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9. Let the function $f_{0}(x)=|x|, f_{1}(x)=\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$, then the area of the closed part of the figure enclosed by the graph of the function $f_{2}(x)$ and the $x$-axis is $\qquad$ | Answer:
As shown in the figure, the area of each small triangle is 1, so the area of the closed part formed by the graph of the function $f_{2}(x)$ and the $x$-axis is 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
【Question 24】
In the figure, color some cells so that any $3 \times 3$ square within the figure contains exactly 4 colored cells. How many cells need to be colored at a minimum? | 【Analysis and Solution】
The problem of extremum.
First, we consider the 4 $3 \times 3$ squares in Figure 1;
To ensure that each of these 4 $3 \times 3$ squares has at least 4 colored cells, and to minimize the number of colored cells; we first consider the overlapping parts of these 4 $3 \times 3$ squares;
The cell at ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Let $P$ be any point on the graph of the function $y=x+\frac{2}{x}(x>0)$, and draw perpendiculars from point $P$ to the line $y=x$ and the $y$-axis, with the feet of the perpendiculars being $A$ and $B$, respectively. Then the value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is $\qquad$ . | Solution: -1.
【Method 1】Let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)$, then the equation of the line $P A$ is
$$
y-\left(x_{0}+\frac{2}{x_{0}}\right)=-\left(x-x_{0}\right) \text {, i.e., } y=-x+2 x_{0}+\frac{2}{x_{0}} \text {. }
$$
From $\left\{\begin{array}{l}y=x, \\ y=-x+2 x_{0}+\frac{2}{x_{0}},\end{array}\right.... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. The two circles in the right figure have only one common point $A$, the diameter of the larger circle is 48 cm, and the diameter of the smaller circle is 30 cm. Two beetles start from point $A$ at the same time, moving in the direction indicated by the arrows at the same speed along the two circles. Question: How m... | 13. 【Solution】The distance between any two points inside a circle is farthest when measured between the endpoints of the diameter. If the beetle crawling along the smaller circle reaches point A, and the beetle crawling along the larger circle reaches point B, the distance between the two beetles is maximized. The circ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16. In $\triangle A B C$, $\angle A B C=70^{\circ}, \angle A C B=30^{\circ}, P$ is a point inside the triangle, $\angle P B C=40^{\circ}, \angle P C B=20^{\circ}$. Prove that: $\frac{C A \cdot A B \cdot B P}{A P \cdot P C \cdot C B}=1$.
(Mathematical Bulletin Problem 1306) | 16. Let $\angle P A C=x$, then $\angle P A B=80^{\circ}-x$. From $\frac{\sin \left(80^{\circ}-x\right)}{\sin x} \cdot \frac{\sin 40^{\circ}}{\sin 30^{\circ}} \cdot \frac{\sin 10^{\circ}}{\sin 20^{\circ}}=1$ and $\sin x=\frac{4 \sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ} \cdot \sin \left(80^{\circ}-x\rig... | 1 | Geometry | proof | Yes | Yes | olympiads | false |
1. $a \in \mathbf{R}$, find the number of non-empty subsets of $M=|x| x \in \mathbf{R}, x^{3}+\left(a^{2}+1\right) x+2 a^{2}+10=0 \mid$.
A. 0
B. 1
C. 2
D. 7 | 1. $\mathrm{B}$
1. Factorize $x^{3}+\left(a^{2}+1\right) x+2 a^{2}+10=(x+2) \cdot\left(x^{2}-2 x+a^{2}+5\right)$ $\Delta=(-2)^{2}-4\left(a^{2}+5\right)<0$ Therefore, there is only 1 real root -2. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
73. As shown in the figure, in rectangle $A B C D$, $A B=3, B C=2$, points $E$ and $G$ are the midpoints of sides $A D$ and $B C$ respectively, point $F$ is a point on $A B$, and points $E$, $G$, and $H$ are collinear. Find the area of the shaded region. | Reference answer: 3 | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16. 2. $47 \star \star$ For some natural numbers $n$, the first digit of the numbers $2^{n}$ and $5^{n}$ is the same. What are these first digits?
Will keep the format and line breaks as requested. | Since $2^{5}=32,5^{5}=3125$, we know that the leading digit can be 3. Clearly, we only need to consider $n>3$. Suppose the leading digit of both $2^{n}$ and $5^{n}$ is $a$, and they are $s+1$ and $t+1$ digits long, respectively, then
$$
a \cdot 10^{s} < 2^{n} < (a+1) \cdot 10^{s}, \quad a \cdot 10^{t} < 5^{n} < (a+1) \... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(7) The minimum value of the function $f(x)=|x-1|+|x-3|+|x-5|+|x-7|$ is | On the number line, points $A, B, C, D$ represent the real numbers $1, 3, 5, 7$ respectively, and point $P$ represents the real number $x$, then
$$
f(x)=|P A|+|P B|+|P C|+|P D|.
$$
Obviously,
$$
|P A|+|P D| \geqslant|A D|=6,
$$
the equality holds if and only if $P$ is on the segment $A D$ (including endpoints);
$$
|P... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
35. Super Wings LeDy is folding a part of a paper airplane. As shown in Figure 1, he first divides a rectangular piece of paper with an area of $60 \mathrm{~cm}^{2}$ into two equal parts using line segment $M N$, then folds the rectangular paper along $M N$ to get Figure 2, and then folds Figure 2 along the axis of sym... | $12$ | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12.94 The equation $x^{n}+(2+x)^{n}+(2-x)^{n}=0$ has a rational solution, the necessary and sufficient condition regarding the positive integer $n$ is what?
(15th Putnam Mathematical Competition, 1955) | [Solution] The necessary and sufficient condition is $n=1$.
When $n=1$, the equation
$$
x^{n}+(2+x)^{n}+(2-x)^{n}=0 .
$$
has a rational solution
$$
x=-4 \text {. }
$$
Next, we prove that $n=1$ is also necessary.
First, because $x, 2+x, 2-x$ cannot all be 0, $n$ cannot be an even number.
When $n$ is an odd number, sup... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. There is a sequence of numbers: $1,1,2,2,4,8,12,96,108, \cdots \cdots$, where the 3rd number is the sum of the two preceding numbers, the 4th number is the product of the two preceding numbers, the 5th number is again the sum of the two preceding numbers, the 6th number is again the product of the two preceding numb... | $0$ | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 21 As shown in Figure 1.4.23, in isosceles $\triangle ABC$, $AB=AC, \angle A=120^{\circ}$, point $D$ is on side $BC$, and $BD=1, DC=2$. Find the length of $AD$. | From the given conditions, we can cut $\triangle A B D$ and move it to the position of $\triangle A C D^{\prime}$. From $\angle A C B=\angle A B C=\angle C A D^{\prime}, A D=D^{\prime} C$, we know that quadrilateral $D C D^{\prime} A$ is an isosceles trapezoid.
Construct $A M \perp C D$ at $M$, then we can find
$$
A M=... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. A two-digit number, when the tens digit and the units digit are swapped, the resulting two-digit number is $\frac{7}{4}$ times the original number. How many such two-digit numbers are there?
(A) 1;
(B) 2;
(C) 4;
(D) infinitely many;
(E) 0.
Answer ( $\quad$ ) | 8. (C)
Let the tens digit of the original number be $a$, and the units digit be $b$, then
$$
\frac{7}{4}(10 a+b)=10 b+a \text {, }
$$
which simplifies to $2 a=b$.
$\because \quad 1 \leqslant a, b \leqslant 9$.
$\therefore a=1,2,3,4$. Then $b=2,4,6,8$.
Therefore, the numbers that meet the condition are $12,24,36,48$, ... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
$4 \cdot 248$ To make the equation
$$
\left[\frac{10^{n}}{x}\right]=1989
$$
have integer solutions, what is the smallest value of the positive integer $n$? | [Solution] From the equation, we get
$$
1989 \leqslant \frac{10^{n}}{x} \cdot 0. \text{ Thus, from (1), we have }
$$
$$
10^{n} \cdot \frac{1}{1990}<x \leqslant 10^{n} \cdot \frac{1}{1989} \text {, }
$$
which means
$$
10^{n} \cdot 0.0005025 \cdots<x \leqslant 10^{n} \cdot 0.0005027 \cdots
$$
When $n=1,2,3, \cdots, 6$,... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 30 (20th Irish Mathematical Olympiad) Find the number of consecutive trailing zeros in 2007! and its last non-zero digit.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $F(k)=\left[\frac{2007}{k}\right]\left(k \in \mathbf{N}_{+}\right)$, then the number of times the prime 5 appears in 2007! is $F(5)+F\left(5^{2}\right)+F\left(5^{3}\right)+\cdots$.
Olympic
Math
Problem
in
Number
Theory
Since $5^{5}>2007$, the above formula is
$F(5)+F\left(5^{2}\right)+F\left(5^{3}\right)+F\left(5^{... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (5 points) From the numbers $1, 2, 3, 4, \cdots, 30$, if you arbitrarily select 10 consecutive numbers, the number of situations where there are exactly 2 prime numbers is:
cases | 4
【Analysis】A natural number, if it has only 1 and itself as factors, such a number is called a prime number; a natural number, if it has other factors besides 1 and itself, such a number is called a composite number; hence the answer. Among 10 prime numbers, if 10 consecutive numbers are chosen from the 30 numbers $1,... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. 40 students form a circle, each numbered from 1 to 40. The teacher randomly selects one student, and this student starts counting 1 to 3 in a clockwise direction. Any student who reports 1 or 2 leaves the circle. This process continues until only one student remains. The number of the last remaining student is 37. ... | 【Analysis】If the initial number of people is $3^{n}$, then starting from person 1, the last person left is the one closest to 40 in the form of $3^{n}$, which is 27. 13 people need to leave, which is an odd number, making it difficult to use. If the initial number of people is $2 \times 3^{n}$, then starting from perso... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The forty-first question: In a square with a side length of 4, if n points are placed arbitrarily, there will always be two points whose distance does not exceed $\sqrt{5}$. Find the minimum value of the integer n.
In a square with a side length of 4, if n points are placed arbitrarily, there will always be two points... | Question 41:
Solution: As shown in the figure, establish a rectangular coordinate system with $AB$ as the $x$-axis and $AD$ as the $y$-axis. Take six points $(2,0)$, $(2,2.75)$, $(0,4)$, $(4,4)$, $(0,1.25)$, and $(4,1.25)$. It can be verified that the distance between any two of these six points is greater than $\sqrt{... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. Define $a \oplus b=(a+2)(b+2)-2$. The calculation result of $1 \times 3 \times 5 \times 7 \times 9 \times 11 \times 13-(1 \bigcirc 30507 \sim 9011)$ is $\qquad$ | 【Analysis】 $a \oplus b \bullet c=[(a+2)(b+2)-2+2](c+2)-2=(a+2)(b+2)(c+2)-2$, thus 1 - 3 - $911=(1+2)(3+2) \cdots(11+2)-2=3 \times 5 \times 7 \times 9 \times 11 \times 13-2$ so the result of the original expression is 2 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\sin y \cdot \cos y+a=0 .
\end{array}\right.
$$
Then $\cos (x+2 y)=$ $\qquad$
(1994, National High School Mathematics Joint Competition) | 【Analysis】This problem appears similar to Example 2, but directly comparing the left sides of the two equations does not yield a suitable function. We can perform a simple transformation on the system of equations:
$$
\left\{\begin{array}{l}
x^{3}+\sin x=2 a, \\
(2 y)^{3}+\sin 2 y=-2 a .
\end{array}\right.
$$
We can c... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
26. If $a, b$ are positive integers, and satisfy $(a+b+3)^{2}=4\left(a^{2}+b^{2}\right)$, then the number of pairs $(a, b)$ that satisfy the equation is ( ).
(A). 0
(B). 2
(C). 4
(D). 6 | Answer: $B$.
Solution: It is clear that $a+b+3$ is an even number, so $a, b$ are one odd and one even, and if $(a, b)$ satisfies the condition, then $(b, a)$ must also satisfy it.
Without loss of generality, assume $a$ is odd and $b$ is even,
Since
$$
(a+b+3)^{2}=4\left(a^{2}+b^{2}\right),
$$
Expanding both sides, we ... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
10. Xiao Ming and Xiao Qiang have the same number of chocolates. They start eating at the same time. Xiao Ming eats 1 piece every 20 minutes and finishes the last piece at 14:40; Xiao Qiang eats 1 piece every 30 minutes and finishes the last piece at 18:00. Then, they started eating their first piece of chocolate at $\... | Answer: 8 | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$1 \cdot 75$ and $\sqrt[3]{5+2 \sqrt{13}}+\sqrt[3]{5-2 \sqrt{13}}$ equal to
(A) $\frac{3}{2}$.
(B) $\frac{\sqrt[3]{65}}{4}$.
(C) $\frac{1+\sqrt[6]{13}}{2}$.
(D) $\sqrt[3]{2}$
(E) None of these.
(31st American High School Mathematics Examination, 1980) | [Solution] Let $a=\sqrt[3]{5+2 \sqrt{13}}, b=\sqrt[3]{5-2 \sqrt{13}}$, and $x=a+b$.
Then $\quad x^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}$,
which means
$$
x^{3}=a^{3}+b^{3}+3 a b(a+b),
$$
Substituting the given values into the equation, we get $x^{3}=10+3 \sqrt[3]{-27} x$,
or
$$
(x-1)\left(x^{2}+x+10\right)=0,
$$
which h... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. (5 points) This year, the sister's age is 3 times the younger sister's age, in 2 years, the sister's age will be 2 times the younger sister's age, so, this year, the sister's age is $\qquad$ years.
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。 | 【Analysis】Let the sister's age this year be $x$ years old, then according to “this year, the older sister's age is 3 times the younger sister's age,” we know that the older sister's age this year is $3 x$ years old. In 2 years, the younger sister's age will be $x+2$ years old, and the older sister will be $3 x+2$ years... | 6 | Logic and Puzzles | other | Yes | Yes | olympiads | false |
76. Rooster one, worth five coins, hen one, worth three coins, three chicks, worth one coin, buy a hundred chickens with a hundred coins. The rooster is at most $\qquad$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
Note: The last sentence is a note and not part of the text to be translated, so it is omitted in the trans... | answer: 12 | 12 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
7. If the complex numbers $\left|z_{i}\right|=1,(i=1,2,3)$, then $\left|\frac{z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}}{z_{1}+z_{2}+z_{3}}\right|=$ | 7. $1 \quad\left|\frac{z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}}{z_{1}+z_{2}+z_{3}}\right|=\left|\frac{\overline{z_{1}}+\overline{z_{2}}+\overline{z_{3}}}{z_{1}+z_{2}+z_{3}}\right|=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. There are 10 young men, each with a different weight and height; for any two young men $\mathbf{A}$ and $\mathbf{B}$, if $\mathbf{A}$ is heavier than $\mathbf{B}$, or $\mathbf{A}$ is taller than $\mathbf{B}$, then we say “$\mathrm{A}$ is not worse than B”; if a young man is not worse than the other 9 people, he is c... | 【Analysis】Let's assume the heights of 10 people are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, and the weights of 10 people are $10, 9, 8, 7, 6, 5, 4, 3, 2, 1$. For any two people $\mathrm{A}$ and $\mathrm{B}$, either $\mathrm{A}$ is taller than $\mathrm{B}$ and $\mathrm{B}$ is heavier than $\mathrm{A}$, or $\mathrm{A}$ is heavi... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) On the clock tower of a railway station, there is an electronic clock. On the boundary of the circular clock face, there is a small colored light at each minute mark. At 9:37:20 PM, there are $\qquad$ small colored lights within the acute angle formed by the minute and hour hands. | 【Analysis】First, analyze the clock face at the hour. When it is 9 o'clock and then it moves 37 minutes and 20 seconds, calculate the distance traveled by the hour hand and the minute hand to find the difference in the number of spaces between them.
【Solution】Solution: According to the problem, we know:
Starting from 9... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
N16 (17-4, USSR) Let $A$ be the sum of the digits of the decimal number $4444^{4144}$, and let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$ (all numbers here are in decimal).
Let's translate the problem and the solution step by step.
### Problem:
N16 (17-4, USSR) Let \( A \) be the sum of th... | Solving: Since
$$
4444^{1444}<10000^{1414}=10^{4 \times 1444},
$$
the number of digits of $4444^{4444}$ does not exceed
$$
4 \times 4444=17776.
$$
$$
\begin{array}{l}
\therefore \quad A \leqslant 17776 \times 9=159984. \\
B \leqslant 1+5 \times 9=46,
\end{array}
$$
so the sum of the digits of $B$, $C \leqslant 4+9=1... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 9 Find the minimum value of the function $y=\sqrt{2 x^{2}-3 x+1}+$ $\sqrt{x^{2}-2 x}$. | Solution: First, find the domain $(-\infty, 0] \cup[2,+\infty)$, note that the functions inside the two square roots are both decreasing on $(-\infty, 0]$ and increasing on $[2,+\infty)$, so the original function is also such. Therefore, $y_{\text {min }}$ $=\min \{f(0), f(2)\}=1$. The minimum value is achieved when $x... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. The common difference of an arithmetic sequence with real terms is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. Such a sequence can have at most $\qquad$ terms. | Let the sequence $\left\{a_{n}\right\}$ have the first term $a_{1}$, and the common difference be $4 \Rightarrow S_{n}=n a_{1}+2 n(n-1)$
$$
\begin{array}{l}
\Rightarrow a_{1}^{2}+S_{n}-a_{1} \leqslant 100 \Rightarrow a_{1}^{2}+(n-1) a_{1}+2 n^{2}-2 n-100 \leqslant 0 \\
\Rightarrow\left(a_{1}+\frac{n-1}{2}\right)^{2}+\f... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. As shown in the right figure, a frog stands at position 1. On the 1st jump, it jumps 1 step to position 2; on the 2nd jump, it jumps 2 steps to position 4; on the 3rd jump, it jumps 3 steps to position 1; ...; on the $\mathbf{n}$th jump, it jumps $\mathbf{n}$ steps. When the frog jumps 20 times in a clockwise direc... | 【Answer】1.
【Analysis】Key point: Periodic problem
The frog jumps a total of: $1+2+3+\ldots+20=210$ (steps), starting from position 1, it returns to position 1 every 6 steps, $210 \div 6=35$ (sets), after jumping 210 steps, it is still at position 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (6 points) $2012 \times 2012 - 2013 \times 2011=$ | 【Solution】Solve: $2012 \times 2012-2013 \times 2011$,
$$
\begin{array}{l}
=2012^{2}-(2012+1) \times(2012-1), \\
=2012^{2}-\left(2012^{2}-1^{2}\right), \\
=2012^{2}-2012^{2}+1, \\
=1 .
\end{array}
$$
Therefore, the answer is: 1 . | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4 \cdot 27$ Given that $a$ is a natural number, a quadratic trinomial with integer coefficients and $a$ as the leading coefficient has two distinct positive roots less than 1. Find the minimum value of $a$.
| [Solution] Let $f(x)=a x^{2}+b x+c$
$$
=a\left(x-x_{1}\right)\left(x-x_{2}\right),
$$
where $0<x_{1}<x_{2}<1$.
From the given information, we know that $f(0)$ and $f(1)$ are both positive integers, so
$$
f(0) \cdot f(1) \geqslant 1 \text {, }
$$
which means $a^{2} x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right) \... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 1 (2005 Russian Mathematical Olympiad) Find the smallest positive integer that cannot be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are all positive integers. | The smallest positive integer sought is 11.
We have:
$1=\frac{4-2}{4-2}, \quad 3=\frac{8-2}{4-2}, \quad 5=\frac{16-1}{4-1}=\frac{2^{5}-2}{2^{3}-2}$,
$7=\frac{16-2}{4-2}, 9=2^{3}+1=\frac{2^{6}-1}{2^{3}-1}=\frac{2^{7}-2}{2^{4}-2}$,
$2=2 \cdot 1=\frac{2^{3}-2^{2}}{2^{2}-2}, \cdots, 10=2 \cdot 5=\frac{2^{6}-2^{2}}{2^{3}-2}... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. (16 points) Let the real-coefficient polynomial
$$
P_{i}(x)=x^{2}+b_{i} x+c_{i}\left(b_{i}, c_{i} \in \mathbf{R}, i=1,2, \cdots, n\right)
$$
be distinct, and for any $1 \leqslant i<j \leqslant n, P_{i}(x)+$ $P_{j}(x)$ has exactly one real root. Find the maximum value of $n$. | 9. The maximum value of $n$ is 3.
On the one hand, when $n=3$, take
$$
\begin{array}{l}
P_{1}(x)=x^{2}-4, P_{2}(x)=x^{2}-4 x+6, \\
P_{3}(x)=x^{2}-8 x+12 .
\end{array}
$$
Then $P_{1}(x)+P_{2}(x)=2(x-1)^{2}$,
$$
\begin{array}{l}
P_{1}(x)+P_{3}(x)=2(x-2)^{2}, \\
P_{3}(x)+P_{2}(x)=2(x-3)^{2},
\end{array}
$$
satisfying th... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Planet Beta has seven countries, each of which has exactly four friendly countries and two enemy countries. There are no three countries that are all enemies with each other. For such a planetary situation, the number of possible alliances of three countries that are all friendly with each other is $\qquad$. | 【Answer】 7
【Explanation】
Use $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{1}$ to represent seven countries, and use dashed lines to connect them to indicate enemy relationships, and solid lines to connect them to indicate friendly relationships. Then each country has 2 dashed lines and 4 solid lines. There are a total... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) As shown in the figure, in the right triangle $A B C$, $\angle A B C=90^{\circ}, A B P A^{\prime} B^{\prime}, B C P B^{\prime} C^{\prime}, A C P A^{\prime}$ $C^{\prime}$, and the distances between the three pairs of parallel lines are all 1. If $A C=10, A B=8, B C=6$, find the maximum distance from a poi... | 【Analysis】Let the point be $P$. If point $P$ is on $A^{\prime} C^{\prime}$, and the distances from $P$ to sides $A B$ and $B C$ are $a$ and $b$ respectively, then the sum of the distances from $P$ to the three sides of triangle $A B C$ is $a+b+1$. Connecting $A P$, $B P$, and $C P$, since the areas of triangles $P A B$... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
46. The distance between $A$ and $B$ is 1000 meters. Person 甲 starts from $A$ and person 乙 starts from $B$, both starting at the same time and moving back and forth between $A$ and $B$. 甲 runs at a speed of 150 meters per minute, and 乙 walks at a speed of 60 meters per minute. Within 30 minutes, 甲 and 乙 meet for the $\... | answer: 3 | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5・11 The area of the figure enclosed by the curve determined by the equation $|x-1|+|y-1|=1$ is
(A) 1 .
(B) 2 .
(C) $\pi$.
(D) 4 .
(China High School Mathematics League, 1982) | ［Solution 1］The given equation can be transformed into the following four equivalent forms
$$
\begin{array}{c}
\left\{\begin{array}{l}
x-1 \geqslant 0, \\
y-1 \geqslant 0, \\
x-1+y-1=1 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x-1 \leqslant 0, \\
y-1 \geqslant 0, \\
1-x+y-1=1 ;
\end{array}\right. \\
\left\{\begin... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
$29 \cdot 169^{1990}$ when divided by 11 leaves a remainder of
(A) 1 .
(B) 3 .
(C) 5 .
(D) 7 .
(China Shanxi Province Taiyuan City High School Mathematics Competition, 1990) | [Solution 1] $9^{1990}-1=9^{5 \cdot 398}-1=\left(9^{5}-1\right)\left[\left(9^{5}\right)^{397}+\cdots+1\right]$,
and
$$
\begin{aligned}
9^{5}-1 & =9^{2} \cdot 9^{2} \cdot 9-1 \\
& =[(7 \cdot 11+4)(7 \cdot 11+4) \cdot 9]-1 \\
& =[11 M+16 \cdot 9]-1 \\
& =\left[11 M^{\prime}+5 \cdot 9\right]-1 \\
& =11 M^{\prime}+44,
\end... | 1 | Number Theory | MCQ | Yes | Yes | olympiads | false |
1. (5 points) The left figure below is the recently discovered Archimedes' "Stomachion" puzzle, dividing a square into 14 polygons: After research, experts found that the 14 pieces can be correctly drawn on a square with a side length of $12 \mathrm{~cm}$, as shown in the figure. Question: The area of the gray piece is... | 【Analysis】According to the characteristics of the gray quadrilateral, reasonably divide the figure into two triangles to find the area.
【Solution】Solution: As shown in the figure, the gray part is quadrilateral $A B C D$, connect $B D$,
then $S$ quadrilateral $A B C D=S_{\triangle A B D}+S_{\triangle B C D}=\frac{1}{2}... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5 Call a number $T$ persistent if the following holds: Whenever $a, b, c, d$ are real numbers different from 0 and 1 such that
$$
a+b+c+d=T
$$
and
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=T
$$
we also have
$$
\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d}=T
$$
(a) If $T$ is persistent, prove that $T$... | 5 (a) Suppose $T$ is persistent. Observe that for any $u$ satisfying $|u| \geq 2$, the equation $x+1 / x=u$ has real solutions (because it can be written in the form $x^{2}-u x+1=0$, which has discriminant $\left.u^{2}-4\right)$. Choose $u$ large enough so that both $u$ and $v=T-u$ have absolute value greater than 2. T... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
31. Given that 2 is both the square root of $x-2$ and the cube root of $2 x-y+1$, then the arithmetic square root of $x^{2}-4 y$ is | Answer: 4
Solution: From the given conditions $\left\{\begin{array}{l}x-2=4 \\ 2 x-y+1=8\end{array}\right.$, we solve to get $\left\{\begin{array}{l}x=6 \\ y=5\end{array}\right.$,
$$
x^{2}-4 y=6^{2}-4 \times 5=16 \text {, }
$$
Therefore, the arithmetic square root of $x^{2}-4 y$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Let $P$ be a moving point on the ellipse $\frac{y^{2}}{4}+\frac{x^{2}}{3}=1$, and let points $A(1,1), B(0,-1)$. Then the maximum value of $|P A|+|P B|$ is $\qquad$ . | Answer: 5. Solution: Take $F(o, 1)$, then $F, B$ are the upper and lower foci of the ellipse, respectively. By the definition of an ellipse, $|P F|+|P B|=4$. Therefore, $|\mathrm{PA}|+|\mathrm{PB}|=4-|\mathrm{PF}|+|\mathrm{PA}| \leqslant 4+|\mathrm{FA}|=4+l=5$.
When $\mathrm{P}$ is at the intersection of the extension ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. If $\sin \alpha+\sin \beta=1, \cos \alpha+\cos \beta=0$, then $\cos 2 \alpha+\cos 2 \beta=$ | 6. 1 . $\sin \alpha=1-\sin \beta, \cos \alpha=-\cos \beta$, squaring and adding the two equations yields $\sin \beta=\frac{1}{2}, \sin \alpha=\frac{1}{2}$, so $\cos 2 \alpha=$ $\cos 2 \beta=1-2 \sin ^{2} \beta=1-2 \sin ^{2} \alpha=\frac{1}{2}$. Therefore, $\cos 2 \alpha+\cos 2 \beta=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Among all the triangles formed by any 3 of the 8 vertices of a rectangular cuboid, the number of acute triangles is
A. 0
B. 6
C. 8
D. 24 | Among the 8 vertices of a cuboid, the triangle formed by 3 vertices is only an acute triangle when the three edges starting from the same vertex are used. Therefore, there are 8 acute triangles in total. So the answer is $C$. | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
$$
\text { 9. } 3333^{8888}+8888^{3333} \equiv
$$
$\qquad$ $(\mod 7)$ | 9. 0 Hint: $3333 \equiv 1(\bmod 7)$, so $3333^{8888} \equiv 1(\bmod 7)$. $8888 \equiv 5(\bmod 7)$, so $8888^{3} \equiv 5^{3}(\bmod 7) \equiv-1(\bmod 7)$. Therefore, $8888^{3333} \equiv-1(\bmod 7)$. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(5) In a regular tetrahedron $ABCD$, $AO \perp$ plane $BCD$, with the foot of the perpendicular being $O$. Let $M$ be a point on the line segment $AO$ such that $\angle BMC=90^{\circ}$. Then $\frac{AM}{MO}=$ $\qquad$ . | (5) 1 Hint: As shown in the figure, connect $O B$. Let the edge length of the regular tetrahedron $A B C D$ be $a$, then
$$
O B=\frac{\sqrt{3}}{3} a, M B=\frac{\sqrt{2}}{2} a .
$$
So $M O=\sqrt{M B^{2}-O B^{2}}=\frac{\sqrt{6}}{6} a=$
$$
\begin{array}{r}
\frac{1}{2} A O=A M . \\
\text { Therefore } \frac{A M}{M O}=1 .
... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15. Let $S=\left\{m \mid m \in \mathbf{N}_{+}, m\right.$ has prime factors no greater than 10$\}$, find the smallest positive integer $n$, such that any $n$-element subset of $S$ always contains 4 distinct numbers whose product is a perfect square. | 15. Let $M$ be any $n$-element subset of $S$. According to the problem, the product of any two numbers $a$ and $b$ ($a \neq b$) in $M$ can be written as $ab = k_{ab}^2 \cdot t_{ab}$, where $k_{ab}$ is a positive integer and $t_{ab} = 2^{\alpha_1} 3^{\alpha_2} 5^{\alpha_3} 7^{\alpha_4}$ ($\alpha_i$ is 0 or 1, $1 \leq i ... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) A deck of playing cards, excluding the joker, has 4 suits totaling 52 cards, with each suit having 13 cards, numbered from 1 to 13. Feifei draws 2 hearts, 3 spades, 4 diamonds, and 5 clubs. If the sum of the face values of the 4 cards Feifei draws is exactly 34, then among them, there are $\qquad$ cards ... | 【Solution】According to the problem:
If there are 5 clubs among the 14 cards, they are all different, and the sum of the smallest 5 numbers is $1+2+3+4+5=15$;
The sum of the smallest 4 diamonds is $1+2+3+4=10$;
The sum of the smallest 3 spades is $1+2+3=6$;
The sum of the smallest 2 hearts is $1+2=3$;
$$
15+10+6+3=34 \t... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. (8 points) December has 31 days. If December 1st of a certain year is a Monday, then December 19th of that year is a Friday $\qquad$ . (Monday to Sunday are represented by the numbers 1 to 7)
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 2. (8 points) December has a total of 31 days. If December 1st of a certain year is a Monday, then December 19th of that year is a
5_. (Monday to Sunday are represented by the numbers 1 to 7)
【Solution】Solution: $19-1=18$ (days)
$18 \div 7=2$ (weeks) $\cdots 4$ (days)
4 days after a Monday is a Friday.
Answer: December... | 5 | Number Theory | proof | Yes | Yes | olympiads | false |
5. In $\triangle A B C$, the lengths of the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $c-a$ equals the height $h$ from vertex $A$ to side $AC$, then the value of $\sin \frac{C-A}{2}+\cos \frac{C+A}{2}$ is
(A) 1
(B) $\frac{1}{2}$
(C) $\frac{1}{3}$
(D) -1 | 5.【Analysis and Solution】As shown in the figure, it is clear that $1=c-a=h=a$. At this time, $\sin \frac{C-\angle A}{2}+\cos \frac{C+\angle A}{2}=\sin 30^{\circ}+\cos 60^{\circ}=1$. | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
9. (16 points) Given real numbers $a, b, c$ satisfy: $a+b+c=5, ab+bc+ca=7$.
Find the maximum value of $abc$.
保留了原文的换行和格式。 | $$
\begin{array}{l}
b(a+c)+c a=7 \Rightarrow b(5-b)+c a=7 \\
\Rightarrow c a=b^{2}-5 b+7 \leqslant\left(\frac{a+c}{2}\right)^{2}=\left(\frac{5-b}{2}\right)^{2} \\
\Rightarrow \frac{1}{3} \leqslant b \leqslant 3 .
\end{array}
$$
Then $a b c=b\left(b^{2}-5 b+7\right)$
$$
\begin{array}{l}
=b^{3}-5 b^{2}+7 b \\
=(b-1)^{2}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. As shown in the figure, $D$ is a point inside $\triangle A B C$, and it satisfies $A B=C D=4$, $\angle A+\angle B D C=180^{\circ}$. Determine the maximum value of $S_{\triangle A B C}-S_{\triangle B D C}$. | As shown in the figure, construct the symmetric point $D^{\prime}$ of $D$ with respect to $BC$. From $\angle A + \angle B D^{\prime} C = 180^{\circ} \Rightarrow A, B, C, D^{\prime}$ are concyclic, and $AB = CD^{\prime} = 4 \Rightarrow AD^{\prime} \parallel BC \Rightarrow ABCD^{\prime}$ is an isosceles trapezoid.
Constr... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Three. (25 points) A total of no more than 30 football teams from the East and West participate in the Super League. The East has 1 more team than the West. Any two teams play exactly one match. Each team earns 3 points for a win, 1 point for a draw, and 0 points for a loss. After the league ends, the statistics show t... | Three, let there be $x$ teams participating in the West, and in the matches between the East and the West, the East wins $y$ games, the West wins $z$ games, and there are $a$ draws.
According to the problem, we have
$$
(y+z): a=2: 1 \text {, }
$$
and $y+z+a=x(x+1)$.
Thus, $a=\frac{1}{3} x(x+1)$,
$$
z=\frac{2}{3} x(x+1... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. (12 points) A natural number that satisfies the following two conditions is called a "lucky number": (1) Starting from the third digit from left to right, each digit is the difference between the two preceding digits (the larger number minus the smaller number); (2) No repeated digits. For example: 132, 871, 54132 ... | 14. (12 points) A natural number that satisfies the following two conditions is called a "lucky number": (1) Starting from the third digit from left to right, each digit is the difference between the two preceding digits (the larger number minus the smaller number). (2) No repeated digits. For example: 132, 871, 54132 ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. (12 points) Cars A and B start from point A to point B at the same time. At the start, car A's speed is 2.5 kilometers per hour faster than car B's. After 10 minutes, car A slows down; 5 minutes later, car B also slows down, at which point car B is 0.5 kilometers per hour slower than car A. Another 25 minutes later... | 13. (12 points) Two cars, A and B, start from point A to point B at the same time. At the start, car A is 2.5 km/h faster than car B. After 10 minutes, car A slows down; 5 minutes later, car B also slows down, at which point car B is 0.5 km/h slower than car A. Another 25 minutes later, both cars arrive at point B at t... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. (8 points) There is a sequence of numbers: $1,3,8,22,60,164,448, \cdots$ where the first number is 1, the second number is 3, and starting from the third number, each number is exactly twice the sum of the previous two numbers. In this sequence, the remainder when the 2000th number is divided by 9 is . $\qquad$ | 【Solution】Solution: By dividing the first few terms of the sequence by 9 and taking the remainder, we get $1, 3, 8, 4, 6, 2, 7, 0, 5, 1, 3, 8$ ... which is a repeating sequence.
$2000 \div 9$ has a remainder of 2. Since "starting from the third number, each number is exactly twice the sum of the previous two numbers," ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
99. In a certain martial arts conference, there are nine levels of experts participating, ranked from highest to lowest as Ranger, Musketeer, Knight, Swordsman, Samurai, Archer, Mage, Hunter, and Priest. For fairness, the grouping rules for the competition are: two or three people form a group. If two people form a gro... | Answer: 9 | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
【Example 2】Using 4 different colors to color the 4 sides of a small square wooden block, with each side painted a different color, how many different coloring patterns can there be? | Here, we not only consider the relative order of the 4 colors, but also the fact that the blocks can be flipped, so there is no distinction between clockwise and counterclockwise relative orders. Therefore, there are
$$
\frac{1}{2 \times 4} A_{4}^{4}=\frac{4 \times 3 \times 2}{2 \times 4}=3
$$
different coloring patte... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. 42 Find the positive root of the equation $x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}$, and prove that there is only one positive root. | [Solution] (1) If $x=2$, then
$$
\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2}}}}=2 \text{.}
$$
$\therefore 2$ is the positive root of the equation.
(2) If $x>2$, let $x=2+\alpha(\alpha>0)$,
then $\because \quad(2+\alpha)^{2}=4+4 \alpha+\alpha^{2}>4+\alpha>4$,
$$
\therefore \quad x^{2}>2+x>4, x>\sqrt{2+x}>2 \text{.}
$$
Similarl... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Curve $C_{1}$ is a part of an ellipse with the origin $O$ as the center and $F_{1}, F_{2}$ as the foci, and curve $C_{2}$ is a part of a parabola with $O$ as the vertex and $F_{2}$ as the focus. Point $A$ is the intersection of curves $C_{1}$ and $C_{2}$, and $\angle A F_{2} F_{1}$ is an obtuse angle. If $\left|A F... | 10. Analysis ~(I) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, then $2 a=\left|A F_{1}\right|+\left|A F_{2}\right|=\frac{7}{2}+\frac{5}{2}=6$, which gives $a=3$. Let $A(x, y), F_{1}(-c, 0), F_{2}(c, 0)$, then $(x+c)^{2}+y^{2}=\left(\frac{7}{2}\right)^{2}$, $(x-c)^{2}+y^{2}=\left(\frac... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.47 Try to find the maximum value of the following expression:
$$
x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}} .
$$ | ［Solution］By Cauchy-Schwarz inequality, we have
$$
\left|x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right|^{2} \leqslant\left(x^{2}+y^{2}\right)\left(2-x^{2}-y^{2}\right) .
$$
Using the AM-GM inequality, we get
$$
\left|x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right| \leqslant \frac{x^{2}+y^{2}+2-x^{2}-y^{2}}{2}=1 \text {. }
$$
If $... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. (15 points) As shown in the figure, $E$ is a point on side $CD$ of parallelogram $ABCD$, and $BD$ intersects $AE$ at point $F$. It is known that the area of triangle $AFD$ is 6, and the area of triangle $DEF$ is 4. Find the area of quadrilateral $BCEF$.
保留源文本的换行和格式，直接输出翻译结果如下：
15. (15 points) As shown in the figu... | 【Solution】Solution: According to the property that the area of a triangle is proportional to its base when the height is constant, we get: $A F: E F=6: 4=3$ : 2;
Since in parallelogram $A B C D$, triangle $A F B$ is similar to triangle $D E F$, we have $A B: D E=A F: E F$ $=3: 2$;
$A B=D C$, so $D C: D E=3: 2$,
Theref... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find all prime numbers $p$ such that
$$
1+2^{p}+3^{p}+\cdots+p^{p}
$$
is a prime number. | 1. When $p=2$, $1+2^{2}=5$ is a prime number.
When $p>2$, $p$ is odd, then
$$
\begin{array}{l}
1+2^{p}+3^{p}+\cdots+p^{p} \\
=\left(1^{p}+(p-1)^{p}\right)+\left(2^{p}+(p-2)^{p}\right)+\cdots+ \\
\quad\left(\left(\frac{p-1}{2}\right)^{p}+\left(\frac{p+1}{2}\right)^{p}\right)+p^{p}
\end{array}
$$
can be divided by $p$,... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. As shown in Figure 3, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through point $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$ respectively. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array... | 14. 2 .
Since $O$ is the midpoint of side $B C$, we have
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} \text {. }
$$
Since points $M, O, N$ are collinear, we have
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 7 (1) Prove: For any natural number $a, a^{4 n+k}$ has the same units digit as $a^{k}$ $(n$, $k \in \mathrm{N}$ );
(2) Find the units digit of $2^{1997}+3^{1997}+7^{1997}+9^{1997}$. | Proof: (1) It needs to be proven that $a^{4 n+k}-a^{k} \equiv 0(\bmod 10)$
Since $a^{4 n+k}-a^{k}=a^{k}\left(a^{4 n}-1\right) \equiv 0\left(\bmod a\left(a^{4}-1\right)\right)$, it is only necessary to prove that $a\left(a^{4}-1\right) \equiv 0(\bmod 10)$.
And $a\left(a^{4}-1\right)=(a-2)(a-1) a(a+1)(a+2)+5(a-1) a(a+1)$... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6 On the surface of a sphere with radius 1, there are four non-coplanar points $A, B, C, D$, and $AB=CD=x, BC=DA=y, CA=BD=z$. Then $x^{2}+y^{2}+z^{2}$ equals ( ).
(A) 2
(B) 4
(C) 8
(D) 16 | $6 \mathrm{C}$ Hint: Construct a rectangular parallelepiped such that the six edges of $A B C D$ are the diagonals of one of its faces. At this time, the length of the space diagonal of the rectangular parallelepiped is $\sqrt{\left(x^{2}+y^{2}+z^{2}\right) / 2}$, which is exactly equal to the diameter of the circumscr... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
9. (16 points) As shown in Figure 1, $A_{1}, A_{2}, A_{3}, A_{4}$ are four distinct points on the parabola $y^{2}=4 x$, and $A_{1}, F, A_{3}$ and $A_{2}, F, A_{4}$ are collinear respectively. Let the lines $A_{1} A_{2}$ and $A_{3} A_{4}$, $A_{1} A_{4}$ and $A_{2} A_{3}$ intersect at points $M$ and $N$, respectively, wh... | (1) Let $A_{1}\left(4 t_{1}^{2}, 4 t_{1}\right), A_{2}\left(4 t_{2}^{2}, 4 t_{2}\right)$, $A_{3}\left(4 t_{3}^{2}, 4 t_{3}\right), A_{4}\left(4 t_{4}^{2}, 4 t_{4}\right)$.
Then $l_{A_{1} A_{3}}:\left(t_{1}+t_{3}\right) y=x+4 t_{1} t_{3}$.
Since the line $A_{1} A_{3}$ passes through the point $F(1,0)$, we get $t_{3}=-\f... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. In the figure below, $A$ is the midpoint of $DE$, the areas of $\triangle D C B$ and $\triangle E B C$ satisfy $S_{\triangle D C B}+S_{\triangle E B C}=12$, then the area of $\triangle A B C$ is $\qquad$ . | $6$ | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16.2.9 * Let $n$ be an integer. If the tens digit of $n^{2}$ is 7, what is the units digit of $n^{2}$? | Let $n=10 x+y, x, y$ be integers, and $0 \leqslant x, y \leqslant 9$, then
$$
n^{2}=100 x^{2}+20 x y+y^{2}=20 A+y^{2}(A \in \mathbf{N}) \text {. }
$$
Since the tens digit of $20 \cdot A$ is even, the tens digit of $y^{2}$ must be odd, which can only be $y^{2}=16$ or 36. Thus, the units digit of $y^{2}$, which is the u... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Let $\varphi(n)$ denote the number of positive integers not exceeding $n$ that are coprime to $n$, and $g(n)$ satisfies that for any $n \in \mathbf{Z}_{+}$, $\sum_{d \| n} g(d)=\varphi(n)$, where $\sum_{d \| n}$ indicates that $d$ takes all positive divisors of $n$. Then $g(50)=$ $\qquad$ | 8. 0 .
It is known that $g(1)=1, g(2)=0$.
From $4=\varphi(5)=g(1)+g(5)$
$$
\Rightarrow g(5)=3 \text {; }
$$
From $4=\varphi(10)=g(1)+g(2)+g(5)+g(10)$
$$
\Rightarrow g(10)=0 \text {; }
$$
From $20=\varphi(25)=g(1)+g(5)+g(25)$
$$
\Rightarrow g(25)=16 \text {. }
$$
Thus $20=\varphi(50)$
$$
\begin{array}{l}
=g(1)+g(5)+... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
132 There is a group of children, all of whose ages are integers. One of them is 10 years old. If the oldest is 13 years old, the sum of the ages of all the children is 50 years, and except for the 10-year-old child, the ages of the other children form an arithmetic sequence in order of size, this group of children has... | 132 5. Removing the 10-year-old child, the sum of the ages of the remaining children is 40, the oldest is 13 years old, $3 \times 13<40$, so there must be at least 4 children. The average age of the oldest and the youngest (at least 1 year old) is $\geqslant 6$, and this average is a divisor of 40, so it can only be 10... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. If the complex numbers $a, b, c$ satisfy
$$
\begin{aligned}
|a|=|b|=|c|=1=a^{2}+b^{2}+c^{2}, \\
\text { then }\left|a^{2020}+b^{2020}+c^{2020}\right|=
\end{aligned}
$$ | $$
\begin{array}{c}
\Rightarrow \cos 2 x+\cos 2 y+\cos 2 z=1, \\
\quad \sin 2 x+\sin 2 y+\sin 2 z=0 .
\end{array}
$$
The above two equations, when transformed using sum-to-product identities, yield
$$
\begin{array}{l}
2 \cos (x+y) \cdot \cos (x-y)=2 \sin ^{2} z, \\
2 \sin (x+y) \cdot \cos (x-y)=-2 \sin z \cdot \cos z ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Let $(1-2 x)^{7}=\sum_{k=0}^{7} a_{k} x^{k}$. Then $2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}+6 a_{6}+7 a_{7}=$ $\qquad$ | 10. 0 .
For $(1-2 x)^{7}=\sum_{k=0}^{7} a_{k} x^{k}$, differentiating both sides with respect to $x$ yields
$$
\begin{aligned}
- & 14(1-2 x)^{6} \\
= & a_{1}+2 a_{2} x+3 a_{3} x^{2}+4 a_{4} x^{3}+5 a_{5} x^{4}+ \\
& 6 a_{6} x^{5}+7 a_{7} x^{6} .
\end{aligned}
$$
Substituting $x=0$ and $x=1$ into the above equation, w... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
25-14 On the $x O y$ plane, the number of distinct common points of the figures: $(x+y-5)(2x-3y+5)=0$ and $(x-y+1)(3x+2y-12)=0$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(G) Infinite.
(21st American High School Mathematics Examination, 1970) | ［Solution］The two equations are equivalent to the following four linear equations:
$$
\begin{array}{ll}
\text { I }: x+y-5=0, & \text { II }: 2 x-3 y+5=0, \\
\text { III: } x-y+1=0, & \text { IV }: 3 x+2 y-12=0 .
\end{array}
$$
It is easy to verify that all four lines pass through the point $(2,3)$.
Since these lines ... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Given $n(n \geqslant 3)$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $\left\{\begin{array}{l}x_{1}+x_{2}+\cdots+x_{n}=n \\ x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=n\end{array}\right.$, then
$$
\frac{x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n-1} x_{n}+x_{n} x_{1}}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}... | $1$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. If $m$ and $n$ are both positive integers, find the minimum value of $\left|2^{m}-181^{n}\right|$.
The above text translated into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. When $m=15, n=2$, we have $\left|2^{m}-181^{n}\right|=\left|2^{15}-181^{2}\right|=7$.
Assume there exist positive integers $m, n$ such that $\left|2^{m}-181^{n}\right|<7$.
Then, by $12^{m}-181^{n} \mid$ being odd, we know $\left|2^{m}-181^{n}\right|=1$ or 3 or 5.
Note that, $2^{m}$ modulo 15 can only be $1, 2, 4, 8$... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
C15 (16-4, Bulgaria) Divide an $8 \times 8$ chessboard into $p$ rectangles, such that the division does not cut through any cell (i.e., it must follow the grid lines of the chessboard), and satisfies the following two conditions:
（1）Each rectangle contains the same number of white and black cells;
（2）If the number of w... | Since the chessboard has 32 white squares, we have
$$
a_{1}+a_{2}+\cdots+a_{p}=32 \text {. }
$$
From (2), $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{p} \geqslant p$, so
$$
\begin{array}{c}
1+2+\cdots+p \leqslant 32, \\
p^{2}+p \leqslant 64,
\end{array}
$$
thus $p \leqslant 7$. That is, the chessboard can be di... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. If vector $\overrightarrow{A B}=(3,-1), \boldsymbol{n}=(2,1), \boldsymbol{n} \cdot \overrightarrow{A C}=7$, then $\boldsymbol{n} \cdot \overrightarrow{B C}=$
A. -2
B. 2
C. -2 or 2
D. 0 | 1. $\mathrm{B}$ Hint: Since $\overrightarrow{B C}=\overrightarrow{A C}-\overrightarrow{A B}$, therefore, $\boldsymbol{n} \cdot \overrightarrow{B C}=\boldsymbol{n} \cdot(\overrightarrow{A C}-\overrightarrow{A B})=\boldsymbol{n} \cdot \overrightarrow{A C}-\boldsymbol{n} \cdot \overrightarrow{A B}=7-(3 \times$ $2-1 \times... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
14. (6th "Hope Cup" Senior High School Competition) The function $y=f(x)$ is an even function and is a periodic function with a period of 2. When $x \in [2,3]$, $f(x)=x-1$. On the graph of $y=f(x)$, there are two points $A$ and $B$, whose y-coordinates are equal ($A$ is to the left of $B$), and their x-coordinates are ... | 14. Since $f(x)$ is a periodic function with a period of 2, and $f(x)=x-1, 2 \leqslant x \leqslant 3$,
when $0 \leqslant x \leqslant 1$, $f(x)=f(x+2)=(x+2)-1=x+1$.
Also, since $f(x)$ is an even function.
when $-1 \leqslant x \leqslant 0$, $f(x)=f(-x)=-x+1$.
Therefore, when $1 \leqslant x \leqslant 2$, $f(x)=f(x-2)=-(x... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(12) (22 points) Suppose a set of planar points $S$ has the properties: (i) no three points are collinear; (ii) the distances between any two points are all different. For two points $A$ and $B$ in $S$, if there exists $C \in S$ such that $|A C|<|A B|<|B C|$, then $A B$ is called a middle edge of $S$. For three points ... | (12) (1) Color all the middle edges of $S$ red, and all other edges blue. When $n \geqslant 6$, according to Ramsey's theorem, there must exist a monochromatic triangle, which must have a middle edge, and thus must be a middle edge triangle.
(2) There exists a 5-element point set with properties (i) and (ii), but it do... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
33. Today is January 18, 2020, and it is displayed on the electronic calendar as 20200118. If there is a multi-digit number $202001188462 \cdots \cdots$, starting from the 9th digit, each digit is the unit digit of the product of the three preceding digits. According to this rule, what is the 2020th digit of this multi... | $2$ | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfy the conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find the l... | From (1), we know that $x=-1$ is the axis of symmetry of $f(x)$, so $-\frac{b}{2a}=-1, b=2a$. Also, $f(1) \geqslant 1$. From (2), we know that $f(1) \leqslant 1$, hence $f(1)=1, a+b+c=1$. Since $f(-1)=0, a-b+c=0$. Therefore, $a=\frac{1}{4}, b=\frac{1}{2}, c=\frac{1}{4}$. Thus, $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\fra... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
57. Given $0<a<1$, and it satisfies $\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18$, then $[10 a]$ $=\ldots$. ([x] represents the greatest integer not exceeding $x$, for example $[2.6]=2,[0.2]=0)$ | Answer: 6 | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given $P(n)=n^{2}+n+1$. For any positive integers $a, b$, if the set
$$
\{P(a), P(a+1), P(a+2), \cdots, P(a+b)\}
$$
has each element not coprime with the product of the other elements, then this set is called "fragrant". Find the minimum number of elements in a fragrant set. | $$
\begin{array}{l}
=\left(n^{2}+n+1, n^{2}+3 n+3\right) \\
=\left(n^{2}+n+1,2 n+2\right) \\
=\left(n^{2}+n+1, n+1\right) \\
=(1, n+1)=1 .
\end{array}
$$
Lemma 2 For $n \neq 2(\bmod 7)$, we have
$$
(P(n), P(n+2))=1 \text {; }
$$
For $n \equiv 2(\bmod 7)$, we have
$$
(P(n), P(n+2))=7 \text {. }
$$
Proof of Lemma 2 Si... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
76. 888 little monkeys obtained 888 treasure chests, and lined them up in a row, numbered $1 \sim 888$. The first little monkey hit all the chests once, the second little monkey hit the chests with numbers that are multiples of 2 once, the third little monkey hit the chests with numbers that are multiples of 3 once... ... | Answer: 8 | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The unit digit of $2+2 \times 3+2 \times 3 \times 3+2 \times 3 \times 3 \times 3+\cdots+2 \times \underbrace{3 \times \cdots \times}_{9 \text { } 3} 3$ is
A. 2
B. 8
C. 4
D. 6 | Analysis: Number Theory, Periodic Digit Problem.
Original expression $=2 \times\left(1+3+3^{2}+3^{3}+\cdots+3^{9}\right), 3^{\mathrm{n}}$'s unit digit appears in a cycle of $3, 9, 7, 1$, 3+9+7+1=20, $9 \div 4=2 \cdots 1$, so the unit digit of the original expression is $2 \times(1+20 \times 2+3) \equiv 2 \times 4=8(\bm... | 8 | Number Theory | MCQ | Yes | Yes | olympiads | false |
6. If $a \neq 0, b \neq 0$, and $\frac{1}{a}+\frac{1}{b}+\frac{2(a+b)}{a^{2}+b^{2}}=0$, then the value of $\frac{a}{b}$ is | answer: -1 | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. From a vertex of a regular $n$-sided polygon, draw $n-3$ diagonals. The sum of all angles (only considering angles less than $180^{\circ}$) formed by any two of these diagonals is $1800^{\circ}$. Then $n=$ $\qquad$ . | $12$ | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. If the sequence of positive numbers $\left\{a_{n}\right\}$ is a geometric sequence, and $a_{2} \cdot a_{6}+2 a_{4} a_{5}+a_{1} a_{9}=25$, then the value of $a_{4}+a_{5}$ is $\qquad$ . | 7. 5 .
Since $a_{2} \cdot a_{6}=a_{4}^{2}, a_{1} \cdot a_{9}=a_{5}^{2}$, we get $a_{4}^{2}+2 a_{4} a_{5}+a_{5}^{2}=25$, thus $\left(a_{4}+a_{5}\right)^{2}=25$, which means $a_{4}+a_{5}=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 4 Given $S=\{1,2,3,4\}$. Let $a_{1}, a_{2}, \cdots, a_{k}$ be a sequence of numbers from $S$, and it includes all permutations of $(1,2,3,4)$ that do not end with 1, i.e., if $\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ is a permutation of $(1,2,3,4)$ and $b_{4} \neq 1$, then there exist $1 \leqslant i_{1}<i_{2}<i... | Since $a_{1}, a_{2}, \cdots, a_{k}$ includes all permutations of $(1,2,3,4)$ where the second number is 1, there exists
$$
1 \leqslant i_{1}<i_{2}<i_{3}<i_{4}<i_{5}<\cdots<i_{m} \leqslant k,
$$
such that $\left(a_{i_{1}}, a_{i_{2}}, a_{i_{3}}\right)$ is a permutation of $(2,3,4)$, $a_{i_{4}}=1$, and $a_{i_{5}}, \cdots... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(4) Given point $O$ is inside $\triangle A B C$, and $3 \overrightarrow{A B}+2 \overrightarrow{B C}+\overrightarrow{C A}=4 \overrightarrow{A O}$, let the area of $\triangle A B C$ be $S_{1}$, and the area of $\triangle O B C$ be $S_{2}$. Then the value of $\frac{S_{1}}{S_{2}}$ is | (4) 2 Hint: From $3 \overrightarrow{A B}+2 \overrightarrow{B C}+\overrightarrow{C A}=4 \overrightarrow{A O}$, we know that
$$
3 \overrightarrow{A B}+2(\overrightarrow{A C}-\overrightarrow{A B})-\overrightarrow{A C}=4 \overrightarrow{A O}.
$$
Therefore, $\overrightarrow{A B}+\overrightarrow{A C}=4 \overrightarrow{A O}$... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
14. Calculate: $(6789+79860+87096+98607+60978) \div(51234+24315+32541+$ $43152+15423)=$ $\qquad$ - | $2$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Teacher Yangyang's city has 4 subway lines, each of which intersects with every other line at exactly one point, serving as the transfer station for those two lines. One day, Teacher Yangyang had a sudden idea to start from the subway station near her home (which is not a transfer station), make at least one transf... | 【Answer】8
【Analysis】First, for 4 lines, each pair has a transfer station, so there are 6 transfer stations in total. Using connection points to represent subway lines and line segments to represent transfer stations, it can be represented as follows:
Then the problem becomes, starting from any point on the graph except... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$$
\underbrace{2 \times 2 \times \ldots \times 2}_{20 \uparrow 2}-1
$$
The result's units digit is $\qquad$ | 【Answer】 5
Analysis: $2 \div 10$ remainder is $2,(2 \times 2) \div 10$ remainder is $4,(2 \times 2 \times 2) \div 10$ remainder is $8,(2 \times 2 \times 2 \times 2) \div 10$ remainder is 6, that is, 4 twos multiplied together form one cycle, $20 \div 4=5$, so $\underbrace{2 \times 2 \times \ldots \times 2}_{20 \text { ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11. (10 points) A three-digit number $N$, when subtracted by 3, added by 4, divided by 5, and multiplied by 6, results in four integers. It is known that the sum of the digits of these four numbers are exactly 4 consecutive natural numbers. How many three-digit numbers $N$ satisfy this condition?
A. 8
B. 6
C. 4
D. 2 | 【Solution】Solution: Considering there will definitely be carrying and borrowing, let the sum of the original number's digits be $a$. Then, $-3, +4$ cannot have a difference of 7, otherwise it cannot form 4 consecutive natural numbers. $\div 5$ indicates the last digit is 0 or 5. When the last digit is 5, $-3, +4$ do no... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
3. As shown in the calendar for March 2006, $A+B+C+D=52$, then, the first Sunday of this month is March $\qquad$.
Translating the question while preserving the format and blank spaces as requested. | answer: 5 | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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