problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
3. The product of the areas of the six faces of a rectangular prism is 14641, then the volume of the rectangular prism is
保留源文本的换行和格式，直接输出翻译结果。 | 【Analysis】Let the length, width, and height of the rectangular prism be $a$, $b$, and $c$,
$$
\begin{aligned}
\left(a^{2} b^{2} c^{2}\right)^{2} & =14641 \\
a^{2} b^{2} c^{2} & =121 \\
(a b c)^{2} & =121 \\
a b c & =11
\end{aligned}
$$ | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 2 Let integers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
|x+y|+|y+z|+|z+x|=4, \\
|x-y|+|y-z|+|z-x|=2 .
\end{array}\right.
$$
Find the value of $x^{2}+y^{2}+z^{2}$. | Let's assume $x \leqslant y \leqslant z$.
Then from equation (2) we get
$$
\begin{array}{l}
(y-x)+(z-y)+(z-x)=2 \\
\Rightarrow z=x+1 \Rightarrow x \leqslant y \leqslant x+1 \\
\Rightarrow y=x \text { or } y=x+1 .
\end{array}
$$
(1) When $y=x$, from equation (1) we get
$$
\begin{array}{l}
|x|+|2 x+1|=2 \\
\Rightarrow x=... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.54 Determine all positive integers $n$ such that $n=[d(n)]^{2}$, where $d(n)$ denotes the number of positive divisors of $n$ (including 1 and itself).
(Canadian Mathematical Olympiad, 1999) | [Solution] Let the standard factorization of $n$ be
$n=p_{1}^{a_{1}} p_{2}^{a_{2} \cdots} p_{k}^{a_{k}}\left(p_{i}\right.$ is a prime, $\left.\alpha_{i} \geqslant 0\right)$.
Then, we have $\quad d(n)=\left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{k}\right)$.
By the problem statement, $n=[d(n)... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (8 points) In a football team, 11 players are on the field, including 1 goalkeeper who does not participate in the formation of defenders, midfielders, and forwards. It is known that the number of defenders is between 3-5, the number of midfielders is between 3-6, and the number of forwards is between 1-3. Therefore... | 【Analysis】The number of defenders is between $3-5$ people, and we can classify and discuss according to the number of defenders, list all possibilities, and thus solve the problem.
【Solution】Solution: Formations are generally arranged in the order of the number of defenders, midfielders, and forwards. When the number ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. As shown in the figure, in the equilateral $\triangle ABC$ with side length 2, $D$ is the midpoint of $BC$, and $E, F$ are moving points on sides $AB, AC$ respectively.
(1) If $\angle EDF=120^{\circ}$, prove that $AE + AF$ is a constant;
(2) If $\angle EDF=60^{\circ}$, is $AE + AF$ a constant? If so, provide a proo... | (1) As shown in the figure, let $A E=x, A F=y$, in $\triangle B E D$,
$$
\frac{2-x}{\sin \alpha}=\frac{1}{\sin \left(120^{\circ}-\alpha\right)} \Rightarrow 2-x=\frac{\sin \alpha}{\sin \left(120^{\circ}-\alpha\right)},
$$
Similarly, in $\triangle C D F$, $\beta=60^{\circ}-\alpha$,
$$
\frac{2-y}{\sin \beta}=\frac{1}{\si... | 3 | Geometry | proof | Yes | Yes | olympiads | false |
7. Given the sets $M=\{x \mid x=3 n, n=1,2,3,4\}, P=\left\{x \mid x=3^{k}, k=\right.$ $1,2,3\}$. If there is a set $S$ that satisfies the condition $(M \cap P) \subseteq S \subseteq(M \cup P)$, then there are $\qquad$ such $S$. | 7. 8 .
$$
M=\{3,6,9,12\}, P=\{3,9,27\}, M \cap P=\{3,9\}, M \cup P=\{3,6,9,12,27\} \text {, }
$$
From the problem, $S=\{3,9\} \cup A$, where $A \subseteq\{6,12,27\}$. There are $2^{3}=8$ (sets). | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Given the set $P=|x| \cdot x^{2}=1 \mid$ and $Q=\{x|m x=1|$, if $Q \subset P$. Then the number of real values that $m$ can take is ( ).
A. 0
B. 1
C. 2
D. 3 | 1. D
Since $P=\{-1,1\}, Q \subset P$, $Q$ could be $\phi, \{-1\}, \{1\}$. When $Q=\phi$, $m=0$,
when $Q=\{-1\}$, $m=-1$,
when $Q=\{1\}$, $m=1$,
thus $m$ has 3 values: $0, -1, 1$. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
6.53 16 students participate in a math competition. All the questions are multiple-choice, with 4 options for each question. After the exam, it is found that any two students have at most 1 question with the same answer. How many questions are there at most? Explain your reasoning. | [Solution 1] Use 16 points to represent 16 students, and connect a line between two points if the two students have the same answer to one question. Thus, there is at most one line between any two points, and the graph can have at most $C_{16}^{2}=120$ lines.
On the other hand, for the $i$-th question, let the number ... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. (7 points) Xiaohong went to the stationery store and bought 4 pencils and 5 notebooks, spending a total of 15 yuan 8 jiao. Xiaoliang bought the same 4 pencils and 7 notebooks, spending a total of 21 yuan 8 jiao. What is the price of 1 notebook? | 【Answer】Solution: $(21.8-15.8) \div(7-5)$
$$
\begin{array}{l}
=6 \div 2 \\
=3 \text { (yuan) }
\end{array}
$$
Answer: Each exercise book costs 3 yuan. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
One, (This question is worth 40 points) Non-negative real numbers $x_{1}, x_{2}, \cdots, x_{2016}$ and real numbers $y_{1}, y_{2}, \cdots, y_{2016}$ satisfy:
(1) $x_{k}^{2}+y_{k}^{2}=1, k=1,2, \cdots, 2016$;
(2) $y_{1}+y_{2}+\cdots+y_{2016}$ is an odd number.
Find the minimum value of $x_{1}+x_{2}+\cdots+x_{2016}$. | Solution: From the given condition (1), we have: $\left|x_{k}\right| \leq 1,\left|y_{k}\right| \leq 1, k=1,2, \cdots, 2016$, thus (note $x_{i} \geq 0$)
$$
\sum_{k=1}^{2016} x_{k} \geq \sum_{k=1}^{2016} x_{k}^{2}=\sum_{k=1}^{2016}\left(1-y_{k}^{2}\right)=2016-\sum_{k=1}^{2016} y_{k}^{2} \geq 2016-\sum_{k=1}^{2016}\left|... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20. The five numbers $a, b, c, d, e$ are all different. The products of each pair of these numbers, arranged in ascending order, are $3, 6, 15, 18, 20, 50, 60, 100, 120, 300$. Then, the five numbers arranged in ascending order, the square of the 2nd number is $\qquad$ . | answer: 10 | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.28 Let \( m \) and \( n \) be natural numbers. Determine the minimum number of distinct prime divisors of the natural number
\[
m(n+9)\left(m+2 n^{2}+3\right)
\]
(53rd Moscow Mathematical Olympiad, 1990) | [Solution] (1) If $m>1$, then $m$ and $m+2 n^{2}+3$ have different parities. In the product $m(n+9)\left(m+2 n^{2}+3\right)$, there must be the even prime number 2 as a divisor. Additionally, there is at least one odd prime number as a divisor.
Therefore, when $m>1$, the product $m(n+9)\left(m+2 n^{2}+3\right)$ has at ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
【Question 4】
A cloth bag contains 10 small balls each of red, yellow, and green colors, all of the same size. The red balls are marked with the number "4", the yellow balls are marked with the number "5", and the green balls are marked with the number "6". Xiao Ming draws 8 balls from the bag, and their sum of numbers ... | 【Analysis and Solution】
If all 8 balls are red, the sum of the numbers is $4 \times 8=32$; replacing a red ball with a yellow ball increases the sum by $5-4=1$; replacing a red ball with a green ball increases the sum by $6-4=2$;
$$
(39-32) \div 2=3 \cdots \cdots 1 ;
$$
Therefore, 3 red balls should be replaced with 3... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5 Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, E $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$, then $\sum_{n=1}^{2008} x_{n}=(\quad)$.
(A) 0
(B) -1
(C) $2+\sqrt{3}$
(D) $-2+\sqrt{3}$ | (5) Since $x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$, let $x_{n}=\tan a_{n}$, thus $x_{n+1}=\tan \left(a_{n}+\frac{\pi}{6}\right)$, $x_{n+6}=x_{n}$. It is easy to calculate that $x_{1}=1, x_{2}=2+\sqrt{3}, x_{3}=-2-\sqrt{3}, x_{4}=-1, x_{5}=$ $-2+\sqrt{3}, x_{6}=2-\sqrt{3}, x_{7}=1, \cdots$, ... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. (7 points) When copying a problem, Lin Lin mistakenly wrote a repeating decimal as 0.123456. If the digits are correct but the dots indicating the repeating section were omitted, there are $\qquad$ possible original decimals. | 【Answer】Solution: According to the analysis,
if there is only one point in the repeating section, then it can only be above the digit "6";
if there are 2 points in the repeating section, then one of them must be above the digit "6"; the other may be above " $1, 2, 3, 4, 5$ ",
so there are $1+5=6$ possibilities in tota... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given the function $f(x)=x^{3}$, the tangent line at the point $\left(a_{k}, f\left(a_{k}\right)\right)\left(k \in \mathbf{N}^{*}\right)$ intersects the $x$-axis at $\left(a_{k+1}, 0\right)$. If $a_{1}=1$, then $\frac{f\left(\sqrt[3]{a_{1}}\right)+f\left(\sqrt[3]{a_{2}}\right)+\cdots+f\left(\sqrt[3]{a_{10}}\right)}{... | The equation of the tangent line to the curve of the function $f(x)$ at the point $\left(a_{k}, f\left(a_{k}\right)\right)$ is $y=3 a_{k}^{2} x-2 a_{k}^{3}$. Let $y=0 \Rightarrow a_{k+1}=\frac{2}{3} a_{k}$, thus $a_{n}=\left(\frac{2}{3}\right)^{n-1}$.
Therefore, $f\left(\sqrt[3]{a_{1}}\right)+f\left(\sqrt[3]{a_{2}}\rig... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. In the race between the tortoise and the hare, the hare's and tortoise's speed ratio at the start was $3: 2$. When the hare reached the pavilion along the way, it found that running another 3 kilometers would complete half the race, so it had a good sleep. When the tortoise passed the pavilion, it increased its spe... | $12$ | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A project needs 100 days to complete. Initially, 10 people worked for 30 days and completed $\frac{1}{5}$ of the entire project. Then, 10 more people were added to complete the project. How many days earlier can the task be completed? | 【Answer】 10
【Analysis】Assuming the work efficiency of each person per day is $a$ units, the total amount of work is $10 a \times 30 \div \frac{1}{5}=1500 a$ (units); after adding 10 more people, the number of days to complete the work is: $(1500 a-30 \times 10 a) \div(10 a+10 a)=60$ (days) ahead of schedule by $100-30-... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let $x y z+y+z=12$. Then the maximum value of $\log _{4} x+\log _{2} y+\log _{2} z$ is $\qquad$ . | 5.3.
From the given conditions, we have
$$
\begin{array}{l}
12=x y z+y+z \geqslant 3 \sqrt[3]{x y^{2} z^{2}} \\
\Rightarrow x y^{2} z^{2} \leqslant 64 .
\end{array}
$$
Then $\log _{4} x+\log _{2} y+\log _{2} z=\log _{4} x y^{2} z^{2}$
$$
\leqslant \log _{4} 64=3 \text {, }
$$
with equality holding if and only if $x=... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (18th "Hope Cup" Senior High School Competition Problem) Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=0, a_{n+1}=\frac{\sqrt{2} a_{n}+\sqrt{6}}{a_{n}-\sqrt{2}}$ $\left(n \in \mathbf{N}^{*}\right)$, then $a_{2007}=$
A. 0
B. $\sqrt{3}$
C. $-\sqrt{3}$
D. $\sqrt{2}$ | 5. A Because $a_{4}=0, a_{2}=\frac{\sqrt{6}}{-\sqrt{2}}=-\sqrt{3}, a_{3}=\frac{\sqrt{2}(-\sqrt{3})+\sqrt{6}}{-\sqrt{3}-\sqrt{2}}=0, a_{4}=-\sqrt{3}$, therefore, in the sequence $\left\{a_{n}\right\}$, all odd terms are 0, and all even terms are $-\sqrt{3}$, so $a_{2007}=0$, choose $\mathrm{A}$. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
15. 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 $a_{m}$ is always divisible by 9, find the minimum value of $n$.
When $m \geqslant n$, $a_{m}$'s value can be divided by 9, find the smallest $n$. | 15. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1)$ $\left(a_{n}-a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots \cdot 4 \cdot 3 \cdot\left(a_{2}-a_{1}\right)=(n+2)!$,
it follows that $a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 7 Determine the smallest natural number $k$, such that for any $a \in[0,1]$ and any $n \in \mathbf{N}$, we have $a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}}$. | Solve: Eliminate one parameter $a$, then find the minimum value of $k$. By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{n+k}=\frac{k}{n+k}. \\
\text { Thus, } a^{k}(1-a)^{n} \leqslant \frac{k^... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
11. (6 points) The average of 6 different non-zero natural numbers is 12. If one of the two-digit numbers $\overline{\mathrm{ab}}$ is replaced by $\overline{\mathrm{ba}}$ (where $a$ and $b$ are non-zero digits), then the average of these 6 numbers becomes 15. Therefore, the number of $\overline{\mathrm{ab}}$ that satis... | 【Solution】Solution: The original sum of the six numbers is: $12 \times 6=72$
The sum of the 6 numbers now is: $15 \times 6=90$
That is, $\overline{b a}-\overline{a b}=90-72=18$
Then, $10 b+a-10 a-b=18$
$$
\begin{array}{r}
9(b-a)=18 \\
b-a=2
\end{array}
$$
Since, the original sum of the six numbers is 72, the smallest ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) Divide 23 cards, each with a number from 1 to 23, into three piles. It is known that the average numbers of the three piles are 13, 4, and 17, respectively. How many cards are there at least in the pile with an average of 13?
【Analysis】According to the problem, let the piles with averages of $13$, $4$, ... | $13 a+4 b+17 c=1+2+3+\cdots+23=276$, substituting $c=23-b-a$ into $13 a+4 b+17 c=276$, we get: $4 a+13 b=115$, based on this, we need to find the minimum number of cards in the pile with an average of 13.
【Solution】Let the piles with averages of $13$, $4$, and $17$ have $a$, $b$, and $c$ cards respectively,
then: $a+b... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. For a finite set $A$, there exists a function $f: \mathbf{N}_{+} \rightarrow A$, with the following property: if $i, j \in \mathbf{N}_{+}$ and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements does the set $A$ have at minimum? | 1. Since the absolute value of the difference between any two of $1,3,6,8$ is a prime number, by the problem's condition, $f(1), f(3), f(6), f(8)$ are distinct elements in $A$, thus $|A| \geqslant 4$. On the other hand, if we let $A=\left\{0,1,2,3 \mid\right.$, and if $x \in \mathbf{N}_{+}, x=4 k+r$, then $f(x)=r$, whe... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
32. Choose several different numbers from $1,2,3,4,5,6,7$, so that the sum of the even numbers equals the sum of the odd numbers. The number of ways that meet the condition is $\qquad$. | answer: 7 | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A. As shown in Figure $1, \odot O$'s diameter $A B$ intersects with chord $C D$ at point $P$, forming an angle of $45^{\circ}$. If $P C^{2}+P D^{2}=8$, then the radius of $\odot O$ is ( ).
(A) $\sqrt{2}$
(B) 2
(C) $2 \sqrt{2}$
(D) 4 | 4. A. B.
Draw $O H \perp C D$, with the foot of the perpendicular at $H$. Without loss of generality, assume point $H$ lies on segment $C P$. Then $\angle O P H=45^{\circ}$.
Thus, $O H=H P=\frac{1}{2}(P C-P D)$.
By $C H=\frac{1}{2}(P C+P D)$, in the right triangle $\triangle O H C$, applying the Pythagorean theorem, w... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
8.164 Suppose there are 3 piles of stones, and A moves 1 stone from one pile to another each time. A can receive payment from B for each move, the amount of which is equal to the difference between the number of stones in the pile where the stone is placed and the number of stones left in the pile where the stone is ta... | [Solution] The only value of the money that A can earn is 0, and of course, the maximum value is also 0. Suppose in each pile of stones, we use line segments to connect the stones in pairs. When A moves a stone, they first have to untie the line segments connecting the stone to all other stones in the pile (one end of ... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. Given the function $f(x)=x+\frac{4}{x}-1$, if there exist $x_{1}, x_{2}, \cdots, x_{n} \in\left[\frac{1}{4}, 4\right]$, such that $f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n-1}\right)=f\left(x_{n}\right)$, then the maximum value of the positive integer $n$ is $\qquad$. | Answer 6.
Analysis From the problem, we have
$$
f(x)=x+\frac{4}{x}-1 \in\left[3,15 \frac{1}{4}\right]
$$
Therefore, when $n$ is as large as possible, the scenario is $3+3+3+3+3 \frac{1}{4}=15 \frac{1}{4}$, at this time $n=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14-22 Solve the equation $\left[x^{3}\right]+\left[x^{2}\right]+[x]=\{x\}-1$.
(Kiev Mathematical Olympiad, 1972) | [Solution] Since the left side of the given equation $\left[x^{3}\right]+\left[x^{2}\right]+[x]$ is an integer, the right side of the equation is also an integer. Also, because
$$
0 \leqslant\{x\}<1,
$$
it must be that
$$
\{x\}=0 .
$$
Thus, $x$ is an integer, and the equation simplifies to
$$
\begin{array}{c}
x^{3}+x... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (5 points) As shown in the figure, there is a cross-shaped shaded area formed by two rectangles in the middle of a rectangle that is 9 cm long and 8 cm wide. If the area of the shaded part is exactly equal to the area of the blank part, then $X=$ $\qquad$ cm.
| 【Analysis】According to the figure, the area of the shaded part is 3 times 8 plus 9 times $x$ minus the overlapping part $3 x$, and the area of the blank part is 9 minus 3 times the difference of 8 minus $x$. Since the area of the shaded part is equal to the area of the blank part, each occupying half of the total area,... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
35. The perimeter of a rectangle is 22 meters. If its length and width are both integers, then there are $\qquad$ possible areas for this rectangle. | answer: 5 | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$19 \cdot 17$ If the sum of the interior angles of a polygon is 3 times the sum of its exterior angles, then the number of sides of this polygon is
(A) 6.
(B) 7.
(C) 8.
(D) 9.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989) | [Solution] The sum of the interior angles of an $n$-sided polygon is $(n-2) \cdot 180^{\circ}$, and the sum of its exterior angles is
$$
n \cdot 180^{\circ}-(n-2) \cdot 180^{\circ}=360^{\circ} \text {. }
$$
According to the problem, we have $(n-2) \cdot 180^{\circ}=3 \cdot 360^{\circ}$, which gives $n=8$. Therefore, t... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
$18 \cdot 110$ As shown, $A B C D$ is an isosceles trapezoid, $A D$ $=B C=5, A B=4, D C=10$, point $C$ is on $D F$, and $B$ is the midpoint of the hypotenuse of the right triangle $\triangle D E F$. Then $C F$ equals
(A) 3.25 .
(B) 3.5 .
(C) 3.75 .
(D) 4.0 .
(E) 4.25 . | [Solution] Draw $B H \perp D F$, then $B H / / E F$. Since $A B C D$ is an isosceles trapezoid,
$$
\therefore \quad C H=\frac{1}{2}(D C-A B)=3,
$$
then
$$
D H=D C-C H=10-3=7 \text {. }
$$
Since $B$ is the midpoint of $D E$, and $\frac{D B}{D E}=\frac{D H}{D F}$,
$$
\therefore D F=14 \text {. }
$$
Then $C F=D F-C D=4$... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. If $\sqrt{24-t^{2}}-\sqrt{8-t^{2}}=2$, then
$$
\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=
$$ | II. 1.8.
From the known equation and using the formula $a+b=\frac{a^{2}-b^{2}}{a-b}$, we get $\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=\frac{24-8}{2}=8$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=1, a_{n+1}=\frac{1}{8} a_{n}^{2}+m\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
If for any positive integer $n$, we have $a_{n}<4$, find the maximum value of the real number $m$. | 15. It is easy to know,
$$
\begin{array}{l}
a_{n+1}-a_{n}=\frac{1}{8} a_{n}^{2}-a_{n}+m \\
=\frac{1}{8}\left(a_{n}-4\right)^{2}+m-2 \geqslant m-2 .
\end{array}
$$
Then $a_{n}=a_{1}+\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right)$
$$
\geqslant 1+(m-2)(n-1) \text {. }
$$
If $m>2$, note that, as $n \rightarrow+\infty$,
$$
(m... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
79. A small boat sails upstream from downstream port $A$, and at a point 4 kilometers away from port $A$, it encounters a log drifting with the current. The boat continues its journey to reach upstream port $B$ and immediately turns around, arriving at port $A$ at the same time as the log. Given that the speed of the c... | Reference answer: 8
Key points: Travel problems - meeting and pursuit, variable speed problems | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
22. A graduating class of 56 people is taking a photo, requiring each row to have 1 more person than the row in front of it, and not to stand in just 1 row. In this case, the first row should have $\qquad$ people. | $5$ | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(16) (This sub-question is worth 20 points)
Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with eccentricity $e=\frac{1}{2}$, the endpoints of the minor axis are $B_{1}, B_{2}$, and the foci are $F_{1}, F_{2}$. The radius of the incircle of the quadrilateral $F_{1} B_{1} F_{2} B_{2}$ is $\frac... | Solution:
(I) As shown in the figure,
Let the point of tangency of the incircle of quadrilateral $F_{1} B_{1} F_{2} B_{2}$ with side $B_{1} B_{2}$ be $G$,
connect $O G,|O G|=\frac{\sqrt{3}}{2}$.
From $S_{\triangle O B_{2} F_{2}}=\frac{1}{2}\left|O B_{2}\right| \cdot\left|O F_{2}\right|=\frac{1}{2}\left|B_{2} F_{2}\rig... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Given $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} \in \mathbf{N}$ and satisfy $a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}$
$$
\begin{array}{c}
a_{1} a_{2} a_{3}=b_{1} b_{2} b_{3} \\
a_{1} a_{2}+a_{1} a_{3}+a_{2} a_{3} \neq b_{1} b_{2}+b_{1} b_{3}+b_{2} b_{3}
\end{array}
$$
then the minimum value of the maximum number among... | 10. 8 .
$a_{1}=2, a_{2}=a_{3}=6, b_{1}=b_{2}=3 \quad b_{3}=8$ satisfies the condition. Let $A=\left\{a_{1}, a_{2}, a_{3}\right\}$, $B=\left\{b_{1}, b_{2}, b_{3}\right\}$. If $\max A \cup B \leqslant 7$, then $A \cap B=\varnothing$. Otherwise, without loss of generality, assume $a_{1}=b_{1}$,
then $\left\{\begin{array}{... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Let $a \in \mathbf{R}$, the equation ||$x-a|-a|=2$ has exactly three distinct roots, then $a=$ | The equation $|x-a|=a \pm 2$ has exactly three distinct roots, then $a+2>0, a-2=0 \Rightarrow a=2$.
| 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. A batch of disaster relief supplies is loaded onto 26 trucks, which travel at a constant speed of $v$ kilometers per hour from a city to the disaster area. If the highway between the two places is 400 kilometers long, and the distance between every two trucks must not be less than $\left(\frac{v}{20}\right)^{2}$ kil... | 8. Let the total time required for all supplies to reach the disaster area be $t$ hours. The first truck takes $\frac{400}{v}$ hours to reach the disaster area, and the distance between the 26th truck and the 1st truck is at least $25 \times\left(\frac{v}{20}\right)^{2}$ kilometers. Therefore,
$t \geqslant \frac{400}{v... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Find the maximum positive integer $m$, such that it is possible to fill each cell of an $m$ row by 8 column grid with one of the letters $C, G, M, O$, and the following property holds: for any two different rows in the grid, there is at most one column where the letters in these two rows are the same. | 6. The largest $m=5$.
First, provide the construction, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline $\mathrm{C}$ & $\mathrm{C}$ & $\mathrm{C}$ & $\mathrm{C}$ & $\mathrm{G}$ & $\mathrm{G}$ & $\mathrm{G}$ & $\mathrm{G}$ \\
\hline $\mathrm{C}$ & $\mathrm{G}$ & $\mathrm{G}$ & $\mathrm{G}$ & $\mat... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. 3 piles of peaches have the numbers $93, 70, 63$, respectively. A monkey moves peaches between the 3 piles. It is known that the monkey can move a maximum of 5 peaches each time, and will eat 1 peach during the move from one pile to another. When the number of peaches in the 3 piles is equal, the monkey has eaten a... | 【Analysis】First, calculate the total number of peaches as $93+70+63=226$. Since they are divided into 3 piles with equal numbers, divide 226 by 3 and make the corresponding calculations based on the result.
【Solution】Solution: $93+70+63=226$,
$$
226 \div 3=75 \cdots 1 \text {, }
$$
Therefore, the number of peaches ea... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Given: $b_{1}, b_{2}, b_{3}, b_{4}$ are positive integers, the polynomial $g(z)=(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1-z^{3}\right)^{b_{3}}\left(1-z^{4}\right)^{b_{4}}$ when expanded and terms higher than 4th degree are omitted, becomes $1-2 z$. Also, $\alpha$ is the largest root of the polynomial $f(x)=x^... | 2. $g(z)=(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1-z^{3}\right)^{b_{3}}\left(1-z^{4}\right)^{b_{4}}$ Comparing the coefficients of $z, z^{2}, z^{3}, z^{4}$, we have:
$$
\begin{array}{l}
b_{1}=2, b_{2}=1, b_{3}=2, b_{4}=3, \\
\therefore \quad f(x)=x^{3}-3 x^{2}+1 .
\end{array}
$$
Let the three roots of $f(x)=0$ ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Fill in the same digit in the $\square$ below to make the equation true.
$$
97+\square \times(19+91 \div \square)=321, \quad \square=
$$
$\qquad$ | 【Analysis】 7
$$
\begin{aligned}
97+\square \times(19+91 \div \square) & =321 \\
97+\square \times 19+\square \times 91 \div \square & =321 \\
97+\square \times 19+91 & =321 \\
\square \times 19 & =133 \\
\square & =7
\end{aligned}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Let $O$ be the center of the base $\triangle A B C$ of the regular tetrahedron $P-A B C$. A moving plane through $O$ intersects the three lateral edges or their extensions of $P-A B C$ at points $Q, R, S$ respectively. Then the sum $\frac{1}{P Q}+\frac{1}{P R}+\frac{1}{P S}$
A. has a maximum value but no minimum val... | Let $D$ be the midpoint of $AB$. Without loss of generality, assume the side length of the equilateral triangular pyramid is 1, then
$$
\overrightarrow{CO}=\frac{2}{3} \overrightarrow{CD} \Rightarrow \overrightarrow{PO}=\frac{2}{3} \overrightarrow{PD}+\frac{1}{3} \overrightarrow{PC}=\frac{1}{3} \overrightarrow{PA}+\fra... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
19.3 $ABCDE$ is a regular pentagon, $AP$, $AQ$, and $AR$ are the perpendiculars from $A$ to the lines $CD$, $CB$ extended, and $DE$ extended, respectively. Let $O$ be the center of the pentagon. If $OP=1$, then $AO + AQ + AR$ equals
(A) 3.
(B) $1+\sqrt{5}$.
(C) 4.
(D) $2+\sqrt{5}$.
(E) 5.
(37th American High School Mat... | [Solution] Connect $A D, A C$, and connect $O D, O C$, and let the side length of the regular pentagon be $a$.
$$
\begin{aligned}
\because S_{\triangle B C D E} & =5 S_{\triangle O C D}=5 \cdot \frac{1}{2} \cdot 1 \cdot a \\
& =\frac{5}{2} a,
\end{aligned}
$$
and
$$
\begin{aligned}
S_{A B C D E} & =S_{\triangle A E D}... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
【Question 10】
Place one black and one white Go stone on the grid points of the square in the figure, but the two stones cannot be on the same line. There are $\qquad$ different ways to do this. (Rotations that result in the same position are considered the same) | 【Analysis and Solution】
Counting.
(1) As shown in Figure 1, when the black stone is placed at the corner, there are 4 different ways to place the white stone; (2) As shown in Figure 2, when the black stone is placed on the edge, there are 4 different ways to place the white stone; (3) As shown in Figure 3, when the bla... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Let $F_{1}, F_{2}$ be the left and right foci of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, respectively. A line passing through point $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$, with $\left|A F_{1}\right|=3\left|B F_{1}\right|$ and $\cos \angle A F_{2} B=\frac{3}{5}$. The line ... | 7. -1
Analysis: Let $\left|F_{1} B\right|=k$, then $k>0$ and $\left|A F_{1}\right|=3 k, |A B|=4 k$. By the definition of an ellipse, we have $\left|A F_{2}\right|=2 a-3 k, \left|B F_{2}\right|=2 a-k$. In $\triangle A B F_{2}$, by the cosine rule, we get $|A B|^{2}=\left|A F_{2}\right|^{2}+\left|B F_{2}\right|^{2}-2\le... | -1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. If the polynomial in $x$
$$
y=x^{3}+3 x^{2}+6 x+c
$$
has three roots that can form a geometric progression, then the constant $c=$ $\qquad$ | 6. 8 .
Let the three roots form a geometric sequence $\alpha, \beta, \gamma$, i.e., $\alpha \gamma=\beta^{2}$ and $\alpha \beta \gamma \neq 0$.
According to Vieta's formulas, we have
$$
\begin{array}{l}
\alpha+\beta+\gamma=-3, \alpha \beta+\beta \gamma+\gamma \alpha=6 \text {. } \\
\text { Hence } 6=\alpha \beta+\beta... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 14 The coefficients $a_{0}, a_{1}, \cdots, a_{n}$ of the polynomial $a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ take values +1 or -1, and the equation $a_{0}+a_{1} x+\cdots+a_{n} x^{n}=0$ has all real roots. Find all polynomials with the above properties.
(29th Putnam Problem) | Obviously, the polynomial we are looking for is neither the zero polynomial nor a zero-degree polynomial. Let the degree of the polynomial we are looking for be $n$.
When $n=1$, the four possible polynomials are: $\pm(x+1), \pm(x-1)$, all of which meet the requirements.
When $n \geqslant 2$, let the roots of equation... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.7 If $\angle A=20^{\circ}, \angle B=25^{\circ}$, then the value of $(1+\operatorname{tg} A)(1+\operatorname{tg} B)$ is
(A) $\sqrt{3}$.
(B) 2 .
(C) $1+\sqrt{2}$.
(D) $2(\operatorname{tg} A+\operatorname{tg} B)$.
(E) None of the above conclusions is correct.
(36th American High School Mathematics Examination, 1985) | [Solution] Since $\operatorname{tg}\left(20^{\circ}+25^{\circ}\right)=\operatorname{tg} 45^{\circ}=1$, and
then
$$
\begin{array}{c}
\operatorname{tg}\left(20^{\circ}+25^{\circ}\right)=\frac{\operatorname{tg} 20^{\circ}+\operatorname{tg} 25^{\circ}}{1-\operatorname{tg} 20^{\circ} \operatorname{tg} 25^{\circ}} \\
\opera... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
42. There are 5 steps in front of the little squirrel's house. If it is stipulated that one step can only climb one or two steps, then there are $\qquad$ different ways to climb these 5 steps. | Answer: 8 | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. (10 points) On a plane, several unit-length sticks can be used to form a pattern of adjacent squares, as shown in the example. Now, using 20 unit-length sticks to form a pattern, it is required that except for the squares in the first row, the squares in the rows below form a rectangle. How many unit squares can su... | 【Analysis】From the above figure, we can see that only when the sides of the small squares are adjacent can we save small sticks, and the closer the shape is to a large square, the more sticks we save. Therefore, this problem can start from a $2 \times 2$ square and a $3 \times 3$ square. From the above figure, we can s... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$$
\begin{array}{l}
A=\{2,0,1,7\}, \\
B=\{x \mid x=a b, a, b \in A\} .
\end{array}
$$
The number of elements in set $B$ is $\qquad$ | $$
\begin{array}{l}
\text { I. 1.7. } \\
\text { It is easy to see that, } B=\{0,1,4,49,2,7,14\} \text {. }
\end{array}
$$
Therefore, $|B|=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. For any real numbers $A, B, C$, the maximum value of the trigonometric expression $\sin ^{2} A \cos ^{2} B+\sin ^{2} B \cos ^{2} C+\sin ^{2} C \cos ^{2} A$ is
$\qquad$ . | 8. 1 Detailed Explanation: $1=\left(\sin ^{2} A+\cos ^{2} A\right)\left(\sin ^{2} B+\cos ^{2} B\right)\left(\sin ^{2} C+\cos ^{2} C\right)=\sin ^{2} A \cos ^{2} B\left(\sin ^{2} C+\cos ^{2} C\right)+$ $\sin ^{2} B \cos ^{2} C\left(\sin ^{2} A+\cos ^{2} A\right)+\sin ^{2} C \cos ^{2} A\left(\sin ^{2} B+\cos ^{2} B\right... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16. A rectangular thin wooden board, 7 meters long and 5 meters wide, has a small rectangular area at one corner damaged by insects, the damaged part is 2 meters long and 1 meter wide. Please design a method to saw off the damaged area in one straight cut, making the area removed as small as possible. What is the minim... | 【Answer】4 square meters
【Solution】As shown in the figure, let $MG=x, EN=y$
Obviously, $\triangle MGF \sim \triangle FEN$, so: $\frac{x}{2}=\frac{1}{y} \Rightarrow xy=2$
$S_{\triangle MGF}=\frac{x}{2}, \quad S_{\triangle AE:}=y$
Find the minimum value of $\frac{x}{2}+y$
$\frac{x}{2}+\frac{2}{x} \leq 2 \times \frac{x}{2}... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. If $P$ is the circumcenter of $\triangle A B C$, and
$$
\overrightarrow{P A}+\overrightarrow{P B}+\lambda \overrightarrow{P C}=\mathbf{0}, \angle C=120^{\circ} \text {. }
$$
then the value of the real number $\lambda$ is $\qquad$ | 5. -1 .
Let the circumradius of $\triangle ABC$ be $R$.
From the given information, we have
$$
\begin{array}{l}
|\overrightarrow{P A}+\overrightarrow{P B}|^{2}=\lambda^{2}|\overrightarrow{P C}|^{2} \\
=|\overrightarrow{P A}|^{2}+|\overrightarrow{P B}|^{2}-2|\overrightarrow{P A}||\overrightarrow{P B}| \cos \frac{C}{2} ... | -1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (3 points) 50 students answered questions $A$ and $B$. Among them, 12 did not answer $A$ correctly, and 30 answered $A$ correctly but did not answer $B$ correctly. Therefore, the number of students who answered both $A$ and $B$ correctly is $\qquad$ people. | 【Answer】Solution: $50-12-30=38-30=8$ (people);
Answer: The number of people who answered both questions $A$ and $B$ correctly is 8.
Therefore, the answer is: 8. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
20. 10 people go to the bookstore to buy books. If it is known that: (1) each person bought three books; (2) any two people have at least one book in common. How many people at least bought the most popular book (the one bought by the most people)? Why? | 20. Let the most popular book be bought by $k$ people, and suppose $A_{1}$ bought books $a, b, c$. Each of the other 9 people bought at least one book that is the same as $a, b, c$, so at least $\left[\frac{9-1}{3}\right]+1=3$ people (denoted as $A_{2}, A_{3}, A_{4}$) also bought the same book among $a, b, c$. Therefor... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (6 points) The calculation result of the expression $\left(1 -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right) \div\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)$ is | $$
\begin{array}{l}
\text { Original expression }=\left(1-\frac{1}{2}-\frac{1}{4}+\frac{1}{5}+\frac{1}{3}-\frac{1}{6}\right) \div\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right) \\
\left(\left(\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right) \div\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)=1\right)\right)
\end{arra... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Let $x, y \in \mathbf{R}$, if $2 x, 1, y-1$ form an arithmetic sequence, and $y+3,|x+1|+|x-1|, \cos (\arccos x)$ form a geometric sequence, then the value of $(x+1)(y+1)$ is $\qquad$. | 1. 4 .
From $2 x, 1, y-1$ forming an arithmetic sequence, we get
$$
y=3-2 x .
$$
From $\cos (\arccos x)=x$ and $-1 \leqslant x \leqslant 1$, we deduce $|x+1|+|x-1|=2$. Therefore, $y+3,2, x$ form a geometric sequence, which gives
$$
x(y+3)=4 \text {. }
$$
Substituting equation (1) into equation (2), we get $2 x^{2}-6... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. 182 For 1 to 1000000000, find the sum of the digits of each number; then for the resulting 1 billion numbers, find the sum of the digits of each number, $\cdots$, until obtaining 1 billion single-digit numbers. Question: Among the resulting numbers, are there more 1s or 2s? | [Solution] 1 is 1 more than 2.
In fact, any number and the sum of its digits have the same remainder when divided by 9. Therefore, in this problem, 1 is obtained from those numbers that give a remainder of 1 when divided by 9: 1, 10, 19, 28, ..., 999999991, 1000000000, while 2 is obtained from those numbers that give a... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The equation $x^{2}-4[x]+3=0, x \in \mathbf{R}$ has
A. 2 solutions
B. 3 solutions
C. 1 solution
D. 4 solutions | 3. B From $x \leqslant[x]<x+1$, we know $2-\sqrt{5}<x<2+\sqrt{5}$. Also, $x^{2}=4[x]-3 \in \mathbf{Z}, x^{2} \leqslant 17, -1 \leqslant[x] \leqslant$ 4, substituting into the equation, we get $x=3, \sqrt{5}$ or 1. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. If 653 divides $\overline{a b 2347}$, then $a+b=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 【Answer】 11
【Solution】Since $653|\overline{a b 2347} \Leftrightarrow 653|(\overline{a b 2347}+653)$, considering $\overline{a b 2347}+653=\overline{a b 3000}$, so $653 \mid \overline{a b 3} \times 1000$. Since 653 is a prime number, and 653 cannot divide 1000, it follows that $653 \mid \overline{a b 3}$, thus leading t... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
17. Given the equation about $x$, $3 x-4=a$ and $\frac{x+a}{3}=1$, if the root of the former is twice the root of the latter, then the value of the constant $a$ is ( ).
(A) 0.
(B) 1.
(C) 2.
(D) 4. | Answer: C.
Solution: Solving the equation $3 x-4=a$, we get
$$
x=\frac{a+4}{3} .
$$
Solving the equation $\frac{x+a}{3}=1$, we get
$$
x=3-a \text {. }
$$
According to the given condition, we have
$$
\frac{a+4}{3}=2(3-a),
$$
Solving this, we get
$$
a=2 \text {. }
$$
Therefore, the correct choice is C. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. Given that $a$, $b$, and $c$ are three distinct real numbers. If any two of the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$
have exactly one common root, find the value of $a^{2}+b^{2}+c^{2}$. | 2. From equations (1) and (2), we know their common root is $p=\frac{b-c}{b-a}$.
Similarly, the common roots of equations (2) and (3), and equations (1) and (3) are $q=\frac{c-a}{c-b}$ and $r=\frac{a-b}{a-c}$, respectively.
Thus, $p q r=-1$.
If any two of $p, q, r$ are equal, assume without loss of generality that $p=... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. On a plane, there are six points, any two of which are at a distance $\geqslant 1$, then the number of pairs of points at a distance of 1 is at most nine. | 10. For each point $P_{i}$, connect it with the point $P_{j}$ that has the minimum distance of 1 to form a graph $G$, and the convex hull of $G$ is $C o G$.
It is not hard to prove: (1) If $P_{i}$ is a vertex of $C o G$, then for the graph $G$, $d\left(P_{i}\right) \leqslant 3$;
(2) If $C o G$ is a convex $m$-gon, $m>1... | 9 | Geometry | proof | Yes | Yes | olympiads | false |
Example 2 In $\triangle A B C$, $\angle A B C=40^{\circ}, \angle A C B=20^{\circ}, N$ is a point inside $\triangle A B C$, $\angle N B C=30^{\circ}$, $\angle N A B=20^{\circ}$. Find the degree measure of $\angle N C B$. | Analysis: According to the Law of Sines, establish a trigonometric equation, and then solve the equation using trigonometric knowledge.
Solution: Let $\angle N C B=\alpha$, then $\angle N C A=20^{\circ}-\alpha$.
In $\triangle A B N$, $\angle A B N=\angle A B C-\angle N B C=10^{\circ}, \angle N A B=20^{\circ}$. By the... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The 34th question: A set of $n$ points $P_{1}, P_{2}, \ldots, P_{n}$ in the plane, denoted as $D$, satisfies that no three points are collinear. A line segment is drawn between any two points, and the lengths of all these line segments are distinct. In a triangle, the side that is neither the longest nor the shortest i... | Question 34:
Solution: First, we prove a lemma: Given any 6 points, with all pairwise distances being distinct, there must be a median-edge triangle.
Proof of the lemma: If an edge is the median edge of a triangle, then color it red; otherwise, color it blue. According to Ramsey's theorem, in a two-colored \( \mathrm{... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
15. (15 points) There are 8 football teams participating in a round-robin tournament. The winning team gets 1 point, the losing team gets 0 points, and in the case of a draw, both teams get 0.5 points. After the tournament, the teams are ranked based on their points from highest to lowest, and it is found that: all tea... | 【Answer】Solution: Each team needs to play 7 matches, so a team that wins all matches gets 7 points,
while the last four teams playing 6 matches among themselves will score at least 6 points in total,
thus the second-place team must score at least 6 points.
If the first-place team wins all matches, then the second-place... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (10 points) Four pieces of paper with 4 squares each, as shown in Figure 1, are used to form the shape shown in Figure 2. If the numbers $1, 3, 5, 7$ (one number per square) are filled into the 16 squares in Figure 2, such that no number is repeated in any row, column, or within any of the four pieces of paper, then... | 【Analysis】As shown in Figure 2,
According to the rule that the four numbers in each cardboard do not repeat, we can get: $A$ $\neq E, A \neq F, B \neq E, B \neq F$, so $A=G, B=H$ or $A=H, B=G$, so $G+H=A+B$, based on this, we need to find the average of the numbers in the four squares $A, B, C, D$.
【Solution】As shown ... | 4 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
$4 \cdot 218$ In a middle school mathematics competition, three problems $A$, $B$, and $C$ were given. Among the 25 students who participated in the competition, each student solved at least one problem. Among those who did not solve $A$, the number of students who solved $B$ is twice the number of students who solved ... | [Solution]. Let the number of people who solved not only $A$ be $x$, only $B$ be $y$, and did not solve $A$ but solved $B$ and $c$ be $z$. Then (refer to the diagram on the previous page),
$$
\left\{\begin{array}{l}
x+x+1+y+z+\frac{y-z}{2}=25, \\
x+1=y+\frac{y-z}{2} .
\end{array}\right.
$$
Solving these equations, we ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 4 Given that the area of quadrilateral $A B C D$ is $9, P$ is a point inside the quadrilateral. Let the centroids of $\triangle A B P, \triangle B C P, \triangle C D P$, and $\triangle D A P$ be $G_{1}, G_{2}, G_{3}, G_{4}$ respectively. Find the area of quadrilateral $G_{1} G_{2} G_{3} G_{4}$. | Let $e=(\overrightarrow{A D} \times \overrightarrow{A B})^{0}$.
Then $18 S_{\text {quadrilateral } G_{1} G_{2} G_{3} G_{4}} e$
$$
\begin{array}{l}
=9\left(G_{3}-G_{1}\right) \times\left(G_{4}-G_{2}\right) \\
=(C+D-A-B) \times(D+A-B-C) \\
=2(C-A) \times(D-B) \\
=4 S_{\text {quadrilateral } A B C D} e .
\end{array}
$$
T... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
(7) Let $a, b, c$ be non-negative real numbers, then the minimum value of $\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}$ is | 7: According to the AM-GM inequality, we get
$$
\begin{aligned}
\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c} & =\frac{c}{a}+\frac{a}{b+c}+\frac{b+c}{c}-1 \\
& \geqslant 3 \sqrt[3]{\frac{c}{a} \cdot \frac{a}{b+c} \cdot \frac{b+c}{c}}-1 \\
& =2 .
\end{aligned}
$$ | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
117 Let the geometric sequence $\left\{z_{n}\right\}$ be such that $z_{1}=1, z_{2}=a+b i, z_{3}=b+a i(a, b \in \mathbf{R}, a>0)$, then the smallest $n$ that satisfies $z_{1}+z_{2}+\cdots+z_{n}=0$ is
A. 6
B. 12
C. 18
D. 24 | 117 B. According to the problem,
$$
(a+b i)^{2}=b+a i \quad(a, b \in \mathbf{R}),
$$
which means
$$
\left\{\begin{array}{l}
a^{2}-b^{2}=b, \\
2 a b=a .
\end{array}\right.
$$
Solving these equations, we get
$$
\left\{\begin{array}{l}
a=\frac{\sqrt{3}}{2}, \\
b=\frac{1}{2},
\end{array}\right.
$$
Therefore,
$$
q=z_{2}=... | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. In $\triangle A B C$, the internal angles $A, B, C$ are opposite the sides $a, b, c$ respectively, and $\sin C \cos \frac{A}{2}=(2-\cos C) \sin \frac{A}{2}$, $\cos A=\frac{3}{5}, a=4$, then the area of $\triangle A B C$ is $\qquad$ . | Solve:
$\cos A=\frac{1-\tan ^{2} \frac{A}{2}}{1+\tan ^{2} \frac{A}{2}}=\frac{3}{5} \Rightarrow \tan \frac{A}{2}=\frac{1}{2}$, thus $\frac{\sin C}{2-\cos C}=\tan \frac{A}{2}=\frac{1}{2}$
$\Rightarrow 2 \sin C+\cos C=2 \Rightarrow 3 \cos ^{2} C-4 \sin C \cos C=0 \Rightarrow C=90^{\circ}$ or $\tan C=\frac{3}{4}$.
When $C=... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. Given: $\sin \alpha=\cos \beta, \cos \alpha=\sin 2 \beta$, then $\sin ^{2} \beta+\cos ^{2} \alpha=$ $\qquad$ | 11. 0 or $\frac{3}{2} \quad \sin \alpha=\cos \beta(1), \cos \alpha=\sin 2 \beta(2)$ By squaring and adding equations (1) and (2), we get $\sin ^{2} 2 \beta=\sin ^{2} \beta$ Therefore, $\sin ^{2} \beta$ $=0$ or $\cos ^{2} \beta=\frac{1}{4}$, then $\sin ^{2} \beta+\cos ^{2} \alpha=0$ or $\frac{3}{2}$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. If for any $x$, we have
$$
\sin ^{n} x+\cos ^{n} x \geqslant \frac{1}{n},
$$
then the maximum value of $n$ is $\qquad$ | 7.8.
Let $x=\frac{\pi}{4}$, then
$$
\begin{array}{l}
\sin ^{n} x+\cos ^{n} x=2\left(\frac{1}{\sqrt{2}}\right)^{n}=\frac{1}{2^{\frac{n-2}{2}}} \geqslant \frac{1}{n} \\
\Rightarrow n \geqslant 2^{\frac{n-2}{2}} .
\end{array}
$$
Notice that, $2^{\frac{9-2}{2}}>9$. Therefore, $n \leqslant 8$.
Also, by the formula $\left(... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. (2000 "Hope Cup" Invitational Competition Question) Let $a>b>c$, and $\frac{1}{a-b}+\frac{1}{b-c} \geqslant \frac{n}{a-c}$ always holds, then what is the maximum value of $n$? | $$
\begin{array}{l}
\text { 6. Let } \vec{a}=\left(\frac{1}{\sqrt{a-b}}, \frac{1}{\sqrt{b-c}}\right), \vec{b}=(\sqrt{a-b}, \sqrt{b-c}), \\
\frac{1}{a-b}+\frac{1}{b-c}=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(1+1)^{2}}{(a-b)+(b-c)}=\frac{4}{a-c} .
\end{array}
$$
Then the maximum ... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. There are 53 books of mathematics and physics on the bookshelf, where no two physics books are placed next to each other, but every mathematics book is adjacent to another mathematics book. Given the following four statements:
(1) There are at least 35 mathematics books;
(2) There are at most 18 physics books;
(3) T... | 5. C.
The scenario with the fewest math books $(S)$ occurs when the number of physics books (W) is at its maximum, as shown below:
Three math books SSS can also appear in other positions simultaneously, as indicated by the arrangements showing (1), (2), and (4) are correct; statement (3) does not hold in the followin... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7$\cdot$70 Alice and Bob came to a hardware store, where colored tassels are sold to be tied on keys to distinguish different keys. Below is a segment of their conversation:
Alice: Are you going to buy some colored tassels to tie on your keys?
Bob: I would like to do that, but the tassels come in only 7 different color... | [Solution] (1) When $n=1$, 1 color of tassel is needed.
(2) When $n=2$, 2 colors of tassels are needed.
(3) When $n \geqslant 3$, only 3 colors of tassels are needed. As shown in the figure, let the keys with blue and green tassels be adjacent, and the rest are tied with red tassels. We just need to remember the number... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $n$ be a natural number. For any real numbers $x, y, z$, it always holds that $\left(x^{2}+y^{2}+z^{2}\right)$ $\leqslant n\left(x^{4}+y^{4}+z^{4}\right)$. Then the minimum value of $n$ is $\qquad$. | 3. Let $a=x^{2}, b=y^{2}, c=z^{2}$, the given inequality becomes
$$
(a+b+c)^{2} \leqslant n\left(a^{2}+b^{2}+c^{2}\right) .
$$
On one hand,
$$
\begin{aligned}
& (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c \\
\leqslant & a^{2}+b^{2}+c^{2}+\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+\left(a^{2}+c^{2}\right) \\... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 81971^{26}+1972^{27}+1973^{28}$ Can it be divisible by 3?
(Kiev Mathematical Olympiad, 1971) | [Solution] $1971 \equiv 0 \quad(\bmod 3)$, so
$$
1971^{26} \equiv 0 \quad(\bmod 3) .
$$
Also, $1972 \equiv 1(\bmod 3)$, so
$$
\begin{aligned}
1972^{27} & \equiv 1^{27} \quad(\bmod 3), \\
1973 & \equiv 2 \quad(\bmod 3), \\
1973^{28} & \equiv 2^{28} \quad(\bmod 3) .
\end{aligned}
$$
Thus, we have
$$
1971^{26}+1972^{27}... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4 Find all functions $f$ from the set of positive integers to the set of real numbers, such that: (1) for any $n \geqslant 1, f(n+1) \geqslant f(n)$; (2) for any $m, n, (m, n)=1$, we have $f(m n)=f(m) f(n) . \quad$ (Supplied by Pan Chengbiao) | Solution: Clearly, $f=0$ is a solution to the problem. Suppose $f \neq 0$, then $f(1) \neq 0$, otherwise for any positive integer $n$ we have $f(n)=f(1) f(n)=0$, which is a contradiction! Thus, we get $f(1)=1$.
From (1), we know $f(2) \geqslant 1$, and there are two cases:
(i) $f(2)=1$, then we can prove
$$
f(n)=1(\for... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11. (5 points) As shown in the figure, two iron rods are vertically placed in a wooden bucket with a flat bottom. After adding water to the bucket, the length of one iron rod above the water is $\frac{1}{3}$ of its total length, and the length of the other iron rod above the water is $\frac{1}{5}$ of its total length. ... | 【Solution】Solution: Let the underwater length of the two iron rods be $x$ cm. According to the problem, we can get the equation:
$$
\begin{array}{l}
x \div \frac{2}{3}+x \div \frac{4}{5}=33 \\
\frac{3}{2} x+\frac{5}{4} x=33 \\
\frac{11}{4} x=33 \\
x=12
\end{array}
$$
Then the difference in length of the two iron rods... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In the same Cartesian coordinate system, there are infinitely many lines with the expression $y=k x+b(k, b$ being real numbers, $k \neq 0)$, (they form a family of lines). Among these lines, no matter how they are selected, it must be ensured that at least two lines pass through exactly the same quadrants. The minim... | 6. $\mathrm{D}$ Hint: Classify by the quadrant the line is in, there are 6 categories. When $k>0$, $\left\{\begin{array}{l}b>0 \\ b=0 ; k>0 \\ b<0 . \text { By the pigeonhole principle, at least } \\ b=0\end{array}\right.$ 7 lines need to be drawn. | 7 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
12. The Smurf loves learning, but she is stuck on the following calculation problem. Can you help her? Calculate: $5.4321 \times 0.5679 - 0.4321 \times 5.5679 + 0.321=$ $\qquad$ . | $1$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In the 99 positive integers $1,2, \cdots, 99$, if any $k$ numbers are taken out, such that there must be two numbers $a, b(a \neq b)$ satisfying $\frac{1}{2} \leqslant \frac{b}{a} \leqslant 2$. Then the smallest possible value of $k$ is
A. 6
B. 7
C. 8
D. 9 | 1. B divides the 99 positive integers from 1 to 99 into 6 groups, such that the ratio of any two numbers in each group is within the closed interval $\left[\frac{1}{2}, 2\right]$, and the number of elements in each group is as large as possible. The groups are as follows: $A_{1}=\{1,2\}, A_{2}=\{3,4,5,6\}, A_{3}=\{7,8,... | 7 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
2. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M, N$ are points on the diagonals $A C$, $B F$ respectively, and $A M=F N$. If $A B=1$, then the maximum value of $M N$ is $\qquad$ . | 2. 1 Analysis: As shown in the figure, draw $M P \perp A B$ at $P$, connect $P N$, it can be proven that $P N \perp A B$, thus $\angle M P N=120^{\circ}$. Let $A M=A N=x$, then, $\frac{M P}{1}=\frac{A M}{\sqrt{2}}$, so $M P=\frac{x}{\sqrt{2}}$, similarly, $P N=\frac{\sqrt{2}-x}{\sqrt{2}}$, in $\triangle M N P$, using t... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12. The system of equations $\left\{\begin{array}{l}x+3 y=3, \\ || x|-| y||=1\end{array}\right.$ has ( ) solutions.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 8 | 12. C.
Graph the functions $x+3 y=3,|| x|-| y||=1$ (as shown in Figure 3), the two functions have three intersection points: $(-3,2)$, $(0,1),\left(\frac{3}{2}, \frac{1}{2}\right)$. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
A1 Determine the positive integer $n$ that satisfies the following equation:
$$
\frac{1}{2^{10}}+\frac{1}{2^{9}}+\frac{1}{2^{8}}=\frac{n}{2^{10}} .
$$ | Solution
Adding the left hand side of the given equation with with a common denominator of $2^{10}$, we have,
$$
\frac{1}{2^{10}}+\frac{1}{2^{9}}+\frac{1}{2^{8}}=\frac{1}{2^{10}}+\frac{2}{2^{10}}+\frac{2^{2}}{2^{10}}=\frac{1+2+4}{2^{10}}=\frac{7}{2^{10}} .
$$
Therefore, $n=7$. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let the complex number $z=\cos \frac{4 \pi}{7}+\mathrm{i} \sin \frac{4 \pi}{7}$. Then the value of $\left|\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}\right|$
is $\qquad$ (answer with a number). | 5.2.
Given that $z$ satisfies the equation $z^{7}-1=0$.
From $z^{7}-1=(z-1) \sum_{i=0}^{6} z^{i}$, and $z \neq 1$, we get
$$
\begin{array}{l}
z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=0 . \\
\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}} \text {, after finding a common denominator, the denominator is } \\
\left... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$31 \cdot 1$ A merchant buys $n$ radios for $d$ dollars, where $d$ is a positive integer. Two of them he sells to a charity sale at half cost, and the rest he sells at a profit of 8 dollars each. If the total profit is 72 dollars, then the smallest possible value of $n$ is
(A) 18 .
(B) 16 .
(C) 15 .
(D) 12 .
(E) 11
(18... | [Solution] Given that the cost of each radio is $\frac{d}{n}$ yuan, according to the problem, we have
$$
(n-2)\left(\frac{d}{n}+8\right)+2 \cdot \frac{d}{2 n}=d+72 \text {, }
$$
The equation can be simplified to $n(n-11)=\frac{d}{8}$, where $n, d$ are positive integers.
If $n \leqslant 11$, then $n(n-11) \leqslant 0$,... | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
9. (5 points) Xiaohong brought 5 RMB notes of 50 yuan, 20 yuan, and 10 yuan denominations, 6 notes of 20 yuan, and 7 notes of 10 yuan. She bought a commodity worth 230 yuan. Therefore, there are $\qquad$ ways to pay. | 【Analysis】To make up 230 using 50, 20, and 10, list all possible ways using the enumeration method.
【Solution】Solution: According to the denomination from largest to smallest, enumerate all possible combinations as shown in the table below:
\begin{tabular}{|c|c|c|}
\hline Denomination & \multicolumn{2}{|c|}{ Number of ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. If $\mathrm{i}$ is the imaginary unit, and the complex number $z=\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}$, then the value of $z^{2016}$ is
A. -1
B. $-\mathrm{i}$
C. i
D. 1 | $$
\begin{array}{l}
z=\cos 30^{\circ}+\mathrm{i} \sin 30^{\circ} \Rightarrow z^{2016}=\cos \left(2016 \cdot 30^{\circ}\right)+\mathrm{i} \sin \left(2016 \cdot 30^{\circ}\right) \\
=\cos \left(168 \cdot 360^{\circ}\right)+\mathrm{i} \sin \left(168 \cdot 360^{\circ}\right)=1, \text{ so the answer is } D .
\end{array}
$$ | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. Let $a, b$ be positive integers, and $a-b \sqrt{3}=(2-\sqrt{3})^{100}$, then the units digit of $a b$ is | Let $a_{n}-b_{n} \sqrt{3}=(2-\sqrt{3})^{n}$, then $a_{n+1}-b_{n+1} \sqrt{3}=(2-\sqrt{3})^{n+1}$
$$
=\left(a_{n}-b_{n} \sqrt{3}\right)(2-\sqrt{3})=2 a_{n}+3 b_{n}-\left(a_{n}+2 b_{n}\right) \sqrt{3} \Rightarrow\left\{\begin{array}{l}
a_{n+1}=2 a_{n}+3 b_{n}, \\
b_{n+1}=a_{n}+2 b_{n} .
\end{array}\right.
$$
Thus, $a_{1}... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 2 Solve the system of simultaneous equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3 .
\end{array}\right.
$$
Find all real or complex roots.
(2nd USA Mathematical Olympiad) | Let $T_{n}=x^{n}+y^{n}+z^{n} (n=1,2, \cdots)$, then $T_{1}=T_{2}=T_{5}=3$.
Let $x, y, z$ be the three roots of the cubic polynomial
$$
f(t)=(t-x)(t-y)(t-z)=t^{3}-b_{1} t^{2}+b_{2} t-b_{3}
$$
Then, by the given conditions, we have $b_{1}=T_{1}=3, b_{2}=\frac{1}{2}\left(T_{1}^{2}-T_{2}\right)=3$. Next, we calculate $b_{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Let $f(x)=\frac{\sin \pi x}{x^{2}}(x \in(0,1))$. Then the minimum value of $g(x)=f(x)+f(1-x)$ is . $\qquad$ | 7.8.
From the given, we have
$$
\begin{aligned}
f^{\prime}(x) & =\frac{\pi x \cos \pi x-2 \sin \pi x}{x^{3}}, \\
f^{\prime \prime}(x) & =\frac{\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x}{x^{3}} .
\end{aligned}
$$
Next, we need to prove
$$
\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x>0.
$... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.