problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
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B1. John walks from home to school with a constant speed, and his sister Joan bikes twice as fast. The distance between their home and school is $3 \mathrm{~km}$. If Joan leaves home 15 minutes after John then they arrive to school at the same time. What is the walking speed (in $\mathrm{km} / \mathrm{h}$ ) of John? | Solution 1: Let $x(\mathrm{~km} / \mathrm{h})$ be the speed of John. Then the speed of Joan is $2 x$. The time John spends for walking to school is $3 / x$ hours. The time Joan spends for biking to school is $3 /(2 x)$ hours. Since she leaves home $0.25 \mathrm{~h}$ after John and arrives to school at the same time, we... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.32 Given a finite set $K_{0}$ in the plane (or in space), add to $K_{0}$ all points obtained by reflecting a point in $K_{0}$ about another point in $K_{0}$, to get the set $K_{1}$. Similarly, we can obtain $K_{2}$ from $K_{1}$, $K_{3}$ from $K_{2}$, and so on.
(1) Suppose the set $K_{0}$ consists of two points $A$ ... | [Solution] (1) It is easy to see that the set $K_{n}$ consists of all points on the line $AB$ that have an integer distance from $A$ and whose distance from the midpoint of the segment $AB$ does not exceed $-\frac{1}{2} \times 3^{n}$, for $n=1,2, \cdots$. Since $\frac{1}{2}\left(3^{6}+1\right)=3651000$, the smallest va... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 11 For non-negative integers $n, f(n)$ is determined by $f(0)=0, f(1)=1, f(n)$ $=f\left(\left[\frac{1}{2} n\right]\right)+n-2\left[\frac{1}{2} n\right]$, find the maximum value of $f(n)$ when $0 \leqslant n \leqslant 1997$. | Solution: Represent $n$ in binary, let $n=\left(a_{k} a_{k-1} \cdots a_{1}\right)_{2}$, where $a_{i}(i=1,2$, $\cdots, k$ ) is 0 or 1, then we have
$$
\left[\frac{1}{2} n\right]=\left(a_{k} a_{k-1} \cdots a_{2}\right)_{2}, n-2\left[\frac{1}{2}\right]=a_{1} \text {. }
$$
Thus, $f(n)=f\left(\left(a_{k} a_{k-1} \cdots a_{... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (1963 24th Putnam Mathematical Competition) Suppose $x^{2}-x+a$ divides $x^{13}+x+90$, determine the value of the positive integer $a$.
untranslated text remains the same as the source text. | 8. Let $x^{13}+x+90=\left(x^{2}-x+a\right) q(x)$, where $a$ is an integer, and $q(x)$ is a polynomial with integer coefficients.
Let $x=-1,0,1$, then
$$
\left\{\begin{array}{l}
(a+2) q(-1)=88, \\
a q(0)=90, \\
a q(1)=92 .
\end{array}\right.
$$
(2)
(3)
From (2) and (3), we get
$a[q(1)-q(0)]=2$, so $a \mid 2$.
Therefore... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1 Calculate $\lg \left[2 \cdot\left(\log _{3} 2 \cdot \log _{2} 3\right)\right]+\lg 5+5^{\log _{5} 3}$.
| (1) 4 | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Given a regular tetrahedron $P-ABC$ with the side length of the base being 6 and the side length of the lateral edges being $\sqrt{21}$. Then the radius of the inscribed sphere of the tetrahedron is $\qquad$. | 7. 1.
Let $P O \perp$ plane $A B C$ at point $O$. Then $O$ is the center of the equilateral $\triangle A B C$. Connect $A O$ and extend it to intersect $B C$ at point $D$, and connect $P D$. Thus, $D$ is the midpoint of $B C$.
It is easy to find that $P D=2 \sqrt{3}, O D=\sqrt{3}, P O=3$.
Let the radius of the inscrib... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
19.18 The ratio of the interior angles of two regular polygons is $3: 2$. The number of such regular polygons is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(E) infinitely many.
(13th American High School Mathematics Examination, 1962) | [Solution] Let $N, n$ represent the number of sides of the two regular polygons, respectively. Therefore, the relationship between their angle measurements can be expressed as
$$
\frac{(N-2) 180^{\circ}}{N}=\frac{3}{2} \cdot \frac{(n-2) 180^{\circ}}{n},
$$
then $N=\frac{4 n}{6-n}$.
Since $n, N$ are both positive integ... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
71. One day, at the entrance of the China Pavilion at the Shanghai World Expo, 945 visitors had already begun waiting to enter. At this time, several people were still coming to the entrance every minute to prepare for entry. In this way, if 4 ticket-checking gates were opened, all waiting visitors could enter in 15 mi... | Reference answer: 11 | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
94. A, B, and C are playing table tennis. Each match involves two players, and the loser is replaced by the third player. This continues. When A has played 9 matches and B has played 6 matches, what is the maximum number of matches C could have played?
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。
94. A, B, and C are playing t... | Answer: 11 | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (8 points) Selling a product with a profit margin of $25 \%$, if you want to increase the profit margin to $40 \%$, then the selling price should be increased by $\qquad$ $\%$.
| 【Answer】Solution: $1+25\% = 125\%$
$$
\begin{array}{l}
1+40\% = 140\% \\
\quad(140\% - 125\%) \div 125\% \\
=15\% \div 125\% \\
=12\%
\end{array}
$$
Answer: The selling price should be increased by $12\%$.
Therefore, the answer is: 12. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$1 \cdot 200$ Write the number $1234567891011 \cdots \cdots 19941995$ on the blackboard, forming the integer $N_{1}$, erase the digits in the even positions of $N_{1}$, the remaining digits form the integer $N_{2}$, erase the digits in the odd positions of $N_{2}$, the remaining digits form the integer $N_{3}$, erase t... | [Solution] Since
$$
9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873,
$$
the integer $N_{1}$ consists of 6873 digits, which are numbered from left to right as 1, $2,3, \cdots, 6873$. We examine the set of these numbers $\{1,2,3, \cdots, 6873\}$.
Erasing the digits at even positions from $N_{1}$ is equivalent to ... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 7 (2003 Belarusian Mathematical Olympiad) Given a convex pentagon $A B C D E$ satisfying $A B=B C, C D=$ $D E, \angle A B C=150^{\circ}, \angle C D E=30^{\circ}, B D=2$. Find the area of pentagon $A B C D E$. | Let $K$ be the point symmetric to $C$ with respect to the line $B D$, then
$$
\begin{array}{l}
B K=B C=B A \\
D K=D C=D E
\end{array}
$$
Construct the angle bisectors of $\angle A B K$ and $\angle E D K$, and let them intersect at point $M$. Thus, $B M$ is the perpendicular bisector of $A K$, and $D M$ is the perpendi... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (5 points) If $\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n}}>\frac{315}{412}$ ( $n$ is a natural number greater than 0), then the smallest value of $n$ that satisfies the condition is . $\qquad$ | 4. (5 points) If $\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{\mathrm{n}}}>\frac{315}{412}$ ( $n$ is a natural number greater than 0), then the smallest value of $n$ that satisfies the condition is $\qquad$
【Solution】Solution: When $n=1$, the left side of the inequality equals $\frac{1}{2}$, which is less than $\frac{31... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9. Given $f(x)=\left\{\begin{array}{ll}x+\frac{1}{2} & \left(0 \leqslant x \leqslant \frac{1}{2}\right) \\ 2(1-x) & \left(\frac{1}{2}<x \leqslant 1\right)\end{array}\right.$, define $f_{1}(x)=f(x), f_{n}(x)=f\left(f_{n-1}(x)\right)$, $n \in \mathbf{N}$, prove: the set $\left\{x \mid f_{15}(x)=x, x \in[0,1]\right\}$ con... | 9. Solution: Let $B=\left\{x \mid f_{15}(x)=x, x \in[0,1]\right\}$, when $x \in A=\left\{\frac{2}{15}, \frac{19}{30}, \frac{11}{15}, \frac{8}{15}, \frac{14}{15}\right\}$, we have $f_{n+5}(x)=$ $f_{n}(x)$, which means $f_{15}(x)=x$, so $x \in B$; when $f(x)=x$, $x=\frac{2}{3}$, in this case, of course, $f_{15}(x)=x$, so... | 9 | Algebra | proof | Yes | Yes | olympiads | false |
1. If $f(x)=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$, then the maximum value of $f(x)$ is $\qquad$ . | $-1.11$.
From the fact that $f(x)$ is defined, we know
$$
0 \leqslant x \leqslant 13 \text {. }
$$
Then $\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
$$
\begin{array}{l}
=\sqrt{\left(6 \sqrt{\frac{x+27}{36}}+2 \sqrt{\frac{13-x}{4}}+3 \sqrt{\frac{x}{9}}\right)^{2}} \\
\leqslant \sqrt{(6+2+3)\left(6 \times \frac{x+27}{36}+2 \times... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7、The denominator is a two-digit number, the numerator is 1, and the fraction can be converted into a finite decimal. There are $\qquad$
The fraction has a two-digit denominator, a numerator of 1, and can be converted into a finite decimal. There are $\qquad$ | 【Analysis】A simplest fraction, if it can be simplified to a simplest fraction, then its denominator can only contain prime factors $2$ and $5$. Fractions with a two-digit denominator and a numerator of 1 are clearly simplest fractions.
Therefore, the denominator of these fractions, when factored into prime factors, s... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14.4.1 * Find all integers $k \geqslant 0$, such that the sequence $k+1, k+2, \cdots, k+10$ contains the maximum number of prime numbers. | When $k=1$, the sequence $k+1, k+2, k+3, \cdots, k+10$ (1) contains 5 prime numbers: 2, 3, 5, 7, 11. When $k=0, k=2$, the sequence (1) only contains 4 prime numbers. If $k \geqslant 3$, then the sequence (1) does not contain the prime number 3. It is known that in any three consecutive odd numbers, one is divisible by ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 5 Find all integers $n$, such that
$$
n^{4}+6 n^{3}+11 n^{2}+3 n+31
$$
is a perfect square.
$(2004$, Western Mathematical Olympiad) | Let $A=n^{4}+6 n^{3}+11 n^{2}+3 n+31$. Then
$$
\begin{array}{l}
A=\left(n^{2}+3 n+1\right)^{2}-3(n-10) \\
=\left(n^{2}+3 n+2\right)^{2}-\left(2 n^{2}+9 n-27\right) .
\end{array}
$$
When $n \geqslant 3$ or $n \leqslant-7$,
$$
\left(n^{2}+3 n\right)^{2}<A<\left(n^{2}+3 n+2\right)^{2} \text {. }
$$
Therefore, $A=\left(n... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A regular tetrahedral box with an edge length of 6 contains a smaller regular tetrahedron. If the smaller tetrahedron can rotate freely within the box, then the maximum edge length of the smaller tetrahedron is $\qquad$ . | 荅靼 2 .
The small regular tetrahedron can rotate freely inside the box, so the maximum edge length of the small regular tetrahedron is when it is inscribed in the sphere that is inscribed in the large regular tetrahedron. Let the radius of the circumscribed sphere of the large regular tetrahedron be $R$, and the radius ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 6 Find all positive integers $k$ such that for any positive numbers $a, b, c$ satisfying the inequality
$k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)$, there must exist a triangle with side lengths $a, b$, and $c$.
(2002 National Girls' Olympiad) | To find the positive integer $k$, we can first determine the upper and lower bounds of $k$, and then prove that the found $k$ satisfies the conditions.
From $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, we know $k>5$. Taking $a, b, c$ that cannot form a triangle, such as $a=$ $b=1, c=2$, then $k(1 \cdot 1+1 \cdot 2+2 \cdo... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Question 33: Let the function $f(x)=a x^{2}+b x+c$ ($a$, $b$, $c \in \mathbb{R}$), satisfying $\max _{x \in[0,1]}|f(x)| \leq 1$. Let $g(x)=$ $cx^{2}+bx+a$. Try to find the maximum possible value of $\max _{x \in[0,1]}|g(x)|$.
保留源文本的换行和格式如下：
Question 33: Let the function $f(x)=a x^{2}+b x+c$ ($a$, $b$, $c \in \mathbb{... | $$
\begin{array}{l}
g(x)=f(0) \cdot x^{2}+\left(4 f\left(\frac{1}{2}\right)-3 f(0)-f(1)\right) x+\left(2 f(1)-4 f\left(\frac{1}{2}\right)+2 f(0)\right) \\
=f(0) \cdot\left(x^{2}-3 x+2\right)+4 f\left(\frac{1}{2}\right)(x-1)-f(1) \cdot(x-2)
\end{array}
$$
Therefore, when $x \in [0,1]$, we have
$$
\begin{array}{l}
|g(x)... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (8 points) The advertising slogan for battery $A$ is “One is stronger than six”. This means that battery $A$ is more durable than other batteries. Let's assume that the power of one $A$ battery is 6 times that of a $B$ battery. There are two clocks with the same power consumption rate. Now, 2 $A$ batteries are insta... | 【Answer】Solution: According to the analysis, since both are operating normally and consuming power at the same rate, the time it takes for Clock A to deplete its battery is 6 times that of Clock B. Therefore, Clock A can operate normally for: $6 \times 2=12$ months, which is $12-2=10$ months longer than Clock B. The an... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. As shown in the figure, the axial section of the reflector on the spotlight bulb of a movie projector is part of an ellipse, with the filament located at focus $F_{2}$, and the distance between the filament and the vertex $A$ of the reflector $|F_{2} A|=1.5$ cm, and the latus rectum of the ellipse $|B C|=5.4$ cm. T... | 14. According to the optical properties of the elliptical mirror, light rays emitted from one focus of the ellipse are reflected and converge at the other focus. Therefore, the lamp should be placed at focus $F_{1}$ to obtain the strongest light. For this, the focal distance needs to be determined.
Assume the equation... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Given that $P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and $(x-1)^{2}+y^{2}=1$ is the incircle of $\triangle P B C$. Then the minimum value of $S_{\triangle P B C}$ is $\qquad$ | 8. 8 .
Let point $P\left(2 t^{2}, 2 t\right)$.
Since it is an incircle, we know $|t|>1$.
Let the slope of the line passing through point $P$ be $k$. Then the equation of the line is
$$
y-2 t=k\left(x-2 t^{2}\right) \text {. }
$$
Since this line is tangent to the circle, we have
$$
d=\frac{\left|2 t^{2} k-2 t-k\right|... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 8 Try to find
$$
\begin{aligned}
p= & (1-1993)\left(1-1993^{2}\right) \cdots\left(1-1993^{1993}\right)+1993\left(1-1993^{2}\right)\left(1-1993^{3}\right) \cdots\left(1-1993^{1993}\right)+ \\
& 1993^{2}\left(1-1993^{3}\right) \cdots\left(1-1993^{1993}\right)+1993^{3}\left(1-1993^{4}\right) \cdots\left(1-1993^{19... | Analysis and Solution Generalization, the $(k+1)$-th term of $p$ has the following form:
$$
p_{k, n}(x)=x^{k}\left(1-x^{k+1}\right)\left(1-x^{k+2}\right) \cdots\left(1-x^{n}\right) \text {. }
$$
Let $\phi_{n}(x)=\sum_{k=0}^{n} p_{k \cdot n}(x)$, then $p=\phi_{1993}$ (1993).
If we can find $\phi_{n}(x)$, the problem ca... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Given that the function $f(x)$ is a monotonic function, and for $x \in$ $(0,+\infty)$, we have $f\left(f(x)+\frac{2}{x}\right)=1$. Then $f(1)=$ ( ).
(A) -3
(B) -2
(C) -1
(D) 0 | 5. D.
Let $f(x)+\frac{2}{x}=k$ (where $k$ is a constant).
$$
\begin{array}{l}
\text { Given } f\left(f(x)+\frac{2}{x}\right)=f(k)=1 \\
\Rightarrow f(x)+\frac{2}{x}=k \Rightarrow f(x)=k-\frac{2}{x} \\
\Rightarrow f(k)=k-\frac{2}{k}=1 .
\end{array}
$$
Given $x \in(0,+\infty)$, solving yields $k=2$.
Thus, $f(x)=2-\frac{... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 8 Find the smallest real number $A$, such that for every quadratic trinomial $f(x)$ satisfying the condition $|f(x)| \leqslant 1(0 \leqslant x \leqslant 1)$, the inequality $f^{\prime}(0) \leqslant A$ holds. | Let the quadratic trinomial be \( f(x) = ax^2 + bx + c \). When \( 0 \leq x \leq 1 \), it satisfies the inequality \( |f(x)| \leq 1 \). Therefore,
\[
|f(0)| \leq 1, \left|f\left(\frac{1}{2}\right)\right| \leq 1, |f(1)| \leq 1.
\]
Since \( f(0) = c, f\left(\frac{1}{2}\right) = \frac{a}{4} + \frac{b}{2} + c, f(1) = a +... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let the sequences $\left\{x_{n}\right\},\left\{y_{n}\right\}$ satisfy $x_{n}+i y_{n}=\left(\frac{\sqrt{3} i-1}{2}\right)^{n}$, find the sum of the first 1994 terms of the sequence $\left\{x_{n}\right\}$, $S_{1994}$. | 5. Let $F=x_{n}+i y_{n}=\left(\frac{\sqrt{3} i-1}{2}\right)^{n}, \quad F^{\prime}=x_{n}-i y_{n}=\left(\frac{-\sqrt{3} i-1}{2}\right)^{n}$, then $F+F^{\prime}=2 x_{n}=\omega^{n}+\bar{\omega}^{n}$. (where $\omega=\frac{-1+\sqrt{3} i}{2}$ ). Thus $2 S_{1994}=-2$, which means $S_{1994}=$ $-1$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms be no less than 3, and the sum of all terms be $97^{2}$. How many such sequences are there? | 9. Let the first term of an arithmetic sequence be $a$, and the common difference be $d$. According to the problem, we have $n a + \frac{n(n-1)}{2} d = 97^2$, which can be rewritten as $[2a + (n-1)d] \cdot n = 2 \times 97^2$. Given that $n$ is a positive integer no less than 3, and 97 is a prime number, the possible va... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. During the middle school math club activities, 11 guests were invited to give specialized lectures. The club paid each guest the same amount, with the total amount being $\overline{1 A 2}$ dollars. Then the digit $A$ in the tens place of the three-digit number is $(\quad$.
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 | 8. D.
From the problem, we know $111 \overline{1 A 2}$.
Thus, $1+2-A=3-A \equiv 0(\bmod 11)$.
Note that, $0 \leqslant A \leqslant 9$.
Then $-6 \leqslant 3-A \leqslant 0 \Rightarrow 3-A=0$
$\Rightarrow A=3$. | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
17. 11 cubic wooden blocks are placed on the ground, first they are painted, then all the blocks are separated. The number of blocks that have exactly three faces painted is $\qquad$ . (Note: The faces in contact with each other and with the ground are not painted) | $7$ | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$$
1 \cdot 2^{\left(0^{(19)}\right)}+\left(\left(2^{0}\right)^{1}\right)^{9}=(\quad) .
$$
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 | 1. C.
$$
\begin{array}{l}
2^{\left(0^{\left(1^{9}\right)}\right)}+\left(\left(2^{0}\right)^{1}\right)^{9}=2^{\left(0^{1}\right)}+\left((1)^{1}\right)^{9} \\
=2^{0}+1=2 .
\end{array}
$$ | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example \ Given $x^{2}+x+1=0$, try to find the value of the rational expression $x^{14}+\frac{1}{x^{14}}$. | Solution: Clearly, $x \neq 1$. From $x^{2}+x+1=\frac{x^{3}-1}{x-1}=0$, we get $x^{3}-1=0$, which means $x$ is a cube root of unity not equal to 1. Therefore, $x^{14}+\frac{1}{x^{14}}=x^{3 \times 4} \times x^{2}+\frac{x}{x^{3 \times 5}}=x^{2}+x=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let the function $f_{0}(x)=|x|, f_{1}(x)=\left|f_{0}(x)-1\right|, f_{2}(x)=$ $\left|f_{1}(x)-2\right|$, then the area of the closed part of the figure enclosed by the graph of the function $y=f_{2}(x)$ and the $x$-axis is $\qquad$. | 3. Draw the graphs of the functions $y=f_{0}(x)$, $y=f_{1}(x)$, $y=f_{2}(x)$ in sequence (as shown in the figure), the required area is 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. A rectangular cuboid (including a cube), all of its edges are integer centimeters, the total length of all edges is 36 centimeters. There are $\qquad$ types of such cuboids. | $7$ | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-35 (1) Find the possible minimum value of the polynomial $P(x, y)=4+x^{2} y^{4}+x^{4} y^{2}-3 x^{2} y^{2}$.
(2) Prove: this polynomial cannot be expressed as a sum of squares of polynomials in variables $x, y$. | ［Solution］(1) By the Arithmetic Mean-Geometric Mean Inequality, we have
$$
1+x^{2} y^{4}+x^{4} y^{2} \geqslant 3 x^{2} y^{2} \text {, }
$$
and the equality holds when $x=y=1$. Therefore, we have
$$
\begin{aligned}
P(x, y) & =3+\left(1+x^{2} y^{4}+x^{4} y^{2}-3 x^{2} y^{2}\right) \\
& \geqslant 3,
\end{aligned}
$$
and... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
21. The formula for the volume of a sphere is $V=\frac{4}{3} \pi r^{3}$, where $r$ is the radius of the sphere. In a cylindrical container, several solid iron spheres with the same radius as the cylinder's base can be placed. When water is poured into the container, the volume of water is six times the volume of one ir... | $12$ | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Let the arithmetic sequence $\left\{a_{n}\right\}$ have all terms as integers, with the first term $a_{1}=2019$, and for any positive integer $n$, there always exists a positive integer $m$ such that $a_{1}+a_{2}+\cdots+a_{n}=a_{m}$. The number of such sequences $\left\{a_{n}\right\}$ is $\qquad$ | Let $\{a_n\}$ be an arithmetic sequence with a common difference $d$, and according to the problem, $d \in \mathbf{Z}$.
There exists a positive integer $k$ such that $a_1 + a_2 = a_k \Rightarrow 2a_1 + d = a_1 + (k-1)d \Rightarrow d = \frac{a_1}{k-2}$.
Thus, $a_1 + a_2 + \cdots + a_n = n a_1 + \frac{n(n-1)}{2} d = a_1 ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy the condition $a+b=2$, then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$. | 1. To find the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$, we can first consider special cases,
When $a=1, b=1, n=1$, $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=1$.
When $a=\frac{1}{2}, b=\frac{3}{2}, n=1$, $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{2}{3}+\frac{2}{5}$.
From this, we can conjecture that $\frac{1}... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. As shown in Figure 1, given the cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, draw a line $l$ through vertex $A_{1}$ in space such that $l$ forms an angle of $60^{\circ}$ with both lines $A C$ and $B C_{1}$. How many such lines $l$ can be drawn?
(A) 4
(B) 3
(C) 2
(D) 1 | 3. (B).
It is easy to know that the angle formed by the skew lines $A C$ and $B C_{1}$ is $60^{\circ}$, so this problem is equivalent to:
Given that the angle formed by the skew lines $a$ and $b$ is $60^{\circ}$, how many lines passing through a fixed point $P$ in space and forming an angle of $60^{\circ}$ with both ... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example 4 Find the largest positive integer $n$, such that the system of equations
$$
(x+1)^{2}+y_{1}^{2}=(x+2)^{2}+y_{2}^{2}=\cdots=(x+k)^{2}+y_{k}^{2}=\cdots=(x+n)^{2}+y_{n}^{2}
$$
has integer solutions $\left(x, y_{1}, y_{2}, \cdots, y_{n}\right)$. | Analysis We use the properties of square numbers to handle the problem.
Solution When $n=3$, it is easy to know that the given system of equations has integer solutions ( $x=-2, y_{1}=0, y_{2}=1, y_{3}=0$ )
When $n=4$,
$$
\left\{\begin{array}{l}
y_{1}^{2}-y_{2}^{2}=2 x+3, \\
y_{2}^{2}-y_{3}^{2}=2 x+5, \\
y_{3}^{2}-y_{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let the set $E=\left\{a_{1}, a_{2}, \cdots, a_{10}\right\}$ have five-element subsets such that: any two elements of $E$ appear in at most two subsets. The maximum number of such five-element subsets is ( ).
A. 252
B. 10
C. 8
D. 18 | 3. C.
Let the number of 5-element subsets that satisfy the condition be $n$. Each element in $E$ appears no more than 4 times, so we have $5n \leqslant 10 \times 4$, which gives $n \leqslant 8$. The construction is as follows:
$$
\begin{array}{l}
\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\},\left\{a_{1}, a_{6}, a_... | 8 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
4.216 * Let $a, b, c$ be positive real numbers. Prove that $\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c} \geqslant 2$. | $$
\begin{array}{r}
\quad \frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}=\frac{c}{a}+\frac{a}{b+c}+\frac{b+c}{c}-1 \\
\geqslant 3 \sqrt[3]{\frac{c}{a} \cdot \frac{a}{b+c} \cdot \frac{b+c}{c}}-1=3-1=2 .
\end{array}
$$
By the AM-GM inequality,
$$
\begin{array}{r}
\quad \frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}=\frac{c}{a}+\frac{a}{b... | 2 | Inequalities | proof | Yes | Yes | olympiads | false |
7. (15 points) Five football teams compete, with each pair of teams playing one match: the winner gets 3 points, the loser gets 0 points, and in the case of a draw, each team gets 1 point. After all the matches are completed, it is found that no team has more than 9 points, and exactly two teams have the same score. Le... | 【Solution】Solution: $5 \times(5-1) \div 2=10$ (matches)
There are a total of 10 matches, with the total score ranging from 20 to 30 points.
The five-digit number $\overline{\mathrm{ABCDE}}$ is exactly a multiple of 15. Using divisibility rules, $E$ can be 0 or 5, considering $E$ is the smallest. If the minimum total sc... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.8 Let $a<b<c<d$ and $(x, y, z, t)$ be any permutation of $(a, b, c, d)$. How many different values can the expression
$$
n=(x-y)^{2}+(y-z)^{2}+(z-t)^{2}+(t-x)^{2}
$$
take? | $$
\begin{aligned}
& \text{[Solution 1]} \text{ Since the expression for } n \text{ is cyclically symmetric with respect to } (x, y, z, t), \text{ we can set } x=a. \\
& \text{Also, since the expression is symmetric with respect to } y \text{ and } t, \text{ for the 6 different permutations of } (b, c, d), n \text{ can... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. If the real numbers $x \cdot y$ 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. $\sqrt{2}$ | 2. B
$(x+5)^{2}+(y-12)^{2}=14^{2}$ is a circle with center $C$ $(-5,12)$ and radius 14. Let $P$ be any point on the circle $L$. Then $|O P|=|C P|-|C C|=14-13=1$.
When points $C$, $O$, and $P$ are collinear, the equality holds, so the minimum value of $P$ to point $O$ is 1, hence the answer is $B$. | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
16. (15 points) Given the functions
$$
f(x)=a \sin x(a \in \mathbf{R}), g(x)=\mathrm{e}^{x} \text {. }
$$
(1) If $00, m0$ always holds, find the minimum value of $k$. | 16. (1) Notice,
$F(x)=a \sin (1-x)+\ln x$,
$F^{\prime}(x)=-a \cos (1-x)+\frac{1}{x}$.
Since $0 < x < 1$,
we have $0 < 1-x < 1$.
Thus, $0 < \cos (1-x) < 1$.
Hence, $F^{\prime}(x) > 0$.
Therefore, $F(x)$ is monotonically increasing in the interval $(0,1)$.
(2) From (1), we know that when $a=1$,
$$
\begin{array}{l}
F(x) <... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
10. Let $f$ be a function from non-negative real numbers to non-negative real numbers, satisfying
$$
f\left(a^{3}\right)+f\left(b^{3}\right)+f\left(c^{3}\right)=3 f(a) f(b) f(c) \text {, (1) }
$$
and $f(1) \neq 1$, where $a, b, c$ are non-negative real numbers. Then $f(2019)=$ $\qquad$ | 10.0 .
Let $a=b=c=1$, we get $3 f(1)=3 f^{3}(1), f(1)=0, \pm 1$.
According to the condition $f(1) \geqslant 0, f(1) \neq 1$, we know $f(1)=0$. Substituting $a=b=1, c=\sqrt[3]{2019}$ into equation (1), we get $f(2019)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2、Define new operations: $A \oplus B=A^{2}+B^{2}, A \otimes B=A$ divided by $\mathbf{B}$'s remainder, then $(2013 \oplus 2014) \otimes 10=$ | 【Answer】5
Analysis: $\left(2013^{2}+2014^{2}\right)$ modulo 10, 2013 $\div$ 10 has a remainder of $3$, 2014 $\div$ 10 has a remainder of 4, which means the remainder of $\left(2013^{2}+2014^{2}\right)$ modulo 10 is the same as the remainder of $\left(3^{2}+4^{2}\right)$ modulo 10, which is 5 | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
35. For any three real numbers $a, b, c$, let $M\{a, b, c\}$ represent the average of these three numbers, and let $\min \{a, b, c\}$ represent the smallest of these three numbers. If
$$
M\{2 x+y+2, x+2 y, 2 x-y\}=\min \{2 x+y+2, x+2 y, 2 x-y\},
$$
then $x+y=$ . $\qquad$ | answer: -4 | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7 Find the prime triplets $(a, b, c)$ that satisfy the following conditions:
(1) $a<b<c<100, a, b, c$ are prime numbers;
(2) $a+1, b+1, c+1$ form a geometric sequence. (Provided by Tao Pingsheng) | Solve according to the conditions,
$$
(a+1)(c+1)=(b+1)^{2},
$$
Let $a+1=n^{2} x, c+1=m^{2} y$, where $x, y$ do not contain square factors greater than 1, then it must be that $x=y$, because according to (1),
$$
(m n)^{2} x y=(b+1)^{2},
$$
then $m n \mid(b+1)$, let $b+1=m n \cdot w$, then (2) becomes,
$$
x y=w^{2} \te... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
16. Given $a+b+c=0$, and $a, b, c$ are all non-zero, then simplify $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ to $\qquad$. | \begin{aligned} \text { 16. Original expression } & =\left(\frac{a}{a}+\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{b}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{c}\right)-\frac{a}{a}-\frac{b}{b}-\frac{c}{c}= \\ \frac{0}{a}+\frac{0}{b}+\frac{0}{c}-3 & =-3 .\end{aligned} | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Let $t=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}$, then the sum of all real solutions of the equation $(t-1)(t-2)(t-3)=0$ with respect to $x$ is $\qquad$ . | 6. 4 Detailed Explanation: Let $f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}$, transform it to $f(x)=\left(\frac{3}{6}\right)^{x}+\left(\frac{4}{6}\right)^{x}+\left(\frac{5}{6}\right)^{x}$, we can see that the function $f(x)$ is a decreasing function on $\mathbf{R}$, $f(3)... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Given $a=2 \sin 45^{\circ}+1, b=2 \cos 45^{\circ}-1$. Then the value of the algebraic expression $\left(\frac{a^{2}+b^{2}}{2 a b}-1\right) \div\left(\frac{a^{2}-b^{2}}{a^{2} b+a b^{2}}\right)$ is | \[\begin{array}{l}\left(\frac{a^{2}+b^{2}}{2 a b}-1\right) \div\left(\frac{a^{2}-b^{2}}{a^{2} b+a b^{2}}\right) \\ =\frac{(a-b)^{2}}{2 a b} \cdot \frac{a b(a+b)}{(a+b)(a-b)} \\ =\frac{a-b}{2}=1\end{array}\] | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. As shown in the figure, the lengths of the base edges of the rectangular prism are $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$, and the height is $6 \mathrm{~cm}$. If a thin string is used to start from point $A$, pass through the four sides and wrap around once to reach point $B$, then the shortest length of the string ... | answer: 10 | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. Let $\alpha, \beta$ be a pair of conjugate complex numbers. If $|\alpha-\beta|=2 \sqrt{3}$, and $\frac{\alpha}{\beta^{2}}$ is a real number, then what is $|\alpha|$? | 9. Let $\alpha=a+b i, \beta=a-b i, a, b \in \mathrm{R}$, then from $|\alpha-\beta|=2 \sqrt{3}$ we get $|b|=\sqrt{3}$. Also, $\frac{\alpha}{\beta^{2}}=\frac{\alpha^{3}}{(\alpha \beta)^{2}}$ is a real number, and $\alpha \beta$ is a real number, so $\alpha^{3}$ is a real number, i.e., $(a+b i)^{3}=\left(a^{3}-3 a b^{2}\r... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. If $x, y$ are both natural numbers, then the number of solutions $(x, y)$ for the equation
$$
[2.018 x]+[5.13 y]=24
$$
is $\qquad$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$). | 2.3.
Since $x$ and $y$ are both natural numbers, and $5.13 \times 5 > 24$, therefore, $y \in \{4,3,2,1,0\}$. At this point, $[5.13 y] = 5 y$.
Similarly, $[2.018 x] = 2 x$.
Thus, the original equation becomes $2 x + 5 y = 24$.
After calculation, $(x, y) = (2,4), (7,2), (12,0)$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. $[a]$ represents the greatest integer not greater than the number $a$. For example
$$
[\sqrt{2}]=1,[-\sqrt{2}]=-2 \text {. }
$$
Then the sum of all roots of the equation $[3 x+1]=2 x-\frac{1}{2}$ is $\qquad$ . | 3. -2
Solution: Let $[3 x+1]$ be an integer $t$. According to the definition of $[a]$, we have
$$
0 \leqslant(3 x+1)-t<1 \text {. }
$$
The original equation is $\quad t=2 x-\frac{1}{2}$,
which means $x=\frac{1}{2} t+\frac{1}{4}$. Substituting this into the inequality above, we get
$$
0 \leqslant\left(\frac{3}{2} t+\f... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example II) Outside an equilateral $\triangle A B C$ with unit area, construct $\triangle A P B, \triangle B Q C, \triangle C R A$ such that $\angle A P B=$ $\angle B Q C=\angle C R A=60^{\circ}$.
(1) Find the maximum area of $\triangle P Q R$;
(2) Find the maximum area of the triangle whose vertices are the incenters ... | Analyzing the problem of the largest area can be transformed into the problem of an equilateral triangle with vertices on a given circle.
Proof (1) Since $\angle A P B=60^{\circ}$.
Therefore, point $P$ lies on the arc $\overparen{A C B}$ symmetrical to $A B$, and this arc is within a circle that is concentric with the ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 4.4.2 Color the 3 vertices of an equilateral triangle with red, blue, and green. How many different schemes are there? If (1) schemes that can be superimposed by rotation are considered the same, (2) schemes that can be superimposed by rotation and reflection are considered the same.
| Solution 1: $D=\{1,2,3\}, R=\{r, b, g\}, C$ has a total of $3^{3}=27$ coloring methods. Using Burnside's formula to calculate:
(1)
$$
\begin{array}{l}
G=\{e,(1,2,3),(1,3,2)\}, \\
N(G, C)=\frac{1}{3}(27+3+3)=11 .
\end{array}
$$
(2)
$$
\begin{array}{l}
G=\{e,(1,2,3),(1,3,2),(1,3),(1,2),(2,3)\}, \\
N(G, C)=\frac{1}{6}\lef... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. In the complex plane $\left\{\begin{array}{l}0 \leqslant \arg (z-1) \leqslant \frac{\pi}{4},(\operatorname{Re}(z) \text { represents the real part of } z) \text { the points that form the corresponding figure have an area of } \\ \operatorname{Re}(z) \leqslant 2\end{array}\right.$ . $\qquad$ | 7. $1 \quad S=\frac{1}{2} \times 2 \times 1=1$ | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 5 As shown in Figure 1.4.6, in quadrilateral $A B C D$, $\angle B=\angle D=90^{\circ}$, $\angle A=60^{\circ}, A B=4, A D=5$, find $B C: C D$. | Extend $A D, B C$ to intersect at point $E$. Since $\angle B=\angle D=90^{\circ}$, we get two right triangles $\triangle A B E$ and $\triangle C D E$. According to the given conditions, $A E=2 A B=8$, $D E=3$.
In the right triangle $\triangle C D E$, we can find $C D=\sqrt{3}, C E=2 \sqrt{3}$. In the right triangle $\... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (8 points) During the New Year, Kangkang serves beer to guests. A bottle of beer can fill 4 glasses. When Qiuqiu pours the beer, there is always half a glass of foam in each glass, and the volume of the beer that turns into foam expands to 3 times its original volume. Therefore, when Qiuqiu pours the beer, a bottle ... | 【Answer】Solution: According to the analysis, it is known that 1 portion of beer can turn into 3 portions of foam. If Qiuqiu pours beer, half of it turns into foam,
then we can divide each cup of beer Qiuqiu pours into 6 portions, so only 4 portions are left when pouring each cup. One bottle of beer can fill 4 cups, wh... | 6 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
6. An exam consists of 6 multiple-choice questions, with the following scoring rules: each person starts with 6 points, gets 4 points for each correct answer, loses 1 point for each wrong answer, and gets 0 points for not answering. Now, 51 students are taking the exam, so at least ( ) people have the same score.
(A) 3... | 【Answer】A
【Question Type】Combination: Drawer Principle
【Analysis】Let the number of correct answers be $x$, and the number of wrong answers be $y$, $x+y \leq 6$; when $x=6$, the score is 30 points;
when $x=5$, $y=0,1$, the corresponding scores are 26, 25;
when $x=4$, $y=0,1,2$, the corresponding scores are $22, 21, 20$;... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. Let $a, b \in [0,1]$, find the maximum and minimum values of $S=\frac{a}{1+b}+\frac{b}{1+a}+(1-a)(1-b)$. | 10. Solution: Since $S=\frac{a}{1+b}+\frac{b}{1+a}+(1-a)(1-b)=\frac{1+a+b+a^{2} b^{2}}{(1+a)(1+b)}=1-\frac{a b(1-a b)}{(1+a)(1+b)} \leqslant 1$, equality holds when $a b=0$ or $a b=1$, so the maximum value of $S$ is 1.
Let $T=\frac{a b(1-a b)}{(1+a)(1+b)}, x=\sqrt{a b}$, then $T=\frac{a b(1-a b)}{1+a+b+a b} \leqslant \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
75. 2 times $a$ plus 2 equals $b$; 2 times $b$ plus 2 equals $c$; 2 times $c$ plus 2 equals $d$; 2 times $d$ plus 2 equals 62. Then, $a=$ ـ. $\qquad$ | Reference answer: 2
Key point: Reverse inference
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Five, $f$ is a mapping from the set of natural numbers $N$ to set $A$. If for $x$, $y \in \mathbf{N}$, $x-y$ is a prime number, then $f(x) \neq f(y)$. How many elements does $A$ have at least? | Five $A$ has at least 4 elements.
Let $f(4 k+i)=i(i=1,2,3,4 . k=0,1,2, \cdots)$, then when $f(x)=f(y)$, $x-y$ is a multiple of 4, and not a prime number. That is, when $x-y$ is a prime number, $f(x) \neq f(y)$. Therefore, the minimum value of $|A| \leqslant 4$.
On the other hand, the 4 numbers $1,3,6,8$, any two of whi... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Let $a=\sum_{i=1}^{24}[\sqrt{i}]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$), and the set $A=\left\{x\left|x \in \mathbf{Z}_{+}, x\right| a\right\}$. Then the number of elements in set $A$ is ( ).
(A) 4
(B) 6
(C) 8
(D) 12 | - 1. C.
Notice,
$$
\begin{aligned}
a= & \sum_{i=1}^{24}[\sqrt{i}] \\
= & ([\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}])+ \\
& ([\sqrt{4}]+[\sqrt{5}]+\cdots+[\sqrt{8}])+ \\
& ([\sqrt{9}]+[\sqrt{10}]+\cdots+[\sqrt{15}])+ \\
& ([\sqrt{16}]+[\sqrt{17}]+\cdots+[\sqrt{24}]) \\
= & 1 \times 3+2 \times 5+3 \times 7+4 \times 9 \\
= & 70=2... | 8 | Number Theory | MCQ | Yes | Yes | olympiads | false |
12. Let $x \in \mathbf{R}$, then the minimum value of the function $f(x)=|2 x-1|+|3 x-2|+|4 x-3|+|5 x-4|$ is | $f(x)=2\left|x-\frac{1}{2}\right|+3\left|x-\frac{2}{3}\right|+4\left|x-\frac{3}{4}\right|+5\left|x-\frac{4}{5}\right|$ consists of 15 terms, the middle term is $\left|x-\frac{3}{4}\right|$, so the minimum value of $f(x)$ is $f\left(\frac{3}{4}\right)=\frac{3}{2}-1+\frac{9}{4}-2+4-\frac{15}{4}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 4.2.4 In an $8 \times 8$ chessboard, what is the minimum number of
shapes that must be placed to achieve a saturated
coverage (no additional shapes can be placed). | Solution: An $8 \times 8$ chessboard can be divided into 16 $2 \times 2$ squares. Each $2 \times 2$ square needs to cover at least 2 squares to possibly be saturated. Therefore, 16 $2 \times 2$ squares require at least 32 squares to be covered, which means at least 11 L-shaped pieces are needed. 11 L-shaped pieces can ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
100. There are 5 different lines on a plane, these 5 lines form $n$ intersection points, then $n$ has $\qquad$ different values. | Answer: 9 | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Let the 20 vertices of a regular 20-sided polygon inscribed in the unit circle in the complex plane correspond to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$. Then the number of different points corresponding to the complex numbers $z_{1}^{1995}, z_{2}^{1995}, \cdots, z_{20}^{1995}$ is
A. 4
B. 5
C. 10
D. 20 | Obviously, $z_{i}=z_{1}^{i}(i=1,2, \cdots, 20)$, where $z_{1}=\cos \frac{\pi}{10}+\mathrm{i} \sin \frac{\pi}{10}$. Thus, $z_{1}^{5} \mid z_{i}^{1995}$ and $z_{1}^{20}=1$, so the different points corresponding to the complex numbers $z_{1}^{1995}, z_{2}^{1995}, \cdots, z_{20}^{1995}$ are $z_{1}^{5}, z_{1}^{10}, z_{1}^{1... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
4. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: x^{2}-\frac{y^{2}}{24}=1$, respectively, and let $P$ be a point on the hyperbola $C$ in the first quadrant. If $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{4}{3}$, then the inradius of $\triangle P F_{1} F_{2}$ is $\qquad$. | 4. 2 .
Let $\left|P F_{1}\right|=4 t$. Then $\left|P F_{2}\right|=3 t$.
Thus $4 t-3 t=\left|P F_{1}\right|-\left|P F_{2}\right|=2$
$$
\Rightarrow t=2,\left|P F_{1}\right|=8,\left|P F_{2}\right|=6 \text {. }
$$
Combining $\left|F_{1} F_{2}\right|=10$, we know that $\triangle P F_{1} F_{2}$ is a right triangle, $P F_{1... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Several different numbers are written on the blackboard. The sum of any three of them is a rational number, while the sum of any two is an irrational number. The maximum number of numbers that can be written on the blackboard is $\qquad$ . | 4.3.
Assume that the number of numbers written on the blackboard is no less than four, denoted as $a, b, c, d$. Then, $a+b+c$ and $b+c+d$ are both rational numbers, which implies that their difference
$$
(b+c+d)-(a+b+c)=d-a
$$
is also a rational number.
Similarly, $b-a$ and $c-a$ are also rational numbers.
Therefore, ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
20. Let $m, n, p, q$ be non-negative integers, and for all $x>0, \frac{(x+1)^{m}}{x^{n}}-1=\frac{(x+1)^{p}}{x^{q}}$ always holds, then $\left(m^{2}+2 n+p\right)^{2 q}=$ $\qquad$ . | Answer: 9 | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) Among the interior angles of seven triangles, there are two right angles and three obtuse angles. Therefore, there are ( ) acute triangles.
A. 1
B. 2
C. 3
D. 4 | 【Analysis】Since a triangle can have at most one right angle or one obtuse angle, according to the problem: there are 2 right-angled triangles, 3 obtuse-angled triangles. As there are a total of 7 triangles, using $7-3-2$ can determine the number of acute-angled triangles.
【Solution】Solution: $7-2-3=2$,
Therefore, ther... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
17.65 In $\triangle A B C$, as shown in the figure, $\frac{C D}{D B}=\frac{3}{1}, \frac{A E}{E B}=\frac{3}{2}$, let $r=\frac{C P}{P E}$, where $P$ is the intersection of $C E$ and $A D$, then $r$ equals
(A) 3.
(B) $\frac{3}{2}$.
(C) 4.
(D) 5.
(E) $\frac{5}{2}$.
(14th American High School Mathematics Examination, 1963) | [Solution] Draw $D F / / A B$, by the property of parallel lines intercepting segments in proportion, we have $\frac{C F}{F E}=\frac{C D}{D B}=\frac{3}{1}, \quad$ and $\quad \frac{F D}{E B}=\frac{C D}{C B}=\frac{3}{4}$.
Thus, $C F=3 F E=3 F P+3 P E$,
and
$$
F D=\frac{3}{4} E B \text {. }
$$
Therefore,
$$
C P=C F+F P=4... | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
Find the maximum value of the function $y=\sec x-\tan x\left(0 \leqslant x<\frac{\pi}{2}\right)$. | Solve $y=\sec x-\tan x=\frac{\sec ^{2} x-\tan ^{2} x}{\sec x+\tan x}=\frac{1}{\sec x+\tan x}\left(0 \leqslant x<\frac{\pi}{2}\right)$,
on $\left[0, \frac{\pi}{2}\right]$, $\sec x+\tan x$ is an increasing function and always greater than 0, so $y=\sec x-\tan x$ is a decreasing function. Therefore, the maximum value must... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
10. The teacher took out 13 cards, with the numbers on the cards ranging from 1 to 13. After keeping one of the cards, the teacher distributed the remaining 12 cards to three students, Jia, Yi, and Bing, with each receiving 4 cards. It is known that the sum of the numbers on the cards in Jia's hand and the sum of the n... | $6$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. When $n$ is a positive integer, the function $f$ satisfies $f(n+3)=\frac{f(n)-1}{f(n)+1}, f(1) \neq 0$ and $f(1) \neq \pm 1$, then the value of $f(8) \cdot f(2018)$ is $\qquad$ | 5. -1 From the given, we have $f(n+6)=-\frac{1}{f(n)}$, so $f(8) \cdot f(2018)=f(8) f(2)=-1$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) The school organizes 482 people to go on an outing, renting 42-seat buses and 20-seat minibuses. If it is required that each person has one seat and each seat is occupied by one person, then there are $\qquad$ rental options. | 【Analysis】Let there be $x$ 42-seat buses and $y$ 20-seat minibuses. According to the problem, we have:
$42 x + 20 y = 482$, we need to find the integer solutions of this equation, which will give us the answer.
【Solution】Let there be $x$ 42-seat buses and $y$ 20-seat minibuses. According to the problem, we have:
$42 x ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. (28th Russian Mathematical Olympiad) In Greek mythology, the "Hydra" god consists of some heads and necks, with each neck connecting two heads. With each strike of a sword, one can sever all the necks connected to a certain head $A$, but head $A$ immediately grows new necks connecting to all the heads it was not pre... | 9. Translating into graph theory language, let the heads be vertices, and the necks be edges, and a strike that cuts the necks connected to head $A$ is called a "reversal" of vertex $A$. It is easy to see that if a vertex $X$ has a degree no greater than 10, then it is only necessary to perform a "reversal" on all vert... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Given real numbers $x, y$ satisfy $x+y=1$. Then, the maximum value of $\left(x^{3}+1\right)\left(y^{3}+1\right)$ is $\qquad$ . | 5.4.
$$
\begin{array}{l}
\text { Given }\left(x^{3}+1\right)\left(y^{3}+1\right) \\
=(x y)^{3}+x^{3}+y^{3}+1 \\
=(x y)^{3}-3 x y+2,
\end{array}
$$
let $t=x y \leqslant\left(\frac{x+y}{2}\right)^{2}=\frac{1}{4}$, then
$$
f(t)=t^{3}-3 t+2 \text {. }
$$
Also, by $f^{\prime}(t)=3 t^{2}-3$, we know that $y=f(t)$ is monoto... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5 There are $n$ people, and it is known that any 2 of them make at most one phone call. The total number of phone calls made among any $n-2$ of them is equal, and it is $3^{k}$ ($k$ is a positive integer). Find all possible values of $n$.
untranslated text remains the same in terms of line breaks and formatti... | Let the total number of calls among $n$ people be $m$. Since $n$ people can form $C_{n}^{n-2}$ groups of $n-2$ people, and the total number of calls among each $n-2$ people is $3^{k}$, the total number of calls in all $n-2$ groups is $C_{n}^{n-2} \cdot 3^{k}$.
On the other hand, in the above counting, each pair of peo... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.48 How many real numbers $a$ are there such that $x^{2}+a x+6 a=0$ has only integer solutions.
(9th American Invitational Mathematics Examination, 1991) | [Solution] Let the equation $x^{2}+a x+6 a=0$ have integer solutions $m, n(m \leqslant n)$, then we have
$$
\begin{array}{c}
m+n=-a, \\
m n=6 a .
\end{array}
$$
Thus, we have
$$
-6(m+n)=m n \text {. }
$$
That is
$$
(m+6)(n+6)=36 \text {. }
$$
Since $36=1 \cdot 36=2 \cdot 18=3 \cdot 12=4 \cdot 9=6 \cdot 6$.
Therefore... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
121 The solutions to the inequality $x(x-1) \leqslant y(1-y)$ can all make $x^{2}+y^{2} \leqslant k$ hold, the minimum value of $k$ is $(\quad)$
A. 0
B. 1
C. 2
D. Greater than 2 | 121 C. The inequality $x(x-1) \leqslant y(1-y)$ is equivalent to $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2} \cdot x^{2}+y^{2}=$ $(\sqrt{k})^{2}$, which represents a circle with radius $\sqrt{k}$. When this circle is internally tangent to $\left(x-\frac{1}{2}\right)^{2}+\left(y-... | 2 | Inequalities | MCQ | Yes | Yes | olympiads | false |
1. Given that $f(x)$ is an odd function defined on $\mathbf{R}$, $f(1)=2, f(2)=3$, then $f(f(-1))=$ | $$
f(-1)=-f(1)=-2, f(-2)=-f(2)=-3 \Rightarrow f(f(-1))=f(-2)=-3
$$ | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(3) Jia writes down the positive integers $1,2, \cdots, 2009$ on the blackboard, then turns his back to the blackboard, and asks Yi to erase some of these numbers and add the remainder when the sum of the erased numbers is divided by 7. After several such operations, only two numbers remain on the blackboard, one of wh... | (3) 5 Hint: Since $1+2+\cdots+2009 \equiv 0(\bmod 7)$, the one-digit number $a$ satisfies $100+a \equiv 0(\bmod 7)$, thus $a \equiv 5(\bmod 7)$, i.e., $a=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Let real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x-y+1 \geqslant 0, \\
y+1 \geqslant 0, \\
x+y+1 \leqslant 0 .
\end{array}\right.
$$
Then the maximum value of $2 x-y$ is | 8. 1 .
When $x=0, y=-1$, $2 x-y$ takes the maximum value 1. | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
12. For a regular tetrahedron $P Q R S$, there are four vertices and six edges, each labeled with a number, making a total of 10 numbers. These 10 numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 11$. Each number is used exactly once, and each number on an edge represents the sum of the numbers at the two vertices it connects. ... | 【Answer】5 (Each vertex is connected by three edges, so the sum of these three edges is equivalent to adding the number at this vertex three times). Considering $S_{\text {edge }}+S_{\text {vertex }}=1+2+3+4+5+6+7+8+9+11=56$, we can deduce that $S_{\text {vertex }}=14$. Possible combinations are: $1+2+3+8$, $1+2+4+7$, $... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. (7 points) Arrange even numbers as shown in the figure below. Ask: 2008 is in the $\qquad$ column.
\begin{tabular}{lllc}
2 & 4 & 6 & 8 \\
16 & 14 & 12 & 10 \\
18 & 20 & 22 & 24 \\
32 & 30 & 28 & 26 \\
$\ldots$ & & &
\end{tabular} | 【Analysis】First, we find that the even numbers in the sequence repeat every 8 numbers, with odd rows increasing from left to right and even rows decreasing from right to left, in reverse order to the previous row; then we determine that 2008 is the $2008 \div 2=1004$th number, and then divide 1004 by 8 to find the rema... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. As shown in the figure, in the right triangle $A B C$, $\angle A C B=\frac{\pi}{2}, A C=B C=2$, point $P$ is a point on the hypotenuse $A B$, and $B P=2 P A$, then $\overrightarrow{C P} \cdot \overrightarrow{C A}+\overrightarrow{C P} \cdot \overrightarrow{C B}=$ $\qquad$ . | Answer 4.
Solution 1: Since
$$
\overrightarrow{C P}=\overrightarrow{C A}+\overrightarrow{A P}=\overrightarrow{C A}+\frac{1}{3} \overrightarrow{A B}=\overrightarrow{C A}+\frac{1}{3}(\overrightarrow{A C}+\overrightarrow{C B})=\frac{2}{3} \overrightarrow{C A}+\frac{1}{3} \overrightarrow{C B}
$$
Therefore,
$$
\overrightar... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Given real numbers $a, b$ satisfy
$$
a+\lg a=10, b+10^{b}=10 \text {. }
$$
Then $\lg (a+b)=$ $\qquad$ | 8. 1 .
Since $\lg a=10-a, 10^{b}=10-b$, therefore, $a$ is the x-coordinate of the intersection point of $y=\lg x$ and $y=10-x$, and $b$ is the y-coordinate of the intersection point of $y=10^{x}$ and $y=10-x$.
Also, $y=\lg x$ and $y=10^{x}$ are symmetric about the line $y=x$, and $y=10-x$ is symmetric about the line ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. If the positive integer $a$ makes the maximum value of the function $y=f(x)=x+\sqrt{13-2 a x}$ also a positive integer, then this maximum value equals
A. 3
B. 4
C. 7
D. 8 | 6. $\mathrm{C}$ Let $\sqrt{13-2 a x}=t,(t \geqslant 0), x=\frac{13-t^{2}}{2 a}, y=-\frac{1}{2 a}(t-a)^{2}+\frac{a^{2}+13}{2 a} \cdot t=a$ when, $y_{\text {max }}=$ $\frac{a^{2}+13}{2 a}$, to make $\frac{a^{2}+13}{2 a}$ a positive integer, $a$ must be an odd number. And $\frac{a^{2}+13}{2 a}=\frac{1}{2}\left(a+\frac{13}... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. Solve the equation $\sqrt{1-\sqrt{x}}+\sqrt[4]{x^{3}}=\sqrt{x+1}$. | 5. According to the Cauchy-Schwarz inequality
$$
\left(\sqrt{1-\sqrt{x}}+\sqrt[4]{x^{3}}\right)^{2} \leqslant\left[(\sqrt{1-\sqrt{x}})^{2}+(\sqrt[4]{x})^{2}\right] \cdot\left[1^{2}+(\sqrt{x})^{2}\right] \text {, }
$$
Since $\sqrt{1-\sqrt{x}}+\sqrt[4]{x^{3}}=\sqrt{x+1}$,
then $\quad\left(\sqrt{1-\sqrt{x}}+\sqrt[1]{x^{3... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let the function be
$$
f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}(x \in[0,+\infty)) .
$$
Then the number of integer points on the graph of the function is $\qquad$ . | 4.3.
It is known that the function $f(x)$ is monotonically decreasing on the interval $[0,+\infty)$, and
$$
f(0)=3, f(1)=2, f(3)=1 .
$$
When $x>3$, we have
$$
0<f(x)<f(3)=1 \text {. }
$$
Therefore, the number of integer points on the graph of the function $y=f(x)(x \in[0,+\infty))$ is 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (10 points) Bunny and Turtle start from location $A$ to the Forest Amusement Park at the same time. Bunny jumps forward 36 meters per minute, and after every 3 minutes of jumping, it plays on the spot. The first time it plays for 0.5 minutes, the second time for 1 minute, the third time for 1.5 minutes, $\cdots$, th... | 【Analysis】First, analyze the time it would take for the rabbit without any rest, then find out the number of rests, and calculate the corresponding time difference which is the time for the tortoise. This will lead to the solution.
【Solution】According to the problem:
The rabbit, without resting, would need $2640 \div ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (2006 College Entrance Examination Mathematics (Shandong Science) Question) Given the odd function $f(x)$ defined on $\mathbf{R}$ satisfies $f(x+2$ ) $=-f(x)$, then the value of $f(6)$ is
A. -1
B. 0
C. 1
D. 2 | 1. B From $f(x+2)=-f(x)$, we can deduce that the function $f(x)$ is a periodic function with a period of 4. Since $f(x)$ is an odd function defined on $\mathbf{R}$, we have $f(6)=f(2)=f(-2)=-f(2)=0$, hence the answer is B. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. Given the function $f(x)=x+\frac{9}{x}$ on the interval $[1,4]$, the maximum value is $M$, and the minimum value is $m$. Then the value of $M-m$ is $\qquad$ . | Ni, 7.4.
Since $f(x)$ is monotonically decreasing on the interval $[1,3]$ and monotonically increasing on the interval $[3,4]$, the minimum value of $f(x)$ is $f(3)=6$.
Also, $f(1)=10, f(4)=\frac{25}{4}$, so the maximum value of $f(x)$ is $f(1)=10$.
Therefore, $M-m=10-6=4$. | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
1. Let $a, b$ be unequal real numbers. If the quadratic function $f(x)=x^{2}+a x+b$ satisfies $f(a)=f(b)$, then the value of $f(2)$ is $\qquad$ . | Solve for the axis of symmetry of $f(x)=x^{2}+a x+b$ which is $x=-\frac{a}{2}=\frac{a+b}{2} \Rightarrow b=-2 a$, thus $f(x)=x^{2}+a x-2 a \Rightarrow f(2)=4+2 a-2 a=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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