problem
stringlengths 2
5.64k
| solution
stringlengths 2
13.5k
| answer
stringlengths 1
43
| problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
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 7. Let $s_{1}, x_{2}, c_{3}, x_{4}$ all be positive real numbers, and $x_{1} + x_{2} + x_{3} - x_{4} = \pi$. Find the minimum value of the expression
\[
\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)
- \left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x_{3}}\right)\left(2 \sin ^{2} x_{4}+\frac{1}{\sin ^{2} x_{4}}\right).
\]
(1991, China National Training Team Test Question)
|
Let $x_{1}+x_{2}$ be a constant. Since
$$
\begin{array}{l}
\sin x_{1} \sin x_{2} \\
=\frac{1}{2}\left[\cos \left(x_{1}-x_{2}\right)-\cos \left(x_{1}+x_{2}\right)\right],
\end{array}
$$
we know that the value of $\sin x_{1} \sin x_{2}$ increases as $\left|x_{1}-x_{2}\right|$ decreases. Let the expression in question be $f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$. If $x_{1}, x_{2}, x_{3}, x_{4}$ are not all equal, without loss of generality, assume $x_{1}>\frac{\pi}{4}>x_{2}$. Let $x_{1}{ }^{\prime}=\frac{\pi}{4}, x_{2}{ }^{\prime}=x_{1}+x_{2}-\frac{\pi}{4}$,
$$
x_{3}{ }^{\prime}=x_{3}, x_{4}{ }^{\prime}=x_{4},
$$
Thus, we have $x_{1}{ }^{\prime}+x_{2}{ }^{\prime}=x_{1}+x_{2}$,
$\left|x_{1}{ }^{\prime}-x_{2}{ }^{\prime}\right|2\left(2 \sin ^{2} x_{1}^{\prime} \sin ^{2} x_{2}{ }^{\prime}+\frac{1}{2 \sin ^{2} x_{1} \sin ^{2} x_{2}{ }^{\prime}}\right) \\
\quad+2\left(\frac{\sin ^{2} x_{1}^{\prime}}{\sin ^{2} x_{2}{ }^{\prime}}+\frac{\sin ^{2} x_{2}^{\prime}}{\sin ^{2} x_{1}{ }^{\prime}}\right) \\
=\left(2 \sin ^{2} x_{1}^{\prime}+\frac{1}{\sin ^{2} x_{1}{ }^{\prime}}\right) \\
\quad \cdot\left(2 \sin ^{2} x_{2}{ }^{\prime}+\frac{1}{\sin ^{\prime} x_{2}^{\prime}}\right) .
\end{array}
$$
Thus,
$$
f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)>f\left(x_{1}{ }^{\prime}, x_{2}{ }^{\prime}, x_{3}{ }^{\prime}, x_{4}{ }^{\prime}\right) .
$$
If $x_{2}{ }^{\prime}, x_{3}{ }^{\prime}, x_{4}{ }^{\prime}$ are not all equal, then we can perform a similar smoothing transformation to prove that when $x_{1}, x_{2}, x_{3}, x_{4}$ are not all equal, we always have
$$
f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)>f\left(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}\right) .
$$
Therefore, the minimum value is $f\left(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}\right)$
$=81$, which is achieved when and only when $x_{1}=x_{2}=x_{3}=x_{4}=\frac{\pi}{4}$.
|
81
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. On a plane, there are $n$ lines, among which $k$ lines are pairwise parallel, and $t$ lines pass through the same point $(k+t<n)$, in addition, any two intersect, and any 3 do not share a point. How many regions do they divide the plane into:
|
The number of intersection points is given by $p_{n}=C_{n}^{2}-C_{k}^{2}-C_{t}^{2}+(t-1)$. According to Theorem 1, the number of regions they divide the plane into is
$$
S_{n}=C_{n}^{2}-C_{1}^{2}-C_{t}^{2}+n+t.
$$
The figure below gives an example F for $n=8, k=4, t=3$. In this case, $S_{8}=C_{8}^{2}-C_{4}^{2}-C_{3}^{2}+8+3=28-6-3+8+3-1=30$.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. In $\triangle ABC$, $G$ is the centroid, and $P$ is a point inside the triangle. The line $PG$ intersects the lines $BC$, $CA$, and $AB$ at $A'$, $B'$, and $C'$, respectively. Prove that: $\frac{A'P}{A'G} + \frac{B'P}{B'G} + \frac{C'P}{C'G} = 3$
(1991, Huanggang Region Junior High School Competition)
|
Prove that by connecting $B G, G C, P B, P C$, and drawing $G G^{\prime} \perp B C$ at $G^{\prime}$, $P P^{\prime} \perp B C$ at $P^{\prime}$, then
$$
\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}
$$
|
3
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. The two roots of the equation $x^{2}-3|x|-4=0$ are $x_{1}$, $x_{2}$, then $x_{1}+x_{2}=$ $\qquad$
|
2. 0
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B-4. $p(x)$ is a non-zero polynomial of degree less than 1992, and $p(x)$ has no common factor with $x^{3}-x$. Let
$$
\frac{d^{1992}}{d x^{1992}}\left(\frac{p(x)}{x^{3}-x}\right)=\frac{f(x)}{g(x)},
$$
where $f(x), g(x)$ are polynomials. Find the smallest possible degree of $f(x)$.
|
The minimum possible degree of $f(x)$ is 3984.
By the division algorithm, we let $p(x) = (x^3 - x)q(x) + r(x)$, where $q(x)$ and $r(x)$ are polynomials, and the degree of $r(x)$ is less than 3, and the degree of $q(x)$ is less than 1989. Then,
$$
\cdot \frac{d^{1992}}{d x^{1992}}\left(\frac{p(x)}{x^3 - x}\right) = \frac{d^{1992}}{d x^{1992}}\left(\frac{r(x)}{x^3 - x}\right).
$$
Since $p(x)$ and $x^3 - x$ have no common factors, $r(x)$ and $x^3 - x$ also have no common factors.
Let $p_0(x)$ be a polynomial that is coprime with $x^3 - x$ and has degree $n < 3$. We will use induction to show that
$$
\frac{d^k}{d x^k}\left(\frac{p_0(x)}{x^3 - x}\right) = \frac{p_k(x)}{(x^3 - x)^{k+1}},
$$
where $p_k(x)$ is a polynomial that is coprime with $x^3 - x$ and $\operatorname{deg}(p_k) = \operatorname{deg}(p_0) + 2k < 3(k+1)$. We have
$$
\begin{array}{l}
\frac{d}{d x}\left(\frac{p_k(x)}{(x^3 - x)^{k+1}}\right) \\
= \frac{(x^3 - x)p_k'(x) - (k+1)(3x^2 - 1)p_k(x)}{(x^3 - x)^{k+2}} \\
= \frac{p_{k+1}(x)}{(x^3 - x)^{k+2}}
\end{array}
$$
The polynomial $p_{k+1}(x)$ is coprime with $x^3 - x$, because the left term is divisible by $x^3 - x$ and is coprime with the right term (the left term refers to $(x^3 - x)p_k'(x)$, and the right term refers to $(k+1)(3x^2 - 1)p_k(x)$ — translator's note). If $p_k(x) = ax^n +$ lower order terms, $a \neq 0$, then the highest degree coefficient of $p_{k+1}(x)$ is $a(n - 3(k+1))$, and since $n \neq 3(k+1)$,
$$
\begin{aligned}
\operatorname{deg}(p_{k+1}) & = \operatorname{deg}(p_k) + 2 = \operatorname{deg}(p_0) + 2k + 2 \\
& < 3 + 3k + 2 < 3(k+2).
\end{aligned}
$$
This completes our induction.
Thus, $\frac{d^{1992}}{d x^{1992}}\left(\frac{r(x)}{x^3 - x}\right) = \frac{f(x)}{g(x)}$,
where $\operatorname{deg}(f) = \operatorname{deg}(r) + 2(1992) \geq 3984$. The minimum value is achieved only when $r(x)$ is a constant.
($\operatorname{deg}(p)$ denotes the highest degree of the polynomial $p(x)$ — translator's note)
|
3984
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. (Question 2 of the 5th Test, *The American Mathematical Monthly*, pages 119 to 121, 1953)
Given a positive integer $n \geqslant 3$, for $n$ complex numbers $z_{1}, z_{2}, \cdots, z_{n}$ with modulus 1, find
$$
\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
|
Example 6. Let $f(x)=\frac{4^{x}}{4^{x}+2}$. Try to find the value of $f\left(\frac{1}{1001}\right)$ $+f\left(\frac{2}{1001}\right)+\cdots+f\left(\frac{1000}{1001}\right)$. (1986, National High School Competition)
|
Let $F=f(x)=\frac{4^{x}}{4^{x}+2}, F^{\prime}=f(y)=$ $\frac{4^{y}}{4^{y}+2}$. If $x+y=1$, then $f(x)+f(y)=1$. In fact, $f(x)+f(y)=f(x)+f(1-x)=\frac{4^{x}}{4^{x}+2}$ $+\frac{4^{1-x}}{4^{1-x}+2}=1$. Therefore, $f\left(\frac{1}{1001}\right)+f\left(\frac{2}{1001}\right)$ $+\cdots+f\left(\frac{999}{1001}\right)+f\left(\frac{1000}{1001}\right)=\left[f\left(\frac{1}{1001}\right)\right.$ $\left.+f\left(\frac{1000}{1001}\right)\right]+\left[f\left(\frac{2}{1001}\right)+f\left(\frac{999}{1001}\right)\right]+\cdots+$ $\left[f\left(\frac{500}{1001}\right)+f\left(\frac{501}{1001}\right)\right]=\underbrace{1+1+\cdots+1}_{500 \uparrow}=500$.
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9. For the set $\{1,2,3, \cdots, n\}$ and any of its non-empty subsets, define the "alternating sum" as follows: rearrange the elements in decreasing order, then alternately add and subtract the subsequent numbers starting from the largest. For example, for $\{1,2,4,6,9\}$, rearranged as $(9,6,4,2,1)$, the alternating sum is $9-6+4-2+1=6$. The alternating sum of $\{5\}$ is 5. For $n=7$, find the total of all alternating sums.
|
Solution: Let's define the alternating sum of the elements of the empty set to be 0. Divide the $2^{n}$ subsets of $\{1,2, \cdots, n\}$ into two categories: one category contains $n$ and has $2^{n-1}$ subsets, and the other category does not contain $n$, also having $2^{n-1}$ subsets. Pair the subset $F=\left\{n, a_{1}, a_{2}, \cdots, a_{i}\right\}$ containing $n$ with the subset $F^{\prime}=\left\{a_{1}, a_{2}, \cdots, a_{i}\right\}$ not containing $n$. Clearly, this pairing is one-to-one, and the sum of the alternating sums of the two subsets in each pair is $n$. Therefore, the total alternating sum $S$ of all elements of $\{1,2, \cdots, n\}$ is $n \cdot 2^{n-1}$. Thus, when $n=7$, the total alternating sum is 448.
|
448
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $E=\{1,2, \cdots, 200\}, G=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$ $\subset E$, and $G$ has the following two properties:
(i) For any $1 \leqslant i \leqslant j \leqslant 100$, it always holds that $a_{i}+a_{j} \neq 201$;
(ii) $\sum_{i=1}^{130} a_{i}=10080$.
Prove: The number of odd numbers in $G$ is a multiple of 4, and the sum of the squares of all numbers in $G$ is a constant. $(1990$, National High School League)
|
6. Divide the elements of $E$ into 100 subsets such that the sum of the two numbers in each subset is 201: $E_{1}=\{1,200\}, E_{2}=\{2,199\}, \cdots, E_{100}=\{100, 101\}$. According to property (i) of $G$, for any $1 \leqslant i \leqslant 100$, the two elements in $E_{i}$ cannot both be in $G$. Since the number of elements in $G$ is exactly equal to the number of elements in $E$, each $E_{i}$ must contain exactly one element of $G$.
Since $2+4+\cdots+200=10100 \neq 10080$, not all elements in $G$ are even. Since 10080 is even, the number of odd elements in $G$ must be even. $G$ can be seen as the result of replacing an even number of even numbers from $\{2,4, \cdots, 200\}$ with odd numbers.
Suppose $2k$ even numbers $2n_{1}, 2n_{2}, \cdots, 2n_{2k}$ are replaced with the odd numbers $201-2n_{1}, 201-2n_{2}, \cdots, 201-2n_{2k}$ to form $G$, then
$$
\sum_{i=1}^{\infty} 2n_{i} - 20 = \sum_{i=1}^{2k} (201 - 2n_{i}).
$$
Thus,
$$
4 \sum_{i=1}^{k} n_{i} = 2k \times 201 + 20.
$$
Since $\sum_{i=1}^{2k} n_{i}$ is an integer, $k$ must be divisible by 2. Let $k = 2l$ (where $l$ is an integer), then the number of odd elements is $4l$.
Since $G$ is obtained by replacing $4l$ even numbers from $\{2,4, \cdots, 200\}$ with odd numbers, let's see how the sum of the squares of these numbers changes:
$$
\begin{array}{l}
\sum_{i=1}^{u} (201 - 2n_{i})^2 \\
= 201^2 \times 4l - 4 \times 201 \sum_{i=1}^{u} n_{i} + \sum_{i=1}^{u} (2n_{i})^2.
\end{array}
$$
Substituting (1) into the right side of the above equation, we get
$$
\begin{array}{l}
\sum_{i=1}^{u} (201 - 2n_{i})^2 = -201 \times 20 + \sum_{i=1}^{u} (2n_{i})^2. \\
\text{Thus, } \sum_{i=1}^{100} a_{i}^2 = (2^2 + 4^2 + \cdots + 200^2) - 20 \times 201 \\
= 1349380 \text{ (a constant).} \\
\end{array}
$$
|
1349380
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example. Given $f(x)=[3 x]+[4 x]+[12 x]+[16 x]$.
(1) Find the number of positive integer values $M$ that $f(x)$ takes when $0<x \leqslant 1$.
(2) Is 97 a value of $f(x)$?
|
$$
\begin{array}{l}
\text { (1) } \sum n_{i}=3+4+12+16=35 . \\
\sum_{1<i<j<1}\left(n_{i}, n_{j}\right)=(3,4)+(3,12)+(3,16)+(4,12)+(4,16)+(12,16)=17. \\
\sum_{1<i<j<k<t}\left(n_{i}, n_{j}, n_{k}\right)=(3,4,12)+(3,4,16)+(3,12,16)+(4,12,16)=7. \\
\left(n_{i}, n_{2}, n_{3}, n_{!}\right)=(3,4,12,16)=1, \\
\end{array}
$$
By Proposition 1, \( M=35-17+7-1=24 \).
$$
\begin{array}{l}
\text { (2) } \because N=9 \cdot \because=2 \times 35+27, r=27, \\
\text { } D=[3,4,12,16]=48, \\
\therefore \frac{D}{\sum_{i=1}^{4} n_{i}} \cdot r=\frac{48}{35} \cdot 27=36.99, \\
\frac{D}{\sum_{i=1}^{4} n_{i}}(r+4-1)=\frac{48}{35} \cdot 30=41.1 .
\end{array}
$$
From \( 36.99 \leqslant t^{\prime}<41.1 \), we know the possible choices are: \( 37,38,39,40,41 \), but only 39 and 40 are multiples of \( \frac{48}{3}=16, \frac{48}{4}=12, \frac{48}{12}=4, \frac{48}{16}=3 \), so we discard \( 37,38,41 \).
\(\because\) Substituting \( x=\frac{39}{48} \) we get:
$$
\begin{array}{l}
f\left(\frac{39}{48}\right)=2+3+9+13=27, \\
\begin{aligned}
\therefore f\left(2 \frac{39}{48}\right)= & f(2)+f\left(\frac{39}{48}\right) \\
= & 2 \times 35+27=97,
\end{aligned}
\end{array}
$$
Thus, 97 is a value of \( f(x)=[3 x]+[4 x]+[12 x]+[16 x] \).
|
97
|
Number Theory
|
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
|
6. Given 95 numbers $a_{1}, a_{2}, \cdots, a_{95}$, each of which can only take one of the two values +1 or -1. Then, the minimum positive value of the sum of their pairwise products $a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{81} a_{23}$ is $\qquad$ .
|
6.13.
Let $N=a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{98} a_{95}$.
Suppose $a_{1}, a_{2}, \cdots, a_{95}$ contain $m \uparrow+1, n \uparrow-1$, then $m+n=95$.
Multiplying (1) by 2 and adding $a_{1}^{2}+a_{2}^{2}+\cdots+a_{95}^{2}=95$, we get
$$
\left(a_{1}+a_{2}+\cdots+a_{95}\right)^{2}=2 N+95 \text{. }
$$
Also, $a_{1}+a_{2}+\cdots+a_{95}=m-n$,
$$
\therefore(m-n)^{2}=2 N+95 \text{. }
$$
The smallest natural number $N=13$, at this time,
$$
\begin{array}{l}
(m-n)^{2}=11^{2}, \text{ i.e., } \\
m-n= \pm 11 .
\end{array}
$$
Solving (2) and (3) simultaneously, we get $m=53, n=42$ or $m=42$, $n=53$. According to this, we can construct the case where $N$ reaches its minimum value.
Thus, the required minimum positive value is 13.
Solution 2: Suppose $a_{1}, a_{2}, \cdots, a_{95}$ contains +1 $n$ times, then -1 appears $95-n$ times. $S=a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}$ has $C_{n}^{2}+C_{95-n}^{2}$ terms of +1, and $C_{95}^{2}-\left(C_{n}^{2}+C_{95-n}^{2}\right)$ terms of -1.
$$
\begin{aligned}
S= & \left(C_{n}^{2}+C_{95-n}^{2}\right)-\left[C_{95}^{2}-\left(C_{n}^{2}+C_{95-n}^{2}\right)\right] \\
= & 2\left(C_{n}^{2}+C_{95-n}^{2}\right)-C_{95}^{2} \\
& =n(n-1)+(95-n)(94-n)-\frac{1}{2} \times 95 \times 94 \\
& =2 n^{2}-(95+94+1) n+\frac{1}{2} \times 95 \times 94 \\
& =2\left[\left(n-\frac{95}{2}\right)^{2}-\frac{95}{4}\right]>0 \\
& \Rightarrow\left|n-\frac{95}{2}\right|>\frac{\sqrt{95}}{2} .
\end{aligned}
$$
Thus, either $n>\frac{95}{2}+\frac{\sqrt{95}}{2}>47.5+4.8=52.3 \Rightarrow n \geqslant 53$ or $n<\frac{95}{2}-\frac{\sqrt{95}}{2}<47.5-4.8=42.7 \Rightarrow n \leqslant 42$.
(4.8 is an underestimate of $\frac{\sqrt{95}}{2}$),
When $n=53$, $S=2 \times 53^{2}-2 \times 95 \times 53+\frac{1}{2} \times 95 \times 94=13$. Similarly, when $n=42$, $S=13$.
From the function graph, the minimum positive value of $S$ is 13.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (This question is worth 25 points) Arrange all positive integers that are coprime with 105 in ascending order. Find the 1000th term of this sequence.
|
$$
\begin{array}{r}
\text { 2. Let } S=\{1,2, \cdots, 105\}, A_{1}=\{x \mid x \in S, \text { and } 3 \mid \\
x\}, A_{2}=\{x \mid x \in S, 45 \mid x\}, A_{3}=\{x \mid x \in S, \text { and } 7 \mid
\end{array}
$$
$$
\begin{array}{l}
\left|\overline{A_{1}} \cap \overline{A_{2}} \cap \overline{A_{3}}\right|=| S \mid-\left(\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|\right) \\
+\left(\left|A_{1} \cap A_{2}\right|+\left|A_{1} \cap A_{3}\right|+\left|A_{2} \cap A_{3}\right|\right) \\
-\left|A_{1} \cap A_{2} \cap A_{3}\right| \text {. } \\
\because|S|=105,\left|A_{1}\right|=35,\left|A_{2}\right|=21,\left|A_{3}\right|=15, \\
\left|A_{1} \cap A_{2}\right|=7,\left|A_{1} \cap A_{3}\right|=5,\left|A_{2} \cap A_{3}\right|=3, \\
\left|A_{1} \cap A_{2} \cap A_{3}\right|=1, \\
\therefore\left|\overline{A_{1}} \cap \overline{A_{2}} \cap \overline{A_{3}}\right|=105-(3 \times 5+3 \times 7+5 \times 7) \\
+(3+5+7)-1 \\
=105-71+15-1=48. \\
\end{array}
$$
Let the sequence of positive integers that are coprime with 105, arranged in ascending order, be $a_{1}, a_{2}, a_{3}, \cdots, a_{n}, \cdots$. Then
$$
a_{1}=1, a_{2}=2, a_{3}=4, \cdots, a_{48}=104.
$$
Let $P=\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{48}\right\}$. On one hand, for $n \geqslant 1$, let $a_{n}=105 k+r(k \geqslant 0,0 \leqslant r \leqslant 104, k, r \in \mathbb{Z})$. Since $(a_{n}, 105)=1$, we have $(r, 105)=1$, hence $r \in P$; on the other hand, for any non-negative integer $k$ and $r \in P$, since $(r, 105)=1$, we have $(105 k+r, 105)=1$. Therefore, there must be an $n$ such that $a_{n}=105 k+r$. This shows that the sequence $\left\{a_{n}\right\}$ consists of and only of numbers of the form $105 k+r(k \geqslant 0, k \in \mathbb{Z}, r \in P)$ arranged in ascending order.
Since the sequence is strictly increasing, and for each fixed non-negative integer $k$, when $r$ takes the values in $P$, there are 48 numbers, and
$$
\begin{array}{c}
1000=48 \times 20+40, \text { so } a_{1000}=105 \times 20+a_{40}. \\
\because a_{48}=104, a_{47}=103, a_{46}=101, a_{45}=97, \\
a_{44}=94, a_{43}=92, a_{42}=89, a_{41}=88, a_{40}=86, \\
\therefore a_{1000}=105 \times 20+86=2186.
\end{array}
$$
|
2186
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (This question is worth 35 points) Given a set of points on a plane $P=\left\{P_{1}\right.$, $\left.P_{2}, \cdots, P_{1004}\right\}$, where no three points in $P$ are collinear. Divide all the points in $P$ arbitrarily into 83 groups, such that each group has at least 3 points, and each point belongs to exactly one group. Then connect any two points in the same group with a line segment, and do not connect points in different groups, thus obtaining a pattern $G$. Different grouping methods result in different patterns. Let the number of triangles in pattern $G$ with vertices from points in $P$ be denoted as $m(G)$.
(1) Find the minimum value $m_{0}$ of $m(G)$;
(2) Let $G^{*}$ be a pattern such that $m\left(G^{*}\right)=m_{0}$. If the line segments (referring to segments with endpoints from points in $P$) in $G^{*}$ are colored using 4 different colors, with each segment colored exactly one color, prove that there exists a coloring scheme such that $G^{*}$, after coloring, contains no triangle with all three sides of the same color.
|
(1) Let $m(G)=m_{0}, G$ be obtained by the partition $X_{1}, X_{2}, \cdots, X_{\text {в3 }}$, where $X_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots$, 83.
$$
\begin{array}{l}
\text { Let }\left|X_{i}\right|=x_{i}(i=1,2, \cdots, 83) \text {, then } x_{1}+x_{2}+\cdots \\
+x_{83}=1994, \text { and } \\
m_{0}=C_{x_{1}}^{3}+C_{2}^{3}+\cdots+C_{x_{83}}^{3} .
\end{array}
$$
We need to prove that when $1 \leqslant i \neq j \leqslant 83$, we have $\left|x_{i}-x_{j}\right| \leqslant 1$.
In fact, if there exist $i, j(1 \leqslant i, j \leqslant 83)$ such that $\left|x_{i}-x_{j}\right| \geqslant 2$, without loss of generality, assume $x_{i}>x_{j}$, then consider the partition of $P$'s points into $Y_{1}, Y_{2}, \cdots, Y_{83}$ (where $Y_{i}$ is the set of points in the $i$-th group, $1 \leqslant i \leqslant 83$), such that
$$
y_{k}=\left|Y_{k}\right|=\left\{\begin{array}{ll}
x_{k}, & \text { when } k \neq i \text { and } k \neq j, \\
x_{i}-1, & \text { when } k=i, \\
x_{j}+1, & \text { when } k=j.
\end{array}\right.
$$
Such a partition clearly exists. Then, for the pattern $G^{\prime}$ obtained by the partition $Y_{1}, Y_{2}, \cdots, Y_{83}$, we have
$$
m\left(G^{\prime}\right)=C_{y_{1}}^{3}+C_{y_{2}}^{3}+\cdots+C_{y_{83}}^{3} .
$$
And $m\left(G^{\prime}\right)-m_{0}=C_{y_{i}}^{3}+C_{y_{j}}^{3}-C_{x_{i}}^{3}-C_{x_{j}}^{3}$
$$
\begin{array}{l}
=C_{x_{i}-1}^{3}+C_{x_{j}+1}^{3}-C_{x_{i}}^{3}-C_{x_{j}}^{3} \\
=C_{x_{i}-1}^{3}+C_{x_{j}}^{3}+C_{x_{j}}^{2}-C_{x_{i}-1}^{3}-C_{x_{i}-1}^{2}-C_{x_{j}}^{3} \\
=C_{x_{j}}^{2}-C_{x_{i}-1}^{2} . \\
\because x_{j}C_{\left\{x_{j} \mid\right.}^{2}.
\end{array}
$$
\therefore C_{\left|x_{j}\right|-1}^{2}-C_{x_{j} \mid}^{2}>0 \text { . }
$$
This contradicts the minimality of $m_{0}$.
$$
\begin{array}{l}
\because 1994=83 \times 24+2=81 \times 24+2 \times 25, \\
\therefore m_{e}=81 \times C_{24}^{3}+2 \times C_{25}^{3}=168544 .
\end{array}
$$
(2) The same as Solution 1.
|
168544
|
Combinatorics
|
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
|
Given 9 points in space, where any 4 points are not coplanar. Connect several line segments among these 9 points so that there is no tetrahedron in the graph. How many triangles can there be at most in the graph?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
First, prove the following conclusion:
In a space graph with $n$ points, if there are no triangles, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$.
Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, where the number of edges from $A_{1}$ is the maximum, say $k$ edges $A_{1} A_{n}, A_{1} A_{n-1}, \cdots, A_{1} A_{n-k+1}$. According to the condition, there are no triangles, so there are no edges between the points $A_{n-k+1}, \cdots, A_{n}$. Thus, each edge in the space graph has at least one endpoint among the points $A_{1}, A_{2}, \cdots, A_{n-k}$, and each $A_{i}$ $(1 \leqslant i \leqslant n-k)$ can have at most $k$ edges, so the total number of edges $\leqslant k(n-k) \leqslant \left(\frac{n^{2}}{4}\right)$.
Next, prove that in a space with 9 points, where no 4 points are coplanar, and several line segments are connected between these 9 points, if the graph contains at least 28 triangles, then there is at least one tetrahedron.
Use proof by contradiction. Assume there is no tetrahedron, and let these 9 points be $A_{1}, A_{2}, \cdots, A_{9}$. By the pigeonhole principle, there must be a point that is a vertex of at least $\left\{\frac{28 \times 3}{9}\right\} = 10$ triangles, where $\left\{\frac{28 \times 3}{9}\right\}$ represents the smallest integer not less than $\frac{28 \times 3}{9}$. Thus, at least 5 edges are drawn from this point, say $A_{1}$.
(1) If 5 edges $A_{1} A_{2}, \cdots, A_{1} A_{6}$ are drawn from $A_{1}$, then by the condition, since there is no tetrahedron, the subgraph formed by the points $A_{2}, A_{3}, \cdots, A_{6}$ has no triangles. By conclusion (*), this subgraph has at most $\left[\frac{5^{2}}{4}\right] = 6$ edges, so the number of triangles with $A_{1}$ as a vertex is at most 6, which is a contradiction.
(2) If 6 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{7}$ are drawn from $A_{1}$, then similarly, the number of triangles with $A_{1}$ as a vertex is at most $\left[\frac{6^{2}}{4}\right] = 9$, which is a contradiction.
(3) If 7 edges $A_{1} A_{2}, \cdots, A_{1} A_{8}$ are drawn from $A_{1}$, since there is no tetrahedron, the subgraph formed by the points $A_{2}, \cdots, A_{8}$ has no triangles. Using conclusion (*), this subgraph has at most $\left[\frac{7^{2}}{4}\right] = 12$ edges. Thus, the number of triangles with $A_{1}$ as a vertex is at most 12, and the number of triangles not having $A_{1}$ as a vertex must have $A_{9}$ as a vertex, and similarly, there are at most 12 such triangles. Therefore, the total number of triangles $\leqslant 12 \times 2 = 24 < 28$, which is a contradiction.
(4) If 8 edges $A_{1} A_{2}, \cdots, A_{1} A_{9}$ are drawn from $A_{1}$, then the subgraph formed by the points $A_{2}, \cdots, A_{9}$ has no triangles. By conclusion (*), this subgraph has at most $\left[\frac{8^{2}}{4}\right] = 16$ edges, so the original graph has at most 16 triangles, which is a contradiction.
Thus, the number of triangles satisfying the conditions is at most 27.
Divide the 9 points $A_{1}, A_{2}, \cdots, A_{9}$ into 3 groups $\left\{A_{1}, A_{2}, A_{3}\right\}$, $\left\{A_{4}, A_{5}, A_{6}\right\}$, $\left\{A_{7}, A_{8}, A_{9}\right\}$, with no edges between points in the same group, and edges between points in different groups. This way, there are $C_{3}^{1} C_{3}^{1} C_{3}^{1} = 27$ triangles. Of course, there is no tetrahedron.
|
27
|
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 {. }
$$
Solving gives $x=1, x=2$ (discard).
$\therefore$ The solution that satisfies the conditions is $x=1, y=1$.
Thus, $x+y+x y=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. From 30 people with distinct ages, select two groups, the first with 12 people and the second with 15 people, such that the oldest person in the first group is younger than the youngest person in the second group. How many ways are there to select these groups?
|
5. 4060.
Assume the ages of these 30 people, from smallest to largest, are
$$
a_{1}<a_{2}<a_{3}<\cdots<a_{29}<a_{30} .
$$
Let the first group of 12 people be $a_{i_{1}}<a_{i_{2}}<\cdots<a_{i_{12}}$,
and the second group of 15 people be $a_{j_{1}}<a_{j_{2}}<\cdots<a_{j_{15}}$,
which are the ages of the two selected groups that meet the requirements.
Thus, $a_{i_{12}}<a_{j_{1}}$.
Therefore, this selection method corresponds one-to-one with
$$
a_{i_{1}}<a_{i_{2}}<\cdots<a_{i_{12}}<a_{j_{1}}<\cdots<a_{j_{15}}
$$
which is a 27-element subset of $\left\{a_{1}, a_{2}, \cdots, a_{30}\right\}$.
Hence, the number of selection methods is
$$
C_{30}^{27}=C_{30}^{3}=4060
$$
|
4060
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (Full marks 30 points) Let $f(x)=x^{2}-(4 a-2) - x-6 a^{2}$ have the minimum value $m$ on the interval $[0,1]$. Try to write the expression for $m$ in terms of $a$, $m=F(a)$. And answer: For what value of $a$ does $m$ achieve its maximum value? What is this maximum value?
---
The translation maintains the original text's format and line breaks as requested.
|
$$
-, 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
|
9.6 On 1000 cards, write down the natural numbers 1 to 1000 (one number per card), and then use these cards to cover 1000 squares in a $1 \times 1994$ rectangle (the rectangle consists of 1994 $1 \times 1$ squares. The size of the cards is also $1 \times 1$). If the square to the right of the square containing the card with $n$ is empty, then the card with $n+1$ can be moved to cover that empty square, which is called one move. Prove: It is impossible to make more than 500,000 such moves.
|
9.6 Notice that the card with 1 written on it will not be moved, so the card with 2 written on it can be moved at most once, the card with 3 written on it can be moved at most twice, and so on. Thus, for any $n \leqslant 1000$, the card with $n$ written on it can be moved at most $n-1$ times (because we can only move it to the right of the card with $n-1$ written on it, and the latter can be moved at most $n-2$ times). This means that the total number of moves that can be made does not exceed $1+2+\cdots+999=999 \cdot \frac{1000}{2}<500000$.
|
500000
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
10. 1 Given? three quadratic trinomials; $P_{1}(x)=x^{2}+p_{1} x$ $+q_{1}, P_{2}(x)=x^{2}-p_{2} x+q_{2}$ and $P_{3}(x)=x^{2}+p_{3} x+q_{3}$. Prove: The equation $\left|P_{1}(x)\right|+\left|P_{2}(x)\right|=\left|P_{3}(x)\right|$ has at most 8 real roots.
|
10.1 Each root of the original equation should be a root of a quadratic trinomial of the form $\pm p_{1}: p_{2} \pm p_{3}$, and with different choices of signs, there are 8 such quadratic trinomials. Since the coefficient of $x^{2}$ has the form $\pm 1 \pm 1 \pm 1$, the coefficient of $x^{2}$ will never be 0 regardless of how the signs are chosen. Therefore, all 8 are genuine quadratic trinomials. However, the roots of two quadratic trinomials with coefficients that are opposite in sign are the same, so the roots of the equation $\left|P_{1}(x)\right|+\mid P_{2}(x) \mid = | P_{3}(x) \mid$ are included among the roots of 4 different quadratic equations. Thus, it has no more than 8 roots.
|
8
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
10. 8 Class
30 students in total, each
student in the class
has the same number of friends.
How many students at most can there be who are
better than the majority of their friends' performance? (Assuming for any two students in the class, their performance can be compared to determine who is better or worse.)
保留源文本的换行和格式，直接输出翻译结果。
|
10.8 䇾案, 25 个.
We refer to a student who performs better than the majority of their friends as a "good student". Let there be $x$ "good students", and each student has $k$ friends.
The best student is better in all $k$ "friend pairs", while each of the other "good students" is better in at least $\left[\frac{k}{2}\right]+1$ (the integer part). Therefore, the "good students" are better in at least $k+$ non-overlapping "friend pairs" in total, which is $\frac{30 k}{2}=15 k$. From this, we know that
$$
x \leqslant 28 \cdot \frac{k}{k+1}+1 .
$$
The number of students is no more than $30-x$. Therefore, we also have
$$
\frac{k+1}{2} \leqslant 30-x \text {. }
$$
The right side of inequality (1) is a decreasing function of $k$, and inequality (2) is equivalent to the condition
$$
k \leqslant 59-2 x \text {. }
$$
Combining (1) and (3), we get
$$
x \leqslant 28 \cdot \frac{59-2 x}{60-2 x}+1 \text {, }
$$
which simplifies to $x^{2}-59 x+856 \geqslant 0$.
The largest integer satisfying (4) and the condition $x \leqslant 30$ is $x=25$, so the number of "good students" does not exceed 25.
Next, we prove that 25 can be achieved.
\begin{tabular}{|c|c|c|c|c|}
\hline 1 & 2 & 3 & 4 & 5 \\
\hline 6 & 7 & 8 & 9 & 10 \\
\hline 11 & 12 & 13 & 14 & 15 \\
\hline 16 & 17 & 18 & 19 & 20 \\
\hline 21 & 22 & 23 & 24 & 25 \\
\hline 26 & 27 & 28 & 29 & 30 \\
\hline
\end{tabular}
35. We number the students in the class from 1 to 30 in the order of their performance, and list these numbers in a $6 \times 5$ table (as shown).
Assume that two students are friends if their numbers in the table fall into one of the following three cases: 1) they are in adjacent rows and different columns; 2) they are in the same column, with one in the bottom row; 3) they are in the top row (i.e., any two students in these three cases are friends). It is not difficult to verify that all the given conditions are satisfied.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a$ be the decimal part of $\sqrt{3}$, $b$ be the decimal part of $\sqrt{2}$: $\frac{a}{(a-b) b}$ has an integer part of $\qquad$
|
$$
\begin{array}{l}
\text { 3. } \frac{a}{(a-b) b}=\frac{\sqrt{3}-1}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{(\sqrt{3}-\sqrt{2})+(\sqrt{2}-1)}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}-\sqrt{2}} \\
=\sqrt{3}+2 \sqrt{2}+1 \\
\approx 5.56 .
\end{array}
$$
The integer part is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $p, q, \frac{2 p-1}{q}, \frac{2 q-1}{p}$ all be integers, and $p>1, q>$
1. Then $p+q=$ $\qquad$ .
|
6. $p+q=8$.
From $\frac{2 q-1}{p}, \frac{2 p-1}{q}$ both being positive integers, we know that one of them must be less than 2. Otherwise, $\frac{2 q-1}{p}>2, \frac{2 p-1}{q}>2$, leading to $2 p-1+2 q-1>2 p+2 q$, which is a contradiction. Assume $0<\frac{2 q-1}{p}<2$, then it must be that $2 q-1=p$. It is easy to see that $q=3, p=5$, thus $p+q=8$.
Note: The original problem statement has a typo in the final result, which should be $p+q=8$ instead of $p+q=3$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, let the three sides of $\triangle A B C$ be $a, b, c$, and all are numbers. If it satisfies $\angle A=3 \angle B$, try to find the minimum value of the perimeter and provide a proof.
|
From the Law of Sines, we have \(\frac{a}{b} = \frac{\sin 3B}{\sin B} = (2 \cos B)^2 - 1\), \(\frac{c}{b} = (2 \cos B)^3 - 4 \cos B\). Since \(2 \cos B = \frac{a^2 + c^2 - b^2}{ac} \in \mathbb{Q}\), there must exist positive integers \(p, q\) with \((p, q) = 1\) such that \(2 \cos B = \frac{p}{q}\). Therefore, \(\frac{a}{b} = \left(\frac{p}{q}\right)^2 - 1 \Leftrightarrow \frac{a}{p^2 - q^2} = \frac{b}{q^2}\); \(\frac{c}{b} = \left(\frac{p}{q}\right)^3 - 2 \cdot \frac{p}{q} \Leftrightarrow \frac{c}{p^3 - 2pq^2} = \frac{b}{q^3}\). Thus, \(\frac{a}{(p^2 - q^2)q} = \frac{b}{q^3} = \frac{c}{p^3 - 2pq^2} = \frac{e}{f}\), where \((e, f) = 1\). This gives \(a = \frac{e}{f}(p^2 q - q^3)\), \(b = \frac{e}{f} q^3\), \(c = \frac{e}{f}(p^3 - 2pq^2)\). To minimize the perimeter, \((a, b, c) = 1\). Since \((e, f) = 1\), it follows that \(f \mid q^3\) and \(f \mid p^3 - 2pq^2\).
We now prove that \(f = 1\). If \(f > 1\), then \(f\) must have a prime factor \(f' > 1 \Rightarrow f' \mid q^3, f' \mid p^3 - 2pq^2 \Rightarrow f' \mid q \Rightarrow f' \mid 2pq^2 \Rightarrow f' \mid p^3 \Rightarrow f' \mid p\). This contradicts \((p, q) = 1\). Hence, \(f = 1\). Since \((a, b, c) = 1\), it follows that \(e = 1\). Therefore, \(a = p^2 q - q^3\), \(b = q^3\), \(c = p^3 - 2pq^2\), with \((p, q) = 1\). On the other hand, from \(0^\circ < A + \mathcal{Q} = 4B < 80^\circ \Rightarrow 0^\circ < B < 45^\circ \Rightarrow \sqrt{2} < 2 \cos B < 2 \Rightarrow \sqrt{2} q < p < 2q\). It is easy to see that the smallest \(p, q\) satisfying this inequality are \(p = 3, q = 2\). At this point, \(c = p(p^2 - 2q^2)\) is minimized when \(p\) is smallest and \(p^2 - 2q^2 = 1\) is also minimized. Hence, \(S = a + b + c = p^2 q + p(p^2 - 2q^2)\). When \(p = 3, q = 2\), we have \(S_{\text{min}} = 18 + 3 = 21\).
|
21
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
There are 1994 points on a circle, which are painted in several different colors, and the number of points of each color is different. Now, take one point from each color set to form a polygon with vertices of different colors inside the circle. To maximize the number of such polygons, how many different colors should the 1994 points be painted, and how many points should each color set contain?
|
Let 1994 points be colored with $n$ colors, and the number of points of each color, in ascending order, is $m_{1}1$. In fact, if $m_{1}=1$, because $m_{1} m_{2} \cdots m_{n}m_{k+1} m_{k}$.
Therefore, when $m^{\prime}{ }_{n+1}, m^{\prime} \star$ replace $m_{\star+1}, m_{\star}$, the value of $M$ increases, leading to a contradiction:
(2) $m_{i+1}-m_{i}=2$ holds for exactly one $i$. For this:
(i) $m_{i+1}-m_{i}=2$ holds for at most one $i$. If otherwise, there exist positive integers $j, k$ such that $1 \leqslant j < k \leq n$ and $m_{j+1} - m_{j} = m_{k+1} - m_{k} = 2$. Then, $m_{j+1} = m_{j} + 2$ and $m_{k+1} = m_{k} + 2$. If we replace $m_{k+1}, m_{j}$ with $m^{\prime}{ }_{k+1}, m^{\prime} j_{j}$, the value of $M$ increases, leading to a contradiction.
(ii) If $m_{i+1}-m_{i}=1$ for all $i=1,2, \cdots, n-1$, then $\sum_{i=1}^{n} m_{i} = n m_{1} + \frac{1}{2} n(n-1) = 1994 = 2 \times 997 \Rightarrow n(2 m_{1} - 1 + n) = 4 \times 997$. Since $n$ and $2 m_{1} - 1 + n$ are one odd and one even, and $2 m_{1} - 1 + n > n$, and 997 is a prime number, the only solution is $n=4$, $2 m_{1} - 1 + n = 997$, giving $m_{1} = 497 = 495 + 22$. If we take $m^{\prime} = m_{k+2} - 2$, then $m_{k} < 2$ and $m_{k+2} > 4$. Hence, $2 m^{\prime} > m_{k+2}$. Replacing $m_{k+2}$ with 2 and $m^{\prime}$, the value of $M$ increases, leading to a contradiction. Therefore, $m_{1} = 2$.
(II) From (I), we can assume the number of points of each color is 2, 3, ..., $k-1$, $k+1$, $k+2$, ..., $n$, $n+1$, $n+2$ ($k \leqslant n-1$).
We have $\frac{1}{2}(n+1)(n+4) - k = 1994$.
This gives $\left\{\begin{array}{l}n^{2} + 5n - 3984 > 0, \\ n^{2} + 3n - 3982 \leqslant 0 .\end{array}\right.$
Solving, we get $61 \leqslant n < 62$.
Taking $n=61$, we have $k = \frac{1}{2}(n+1)(n+4) - 1994 = 31 \times 65 - 1994 = 21$. Therefore, 1994 points can be colored with 61 colors, with the number of points of each color being $2, 3, \cdots, 19, 20, 22, 23, \cdots, 61, 62, 63$. In this case, the resulting polygon is a 61-sided polygon, and the number of such polygons is the maximum. (Xu Jingling, Jiugu Middle School, Susong County, Anhui Province)
|
61
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. As shown in the figure, in pentagon $A B C D E$, $\angle B=\angle A E D=$ $90^{\circ}, A B=C D=A E=B C+D E=1$. Find the area of this pentagon. (1992, Beijing Junior High School Grade 2 Mathematics Competition)
|
Analysis We immediately think of connecting $A C, A D$, because there are two right-angled triangles. But we also find that it is not easy to directly calculate the area of each triangle. Given the condition $B C+D E=1$, we wonder if we can join $B C$ to one end of $D E$, making $E F=B C$. Connect $A F$. Then we find $\triangle A B C \cong$ $\triangle A E F$. So $F D=1$, and $=A C \cdot \triangle E=A B, \triangle A D F$ are triangles with bases of 1, each with an area of $\frac{1}{2}$, and $\triangle A C D$ is congruent to $\triangle A D F$, thus the total area is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. As shown in the figure, let $D$ and $E$ be on the sides $AC$ and $AB$ of $\triangle ABC$, respectively. $BD$ and $CE$ intersect at $F$, $AE = EB$, $\frac{AD}{DC} = \frac{2}{3}$, and $S_{\triangle ABC} = 40$. Find $S_{\text{quadrilateral AEFD}}$. (6th National Junior High School Mathematics Correspondence Competition)
|
Analyzing, first connect $A F$ and then set:
$$
S_{\triangle A E P}=x, S_{\triangle A P D}=y, S_{\triangle C D F}=z, S_{\triangle B E P}=t\left(S_{\triangle B C F}\right.
$$
$=u$ ), to find $S_{\text {quadrilateral } A E F D}$, we only need to find the values of $x$ and $y$.
We easily get:
$$
\begin{array}{l}
x+y+z=20, \\
x+y+t=40 \times \frac{2}{5}=16, \\
x=t, \\
\frac{y}{z}=\frac{2}{3} .
\end{array}
$$
Solving these, $x=5, y=6$.
Thus, $S_{\text {quadrilateral } A E F D}=11$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $S=\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$ that satisfy: (1) $\left|A_{i}\right|=5, i=1,2, \cdots, k_{i}$ (2) $\mid A_{i} \cap$ $A_{j} \mid \leqslant 2, i, j=1,2, \cdots, k, i \neq j$. Find the maximum value of $k$. (6th test, 1st question, provided by Wu Chang)
|
First, prove that each element in $S$ belongs to at most 3 of the sets $A_{1}, A_{2}, \cdots, A_{k}$.
By contradiction, suppose there is an element in $S$ that belongs to at least 4 subsets. Without loss of generality, assume $1 \in A_{1}, A_{2}, A_{3}, A_{4}$.
(1) If each of the other elements in $S$ belongs to at most 2 of the sets $A_{1}, A_{2}, A_{3}, A_{4}$, each $A_{i} (i=1,2,3,4)$ has 5 positive integers. Removing 1, there are 4 positive integers left. Thus, excluding 1, $A_{1}, A_{2}, A_{3}, A_{4}$ should have $4 \times 4 = 16$ positive integers, some of which are repeated. However, $S - \{1\}$ consists of 9 different positive integers, so there should be $16 - 9 = 7$ repeated positive integers. Given the initial assumption, 7 positive integers belong to exactly 2 of these 4 sets. Therefore, using the problem's condition, we have
$$
2 C_{4}^{2} \geqslant \sum_{1<i<j<4}\left|A_{i} \cap A_{j}\right| \geqslant C_{4}^{2} + 7,
$$
but $2 C_{4}^{2} = 12$ and $C_{4}^{2} + 7 = 13$, which is a contradiction.
(2) If there is another element in $S$ that belongs to at least 3 of the sets $A_{1}, A_{2}, A_{3}, A_{4}$, without loss of generality, assume $2 \in A_{1} \cap A_{2} \cap A_{3}$. Since each of $A_{1}, A_{2}, A_{3}$ has 5 positive integers, removing 1 and 2, each $A_{i} (i=1,2,3)$ has 3 positive integers left. Thus, excluding 1 and 2, $A_{1}, A_{2}, A_{3}$ should have $3 \times 3 = 9$ positive integers, but $S - \{1, 2\}$ consists of 8 different positive integers. Therefore, $9 - 8 = 1$, meaning there is at least one positive integer, which is neither 1 nor 2, belonging to 2 of the sets $A_{1}, A_{2}, A_{3}$. Thus, we have
$$
6 = 2 C_{3}^{2} \geqslant \sum_{1<i<j<3}\left|A_{i} \cap A_{j}\right| \geqslant 2 \times 3 + 1 = 7,
$$
which is a contradiction.
Thus, $k \leqslant \frac{3 \times 10}{5} = 6$.
On the other hand, let
$$
\begin{array}{l}
A_{1}=\{1,2,3,4,5\}, \quad A_{2}=\{1,2,6,7,10\} \\
A_{3}=\{1,3,8,9,10\}, \quad A_{4}=\{2,4,5,8,9\} \\
A_{5}=\{3,5,6,7,8\}, \quad A_{6}=\{4,5,7,9,10\}.
\end{array}
$$
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. It is known that a total of 12 theater groups are participating in a 7-day drama festival. Each day, some of the groups perform on stage, while the groups not performing watch from the audience. It is required that each group must see the performances of all other groups. What is the minimum number of performances needed? (From the 7th Partner School Test, Question 2, provided by Li Chengzhang)
|
Certainly, here is the translation of the provided text into English, preserving the original formatting:
```
It is impossible for 3 troupes to perform only for 2 days and meet the conditions of the problem. By the pigeonhole principle, on one day, at least two troupes must perform, and on that day, these two troupes cannot observe each other. They must observe each other on the remaining day, which is clearly impossible.
Next, to meet the conditions of the problem, 8 troupes cannot perform only for 3 days. Using the pigeonhole principle, on one day, at least 3 troupes must perform, and on that day, these 3 troupes cannot observe each other. Therefore, they must observe each other's performances on the remaining 2 days, which is known to be impossible.
Now, we prove that 9 troupes cannot meet the conditions of the problem by performing only for 4 days. If one troupe performs only one day, then on that day, the other 8 troupes must be in the audience. With the remaining 3 days, the 8 troupes must observe each other's performances, which has already been proven impossible.
Of course, one troupe cannot perform for 4 days, otherwise, that troupe cannot observe the performances of other troupes. Therefore, each troupe performs at most 3 days. If one troupe performs for 3 days, it must be in the audience on the 4th day. On that day, the remaining 8 troupes must all perform, so in the first 3 days, the 8 troupes must observe each other's performances, which is known to be impossible. Therefore, each troupe performs at most 2 days, and in 4 days, choosing 2 days, from \( C_{\mathrm{I}}^{n=6} \), these troupes cannot observe each other's performances, as originally stated.
Now, let's solve this problem. If there are \( k \) troupes that perform only once, and the remaining \( 12 - k \) troupes each perform at least twice. Then these \( k \) troupes must be scheduled to perform on \( k \) days, and the other troupes can only be in the audience on those \( k \) days. The \( 12 - k \) troupes must observe each other's performances in the remaining \( 7 - k \) days. When \( k = 7, 6 \), it is clearly impossible. When \( k = 5, 4, 3 \), it has already been proven to be impossible. Therefore, \( k \leq 2 \).
When \( k = 1 \), at least \( 1 + 2 \times 11 = 23 \) performances are required. When \( k = 2 \), at least \( 2 + 2 \times 10 = 22 \) performances are required.
Below is an example of a total of 22 performances. Use \( A_{3} = \{3,4\} \) to indicate that the 3rd troupe performs on the 3rd and 4th days, \( A_{4} = \{3, 5\} \) to indicate that the 4th troupe performs on the 3rd and 5th days, etc.
\[
\begin{array}{l}
A_{1} = \{1\}, A_{2} = \{2\}, A_{3} = \{3,4\}, A_{4} = \{3,5\}, A_{5} = \\
\{3,6\}, A_{6} = \{3,7\}, A_{7} = \{4,5\}, A_{8} = \{4,6\}, A_{9} = \{4, \\
7\}, A_{10} = \{5,6\}, A_{11} = \{5,7\}, A_{12} = \{6,7\},
\end{array}
\]
```
|
22
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(1) (This question is worth 35 points) Let the set $A=\{1,2,3, \cdots, 366\}$. If a binary subset $B=\{a, b\}$ of $A$ satisfies 17|(a $+b$), then $B$ is said to have property $P$.
(1) Find the number of all binary subsets of $A$ that have property $P$;
(2) Find the number of all pairwise disjoint binary subsets of $A$ that have property $P$.
|
(1) $a+b$ is divisible by 17 if and only if $a+b \equiv 0 \pmod{17}$, i.e., $\square$
$$
\begin{array}{l}
a \equiv k \pmod{17}, b \equiv 17-k \pmod{17}, \\
k=0,1,2, \cdots, 16 .
\end{array}
$$
Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: [0], [1], ..., [16]. Since $366=17 \times 21+9$, there are 21 numbers in [0], 22 numbers in [1], [2], ..., [9], and 21 numbers in [10], ..., [16].
Let $17 \mid a+b$, if and only if $a, b \in [0]$ or $a \in [k], b \in [17-k]$.
When $a, b \in [0]$, the number of subsets with property $P$ is $C_{21}^{2}=210$.
When $a \in [k], b \in [17-k], k=1,2, \cdots, 7$, the number of subsets with property $P$ is $C_{22}^{1} \times C_{21}^{1}=462$ each.
When $a \in [8], b \in [9]$, the number of subsets with property $P$ is $C_{22}^{1} \times C_{22}^{1}=484$.
Therefore, the total number of two-element subsets of $A$ with property $P$ is
$$
210+462 \times 7+484=3928 \text{ (sets). }
$$
(2) To ensure that the two-element subsets are pairwise disjoint, we can pair them as follows:
$a \in [0], b \in [0]$, there are 10 subsets;
$a \in [k], b \in [17-k], k=1,2, \cdots, 7$, there are 21 subsets;
$a \in [8], b \in [9]$, there are 22 subsets.
Thus, the total number of two-element subsets of $A$ with property $P$ that are pairwise disjoint is
$$
10+21 \times 7+22=179 \uparrow .
$$
|
3928
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. From the numbers $1,2, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, and $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways of selection are there that meet the conditions? (1989, National High School Competition)
|
Destruct the following model:
Take 10 identical white balls and arrange them in a row, take 5 identical black balls and divide them into 3 groups in the order of $2,2,1$, then insert them into the 10 gaps between the white balls, from left to right, excluding the left end but including the right end. There are $C_{10}^{3}$ ways to insert. Each insertion method corresponds to an unordered set. As shown in the figure: $\square$
etc. Therefore, the number of different ways to draw is $C_{10}^{3}=120$.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Among the 95 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 95^{2}$, the numbers with an odd digit in the tens place total $\qquad$.
untranslated part: $\qquad$
|
In $1^{2}, 2^{2}, \cdots, 10^{2}$, it is found through calculation that the tens digit is odd only for $4^{2}=16, 6^{2}=36$. A two-digit square number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2},
$$
Therefore, $b$ can only be 4 or 6. That is, in every 10 consecutive numbers, there are two numbers whose tens digit is odd, hence it is deduced that in $1^{2}, 2^{2}, 3^{2}, \cdots, 95^{2}$, the number of squares with an odd tens digit is 19.
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $\alpha$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$, then the value of $\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}$ is. $\qquad$ .
|
2. 20 .
Factorization yields
$$
\begin{array}{l}
\alpha^{3}-1=(\alpha-1)\left(\alpha^{2}+\alpha+1\right), \\
\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}=\alpha^{2}(\alpha-1)(\alpha+1)^{2} .
\end{array}
$$
$\alpha$ satisfies $\alpha^{2}+\alpha-\frac{1}{4}=0$, hence $\alpha \neq 1$.
$$
\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}=\frac{\alpha^{2}+\alpha+1}{\alpha^{2}(\alpha+1)^{2}}=\frac{\frac{1}{4}+1}{\left(\frac{1}{4}\right)^{2}}=20 .
$$
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ .
|
3. 1.
The formula yields
$$
\begin{aligned}
y & =(x-1)^{2}+x+\frac{1}{x}-1 \\
& =(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1 .
\end{aligned}
$$
When $x=1$, both $(x-1)^{2}$ and $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}$ simultaneously take the minimum value of 0, so the minimum value of $y=x^{2}-x+\frac{1}{x}$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The "Monkey Cycling" performance in the circus is carried out by 5 monkeys using 5 bicycles, with each monkey riding at least once, but no monkey can ride the same bicycle more than once. After the performance, the 5 monkeys rode the bicycles $2, 2, 3, 5, x$ times respectively, and the 5 bicycles were ridden $1, 1, 2, 4, y$ times respectively. Therefore, $x+y$ equals ( ).
(A) 5
(B) 6
(C) 7
(D) 8
|
,- 1 (B).
Every time the monkey rides a bike, the monkey and the bike each get one count, so the number of times the monkey rides a bike = the number of times the bike is ridden, i.e., $2+2+3+5+x=1+1+2+4+y$. Simplifying, we get $x+4=y$. From $x \geqslant 1$, we get $y \geqslant 5$. Also, from the problem, $y \leqslant 5$, so $y=5, x=1$. Therefore, $x+y=6$.
|
6
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
For each natural number $n$, the parabola $y=\left(n^{2}+n\right) x^{2}$
$-(2 n+1) x+1$ intersects the $x$-axis at points $A_{n}, B_{n}$. Let $\left|A_{n} B_{n}\right|$ denote the distance between these two points. Then $\left|A_{1} B_{1}\right|+\left|A_{2} B_{2}\right|+\cdots$ $+\left|A_{1995} B_{1995}\right|$ is $\qquad$.
|
$$
\begin{aligned}
& \text { II.1. } 1995 \\
& \because y=(n x-1)[(n+1) x-1], \\
x_{1}= & \frac{1}{n}, x_{2}=\frac{1}{n+1}, \\
& \therefore\left|A_{n} B_{n}\right|=\frac{1}{n}-\frac{1}{n+1} .
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
& \left|A_{1} B_{1}\right|+\left|A_{2} B_{2}\right|+\cdots+\left|A_{1005} B_{1005}\right| \\
= & \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{1995}-\frac{1}{1996}\right) \\
=1- & \frac{1}{1996}=\frac{1995}{1996} .
\end{aligned}
$$
|
1995
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The ice numbers $a_{1}, u_{2}, \cdots, a_{9}$ can only take two different values, +1 or -1. Then, the maximum value of the expression $a_{1} a_{5} a_{9}-a_{1} a_{6} a_{8}+a_{2} a_{8} a_{7}-$ $a_{2} a_{4} a_{9}+a_{3} a_{4} a_{8}-a_{3} a_{5} a_{7}$ is $\qquad$.
|
4. 4 .
Since each term in the sum can only be 11 or -1, the sum is even (the parity of $a-b$ and $a+b$ is the same), so the maximum value of the sum is at most 6. It is easy to see that the sum cannot be 6, because in this case $a_{1} a_{5} a_{9}, a_{2} a_{6} a_{7}, a_{3} a_{6} a_{8}$ should all be +1, making their product equal to +1. However, $a_{1} a_{8} a_{8}, a_{2} a_{4} a_{9}, a_{3} a_{5} a_{8}$ should all be -1, making their product equal to -1.
But in fact, these two products should be equal, so the sum cannot be 6. However, the sum can be 4. For example, when $a_{1}=a_{2}=a_{3}=a_{5}=a_{8}=a_{0}=1, a_{4}=a_{6}=a_{7}=-1$, the sum is equal to 4. In conclusion, the maximum value of the sum is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let the function $f(x)$ have a domain and range both equal to $R$, and for any $a, b \in R$ there is $f[a f(b)]=a b$. Then the value of $|f(1995)|$ is $\qquad$
|
Ni, i. 1995.
For the known functional equation, substituting $a=1, a=b, a=f(b)$, we have $f(f(b))=b, f(b f(b))=b^{2}, f\left[f^{2}(b)\right]=b f(b)$. Applying the function $f$ to the third equation, $f\left\{f\left[f^{2}(b)\right]\right\}=f(b f(b))=b^{2}$. Thus, from $f(J(b))=b$, we have $f^{2}(b)=b^{2}$, i.e., $|f(b)|=b$. Therefore, $|f(1995)|=1995$.
|
1995
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $f(x)=a x^{2}+b x+c(a, b, c \in R, a \neq 0)$. If for $|x| \leqslant 1$, $|f(x)| \leqslant 1$, then for $|x| \leqslant 1$, the maximum value of $|2 a x+b|$ is $\qquad$
|
4. 4 .
Given $f(0)=c, f(1)=a+b+c, f(-1)=a-b+c$,
we solve to get $a=\frac{f(1)+f(-1)-2 f(0)}{2}, b=\frac{f(1)-f(-1)}{2}$.
For $|x| \leqslant 1$, $|f(x)| \leqslant 1$, we have
$$
\begin{array}{l}
|2 a x+b| \\
=\left|[f(1)+f(-1)-2 f(0)] x\right. \\
\left.+\frac{f(1)-f(-1)}{2} \right| \\
=\left|\left(x+\frac{1}{2}\right) f(1)+\left(x-\frac{1}{2}\right) f(-1)-2 x f(0)\right| \\
\leqslant\left|x+\frac{1}{2}\right||f(1)|+\left|x-\frac{1}{2}\right||f(-1)| \\
\quad \quad+|2 x||f(0)| \\
\leqslant\left|x+\frac{1}{2}\right|+\left|x-\frac{1}{2}\right|+2 \leqslant 4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(4) (Total 20 points) In the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $M$ is a point on $AB$, and $AM^2 + BM^2 + CM^2 = 2AM + 2BM + 2CM - 3$. If $P$ is a moving point on the segment $AC$, and $\odot O$ is the circle passing through points $P, M, C$, and a line through $P$ parallel to $AB$ intersects $\odot O$ at point $D$.
(1) (8 points) Prove that $M$ is the midpoint of $AB$;
(2) (12 points) Find the length of $PD$.
|
(1) From the given, we have
$$
(A M-1)^{2}+(B M-1)^{2}+(C M-1)^{2}=0,
$$
which implies $A M=B M=C M=1$, meaning $M$ is the midpoint of $A B$.
(2) From $\left\{\begin{array}{l}M A=M C \Rightarrow \angle A=\angle M C A, \\ P D / / A B: \angle A=\angle C P D\end{array}\right.$
$\Rightarrow \angle M C A=\angle C P D$, thus $\overparen{P} M=\overparen{C D}$, and $P C / / M D$. Since $A P \| A$, then $A P D M$ is a parallelogram, hence $P D=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, (20 points) A mall installs an escalator from the first floor to the second floor, moving at a uniform speed that is twice the speed of a child. It is known that a boy passed 27 steps to reach the top of the escalator, while a girl walked 18 steps to reach the top. (Assume the boy and girl each step on one step at a time)
(1) (10 points) How many steps are visible on the escalator?
(2) (10 points) If there is a staircase next to the escalator leading from the second floor to the first floor, with the same number of steps as the escalator, and the children each return to the top of the escalator at their original speeds, then walk down the stairs, and ride the escalator again (ignoring the distance between the escalator and the stairs), how many steps will the boy have walked when he catches up to the girl for the first time?
|
(1) Let the girl's speed be $x$ levels/min, the escalator's speed be $y$ levels/min, the boy's speed be $2x$ levels/min, and the stairs have $s$ levels, then
$$
\left.\begin{array}{l}
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y}
\end{array}\right. \\
\Rightarrow \frac{13.5}{18}=\frac{s-27}{s-18}
\end{array}\right\}
$$
(2) Let the boy pass the girl once after walking the escalator $m$ times and the stairs $n$ times, then the girl has walked the escalator $(m-1)$ times and the stairs at a speed of $x$ levels/min.
$$
\begin{array}{l}
\frac{54 m}{4 x}+\frac{54 n}{2 x}=\frac{54(m-1)}{3 x}+\frac{54(n-1)}{x}, \\
\frac{m}{4}+\frac{n}{2}=\frac{m-1}{3}+\frac{n-1}{1}, \\
6 n+m=16 .
\end{array}
$$
One of $m, n$ must be a positive integer, and $0 \leqslant|m-n| \leqslant 1$.
\begin{tabular}{c|ccccc}
$m$ & 1 & 2 & 3 & 4 & 5 \\
\hline$n=\frac{16-m}{6}$ & $2 \frac{1}{2}$ & $2 \frac{1}{3}$ & $2 \frac{1}{6}$ & 2 & $1 \frac{5}{6}$
\end{tabular}
\begin{tabular}{c|cc}
$n$ & 1 & 2 \\
\hline$m=16-6 n$ & 10 & 4
\end{tabular}
Obviously, only $m=3, n=2 \frac{1}{6}$,
So the boy walked the escalator 3 times and the stairs $2 \frac{1}{6}$ times.
Therefore, the total number of steps the boy walked $=3 \times 27+2 \frac{1}{6} \times 54=198$
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In the table below, the average of any three adjacent small squares in the upper row is $1(x \neq 0)$, and the square of any four adjacent small squares in the lower row is also 1. Then the value of $\frac{x^{2}+y^{2}+z^{2}-32}{y z+16}$ is $ـ$. $\qquad$
|
ii. 3 .
From the problem, we know that in the above row, the numbers in every two small squares separated by one are equal, thus we can deduce
$$
\frac{x+7-y}{3}=1 \text {. }
$$
Similarly, for the row below, we get
$$
\begin{array}{l}
\frac{z-x+9.5-1.5}{4}=1 . \\
\begin{array}{l}
\therefore y=x+4, z=x-4, \text { then } \\
\frac{x^{2}+y^{2}+z^{2}-32}{y z+16}=\frac{x^{2}+\cdot(x+4)^{2}+(x-4)^{2}-32}{(x+4)(x-4)+16} \\
=\frac{3 x^{2}}{x^{2}}=3 .
\end{array}
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12. Is there a north-facing 100 petals arrangement, such that they exactly have 1985 intersections.
|
Solution 1: Since $x$ parallel lines and another set of $100 - x$ parallel lines can yield $x(100-x)$ intersection points, and
$$
x(100-x)=1985
$$
has no integer solutions, we can consider a grid formed by 99 lines. Since
$$
x(99-x)<1955
$$
has solutions $x \leqslant 26$ or $x \geqslant 73, x \in \mathbb{N}$, and
$$
1985=73 \times 20+90-12,
$$
we can make the following selection:
(1) 73 horizontal lines and 26 vertical lines
$$
\begin{array}{l}
x=k, k=1,2, \cdots, 73, \\
y=k, k=1,2, \cdots, 26 .
\end{array}
$$
These 99 lines can yield $73 \times 26$ intersection points.
(2) Draw the line $y=x+14$ intersecting with the above 99 lines, yielding 99 intersection points, among which 12 points $(1,15),(2,15), \cdots,(12,26)$ are excluded, leaving $99-12$ intersection points.
From (1) and (2), 100 lines can yield $73 \times 26 + 99 - 12 = 1985$ intersection points.
Solution 2: Among 100 lines, if no two lines are parallel and no three lines are concurrent, then there can be $C_{100}^{2}=4950$ intersection points. Now, using concurrent lines to reduce the number of intersection points, each set of $n$ concurrent lines reduces the number of intersection points by $C_{n}^{2}-1$.
Suppose there are $k$ sets of concurrent lines, with the number of lines in each set being $n_1, n_2, \cdots, n_k$. Then we have
$$
\begin{array}{l}
n_1 + n_2 + \cdots + n_k \leqslant 100, \\
C_{n_1}^{2}-1 + C_{n_2}^{2}-1 + \cdots + C_{n_k}^{2}-1 = 2965 \text{ (group 1) } \\
C_{100}^{2} - 2965 \text{ (group 2) }.
\end{array}
$$
To satisfy
$$
C_{n_1}^{2}-1 < 2965,
$$
the largest integer $n_1 = 77$, at which point
$$
C_{77}^{2}-1 = 2925.
$$
Therefore, we can construct a set of 77 concurrent lines, which still needs to reduce 40 intersection points. To satisfy
$$
C_{n_2}^{2}-1 < 40,
$$
the largest integer $n_2 = 9$, at which point
$$
C_{9}^{2}-1 = 35.
$$
Therefore, we can construct another set of 9 concurrent lines, which still needs to reduce 5 intersection points.
Since $C_{4}^{2}-1 = 5$, we can construct a set of 4 concurrent lines.
Since $77 + 9 + 4 = 90 < 100$, the 100 lines can be constructed as 77, 9, and 4 concurrent lines, with the remaining 10 lines left unchanged.
|
1985
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In a tennis tournament, $n$ women and $2 n$ men participate, and each player plays against all other players exactly once. If there are no ties, the ratio of the number of games won by women to the number of games won by men is 7:5. Then $n=$
|
6. $n=3$.
Let $k$ represent the number of times the woman wins over Luo Ziyue.
Here, we set $\frac{k+\mathrm{C}_{0}^{2}}{\mathrm{C}_{\mathrm{jn}}^{2}}=\frac{7}{12}, k=\frac{7}{12} \mathrm{C}_{3 n}-\mathrm{C}_{n}^{2}$. From $k \leq 2 n^{2}$, we can solve to get $n \leq 3$.
When $n=1$, we get $k=\frac{7}{4}$; when $n=2$, $k=\frac{31}{4}$; when $n=3$, $k=18$. Thus, $n=3$.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at $D$, $P$ is the midpoint of $AD$, $BP$ intersects $AC$ at $E$, $EF \perp BC$ at $F$. Given $AE=3$, $EC=$ 12. Find the length of $EF$.
|
Analysis: Consider $\triangle A B C$ as its machine triangle, given that $E$, $P$ are the fixed ratio points of $A C, A D$ respectively, hence from (*) we get $\left(1+\frac{C D}{D B}\right) \cdot \frac{3}{12}=1$, i.e., $\frac{C D}{D B}=3$.
From the projection theorem, we can find $B D=\frac{5}{2} \sqrt{3}, D C$ $=\frac{15}{2} \sqrt{3}$, thus $A D=\frac{15}{2}$.
By $\triangle C A D C \sim \triangle C E F$ we get $E F=6$.
Using the above basic figure and (*) formula, prove the following.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Among the 35 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 35^{2}$, the numbers with an odd digit in the tens place are $\qquad$ in total.
|
3. 7. Among $1^{2}, 2^{2}, \cdots, 9^{2}$, those with an odd tens digit are only $4^{2}=16, 6^{2}=36$.
The square of a two-digit number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2} \text {. }
$$
It is evident that, when the unit digit is $A$ and $\hat{C}$, the tens digit of the square is odd. That is, only $4^{2}, 6^{2}, 14^{2}, 16^{2}, 24^{2}, 26^{2}, 34^{2}$ have
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\frac{a}{a^{3}+a^{2} b+6 b^{2}+b^{3}}+\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}} \\
+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$
|
\[
\begin{aligned}
=、 \text { Original expression } & =\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\
& +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\
= & \frac{a^{2}+b^{2} .}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\
& =0 .
\end{aligned}
\]
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six, a boat sails between
$A, B$ two docks. When sailing downstream, it takes 40 minutes to be 4 kilometers away from the destination, and when sailing upstream, it needs $1 \frac{1}{3}$ hours to arrive. It is known that the upstream speed is 12 kilometers per hour. Find the speed of the boat in still water.
|
Six, let the speed of the current be $x$ kilometers/hour. Then
$$
\frac{2}{3} x+4=1 \frac{1}{3} \times 12 \text {. }
$$
Solving for $x$ gives $x=18$.
Therefore, the speed of the boat in the stream is
$$
(18+12) \div 2=15 \text { (kilometers/hour). }
$$
Answer: The speed of the boat in still water is 15 kilometers/hour.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the equation $2 x^{2}+2 k x-13 k+1=0$ has two real roots whose squares sum to 13. Then $k=$ $\qquad$ .
|
1.1. By the relationship between roots and coefficients, we solve to get $k=1$ or $k=$ -14. Substituting these values back into the original equation, we find that $k=-14$ does not satisfy the conditions, so it is discarded; $k=1$ satisfies the conditions.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A four-digit number has the following property: dividing this four-digit number by its last two digits yields a perfect square (if the tens digit is zero, then divide by the units digit), and this perfect square is exactly the square of the first two digits plus 1. For example, $4802 \div 2$ $=2401=49^{2}=(48+1)^{2}$. Then the smallest four-digit number with the above property is $\qquad$
|
5. 1085 Let this four-digit number be $100 c_{1}+c_{2}$, where $10 \leqslant c_{1} \leqslant 99, 1 \leqslant c_{2} \leqslant 99$. According to the problem, we have
$$
\begin{array}{l}
100 c_{1}+c_{2}=\left(c_{1}+1\right)^{2} c_{2}=c_{1}^{2} c_{2}+2 c_{1} c_{2}+c_{2}, \\
\therefore 100 c_{1}=c_{1} c_{2}\left(c_{2}+2\right),
\end{array}
$$
Therefore, $\left(c_{1}+2\right) \mid 100$, and $10 \leqslant c_{1} \leqslant 99$,
then $c_{1}=\{18,23,48,98\}, c_{2}=\{5,4,2,1\}$.
Thus, the four-digit numbers are $1805, 2504, 4802, 280 \%$, and the smallest four-digit number among them is 1805.
|
1085
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (30 points) Given positive numbers $m, n$ are the roots of the quadratic equation $x^{2}+$ $p x+q=0$, and $m^{2}+n^{2}=3, m n=1$. Find the value of the polynomial $x^{3}-(\sqrt{5}-1) x^{2}-(\sqrt{5}-1) x+1994$.
|
$$
=, \because m, n>0, \therefore m+n=5 .
$$
Thus, $p=-(m+n)=-\sqrt{5}, q=mn=1$,
$$
\therefore x^{2}-\sqrt{5} x+1=0 \text {. }
$$
$$
\text { Also } \begin{array}{l}
x^{3}-(\sqrt{5}-1) x^{2}-(\sqrt{5}-1) x+1994 \\
=\left(x^{3}+1\right)-(\sqrt{5}-1) x(x+1)+1993 \\
=(x+1)\left[\left(x^{2}-x+1\right)-(\sqrt{5}-1) x\right] \\
\quad+1993 \\
=(x+1)\left(x^{2}-\sqrt{5} x+1\right)+1993 \\
= 1993 .
\end{array}
$$
|
1993
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that the area of $\triangle A B C$ is $1, D$ is the midpoint of $B C, E$, $F$ are on $A C, A B$ respectively, and $S_{\triangle B D P}=\frac{1}{5}, S_{\triangle C D E}=\frac{1}{3}$. Then $S_{\triangle D S P}=$ $\qquad$
|
(7)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. You asked me to translate a text, but the text you provided is just a number in parentheses. If you need a translation, please provide the actual text you want translated.
For the given input, the translation would simply be:
(7)
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The square of the first and last digits of a four-digit number adds up to 13, and the square of the middle two digits adds up to 85. If 1089 is subtracted from this number, the result is a four-digit number written with the same digits but in reverse order. The original four-digit number is $\qquad$
|
8. 3762
|
3762
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (18 points) In a round-robin football tournament with $n$ teams (i.e., 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 case of a draw, and 0 points to the losing team. The result is that one team has more points than any other team, yet has fewer wins than any other team. Find the smallest possible value of $n$.
|
Let the team with the highest score be Team $A$, and assume Team $A$ wins $k$ games, draws $m$ games, and thus Team $A$'s total points are $2k + m$.
From the given conditions, every other team must win at least $k+1$ games, and their points must be no less than $2(k+1)$. Therefore, we have $2k + m > 2(k+1)$, which simplifies to $m > 2$. Hence, $m \geqslant 3$.
Thus, there is a team that draws with Team $A$, and this team's score is no less than $2(k+1) + 1$. Therefore,
$$
2k + m > 2(k+1) + 1, \quad m > 3.
$$
This implies $m \geqslant 4$.
From this, we derive $n \geqslant m + 1 = 5$.
If $n = 5$, then $m = 4$, $k = 0$, and Team $A$ scores 4 points. However, in this case, 5 teams play a total of 10 games, and the total points of all teams sum up to 20 points. According to the problem, Team $A$'s score $>\frac{20}{5} = 4$, leading to a contradiction.
Therefore, $n \geqslant 6$.
When $n = 6$, there exists a match that meets the conditions (see the table below), so the minimum possible value of $n$ is 6.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $m$ satisfies the equation
$$
\begin{array}{l}
\sqrt{3 x+5 y-2-m}+\sqrt{2 x+3 y-m} \\
=\sqrt{x-199+y} \cdot \sqrt{199-x-y},
\end{array}
$$
try to determine the value of $m$.
|
Analyzing this problem, we have one equation with three unknowns, so we can only solve it cleverly. Observing the characteristics of each algebraic expression in the original equation, we know that
$$
x-199+y \geqslant 0,
$$
and $199-x-y \geqslant 0$.
Thus, $x+y \geqslant 199$,
and $x+y \leqslant 199$.
Therefore, $x+y=199$.
Substituting $x=199-y$ into the original equation, we get
$$
\sqrt{2 y-m+595}+\sqrt{y-m+398}=0 \text {. }
$$
From (2), we know that it can only be $2 y-m+595=0$ and $y-m+398=0$. Eliminating $y$, we get $m=201$.
|
201
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) There are both coins and banknotes for 1 fen, 2 fen, and 5 fen. To exchange one jiao into several 1 fen, 2 fen, and 5 fen coins or banknotes, how many different methods are there?
|
Three, suppose one dime can be exchanged for $x, y, z$ coins of 1 cent, 2 cents, and 5 cents respectively. Then we have
$$
x+2 y+5 z=10 \text {. }
$$
Solving this, we get $(x, y, z)=(0,0,2),(0,5,0),(1,2$,
$$
\begin{array}{l}
1),(2,4,0),(3,1,1)(4,3,0),(5,0,1),(6, \\
2,0),(8,1,0),(10,0,0) .
\end{array}
$$
Suppose $m$ coins are paid using $x$ coins and $x^{\prime}$ notes, with $m=0^{\prime}+m=1^{\prime}+(m-1)=\cdots=m^{\prime}+$ 0, then there are $(m+1)$ ways to pay $m$ coins using coins and notes. Therefore, the number of different ways to exchange one dime is
$$
\begin{array}{l}
1 \times 1 \times 3+1 \times 6 \times 1+2 \times 3 \times 2+3 \times 5 \times \\
1+4 \times 2 \times 2+5 \times 4 \times 1+6 \times 1 \times 2+7 \times 3 \times 1 \\
+9 \times 2 \times 1+11 \times 1 \times 1=134 \text { ways. }
\end{array}
$$
(Provided by Yang Shilin, Changlin High School, Xichong County, Sichuan Province)
|
134
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, here $a, b, c, d$ are real numbers. It is known that $f(1)=5, f(2)=10$, $f(3)=15$, then $f(8)+f(-4)$ is $(\quad)$.
(A) 2500
(B) uncertain
(C) 2540
(D) 860
|
4. (C).
$$
\begin{array}{c}
f(x)-5 x \\
=(x-1)(x-2)(x-3)(x+h), \\
f(8)-40=210(8+h), \\
f(-4)+20=-210(h-4), \\
\text { then, } f(8)+f(-4)=2540 .
\end{array}
$$
|
2540
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12. Let $M, x, y$ be positive integers, and $\sqrt{M-\sqrt{28}}=\sqrt{x}-\sqrt{y}$. Then the value of $x+y+M$ is ( ).
(1994, Hope Cup Mathematics Competition)
|
Given $M-\sqrt{28}=(x+y)$ $-\sqrt{4 x y}$. Therefore, $x+y=M$, and $4 x y=28$. Since $x, y$ are positive integers, and it is specified that $x \geqslant y$, from $x y=7$, we get $x=7, y=1, M=8, x+y+M=16$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Find the smallest positive integer $n$, such that in any 9-vertex and $n$-edge graph with edges colored in two colors, there must exist a monochromatic triangle.
(33rd IMO, Problem 3)
|
Since $K_{2}$ can have one edge removed and be colored with 0 colors to ensure it contains no monochromatic triangles, and $K_{5}$ can be colored with 2 colors to ensure it contains no monochromatic triangles. $K$ can be divided into 4 $K_{2}$'s and 1 $K_{1}$ with no common vertices, so in $K$, 4 edges can be removed and then colored with $2+0=2$ colors to ensure it contains no monochromatic triangles. The specific coloring method is shown in the figure. Therefore, to ensure that $K$ contains a monochromatic triangle after removing $m$ edges and any 2-coloring, it must be that $m \leqslant 3$.
When 3 edges are removed from $K$, it must contain a complete graph $K_{6}$ (each removed edge removes one endpoint, leaving 6 vertices to form $K_{6}$). Therefore, in $K_{9}$, after removing any 3 edges and 2-coloring, it must contain a monochromatic triangle. Thus, the smallest positive integer
$$
n=C_{9}^{2}-3=33 .
$$
|
33
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
The last digit is not 0, after removing one digit (but not the first digit) from it, it becomes a smaller number that is an integer multiple of the resulting number (i.e., the original number is an integer multiple of the resulting number -- translator's note).
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
57. 19. Let the original number be $b$ times the later number, i.e.,
$$
\begin{array}{l}
\overline{a_{k} \cdots a_{l+1}} \cdot 10^{t}+a_{t} \cdot 10^{t-1}+\overline{a_{i-1} \cdots a_{1}} \\
=b\left(\overline{a_{k} \cdots a_{l+1}} \cdot 10^{t-1}+\overline{a_{t-1} \cdots a_{1}}\right) .
\end{array}
$$
It is easy to see that $110$, then $l=k-1$, and from (1) we get
$$
\begin{array}{l}
{\left[a_{k-1}-(b-10) a_{k}\right] \cdot 10^{4-2}} \\
=(b-1) \cdot \overline{a_{k-2} \cdots a_{k}} .
\end{array}
$$
$$
u_{k-1}>(b-10) a_{k} \text {. }
$$
At this point, (2) becomes
$$
\begin{array}{l}
(b-1) \cdot \overline{a_{k-2} \cdots a_{1}} \\
=2^{k-2} \cdot 5^{k-2} \cdot\left[a_{k-1}-(b-10) a_{k}\right] .
\end{array}
$$
Since $a_{1} \neq 0$, $\overline{a_{1-2} \cdots a_{1}}$ cannot be both a multiple of 2 and a multiple of 5. When it is a multiple of 2, $b-1$ should be a multiple of $5^{1-2}$. But since $b-17 a_{6} .
$$
Since $a_{6} \geqslant 1$, the only suitable values for $a_{5}$ and $a_{6}$ in (5) are:
$a_{5}=8, a_{6}=1$ or $a_{5}=9, a_{6}=1$.
Substituting $a_{5}=8, a_{6}=1, b-1=16$ into (4), we get
$$
\overline{a_{4} a_{3} a_{2} a_{1}}=\frac{10000}{16}=625 .
$$
This indicates that the original number is
$$
\overline{a_{6} a_{5} a_{4} a_{3} a_{2} a_{1}}=180625 \text {. }
$$
In fact, $180625=17 \times 10625$. This meets the requirement.
Substituting $a_{5}=9, a_{6}=1, b-1=16$ into (4), we get
$$
\overline{a_{4} a_{3} a_{2} a_{1}}=\frac{20000}{16}=1250 .
$$
Since $a_{i}=0$ in this case, it does not meet the requirement.
Next, we discuss $b1$. After similar discussions, we find that when $b-1=$ $8, l-1=3, \overline{a_{3} a_{2} a_{1}}=875$, the right side of (7) is maximized, and we have
$$
\overline{a_{4} \cdots a_{5}}+a_{4}=7 .
$$
Thus, $k=5, a_{5} \leqslant 7$. Therefore, the largest positive integer in this case is 70875.
Since $70875=9 \times 7875$. And $70875<180625$, the largest positive integer that meets the requirement is 180625.
|
180625
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. During the New Year festival, the school bought a chocolate bar for each student. It is known that if the chocolate bars are packed 20 to a box, then 5 more boxes are used compared to when they are packed 24 to a box. How many students are there?
|
3. If each tray contains 20 pieces and requires $x$ boxes per week, then, $20 x=24(x-5)$. Solving for $x$ gives $x=30$. Therefore, the number of students is $20 \times 30=600$.
|
600
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. On the diagonal $BD$ of square $ABCD$, take two points $E$ and $F$, such that the extension of $AE$ intersects side $BC$ at point $M$, and the extension of $AF$ intersects side $CD$ at point $N$, with $CM = CN$. If $BE = 3$, $EF = 4$, what is the length of the diagonal of this square?
|
3. From the condition, we get
$\triangle A^{\prime} B M I \triangle \triangle A C M$.
$\therefore \angle B E=\angle D A E$.
Also, $A B=A D$,
$\angle A B E=\angle A D F$,
$\therefore \triangle A B E \simeq \triangle A D F$.
Therefore, $D F=B E=3$.
$\therefore A C=B D=10$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8. Calculate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}$
$$
\begin{array}{l}
+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7}) \\
\cdot(-\sqrt{5}+\sqrt{6}+\sqrt{7}) .
\end{array}
$$
|
Analyze the repeated application of the difference of squares formula to get the result as 104.
|
104
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. For positive $x$, we have $x^{2}+\frac{1}{x^{2}}=7$. Prove that $x^{5}+\frac{1}{x^{5}}$ is a number, and calculate the result.
|
4. From the equation $\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2$ and $x+\frac{1}{x}=3$.
We get
$$
\begin{array}{l}
27:=\left(x+\frac{1}{x}\right)^{3} \\
=x^{3}+3 x^{2} \cdot \frac{1}{x}+3 x \cdot \frac{1}{x^{2}}+\frac{1}{x^{3}} \\
=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right) .
\end{array}
$$
Thus, $x^{3}+\frac{1}{x^{3}}=18$.
$$
\begin{array}{l}
\therefore \quad 7 \times 18=\left(x^{2}+\frac{1}{x^{2}}\right) \times\left(x^{3}+\frac{1}{x^{3}}\right) \\
=x^{5}+x^{2} \cdot \frac{1}{x^{3}}+\frac{1}{x^{2}} \cdot x^{3}+\frac{1}{x^{5}} \\
=x^{5}+\frac{1}{x^{5}}+x+\frac{1}{x} . \\
\therefore x^{5}+\frac{1}{x^{5}}=126-3=123 .
\end{array}
$$
|
123
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Calculate $(\sqrt{1300}]+[\sqrt{1901}]+\cdots$ $+[\sqrt{1993}]=(\quad)$.
(1993, Hope Cup Mathematics Competition)
|
$$
\begin{array}{l}
\text { Analysis } \because 43<\sqrt{1900}, \sqrt{1901}, \cdots, \\
\sqrt{1935}<44,44 \leqslant \sqrt{1936}, \sqrt{1937}, \cdots, \\
\sqrt{1993}<45, \\
\therefore[\sqrt{1900}],[\sqrt{1901}], \cdots,[\sqrt{1935}] \text { are all } 43, \text { a total of 36, } [\sqrt{1936}],[\sqrt{1937}], \cdots, [\sqrt{1993}] \text { are all } 44, \text { a total of 58. } \\
\text { Therefore, the required value is } 43 \times 36+44 \times 58=4100.
\end{array}
$$
|
4100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3. 3 .
From $[\lg x] \leqslant \lg x$, we get $\lg ^{2} x-\lg x-2 \leqslant 0$, which means $-1 \leqslant \lg x \leqslant 2$.
When $-1 \leqslant \lg x<0$, we have $[\lg x]=-1$. Substituting into the original equation, we get $\lg x= \pm 1$, but $\lg x=1$ is not valid, so $\lg x=-1, x_{1}=\frac{1}{10}$.
When $0 \leqslant \lg x<1$, we have $[\lg x]=0$, substituting into the original equation, we get $\lg x$ $= \pm \sqrt{2}$, both of which are not valid.
When $1 \leqslant \lg x<2$, we have $[\lg x]=1$, substituting into the original equation, we get $\lg x$ $= \pm \sqrt{3}$. But $\lg x=-\sqrt{3}$ is not valid, so $\lg x=\sqrt{3}, x_{2}$ $=10 \sqrt{3}$.
When $\lg x=2$, we get $x_{3}=100$.
Therefore, the original equation has
3 real roots.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In the Cartesian coordinate plane, the number of integer points (i.e., points with both coordinates as integers) that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y \leqslant 3 x, \\
y \geqslant \frac{1}{3} x, \\
x+y \leqslant 100
\end{array}\right.
$$
is
|
$$
1+2+3+\cdots+101=\frac{102 \times 101}{2}=5151 \uparrow .
$$
The region enclosed by $y=\frac{1}{3} x, x+y=100$, and the $x$-axis (excluding the boundary $y=\frac{1}{3} x$) contains
$$
\begin{aligned}
& (1+1+1+1)+(2+2+2+2)+\cdots \\
& +(25+25+25+25) \\
= & 4 \times(1+2+\cdots+25)=1300 \text { points. }
\end{aligned}
$$
The region enclosed by $y=3 x, x+y=100$, and the $y$-axis (excluding the boundary $y=3 x$) also contains 1300 points. Therefore, the total number of points satisfying the given conditions is $5151 - 2 \times 1300 = 2551$.
4. 2551 .
The region enclosed by the two coordinate axes and the line $x+y=100$ (including the boundaries) contains
|
2551
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Color each vertex of a square pyramid with one color, and make the endpoints of the same edge have different colors. If only 5 colors are available, the total number of different coloring methods is $\qquad$
|
$5,4,20$.
Suppose, for the pyramid $S-ABCD$, the vertices $S, A, B$ are colored with different colors, and they have a total of $5 \times 4 \times 3=60$ coloring methods.
When $S, A, B$ are already colored, assume their colors are $1, 2, 3$ respectively; if $C$ is colored with color 2, then $D$ can be colored with one of the colors $3, 4, 5$, which gives 3 coloring methods; if $C$ is colored with color 4, then $D$ can be colored with color 3 or 5, which gives 2 coloring methods; if $C$ is colored with color 5, then $D$ can be colored with color 3 or 4, which also gives 2 coloring methods. Therefore, when $S, A, B$ are already colored, $C$ and $D$ have 7 coloring methods.
Thus, the total number of coloring methods is $60 \times 7=420$.
|
420
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $M=\{1,2,3, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: if $x \in A$, then $15 x \notin A$. Then the maximum number of elements in $A$ is $\qquad$ .
|
6. 1870.
Let $n(A)$ denote the number of elements in the set $A$.
By the problem's setup, at least one of the numbers $k$ and $15k$ (where $k=9,10, \cdots, 133$) does not belong to $A$, so at least $125(=133-9+1)$ numbers do not belong to $A$, i.e., $n(A) \leqslant 1995-125=1870$. On the other hand, we can take $A=\{1,2, \cdots, 8\} \cup\{134,135, \cdots, 1995\}$, which satisfies the problem's conditions, and in this case, $n(A)=1870$. Therefore, the maximum value of $n(A)$ is 1870.
|
1870
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (This question is worth 25 points) Find all real numbers $p$ such that the cubic equation $5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p$ has three roots that are all natural numbers.
|
From observation, it is easy to see that $x=1$ is a root of the original cubic equation. By synthetic division, the original cubic equation can be reduced to a quadratic equation:
$$
5 x^{2}-5 p x+66 p-1=0 \text {. }
$$
The three roots of the original cubic equation are natural numbers, and the two roots of the quadratic equation (*) are also natural numbers. Let $\alpha, \nu(2: \div \leqslant v)$ be the two roots of equation (*), then by Vieta's formulas, we have:
$$
\left\{\begin{array}{l}
u+v=p, \\
u v=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$
Substituting (1) into (2) gives $5 u v=66(u+v)-1$. It can be known that $u, v$ cannot be divisible by $2,3,11$.
From (3), we get $v=\frac{66 u-1}{5 u-66}$.
Since $u, v$ are natural numbers, from (4) we know $u>\frac{66}{5}$, i.e., $u \geqslant 14$.
Also, since $2 \nmid u$ and $3 \nmid u$, we know $u \geqslant 17$.
From $v \geqslant u$ and (4), we get $\frac{66 u-1}{5 u-66} \geqslant u$, i.e., $5 u^{2}-132 u+1 \leqslant 0$.
Thus, $u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}<\frac{132}{5}$.
$\therefore 17 \leqslant u \leqslant 26$.
Furthermore, since $2 \nmid u$ and $3 \nmid u$, $u$ can only be $17,19,23,25$.
When $u=17$, from (4) we get $v=\frac{1121}{19}=59$.
When $u=19$, from (4) we get $v$ is not a natural number, so it should be discarded.
When $u=23$, from (4) we get $v$ is not a natural number, so it should be discarded.
When $u=25$, from (4) we get $v$ is not a natural number, so it should be discarded.
Therefore, only when $p=u+v=76$, the two roots of equation (*) are natural numbers, and the three roots of the original equation are natural numbers.
|
76
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Find the smallest prime $p$ that cannot be expressed as $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers.
|
It has been verified that $2,3,5,7,11,13,17,19,23,29,31$, and 37 can all be expressed in the form $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers:
$$
\begin{array}{l}
2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, \\
7=2^{3}-3^{0}, 11=3^{3}-2^{4}, 13=2^{4}-3^{1}, \\
17=3^{4}-2^{6}, 19=3^{3}-2^{3}, 23=3^{3}-2^{2}, \\
29=2^{5}-3^{1}, 31=2^{5}-3^{0}, 37=2^{6}-3^{3} .
\end{array}
$$
Pig 41 is the smallest prime number that cannot be expressed in this form. To verify this conjecture, we examine the following two indeterminate equations:
( I ) $2^{u}-3^{v}=41 ;$
( II ) $3^{x}-2^{y}=41$.
Let $(u, v)$ be a non-negative integer solution to equation (I), then $2^{u}>41, u \geqslant 6$. Therefore,
$$
-3^{v} \equiv 1 \text { (mod } 8 \text { ). }
$$
However, the residue of $3^{v}$ modulo 8 can only be 1 or 3, so equation (I) has no non-negative integer solutions.
Let $(x, y)$ be a non-negative integer solution to equation (II), then $3^{x}>41, x \geqslant 4$. Therefore,
$$
2^{y} \equiv 1(\bmod 3) \text {. }
$$
Thus, $y$ must be even. Let $y=2t$, and we get
$$
3^{x} \equiv 1(\bmod 4) .
$$
This implies that $x$ must also be even. Let $x=2s$. Therefore,
$$
41=3^{x}-2^{y}=3^{2s}-2^{2t}=\left(3^{s}+2^{t}\right)\left(3^{s}-2^{t}\right) .
$$
For this equation to hold, it must be that
$$
\left\{\begin{array}{l}
3^{s}+2^{t}=41, \\
3^{s}-2^{t}=1 .
\end{array}\right.
$$
This implies $3^{s}=21, 2^{t}=20$.
But this is impossible. Therefore, equation (II) also has no non-negative integer solutions.
In conclusion, we find that the smallest prime number that cannot be expressed in the form $\left|3^{a}-2^{b}\right|$ (where $a$ and $b$ are non-negative integers) is 41.
|
41
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. 21 people participate in an exam, with the test paper containing 15 true/false questions. It is known that for any two people, there is at least one question that both of them answered correctly. What is the minimum number of people who answered the most questions correctly? Please explain your reasoning.
|
For the $i$-th problem, $a_{i}$ people answered correctly, thus there are exactly
$$
b_{i}=C_{s_{i}}^{z}
$$
pairs of people who answered the $i$-th problem correctly $(i=1,2, \cdots, 15)$.
Below, we focus on the sum $\sum_{i=1}^{15} b_{i}$.
Let $a=\max \left\{a_{1}, a_{2}, \cdots, a_{15}\right\}$, then we have
$$
15 C_{a}^{2} \geqslant \sum_{i=1}^{15} b_{i} \geqslant C_{21}^{2}, \quad a(a-1) \geqslant \frac{420}{15}=28,
$$
$a \geqslant 6$.
We will show that $a=6$ is impossible, hence, the problem with the most correct answers must have at least 7 people answering correctly.
Assume $a=6$, meaning no more than 6 people answered any question correctly. We will derive a contradiction. If someone answered only three questions correctly, then for each of these questions, they could only share the correct answer with at most 5 other people, meaning they could share correct answers with at most 15 other people, which contradicts the assumption. Therefore, each person must answer at least four questions correctly.
Since $21 \times 5 > 6 \times 15$,
it is impossible for everyone to answer five or more questions correctly, so there must be at least one person who answers exactly four questions correctly, and these four questions form four sets that have no common members except for this person, for example:
$$
\begin{array}{l}
S_{1}=\{1,2,3,4,5,6\}, \\
S_{2}=\{1,7,8, \cdots, 11\}, \\
S_{3}=\{1,12,13, \cdots, 16\}, \\
S_{4}=\{1,17,18, \cdots, 21\},
\end{array}
$$
On the other hand, among all 15 questions, at least 12 questions must have 6 people answering correctly, otherwise it would lead to a contradiction:
$$
C_{21}^{2} \leqslant \sum_{i=1}^{15} b_{i} \leqslant 11 C_{6}^{2}+4 C_{5}^{2}=205 .
$$
Excluding the aforementioned four questions, there are still eight questions with 6 people answering correctly. Consider the 6 people who answered one of these eight questions correctly. Among them, either more than three people belong to the same $S_{j}$, or two people belong to some $S_{j}$, and two others belong to $S_{k}(j, k \in\{1,2,3,4\}, j \neq k)$. In any case, when calculating $\sum_{i=1}^{15} b_{i}$, each of these eight questions will result in at least two duplicate counts. This leads to
$$
C_{21}^{2} \leqslant \sum_{i=1}^{15} b_{i}-8 \times 2 \leqslant 15 C_{6}^{2}-16=209 .
$$
The resulting contradiction shows that the case $a=6$ is impossible.
Based on the above discussion, we confirm that $a \geqslant 7$.
Next, we construct an example to show that the case $a=7$ is possible. For this, we first number the participants from 1 to 21 and define the following sets of participants:
$$
\begin{array}{l}
P_{i}=\{2 i-1,2 i\}, i=1, \cdots, 63 \\
P_{1}=\{13,14,15\}, \\
P_{8}=\{16,17,18\}, \\
P_{0}=\{19,20,21\},
\end{array}
$$
Using these notations, we construct the following table to indicate the participants who answered each question correctly. It is easy to verify that in the scenario shown in the table, every pair of the 21 participants has at least one question they both answered correctly, and no more than 7 people answered the same question correctly.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $S=\left\{A=\left(a_{1}, \cdots, a_{8}\right) \mid a_{i}=0\right.$ or $1, i=1, \cdots$, 8\}. For two elements $A=\left(a_{1}, \cdots, a_{8}\right)$ and $B=\left(b_{1}\right.$, $\cdots, b_{8}$ ) in $S$, denote
$$
d(A, B)=\sum_{i=1}^{\delta}\left|a_{i}-b_{i}\right|,
$$
and call it the distance between $A$ and $B$. What is the minimum number of elements that can be selected from $S$ such that the distance between any two of them is $\geqslant 5$?
|
Solution One
(I) First, we point out that the sum of the weights of any two codewords in $\mathscr{D}$ does not exceed 11; otherwise, if
$$
\omega(X)+\omega(Y) \geqslant 12,
$$
then because $12-8=4$, these two codewords must share at least four positions with 1s, and the distance between them
$$
a^{\prime}\left(X, y^{\prime}\right) \leq 8-4=4.
$$
Therefore, $\mathscr{D}$ can contain at most one codeword with a weight $\geqslant 6$. Furthermore, if $\mathscr{D}$ contains a codeword with a weight $\geqslant 7$, then $\mathscr{D}$ can only contain that codeword and the all-zero codeword.
(I) Noting that $5+5-8=2$, we conclude: if $\mathscr{D}$ contains two codewords $A$ and $B$ with a weight of 5, then these two codewords must share at least two positions with 1s. We can further conclude: these two codewords share exactly two positions with 1s. Otherwise, if they share three positions with 1s, then they must also share at least one position with 0s, and the distance between the two codewords
$$
d(A, B) \leqslant 4.
$$
Based on the above discussion, we can conclude: $\mathscr{D}$ can contain at most two codewords with a weight of 5 (otherwise, the third codeword with a weight of 5 must share at least three positions with 1s with either $A$ or $B$).
(1) In summary, $\mathscr{D}$ can contain at most four codewords, one with a weight of 0, one with a weight of 6, and two with a weight of 5. The following example shows that a $\mathscr{D}$ consisting of four codewords can be realized:
$$
\begin{array}{l}
(0,0,0,0,0,0,0,0) \\
(1,1,1,0,0,1,1,1) \\
(1,1,1,1,1,0,0,0) \\
(0,0,0,1,1,1,1,1).
\end{array}
$$
Solution Two First, consider an easier problem: what is the maximum number of codewords of length 7, where the distance between any two is $\geqslant 5$? The answer is: at most two. First, assume $(0,0,0,0,0,0,0)$ is one of the codewords. Then it is impossible to have two more codewords. Otherwise, each of these two codewords would have a weight $\geqslant 5$. However,
$$
5+5-7=3.
$$
So, these two codewords must share at least three positions with 1s. The distance between these two codewords is no more than $7-3=4$.
Now, let's return to the original problem. For codewords of length 8, the first bit can be 0 or 1. Codewords with the same first bit $(a \in\{0,1\})$, the remaining bits form a codeword of length 7, and thus there can be at most two such codewords with a distance $\geqslant 5$. Therefore, there can be at most four codewords of length 8, where the distance between any two is $\geqslant 5$.
An example of "at most four" being achievable is the same as in Solution One.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}(n>4)$ where all interior angles are integer multiples of $15^{\circ}$, and $\angle A_{1}+\angle A_{2}+\angle A_{3}=$ $285^{\circ}$. Then, $n=$
|
2. 10 (Hint: The sum of the $n-3$ interior angles other than the male one is $(n-2) \cdot$ $180^{\circ}-285^{\circ}$, it can be divided by $n-3$, and its quotient should also be an integer multiple of $15^{\circ}$.）
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A football made of leather in black and white colors is shown as follows. It is known that this football has 12 black leather pieces. Therefore, this football has white leather pieces.
保留源文本的换行和格式，直接输出翻译结果。
|
4. 20 (Hint: Each black piece is a pentagon, and the total number of edges for 12 pieces is: $5 \times 12=60$ (edges). Each white piece is a hexagon, and three of its edges are adjacent to the edges of black pieces. That is, three edges of black pieces determine one white piece. Therefore, the number of white pieces is $60 \div 3=20$ (pieces).)
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. For real numbers $x, y$, define a new operation: $x * y=a x+b y+$ $c$, where $a, b, c$ are constants, and the right side of the equation is the usual addition and multiplication operations. It is known that $3 * 5=15,4 * 7=28$. Then, 1*1= $\qquad$
|
7. -11
Brief solution: According to the definition, we know that $3 * 5=3a+5b+c=15, 4 * 7$ $=4a+7b+c=28$, and $1 * 1=a+b+c=3(3a+5b+$c) $-2(4a+7b+c)=3 \times 15-2 \times 28=-11$.
|
-11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (12 points) Given points $P_{1}\left(x_{1}, 1994\right), P_{2}\left(x_{2}, 1994\right)$ are two points on the graph of the quadratic function $y=a x^{2}+b x+7(a \neq 0)$. Try to find the value of the quadratic function $y=x_{1}+x_{2}$.
|
Three, since $P_{1}, P_{2}$ are two points on the graph of a quadratic function, we have
$$
\begin{array}{l}
a x_{1}^{2}+b x_{1}+7=1994, \\
a x_{2}^{2}+b x_{2}+7=1994 .
\end{array}
$$
Subtracting the two equations and simplifying, we get
$$
\left(x_{1}-x_{2}\right)\left[a\left(x_{1}+x_{2}\right)+b\right]=0 .
$$
Since $P_{1}, P_{2}$ are two distinct points, i.e., $x_{1} \neq x_{2}$, we have
$$
x_{1}+x_{2}=-\frac{b}{a},
$$
Thus, when $x=x_{1}+x_{2}$, the value of $y$ is 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. (India) The sequence of positive integers $\left\{f_{n}\right\}_{m=1}^{m}$ is defined as follows:
$f(1)=1$, and for $n \geqslant 2$,
$$
f(n)=\left\{\begin{array}{l}
f(n-1)-n, \text { when } f(n-1)>n ; \\
f(n-1)+n, \text { when } f(n-1) \leqslant n .
\end{array}\right.
$$
Let $S=\{n \in \mathbb{N} \mid f(n)=1993\}$.
(1) Prove that $S$ is an infinite set;
(2) Find the smallest positive integer in $S$;
(3) If the elements of $S$ are arranged in increasing order as $n_{1}<n_{2}$ $<n_{3}<\cdots$, then $\lim _{k \rightarrow \infty} \frac{n_{k+1}}{n_{k}}=3$.
|
(i) We point out that if $f(n)=1$, then $f(3n+3)=1$. From $f(n)=1$, we can sequentially obtain according to the definition:
$$
\begin{array}{l}
f(n+1)=n+2, f(n+2)=2n+4, \\
f(n+3)=n+1, f(n+4)=2n+5, \\
f(n+5)=n, \quad f(n+6)=2n+6 \\
f(n+7)=n-1, \cdots
\end{array}
$$
It is evident that the sequence $f(n+1), f(n+3), f(n+5), \cdots, f(3n+3)$ has values exactly as the sequence $n+2, n+1, n, \cdots, 1$, and the sequence $f(n+2), f(n+4), f(n+6), \cdots, f(3n+2)$ has values exactly as the sequence $2n+4, 2n+5, 2n+6, \cdots, 3n+4$. Therefore, $f(3n+3)=1$ and when $n \geq 1993$, from the above argument, among all natural numbers $n$ such that $b_i < n \leq b_{i+1}$, there is exactly one that makes $f(n)=1993$. Thus, $S$ is an infinite set.
(iv) Similarly to (iii), when $i \geq 8$, we have
$$
f(b_i + 2j - 1) = b_i + 3 - j = 1993
$$
Solving for $j$ gives $j = b_i - 1990$. Therefore, the value of $n_i$ that makes $f(n_i) = 1393$ is
$$
\begin{aligned}
a_1 & = b_i + 2j - 1 = 3 \cdot 3^{i-2} - 3981 \\
& = \frac{9}{2}(5 \cdot 3^{i-2} - 1) - 3982; \quad i = 8, 9, \cdots.
\end{aligned}
$$
Thus, we have
$$
\lim_{k \to \infty} \frac{n_{k+1}}{n_k} = \lim_{k \to \infty} \frac{\frac{9}{2}(5 \cdot 3^{k-1} - 1) - 3981}{\frac{9}{2}(5 \cdot 3^{k-2} - 1) - 3981} = 3.
$$
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Nephew 8. Find the number of lattice points inside and on the boundary of the triangle formed by the line $y=\frac{2}{3} x-\frac{1}{2}, x=10$ and the $x$-axis.
|
Given the function $y=\frac{2}{3} x-\frac{1}{2}$, when $x$ takes the values $1,2, \cdots$, 10 (note: when $x=\frac{3}{4}$, $y=0$), the corresponding $y$ values are $\frac{1}{6}, \frac{5}{6}, \frac{3}{2}, \frac{13}{6}, \frac{17}{6}, \frac{7}{2}, \frac{25}{6}, \frac{29}{6}, \frac{11}{2}, \frac{37}{6}$. The number of integer points on the known triangle (including the boundary) is equal to the sum of the integer parts of the $y$-coordinates plus the 10 integer points on the $x$-axis, i.e.,
$$
\begin{array}{c}
{\left[\frac{1}{6}\right]+\left[\frac{5}{6}\right]+\left[\frac{3}{2}\right]+\left[\frac{13}{6}\right]+\left[\frac{17}{6}\right]+} \\
{\left[\frac{7}{2}\right]+\left[\frac{25}{6}\right]+\left[\frac{29}{6}\right]+\left[\frac{11}{2}\right]+\left[\frac{37}{6}\right]+10=37}
\end{array}
$$
Therefore, there are 37 integer points inside and on the boundary of the triangle.
|
37
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In $\triangle A B C$, $\angle C=3 \angle A, a=27, c=48$. Then $b$ $=$ ـ $\qquad$
|
4. 35 .
As shown. Draw the trisectors of $\angle C$ as $C D$ and $C E$.
$$
\begin{array}{l}
\text { Let } C D=1, A N=x, \\
A E=E C=y .
\end{array}
$$
From $\triangle C D B \sim \triangle A C B$, we have
$$
\begin{array}{l}
\frac{a}{c}=\frac{t}{b}=\frac{c-x}{a}, \\
t=\frac{a b}{c}, x=\frac{c^{2}-a^{2}}{c} .
\end{array}
$$
Also, from $\triangle C E D \sim \triangle A C D$, we have
$$
\frac{t}{x}-\frac{y}{b}=\frac{x-y}{t} .
$$
Thus, $\frac{x}{b+t}=\frac{t}{x}, x^{2}=t(b+t)$.
(1) Substituting (1) into (2) and simplifying, we get
$$
b^{2}=\frac{1}{a}(c+a)(c-a)^{2} .
$$
Therefore, $b=35$.
|
35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, from the 1995 natural numbers $1,2,3, \cdots$, 1995, remove some numbers so that in the remaining numbers, no number is equal to the product of any two other numbers. How many numbers must be removed at a minimum to achieve this?
|
(1) First, prove: When any 42 numbers are removed, there will be three numbers among the remaining numbers, such that the product of two of them equals the third.
We consider the array $(2,87,2 \times 87),(3,86,3 \times 86)$, $\cdots,(44,5,44 \times 45)$,
i.e., $(x, 89-x, x(89-x))(2 \leqslant x \leqslant 44)$.
$$
\begin{array}{l}
\because x(89-x)=89 x-x^{2}=44.5^{2}-(44.5-x)^{2}, \\
\therefore \text { when } x \text { takes } 2,3, \cdots, 44 \text {, } x(89-x) \text { are all different, and } \\
x(89-x) \leqslant 44.5^{2}-(44.5-44)^{2} \\
=44 \times 45=19801995.
\end{array}
$$
Combining the above, we know that at least 43 numbers need to be removed.
|
43
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $[x]$ denotes the greatest integer not exceeding $x$, $a, b, c \in$ $R^{+}, a+b+c=1$, let $M=\sqrt{3 a+1}+\sqrt{3 b+1}+$ $\sqrt{3 c+1}$. Then the value of $[M]$ is $($.
|
5. B.
Given $a \in(0,1), b \in(0,1), c \in(0,1) \Rightarrow a^{2} & \sqrt{a^{2}+2 a+1}+\sqrt{b^{2}+2 b+1} \\
& +\sqrt{c^{2}+2 c+1} \\
= & (a+b+c)+3=4 .
\end{aligned}
$$
And $M=\sqrt{(3 a+1) \cdot 1}+\sqrt{(3 b+1) \cdot 1}$
$$
\begin{aligned}
& +\sqrt{(3 c+1) \cdot 1} \\
& \frac{(3 a+1)+1}{2}+\frac{(3 b+1)+1}{2}+\frac{(3 c+1)+1}{2} \\
= & 3+\frac{3}{2}(a+b+c)=4.5, \\
\therefore & 4<M<4.5,[M]=4 .
\end{aligned}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given that there are exactly 600 triangles with integer sides, all of different lengths, and the longest side is exactly $n$. Find $n$.
The longest side of the triangle is $n$, and the other two sides are $a$ and $b$ with $a < b < n$. The triangle inequality theorem states that:
1. $a + b > n$
2. $a + n > b$
3. $b + n > a$
Since $a < b < n$, the second and third inequalities are always satisfied. Therefore, the only relevant inequality is $a + b > n$.
To find the number of such triangles, we need to count the number of pairs $(a, b)$ that satisfy $a < b < n$ and $a + b > n$.
For a fixed $b$, $a$ can take values from $b - n + 1$ to $b - 1$. The number of valid $a$ values for a given $b$ is $b - \left\lceil \frac{n + 1}{2} \right\rceil + 1$.
Summing over all possible values of $b$ from $\left\lceil \frac{n + 1}{2} \right\rceil$ to $n - 1$, we get the total number of triangles:
\[
\sum_{b=\left\lceil \frac{n + 1}{2} \right\rceil}^{n-1} (b - \left\lceil \frac{n + 1}{2} \right\rceil + 1)
\]
This sum can be simplified to:
\[
\sum_{k=1}^{n - \left\lceil \frac{n + 1}{2} \right\rceil} k = \frac{(n - \left\lceil \frac{n + 1}{2} \right\rceil)(n - \left\lceil \frac{n + 1}{2} \right\rceil + 1)}{2}
\]
Given that this sum equals 600, we can solve for $n$:
\[
\frac{(n - \left\lceil \frac{n + 1}{2} \right\rceil)(n - \left\lceil \frac{n + 1}{2} \right\rceil + 1)}{2} = 600
\]
By trial and error or solving the quadratic equation, we find that $n = 49$ satisfies this equation. Therefore, the value of $n$ is:
\[
\boxed{49}
\]
|
Analysis: Since one side of the triangle is fixed at length $n$, and the three side lengths $x, y, n$ are all integers, the number of such triangular shapes corresponds one-to-one with the lattice points $(x, y)$ in the coordinate plane. Therefore, the lattice point method can be used to solve this problem.
Let the lengths of the other two sides of the triangle be $x, y$. Thus, the number of positive integer solutions satisfying $x+y>n, x<y<n$ is the number of triangles.
When $n$ is odd, the number of triangles is the number of lattice points within $\triangle B C D$ in Figure 1. That is,
$$
\begin{array}{l}
2+4+6+\cdots+(n-3) \\
=\frac{1}{4} n(n-3) .
\end{array}
$$
When $n$ is even, the number of triangles is the number of lattice points within $\triangle B C D$ in Figure 2, which is
$1+3+5+\cdots+(n-3)=\frac{1}{4}(n-2)^{2}$.
When $-\frac{1}{4} n(n-3)=600$, solving gives $n=51$:
When $\frac{1}{4}(n-2)^{2}=600$, there is no integer solution.
Therefore, the required $n=51$.
|
51
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the functions $y=2 \cos \pi x(0 \leqslant x \leqslant 2)$ and $y=2(x \in$ $R$ ) whose graphs enclose a closed plane figure. Then the area of this figure is $\qquad$ .
|
2. As shown in the figure, by symmetry, the area of $CDE$ = the area of $AOD$ + the area of $BCF$, which means the shaded area is equal to the area of square $OABF$, so the answer is 4.
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Now we have 1 one-tenth yuan note, 1 one-tenth yuan note, 1 six-tenth yuan note, 5 one yuan notes, and 2 two yuan notes. The number of different amounts (excluding the case of not paying) that can be made is $\qquad$
|
6. Using: Jiao Shi Ji to match 0 Jiao, 1 Jiao, 2 Zhong different denominations. That is, $(1+i)$ kinds.
Using 5 Jiao can match 0 Jiao, 5 Jiao, which is $(1+1)$ kinds; Yuan. That is, $(5+1)$ kinds;
Using 5 Yuan can match 0 Yuan, 5 Yuan, 10 Yuan, which is $(2+1)$ kinds.
Considering the amounts to be paid are 5.5, 1.5, 2.5, 3.5, 5, 6.5, 7.5, 8.5, 10, 10.1, 10.2, 10.3, 10.5, 10.6, 10.7, 10.8 Yuan, each has 2 methods, so the total number of ways to pay is
$$
\begin{array}{l}
(1+1)(1+1)(1+1)+(5+1)(2+1)-16-1 \\
=127 \text { ( )} .
\end{array}
$$
|
127
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the natural numbers $a, x, y$ satisfy $\sqrt{a-2 \sqrt{6}}=\sqrt{x} -\sqrt{y}$, then the maximum value of $a$ is $\qquad$.
|
$$
=, 1, a_{\max }=7 \text {. }
$$
Squaring both sides of the known equation yields $a-2 \sqrt{6}=x+y-$ number, it must be that $x+y=\alpha$ and $x y=6$. From the known equation, we also know that $x>y$, so it can only be $x=6, y=1$ or $x=3, y=2$. Therefore, $a=7$ or 2, the maximum value is 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For the quadratic function $y=x^{2}-4 a x+5 a^{2}$ $-3 a$ in terms of the independent variable $x$, the minimum value $m$ is a function of $a$, and $a$ satisfies the inequality $0 \leqslant$ $a^{2}-4 a-2 \leqslant 10$. Then the maximum value of $m$ is $\qquad$ $ـ$
|
2. 18 .
Solving $0 \leqslant a^{2}-4 a-2 \leqslant 10$ yields $-2 \leqslant a \leqslant 2-\sqrt{6}$ or $2+\sqrt{6} \leqslant a \leqslant 6$. Also, $m=a^{2}-3 a$, considering the quadratic function, $m=a^{2}-3 a$ when $-2 \leqslant a \leqslant 2-\sqrt{6}$ and $2+\sqrt{6} \leqslant a \leqslant 6$, we can see that the maximum value occurs when $a=6$, thus $m=6^{2}-3 \times 6=18$
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Person A and Person B start from the same point $A$ on a circular track at the same time and run in opposite directions. Person A's speed is $5 \mathrm{~m}$ per second, and Person B's speed is $7 \mathrm{~m}$ per second. They stop running when they meet again at point $A$ for the first time. During this period, they meet a total of $n$ times, so $n=$ $\qquad$
|
3. 12 .
Two $\lambda \mu A$ start from $A$ and meet once (not at point $A$). At the moment of their first meeting, the two have run a certain distance. From the ratio of their speeds, it can be deduced that person A has run $\frac{5}{12}$ of a lap, and person B has run $\frac{7}{12}$ of a lap. Therefore, after the $n$-th meeting, person A has run $\frac{5}{12} n$ laps, and person B has run $\frac{7}{12} n$ laps. When they meet again at point $A$, both person A and person B have run an integer number of laps, so $\frac{5}{12} n$ and $\frac{7}{12} n$ are both integers. The smallest value of $n$ is 12, thus they have met 12 times.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
ニ、(Full score 25 points) Write 1995 equations of the form $A_{i} x^{2}+B_{i} x+C_{i}=0(i=1,2, \cdots, 1995)$ on the blackboard. Two people, A and B, take turns to play a game. Each time, only one non-zero real number is allowed to replace one of the $A_{i}, B$ or $C$ in the $n$ equations $(i=1,2, \cdots, 1995)$. Once a coefficient in the equation is filled, it cannot be changed. After $3 \times 1995$ steps, 1995 quadratic equations with specific real coefficients are obtained. A always tries to make as many quadratic equations as possible have no roots, while B always tries to prevent A from achieving his goal. If A fills first, how many rootless equations can A get at most, regardless of what B does?
|
Second, the answer is that Jia can get at most 998 rootless equations.
If Jia first fills in the coefficient of the linear term $B_{1}$ with 1, and tries to fill in the linear term coefficient of each equation with 1 as much as possible, at this time, Jia has at least $\frac{1995+1}{2}=998$ chances.
If Yi fills in the coefficient of the $x^{2}$ term (or the constant term) with $a$ in the equations where Jia has already filled in the coefficient of the linear term, then Jia will fill in the third position with $\frac{1}{a}$, resulting in the equation $a x^{2}+x+\frac{1}{a}=0$, or $\frac{1}{a} x^{2}+x+a=0$. At this point, $\Delta=1-4 \cdot \frac{1}{a} \cdot a=-3<0$, so the equation has no roots. Therefore, Jia can get at least 998 rootless equations.
|
998
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The Chinese Football League A has 12 football clubs participating, with a home and away double round-robin system, meaning any two teams play each other twice, once at home and once away. A win earns 3 points, a draw earns 1 point each, and a loss earns 0 points. At the end of the league, teams are ranked according to their points. The maximum points difference between two teams with adjacent positions on the points table can be $\qquad$ points.
|
$=1.46$ points.
Suppose the score difference between the $k$-th and the $(k+1)$-th is the largest. Since the total score of each match between any two teams is at most 3 points and at least 2 points, the maximum total score of the first $k$ teams playing against each other is $3 \cdot 2 C_{k}^{2}=6 C_{k}^{2}$. Additionally, the maximum total score of the first $k$ teams playing against the remaining $12-k$ teams is $3 \cdot 2 k(12-\dot{k})$. Therefore, the maximum total score of the first $k$ teams is $6 C_{k}^{2}+6 k(12-k)$. Thus, the maximum score of the $k$-th team is $\frac{6 C_{k}^{2}+6 k(12-k)}{k}=69-3 k$ points.
The minimum total score of the remaining $12-k$ teams playing against each other is $2(12-k)(11-k)$ points. Therefore, the minimum score of the $(k+1)$-th team is $2(11-k)$ points.
Thus, the score difference between the $k$-th and the $(k+1)$-th team is $(69-3 k) - 2(11-k) = 47 - k$ points, and its maximum value is 46 points when $k=1$.
In fact, if the first team wins all 22 matches and scores 66 points, while the remaining 11 teams draw against each other and each scores 20 points, the difference is exactly 46 points.
|
46
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the plane, a parabola divides the plane into two parts, two parabolas can divide the plane into at most seven parts. Then 10 parabolas can divide the plane into $\qquad$ parts.
|
3. 191 .
Let $n$ parabolas divide the plane into $a_{n}$ regions. When adding another parabola to maximize the number of regions, this new parabola intersects each of the $n$ given parabolas at 4 points, thus the $n+1$ parabolas are divided into $4n+1$ segments, each of which splits an original region into two. This leads to the recurrence relation $a_{n+1}=a_{n}+4n+1$, and $a_{1}=2$. From this, we can derive $a_{n}=2n^{2}-n+1$. For $n=10$, $a_{10}=191$.
|
191
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Consider the hyperbola \((x-2)^{2}-\frac{y^{2}}{2}=1\). A line \(l\) is drawn through the right focus of the hyperbola, intersecting the hyperbola at points \(A\) and \(B\). If \(|A B|=4\), then the number of such lines is \(\qquad\).
|
4. 3 lines.
The coordinates of the right focus are $(2+\sqrt{3}, 0)$. The length of the chord perpendicular to the real axis through the right focus is exactly 4. The length of the chord $d$ formed by passing through the right focus and intersecting both branches of the hyperbola has the range $d>2$, so there are two more chords of length 4, making a total of 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.