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11.16 Expand $\left(x^{2}-2 x y+y^{2}\right)^{7}$, the sum of its coefficients is
(A) 0 .
(B) 7 .
(C) 14 .
(D) 128 .
(E) $128^{2}$.
(16th American High School Mathematics Examination, 1965) | [Solution] From $\left(x^{2}-2 x y+y^{2}\right)^{7}=\left[(x-y)^{2}\right]^{7}=(x-y)^{14}$, in its expanded form, let $x=y=1$ which gives the sum of all coefficients.
And $(1-1)^{14}=0$.
Therefore, the answer is $(A)$. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. Given $z \in \mathrm{C}$, if the equation $x^{2}-2 z x+\frac{3}{4}+\mathrm{i}=0$ (i being the imaginary unit) has real roots, then the minimum value of the modulus of the complex number $z$ $|z|$ is $\qquad$ . | $$
2 z x=x^{2}+\frac{3}{4}+\mathrm{i} \Rightarrow 2 z=x+\frac{3}{4 x}+\frac{1}{x} \mathrm{i} \Rightarrow 4|z|^{2}=\left(x+\frac{3}{4 x}\right)^{2}+\frac{1}{x^{2}}=x^{2}+\frac{25}{16 x^{2}}+\frac{3}{2} \geqslant 4 \Rightarrow|z| \geqslant 1 \text {, }
$$
When the equality holds, $x= \pm \frac{\sqrt{5}}{2}$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.44 A finite set of polygons in the plane is said to be properly placed if for any two of them, there is a line through the origin that intersects both.
Find the smallest natural number $m$ such that for any properly placed set of polygons, one can draw $m$ lines through the origin so that each polygon in the set int... | ［Solution］In the case of 3 polygons shown in the right figure, any line passing through the origin intersects at most two of them, hence we know that $m \geqslant 2$.
Since there are only a finite number of polygons, there must be one polygon that subtends the smallest angle at the origin. Let this polygon be $M$ and ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. A two-digit number, after swapping the tens and units, generates a new two-digit number. The sum of the new number and the original number is 132. How many two-digit numbers satisfy this condition?
(A) 5
(B) 7
(C) 9
(D) 11
(E) 12 | 11. B.
Let the two-digit number that satisfies the condition be $10 a+b(a, b \in \mathbf{Z}_{+}, 1 \leqslant a, b \leqslant 9)$. Then, the new number obtained by swapping the tens and units digits is $10 b+a$.
$$
\begin{array}{c}
\text { Then }(10 a+b)+(10 b+a)=132 \\
\Rightarrow a+b=12 . \\
\text { Therefore, }(a, b)... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
19. A batch of feed can supply 10 ducks and 15 chickens for 6 days, or 12 ducks and 6 chickens for 7 days. Then this batch of feed can supply $\qquad$ ducks for 21 days. | 【Solution】Solution: Let 1 duck eat $x$ feed per day, and 1 chicken eat $y$ feed per day. According to the problem, we have:
$$
\begin{aligned}
(10 x+15 y) \times 6 & =(12 x+6 y) \times 7, \\
60 x+90 y & =84 x+42 y \\
24 x & =48 y \\
x & =2 y,
\end{aligned}
$$
Substituting $2 y=x$ into: $(12 x+6 y) \times 7=(12 x+3 x) ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$32 \cdot 35$ 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.
(China Junior High School Mathematics League, 1984) | [Solution] Let the tens digit of the original number be $a$, and the units digit be $b$. According to the problem, we have $-\frac{7}{4}(10 a+$
$b)=10 b+a$, which simplifies to $2 a \doteq b$.
$\because \quad 1 \leqslant a, \quad b \leqslant 9$,
$\therefore a=1,2,3,4$, then $b=2,4,6,8$.
The numbers that meet the condit... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
3. In the spatial rectangular coordinate system, given $O(0,0,0), A(1,0,0), B(0,1,0), C(0,0,1)$, the number of points that are equidistant from the planes $O A B$, $O B C$, $O A C$, and $A B C$ is ( )
A: 1
B: 4
C: 5
D: infinitely many | Huijun $\mathrm{C}$.
Analysis The incenter of the tetrahedron $O A B C$ and the centers of the four "exspheres" are equidistant from the faces $O A B$, $O B C$, $O A C$, and $A B C$, so the number of points that satisfy the condition is 5. | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
10.41 Given any 5 points in the plane, where no three points are collinear and no four points are concyclic. If a circle passes through three of these points and the other two points are one inside and one outside the circle, it is called a "good circle". If the number of good circles is denoted as $n$, find all possib... | [Solution] Choose any two points $A, B$ from 5 points and draw a line through these two points. If the other three points $C, D, E$ are on the same side of line $AB$, then consider $\angle ACB, \angle ADB, \angle AEB$. Without loss of generality, assume $\angle ACB < 180^{\circ}$. Then the circle $ADB$ is the only good... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (8 points) Jia, Yi, and Bing are sawing wooden sticks of the same thickness. They are given 8 meters, 10 meters, and 6 meters of wooden sticks respectively, and are required to saw them into 2-meter segments. After the work is done, Jia, Yi, and Bing have sawn 24, 25, and 27 segments respectively. The fastest sawing... | 【Solution】Solution: For A: $8 \div 2=4$ (segments)
$$
\begin{array}{l}
4-1=3 \text { (times) } \\
3 \times(24 \div 4) \\
=3 \times 6 \\
=18 \text { (times) } \\
\text { For B: } 10 \div 2=5 \text { (segments) } \\
5-1=4 \text { (times) } \\
4 \times(25 \div 5) \\
=4 \times 5 \\
=20 \text { (times) } \\
\end{array}
$$
... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In $\triangle A B C$, the sides $a, b, c$ are opposite to the angles $\angle A, \angle B, \angle C$ respectively. If
$$
2 \sin ^{2} B+3 \sin ^{2} C=2 \sin A \cdot \sin B \cdot \sin C+\sin ^{2} A \text {, }
$$
then $\tan A=$ $\qquad$ | 5. -1 .
From the Law of Sines, we have
$$
2 b^{2}+3 c^{2}=2 b c \sin A+a^{2} \text {. }
$$
From the Law of Cosines, we have
$$
\begin{array}{l}
2 b^{2}+3 c^{2}=2 b c \sin A+b^{2}+c^{2}-2 b c \cos A \\
\Rightarrow \frac{b^{2}+2 c^{2}}{2 b c}=\sin A-\cos A .
\end{array}
$$
Since $\frac{b^{2}+2 c^{2}}{2 b c} \geqslant ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(1) If the real number $x$ satisfies: for any positive number $a$, $x^{2}<1+a$, then the minimum value of $x$ is ( ).
(A) 0
(B) 1
(C) -1
(D) Does not exist | (1) $\mathrm{C}$ Hint: Since $a>0$, then $a+1>1, x^{2}<1+a$ is equivalent to $x^{2} \leqslant 1, -1 \leqslant x \leqslant 1$. | -1 | Inequalities | MCQ | Yes | Yes | olympiads | false |
2. Given non-zero real numbers $x, y, z$ form an arithmetic sequence, $x+1, -y, z$ and $x, y, z+2$ each form a geometric sequence, then the value of $y$ is ( ).
A. 8
B. 16
C. 12
D. 20 | 2. The system of equations $\left\{\begin{array}{l}2 y=x+z \\ y^{2}=z(x+1) \\ y^{2}=x(z+2)\end{array}\right.$
From (2)-(3) we get $z=2 x$, substituting into (1) gives $x=\frac{2}{3} y, \therefore z=\frac{4}{3} y$.
Substituting back into (2) yields the valid value of $y$ as $y=12$. Therefore, the answer is C. | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
(3) (2006 - Shandong) Given an odd function $f(x)$ defined on $\mathbf{R}$ that satisfies $f(x+2)=-f(x)$, then the value of $f(6)$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2 | Solve $f(6)=f(4+2)=-f(4)=-f(2+2)=-[-f(2)]=f(2)$, for an odd function defined on $\mathbf{R}$, we always have $f(0)=0$, thus, $f(0+2)=-f(0)$, which means $f(2)=0$, so, $f(6)=0$, the answer is B. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
60. Let the number of positive divisors of 3600 be $m$, and the number of positive divisors of 36 be $n$, then $\frac{m}{n}=$ | Answer: 5.
Solution: Since
$$
3600=2^{4} \times 3^{2} \times 5^{2},
$$
Therefore, the number of positive divisors of 3600
$$
m=(4+1)(2+1)(2+1)=5 \times 3 \times 3=45.
$$
Also, since
$$
\begin{array}{c}
36=2^{2} \times 3^{2}, \\
27
\end{array}
$$
Therefore, the number of positive divisors of 36
$$
\begin{array}{c}
n=... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Given two lines with slopes of $2$ and $\frac{1}{2}$ intersecting at the point $(2,2)$. The area of the triangle formed by these lines and a third line $x+y=10$ is ( ).
(A) 4
(B) $4 \sqrt{2}$
(C) 6
(D) 8
(E) $6 \sqrt{2}$ | 7. C.
From the problem, we can set the equations of the two lines as
$$
l_{1}: y=2 x+b_{1}, l_{2}: y=\frac{1}{2} x+b_{2} \text {. }
$$
Substituting the point $(2,2)$, we get
$$
l_{1}: y=2 x-2, l_{2}: y=\frac{1}{2} x+1 \text {. }
$$
Thus, the intersections with the line $l_{3}: y=-x+10$ are at points
$$
B(4,6), C(6,4... | 6 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. Given the quadratic function
$$
f(x)=a x^{2}+\left(16-a^{3}\right) x-16 a^{2}(a>0)
$$
the graph of which intersects the $x$-axis at points $A$ and $B$. Then the minimum length of segment $A B$ is $\qquad$. | -1.12 .
The factorization yields $f(x)=\left(x-a^{2}\right)(a x+16)$. Thus, $A\left(a^{2}, 0\right), B\left(-\frac{16}{a}, 0\right)$.
Therefore, the length of segment $A B$ is $a^{2}+\frac{16}{a}$. Let $F(a)=a^{2}+\frac{16}{a}$, then
$$
F^{\prime}(a)=2 a-\frac{16}{a^{2}}=\frac{2\left(a^{3}-8\right)}{a^{2}} .
$$
Hence,... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
88. A box contains ten balls numbered from 1 to 10. Xiao Ming takes nine balls from the box in three successive draws. If starting from the second draw, the sum of the numbers on the balls drawn each time is twice the sum of the numbers from the previous draw, then the number of the ball not drawn is $\qquad$ | answer: 6 | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Question 211, Find the largest integer $k$, such that $\left[\frac{n}{\sqrt{3}}\right]+1>\frac{\mathrm{n}^{2}}{\sqrt{3 \mathrm{n}^{2}-k}}$ holds for all positive integers $n \geq 2$.
---
The translation maintains the original text's format and line breaks. | Question 211, Solution Method One: We first test the first few values.
When $\mathrm{n}=2$, we require $2>\frac{4}{\sqrt{12-k}}$, which gives $\mathrm{k} \leq 7$;
When $n=3$, we require $2>\frac{9}{\sqrt{27-k}}$, which gives $k \leq 6$;
When $n=4$, we require $3>\frac{16}{\sqrt{48-k}}$, which gives $k \leq 5$;
When $n=... | 5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Five. (15 points) Given an $n$-element set of positive integers
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\},
$$
for any $i \in\{1,2, \cdots, n\}$, the set $A$ with the element $a_{i}$ removed, denoted as $A_{i}$, can be divided into the union of two disjoint subsets, and the sums of the elements in these two subse... | (1) Let $M=a_{1}+a_{2}+\cdots+a_{n}$.
By the problem statement, the sum of the elements in the set $A_{i}(i=1,2, \cdots, n)$ is $M-a_{i}$, and $M-a_{i}$ is even. Therefore, $a_{i}$ and $M$ have the same parity.
(i) If $M$ is odd, then $a_{i}(i=1,2, \cdots, n)$ is also odd, and $n$ is odd.
(ii) If $M$ is even, then $a_{... | 7 | Number Theory | proof | Yes | Yes | olympiads | false |
15. (12 points) Chen Chen, a little kid, found that he has a total of 20 coins of 1 jiao and 5 jiao, and the total amount of money is 8 yuan. How many 1 jiao coins does he have? | 【Answer】Solution: 8 yuan $=80$ jiao,
Assuming all are 5 jiao coins, then the 1 jiao ones are:
$$
\begin{aligned}
& (5 \times 20-80) \div(5-1) \\
= & 20 \div 4 \\
= & 5 \text { (pieces); }
\end{aligned}
$$
Answer: There are 5 pieces of 1 jiao coins. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. The sum of the integer parts of all positive real numbers $x$ that satisfy $x^{4}-x^{3}-2 \sqrt{5} x^{2}-7 x^{2}+\sqrt{5} x+3 x+7 \sqrt{5}+17=0$ is | $$
\begin{array}{l}
\text { Let } x^{4}-x^{3}-7 x^{2}+3 x+12=\left(x^{2}+a x-3\right)\left(x^{2}+b x-4\right) \\
\Rightarrow\left\{\begin{array} { l }
{ a + b = - 1 , } \\
{ 4 a + 3 b = - 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=0, \\
b=-1 .
\end{array} \Rightarrow x^{4}-x^{3}-7 x^{2}+3 x+12=\left(x^{2}-3... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. In the Cartesian coordinate system $x O y$, the coordinates of point $F$ are $(1,0)$, and points $A, B$ are on the parabola $y^{2}=4 x$, satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B}=-4,|\overrightarrow{F A}|-|\overrightarrow{F B}|=4 \sqrt{3}$, find the value of $\overrightarrow{F A} \cdot \overrighta... | Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, according to the problem, we have $x_{1}=\frac{y_{1}^{2}}{4}, x_{2}=\frac{y_{2}^{2}}{4}$. Then $\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2}=\frac{y_{1}^{2} y_{2}^{2}}{16}+y_{1} y_{2}=-4$
$$
\begin{array}{l}
\Rightarrow y_{1}^{2} y... | -11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16. Elsa makes ice sculptures with ice blocks, 1 ice block can make 1 small ice sculpture, 3 ice blocks can make 1 large ice sculpture. The leftover shavings from making 2 small ice sculptures or 1 large ice sculpture are just enough to make 1 ice block. Given 30 ice blocks, to make the number of small ice sculptures m... | $11$ | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A1. Shawn's password to unlock his phone is four digits long, made up of two 5 s and two 3s. How many different possibilities are there for Shawn's password? | Solution: This is the number of ways to choose 2 spots out of 4 , thus there are $\binom{4}{2}=6$ possibilities. Alternatively one can list all the possibilities: $3355,3535,3553,5335,5353,5533$.
Answer: 6 . | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Calculate: $(70 \div 4+90 \div 4) \div 4=$ | 【Analysis】We can start the calculation from inside the parentheses, and the two divisions within the parentheses can be converted into fractional forms, and then calculate the final result.
【Solution】According to the analysis, the original expression is $(70 \div 4+90 \div 4) \div 4$
$$
\begin{array}{l}
=(70+90) \div ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$12 \cdot 10$ If $x=\frac{1-i \sqrt{3}}{2}$ where $i=\sqrt{-1}$, then $\frac{1}{x^{2}-x}$ equals
(A) -2 .
(B) -1 .
(C) $1+i \sqrt{3}$.
(D) 1 .
(E) 2 .
(23rd American High School Mathematics Examination, 1972) | [Solution] From $x^{2}-x$
$$
\begin{array}{l}
=\frac{1}{4}(1-i \sqrt{3})^{2}-\frac{1}{2}(1-i \sqrt{3}) \\
=\frac{1}{4}(-2-2 i \sqrt{3})-\frac{1}{2}(1-i \sqrt{3}) \\
=-1,
\end{array}
$$
Therefore, the reciprocal of $x^{2}-x$ is still -1. Hence, the correct choice is $(B)$. | -1 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. Given the equation $|x-20|+|x+20|=2020$ with two solutions $x_{1} 、 x_{2}$. Then $x_{1}+x_{2}=(\quad)$.
(A) 2020
(B) 1010
(C) 0
(D) -2020 | -,1. C.
From the symmetry of the function about the $y$-axis, we know that $x_{1}+x_{2}=0$. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. Let point $O$ be inside a regular tetrahedron $A B C D$, and $G$ be the center of the regular tetrahedron. The line $O G$ intersects the faces of the tetrahedron at points $A_{1}, B_{1}, C_{1}, D_{1}$. Then the value of $\frac{A_{1} O}{A_{1} G}+\frac{B_{1} O}{B_{1} G}+\frac{C_{1} O}{C_{1} G}+\frac{D_{1} O}{D_{1} G}$... | 2. D Let the distance from point $G$ to each face of the tetrahedron be $r$, and the distances from point $O$ to the planes $BCD$, $CDA$, $DAB$, $ABC$ be $h_{1}$,
$$
\begin{array}{c}
h_{2}, h_{3}, h_{4} \text {, then } \frac{V_{O-BCD}}{V_{G-BCD}}=\frac{h_{1}}{r}=\frac{A_{1} O}{A_{1} G}, \frac{V_{O}-CDA}{V_{G-CDA}}=\fra... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
8.16 The number of positive integer pairs $(x, y)$ that satisfy the equation $7^{x}-3 \cdot 2^{y}=1$ is
(A) 0.
(B) 1.
(C) 2.
(D) Infinitely many.
(High School Mathematics Competition in Fuzhou City, Fujian Province, China, 1993) | [Solution] The given equation can be transformed into
$\frac{7^{x}-1}{7-1}=2^{y-1}$, which is $7^{x-1}+7^{x-2}+\cdots+1=2^{y-1}$.
If $y=1$, then $x=1$, i.e., $(1,1)$ satisfies the equation.
If $y>1$, since the right side of (1) is even, the left side must be an even number of odd numbers, which means $x$ is even, so th... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
2. Let $x_{i} \geqslant 0(i=1,2,3, \cdots, n)$, and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} x_{k} x_{j}=1$, find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$. | Since $\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} x_{k} x_{j} \geqslant 1$, equality holds if and only if there exists $i$ such that $x_{i}=1$ and the rest $x_{j}=0, j \neq i$, so the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1.
Let $x_{k}=\sqrt{k} y_{k}(1 \leqs... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.37 If $n, p$ are two natural numbers, let
$$
S_{n, p}=1^{p}+2^{p}+\cdots+n^{p} .
$$
Determine the natural number $p$, such that for any natural number $n, S_{n, p}$ is a perfect square.
(French Mathematical Olympiad, 1991) | [Solution] Since for any natural number $n, S_{n, p}$ is a perfect square, in particular, taking $n=2, \quad S_{2, p}$ is also a perfect square. That is, there exists a natural number $x$ satisfying the following equation:
So
$$
\begin{array}{c}
x^{2}=1+2^{p} \quad(x>1) . \\
x^{2}-1=2^{p}, \\
(x-1)(x+1)=2^{p} .
\end{ar... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) There are 8 children, each wearing a red hat or a blue hat. If a child sees 3 or more other children wearing red hats, they take a red balloon; otherwise, they take a blue balloon. In the end, some children have red balloons and some have blue balloons. How many children are wearing red hats? $\qquad$ | 【Answer】Solution: If there is 1 or 2 children wearing red hats, then all the children will take out blue balloons;
If there are 3 children wearing red hats, then the children wearing red hats will all take out blue balloons, and the children wearing blue hats will take out red balloons; this meets the requirements of t... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
12. The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{0}=1, a_{1}=2, a_{n+2}=a_{n}+a_{n+1}^{2}$, then $a_{2006} \equiv$ $\qquad$ $(\bmod 7)$ | 12. 6
Hint: $a_{0} \equiv 1(\bmod 7), a_{1} \equiv 2(\bmod 7)$,
$$
\begin{array}{l}
a_{2} \equiv 2^{2}+1 \equiv 5(\bmod 7), a_{3} \equiv 6(\bmod 7), \\
a_{4} \equiv 6(\bmod 7), a_{5} \equiv 0(\bmod 7), \\
a_{6} \equiv 6(\bmod 7), a_{7} \equiv 1(\bmod 7), \\
a_{8} \equiv 2(\bmod 7) .
\end{array}
$$
Therefore, $a_{n+7}... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. If $M=\left\{(x, y)|| \tan \pi y \mid+\sin ^{2} \pi x=0\right\}, N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant 2\right\}$, then $|M \cap N|=(\quad)$
A. 4
B. 5
C. 8
D. 9 | 2. D From $\tan \pi y=0$ we get $y=k, k \in \mathbf{Z}$; from $\sin \pi x=0$ we get $x=k', k' \in \mathbf{Z}$; also, $x^{2}+y^{2} \leqslant 2$, so $k=0$, $-1,1 ; k'=0,-1,1$. Therefore, $M \cap N$ contains a total of 9 points. | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Let $x \geqslant 0, y \geqslant 0$, and $x+2 y=\frac{1}{2}$, then the maximum value of the function $u=\log \frac{1}{2}\left(8 x y+4 y^{2}+1\right)$ is
A. $\log \frac{1}{2} \frac{4}{3}$
B. 0
C. 1
D. $\log _{\frac{1}{2}} \frac{3}{4}$ | 4. B $x+2 y=\frac{1}{2}, y=\frac{1}{4}-\frac{1}{2} x$, from $x \geqslant 0$ we know, $y \leqslant \frac{1}{4}$. Substituting $x=\frac{1}{2}-2 y$ into $u$, $u=\log \frac{1}{2}\left(-12 y^{2}+\right.$ $4 y+1)$. Let $-12 y^{2}+4 y+1>0$, thus $y \in\left[0, \frac{1}{4}\right]$. $-12 y^{2}+4 y+1 \in\left[1, \frac{4}{3}\righ... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
11. Given that vectors $\alpha, \beta$ are two mutually perpendicular unit vectors in a plane, and
$$
(3 \boldsymbol{\alpha}-\boldsymbol{\gamma}) \cdot(4 \boldsymbol{\beta}-\boldsymbol{\gamma})=0 .
$$
Then the maximum value of $|\boldsymbol{\gamma}|$ is $\qquad$ | 11. 5 .
As shown in Figure 2, let
$$
\begin{array}{l}
\overrightarrow{O A}=3 \alpha, \\
\overrightarrow{O B}=4 \boldsymbol{\beta}, \\
\overrightarrow{O C}=\gamma .
\end{array}
$$
From the given information, $\overrightarrow{A C} \perp \overrightarrow{B C}$. Therefore, point $C$ lies on the circle with $A B$ as its di... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. (3 points) The greatest common divisor and the least common multiple of two numbers are 3 and 135, respectively. Find the minimum difference between these two numbers is $\qquad$ . | 【Solution】Solution: Since $135 \div 3=45, 45$ can be decomposed into two coprime numbers in two ways, which are 1 and $45$, 9 and 5, so the pair with the smallest difference is: 9 and 5,
Therefore, these two numbers are:
$$
\begin{array}{l}
9 \times 3=27 \\
5 \times 3=15 \\
27-15=12
\end{array}
$$
Answer: The smalles... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Let $a_{i}(i=1,2,3,4)$ be real numbers. If the sum of the elements of all non-empty proper subsets of the set $\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$ is
28, then $a_{1}+a_{2}+a_{3}+a_{4}=$ $\qquad$ | Answer: 4
Explanation: The number of non-empty proper subsets containing the element $a_{i}(i=1,2,3,4)$ is 7, so the sum of the elements of all non-empty proper subsets of $\mathrm{I}=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$ is $7\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=28$, thus $a_{1}+a_{2}+a_{3}+a_{4}=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the Cartesian coordinate system $x O y$, points with both coordinates as integers are called integer points. It is known that $k$ is a real number. When the intersection point of the two different lines $y=4 k x-\frac{1}{k}$ and $y=\frac{1}{k} x+2$ is an integer point, the number of values that $k$ can take is ( ... | 3. C.
When $x=0$, $y=2$, so
$$
2=-\frac{1}{k} \Rightarrow k=-\frac{1}{2} \text {. }
$$
At this point, the two lines coincide.
Therefore, $x \neq 0$, and $k \neq-\frac{1}{2}$.
Obviously, $k \neq 0$, then
$4 k x-\frac{1}{k}=\frac{1}{k} x+2$
$\Rightarrow k=\frac{x+1}{2 x}$
$\Rightarrow y=\frac{2}{x+1}+2 x$.
Since $x$ an... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. (10 points) From the integers 1 to 10, at least ( ) numbers must be taken to ensure that there are two numbers whose sum is 10.
A. 4
B. 5
C. 6
D. 7 | 【Analysis】10 natural numbers are: $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$; those that sum to 10 are $(1,9), (2, 8); (3, 7); (4, 6)$; the sum of the two numbers in these four pairs is 10. According to the pigeonhole principle, consider the worst-case scenario: taking out 6 numbers, which are the numbers $5, 10$ and one number f... | 7 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
5.9 Expand the expression $\left(x^{7}+x^{5}+1\right)^{20}$ and combine like terms, to get a polynomial. Find the coefficients of $x^{17}$ and $x^{18}$ in this polynomial. | [Solution] Since $17=5 \times 2+7 \times 1$, the coefficient of $x^{17}$ is
$$
\frac{20!}{2!(20-2-1)!}=\frac{20!}{2!\cdot 17!}=\frac{20 \cdot 19 \cdot 18}{2}=3420 \text {. }
$$
Since 18 cannot be expressed as the sum of several 5s and 7s, this polynomial does not contain a term of $x^{18}$, or in other words, the coef... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8、A number lies between 2013 and 2156, and when divided by $5, 11, 13$ the remainders are the same. The largest possible remainder is $\qquad$ - | 【Analysis】This number minus the remainder results in a common multiple of $5, 11, 13$. Since the divisor includes 5, the remainder can be at most 4. Therefore, this number minus the remainder is at least 2009 and at most 2156 (i.e., when the remainder is 0). $[5,11,13]=715$, and $715 \times 3=2145$. Thus, 2149 is betwe... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
33.34 Let $[x]$ denote the greatest integer not exceeding $x$. Then the number of positive integer solutions to the equation $[1.9 x]+[8.8 y]=36$ is
(A) 5.
(B) 4.
(C) 3.
(D) 2.
(China Guangzhou, Wuhan, Fuzhou, and other five cities Junior High School Mathematics League, 1988) | ［Solution］Since $x, y$ are positive integers, then first $y \geqslant 1$.
Also, $[8.8 y]<36$, so $y \leqslant 4$. That is, $1 \leqslant y \leqslant 4$.
By substituting $y=1,2,3,4$ into the equation, we find that $(15,1),(10,2),(1,4)$ are its solutions. Therefore, the answer is $(C)$. | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
40. At 8:00 AM, the Knight and the Warrior set off from ports $A$ and $B$ respectively, heading towards each other. The Knight arrives at the downstream port $B$, and the Warrior arrives at the upstream port $A$, and both immediately turn around and return. At 10:00 AM, both ships return to their respective starting po... | \begin{tabular}{r}
$\mathbf{4 0}$ \\
\hline \\
6
\end{tabular} | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. When $s, t$ take all real numbers, then the minimum value that $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ can reach is | 9.2 Hint: $(s+5, s)$ is a point on the line, and $\left\{\begin{array}{l}x=3|\cos t|, \\ y=2|\sin t|\end{array}\right.$ is a point on the elliptical arc. Clearly, the point $(3,0)$ on it has the minimum square of the distance to the line, which is 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. "24 Game" is a familiar math game to many people, the game process is as follows: arbitrarily draw 4 cards from 52 cards (excluding the joker), use the numbers on these 4 cards $(A=1, J=11, Q=12, K=13)$ to get 24 through addition, subtraction, multiplication, and division. The first person to find the algorithm wins... | $4$ | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Solve the equation
$$
\frac{x}{2+\frac{x}{2+} \cdot \cdot 2+\frac{x}{1+\sqrt{1+x}}}=1 .
$$
(The expression on the left side of the equation contains 1985 twos)
(1985 All-Soviet Union Mathematical Olympiad) | Solve: Since $\frac{x}{\sqrt{x+1}+1}=\sqrt{x+1}-1$, let $f(x)=2+\sqrt{x+1}-1$ $=\sqrt{x+1}+1$, then the original equation is clearly an iterative equation
$$
\begin{array}{c}
f^{[n]}(x)=2+\frac{f^{[n-1]}(x)}{\sqrt{f^{[n-1]}(x)+1}+1}, \\
f^{[n]}(x)=\sqrt{f^{[n-1]}(x)+1}+1 .
\end{array}
$$
The fixed points of $f(x)$ are ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Four schools each send 3 representatives to form $n$ groups to participate in social practice activities (each representative can participate in several groups). The rules are: (1) representatives from the same school are not in the same group; (2) any two representatives from different schools are in exactly one gr... | 9.9 Group. First consider the representatives of cities A and B, let the representatives of city A be $A_{1}, A_{2}, A_{3}$, and the representatives of city B be $B_{1}, B_{2}, B_{3}$. They can form 9 pairs: $A_{1} B_{1}, A_{1} B_{2}, A_{1} B_{3}, A_{2} B_{1}, A_{2} B_{2}, A_{2} B_{3}, A_{3} B_{1}, A_{3} B_{2}, A_{3} B... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. (15 points) Math Competition, 8 fill-in-the-blank questions, 4 points for each correct answer, 0 points for each incorrect answer; 6 short-answer questions, 7 points for each correct answer, 0 points for each incorrect answer, 400 participants, what is the minimum number of people who have the same total score? | 【Analysis】First, find out how many different outcomes there are, then divide 400 by the number of groups to see if there is any remainder, which is the maximum value of the average distribution.
【Solution】Solution: Method one: Let the number of 4-point questions answered correctly be $a$, and the number of 7-point que... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Example 5.2.4] Let $x, y, z, w$ be real numbers, and satisfy:
(1) $x+y+z+w=0$;
(2) $x^{7}+y^{7}+z^{7}+w^{7}=0$.
Find the value of $w(w+x)(w+y)(w+z)$. | If we take $x, y, z, w$ as the main variables, the two given conditions are not sufficient to directly determine the values of $x, y, z, w$. However, the problem only requires us to find the value of the product of four factors: $w(w+x)$, $(w+y)(w+z)$. Therefore, we still divide $x, y, z, w$ into two camps: main and au... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
19. Xiaoling's father takes the subway or bus to and from work every morning and afternoon without transferring. In the 10 working days, he took the subway 9 times, and the bus in the morning 5 times, so he took the bus in the afternoon ( ) times. | answer : 6 | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. (5 points) In the third grade, there are two classes, Class A and Class B. If 4 students are transferred from Class A to Class B, the number of students in both classes will be equal. Class A has $\qquad$ more students than Class B. | 【Answer】Solution: $4 \times 2=8$ (people) Answer: Class A has 8 more people than Class B. Therefore, the answer is: 8. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$33 \cdot 60$ denotes $25_{b}$ as a two-digit number in base $b$. If $52_{b}$ is twice $25_{b}$, then $b$ equals
(A) 7.
(B) 8.
(C) 9.
(D) 11.
(E) 12.
(16th American High School Mathematics Examination, 1965) | [Solution] From the given, $25_{b}=2 b^{1}+5 b^{0}, 52_{b}=5 b^{1}+2 b^{0}$, and by the problem statement $52_{b}=2\left(25_{b}\right)$, which means $5 b+2=2(2 b+5)$, solving this gives $b=8$. Therefore, the answer is $(B)$. | 8 | Algebra | MCQ | Yes | Yes | olympiads | false |
12. Dora the Explorer and her monkey friend Boots encountered a math problem during one of their adventures:
$$
2020 \times 2019 - 2021 \times 2018
$$
Seeing Boots looking puzzled, Dora said: “Don’t worry, let’s think together if there’s a simpler way?” So, the result of this calculation is $\qquad$ . | $2$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
17・106 Through any point $P$ inside $\triangle ABC$, draw $IF \parallel BC, EH \parallel AC, DG \parallel AB$. Then the value of $\frac{IF}{BC} + \frac{EH}{AC} + \frac{DG}{AB}$ is
(A) $\frac{3}{2}$.
(B) 2.
(C) $\frac{4}{3}$.
(D) $\frac{5}{3}$.
(Chinese Jiangsu Province Suzhou Junior High School Mathematics Competition,... | [Solution 1] Let the three sides of $\triangle ABC$ be $a, b, c$, and let $DE=a^{\prime}, \quad GF=b^{\prime}$, $HI=c^{\prime}$. From the conclusion of the previous problem, we have
$$
\frac{IF}{BC}=\frac{a-a^{\prime}}{a}=1-\frac{a^{\prime}}{a}=1-\frac{S_{\triangle PBC}}{S_{\triangle ABC}},
$$
Similarly,
$$
\begin{arr... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
6、Taking the train from City A to City B, it initially took 19.5 hours at the beginning of 1998. In 1998, the train's speed was increased by $30\%$ for the first time, in 1999 it was increased by $25\%$ for the second time, and in 2000 it was increased by $20\%$ for the third time. After these three speed increases, th... | 6. Solution: According to the problem, when the distance is constant, speed and time are inversely proportional. $19.5 \div(1+30 \%) \div(1+25 \%) \div(1+20 \%)=19.5 \times \frac{100}{130} \times \frac{100}{125} \times \frac{100}{120}=10$ (hours) Answer: It takes only 10 hours to travel from City A to City B by train. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Given the following two equations in $x$: $6(x+8)=18 x ; 6 x-2(a-x)=2 a+x$ have the same solution, then $a=(\quad)$. | 【Analysis】
$$
\begin{aligned}
6(x+8) & =18 x \\
6 x+48 & =18 x \\
12 x & =48 \\
x & =4
\end{aligned}
$$
Substitute into the second equation, solve the equation
$$
\begin{aligned}
6 \times 4-2(a-4) & =2 a+4 \\
24-2 a+8 & =2 a+4 \\
4 a & =28 \\
a & =7
\end{aligned}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
SKET $k$ Given: On the parabola $y=x^{2}$, there are three vertices $A, B, C$ of a square. Find: The minimum value of the area of such a square.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Assume that among the three vertices of the square on the parabola, two are on the right side of the $y$-axis (including the $y$-axis), and let the coordinates of points $A$, $B$, and $C$ be $\left(x_{i}, y_{i}\right), i=1,2,3$. The slope of $BC$ is $k (k > 0)$. Since the points lie on the parabola $y=x^{2}$, we have $... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. The Yangtze Farm has a pasture, where grass grows uniformly every day. If 24 cows are grazed on the pasture, they will finish the grass in 6 days; if only 21 cows are grazed, it will take 8 days to finish the grass. How many days will it take to finish the grass if 36 cows are grazed? | Assuming one cow eats 1 unit of grass per day, the new grass grows daily: $(21 \times 8 - 24 \times 6) \div (8 - 6) = 12$
Original grass amount: $(21 - 12) \times 8 = 72$
Therefore, $72 \div (36 - 12) = 3$ (days) | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. $\frac{1}{1-\log _{2} 20}+\frac{1}{1-\log _{5} 50}$ The value is $\qquad$ | $$
\frac{1}{1-\log _{2} 20}+\frac{1}{1-\log _{5} 50}=-\frac{1}{\log _{2} 10}-\frac{1}{\log _{5} 10}=-\lg 2-\lg 5=-1 .
$$
Translate the above text into English, keeping the original text's line breaks and format:
$$
\frac{1}{1-\log _{2} 20}+\frac{1}{1-\log _{5} 50}=-\frac{1}{\log _{2} 10}-\frac{1}{\log _{5} 10}=-\lg 2... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The number of real roots of the equation $\sin x=\lg x$ is | 2. 3 .
According to the graphs of the functions $y=\sin x$ and $y=\lg x$, in the interval $x \in(0,10)$, it can be seen that their graphs have 3 intersection points, and the equation $\sin x=\lg x$ has three roots. | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
15. In the 6 circles below, fill in $0,1,2,3,4,5$ with each number used only once, so that the sum of the three numbers on each side is equal. | 【Key Point】Number Array Diagram【Difficulty】Adjacent
【Answer】
【Analysis】Let the 3 vertices be $a, b, c, 0+1+2+3+4+5+a+b+c=15+a+b+c$ is the sum of the 3 edges, $a+b+c$ is a multiple of 3. When $a+b+c=0+1+2=3$, the sum of each edge is $(15+3) \div 3=6$. By rotating the two numbers above, 4 more results can be obtained. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. The number of ordered pairs $(a, b)$ that satisfy $(a+b \mathrm{i})^{6}=a-b \mathrm{i}(a, b \in \mathbf{R})$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 8 .
Let $z=a+b$ i. Then
$$
z^{6}=\bar{z} \Rightarrow|z|^{6}=|\bar{z}|=|z| \text {. }
$$
Thus, $|z|=0$ or 1.
When $|z|=0$, $z=0$, i.e., $a=b=0$. Clearly, $(0,0)$ satisfies the condition;
When $|z|=1$, from $z^{6}=\bar{z}$, we know
$$
z^{7}=z \cdot \bar{z}=|z|^{2}=1 \text {. }
$$
Note that, $z^{7}=1$ has 7 complex ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $m$ and $n$ be positive integers, and $m>n$. If the last two digits of $9^{m}$ and $9^{n}$ are the same, then the minimum value of $m-n$ is $\qquad$ | 3. 10 .
From the problem, we know that $9^{m}-9^{n}=9^{n}\left(9^{m-n}-1\right)$ is a multiple of 100. Therefore, $9^{m-n}-1$ is also a multiple of 100. Hence, the last two digits of $9^{m-n}$ are 01.
Clearly, $m-n$ is an even number. Let $m-n=2 t$ (where $t$ is a positive integer). Then $9^{m-n}=9^{2 t}=81^{t}$.
It... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
21. As shown in the figure, curve $C_{1}$ is part of an ellipse centered at the origin $O$ with foci $F_{1}, F_{2}$, and curve $C_{2}$ is part of a parabola with vertex at $O$ and focus at $F_{2}(1,0)$. Point $A\left(\frac{3}{2}, \sqrt{6}\right)$ is the intersection of curves $C_{1}$ and $C_{2}$.
(1) Find the equations... | (1) According to the problem, for the ellipse we have $\left\{\begin{array}{l}a^{2}-b^{2}=1, \\ \frac{9}{4 a^{2}}+\frac{6}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=9, \\ b^{2}=8 .\end{array}\right.\right.$
Thus, the equation of the ellipse is $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$, and the equation of ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
(2) Divide the sides of the equilateral $\triangle A B C$ with side length 1 into $n$ equal parts, and draw lines parallel to the other two sides through each division point, dividing the triangle into smaller triangles. The vertices of these small triangles are called nodes. A real number is placed at each node. It is... | Without loss of generality, let $a \leqslant b \leqslant c$. We refer to property (II) as the "diamond property". As shown in Figure 1, consider three adjacent nodes $D, E, F$ on the same line, with numbers $d, e, f$ placed on them; then take node $G$ adjacent to $D, E$, and node $H$ adjacent to $E, F$, with numbers $g... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
20.26 In the attached figure, $T P$ and $T^{\prime} Q$ are two parallel tangent lines of this circle with radius $r$, $T$ and $T^{\prime}$ are the points of tangency. $P T^{\prime \prime} Q$ is a third tangent line with the point of tangency $T^{\prime \prime}$.
If $T P=4, \quad T^{\prime} Q=9$, then $r$ is
(A) $\frac{... | [Solution] Connect $O P, O Q, O T, O T^{\prime}$, then by $T P=T^{\prime \prime} P, O T=O T^{\prime \prime}=r$, and $\angle P T^{\prime \prime} O=\angle P T O=90^{\circ}$, thus $\triangle O T P \cong \triangle O T^{\prime \prime} \mathrm{P}$.
Similarly, we can prove $\triangle O T^{\prime} Q \cong \triangle O T^{\prime... | 6 | Geometry | MCQ | Yes | Yes | olympiads | false |
$33 \cdot 5$ and $3^{1991}+1991^{3}$, when expressed in decimal form, have their last digit as
(A) 8.
(B) 4.
(C) 2.
(D) 0.
(3rd "Five Sheep Cup" Junior High School Mathematics Competition, 1991) | [Solution] $3^{1091}=3^{4 \cdot 477+3}=81^{497} \cdot 3^{3}$, its unit digit is 7. Also, the unit digit of $1991^{3}$ is 1, so the last digit of $3^{1991}+1991^{3}$ is 8. Therefore, the answer is $(A)$. | 8 | Number Theory | MCQ | Yes | Yes | olympiads | false |
(3) For the number $(3+\sqrt{8})^{2 n}\left(n \in \mathbf{N}^{*}\right.$ and $\left.n \geqslant 2009\right)$, let $[x]$ be the integer part of $x$. Then the remainder when $\left[(3+\sqrt{8})^{2 n}\right]$ is divided by 8 is $(\quad)$.
(A) 1
(B) 3
(C) 4
(D) 7 | (3) $\mathrm{A}$ Hint: $x=(3+\sqrt{8})^{2 n}$
$$
\begin{aligned}
= & \mathrm{C}_{2 n}^{0} \cdot 3^{2 n}+\mathrm{C}_{2 n}^{1} \cdot 3^{2 n-1} \cdot(\sqrt{8}) \\
& +\mathrm{C}_{2 n}^{2} \cdot 3^{2 n-2} \cdot(\sqrt{8})^{2}+\cdots \\
= & A+\sqrt{8} B\left(A, B \in \mathbf{N}^{*}\right),
\end{aligned}
$$
$$
\begin{aligned}
... | 1 | Number Theory | MCQ | Yes | Yes | olympiads | false |
8. (3 points) For a certain product, if the purchase price decreases by 10%, and the selling price remains unchanged, then the gross profit margin (Gross Profit Margin = $\frac{\text{Selling Price - Purchase Price}}{\text{Purchase Price}} \times 100 \%$) can increase by $12 \%$, then the original gross profit margin fo... | 8. (3 points) For a certain product, if the purchase price decreases by $10 \%$, and the selling price remains unchanged, then the gross profit margin (gross profit margin = $\frac{\text { selling price - purchase price }}{\text { purchase price }} \times 100 \%$) can increase by $12 \%$, then the original gross profit... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. If $z$ is a complex number with a non-zero real part, then the minimum value of $\frac{\operatorname{Re}\left(z^{4}\right)}{(\operatorname{Re}(z))^{4}}$ is $\qquad$ . (Wang Yongxi)
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result dire... | 7. -8 .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}, a \neq 0)$, then it is easy to know that $\operatorname{Re}\left(z^{4}\right)=a^{4}-6 a^{2} b^{2}+b^{4}$. Therefore,
$$
\frac{\operatorname{Re}\left(z^{4}\right)}{(\operatorname{Re}(z))^{4}}=\frac{a^{4}-6 a^{2} b^{2}+b^{4}}{a^{4}}=t^{2}-6 t-1=(t-3)^{2}-8 \geqslant 8,
... | -8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
3. In the equation $A \times B \times \overline{B B C}=\overline{A D A A}$, the same letters represent the same digits, different letters represent different digits, and $D$ is the smallest natural number, $A$ is the smallest prime number. Using the digits $A$, $B$, and $C$ to form natural numbers without repeated digi... | $6$ | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Let $[t]$ denote the greatest integer not exceeding the real number $t$, and let $\{t\}=t-[t]$. Given that the real number $x$ satisfies $x^{3}+\frac{1}{x^{3}}=18$. Then $\{x\}+\left\{\frac{1}{x}\right\}=(\quad)$.
(A) $\frac{1}{2}$
(B) $3-\sqrt{5}$
(C) $\frac{3-\sqrt{5}}{2}$
(D) 1 | 5. D.
Let $x+\frac{1}{x}=a$. Then
$$
\begin{array}{l}
x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}-1\right) \\
=\left(x+\frac{1}{x}\right)\left[\left(x+\frac{1}{x}\right)^{2}-3\right]=a\left(a^{2}-3\right) .
\end{array}
$$
Thus $a\left(a^{2}-3\right)=18$
$$
\begin{array}{l}
\Rightarrow(... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 5 (2003 Hunan Province Competition Question) Let $x, y \in \mathbf{R}$, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{2003}+2002(x-1)=-1, \\
(y-2)^{2003}+2002(y-2)=1 .
\end{array}\right.
$$
Then $x+y=$ $\qquad$ | Solution: Fill in 3. Reason: Construct the function $f(t)=t^{2003}+2002 t$, it is easy to know that $f(t)$ is an odd function and a monotonic function on $\mathbf{R}$. From this, we can get $f(x-1)=-f(y-2)$, which means $f(x-1)=f(2-y)$. Therefore, $x-1=2-y$, which leads to $x+y=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4 \cdot 25$ Introduce a relation “ $\rightarrow$ ” among the elements of set $S$, such that
(1) For any $a, b \in S$, either $a \rightarrow b$ or $b \rightarrow a$, but not both;
(2) For any $a, b, c \in S$, if $a \rightarrow b, b \rightarrow c$, then $c \rightarrow a$.
How many elements can set $S$ contain at most? | [Solution] Clearly, when $S=\{a, b, c\}$ and the relationships among them are $a \rightarrow b, b \rightarrow c$, $c \rightarrow a$, the conditions of the problem are satisfied. Therefore, the maximum number of elements sought is no less than 3.
If $S$ contains 4 elements $a, b, c, d$, then by the pigeonhole principle... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) Use 40 yuan to buy three types of exercise books priced at 2 yuan, 5 yuan, and 11 yuan each. You must buy at least one of each type, and the money must be exactly spent. How many different ways are there to buy the exercise books? $\qquad$ kinds. | 【Analysis】Subtract 1 from each type, leaving $40-2-5-11=22$ yuan. Then, based on the remaining money, classify and solve the problem.
【Solution】Solution: Subtract 1 from each type, leaving $40-2-5-11=22$ yuan. If you buy 2 more 11-yuan books, it will be exactly used up, counting 1 method; If you buy 1 more 11-yuan boo... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13. In the Cartesian coordinate system $x O y$, point $P(a, b)$, after a certain transformation, gets the corresponding point $P^{\prime}(3 a+ 2,3 b-1)$. Given that $A, B, C$ are three non-collinear points, and after this transformation, their corresponding points are $A^{\prime}, B^{\prime}, C^{\prime}$, respectively.... | $9$ | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Point $P\left(x_{0}, y_{0}\right)$ is any point on the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{2}=1$, and the line $y=k x+m(k m \neq 0)$ intersects the ellipse $C_{1}: \frac{x^{2}}{4}+y^{2}=1$ to form a chord $M N$ which is bisected by the line $O P$ and satisfies $m y_{0}>-1$. The area of $\triangle P M N$ is 1. Det... | 10. Analysis: Substituting \( y = kx + m \) into \( \frac{x^2}{4} + y^2 = 1 \), we get: \( \left(4k^2 + 1\right)x^2 + 8kmx + 4m^2 - 4 = 0 \), hence \( x_1 + x_2 = -\frac{8km}{4k^2 + 1}, x_1 x_2 = \frac{4m^2 - 4}{4k^2 + 1} \), then
$$
|MN| = \sqrt{k^2 + 1} |x_1 - x_2| = \sqrt{k^2 + 1} \sqrt{\frac{64k^2m^2}{(4k^2 + 1)^2}... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given that $x, y, z$ are 3 real numbers greater than or equal to 1, then
$$
\left(\frac{\sqrt{x^{2}(y-1)^{2}+y^{2}}}{x y}+\frac{\sqrt{y^{2}(z-1)^{2}+z^{2}}}{y z}+\frac{\sqrt{z^{2}(x-1)^{2}+x^{2}}}{z x}\right)^{2}
$$
the sum of the numerator and denominator of the minimum value written as a simplified fraction is $\... | Solve:
$$
\frac{\sqrt{x^{2}(y-1)^{2}+y^{2}}}{x y}=\sqrt{\left(1-\frac{1}{y}\right)^{2}+\frac{1}{x^{2}}} \leqslant \frac{\sqrt{2}}{2}\left[\left(1-\frac{1}{y}\right)+\frac{1}{x}\right] \text {, }
$$
Similarly, we get $\frac{\sqrt{y^{2}(z-1)^{2}+z^{2}}}{y z} \leqslant \frac{\sqrt{2}}{2}\left[\left(1-\frac{1}{z}\right)+\... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Given the parabola $\Gamma: y^{2}=8 x$ with focus $F$, a line $l$ through $F$ intersects the parabola $\Gamma$ at points $A, B$. Tangents to the parabola $\Gamma$ at $A, B$ intersect the $y$-axis at points $P, Q$ respectively. Find the minimum value of the area of quadrilateral $A P Q B$.
Translate the above text i... | 9 Analysis: It is known that the focus of the parabola is $F(2,0)$. By symmetry, let's assume $A\left(2 x_{1}^{2}, 4 x_{1}\right)\left(x_{1}>0\right), B\left(2 x_{2}^{2}, 4 x_{2}\right)\left(x_{2}<0\right)$, then $\frac{4 x_{1}-0}{2 x_{1}^{2}-2}=\frac{4 x_{2}-0}{2 x_{2}^{2}-2}$, simplifying gives $x_{1} x_{2}=-1 \quad(... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. Real numbers $a, b, c$ satisfy $5^{a}=2^{b}=\sqrt{10^{c}}$, and $a b \neq 0, \frac{c}{a}+\frac{c}{b}$ has the value of $\qquad$ . | 9. 2 .
Let $5^{a}=2^{b}=\sqrt{10^{c}}=k>0$, then $a \lg 5=b \lg 2=\frac{c}{2}=\lg k$, so $\frac{c}{a}+\frac{c}{b}=2 \lg 5+2 \lg 2=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(5) Let two regular tetrahedra $P-ABC$ and $Q-ABC$ be inscribed in the same sphere. If the dihedral angle between a lateral face and the base of the regular tetrahedron $P-ABC$ is $45^{\circ}$, then the tangent value of the dihedral angle between a lateral face and the base of the regular tetrahedron $Q-ABC$ is $\qquad... | 5 . 4 Hint: As shown in the figure, connect $P Q$, then $P Q \perp$ plane $A B C$, with the foot of the perpendicular $H$ being the center of the equilateral $\triangle A B C$, and $P Q$ passing through the center of the sphere $O$. Connect $C H$ and extend it to intersect $A B$ at point $M$, then $M$ is the midpoint o... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example II Construct $\triangle A P B, \triangle B Q C, \triangle C R A$ outside an equilateral $\triangle A B C$ with unit area, such that $\angle A P B = \angle B Q C = \angle C R A = 60^{\circ}$.
(1) Find the maximum area of $\triangle P Q R$;
(2) Find the maximum area of the triangle whose vertices are the incenter... | Analyzing the problem of the largest area can be transformed into the problem of an equilateral triangle with vertices on a given circle.
Proof (1) Since $\angle A P B=60^{\circ}$.
Therefore, point $P$ lies on the arc symmetrical to $A C B$ about $A B$, and this arc is within a circle that is concentric with the circum... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (6 points) Convert $\frac{13}{999}$ to a decimal, the digit in the 2015th position after the decimal point is
| 【Solution】Solution: $\frac{13}{999}=13 \div 999=0.013013013013013013013013013013013 \cdots$
$$
2015 \div 3=671 \cdots 2
$$
Therefore, the digit at the 2015th position in the decimal part is: 1.
The answer is: 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find the smallest natural number $n$, such that in any two-coloring of $K_{n}$ there are always two monochromatic triangles that share exactly one vertex.
---
The translation maintains the original text's format and line breaks as requested. | When $n=8$, divide 8 points into two groups, each with 4 points, connect red edges between points in the same group, and blue edges between points in different groups. In this coloring of $K_{8}$, there are no two monochromatic triangles that share exactly one vertex.
When $n \geqslant 9$, consider a $K_{9}$ formed by... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
28.2.11 ** Let $M$ be a set of $n$ points in the plane, satisfying:
(1) $M$ contains 7 points which are the 7 vertices of a convex heptagon;
(2) For any 5 points in $M$, if these 5 points are the 5 vertices of a convex pentagon, then this convex pentagon contains at least one point from $M$ inside it.
Find the minimum ... | Parse the proof that $n \geqslant 11$.
Consider a convex heptagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}$ with vertices in $M$. Connect $A_{1} A_{5}$. By condition (1), in the convex pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$, there is at least one point in $M$, denoted as $P_{1}$. Connect $P_{1} A_{1}$ and $P_{1} A_... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 4: Let $\overline{2 x 3}+326=\overline{5 y 9}$, and $9 \mid \overline{5 y 9}$. Find $x+y$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve:
$$
9|\overline{5 y 9} \Rightarrow 9|(5+y+9) \Rightarrow 9 \mid(5+y) \text {. }
$$
But $0 \leqslant y \leqslant 9$, then $y=4$.
$$
\begin{array}{l}
\text { Hence } \overline{2 x 3}+326=549 \Rightarrow \overline{2 x 3}=223 \\
\Rightarrow x=2 \Rightarrow x+y=6 .
\end{array}
$$ | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The teacher asked the students: "How many of you reviewed math yesterday?"
Zhang: "No one."
Li: "One person."
Wang: "Two people."
Zhao: "Three people."
Liu: "Four people."
The teacher knew that some of them reviewed yesterday afternoon, and some did not. Those who reviewed spoke the truth, and those who did not revi... | Analysis: Only 1 of these 5 statements is true, so among these 5 people, 1 person studied math. Therefore, the answer is B. | 1 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
(4) Let $r>1$ be a real number. If a moving point $z$ in the complex plane satisfies $|z|=r$, then the trajectory of the moving point $\omega=z+\frac{1}{z}$ is ( ).
(A) An ellipse with a focal distance of 4
(B) An ellipse with a focal distance of $\frac{4}{r}$
(C) An ellipse with a focal distance of 2
(D) An ellipse wi... | (4) A Hint: Let $z=r(\cos \theta+\mathrm{i} \sin \theta), \theta \in[0,2 \pi)$, then
$$
\begin{aligned}
\omega & =z+\frac{1}{z}=r(\cos \theta+\mathrm{i} \sin \theta)+\frac{1}{r(\cos \theta+\mathrm{i} \sin \theta)} \\
& =\left(r \cos \theta+\frac{1}{r} \cos \theta\right)+\left(r \sin \theta-\frac{1}{r} \sin \theta\right... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. Given the imaginary number $z$ satisfies $z^{3}+1=0$, then $\left(\frac{z}{z-1}\right)^{2018}+\left(\frac{1}{z-1}\right)^{2018}=$ | 荅郪 -1 .
Solving $z^{3}+1=0 \Rightarrow(z+1)\left(z^{2}-z+1\right)=0 \Rightarrow z^{2}-z+1=0$, so
$$
\left(\frac{z}{z-1}\right)^{2018}+\left(\frac{1}{z-1}\right)^{2018}=\frac{z^{2018}+1}{\left(z^{2}\right)^{2018}}=\frac{\left(z^{3}\right)^{672} z^{2}+1}{\left(z^{3}\right)^{1345} z}=\frac{z^{2}+1}{-z}=-1
$$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (6 points) 12 question setters make guesses about the answer to this question, with their guesses being “not less than 1”, “not greater than 2”, “not less than 3”, “not greater than 4”, “not less than 11”, “not greater than 12” (“not less than” followed by an odd number, “not greater than” followed by an even number... | 【Answer】Solution: According to the analysis, since there are only 12 teachers in total, "no more than 12" is correct;
"no more than 2" and "no less than 3" are mutually exclusive, similarly "no more than 4" and "no less than 5",
"no more than 6" and "no less than 7", "no more than 8" and "no less than 9",
"no more than... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Given $a$ is a real number. Then, the number of subsets of the set $M=|x| x^{2}-3 x-a^{2}+2=0, x \in$ $\mathbf{R}$ : is ( ).
A. 1
B. 2
C. 4
J). Uncertain | 1. $\mathrm{C}$
From the discriminant $\Delta=1+4 a^{2}>0$ of the equation $x^{2}-3 x-a^{2}+2=0$, we know that the equation has two distinct real roots. Therefore, $M$ has 2 elements, and the set $M$ has $2^{2}=4$ subsets. | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
12. The figure below shows the number of days required for three individuals, A, B, and C, to complete a certain project individually. According to the information in the figure, if A works for 2 days first, followed by B and C working together for 4 days, and the remaining work is completed by C alone, then the total ... | 12. The figure shows the number of days required for three individuals, A, B, and C, to complete a certain project individually. According to the information in the figure, if A works for 2 days first, then B and C work together for 4 days, and finally, the remaining work is completed by C alone, the total time to comp... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (8 points) Xiao Zhang has 200 pencils, and Xiao Li has 20 fountain pens. Each time Xiao Zhang gives Xiao Li 6 pencils, Xiao Li gives Xiao Zhang 1 fountain pen in return. After $\qquad$ such exchanges, the number of pencils Xiao Zhang has is 11 times the number of fountain pens Xiao Li has. | 【Answer】Solution: According to the problem, we have:
After the first time, Xiao Zhang has 194 pencils left, and Xiao Li has 19 pens left.
After the second time, Xiao Zhang has 188 pencils left, and Xiao Li has 18 pens left.
After the third time, Xiao Zhang has 182 pencils left, and Xiao Li has 17 pens left.
After the f... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3 (China Girls Mathematical Olympiad) Find all positive real numbers $a$ such that there exists a positive integer $n$ and $n$ pairwise disjoint infinite sets $A_{1}, A_{2}, \cdots, A_{n}$ satisfying $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=\mathbf{Z}$, and for any two numbers $b>c$ in each $A_{i}$, we have $b-... | Analysis We look for the most familiar model, and find that classifying integers into odd and even, and then further classifying the even set based on the odd or even nature after dividing by 2, can be extended similarly. Although it requires infinite classification to meet the condition, it provides useful information... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4.2. 12 * Given that $x, y, z$ are positive numbers, and satisfy $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$.
保留源文本的换行和格式，翻译结果如下：
4.2. 12 * Given that $x, y, z$ are positive numbers, and satisfy $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$. | From the arithmetic mean being greater than or equal to the geometric mean, we get
$$
(x+y)(x+z)=y z+x(x+y+z) \geqslant 2 \sqrt{y z \cdot x(x+y+z)}-2,
$$
and when $x=\sqrt{2}-1, y=z=1$, the above inequality holds with equality. Therefore, the minimum value of the expression $(x+y)(x+z)$ is 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Given that the radius $O D$ of $\odot O$ is perpendicular to the chord $A B$, and intersects $A B$ at point $C$. Connect $A O$ and extend it to intersect $\odot O$ at point $E$. If $A B=8, C D=2$, then the area of $\triangle B C E$ is ( ).
(A) 12
(B) 15
(C) 16
(D) 18 | 5. A.
As shown in Figure 4, let $O C=x$.
Then $O A=O D=x+2$.
In the right triangle $\triangle O A C$, by
the Pythagorean theorem, we have
$$
\begin{array}{l}
O C^{2}+A C^{2}=O A^{2} \\
\Rightarrow x^{2}+4^{2}=(x+2)^{2} \\
\Rightarrow x=3 .
\end{array}
$$
Since $O C$ is the midline of $\triangle A B E$, we have
$$
B E... | 12 | Geometry | MCQ | Yes | Yes | olympiads | false |
5. (10 points) Class 1 of Grade 4 used the class fund to purchase three types of stationery, A, B, and C, with unit prices of 3 yuan, 2 yuan, and 1 yuan, respectively.
It is known that the number of B type stationery purchased is 2 less than the number of A type stationery purchased, and the cost of A type stationery d... | 5. (10 points) Class 1 of Grade 4 used the class fund to purchase three types of stationery, A, B, and C, with unit prices of 3 yuan, 2 yuan, and 1 yuan, respectively. It is known that the number of B stationery purchased is 2 less than the number of A stationery purchased, and the cost of purchasing A stationery does ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
28. As shown in the figure, robots $A$ and $B$ start from point A at the same time, moving in opposite directions on a circular path with a diameter of an integer number of meters. They meet 8 times in 18 minutes; if $A$'s speed is increased by 6 meters per minute, then $A$ and $B$ meet 10 times in 18 minutes. The mini... | $12$ | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Let the function $f(x)=\frac{(x+1)^{2}+\ln \left(\sqrt{x^{2}+1}+x\right)}{x^{2}+1}$ have a maximum value of $M$ and a minimum value of $N$. Determine the value of $M+N$. | $$
f(x)=\frac{x^{2}+1+2 x+\ln \left(\sqrt{x^{2}+1}+x\right)}{x^{2}+1}=1+\frac{2 x+\ln \left(\sqrt{x^{2}+1}+x\right)}{x^{2}+1} \text {, }
$$
Notice that $g(x)=\frac{2 x+\ln \left(\sqrt{x^{2}+1}+x\right)}{x^{2}+1}$ is an odd function, so $g(x)_{\text {min }}=-g(x)_{\max }$,
thus $M=1+g(x)_{\max }, N=1-g(x)_{\max } \Righ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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