problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
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5. In the spatial quadrilateral $ABCD$, it is known that $AB=2, BC=3, CD=4, DA=5$, then $\overrightarrow{AC} \cdot \overrightarrow{BD}=$ | $$
\begin{array}{l}
\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A D}+\overrightarrow{D C}, \overrightarrow{B D}=\overrightarrow{B A}+\overrightarrow{A D}=\overrightarrow{B C}+\overrightarrow{C D}, \\
\text { then } 4 \overrightarrow{A C} \cdot \overrightarrow{B D}=(\overrightarrow{A B... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 3.3.4 There are 2005 same-sized circular paper pieces placed on a table without overlapping. How many different colors are needed at least to ensure that each piece can be colored with one color, so that any two touching circular paper pieces are colored differently? | Solution: Let the number of colors that can ensure the requirement be $k$, then $k>3$.
That is, three colors cannot guarantee the requirement. For example, in the figure, $A, B, E$ can only be colored with 1 and 3, and $A$ is one of them, while $B, E$ are the other. At this point, $C, D$ can only be colored with 2, wh... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. The plane $\alpha$ forms angles $\theta_{i}(i=1,2, \cdots, 6)$ with the six faces of a rectangular prism, then the value of $\sum_{i=1}^{6} \sin ^{2} \theta_{i}$ is
A. 2
B. 3
C. 4
D. 6 | Consider the normal vector $\overrightarrow{A C_{1}}$ of $\alpha$, and construct a rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ with faces parallel to the original rectangular prism. Then, the sum of the cosines of the angles formed by $A C_{1}$ and the faces of the rectangular prism $A B C D-A_{1} B_{1} C_{1} D... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
19. (6 points) Use small cubes with edge length $m$ to form a large cube with edge length 12. Now, paint the surface (6 faces) of the large cube red, where the number of small cubes with only one red face is equal to the number of small cubes with only two red faces. Then $m=$ $\qquad$ . | 【Solution】Solution: According to the problem, there are $12 \div m$ small cubes on each edge of the large cube. Let $12 \div m=n$, which means there are $n$ small cubes on each edge of the large cube,
$$
\begin{array}{c}
6(n-2)^{2}=12(n-2) \\
(n-2)^{2}=2(n-2) \\
n-2=2 \\
n=4
\end{array}
$$
Since $12 \div m=4$
Therefor... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$1 \cdot 15$ Given 5 distinct positive numbers can be divided into two groups such that the sums of the numbers in each group are equal, how many different ways are there to divide these numbers into such groups?
The above text translated into English, keeping the original text's line breaks and format, is as follow... | [Solution] (1) If the number of elements in the two groups with equal sums are 1 and 4, respectively, i.e., the largest number among the 5 numbers equals the sum of the other 4 numbers, then there is obviously only 1 way to divide them.
(2) Suppose the number of elements in the two groups with equal sums are 2 and 3, r... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12. For a positive integer $n, n!=1 \times 2 \times \cdots \times n$. If a positive integer $N$ satisfies $5! \times 9! = 12 \times N!$, then the value of $N$ is ( ).
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14 | 12. A.
Notice that, $5!=120$.
Then $120 \times 9!=12 \times N$ !
$$
\begin{array}{l}
\Rightarrow N!=10 \times 9!=10! \\
\Rightarrow N=10 .
\end{array}
$$ | 10 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. Let the function $f_{0}(x)=|x|, f_{1}(x)=\left|f_{0}(x)-1\right|, f_{2}(x)=\mid f_{1}(x)$ $-2 \mid$, then the area of the closed part of the figure enclosed by the graph of the function $y=f_{2}(x)$ and the $x$-axis is $\qquad$. | 3. Draw the graphs of the functions $y=f_{0}(x), y$ $=f_{1}(x), y=f_{2}(x)$ in sequence (as shown in the figure). The required area is 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy $a+b=2$, then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$ . | $2=a+b \geqslant 2 \sqrt{a b} \Rightarrow a b \leqslant 1 \Rightarrow a^{n} b^{n} \leqslant 1$, equality holds when $a=b=1$.
Thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{2+a^{n}+b^{n}}{\left(1+a^{n}\right)\left(1+b^{n}\right)}=\frac{2+a^{n}+b^{n}}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1$,
therefore, the minimum value... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate: $\frac{9 \times 11+31}{2015} \times \frac{31+6 \frac{1}{5}+4 \frac{3}{7}}{1+\frac{1}{5}+\frac{1}{7}}=$ | 【Answer】2
【Solution】
$$
\begin{array}{l}
\frac{9 \times 11+31}{2015} \times \frac{31+6 \frac{1}{5}+4 \frac{3}{7}}{1+\frac{1}{5}+\frac{1}{7}} \\
=\frac{99+31}{2015} \times \frac{31+\frac{31}{5}+\frac{31}{7}}{1+\frac{1}{5}+\frac{1}{7}} \\
=\frac{130}{5 \times 13 \times 31} \times \frac{31 \times\left(1+\frac{1}{5}+\frac... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
21. If the polynomial $x^{2}-(a+4) x+3 a-4$ (where $a>0$) can be factored into the product of two linear factors $(x+b),(x+c)$ (where $b, c$ are integers), then $a=$ $\qquad$ | $8$ | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Choose four different digits from 0 to 9 to fill in the four parentheses in the box, there are ( ) ways to make the statement in the box correct.
This sentence contains ( ) numbers greater than 1, ( ) numbers greater than 2, ( ) numbers greater than 3, and ( ) numbers greater than 4.
(A) 1
(B) 2
(C) 3
(D) 4 | (1) Let the numbers filled in the four parentheses be $a$, $b$, $c$, and $d$ in sequence. There are 8 numbers in this sentence, and obviously $a > b > c > d \geqslant 0$.
(2) Since the four numbers in the parentheses are different, and only $0$ and $1$ are not greater than 1 (including the given 1), so $\mathrm{a} \geq... | 2 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
2. Solve the system of equations $\left\{\begin{array}{l}x+y+z=3 \\ x^{2}+y^{2}+z^{2}=3 \\ x^{5}+y^{5}+z^{5}=3\end{array}\right.$. | 2. Let $s_{n}=x^{n}+y^{n}+z^{n}(n=1,2, \cdots)$. Then $s_{1}=s_{2}=s_{5}=3$. Suppose $x, y, z$ are the three roots of the cubic polynomial $f(t)=(t-x)(t-y)(t-z)=t^{3}-\sigma_{1} t^{2}+\sigma_{2} t-\sigma_{3}$, then by the given conditions $\sigma_{1}=s_{1}=3, \sigma_{2}=\frac{1}{2}\left(s_{1}^{2}-s_{2}\right)=3$. Next,... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Let the side length of the regular hexagon $A B C D E F$ be 1. Then $(\overrightarrow{A B}+\overrightarrow{D C}) \cdot(\overrightarrow{A D}+\overrightarrow{B E})=$ $\qquad$ . | $$
-1 .-3 \text {. }
$$
As shown in Figure 1, establish a Cartesian coordinate system, and set $C(1,0)$.
Thus, $B\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), A\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$,
$$
\begin{array}{l}
D\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right), E\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\righ... | -3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
33. Given $a-b=-1, c-a=2$, then $(a-b)^{2}+(c-b)^{2}+(c-a)^{2}=$ | Answer: 6.
Solution: Since $c-b=(c-a)+(a-b)=2+(-1)=1$, therefore $(a-b)^{2}+(c-b)^{2}+(c-a)^{2}=(-1)^{2}+1^{2}+2^{2}=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The area of the large square $\mathrm{ABCD}$ is 18 square centimeters, and the side $\mathrm{MN}$ of the gray square $\mathrm{MNPQ}$ lies on the diagonal $\mathrm{BD}$, with vertex $\mathrm{P}$ on side $\mathrm{BC}$ and $\mathrm{Q}$ on side $\mathrm{CD}$. What is the area of the gray square $\mathrm{MNPQ}$ in square ce... | 10. Answer: 4
Analysis: Connect $\mathrm{AC}$ intersecting BD at 0, and construct the circumscribed square $\mathrm{EFGH}$ of the larger square $\mathrm{ABCD}$, as shown in the figure. Then, the area of square $\mathrm{EFGH}$ is 36 square centimeters. Therefore, $\mathrm{DB}=\mathrm{AC}=6$ centimeters.
It is easy to se... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
I. Multiple Choice Questions (10 points per question, full score 60 points. Among the four options for each question below, only one is correct. Please write the English letter representing the correct answer in the parentheses of each question.)
1. $2012.25 \times 2013.75-2010.25 \times 2015.75=($ ).
A. 5
B. 6
C. 7
D.... | $$
\begin{aligned}
\text { Original expression } & =(2010.25+2) \times(2015.75-2)-2010.25 \times 2015.75 \\
& =2015.75 \times 2-2010.25 \times 2-4 \\
& =7
\end{aligned}
$$
The answer is $C$. | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 4 A square blackboard is divided into $n^{2}$ small squares of side length 1 by a grid of horizontal and vertical lines. For what maximum positive integer $n$ is it always possible to select $n$ small squares such that any rectangle of area at least $n$ contains at least one of the selected small squares (the s... | From the problem, it is evident that if $n$ small squares are selected to meet the problem's conditions, then in each row and each column, there is exactly one selected small square.
Call the row in which the selected square in the first column is located as $A$. If $A$ is the first row, then call the second and third... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Let $a$ be an integer such that $x^{2}-x+a$ divides $x^{13}+x+90$. Find the value of $a$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Analysis: This problem can be solved by dividing $x^{13}+x+90$ by $x^{2}-x+a$, and setting the coefficients of the resulting remainder (at most a linear term) to 0 to find $a$. Alternatively, the roots of $x^{2}-x+a=0$, $x=\frac{1 \pm \sqrt{1-4 a}}{2}$, can be substituted into $x^{13}+x+90=0$ to find $a$. However, both... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$13 \cdot 41$ Given that there are 64 light bulbs on the signal panel, each controlled by 64 buttons. Each light bulb is controlled by its own button. Each time it is activated, any group of buttons can be pressed, and the on/off state of each light bulb can be recorded. To figure out which button controls each light b... | [Solution] Number 64 buttons from 0 to 63, and represent these numbers in binary, denoted as $000000,000001, \cdots, 111111$. On the $k$-th start, press the buttons whose $k$-th digit is 1 $(k=1,2, \cdots, 6)$, and record the $k$-th digit as 1 for the lights that turn on, and 0 for those that remain off. After 6 starts... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$13 \cdot 39$ For all real numbers $x, y$, and $z$, then
$$
\left(x^{2}+y^{2}+z^{2}\right)^{2} \leqslant n\left(x^{4}+y^{4}+z^{4}\right) \text {. }
$$
the smallest integer $n$ is
(A) 2 .
(B) 3 .
(C) 4 .
(D) 6 .
(E) There is no such integer $n$.
(28th American High School Mathematics Examination, 1977) | [Solution] Let $a=x^{2}, \quad b=y^{2}, \quad c=z^{2}$, then
we have
$$
0 \leqslant(a-b)^{2}+(b-c)^{2}+(c-a)^{2},
$$
or
$$
\begin{array}{c}
a b+b c+c a \leqslant a^{2}+b^{2}+c^{2}, \\
\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \leqslant 1,
\end{array}
$$
Multiplying both sides of the inequality by 2 and adding 1, we get
... | 3 | Inequalities | MCQ | Yes | Yes | olympiads | false |
9. (1995 China National Team Selection Test for IMO) 21 people participate in an exam, which consists of 15 true/false questions. It is known that for any two people, there is at least one question that both of them answered correctly. What is the minimum number of people who answered the most answered question correct... | 9. Let the number of people who correctly answer the $i$-th question be $a_{i}$, then the number of pairs who answer the $i$-th question correctly is
$$
b_{i}=C_{u_{i}}^{2}
$$
for $(i=1,2, \cdots, 15)$. We will focus on the sum $\sum_{i=1}^{15} b_{i}$.
Let $a=\max \left\{a_{1}, a_{2}, \cdots, a_{15}\right\}$, then we ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. If the function $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a sum of its maximum and minimum values equal to 4, then $a+b=$ $\qquad$ . | 5.3.
Given $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a maximum or minimum value, we know $b=0$.
Then $y=\frac{a+\sin x}{2+\cos x}$
$\Rightarrow \sin x-y \cos x=a-2 y$
$\Rightarrow \sin (x+\alpha)=\frac{a-2 y}{\sqrt{1+y^{2}}}$
$\Rightarrow|a-2 y| \leqslant \sqrt{1+y^{2}}$.
By Vieta's formulas, $\frac{4 a}{3}=4 \Lef... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (3 points) A student, while solving a calculation problem, was supposed to divide by 20 in the last step, but mistakenly added 20, thus obtaining an incorrect result of 180. What should the correct answer to this calculation problem be? $\qquad$ . | 2. (3 points) A student, while solving a calculation problem, should have divided by 20 in the last step, but mistakenly added 20, resulting in an incorrect answer of 180. What should the correct answer to this calculation problem be? $\qquad$ .
【Solution】Let the result before the last step be $a$,
$$
a+20=180 \text {... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Given a positive integer $n$ does not exceed 2000, and can be expressed as the sum of no less than 60 consecutive positive integers. Then, the number of such $n$ is $\qquad$ . | Since $1+2+\cdots+62=63 \times 31=1953$, we have $n=\sum_{k=1}^{60} k, \sum_{k=1}^{61} k, \sum_{k=1}^{62} k, \sum_{k=2}^{61} k, \sum_{k=2}^{62} k$,
$$
\sum_{k=3}^{62} k \text { for a total of } 6 \text { terms. }
$$ | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. In Lion Tiger Mountain, there is a group of lions and tigers, and the $\frac{3}{5}$ of the number of lions is equal to the $\frac{2}{3}$ of the number of tigers. Therefore, Lion Tiger Mountain has at least $\qquad$ lions. | $10$ | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For any positive integer $n$, let $f_{1}(n)$ denote the square of the sum of the digits of $n$ plus $r+1$, where $r$ is the integer such that $n=3q+r, 0 \leqslant r<3$. For $k \geqslant 2$, let $f_{k}(n)=f_{1}\left(f_{k-1}(n)\right)$. Therefore, $f_{1990}(2345)$ $=$ $\qquad$. | 55 3. It is easy to calculate $f_{1}(2345)=199, f_{2}(2345)=363, f_{3}(2345)=145$,
$$
\begin{array}{l}
f_{4}(2345)=102, f_{5}(2345)=10, f_{6}(2345)=3, f_{7}(2345)=10, \\
f_{8}(2345)=3, \cdots, f_{1990}(2345)=3 .
\end{array}
$$ | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$20 \cdot 104$ Two circles with radii 13 and 5 intersect, and the distance between their centers is 12. Then the length of the common chord of the two circles is
(A) $3 \sqrt{11}$.
(B) $\frac{65}{6}$.
(C) $4 \sqrt{6}$.
(D) 10 .
(E) None of the above results is correct.
(Tianjin, China Junior High School Mathematics Co... | [Solution] As shown in the figure, let the centers of the two circles be $O, O_{1}$, and the intersection points of the two circles be $A, B$.
Connect $O_{1} A$. In $\triangle O O_{1} A$, $O A=13, O O_{1}=$ $12, O A=5$.
Then, by $5^{2}+12^{2}=13^{2}$, we know that $\angle A O_{1} O$ is a right angle.
Thus, the chord $... | 10 | Geometry | MCQ | Yes | Yes | olympiads | false |
Exercise 1 Let $a, b, c, d$ be distinct real numbers, and $a+b+c+d=3, a^{2}+b^{2}+c^{2}+d^{2}=45$. Find the value of the algebraic expression
$$
\begin{array}{l}
\frac{a^{5}}{(a-b)(a-c)(a-d)}+\frac{b^{5}}{(b-a)(b-c)(b-d)} \\
+\frac{c^{5}}{(c-a)(c-b)(c-d)}+\frac{d^{5}}{(d-a)(d-b)(d-c)} .
\end{array}
$$ | Let \( S_{n}=\frac{a^{n}}{(a-b)(a-c)(a-d)} \)
\[
\begin{array}{l}
+\frac{b^{n}}{(b-a)(b-c)(b-d)} \\
+\frac{c^{n}}{(c-a)(c-b)(c-d)} \\
+\frac{d^{n}}{(d-a)(d-b)(d-c)} .
\end{array}
\]
By the Lagrange interpolation formula, we know
\[
\begin{aligned}
& \frac{(x-b)(x-c)(x-d)}{(a-b)(a-c)(a-d)} a^{3}+\frac{(x-a)(x-c)(x-d)}{... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given $A=\left\{x \mid x^{2}-m x+m^{2}-19=0\right\}, B=\left\{x \mid \log _{2}\left(x^{2}-5 x+\right.\right.$ $8)=1\}, C=\left\{x \mid x^{2}+2 x-8=0\right\}$, and $A \cap B \neq \varnothing, A \cap C=\varnothing$, find the value of $m$. | 3. From the given, $B=\{2,3\}, C=\{2,-4\}$, then $3 \in A$, so $3^{2}-3 m+m^{2}-19=$ 0 , solving for $m$ yields $m=5$ or -2. Therefore, upon verification, $m=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. Let $\mathrm{AB}$ be the major axis of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{6}=1$. The moving chord $\mathrm{PQ}$ of this ellipse passes through $\mathrm{C}(2,0)$, but does not pass through the origin. The lines $\mathrm{AP}$ and $\mathrm{QB}$ intersect at point $\mathrm{M}$, and the lines $\mathrm{PB}$ and $... | Solution: The ellipse is: $3 x^{2}+8 y^{2}=48$, let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, from $A(-4,0), B(4,0)$ we get the lines $A P, Q B$ as $\left\{\begin{array}{l}y\left(x_{1}+4\right)=y_{1}(x+4) \\ y\left(x_{2}-4\right)=y_{2}(x-4)\end{array}\right.$, eliminating $\mathrm{y}$ we get $x_{M}\left... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. If $\sin ^{3} x+\cos ^{3} x=1$, then $\sin x+\cos x=$ $\qquad$ . | - 1.1.
Notice,
$$
1=\sin ^{2} x+\cos ^{2} x \geqslant \sin ^{3} x+\cos ^{3} x=1 \text {. }
$$
Therefore, $\sin x+\cos x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. As shown in the figure, in rectangle $A B C D$, $\triangle A E D$ and $\triangle B F C$ are both isosceles right triangles, $E F=A D=2$. Then the area of rectangle $A B C D$ is $\qquad$ . | 【Answer】 8
【Solution】The rectangle can be cut as shown in the figure below. Since $E F=A D=2 A G$, the area of the entire rectangle is 8 times the area of the small square. Since the area of one small square is 1, the area of the rectangle is 8. | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
【Question 10】
A grade has three classes: Class A, Class B, and Class C. Class A has 4 more girls than Class B, and Class B has 1 more girl than Class C. If the first group of Class A is moved to Class B, the first group of Class B is moved to Class C, and the first group of Class C is moved to Class A, then the number ... | 【Analysis and Solution】
Average number problem.
Taking the "original number of girls in Class C" as the benchmark, Class B has 1 more girl than Class C, and Class A has $4+1=5$ more girls than Class C; Class C has increased by $(1+5) \div 3=2$ girls compared to the original, Class B has increased by 2-1=1 girl compared... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (8 points) The units digit of 2010 2009s is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 【Solution】Solution: The unit digit is determined only by the unit digit of a multi-digit number, that is, the unit digit is determined by the product of 2010 nines,
According to the cycle of $9, 1$ every 2 digits,
$$
2010 \div 2=1005
$$
There is no remainder,
Therefore, the unit digit of the product of 2010 nines is ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
15. Suppose there are 128 ones written on a blackboard. In each step, you can erase any two numbers \(a\) and \(b\) and write \(ab + 1\). After performing this operation 127 times, only one number remains. Let the maximum possible value of this remaining number be \(A\). Determine the last digit of \(A\).
(1992 Saint P... | 15. First, prove that as long as each step involves the operation on the smallest two numbers on the blackboard, the final remaining number can reach its maximum value. We denote the operation on numbers \(a\) and \(b\) on the blackboard as \(a * b = ab + 1\). Suppose at some step, the operation is not performed on the... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$12.2 .1 *$ As shown in the figure, for the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, a line $l$ is drawn through vertex $A_{1}$ such that $l$ forms an angle of $60^{\circ}$ with both lines $A C$ and $B C_{1}$. How many such lines $l$ are there?
(A) 1
(B) 2
(C) 3
(D) More than 3 | Obviously, $A D_{1} // B C_{1}$, the number of lines passing through point $A$ that form a $60^{\circ}$ angle with both lines $A C$ and $A D_{1}$ is 3. Therefore, the number of lines $l$ passing through point $A_{1}$ that satisfy the conditions is also 3. Hence, the answer is C. | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
8. Determine $\left(2017^{\frac{1}{\log _{2} 2017}} \times 2017^{\frac{1}{\log _{4} 2017}} \times 2017^{\frac{1}{\log _{8} 2017}} \times 2017^{\frac{1}{\log _{16} 2017}} \times 2017^{\frac{1}{\log _{32} 2017}}\right)^{\frac{1}{5}}$.
| $$
\begin{array}{l}
\text { Original expression }=2017^{\frac{1}{5}}\left(\frac{1}{\log _{2} 2017}+\frac{1}{\log _{4} 2017}+\frac{1}{\log _{8} 2017}+\frac{1}{\log _{16} 2017}+\frac{1}{\log _{32} 2017}\right) \\
=2017^{\frac{1}{5}}\left(\log _{2017} 2+\log _{2017} 4+\log _{2017} 8+\log _{2017} 16+\log _{2017} 32\right)=... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. $\left(2 \sin 36^{\circ} \cdot 2 \sin 72^{\circ}\right)^{2}=$ | 3.5.
$$
\begin{array}{l}
\text { Original expression }=4 \times 2 \sin ^{2} 36^{\circ} \times 2 \sin ^{2} 72^{\circ} \\
=4\left(1-\cos 72^{\circ}\right)\left(1-\cos 144^{\circ}\right) .
\end{array}
$$
And $\cos 72^{\circ}+\cos 144^{\circ}$
$$
\begin{array}{l}
=\frac{1}{2 \sin 36^{\circ}}\left(2 \sin 36^{\circ} \cdot \... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let $x, y, z$ be positive real numbers, $M=\max \left\{x y+\frac{2}{z}, z+\frac{2}{y}, y+z+\frac{1}{x}\right\}$. Then the minimum value of $M$ is | $3 M \geqslant x y+\frac{2}{z}+z+\frac{2}{y}+y+z+\frac{1}{x}=x y+\frac{1}{z}+\frac{1}{z}+z+\frac{1}{y}+\frac{1}{y}+y+z+\frac{1}{x}$ $\geqslant 9 \sqrt[9]{x y \cdot \frac{1}{z} \cdot \frac{1}{z} \cdot z \cdot \frac{1}{y} \cdot \frac{1}{y} \cdot y \cdot z \cdot \frac{1}{x}}=9 \Rightarrow M \geqslant 3$, equality holds wh... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
12. Define a new operation: $x \odot y=18+x-a \times y$, where $a$ is a constant. For example:
$$
1 \odot 2=18+1-a \times 2 \text {. }
$$
If 2๑3=8, then $3 \odot 5=$ $\qquad$ , $5 \odot 3=$ $\qquad$ | Reference answer: 1,11
Key points: Calculation - Defining new operations | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 2 Let $S=\{1,2,3,4\}, n$ terms of the sequence: $a_{1}, a_{2}, \cdots, a_{n}$ have the following property: for any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted by। $B$ ।), there are adjacent। $B$ ।terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997 ... | The minimum value of $n$ is 8.
First, prove that each number in $S$ appears at least 2 times in the sequence $a_{1}, a_{2}, \cdots, a_{n}$.
In fact, if a number in $S$ appears only once in this sequence, since there are 3 subsets containing this number, but in the sequence, the adjacent pairs containing this number can... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14.4.27 ** Find all prime numbers that can be expressed both as the sum of two prime numbers and as the difference of two prime numbers. | The prime number $r$ that satisfies the condition is only one, which is 5. In fact, the sum and difference of two odd primes are even numbers, so $r=p+2=q-2$, where $p$ and $q$ are odd primes. Thus, $p, r=p+2, q=r+2$ are three consecutive odd primes. At this point, there is only one unique set: $3,5,7$ (since among any... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. (25 points) Given the functions
$$
f(x)=x-\ln x-2, \quad g(x)=x \ln x+x \text {. }
$$
(1) Prove that $f(x)$ has a zero in the interval $(3,4)$;
(2) If $k \in \mathbf{Z}$, and $g(x)>k(x-1)$ for all $x>1$, find the maximum value of $k$. | 14. (1) Let $f(x)=x-\ln x-2$. Then
$$
f^{\prime}(x)=1-\frac{1}{x} \text {. }
$$
Therefore, $f(x)$ is monotonically increasing on the interval $(1,+\infty)$.
$$
\text { Also, } f(3)=1-\ln 30 \text {, }
$$
Thus, $f(x)$ has a zero point $x_{0} \in(3,4)$.
(2) The unique zero point $x_{0}$ of $f(x)$ clearly satisfies
$$
x... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the figure, the five Chinese characters “华罗庚金杯” represent five different digits from $1-5$. By adding the numbers at the endpoints of each line segment, five sums are obtained. There are $\qquad$ ways to make these five sums exactly five consecutive natural numbers. | 【Analysis】According to “the sum of the numbers at the endpoints of each line segment is exactly 5 consecutive natural numbers,” it can be seen that these 5 sums are somewhat larger than the original $1, 2, 3, 4, 5$; the numbers at the 5 vertices of the pentagram are counted twice, so the total sum of the 5 sums can be ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
26. There is a natural number $x$. When divided by 3, the remainder is 2; when divided by 5, the remainder is 3. The remainder when $x$ is divided by 15 is $\qquad$ | Reference answer: 8 | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.1.17 $\underset{\star}{\star \star}$ (1) Let $n$ be a prime number greater than 3. Find the value of $\left(1+2 \cos \frac{2 \pi}{n}\right)(1+$ $\left.2 \cos \frac{4 \pi}{n}\right)\left(1+2 \cos \frac{6 \pi}{n}\right) \cdots \cdots \cdot\left(1+2 \cos \frac{2 n \pi}{n}\right)$;
(2) Let $n$ be a natural number greater... | (1) Let $w=e^{\frac{2 x i}{n}}$. Clearly, $w^{n}=1, w^{-\frac{n}{2}}=-1, w^{k}+w^{-k}=2 \cos \frac{2 k \pi}{n}$. Therefore, we have
$$
\begin{aligned}
\prod_{k=1}^{n}\left(1+2 \cos \frac{2 k \pi}{n}\right) & =\prod_{k=1}^{n}\left(1+w^{k}+w^{-k}\right)=\prod_{k=1}^{n} w^{-k}\left(w^{k}+w^{2 k}+1\right) \\
& =w^{-\frac{n... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$16 \cdot 19$ On a plane, given three non-collinear points $A, B, C$, construct a line $l$ such that the distances from points $A, B, C$ to line $l$ are in the ratio $1: 1: 2$ or $1: 2: 1$ or $2: 1: 1$. The number of such lines is
(A) 3.
(B) 6.
(C) 9.
(D) 12.
(E) 15.
(China Junior High School Mathematics Correspondence... | [Solution] Connect $A B, B C, C A$, extend $A B$, take $B E=$ $A B$; extend $A C$, so that $C F$
$=A C$. Take the midpoint $D$ of $B C$. Connect $E D, F D$, and extend them respectively to intersect $A C$ at $H$, and $A B$ at $G$.
Then the lines $E H, F G, G H, E F$ are all at a distance ratio of $1: 1: 2$ from point... | 12 | Geometry | MCQ | Yes | Yes | olympiads | false |
$29 \cdot 54$ If $a, b$ are natural numbers, and $a$ divided by 7 leaves a remainder of $2, b$ divided by 7 leaves a remainder of 5. When $a^{2}>3 b$, the remainder of $a^{2}-3 b$ divided by 7 is
(A) 1 .
(B) 3 .
(C) 4 .
(D) 6 .
(China Jiangxi Province Nanchang Junior High School Mathematics Competition, 1989) | [Solution] Let $a=7 m+2, b=7 n+5 \quad(m, n \in \mathbb{N})$, then
$$
\begin{aligned}
a^{2}-3 b= & (7 m+2)^{2}-3(7 n+5) \\
& =7\left(7 m^{2}+4 m-3 n\right)-11 \\
& =7\left(7 m^{2}+4 m-3 n\right)-14+3,
\end{aligned}
$$
we know that $a^{2}-3 b$ leaves a remainder of 3 when divided by 7.
Therefore, the answer is $(B)$. | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
* Given the quadratic equation in $x$: $a(1+\mathrm{i}) x^{2}+\left(1+a^{2} \mathrm{i}\right) x+a^{2}+\mathrm{i}=0$ has real roots, find the value of the real number $a$. | Suppose the original equation has a real root $x_{0}$, substituting into the original equation gives $a(1+\mathrm{i}) x_{0}^{2}+\left(1+a^{2} \mathrm{i}\right) x_{0}+a^{2}+\mathrm{i}=0$, which is $\left(a x_{0}^{2}+x_{0}+a^{2}\right)+\mathrm{i}\left(a x_{0}^{2}+a^{2} x_{0}+1\right)=0$. According to the necessary and su... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
56. In the figure, the two circles 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 ants start from point $A$ at the same time and move in the direction indicated by the arrows at the same speed, each along one of the circles. When the ant ... | Answer: 4 | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The line $l$ intersects the curve $C: x^{2}-y^{2}=1(x>0)$ at points $A$ and $B$. Then the minimum value of $f=\overrightarrow{O A} \cdot \overrightarrow{O B}$ is $\qquad$ | 4. 1 .
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
\begin{aligned}
f & =x_{1} x_{2}+y_{1} y_{2} \\
& =\frac{1}{2}\left[\left(x_{1}+y_{1}\right)\left(x_{2}+y_{2}\right)+\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\right] \\
& \geqslant 1 .
\end{aligned}
$$
When points $A$ and $B$ are symm... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) Xiao Ming marks several points on the sides of a square, with exactly 3 on each side. The minimum number of points marked is
( ) .
A. 12
B. 10
C. 8
D. 6 | 【Analysis】To minimize the number of points marked on each edge, points must be marked at all four vertices of the square. Therefore, using the formula for the number of points around the perimeter of the square: number of points on each side $\times 4 - 4$, we can find the minimum number of points to be marked.
【Solut... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example 7 Let the roots of the odd-degree real-coefficient equation $f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}=0$ all lie on the unit circle, and $-a_{n}=a_{0} \neq 0$, find $a_{0}+a_{1}+\cdots+a_{n}$. | Let $f(c)=a_{0} c^{n}+a_{1} c^{n-1}+\cdots+a_{n-1} c+a_{n}=0$, i.e., the complex number $c$ is a root of $f(x)=0$, then the conjugate complex number $\bar{c}$ is also a root of $f(x)=0$. However, $f(x)=0$ is an equation of odd degree, so apart from the conjugate pairs of imaginary roots, it must have an odd number of r... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$3 、 A$ number $\mathrm{A}$ is a prime number, and $\mathrm{A}+14 、 \mathrm{~A}+18 、 \mathrm{~A}+32 、 \mathrm{~A}+36$ are also prime numbers. What is the value of $\mathrm{A}$? | 【Analysis】14 divided by 5 leaves a remainder of 4, 18 divided by 5 leaves a remainder of 3, 32 divided by 5 leaves a remainder of 2, 36 divided by 5 leaves a remainder of 1, so among A, A+14, $\mathrm{A}+18$, $\mathrm{A}+32$, $\mathrm{A}+36$, there must be one that is a multiple of 5, and also a prime number, so it can... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
178 In $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $c-a$ equals the height $h$ from $A C$, then $\left(\cos \frac{A}{2}-\sin \frac{A}{2}\right) \cdot\left(\sin \frac{C}{2}+\cos \frac{C}{2}\right)=$ $\qquad$ . | 1781 . Since $c-a=h$, $A$ is an acute angle, $\cos \frac{A}{2}>\sin \frac{A}{2}$. And $c-a=c \sin A$, that is
$$
\begin{aligned}
& \sin C-\sin A=\sin C \sin A . \\
& \left(\cos \frac{A}{2}-\sin \frac{A}{2}\right)^{2}\left(\sin \frac{C}{2}+\cos \frac{C}{2}\right)^{2} \\
= & (1-\sin A)(1+\sin C) \\
= & 1-\sin A+\sin C-\s... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
18.4.5 $\star \star$ Given that $m, n$ are positive integers, find the minimum value of $\left|12^{m}-5^{n}\right|$, and prove it. | When $m=n=1$, $\left|12^{m}-5^{n}\right|=7$.
Since $12^{m} \equiv 0(\bmod 2), 5^{n} \neq 0(\bmod 2)$, so $\left|12^{m}-5^{n}\right| \neq 0,2,4,6$.
Since $3 \mid 12^{m}, 3 \nmid 5^{n}$, so $\left|12^{m}-5^{n}\right| \neq 3$. Since $5 \nmid 12^{m}, 5 \mid 5^{n}$, so $\left|12^{m}-5^{n}\right| \neq 5$. Assume $\left|12^{m... | 7 | Number Theory | proof | Yes | Yes | olympiads | false |
2. (6 points) Buying 5 pounds of cucumbers costs 11 yuan 8 jiao, which is 1 yuan 4 jiao less than buying 4 pounds of tomatoes. Therefore, the price per pound of tomatoes is $\qquad$yuan $\qquad$jiao. | 【Analysis】First, based on the fact that buying 5 pounds of cucumbers costs 11 yuan 8 jiao, which is 1 yuan 4 jiao less than buying 4 pounds of tomatoes, find out the cost of buying tomatoes, then use the unit price $=$ total price $\div$ quantity to solve the problem.
【Solution】Solution: 11 yuan 8 jiao $=11.8$ yuan, 1... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.54 Suppose the radius of the small circle is $\frac{r}{2}$, and the radius of the large circle is $r$. How many small circles are needed at minimum to completely cover the area of the large circle? | [Solution] (1) A circular surface with radius $r$ can be covered by 7 small circular discs with radius $\frac{r}{2}$. Construct a regular hexagon inscribed in the large circle $O$, and then, with the midpoints of each side of the hexagon and point $O$ as centers, and $\frac{r}{2}$ as the radius, draw 7 small circles. T... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Given that $q$ is a positive rational number. There are two ants starting from a point $X$ on a plane at the same time. Each ant chooses one of the four directions (east, south, west, north) and moves $q^{n}$ meters in the $n$-th minute $(n=1,2, \cdots)$. Suppose they meet again after some integer minutes, but their... | 2. Establish a Cartesian coordinate system with $X$ as the origin, the positive east and north directions as the positive directions of the $x$-axis and $y$-axis, respectively.
Let the sum of the horizontal and vertical coordinates of the first ant after $n$ minutes be $a_{n}$, and the corresponding coordinate sum of ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Let $a, b, c$ be positive numbers, and $abc=1$, find the minimum value of $\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$. | Solution: Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}^{+}\right)$, then $\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1}=\frac{y}{y+2 x}+\frac{z}{z+2 y}+$ $\frac{x}{x+2 z}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{c}
{[y(y+2 x)+z(z+2 y)+x(x+2 z)] \cdot} \\
\left(\frac... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
28. Given non-negative numbers $a, b, c, d, e$ satisfy the equation $a+b+c+d+e=2$. If the maximum value of $a+b+c, b+c+d$, $c+d+e$ is $M$, then the minimum value of $M$ is $\qquad$ . | $1$ | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
18.18 Extend the sides $AD, BC$ of the convex quadrilateral $ABCD$ through $D, C$ to meet at point $E$. Let $s=\angle CDE + \angle DCE$, and $s'=\angle BAD + \angle ABC$. If $r=\frac{s}{s'}$, then
(A) either $r=1$, or $r>1$.
(B) either $r=1$, or $r1$.
(E) $r=1$.
(19th American High School Mathematics Examination, 1968) | [Solution] In $\triangle EDC$, we have
$$
\angle E + \angle CDE + \angle DCE = \angle E + s = 180^{\circ}.
$$
In $\triangle EAB$, we also have
$$
\angle E + \angle BAE + \angle ABE = \angle E + s' = 180^{\circ}.
$$
Thus, $s = s' = 180^{\circ} - \angle E$.
Therefore, the answer is $(E)$. | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
(3) In rectangle $A B C D$, $A B=2, B C=3, E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. With $E F$ as the axis of rotation, rotate $\triangle F A B$ in space by $90^{\circ}$ to $\triangle F A^{\prime} B^{\prime}$. Then the volume of the tetrahedron $A^{\prime} B^{\prime} C D$ is $\qquad$. | 3 2 Hint: $E F=B C=3, \triangle F A^{\prime} B^{\prime}$ divides the tetrahedron $A^{\prime} B^{\prime} C D$ into two tetrahedrons $C F A^{\prime} B^{\prime}$ and $D F A^{\prime} B^{\prime}$ of equal volume, with heights $C F=D F=$ 1, and base area $S_{\triangle F A^{\prime} B^{\prime}}=3$, then the volume
$$
V_{A^{\pr... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
74. In a rectangular coordinate system, there are four points $A(-1,1)$, $B(5,0)$, $C(3,-3)$, and $D(-3,-2)$. After moving point $A$ down by one unit and moving point $C$ up by one unit, the area of the new quadrilateral $ABCD$ is . $\qquad$ | Reference answer: 12 | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Given a positive integer that has exactly 2017 positive divisors and is not coprime with 2018. What is the remainder when the sum of all such positive integers is divided by 2019? | 7.2.
Notice,
$$
2018=2 \times 1009, 2019=3 \times 673 \text {, }
$$
and $673$, $1009$, and $2017$ are all prime numbers.
By the problem, a positive integer with exactly 2017 positive divisors must be the 2016th power of some prime number.
Thus, the sum to consider is $2^{2016} + 1009^{2016}$. At this point,
$$
\begin... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
【Question 14】
If there are 5 medicine boxes, and every 2 boxes contain one kind of the same medicine, with each kind of medicine appearing in exactly 2 boxes, then there are $\qquad$ kinds of medicine. | 【Analysis and Solution】 $C_{5}^{2}=\frac{5 \times 4}{2 \times 1}=10$ kinds of medicine. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. A test paper consists of four multiple-choice questions, each with three options $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$. Several students take the exam, and after grading, it is found that any three students have different answers to at least one question, and no student leaves any question unanswered. Then th... | 7.9.
If 10 people take the exam, then for the first question, at least 7 people choose two options; for the second question, at least 5 people among these 7 choose two options; for the third question, at least 4 people among these 5 choose two options; for the fourth question, at least 3 people among these 4 choose tw... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.49 John scored 93 points in this year's AHSME. If the old scoring method were still in effect, he would have scored only 84 points based on the same answers. The number of questions he left unanswered is (the new scoring system awards 5 points for a correct answer, 0 points for a wrong answer, and 2 points for an una... | ［Solution］Let John answer $x$ questions correctly, $y$ questions incorrectly, and leave $z$ questions unanswered. According to the problem, we can set up the following system of equations:
$$
\left\{\begin{array}{l}
5 x+0 \cdot y+2 z=93, \\
30+4 x-y+0 \cdot z=84, \\
x+y+z=30 .
\end{array}\right.
$$
Adding (2) and (3) g... | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
If $p, q, \frac{2 p-1}{q}, \frac{2 q-1}{p}$ are all integers, and $p>1$, $q>1$, find the value of $p+q$. | Solution 1 First, $p \neq q$.
Indeed, if $p=q$, then
\[
\frac{2 p-1}{q}=\frac{2 p-1}{p}=2-\frac{1}{p} \quad \text{, since } p>1, \quad \frac{2 p-1}{q} \text{ is not an integer, contradicting the problem statement.}
\]
By symmetry, assume without loss of generality that $p>q$, and let $\frac{2 q-1}{p}=m$, then $m$ is a... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$3 \cdot 25$ On a two-lane road, to overtake car $B$, car $A$ must pass and maintain a distance of 30 feet. At this moment, car $C$ is coming from the opposite direction, 210 feet away from $A$ (as shown in the figure). If $B$ and $C$ maintain their original speeds of $40 \mathrm{mph}$ and $50 \mathrm{mph}$ respectivel... | [Solution] Let the increased speed of car $A$ be $r$, and the distance it travels to safely overtake car $B$ be $d$. Then, when car $A$ safely overtakes car $B$,
$$
\frac{d}{50+r}=\frac{d-\frac{30}{5280}}{40},
$$
and when car $A$ safely passes car $C$,
$$
\frac{d}{50+r}=\frac{\frac{210}{5280}-d}{50} \text {. }
$$
Sol... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
5 In a rectangle $A B C D$ with an area of 1 (including the boundary) there are 5 points, no three of which are collinear. Find the minimum number of triangles, formed by these 5 points as vertices, that have an area not greater than $\frac{1}{4}$. | The proof of this problem requires the use of the following common conclusion, which we will use as a lemma: the area of any triangle within a rectangle is no more than half the area of the rectangle.
In rectangle \(ABCD\), if a triangle formed by any three points has an area no more than \(\frac{1}{4}\), then these t... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
(9) Let $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, satisfying $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$. Then the value of $S_{20}$ is $\qquad$. | (9) 0 Hint: From the condition, we know $a_{1}=\frac{\left(a_{1}+1\right)^{2}}{4}$, solving this gives $a_{1}=1$.
When $n \geqslant 2$, from $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$ we know $S_{n-1}=\frac{\left(a_{n-1}+1\right)^{2}}{4}$, so
$$
a_{n}=\frac{1}{4}\left(a_{n}+1\right)^{2}-\frac{1}{4}\left(a_{n-1}+1\right... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3 A school is hosting a football round-robin tournament, where each participating team plays one match against every other team. A team earns 2 points for a win, 1 point for a draw, and 0 points for a loss. It is known that only one team has the highest points, but they have the fewest wins. How many teams must... | The team with the most points is called the champion. Suppose the champion wins $n$ games and draws $m$ games, then they have a total of $2n + m$ points. According to the assumption, the other teams win no less than $n+1$ games, meaning their points are no less than $2(n+1)$. From $2n + m > 2(n+1)$, we get $m \geqslant... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$128 \lim _{n \rightarrow \infty} \frac{1}{\sqrt[3]{n}}\left(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+\frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+\cdots+\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}}\right)=$ | 128 1. Since
$$
\begin{aligned}
& \frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+\cdots+\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}} \\
= & \frac{\sqrt[3]{2}-1}{2-1}+\cdots+\frac{\sqrt[3]{n}-\sqrt[3]{n-1}}{n-(n-1)}=\sqrt[3]{n}-1
\end{aligned}
$$
Therefore, the required limit is $\lim _{n \rightarrow \infty} \fr... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) Calculate: $\left[\left(6.875-2 \frac{1}{2}\right) \times 25 \%+\left(3 \frac{23}{24}+1 \frac{2}{3}\right) \div 4\right] \div 2.5=$ $\qquad$ | 【Solution】Solve: [ $\left.\left.6.875-2 \frac{1}{2}\right) \times 25 \%+\left(3 \frac{23}{24}+1 \frac{2}{3}\right) \div 4\right] \div 2.5$,
$$
\begin{array}{l}
=\left[\left(6 \frac{7}{8}-2 \frac{1}{2}\right) \times \frac{1}{4}+\left(3 \frac{23}{24}+1 \frac{16}{24}\right) \times \frac{1}{4}\right] \div 2.5, \\
=\left[\f... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.67 Given that a $2 \times 2$ square paper piece covers no fewer than 7 nodes on a grid paper with a side length of 1, how many nodes does it cover exactly? | ［Solution］If a square piece of paper covers two nodes that are $2 \sqrt{2}$ apart, then the square exactly covers 4 grid squares, thus covering 9 nodes.
According to the given information, the square piece of paper covers at least 7 nodes. If it is not the aforementioned case, then the 7 nodes must be 3 nodes each in ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. For any $\alpha, \beta \in\left(0, \frac{2 \pi}{3}\right)$, we have
$$
\begin{array}{l}
4 \cos ^{2} \alpha+2 \cos \alpha \cdot \cos \beta+4 \cos ^{2} \beta- \\
3 \cos \alpha-3 \cos \beta-k<0 .
\end{array}
$$
Then the minimum value of $k$ is $\qquad$ | 5.6.
Substitution $\left(x_{1}, x_{2}\right)=(2 \cos \alpha, 2 \cos \beta)$.
Then the given inequality becomes
$$
2 x_{1}^{2}+x_{1} x_{2}+2 x_{2}^{2}-3 x_{1}-3 x_{2}-2 k<0 \text {. }
$$
Since $x_{1}, x_{2} \in(-1,2)$, we have
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{1}-2\right)+\left(x_{2}+1\right)\left(x_{2}... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
$6 \cdot 84$ For a finite set $A$, there exists a function $f: N \rightarrow A$ with the following property: if $|i-j|$ is a prime number, then $f(i) \neq f(j), N=\{1,2, \cdots\}$, find the minimum number of elements in the finite set $A$.
| [Solution] Starting from 1, the smallest natural number whose difference with it is a prime number is 3; the smallest natural number whose difference with the previous two is a prime number is 6; the smallest natural number whose difference with the first three is a prime number is 8. This means that for the set $M=\{1... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(20) Given the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$, a line through its left focus $F_{1}$ intersects the ellipse at points $A$ and $B$. Point $D(a, 0)$ is to the right of $F_{1}$. Lines $A D$ and $B D$ intersect the left directrix of the ellipse at points $M$ and $N$, respectively. If the circle with di... | (20) $F_{1}(-3,0)$, the left directrix equation is $x=-\frac{25}{3}$; $A B$ equation is $y=k(x+3)$ ( $k$ is the slope).
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then
$$
\left\{\begin{array}{l}
y=k(x+3), \\
\frac{x^{2}}{25}+\frac{y^{2}}{16}=1,
\end{array} \Rightarrow\left(16+25 k^{2}\right) x^{2}+15... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. As shown in Figure 2, point $A$ is on the positive $y$-axis, point $B$ is on the positive $x$-axis, $S_{\triangle A O B}=9$, segment $A B$ intersects the graph of the inverse proportion function $y=\frac{k}{x}$ at points $C$ and $D$. If $C D=$ $\frac{1}{3} A B$, and $A C=B D$, then $k=$ $\qquad$ . | 8. 4 .
Let point $A\left(0, y_{A}\right), B\left(x_{B}, 0\right)$.
Given $C D=\frac{1}{3} A B, A C=B D$, we know that $C$ and $D$ are the trisection points of segment $A B$.
Thus, $x_{C}=\frac{1}{3} x_{B}, y_{C}=\frac{2}{3} y_{A}$.
Therefore, $k=x_{C} y_{C}=\frac{2}{9} x_{B} y_{A}=\frac{4}{9} S_{\triangle A O B}=4$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
(11) The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is | 119 Prompt: Since $(\sqrt{2}+\sqrt{3})^{2010}+(\sqrt{2}-\sqrt{3})^{2010}$ is an integer, then the fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is $1-(\sqrt{2}-\sqrt{3})^{2010}$, and because
$$
0<(\sqrt{2}-\sqrt{3})^{2010}<0.2^{1005}<(0.008)^{300},
$$
therefore
$$
0.9<1-(\sqrt{2}-\sqrt{3})^{2010}<1 \text {, }
$$
it... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. Satisfy:
$$
x+y^{2}=z^{3}, x^{2}+y^{3}=z^{4}, x^{3}+y^{4}=z^{5}
$$
the number of ordered triples of real numbers $(x, y, z)$ is $\qquad$. | 12.7.
From the problem, we have
$$
\begin{array}{l}
\left(x+y^{2}\right)\left(x^{3}+y^{4}\right)=z^{8}=\left(x^{2}+y^{3}\right)^{2} \\
\Rightarrow x y^{2}(x-y)^{2}=0 .
\end{array}
$$
If $x=0$, then $y^{2}=z^{3}, y^{3}=z^{4}$. In this case, the only solutions are $(x, y, z)=(0,0,0)$ or $(0,1,1)$.
If $x \neq 0, y=0$, t... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
II. Fill-in-the-blank Questions (Full marks 54 points, each question 9 points)
1. Given a positive integer $n$ does not exceed 2000, and can be expressed as the sum of at least 60 consecutive positive integers, then the number of such $n$ is $\qquad$. | 6
1. 【Analysis and Solution】The sum of $k$ consecutive positive integers with the first term $a$ is $S_{k}=k a+\frac{k(k+1)}{2} \geqslant \frac{k(k+1)}{2}$. From $S_{k} \leqslant 2000$, we can get $60 \leqslant k \leqslant 62$.
When $k=60$, $S_{k}=60 a+30 \times 59$, from $S_{k} \leqslant 2000$, we can get $a \leqslant... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Given $\boldsymbol{a}=\left(\cos \frac{2}{3} \pi, \sin \frac{2}{3} \pi\right), \overrightarrow{O A}=\boldsymbol{a}-\boldsymbol{b}, \overrightarrow{O B}=\boldsymbol{a}+\boldsymbol{b}$, if $\triangle O A B$ is an isosceles right triangle with $O$ as the right-angle vertex, then the area of $\triangle O A B$ is $\qquad... | 4. 1 Detailed Explanation: Let vector $\boldsymbol{b}=(x, y)$, then $\left\{\begin{array}{l}(\boldsymbol{a}+\boldsymbol{b})(\boldsymbol{a}-\boldsymbol{b})=0 \\ |\boldsymbol{a}+\boldsymbol{b}|=|\boldsymbol{a}-\boldsymbol{b}|\end{array}\right.$,
$$
\begin{array}{l}
\text { i.e., }\left\{\begin{array} { l }
{ ( x - \frac... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. Given that $m, n$ are positive integers, where $n$ is odd, find the greatest common divisor of $2^{m}+1$ and $2^{n}-1$.
untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This part is a note and not part of the problem statement, so it is not translated. | Given $n$ is odd $\Rightarrow\left(2^{m}+1\right)\left|\left(2^{m n}+1\right),\left(2^{n}-1\right)\right|\left(2^{m n}-1\right)$, then $\left(2^{m}+1,2^{n}-1\right) \mid\left(2^{m n}+1,2^{m n}-1\right)=\left(2,2^{m n}-1\right)=1$, so $\left(2^{m}+1,2^{n}-1\right)=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. If $M=\left\{z \left\lvert\, z=\frac{t}{1+t}+\mathrm{i} \frac{1+t}{t}\right., t \in \mathbf{R}, t \neq-1, t \neq 0\right\}$, $N=\{z|z=\sqrt{2} \cdot[\cos (\arcsin t)+\mathrm{i} \cos (\arccos t)], t \in \mathbf{R}| t \mid, \leqslant 1\}$, then the number of elements in $M \cap N$ is
A. 0
B. 1
C. 2
D. 4 | $$
M=\{(x, y) \mid x y=1, x \neq 1\}, N=\left\{(x, y) \mid x^{2}+y^{2}=2, x \geqslant 0\right\} \text {, }
$$
From the function graphs, we can see that $M \cap N=\varnothing$. Therefore, the answer is $A$. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. Let the set $T=\left\{x_{1}, x_{2}, \cdots, x_{10}\right\}$ have five-element subsets such that any two elements of $T$ appear in at most two subsets. The maximum number of such subsets is $\qquad$ . | 8. 8 Detailed Explanation: An element has no more than two opportunities to be in the same subset with other 9 elements, so the number of binary pairs it can form does not exceed 18 times. Since it is a 5-element subset, the number of times this element appears does not exceed 4 times. There are a total of $5 n$ elemen... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. If $a \geqslant b \geqslant c, a+b+c=0$, and $x_{1}, x_{2}$ are the two real roots of the quadratic equation $a x^{2}+b x+c=0$, then the sum of the maximum and minimum values of $\left|x_{1}^{2}-x_{2}^{2}\right|$ is $\qquad$. | $$
\begin{array}{l}
\left|x_{1}^{2}-x_{2}^{2}\right|=\left|\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)\right|=\left|\frac{b}{a} \cdot \frac{\sqrt{b^{2}-4 a c}}{a}\right|=\left|\frac{b \sqrt{b^{2}+4 a(a+b)}}{a^{2}}\right| \\
=\left|\frac{b(2 a+b)}{a^{2}}\right|=\left|\left(\frac{b}{a}\right)^{2}+2 \cdot \frac{b}{a}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. The Youth Palace is recruiting for spring calligraphy, art, and music instrument classes. The calligraphy class has enrolled 29 students, among whom 13 are enrolled in both calligraphy and art, 12 are enrolled in both calligraphy and music instruments, and 5 are enrolled in all three subjects. Therefore, the number... | $9$ | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Multiply the consecutive natural numbers from 1 to 25, which is $1 \times 2 \times 3 \times \cdots \times 25$, denoted as 25! (read as 25 factorial). When 25! is divided by 3, it is clear that 25! is divisible by 3, resulting in a quotient; then divide this quotient by 3, ..., continue dividing by 3 until the quotie... | Equivalent to calculating how many factors of 3 are in 25!;
Multiples of 3: $[25 \div 3]=8$ (numbers), multiples of 9: $[8 \div 3]=2$ (numbers)
Total factors of 3: $8+2=10$ (numbers), so 3 divides exactly 10 times. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 7 (2005 Fujian Province College Entrance Examination Question) $f(x)$ is a function defined on $\mathbf{R}$ with a period of 3 and is an odd function, and $f(2)=0$. Then the minimum number of solutions to the equation $f(x)=0$ in the interval $(0,6)$ is
A. 2
B. 3
C. 4
D. 7 | Solve: From $f(-x)=-f(x), f(x+3)=f(x)$, we know that the graph of $f(x)$ is symmetric about the point $\left(\frac{3}{2}, 0\right)$, i.e., $\rho\left(\frac{3}{2}\right)=0$. Also, $f(2)=0$. By symmetry, $f(1)=0$. And $f(3)=f(0)=0$, so within one period segment $(0,3]$, the solutions to the equation $f(x)=0$ are four: $x... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Let the maximum value of the function $f(x)=\sqrt{2-x}+\sqrt{3 x+12}$ be $M$, and the minimum value be $m$. Then the value of $\frac{M}{m}$ is ( ).
(A) $\frac{\sqrt{6}}{2}$
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) 2 | 4. D.
From the conditions, the domain of the function $f(x)$ is $[-4,2]$. When $x \in(-4,2)$,
$$
\begin{array}{l}
f^{\prime}(x)=\frac{-1}{2 \sqrt{2-x}}+\frac{3}{2 \sqrt{3 x+12}} . \\
\text { By } f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2} . \\
\text { Also, } f(-4)=\sqrt{6}, f(2)=3 \sqrt{2}, f\left(\frac{1}{2}\right)=2... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. (6 points) Dad is 24 years older than his son. This year, Dad's age is five times that of his son. $\qquad$ years later, Dad's age will be
three times that of his son. | 【Answer】Solution: According to the problem, using the difference multiple formula, we get:
This year, when the father's age is five times the son's age, the son's age is: $24 \div(5-1)=6$ (years);
When the father's age is three times the son's age, the son's age is: $24 \div(3-1)=12$ (years); $12-6=6$ (years).
Answer:... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A city has $n$ high schools, the $i$-th high school sends $C_{i}\left(1 \leqslant C_{i} \leqslant 39,1 \leqslant i \leqslant n\right)$ students to the stadium to watch a ball game, the total number of all students is $\sum_{i=1}^{n} C_{i}=1990$, each row in the stand has 199 seats, and it is required that students f... | Since $199=5 \cdot 39+4$, at least 5 schools' students can be seated in each row.
Now, all students are seated by school, with no empty seats, and a total of 10 rows are arranged. In this way, the situation where students from the same school are not seated together can only occur at the end of one row and the beginnin... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. John scored 93 points in a certain AHSME competition. If the old scoring method were still valid, he would have scored only 84 points with the same answers. The number of questions he did not answer is (the new scoring standard is 5 points for a correct answer, 0 points for a wrong answer, and 2 points for unanswere... | 4. B Let John answer $x$ questions correctly, $y$ questions incorrectly, and leave $z$ questions unanswered, then we have $5 x+0 y+2 z=93,30+4 x-y+0 z=84, x+y$ $+z=30$, solving these equations yields $z=9$. | 9 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
7. (10 points) The school organizes 1511 people to go on an outing, renting 42-seat buses and 25-seat minibuses. If it is required that each person has exactly one seat and each seat is occupied by one person, then there are $\qquad$ rental options.
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 ... | 7. (10 points) The school organizes 1511 people to go on an outing, renting 42-seat buses and 25-seat minibuses. If it is required that each person has exactly one seat and each seat is occupied by one person, then there are $\qquad$ rental schemes.
【Analysis】All seats are fully occupied. Let the number of buses be $x... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7、A and B read a 120-page book, starting on October 1st. A reads 8 pages every day; B reads 13 pages every day, but he takes a break every two days. By the end of the long holiday on October 7th, A and B $\qquad$ compared $\qquad$ who read more, and by how many pages. | 【Analysis】Two people read books together for 7 days, A read $7 \times 8=56$ pages.
In 7 days, since $7 \div 3=2 \cdots 1$, B rested for 2 days, read books for 5 days, and read $5 \times 13=65$ pages.
Therefore, B read more than A, 9 pages more. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Non-negative real numbers $x_{1}, x_{2}, \cdots, x_{2016}$ and real numbers $y_{1}, y_{2}, \cdots, y_{2016}$ satisfy:
(1) $x_{k}^{2}+y_{k}^{2}=1, k=1,2, \cdots, 2016$;
(2) $y_{1}+y_{2}+\cdots+y_{2016}$ is an odd number.
Find the minimum value of $x_{1}+x_{2}+\cdots+x_{2016}$. | From condition (1), we get: $\left|x_{k}\right| \leqslant 1,\left|y_{k}\right| \leqslant 1, k=1,2, \cdots, 2016$, thus (note $x_{k} \geqslant 0$)
$$
\sum_{k=1}^{2016} x_{k} \geqslant \sum_{k=1}^{2016} x_{k}^{2}=\sum_{k=1}^{2016}\left(1-y_{k}^{2}\right)=2016-\sum_{k=1}^{2016} y_{k}^{2} \geqslant 2016-\sum_{k=1}^{2016}\l... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Color the positive integers $1,2, \cdots, 15$ in blue or red. Find the number of coloring methods that satisfy the following conditions:
(1) The positive integer 15 is red;
(2) If the positive integers $x$ and $y$ are of different colors, and $x+y \leqslant 15$, then $x+y$ is colored blue;
(3) If the positive intege... | 2. First, consider the case where 1 is red. Then from $2 \sim 15$ are all red.
Otherwise, assume some number $k$ is blue. By condition (3), we know $1 \cdot k=k$ is red, which is a contradiction. Thus, $1,2, \cdots, 15$ are all red, satisfying the conditions.
Now consider the case where 1 is blue.
$$
\begin{array}{l}
... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Student A, B, C, and D participated in a science mini-test. The test only has 5 true or false questions, with 1 point awarded for each correct answer. The table below provides the answers of each student and the scores of A, B, and C. How many points did D get?
The translation maintains the original text's line brea... | Logical reasoning.
Comparing Yi and Bing:
Yi and Bing only differ on the fourth question, and Yi's score is 1 point lower than Bing's; therefore, the correct answer to the fourth question is “Yes”;
Comparing Jia and Ding:
Jia and Ding have opposite answers for the first, second, third, and fifth questions;
Thus, for th... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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