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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., $\frac{1}{2} m_{\mathrm{e}} \times \frac{2}{3} m_{\mathrm{c}}=\left(\frac{1}{2} b\right)^{2}$.
Here, the median
$$
m_{c}^{2}=\frac{1}{2} a^{2}+\frac{1}{2} b^{2}-\frac{1}{4} c^{2} .
$$
From (2) and (3), we get
$$
2 a^{2}=b^{2}+c^{2} \text {. }
$$
Thus, by the cosine rule and (4), we have
$$
b^{2}+c^{2}=4 b c \cos \angle B A C \text {. }
$$
Let $\frac{b}{c}=\lambda$, then
$$
\lambda^{2}+1=4 \lambda \cos \angle B A C .
$$
In (6), there are real solutions if and only if $\angle B A C \leqslant 60^{\circ}$. The solutions $\lambda_{1}, \lambda_{2}$ are reciprocals of each other, thus generating similar triangles. Therefore, for $\angle B A C \leqslant 60^{\circ}$, there is only one non-similar $\triangle A B C$ that satisfies the conditions. When $\angle B A C > 60^{\circ}$, there are no $\triangle A B C$ that satisfy the conditions.
|
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 at most? Prove your conclusion.
|
. 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 $\triangle PAB \cong \triangle PAC$ in area, line $AP$ must pass through the midpoint of diagonal $BD$. Similarly, line $CP$ must also pass through the midpoint of $BD$. Therefore, $P$ is the midpoint of $BD$. Clearly, in this case, $\triangle ABD$ and $\triangle BCD$ have equal areas. This means that if there is a point $P$ inside the quadrilateral that meets the requirements, then the quadrilateral $ABCD$ is divided into two equal areas by one of its diagonals. It is easy to see that this condition is also sufficient.
(ii) Next, consider the case where point $P$ is outside the quadrilateral.
Since $\triangle APB \cong \triangle APD$ in area, $AP \parallel BD$. Similarly, $AC \parallel PD$. Therefore, quadrilateral $AEDP$ is a parallelogram. At this time, $S_{\triangle BVP} = S_{\triangle DVP} = \frac{1}{2} S_{\triangle BCD}$. Additionally, it is easy to see that any point $Q$ in the angle domain formed by extending two adjacent sides of the quadrilateral, such as extending $BA$ and $DA$, does not meet the requirements. This means that if a point $P$ outside the quadrilateral meets the requirements, then one of the four triangles formed by the two diagonals of the quadrilateral must have an area equal to half the area of the quadrilateral. We will now prove that this condition is also sufficient.
Assume $S_{\triangle BDP} = \frac{1}{2} S_{\triangle BCD}$. Draw a line through $A$ parallel to $BD$ and a line through $D$ parallel to $CA$, and let these two lines intersect at point $P$. Then quadrilateral $AEDP$ is a parallelogram and $S_{\triangle PAD} = \frac{1}{2} S_{ABCD}$. Since $S_{\triangle BE} < S_{\triangle ED}$, it follows that $BE < ED$. Therefore, $PB$ is inside the pentagon $PABCD$ and will not be outside the quadrilateral. Similarly, $PC$ is also inside the quadrilateral. Thus, $S_{\triangle APB} = S_{\triangle PBD} = S_{\triangle PDC}$. Consequently, $S_{\triangle PBC} = S_{PABCD} - S_{\triangle PAB} - S_{\triangle PCD} = S_{\triangle PAD}$. This proves that point $P$ meets the requirements of the problem.
(iii) From (i) and (ii), we know that there is at most one point $P$ outside the quadrilateral $ABCD$ that meets the requirements, and the interior and exterior of the quadrilateral cannot simultaneously have points that meet the requirements. Furthermore, if both diagonals of the quadrilateral divide the quadrilateral into equal areas, it is easy to prove that the quadrilateral is a parallelogram, so there is at most one point $P$ inside the quadrilateral that meets the requirements. In summary, the point $P$ that satisfies (1) can have at most one point.
|
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 on a circle. How many birds are there on the circle with the most birds, at minimum?
|
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 (hereinafter referred to as a four-point circle), which can form 252 four-point circles (including repeated counts). Each four-point circle belongs to 6 different groups of 5 points, so there should be 42 different four-point circles. Each four-point circle has four known points, totaling 168 points. Since there are 10 known points, there is one point $A$ on 17 different four-point circles. On these 17 four-point circles, apart from $A$, there are 3 other known points, totaling 51 points. Apart from $A$, there are 9 known points, so there is a point $B$ different from $A$, passing through $A$: 17 circles at least 6 pass through $B$, meaning there are at least two different four-point circles passing through $A$ and $B$, which have 12 points and they do not overlap, which is impossible.
(2) Suppose 5 known points $A_{1}, A_{2}, A_{3}, A_{4}$, $A_{8}$ are all on circle $S_{1}$, then the other 5 points can have at most one point not on $S_{1}$. If not, let $B, C$ not be on $S_{1}$.
Consider the 5-point group $A_{1}, A_{2}, A_{3}, B, C$. According to the given, 4 points among the 5 points must be concyclic on $S_{2}$. Since $B, C$ are not on $S_{1}$, $A_{1}, A_{2}, A_{3}$ cannot be on $S_{2}$, otherwise $S_{2}$ and $S_{1}$ would coincide. Assume $A_{1}, A_{2}$ are on $S_{2}$, then $A_{3}, A_{4}, A_{8}$ are not on $S_{2}$.
Consider the 5-point group $A_{3}, A_{4}, A_{5}, B, C$. Assume $A_{3}, A_{4}, B, C$ are concyclic on $S_{3}$, then $A_{1}, A_{2}, A_{5}$ are not on $S_{3}$.
Finally, consider the 5-point group $A_{1}, A_{3}, A_{8}, B, C$. Since $A_{1}, A_{3}, A_{8}$ are on $S_{1}$ and $B, C$ are not on $S_{1}$, $A_{1}, A_{3}, A_{8}, B$ and $A_{1}, A_{3}, A_{8}, C$ are not concyclic. Since $A_{1}, B, C$ are on $S_{2}$ and $A_{3}, A_{5}$ are not on $S_{2}$, $A_{1}, A_{3}, B, C$ and $A_{1}, A_{5}, B, C$ are not concyclic; similarly, it can be proven that $A_{3}, A_{5}, B, C$ are not concyclic, meaning any 4 points in this 5-point group are not concyclic, which contradicts the given.
In summary, the circle containing the most points must have at least 9 known points. Additionally, if 9 known points are on one circle and the other point is not on this circle, these 10 points clearly meet the requirements of the problem. Therefore, the circle with the most birds must have at least 9 birds.
|
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 coefficient $p > 0$ of $f(x) = p x^{2} - q x + r$, and its roots are within $(0,1)$, we have $f(1) = (p - q + r) = 0$, and $r(p - q + r) > 0$. Given that $p, q, r$ are positive integers, we have $r(p - q + r) \geq 1$. Thus, from the inequality, we get
$$
p^{2} > 16 r(p - q + r) = 16.
$$
Therefore, $p > 1$, i.e., $p > 5$.
When $p = 5$, the quadratic equation $5 x^{2} - 5 x + 1 = 0$ has two roots $\alpha = \frac{5 + \sqrt{5}}{10}, \beta = \frac{5 - \sqrt{5}}{10}$, both of which are within $(0,1)$. Hence, $p_{10} = 5$.
|
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 $f(0)=3$.
$$
\text { Let } F(n)=\frac{f(f(f(f \cdots(f(n)) \cdots)))}{10 \uparrow \bar{f}} \text {. }
$$
Prove: $F[f(1990)+f(5)+f(13)]$
$$
=F[f(1990) f(3) f(25)] .
$$
|
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., $m_{\mathrm{k}+1}=n_{\mathrm{k}}, n_{\mathrm{k}+1}=m_{\mathrm{k}}+n_{\mathrm{b}}$.
Notice that $\frac{m_{1}}{n_{1}}=\frac{1}{1}, m_{1}=1, n_{1}=1$, comparing it with the Fibonacci sequence: $f_{1}=1, f_{2}=1, f_{k+1}$ $=f_{k}+f_{k+1}(k \geqslant 1)$, we can see that,
$$
m_{1}=f_{\mathrm{k}}, \quad n_{\mathrm{k}}=f_{\mathrm{k}+1} \text {. }
$$
Thus, $m^{2}+m n-n^{2}$
$$
\begin{array}{l}
=f_{1988}^{2}+f_{1988} f_{1989}-f_{1989}^{2} \\
=f_{1988}^{2}+f_{1988}\left(f_{1988}+f_{1987}\right) \\
-\left(f_{1987}^{2}+2 f_{1988} f_{1987}+f_{1988}^{2}\right) \\
=-\left(f_{1937}^{2}+f_{1987} f_{1988}-f_{1988}\right) \\
=f_{1980}^{2}+f_{1986} f_{1087}-f_{1987}^{2} \\
=\cdots \cdots
\end{array}
$$
$$
\begin{array}{l}
=f_{2}^{2}+f_{2} f_{3}-f_{3}^{2} \\
=1^{2}+1 \cdot 2-2^{2}=-1 .
\end{array}
$$
|
-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 until $O$ touches the $AB$ side, completing one "left move". Then, with $O$ touching the $AB$ side as the pivot, the triangular colored plate rotates to the left again, making the second "left move". This continues until the 1990th "left move" is completed and the plate stops. How far is the red vertex $O$ from point $A$ at this time? Explain your reasoning.
|
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 and its red vertex completely return to their original positions.
Since $1990=48 \times 41+22$,
1990 left advances are equivalent to 22 left advances. After 16 left advances, O moves to the F position. After another 6 left advances, the vertex moves to the midpoint P on the AB side, at which point $PA=3$.
|
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 construct a function $f$ that satisfies the given conditions for a four-element set.
$$
\begin{array}{c}
\text { Let } A=\left\{a_{0}, a_{1}, a_{2}, a_{3}\right\} \text { and } N_{k} \\
=\{x \mid x=k+4 n, n \in Z\}(k=0,1,2,3) .
\end{array}
$$
For any $x \in N$, define $f(x)=a_{k}$ if $x \in N_{1}$. In this case, for any $i, j \in N$, if $f(i) = f(j)$, by the definition of the function, $i$ and $j$ must be elements of the same set $N_{1}$, so $i-j$ is divisible by 4, and thus $|i-j|$ cannot be a prime number. This means that for a four-element set $A$, a function $f$ that satisfies the conditions exists. In summary, the minimum number of elements in set $A$ is 4.
|
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
$$
\begin{array}{l}
(-42,-7),(-24,-8),(-18,-9), \\
(-15,-10),(-12,-12),(-5,30), \\
(-4,12),(-3,6),(-2,3),(0,0) .
\end{array}
$$
The corresponding values of $a=-(m+n)$ are $49,32,27,25$,
$$
{ }^{2} 4,-25,-8,-3,-1,0 \text {. A total of 10. }
$$
|
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
$$
\begin{array}{l}
\left(\sum_{k=1}^{n} a_{b}\right)^{2}+\left(\sum_{k=1}^{n}(2 h-1)\right)^{2} \\
=\sqrt{17^{2}+n^{4}}
\end{array}
$$
Let $\sum_{k=1}^{a} t_{i}=\sqrt{17^{2}+n^{4}}$. And we can choose $a_{h}$ such that $S_{n}=\sqrt{17^{2}+n^{4}}$. When $S_{n}$ is an integer, then
$$
\begin{array}{l}
17^{2}=S_{u}^{2}-n^{4}=\left(S_{n}-n^{2}\right)\left(S_{u}+n^{2}\right) . \\
\text { By }\left\{\begin{array}{l}
S_{u}+n^{2}=17^{2} \\
S_{n}-n^{2}=1,
\end{array} \text { we solve to get } S_{n}=1+5, n=12\right. \text {. }
\end{array}
$$
|
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 $n=3 k+1$,
$$
\begin{array}{l}
x^{\mathrm{n}}+x^{2 \mathrm{n}}=\left(x^{3}\right)^{k} \cdot x+\left(x^{3}\right)^{2 k} \cdot x^{2} \\
=x+x^{2}=-1 ;
\end{array}
$$
When $n=3 k+2$,
$$
\begin{array}{l}
x^{n}+x^{20} \\
=\left(x^{3}\right)^{1} \cdot x^{2}+\left(x^{3}\right)^{2 k} \cdot x^{3} \cdot x \\
=x^{2}+x=-1
\end{array}
$$
|
-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 remainder of $1,3,6,5$ : $3,6,2,5,3$ when divided by 7. That is, $n$ cannot have a remainder of $1,2,3,5,6$ when divided by 7. Therefore, the remainder when $n$ is divided by 7 can only be 0 or 4. Since $n$ is a natural number, $n_{1}=4, n_{2}=7$, $n_{1} n_{2}=28$.
|
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
$$
\begin{array}{l}
12^{2}=13^{2} \\
\cdot \frac{x}{5+x}+15^{2} \cdot \frac{5}{5+x}-5 x,
\end{array}
$$
Solving this equation yields $x_{1}=9, x_{2}=-9$ (discard). Therefore, $D C=9$.
|
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 $k \geqslant 0$, $x_1 = -k^2$, $x_2 = -(k + 1)^2$, then $\sqrt[4]{x_1^2} - \sqrt[4]{x_2^2} = -1$,
When $k < 0$, $x_1 = -(k + 1)^2$, $x_2 = -k^2$, then $\sqrt[4]{x_1^2} - \sqrt[4]{x_2^2} = -1$.
|
-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 gives $C E=2, C E=-8$ (discard),
$$
\begin{array}{l}
\therefore B E \cdot D E=A E \cdot C E=6 \times 2=12 . \\
\text { Also } B D<B C+C D=4+4=8, \\
\therefore\left\{\begin{array} { l }
{ B E = 4 , } \\
{ D E = 3 ; }
\end{array} \text { or } \left\{\begin{array}{l}
B E=3, \\
D E=4 .
\end{array}\right.\right. \\
\therefore B D=7 .
\end{array}
$$
|
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 there are at most $2+4=6$ points.
If there is no point at the South Pole, it can be proven that the number of points does not exceed 5. Otherwise, there would be at least 5 points $A_{1}$, $A_{2}, \cdots, A_{5}$ in the Southern Hemisphere (including the equator). Let $A^{\prime}$ be the South Pole, and the arcs $A^{\prime} A_{1}(i=1,2, \cdots, 5)$ intersect the equator at $A_{1}^{\prime}$. Then, among $\angle A_{1}^{\prime} A^{\prime} A_{\mathrm{j}}^{\prime}(1 \leqslant i \neq j \leqslant 5)$, at least one angle is less than or equal to $72^{\circ}$. Suppose $\angle A_{1}^{\prime} A^{\prime} A_{2}^{\prime} \leqslant 72^{\circ}$, then in the spherical triangle $\triangle A_{1}^{\prime} A^{\prime} A_{2}^{\prime}$, the distance between any two points is less than $\sqrt{2}$, which is a contradiction. Therefore, the maximum number of points on the sphere is 6. (2) The maximum number of points is 4. The proof is similar to (1).
|
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 coins?
(1) Assuming $a$ and $b$ are known, solve this problem.
(2) Assuming $a$ and $b$ are unknown, solve this problem.
|
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 is $a+b$, take one out and weigh it again, then take another one from the remaining pile with the fake coin and weigh it, and the other fake coin can be found.
(2) At least 4 weighings are required.
First, take two coins from one pile and weigh them separately. (1) If the two coins weigh the same, then they are real coins. The remaining one is a fake coin, and $b$ is known. Take two more coins from the other pile and weigh them separately, and the fake coins can be found; (2) If the two coins weigh differently, one weighs $a$ and the other weighs $b$, at this point, take two coins from the other pile and weigh them together with the remaining one from this pile. If the weight is $3b$, then the two fake coins are found; if the weight is $a+2b$, take one of the two coins already taken from the other pile and weigh it again, and the two fake coins can also be found.
|
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 $[0, 1]$. Each flea can independently jump once or stay in place, which counts as one step. How many steps are needed at minimum to ensure that the two fleas can always jump to the same small segment divided by the marked points on $[0, 1]$?
|
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 be 1.
The following proof shows that, regardless of the number of fixed points and how they are placed on the segment $[0,1]$, it is always possible to make both fleas jump into the longest segment after two jumps (if there are multiple such segments, we can choose one of them). By symmetry, it is sufficient to prove this for the flea starting at point 0.
Let the length of the chosen longest segment be $s$, and $\alpha$ be one of its endpoints.
If $\alpha < s$ (see the following figure), the flea can jump over the fixed point $\alpha$ in one step and land in the selected segment $[\alpha, \alpha+s]$ (if $\alpha=0$, the flea is already in the selected segment and does not need to jump at all). If $\alpha \geqslant s$, consider the interval $\left[\frac{\alpha-s}{2}, \frac{\alpha+s}{2}\right]$, which has a length of $s$ (see the following figure). This interval must contain at least one fixed point $\beta$. Otherwise, the length of the segment containing this interval would be greater than $s$, which is impossible. By jumping over $\beta$, the flea will land at point $2 \beta \in [\alpha-s, \alpha+s]$. If $2 \beta \notin [\alpha, \alpha+s]$, then by jumping over point $\alpha$ again, the flea will definitely land in the selected segment $[\alpha, \alpha+s]$.
|
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 has won fewer matches than any other team. How many teams must have participated at a minimum?
|
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. Therefore,
$$
\begin{array}{l}
2n + m > 2(n + 1), \\
m \geq 3.
\end{array}
$$
Thus, there must be a team that draws with the champion team, and this team's score should be no less than $2(n + 1) + 1$ points. Therefore,
$$
\begin{array}{l}
2n + m > 2(n + 1) + 1, \\
m \geq 4.
\end{array}
$$
Let there be $s$ teams in total. The champion must win at least one match; otherwise, its score would not exceed $s - 1$ points. Any other team would score strictly less than $s - 1$ points, and the total score of all participating teams would be less than $s(s - 1)$ points, while the total score of $s$ teams is $s(s - 1)$ points, leading to a contradiction.
Thus, $m > 4, n \geq 1$, meaning the champion team $A$ must play at least 5 matches, implying there are at least 6 teams in the competition.
A score table for 6 teams can always be constructed to meet the given conditions, where the champion team $A$ wins the minimum number of matches:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & Score \\
\hline$A$ & & 1 & 1 & 1 & 1 & 2 & 6 \\
\hline$B$ & 1 & & 2 & 0 & 0 & 2 & 5 \\
\hline$C$ & 1 & 0 & & 0 & 2 & 2 & 5 \\
\hline$D$ & 1 & 2 & 2 & & 0 & 0 & 5 \\
\hline$E$ & 1 & 2 & 0 & 2 & & 0 & 5 \\
\hline$F$ & 0 & 0 & 0 & 2 & 2 & & 4 \\
\hline
\end{tabular}
|
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 possible values of $n$.
---
Note: The translation preserves the original text's line breaks and formatting.
|
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 circle; if $\angle A E B + \angle A C B = 180^{\circ}$, then circles $A C B, A D B, A E B$ are all good circles. This means that through two fixed points, there is either one or three good circles.
From 5 points, a total of 10 point pairs can be formed. Through each point pair, there is at least one good circle, so there are at least 10 good circles (including repeated counts). Each good circle passes through 3 point pairs, so there are at least 4 different good circles, and $n \geqslant 1$.
Connecting each pair of points among the 5 points with a line segment, each line segment is either a chord of a good circle or a common chord of 3 good circles. If there are at least 5 good circles, then they have at least 15 chords. Since there are only 10 line segments in total and each line segment contributes 1 or 3 to the count, the sum of 10 odd numbers is even, which cannot be 15, so there must be at least 6 different good circles. This means that at least 4 line segments are chords of good circles: 4 line segments have 8 endpoints, so there must be at least two line segments sharing a common endpoint, say $A B, A C$. Thus, $A B D$ and $A C D$ are both good circles, so there are at least three good circles through $A, D$.
Assume $A B, A C, A D, A E$ with the shortest one being $A B$, then $\angle A C B, \angle A D B, \angle A E B$ are all acute angles. If $C, D, E$ are on the same side of line $A B$, then there is only one good circle through $A, B$, so $C, D, E$ must be on different sides of line $A B$. In this case, since the sum of any two angles among $\angle A C B, \angle A D B, \angle A E B$ is less than $180^{\circ}$, there can only be one good circle through $A, B$, which is a contradiction.
In summary, the number of good circles $n$ must be 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For $a \geqslant 1$, calculate the sum of the infinite series
$$
\begin{array}{l}
\frac{a}{a+1}+\frac{a^{2}}{(a+1)\left(a^{2}+1\right)} \\
+\frac{a^{4}}{\left.(a+1) a^{2}+1\right)\left(a^{4}+1\right)} \\
+\frac{a^{8}}{(a+1)\left(a^{2}+1\right)\left(a^{4}+1\right)\left(a^{8}+1\right)}+\cdots
\end{array}
$$
|
$\begin{array}{c}\text { I. When } a=1 \text {, } \Sigma=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \\ =\frac{\frac{1}{2}}{1-\frac{1}{2}}=1 \text {. When } a>1 \text {, we have } \Sigma_{1}=\frac{a^{2}-a}{a^{2}-1}, \\ \Sigma_{2}=\frac{a^{4}-a}{a^{4}-1}, \Sigma_{3}=\frac{a^{8}-a}{a^{8}-1}, \cdots, \Sigma_{n}=\frac{a^{2}-a}{a^{2^{n}}-1} \\ \text { Therefore, } \Sigma=\lim _{n \rightarrow \infty} \Sigma_{n}=\lim _{n \rightarrow \infty} \frac{a^{2^{n}}-a}{a^{2^{n}}-1}=1 .\end{array}$
|
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
$$
\begin{array}{l}
\Sigma=\left[\left(\frac{\cos \theta}{\sqrt{3}}-1\right)^{2}+\left(\frac{1}{\sqrt{3}}+\frac{\sin \theta}{\sqrt{3}}\right)^{2}\right] \\
+\left[\left(\frac{\cos \theta}{\sqrt{3}}\right)^{2}+\left(\frac{\sin \theta}{\sqrt{3}}-\frac{2}{\sqrt{3}}\right)^{2}\right] \\
+\left[\left(\frac{\cos \theta}{\sqrt{3}}+1\right)^{2}+\left(\frac{1}{\sqrt{3}}+\frac{\sin \theta}{\sqrt{3}}\right)^{2}\right] \\
=\cos ^{2} \theta+\sin ^{2} \theta+4+\left(-\frac{2}{\sqrt{3}}\right. \\
\left.+\frac{2}{\sqrt{3}}\right) \cos \theta+\left(\frac{2}{3}-\frac{4}{3}+\frac{2}{3}\right) \\
\sin ^{2} \theta=5 \text{ (constant)}
\end{array}
$$
|
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 $1 \leqslant i_{1}<i_{2}<i_{3}<i_{4} \leqslant k$, and $\left(a_{i_{1}}, a_{i_{2}}, a_{i_{3}}, a_{i_{4}}\right)=\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$. Find the minimum value of the number of terms $k$ in such a sequence.
|
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 any permutation of $\{1,2,3,4\}$.
Direct verification shows that the sequence 1, 2, 3, 4, 1, 2, 3, 1, 4, 2, 3, 1 contains any permutation of $S$, thus 12 is the minimum value.
Returning to the problem itself. If there is a sequence with fewer than 11 terms that contains any permutation of $S$ not ending in 1, then adding one more positive term to this sequence would result in a sequence containing all permutations of $S$, but with fewer than 12 terms, which is a contradiction. Additionally, 1, 2, 3, 4, 1, 2, 3, 1, 4, 2, 3 contains any permutation of $S$ not ending in 1. Therefore, the minimum value of $k$ is 11.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Find the minimum value of the function $f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \quad(x \in R)$.
|
$$
\begin{array}{l}
\text { Sol } \because\left(x^{2}+1\right)-\cos x \\
=x^{2}+(1-\cos x) \geqslant 0, \\
\left(x^{2}+1\right)-2 x=(x-1)^{2} \geqslant 0, \\
\therefore \quad x^{2}+1 \geqslant \cos x, x^{2}+1 \geqslant 2 x \\
\text { Therefore, } f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \\
=x^{2}+1 \geqslant 1,
\end{array}
$$
where the equality holds when $x=0$, hence the minimum value of $f(x)$ is 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 700. Therefore, the digit at the required position is 0.
|
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 number of alliances that can be formed to meet the requirements?
(b) If an additional restriction is added, that the union of any two alliances contains no more than 80 countries, how would you answer the above question?
|
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 $a_{1}, a_{2}, a_{8}, a_{4}, a_{8}, b_{1}, b_{2}, b_{8}, b_{4}$, $b_{B}$, each with 10 countries, and forming the following alliances:
$$
\begin{array}{l}
a_{1} \cup b_{1} \cup b_{2} \cup b_{8} \cup a_{3}, \\
a_{2} \cup b_{2} \cup b_{3} \cup b_{4} \cup a_{4}, \\
a_{3} \cup b_{8} \cup b_{4} \cup b_{8} \cup a_{5}, \\
a_{1} \cup b_{4} \cup b_{8} \cup b_{1} \cup a_{1}, \\
a_{3} \cup b_{5} \cup b_{1} \cup b_{2} \cup a_{2}, \\
a_{1} \cup a_{2} \cup a_{9} \cup a_{4} \cup a_{8},
\end{array}
$$
|
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 $B \subset \{1,2, \cdots, n-1\}$, let $f(B) = B \cup \{n\} = A$. Then $S_{B} + S_{f(B)} = (-1)^{|B|} \cdot n$. By Lemma 4, there are $2^{n-2}$ sets $B$ such that $2 \mid |B|$, hence there are also $2^{n-2}$ sets $f(B)$ such that $2 \mid |f(B)|$.
There are $2^{n-2}$ sets $B$ such that $2 \mid |B|$, hence there are also $2^{n-2}$ sets $f(B)$ such that $2 \mid |f(B)|$.
$$
\begin{array}{l}
\therefore \sum_{n \in A \subset S} S_{A} + \sum_{\substack{B \subset S-\{ \\ 2+|B|}} S_{B} = n \cdot 2^{n-2}, \\
\sum_{\substack{n \in A \subset S \\ 2+|A|}} S_{A} + \sum_{\substack{B \subset S \\ 2 \mid |B|-\{n\}}} S_{B} = \cdots n \cdot 2^{n-2}, \\
\end{array}
$$
Finally, $\sum_{A \subset S} S_{A} = n \cdot 2^{n-2} + (-n) \cdot 2^{n-2} = 0$.
|
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 inequality true when $0 \leqslant x \leqslant 1$?
|
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 $x$ approaches 0, the above expression tends to infinity. Therefore, for any fixed $q$, there exists an $x$ value sufficiently close to 0 such that the required inequality does not hold. For $t=2$, when $x=0$, the right-hand side of the above expression equals 4. Thus, $q \geqslant 4$. For $0<x \leqslant 1$, since $\sqrt{1+x}+\sqrt{1-x}<1$ and $\sqrt{1-x^{2}}<1$, when $t=2$, the minimum value of $q$ is 4.
|
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 cases.
|
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
$$
(d + k y)^{2} + y^{2} = 1.
$$
Thus, $\left(k^{2} + 1\right) y^{2} + 2 d k y + \left(d^{2} - 1\right) = 0$.
As shown in the figure, let $B\left(x_{1}, y_{1}\right)$ and $D\left(x_{2}, y_{2}\right)$, then $y_{1} > y_{2}$ are the roots of (5), and $y_{1} > 0, y_{2} < 0$, so
$$
\begin{array}{l}
s_{1} - s_{2} = -2 y_{1} - 2 y_{2} = \frac{4 d k}{k^{2} + 1}, \\
\frac{s_{1} - s_{2}}{d} = \frac{4 k}{k^{2} + 1} \leqslant 2.
\end{array}
$$
The upper bound 2 in (6) is reached if and only if $k = 1$, which means that line $BD$ forms an angle of $\frac{\pi}{4}$ with the $x$-axis. By symmetry, this also means that $AC \perp BD$ (but the intersection point $E$ does not coincide with point $O$) when it is reached.
|
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 $1,2,3, \cdots, 9,10$. After checking the perfect squares, the answer to the original problem is 2.
|
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 $Q_{1}, Q_{2}$, $Q_{3}$. Prove:
$$
\frac{P_{1} P}{P Q_{1}}+\frac{P_{2} P}{P Q_{2}}+\frac{P_{3} P}{P Q_{3}} \geqslant 6 "
$$
|
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 $S_{1}, S_{2}, S_{3}$, respectively. Then,
$$
\begin{array}{c}
\frac{P_{1} P}{P Q_{1}}=\frac{S_{\triangle P_{1} P_{2} P}}{S_{\triangle P P_{2} Q_{1}}}=\frac{S_{\triangle P_{3} P_{1} P}}{S_{\triangle P Q_{1} P_{3}}} \\
\quad=\frac{S_{\triangle P_{1} P_{2} P}+S_{\triangle P_{3} P_{1} P}}{S_{\triangle P P_{2} Q_{1}}+S_{\triangle P Q_{1} P_{3}}} \\
=\frac{S_{2}+S_{3}}{S_{1}} .
\end{array}
$$
Similarly, we have
$$
\frac{P_{2} P}{P_{2} Q_{2}}=\frac{S_{3}+S_{1}}{S_{2}}, \frac{P_{3} P}{P Q_{3}}=\frac{S_{1}+S_{2}}{S_{3}} \text {. }
$$
Therefore, $\frac{P_{1} P}{P Q_{1}}+\frac{P_{2} P}{P Q_{2}}+\frac{P_{3} P}{P Q_{3}}$
$$
\begin{array}{l}
=\frac{S_{2}+S_{3}}{S_{1}}+\frac{S_{3}+S_{1}}{S_{2}}+\frac{S_{1}+S_{2}}{S_{3}} \\
=\left(\frac{S_{2}}{S_{3}}+\frac{S_{3}}{S_{2}}\right)+\left(\frac{S_{3}}{S_{1}}+\frac{S_{1}}{S_{3}}\right) \\
+\left(\frac{S_{1}}{S_{2}}+\frac{S_{2}}{S_{1}}\right) \geqslant 2+2+2=6 .
\end{array}
$$
|
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 people can buy 甲, two more can buy 乙, and two more can buy 丙. Therefore, at most 7 people can buy 甲, 乙, and 丙, and the remaining 3 people cannot share a book with $A$, which is a contradiction.
If $n=4$. In this case, we can prove that each book is bought by exactly 4 people. Because otherwise, at least one book is bought by at most three people. Assume the 甲 book bought by $A$ is bought by at most three people. That is, apart from $A$, 甲 is bought by at most two more people, and 乙 and 丙 are each bought by at most three more people. Thus, 甲, 乙, and 丙 are bought by at most $1+2+3+3=9$ people, and the remaining one cannot share a book with $A$, which is a contradiction.
The above argument shows that when $n=4$, each book is bought by exactly 4 people. However, 4 does not divide 30, which is a contradiction. Therefore, $n \geqslant 5$. In fact, $n=5$.
Below is an example for $n=5$: Assume there are 1, 2, ..., 7 seven types of books. The 10 people's book purchase situation is as follows:
\begin{tabular}{|c|c|c|c|c|}
\hline 2 & $3),(1$ & 4 & $5),(2$ & 4 \\
\hline 4 & $6),(2$ & 5 & $7),(2$ & 5 \\
\hline 4 & $7),(3$ & 4 & $7),(3$ & 5 \\
\hline
\end{tabular}
Solution 2: Using numbers to represent the book types, then when 10 people buy books as follows:
$$
\begin{array}{l}
(123),(134),(145),(156),(126), \\
\text { (235), (245), (246), (346), (356) }
\end{array}
$$
it satisfies the requirements of the problem and each type of book is bought by exactly 5 people, so the minimum value is no more than 5.
Assume the minimum value is 4, and 10 people buy a total of $n$ types of books, with the $i$-th type of book being bought by $m_{i}$ people. Thus, $m_{i} \leqslant 4$ and $m_{1} + m_{2} + \cdots + m_{n} = 30$. When two people buy the same book, it is called a "book pair". By the given condition, each pair of people has at least one book pair, so there are at least $C_{10}^{2} = 45$ book pairs. On the other hand, the number of book pairs formed by the $i$-th type of book is $C_{m_{i}}^{2}$, so there are a total of $C_{m_{1}}^{2} + C_{m_{2}}^{2} + \cdots + C_{m_{n}}^{2}$ book pairs. Therefore, there must be
$$
C_{m_{1}}^{2} + C_{m_{2}}^{2} + \cdots + C_{m_{n}}^{2} \geqslant 45.
$$
However, since $C_{4}^{2} = 6$, $C_{3}^{2} = 3$, and $C_{2}^{2} = 1$, we also know
$$
C_{m_{1}}^{2} + C_{m_{2}}^{2} + \cdots + C_{m_{n}}^{2} \leqslant 7 C_{4}^{2} + C_{2}^{2} = 43.
$$
Since (1) and (2) are contradictory, the minimum value must be 5.
|
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 .
|
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 $\triangle O N D \backsim$
$\triangle O M C$, then
$$
\frac{O D}{O C}=\frac{D N}{C M} \text {, i.e. } \frac{6-y}{12}=\frac{y}{6} .
$$
Solving for $y$ gives $y=2$, i.e., the radius of $\odot D$ is 2.
|
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 $S$. Find the maximum value of $S$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
$$
\begin{array}{l}
\text { Three, Solution: As shown in the figure, } \because P A / / \text { plane } D F G H L, \\
\therefore P A / / E F, P A / / H L, P A / / Q G . \\
\text { Also, } \because B D / / \text { plane } D F G H L, \\
\therefore B D / / E L, B D / / F H . \\
\text { Therefore, } \frac{E F}{P A}=\frac{B E}{B A}=\frac{D L}{D A}=\frac{H L}{P A} . \\
\text { Hence, } E F=H L . \\
\text { Also, } \because P O \perp \perp \\
A B C D, B D \perp A C, \\
\therefore P A \perp B D, \\
E F \perp E L . \\
\quad \text { Also, } \because F H / / \\
B D, P-A B C D \text { is a }
\end{array}
$$
regular quadrilateral pyramid.
$$
\therefore P H=P F,
$$
thus $\angle P F G \cong \triangle P I I G$.
Therefore, $G F=G H$, so the pentagonal section $E F G K L$ is composed of two congruent right trapezoids.
Since $E L / / B D$, $\triangle A E L$ is an isosceles right triangle. Let $E Q=x$. Then $Q L=x$.
So, $\frac{E F}{P A}=\frac{B E}{B A}=\frac{O Q}{O A}=\frac{\frac{3 \sqrt{2}}{2}-x}{\frac{3 \sqrt{2}}{2}}$.
$$
E F=\left(1-\frac{\sqrt{2}}{3} x\right) P A \text {. }
$$
Similarly, $Q G=\left(1-\frac{\sqrt{2}}{6} x\right) P A$.
Also, $P A=\sqrt{P O^{2}+O A^{2}}=\frac{9}{2} \sqrt{2}$.
Therefore, $S=(E F+Q G) E Q$
$$
=\left(2-\frac{\sqrt{2}}{2} x\right) \cdot \frac{\sqrt{2}}{2} x \text {. }
$$
Thus, $S=-\frac{9}{2} x^{2}+9 \sqrt{2} x$.
When and only when $x=\sqrt{2}$, $S$ reaches its maximum value of 9.
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|
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 all sets that satisfy the above condition.
Where the notation $\min (a, b, c)$ denotes the minimum of $a, b, c$, and $\max (a, b, c)$ denotes the maximum of $a, b, c$.
```
|
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 $z_{0}=0$, satisfy the given conditions for a total of 7 points.
If $n>7$, we prove that there must be two points $z^{\prime}, z^{\prime \prime}$ such that
$$
\left|z^{\prime}-z^{\prime \prime}\right|7$, then at least one of the sectors must contain two points $Z^{\prime}, Z^{\prime \prime}$. Without loss of generality, assume
$$
\left|z^{\prime}\right| \leqslant\left|z^{\prime \prime}\right|, \angle Z^{\prime} Z_{0} Z^{\prime \prime}=\varphi0$.
|
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 $\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z} \Rightarrow \sqrt{3}$,
with equality holding when $x=y=z=\frac{1}{\sqrt{3}}$.
|
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 \( T(x, x), A(0,1), B\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right), C\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right) \), we have \( y = TA + TB + TC \). It is easy to verify that \( \triangle ABC \) is an equilateral triangle centered at \( O(0, 0) \) with side length 1. According to the Fermat point principle, when \( T \) coincides with \( O \), \( TA + TB + TC \) reaches its minimum value. Therefore, the minimum value of \( y \) is obtained when \( T(0,0) \), which is \( y(0) = 3 \).
|
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 squares are parallel or perpendicular to the sides of $R$;
(2) Including $T$, no two small squares have any common points;
(3) For any two small squares including $T$, there must be a straight line passing through them (including edges) that is parallel to one of the sides of $R$.
How many more small squares can be placed in $R$ besides $T$ while satisfying the requirements? Prove your conclusion.
|
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 placed on one of the 8 larger squares of side length 2, which are outside $T$ and bounded by dashed lines (such as $A B C D$).
Obviously, each larger square can contain at most one small cube that meets the requirements.
c) On any two larger squares that share a common vertex with $T$ (such as $A B C D$ and $C E F G$), two small cubes cannot be placed simultaneously, otherwise condition (3) would not be satisfied. The 8 larger squares can be divided into 4 pairs in this manner, and each pair of larger squares can contain at most one small cube that meets the requirements, meaning that the number of small cubes placed does not exceed 4.
d) As shown in the figure, placing 4 small cubes, it is easy to see that the conditions are met.
Therefore, the maximum number of small cubes that can be placed is 4.
(Maoyoude provided)
|
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 with the above process, we can reduce the target expression to a lower-degree algebraic expression in terms of known values.
$$
\begin{array}{l}
m^{4}+n^{4}=\left(m^{2}+n^{2}\right)^{2}-2 m^{2} n^{2} \\
=9-2 \times 1=7, \\
m^{5}+n^{5}=\left(m^{2}+n^{2}\right)\left(m^{3}+n^{3}\right)-m^{2} n^{2}(m+ \\
n)=11 .
\end{array}
$$
|
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
$$
x_{1}+x_{2}=-\frac{b}{a} \text {. }
$$
Therefore,
$$
\begin{array}{l}
a\left(x_{1}+x_{2}\right)^{2}+b\left(x_{1}+x_{2}\right) \\
=a\left(-\frac{b}{a}\right)^{2}+b\left(-\frac{b}{a}\right)=0 .
\end{array}
$$
|
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 $x_{1}+x_{2}+\cdots+x_{2 n}=0$,
and. $A_{i}=x_{i}+x_{i+1}+x_{i+2}$
$$
=\left\{\begin{array}{ll}
-3, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { are all odd, } \\
-1, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { exactly 2 are odd, } \\
+1, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { exactly 1 is odd, } \\
+3, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { are all even. }
\end{array}\right.
$$
where $x_{2 n+i}=x_{i}$.
$$
\text { By } \begin{aligned}
0 & =3\left(x_{1}+x_{2}+\cdots+x_{2 n}\right) \\
& =A_{1}+A_{2}+\cdots+A_{2 n} \\
& =(-3 a)-b+\cdots+3 d) \\
& =-3(a-c)-(b \cdots c)
\end{aligned}
$$
we get $\quad \frac{b-c}{a-d}=-3$.
|
-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}
\begin{tabular}{l}
$5+1$ \\
$4+2$ \\
$3+3$
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
$4+1+1$ \\
$3+2+1$ \\
$2+2+2$
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
$3+1+1+1$ \\
$2+2+1+1$
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{c}
$2+1+1$ \\
$+1+1$
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{c}
$1+1+1$ \\
$+1+1+1$
\end{tabular}
\end{tabular} \\
\hline \begin{tabular}{l}
\begin{tabular}{c}
$P(6$, \\
$k)$
\end{tabular}
\end{tabular} & 1 & 3 & 3 & 2 & 1 & 1 \\
\hline
\end{tabular}
Thus, $P(6)=\sum_{i=1}^{6} P(6, i)=11$.
|
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 taken, so the parity of $\sum_{k=9}^{23} a_{k}$ is known. From the three parities, it is easy to determine the parity of $a_{1}+2 \cdot \sum_{k=2}^{23} a_{k}$. If it is odd, then $a_{1}$ is odd; otherwise, $a_{1}$ is even.
Clearly, asking only once is not enough. Asking only twice is also not enough.
It is easy to see that in the two questions, $a_{1}$ must be included exactly once. Suppose $a_{1}$ and $a_{2}$ are included in the first question. We then consider two cases:
(1) If $a_{2}$ is not included in the second question, then changing the parity of both $a_{1}$ and $a_{2}$ simultaneously does not affect the answer, so the parity of $a_{1}$ cannot be determined.
(2) If $a_{2}$ is included in the second question, since $a_{1}$ is not included in the second question, the second question must include some $a_{3}$, which does not appear in the first question. Then, changing the parity of $a_{1}$, $a_{2}$, and $a_{3}$ simultaneously does not affect the answer, so the parity of $a_{1}$ cannot be determined.
In summary, at least three questions are required.
|
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 & 3 & 4 \\
\hline 2 & 5 & 6 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 1 & 3 & 5 \\
\hline 2 & 4 & 6 \\
\hline
\end{tabular}
In summary, there are 5 ways to fill.
(Provided by Lai Guangfa, Hefei No. 46 Middle School, 230061)
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 3. Let } x>\frac{1}{4} \text {, simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$
|
$\begin{array}{l}\text { 3. } \because \sqrt{x+\frac{1}{2} \pm \frac{1}{2} \sqrt{4 x+1}} \\ =\sqrt{\frac{2 x+1 \pm \sqrt{4 x+1}}{2}} \\ =\frac{1}{2} \sqrt{4 x+2 \pm 2 \sqrt{4 x+1}} \\ =\frac{1}{2} \sqrt{(\sqrt{4 x+1} \pm 1)^{2}} \\ =\frac{1}{2}|\sqrt{4 x+1} \pm 1|, \\ \therefore \text { when } x>\frac{1}{4} \text {, the original expression }=\frac{1}{2}(\sqrt{4 x+1}+1) \\ -\frac{1}{2}(\sqrt{4 x+1}-1)=1 . \\\end{array}$
|
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 $1992^{1992}$ is divided by 1993 is 1.
|
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.
If there is a 4th point $D \in M$, then there must be a point $D' \in M$, such that $D'$, $A$, $D$ form an equilateral triangle, $\angle DAD' = 60^\circ$, so one of the points must be outside $\triangle ABC$. Therefore, without loss of generality, assume $D$ is within the segment $AC$, and $BD$ intersects the chord $AC$ at $K$. Also, without loss of generality, assume $D'$ is within the segment $BC$. Rotating $\triangle ABD$ by $60^\circ$ around $B$ to coincide with $\triangle CBD'$, then $K$ coincides with a point $K'$ on $BD'$. $\angle BCK' = \angle BAK = 60^\circ > 30^\circ$. Thus, $K'$ is not within the segment $BC$, and $D'$ is not within the segment $BC$, leading to a contradiction. Therefore, $M$ cannot contain a 4th point.
Therefore, $M$ contains exactly three points, which form an equilateral triangle.
|
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 $y$, then before $x$ and $y$ "meet", they are combined with $x$ and $y$. That is, we have to operate on two numbers of the form $\left(\left(\cdots\left(\left(x * a_{1}\right) * a_{2}\right) \cdots\right) * a_{1}\right)$ and $\left(\left(\cdots\left(\left(y * b_{1}\right) * b_{2}\right) \cdots\right) *\right.$ $\left.b_{n}\right)$, which are respectively equal to
$$
\begin{aligned}
X & =a_{1} a_{2} \cdots a_{k}\left(x+\frac{1}{a_{1}}+\frac{1}{a_{1} a_{2}}+\cdots+\frac{1}{a_{1} a_{2} \cdots a_{k}}\right) \\
& =a_{2} \cdots a_{k}\left(a_{1} x+1+u\right), \\
Y & =b_{1} b_{2} \cdots b_{n}\left(y+\frac{1}{b_{1}}+\frac{1}{b_{1} b_{2}}+\cdots+\frac{1}{b_{1} b_{2} \cdots b_{n}}\right) \\
& =b_{1} b_{2} \cdots b_{n}(y+v),
\end{aligned}
$$
where $u>0$ and $v>0$ represent the sum of the remaining numbers in the respective parentheses. We can assume $b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n}$, if there is $b_{j}>b_{j+1}$, then by swapping $b_{j}$ and $b_{j+1}$, we can increase $Y$.
Now, let's swap the positions of $a_{1}$ and $y$, obtaining
$$
\begin{array}{c}
X^{\prime}=a_{2} \cdots a_{k}(x y+1+u), \\
Y^{\prime}=b_{1} b_{2} \cdots b_{n}\left(a_{1}+v\right), \text { then we have } \\
\left(X^{\prime} Y^{\prime}-1\right)-(X Y-1)=X^{\prime} Y^{\prime}-X Y \\
=a_{2} \cdots a_{k} b_{1} b_{2} \cdots b_{n}\left[(x y+1+u)\left(a_{1}+v\right)\right. \\
\left.-\left(a_{1} x+1+u\right)(y+v)\right] \\
=c\left(a_{1}-y\right)(u+1-v x),
\end{array}
$$
where $c>0, a_{1}>y, v \leqslant \frac{1}{b_{1}}+\frac{1}{b_{1}^{2}}+\cdots \leqslant \frac{1}{b_{1}-1} \leqslant \frac{1}{x}$, i.e., $u+1-v x>0$, thus $X^{\prime} Y^{\prime}-X Y>0$. This means that by the above adjustment, we can increase the intermediate result, which contradicts our assumption about the optimality of the operation process.
Now, we can follow the optimal procedure we have proven to obtain the maximum possible value $A$, and it is easy to see that its last digit is 2.
|
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 there are at least $n$ teams participating, then $n=$ $\qquad$ .
|
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,
$2n + m > 2(n+1)$. This implies $m \geqslant 3$.
Thus, there exists a team that draws with team $A$ and scores no less than $2(n+1) + 1$, so we have
$2n + m > 2(n+1) + 1$, which implies $m \geqslant 4$.
Clearly, team $A$ must win at least one game, i.e., $n \geqslant 1$.
Therefore, team $A$ must play against at least 5 teams, meaning there are at least 6 teams participating in the competition. A tournament schedule for 6 teams that meets the conditions of the problem can be designed (omitted).
|
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 $A P_{1}=A P_{2}$, $B P_{2}=B P_{3}, C P_{3}=C P_{4}, A P_{4}=A P_{5}, B P_{5}=$ $B P_{6}, \cdots$. Then $\left|P_{2} P_{1994}\right|=$ $\qquad$
|
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 point. We have
$$
\begin{array}{l}
P_{2}-A=e^{A i}\left(P_{1}-A\right), \\
P_{5}-A=e^{A i}\left(P_{4}-A\right) .
\end{array}
$$
Subtracting, $P_{5}-P_{2}=e^{A i}\left(P_{4}-P_{1}\right)$.
Similarly, $P_{6}-P_{3}=e^{R i}\left(P_{5}-P_{2}\right)$
$$
=e^{(A-B) i}\left(P_{4}-P_{1}\right),
$$
$$
\begin{aligned}
P_{7}-P_{4} & =e^{C_{i}}\left(P_{6}-P_{3}\right) \\
& =e^{(A+B+C) i}\left(P_{4}-P_{1}\right) \\
& =e^{\pi i}\left(P_{4}-P_{1}\right)=-\left(P_{4}-P_{1}\right),
\end{aligned}
$$
we get $P_{7}=P_{1}$:
Similarly, $P_{n+6}=P_{n}$.
Since $1994=6 \times 332+2$,
we know $P_{1994}=P_{2}$.
Thus, $\left|P_{2} P_{1994}\right|=0$.
|
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 $4x$.
$$
\begin{aligned}
V & =\frac{1}{4} \cdot 4 x(80-2 x)(50-2 x) \\
& \leqslant \frac{1}{4} \cdot\left(\frac{4 x+80-2 x+50-2 x}{3}\right)^{3} \\
& =\frac{130^{3}}{108} .
\end{aligned}
$$
However, the condition for equality to hold should be $4 x=80 - 2 x=50-2 x$, which is impossible. Thus, this solution is incorrect. To achieve equality and find the extremum, the correct approach is
$$
\begin{array}{l}
V=\frac{1}{12} \cdot 6 x \cdot(80-2 x) \cdot 2(50-2 x) \\
\leqslant \frac{1}{12} \cdot\left[\frac{6 x+(80-2 x)+(100-4 x)}{3}\right]^{3} \\
=18000 .
\end{array}
$$
Equality holds when and only when $6 x=80-2 x=100-4 x$, i.e., $x=10$, at which $V$ reaches its maximum value.
This solution is concise, but why is it transformed this way?
This can be explained by introducing parameters into the inequality to obtain the answer.
$$
\begin{array}{l}
V=\frac{1}{\alpha \beta} \cdot \alpha x \cdot \beta(50-2 x) \cdot(80-2 x) \\
\leqslant \frac{1}{\alpha \beta}\left(\frac{\alpha x+50 \beta-2 \beta x+80-2 x}{3}\right)^{3} \\
=\frac{1}{\alpha \beta}\left[\frac{(50 \beta+80)+(\alpha-2 \beta-2) x}{3}\right]^{3} .
\end{array}
$$
To make (1) independent of $x$, it is necessary that
$$
\alpha-2 \beta-2=0 .
$$
For equality to hold, it must also be true that
$$
\alpha x=50 \beta-2 \beta x=80-2 x .
$$
From (2) and (3), we can obtain $\alpha=6, \beta=2$.
|
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 destination simultaneously?
|
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 reach the destination is at least 12 hours.
|
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 straight every time he reaches point $B$. The father runs 100 meters in 20 seconds, and the son runs 100 meters in 19 seconds. If they run at these speeds, how many times will they meet in 35 minutes from the start?
|
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
$$
\frac{250 N_{1}+x}{100 / 20}=\frac{400 N_{2}+x}{100 / 19} .
$$
This simplifies to $\frac{x}{200}=38 N_{2}-25 N_{1}$ being an integer. Since $0 \leqslant \frac{x}{200} \leqslant 1$, it can only be that $x=0$ or $x=200$. Therefore, their meeting points are at $A$ or $B$.
$\therefore 38 N_{2}-25 N_{1}-1$ or $38 N_{2}-25 N_{1}=1$.
Solving $38 N_{2}-25 N_{1}=0$ gives $N_{1}=38 k, N_{2}=$ $25 k .(k \in \mathbb{Z})$
Solving $38 N_{2}-25 N_{1}=1$ gives $N_{1}=38 k_{1}+3$, $N_{2}=25 k_{1}+2 .(k \in \mathbb{Z})$
Given that the father and son each run one lap in 50 and 76 seconds, respectively.
And $\left[\frac{2100}{50}\right)=42$ (laps), $\left[\frac{2100}{76}\right]=27$ (laps).
Their meeting at point $A$ can be determined by the solution to $38 N_{1}-25 N_{2}=0$.
Therefore, for $N_{1}=38 k, N_{2}=25 k$, we can take $k=$ 0,1 .
Similarly, for $N_{1}=38 k_{1}+3, N_{2}=25 k_{1}+2$, we can take $k_{1}=0,1$.
Thus, excluding the starting point, they meet 3 times. Answer: From the start of the race, they meet 3 times in 35 minutes.
|
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 exists a circle that can cover all the points, and therefore, there must also exist a smallest circle that can cover all the points. Let this circle be $\odot C$, with radius $r$. By the problem statement, $r \geqslant 1$.
We now prove that $r \leqslant 1$.
If there is only one point or no points on the circumference of $\odot C$ among the eight points, then there must be a circle with a radius smaller than $r$ that can cover all the points. $\odot C$ would not be the smallest circle covering all the points, which is impossible. Therefore, the circumference of $\odot C$ must have at least two points.
Assume that $\odot C$ has exactly two points, then these two points must be at the ends of a diameter, otherwise $\odot C$ can still be reduced. Since the distance between the two points is no more than 2, the radius $r$ of $\odot C$ is $\leqslant 1$.
If $\odot C$ has more points, $a_{1}, a_{2}, \cdots, a_{k}(3 \leqslant k \leqslant n)$, then these $k$ points cannot all be on a minor arc, otherwise $\odot C$ can still be reduced. This means that at least three points are on a major arc, and the triangle formed by these three points does not contain an obtuse angle. $\odot C$ is the smallest circle containing these three points. Since a circle with a radius of 1 that covers these three points can completely cover $\odot C$, it follows that $r \leqslant 1$.
Since $r \geqslant 1, r \leqslant 1$, thus $r=1$.
|
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 $P Q R S$ as their bases.)
|
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 that any two numbers selected as side lengths uniquely determine a right-angled triangle. Try to find the maximum value of $k$.
|
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 $2 a_{i} \leqslant a_{j}$. Therefore, any two numbers $a_{i}, a_{j}$ can uniquely determine an isosceles triangle. Hence, $k \geqslant 11$.
(2) Suppose $a_{1}<a_{2}<a_{3}<\cdots<a_{12}$ are any 12 numbers selected from $A$, and any two of these numbers can uniquely determine an isosceles triangle. Then, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{3} \geqslant 4, a_{4} \geqslant 8, \cdots, a_{11} \geqslant 1024, a_{12} \geqslant 2048$. This contradicts $a_{12} \leqslant 1994$. This indicates that at most 11 numbers can be selected from $A$ such that any two of them can uniquely determine an isosceles triangle. Therefore, the maximum value of $k$ is 11.
(Tongtan Town, Huxian County, Sichuan, 646111, Answered by Zhong Dingji)
|
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 all the stones away?
|
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; second, take away $2^{9}=$ 512 stones from each pile that has enough; …… finally, take away the stones from the piles that have only $2^{0}=1$ stone left. This way, a total of 11 operations are used.
Since there are piles with only one stone, there must be an operation to take one stone; if the remaining operations all take more than 2 stones, then the piles with exactly two stones cannot be taken away. Therefore, 2 operations can take at most $1+2=3$ stones, ..., 10 operations can take at most $1+2+$ $2^{2}+\cdots+2^{9}=1023$ stones. Thus, 1950 piles of stones require at least 11 operations.
|
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.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If natural numbers $m, n$ satisfy $m+n>m n$, then the value of $m$ $+n-m n$ is $\qquad$
|
2. 1. From the given, we have $\frac{1}{n}+\frac{1}{m}>1$. If $n, m \geqslant 2$, then $\frac{1}{n}$ $+\frac{1}{m} \leqslant \frac{1}{2}$, which is a contradiction. Therefore, at least one of $m, n$ must be 1. Without loss of generality, let $m=1$, then $m+n-m n=1+n-1 \cdot n=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If the complex number $z$ satisfies $3 z^{6}+2 i \cdot z^{5}-2 z-3 i=$ 0 . Then $|z|=$ $\qquad$ .
|
6. $|z|=1$.
The equation is $z^{5}=\frac{2 z+3 i}{3 z+2 i}$. Let $z=a+b i$.
$$
\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 {. }
$$
If $a^{2}+b^{2}>1$, then the left side of (1) $=|z|^{5}=\left(\sqrt{a^{2}+b^{2}}\right)^{5}>$ 1. But the right side has both the numerator and denominator positive, and the difference between the denominator and the numerator is $5\left(a^{2}+\right.$ $\left.b^{2}\right)-5>0$, making the fraction value less than 1. The equation cannot hold.
If $a^{2}+b^{2}<0$, similarly, it can be deduced that the equation does not hold.
Only when $a^{2}+b^{2}=1,|z|=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11. Let $a_{n}=6^{n}-8^{n}$. Find the remainder when $a_{94}$ is divided by 49. (Adapted from the first American Mathematical Invitational Competition)
|
$$
\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}
$$
$\therefore a_{94}$ when divided by 49 leaves a remainder of 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.7. A square wooden board is divided into $n^{2}$ unit squares by horizontal and vertical lines. Mark $n$ squares so that any rectangle with an area of at least $n$ and whose sides lie along the grid lines contains at least one marked square. Find the largest $n$ that satisfies this condition.
|
10.7. 7 .
Obviously, if $n$ marked cells satisfy the conditions of the problem, then in each row and each column there is exactly one marked cell. Let $n \geqslant 3$ (obviously, $n=2$ is not the maximum), take the first row with a marked cell as $A$, the row adjacent to $A$ as $B$, and take either a row adjacent to $A$ (and not coinciding with $B$) or a row adjacent to $B$ (and not coinciding with $A$) as $C$.
Let $b$ be the number of the marked cell in row $B$. If $b \leqslant n-\left[\frac{n}{2}\right]$ or $b>\left[\frac{n}{2}\right]+1$, then in rows $A$ and $B$ there exists a rectangle with an area of at least $n$ that does not contain any marked cells, so,
$$
n-\left[\frac{n}{2}\right]7$, then the area of both rectangles is not less than $n$, but row $C$ has only one marked cell, meaning one of these rectangles does not contain a marked cell.
Thus, we have proved that $n \leqslant 7$. A $7 \times 7$ board that meets the conditions is shown above.
|
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
$$
\begin{array}{l}
C D^{2}=A D \cdot B D . \\
\therefore \frac{A D}{B D} \operatorname{tg}^{2} \frac{\theta}{2}=\frac{A D}{B D} \cdot \frac{C D^{2}}{A D^{2}} \\
=\frac{A D}{B D} \cdot \frac{A D \cdot B D}{A D^{2}}=1 .
\end{array}
$$
|
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
$$
\begin{array}{c}
P P^{\prime} / / C G^{\prime}, \frac{P P^{\prime}}{G G^{\prime}}=\frac{A^{\prime} P^{2}}{A^{\prime} G} . \\
\text { Also, } \frac{S_{\triangle P B C}}{S_{\triangle G B C}}=\frac{P F^{\prime}}{G G^{\prime}}, \\
\text { hence } \frac{S_{\triangle P B C}}{S_{\triangle G B C}}=\frac{A^{\prime} P}{A^{\prime} G} .
\end{array}
$$
Since $G$ is the centroid, we have
$$
S_{\triangle G A B}=S_{\triangle G B C}=S_{\triangle G C A}=\frac{1}{3} S_{\triangle A B C} .
$$
From (1) + (2) + (3) and (4), we get
$$
\begin{array}{l}
\frac{A^{\prime} P}{A^{\prime} G}+\frac{B^{\prime} P}{B^{\prime} G}+\frac{C^{\prime} P}{C^{\prime} G} \\
=\frac{3 S_{\triangle P B C}}{S_{\triangle A B C}}+\frac{3 S_{\triangle P C A}}{S_{\triangle A B C}}+\frac{3 S_{\triangle P A B}}{S_{\triangle A B C}}=3 .
\end{array}
$$
|
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
$$
\min _{x_{1}, z_{2} \cdots, s_{n}}\left[\max _{\omega \in C,|\omega|=1} \prod_{j=1}^{n}\left|\omega-z_{j}\right|\right] \text {, }
$$
and discuss the conditions that the complex numbers $z_{1}$, $z_{2}, \cdots, z_{n}$ satisfy when the minimum of the maximum value is achieved.
|
$$
\begin{array}{l}
f(u)=\prod_{j=1}^{n}\left(u-x_{j}\right) \\
=u^{n}+C_{n-1} u^{n-1}+\cdots+C_{1} u+C_{0} .
\end{array}
$$
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, $\varepsilon_{k}=e^{i \frac{2 x+\theta}{n}}, k=1,2, \cdots, n$ (Note: for any angle $\psi, e^{i \phi}=\cos \psi+i \sin \psi$). Then, $\varepsilon_{i}^{i}=e^{i \theta}=C_{0}$, and when $1 \leqslant m \leqslant n-1$, since $\left(e^{i \frac{2 \pi}{n}}\right)^{m} \neq 1$, we have
Thus, we have
$$
\begin{array}{l}
\sum_{k=1}^{n}\left|f\left(\varepsilon_{k}\right)\right| \geqslant\left|\sum_{k=1}^{n} f\left(\varepsilon_{k}\right)\right| \\
=\left|\sum_{k=1}^{n} \varepsilon_{i}^{n}+\sum_{i=1}^{n-1} C_{i} \sum_{k=1}^{n} \varepsilon_{k}^{k}+n C_{0}\right| \\
=\left|2 n C_{0}\right|=2 n .
\end{array}
$$
From (2), there is at least one $\varepsilon_{k}$ such that $\left|f\left(\varepsilon_{k}\right)\right| \geqslant 2$, i.e.,
$$
\max _{\{u \in \mathbb{C} \mid |u|=1\}} \prod_{j=1}^{n}\left|u-z_{j}\right| \geqslant 2.
$$
If (3) holds with equality, then from (2) we know that for $k=1,2, \cdots, n, f\left(\varepsilon_{k}\right)$ not only have the same modulus 2, but also the same argument. Using (2) again, we have $\sum_{k=1}^{n} f\left(\varepsilon_{k}\right)=2 n C_{0}$, so $f\left(\varepsilon_{k}\right)=2 C_{0}$, for $k=1, 2, \cdots, n$. Using (1), we can see that the $(n-1)$-degree polynomial $C_{n-1} u^{n-1} + C_{n-2} u^{n-2} + \cdots + C_{1} u$ is zero when $u=\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n}$, thus this polynomial is identically zero. Therefore,
$$
f(u)=u^{n}+C_{0}.
$$
Since $f\left(z_{k}\right)=0, k=1,2, \cdots, n$, we can conclude that $x_{1}, z_{1}, \cdots, z_{n}$ are the $n$-th roots of $-C_{0}$. When the minimum of the maximum value is reached, in the complex plane, the points corresponding to the complex numbers $z_{1}, z_{2}, \cdots, z_{n}$ are the $n$ vertices of a regular $n$-sided polygon inscribed in the unit circle.
|
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 in this group at minimum?
|
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 this group of students has only 7 people, then by the pigeonhole principle, there exists an $i, 1 \leqslant i \leqslant 4$, such that the number of people whose answer to the first question is $i$ is at most 1, let this person's answer be $(i, j, k)$. Thus, among the 16 different answers where the first question is $i$, it can only cover 7 of $(i, j, x)$ and $(i, x, k)$, the remaining 9 must be covered by other answer sets, which requires 9 more sets, leading to a contradiction. Therefore, this group of students must have at least 8 people.
|
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 $3 * x=3^{2}+x-3=6+x \leqslant 6+\frac{3}{2}$ $<8$, which does not satisfy $3 * x=8$.
Therefore, the $x$ that satisfies $3 * x=8$ is $x=\frac{1+\sqrt{21}}{2}$, which satisfies $x^{2}-x-5=0$.
$$
\begin{array}{l}
\text { So, } x^{3}-4 x=x\left(x^{2}-x-5\right)+x^{2}+x=x^{2}+x \\
\quad=x^{2}-x-5+2 x+5=2 x+5 . \\
{\left[x^{3}-4 x\right]=[2 x+5]=[2 x]+5} \\
=[1+\sqrt{21}]+5=10 \text { is the answer. }
\end{array}
$$
|
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 $\left\{x_{n}\right\}$, $S_{1994}$.
|
$\begin{array}{l}\text { Let } \begin{array}{l} \text { } F:=x_{n}+y_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n}, F^{\prime} \\ =x_{h}-j_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n} . \text { Then } F+F^{\prime}=2 x_{n} \\ =\omega^{n}+(\bar{\omega})^{n}\left(\text { where } \omega=\frac{-1+\sqrt{3} i}{2}\right) . \text { Therefore, } \\ 2 S_{1994}= \omega+\omega^{2}+\cdots+\omega^{1994}+\bar{\omega}+(\bar{\omega})^{2} \\ +\cdots+(\bar{\omega})^{1994} \\ = \frac{\omega\left(\omega^{1994}-1\right)}{\omega-1}+\frac{\bar{\omega}\left[(\bar{\omega})^{1994}-1\right]}{\bar{\omega}-1} \\ = \frac{\omega\left(\omega^{2}-1\right)}{\omega-1}+\frac{\bar{\omega}\left[(\bar{\omega})^{2}-1\right]}{\bar{\omega}-1} \\ = \omega^{2}+\omega+(\bar{\omega})^{2}+\bar{\omega}=-2 .\end{array}\end{array}$
Thus, $S_{1994}=-1$.
|
-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 coordinates as integers) in the point set $A \cap B$ is
|
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-coordinate can be 2, 3, 4, 5. The axis of symmetry $E F$ passes through the crescent shape $S$, and it is found that it passes through only one integer point $C_{4}(2,3)$. The integer points in $S$ with an x-coordinate of 1 are 3:
$$
C_{1}(1,5), C_{2}(1,4), C_{3}(1,3).
$$
The axis-symmetric points of these three points are, in order,
$$
C_{5}(2,2), C_{0}(3,2), C_{7}(4,2).
$$
Therefore, there are a total of 7 integer points.
|
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。
Through a non-central point inside a regular dodecagon,
what is the maximum number of different diagonals that can be drawn?
|
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 $\overparen{A_{5} A_{8}}=\overparen{A_{8} A_{11}}$, $\overparen{A_{11} A_{1}}=\overparen{A_{1} A_{3}}$, and $\overparen{A_{3} A_{4}}=\overparen{A_{4} A_{5}}$, the lines $A_{3} A_{8}$, $A_{5} A_{1}$, and $A_{11} A_{1}$ are the angle bisectors of $\triangle A_{3} A_{5} A_{11}$. Therefore, $A_{1} A_{5}$ passes through the intersection point $K$ of the diagonals $A_{3} A_{8}$ and $A_{4} A_{11}$. Similarly, the lines $A_{2} A_{6}$, $A_{4} A_{11}$, and $A_{8} A_{3}$ are the angle bisectors of $\triangle A_{2} A_{4} A_{8}$, so the diagonal $A_{2} A_{6}$ also passes through point $K$. Thus, these 4 diagonals intersect at point $K$. Point $K$ lies on the chord $A_{1} A_{5}$, and $K$ is different from the center $O$.
Next, prove that in a regular dodecagon, there do not exist 5 diagonals intersecting at a non-central point inside the dodecagon.
Use proof by contradiction. Assume that such 5 diagonals exist. Then, the 10 endpoints of these 5 diagonals are all distinct, and any one of these diagonals divides the circumcircle of the dodecagon into two arcs, each containing 4 endpoints of the other 4 diagonals. This implies that if these 10 endpoints are given, then the 5 diagonals are also determined.
By the symmetry of the vertices of the regular dodecagon, without loss of generality, assume that among the 12 vertices, excluding $A_{1}$ and one of $A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$, these 10 endpoints are given. Now, consider the cases for this one point.
(1) If this point is $A_{2}$, since $\overparen{A_{1} A_{7}}=\overparen{A_{9} A_{12}}$, the intersection point $M$ of the diagonals $A_{4} A_{9}$ and $A_{7} A_{12}$ lies on the line through the center $O$ and point $A_{8}$. This can be seen by noting that $\triangle M A_{4} A_{7}$ is congruent to $\triangle M A_{12} A_{9}$, and $\triangle O M A_{9}$ is congruent to $\triangle O M A_{7}$. Thus, the intersection point $M$ does not lie on the other diagonal $A_{3} A_{\varepsilon}$, leading to a contradiction.
(2) If this point is $A_{3}$, since $\overparen{A_{4} A_{5}}=\overparen{A_{9} A_{10}}$, the intersection point $M$ of the diagonals $A_{4} A_{9}$ and $A_{5} A_{10}$ lies on the line through the center $O$ and point $A_{7}$. The point $M$ does not lie on the other diagonal $A_{2} A_{0}$, also leading to a contradiction.
(3) If this point is one of $A_{6}, A_{5}, A_{5}, A_{7}$, then the diagonals $A_{2} A_{8}$ and $A_{3} A_{9}$ are diameters and must intersect at the center $O$, leading to a contradiction.
In conclusion, the answer is 4.
|
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, $x\left(e^{y}-1\right) y\left(e^{x}-1\right)>0 \Leftrightarrow x^{2} y^{2}>0$. That is, when the two terms on the left side of (*) are not 0, these two terms have the same sign.
$$
\begin{array}{l}
\left.\therefore x\left(e^{y}-1\right)+y\left(e^{x}-1\right)=1\right) \Leftrightarrow\left\{\begin{array}{l}
x\left(e^{y}-1\right)=0, \\
y\left(e^{x}-1\right)=0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array} { l }
{ x = 0 , y \in R , } \\
{ y = 0 , x \in R }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=0, \\
y=0 .
\end{array}\right.\right.
\end{array}
$$
Also, $y=0$ means $x^{2}-1=0$, giving $x= \pm 1$.
Therefore, the original equation has roots $0,1,-1$, i.e., $A=\{0,-1$, $1\}$. The sum of the squares of its elements is 2.
|
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
$$
2 x^{2}-6 x+4=0 \text {. }
$$
Solving gives $x=1, x=2$ (discard).
$\therefore$ The solution that satisfies the conditions is $x=1, y=1$.
Thus, $x+y+x y=3$.
|
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 original text's format and line breaks as requested.
|
$$
-, f(x)=[x-(2 a-1)]^{2}-10 a^{2}+4 a-1 .
$$
(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., $\frac{1}{2}<a<1$, the minimum value is obtained at the vertex of the parabola. At this time, $m=-10 a^{2}+4 a-1$.
Combining the above, we get the piecewise function for $m$
$$
m=F(a)=\left\{\begin{array}{l}
-6 a^{2}, \quad a \leqslant \frac{1}{2}, \\
-10 a^{2}+4 a-1, \frac{1}{2}<a<1, \\
-6 a^{2}-4 a+3, \quad a \geqslant 1 .
\end{array}\right.
$$
Next, consider finding the maximum value of $m$. We need to discuss $F(a)$ in segments.
When $a \leqslant \frac{1}{2}$, $m=-6 a^{2} \leqslant 0$, equality holds only when $a=0$, at which time the maximum value is 0.
When $-\frac{1}{2}<a<1$, $m=-10\left(a-\frac{1}{5}\right)^{2}-\frac{3}{5}$ $\leqslant-10\left(\frac{1}{2}-\frac{1}{5}\right)^{2}-\frac{3}{5}=-\frac{3}{2}$. At this time, the maximum value is $\frac{3}{2}$.
When $a \geqslant 1$, $m=-6\left(a+\frac{1}{3}\right)^{2}+\frac{11}{3}$ $\leqslant-6\left(1+\frac{1}{3}\right)^{2}+\frac{11}{3}=-7$. At this time, the maximum value is -7.
In summary, when $a=0$, $m$ achieves its maximum value of 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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