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Example 1. A pile of toothpicks 1000 in number, two people take turns to pick any number from it, but the number of toothpicks taken each time must not exceed 7. The one who gets the last toothpick loses. How many toothpicks should the first player take on the first turn to ensure victory? (New York Math Competition) | From $1000=125 \times 8$, we know that one should first dare to take 7 moves, so that the latter can achieve a balanced state: $124 \times(7+1)+1$. | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. In $\triangle A B C$, $D, E, F$ are the midpoints of $B C$, $C A, A B$ respectively, and $G$ is the centroid. For each value of $\angle B A C$, how many non-similar $\triangle A B C$ are there such that $A E G F$ is a cyclic quadrilateral? | 1. From $A, E, G, F$ being concyclic, we get
$$
\begin{aligned}
\angle C G_{E} & =\angle B A C \\
= & \angle C E D .
\end{aligned}
$$
Let $C G$ intersect $D E$ at $M$. Then from (1),
it is easy to derive
$$
C M \times C G=C E^{2},
$$
i.e., $\frac{1}{2} m_{\mathrm{e}} \times \frac{2}{3} m_{\mathrm{c}}=\left(\frac{1}{2... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
On a plane, there is a convex quadrilateral $A B C D$.
(1) If there exists a point $P$ on the plane such that the areas of $\triangle A B P, \triangle B C P, \triangle C D P, \triangle D A P$ are all equal, what condition should the quadrilateral $A B C D$ satisfy?
(2) How many such points $P$ can there be on the plane... | . Solving (i) First consider the case where $P$ is inside the quadrilateral.
If points $A, P, C$ are collinear, and points $B, P, D$ are also collinear, then quadrilateral $ABCD$ is a parallelogram, and $P$ is the intersection of the two diagonals.
If points $A, P, C$ are not collinear, by the fact that $\triangle PAB... | 1 | Geometry | proof | Yes | Yes | cn_contest | false |
Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are on a circle. How many birds are there on the circle with the most birds, at minimum?
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。
---
Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are... | We use 10 points to represent 10 birds.
(1) Among the 10 points, there must be 5 points that are concyclic. If not, then any 5 points among the 10 points are not concyclic. The 10 known points can form $C_{10}^{5}=252$ groups of 5 points. According to the given, each group of 5 points has four points that are concyclic... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. (Shanghai $\cdot$ Senior High School) If positive integers $p, q, r$ make the quadratic equation $p x^{2}-{ }_{y} x+r=0$ have two distinct real roots in the open interval $(0,1)$, find the minimum value of $p$.
| Let $\alpha, \beta$ be the two roots of the quadratic equation $p x^{2}-q x+\gamma=0$, $\alpha \neq \beta, 0<\alpha, \beta<1$. Then
$$
\alpha \beta(1-\alpha)(1-\beta) = \frac{r}{p} \left[1 - \frac{q}{p} + \frac{r}{p}\right] = \frac{r}{p^{2}}(p-q+r).
$$
Therefore, $p^{2} > 16 r(p-q+r)$.
Since the quadratic term coeffic... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. (First State "Killing Hope Cup" - Grade 10) Let the function $f(n)=k$, where $n$ is a natural number, and $k$ is the $n$-th digit after the decimal point of the irrational number $\pi=3.1415926535 \cdots$. It is also defined that $f(0)=3$.
$$
\text { Let } F(n)=\frac{f(f(f(f \cdots(f(n)) \cdots)))}{10 \uparrow \bar{... | It can be easily proven that for non-negative integers $n$, $F(n)=1$ always holds.
Below $f(1990)+f(5)+f(13)$ and $f(1990) f(3)$
- $f(25)$ are all non-negative integers, so $F(f(1990)+f(5)$
$$
+f(13) \equiv 1 \equiv F[(1990) f(3) f(25)] .
$$ | 1 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 1. (14th All-Russian Mathematics Competition) Let
$$
\frac{1}{1+\frac{1}{1+\frac{1}{1+}}+\frac{1}{1}}=-\frac{m}{n},
$$
where $m$ and $n$ are coprime natural numbers, and the left side of the equation has 1988 fraction lines. Calculate $m^{2}+m n-n^{2}$. | Let the value of the complex fraction with $k$ layers of fraction lines be $\frac{m_{k}}{n_{k}},\left(m_{k}, n_{k}^{*}\right)=1$, then
$$
\frac{m_{\mathrm{k}+1}}{n_{\mathrm{k}+1}}=\frac{1}{1+\frac{m_{\mathrm{k}}}{n_{\mathrm{k}}}}=\frac{n_{\mathrm{k}}}{m_{\triangle}+n_{\mathrm{k}}},
$$
i.e., $m_{\mathrm{k}+1}=n_{\mathr... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
19. (Wuhu City)
In a square plate $ABCD$ with a side length of 8 units, there is an equilateral triangular colored plate $OEF$ (vertex $O$ is painted red, $E, F$ are white) placed as shown (E coincides with D, and $EF$ coincides with $AD$). Now, the triangular colored plate rotates to the left with $F$ as the pivot un... | The "left advance" rule for the triangular color board is:
?.
(1) Every 3 left advances, the red vertex is inside the disk;
(2) Every 16 left advances, the triangular board returns to its original starting position, but the red vertex moves to the point F on the AD side;
(3) Every 48 left advances, the triangular board... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. (Lokni Army) Let $A$ be a finite set, $N=\{1, 2, 3, \cdots\}$. If there exists a function $f: N \rightarrow A$ with the following property: if $|i-j|$ is a prime number, then $f(i) \neq f(j)$. Try to find the minimum number of elements in $A$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Consider in $N$ taking $i=1,3,6,8$. The absolute value of the difference between any two of these numbers is a prime number, so their images $f(1), f(3), f(6), f(8)$ are the same, meaning that the set $A$ must have at least 4 elements.
Below, we prove that the minimum number of elements in set $A$ is 4. For this, we c... | 4 | Logic and Puzzles | other | Yes | Yes | cn_contest | false |
8. How many real numbers $a$ are there such that $x^{2}+a x+6 a$ $=0$ has only integer solutions?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note at the end is not part of the translation but is provided to clarify that the format and structure of the original text have been maintained in the translation. | 8. Let $x^{2}+a x+6 a=$ O have integer solutions $m, n$ $(m \leqslant n)$. Then we have
$$
a=-(m+n), \quad 6 a=m n .
$$
Since $a$ must be an integer, we also have
$$
-6(m+n)=m n \text { . }
$$
H $(m+6)(n+6)=36$. Since $36=36 \times 1=18 \times 2=12 \times 3=6 \times 6$, the solutions satisfying $m \leqslant n$ are
$$... | 10 | Other | math-word-problem | Yes | Yes | cn_contest | false |
15. For a positive integer $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value. Where $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ are positive integers, and their sum is 17. There is a unique value of $n$ for which $S_{n}$ is an integer, find $n$. | 15. Consider each $l_{i}=\sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the hypotenuse of a right-angled triangle, with the two legs being $2 k-1$ and $a_{b}$. When these right-angled triangles are placed together to form a ladder, let $A, B$ be the starting and ending points, respectively.
The distance from $A$ to $B$ is
$$
\begin{... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Let $x$ be a cube root of 1 different from 1, find the value of $x^{\text {D}}$ $+x^{2}$. $(n \in N)$ | Solve: From $x^{3}=1$, i.e., $x^{3}-1=0$, which is also $(x-1)\left(x^{2}+x+1\right)=0$.
Since $x \neq 1$, it follows that $x^{2}+x=-1$.
When $n=3k$,
$$
x^{\mathrm{n}}+x^{2 n}=\left(x^{3}\right)^{k}+\left(x^{3}\right)^{2 k}=2 \text {; }
$$
When $n=3 k+1$,
$$
\begin{array}{l}
x^{\mathrm{n}}+x^{2 \mathrm{n}}=\left(x^{3}... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If two real-coefficient quadratic equations in $x$, $x^{2}+x+a=0$ and $x^{2}+a x+1=0$, have at least one common real root, then $a=$ $\qquad$ | 1. $-2 ;$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Among all possible four-digit numbers formed using the digits $1,9,9,0$, for each such four-digit number and a natural number $n$, their sum when divided by 7 does not leave a remainder of 1. List all such natural numbers $n$ in descending order.
$$
n_{1}<n_{2}<n_{3}<n_{4}<\cdots \cdots,
$$
Find: the value of $n_{1... | 3. Solution
$1,0,9,0$ digits can form the following four-digit numbers: 1099, 1909, 1990, 9019, 9091, 9109, $9190,9901,9910$ for a total of nine. The remainders when these nine numbers are divided by 7 are $0,5,2,3,5,2,6$, 3 , 5 . Since $n$ and their sum cannot be divisible by 7 with a remainder of 1, $n$ cannot have a... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-c-a}{b}$ $=3$, and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$. Then $x-a-b-c=$ | 5. 0 ;
The above text has been translated into English, maintaining the original text's line breaks and format. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. In $\triangle A B C$, $D$ is a point on side $B C$. It is known that $A B=13, A D=12, A C=15, B D=5$. What is $D C$? (3rd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition) | According to Stewart's Theorem, we have
$$
A D^{2}=A B^{2} \cdot \frac{C D}{B C}+A C^{2} \cdot \frac{B D}{B C}-B D \cdot D C \text {. }
$$
Let $D C=x$,
then $B C=5+x$.
Substituting the known data
into the above equation, we get
$$
\begin{array}{l}
12^{2}=13^{2} \\
\cdot \frac{x}{5+x}+15^{2} \cdot \frac{5}{5+x}-5 x,
\e... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, given that $a$ is an integer, the equation $x^{2}+(2 a+1) x$ $+a^{2}=0$ has integer roots $x_{1}, x_{2}, x_{1}>x_{2}$. Try to find the value of $\sqrt[4]{x_{1}^{2}}-\sqrt[4]{x_{2}^{2}}$. | Three, Solution: From the problem, we know that the discriminant $4a + 1$ is a perfect square, so $a \geqslant 0$. Since $4a + 1$ is odd, we can set $(2k + 1)^2 = 4a + 1$, solving for $a = k(k + 1)$.
$$
x_{1,2} = \frac{1}{2} \left[ \left( -2k^2 - 2k - 1 \right) \pm \sqrt{(2k + 1)^2} \right].
$$
When $k \geqslant 0$, $... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. As shown in the figure, points $A, B, C, D$ lie on the same circle, and $BC=DC=4, AE=6$. The lengths of segments $BE$ and $DE$ are both positive integers. What is the length of $BD$? (1988
National Junior High School Mathematics Competition) | Solve in $\triangle B C D$, $B C=D C$, H
Inference 1 gives
$$
C E^{2}=B C^{2}-B E \cdot D E,
$$
$\because A, B, C, D$ are concyclic,
$$
\begin{array}{l}
\therefore B E \cdot D E=A E \cdot C E, \\
\therefore C E^{2}=B C^{2}-A E \cdot C E=4^{2}-6 \times C E,
\end{array}
$$
which is $C E^{2}+6 \cdot C E-16=0$.
Solving gi... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Factorize: $a^{3}+2 a^{2}-12 a+15$ $=$ $\qquad$ - If $a$ is a certain natural number, and the above expression represents a prime number, then this prime number is | 3. $\left(a^{2}-3 a+3\right)(a+5), 7$; | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Place several points on the unit sphere such that the distance between any two points is (1) at least $\sqrt{2}$; (2) greater than $\sqrt{2}$. Determine the maximum number of points and prove your conclusion. | 13. (1) The maximum number of points is 6. If $A$ is one of these points, let's assume $A$ is at the North Pole, then the remaining points must all be in the Southern Hemisphere (including the equator). If there is only one point $B$ at the South Pole, then the rest of the points are all on the equator, in which case t... | 6 | Geometry | proof | Yes | Yes | cn_contest | false |
21. Let the weight of the counterfeit coin be $a$, and the weight of the genuine coin be $b$ $(a \neq b)$. There are two piles of three coins each, and it is known that each pile contains exactly one counterfeit coin. How many times at least must a precise scale (not a balance) be used to find these two counterfeit coi... | 21. (1) At least 3 weighings are required.
First, take one coin from each of the two piles, (1) if the weight is $2a$, then the real coins have been found; (2) if the weight is $2b$, then take one coin from the remaining two in each pile and weigh them separately, and the two fake coins can be found; (3) if the weight... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
4. There are two fleas at the two endpoints of the line segment $[0, 1]$. Some points are marked within the line segment. Each flea can jump over the marked points such that the positions before and after the jump are symmetric about the marked point, and they must not jump out of the range of the segment $[0, 1]$. Eac... | 4. The segment [0, 1] is divided into smaller segments by certain fixed points, with lengths $\frac{17}{23}$ and $\frac{19}{23}$. Therefore, the two fleas cannot land on the same segment after each taking one step (see the following supplement). From this, we can conclude that the minimum number of steps required must ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
18. (16th All-Russian Mathematics Competition) In the school football championship, each team must play a match against every other team. Each match awards 2 points to the winning team, 1 point to each team in the event of a draw, and 0 points to the losing team. It is known that one team has the highest score, but it ... | Let the team with the highest score, denoted as team $A$, be the champion. Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.
From the given conditions, every other team must win at least $n + 1$ matches, meaning their score is no less than $2(n + 1)$ points. ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, given any 5 points on a plane, where no three points are collinear and no four points are concyclic. If a circle passes through three of these points, and the other two points are respectively inside and outside the circle, then it is called a "good circle".
Let the number of good circles be $n$, find all possib... | Three, among: 5 points, take any two points $A, B$ and draw the line through $A$, $B$. If the other three points $C, D, E$ are on the same side of line $A B$, then consider $\angle A C B, \angle A D B, \angle A E B$. Without loss of generality, if $\angle A C B < 180^{\circ}$, then circle $A D B$ is the unique good cir... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
For $a \geqslant 1$, calculate the sum of the infinite series
$$
\begin{array}{l}
\frac{a}{a+1}+\frac{a^{2}}{(a+1)\left(a^{2}+1\right)} \\
+\frac{a^{4}}{\left.(a+1) a^{2}+1\right)\left(a^{4}+1\right)} \\
+\frac{a^{8}}{(a+1)\left(a^{2}+1\right)\left(a^{4}+1\right)\left(a^{8}+1\right)}+\cdots
\end{array}
$$ | $\begin{array}{c}\text { I. When } a=1 \text {, } \Sigma=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \\ =\frac{\frac{1}{2}}{1-\frac{1}{2}}=1 \text {. When } a>1 \text {, we have } \Sigma_{1}=\frac{a^{2}-a}{a^{2}-1}, \\ \Sigma_{2}=\frac{a^{4}-a}{a^{4}-1}, \Sigma_{3}=\frac{a^{8}-a}{a^{8}-1}, \cdots, \Sigma_{n}=\frac{a^{2}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, let $A B C$ be an equilateral triangle, and $P$ a point on its incircle. Prove that $P A^{2}+P B^{2}+P C^{2}$ is a constant. | Four, as shown in the figure, establish
a rectangular coordinate system,
the coordinates of point $P$ are
$$
\begin{array}{l}
\left(\frac{\sqrt{3}}{3} \cos \theta,\right. \\
\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3} \\
\cdot \sin \theta) .
\end{array}
$$
Let $\Sigma=P A^{2}+P B^{2}+P C^{2}$, then
$$
\begin{array}{l}
\Sigm... | 5 | Geometry | proof | Yes | Yes | cn_contest | false |
Let $S=\{1,2,3,4\}, a_{1}, a_{2}, \cdots$, be any permutation ending with 1, i.e., for any permutation $\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ of the four numbers in $S$ that does not end with 1, $\left(b_{4} \neq 1\right)$, there exist $i_{1}, i_{2}, i_{3}, i_{4}$, such that $1 \leqslant i_{1}<i_{2}<i_{3}<i_{4} \leq... | II. Hint: First consider the minimum value of the number of terms $k$ in any sequence containing a permutation of $S$.
1. For $S=\{1,2,3\}$, it can be proven that a sequence with only 6 terms cannot contain any permutation of $\{1,2,3\}$.
2. For $S=\{1,2,3,4\}$, prove that a sequence with only 11 terms cannot contain a... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Find the minimum value of the function $f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \quad(x \in R)$. | $$
\begin{array}{l}
\text { Sol } \because\left(x^{2}+1\right)-\cos x \\
=x^{2}+(1-\cos x) \geqslant 0, \\
\left(x^{2}+1\right)-2 x=(x-1)^{2} \geqslant 0, \\
\therefore \quad x^{2}+1 \geqslant \cos x, x^{2}+1 \geqslant 2 x \\
\text { Therefore, } f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \\
=x^{2}+1 \geqslant 1,
\... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. $A, B$ are two fixed points on a plane, find a point $C$ on the plane such that $\triangle A B C$ forms an isosceles triangle. There are $\qquad$ such points $C$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The blank space represented by $\qquad$ in the original text is kept as is in the translation. | 4.6.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 6 | Logic and Puzzles | other | Yes | Yes | cn_contest | false |
Sure, here is the translation:
---
One, starting from the natural number 1, write down in sequence to form the following series of numbers: $12345678910111213 \cdots$. With each digit occupying one position, determine the digit at the 1992nd position. | In the following sequence of digits, there are 9 single-digit numbers, $2 \times 90$ two-digit numbers, and $3 \times 900$ three-digit numbers. From $(1992-9-2 \times 90) \div 3$ $=601$, we know that the 1992nd position is in the 601st three-digit number starting from 100, which is the unit digit of the natural number ... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Given $-\frac{x-b}{c}+\frac{x-b-c}{a}$ $+\frac{x-c-a}{b}=3$,
and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$.
Then $x-a-b-c=$ $\qquad$ (8th Jincheng
Mathematics Competition) | From the known, we get $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(x-a-b-c)=0$, so we should fill in “0”. This is transforming the known to the unknown, the other is transforming the unknown to the known.
| 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. If $x^{3}-x^{2}+x-2=0$. Then $x^{4}+2 x^{3}-2 x^{2}+x-1=$ $\qquad$ . (1991, Hubei Huanggang Region Mathematics Competition). | Left= $\begin{aligned} & \left(x^{4}-x^{3}+x^{2}-2 x\right) \\ & +\left(3 x^{3}-3 x^{2}+3 x-6\right)+5 \\ = & x \cdot 0+3 \cdot 0+5=5 .\end{aligned}$ | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $a, b$ are integers, $a$ divided by 7 leaves a remainder of 3, and $b$ divided by 7 leaves a remainder of 5. When $a^{2}>4 b$, find the remainder when $a^{2}-4 b$ is divided by 7. | 4. Slightly explained: Let $a=$
$$
7 m+3, b=7 n+5 \text {, }
$$
where $m, n$ are integers, then
$$
\begin{aligned}
a^{2}-4 b & =(7 m+3)^{2}-4(7 n+5) \\
& =7\left(7 m^{2}+6 m-4 n-2\right)+3 .
\end{aligned}
$$
Since $7 m^{2}+6 m-4 n-2$ is an integer,
thus $a^{2}-4 b$ leaves a remainder of 3 when divided by 7. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. On the planet, there are 100 mutually hostile countries. To maintain peace, they decide to form several alliances, with the requirement that each alliance includes no more than 50 countries, and any two countries must be in at least one alliance. How many alliances are needed at a minimum?
(a) What is the minimum nu... | 6. 6 alliances. Each country should join no less than 3 alliances, so the number of alliances is at least 6.
(a) Only by dividing 100 countries into 4 groups, each with 25 countries, and then through combinations, we get $C{ }_{4}^{2}=6$ alliances.
(b) Only by dividing 100 countries into 10 groups $a_{1}, a_{2}, a_{8},... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For every $A \subset S$, let
$$
S_{\mathrm{A}}=\left\{\begin{array}{ll}
(-)^{\mid \mathrm{A}} \mid \sum_{\mathbf{a} \in \mathrm{A}} a, & A \neq \varnothing, \\
0, & A=\varnothing .
\end{array}\right.
$$
Find $\sum_{\mathrm{A} \subset \mathrm{S}} S_{\mathrm{A}}$. | Solve: There are $2^{n-1}$ sets $A$ satisfying $n \in A \subset S$, and there are also $2^{n-1}$ sets $B$ satisfying $n \notin B \subset S$. Note that $B$ covers all subsets of $\{1,2, \cdots, n-1\}$, and each $A$ satisfying $n \in A \subset S$ can be obtained by adding the element $n$ to some $B$. For each $B \subset ... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
21. Let when $0 \leqslant x \leqslant 1$, there exists a positive number $q$ such that $\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{\mathrm{t}}}{q}$ holds, find the smallest positive number $t$ that makes the above inequality true. For this smallest $t$ value, what is the minimum value of $q$ that makes the above inequa... | 21. Since $y=\sqrt{x}$ is strictly concave, for $0<x<1$, we have
$$
q \geqslant 2-(\sqrt{1+x}+\sqrt{1-x}).
$$
After two constant transformations, we get
$$
\begin{array}{l}
q \geqslant x^{1-2}\left(1+\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) \\
\cdot\left(1+\sqrt{1-x^{2}}\right).
\end{array}
$$
If $t<2$, then as $x$ app... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Test $B-5$. Let $T$ be the inscribed trapezoid $ABCD$ (counterclockwise) in the unit circle $O$, $ABCDI, AB=s_{1}$, $CD=s_{2}, OE=d, E$ is the intersection point of the heights. When $d \neq 0$, determine the minimum upper bound of $\frac{s_{1}-s_{2}}{d}$. If the minimum upper bound can be achieved, determine all such ... | Solve: As shown in the figure, establish a rectangular coordinate system with $O$ as the origin, $AB \perp x$-axis, and the coordinates of $E$ are $(d, 0)$. The equation of $BD$ can be set as $x - d = k y$, where $k^{-1}$ is the slope of line $BD$. Since $B$ and $D$ are on the unit circle, we have
$$
(d + k y)^{2} + y^... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. The number of natural numbers $n$ that make $n^{2}-19 n+91$ a perfect square is?
Will the above text be translated into English, please keep the original text's line breaks and format, and output the translation result directly. | Notice that $n^{2}-19 n+91=(n-9)^{2}+(10-$ $n$ ). When $n>10$,
$$
(n-10)^{2}<(n-9)^{2}+(10-n)<(n-9)^{2} .
$$
The integers between two consecutive perfect squares cannot be perfect squares, therefore the natural numbers $\pi$ that make $n^{2}-10 n+91$ a perfect square can only be from the ten numbers $1,2,3, \cdots, 9,... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The graph of a quadratic function passes through $(1,0),(5, 0)$, the axis of symmetry is parallel to the $y$-axis, but does not pass through points above the line $y=2x$. Then the product of the maximum and minimum values of the coordinates of its vertex is $\qquad$ . | 1. 4.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
In the letter摘登 of the sixth issue of "Middle School Mathematics" in 1992, Comrade Yang Xuezhi from Fuzhou No. 24 High School mentioned a conjecture by a senior high school student: "Given $\triangle P_{1} P_{2} P_{3}$ and a point $P$ inside it. The lines $P_{1} P, P_{2} P, P_{3} P$ intersect the opposite sides at $Q_{... | First, it should be noted that the above "conjecture" has long been proposed as a correct proposition and has been proven. Now, we introduce a proof method as follows (figure omitted).
Proof: Let the areas of $\triangle P_{2} P P_{3}, \triangle P_{3} P P_{1}, \triangle P_{1} P P_{2}$ be $S_{1}, S_{2}, S_{3}$, respecti... | 6 | Inequalities | proof | Yes | Yes | cn_contest | false |
Five, 10 people go to the bookstore to buy books, 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 most bought the most purchased book? Explain your reasoning.
(Na Chengzhang provided the question) | Solution 1: Let's assume that each person buys at most one copy of the same book. Thus, by condition (1), 10 people buy a total of 30 books.
Let the minimum number of people who buy the most popular book be $n$. Assume $A$ buys books 甲, 乙, and 丙.
Clearly, $n > 3$. Because if $n \leqslant 3$, then at most two more peo... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $A, B, C$ be the three interior angles of $\triangle ABC$, then the imaginary part of the complex number
$$
\frac{(1+\cos 2B+i \sin 2 B)(1+\cos 2 C+i \sin 2 C)}{1+\cos 2 A-i \sin 2 A}
$$
is . $\qquad$ | 3. 0 .
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "3. 0 ." is already in a numerical and punctuation format that is universal and does not require translation, the output remains the same:
3. ... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $x, y$ be coprime natural numbers, and $xy=$ 1992. Then the number of different ordered pairs $(x, y)$ is $\qquad$ | 5. 8.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. For the right square prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ with a base edge length of 1. If the dihedral angle $A-B D_{1}-C$ is $\frac{2 \pi}{3}$, then $A A_{1}=$ $\qquad$ | 15. 1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. The chord $AB=18$ of the sector $OAB$, a circle $C$ with radius 6 is exactly tangent to $OA$, $OB$, and the arc $\widehat{AB}$. Another circle $D$ is tangent to circle $C$, $OA$, and $OB$ (as shown in the figure). Then the radius of circle $D$ is $\qquad$. | 4. Solution As shown in the figure, let the radii of $\odot O, \odot D$ be $x, y$ respectively. Then $\alpha C=x-6$.
Since $\triangle O L B \backsim \triangle O M C$.
Then
$$
\frac{O B}{O C}=\frac{L B}{C M},
$$
which means $\frac{x}{x-6}=\frac{9}{6}$.
Solving for $x$ gives $x=18, Q C=12$, $O D=$ $6-y$.
Also, since $\... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (This question is worth 20 points) In the regular quadrilateral pyramid $P$ $A B C D$, $A B=3, O$ is the projection of $P$ on the base, $P O=6, Q$ is a moving point on $A O$, and the section passing through point $Q$ and parallel to $P A, B D$ is a pentagon $D F G H L$, with the area of the section being $S$. Fi... | $$
\begin{array}{l}
\text { Three, Solution: As shown in the figure, } \because P A / / \text { plane } D F G H L, \\
\therefore P A / / E F, P A / / H L, P A / / Q G . \\
\text { Also, } \because B D / / \text { plane } D F G H L, \\
\therefore B D / / E L, B D / / F H . \\
\text { Therefore, } \frac{E F}{P A}=\frac{B... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Sure, here is the translated text:
```
II. (This question is worth 35 points) Let the set $Z=\left\{z_{1}\right.$, $\left.z_{2}, \cdots, z_{n}\right\}$ satisfy the inequality
$$
\min _{i \neq j}\left|z_{i}-z_{j}\right| \geqslant \max _{i}\left|z_{i}\right| \text {. }
$$
Find the largest $n$, and for this $n$, find al... | Let $\left|z_{m}\right|=\max \left|z_{i}\right|$.
Thus, all points $z_{i}$ on the plane are distributed within a circle of radius $\left|z_{m}\right|=R$ centered at $z_{0}=0$.
It can be seen that on the circumference $R$, the six vertices $z,(j=1,2,3,4,5,6)$ of any inscribed regular hexagon, along with $z_{0}=0$, sati... | 7 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. If $n$ is a natural number, and $n^{3}+2 n^{2}+9 n+8$ is the cube of some natural number, then $n=$ $\qquad$ . | 2. 7 .
Since $n \in N$,
$$
\begin{aligned}
n^{3} & <n^{3}+2 n^{2}+9 n+8 \\
& <(n+2)^{3}=n^{3}+6 n^{2}+12 n+8,
\end{aligned}
$$
thus, only
$$
\begin{array}{l}
n^{3}+2 n^{2}+9 n+8 \\
=(n+1)^{3}=n^{3}+3 n^{2}+3 n+1 .
\end{array}
$$
Solving for the positive root $n=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. If $x>0, y>0, c>0$, and $x^{2}+y^{2}+$ $z^{2}=1$. Then the minimum value of the expression $\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z}$ is
| 6. 3 .
Let $\frac{yz}{x}=a, \frac{xz}{y}=b, \frac{xy}{z}=c$, then
$$
x^{2}+y^{2}+z^{2}=1 \Leftrightarrow ab+bc+ca=1 .
$$
Therefore, $a^{2}+b^{2}+c^{2} \geqslant ab+bc+ca=1$.
Thus, $(a+b+c)^{2}$
$$
=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) \geqslant 3,
$$
which implies $\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z} \Rightarrow \sqrt{3... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{aligned}
y=\sqrt{2 x^{2}} & -2 x+1 \\
& +\sqrt{2 x^{2}-(\sqrt{3}-1) x+1} \\
& +\sqrt{2 x^{2}-(\sqrt{3}+1) x+1}
\end{aligned}
$$
Find the minimum value of the function above. | Obviously, since
\[
\begin{aligned}
y=\sqrt{x^{2}}+ & (x-1)^{2} \\
& +\sqrt{\left(x-\frac{\sqrt{3}}{2}\right)^{2}+\left(x+\frac{1}{2}\right)^{2}} \\
+ & \sqrt{\left(x-\frac{\sqrt{3}}{2}\right)^{2}+\left(x-\frac{1}{2}\right)^{2}},
\end{aligned}
\]
then for the points \( T(x, x), A(0,1), B\left(\frac{\sqrt{3}}{2}, -\frac... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, as shown in the figure, $R$ is a square grid composed of 25 small squares with a side length of 1. Place a small square $T$ with a side length of 1 on the center square of $R$, and then place more small squares with a side length of 1 in $R$ according to the following requirements:
(1) The sides of these small sq... | Four, Answer: The maximum number of small cubes that can be placed is 4.
Proof As shown in the figure,
a) The small cubes that satisfy condition (3) can only be placed within the figure formed by removing a small square of side length 1 from each corner of $R$.
b) The small cubes that satisfy condition (3) must also be... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ (Math Competition) | By analyzing, we compare $E_{1}$ with $x^{2}-x-1=0$ and find that $m, n$ are the two distinct real roots of the equation. By Vieta's formulas, we get
$$
\begin{aligned}
m+n= & 1, m n=-1 \\
m^{2}+n^{2} & =(m+n)^{2}-2 m n=3 \\
m^{3}+n^{3} & =(m+n)\left(m^{2}+n^{2}\right)-m n(m+n) \\
& =4 .
\end{aligned}
$$
By analogy wi... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For the quadratic function $y=a x^{2}+b x(a b \neq 0)$, when $x$ takes $x_{1}, x_{2}\left(x_{1} \neq x_{2}\right)$, the function values are equal. Then, when $x$ takes $x_{1}+x_{2}$, the function value is | 2. Let $x$ take $x_{1}, x_{2}$, the function values are $-c$, then we have
$$
\left\{\begin{array}{l}
a x_{1}^{2}+b x_{1}=-c, \\
a x_{2}^{2}+b x_{2}=-c .
\end{array}\right.
$$
That is, $x_{1}, x_{2}$ are the two distinct roots of the quadratic equation
$$
a x^{2}+b x+c=0
$$
and
$$
x_{1}+x_{2}=-\frac{b}{a} \text {. }
... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Place the natural numbers $1,2,3,4, \cdots, 2 n$ in any order on a circle. It is found that there are $a$ groups of three consecutive numbers that are all odd, $b$ groups where exactly two are odd, $c$ groups where exactly one is odd, and $d$ groups where none are odd. Then $\frac{b-c}{a-d}=$ $\qquad$ . | 4. -3 .
Let the numbers on a circle be recorded in reverse order as $x_{1}, x_{2}$, $\cdots, x_{2 n}$, and satisfy
$$
x_{i}=\left\{\begin{array}{l}
-1, \text { when } x_{i} \text { represents an odd number, } \\
+1, \text { when } x_{i} \text { represents an even number. }
\end{array}\right.
$$
Then $x_{1}+x_{2}+\cdo... | -3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. When $n=6$, find the value of $P(6)$.
| Solve: Make the partition table for 6
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline \begin{tabular}{l}
\begin{tabular}{l}
$k-$ \\
partition \\
\hline
\end{tabular}
\end{tabular} & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline \begin{tabular}{l}
partition \\
ways
\end{tabular} & \begin{tabular}{l}
6
\end{tabular} & \begin{tabular}{l}
\be... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. $p, q$ are natural numbers, and there are $11 q-1 \geqslant 15 \psi, 10 p \geqslant 7 q$ $+1, \frac{7}{10}<\frac{q}{p}<\frac{11}{15}$. Then the smallest $q=$ $\qquad$ . | 2. $q=7$.
Solve $\left.\begin{array}{l}11 q-1 \geqslant 15 p, \\ 10 p \geqslant 7 q+1\end{array}\right\} \Rightarrow 11 q-1 \geqslant \frac{3}{2}(7 q+1) \Rightarrow q$
$\geqslant 5$. When $q=5,6$, check and find it does not meet the requirements, $q=7$ when, $p=5$. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
*Five, (20 points) 100 matchboxes, numbered 1 to 100. We can ask whether the total number of matches in any 15 boxes is odd or even. What is the minimum number of questions needed to determine the parity (odd or even) of the number of matches in box 1? | Let $a_{i}$ represent the number of matches in the $i$-th box.
The first time, boxes 1 to 15 are taken, so the parity of $\sum_{k=1}^{15} a_{k}$ is known;
The second time, boxes 2 to 8 and 16 to 23 are taken, so the parity of $\sum_{k=2}^{8} a_{k}+\sum_{k=16}^{23} a_{k}$ is known;
The third time, boxes 9 to 23 are ta... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) Fill in the right table with $1,2,3,4,5,6$ respectively, so that in each row, the number on the left is less than the number on the right, and in each column, the number on top is less than the number below. How many ways are there to fill the table? Provide an analysis process. | Five, as shown in the right figure, from the known information we can get: $a$ is the smallest, $f$ is the largest, so $a=1$, $f=6$. From $bd$, we need to discuss the following two cases:
(1) When $bd$, then $b=3, d=2, c=4$ or
5.
In this case, there are the following two ways to fill:
\begin{tabular}{|l|l|l|}
\hline 1 ... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 3. Let } x>\frac{1}{4} \text {, simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$ | $\begin{array}{l}\text { 3. } \because \sqrt{x+\frac{1}{2} \pm \frac{1}{2} \sqrt{4 x+1}} \\ =\sqrt{\frac{2 x+1 \pm \sqrt{4 x+1}}{2}} \\ =\frac{1}{2} \sqrt{4 x+2 \pm 2 \sqrt{4 x+1}} \\ =\frac{1}{2} \sqrt{(\sqrt{4 x+1} \pm 1)^{2}} \\ =\frac{1}{2}|\sqrt{4 x+1} \pm 1|, \\ \therefore \text { when } x>\frac{1}{4} \text {, th... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$4.1992^{1002}$ when divided by 1993 leaves a remainder of | $$
\begin{aligned}
4 \cdot 1992^{1002} & =(1993-1)^{2 \cdot 2 \cdot 2 \cdot 3 \cdot 83} \\
& =\left(1993^{2}-2 \cdot 1993+1\right)^{2 \cdot 2 \cdot 3 \cdot 83}
\end{aligned}
$$
Continuing the calculation in a similar manner, the first term that appears is 1993, and the last term is 1. Therefore, the remainder when $19... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If the inequality $|x-2|+|x-1| \geqslant a$ holds for all real numbers $x$, then the maximum value of $\boldsymbol{a}$ is $\qquad$ . | 3. 1.
Make $y=|x-2|+|x-1|$ $=\left\{\begin{array}{l}3-2 x, \quad x<1 \\ 1, \quad 1 \leqslant x \leqslant 2 \\ 2 x-3, \quad x>2\end{array}\right.$ Obviously, $|x-2|+|x-1| \geqslant 1$. Therefore, the maximum value of $a$ is 1. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. For a finite set of points $M$ in the plane, it has the property: for any two points $A, B$ in $M$, there must exist a third point $C \in M$ such that $\triangle A B C$ is an equilateral triangle. Find the maximum value of $|M|$. | 2. Let $A, B \in M$ and $AB$ be the longest, by the given condition there exists $C \in M$, such that $\triangle ABC$ is an equilateral triangle. Clearly, the point set $M$ is entirely within the union of the three segments $CAB$, $ABC$, $BCA$ with $A$, $B$, $C$ as centers and $AB$ as the radius, as shown in the figure... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. There are 128 ones written on the blackboard. In each step, you can erase any two numbers $a$ and $b$ on the blackboard, and write the number $a \cdot b + 1$. After doing this 127 times, only one number remains. Denote the maximum possible value of this remaining number as $A$. Find the last digit of $A$.
| 8. Let's prove that by operating on the smallest two numbers on the board at each step, we can achieve the maximum final number. For convenience, we use $a * b$ to denote the operation of removing $a$ and $b$, and immediately adding $a b + 1$. Suppose at some step we do not operate on the smallest two numbers $x$ and $... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Find the minimum value of the function $f(u, v)=(u-v)^{2}+\left(\sqrt{2-u^{2}}\right.$ $\left.-\frac{9}{v}\right)^{2}$. (1983 Putnam Competition) | Analysis: The problem is equivalent to finding the shortest distance (squared) between points on two curves. A sketch reveals that $[f(u, v)]_{\min }=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
*5. In a football championship, each team is required to play a match against all other teams. Each match, the winning team gets 2 points, a draw results in 1 point for each team, and the losing team gets 0 points. It is known that one team has the highest score, but it has won fewer matches than any other team. If the... | 5. $n=6$. We call the team $A$ with the highest score the winning team. Suppose team $A$ wins $n$ games and draws $m$ games, then the total score of team $A$ is $2n + m$ points.
From the given conditions, each of the other teams must win at least $n+1$ games, i.e., score no less than $2(n+1)$ points. Therefore,
$2n + ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. $A, B, C$ are three points on line $l$, and $A B=B C=5$, and $P$ is a point outside line $l$, $\angle A P B=\frac{\pi}{2}, \angle B P C=\frac{\pi}{4}$. Then the distance from $P$ to line $l$ is $\qquad$ | 3. 2
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
* 2. On the three sides of $\triangle A B C$, take points $P_{1}, P_{2}$, $P_{3}, P_{4}, P_{5}, P_{6}, \cdots$, such that $P_{1}, P_{4}, P_{7}, \cdots$ are on $A C$, $P_{2}, P_{5}, P_{8}, \cdots$ are on $A B$, and $P_{3}, P_{6}, P_{9}, \cdots$ are on $B C$, and $A P_{1}=A P_{2}$, $B P_{2}=B P_{3}, C P_{3}=C P_{4}, A P_... | 2. 0
Construct the incircle of $\triangle ABC$. By the property of tangent segments, we can prove that $P_{n+6}=P_{6}$. Therefore, $P_{1994}=P_{2}$, which gives $P_{2} P_{1994}=$ 0. Now, let's handle this using complex numbers.
Set up the complex plane, and let each point's letter represent the complex number at that... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8. There is a rectangular sheet of size $80 \times 50$. Now, we need to cut off a square of the same size from each corner and then make it into an open box. What should be the side length $y$ of the square to be cut off so that the volume of this open box is maximized? | Let the side length of the cut-out square be $x$, then the volume of the box made is
$$
V=x(80-2 x)(50-2 x) .
$$
To find the maximum value of $V$, a common approach is to try to make the product a constant, which involves converting $x$ into $4x$.
$$
\begin{aligned}
V & =\frac{1}{4} \cdot 4 x(80-2 x)(50-2 x) \\
& \leq... | 10 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. 1000 teachers and students of a school are to visit a place 100 km away from the school. There are five cars available, each capable of carrying 50 people, with a speed of 25 km/h, while the walking speed of a person is 5 km/h. How much time is required for all teachers and students to arrive at the destinat... | Let's assume they walk $3 x$ kilometers throughout the journey. According to the problem, we have
$$
\frac{100-3 x+100-4 x}{25}=\frac{x}{5} .
$$
Thus, $\frac{200-7 x}{5}=x$, which means $3 x=50$.
The total time spent walking and riding is
$$
\frac{100-3 x}{25}+\frac{3 x}{5}=12 \text{. }
$$
Answer: The total time to r... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. In the figure, the large circle is a 400-meter track, and the track from $A$ to $B$ is 200 meters long. The straight-line distance is 50 meters. A father and son start running counterclockwise from point $A$ along the track for a long-distance run. The son runs the large circle, while the father runs straigh... | Let the number of laps the father and son run when they meet be $N_{1}, N_{2}\left(N_{1}, N_{2}\right.$ are positive integers), and let the distance from point $A$ to the meeting point on the left half-circle be $x$ meters $(0 \leqslant x \leqslant 200)$. According to the problem, we have
$$
\frac{250 N_{1}+x}{100 / 20... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. There are $n$ points on a plane, where any three points can be covered by a circle of radius 1, but there are always three points that cannot be covered by any circle of radius less than 1. Find the minimum radius of a circle that can cover all $n$ points. | We prove that the radius of the smallest circle covering $n$ points is 1.
Since any three points can be covered by a circle with a radius of 1, the distance between any two points is no more than 2.
Thus, a circle with a radius of 2 centered at any one of the points can cover all the points. This implies that there e... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. In trapezoid $A B C D$, $A D / / B C, \angle B=$ $30^{\circ}, \angle C=60^{\circ}, E, M, F, N$ are the midpoints of $A B, B C$. $C D, D A$ respectively. Given $B C=7, M N=3$. Then $E F$ | 4. 4(Extend $B A, C D$ to meet at $H$, then $\triangle B H C$ is a right triangle.) | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the equation $a x^{2}+b x+c=0(a \neq 0)$, the sum of the two roots is $S_{1}$, the sum of the squares of the two roots is $S_{2}$, and the sum of the cubes of the two roots is $S_{3}$. Then the value of $a S_{1}+b S_{2}+c S_{3}$ is $\qquad$ | 2. 0
(Hint: Let the two roots of the equation be $x_{1}, x_{2}$. It is easy to see that $a x_{1}^{3}+b x_{1}^{2}+c x_{1}=$ $\left.0, a x_{2}^{3}+b x_{2}^{2}+c x_{2}=0.\right)$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x=1-\sqrt{3}$. Then $x^{5}-2 x^{4}-2 x^{3}$ $+x^{2}-2 x-1$ is
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 1 (Hint: Original expression $=x^{3}\left(x^{2}-\right.$
$$
\left.2 x-2)+\left(x^{2}-2 x-2\right)+1\right)
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in the figure, in isosceles $\triangle ABC$, $AB=AC$, $\angle A=120^{\circ}$, point $D$ is on side $BC$, and $BD=1$, $DC=2$, then $AD=$ | 5. 1 (Hint: Draw $D N \perp A B$, and draw the median $A M$ of side $B C$. Thus, we can prove $\triangle A D N \cong \triangle A D M$, getting $A D=$ $2 D N=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. $M$ is a point inside the convex quadrilateral $A B C D$, and the points symmetric to $M$ with respect to the midpoints of the sides are $P, Q, R, S$. If the area of quadrilateral $A B C D$ is 1, then the area of quadrilateral $P Q R S$ is equal to | 6. 2 (Hint: Connect the midpoints of each pair of adjacent sides of quadrilateral $A B C D$, to get a parallelogram, whose area is easily known to be $\frac{1}{2}$. On the other hand, these four sides are precisely the midlines of four triangles that share $M$ as a common vertex and have the four sides of quadrilateral... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. If real numbers $x, y, z$ satisfy the equation
$$
\sqrt{x+5+\sqrt{x-4}}+\frac{|x+y-z|}{4}=3 \text {, }
$$
then the last digit of $(5 x+3 y-3 z)^{1994}$ is | 1. 4 | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
* 5. For a positive integer $m$, its unit digit is denoted by $f(m)$, and let $a_{n}=f\left(2^{n+1}-1\right)(n=1,2, \cdots)$. Then $a_{1994}=$ $\qquad$. | 5. 7
$$
\begin{aligned}
1994 & =4 \times 498+2, \\
a_{1994} & =f\left(2^{4 \times 198+9}-1\right)=f\left(2^{4 \times 198} \times 8-1\right) \\
& =f\left(16^{448} \times 8-1\right)=f(6 \times 8-1)=7 .
\end{aligned}
$$ | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Initially 17. M $1,2,3,4, \cdots, 1994$ are 1994 numbers from which $k$ numbers are arbitrarily selected, such that any two numbers selected as side lengths uniquely determine a right-angled triangle. Try to find the maximum value of $k$.
In the 1994 numbers $1,2,3,4, \cdots, 1994$, arbitrarily select $k$ numbers, so ... | Solve for the maximum value of $k$ being 11.
(1) From $A=\{1,2,3,4, \cdots, 1994\}$, select $\{1,2,4,8,16,32,64,128,256,512,1024\}$, a total of 11 numbers, using the recursive formula: $a_{1}=1, a_{n+1}=2 a_{n}$. Any two numbers $a_{i}, a_{j}$ $(i<j)$ in this set satisfy $a_{i}<a_{j}$, and $2 a_{i} \leqslant a_{j}$. Th... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
The 24th All-Union Mathematical Olympiad has a problem:
There are 1990 piles of stones, with the number of stones being $1, 2, \cdots$, 1990. The operation is as follows: each time, you can choose any number of piles and take the same number of stones from each of them. How many operations are needed at least to take ... | Solve the confusion of
$$
\begin{array}{l}
1990=: 2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+0 \cdot 2^{5} \\
+0 \cdot 2^{4}+0 \cdot 2^{3}+2^{2}+2^{1}+0 \cdot 2^{0},
\end{array}
$$
and write $1,2, \cdots, 1989$ in binary form. The operation is as follows:
First, take away $2^{10}=1024$ stones from each pile that has enough; seco... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}-0.25^{2} \div\left(-\frac{1}{2}\right)^{4} \times(-2)^{3}+\left(1 \frac{3}{8}+\right. \\ \left.2 \frac{1}{3}-3.75\right) \times 24\end{array}$ | One, 7.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Ten, (10 points) If $a>0, b>0$, and $\sqrt{a}(\sqrt{a} + \sqrt{b}) = 3 \sqrt{b}(\sqrt{a} + 5 \sqrt{b})$. Find the value of $\frac{2a + 3b + \sqrt{ab}}{a - b + \sqrt{ab}}$. | $$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If natural numbers $m, n$ satisfy $m+n>m n$, then the value of $m$ $+n-m n$ is $\qquad$ | 2. 1. From the given, we have $\frac{1}{n}+\frac{1}{m}>1$. If $n, m \geqslant 2$, then $\frac{1}{n}$ $+\frac{1}{m} \leqslant \frac{1}{2}$, which is a contradiction. Therefore, at least one of $m, n$ must be 1. Without loss of generality, let $m=1$, then $m+n-m n=1+n-1 \cdot n=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. If the complex number $z$ satisfies $3 z^{6}+2 i \cdot z^{5}-2 z-3 i=$ 0 . Then $|z|=$ $\qquad$ . | 6. $|z|=1$.
The equation is $z^{5}=\frac{2 z+3 i}{3 z+2 i}$. Let $z=a+b i$.
$$
\left|z^{5}\right|=\frac{|2 z+3 i|}{|3 z+2 i|}=\sqrt{\frac{4\left(a^{2}+b^{2}\right)+12 b+9}{9\left(a^{2}+b^{2}\right)+12 b+4}} \text {. }
$$
If $a^{2}+b^{2}>1$, then the left side of (1) $=|z|^{5}=\left(\sqrt{a^{2}+b^{2}}\right)^{5}>$ 1. B... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11. Let $a_{n}=6^{n}-8^{n}$. Find the remainder when $a_{94}$ is divided by 49. (Adapted from the first American Mathematical Invitational Competition) | $$
\begin{aligned}
a_{94} & =6^{94}-8^{94}=(7-1)^{94}-(7+1)^{94} \\
= & -2\left(C_{94}^{1} \cdot 7^{93}+C_{94}^{3} \cdot 7^{91}+\cdots\right. \\
& \left.+C_{94}^{91} \cdot 7^{3}+C_{94}^{93} \cdot 7\right) \\
= & 49 k-2 \cdot 94 \cdot 7 \\
= & 49(k-27)+7 .(k \in Z)
\end{aligned}
$$
$\therefore a_{94}$ when divided by 49... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10.7. A square wooden board is divided into $n^{2}$ unit squares by horizontal and vertical lines. Mark $n$ squares so that any rectangle with an area of at least $n$ and whose sides lie along the grid lines contains at least one marked square. Find the largest $n$ that satisfies this condition. | 10.7. 7 .
Obviously, if $n$ marked cells satisfy the conditions of the problem, then in each row and each column there is exactly one marked cell. Let $n \geqslant 3$ (obviously, $n=2$ is not the maximum), take the first row with a marked cell as $A$, the row adjacent to $A$ as $B$, and take either a row adjacent to $... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in the figure, $AB$ is the diameter of the semicircle $\odot O$, $CD \perp AB$. Let $\angle COD = \theta$, then $\frac{AD}{BD}$ - $\tan^2 \frac{\theta}{2}=$ $\qquad$ | 4. 1.
Connect $A C$, then $\angle C A D=\frac{1}{2} \angle \theta$, and $\operatorname{tg} \frac{\theta}{2}=\operatorname{tg} \angle C A D=\frac{C D}{A D}$.
According to the projection theorem, we know
$$
\begin{array}{l}
C D^{2}=A D \cdot B D . \\
\therefore \frac{A D}{B D} \operatorname{tg}^{2} \frac{\theta}{2}=\fra... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. In $\triangle ABC$, $G$ is the centroid, and $P$ is a point inside the triangle. The line $PG$ intersects the lines $BC$, $CA$, and $AB$ at $A'$, $B'$, and $C'$, respectively. Prove that: $\frac{A'P}{A'G} + \frac{B'P}{B'G} + \frac{C'P}{C'G} = 3$
(1991, Huanggang Region Junior High School Competition) | Prove that by connecting $B G, G C, P B, P C$, and drawing $G G^{\prime} \perp B C$ at $G^{\prime}$, $P P^{\prime} \perp B C$ at $P^{\prime}$, then
$$
\begin{array}{c}
P P^{\prime} / / C G^{\prime}, \frac{P P^{\prime}}{G G^{\prime}}=\frac{A^{\prime} P^{2}}{A^{\prime} G} . \\
\text { Also, } \frac{S_{\triangle P B C}}{S... | 3 | Geometry | proof | Yes | Yes | cn_contest | false |
2. The two roots of the equation $x^{2}-3|x|-4=0$ are $x_{1}$, $x_{2}$, then $x_{1}+x_{2}=$ $\qquad$ | 2. 0
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. (Question 2 of the 5th Test, *The American Mathematical Monthly*, pages 119 to 121, 1953)
Given a positive integer $n \geqslant 3$, for $n$ complex numbers $z_{1}, z_{2}, \cdots, z_{n}$ with modulus 1, find
$$
\min _{x_{1}, z_{2} \cdots, s_{n}}\left[\max _{\omega \in C,|\omega|=1} \prod_{j=1}^{n}\left|\omega-z_{j}\... | $$
\begin{array}{l}
f(u)=\prod_{j=1}^{n}\left(u-x_{j}\right) \\
=u^{n}+C_{n-1} u^{n-1}+\cdots+C_{1} u+C_{0} .
\end{array}
$$
Here, $\left|C_{0}\right|=\left|z_{1} z_{2} \cdots z_{n}\right|=1$. Let $C_{0}=e^{i\theta}, 0 \leqslant \theta<2 \pi$. Take $n$ complex numbers of modulus 1, $\varepsilon_{k}=e^{i \frac{2 x+\the... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (35 points) A group of students took an exam with 3 multiple-choice questions, each with four options. It is known that any two students in this group have at most one answer in common, and if one more student is added, regardless of their answers, the above property no longer holds. How many students are there ... | Three, this group of students must have at least 8 people.
First, consider if this group of students has 8 people, then it can meet the conditions of the problem.
Let the answers of these 8 people be $(1,1,1),(1,2,2),(2,2,1)$, $(2,1,2),(3,4,3),(3,3,4),(4,3,3),(4,4,4)$, which will suffice.
On the other hand, assume thi... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) For positive numbers $a, b$, define the operation * as follows: when $a \leqslant b$, $a * b=a b$; when $b<a<2 b$, $a * b=b^{2}+a-b$; when $a \geqslant 2 b$, $a * b=a^{2}+b-a$. If $3 * x=8$, find the value of $\left[x^{3}-4 x\right]$ (where $[x]$ represents the greatest integer not exceeding $x$). | If $3 \leqslant x$, then $3 * x=3 x \geqslant 3 \cdot 3>8$. This does not satisfy $3 * x=8$.
If $x<3<2 x$, then $3 * x=x^{2}+3-x=8$, which means $x^{2}-x-5=0$.
Thus, $x=\frac{1+\sqrt{21}}{2}$ (taking the positive root) (since $\sqrt{21}<5$ satisfies $x<3<2 x$).
If $3 \geqslant 2 x$, then $3 * x=3^{2}+x-3=6+x \leqslan... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. Let the sequences $\left\{x_{n}\right\},\left\{y_{n}\right\}$ satisfy $x_{n}+y_{n} i=\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}$. | $\begin{array}{l}\text { Let } \begin{array}{l} \text { } F:=x_{n}+y_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n}, F^{\prime} \\ =x_{h}-j_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n} . \text { Then } F+F^{\prime}=2 x_{n} \\ =\omega^{n}+(\bar{\omega})^{n}\left(\text { where } \omega=\frac{-1+\sqrt{3} i}{2}\right) . \t... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\sin y \cos y+a=0$. Then $\cos (x+2 y)=$ $\qquad$ . | 2.1.
Given that $x^{3}+\sin x=2 a=(-2 y)^{3}+\sin (-2 y)$. Let $f(t)=t^{3}+\sin t$, then $f(x)=f(-2 y)$. The function $f(t)=t^{3}+\sin t$ is strictly increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, $x=-2 y, x + 2 y=0$. Hence, $\cos (x+2 y)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the point sets
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(x-3)^{2}+(y-4)^{2} \leqslant\left(\frac{5}{2}\right)^{2}\right.\right\}, \\
B=\left\{(x, y) \left\lvert\,(x-4)^{2}+(y-5)^{2}>\left(\frac{5}{2}\right)^{2}\right.\right\} .
\end{array}
$$
Then the number of integer points (i.e., points with both c... | 3. 7.
As shown in the figure, circles $E$ and $F$ intersect at points $M$ and $N$. The entire figure is symmetric about the line connecting the centers $E F$. Among them, $A \cap B$ is a crescent shape $S$ near the origin $O$ on the lower left. The x-coordinate of the points in $S$ can be 1, 2, 3, 4, and the y-coordin... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Through a non-central point inside a regular dodecagon, what is the maximum number of different diagonals that can be drawn?
保留源文本的换行和格式，这里的翻译结果应该是：
Through a non-central point inside a regular dodecagon, what is the maximum number of different diagonals that can be drawn?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Throug... | Let the regular dodecagon be $A_{1} A_{2} \cdots A_{12}$.
First, prove that the 4 diagonals $A_{1} A_{5}, A_{2} A_{6}, A_{3} A_{8}, A_{4} A_{11}$ intersect at a non-central point inside the regular dodecagon.
Since $\overparen{A_{5} A_{8}}=\overparen{A_{8} A_{11}}$, $\overparen{A_{11} A_{1}}=\overparen{A_{1} A_{3}}$,... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. The equation $x^{2}+m x+1=0$ and the equation $x^{2}-x$ $-m=0$ have one common root, then $m=$ $\qquad$ | Two equations have a common root $x=-1$. Substituting this root into either equation, we get $m=2$.
| 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The solution set of the equation $x^{2}+x-1=x e^{x^{2}-1}+\left(x^{2}-1\right) e^{x}$ is $A$ (where, $e$ is an irrational number, $e=2.71828$ $\cdots$). Then the sum of the squares of all elements in $A$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 5. (C)
Let $y=x^{2}-1$. The original equation becomes
$$
x+y=x e^{y}+y e^{x} \text {, }
$$
which is $x\left(e^{y}-1\right)+y\left(e^{x}-1\right)=0$.
Since $x>0$ implies $e^{x}-1>0$, i.e., $x$ and $e^{x}-1$ always have the same sign, $y$ and $e^{y}-1$ also always have the same sign.
Therefore, $x\left(e^{y}-1\right) ... | 2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. The numbers $2 x, 1, y-1$ form an arithmetic sequence in the given order, and $y+3,|x+1|+|x-1|$, $\cos (\arccos x)$ form a geometric sequence in the given order. Then $x+y+x y=$ | 4. 3.
Since $2 x, 1, y-1$ form an arithmetic sequence, we have
$$
2 x+y-1=2 \Rightarrow y=3-2 x \text {. }
$$
And $\cos (\arccos x)=x,-1 \leqslant x \leqslant 1$, it follows that,
$$
|x+1|+|x-1|=2 \text {. }
$$
From the second condition, we get $\quad(y+3) x=4$.
Substituting (1) into (2) yields
$$
2 x^{2}-6 x+4=0 \t... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (Full marks 30 points) Let $f(x)=x^{2}-(4 a-2) - x-6 a^{2}$ have the minimum value $m$ on the interval $[0,1]$. Try to write the expression for $m$ in terms of $a$, $m=F(a)$. And answer: For what value of $a$ does $m$ achieve its maximum value? What is this maximum value?
---
The translation maintains the origin... | $$
-, f(x)=[x-(2 a-1)]^{2}-10 a^{2}+4 a-1 .
$$
(i) When $2 a-1 \leqslant 0$, i.e., $a \leqslant \frac{1}{2}$, $f(x)$ is \. on $[0,1]$. At this time, $m=f(0)=-6 a^{2}$.
(ii) When $2 a-1 \geqslant 1$, i.e., $a \geqslant 1$, $f(x)$ is L. on $[0,1]$. At this time, $m=f(1)=-6 a^{2}-4 a+3$.
(iii) When $0<2 a-1<1$, i.e., $\fr... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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