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1. If real numbers $x, y$ satisfy $(x+5)^{2}+(y-12)^{2}=14^{2}$, then the minimum value of $x^{2}+y^{2}$ is
A. 2
B. 1
C. $\sqrt{3}$
D. $\sqrt{2}$
|
1. B Hint: Combine geometric and algebraic thinking, notice that $(x+5)^{2}+(y-12)^{2}=14^{2}$ is a circle with center $C(-5,12)$ and radius 14. Let $P$ be any point on the circle, then $|O P| \geqslant|C P|-|O C|=14-13=1$.
|
1
|
Geometry
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
Example 1 Let $x_{1}, x_{2}, x_{3}, \cdots$ be a decreasing sequence of positive numbers such that for any natural number $n$,
$$
x_{1}+\frac{x_{4}}{2}+\frac{x_{9}}{3}+\cdots+\frac{x_{n}^{2}}{n} \leqslant 1 \text {. }
$$
Prove that for any natural number $n$, $x_{1}+\frac{x_{2}}{2}+\frac{x_{3}}{3}+\cdots+\frac{x_{n}}{n}<3$.
(13th All-Soviet Union Olympiad Problem)
|
Prove that since $x_{1}, x_{2}, x_{3}, \cdots$ is a decreasing sequence of positive numbers, for any natural number $k$, we have $\frac{x_{k}^{2}}{k^{2}}+\frac{x_{k^{2}+1}}{k^{2}+1}+\cdots+\frac{x_{(k+1)^{2}-1}}{(k+1)^{2}-1}<(2 k+1) \frac{x_{k^{2}}}{k^{2}} \leqslant 3 \frac{x_{k}{ }^{2}}{k}$. For any natural number $n$, there exists a natural number $k$ such that $k^{2} \leqslant n \leqslant(k+1)^{2}-1$, thus $x_{1}+\frac{x_{2}}{2}+\cdots+\frac{x_{n}}{n}<3\left(x_{1}+\frac{x_{4}}{2}+\cdots+\frac{x_{k}{ }^{2}}{k}\right) \leqslant 3$.
|
3
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
20 Find the real solutions of the equation $\left(x^{2008}+1\right)\left(1+x^{2}+x^{4}+\cdots+x^{2006}\right)=2008 x^{2007}$.
|
(20 Divide both sides of the equation by $x^{2007}$, we get
$$
\begin{array}{c}
\left(x+\frac{1}{x^{2007}}\right)\left(1+x^{2}+x^{4}+\cdots+x^{2006}\right)=2008 \\
x+x^{3}+x^{5}+\cdots+x^{2007}+\frac{1}{x^{2007}}+\frac{1}{x^{2005}}+\cdots+\frac{1}{x}=2008, \\
2008=x+x^{3}+x^{5}+\cdots+x^{2007}+\frac{1}{x^{2007}}+\frac{1}{x^{2005}}+\cdots+\frac{1}{x} \\
\geqslant 2 \times 1004 \\
=2008
\end{array}
$$
The equality holds if and only if $x=\frac{1}{x}, x^{3}=\frac{1}{x^{3}}, \cdots, x^{2007}=\frac{1}{x^{2007}}$, which means $x=\pm 1$. However, when $x=-1$, it does not satisfy the original equation. Therefore, $x=1$ is the only real solution to the original equation.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
One. (20 points) Given the function
$$
f(x)=2 \cos x(\cos x+\sqrt{3} \sin x)-1(x \in \mathbf{R}) \text {. }
$$
(1) Find the intervals where the function $f(x)$ is monotonically increasing;
(2) Let points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$ all lie on the graph of the function $y=f(x)$, and satisfy
$$
x_{1}=\frac{\pi}{6}, x_{n+1}-x_{n}=\frac{\pi}{2}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the value of $y_{1}+y_{2}+\cdots+y_{2018}$.
|
(1) From the problem, we have
$$
f(x)=\cos 2 x+\sqrt{3} \sin 2 x=2 \sin \left(2 x+\frac{\pi}{6}\right) \text {. }
$$
Therefore, the monotonic increasing interval of $f(x)$ is
$$
\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right](k \in \mathbf{Z}) \text {. }
$$
(2) Let $t_{n}=2 x_{n}+\frac{\pi}{6}, t_{1}=2 x_{1}+\frac{\pi}{6}=\frac{\pi}{2}$.
Then $t_{n+1}-t_{n}=\pi \Rightarrow t_{n}=\left(n-\frac{1}{2}\right) \pi$.
Thus, $y_{n}=2 \sin \frac{2 n-1}{2} \pi$
$$
=\left\{\begin{array}{ll}
2, & n=2 k-1 ; \\
-2, & n=2 k
\end{array}(k \in \mathbf{Z}) .\right.
$$
Therefore, $y_{1}+y_{2}+\cdots+y_{2018}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 5.7 How many 35-complete partitions with the smallest number of parts are there? What are their part numbers?
|
Solution: $35+1=36=2^{2} \cdot 3^{2}$. By Theorem 5.15, the number of 35-complete partitions with the smallest number of parts is $\frac{(2+2)!}{2!2!}=6$, and their part numbers are
$$
2(2-1)+2(3-1)=6 .
$$
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
$28.816 \div(0.40+0.41+0.42+\cdots+0.59)$ The integer part of the value is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
|
[Solution] The original expression $>16 \div(0.6 \cdot 20)=16 \div 12>1$, and $\quad$ the original expression $<16 \div(0.4 \cdot 20)=16 \div 8=2$. Therefore, the integer part of the original expression is 1. Hence, the answer is $(A)$.
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
16. Anna and Bonnie are racing on a 400-meter oval track. They start at the same time, but Anna is ahead because she is 25% faster than Bonnie. Then Anna will have run ( ) laps when she first overtakes Bonnie.
(A) $\frac{5}{4}$
(B) $\frac{10}{3}$
(C) 4
(D) 5
(E) 25
|
16. D.
Since Anna is 25% faster than Bonnie, when Bonnie finishes the first lap, Anna has run an additional $\frac{1}{4}$ lap. Thus, when Bonnie completes 4 laps, Anna has run an extra full lap. This means that when Anna completes 5 laps, she overtakes Bonnie for the first time.
|
5
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
13. (15 points) There are 60 students in the class, and their birthdays are recorded as a certain month and day. Ask each student the same two questions: How many students in the class have the same birth month as you? How many students in the class have the same birth day as you (for example, the birth day of January 12 and December 12 is the same). It is found that the answers obtained include all integers from 0 to 14. Then, what is the minimum number of students who have the same birthday?
|
【Analysis】The answers for the same month and the same day number cover from 0 to 14, meaning the number of people with the same month and the same day number ranges from 1 to 15, thus leading to the analysis and solution.
【Solution】The answers include all integers from 0 to 14, which means each answer contains a number of students ranging from 1 to 15.
Since $1+2+3+\cdots+15=120=2 \times 60$,
it follows that whether the answer is for the same month or the same day number, the number of people with the same month or the same day number will not repeat (for example, if there are 3 people born in a certain month, there will not be 3 people born on a certain day). Therefore, when counting the number of people with the same month or the same day number, each of the 15 numbers from 1 to 15 appears only once. To minimize the number of people with the same month and day, the distribution of months or day numbers should be as dispersed as possible. For example, 60 can be broken down as: $60=1+2+3+4+5+7+8+9+10+11$ for the number of people in the same month,
and another case: $60=6+12+13+14+15$ for the number of people with the same day number.
Analyzing the largest number 15, if 15 people with the same day number are distributed among the above 10 months, it is known that there will be at least two people with the same month and day.
Therefore: The number of people in the class with the same birthday is at least 2.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (10 points) Among all the factors of the four-digit number Shuangcheng Shuangcheng, 3 are prime numbers, and the other 39 are not prime. Then,
the four-digit number Shuangcheng Shuangcheng has $\qquad$ factors.
|
【Solution】Solution: First, according to the sum of the digits in odd and even positions being equal, it must be a multiple of 11. The total number of factors is $3+39=42$ (individuals), decomposing 42 into the product of 3 numbers $42=2 \times 3 \times 7$.
$\overline{\text { double become become double }}=a \times b^{2} \times c^{6}$.
If it is $11 \times 5^{2} \times 2^{6}=17600$ (not a four-digit number, does not meet the condition). Let's check if the smallest number is $\overline{\text { double become become double }}=11 \times 3^{2} \times 2^{6}=6336$.
Become double double become $=3663=11 \times 37 \times 3^{2}$. The number of factors is $2 \times 2 \times 3=12$ (individuals).
Therefore, the answer is: 12 individuals.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Question 140, Find the smallest real number $\mathrm{r}$, such that there exists a sequence of positive real numbers $\left\{\mathrm{x}_{\mathrm{n}}\right\}$, satisfying for any $\mathrm{n} \in \mathrm{N}^{+}$, we have $\sum_{\mathrm{i}=1}^{\mathrm{n}+1} \mathrm{x}_{\mathrm{i}} \leq \mathrm{r} \cdot \mathrm{x}_{\mathrm{n}}$.
|
Question 140, Solution: Clearly, $r \in R^{+}$.
Let $S_{n}=\sum_{i=1}^{n} x_{1}$, then $\left\{S_{n}\right\}$ is a strictly increasing sequence, and $S_{n+1} \leq r \cdot\left(S_{n}-S_{n-1}\right)$. Therefore, we know that $r \cdot S_{n} \geq S_{n+1}+r \cdot S_{n-1} \geq 2 \sqrt{S_{n+1} \cdot r \cdot S_{n-1}}$, rearranging gives $S_{n+1} \leq \frac{r}{S_{n}} \leq \frac{S_{n}}{S_{n-1}}$, thus $\frac{S_{n+1}}{S_{n}} \leq\left(\frac{r}{4}\right)^{n-1} \cdot \frac{S_{2}}{S_{1}}$. If $r<4$, then as $n \rightarrow+\infty$, $\left(\frac{r}{4}\right)^{n-1} \cdot \frac{s_{2}}{s_{1}} \rightarrow 0$, so for sufficiently large $n$, there must be $\frac{s_{n+1}}{s_{n}}<1$, which contradicts the fact that $\left\{S_{n}\right\}$ is a strictly increasing sequence. Hence, $r \geq 4$.
On the other hand, when $r=4$, taking $x_{1}=2^{i-1}$, we have $\sum_{i=1}^{n+1} x_{1}=4 x_{n}$.
In conclusion, the minimum value of $r$ that satisfies the condition is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (6 points) $A, B, C, D$ four people stay in four rooms numbered $1, 2, 3, 4$, with one person per room; so that $B$ does not stay in room 2, and $B, C$ two people require to stay in adjacent numbered rooms. The number of ways to arrange this is. $\qquad$
|
【Solution】Solution: According to the problem, $B$ does not stay in room 2, so $B$ can stay in rooms $1, 3, 4$. If $B$ stays in room 1, then $C$ can stay in room 2, and the remaining 2 people can be arranged in the other 2 rooms. In this case, there are $A_{2}^{2}=2$ situations.
If $B$ stays in room 3, then $C$ can stay in rooms 2 or 4, which gives 2 situations, and the remaining 2 people can be arranged in the other 2 rooms,
in this case, there are $2 \times A_{2}^{2}=4$ situations.
If $B$ stays in room 4, then $C$ can stay in room 3, and the remaining 2 people can be arranged in the other 2 rooms. In this case, there are $A_{2}^{2}=2$ situations,
There are a total of $2+4+2=8$ situations,
Therefore, the answer is 8.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (19th "Hope Cup" Senior High School Competition Question) The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n+1}=-\frac{1}{a_{n}+1}$, then $a_{2008}$ equals
A. $-\frac{4}{3}$
B. $-\frac{1}{4}$
C. 3
D. -3
|
4. $\mathrm{C} \quad a_{1}=3, a_{2}=-\frac{1}{a_{1}+1}=-\frac{1}{4}, a_{3}=-\frac{1}{a_{2}+1}=-\frac{4}{3}, a_{4}=-\frac{1}{a_{3}+1}=3, \cdots$
It can be seen that the sequence $\left\{a_{n}\right\}$ has a period of 3. Therefore, from
$$
2008=669 \times 3+1
$$
we get $a_{2008}=a_{1}=3$, hence the answer is $\mathrm{C}$.
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
11. 6 teams participate in a round-robin tournament (each team plays one match against each of the other teams), with 3 points awarded for a win, 1 point for a draw, and 0 points for a loss. In the end, the 6 teams have different points, the first and second place teams differ by 4 points, the fourth and fifth place teams also differ by 4 points, and the third place team won 2 matches, while the sixth place team drew 2 matches. Therefore, the points of the third place team are $\qquad$ points.
|
$8$
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. (12 points) 2014 students line up in a row, and report numbers according to the following rules: If a student reports a single-digit number, the next student should report the sum of this number and 8; If a student reports a two-digit number, the next student should report the sum of the unit digit of this number and 7. Now, let the first student report 1, then the number reported by the last student is $\qquad$
|
14. (12 points) 2014 students line up from front to back, and report numbers according to the following rules: If a student reports a single-digit number, the next student should report the sum of this number and 8; if a student reports a two-digit number, the next student should report the sum of the unit digit of this number and 7. Now, the first student reports 1, so the last student reports the number $\qquad$.
【Solution】Solution: Since the cycle starts from the second student, each cycle involves 11 students, reporting numbers as $9, 17, 14, 11, 8, 16, 13, 10, 7, 15, 12$,
So $(2014-1) \div 11=183$,
the corresponding number in one cycle is the 11th number, which is 12, hence the answer is 12.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (26th Russian Mathematical Olympiad) Let $a$, $b$, $c$ be three distinct real numbers such that the equations $x^{2} + a x + 1 = 0$ and $x^{2} + b x + c = 0$ have a common real root, and the equations $x^{2} + x + a = 0$ and $x^{2} + c x + b = 0$ also have a common real root. Find $a + b + c$.
|
5. If $x_{1}^{2}+a x_{1}+1=0$ and $x_{1}^{2}+b x_{1}+c=0$, then $(a-b) x_{1}+(1-c)=0$. Therefore, $x_{1}=\frac{c-1}{a-b}$. Similarly, from the equations $x_{2}^{2}+x_{2}+a=0$ and $x_{2}^{2}+c x_{2}+b=0$, we derive $x_{2}=\frac{a-b}{c-1}$ (obviously $c \neq 1$). Hence, $x_{2}=\frac{1}{x_{1}}$.
On the other hand, by Vieta's formulas, $\frac{1}{x_{1}}$ is a root of the first equation. This indicates that $x_{2}$ is a common root of the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$. Therefore, $(a-1)\left(x_{2}-1\right)=0$. However, when $a=1$, these two equations have no real roots, so it must be that $x_{2}=1$, and thus $x_{1}=1$. Consequently, $a=-2, b+c=-1$, so $a+b+c=-3$.
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3 Given $(\sqrt{3}+1) x=\sqrt{3}-1$. Then $x^{4}-5 x^{3}+6 x^{2}-5 x+4=$ $\qquad$
(2013, Huanggang City, Hubei Province Mathematics Competition)
|
Solve: From $x=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$, we get $x-2=\sqrt{3}$. Squaring it, we obtain
$$
x^{2}-4 x+1=0 \text {. }
$$
Use equation (1) to reduce the power of the target algebraic expression. Performing polynomial division, we get
$$
\begin{array}{l}
x^{4}-5 x^{3}+6 x^{2}-5 x+4 \\
=\left(x^{2}-4 x+1\right)\left(x^{2}-x+1\right)+3=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (10 points) $\left[x-\frac{1}{2}\right]=3 x-5$, here $[x]$ represents the greatest integer not exceeding $x$, then $x=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
【Analysis】According to the problem, for the original equation to hold, then $\left[x-\frac{1}{2}\right] \leqslant x-\frac{1}{2}, \Rightarrow 3 x-5 \leqslant x-\frac{1}{2}$, and $3 x-5$ is an integer, it is not difficult to find that $x=2$.
【Solution】Solution: According to the analysis, for the original equation to hold, then $\left[x-\frac{1}{2}\right] \leqslant x-\frac{1}{2}, \Rightarrow 3 x-5 \leqslant x-\frac{1}{2}, \Rightarrow x \leqslant \frac{9}{4}$, $\because 3 x-5 \geqslant 0 \therefore x=2$
Since $3 x-5$ is an integer, it is not difficult to find that $x=2$.
Therefore, the answer is: 2
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
16. (3 points) $A, B, C, D$ four boxes contain $8, 6, 3, 1$ balls respectively. The first child finds the box with the fewest balls, then takes one ball from each of the other boxes and puts it into this box; the second child also finds the box with the fewest balls, and then also takes one ball from each of the other boxes and puts it into this box, ..., after the 50th child has finished, the number of balls in box $A$ is $\qquad$ .
|
A $\cdots)$, the 6th child repeats with the 2nd, meaning a cycle of 4 groups; thus, by analogy: $(50-1) \div 4=12 \cdots 1$ (times);
That is: Excluding the first irregular group, there should be 49 repeated groups, with one remaining, so after the 50th child takes, the four boxes $A B C D$ should contain: $6,4,5,3$ balls respectively.
【Solution】Solution: As analyzed: The 6th child repeats with the 2nd, meaning a cycle of 4 groups; thus, by analogy: $(50-1) \div 4=12 \cdots 1$ (times);
After the 50th child takes, the four boxes $A B C D$ should contain: $6,4,5,3$ balls respectively;
Answer: After 50 children have placed their balls, the number of balls in box $A$ is 6;
Therefore, the answer is: 6.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a>2, a_{2017}=$ 2017, and for any positive integer $n, a_{n+1}=a_{n}^{2}-2$. Then $\left[\frac{\sqrt{a-2}}{10^{6}} a_{1} a_{2} \cdots a_{2017}\right]=$ $\qquad$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
|
5.2.
Let $a_{1}=a=2+\varepsilon(\varepsilon>0)$.
Then $a_{2}=a_{1}^{2}-2=(2+\varepsilon)^{2}-2$
$$
\begin{array}{l}
=2+4 \varepsilon+\varepsilon^{2}>2+4 \varepsilon, \\
a_{3}=a_{2}^{2}-2>(2+4 \varepsilon)^{2}-2 \\
=2+16 \varepsilon+16 \varepsilon^{2}>2+16 \varepsilon .
\end{array}
$$
Using mathematical induction, we can prove: for any positive integer $n$,
$$
a_{n} \geqslant 2+4^{n-1} \varepsilon \text {. }
$$
Thus $2017=a_{2017}>2+4^{2016} \varepsilon$
$$
\Rightarrow 02$, then
$$
\begin{array}{l}
a_{1} a_{2} \cdots a_{2017}=2017 \sqrt{\frac{2017^{2}-4}{a^{2}-4}} . \\
\text { Hence }\left[\frac{\sqrt{a-2}}{10^{6}} a_{1} a_{2} \cdots a_{2017}\right] \\
=\left[\frac{2017}{10^{6}} \sqrt{\frac{2017^{2}-4}{a+2}}\right] .
\end{array}
$$
Substituting $a=2+\varepsilon$ and using equation (2), we get
$$
\left[\frac{\sqrt{a-2}}{10^{6}} a_{1} a_{2} \cdots a_{2017}\right]=2
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Given that $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are two rhombuses with side lengths of $\sqrt{3}+1$. If $A C \perp A^{\prime} C^{\prime}$, $\angle A B C=\angle A^{\prime} B^{\prime} C^{\prime}=120^{\circ}$, then the perimeter of the shaded part is $\qquad$
|
$8$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
$35 \cdot 22$ A supermarket has 128 boxes of apples. Each box contains between 120 and 144 apples. Boxes with the same number of apples are called a group. What is the minimum number $n$ of boxes in the largest group?
(A) 4 .
(B) 5 .
(C) 6 .
(D) 24 .
(E) 25 .
(27th American High School Mathematics Examination, 1976)
|
[Solution] The number of apples that can be packed in 1 box is $120, 121, \cdots, 144$, which gives 25 different ways. If each way is used to pack 5 boxes, then at most $25 \cdot 5 = 125$ boxes.
Since there are 128 boxes now, we have $n \geqslant 6$.
Because there is the following possibility: boxes containing $k$ apples $(k=120,121,122)$ each have 6 boxes, and boxes containing $l$ apples $(123 \leqslant l \leqslant 144)$ each have 5 boxes. This way, $n \leqslant 6$.
Therefore, $n=6$.
So the answer is $(C)$.
Note: If the total number of boxes is 126, it is a problem from the 2nd Tianjin "Chinese Youth Forest" Mathematics Invitational Competition (1987).
And 126 is the minimum number of boxes to ensure the conclusion of the proposition holds.
|
6
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
5. Let $S$ be a set composed of positive integers. If for any three distinct $x, y, z \in S$, at least one of $x, y, z$ is a divisor of $x+y+z$, then $S$ is called "beautiful". Prove that there exists a positive integer $N$ such that the following relationship holds: for any beautiful set $S$, there exists a positive integer $n_{S} \geq 2$, such that the number of elements in $S$ that are not divisible by $n_{S}$ is at most $N$. Determine the minimum value of $N$.
|
5. The minimum value of $N$ is 6.
If a beautiful set $S$ satisfies that any two different elements are coprime, then this set is called "extremely beautiful."
First, prove that $N \leqslant 5$ does not meet the requirements.
Let $a_{1}$ and $a_{2}$ be two odd numbers not less than 3, and $a_{1}$ and $a_{2}$ are coprime. By the Chinese Remainder Theorem, there exists an odd number $a_{3}$ not less than 3, satisfying:
\[
\left\{\begin{array}{l}
a_{3} = a_{2} \pmod{a_{1}}, \\
a_{3} = -a_{1} \pmod{a_{2}}.
\end{array}\right.
\]
Since $a_{1}$ and $a_{2}$ are coprime, $a_{3}$ is coprime with $a_{1}$ and $a_{2}$.
Similarly, there exist odd numbers $a_{4}$ and $a_{5}$ not less than 3, satisfying:
\[
\left\{\begin{array}{l}
a_{4} = -a_{2} \pmod{a_{1}}, \\
a_{4} \equiv -a_{1} \pmod{a_{2}}, \\
a_{4} = -a_{2} \pmod{a_{3}};
\end{array}\right.
\]
\[
\left\{\begin{array}{l}
a_{5} = -a_{2} \pmod{a_{1}}, \\
a_{5} \equiv a_{1} \pmod{a_{2}}, \\
a_{5} = a_{2} \pmod{a_{3}}, \\
a_{5} = -a_{1} \pmod{a_{4}}.
\end{array}\right.
\]
Clearly, any two numbers in $a_{1}, a_{2}, \cdots, a_{5}$ are coprime. Since they are all not less than 3, they are also all different.
If we take $S = \{1, 2, a_{1}, a_{2}, \cdots, a_{5}\}$, then it is clear that $S$ is a beautiful set. Clearly, no matter how $n_{s}$ is chosen, at most one number in $s$ is a multiple of $n_{s}$, thus, $N \leqslant 5$, which does not meet the requirements.
Next, prove that $N=6$ meets the requirements.
Lemma 1 If positive integers $x, y, z$ satisfy
$x < z, y < z$ and $z \mid (x + y + z)$,
then $x + y = z$.
Proof of Lemma 1: Note that,
$z \mid (x + y)$, and $0 < \frac{x + y}{z} < 2$.
Thus, $\frac{x + y}{z} = 1 \Rightarrow x + y = z$.
Lemma 2 If $x, y, z$ are positive odd numbers, and $x < z, y < z$,
then $z \mid (x + y + z)$ does not hold.
Proof of Lemma 2: If $z \mid (x + y + z)$, by Lemma 1, we know $z = x + y$, which contradicts the fact that $x, y, z$ are all odd numbers.
Lemma 3 Given that $S$ is an extremely beautiful set, and $S$ does not contain 1 and even numbers, if $x, y \in S$ and $x < y$, then there is at most one $z \in S$ such that $z < x$ and $z$ is not a factor of $x + y$.
Proof of Lemma 3: Let $z$ be an odd element in set $S$, and $z < x$, then $z < x < y$.
By Lemma 2, we know $y$ is not a factor of $x + y + z$.
Since $x, y, z$ are different elements in $S$, then
$x \mid (x + y + z)$ or $z \mid (x + y + z)$.
Assume there are two or more elements $z \in S$, and not a factor of $x + y$. Then $z$ is also not a factor of $x + y + z$.
Thus, $x \mid (x + y + z)$.
Take two $z$ that satisfy the assumption and denote them as $z_{1}, z_{2}$, and $z_{1} < z_{2}$,
then $z_{2} - z_{1} = (z_{2} + y) - (z_{1} + y)$ must also be a positive integer multiple of $x$. But since $1 \leq z_{2} - z_{1} < z_{2}$, therefore, $z_{2} - z_{1} < x$,
which is a contradiction.
Lemmas 1 to 3 are proved.
Next, prove that an extremely beautiful set has at most seven elements.
Since this set contains at most one even number,
it is sufficient to prove that an extremely beautiful set that does not contain 1 and even numbers has at most five elements.
Assume such a set has six elements, and we will prove the contradiction.
Let these six elements be $x_{1}, x_{2}$, in ascending order,
If for some $k (k=1,2,3), x_{k}$ divides $y_{4}$,
$y_{5}, y_{6}$, then $x_{k}$ divides $2 x_{4} = y_{5} + y_{6} - y_{4}$, which contradicts the fact that $x_{k}, x_{4}$
are both odd numbers not less than 3 and $x_{k}, x_{4}$ are coprime:
By Lemma 3, we know that for $l=4,5,6$, there are more than two
$k (k=1,2,3)$ such that $x_{k} \mid y_{k}$. Then there are exactly six pairs
$(k, l)$ such that $x_{k} \mid y_{l}$, and each $y_{l}$ is divisible by exactly 2
$x_{k}$.
By Lemma 3, we know $y_{4}$ is divisible by at least three of $x_{1}, x_{2}, x_{3}, x_{4}$, thus, $y_{4}$ is divisible by $x_{4}$.
If $x_{i}$ is not a factor of $y_{5}$, $x_{j}$ is not a factor of $y_{6} y_{6}$
$(i, j=1,2,3)$, then by Lemma 2 we get
$x_{4} \mid (x_{6} + x_{i}), x_{4} \mid (x_{5} + x_{j})$.
Thus, the number of elements in the set that satisfy the previous discussion is at most 5. Therefore, an extremely beautiful set has at most seven elements.
Let $S$ be an extremely beautiful set, $n_{s}=2$. Then there are at most six elements that are not multiples of $n_{s}$. If $S$ is not extremely beautiful, then for a positive integer $n$, if $|S| \geqslant n$, let $S_{n}$ denote the set of the smallest $n$ elements in $S$. Take the smallest $n$ such that $S_{n}$ is not an extremely beautiful set, then $|S_{n}| \leqslant 8$.
Thus, there must exist a prime $p$, such that $p$ is a factor of two elements in $S_{n}$. Without loss of generality, let these two numbers be $a, b$. If $m \in S$ but $m \notin S_{n}$. At this time, $a \mid (a + b + m)$ or $b \mid (a + b + m)$.
Thus, $p \mid (a + b + m)$.
Therefore, $p \mid m$.
Similarly, if $m \mid (a + b + m)$, by Lemma 1, we get
$m = a + b$, then $p \nmid m$. Thus, taking $n_{s} = p$, we get that the elements in $S$ that are not multiples of $n_{s}$ are the elements in $S_{n}$ different from $a$ and $b$, and there are at most six elements.
Therefore, $N=6$ meets the requirements and is also the minimum.
|
6
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
A heptagon with all interior angles less than 180 degrees has at least ( ) obtuse angles.
(A) 6
(B) 5
(C) 4
(D) 3
|
Solution: The following solution tries to use elementary school level thinking as much as possible:
The sum of the interior angles of a heptagon is $180 \times(7-2)=900^{\circ}$
Try to have as many angles as possible be $90^{\circ}$,
1) If 6 angles are $90^{\circ}$, then the last 1 angle is greater than $180^{\circ}$
2) If 5 angles are $90^{\circ}$, then the sum of the other 2 angles is $900-5 \times 90=450^{\circ}$, $450^{\circ}>2 \times 180^{\circ}$, there must be an angle greater than $180^{\circ}$
3) If 4 angles are $90^{\circ}$, then the sum of the other 3 angles is $900-4 \times 90=540^{\circ}$, $540^{\circ}=3 \times 180^{\circ}$, there must be an angle greater than or equal to $180^{\circ}$
4) If 3 angles are $90^{\circ}$, then the sum of the other 4 angles is $900-3 \times 90=630^{\circ}$, $640<4 \times 180^{\circ}$, so at least 4 angles are obtuse
|
4
|
Geometry
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
4. Let $f(x)=a x+b$ (where $a, b$ are real numbers), $f_{1}(x)=f(x), f_{n+1}(x)=f\left(f_{n}(x)\right), n=1$, $2,3, \cdots$, If $2 a+b=-2$, and $f_{k}(x)=-243 x+244$, then $k=$
|
4. 5 Detailed Explanation: We can first find $f_{1}(x), f_{2}(x), f_{3}(x)$, and by induction we get $f_{k}(x)=a^{k} x+\frac{1-a^{k}}{1-a} \cdot b$. Substituting the known values, we can solve for the result.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The area of the figure enclosed by the curve determined by the equation $|x-1|+|y-1|=1$ is
(A) 1
(B) 2
(C) $\pi$
( D ) 4
|
(B)
4.【Analysis and Solution】For expressions containing absolute values, discuss in cases to remove the absolute value symbol, obtaining four line segments.
They determine a curve that is a square with vertices at $(1,0),(2,1),(1,2),(0,1)$, with a side length of $\sqrt{2}$, thus the area is 2.
|
2
|
Geometry
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
7. The right figure is composed of 4 regular hexagons, each with an area of 6. Using the vertices of these 4 hexagons as vertices, the number of equilateral triangles that can be formed with an area of 4 is $\qquad$ .
|
8,
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Find the smallest natural number $n$, such that in any two-coloring of $K_{n}$ there exist 5 edge-disjoint monochromatic triangles.
|
Solution: $n_{\min }=11$.
(1) When $n \leqslant 10$. We can take a part or all of the following graph, where there do not exist 5 monochromatic triangles without common edges (solid lines represent red, no lines represent blue).
(2) When $n=11$, establish the following three lemmas:
Lemma 1: In a two-coloring of $K_{6}$, there must be two monochromatic triangles.
Lemma 2: In a two-coloring of $K_{7}$, there must be two monochromatic triangles without common edges.
Lemma 3: In a two-coloring of $K_{11}$, there must be two monochromatic triangles with exactly one common vertex.
Lemma 1 is a well-known structure. For the proof of Lemma 2, by Lemma 1, there are two monochromatic triangles. If they have no common edges, the lemma is proved. If they do, assume without loss of generality that $\triangle A_{1} A_{2} A_{3}$ and $\triangle A_{1} A_{2} A_{4}$ are both red triangles. Among them, there must be one different from $\triangle A_{2} A_{3} A_{4}$, which must be without common edges with $\triangle A_{1} A_{2} A_{3}$ or $\triangle A_{1} A_{2} A_{4}$ (drawing a diagram, it is somewhat complex).
Proof of Lemma 3: By Lemma 2, assume there exist monochromatic triangles $\triangle A_{1} A_{2} A_{3}$, $\triangle A_{4} A_{5} A_{6}$, $\triangle A_{7} A_{8} A_{9}$ (using proof by contradiction). Take 7 points $A_{1}, A_{2}, A_{3}, A_{4}, A_{7}, A_{10}, A_{11}$, among which there will be two monochromatic triangles without common edges and no common vertices. Among them, there must be one with a common vertex with $\triangle A_{1} A_{2} A_{3}$, $\triangle A_{4} A_{5} A_{6}$, or $\triangle A_{7} A_{8} A_{9}$, leading to a contradiction!
Next, we prove that $n=11$ satisfies the condition, using proof by contradiction, assume it does not hold.
First, by Lemma 3, we can assume $\triangle A_{1} A_{2} A_{3}$ and $\triangle A_{1} A_{4} A_{5}$ are monochromatic triangles.
Consider the 6 points $A_{6}, A_{7}, A_{8}, A_{9}, A_{10}, A_{11}$. If the edges between them are all the same color, then $\triangle A_{6} A_{7} A_{8}$, $\triangle A_{6} A_{10} A_{11}$, $\triangle A_{8} A_{9} A_{10}$, $\triangle A_{1} A_{2} A_{3}$, and $\triangle A_{1} A_{4} A_{5}$ are 5 monochromatic triangles without common edges, leading to a contradiction!
If there is a monochromatic triangle $T$ that has no common edges with $\triangle A_{1} A_{2} A_{3}$ and $\triangle A_{1} A_{4} A_{5}$, and shares a vertex with at least one of them, assume without loss of generality that $A_{2} \notin T$, $A_{4} \notin T$, $A_{6} \in T$. Consider the 7 points $A_{2}, A_{4}, A_{6}, A_{8}, A_{9}, A_{10}, A_{11}$, among which there are two monochromatic triangles without common edges (assume without loss of generality that $A_{8}, A_{9}, A_{10}, A_{11} \notin T$), leading to a contradiction!
By (1), assume the edges between $A_{6}, A_{7}, A_{8}, A_{9}$ are not all the same color. Among $A_{2}, A_{4}, A_{6}, A_{7}, A_{8}, A_{9}$, there are two monochromatic triangles, and these two triangles do not contain $A_{2}, A_{4}$. Assume without loss of generality that they are $\triangle A_{6} A_{8} A_{9}$ and $\triangle A_{7} A_{8} A_{9}$, and are red, then $A_{6} A_{7}$ is blue. Consider the 6 points $A_{2}, A_{4}, A_{6}, A_{7}, A_{k}, A_{i} (k=8,9; i=10,11)$. Similarly, we get: $A_{i} A_{k}, A_{i} A_{6}, A_{i} A_{7}$ are red.
Similarly, we get $A_{10} A_{11}$ is red. This way, between $A_{6}, A_{7}, A_{8}, A_{8}, A_{10}, A_{11}$, only $A_{6} A_{7}$ is blue. Consider $\triangle A_{1} A_{2} A_{3}$, $\triangle A_{1} A_{4} A_{5}$, $\triangle A_{7} A_{10} A_{11}$, $\triangle A_{8} A_{6} A_{11}$, $\triangle A_{8} A_{9} A_{10}$, a total of 5 monochromatic triangles, leading to a contradiction!
Thus, the assumption is false, $n=11$ satisfies the condition.
In conclusion, $n_{\min }=11$.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
19. Given that the three sides of $\triangle A B C$ are all integers, $\cos A=\frac{11}{16}, \cos B=\frac{7}{8}, \cos C=-\frac{1}{4}$. Then the minimum possible value of the perimeter of $\triangle A B C$ is ( ).
(A) 9
(B) 12
(C) 23
(D) 27
(E) 44
|
19. A.
From the problem, we know
$$
\begin{array}{l}
\cos A=\frac{11}{16}, \sin A=\frac{3 \sqrt{15}}{16}, \\
\cos B=\frac{7}{8}, \sin B=\frac{\sqrt{15}}{8}, \\
\cos C=-\frac{1}{4}, \sin C=\frac{\sqrt{15}}{4} .
\end{array}
$$
By the Law of Sines, we get
$$
a: b: c=\sin A: \sin B: \sin C=3: 2: 4 \text {. }
$$
Therefore, the minimum possible perimeter of $\triangle A B C$ is
$$
3+2+4=9 \text {. }
$$
|
9
|
Geometry
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
Example 3 Given $f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x$.
(1) Solve the equation $f(x)=0$;
(2) Find the number of subsets of the set $M=\left\{n \mid f\left(n^{2}-214 n-1998\right)\right.$ $\geqslant 0, n \in \mathbf{Z}\}$.
|
$$
\begin{array}{l}
\text { (1) For any } 0<x_{1}<x_{2}, \text { we have } \frac{x_{1}+1}{x_{2}+1}>\frac{x_{1}}{x_{2}}, \therefore \lg \frac{x_{1}+1}{x_{2}+1}>\lg \frac{x_{1}}{x_{2}} . \\
\therefore f\left(x_{1}\right)-f\left(x_{2}\right)>\lg \frac{x_{1}+1}{x_{2}+1}-\log _{9} \frac{x_{1}}{x_{2}} \\
\quad=\lg \frac{x_{1}}{x_{2}}-\frac{\lg \frac{x_{1}}{x_{2}}}{\lg 9} \\
\because 0<\frac{x_{1}}{x_{2}}<1, \therefore \lg \frac{x_{1}}{x_{2}}<0, \therefore \lg \frac{x_{1}}{x_{2}}-\frac{\lg \frac{x_{1}}{x_{2}}}{\lg 9} \\
\quad>\lg \frac{x_{1}}{x_{2}}-\lg \frac{x_{1}}{x_{2}}=0 \\
\therefore f(x) \text { is a decreasing function on }(0,+\infty) \text {. }
\end{array}
$$
$\therefore f(x)$ is a decreasing function on $(0,+\infty)$.
Notice that $f(9)=0, \therefore$ when $x>9$, $f(x)f(9)=0$,
$\therefore f(x)=0$ has and only has one root $x=9$.
（2）From $f\left(n^{2}-214 n-1998\right) \geqslant 0 \Rightarrow f\left(n^{2}-\right.$
$$
\begin{array}{l}
214 n-1998) \geqslant f(9), \\
\therefore\left\{\begin{array}{l}
n^{2}-214 n-1998 \leqslant 9 \\
n^{2}-214 n-1998>0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
n^{2}-214 n-2007 \leqslant 0 \\
n^{2}-214 n-1998>0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
(n-223)(n+9) \leqslant 0 \\
(n-107)^{2}>1998+107^{2}=13447>115^{2}
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
-9 \leqslant n \leqslant 223 \\
n>222 \text { or } n<-8
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
-9 \leqslant n \leqslant 223 \\
n \geqslant 223 \text { or } n \leqslant-9
\end{array}\right.
\end{array}
$$
$$
\therefore n=223 \text { or } n=-9, \therefore M=\{-9,223\} \text {, }
$$
The number of subsets of $M$ is 4.
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4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
35. Given $a=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $a^{4}-5 a^{3}+10 a^{2}-11 a+4$ is
|
Answer: 1
Solution: Given $a=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{3-\sqrt{5}}{2}$, we get $\quad 3-2 a=\sqrt{5}$,
Therefore,
$$
(2 a-3)^{2}=5 \text {, }
$$
which means
$$
4 a^{2}-12 a+4=0 \text {, }
$$
or equivalently
$$
a^{2}-3 a+1=0,
$$
Thus, $a^{4}-5 a^{3}+10 a^{2}-11 a+4=a^{2}\left(a^{2}-3 a+1\right)-2 a\left(a^{2}-3 a+1\right)+3\left(a^{2}-3 a+1\right)+1=1$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Teacher Wang has 45 candies, and he decides to eat some every day. Since the candies are very delicious, starting from the second day, the number of candies he eats each day is 3 more than the previous day, and he finishes all the candies in 5 days. So, how many candies did Teacher Wang eat on the second day? ( )
|
【Analysis】Calculation, Arithmetic Sequence
Since the number of candies eaten each day is 3 more than the previous day, it forms an arithmetic sequence with a common difference of 3, and there are 5 terms. Directly find the middle term, the third day $45 \div 5=9$, so the second day is $9-3=6$ (candies).
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, with the sum of its first $n$ terms being $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-b\right|<\frac{1}{125}$ is what?
|
2. $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right), a_{1}-1=8$, it can be known that the sequence $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore, $\left|S_{n}-n-6\right|=6 \times\left(\frac{1}{3}\right)^{n}250>3^{5}$, from which we can deduce that $n-1>5, n>6$, hence $n \geqslant 7$, so the smallest integer $n=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. Place the numbers $1,2,3,4,5$ in a circle. We call a placement method a "central ring" placement if for some number $n$ in the range 1 to 15, it is impossible to select several consecutive numbers on the circle such that their sum is $n$. If two placement methods are the same after rotation or reflection, they are considered the same method. How many "central ring" placement methods are there in total?
(Explanation: If placed as shown in the figure below, then $1-5$ can be taken as a single number, $6=5+1,7=3+4,8=5+1+2$, $9=2+3+4,10=1+2+3+4,11=5+1+2+3,12=4+5+1+2,13=3+4+5+1,14=2+3+4+5,15=1+2+3+4+5$, so all numbers from 1 to 15 are obtained, and this is not a "central ring" placement method.)
|
【Analysis】 $1,2,3,4,5$ and $15,14,13,12,11,10$ can definitely be taken; if 6 can be taken, then $15-6=9$, which cannot be taken; if 7 can be taken, then $15-7=8$, which cannot be taken; any of the above conditions being met is a valid placement method.
$$
\begin{array}{l}
6=1+5=2+4=1+2+3 ; \\
7=2+5=3+4=1+2+4 ;
\end{array}
$$
Without 6, (1,5 separated, 2,4 separated, and 1,2,3 not together), 1 way; without 7, (2,5 separated, 3,4 separated, and 1,2,4 not together), 1 way; in total, two ways.
As shown in the figure below
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
$29 \cdot 33$ How many integers $n$ between 1 and 100 allow $x^{2}+x-n$ to be factored into the product of two linear factors with integer coefficients
(A) 0 .
(B) 1 .
(C) 2 .
(D) 9 .
(E) 10 .
(40th American High School Mathematics Examination, 1989)
|
[Solution] Let $x^{2}+x+n=(x-a)(x-b)$,
then $a+b=-1$, and $ab=-n$.
That is, the absolute values of $a$ and $b$ differ by 1, and the absolute value of the negative integer is larger. Also, $ab=-n$, with $n$ ranging from 1 to 100. Thus, we have the table:
\begin{tabular}{r|rrrrrrrrr}
\hline$a$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline$b$ & -2 & -3 & -4 & -5 & -6 & -7 & -8 & -9 & -10 \\
\hline$n$ & 2 & 6 & 12 & 20 & 30 & 42 & 56 & 72 & 90 \\
\hline
\end{tabular}
Thus, there are 9 values of $n$ that satisfy the conditions.
Therefore, the answer is $(D)$.
|
9
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
(1) Given the function $f(x)=x^{3}+a x^{2}+x+1(a \in \mathbf{R})$ is decreasing in the interval $\left(-\frac{2}{3},-\frac{1}{3}\right)$ and increasing in the interval $\left(-\frac{1}{3},+\infty\right)$, then $a=$
|
(2) Hint: From the given, we know $f^{\prime}(x)=3 x^{2}+2 a x+1$, and $x=-\frac{1}{3}$ is an extremum point of the function $f(x)$, i.e.,
$$
f^{\prime}\left(-\frac{1}{3}\right)=-\frac{2}{3} a+\frac{4}{3}=0,
$$
Solving for $a$ yields $a=2$.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given 3 feet $=1$ yard. If there is a rectangular field that is 12 feet long and 9 feet wide, then it requires ( ) square yards of grass to cover it completely.
(A) 12
(B) 36
(C) 108
(D) 324
(E) 972
|
1. A.
$$
\frac{12}{3} \times \frac{9}{3}=12 \text{.}
$$
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
7. In the Cartesian coordinate system, if the circle with center $(r+1,0)$ and radius $r$ has a point $(a, b)$ satisfying $b^{2} \geq 4 a$, then the minimum value of $r$ is $\qquad$ .
|
Answer: 4.
Solution: From the condition, we have $(a-r-1)^{2}+b^{2}=r^{2}$, hence
$$
4 a \leq b^{2}=r^{2}-(a-r-1)^{2}=2 r(a-1)-(a-1)^{2}.
$$
That is, $a^{2}-2(r-1) a+2 r+1 \leq 0$.
The above quadratic inequality in $a$ has a solution, so the discriminant
$$
(2(r-1))^{2}-4(2 r+1)=4 r(r-4) \geq 0 \text{, }
$$
Solving this, we get $r \geq 4$.
Upon verification, when $r=4$, $(a, b)=(3,2 \sqrt{3})$ satisfies the condition. Therefore, the minimum value of $r$ is 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (2002 Hunan Province Competition Question) Given $a_{1}=1, a_{2}=3, a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n}$. If for $m \geqslant n$, the value of $a_{m}$ can always be divided by 9, find the minimum value of $n$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
8. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1)\left(a_{n}-\right.$ $\left.a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots 4 \cdot 3 \cdot\left(a_{2}-a_{1}\right)=(n+2)!$,
it follows that $a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{n}-a_{n-1}\right)=1+2!+3!+\cdots+n!(n \geqslant 1)$.
Given $a_{1}=1, a_{2}=3, a_{3}=9, a_{1}=33, a_{5}=153$, at this point 153 is divisible by 9.
When $m \geqslant 5$, $a_{n}=a_{5}+\sum_{k=6}^{m} k!$, and for $k \geqslant 6$, $k!$ is divisible by 9.
Therefore, when $m \geqslant 5$, $a_{n}$ is divisible by 9, so the smallest value of $n$ is 5.
|
5
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
1. Solution: 3 From $[\lg x] \leqslant \lg x$ we get $\lg ^{2} x-\lg x-2 \leqslant 0$ which means $-1 \leqslant \lg x \leqslant 2$.
When $-1 \leqslant \lg x<0$, we have $[\lg x]=-1$, substituting into the original equation gives $\lg x= \pm 1$, but $\lg x=1$ is not valid, so $\lg x=-1, x_{1}=\frac{1}{10}$.
When $0 \leqslant \lg x<1$, we have $[\lg x]=0$, substituting into the original equation gives $\lg x= \pm \sqrt{2}$, both are not valid.
When $1 \leqslant \lg x<2$, we have $[\lg x]=1$, substituting into the original equation gives $\lg x= \pm \sqrt{3}$, but $\lg x=-\sqrt{3}$ is not valid, so $\lg x=\sqrt{3}, x_{2}=$ $10^{\sqrt{3}}$.
When $\lg x=2$, we get $x_{3}=100$. Therefore, the original equation has 3 real roots.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given point $G$ is the centroid of $\triangle A B C$, and $\overrightarrow{A B}=\boldsymbol{a}, \overrightarrow{A C}=\boldsymbol{b}$. If $P Q$ passes through point $G$ and intersects $A B, A C$ at points $P, Q$, and $\overrightarrow{A P}=m \boldsymbol{a}, \overrightarrow{A Q}=n \boldsymbol{b}$, then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$ .
|
1. 3 Detailed Explanation: Special Value Method. When $P Q // B C$, $m=n=\frac{2}{3}$, thus $\frac{1}{m}+\frac{1}{n}=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
13. Given the sequence $a_{1}=1, a_{n}=1-\frac{1}{4 a_{n-1}}$, and $b_{n}=\frac{2}{2 a_{n}-1}$.
(1) Prove: For $n \geqslant 1$, it always holds that $\frac{1}{2}<a_{n} \leqslant 1$;
(2) Find the general term formula for the sequence $\left\{b_{n}\right\}$;
(3) Find the value of $\lim _{n \rightarrow+\infty} \frac{a_{n}}{b_{n}}$.
|
13. (1) Prove the following by mathematical induction:
(1) When $n=1$, $a_{1}=1$, we have $\frac{1}{2}<a_{1} \leqslant 1$;
(2) Assume that when $n=k$, the proposition holds, i.e., $\frac{1}{2}<a_{k} \leqslant 1$, then when $n=k+1$, since $a_{k+1}=1-\frac{1}{4 a_{k}}$, and because $\frac{1}{2}<a_{k} \leqslant 1$, we have $-\frac{1}{2}<-\frac{1}{4 a_{k}} \leqslant-\frac{1}{4}$, so $1-\frac{1}{2}<1-\frac{1}{4 a_{k}} \leqslant 1-\frac{1}{4}<1$, i.e., $\frac{1}{2}<a_{k+1}<1$, which means that when $n=k+1$, the proposition also holds.
From (1) and (2), we know that for all $n \geqslant 1$, $\frac{1}{2}<a_{n} \leqslant 1$ always holds;
(2) $b_{1}=\frac{2}{2 a_{1}-1}=\frac{2}{2-1}=2$, so $a_{k}=1-\frac{1}{4 a_{k-1}}$.
Therefore, when $k \geqslant 2$, $b_{k}-b_{k-1}=\frac{2}{2 a_{k}-1}-\frac{2}{2 a_{k-1}-1}=\frac{2}{2\left(1-\frac{1}{4 a_{k-1}}\right)-1}-\frac{2}{2 a_{k-1}-1}=$ $\frac{4 a_{k-1}}{2 a_{k-1}-1}-\frac{2}{2 a_{k-1}-1}=\frac{2\left(2 a_{k-1}-1\right)}{2 a_{k-1}-1}=2$, so $\left\{b_{n}\right\}$ is an arithmetic sequence with the first term 2 and common difference 2;
(3) From $b_{n}=\frac{2}{2 a_{n}-1}$, we get $a_{n}=\frac{1}{b_{n}}+\frac{1}{2}$, and $b_{n}=2+2(n-1)=2 n$, so $a_{n}=\frac{1}{2 n}+\frac{1}{2}$. Therefore, $\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\lim _{n \rightarrow \infty} a_{n} \cdot \lim _{n \rightarrow \infty} \frac{1}{b_{n}}=\lim _{n \rightarrow \infty}\left(\frac{1}{2 n}+\frac{1}{2}\right) \cdot \lim _{n \rightarrow \infty} \frac{1}{2 n}=\frac{1}{2} \times 0=0$.
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
6. In a regular quadrilateral pyramid $P-A B C D$, $G$ is the centroid of $\triangle P B C$. Then $\frac{V_{G-P A D}}{V_{G-P A B}}=$ $\qquad$
|
As shown in the figure, $M, N, F$ are the midpoints of the corresponding sides, $M E \perp P N$, $O H \perp P F$. Since $P-A B C D$ is a regular quadrilateral pyramid, we have $\angle M N E=\angle O F H=\alpha$, thus
$$
M E=M N \sin \alpha=2 O F \sin \alpha=2 O H \text {. }
$$
This means the distance from point $M$ to plane $P A D$ is twice the distance from point $M$ to plane $P A B$, hence the distance from point $G$ to plane $P A D$ is also twice the distance from point $G$ to plane $P A B$.
Therefore, $\frac{V_{G-P A D}}{V_{G-P A B}}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Let $a, b, c \in \mathbf{R}^{+}$. Prove that $\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+a+c)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8$.
(2003 USA Olympiad Problem)
|
7. Let $a+b+c=s, \frac{a}{a+b+c}=x_{1}, \frac{b}{a+b+c}=x_{2}, \frac{c}{a+b+c}=x_{3}$, then the original proposition is transformed into: Given positive real numbers $x_{1}, x_{2}, x_{3}$ satisfying $\sum_{i=1}^{3} x_{i}=1$, prove that $\sum_{i=1}^{3} \frac{\left(1+x_{i}\right)^{2}}{2 x_{i}^{2}+\left(1-x_{i}\right)^{2}} \leqslant 8$.
Let $f(x)=\frac{(x+1)^{2}}{3 x^{2}-2 x+1}, x \in(0,1)$, the tangent line function at the mean $x=\frac{1}{3}$ is $y=\frac{12 x+4}{3}$.
And $\frac{(x+1)^{2}}{3 x^{2}-2 x+1}-\frac{12 x+4}{3}=\frac{-(4 x+1)(3 x-1)^{2}}{3\left(3 x^{2}-2 x+1\right)} \leqslant 0$, and when $x \in(0,1)$. Therefore, $\sum_{i=1}^{3} \frac{\left(x_{i}+1\right)^{2}}{3 x_{i}^{2}-2 x_{i}+1} \leqslant \sum_{i=1}^{3} \frac{12 x_{i}+4}{3}=\frac{12+12}{3}=8$.
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
3. The faces of a hexahedron and the faces of a regular octahedron are all equilateral triangles with side length $a$. The ratio of the radii of the inscribed spheres of these two polyhedra is a reduced fraction $\frac{m}{n}$. Then, the product $m \cdot n$ is $\qquad$.
|
6
3.【Analysis and Solution】Solution 1 Let the radius of the inscribed sphere of this hexahedron be $r_{1}$. On one hand, the volume of this hexahedron can be calculated by multiplying the radius of the inscribed sphere by the surface area and then dividing by $\frac{1}{3}$. On the other hand, it can also be calculated as the volume of two regular tetrahedrons with edge length $a$. Therefore, we have
$$
\frac{1}{3} \times 6 \times \frac{\sqrt{3}}{4} a^{2} \times r_{1}=2 \times \frac{1}{3} \times \frac{\sqrt{3}}{4} a^{2} \cdot \sqrt{a^{2}-\left(\frac{\sqrt{3} a}{3}\right)^{2}} .
$$
Let the radius of the inscribed sphere of the regular octahedron be $r_{2}$. On one hand, the volume of this regular octahedron can be calculated by multiplying the radius of the inscribed sphere by the surface area and then dividing by $\frac{1}{3}$. On the other hand, it can also be calculated as the volume of two regular square pyramids with edge length $a$. Therefore, we have
84
$$
\frac{1}{3} \times 8 \times \frac{\sqrt{3}}{4} a^{2} \cdot r_{2}=2 \times \frac{1}{3} \times a^{2} \times \sqrt{a^{2}-\left(\frac{\sqrt{2} a}{2}\right)^{2}} .
$$
From equations (1) and (2), we get $\frac{r_{1}}{r_{2}}=\frac{2}{3}$.
Therefore, $\frac{m}{n}=\frac{2}{3}, m \cdot n=6$.
Solution 2 As shown in Figure $\mathrm{a}$, the radius of the inscribed sphere of the hexahedron is the perpendicular distance $O R$ from the center $O$ of the equilateral $\triangle B C D$ to one of the faces $A B D$, and $O R$ is also the altitude from the hypotenuse $A M$ of the right $\triangle A O M$.
$$
\therefore r_{1}=O R=\frac{A O \cdot O M}{A M} \text {. }
$$
But $A M=\frac{\sqrt{3}}{2} a, O M=\frac{\sqrt{3}}{6} a$,
$$
\therefore A O=\frac{\sqrt{6}}{3} a \text {. }
$$
Substituting into (3) gives $r_{1}=\frac{\sqrt{6} a}{9}$.
In Figure $\mathrm{b}$, the center of the inscribed sphere of the regular octahedron is the center $O$ of the square $B C D E$, and the perpendicular distance $O R$ from $O$ to one of the faces $A B C$ is the radius $r_{2}$ of the inscribed sphere of the regular octahedron, and $O R$ is also the altitude from the hypotenuse of the right $\triangle A O M$.
$$
\therefore r_{2}=O R=\frac{A O \times O M}{A M} \text {. }
$$
But $A M=\frac{\sqrt{3}}{2} a, O M=\frac{a}{2}$,
$$
\therefore A O=\frac{\sqrt{2}}{2} a \text {. }
$$
Substituting into (4) gives $r_{2}=\frac{\sqrt{6}}{6} a$.
$$
\begin{array}{l}
\therefore \frac{r_{1}}{r_{2}}=\frac{\frac{\sqrt{6} a}{9}}{\frac{\sqrt{6}}{6} a}=\frac{2}{3} \cdot 2,3 \text { are coprime, } \\
\therefore m=2, n=3, m n=2 \times 3=6 .
\end{array}
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
$\qquad$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
【Analysis】Since $\left(101010_{2}=(42)_{10}\right.$ is divisible by 7, every 6 bits are exactly divisible by 7.
$2014 \div 6=335 \cdots 4$, so ( $\underbrace{201010 \cdots 10}_{2014 \text { years }}$ when divided by 7 has the same remainder as $\left(1010_{2}\right.$ when divided by 7.
And $(1010)_{2}=10_{10}$, the remainder is 3
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
$31 \cdot 28$ satisfies the equation $x^{4} y^{4}-10 x^{2} y^{2}+9=0$ for the number of different pairs $(x, y)$ (where $x, y$ are positive integers) is
(A) 0.
(B) 3.
(C) 4.
(D) 12.
(E) infinite.
(20th American High School Mathematics Examination, 1969)
|
［Solution］The given equation is equivalent to
$$
\left(x^{2} y^{2}-1\right)\left(x^{2} y^{2}-9\right)=0 \text {, }
$$
Therefore,
$$
x y= \pm 1 \text { or } \pm 3 \text {. }
$$
Thus, the different pairs of positive integers that satisfy the equation are $(1,1),(1,3)$ or $(3,1)$. Hence, the answer is $(B)$.
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
2. There are 3 logs, each cut into 3 pieces, it took a total of 18 minutes, how many minutes does it take to saw each time?
|
2. $(3-1) \times 3=6$ (times) $18 \div 6=3$ (minutes)
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. On the blackboard, there are $N(N \geqslant 9)$ distinct non-negative real numbers less than 1. It is known that for any eight numbers on the blackboard, there exists another number on the blackboard such that the sum of these nine numbers is an integer. Find all possible values of $N$.
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。
|
8. $N=9$.
An example for $N=9$ exists. It is sufficient for the sum of these nine numbers to be an integer.
Now, let $N>9$. Let $S$ denote the sum of all numbers.
Take any seven numbers $a_{1}, a_{2}, \cdots, a_{7}$, and let $T$ denote their sum, and $A$ denote the complement of the set formed by these seven numbers.
Take any $c \in A$, there exists $d \in A$ such that $T+c+d$ is an integer.
It is easy to see that such a $d$ is unique.
In fact, if $T+c+d$ and $T+c+d^{\prime}$ are both integers, then
$$
\left|d-d^{\prime}\right|=\left|(T+c+d)-\left(T+c+d^{\prime}\right)\right|
$$
is an integer.
Thus, $d=d^{\prime}$.
The above $d$ is paired with $c$. Thus, the numbers in set $A$ are paired as follows:
$$
\left(c_{1}, d_{1}\right),\left(c_{2}, d_{2}\right), \cdots,\left(c_{l}, d_{l}\right),
$$
where $l=\frac{N-7}{2}>1$.
Consider the sum
$$
S_{1}=\sum_{i=1}^{l}\left(T+c_{i}+d_{i}\right)=l T+(S-T)
$$
which is an integer.
Thus, $T=\frac{S_{1}-S}{l-1}$.
Now, take the numbers $a_{2}, a_{3}, \cdots, a_{8}$ on the board. Let their sum be $T^{\prime \prime}$.
As before, we get $T^{\prime}=\frac{S_{1}^{\prime}-S}{l-1}$.
Therefore, $a_{1}-a_{8}=\frac{S_{1}-S}{l-1}-\frac{S_{1}^{\prime}-S}{l-1}=\frac{S_{1}-S_{1}^{\prime}}{l-1}$.
Since $a_{1}$ and $a_{8}$ can be any two numbers on the board, we conclude that the difference between any two numbers on the board has the form $\frac{k}{l-1}(k \in \mathbf{Z})$.
Let $a$ be the smallest number on the board. Then the numbers on the board belong to the set
$$
\left\{a, a+\frac{1}{l-1}, a+\frac{2}{l-1}, \cdots, a+\frac{l-2}{l-1}\right\}
$$
which contains at most $l-1$ numbers.
This contradicts the fact that the number of numbers is $N=7+2 l>l$.
|
9
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8 If the equation $z^{2009}+z^{2008}+1=0$ has roots of modulus 1, then the sum of all roots of modulus 1 is $\qquad$ .
|
8 Let $z$ be a root satisfying the condition, then the original equation is equivalent to $z^{2008}(z+1)=-1$. Taking the modulus on both sides, we get $|z+1|=1$. Since $|z|=1$, all roots with modulus 1 can only be the complex numbers corresponding to the intersection points of the circle with center $(-1,0)$ and radius 1, and the circle with center $(0,0)$ and radius 1 on the complex plane, so $z=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$. Upon verification, these two roots are both roots of the original equation, thus the sum of all roots with modulus 1 is -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. For any $x \in \mathbf{R}$, the function $f(x)$ satisfies the relation $f(x+2008)=f(x+2007)+f(x+2009)$, and $f(1)=\lg \frac{3}{2}, f(2)=\lg 15$, then the value of $f(2007)$ is $\qquad$ .
|
10. 1. By substituting $x-2007$ for $x$ in the given relation, we get $f(x)=f(x-1)+f(x+1)$.
Replacing $x$ with $x+1$, we get $f(x+1)=f(x)+f(x+2)$.
Adding (1) and (2) yields $f(x-1)+f(x+2)=0$.
Replacing $x$ with $x+1$, we get $f(x)+f(x+3)=0$, so $f(x+3)=-f(x), f(x+6)=f[(x+3)+3]=-f(x+3)=f(x)$. Therefore, $f(x)$ is a periodic function, and 6 is one of its periods. Thus, $f(2007)=f(6 \times 334+3)=f(3)$.
Also, $f(2)=f(1)+f(3), f(3)=f(2)-f(1)=\lg 15-\lg \frac{3}{2}=1$. Therefore, $f(2007)=f(3)=1$.
Note: The conclusion of this problem can be generalized as follows: If the function $y=f(x)(x \in \mathbf{R})$ satisfies $f(x)=f(x-a)+f(x+a)$, then the function $y=f(x)$ is a periodic function.
Proof: Since for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(x)=f(x-a)+f(x+a), \\
f(x-a)=f(x-2 a)+f(x) .
\end{array}
$$
Adding (1) and (2) gives
$$
f(a+x)=-f(x-2 a), f(x)=-f(x-3 a) .
$$
Thus,
$$
f(x+6 a)=-f(x+6 a-3 a)=-f(x+3 a)=f(x+3 a-3 a)=f(x) \text {. }
$$
Therefore, the function $y=f(x)(x \in \mathbf{R})$ is a periodic function, and $6 a$ is one of its periods.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the Land of Dwarfs, there are 1 yuan, 5 yuan, 10 yuan, and 50 yuan gold coins. To buy a sheep priced at 168 yuan without receiving change, the minimum number of gold coins needed is $\qquad$.
|
$8$
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5 Let $a, b$ be two positive numbers, and $a>b$. Points $P, Q$ are on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the line connecting point $A(-a, 0)$ and $Q$ is parallel to the line $O P$, and intersects the $y$-axis at point $R, O$ being the origin, then $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=$ $\qquad$.
|
5 2. Let $A Q:\left\{\begin{array}{l}x=-a+t \cos \theta \\ y=t \sin \theta\end{array}\right.$ ( $t$ is a parameter)
Substitute (1) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$$
t=\frac{2 a b^{2} \cos \theta}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} .
$$
Therefore,
$$
|A Q|=\frac{2 a b^{2}|\cos \theta|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} .
$$
In (1), let $x=0$, we get $t=\frac{\alpha}{\cos \theta}$. So $|A R|=\frac{\alpha}{|\cos \theta|}$.
Also, let
$$
O P:\left\{\begin{array}{l}
x=t^{\prime} \cos \theta \\
y=t^{\prime} \sin \theta
\end{array} \text { ( } t^{\prime} \text { is a parameter }\right)
$$
Substitute (2) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$$
t^{\prime 2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} .
$$
Therefore,
$$
|O P|^{2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} .
$$
Hence,
$$
\frac{|A Q| \cdot|A R|}{|O P|^{2}}=\cdots=2 .
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. If $\frac{1-\cos \theta}{4+\sin ^{2} \theta}=\frac{1}{2}$, then $\left(4+\cos ^{3} \theta\right) \cdot\left(3+\sin ^{3} \theta\right)=$
|
2. 9 Detailed Explanation: From the condition, we get $2-2 \cos \theta=4+\sin ^{2} \theta, \Rightarrow(\cos \theta-1)^{2}=4, \Rightarrow \cos \theta=-1$, then $\left(4+\cos ^{3} \theta\right) \cdot\left(3+\sin ^{3} \theta\right)=9$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Given the vector $\boldsymbol{a}=(\cos \theta, \sin \theta)$, vector $\boldsymbol{b}=(\sqrt{3},-1)$, then the maximum value of $|2 \boldsymbol{a}-\boldsymbol{b}|$ is
|
5. 4 Hint: $2 \boldsymbol{a}-\boldsymbol{b}=(2 \cos \theta-\sqrt{3} \cdot 1+2 \sin \theta) \quad$ so $\quad|2 \boldsymbol{a}-\boldsymbol{b}|=\sqrt{8-8 \cos \left(\theta+\frac{\pi}{6}\right)}$ Therefore, when $\cos \left(\theta+\frac{\pi}{6}\right)=-1$, $|2 \boldsymbol{a}-\boldsymbol{b}|_{\text {max }}=4$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 10 Let real numbers $s, t$ satisfy $19 s^{2}+99 s+1=0, t^{2}+99 t+19=0$, and $s t \neq 1$. Find the value of $\frac{s t+4 s+1}{t}$.
|
Solve: Given $19 s^{2}+99 s+1=0, t^{2}+99 t+19=0$, we have
$$
19 s^{2}+99 s+1=0, 19\left(\frac{1}{t}\right)^{2}+99 \cdot\left(\frac{1}{t}\right)+1=0 \text {. }
$$
Since $s t \neq 1$, we know $s \neq \frac{1}{t}$. Therefore, $s, \frac{1}{t}$ are the two distinct real roots of the equation $19 x^{2}+99 x+1=0$.
By the relationship between roots and coefficients, we have $s+\frac{1}{t}=-\frac{99}{19}, s \cdot \frac{1}{t}=\frac{1}{19}$.
$$
\text { Thus } \frac{s t+4 s+1}{t}=s+4\left(\frac{s}{t}\right)+\frac{1}{t}=\left(s+\frac{1}{t}\right)+4 \cdot s \cdot \frac{1}{t}=-\frac{99}{19}+\frac{4}{19}=-5 \text {. }
$$
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
$18.4 .7 * \star$ Among numbers of the form $36^{k}-5^{l}$, where $k$ and $l$ are natural numbers. Taking the minimum value in absolute value, prove: the number obtained in this way is certainly the smallest positive number.
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
$18.4 .7 * \star$ Among numbers of the form $36^{k}-5^{l}$, where $k$ and $l$ are natural numbers. Taking the minimum value in absolute value, prove: the number obtained in this way is certainly the smallest positive number.
|
The units digit of the numbers $36^{k}$ and $5^{l}$ are 6 and 5, respectively. Therefore, the units digit of $\left|36^{k}-5^{l}\right|$ is either 1 or 9.
However, $36^{k}-5^{l}=1$ and $5^{l}-36^{k}=9$ cannot both hold. Otherwise, $36^{k}-5^{l}=1$ can be written as $5^{l} = \left(6^{k}-1\right)\left(6^{k}+1\right)$. Clearly, $5 \nmid 6^{k}+1$. Therefore, $\left(6^{k}-1\right)\left(6^{k}+1\right)$ cannot be $5^{l}$. Similarly, in $5^{l}-36^{2k}=9$, $3 \nmid 5^{l}$, so the equation clearly does not hold.
When we take $k=1, l=2$, we get $36-5^{2}=11$, which is the smallest positive number that satisfies the condition.
|
11
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
5. In $\triangle A B C$, the lengths of the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $c-a$ equals the height $h$ from vertex $A$ to side $AC$, then the value of $\sin \frac{C-A}{2}+\cos \frac{C+A}{2}$ is
(A) 1 ;
(B) $\frac{1}{2}$;
(C) $\frac{1}{3}$;
(D) -1 .
|
5. (A)
As shown in the right figure,
$$
1=c-a=h=a .
$$
At this moment,
$$
\begin{aligned}
& \sin \frac{C-A}{2}+\cos \frac{C+A}{2} \\
= & \sin 30^{\circ}+\cos 60^{\circ}=1
\end{aligned}
$$
|
1
|
Geometry
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
4. (42nd IMO Shortlist) Let $\triangle ABC$ be an acute-angled triangle. Construct isosceles triangles $\triangle DAC$, $\triangle EAB$, and $\triangle FBC$ outside $\triangle ABC$ such that $DA = DC$, $EA = EB$, $FB = FC$, $\angle ADC = 2 \angle BAC$, $\angle BEA = 2 \angle ABC$, and $\angle CFB = 2 \angle ACB$. Let $D'$ be the intersection of line $DB$ and $EF$, $E'$ be the intersection of line $EC$ and $DF$, and $F'$ be the intersection of line $FA$ and $DE$. Find the value of $\frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'}$.
|
4. Since $\triangle A B C$ is an acute triangle
Therefore, $\angle A D C, \angle B E A, \angle C F B < \pi$
So $\angle D A C = \frac{\pi}{2} - \frac{1}{2} \angle A D C = \frac{\pi}{2} - \angle B A C$
$$
\angle B A E = \frac{\pi}{2} - \frac{1}{2} \angle B E A = \frac{\pi}{2} - \angle A B C
$$
So $\angle D A E = \angle D A C + \angle B A C + \angle B A E = \pi - \angle A B C < \pi$
Similarly, $\angle E B F < \pi, \angle F C D < \pi$
Therefore, hexagon $A B C D E F$ is a convex hexagon, and
$$
\angle A D C + \angle B E A + \angle C F B = 2(\angle B A C + \angle A B C + \angle A C B) = 2 \pi
$$
Let $\omega_{1}, \omega_{2}, \omega_{3}$ be the circles with centers at $D, E, F$ and radii $D A, E B, C F$ respectively.
Since $\angle A D C + \angle B E A + \angle C F B = 2 \pi$
Therefore, the circles $\omega_{1}, \omega_{2}, \omega_{3}$ intersect at a common point.
Let this intersection point be $O$, then point $O$ is the reflection of $C$ over the line $D F$, and also the reflection of $A$ over the line $D E$ and the reflection of $B$ over the line $E F$.
So $\frac{D B}{D D^{\prime}} = \frac{D D^{\prime} + D^{\prime} B}{D D^{\prime}} = 1 + \frac{D^{\prime} B}{D D^{\prime}} = 1 + \frac{S_{\triangle E F B}}{S_{\triangle E F D}} = 1 + \frac{S_{\triangle E F O}}{S_{\triangle E F D}}$
Similarly, $\frac{E C}{E E^{\prime}} = 1 + \frac{S_{\triangle D F O}}{S_{\triangle E F D}}, \frac{F A}{F F^{\prime}} = 1 + \frac{S_{\triangle D E O}}{S_{\triangle E F D}}$
Therefore, $\frac{D B}{D D^{\prime}} + \frac{E C}{E E^{\prime}} + \frac{F A}{F F^{\prime}} = 3 + \frac{S_{\triangle E F O} + S_{\triangle D F O} + S_{\triangle D E O}}{S_{\triangle E F D}} = 4$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, with the sum of its first $n$ terms being $S_{n}$. Then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is $\qquad$.
|
3. $n=7$ Detailed Explanation: From the recurrence relation, we get: $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right)$, thus $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore, $S_{n}-n=\left(a_{1}-1\right)+\left(a_{2}-1\right)+\cdots+\left(a_{n}-1\right)=\frac{8\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}=6-6 \times\left(-\frac{1}{3}\right)^{n}$, $\therefore\left|S_{n}-n-6\right|=6 \times\left(\frac{1}{3}\right)^{n}250, \therefore$ the smallest integer $n$ that satisfies the condition is $n=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. If real numbers $\alpha, \beta$ satisfy:
$$
\alpha^{3}-3 \alpha^{2}+5 \alpha=1, \beta^{3}-3 \beta^{2}+5 \beta=5 \text {, }
$$
then $\alpha+\beta=$ $\qquad$ .
|
4. 2 .
Construct the function $f(x)=x^{3}+2 x$.
According to the problem, we know that $f(\alpha-1)=-2, f(\beta-1)=2$.
By the odd-even property and monotonicity of $f(x)$, we get
$$
(\alpha-1)+(\beta-1)=0 \text {. }
$$
Therefore, $\alpha+\beta=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
$4 \cdot 36$ Find the smallest positive integer $n$, such that in any two-coloring of $K_{n}$, there exist 3 monochromatic triangles, each pair of which has no common edge.
untranslated text preserved the line breaks and format.
|
[Solution] In the given two-colored $K_{8}$, there are two blue $K_{4}$, and the rest of the unshown edges are all red. It is easy to see that there are no red triangles and exactly 8 blue triangles in the graph. However, among any 3 blue triangles, there are always 2 triangles in the same blue $K_{4}$, sharing 1 common edge. Therefore, the smallest positive integer $n \geqslant 9$.
For a two-colored $K_{9}$, let $\triangle A_{1} A_{2} A_{3}$ be one of its monochromatic triangles. Consider the complete subgraph $K_{6}$ with vertices $\left\{A_{4}, A_{5}, A_{6}, A_{7}, A_{8}, A_{9}\right\}$, which must contain a monochromatic triangle. Suppose $\triangle A_{7} A_{8} A_{9}$ is a blue triangle. Now consider the $K_{5}$ with vertices $\left\{A_{3}, A_{4}, A_{5}, A_{6}, A_{7}\right\}$. If it contains a monochromatic triangle, then together with the previous two, there are 3 monochromatic triangles with no common edges. If it does not contain a monochromatic triangle, it can be decomposed into one red and one blue cycle, each with 5 edges (see figure). Similarly, in the $K_{5}$ with vertices $\left\{A_{3}, A_{4}, A_{5}, A_{6}, A_{8}\right\}$ and $\left\{A_{3}, A_{4}, A_{5}, A_{6}, A_{9}\right\}$, if either $K_{5}$ contains a monochromatic triangle, the problem is solved. Otherwise, both $K_{5}$ can be decomposed into one red and one blue cycle, each with 5 edges. Since the colors of the $K_{4}$ with vertices $\left\{A_{3}, A_{4}, A_{5}, A_{6}\right\}$ are already determined, it must be that $A_{3} A_{8}, A_{3} A_{9}, A_{6} A_{8}, A_{6} A_{9}$ are all blue edges. Thus, $\triangle A_{3} A_{7} A_{9}$ and $\triangle A_{6} A_{7} A_{8}$ are both blue triangles. Adding the initial monochromatic triangle $A_{1} A_{2} A_{3}$ meets the requirements of the problem.
In summary, the smallest positive integer $n=9$.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
$32 \cdot 25$ Among the 101 integers $1,2,3, \cdots, 101$, the numbers that can be divided by both 3 and 5 and leave a remainder of 1 are
(A) 6.
(B) 7.
(C) 8.
(D) 9.
(2nd "Jinyun Cup" Junior High School Mathematics Invitational, 1985)
|
〔Solution〕 Obviously, numbers that leave a remainder of 1 when divided by both 3 and 5 (note that 3 and 5 are both prime numbers) can be expressed as
$$
3 \cdot 5 k+1=15 k+1 \quad(k=0,1,2 \cdots),
$$
Thus, among the first 101 natural numbers, there are $1,16,31,46,61,76,91$, a total of 7 numbers that meet the condition. Therefore, the answer is $(B)$.
|
7
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
$17 \cdot 102$ In the figure, $D$ is the midpoint of side $BC$ of $\triangle ABC$, $E$ is an internal point on side $AC$, $AC=3CE$, $BE$ and $AD$ intersect at $G$. Then $AG: GD$ is
(A) 2 .
(B) 3 .
(C) 3 or 4 .
(D) 4 .
|
$\begin{array}{l}\text { [Solution] Draw } D F / / B E \text { intersecting } A C \text { at } F \text {. } \\ \text { Since } D \text { is the midpoint of } B C \text {, then } F \text { is the midpoint of } E C \text {. } \\ \text { Given } A C=3 C E \text {, we have } A E=2 C E=4 E F \text {, } \\ \text { and } D F / / B E, \\ \text { thus } A G: G D=A E: E F=4: 1=4 . \\ \text { Therefore, the answer is }(D) .\end{array}$
|
4
|
Geometry
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
52 Given $\left(a x^{4}+b x^{3}+c x^{2}+d x+e\right)^{5} \cdot\left(a x^{4}-b x^{3}+c x^{2}-d x+e\right)^{5}=a_{0}+a_{1} x+$ $a_{2} x^{2}+\cdots+a_{41} x^{10}$, then $a_{1}+a_{3}+a_{5}+\cdots+a_{39}=$ $\qquad$.
|
520 . Let $F(x)=\left(a x^{4}+b x^{3}+c x^{2}+d x+e\right)^{5} \cdot\left(a x^{4}-b x^{3}+c x^{2}-d x+e\right)^{5}$, from $F(-x)=F(x)$ we know that $F(x)$ is an even function, so the expansion of $F(x)$ does not contain odd powers of $x$, i.e.,
$$
a_{1}=a_{3}=\cdots=a_{39}=0,
$$
thus
$$
a_{1}+a_{3}+a_{5}+\cdots+a_{39}=0 .
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Pack 35 egg yolk mooncakes, there are two packaging specifications: large bags with 9 mooncakes per pack, and small bags with 4 mooncakes per pack. It is required that no mooncakes are left over, so a total of _ $\qquad$ packs were made.
|
【Answer】 5
Analysis: Let the number of large bags be $\mathrm{x}$, and the number of small bags be $\mathrm{y}$, (both $\mathrm{x}$ and $\mathrm{y}$ are integers) so $9 \mathrm{x}+4 \mathrm{y}=35$. It is easy to get $\left\{\begin{array}{l}x=3 \\ y=2\end{array}\right.$, so a total of $2+3=5$ bags were packed.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. As shown in the figure, in $\triangle A B C$, $\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}, \overrightarrow{A H} \cdot \overrightarrow{B C}=0, \overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$, then the eccentricity of the hyperbola passing through point $C$, with $A, H$ as the two foci is $\qquad$ .
|
Given, $\cos C=2 \cos ^{2} \frac{C}{2}-1=\frac{3}{5}, \sin C=\frac{4}{5}$, let's assume $A C=B C=5$, then $A H=4, C H=3$. Therefore, for the hyperbola, $2 a=2,2 c=4 \Rightarrow e=\frac{c}{a}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.3.11 ** Given $\triangle A B C$ with corresponding side lengths $a, b, c$, and $P$ is a point in the plane of $\triangle A B C$. If $P A=p, P B=q, P C=r$, then $\frac{p q}{a b}+\frac{q r}{b c}+\frac{r p}{a c} \geqslant 1$.
|
Let $A$, $B$, $C$ correspond to the complex numbers $t_{1}$, $t_{2}$, $t_{3}$, respectively, and let $P$ correspond to the origin $O$. Then $\frac{p q}{a b}=\frac{\left|t_{1} t_{2}\right|}{\left|\left(t_{3}-t_{1}\right)\left(t_{3}-t_{2}\right)\right|}$, and similarly for the other two expressions. Therefore,
$$
\frac{p q}{a b}+\frac{q r}{b c}+\frac{r p}{a c} \geqslant\left|\frac{t_{1} t_{2}}{\left(t_{3}-t_{1}\right)\left(t_{3}-t_{2}\right)}+\frac{t_{2} t_{3}}{\left(t_{1}-t_{2}\right)\left(t_{1}-t_{3}\right)}+\frac{t_{3} t_{1}}{\left(t_{2}-t_{3}\right)\left(t_{2}-t_{1}\right)}\right|=1 .
$$
|
1
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
3. Given a positive real number $x$ satisfies
$$
x^{3}+x^{-3}+x^{6}+x^{-6}=2754 \text {. }
$$
Then $x+\frac{1}{x}=$ $\qquad$ .
|
3. 4 .
Transform the given equation into
$$
\begin{array}{l}
\left(x^{3}+x^{-3}\right)^{2}+\left(x^{3}+x^{-3}\right)-2756=0 \\
\Rightarrow\left(x^{3}+x^{-3}+53\right)\left(x^{3}+x^{-3}-52\right)=0 .
\end{array}
$$
Notice that, $x^{3}+x^{-3}+53>0$.
Thus, $x^{3}+x^{-3}=52$.
Let $b=x+x^{-1}$.
$$
\begin{array}{l}
\text { Then } x^{3}+x^{-3}=\left(x+x^{-1}\right)\left[\left(x+x^{-1}\right)^{2}-3\right] \\
\Rightarrow b\left(b^{2}-3\right)=52 \\
\Rightarrow(b-4)\left(b^{2}+4 b+13\right)=0 \\
\Rightarrow b=4 .
\end{array}
$$
Therefore, $x+\frac{1}{x}=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In space, there are $n(n \geqslant 3)$ planes, among which any three planes do not have a common perpendicular plane. There are the following four conclusions:
(1) No two planes are parallel to each other;
(2) No three planes intersect in a single line;
(3) Any two lines of intersection between planes are not parallel;
(4) Each line of intersection between planes intersects with $n-2$ planes.
Among these, the number of correct conclusions is . $\qquad$
|
3. 4 .
If two planes $\alpha_{1}, \alpha_{2}$ are parallel, then the common perpendicular plane of $\alpha_{1}, \alpha_{3}$ is also perpendicular to $\alpha_{2}$, which contradicts the condition that any three planes have no common perpendicular plane. Conclusion (1) is true.
If three planes intersect at a line, then the plane perpendicular to this line is the common perpendicular plane of the three planes, which contradicts the given condition. Conclusion (2) is true.
If two lines of intersection are parallel, then the plane perpendicular to these two lines is at least the common perpendicular plane of three planes, which contradicts the given condition. Conclusion (3) is true.
If the line of intersection $l$ of $\alpha_{1}$ and $\alpha_{2}$ does not intersect with plane $\alpha_{3}$, then there are two possibilities: $l$ lies on $\alpha_{3}$ (where $\alpha_{1}, \alpha_{2}, \alpha_{3}$ intersect at $l$), or $l$ is parallel to $\alpha_{3}$ (where $\alpha_{3}$ is parallel to either $\alpha_{1}$ or $\alpha_{2}$, or $\alpha_{3}$ intersects $\alpha_{1}, \alpha_{2}$ at three parallel lines). Both of these possibilities contradict the given conditions. Conclusion (4) is true.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
21. Let $a$, $b$, $c$ be non-zero real numbers, and $a+b+c=0$. Then the value of $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b c}{|a b c|}$ could be ( ).
(A) 0
(B) $\pm 1$
(C) $\pm 2$
(D) 0 or $\pm 2$
(E) 0 or $\pm 1$
|
21. A.
Assume $a \leqslant b \leqslant c$.
Since $a, b, c$ are non-zero real numbers, and $a+b+c=0$, thus, $a<0$, at this time,
$$
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b c}{|a b c|}=0 \text { ; }
$$
When $b>0$, $a b c<0$, at this time,
$$
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b c}{|a b c|}=0 .
$$
|
0
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
8. In the plane region
$$
M=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}
0 \leqslant y \leqslant 2-x, \\
0 \leqslant x \leqslant 2
\end{array}\right\}\right.\right.
$$
take any $k$ points, it is always possible to divide these $k$ points into two groups $A$ and $B$, such that the sum of the x-coordinates of the points in group $A$ is no more than 6, and the sum of the y-coordinates of the points in group $B$ is no more than 6. Then the maximum value of the positive integer $k$ is . $\qquad$
|
8. 11.
Since there exist 12 points in the plane region $M$:
$$
\begin{array}{l}
P_{t}\left(1-10^{-4}, 1+10^{-t-3} t\right)(t=1,2, \cdots, 5), \\
P_{t}\left(1+6 \times 10^{-4}, 1-10^{-4} t\right)(t=6,7, \cdots, 12),
\end{array}
$$
these points cannot be divided into two groups as required by the problem, so $k \leqslant 11$.
Next, we prove that $k=11$ satisfies the problem's conditions.
Take any 11 points $P_{i}\left(x_{i}, y_{i}\right)$ $(i=1,2, \cdots, 11)$ in the plane region $M$. By the problem's conditions, we know $y_{i} \leqslant 2-x_{i}(i=1, 2, \cdots, 11)$. Without loss of generality, assume $0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{11} \leqslant 2$, and consider the following two cases.
(1) If $\sum_{i=1}^{11} x_{i} \leqslant 6$, then take $A=\left\{P_{1}, P_{2}, \cdots, P_{8}\right\}$, $B=\left\{P_{9}, P_{10}, P_{11}\right\}$.
(2) If $\sum_{i=1}^{11} x_{i}>6$, then there exists a unique positive integer $k \in \{4,5, \cdots, 10\}$ such that $\sum_{i=1}^{k} x_{i} \leqslant 6$.
But $\sum_{i=1}^{k+1} x_{i}>6$, so $x_{k+1}>\frac{6}{k+1}$, and thus,
$$
\sum_{i=k+1}^{11} y_{i} \leqslant \sum_{i=k+1}^{11}\left(2-x_{i}\right)
$$
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.78 On the blackboard, all natural numbers from 1 to 1988 are written. Operations $A$ and $B$ are alternately performed on these numbers, i.e., first $A$, then $B$, then $A$ again, then $B$, and so on. Operation $A$ involves subtracting the same natural number from each number on the blackboard (the number subtracted can be different in different $A$ operations); Operation $B$ involves erasing two numbers from the blackboard and writing down their sum. The process continues until after a certain $B$ operation, only one number remains on the blackboard. Given that this number is non-negative, find this number.
|
［Solution］Since operation $A$ does not decrease the number of numbers on the blackboard, while operation $B$ reduces the number of numbers by 1 each time, after operation $A$ and operation $B$ are each performed 1987 times, only 1 number remains on the blackboard.
Let $d_{k}$ be the natural number subtracted during the $k$-th operation $A$, where $k=1,2, \cdots$, 1987. Since in the $k$-th operation $A$, the sum of all numbers on the blackboard will decrease by $(1989 - k) d_{k}$, and operation $B$ does not change this sum, after alternating 1987 times of operation $A$ and operation $B$, the number written on the blackboard should be
$$
x=\sum_{k=1}^{1988} k-\sum_{k=1}^{1987}(1989-k) d_{k}=\sum_{k=1}^{1987}(1989-k)\left(1-d_{k}\right)+1 .
$$
Since $d_{k}$ is a natural number, $1-d_{k} \leqslant 0, k=1,2, \cdots, 1987$. Also, since $x \geqslant 0$, it must be that $d_{k}$ $=1, k=1,2, \cdots, 1987$. Thus, we get $x=1$, meaning the last remaining number on the blackboard is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Given positive numbers $a, b, x, y, z$ satisfying $a \leqslant x, y, z \leqslant b$, and the difference between the maximum and minimum values of $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is $\frac{9}{5}$, then $\frac{2 b}{a}=$ $\qquad$ .
|
$5$
|
5
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Let $\triangle A B C$ have internal angles $A, B, C$ with opposite sides $a, b, c$ respectively, and satisfy $a \cos B - b \cos A = \frac{3}{5} c$. Then the value of $\frac{\tan A}{\tan B}$ is
|
Solution: 4.
[Method 1] From the given and the cosine rule, we have
$$
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c \text {, i.e., } a^{2}-b^{2}=\frac{3}{5} c^{2} \text { . }
$$
Thus, $\frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\sin B \cos A}=\frac{a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}}{b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}}=\frac{c^{2}+a^{2}-b^{2}}{b^{2}+c^{2}-a^{2}}=\frac{\frac{8}{5} c^{2}}{\frac{2}{5} c^{2}}=4$.
[Method 2] As shown in Figure 2, draw $C D \perp A B$, with the foot of the perpendicular at $D$, then
$$
a \cos B=D B, b \cos A=A D .
$$
From the given, we have $D B-A D=\frac{3}{5} c$.
Also, $D B+D A=c$.
Solving these equations simultaneously, we get $A D=\frac{1}{5} c, D B=\frac{4}{5} c$.
Thus, $\frac{\tan A}{\tan B}=\frac{\frac{C D}{A D}}{\frac{A D}{D B}}=\frac{D B}{A D}=4$.
[Method 3] From the projection theorem, we have $a \cos B+b \cos A=c$.
Also, $a \cos B-b \cos A=\frac{3}{5} c$.
Solving these equations simultaneously, we get $a \cos B=\frac{4}{5} c, b \cos A=\frac{1}{5} c$.
Thus, $\frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\sin B \cos A}=\frac{a \cos B}{b \cos A}=\frac{\frac{4}{5} c}{\frac{1}{5} c}=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3 Let $S=\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$ and satisfy (1) $\left|A_{i}\right|=5, i=1,2, \cdots, k$; (2) $\left|A_{i} \cap A_{j}\right| \leqslant 2,1 \leqslant i<j \leqslant k$. Find the maximum value of $k$.
|
Solution: Let $k$ be the number of subsets satisfying conditions (1) and (2), and let $i$ belong to $x_{i}$ of these $k$ subsets, where $i=1,2, \cdots, 10$. If $i \in A_{j}, i \in A_{k}, j \neq k$, then $i$ is called a repeated pair. Thus, the number of repeated pairs caused by the number $i$ is $\mathrm{C}_{x_{i}}^{2}$. The total number of repeated pairs caused by the 10 elements in $S$ is $\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{10}}^{2}, x_{1}+x_{2}+\cdots+x_{10}=5 k$.
On the other hand, there are at most two repeated pairs between any two subsets, so there are at most $2 \mathrm{C}_{k}^{2}$ repeated pairs among the $k$ subsets. Therefore, we have
$$
\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \leqslant 2 \mathrm{C}_{k}^{2}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \\
=\frac{1}{2}\left\{x_{1}\left(x_{1}-1\right)+x_{2}\left(x_{2}-1\right)+\cdots+x_{10}\right. \\
\left.\left(x_{10}-1\right)\right\} \\
=\frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}\right)-\frac{1}{2}\left(x_{1}+x_{2}+\cdots\right. \\
\left.+x_{10}\right) \\
=\frac{1}{2}\left(x_{2}^{1}+x_{2}^{2}+\cdots+x_{10}^{2}\right)-\frac{5}{2} k \\
\geqslant \frac{1}{20}(5 k)^{2}-\frac{5}{2} k=\frac{5}{4} k(k-2) \\
\end{array}
$$
From (1) and (2), we get
$$
\frac{5}{4}(k-2) \leqslant k-1
$$
Solving (3) yields $k \leqslant 6$. This indicates that there are at most 6 subsets.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
(9) (14 points) Let the line $l: y=k x+m$ (where $k, m$ are integers) intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ at two distinct points $A, B$, and intersect the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ at two distinct points $C, D$. Does there exist a line $l$ such that the vector $\overrightarrow{A C}+\overrightarrow{B D}=0$? If it exists, how many such lines are there? If not, please explain the reason.
|
9 From $\left\{\begin{array}{l}y=k x+m, \\ \frac{x^{2}}{16}+\frac{y^{2}}{12}=1,\end{array}\right.$ eliminating $y$ and simplifying, we get
$$
\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0 .
$$
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, then $x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}$.
$$
\Delta_{1}=(8 k m)^{2}-4\left(3+4 k^{2}\right)\left(4 m^{2}-48\right)>0 \text {. }
$$
From $\left\{\begin{array}{l}y=k x+m, \\ \frac{x^{2}}{4}-\frac{y^{2}}{12}=1,\end{array}\right.$ eliminating $y$ and simplifying, we get
$$
\left(3-k^{2}\right) x^{2}-2 k m x-m^{2}-12=0 \text {. }
$$
Let $C\left(x_{3}, y_{3}\right)$ and $D\left(x_{4}, y_{4}\right)$, then $x_{3}+x_{4}=\frac{2 k m}{3-k^{2}}$.
$$
\Delta_{2}=(-2 k m)^{2}+4\left(3-k^{2}\right)\left(m^{2}+12\right)>0 .
$$
Since $\overrightarrow{A C}+\overrightarrow{B D}=0$, we have $\left(x_{4}-x_{2}\right)+\left(x_{3}-x_{1}\right)=0$, and at this time $\left(y_{4}-y_{2}\right)+\left(y_{3}-y_{1}\right)=0$. From $x_{1}+x_{2}=x_{3}+x_{4}$, we get
$$
-\frac{8 k m}{3+4 k^{2}}=\frac{2 k m}{3-k^{2}} \text {. }
$$
Thus, $2 k m=0$ or $-\frac{4}{3+4 k^{2}}=\frac{1}{3-k^{2}}$. Solving the above equation, we get $k=0$ or $m=0$. When $k=0$, from (1) and (2) we get $-2 \sqrt{3}<m<2 \sqrt{3}$. Since $m$ is an integer, the values of $m$ are $-3,-2,-1,0,1,2,3$. When $m=0$, from (1) and (2) we get $-\sqrt{3}<k<\sqrt{3}$. Since $k$ is an integer, the values of $k$ are $-1,0,1$. Therefore, there are 9 lines that satisfy the conditions.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given that point $P(x, y)$ lies on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$, find the maximum value of $2 x-y$.
untranslated text remains the same as the source text.
|
4. Let $\vec{a}=\left(\frac{x}{2}, \frac{y}{3}\right), \vec{b}=(4,-3)$, then
$$
1=\frac{x^{2}}{4}+\frac{y^{2}}{9}=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(2 x-y)^{2}}{4^{2}+(-3)^{2}}=\frac{(2 x-y)^{2}}{25} \text {. }
$$
$(2 x-y)^{2} \leqslant 25, 2x-y$'s maximum value is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Given: $\left\{\begin{array}{l}5 x+10 y \leqslant 30 \\ 2 x-y \leqslant 3 \\ x \cdot y \in N\end{array}\right.$ Then the maximum value of $x+y$ is ().
A. 6
B. 5
C. 4
D. 3
|
6. C
From $\left\{\begin{array}{l}5 x+10 y=30 \\ 2 x-y=3\end{array}\right.$ we get the intersection point $D\left(\frac{12}{5}, \frac{9}{5}\right)$. Substituting into $x+y$ gives: $x^{1}+y^{1}=\frac{21}{5}=4 \frac{1}{5}$, so $x+y \leqslant 4 \frac{1}{5}$. Since $x, y \in N$, we have $x+y \leqslant 4$. When $x=y=2$, $(x+y)_{\max }=4$ (Note: $x+y$ taking 4 may sometimes have more than one set of $x, y$).
|
4
|
Inequalities
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
Example 7: In $1^{2}, 2^{2}, 3^{2}, \cdots, 2005^{2}$, add a “+” or “-” sign before each number to make their algebraic sum the smallest non-negative number, and write out the equation.
|
Analysis: First, we explore the general pattern.
Solution: Since there are 1003 odd numbers among the 2005 numbers from $1^{2}$ to $2005^{2}$, the sum of these 2005 numbers is odd. Changing some of the “+” signs to “-” signs does not change the parity of the result, i.e., the algebraic sum of these 2005 numbers is odd. Therefore, the smallest non-negative number sought cannot be 0.
Since $n^{2}-(n+1)^{2}-(n+2)^{2}+(n+3)^{2}=4$, by placing “+” signs before the 1st, 4th, 6th, and 7th numbers and “-” signs before the 2nd, 3rd, 5th, and 8th numbers in a sequence of 8 consecutive perfect squares, the algebraic sum of these 8 numbers is 0. Given $2005=8 \times 250+5$, and $-1^{2}+2^{2}-3^{2}+4^{2}-5^{2}=-15$, we can take
$$
\begin{array}{l}
\left(6^{2}-7^{2}-8^{2}+9^{2}\right)+\left(10^{2}-11^{2}-12^{2}+13^{2}\right)+ \\
\left(14^{2}-15^{2}-16^{2}+17^{2}\right)+\left(18^{2}-19^{2}-20^{2}+21^{2}\right)=16 .
\end{array}
$$
From $22^{2}$ to $2005^{2}$, we can divide into 248 groups (each group consisting of 8 numbers), with the algebraic sum of each group of 8 numbers being 0. At this point, the algebraic sum of these 2005 numbers can be 1. That is, the smallest non-negative integer value that the algebraic sum can take is 1.
!Note: Using parity analysis can handle the problem as a whole.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Test D Let $M$ be the set of all polynomials of the form
$$
P(x)=a x^{3}+b x^{2}+c x+d \quad(a, b, c, d \in \mathbf{R}),
$$
and which satisfy $|p(x)| \leqslant 1$ for $x \in[-1,1]$. Prove: there must be a number $k$, such that for all $p(x) \in M$, we have $a \leqslant k$, and find the smallest $k$.
|
Prove that for any four values $x_{0}, x_{1}, x_{2}, x_{3}$ chosen in the interval $[-1,1]$ satisfying the condition
$$
0<\left|x_{i}-x_{j}\right| \leqslant 1 \quad(0 \leqslant i \leqslant j \leqslant 3)
$$
then by Lemma 1, we have
$$
|a| \leqslant \max _{0 \leqslant 3 \leqslant 3}\left\{\left|p\left(x_{i}\right)\right|\right\}\left(\sum_{i=1}^{3}\left|a_{i}\right|\right) \leqslant \sum_{i=1}^{3}\left|a_{i}\right| \text {. }
$$
Let $k=\sum_{i=1}^{3}\left|a_{i}\right|$. Under condition (1), for each $0 \leqslant i \leqslant 3$, we have
$$
\left|a_{i}\right|=\prod_{\substack{0 \leq i \leq 3 \\ j \neq i}}\left(x_{i}-x_{j}\right)^{-1} \geqslant 1 .
$$
Thus, $k \geqslant 4$, so $|a| \leqslant 4$. Furthermore, since $p_{0}(x)=4 x^{3}-3 x$ satisfies $p_{0}(-1)=-1, p_{0}(1)=1$, and at its critical points,
$$
p_{0}\left(-\frac{1}{2}\right)=1, p_{0}\left(\frac{1}{2}\right)=-1 \text {. }
$$
Therefore, $p_{0}(x) \in M$, and thus $|a|$ can reach 4, so $\min k=4$.
Clearly, when $x \in[-1,1]$, let $x=\sin \theta, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then $\left|p_{0}(x)\right|=|\sin 3 \theta| \leqslant 1$, hence $p_{0}(x) \in M$.
Clearly, $p_{0}(x)$ is the famous Tchebychev polynomial. For the $n$-th Tchebychev polynomial $T_{n}(x)$, when $-1 \leqslant x \leqslant 1$, we have $\left|T_{n}(x)\right| \leqslant 1$.
|
4
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
$32 \cdot 32$ A two-digit natural number has a one-digit number (including 0) inserted in the middle, turning it into a three-digit number. Some two-digit numbers, when a one-digit number is inserted in the middle, become a three-digit number that is exactly 9 times the original two-digit number. The number of such two-digit numbers is
(A) 1 .
(B) 4 .
(C) 10 .
(D) more than 10 .
("Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1993)
|
[Solution] If a "three-digit number" is 9 times a "two-digit number", then the sum of such a "three-digit number" and a "two-digit number" should be 10 times the "two-digit number".
Therefore, the unit digit of such a "two-digit number" can only be 0 and 5.
Also, because the inserted digit should be less than the unit digit of the "two-digit number", the unit digit of the "two-digit number" can only be 5, and the inserted unit digit can only be $0,1,2,3,4$.
And a "three-digit number" is a multiple of 9 $\Longleftrightarrow$ the sum of its digits can be divided by 9.
Thus, the inserted digit can only be $0,1,2,3$, and the corresponding "two-digit number" is $45,35,25$ and 15, four in total.
Hence, the answer is $(B)$.
|
4
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
3. Let the complex numbers $z_{1}=(2-a)+(1-b) \mathrm{i}, z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, z_{3}=(3-a)+(3-2 b) \mathrm{i}$, where $a, b \in \mathbf{R}$. When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ is minimized, $3 a+4 b=$ $\qquad$ .
|
3. 12 .
It is easy to find that $z_{1}+z_{2}+z_{3}=8+6 \mathrm{i}$, thus $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right|=10$. $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ achieves its minimum value if and only if
$$
\frac{2-a}{1-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{8}{6}
$$
Solving this, we get $a=\frac{7}{3}, b=\frac{5}{4}$. Therefore, $3 a+4 b=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. The equation
$$
\lg ^{2} x-[\lg x]-2=0
$$
|
3. From $[\lg x] \leqslant \lg x$ we get $\lg ^{2} x-\lg x-2 \leqslant 0$ which means $-1 \leqslant \lg x \leqslant 2$
When $-1 \leqslant \lg x<0$, we have $[\lg x]=-1$, substituting into the original equation gives $\lg x= \pm 1$, but $\lg x=1$ is not valid, $\therefore \lg x=-1, x_{1}=\frac{1}{10}$.
When $0 \leqslant \lg x<1$, we have $[\lg x]=0$, substituting into the original equation gives $\lg x= \pm \sqrt{2}$, both are not valid.
When $1 \leqslant \lg x<2$, we have $[\lg x]=1$, substituting into the original equation gives $\lg x= \pm \sqrt{3}$, but $\lg x=-\sqrt{3}$ is not valid, $\therefore \lg x=\sqrt{3}, x_{2}=10^{\sqrt{3}}$.
When $\lg x=2$, we get $x_{3}=100$.
$\therefore$ The original equation has 3 real roots.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. Given a function $f(x)$ defined on $\mathbf{R}$ whose graph is centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, and satisfies $f(x)=-f\left(x+\frac{3}{2}\right), f(-1)=1, f(0)=-2$, then the value of $f(1)+f(2)+\cdots+f(2008)$ is $\qquad$
|
12. From the fact that the graph of the function $f(x)$ is centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, we know that $f(x)=-f\left(-x-\frac{3}{2}\right)$.
Given that $f(x)=-f\left(x+\frac{3}{2}\right)$, it follows that $f\left(-x-\frac{3}{2}\right)=f\left(x+\frac{3}{2}\right)$.
Therefore, $f(-x)=f(x)$.
Hence, $f(x+3)=f\left[\left(x+\frac{3}{2}\right)+\frac{3}{2}\right]=-f\left(x+\frac{3}{2}\right)=f(x)$.
Thus, $f(x)$ is an even function with a period of 3.
Consequently, $f(1)=f(-1)=1, f(2)=f(-1+3)=f(-1)=1, f(3)=f(0)=-2$.
Therefore, $f(1)+f(2)+\cdots+f(2008)=669[f(1)+f(2)+f(3)]+f(2008)=f(2008)=f(1)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. Let the quadratic equation in $x$ be $2 x^{2}-t x-2=0$ with two roots $\alpha$ and
(Problem 13 interval) $\beta(\alpha<\beta)$.
(1) If $x_{1} 、 x_{2}$ are two different points in the interval $[\alpha, \beta]$, prove: $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4<0$;
(2) Let $f(x)=\frac{4 x-t}{x^{2}+1}$, and the maximum and minimum values of $f(x)$ in the interval $[\alpha, \beta]$ be $f_{\max }$ and $f_{\min }$, respectively. Let $g(t)=f_{\max }-f_{\min }$, find the minimum value of $g(t)$.
|
14. (I) From the conditions, we have: $\alpha+\beta=\frac{t}{2}, \alpha \beta=-1$.
Assume $x_{1}4\left(x_{1}-\alpha\right)\left(x_{2}-\beta\right)=4 x_{1} x_{2}-4\left(\alpha x_{2}+\beta x_{1}\right)+4 \alpha \beta=4 x_{1} x_{2}-2(\alpha+\beta)\left(x_{1}+\right.$ $\left.x_{2}\right)+4 \alpha \beta+2(\alpha-\beta)\left(x_{1}-x_{2}\right)>4 x_{1} x_{2}-2(\alpha+\beta)\left(x_{1}+x_{2}\right)+4 \alpha \beta=4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4$.
Thus, $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-40, f_{\min }=f(\alpha)<0$.
$$
g(t)=f(\beta)-f(\alpha)=|f(\beta)+| f(\alpha) \mid \geqslant 2 \sqrt{|f(\beta) \cdot| f(\alpha) \mid}=2 .
$$
Equality holds if and only if $f(\beta)=-f(\alpha)=2$, i.e., $\frac{8}{\sqrt{t^{2}+16}+t}=2$, which simplifies to $t=0$.
Therefore, the minimum value of $g(t)$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. When $s$ and $t$ take all real values, the minimum value that $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ can reach is $\qquad$
|
Solve the system $\left\{\begin{array}{l}x=s+5, \\ y=s\end{array} \Rightarrow x-y-5=0,\left\{\begin{array}{l}x=3|\cos t|, \\ y=2|\sin t|\end{array} \Rightarrow \frac{x^{2}}{9}+\frac{y^{2}}{4}=1(x, y \geqslant 0)\right.\right.$, thus the original expression represents the square of the minimum distance from points on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ in the first quadrant to the line $x-y-5=0$. Let $P(3 \cos t, 2 \sin t), t \in\left[0, \frac{\pi}{2}\right]$,
then $d=\frac{|3 \cos t-2 \sin t-5|}{\sqrt{2}}=\frac{|\sqrt{13} \cos (t+\varphi)-5|}{\sqrt{2}}=\frac{5-\sqrt{13} \cos (t+\varphi)}{\sqrt{2}}$, where $\cos \varphi=\frac{3}{\sqrt{13}}, \sin \varphi=\frac{2}{\sqrt{13}}, y=\cos x$ is monotonically decreasing on $\left[\varphi, \frac{\pi}{2}+\varphi\right]$, thus when $t=0$, $d$ achieves its minimum value $\sqrt{2}$.
Therefore, the minimum value of $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
16. The function $f(x)=a x^{2}+b x+c, a, b, c \in \mathbf{Z}$, and $f(x)$ has two distinct roots in $(0,1)$. Find the smallest positive integer $a$ that satisfies the above conditions.
|
16. Let the roots of the equation $f(x)=0$ be $r$ and $s(0<s<1<r)$. Then, $f(x)=a(x-r)(x-s)=a[x^2-(r+s)x+rs]$. Since $0<s<1<r$, we have $f(0)=c=ars>0, f(1)=a+b+c=a(1-r)(1-s)>$ 0. Given $0<s<1<r, c \in \mathbf{Z}$, then $c \geqslant 1$, i.e., $f(0) \geqslant 1$. Also, $a+b+c \geqslant 1$, i.e., $f(1) \geqslant 1$, so $1 \leqslant f(0) f(1)=a^{2} r(1-r) \cdot s(1-s) \leqslant \frac{a^{2}}{16}$. Since $r \neq s$, $r$ and $s$ cannot both be $\frac{1}{2}$, so the equality in the above inequality does not hold. Therefore, $a^{2}>16$, and the smallest positive integer $a=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. As shown in the figure, the graph of the linear function $y=k x+b$ passes through the point $P(1,3)$, and intersects the positive halves of the $x$-axis and $y$-axis at points $A, B$, respectively. Point $O$ is the origin. The minimum value of the area of $\triangle A O B$ is $\qquad$
|
$6$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (10 points) Four people went to a bookstore to buy books. Each person bought 4 different books, and every two people have exactly 2 books in common. Therefore, these 4 people bought at least $\qquad$ kinds of books.
|
【Analysis】From simple properties: If there are only two people, to meet the requirements of the problem, there must be 6 different books. Number these 6 books: $1, 2, 3, 4, 5, 6$. Suppose A buys $1, 2, 3, 4$, and B buys $1, 2, 5, 6$. At this point, C arrives. To meet the requirements, C can choose not to buy any other books, and he buys the books numbered $3, 4, 5, 6$, which also meets the requirements. Note that when three people only buy 6 books, after A and B's book choices are determined, C's book choices are unique. C cannot buy the books that both A and B have bought, so he must buy one book that is "unique" to him. To make the problem easier to understand, use a table (see the solution section).
【Solution】If there are only two people, to meet the requirements, there must be 6 different books, numbered $1, 2, 3, 4, 5, 6$. Suppose A buys $1, 2, 3, 4$, and B buys $1, 2, 5, 6$. At this point, C arrives. To meet the requirements, C can choose not to buy any other books, and he buys the books numbered $3, 4, 5, 6$, which also meets the requirements. Note that when three people only buy 6 books, after A and B's book choices are determined, C's book choices are unique. C cannot buy the books that both A and B have bought, so he must buy one book that is "unique" to him. Use a table:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline A & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ & & \\
\hline B & $\checkmark$ & $\checkmark$ & & & $\checkmark$ & $\checkmark$ \\
\hline C & & & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ \\
\hline
\end{tabular}
At this point, D arrives. From C's perspective, he must have at least one "unique" book, so when there are 4 people, there must be at least 7 books.
That is, A buys the books numbered $(1, 2, 3, 4)$, B buys the books numbered $(1, 2, 5, 6)$, C buys the books numbered $(3, 4, 5, 6)$, and D can buy $(1, 3, 5, 7)$. Therefore, when there are 4 people, they must buy at least $(1, 2, 3, 4, 5, 6, 7)$, which is 7 different books.
Thus, the answer is: 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The number of solutions for $x$ in the interval $(-\pi, \pi)$ of the equation $(a-1)(\sin 2 x+\cos x)+(a-1)(\sin x-\cos 2 x)=0(a<0)$ is ( ).
A. 2
B. 3
C. 4
D. 5 or more than 5
|
3. C. Reason: The original equation can be transformed into
$$
\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)\left[\cos \left(\frac{3 x}{2}-\frac{\pi}{4}\right)+\frac{a+1}{a-1} \sin \left(\frac{3 x}{2}-\frac{\pi}{4}\right)\right]=0 .
$$
(1) The equation $\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)=0$ has a solution $x=-\frac{\pi}{2}$ in $(-\pi, \pi)$.
(2) For the equation $\cos \left(\frac{3 x}{2}-\frac{\pi}{4}\right)+\frac{a+1}{a-1} \sin \left(\frac{3 x}{2}-\frac{\pi}{4}\right)=0$:
When $a=-1$, it becomes $\cos \left(\frac{3 x}{2}-\frac{\pi}{4}\right)=0$, yielding $x=-\frac{5 \pi}{6},-\frac{\pi}{6}, \frac{\pi}{2}$;
When $a \neq-1$, it can be transformed into $\tan \left(\frac{3 x}{2}-\frac{\pi}{4}\right)=\frac{1-a}{a+1}$. * When $a<0$ and $a \neq 1$, $\frac{a-1}{a+1} \in(-\infty,-1) \cup$ $(1,+\infty)$. And $y=\tan \left(\frac{3 x}{2}-\frac{\pi}{4}\right)$ has 3 periods in $(-\pi, \pi)$, so equation * always has 3 solutions, none of which is $-\frac{\pi}{2}$. Therefore, when $a<0$, the original equation has 4 solutions.
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
8. There is a pair of tightly adjacent rubber drive wheels, each with a mark line passing through the center (as shown in the figure below).
The radius of the driving wheel is 105 cm, and the radius of the driven wheel is 90 cm. When they start to rotate, the mark lines on both wheels are in a straight line. Question: How many turns does the driving wheel need to make at least so that the mark lines on both wheels are in a straight line again?
|
8. The driving gear has rotated 3 turns
8.【Solution】The least common multiple of 105 and 90 is $630. 630 \div 105=6$,
So the driving gear has rotated 6 half-turns, which is 3 turns, and the marks on both gears are aligned again
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1.15 ** In $\triangle A B C$, it is known that $\sin A \cdot \cos ^{2} \frac{C}{2}+\sin C \cdot \cos ^{2} \frac{A}{2}=\frac{3}{2} \sin B$. Find the value of $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}$.
|
From the given, we have
$$
\begin{array}{c}
\sin A \cdot \frac{1+\cos C}{2}+\sin C \cdot \frac{1+\cos A}{2}=\frac{3}{2} \sin B \\
\sin A+\sin C+\sin A \cdot \cos C+\cos A \cdot \sin C=3 \sin B \\
\sin A+\sin C+\sin (A+C)=3 \sin B \\
\sin A+\sin C=2 \sin B \\
2 \sin \frac{A+C}{2} \cdot \cos \frac{A-C}{2}=4 \sin \frac{B}{2} \cdot \cos \frac{B}{2}
\end{array}
$$
Therefore, $\cos \frac{A-C}{2}=2 \sin \frac{B}{2}$, which means $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. Let the left focus of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ be $F$, and let the line $l$ passing through $(1,1)$ intersect the ellipse at points $A$ and $B$. When the perimeter of $\triangle FAB$ is maximized, the area of $\triangle FAB$ is $\qquad$
|
11.3.
Let $F_{1}$ be the right focus.
Notice,
$$
\begin{array}{l}
F A+F B+A B \\
=2 a-F_{1} A+2 a-F_{1} B+A B \\
=4 a-\left(F_{1} A+F_{1} B-A B\right) \leqslant 4 a,
\end{array}
$$
The equality holds when $l$ passes through point $F_{1}$.
Thus, $l$ is perpendicular to the $x$-axis, at this moment, $S_{\triangle F A B}=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. For positive integer $a$ and integers $b, c$, in the rectangular coordinate system $O-xyz$, points $O(0,0,0)$, $A(a, b, c)$, and $B\left(x^{2}, x, 1\right)$ satisfy $=\frac{\pi}{2}$. The real number $x$ has exactly two distinct real solutions $x_{1}, x_{2} \in (0,1)$. The minimum value of $a$ is $\qquad$.
|
4. 5 Detailed Explanation: For the equation $a x^{2}+b x+c=0$ to have two distinct real roots $x_{1}, x_{2}$ in the open interval $(0,1)$, assume $0<c>0, a \geqslant 2$; from $x_{1}+x_{2}=-\frac{b}{a}>0$ we know $b<0$, and $x_{1} x_{2}=\frac{c}{a}>0$, thus $c>0$. Therefore, $4 a c+1 \leqslant(a+c-1)^{2}$, which simplifies to $a^{2}+c^{2}-2 a c-2 a-2 c \geqslant 0$. If $a=2$, then $c^{2}-6 c \geqslant 0$, leading to $c \geqslant 6$, a contradiction! If $a=3$, then $c^{2}-8 c+3 \geqslant 0$, leading to $c \geqslant 8$, a contradiction! If $a=4$, then $c^{2}-10 c+8 \geqslant 0$, leading to $c \geqslant 10$, a contradiction! If $a=5$, then $c^{2}-12 c+15 \geqslant 0$, leading to $c=1$, or $c \geqslant 11$, so we can take $c=1, b=-5$, solving the equation $5 x^{2}-5 x+1=0$, we get $\left(x_{1}, x_{2}\right)=\left(\frac{5+\sqrt{5}}{10}, \frac{5-\sqrt{5}}{10}\right)$. Therefore, $a_{\min }=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. (5 points) A team of 23 people participated in the 16th Guangzhou Asian Games. They were arranged in descending order of height. The average height of the top 5 players is 3 cm more than the average height of the top 8 players; the average height of the last 15 players is 0.5 cm less than the average height of the last 18 players. How much more is the average height of the top 8 players compared to the average height of the last 15 players?
|
【Answer】Solution: The average height of positions $6 \sim 8$ is less than the first 5 positions by: $3 \div 3 \times 8=8$ (cm); therefore, the average height of positions $6 \sim 8$ is more than the average height of positions $9 \sim 23$ by $0.5 \div 3 \times 18=3$ cm. The average height of the first 8 positions is more than the average height of positions $9 \sim 23$ by $8-3+3=8$ cm.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. (20 points) Given a function $f(x)$ defined on $[0,1]$, $f(0)=0, f(1)=1$, and satisfying the following conditions:
(a) For any $x \in[0,1]$, $f(x) \geqslant 0$;
(b) For any two real numbers $x_{1}, x_{2}$ satisfying $x_{1} \geqslant 0, x_{2} \geqslant 0, x_{1}+x_{2} \leqslant 1$, $f\left(x_{1}+x_{2}\right) \geqslant$ $f\left(x_{1}\right)+f\left(x_{2}\right)$.
Find the smallest positive number $c$, such that for any function $f(x)$ satisfying the above conditions and any real number $x \in[0,1]$, $f(x) \leqslant c x$.
(Feng Yuefeng)
|
11. For any $0 \leqslant x \leqslant y \leqslant 1$, we have $0 \leqslant y-x \leqslant 1$. By condition (a), we have $f(y-x) \geqslant 0$. Therefore, by condition (b), we have
$$
f(y)=f(y-x+x) \geqslant f(y-x)+f(x) \geqslant f(x) .
$$
Thus, for any $0 \leqslant x \leqslant y \leqslant 1$, we have
$$
f(x) \leqslant f(y) \leqslant f(1) \leqslant 1 .
$$
Moreover, by condition (b), for any $x \in\left[0, \frac{1}{2}\right]$, we have
$$
f(2 x) \geqslant f(x)+f(x)=2 f(x),
$$
Combining this with mathematical induction, we can see that for any positive integer $k$ and $x \in\left[0, \frac{1}{2^{k}}\right]$, we have
$$
f\left(2^{k} x\right) \geqslant 2^{k} f(x) \text {. }
$$
Next, we prove that for any $x \in[0,1]$, we have $f(x) \leqslant 2 x$.
First, when $x=0$, $f(0)=0 \leqslant 2 \times 0$. So when $x=0$, $f(x) \leqslant 2 x$ holds.
Assume there exists $x_{0} \in(0,1]$, such that $f\left(x_{0}\right)>2 x_{0}$. Since $x_{0} \neq 0$, there must exist a positive integer $k$ such that $2^{k-1} x_{0} \leqslant 1$, but $2^{k} x_{0}>1$. For such $k$, by equation (2), we have
$$
f\left(2^{k-1} x_{0}\right) \geqslant 2^{k-1} f\left(x_{0}\right)>2^{k-1}\left(2 x_{0}\right)=2^{k} x_{0}>1 .
$$
But $2^{k-1} x_{0} \leqslant 1$, by equation (1), we have
$$
f\left(2^{k-1} x_{0}\right) \leqslant f(1)=1,
$$
which is a contradiction.
Therefore, for any $0 \leqslant x \leqslant 1$, $f(x) \leqslant 2 x$, so $c=2$ is valid.
Next, we prove that $c \geqslant 2$.
Assume $c < 2$.
Notice that the range of $f(x)$ is $[0,1]$. To make $f(t)$ as large as possible, we can take $f(t)=1$, which by the definition of $f(x)$, must mean $\frac{1}{2} c t < 1$, i.e., $t < \frac{2}{c}$. Since $c < 2$, we have $t < 1$. Thus, $x=t$ does not satisfy $f(x) \leqslant c x$, which is a contradiction.
In conclusion, the minimum value of $c$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 14 Given $n$ points $p_{1}, p_{2}, \cdots, p_{n}$ in a plane, no three of which are collinear, each point $p_{i}(1 \leqslant i \leqslant n)$ is arbitrarily colored red or blue. Let $S$ be the set of triangles whose vertex set is $\left\{p_{1}, p_{2}, \cdots, p_{n}\right\}$, and has the property: for any two line segments $p_{i} p_{j}$ and $p_{u} p_{v}$, the number of triangles in $S$ that have $p_{i} p_{j}$ as a side is the same as the number of triangles in $S$ that have $p_{u} p_{v}$ as a side. Find the smallest $n$ such that $S$ always contains two triangles, each of whose vertices have the same color.
(48th IMO China National Training Team Test Question)
|
Let the number of triangles in $S$ with $p_{i} p_{j}$ as an edge be $k$, then $k$ is a positive integer and is independent of $i, j$. Since there are $C_{n}^{2}$ edges $p_{i} p_{j}$, there are $k \cdot C_{n}^{2}$ triangles (including repeated counts). Each triangle has three edges, and each triangle is counted 3 times in the above count, so $|S|=\frac{1}{3} k C_{n}^{2}$. Let the number of triangles in $S$ with vertices of the same color be $x$, then the number of triangles in $S$ with vertices not all the same color is $\frac{1}{3} k C_{n}^{2}-x$. Each triangle with vertices not all the same color produces two edges with different colored endpoints (referred to as different colored edges), so $S$ has $2\left(\frac{1}{3} k C_{n}^{2}-x\right)$ different colored edges.
On the other hand, let $p_{1}, p_{2}, \cdots, p_{n}$ have $n_{1}$ red points and $n_{2}$ blue points, with $n_{1}+n_{2}=n$. By assumption, each different colored edge appears $k$ times in the triangles of $S$, so there are $k n_{1} n_{2}$ different colored edges, thus
$$
2\left(\frac{1}{3} k C_{n}^{2}-x\right)=k n_{1} n_{2} .
$$
Solving for $x$ gives
$$
\begin{aligned}
x & =\frac{k}{6}\left(2 C_{n}^{2}-3 n_{1} n_{2}\right)=\frac{k}{6}\left(n^{2}-n-3 n_{1} n_{2}\right) \\
& \geqslant \frac{k}{6}\left[n^{2}-n-3\left(\frac{n_{1}+n_{2}}{2}\right)^{2}\right]=\frac{k}{6} n\left(\frac{n}{4}-1\right)=\frac{k}{24} n(n-4) .
\end{aligned}
$$
Therefore, when $n \geqslant 8$, since $k \geqslant 1$, we have
$$
x \geqslant \frac{n}{24}(n-4) \geqslant \frac{8 \times 4}{24}=\frac{4}{3}>1 \text {. }
$$
Thus, $x \geqslant 2$, so $n \geqslant 8$.
When $n=7$, the conclusion is not necessarily true. For example, if points $1,2,4$ are colored red and points $3,5,6,7$ are colored blue, then the set of triangles $\{1,2,4\},\{2,3,5\},\{3,4,6\},\{4,5,7\},\{5,6,1\},\{6,7,2\},\{7,1,3\}$ meets the requirement (each edge appears in one triangle), but there is only one same-colored triangle $\{1,2,4\}$.
Therefore, the smallest positive integer $n=8$.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. Let the plane point sets be
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(y-x)\left(y-\frac{18}{25 x}\right) \geqslant 0\right.\right\}, \\
B=\left\{(x, y) \mid(x-1)^{2}+(y-1)^{2} \leqslant 1\right\} .
\end{array}
$$
If $(x, y) \in A \cap B$, find the minimum value of $2 x-y$.
|
Let $z=2 x-y$. Then $y=2 x-z,-z$ represents the y-intercept of the line $y=2 x-z$.
It is easy to know that when the line $y=2 x-z$ passes through point $P$ in the shaded area, $z=2 x-y$ reaches its minimum value.
Point $P$ is on the circle $(x-1)^{2}+(y-1)^{2}=1$, with coordinates $(1+\cos \theta, 1+\sin \theta)$. From the graph, we know $\theta \in\left(\frac{\pi}{2}, \pi\right)$.
Since point $P$ is on the curve $y=\frac{8}{25 x}$, we have,
$$
\begin{array}{l}
(1+\cos \theta)(1+\sin \theta)=\frac{18}{25} \\
\Rightarrow \sin \theta \cdot \cos \theta+\sin \theta+\cos \theta+\frac{7}{25}=0 .
\end{array}
$$
Let $\sin \theta+\cos \theta=t$. Then
$$
\sin \theta \cdot \cos \theta=\frac{1}{2}\left(t^{2}-1\right) \text {. }
$$
Substituting into equation (1) we get
$$
\begin{array}{l}
\frac{1}{2}\left(t^{2}-1\right)+t+\frac{7}{25}=0 \\
\Rightarrow t=\frac{1}{5} \text { or } t=-\frac{11}{5}(\text { discard }) \\
\Rightarrow \sin \theta+\cos \theta=\frac{1}{5} \\
\Rightarrow \sin \theta=\frac{4}{5}, \cos \theta=-\frac{3}{5} \\
\Rightarrow P\left(\frac{2}{5}, \frac{9}{5}\right), z_{\min }=2 \times \frac{2}{5}-\frac{9}{5}=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
62. Arrange 7 small cubes to form a "T" shape, then paint the surface red, and finally separate the small cubes. The number of small cubes with 3 faces painted red is $\qquad$.
Arrange 7 small cubes to form a "T" shape, then paint the surface red, and finally separate the small cubes. The number of small cubes with 3 faces painted red is $\qquad$.
|
Answer: 1
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{n}=(\sqrt{2}+1)^{n}-(\sqrt{2}-1)^{n}(n \in \mathbf{N})$, and $[x]$ represents the greatest integer not exceeding the real number $x$, then the unit digit of $\left[a_{2017}\right]$ is
A. 2
B. 4
C. 6
D. 8
|
Let $b_{n}=(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n} \Rightarrow b_{1}=2, b_{2}=6$, and $x=1+\sqrt{2} \Rightarrow(x-1)^{2}=2$ $\Rightarrow x^{2}-2 x-1=0$, thus $b_{n+2}=2 b_{n+1}+b_{n}$. The pattern of the last digits is $2,6,4,4,2,8,8,4,6,6,8,2$, $2,6, \cdots$, with a period of 12, hence the last digit of $b_{2017}$ is 2, so the answer is $A$.
|
2
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
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