\left\{\begin{array}{l}
2 x+y+3 z=23, \\
x+4 y+5 z=36
\end{array}\right.
$$
From (1) - (2) $\times 2$, we get $y=7-z$.
Substituting $y$ into (2), we get $x=8-z$.
Substituting $x, y$ into $x+2 y+3 z=8-z+14-2 z+3 z=22$ (meters).
|
22
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. There are several warriors, forming a rectangular formation that is exactly eight columns wide. If 120 more people are added to or 120 people are removed from the formation, a square formation can be formed in both cases. How many warriors are there in the original rectangular formation?
|
10. Solution: Let the original number of soldiers be $8x$ people,
From the given, $8x+120$ and $8x-120$ are both perfect squares, then we have
$$
\left\{\begin{array}{l}
8 x+120=m^{2} \\
8 x-120=n^{3}
\end{array}\right.
$$
( $m, n$ are positive integers).
By (1) - (2) we get $m^{2}-n^{2}=240$,
which is $(m+n)(m-n)=240$.
From equations (1) and (2), we know that $m, n$ can be divided by 4, so $m+n$ and $m-n$ can be divided by 4.
$$
\begin{array}{l}
\text { Now (1) }\left\{\begin{array}{l}
m+n=60, \\
m-n=4,
\end{array}\right. \\
\text { ( } \mathbb{R} \text { ) }\left\{\begin{array}{l}
m+n=20, \\
m-n=12,
\end{array}\right.
\end{array}
$$
From (I) we get $m=32, n=28$, then
$$
8 x=32^{2}-120=904 \text {. }
$$
From (II) we get $m=16, n=4$, then
$$
8 x=16^{2}-120=136
$$
|
904
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, in a $3 \times 3$ square grid, fill in nine different natural numbers such that the product of the three numbers in each row and the product of the three numbers in each column are all equal (we denote this product by $P$).
(1) Prove that this arrangement of numbers is possible.
(2) Determine which of the six numbers 1990, 1991, 1992, 1993, 1994, 1995 can be the value of $P$.
(3) Find the minimum value of $P$ and explain your reasoning.
|
Three, (1) Proof It is easy to prove the following filling method 1 As shown in the figure, fill in 1, 2, 3, 4, 5, 6, $8, 15, 20$ nine different natural numbers, then the product of the three numbers in each row and the product of the three numbers in each column are all equal to 120.
Therefore, the filling method required by the problem can be achieved.
(2) Solution Obviously, the nine natural numbers to be filled should be nine different divisors of $P$.
Also, the divisor $P$ itself cannot be filled into the table,
Therefore, if the filling method can be achieved, the number of different divisors of $P$ must be greater than or equal to 10.
Among 1990, 1991, 1992, 1993, 1994, 1995, $1990=2 \times 5 \times 199$ has a total of eight different divisors. $1991=11 \times 181$ has a total of four different divisors, $1992=2^{8} \times 3 \times 83$ has a total of sixteen different divisors, $1993=1 \times 1993$ has a total of two different divisors, $1994=2 \times 997$ has a total of four different divisors, $1995=3 \times 5 \times 7 \times 19$ has a total of sixteen different divisors.
Obviously, $P$ cannot take 1990, 1991, 1993, 1994, $P$ may take the values 1992, 1995.
Also, for $P=1992$, there is 1.
$$
P=1995 \text { has } 2 \text {. }
$$
Therefore, among 1990, 1991, 1992, 1993, 1994, 1995, the possible values of $P$ are 1992 and 1995. (3) Solution In general, in the $2 \times 2$ square in the upper left corner, fill in four different natural numbers $a$, $b$, $c$, $d$. In the lower right corner, fill in a natural number $e$ that is different from them, and in the remaining four cells, fill in $-\frac{d}{e}$, $-\frac{a b}{e}$, $\frac{b d}{e}$, $\frac{a c}{e}$, so that $P=\frac{a b c d}{e}$. In this case, we only need to select $a$, $b$, $c$, $d$ such that $\frac{c d}{e}$, $\frac{a b}{e}$, $\frac{b d}{e}$, $\frac{a c}{e}$ are all natural numbers, and they are all different from each other.
In general, it can be assumed that $e$ is smaller than $a$, $b$, $c$, $d$. If not, appropriately swap rows and columns to move the smallest number to the lower right corner, without affecting the value of $P$. When $e=1$, the product $P=\frac{a b c d}{e}$ takes the minimum value, which is clearly 120, as shown in the table filled in (1).
When $e=2$, $P \geqslant 180$,
When $e=3$, $P \geqslant 280$,
When $e=4$, $P \geqslant 420$.
Finally, when $e>5$, $\frac{a b c d}{e}>\frac{e^{4}}{e}>125>120$. Therefore, the minimum value that $P$ can take is 120.
|
120
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that a certain four-digit number is exactly equal to the fourth power of the sum of its digits, then this four-digit number is $\qquad$
|
9. 2401 .
The above text has been translated into English, maintaining the original text's line breaks and format.
|
2401
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In isosceles $\triangle A B C$, it is known that $A B=A C$ $=k B C$, where $k$ is a natural number greater than 1. Points $D$ and $E$ are on $A B$ and $A C$ respectively, and $D B=B C=C E$. $C D$ and $B E$ intersect at $O$.
Find the smallest natural number $k$ such that $\frac{O C}{B C}$ is a rational number.
|
II. Solution: As shown, connect $D E$, it is easy to know that $B C E D$ is an isosceles trapezoid. Also, from the given conditions, we have
$$
\angle 2=\angle 1=\angle 3 \text {, }
$$
thus $\triangle O B C \cos \triangle B C D$,
which means $O C \cdot C D=B C^{2}$.
Also, $\frac{C O}{O D}={ }_{D E}^{B C}=\frac{A B}{A D}$
$$
=\frac{A B}{A B-D B}=\stackrel{k}{k-1} \text {, }
$$
$\frac{C O}{C D}=\frac{k}{2 k-1}$.
(1) $\times$ ( 2 ), we get
$$
{ }_{B C} O C=\sqrt{\frac{k}{2 k-1}} .
$$
Since $\sqrt{\frac{k}{2 k-1}}$ is a rational number, and $k$ and $2 k-1$ are coprime, both $k$ and $2 k-1$ must be perfect squares.
$\cdots$ when $k=4,9,16$, $2 k-1=7,17$,
are not perfect squares.
When $k=25$, $2 k-1=49$ is a perfect square.
In summary, the smallest natural number $k$ that makes $\frac{O C}{B C}$ a rational number is $k=25$.
|
25
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, find all such three-digit numbers. If a two-digit number itself increases by 3, then, the sum of the digits of the resulting number is equal to one third of the sum of the digits of the original three-digit number.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Four, find all such three-digit numbers. If a two-digit number itself increases by 3, then, the sum of the digits of the resulting number is equal to one third of the sum of the digits of the original three-digit number.
|
Let the three-digit number $abc$ satisfy the given conditions. Then
(1) $c \geqslant 7$. In fact, if $c \leqslant b$, then the sum of the digits of $\overline{abc}+3$ will increase by 3, which does not meet the conditions.
(2) $b \neq 9$. This is because if $c \geqslant 7, b=9, a=9$, then
$$
\overline{abc}+3=1000+(c+3-10)
$$
the sum of the digits is
$$
\begin{array}{l}
1+(c+3-10)=c-6 \neq \frac{1}{3}(9+9+c) \\
\text { If } c \geq 7, \quad b=9, a<9, \text { then } \\
\overline{ab} \bar{c}+3=(a+1) \cdot 100+(c+3-10)
\end{array}
$$
the sum of the digits is
$$
(a+1)+(c+3-10)=a+c-b \text {. }
$$
Thus, $a+c-6=\frac{1}{3}(a+9+c)$ leads to $2a+2c=27$, which is impossible.
From (1) and (2), we have
$$
\begin{aligned}
a \vec{b} c+3 & =a \cdot 100+(b+1) \cdot 10 \\
& +(c+3-10),
\end{aligned}
$$
the sum of the digits is
$$
\begin{array}{l}
a+(b+1)+(c+3-10) \\
=a+b+c-6=\frac{1}{3}(a+b+c) .
\end{array}
$$
Thus, $a+b+c=9$.
Since $a \geqslant 1, c \geqslant 7$, the numbers that satisfy the conditions are:
$$
117,108,207 .
$$
Their sum is $117+108+207=432$.
|
432
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The expression $\frac{\left(2^{4}+\frac{1}{4}\right)\left(4^{4}+\frac{1}{4}\right)\left(6^{4}+\frac{1}{4}\right)}{\left(1^{4}+\frac{1}{4}\right)\left(3^{4}+\frac{1}{4}\right)\left(5^{4}+\frac{1}{4}\right)}$ $\times \frac{\left(8^{4}+\frac{1}{4}\right)\left(10^{4}+\frac{1}{4}\right)}{\left(7^{4}+\frac{1}{4}\right)\left(9^{4}+\frac{1}{4}\right)}$ represents a positive integer. This positive integer is
|
$221$
|
221
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The denominator of a reduced fraction is 30, find the sum of all such positive rational numbers less than 10.
untranslated text remains unchanged:
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。
|
1. Xiao Guang 10 points: If a fraction can be simplified to the form $\frac{30 n+r}{30}$, where $n$ and $r$ are integers satisfying $0 \leqslant n \leqslant 9$ and $0 \leqslant r<30$.
When and only when $r$ is coprime with 30, $\frac{30 n+r}{30}$ is a simplified fraction, thus, $r \in\{1,7,11,13,17,19,23,29\}$.
Since $n$ has 10 possible values and $r$ has 8 possible values, and each value of $\frac{30 n+r}{30}$ is unique, the total number of such fractions is $8 \times 10=80$. $\because \quad 10-\frac{k}{30}=\frac{300-k}{30}$,
$\therefore$ the fractions $\frac{k}{30}$ can be paired for calculation.
Thus, the sum of all such fractions is
$$
\begin{array}{c}
\left(\frac{1}{30}+\frac{300-1}{30}\right)+\left(\frac{7}{30}+\frac{300-7}{30}\right)+ \\
\cdots+\left(\frac{149}{30}+\frac{300-149}{30}\right)=40 \times 10=400
\end{array}
$$
|
400
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. A decimal positive integer of at least two digits, where each digit is smaller than the digit to its right, is called an "ascending number." How many ascending numbers are there?
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。
|
2. An "ascending number" must have at least two non-zero digits. It is a subset of the set $S=\{1,2,3,4,5,6,7, 8, 9\}$ consisting of two or more elements. The set determined by all elements is unique. Therefore, the number of ascending numbers is the number of subsets of $S$ consisting of two or more elements. Since $S$ has 9 elements, the total number of subsets is
$$
C_{0}^{0}+C_{0}^{1}+C_{9}^{2}+\cdots+C_{9}^{9}=2^{9}=512 .
$$
Since $C_{8}^{0}+C_{9}^{1}=10$ subsets have fewer than 2 elements, the number of ascending numbers is $512-10=502$.
|
502
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
5. $S$ is the set of rational numbers $r$ where $0<r<1$, and $r$ has a repeating decimal expansion of the form $0 . a b c a b c a b c \cdots$ $=0 . \dot{a} b \dot{c}, a, b, c$ are not necessarily distinct. Among the elements of $S$ that can be written as a fraction in simplest form, how many different numerators are there?
|
5. Since $0 . \dot{a} b \dot{c}=\frac{a b c}{999}$, and $999=3^{8} \cdot 37$. If $a b c$ is neither divisible by 3 nor by 37, then the fraction is in its simplest form. By the principle of inclusion-exclusion, we have
$$
999-\left(\frac{999}{3}+\frac{999}{37}\right)+\frac{999}{3 \cdot 37}=648
$$
such numbers.
In addition, there are fractions of the form $\frac{k}{37}$, where the natural number $k$ is less than 37 and is a multiple of 3. Such $k$ are $3, 6, 9, \cdots, 36$, a total of 12.
Therefore, the number of fractions that satisfy the condition is $648+12=660$.
|
660
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In $\{1000,1001, \cdots, 2000\}$, how many pairs of consecutive integers can be added without carrying over a digit?
|
6 . Let $n$ have the decimal representation $1 a b c$. If one of $a, b$, $c$ is $5, 6, 7$ or 8, then adding $n$ and $n+1$ requires carrying 3. If $b=9, c \neq 9$ or $a=9, b$ and $c$ are not 9, then adding $n$ and $n+1$ requires carrying.
If $n$ is not as described above, then it must be of one of the following forms:
$1 a b c, 1 a b 9,1 a 99,1999$,
where $a, b, c \in\{0,1,2, 3, 4\}$. For such $n$, adding $n$ and $n+1$ does not require carrying. Therefore, there are $5^{3}+5^{2}+5+1=156$ such $n$.
|
156
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. In trapezoid $A B C D$, $A B \| C D$, and $A B$ $=92, B C=50, C D=19, A D=70$. A circle with center $P$ on $A B$ is tangent to sides $B C$ and $A D$. If $A P=\frac{m}{n}$, where $m, n$ are coprime positive integers. Find $m+n_{0} \quad$
|
9. As shown in the figure, extend $A D$ and $B C$ to intersect at point $Q$. Since the distances from $P$ to $A Q$ and $B Q$ are equal, $P$ lies on the angle bisector of $\angle A Q B$, thus we have
$$
\frac{A P}{B P}=\frac{A Q}{B Q}
$$
Since $A B \parallel C D$, we have
$$
\frac{A Q}{B Q}=\frac{A D}{B C}=\frac{7}{5}.
$$
From (1) and (2), we get $\frac{A P}{B P}=\frac{7}{5}$.
$$
\because \quad A P+P B=92.
$$
Substituting (4) into (3), we get $A P=\frac{161}{3}$. Therefore,
$$
m+n=161+3=164
$$
|
164
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Lines $l_{1}, l_{2}$ both pass through the origin, and in the first quadrant, they form angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ with the positive $x$-axis, respectively. Define $R(l)$ as the line obtained by reflecting $l$ about $l_{1}$ and then reflecting the result about $l_{2}$. Let $R^{(1)}(l)=R(l), R^{\mathrm{a}}(l)$ $=R\left(R^{(n-1)}(l)\right)(n \geqslant 2)$. If the equation of line $l$ is $y=\frac{19}{92} x$, find the smallest positive integer $m_{0}$ such that $R^{(m)}(l)=l$.
|
A line passing through the origin and forming angles $\theta_{0}, \theta, \lambda$ with the positive x-axis. The symmetric line $\lambda^{\prime}$ about $\lambda$ forms an angle with the x-axis of $\theta_{0} + (\theta_{0} - \theta) = 2 \theta_{0} - \theta$.
Therefore, the symmetric line $\lambda_{1}$ about $l_{1}$ passes through the origin and forms an angle with the positive x-axis of $2 \cdot \frac{\pi}{70} - \theta$. The symmetric line $\lambda_{2}$ about $l_{2}$ passes through the origin and forms an angle with the positive x-axis of
$$
2 \cdot \frac{\pi}{54} - \left(2 \cdot \frac{\pi}{70} - \theta\right) = -\frac{8 \pi}{945} + \theta.
$$
Thus, $R(\lambda)$ can be obtained by rotating $\lambda$ by $-\frac{8 \pi}{945}$. Therefore, $R^{(m)}(\lambda)$ can be obtained by rotating $\lambda$ by $-\frac{8 m}{945}$.
If $\left.R^{(m)}\right)(\lambda) = \lambda$, then $\frac{8 m}{945}$ must be an integer. Therefore, the smallest positive integer $m$ is 945.
|
945
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. In a game, two players take turns to "eat squares" from a $5 \times 7$ grid chessboard. To "eat a square," a player selects an uneaten square and moves the piece to that square, then all the squares in the quadrant formed (along the left edge of the square upwards, and along the bottom edge of the square to the right) are eaten. For example, in the right figure, moving the piece to the shaded square results in the shaded square and the four squares marked with $\times$ being eaten (the squares with lines in them were previously eaten). The goal of the game is to make the opponent eat the last square. The figure above shows a situation that can occur during the game. How many different situations can appear at most during the game process?
|
12. In this game, the height of the table formed by the un-eaten squares from left to right is not constant (as shown in the right figure). It is easy to prove that this situation is both sufficient and necessary. Moreover, each shape can be completely described by a 12-step zigzag line. This zigzag line starts from the top-left corner of the original shape to the bottom-right corner, following the boundary formed by the eaten and uneaten squares. This zigzag line can be described by a sequence of 12 letters, consisting of 7 $H$s (where $H$ represents the top edge of an uneaten square or the bottom edge of an eaten square) and 5 $V$s (where $V$ represents moving vertically down one unit from the top-right vertex of an uneaten square to the vertex of the adjacent lower square). For example, the situation given in the original figure can be described as $H H H^{2} H^{V} V H H H V \square$, and the situation shown above can be described as $V V_{H} H H H V H V Y H H$. Therefore, there are $\frac{12!}{7!5!}=792$ possible sequences of $H$ and $V$. These include the sequences $H H H H H H H V V V V V$ and $V V V V V H H H H H H H$ (i.e., the full and empty figures).
|
792
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. In $\triangle A B C$, it is known that $A B=9, B C: C A$ $=40: 41$. Find the maximum value of the area of $\triangle A B C$.
|
13. Let $AB = c$, $AC = br$, $BC = ar$, $(a < b)$. Establish a Cartesian coordinate system with point $A$ as the origin and the line $AB$ as the x-axis (as shown in the figure), set $A(0,0)$, $B(c, 0)$, $C(x, y)$.
From $\frac{BC}{AC} = \frac{a}{b}$, we get
$$
\frac{\sqrt{(x-c)^{2} + y^{2}}}{\sqrt{x^{2} + y^{2}}} = \frac{a}{b},
$$
Squaring both sides and simplifying, we get
$$
\left(x - \frac{b^{2} c}{b^{2} - a^{2}}\right)^{2} + y^{2} = \frac{a^{2} b^{2} c^{2}}{\left(b^{2} - a^{2}\right)^{2}}.
$$
This circle is called the Apollonian circle.
When the height of the triangle is the radius of the Apollonian circle, the area of the triangle is maximized. That is,
$$
S_{\triangle ABC} = \frac{1}{2} c \left(\frac{a b c}{b^{2} - a^{2}}\right).
$$
Substituting $\frac{a}{b} = \frac{40}{41}$ and $c = 9$ into the above formula, we get $S_{\triangle ABC} = 820$.
|
820
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. In $\triangle A B C$, points $A^{\prime}, B^{\prime}, C^{\prime}$ are on sides $B C, A C, A B$ respectively, $A A^{\prime}, B B^{\prime}, C C^{\prime}$ intersect at point $O$, and $\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}=92$. Find the value of $\frac{A O}{O A^{\prime}} \cdot \frac{B O}{O B^{\prime}} - \frac{C O}{O C^{\prime}}$.
|
14. As shown in the figure, since $\triangle A O B$ and $\triangle A^{\prime} O B$ have the same height, therefore,
$$
\begin{aligned}
\frac{A O}{O A^{\prime}} & =\frac{S_{\triangle A O B}}{S_{\triangle A^{\prime} O B}} \\
& =\frac{S_{\triangle C O A}}{S_{\triangle C O A^{\prime}}}+S_{\triangle C O A} \\
& =\frac{S_{\triangle A O B}+S}{S_{\triangle A^{\prime} O B}+S_{\triangle C O A^{\prime}}} \\
& =\frac{S_{\triangle B O C}+S}{S_{\triangle B O C}}.
\end{aligned}
$$
Let $x=S_{\triangle B O C}, y=S_{\triangle C O A}$,
$z=S_{\triangle A O B}$, then
$$
\frac{A O}{O A^{\prime}}=\frac{z+y}{x}, \frac{B O}{O B^{\prime}}=\frac{x+z}{y}, \frac{C O}{O C^{\prime}}=\frac{y+x}{z} \text {. }
$$
Therefore,
$$
\begin{array}{l}
\frac{A O}{O A^{\prime}} \cdot \frac{B O}{O B^{\prime}} \cdot \frac{C O}{O C^{\prime}} \\
=\frac{(z+y)(x+z)(y+x)}{x y z} \\
=\frac{z+y}{x}+\frac{x+z}{y}+\frac{y+x}{z}+2 \\
=\left(\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}\right)+2 \\
=92+2=94 \text {. } \\
\end{array}
$$
|
94
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. If there exists a positive integer $m$ such that $m!$ ends with exactly $n$ zeros, then the positive integer $n$ is called a "factorial tail number." How many non-"factorial tail number" positive integers are there less than 1992?
|
15. Let $f(m)$ be the number of trailing zeros in the number $m!$. Clearly, $f(m)$ is a non-decreasing function of $m$, and when $m$ is a multiple of 5, we have
$$
\begin{aligned}
f(m) & =f(m+1)=f(m+2) \\
& =f(m+3)=f(m+4) \\
& 4 \times 1991=7964. \text{ Using (2), it is easy to get }
$$
$$
f(7965)=1988, f(7975)=1991 .
$$
If we take the sequence (1) from the first term to $f(7979) = 1991$, we get
$$
\begin{array}{l}
\quad 0,0,0,0,0,1,1,1,1,1, \cdots, 1989, \\
1991,1991,1991,1991,1991 .
\end{array}
$$
Sequence (3) contains a total of 7980 numbers. Since each different integer appears exactly 5 times, there are 7980 / 5 = 1596 different integers in sequence (3), and they are all elements of the set $\{0,1, 2, \ldots, 1991\}$. Therefore, there are 1992 - 1596 = 396 integers that do not appear in (3). Thus, there are 396 positive integers less than 1992 that are not "factorial trailing zero numbers". (Solved by Liu Zhi, a student at Peking University)
|
396
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A lottery ticket has 50 spaces arranged in sequence, and each participant should fill in the numbers 1 to 50 without repetition on their ticket, with the order determined by themselves. The host also fills in these 50 numbers on a ticket as the answer key. If a participant's number in any space matches the number in the corresponding space on the answer key, they win the lottery. How many tickets should a participant fill out at least to ensure they will definitely win (i.e., no matter how the answer key is filled, they will have at least one winning ticket)?
|
4. 26 tickets. To ensure a win, for example, 26 tickets can be filled out as follows:
$$
\begin{array}{l}
1,2,3, \cdots, 25,26,27, \cdots, 50 \\
2,3,4, \cdots, 26,1,27, \cdots, 50 \\
3,4,5, \cdots, 1,2,27, \cdots, 50 \\
\cdots \cdots
\end{array}
$$
$$
\begin{array}{l}
25,26,1, \cdots, 23,24,27, \cdots, 50 \\
26,1,2, \cdots, 24,25,27, \cdots, 50
\end{array}
$$
At least one number will appear in the first 26 positions (since there are only 24 positions in the back). Therefore, this number must match one of the 26 tickets filled out. To prove that filling out 25 tickets may not guarantee a win, place the 25 filled tickets and one blank ticket one by one. Let's see how to fill the blank ticket so that the numbers filled in do not match any of the 25 filled tickets.
First, it is clear that 1 can be filled in an appropriate position so that its position does not match the 1 on any of the 25 tickets. Assuming that 1, 2, ..., $a-1$ have already been filled, let's see how to fill $a$. If it can be filled and is in a different position from the $a$ on the 25 tickets, then do so. If it cannot be filled, it means that all suitable positions for $a$ are already occupied. Originally, there should be no fewer than 25 suitable positions for $a$; this is because each ticket can only create one "forbidden position." Suppose the suitable positions for $a$ on the blank ticket have already been filled with $x_{1}, x_{2}, \cdots, x_{28}$. Take any empty position; clearly, there are no fewer than 25 suitable numbers for that position, so at least one of $x_{1}, x_{2}, \cdots, x_{28}$ and $a$ can be filled in that position. Since $a$ is not suitable for that position, the position is suitable for filling some number $x_{1}$. Fill $x$ in that position, and then fill $a$ in the original position of $x_{1}$ ${ }^{1}$. Continue this process until the end, and you will get a blank ticket that ensures no ticket can win.
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. On the planet, there are 100 mutually hostile countries. To maintain peace, they decide to form several alliances, with the requirement that each alliance includes no more than 50 countries, and any two countries must be in at least one alliance. How many alliances are needed at a minimum?
(a) What is the minimum number of alliances that can be formed to meet the requirements?
(b) If an additional restriction is added, that the union of any two alliances contains no more than 80 countries, how would you answer the above question?
|
6. 6 alliances. Each country should join no less than 3 alliances, so the number of alliances is at least 6.
(a) Only by dividing 100 countries into 4 groups, each with 25 countries, and then through combinations, we get $C{ }_{4}^{2}=6$ alliances.
(b) Only by dividing 100 countries into 10 groups $a_{1}, a_{2}, a_{8}, a_{4}, a_{8}, b_{1}, b_{2}, b_{8}, b_{4}$, $b_{B}$, each with 10 countries, and forming the following alliances:
$$
\begin{array}{l}
a_{1} \cup b_{1} \cup b_{2} \cup b_{8} \cup a_{3}, \\
a_{2} \cup b_{2} \cup b_{3} \cup b_{4} \cup a_{4}, \\
a_{3} \cup b_{8} \cup b_{4} \cup b_{8} \cup a_{5}, \\
a_{1} \cup b_{4} \cup b_{8} \cup b_{1} \cup a_{1}, \\
a_{3} \cup b_{5} \cup b_{1} \cup b_{2} \cup a_{2}, \\
a_{1} \cup a_{2} \cup a_{9} \cup a_{4} \cup a_{8},
\end{array}
$$
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For every $A \subset S$, let
$$
S_{\mathrm{A}}=\left\{\begin{array}{ll}
(-)^{\mid \mathrm{A}} \mid \sum_{\mathbf{a} \in \mathrm{A}} a, & A \neq \varnothing, \\
0, & A=\varnothing .
\end{array}\right.
$$
Find $\sum_{\mathrm{A} \subset \mathrm{S}} S_{\mathrm{A}}$.
|
Solve: There are $2^{n-1}$ sets $A$ satisfying $n \in A \subset S$, and there are also $2^{n-1}$ sets $B$ satisfying $n \notin B \subset S$. Note that $B$ covers all subsets of $\{1,2, \cdots, n-1\}$, and each $A$ satisfying $n \in A \subset S$ can be obtained by adding the element $n$ to some $B$. For each $B \subset \{1,2, \cdots, n-1\}$, let $f(B) = B \cup \{n\} = A$. Then $S_{B} + S_{f(B)} = (-1)^{|B|} \cdot n$. By Lemma 4, there are $2^{n-2}$ sets $B$ such that $2 \mid |B|$, hence there are also $2^{n-2}$ sets $f(B)$ such that $2 \mid |f(B)|$.
There are $2^{n-2}$ sets $B$ such that $2 \mid |B|$, hence there are also $2^{n-2}$ sets $f(B)$ such that $2 \mid |f(B)|$.
$$
\begin{array}{l}
\therefore \sum_{n \in A \subset S} S_{A} + \sum_{\substack{B \subset S-\{ \\ 2+|B|}} S_{B} = n \cdot 2^{n-2}, \\
\sum_{\substack{n \in A \subset S \\ 2+|A|}} S_{A} + \sum_{\substack{B \subset S \\ 2 \mid |B|-\{n\}}} S_{B} = \cdots n \cdot 2^{n-2}, \\
\end{array}
$$
Finally, $\sum_{A \subset S} S_{A} = n \cdot 2^{n-2} + (-n) \cdot 2^{n-2} = 0$.
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 5 Let the set $S=\{1,2, \cdots, 1000\}$. Now for any non-empty subset $A$ of $S$, let $\alpha_{\mathrm{A}}$ denote the sum of the largest and smallest numbers in $A$, then, the arithmetic mean of all such $\alpha_{\Delta}$ is $\qquad$
|
The answer is 1001.
Let $S=\{1,2, \cdots, n\}, n \in N$. Let $m_{A}, M_{A}$ represent the minimum and maximum elements of a non-empty subset $A$ of $S$, respectively. Then the arithmetic mean of all $\alpha_{A}$ is
$$
\frac{1}{2^{n}-1} \sum_{\substack{A \subset S \\ A \neq \varnothing}}\left(m_{A}+M_{A}\right).
$$
Since for $1 \leqslant k \leqslant n$, there are $2^{n-k}$ subsets $A$ satisfying $m_{A}=k$, we have
$$
\begin{array}{l}
\sum_{\substack{A \subset S \\ A \neq \infty}} m_{A}=1 \cdot 2^{n-1}+2 \cdot 2^{n-2}+\cdots+ \\
k \cdot 2^{n-1}+\cdots+n \cdot 2^{0}.
\end{array}
$$
And for $1 \leqslant k \leqslant n$, there are $2^{k-1}$ subsets $A$ such that $M_{A}=k$, hence
$$
\begin{array}{l}
\sum_{\substack{A}} M_{A}=n \cdot 2^{n-1}+(n-1) 2^{n-2}+\cdots \\
\quad+(n-k+1) 2^{n-k}+\cdots+1 \cdot 2^{0}.(3)
\end{array}
$$
Adding (2) and (3) gives
$$
\begin{array}{l}
\sum_{A \subseteq S}\left(m_{A}+M_{A}\right)=(n+1)\left(2^{n-1}+2^{n-2}\right. \\
A \neq \varnothing \\
+\cdots+2+1)=(n+1)\left(2^{n}-1\right).
\end{array}
$$
Combining (1), we know that the arithmetic mean of all such $\alpha_{A}$ is $n+1$.
From (2), we get
$$
\begin{array}{l}
2 \sum_{A \subset S} m_{A}=2^{n}+2 \cdot 2^{n-1}+\cdots \\
A \neq \varnothing \\
\quad+(t+1) 2^{n-I}+\cdots+n \cdot 2.
\end{array}
$$
Subtracting (2) from (A), we get
$$
\begin{aligned}
\sum_{\substack{A \subset S}}^{n_{A} \neq \varnothing} & =2^{n}+2^{n-1}+\cdots+2^{n-x}+2-n \\
& =2^{u+1}-n-2.
\end{aligned}
$$
Similarly, we have
$$
\sum_{\substack{A \subset S \\ A \neq \varnothing}} M_{\triangle}=(n-1) 2^{\mathbb{n}}+1
$$
Finally, let $1 \leqslant i<i \leqslant n$, then the number of subsets $A \subset S$ with $i$ and $1$ as the minimum and maximum elements, respectively, is $2^{i-1-1}$, and in this case, $\alpha_{A}=i+j$. Noting that $\alpha_{\{1\}}=2 i, 1 \leqslant i \leqslant n$. Therefore, we have the identity
$$
\begin{array}{l}
\mathbf{2} \sum_{\mathbf{1}=1}^{\mathbf{n}} i+\sum_{1 \leqslant \mathrm{i}<\mathbf{j} \leqslant \mathbf{n}}(i+j) 2^{i-1-1} \\
=(n+1) \cdot\left(2^{\mathbf{n}}-1\right),
\end{array}
$$
or $\sum_{1 \leqslant \mathrm{i}<\mathrm{j}=\mathrm{n}}(i+1) 2^{\mathrm{j}-1}=2(n+1)\left(2^{\mathrm{n}}-n-1\right)$ (8)
|
1001
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
19. For $0<x<\frac{\pi}{2}$, let $\frac{\operatorname{ctg}^{3} x}{\operatorname{ctg} 3 x}$ take all real numbers except those in the open interval $(a, b)$. Find $a+b$.
---
Note: The function $\operatorname{ctg} x$ is the cotangent function, often denoted as $\cot x$ in English.
|
19. Notice that $\operatorname{ctg} 3 x=\frac{\operatorname{ctg}^{3} x-3 \operatorname{ctg} x}{3 \operatorname{ctg}^{2} x-1}$. Let $y=\frac{\operatorname{ctg}^{3} x}{\operatorname{ctg} 3 x}, t=\operatorname{ctg}^{2} x$, then
$$
3 t^{2}-(y+1)t+3 y=0 \text {. }
$$
Thus this equation has real roots, hence
$$
\Delta=(y+1)^{2}-36 y=y^{2}-34 y+1
$$
must be non-negative. And the roots of $y^{2}-34 y+1=0$ are
$$
a=17-12 \sqrt{2} \text { and } b=17+12 \sqrt{2} \text {. }
$$
Therefore, $\Delta \geqslant 0$ if and only if $y \leqslant a$ or $y \geqslant b$. Every $y$ value outside of $(a, b)$ is attainable, and thus, we have
$$
a+b=34 \text {. }
$$
|
34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
21. Let when $0 \leqslant x \leqslant 1$, there exists a positive number $q$ such that $\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{\mathrm{t}}}{q}$ holds, find the smallest positive number $t$ that makes the above inequality true. For this smallest $t$ value, what is the minimum value of $q$ that makes the above inequality true when $0 \leqslant x \leqslant 1$?
|
21. Since $y=\sqrt{x}$ is strictly concave, for $0<x<1$, we have
$$
q \geqslant 2-(\sqrt{1+x}+\sqrt{1-x}).
$$
After two constant transformations, we get
$$
\begin{array}{l}
q \geqslant x^{1-2}\left(1+\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) \\
\cdot\left(1+\sqrt{1-x^{2}}\right).
\end{array}
$$
If $t<2$, then as $x$ approaches 0, the above expression tends to infinity. Therefore, for any fixed $q$, there exists an $x$ value sufficiently close to 0 such that the required inequality does not hold. For $t=2$, when $x=0$, the right-hand side of the above expression equals 4. Thus, $q \geqslant 4$. For $0<x \leqslant 1$, since $\sqrt{1+x}+\sqrt{1-x}<1$ and $\sqrt{1-x^{2}}<1$, when $t=2$, the minimum value of $q$ is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Test Question $A-1$. How many evil numbers have a 10-adic representation that is exactly 0,1 alternating and ends with 1?
|
Let $P_{\mathrm{k}}=1010 \cdots 101$ ( $k$ zeros) be a prime number in decimal notation, $k \geqslant 1$. Clearly, $P_{1}=101$ is a prime number. When $k>2$, if $k$ is odd, then 101 clearly divides $P_{\times 3}$. If $k$ is even, then
$$
\begin{array}{l}
11 P_{k}=11 \cdots 1(2 k+2 \text { ones}) \stackrel{\text { denote as }}{=} Q_{2 k+2}, \\
\text { hence } 11 P_{k}=Q_{k+1} \cdot R,
\end{array}
$$
$R$ is $100 \cdots 01(k$ zeros$)$, and 11 divides $R$, so $Q_{\mathrm{x}+1}$ divides $P_{\mathrm{k}}$.
Therefore, only the prime number (101) satisfies the problem's requirements.
|
101
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Test $B-5$. Let $T$ be the inscribed trapezoid $ABCD$ (counterclockwise) in the unit circle $O$, $ABCDI, AB=s_{1}$, $CD=s_{2}, OE=d, E$ is the intersection point of the heights. When $d \neq 0$, determine the minimum upper bound of $\frac{s_{1}-s_{2}}{d}$. If the minimum upper bound can be achieved, determine all such cases.
|
Solve: As shown in the figure, establish a rectangular coordinate system with $O$ as the origin, $AB \perp x$-axis, and the coordinates of $E$ are $(d, 0)$. The equation of $BD$ can be set as $x - d = k y$, where $k^{-1}$ is the slope of line $BD$. Since $B$ and $D$ are on the unit circle, we have
$$
(d + k y)^{2} + y^{2} = 1.
$$
Thus, $\left(k^{2} + 1\right) y^{2} + 2 d k y + \left(d^{2} - 1\right) = 0$.
As shown in the figure, let $B\left(x_{1}, y_{1}\right)$ and $D\left(x_{2}, y_{2}\right)$, then $y_{1} > y_{2}$ are the roots of (5), and $y_{1} > 0, y_{2} < 0$, so
$$
\begin{array}{l}
s_{1} - s_{2} = -2 y_{1} - 2 y_{2} = \frac{4 d k}{k^{2} + 1}, \\
\frac{s_{1} - s_{2}}{d} = \frac{4 k}{k^{2} + 1} \leqslant 2.
\end{array}
$$
The upper bound 2 in (6) is reached if and only if $k = 1$, which means that line $BD$ forms an angle of $\frac{\pi}{4}$ with the $x$-axis. By symmetry, this also means that $AC \perp BD$ (but the intersection point $E$ does not coincide with point $O$) when it is reached.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Try to find the four-digit number $\overline{x x y y}$, such that it is a perfect square.
|
Solution: Since the required square number $\overline{x x y y}=11 \cdot(100 x + y)$, therefore, $100 x+y$ can be divided by 11, i.e., $11 | (x + y)$. Noting that $x$ and $y$ are both digits, thus $x+y=11$. After verification, only $x=7, y=4$ satisfies the condition, hence the required four-digit square number is 7744.
|
7744
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Five monkeys found Wang Yiguan's peaches. How can they be fairly divided when one monkey secretly gets up, eats one peach, and the remaining peaches are exactly divided into five equal parts, after which it hides its share and goes back to sleep. The second monkey gets up, also eats one peach, and the remaining peaches are again exactly divided into five equal parts, after which it also takes its share and goes back to sleep. The third, fourth, and fifth monkeys all do the same. How many peaches were there at the very beginning?
|
This problem has many solutions, for example, it can first be transformed into an indeterminate equation, and then its positive integer solutions can be sought. However, there is a simpler solution: Imagine that when Sun Wukong, the Great Sage Equal to Heaven, found his subordinates facing such a difficult problem, he had a kind heart and pulled out four hairs from his sleeve to turn into four monkeys, who added to a pile of peaches while the Monkey King was sleeping. Thus, when the first monkey came, it could exactly divide the pile of peaches into five equal parts. Suppose the part it took did not contain any peaches transformed from the hairs, and the quantity it obtained was the same as what the others would get (five equal parts). This scenario can be deduced further. Because the integer 4, plus the four hair-transformed monkeys, the number of peaches in the pile can be divided by 5 five times in succession, which means: the original pile of peaches had a minimum number of
$$
5^{5}-4=3121.
$$
The mathematical puzzle in the above example 4 can actually serve as a math competition problem for elementary school students. We can see closely that such problems can also fascinate the greatest scholars. This fully demonstrates that good mathematical puzzles are the wonders in the garden of wisdom that all those who are diligent in thinking can enjoy.
|
3121
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. The numbers 1447, 1005, and 1231 have something in common. That is, each number is a four-digit number starting with $\mathrm{i}$, and in each four-digit number, exactly two digits are the same. How many such four-digit numbers are there? (1st AIME)
|
Solve for the four-digit numbers that must contain one 1 or two 1s.
(i) The case with two $\hat{\mathrm{i}} 1$s
Choose two numbers from the remaining 9 digits (excluding 1), which has $C_{9}^{2}$ ways. Then, form any permutation of a three-digit number with these two numbers and one 1, which has $P_{3}^{3}$ ways. Thus, the four-digit numbers starting with 1 are $N_{1}=C_{9}^{2} \cdot P_{3}^{2}$ (numbers).
(ii) The case with only one 1
Choose two numbers from the remaining 9 digits $\left(C_{9}^{2}\right)$, and one of these numbers is repeated $\left(C_{2}^{1}\right)$. The three digits can form a three-digit number in $\frac{P_{3}^{3}}{2}$ ways. Thus, the four-digit numbers starting with 1 are $N_{2}=\frac{C_{9}^{2} \cdot C_{2}^{1} \cdot P_{3}^{3}}{2}$ (numbers).
Therefore, the total number of four-digit numbers that meet the criteria is
$$
N=N_{1}+N_{2}=432 \text { (numbers). }
$$
|
432
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Use 6 white beads, 8 blue beads, and 1 red bead to form a string. How many different ways can this be done?
保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a repetition of the instruction and should not be part of the translation. Here is the corrected version:
Example 2. Use 6 white beads, 8 blue beads, and 1 red bead to form a string. How many different ways can this be done?
|
This is a circular permutation problem. If we fix the red bead, it turns into a linear permutation problem. By the principle of the whole, apart from the red bead, the remaining 14 beads have $N_{1}=\frac{P_{14}^{14}}{P_{6}^{6} \cdot P_{8}^{8}}=3003$ ways to be strung.
Below, we discuss two cases based on whether the 14 beads are symmetrical or asymmetrical relative to the red bead.
(1) The number of ways to string the beads symmetrically about the red bead is
$$
N_{2}=\frac{P_{7}^{7}}{P_{3}^{3} \cdot P_{4}^{4}}=35 \text { (ways). }
$$
(2) The number of ways to string the beads asymmetrically about the red bead is
$$
N^{\prime}=N_{1}-N_{2}=2963 \text { (ways). }
$$
If only beads of the same color are swapped, the resulting stringing methods are considered the same, so the number of asymmetrical stringing methods relative to the red bead is
$$
N_{2}=\frac{1}{2} N^{\prime}=1484 \text { (ways). }
$$
Therefore, the total number of different stringing methods is $N=N_{2}+1-N_{3}=1519$ (ways).
Note: When solving such problems, the principle of classification often starts from the process of the problem, while the principle of the whole often starts from the essence of the problem. They observe the problem from different angles. Therefore, when applying them, one should not emphasize one aspect while neglecting the other.
|
1519
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
List 3. From $1,2,3, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, such that $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways are there to select the numbers that meet the above requirements? (1389, National High School League)
|
Solution: Obviously, $a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+(14-$ $\left.a_{3}\right)=14$, where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-$ $a_{3} \geqslant 0$.
Transform the above equation to
$$
\left(a_{1}-1\right)+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+(14
$$
$\left.-a_{3}\right)=7$.
At this point, $a_{1}-1, a_{2}-a_{1}-3, a_{3}-a_{2}-3,14-c_{3}$ are all non-negative integers. It is known that this indeterminate equation has $C_{i+4-1}^{\gamma}=$ $C_{10}^{?}$ different sets of non-negative integer solutions, therefore, the number of different ways that meet the requirements is $C_{10}^{3}=120$ (using the combination formula).
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
List 5. Arrange 8 cards $A A B B C D E F$ in a row, the number of arrangements where the same letter cards are allowed to be adjacent is how many?
Arrange 8 cards $A A B B C D E F$ in a row, with the same letter cards allowed to be adjacent. How many such arrangements are there?
|
Let the number of ways to arrange 8 cards in a row be $S$, then $|S|=\frac{P_{8}^{8}}{P_{2}^{2} \times \bar{P}_{2}^{2}}$; let the situation where the two cards with the letter $\Lambda$ are adjacent be $A_{1}$, and the situation where the two cards with the letter $B$ are adjacent be $A_{2}$, then $\left|A_{1}\right|:=\left|A_{2}\right|=\frac{P_{i}^{\rangle}}{P_{i}^{?}}$.
If both the two $A$s and the two $B$s are adjacent, the situation is $A_{1} \cap A_{2}$, and $\left|A_{1} \cap A_{2}\right|=D_{6}^{6}$, and the situation where no identical letters are adjacent is $\overline{A_{1}} \cap \overline{A_{2}}$. By the principle of inclusion-exclusion,
$$
\begin{array}{l}
=5760 \text { (ways). } \\
\end{array}
$$
That is, there are 5760 arrangements that meet the requirements of the problem.
|
5760
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. A person wrote 6 letters to 6 different people and prepared 6 envelopes with the recipients' addresses written on them. How many ways are there to place the letters into the envelopes so that no letter matches the recipient on the envelope? (Polish Competition Question)
|
Let's denote the set of all ways to place 6 letters into 5 different envelopes (i, $2, \cdots, 6$) such that one letter is left out, and the remaining letters are placed into the 5 envelopes in any order, as $\Lambda_{i}$. Then, $\left|A_{i}\right|=P_{5}^{5}=5!$.
Similarly,
$$
\begin{array}{l}
\left|A_{i} \cap A_{j}\right|=P_{4}^{4}=4!, \\
\cdots \cdots \cdots \\
\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \cap A_{5} \cap A_{6}\right|=P_{1}^{1}=1!.
\end{array}
$$
The scenario where none of the 6 letters match the recipient on the envelopes is $\overline{A_{1}} \cap \overline{A_{2}} \cap \cdots \cap \overline{A_{6}}$.
By the principle of inclusion-exclusion,
$$
\begin{array}{l}
\left|\overline{A_{1}} \cap \overline{\Lambda_{2}} \cap \cdots \cap \overline{A_{5}}\right| \\
=|S|-\sum_{1 \leq i \leq 6}\left|A_{i}\right|+\sum_{1 \leq i<j \leq 6}\left|A_{1} \cap A_{1}\right|-\cdots+ \\
(-1)^{6}\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \cap A_{5} \cap A_{5}\right| \\
=6!-C_{6}^{1} \cdot 5!+C_{6}^{2} \cdot 4!-C_{6}^{3} \cdot 31 \\
+C_{6}^{4} \cdot 2!-C_{6}^{5} \cdot 1!+C_{6}^{5} \\
=265 .
\end{array}
$$
Thus, the number of ways to place the letters such that none of the letters match the recipient on the envelopes is 265.
|
265
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. There are exactly 35 consecutive natural numbers whose integer parts of the arithmetic square roots are the same. Then, what is this identical integer?
---
The translation maintains the original text's format and line breaks as requested.
|
Analyzing to determine the integer part of a real number $a$ actually means finding two adjacent integers $n$ and $n+1$, such that $n \leqslant a < n+1$, at this point the integer part of $a$ is $n$. Using the property $2^{\circ}$ above, the answer to this question is 17.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. The number of natural numbers $n$ that make $n^{2}-19 n+91$ a perfect square is?
Will the above text be translated into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Notice that $n^{2}-19 n+91=(n-9)^{2}+(10-$ $n$ ). When $n>10$,
$$
(n-10)^{2}<(n-9)^{2}+(10-n)<(n-9)^{2} .
$$
The integers between two consecutive perfect squares cannot be perfect squares, therefore the natural numbers $\pi$ that make $n^{2}-10 n+91$ a perfect square can only be from the ten numbers $1,2,3, \cdots, 9,10$. After checking the perfect squares, the answer to the original problem is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The graph of a quadratic function passes through $(1,0),(5, 0)$, the axis of symmetry is parallel to the $y$-axis, but does not pass through points above the line $y=2x$. Then the product of the maximum and minimum values of the coordinates of its vertex is $\qquad$ .
|
1. 4.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Toss a coin, if it lands heads, point $P$ moves +1 on the number line, if it lands tails, it does not move. The coin is tossed no more than 12 times, and if point $P$ reaches coordinate +10, no more tossing occurs. Then the total number of different ways for point $P$ to reach coordinate +10 is .
|
3. 66.
Obviously, the last coin toss must be heads.
(i) To get 10 points with 10 heads, there is 1 way.
(ii) For 11 tosses, since the last one is heads, the first 10 must include 1 tails, which has $C_{10}^{1}=10$ ways.
(iii) For 12 tosses, since the last one is heads, the first 11 must include 2 tails, which has $C_{10}^{2}=55$ ways.
In total, there are $1+10+55=66$ ways.
|
66
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $f(x)=(\sin x+4 \sin \theta+4)^{2}+(\cos x$ $-5 \cos \theta)^{2}$, the minimum value of $f(x)$ is $g(\theta)$. Then the maximum value of $g(\theta)$ is
|
6. 49.
For $f(x)$, expanding and rearranging yields
$$
\begin{aligned}
f(x) & =8(1+\sin \theta) \sin x-10 \cos \theta \cdot \cos x \\
& -9 \sin ^{2} \theta+32 \sin \theta+42 \\
& =\sqrt{64(1+\sin \theta)^{2}+100 \cos ^{2} \theta} \\
& \cdot \sin (x+\varphi)-9 \sin ^{2} \theta+32 \sin \theta \\
& +42 . \\
g(\theta) & =-\sqrt{6} 4(1+\sin \theta)^{2}+100 \cos ^{2} \theta \\
& \cdots-9 \sin ^{2} \theta+32 \sin \theta+42 \\
& =-2 \sqrt{-9 \sin \theta+32 \sin \theta+41} \\
& -9 \sin ^{2} \theta+32 \sin \theta+42 .
\end{aligned}
$$
Let $t=\sqrt{-9 \sin ^{2} \theta+32 \sin \theta+41}$,
then $0 \leqslant t \leqslant 8$.
$$
g(\theta)=t^{2}-2 t+1=(t-1)^{2}, 0 \leqslant t \leqslant 8 .
$$
Thus, $g(\theta)_{\max }=(8-1)^{2}=49$.
|
49
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In the letter摘登 of the sixth issue of "Middle School Mathematics" in 1992, Comrade Yang Xuezhi from Fuzhou No. 24 High School mentioned a conjecture by a senior high school student: "Given $\triangle P_{1} P_{2} P_{3}$ and a point $P$ inside it. The lines $P_{1} P, P_{2} P, P_{3} P$ intersect the opposite sides at $Q_{1}, Q_{2}$, $Q_{3}$. Prove:
$$
\frac{P_{1} P}{P Q_{1}}+\frac{P_{2} P}{P Q_{2}}+\frac{P_{3} P}{P Q_{3}} \geqslant 6 "
$$
|
First, it should be noted that the above "conjecture" has long been proposed as a correct proposition and has been proven. Now, we introduce a proof method as follows (figure omitted).
Proof: Let the areas of $\triangle P_{2} P P_{3}, \triangle P_{3} P P_{1}, \triangle P_{1} P P_{2}$ be $S_{1}, S_{2}, S_{3}$, respectively. Then,
$$
\begin{array}{c}
\frac{P_{1} P}{P Q_{1}}=\frac{S_{\triangle P_{1} P_{2} P}}{S_{\triangle P P_{2} Q_{1}}}=\frac{S_{\triangle P_{3} P_{1} P}}{S_{\triangle P Q_{1} P_{3}}} \\
\quad=\frac{S_{\triangle P_{1} P_{2} P}+S_{\triangle P_{3} P_{1} P}}{S_{\triangle P P_{2} Q_{1}}+S_{\triangle P Q_{1} P_{3}}} \\
=\frac{S_{2}+S_{3}}{S_{1}} .
\end{array}
$$
Similarly, we have
$$
\frac{P_{2} P}{P_{2} Q_{2}}=\frac{S_{3}+S_{1}}{S_{2}}, \frac{P_{3} P}{P Q_{3}}=\frac{S_{1}+S_{2}}{S_{3}} \text {. }
$$
Therefore, $\frac{P_{1} P}{P Q_{1}}+\frac{P_{2} P}{P Q_{2}}+\frac{P_{3} P}{P Q_{3}}$
$$
\begin{array}{l}
=\frac{S_{2}+S_{3}}{S_{1}}+\frac{S_{3}+S_{1}}{S_{2}}+\frac{S_{1}+S_{2}}{S_{3}} \\
=\left(\frac{S_{2}}{S_{3}}+\frac{S_{3}}{S_{2}}\right)+\left(\frac{S_{3}}{S_{1}}+\frac{S_{1}}{S_{3}}\right) \\
+\left(\frac{S_{1}}{S_{2}}+\frac{S_{2}}{S_{1}}\right) \geqslant 2+2+2=6 .
\end{array}
$$
|
6
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Five, 10 people go to the bookstore to buy books, it is known that
(1) Each person bought three books;
(2) Any two people have at least one book in common.
How many people at most bought the most purchased book? Explain your reasoning.
(Na Chengzhang provided the question)
|
Solution 1: Let's assume that each person buys at most one copy of the same book. Thus, by condition (1), 10 people buy a total of 30 books.
Let the minimum number of people who buy the most popular book be $n$. Assume $A$ buys books 甲, 乙, and 丙.
Clearly, $n > 3$. Because if $n \leqslant 3$, then at most two more people can buy 甲, two more can buy 乙, and two more can buy 丙. Therefore, at most 7 people can buy 甲, 乙, and 丙, and the remaining 3 people cannot share a book with $A$, which is a contradiction.
If $n=4$. In this case, we can prove that each book is bought by exactly 4 people. Because otherwise, at least one book is bought by at most three people. Assume the 甲 book bought by $A$ is bought by at most three people. That is, apart from $A$, 甲 is bought by at most two more people, and 乙 and 丙 are each bought by at most three more people. Thus, 甲, 乙, and 丙 are bought by at most $1+2+3+3=9$ people, and the remaining one cannot share a book with $A$, which is a contradiction.
The above argument shows that when $n=4$, each book is bought by exactly 4 people. However, 4 does not divide 30, which is a contradiction. Therefore, $n \geqslant 5$. In fact, $n=5$.
Below is an example for $n=5$: Assume there are 1, 2, ..., 7 seven types of books. The 10 people's book purchase situation is as follows:
\begin{tabular}{|c|c|c|c|c|}
\hline 2 & $3),(1$ & 4 & $5),(2$ & 4 \\
\hline 4 & $6),(2$ & 5 & $7),(2$ & 5 \\
\hline 4 & $7),(3$ & 4 & $7),(3$ & 5 \\
\hline
\end{tabular}
Solution 2: Using numbers to represent the book types, then when 10 people buy books as follows:
$$
\begin{array}{l}
(123),(134),(145),(156),(126), \\
\text { (235), (245), (246), (346), (356) }
\end{array}
$$
it satisfies the requirements of the problem and each type of book is bought by exactly 5 people, so the minimum value is no more than 5.
Assume the minimum value is 4, and 10 people buy a total of $n$ types of books, with the $i$-th type of book being bought by $m_{i}$ people. Thus, $m_{i} \leqslant 4$ and $m_{1} + m_{2} + \cdots + m_{n} = 30$. When two people buy the same book, it is called a "book pair". By the given condition, each pair of people has at least one book pair, so there are at least $C_{10}^{2} = 45$ book pairs. On the other hand, the number of book pairs formed by the $i$-th type of book is $C_{m_{i}}^{2}$, so there are a total of $C_{m_{1}}^{2} + C_{m_{2}}^{2} + \cdots + C_{m_{n}}^{2}$ book pairs. Therefore, there must be
$$
C_{m_{1}}^{2} + C_{m_{2}}^{2} + \cdots + C_{m_{n}}^{2} \geqslant 45.
$$
However, since $C_{4}^{2} = 6$, $C_{3}^{2} = 3$, and $C_{2}^{2} = 1$, we also know
$$
C_{m_{1}}^{2} + C_{m_{2}}^{2} + \cdots + C_{m_{n}}^{2} \leqslant 7 C_{4}^{2} + C_{2}^{2} = 43.
$$
Since (1) and (2) are contradictory, the minimum value must be 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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