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 |
|---|---|---|---|---|---|---|---|---|
Example 6 There are two rivers, A and B, each with a flow rate of $300 \mathrm{~m}^{3} / \mathrm{s}$, which converge at a certain point and continuously mix. Their sediment contents are $2 \mathrm{~kg} / \mathrm{m}^{3}$ and $0.2 \mathrm{~kg} / \mathrm{m}^{3}$, respectively. Assume that from the convergence point, there are several observation points along the shore. During the flow between two adjacent observation points, the mixing effect of the two streams is equivalent to the exchange of $100 \mathrm{~m}^{3}$ of water between the two streams in 1 second, i.e., $100 \mathrm{~m}^{3}$ of water flows from stream A to stream B, mixes, and then $100 \mathrm{~m}^{3}$ of water flows from stream B back to stream A, and mixes together. Question: From which observation point onwards will the difference in sediment content between the two streams be less than $0.01 \mathrm{~kg} / \mathrm{m}^{3}$ (ignoring sedimentation)?
(1999, Comprehensive Test for University Admission of Outstanding Students)
|
Solution: Let the sand content of two water flows be $a \, \text{kg} / \text{m}^{3}$ and $b \, \text{kg} / \text{m}^{3}$, and the water volumes flowing through in a unit time be $p \, \text{m}^{3}$ and $q \, \text{m}^{3}$, respectively. Then the sand content after mixing is
$$
\frac{a p + b q}{p + q} \, \text{kg} / \text{m}^{3}.
$$
At the $n$-th observation point, the sand content of water flow $A$ is $a_{n} \, \text{kg} / \text{m}^{3}$, and the sand content of water flow $B$ is $b_{n} \, \text{kg} / \text{m}^{3}$, where $n = 1, 2, \cdots$. Then
$$
\begin{aligned}
a_{1} & = 2, b_{1} = 0.2, \\
b_{n} & = \frac{1}{400} \left(300 b_{n-1} + 100 a_{n-1}\right) \\
& = \frac{1}{4} \left(3 b_{n-1} + a_{n-1}\right). \\
a_{n} & = \frac{1}{300} \left(200 a_{n-1} + 100 b_{n}\right) \\
& = \frac{1}{4} \left(3 a_{n-1} + b_{n-1}\right).
\end{aligned}
$$
Thus, $a_{n} - b_{n} = \frac{1}{2} \left(a_{n-1} - b_{n-1}\right)$
$$
= \cdots = \frac{1}{2^{n-1}} \left(a_{1} - b_{1}\right) = \frac{1}{2^{n-1}}.
$$
$$
2^{n-1} > 180, \quad n \geqslant 9.
$$
From this, we can see that starting from the 9th observation point, the difference in sand content between the two water flows is less than $0.01 \, \text{kg} / \text{m}^{3}$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given 19 drawers arranged in a row, place $n$ identical balls into them such that the number of balls in each drawer does not exceed the number of balls in the drawer to its left. Let the number of such arrangements be denoted by $F_{m, n}$.
(1) Find $F_{1, n}$;
(2) If $F_{m, 0}=1$, prove:
$$
F_{m, n}=\left\{\begin{array}{ll}
F_{n, n}, & (m>n \geqslant 1) \\
F_{m-1, n}+F_{m, n-m} . & (1<m \leqslant n)
\end{array}\right.
$$
(3) Calculate $F_{3,8}$.
(1991, Jiangsu Province Mathematical Summer Camp)
|
Solution: (1) When $m=1$, there is obviously only one way, i.e., $F_{1, n}=1$.
(2) When $m>n \geqslant 1$, for any method that meets the conditions, there will be no balls in all the drawers to the right of the $n$-th drawer, so $F_{m, n}=F_{n, n}$.
When $1$ 0, if we reduce the number of balls in each drawer by 1, we clearly get a method of placing $n-m$ balls into $m$ drawers. Conversely, for any method of placing $n-m$ balls into $m$ drawers, if we increase the number of balls in each drawer by 1, we get a method of placing $n$ balls into $m$ drawers such that $x_{m}>0$. Therefore, $F_{m, n}=F_{m-1, n}+F_{m, n-m}$.
(3) From the result in (2), it is not difficult to deduce that $F_{3,8}=10$.
Example 11 Let $P_{1}$ be a point on the side $AB$ of the equilateral $\triangle ABC$. Draw a perpendicular from $P_{1}$ to the side $BC$, with the foot of the perpendicular being $Q_{1}$. Draw a perpendicular from $Q_{1}$ to the side $CA$, with the foot of the perpendicular being $R_{1}$. Draw a perpendicular from $R_{1}$ to the side $AB$, with the foot of the perpendicular being $P_{2}$, and so on, to get points $Q_{2}, R_{2}, P_{3}, Q_{3}, R_{3}, \cdots$. When $n$ $\rightarrow \infty$, which point does $P_{n}$ approach infinitely?
Solution: As shown in Figure 2, let $B P_{n}=x_{n}, A B=a$, then $B P_{n+1}=x_{n+1}$. Find the relationship between $x_{n}$ and $x_{n+1}$:
$$
\begin{array}{l}
B P_{n}=x_{n} \\
\rightarrow B Q_{n}=\frac{1}{2} x_{n} \\
\rightarrow Q_{n} C=a-\frac{1}{2} x_{n} \\
\rightarrow C R_{n}=\frac{1}{2}\left(a-\frac{1}{2} x_{n}\right) \\
\rightarrow A R_{n}=a-C R_{n}=\frac{1}{2} a+\frac{1}{4} x_{n} \\
\rightarrow A P_{n+1}=\frac{1}{2} A R_{n}=\frac{1}{4} a+\frac{1}{8} x_{n} \\
\rightarrow B P_{n+1}=x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a, \\
\text { hence } x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a .(n \geqslant 1)
\end{array}
$$
Thus, $x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a \cdot(n \geqslant 1)$
It is easy to get $x_{n}=\frac{2}{3} a+\left(x_{1}-\frac{2}{3} a\right)\left(-\frac{1}{8}\right)^{n-1}$
Therefore, $\lim x_{n}=\frac{2}{3} a$, which means that point $P_{n}$ approaches infinitely close to the point that divides $AB$ in the ratio $1:2$.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-$ $a)$, and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
(1999, National Junior High School Mathematics Competition)
|
Solution: According to the characteristics of the required expression, the given equation can be regarded as a quadratic equation in $a$ or $b+c$. Therefore, there are two methods of solution.
Solution One: Transform the given equation into a quadratic equation in $a$
$$
a^{2}-(b+c) a+\frac{1}{4}(b+c)^{2}=0 .
$$
Its discriminant: $\Delta=(b+c)^{2}-4 \times \frac{1}{4}(b+c)^{2}=0$. Therefore, equation (1) has two equal roots $a=\frac{b+c}{2}$. Thus, $\frac{b+c}{a}=2$.
Solution Two: Transform the given equation into a quadratic equation in $(b+c)$
$$
(b+c)^{2}-4 a(b+c)+4 a^{2}=0 .
$$
Its discriminant $\Delta=16 a^{2}-4 \times 4 a^{2}=0$. Therefore, equation (2) has two equal roots $b+c=2 a$. Thus, $\frac{b+c}{a}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $a_{n}$ be the coefficient of the $x$ term in the expansion of $(3-\sqrt{x})^{n}$ $(n=2,3,4, \cdots)$. Then
$$
\lim _{n \rightarrow \infty}\left(\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{n}}{a_{n}}\right)=
$$
$\qquad$
|
8.18
By the binomial theorem, $a_{n}=\mathrm{C}_{n}^{2} \cdot 3^{n-2}$. Therefore,
$$
\begin{array}{l}
\frac{3^{n}}{a_{n}}=\frac{3^{2} \times 2}{n(n-1)}=18\left(\frac{1}{n-1}-\frac{1}{n}\right) . \\
\lim _{n \rightarrow \infty}\left(\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{n}}{a_{n}}\right) \\
=\lim _{n \rightarrow \infty} 18\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n}\right) \\
=\lim _{n \rightarrow \infty} 18\left(1-\frac{1}{n}\right)=18 .
\end{array}
$$
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. If:
(1) $a, b, c, d$ all belong to $\{1,2,3,4\}$;
(2) $a \neq b, b \neq c, c \neq d, d \neq a$;
(3) $a$ is the smallest value among $a, b, c, d$.
Then, the number of different four-digit numbers $\overline{a b c d}$ that can be formed is
$\qquad$
|
12.28.
When $\overline{a b c d}$ has exactly 2 different digits, it can form $C_{4}^{2}=6$ different numbers.
When $\overline{a b c d}$ has exactly 3 different digits, it can form $\mathrm{C}_{3}^{\mathrm{l}} \mathrm{C}_{2}^{\mathrm{d}} \mathrm{C}_{2}^{1}+$ $\mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1}=12+4=16$ different numbers.
When $\overline{a^{2} d d}$ has exactly 4 different digits, it can form $\mathrm{P}_{3}^{3}=6$ different numbers.
Therefore, the total number of numbers that meet the requirements is $6+16+6=28$.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given positive integers $k, m, n$, satisfying $1 \leqslant k \leqslant m \leqslant n$. Try to find
$$
\sum_{i=0}^{n}(-1)^{i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}
$$
and write down the derivation process.
(Xu Yichao, provided)
|
II. The answer to this question is 0. Below, we will prove this combinatorial identity by constructing polynomials and using interpolation and difference methods.
Method 1: Construct the polynomial
$$
\begin{aligned}
f(x) & =\sum_{i=0}^{n} a_{i} x(x+1) \cdots(x+i-1)(x+i+1) \\
& \cdots(x+n)-(x-m-1) \cdots(x-m-n) .
\end{aligned}
$$
We will appropriately choose the coefficients $a_{i}(0 \leqslant i \leqslant n)$ such that
$$
f(x) \equiv 0 .
$$
Notice that $f(x)$ is a polynomial of degree at most $n$, so $f(x) \equiv 0$ if and only if $f(x)$ has $n+1$ roots $0, -1, -2, \cdots, -n$. Therefore, for $0 \leqslant i \leqslant n$, we should have
$$
\begin{aligned}
0= & f(-i) \\
= & a_{i}(-i)(-i+1) \cdots(-i+i-1)(-i+i+ \\
& 1) \cdots(-i+n)-(-i-m-1) \cdots(-i \\
& -m-n) \cdot \\
= & (-1)^{i} i!(n-i)!a_{i}-(-1)^{n} \\
& \cdot \frac{(m+n+i)!}{m+i)!},
\end{aligned}
$$
which implies $a_{i}:=(-1)^{n+i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!}$. Thus, we obtain the algebraic identity
$$
\begin{array}{l}
\sum_{i=0}^{n}(-1)^{n+i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!} x(x+1) \cdots \\
\quad \cdot(x+i-1)(x+i+1) \cdots(x+n) \\
=(x-m-1) \cdots(x-m-n) .
\end{array}
$$
In particular, substituting $x=n+k$ in the above equation, and given $1 \leqslant k \leqslant m \leqslant n$, we have $m+1 \leqslant n+k \leqslant m+n$, so the right-hand side of the equation is 0. Therefore,
$$
\begin{array}{l}
\sum_{i=0}^{n}(-1)^{n+i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!} \\
\cdot k(k+1) \cdots(n+k+i-1)(n+k+i) \\
\cdot(n+k+i+1) \cdots(2 n+k)=0 .
\end{array}
$$
By canceling the factor independent of $i$,
$$
(-1)^{n} k(k+1) \cdots(2 n+k)
$$
we obtain the desired result.
Method 2: Let
$$
g(x)=\frac{(x+m+1)(x+m+2) \cdots(x+m+n)}{x+n+k} .
$$
Given $1 \leqslant k \leqslant m \leqslant n$, we have $m+1 \leqslant n+k \leqslant m+n$. Therefore, $g(x)$ is a polynomial of degree $n-1$.
By the difference formula, the $n$-th difference of $g(x)$ at 0 is
$$
\begin{array}{l}
\Delta^{n} g(0)=\sum_{i=0}^{n}(-1)^{i} \mathrm{C}_{\mathrm{m}}^{i} g(i) \\
=\sum_{i=0}^{n}(-1)^{i} \frac{n!}{i!(n-i)!} \cdot \frac{(m+n+i)!}{(m+i)!} \cdot \frac{1}{n+k+i} .
\end{array}
$$
We know that the $n$-th difference of an $(n-1)$-degree polynomial is always 0, hence
$$
\begin{array}{l}
\sum_{i=0}^{n}(-1)^{i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!} \cdot \frac{1}{n+k+i} \\
=\frac{\Delta^{n} g(0)}{n!}=0 .
\end{array}
$$
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
'. Six, let $n$ be a positive integer, and let the set
$M=\{(x, y) \mid x, y$ are integers, $1 \leqslant x, y \leqslant n\}$.
Define a function $f$ on $M$ with the following properties:
(a) $f(x, y)$ takes values in the set of non-negative integers;
(b) When $1 \leqslant x \leqslant n$, we have $\sum_{y=1}^{n} f(x, y)=n-1$;
(c) If $f\left(x_{1}, y_{1}\right) f\left(x_{2}, y_{2}\right)>0$, then
$\left(x_{1}-x_{2}\right)\left(y_{1}-y_{2}\right) \geqslant 0$.
Determine the number of such functions $f$, denoted by $N(n)$, and find the specific value of $N(4)$.
(Supplied by Yao Jiangan)
|
Six, we first prove the following general lemma.
Lemma In each cell of an $m$-row $n$-column grid, fill in a non-negative integer, the number filled in the cell at the $i$-th row and $j$-th column is denoted by $a_{ij}$, $r_i (1 \leqslant i \leqslant m)$ and $s_j (1 \leqslant j \leqslant n)$ are non-negative integers, satisfying $\sum_{i=1}^{m} r_i = \sum_{j=1}^{n} s_j$. Then, there exists a unique way to fill the numbers that satisfies the following properties:
(1) $\sum_{j=1}^{n} a_{ij} = r_i (1 \leqslant i \leqslant m)$;
(2) $\sum_{i=1}^{m} a_{ij} = s_j (1 \leqslant j \leqslant n)$;
(3) If $a_{ij} a_{kl} > 0$, then $(k-i)(l-j) \geqslant 0$.
Proof of the lemma: We use induction on $m+n$: When $m=n=1$, it is clear that $a_{11} = r_1 = s_1$. The proposition holds.
Assume the proposition holds for grids with the sum of rows and columns less than $m+n$, and consider the filling method for an $m$-row $n$-column grid.
As shown in Figure 4, the numbers in block $A_1$ have the form $a_{1j} (j \geqslant 2)$, and the numbers in block $A_2$ have the form $a_{i1} (i \geqslant 2)$. Note that $(i-1)(1-j) < 0$, so by condition (3), either all numbers in $A_1$ or $A_2$ are zero.
Also, the sum of the numbers in $A_1$ plus $a_{11} = r_1$, and the sum of the numbers in $A_2$ plus $a_{11} = s_1$, so it must be that $a_{11} = \min(r_1, s_1)$. By symmetry, assume $r_1 \leqslant s_1$, then all numbers in $A_1$ are zero.
Consider the sub-grid formed by the last $n-1$ rows of this grid, with the sums of the rows being $r_2, \cdots, r_m$ and the sums of the columns being $s_1 - r_1, s_2, \cdots, s_n$, all of which are non-negative integers, and
$$
r_2 + \cdots + r_m = (s_1 - r_1) + s_2 + \cdots + s_n.
$$
Thus, applying the induction hypothesis to this sub-grid, we know that the filling method for this part is unique.
The numbers in the first row are already determined, with only $a_{11}$ being non-zero, and the inequality $(i-1)(j-1) \geqslant 0$ holds for any $a_{ij}$. Therefore, the filling method for the entire grid also satisfies condition (3), and the proposition is proved.
We now use the lemma to solve the original problem.
The set $M$ can be viewed as an $n$-row $n$-column grid, and each function $f$ on $M$ corresponds to a filling method for the grid. Now, the column sums of the numbers in the grid are given, and the sum of all numbers in the grid is $n(n-1)$. Hence, by the lemma, any selection of $n$ non-negative integers that sum to $n(n-1)$ uniquely corresponds to a function $f$. Therefore, the answer to this problem is the number of non-negative integer solutions to the system of equations
$$
b_1 + b_2 + \cdots + b_n = n(n-1)
$$
which is
$$
N(n) = \mathrm{C}_{n(n-1) + (n-1)}^{n-1} = \mathrm{C}_{n^2 - 1}^{n-1}
$$
In particular, $N(4) = \mathrm{C}_{15}^{3} = 455$.
|
455
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. $A B$ is a chord of $\odot O$, $P$ is a point outside $\odot O$, $P B$ is tangent to $\odot O$ at $B$, $P A$ intersects $\odot O$ at $C$, and $A C = B C$, $P D \perp A B$ at $D$, $E$ is the midpoint of $A B$, $D E = 1000$. Then $P B=$
|
2.2000 .
As shown in Figure 5, take the midpoint $F$ of $PA$, and connect $DF$, $EF$. Then $EF \parallel PB$. Also, $DF = \frac{1}{2} PA$, so $DF = AF$. Since $AC = BC$, $\triangle AFD$ is an isosceles triangle. Therefore, $\angle 2 = \angle 4 = \angle 1$. Since $EF \parallel PB$ and $DF \parallel BC$, $\angle 3 = \angle 5$. From the fact that $PB$ is a tangent, we know $\angle 5 = \angle 2$. Thus, $\angle 3 = \angle 4$, and $DE = L_i$, which means $PB = 2EF = 2DE = 2000$.
|
2000
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Given real numbers $x, y, z$ satisfy $x+y=5$, $z^{2}=x y+y-9$. Then, $x+2 y+3 z=$ $\qquad$
(1995, Zu Chongzhi Cup Junior High School Mathematics Invitational Competition)
|
Solution: The problem provides two equations with three unknowns. Generally speaking, when the number of unknowns exceeds the number of equations, it is impossible to solve for each unknown individually. However, under the condition that $x$, $y$, and $z$ are all real numbers, a solution may still be possible. In fact, from $x+y=5$, we get $x=5-y$, substituting this into the second equation, we have
$$
z^{2}=y(5-y)+y-9=-(y-3)^{2}.
$$
The left side of the equation is a non-negative real number, so the right side must also be a non-negative real number. However, $-(y-3)^{2}$ is a non-positive real number. Therefore, $y-3=0$. Thus, $y=3$. We can then find
$$
\begin{array}{l}
x=5-y=2, z^{2}=0, z=0 . \\
\text { Hence } x+2 y+3 z=2+2 \times 3+3 \times 0=8 .
\end{array}
$$
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a<b<c<d$. If variables $x, y, z, t$ are a permutation of $a, b, c, d$, then the expression
$$
\begin{array}{l}
n(x, y, z, t)=(x-y)^{2}+(y-z)^{2} \\
\quad+(z-t)^{2}+(t-x)^{2}
\end{array}
$$
can take different values.
|
$=、 1.3$.
If we add two more terms $(x-z)^{2}$ and $(y-t)^{2}$ to $n(x, y, z, t)$, then $n(x, y, z, t)+(x-z)^{2}+(y-t)^{2}$ becomes a fully symmetric expression in terms of $x, y, z, t$. Therefore, the different values of $n(x, y, z, t)$ depend only on the different values of $(x-z)^{2}+(y-t)^{2}=\left(x^{2}+y^{2}+z^{2}+t^{2}\right)-2(x z+y t)$. The first term on the right side, $\left(x^{2}+y^{2}+z^{2}+t^{2}\right)$, is also fully symmetric, so $n$ takes different values only depending on $x z+y t$, which has exactly three different values: $a b+c d, a c+b d, a d+b c$. In fact,
$$
\begin{array}{l}
(a b+c d)-(a c+b d)=a(b-c)+d(c-b) \\
=(b-c)(a-d)>0 .
\end{array}
$$
Thus, $a b+c d>a c+b d$.
Similarly, $a c+b d>a d+b c$.
Therefore, $n(x, y, z, t)$ can take three different values.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $a$, $b$, and $c$ be the sides of $\triangle ABC$, and $a^{2}+b^{2}=m c^{2}$. If $\frac{\cot C}{\cot A+\cot B}=999$, then $m$ $=$ $\qquad$
|
2.1999 .
$$
\begin{array}{l}
\because \frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}} \\
\quad=\frac{\sin A \cdot \sin B \cdot \cos C}{\sin C(\cos A \cdot \sin B+\sin A \cdot \cos B)} \\
\quad=\cos C \cdot \frac{\sin A \cdot \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^{2}} \\
\quad=\frac{m-1}{2}=999, \\
\therefore m=2 \times 999+1=1999 .
\end{array}
$$
|
1999
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the positive direction of the $y$-axis at point $B$. Find the number of isosceles right triangles inscribed in the ellipse with point $B$ as the right-angle vertex.
---
Please note that the translation preserves the original format and line breaks.
|
Let the right-angled isosceles triangle inscribed in the ellipse be $\triangle A B C$, and suppose the equation of $A B$ is
\[
\left\{\begin{array}{l}
x=t \cos \alpha, \\
y=b+t \sin \alpha .
\end{array}\right.
\]
(t is a parameter)
Substituting (1) into $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, we get
\[
\left(b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha\right) t^{2}+2 a^{2} b t \sin \alpha=0.
\]
$t=0$ corresponds to point $B$, which we discard. Therefore,
\[
|A B|=-t=\frac{2 a^{2} b \sin \alpha}{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}.
\]
Suppose the equation of $B C$ is
\[
\left\{\begin{array}{l}
x=t \cos \left(\alpha+90^{\circ}\right)=-t \sin \alpha, \\
y=b+t \sin \left(\alpha+90^{\circ}\right)=b+t \cos \alpha .
\end{array}\right.
\]
(t is a parameter)
Substituting (3) into $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, we get
\[
\left(b^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha\right) t^{2}+2 a^{2} b t \cos \alpha=0.
\]
$t=0$ corresponds to point $B$, which we discard. Therefore,
\[
|B C|=-t=\frac{2 a^{2} b \cos \alpha}{b^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha}.
\]
Since $|B A|=|B C|$, from (2) and (4) we have
\[
\frac{2 a^{2} b \sin \alpha}{b^{2} \cdot \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}=\frac{2 a^{2} b \cos \alpha}{b^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha}.
\]
Rearranging the above equation, we get
\[
(\tan \alpha-1)\left[b^{2} \tan ^{2} \alpha+\left(b^{2}-a^{2}\right) \tan \alpha+b^{2}\right]=0.
\]
From $\tan \alpha-1=0$, we have $\alpha=45^{\circ}$, indicating that $\triangle A B C$ is an isosceles right-angled triangle symmetric about the y-axis;
From $b^{2} \tan ^{2} \alpha+\left(b^{2}-a^{2}\right) \tan \alpha+b^{2}=0$, the discriminant $\Delta=\left(b^{2}-a^{2}\right)^{2}-4 b^{4}=\left(a^{2}-3 b^{2}\right)\left(a^{2}+b^{2}\right)$. When $\Delta>0$, i.e., $a>\sqrt{3} b$, there are two such triangles; when $\Delta=0$, i.e., $a^{2}=3 b^{2}$, we have $\tan \alpha=1$.
Therefore, when $a>\sqrt{3} b$, there are two such triangles; when $b<a \leqslant \sqrt{3} b$, there is one such triangle.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 50 points) On a straight ruler of length $36 \mathrm{~cm}$, mark $n$ graduations so that the ruler can measure any integer $\mathrm{cm}$ length in the range $[1,36]$ in one go. Find the minimum value of $n$.
|
Three, if the ruler is marked with 7 (or fewer than 7) graduations, we can prove that it is impossible to measure any integer length in the range [1, 6] cm in a single measurement.
In fact, 7 graduations, including the two end lines of the ruler, total 9 lines, which have 36 different combinations. Therefore, 7 graduations can measure at most 36 different lengths.
If there are two segments of the same length among the 8 segments created by 7 graduations, the number of different lengths that can be measured will be less than 36, clearly failing to meet the measurement requirement.
If none of the 8 segments created by 7 graduations have the same length, note that \(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36\). It is easy to see that at this point, there is one segment of each length from \(1 \mathrm{~cm}\) to \(8 \mathrm{~cm}\).
\(1^{\circ}\) If the segment of length \(1 \mathrm{~cm}\) is not at one end of the ruler, then it cannot measure \(35 \mathrm{~cm}\); if the segment of length \(2 \mathrm{~cm}\) is not at one end of the ruler, then it cannot measure \(34 \mathrm{~cm}\).
\(2^{\circ}\) If the segment of length \(1 \mathrm{~cm}\) is at one end of the ruler and the segment of length \(2 \mathrm{~cm}\) is at the other end, then if the segment of length \(3 \mathrm{~cm}\) is not adjacent to the segment of length \(1 \mathrm{~cm}\), it cannot measure \(32 \mathrm{~cm}\); if the segment of length \(3 \mathrm{~cm}\) is adjacent to the segment of length \(1 \mathrm{~cm}\), it cannot measure \(31 \mathrm{~cm}\).
From this, it is evident that to meet the measurement requirement, the number of graduations on the ruler must be at least 8.
8 graduations divide the ruler into 9 segments. When the lengths of these 9 segments are \(1, 2, 3, 7, 7, 7, 4, 4, 1\), it is known that it can achieve the measurement of any integer length in the range \([1, 36]\) cm in a single measurement.
Therefore, the conclusion of this problem is: the minimum value of \(n\) is 8, meaning that at least 8 graduations must be marked on a 36 cm ruler to possibly achieve (the given graduation distribution is one method) the measurement of any integer length in the range \([1, 36]\) cm in a single measurement.
(Contributor: Liu Yue, Tongling No. 3 Middle School, Anhui Province, 244000)
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given that the bases of two congruent regular triangular pyramids are glued together, exactly forming a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$
(1996, National High School League)
|
Analysis: There are two difficulties in this problem: one is to determine which of $AC$ and $CD$ is 2 in length; the other is to determine which of the line segments $AC$, $CD$, and $AB$ represents the distance between the farthest two vertices.
Solution: As shown in Figure 6, construct $CE \perp AD$, and connect $EF$. It is easy to prove that $EF \perp AD$, so $\angle CEF$ is the plane angle of the dihedral angle formed by plane $ADF$ and plane $ACD$.
Let point $G$ be the midpoint of $CD$. Similarly, $\angle AGB$ is the plane angle of the dihedral angle formed by plane $ACD$ and plane $BCD$. Given that $\angle CEF = \angle AGB$. Let the side length of the base $\triangle CDF$ be $2a$, and the length of the side edge $AC$ be $b$. In $\triangle ACD$, $CE \cdot b = AG \cdot 2a$, thus
$$
CE = \frac{AG \cdot 2a}{b} = \frac{\sqrt{b^2 - a^2} \cdot 2a}{b}.
$$
In $\triangle ABC$, it is easy to find that
$$
\begin{aligned}
AB & = 2 \sqrt{b^2 - \left(\frac{2}{3} \sqrt{3} a\right)^2} \\
& = 2 \sqrt{b^2 - \frac{4}{3} a^2}.
\end{aligned}
$$
From $\triangle CEF \sim \triangle AGB$, we get
$$
\frac{AB}{CF} = \frac{AG}{CE},
$$
which means $\frac{2 \sqrt{b^2 - \frac{4}{3} a^2}}{2a} = \frac{\sqrt{b^2 - a^2}}{\frac{\sqrt{b^2 - a^2} \cdot 2a}{b}}$.
Thus, $b = \frac{4}{3} a = \frac{2}{3} \times 2a$.
Therefore, $AC < CD$.
Given that $b = 2$, we have $2a = 3$, and $AB = 2$.
Thus, $AC = AB < CD$.
Hence, the distance between the farthest two vertices is 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Select $k$ edges and face diagonals from a cube such that any two line segments are skew lines. What is the maximum value of $k$?
|
(提示:考察线段 $A C 、 B C_{1} 、 D_{1} B_{1} 、 A_{1} D$, 它们所在的直线两两都是异面直线. 若存在 5 条或 5 条以上满足条件的线段, 则它们的端点相异, 且不少于 10 个, 这与正方体只有 8 个端点矛盾, 故 $k$ 的最大值是 4.)
(Translation: Consider the line segments $A C, B C_{1}, D_{1} B_{1}, A_{1} D$, the lines on which they lie are pairwise skew lines. If there are 5 or more line segments that satisfy the condition, then their endpoints are distinct and there are at least 10 of them, which contradicts the fact that a cube has only 8 vertices. Therefore, the maximum value of $k$ is 4.)
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.5. Let $M$ be a finite set of numbers. It is known that from any 3 elements of it, two numbers can be found whose sum belongs to $M$. How many elements can $M$ have at most?
|
10.5.7.
An example of a set of numbers consisting of 7 elements is: $\{-3,-2, -1,0,1,2,3\}$.
We will prove that for $m \geqslant 8$, any set of numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ does not have the required property. Without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>\cdots>a_{m}$ and $a_{4}>0$ (since multiplying each number by -1 does not change our property). Thus, $a_{1}+a_{2}>a_{1}+a_{3}>a_{1}+a_{4}>a_{1}$, so the sums $a_{1}+a_{2}, a_{1}+a_{3}, a_{1}+a_{4}$ do not belong to the set $A$. Furthermore, the sums $a_{2}+a_{3}$ and $a_{2}+a_{4}$ cannot both belong to the set $A$, because $a_{2}+a_{3}>a_{2}, a_{2}+a_{4}>a_{2}$, and $a_{2}+a_{3} \neq a_{2}+a_{4}$. Therefore, for the tuples $\left(a_{1}, a_{2}, a_{3}\right)$ and $\left(a_{1}, a_{2}, a_{4}\right)$, at least one of the tuples has the property that the sum of any two of its elements is not an element of $A$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A biologist
observed a chameleon
for a while. During the rest periods, the chameleon can immediately catch a fly. Time:
(1) How many flies did the chameleon catch before its first 9-minute rest?
(2) How many minutes later did the chameleon catch the 98th fly?
(3) After 1999 minutes, how many flies did the chameleon catch in total?
|
Solution: Let the time the chameleon rests before catching the $m$-th fly be $r(m)$. Then $r(1)=1, r(2m)=r(m), r(2m+1)=r(m)+1$. This indicates that $r(m)$ is equal to the number of 1s in the binary representation of the number $m$.
Let $i(m)$ be the moment the chameleon catches the $m$-th fly, and $f(n)$ be the total number of flies the chameleon catches after $n$ minutes. For each natural number $m$, we have:
$$
t(m)=\sum_{i=1}^{m} r(i), f(t(m))=m .
$$
Thus, the following recursive formulas can be derived:
$$
\begin{array}{l}
t(2m+1)=2t(m)+m+1, \\
t(2m)=2t(m)+m-r(m), \\
t\left(2^{p}m\right)=2^{n}t(m)+pm2^{2^{p}}-\left(2^{n}-1\right)r(m) .
\end{array}
$$
The first formula is derived from the following equations:
$$
\begin{array}{l}
\sum_{i=1}^{m} r(2i)=\sum_{i=1}^{m} r(i)=t(m), \\
\sum_{i=0}^{m} r(2i+1)=1+\sum_{i=1}^{m}[r(i)+1] \\
=t(m)+m+1
\end{array}
$$
By adding them together. The second formula is derived from the following equation:
$$
\begin{array}{l}
t(2m)=t(2m+1)-r(2m+1) \\
=2t(m)+m+1-[r(m)+1] \\
=2t(m) \div m-r(m) .
\end{array}
$$
(i) To find $m$ such that $r(m+1)=9$. In binary, the smallest number with 9 ones is $\frac{11 \cdots 1}{9 \uparrow 1}=2^{9}-1=511$. Therefore, $m=510$.
(2) For
$$
\begin{array}{l}
t(98)-2r(49)+49-r(49), \\
t(49)=2t(24)+25 . \\
r(24)=2^{3}t(3)+3 \times 3 \times 2^{2}-\left(2^{3}-1\right)(3), \\
r(1)=r(2)=1, r(3)=2 . \\
r(49)=r\left(110001_{2}\right)=3,
\end{array}
$$
Therefore, $t(3)=4, t(24)=54, t(49)=133$,
$t(98)=312$.
(3) Since $f(n)=m$ if and only if $n \in [t(m), t(m+1)]$, we need to find $m_{0}$ such that $t\left(m_{n}\right) \leqslant 1990 < t\left(m_{0}+1\right)$.
Given
$$
\begin{array}{l}
t\left(2^{p}-1\right)=t\left(2\left(2^{p-1}-1\right)+1\right) \\
=2t\left(2^{p-1}-1\right)+2^{p-1} .
\end{array}
$$
Therefore, $t\left(2^{p}-1\right)=p2^{p-1}$.
And $t\left(2^{p}\right)=2^{\prime \prime}t(1)+p2^{p \prime}-\left(2^{\prime \prime}-1\right)r(1)$
$=p \cdot 2^{p-1}+1$.
$t\left(11 \cdots 100 \cdots 0_{2}\right)=t\left(2^{\prime \prime}\left(2^{q}-1\right)\right)$
$q \uparrow 1 p$ zeros
$$
=2^{p}t\left(2^{q}-1\right)+p\left(2^{q}-1\right)2^{p-1}-\left(2^{p}-1\right)r\left(2^{q}\right.
$$
$$
\begin{array}{l}
=2^{p} \cdot q \cdot 2^{q-1}+p\left(2^{q}-1\right)2^{p}-\left(2^{p}-1\right)q \\
=(p+q)2^{n+q}1-p \cdot 2^{p-1}-q \cdot 2^{p}+q .
\end{array}
$$
From
$$
t\left(2^{8}\right)=8 \times 2^{7}+1<1999<9 \times 2^{8}+1=t\left(2^{9}\right)
$$
We get
$2^{8}<m_{0}<2^{9}$.
Thus, the binary representation of $m_{0}$ has 9 digits. Let $q=3, p=6$ and $q=4, p=5$, then we have
$$
\begin{aligned}
\iota\left(\overline{111000000_{2}}\right) & =9 \times 2^{8}-6 \times 2^{5}-3 \times 2^{6}+3 \\
& =1923, \\
\iota\left(\overline{111100000_{2}}\right) & =9 \times 2^{8}-5 \times 2^{4}-4 \times 2^{5}+4 \\
& =2100 .
\end{aligned}
$$
Therefore, the first 4 digits of the binary representation of $m_{0}$ are 1110.
Since $t\left(111010000_{2}\right)=2004$,
$$
\begin{array}{l}
t\left(\overline{111001111_{2}}\right)=2900, \\
\because\left(11 \overline{100110_{2}}\right)=1992,
\end{array}
$$
Therefore, $f(1999)=\overline{1} \overline{1} 001110_{2}=462$.
|
462
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1.100 people share 1000 RMB, and the money of any 10 people does not exceed 190 RMB. Then, the most one person can have is ( ) RMB.
(A) 109
(B) 109
(C) 118
(D) 119
|
-1 . (B).
Let the person with the most money have $x$ yuan, and the rest are divided into 11 groups, each with 9 people, and this person together with each group of 9 people does not exceed 190 yuan. Therefore,
$$
190 \times 11 \geqslant 1000-x+11 x .
$$
This gives $x \leqslant 109$.
So, one person can have at most 109 yuan, and the rest each have 9 yuan, which meets the requirements.
|
109
|
Inequalities
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the two real roots of $x^{2}-p x+q=0$ be $\alpha, \beta$, and the quadratic equation with $\alpha^{2}, \beta^{2}$ as roots is still $x^{2}-p x+q=0$. Then the number of pairs $(p, q)$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 0
|
5. (B).
From the problem,
$\left\{\begin{array}{l}\alpha+\beta=p, \\ \alpha \beta=q\end{array}\right.$ and $\left\{\begin{array}{l}\alpha^{2}+\beta^{2}=p, \\ \alpha^{2} \beta^{2}=q .\end{array}\right.$
Thus, $q^{2}=q$. Therefore, $q=0$ or $q=1$.
And we have $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$, which means $p^{2}=p+2 q$.
(1) When $q=1$, $p=2$ or $p=-1$;
(2) When $q=0$, $p=0$ or $p=1$.
But $\alpha^{2}+\beta^{2}=p \geqslant 0$, so $p=-1$ (discard).
Therefore, the pairs $(p, q)$ that satisfy the problem can be $(2,1),(0,0)$, $(1,0)$, a total of 3 pairs.
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Real numbers $a, b, c$ are all non-zero, and $a+b+c=$
0. Then
$$
=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)
$$
|
\begin{aligned} \text { II.1. } & -3 . \\ \text { Original expression }= & a\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)+b\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right) \\ & +c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3 \\ = & \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(a+b+c)-3=-3 .\end{aligned}
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 2,
Square $A B C D$ has a side length of $1, E$ is a point on the extension of $C B$, connect $E D$ intersecting $A B$ at $P$, and $P E$ $=\sqrt{3}$. Then the value of $B E-P B$ is $\qquad$
|
4.1.
Let $B E=x, P B=y$, then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=(\sqrt{3})^{2}, \\
\frac{y}{1}=\frac{x}{x+1} .
\end{array}\right.
$$
From (2), we have $x-y=x y$.
From (1) and (3), we have
$$
\begin{array}{l}
(x-y)^{2}+2(x-y)-3=0, \\
(x-y+3)(x-y-1)=0 .
\end{array}
$$
Clearly, $x>y, x-y+3>0$.
Therefore, $x-y-1=0, x-y=1$.
Thus, the value of $B E-P B$ is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $a_{n}=\log _{n}(n+1)$, let $\sum_{n=2}^{1023} \frac{1}{\log _{100} a_{n}}$ $=\frac{q}{p}$, where $p, q$ are integers, and $(p, q)=1$. Then $p+q=(\quad)$.
(A) 3
(B) 1023
(C) 2000
(D) 2001
|
-1 (A).
$$
\begin{array}{l}
\sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\sum_{n=2}^{1023} \log _{100} a_{n} \\
=\log _{100}\left(a_{2} a_{3} \cdots a_{1023}\right) . \\
\text{Since } a_{n}=\log _{n}(n+1)=\frac{\lg (n+1)}{\lg n}, \\
\text{therefore } a_{2} a_{3} \cdots a_{1023}=\frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\lg 3} \cdots \cdots \frac{\lg 1024}{\lg 1023}=\frac{\lg 1024}{\lg 2} \\
=\log _{2} 1024=10 . \\
\text{Hence } \sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\log _{100} 10=\frac{1}{2}, \text{ i.e., } \frac{q}{p}=\frac{1}{2} . \\
\text{Since } (p, q)=1, \\
\text{therefore } p=2, q=1 .
\end{array}
$$
Thus $p+q=3$.
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given that the polynomial $3 x^{3}+a x^{2}+b x+1$ can be divided by $x^{2}+1$, and the quotient is $3 x+1$. Then, the value of $(-a)^{i}$ is $\qquad$
（Dinghe Hesheng Junior High School Math Competition）
|
Solution: According to the polynomial identity, we have
$$
3 x^{3}+a x^{2}+b x+1=\left(x^{2}+1\right)(3 x+1) \text {. }
$$
Taking $x=1$ gives $a+b+4=8$.
Taking $x=-1$ gives $a-b-2=-4$.
Solving these, we get $a=1, b=3$.
$$
\therefore(-a)^{b}=(-1)^{3}=-1 \text {. }
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Height 97 Find the last three digits of $2^{2^{2} 1100}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
$1000=2^{3} \times 5^{3}$ divided by the remainder.
Obviously, $x==1\left(\bmod ^{1} 2^{1}\right)$.
Next, consider the remainder of $x$ divided by $5^{3}$.
By Euler's theorem, $2^{2 / 5^{3}} \equiv \mathrm{I}\left(\bmod 5^{3}\right)$.
That is, $2^{+5^{2}}=1\left(\operatorname{mxl} 55^{3}\right)$.
Thus, $16^{35}=1\left(\bmod 5^{3}\right)$.
From equation (1), to find the remainder of $2^{1 \text { WNF }}$ divided by 25, by the binomial theorem we get
$$
\begin{array}{l}
2^{1} \text { (2)44 }=4^{454}=(5-1)^{4444} \\
\equiv(-1)^{x, 94}+\mathrm{C}_{494}^{1} \times 5 \times(-1)^{495} \\
=-115 \times 999=-115 \times(1000-1) \\
=-6+25 \times 2000=19(\bmod 25) \text {. } \\
\end{array}
$$
Therefore, $16 x=16^{256+24}=16^{211} \times\left(16^{25}\right)^{k}$
$$
\begin{array}{l}
=16^{24} \equiv(25-9)^{211} \equiv(\cdots 9)^{29} \equiv(10-1)^{20} \\
\equiv(-1)^{20}+C_{21}^{1} \times(-1)^{14} \times 10=1-200 \\
\equiv-199 \equiv 250-199 \equiv 51\left(\bmod 5^{3}\right) .
\end{array}
$$
That is, $16 . x=51(\bmod 125)$.
So, $16 x \equiv 51+125 \equiv 176=16 \times 11(\bmod 125)$. Hence $x \equiv 11(\bmod 125)$ (since 16 and 125 are coprime).
Where $k$ represents some positive integer.
From $x \equiv 0(\bmod 8)$ and $x \equiv 11(\bmod 125)$ we get
$$
\left\{\begin{array}{l}
125 x \equiv 0(\bmod 8 \times 125) \\
8 x \equiv 88(\bmod 8 \times 125)
\end{array} .\right.
$$
Therefore, $125 x-16 \times 8 x=0-16 \times 88(\bmod 1000)$,
That is, $-3 . x \equiv-1408(\bmod 1000)$.
$$
3 x \equiv 408 \cong 3 \times 136(\bmod 1000) \text {, }
$$
Hence $x \equiv 136(\bmod 1000)$ (since 3 and 1000 are coprime),
That is, the last three digits of $x=2^{2^{2} \text { (1001 }}$ are 136.
|
136
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Third question: There are $n$ people, and it is known that any two of them make at most one phone call. Any $n-2$ of them have the same total number of phone calls, which is $3^{k}$ times, where $k$ is a natural number. Find all possible values of $n$.
---
The translation maintains the original text's format and line breaks.
|
Solution 1: Clearly, $n \geqslant 5$. Let the $n$ people be $n$ points $A_{1}, A_{2}, \cdots, A_{n}$. If $A_{i}$ and $A_{j}$ make a phone call, then connect $A_{i} A_{j}$. Therefore, there must be line segments among these $n$ points. Without loss of generality, assume there is a line segment between $A_{1} A_{2}$:
If there is no line segment between $A_{1} A_{3}$, consider the $n-2$ points $A_{1} A_{4} A_{5} \cdots A_{n}$, $A_{2} A_{4} A_{5} \cdots A_{n}$, and $A_{3} A_{4} A_{5} \cdots A_{n}$. By the problem's condition, the total number of line segments between $A_{1}$, $A_{2}$, and $A_{3}$ and $A_{4}$, $A_{5}$, $\cdots$, $A_{n}$ are all equal, denoted as $m$.
Adding $A_{2}$ to $A_{1} A_{4} A_{5} \cdots A_{n}$, the total number of line segments among these $n-1$ points is $s=3^{k}+m+1$. Removing any one point from these $n-1$ points, the total number of line segments among the remaining $n-2$ points is $3^{k}$, so each point is connected to the other $n-2$ points by $m+1$ line segments. Thus,
$$
s=\frac{1}{2}(n-1)(m+1).
$$
Adding $A_{3}$ to $A_{1} A_{4} A_{5} \cdots A_{n}$, the total number of line segments among these $n-1$ points is $t=3^{k}+m$. Similarly, we get $t=\frac{1}{2}(n-1) m$.
Therefore, from $s=t+1$ we have
$\frac{1}{2}(n-1)(m+1)=\frac{1}{2}(n-1) m+1$, which implies $n=3$, a contradiction. So there must be a line segment between $A_{1} A_{3}$.
Similarly, there must be a line segment between $A_{2} A_{3}$. Therefore, $A_{1}$ and $A_{2}$ are connected to all $A_{i}(i=3,4, \cdots, n)$.
For $A_{i}$ and $A_{j}(i \neq j)$, since $A_{i}$ is connected to $A_{1}$, similarly $A_{i}$ is connected to $A_{j}$. Therefore, $A_{i}$ and $A_{j}$ $(i, j=1,2, \cdots, n)$ are all connected.
Thus, $3^{k}=\frac{1}{2}(n-2)(n-3)$. Hence, $n=5$.
(Provided by Class 1, Senior 3, Xinhua High School, Tianjin, 300201, Tutor: Wang Hao)
Solution 2: Clearly, $n \geqslant 5$. Let the $n$ people be $A_{1}$, $A_{2}, \cdots, A_{n}$. Let the number of calls made by $A_{i}$ be $d\left(A_{i}\right), i=1,2, \cdots, n$.
If $d\left(A_{1}\right)=0$, among $A_{3}, A_{4}, \cdots, A_{n}$, there must be one person, say $A_{3}$, who has called at least one of $A_{4}$, $A_{5}, \cdots, A_{n}$. Thus, the total number of calls between $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$ and $A_{3}, A_{4}, \cdots, A_{n}$ are not equal, a contradiction. Therefore, $d\left(A_{i}\right) \neq 0(i=1,2, \cdots, n)$.
If $d\left(A_{1}\right) \neq d\left(A_{2}\right)$, assume without loss of generality that $d\left(A_{1}\right)<d\left(A_{2}\right)$. Among $A_{3}, A_{4}, \cdots, A_{n}$, if there is one person, say $A_{3}$, who has called both $A_{1}$ and $A_{2}$ or neither, then the total number of calls between $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$ and $A_{2}, A_{4}, A_{5}, \cdots, A_{n}$ are not equal, a contradiction. If each person in $A_{3}, A_{4}, \cdots, A_{n}$ either calls $A_{1}$ but not $A_{2}$ or calls $A_{2}$ but not $A_{1}$, when all $A_{3}, A_{4}, \cdots, A_{n}$ call $A_{2}$, $A_{1}$ can only call $A_{2}$, so the total number of calls between $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$ is less than that between $A_{2}, A_{4}, \cdots, A_{n}$. When there is one person, say $A_{3}$, who calls $A_{1}$ but not $A_{2}$, the total number of calls between $A_{1}, A_{4}, \cdots, A_{n}$ is still less than that between $A_{2}, A_{4}, A_{5}, \cdots, A_{n}$, a contradiction. Therefore, $d\left(A_{i}\right)=d\left(A_{j}\right), i \neq j, i, j=1,2, \cdots, n$. Let this common value be $d$.
If $1 \leqslant a \leqslant n-2$, then there exist $A_{i}, A_{j}(i \neq j)$ who have not called each other, and there exist $A_{s}, A_{t}(s \neq t)$ who have called each other. The total number of calls among the $n-2$ people excluding $A_{i}, A_{j}$ is $\frac{1}{2} n d-2 d$, and the total number of calls among the $n-2$ people excluding $A_{s}, A_{t}$ is $\frac{1}{2} n d-2 d+1$, a contradiction. Therefore, $d=n-1$.
Thus, $\frac{1}{2} n(n-1)-2(n-1)+1=3^{k}$, i.e., $(n-2)(n-3)=2 \times 3^{k}$. Hence, $n=5$.
(Provided by Zhu Hengjie, Zibo Teaching Research Office, Shandong Province, 255033)
Solution 3: Clearly, $n \geqslant 5$. Since $n$ people can form $\mathrm{C}_{n}^{n-2}=\mathrm{C}_{n}^{2}$ "groups of $n-2$ people", the total number of calls among $n$ people, counted with multiplicity, is $3^{k} C_{n}^{2}$.
If two people make a call, this call is counted once in $\mathrm{C}_{n-2}^{n-4}=\mathrm{C}_{n-2}^{2}$ "groups of $n-2$ people". Therefore, the multiplicity of each call is $\mathrm{C}_{n-2}^{2}$. Hence, the actual total number of calls among $n$ people is $l=\frac{C_{n}^{2} 3^{k}}{\mathrm{C}_{n-2}^{2}}$, where $l$ is a natural number, i.e.,
$$
\frac{(n-2)(n-3)}{2} l=\frac{n(n-1)}{2} 3^{k}.
$$
Since one of $n$ and $n-2$ or $n-1$ and $n-3$ is a multiple of 4 and the other is an even number but not a multiple of 4, $\frac{1}{2}(n-2)(n-3)$ and $\frac{1}{2} n(n-1)$ are one odd and one even.
When $n_{1}$ is odd, $\frac{n_{1}-1}{2}=1, 3 n_{1}-2=7$, $7 \mid 3^{k}$, a contradiction. When $n_{1}$ is even, $n_{1}-1=1$, $\frac{1}{2}\left(3 n_{1}-2\right)=2,2 \mid 3^{k}$, a contradiction. Hence, $d=1$, and we have
$$
\left.\frac{1}{2}(n-2)(n-3) \right\rvert\, 3^{k}.
$$
If $n$ is even, since $\left(\frac{n-2}{2}, n-3\right)=1$, we have $\frac{n-2}{2}=1, n=4$, a contradiction.
If $n$ is odd, we have $\frac{n-3}{2}=1, n=5$.
When $n=5$, the condition is satisfied.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy $a+b=2$. Then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$
(1990, National High School Mathematics Competition)
|
Solution: $\because a, b>0$,
$$
\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1 \text {. }
$$
Thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}}$ $\geqslant 1$.
When $a=b=1$, the above expression $=1$, hence the minimum value is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 When $s$ and $t$ take all real numbers, then the minimum value that $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ can achieve is $\qquad$
(1989, National High School Mathematics Competition)
|
Solution: As shown in Figure 1, the distance squared between any point on the line
$$
\left\{\begin{array}{l}
x=s+5, \\
y=s
\end{array}\right.
$$
and any point on the elliptical arc
$$
\left\{\begin{array}{l}
x=3|\cos t|, \\
y=2|\sin t|
\end{array}\right.
$$
is what we are looking for.
Indeed, the shortest distance is the perpendicular distance from point $A$ to the line, i.e., the distance squared from the point $(-3,0)$ to the line. Its value is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $X=\{1,2, \cdots, 2001\}$. Find the smallest positive integer $m$, such that for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ (where $u$ and $v$ can be the same), such that $u+v$ is a power of 2.
（Supplied by Zhang Zhusheng）
|
To divide $X$ into the following 5 subsets for examination:
$$
\begin{array}{l}
2001=1024+977 \geqslant x \geqslant 1024-977=47, \\
46=32+14 \geqslant x \geqslant 32-14=18, \\
17=16+1 \geqslant x \geqslant 16-1=15, \\
14=8+6 \geqslant x \geqslant 8-6=2, \\
x=1 .
\end{array}
$$
To construct an example that does not satisfy the requirements of the problem and yet contains the maximum number of elements, this subset must not contain any power of 2, and in each pair $\left\{2^{r}+a, 2^{r}-a\right\}$, only one can be included in the set. Let
$$
Y=\{2001,2000, \cdots, 1025\} \cup\{46,45, \cdots, 33\}
$$
$\cup\{17\} \cup\{14,13, \cdots, 9\} \cup\{1\}$,
then $|Y|=998$ and for any $u, v \in Y, u+v$ is not a power of 2.
In fact, when $u, v \in Y$, assume without loss of generality that $u \geqslant v$ and there is $2^{r}<u \leqslant 2^{r}+a<2^{r+1}$, where when $r$ takes the values $10,5,4,3$ respectively, the corresponding $a$ values are $977,14,1,6$.
(1) If $2^{r}<v \leqslant u$, then
$$
2^{r+1}<u+v<2^{r+2} \text {. }
$$
$u+v$ cannot be a power of 2.
(2) If $1 \leqslant v<2^{r}$, then when $2^{r}<u \leqslant 2^{r}+a, 1 \leqslant a<2^{r}$, we have $1 \leqslant v<2^{r}-a$. Thus
$$
2^{r}<u+v<2^{r+1}
$$
This shows that $u+v$ cannot be a power of 2. Therefore, the sum of any two numbers in the subset $Y$ is not a power of 2.
Hence, the smallest positive integer $m \geqslant 999$.
$$
\begin{array}{l}
A_{i}:=\{1524-i, 1024+i\} ; i=1,2, \cdots, 977, \\
B=\{32-j, 32+j\}, j=1,2, \cdots, 14, \\
C=\{15,17\}, \\
D_{k}=\{8-k, 8+k\}, k=1,2, \cdots, 6, \\
E=\{1,8,16,32,1024\} .
\end{array}
$$
For any 999-element subset $W$ of $S$, if $W \cap E \neq \varnothing$, then taking any element from it, its double is a power of 2; if $W \cap E=\varnothing$, then the 999 elements of $W$ belong to the preceding 998 two-element subsets. By the pigeonhole principle, there must be different $u$ and $v$ in $W$ belonging to the same subset. Clearly, $u+v$ is a power of 2.
In summary, the smallest positive integer $m=999$.
|
999
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $a$, $b$, $c$, $a+b-c$, $a+c-b$, $b+c-a$, and $a+b+c$ be 7 distinct prime numbers, and suppose that the sum of two of $a$, $b$, and $c$ is 800. Let $d$ be the difference between the largest and smallest of these 7 prime numbers. Find the maximum possible value of $d$.
(Liang Darong, problem contributor)
|
$$
\begin{array}{l}
\text { Let's assume } a<b<c<d \text {, and } a, b, c, d \text { are all prime numbers. } \\
\therefore c<a+b<a+c<b+c .
\end{array}
$$
Also, since one of $a+b$, $a+c$, and $b+c$ is 800,
$$
\therefore c<800 \text {. }
$$
Since $799=17 \times 47$ and 798 are not prime numbers, but 797 is a prime number, we have
$$
\begin{array}{l}
c \leqslant 797, d \geqslant 1594 . \\
\text { In another case, when } a+b=1300 \text {, note that } \\
a=5, b=795, \\
a=7, b=793=13 \times 61, \\
a=11, b=789=3 \times 263
\end{array}
$$
are not all prime numbers, thus cannot meet the requirements of the problem.
However, when $a=13, b=787$, both are prime numbers, and at this time, $a+b-c=3, a+c-b=23$ are also prime numbers. It is easy to verify that $b+c-a=1571$ and $a+b+c=1597$ are also prime numbers.
In summary, the maximum possible value of $d$ is 1594.
|
1594
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, divide a circle with a circumference of 24 into 24 equal segments, and select 8 points from the 24 points; such that the arc length between any two points is not equal to 3 and 8. How many different ways are there to select such 8-point groups? Explain your reasoning.
（Supplied by Li Chengzhang）
|
Solution 1: Number the 24 points as $1,2, \cdots, 24$, and arrange them in a $3 \times 8$ number table according to their "bad relationship":
$1,4,7,10,13,16,19,22$,
$9,12,15,18,21,24,3, \quad 6$,
$17,20,23, \quad 2, \quad 5, \quad 8, \quad 11, \quad 14$.
It is easy to see that the arc length between two points represented by adjacent numbers in each row is 3, and the arc length between two points represented by adjacent numbers in each column is 8 (the first and last numbers are also considered adjacent). Thus, the requirement for the 8 points in the problem is transformed into the requirement that the numbers of the 8 points selected in the table are not adjacent. Therefore, exactly one number is taken from each column, and at most 4 non-adjacent numbers are taken from each row. Thus, the number of numbers taken from the 3 rows can only be 4 different cases:
$$
\{4,4,0\},\{4,3,1\},\{4,2,2\},\{3,3,2\} \text {. }
$$
(1) $\{4,4,0\}$. Choose one row from the 3 rows to not take any numbers, there are 3 different ways to do this. In the other two rows, the first row takes 4 non-adjacent numbers, there are 2 different ways to do this. The remaining 4 columns are uniquely determined as the columns for the 4 numbers taken from the other row. By the multiplication principle, there are 6 different ways in this case.
(2) $\{4,3,1\}$. The number of numbers taken from the 3 rows are 4, 3, and 1, respectively, there are $3!=6$ different arrangements. In one row, take 4 non-adjacent numbers, there are 2 different ways to do this. In another row and the remaining 4 columns, choose 1 number, there are 4 different ways to do this. Finally, in the third row, choose 1 number from the remaining 3 columns, the selection is uniquely determined. By the multiplication principle, there are $6 \times 2 \times 4=48$ different ways in this case.
(3) $\{4,2,2\}$. Choose one row from the 3 rows to take 4 non-adjacent numbers, there are 3 different ways to choose the row and 2 different ways to take the numbers, a total of 6 different ways. The remaining 4 columns are non-adjacent. From the 4 columns, choose 2 columns for the second row, there are $\mathrm{C}_{4}^{2}=6$ different ways. By the multiplication principle, there are $6 \times 6=36$ different ways in this case.
(4) $\{3,3,2\}$. From the 3 rows, choose one row to take 2 non-adjacent numbers, there are 3 different ways to choose the row and $\mathrm{C}_{7}^{2}-1=20$ different ways to take the numbers (subtract 1 to exclude the case where the two numbers are in the 1st and 8th columns, respectively).
After choosing the two numbers, the remaining 6 columns are divided into two parts. There are 3 different segment scenarios: $\{1,5\},\{2,4\}$, and $\{3,3\}$. The number of different ways for these segments are 8, 8, and 4, respectively. It is easy to see the different ways:
(i) For the $\{1,5\}$ segment, choose 3 non-adjacent columns, there are 8 different ways;
(ii) For the $\{2,4\}$ segment, choose 3 non-adjacent columns, there are 4 different ways;
(iii) For the $\{3,3\}$ segment, choose 3 non-adjacent columns, there are 2 different ways.
Therefore, the number of different ways in this case is
$3 \times\{8 \times 2+8 \times 4+4 \times 2\}=168$.
In summary, the number of different ways that meet the requirements of the problem is
$$
6+48+36+168=258 \text {. }
$$
Solution 2: As in Solution 1, write the 24 numbers in a $3 \times 8$ number table. Thus, when selecting 8 numbers, exactly one number is taken from each column. At this time, taking 1 number from the 1st column, there are 3 different ways. After the 1st column is determined, the number taken from the 2nd column cannot be in the same row as the number taken from the 1st column, so there are only 2 different ways. For each subsequent column, there are 2 different ways, a total of $3 \times 2^{7}$ different ways. But among these, the ways where the number taken from the 1st column and the number taken from the 8th column are in the same row do not meet the requirements.
If we denote the number of different ways to take one number from each column of a $3 \times n$ number table, and any two adjacent columns (including the $n$-th column and the 1st column) do not take numbers from the same row, as $\dot{x}_{n}$, then the above conclusion can be written as
$$
x_{8}+x_{7}=3 \times 2^{7} \text {. }
$$
Similarly, we can get
$$
x_{n}+x_{n-1}=3 \times 2^{n-1} \text {. }
$$
From this recurrence relation, we get
$$
\begin{aligned}
x_{8} & =3 \times 2^{7}-x_{7} \\
& =3 \times 2^{7}-\left(3 \times 2^{6}-x_{6}\right) \\
& =3 \times\left(2^{7}-2^{6}\right)+x_{6} \\
& =\cdots \cdots \\
& =3 \times\left(2^{7}-2^{6}+2^{5}-2^{4}+2^{3}-2^{2}+2\right) \\
& =3 \times 86=258,
\end{aligned}
$$
i.e., the number of different ways that meet the requirements of the problem is 258.
|
258
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Let $x_{1}, x_{2}, \cdots, x_{7}$ all be integers, and
$$
\begin{array}{l}
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}+25 x_{5}+36 x_{6} \\
+49 x_{7}=1, \\
4 x_{1}+9 x_{2}+16 x_{3}+25 x_{4}+36 x_{5}+49 x_{6} \\
+64 x_{7}=12, \\
9 x_{1}+16 x_{2}+25 x_{3}+36 x_{4}+49 x_{5}+64 x_{6} \\
+81 x_{7}=123 .
\end{array}
$$
Find $16 x_{1}+25 x_{2}+36 x_{3}+49 x_{4}+64 x_{5}+$
$$
81 x_{6}+100 x_{7}=
$$
. $\qquad$
|
Solution: Since the coefficients of the same letters in the four equations are the squares of four consecutive natural numbers, i.e., $n^{2}, (n+1)^{2}, (n+2)^{2}, (n+3)^{2}$, to find the value of (4), it is necessary to express (4) using (1), (2), and (3), i.e., to express $(n+3)^{2}$ using $n^{2}, (n+1)^{2}, (n+2)^{2}$. Therefore, we can set
$$
(n+3)^{2}=a(n+2)^{2}+b(n+1)^{2}+c^{2} \text {. }
$$
$$
\text { Let } n=0 \text {, we get } 9=4 a+b \text {; }
$$
Taking $n:=-1$, we get $4=a \div c$;
Taking $n:=\cdots$, we get $1=b+4 c$.
Solving, we get $a=3, b=-3, c=1$.
Thus, $(n+3)^{2}=3(n+2)^{2}-3(n+1)^{2}+n^{2}$.
Therefore, $16 x_{1}+25 x_{2}+36 x_{3}+49 x_{4}+64 x_{5}+$ $81 x_{6}+100 x_{7}$
$$
=3 \times 123-3 \times 12+1=334 \text {. }
$$
|
334
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (Problem 16) The permutation $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ of $1,2,3,4,5$ has the following property: for $1 \leqslant i \leqslant 4$, $a_{1}$, $a_{2} \cdots, a_{i}$ does not form a permutation of $1,2, \cdots, i$. Find the number of such permutations.
|
Obviously, $a_{1} \neq 1$.
When $a_{1}=5$, all $4!$ permutations meet the requirements.
When $a_{1}=4$, the $3!$ permutations where $a_{5}=5$ do not meet the requirements, so the number of permutations that meet the requirements is $(4!-3!)$.
When $a_{1}=3$, permutations in the form of $3 \times \times \times 5$ and $3 \times \times 54$ do not meet the requirements, so the number of permutations that meet the requirements is $(4!-3!-2!)$.
When $a_{1}=2$, permutations in the form of $2 \times \times \times 5, 2 \times \times 54, 2 \times 534, 2 \times 453, 2 \times 543$ do not meet the requirements, (continued on page 43)
(continued from page 36) so the number of permutations that meet the requirements is $(4! - 3! - 2! - 3 \times 1! - 1! - 1!)$.
In summary, the number of permutations that meet the requirements is
$$
\begin{array}{l}
4!+(4!-3!)+(4!-3!-2!)+(4!-3!-2! \\
-3 \times 1!)=71
\end{array}
$$
|
71
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (This question is worth 16 points) How many ordered pairs of positive integers $(x, y)$ have the following properties:
$y<x \leqslant 100$, and $\frac{x}{y}$ and $\frac{x+1}{y+1}$ are both integers?
|
Three, let $\frac{x+1}{y+1}=m(m \in \mathbf{N}, m>1)$, then $x=m y+(m-1)$.
Since $y \mid x$, it follows that $y \mid(m-1)$.
Let $m-1=k y(k \in \mathbf{N})$, then $m=k y+1$,
Substituting into (1), we get
$$
\begin{array}{l}
x=(k y+1) y+k y=k y(y+1)+y \leqslant 100, \\
k \leqslant \frac{100-y}{y(y+1)} .
\end{array}
$$
Thus, there are $\left[\frac{100-y}{y(y+1)}\right]$ positive integers $k$ that satisfy the above inequality.
Conversely, for each positive integer $y$, equation (2) shows a one-to-one correspondence between $k$ and the ordered pair $(x, y)$, and makes
$$
\frac{x}{y}=k(y+1)+1, \frac{x+1}{y+1}=k y+1 .
$$
Notice that, when $10 \leqslant y \leqslant 99$,
$$
y(y+1) \geqslant 110>100-y,\left[\frac{100-y}{y(y+1)}\right]=0,
$$
The number of ordered pairs is
$$
\begin{array}{l}
\sum_{y=1}^{99}\left[\frac{100-y}{y(y+1)}\right]=\sum_{y=1}^{9}\left[\frac{100-y}{y(y+1)}\right] \\
=49+16+8+4+3+2+1+1+1=85 .
\end{array}
$$
|
85
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The ground floor of a hotel has 5 fewer rooms than the second floor. A tour group has 48 people. If all are arranged to stay on the ground floor, each room can accommodate 4 people, but there are not enough rooms; if each room accommodates 5 people, some rooms are not fully occupied. If all are arranged to stay on the second floor, each room can accommodate 3 people, but there are not enough rooms; if each room accommodates 4 people, some rooms are not fully occupied: The ground floor of this hotel has ( ) rooms.
(A) 9
(B) 10
(C) 11
(D) 12
|
5. (B).
Let the number of guest rooms on the ground floor be $x$, then the number of rooms on the second floor is $x+5$. According to the problem, we have the system of inequalities
$$
\left\{\begin{array}{l}
\frac{48}{5}<x<\frac{48}{4}, \\
\frac{48}{4}<x+5<\frac{48}{3} .
\end{array}\right.
$$
which simplifies to $\left\{\begin{array}{l}9.6<x<12, \\ 7<x<11\end{array} \Rightarrow 9.6<x<11\right.$.
Solving this, we get $x=10$.
|
10
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 1, in the acute triangle $\triangle ABC$, the three altitudes $AD$, $BE$, and $CF$ intersect at $H$. Connect $DE$, $EF$, and $FD$. Then the number of triangles in the figure is ( ).
(A) 63
(B) 47
(C) 45
(D) 40
|
6. (B).
The number of triangles with $A$ as a vertex is 13 (excluding $\triangle A B C$); similarly, the number of triangles with $B, C$ as vertices is also 13 each. In total, there are 39.
In $\triangle D E F$, the number of triangles with $D$ as a vertex is 8 (excluding $\triangle D E F$); similarly, the number of triangles with $E, F$ as vertices is also 8 each. In total, there are 24.
Among them, the number of triangles with $A B$ as a side is 3; the number of triangles with $B C, C D$ as sides is also 3 each, totaling 9. In $\triangle D E F$, the number of triangles with $D E, E F, F D$ as sides is 9. These triangles are counted twice and should be subtracted.
Add $\triangle A B C, \triangle D E F$.
The total number of triangles in the figure is
$$
39+24-9 \times 2+2=47 \text { (triangles). }
$$
|
47
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. In space, there are 4 non-coplanar fixed points. The number of parallelepipeds that can be formed with these 4 points as vertices is ( ).
(A) 20
(B) 32
(C) 25
(D) 29
|
6. (D).
Each parallelepiped is uniquely determined by the designated 1 vertex and 3 median planes (each of these median planes is equidistant from all vertices of the parallelepiped). For any given 4 non-coplanar points, there exist 7 planes equidistant to these 4 points. Choosing any 3 out of these 7, there are $\mathrm{C}_{7}^{3}=35$ ways. However, now we need 3 planes that intersect at one point (the center of the parallelepiped), so we must exclude the groups of three planes that are parallel to the same line, and there are 6 such groups. Therefore, the total number of parallelepipeds that can be determined is 29.
|
29
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) A conference was attended by $12 k$ people $(k \in$ $\mathbf{N}$ ), where each person has greeted exactly $3 k+6$ other people. For any two people, the number of people who have greeted both of them is the same. How many people attended the conference?
|
Three, connecting the points of people and their acquaintances, each point forms $\mathrm{C}_{3 k+6}^{2}$ angles, each simple connection corresponds to a vertex, so the graph has a total of $12 k \cdot C_{3 k+6}^{2}$ angles. On the other hand, suppose for any two people, there are $n$ people who know both of them, thus for each pair, there are $n$ angles, each angle uniquely corresponding to a pair of people, and for each pair, there are exactly $n$ angles, from this perspective, the total number of angles is $n \cdot \mathrm{C}_{12 k}^{2}$.
Of course, $n \cdot \mathrm{C}_{12 k}^{2}=12 k \cdot \mathrm{C}_{3 k+6}^{2}$, simplifying gives $9 k^{2}+33 k -12 n k+n+30=0$, where $n, k$ are integers. It is easy to see that $3 \mid n$, let $n=3 m$, further simplifying gives $3 k^{2}+11 k-12 k m+m+10=0$, i.e., $4 m=\frac{k+3+(9 k+43)}{12 k-1}$. When $k \geqslant 15$, $12 k -1>9 k+43, 4 m$ cannot be an integer. For $1 \leqslant k \leqslant 14$, only when $k=3$, $\frac{9 k+43}{12 k-1}$ is an integer. Therefore, it can only be 36 people.
Below is a construction to show that the proposition holds.
\begin{tabular}{|c|c|c|c|c|c|}
\hline$R$ & $O$ & $Y$ & $G$ & $B$ & $V$ \\
\hline$V$ & $R$ & $O$ & $Y$ & $G$ & $B$ \\
\hline$B$ & $V$ & $R$ & $O$ & $Y$ & $G$ \\
\hline$G$ & $B$ & $V$ & $R$ & $O$ & $Y$ \\
\hline$Y$ & $G$ & $B$ & $V$ & $R$ & $O$ \\
\hline$O$ & $Y$ & $G$ & $B$ & $V$ & $R$ \\
\hline
\end{tabular}
In the table, $R, O, Y, G, B, V$ represent 6 colors. Color the people according to the above arrangement, each person knows exactly the people sitting in the same row, the same column, or having the same color as them, thus ensuring that each person knows exactly 15 people. It only remains to prove that this construction satisfies the second condition.
Let $P, Q$ be any two attendees. If they are in the same row, the people who know both of them are the other 4 people in this row, plus the person in the same column as $P$ and the same color as $Q$, and the person in the same column as $Q$ and the same color as $P$. If $P, Q$ are of the same color or in the same column, the analysis is similar. Assuming $P$ and $Q$ are not in the same row, not in the same column, and have different colors, the 6 people they both know are: the person in $P$'s row and $Q$'s column (different in color from both $P$ and $Q$), the person in $P$'s column and $Q$'s row, the 2 people in $P$'s row and column and the same color as $Q$, and the 2 people in $Q$'s row and column and the same color as $P$.
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Cut a wire of length $143 \mathrm{~cm}$ into $n$ small segments $(n \geqslant 3)$, with each segment no less than $1 \mathrm{~cm}$. If no three segments can form a triangle, the maximum value of $n$ is
|
2.1.10.
Each segment should be as small as possible, and the sum of any two segments
should not exceed the third segment. A $143 \mathrm{~cm}$ wire should be divided into 10
such segments:
$1 \mathrm{~cm}, 1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}, 8 \mathrm{~cm}$,
$13 \mathrm{~cm}, 21 \mathrm{~cm}, 34 \mathrm{~cm}, 55 \mathrm{~cm}$
This is acceptable.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a$, $b$, $c$, $d$ are the thousands, hundreds, tens, and units digits of a four-digit number, respectively, and the digits in lower positions are not less than those in higher positions. When $|a-b|+|b-c|+|c-d|+|d-a|$ takes the maximum value, the maximum value of this four-digit number is $\qquad$ .
|
$$
\begin{array}{l}
\text { 3. } 1999 . \\
\because a \leqslant b \leqslant c \leqslant d, \\
\therefore a-b \leqslant 0, b-c \leqslant 0, c-d \leqslant 0, d-a \geqslant 0 . \\
\therefore|a-b|+|b-c|+|c-d|+|d-a| \\
=b-a+c-b+d-c+d-a \\
=2(d-a) .
\end{array}
$$
When $d=9, a=1$,
$$
|a-b|+|b-c|+|c-d|+|d-a|
$$
has a maximum value of 16. At this time, $b$ and $c$ can be any numbers.
Therefore, the maximum value of this four-digit number is 1999.
$$
|
1999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A monkey is climbing an 8-rung ladder, each time it can climb one rung or jump two rungs, and at most jump three rungs. From the ground to the top rung, there are $\qquad$ different ways to climb and jump.
|
4.81.
Starting with simple cases:
(1) If there is 1 step, then there is only one way to climb. That is, $a_{1}=1$.
(2) If there are 2 steps, then there are 2 ways to climb:
(1) Climb one step at a time;
(2) Leap two steps at once, i.e., $a_{2}=2$.
(3) If there are 3 steps, then there are 4 ways to climb:
(1) Climb one step at a time;
(2) Climb one step the first time and leap two steps the second time;
(3) Leap two steps the first time and climb one step the second time;
(4) Leap three steps at once, i.e., $a_{3}=4$.
(4) If there are 4 steps, then we can categorize the ways based on the number of steps taken the first time.
(1) Climb one step the first time, leaving three steps, which according to the previous results, has $a_{3}=4$ ways;
(2) Leap two steps the first time, leaving two steps, which according to the previous results, has $a_{2}=2$ ways;
(3) Leap three steps the first time, leaving one step, which has $a_{1}=1$ way.
Thus, $a_{4}=a_{3}+a_{2}+a_{1}=4+2+1=7$ (ways).
Similarly, $a_{5}=a_{4}+a_{3}+a_{2}=13$ (ways);
$a_{6}=a_{5}+a_{4}+a_{3}=24$ (ways);
$a_{7}=a_{6}+a_{5}+a_{4}=44$ (ways);
$a_{8}=a_{7}+a_{6}+a_{5}=81$ (ways).
Therefore, there are 81 ways to climb.
|
81
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, and $x_{n+1}=$ $\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$, then $x_{1999}-x_{601}=$ $\qquad$ .
|
3.0 .
From $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$ we get $x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$.
Let $x_{n}=\tan \alpha_{n}$, then
$$
x_{n+1}=\tan \alpha_{n+1}=\tan \left(\alpha_{n}+\frac{\pi}{6}\right) \text {. }
$$
Therefore, $x_{n+6}=x_{n}$,
which means the sequence $\left\{x_{n}\right\}$ is a periodic sequence.
From $1999=333 \times 6+1,601=100 \times 6+1$ we get
$$
x_{1999}-x_{601}=0 \text {. }
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the surface area and volume of a circular cone are divided into upper and lower parts by a plane parallel to the base in the ratio $k$, then the minimum value of $c$ that makes $k c>1$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
4.7.
As shown in Figure 3, let $C O_{1}=r_{1}, A O=r_{2}$, the surface area of the smaller cone is $S_{1}$. The lateral surface area of the frustum is $S_{2}$, then
$$
\frac{S_{2}+S_{\text {base }}}{S_{1}}=\frac{1}{k},
$$
which means
$$
\frac{S_{1}}{S_{\text {lateral }}+S_{\text {base }}}=\frac{k}{k+1} \text {. }
$$
Given $S_{1}=\frac{\pi r_{1}^{2}}{\sin \theta}, S_{\text {lateral }}=\frac{\pi r_{2}^{2}}{\sin \theta}, S_{\text {base }}=\pi r_{2}^{2}$, we have
$$
\frac{k}{k+1}=\frac{r_{1}^{2}}{r_{2}^{2}} \cdot \frac{1}{1+\sin \theta} \text {. }
$$
Also, $\frac{V_{\text {frustum }}}{V_{\text {small cone }}}=\frac{1}{k}$, then
$$
\frac{V_{\text {cone }}}{V_{\text {small cone }}}=\frac{1+k}{k} \text {, }
$$
which means $\frac{r_{1}^{3}}{r_{2}^{3}}=\frac{k}{k+1}$.
Thus,
$$
\begin{array}{l}
\left(\frac{k}{k+1}(1+\sin \theta)\right)^{\frac{1}{2}} \\
=\frac{r_{1}}{r_{2}}=\left(\frac{\dot{k}}{k+\dot{\mathrm{i}}}\right)^{\frac{1}{3}} .
\end{array}
$$
Therefore, $\frac{1}{k}=(1+\sin \theta)^{3}-1\frac{1}{k}=(1+\sin \theta)^{3}-1$.
Hence, the minimum value of $c$ is 7.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6.12 friends have a weekly dinner together, each week they are divided into three groups, each group 4 people, and different groups sit at different tables. If it is required that any two of these friends sit at the same table at least once, then at least how many weeks are needed.
|
6.5.
First, for any individual, sitting with 3 different people each week, it would take at least 4 weeks.
Second, there are $C_{12}^{2}=66$ pairs among 12 people. Each table has $\mathrm{C}_{4}^{2}=6$ pairs, so in the first week, $3 \times 6=18$ pairs get to know each other.
Since 4 people sit at 3 tables, after the first week, at least two people at each table must have sat together in the first week. Therefore, the maximum number of new pairs that can form each week is $6-1=5$ pairs per table, totaling 15 pairs.
Since $18+15+15+15=63$, it is impossible to have all 66 pairs sit together in 4 weeks, so it must take 5 weeks.
Using 5 weeks is feasible. For example, 18 pairs in the first week, and 12 pairs each in the remaining 4 weeks, totaling $18+12 \times 4=66$ pairs. Below is a specific seating arrangement:
\begin{tabular}{crrrrrrrrrrrrrr}
Week & & \multicolumn{5}{c}{ Table 1} & & & & Table 2 & & & \multicolumn{5}{c}{ Table 3} & \\
1 & 1 & 2 & 3 & 4 & 5 & 6 & 9 & 10 & 7 & 8 & 11 & 12 \\
2 & 1 & 2 & 5 & 6 & 3 & 4 & 7 & 8 & 9 & 10 & 11 & 12 \\
3 & 1 & 2 & 7 & 8 & 3 & 4 & 9 & 10 & 5 & 6 & 11 & 12 \\
4 & 1 & 2 & 9 & 10 & 3 & 4 & 11 & 12 & 5 & 6 & 7 & 8 \\
5 & 1 & 2 & 11 & 12 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10
\end{tabular}
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Arrange all powers of 3 and the sums of distinct powers of 3 in an increasing sequence:
$$
1,3,4,9,10,12,13, \cdots \text {. }
$$
Find the 100th term of this sequence.
|
Analysis: Write the known sequence in the form of the sum of powers of 3:
$$
\begin{array}{l}
a_{1}=3^{0}, a_{2}=3^{1}, a_{3}=3^{1}+3^{0}, \\
a_{4}=3^{2}, a_{5}=3^{2}+3^{0}, a_{6}=3^{2}+3^{1}, \\
a_{7}=3^{2}+3^{1}+3^{0}, \cdots
\end{array}
$$
It is easy to find that the terms correspond exactly to the binary representation of the natural number sequence:
$$
\begin{array}{l}
1=2^{0}, 2=2^{1}, 3=2^{1}+2^{0}, \\
4=2^{2}, 5=2^{2}+2^{0}, 6=2^{2}+2^{1}, \\
7=2^{2}+2^{1}+2^{0}, \cdots
\end{array}
$$
Since $100=(1100100)_{2}=2^{6}+2^{5}+2^{2}$,
the 100th term of the original sequence is
$$
3^{6}+3^{5}+3^{2}=981 \text {. }
$$
Explanation: This example demonstrates the superiority of base $p$ $(p \neq 0)$. Generally, each base system has its strengths and weaknesses. Depending on the specific problem, flexibly choosing the base system often yields surprising results. Please see another example.
|
981
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Find the smallest positive integer $n$, such that every $n$-element subset of $S=\{1,2, \cdots, 150\}$ contains 4 pairwise coprime numbers (it is known that $S$ contains a total of 35 prime numbers).
|
Solution: Consider the number of multiples of 2, 3, or 5 in $S$. We have
$$
\begin{array}{l}
{\left[\frac{150}{2}\right]+\left[\frac{150}{3}\right]+\left[\frac{150}{5}\right]-\left[\frac{150}{2 \times 3}\right]-\left[\frac{150}{2 \times 5}\right]} \\
-\left[\frac{150}{3 \times 5}\right] +\left[\frac{150}{2 \times 3 \times 5}\right]=110 .
\end{array}
$$
When $n=110$, all numbers can be multiples of 2, 3, or 5, so in this subset, it is impossible to find 4 numbers that are pairwise coprime. Therefore, $n \geqslant 111$.
Next, we prove that $n=111$ satisfies the requirement.
To do this, construct the following 6 sets:
$$
\begin{array}{l}
A_{1}=\{1\} \cup\{35 \text{ primes in } S\}, \\
A_{2}=\{2 \times 67,3 \times 43,5 \times 17,7 \times 19,11 \times 13\}, \\
A_{3}=\{2 \times 53,3 \times 47,5 \times 23,7 \times 17\} . \\
A_{4}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}\right\}, \\
A_{5}=\left\{2 \times 19,3^{3}, 5 \times 13,7 \times 11\right\}, \\
A_{6}=\left\{2^{3}, 3 \times 23,5 \times 11,7 \times 13\right\} . \\
\text{Let } A=A_{1} \cup A_{2} \cup A_{3} \cup \cdots \cup A_{6} .
\end{array}
$$
Each of the above $A_{i}(i=1,2, \cdots, 6)$ must contain at least 4 elements that are pairwise coprime, and $A_{i} \cap A_{j}=\varnothing(1 \leqslant i<j \leqslant 6)$, to prove the conclusion of this problem, it is necessary to ensure that $|A|=58$.
Thus, if 111 numbers are selected from $S$, at least 19 numbers must be selected from $A$, and by the pigeonhole principle, there must be some $A_{i}(i=1,2, \cdots, 6)$ from which 4 numbers are selected, and these four numbers are pairwise coprime.
(Yuan Jin, Anhui Normal University Affiliated High School, 241001)
|
111
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Does there exist a prime number that remains prime when 16 and 20 are added to it? If so, can the number of such primes be determined?
|
Solution: Testing with prime numbers $2, 3, 5, 7, 11, 13 \cdots$, we find that 3 meets the requirement. Below, we use the elimination method to prove that apart from 3, no other prime number satisfies the requirement.
Divide the natural numbers into three categories: $3n, 3n+1, 3n+2$ $(n \in \mathbb{N})$,
$\because 2n+1+20 = 2(n+7)$ is a composite number,
$3r+2+16 = 3(n+6)$ is a composite number,
thus numbers of the form $3n+1$ and $3n+2$ do not become prime numbers when 16 and 20 are added.
Among the remaining numbers of the form $3n$, it is clear that except for 3, when $n>1$, they are all composite numbers.
Therefore, the only prime number that meets the condition is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 What is the largest even number that cannot be written as the sum of two odd composite numbers?
The largest even number that cannot be written as the sum of two odd composite numbers is 38.
Note: The original text did not provide the answer, this is based on known mathematical results. However, if you need the exact answer as per the original text, please clarify.
(Translation of the provided text only, without additional content)
Example 7 What is the largest even number that cannot be written as the sum of two odd composite numbers?
|
Solution: First, prove that even numbers $n$ greater than 38 can all be written as the sum of two odd composite numbers.
Obviously, if the unit digit of $n$ is 0, then $n=5k+15$; if the unit digit of $n$ is 2, then $n=5k+27$; if the unit digit of $n$ is 4, then $n=5k+9$; if the unit digit of $n$ is 6, then $n=5k+21$; if the unit digit of $n$ is 8, then $n=5k+33$, where $k$ is an integer greater than 2.
Next, find the largest even number that meets the condition.
Among the even numbers less than 40, remove those that can be expressed as the sum of two odd composite numbers.
For numbers with a unit digit of 0, it is easy to verify that only 10 and 20 are qualified.
Similarly, for numbers with a unit digit of 2, 2, 12, 22, and 32 are all qualified; for numbers with a unit digit of 4, only 4 and 14 are qualified; for numbers with a unit digit of 6, only 6, 16, and 26 are qualified; for numbers with a unit digit of 8, 8, 18, 28, and 38 are all qualified.
After elimination, the largest even number less than 40 that meets the condition is 38.
$$
\begin{aligned}
38 & =1+37=3+35=5+33=7+31 \\
& =9+29=11+27=13+25 \\
& =15+23=17+21=19+19
\end{aligned}
$$
None of these are the sum of two odd composite numbers.
None of these are the sum of two odd composite numbers.
|
38
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given $a_{1}=1, a_{2}=\frac{5}{2}, a_{n+1}=$ $\frac{a_{n}+b_{n}}{2}, b_{n+1}=\frac{2 a_{n} b_{n}}{a_{n}+b_{n}}$. Prove:
$$
\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .
$$
|
Proof: From the characteristic of the simultaneous recurrence relations, we know that $\left\{a_{n}\right\}$ is an arithmetic mean sequence, and $\left\{b_{n}\right\}$ is a harmonic mean sequence. By considering the inequality of means, we can insert the geometric mean between $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$. Therefore, we construct the auxiliary sequence $c_{n}=a_{n} b_{n}$.
From the simultaneous recurrence relations, we have
$$
\begin{aligned}
c_{n+1} & =a_{n+1} b_{n+1}=\frac{a_{n}+b_{n}}{2} \cdot \frac{2 a_{n} b_{n}}{a_{n}+b_{n}} \\
& =a_{n} b_{n}=c_{n} .
\end{aligned}
$$
$\therefore\left\{c_{n}\right\}$ is a constant sequence, with $c_{n}=c_{1}=a_{1} b_{1}$. From the initial conditions,
$$
\frac{5}{2}=a_{2}=\frac{a_{1}+b_{1}}{2}=\frac{1+b_{1}}{2},
$$
we get $b_{1}=4, a_{1}=1$.
$$
\begin{array}{l}
\therefore c_{n}=4, b_{n}=\frac{c_{n}}{a_{n}}=\frac{4}{a_{n}}, a_{n+1}=\frac{a_{n}^{2}+4}{2 a_{n}} . \\
\therefore \frac{a_{n+1}+2}{a_{n+1}-2}=\left(\frac{a_{n}+2}{a_{n}-2}\right)^{2}=\left(\frac{a_{n-1}+2}{a_{n-1}-2}\right)^{2^{2}} \\
=\cdots=\left(\frac{a_{1}+2}{a_{1}-2}\right)^{2^{n}}=\left(\frac{1+2}{1-2}\right)^{2^{n}}=3^{2^{n}} .
\end{array}
$$
Thus, $a_{n}+2=3^{2^{n-1}}\left(a_{n}-2\right)$.
Solving for $a_{n}$, we get $a_{n}=2 \frac{3^{2^{n-1}}+1}{3^{2^{n-1}}-1}$,
$$
\begin{array}{l}
b_{n}=\frac{4}{a_{n}}=2 \frac{3^{2^{n-1}}-1}{3^{2^{n-1}}+1} . \\
\therefore \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} 2 \frac{3^{2^{n-1}}+1}{3^{2^{n-1}}-1}=2, \\
\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} 2 \frac{3^{2^{n-1}}-1}{3^{2^{n-1}}+1}=2 .
\end{array}
$$
Therefore, $\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}$.
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $x$ and $y$ are positive integers, and $xy + x + y$ $=23, x^2y + xy^2=120$. Then $x^2 + y^2=$ $\qquad$
|
3.34 .
Let $x+y=s, xy=t$. Then $s+t=23, st=120$. We get $s=8, t=15$ or $s=15, t=8$ (discard). $x^{2}+y^{2}=(x+y)^{2}-2xy=s^{2}-2t=34$.
|
34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A positive integer, if added to 100 and 168 respectively, can result in two perfect squares. This positive integer is $\qquad$
|
4. $n=156$.
Let this number be $n$, and $n+168=a^{2}, n+100=b^{2}$. Then $a^{2}-b^{2}=68=2^{2} \times 17$,
which means $(a+b)(a-b)=2^{2} \times 17$.
However, $a+b$ and $a-b$ have the same parity,
so $a+b=34, a-b=2$.
Thus, $a=18$, and hence $n=156$.
|
156
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
As shown in Figure $3, D$ and $E$ are points on side $BC$ of $\triangle ABC$, and $F$ is a point on the extension of $BC$. $\angle DAE = \angle CAF$.
(1) Determine the positional relationship between the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$, and prove your conclusion.
(2) If the radius of the circumcircle of $\triangle ABD$ is twice the radius of the circumcircle of $\triangle AEC$, and $BC=6, AB=4$, find the length of $BE$.
|
Three, (1) Two circles are externally tangent.
Draw the tangent line $l$ of $\odot A B D$, then $\angle 1=\angle B$.
$$
\begin{array}{l}
\because \angle 3=\angle B+\angle C, \\
\therefore \angle 3=\angle 1+\angle C . \\
\because \angle 1+\angle 2=\angle 3=\angle 1+\angle C, \\
\therefore \angle 2=\angle C .
\end{array}
$$
Draw $A P \perp l$ through $A$, intersecting $\odot A E C$ at point $P$, and connect $P E$.
$\because \angle P=\angle A C E$, then $\angle 2=\angle P$.
$$
\therefore \angle P A E+\angle P=90^{\circ} \text {. }
$$
Thus, $\angle A E P=90^{\circ}$, so $A P$ is the tangent line of $\odot A E C$, i.e., the circle is tangent at point $A$.
(2) Extend $D A$ to intersect $\odot A E C$ at $G$ (assuming $F$ is on $\odot A E C$), and connect $(S F$.
$$
\text { Since } \angle 4=\angle D A E+\angle A E D=\angle 3+\angle A F C \text {, }
$$
we have $\angle 4+\angle 5=180^{\circ}$, then $\angle 4=\angle A G F$.
$\therefore \triangle A D B \backsim \triangle A G F$.
$\therefore A B: A F=2$ (i.e., equal to the ratio of the radii of the two circles).
But $A B=4$,
$\therefore A F=2$. (This can also be done using the sine rule)
$\because B A \cdot B F=B E \cdot B C$,
$\therefore B E=4$.
(Provided by Yu Qiao)
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $m$ and $n$ are integers, the equation
$$
x^{2}+(n-2) \sqrt{n-1} x+m+18=0
$$
has two distinct real roots, and the equation
$$
x^{2}-(n-6) \sqrt{n-1} x+m-37=0
$$
has two equal real roots. Find the minimum value of $n$, and explain the reasoning.
|
1. $\left\{\begin{array}{l}n \geqslant 1, \\ (n-2)^{2}(n-1)-4(m+18)>0, \\ (n-6)^{2}(n-1)-4(m-37)=0 .\end{array}\right.$
(2) - (3), and rearrange to get
$$
(n-4)(n-1)>27.5 \text {, }
$$
then $n \geqslant 8$.
When $n=8$, $m=44$, which satisfies the given conditions. Therefore, the minimum value of $x$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $f(x)=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x$ is an even function, $g(x)=2^{x}+\frac{a+b}{2^{x}}$ is an odd function, where $a 、 b \in \mathbf{C}$. Then $a+a^{2}+a^{3}+\cdots+a^{2000}+b+b^{2}$ $+b^{3}+\cdots+b^{2000}=$
|
3. -2 .
From the known $f(-x)=f(x), g(-x)=-g(x)$ we get
$$
\begin{array}{l}
\log _{\frac{1}{3}}\left(3^{-x}+1\right)-\frac{1}{2} a b x \\
=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x . \\
2^{-x}+\frac{a+b}{2^{-x}}=-\left(2^{x}+\frac{a+b}{2^{x}}\right) .
\end{array}
$$
Simplifying, from equation (1) we get $a b=1$, and from equation (2) we get $a+b=-1$. Therefore, $a$ and $b$ are a pair of conjugate complex roots of the equation $x^{2}+x+1=0$, which are the imaginary cube roots of 1. Then
$$
\begin{aligned}
a & +a^{2}+a^{3}+\cdots+a^{1999}+a^{2000}+b+b^{2} \\
& +b^{3}+\cdots+b^{1999}+b^{2000} \\
= & a^{1999}+a^{2000}+b^{1999}+b^{2000} . \\
= & a+a^{2}+b+b^{2}=-1-1=-2 .
\end{aligned}
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Find the smallest positive integer $n$ that has exactly 144 different positive divisors, and among them, there are 10 consecutive integers.
(26th IMO Shortlist)
|
Analysis: According to the problem, $n$ is a multiple of the least common multiple of 10 consecutive integers, and thus must be divisible by $2, 3, \cdots, 10$. Since $8=2^{3} ; 9=3^{2}, 10=2 \times 5$, its standard factorization must contain at least the factors $2^{3} \times 3^{2} \times 5 \times 7$. Therefore, we can set
$$
n=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \cdots \cdot p_{s}^{\alpha_{5}} .
$$
where $\alpha_{1} \geqslant 3, \alpha_{2} \geqslant 2, \alpha_{3} \geqslant 1, \alpha_{4} \geqslant 1$.
Since the number of divisors of $n$ is
$$
\begin{aligned}
d(n) & =\left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{s}\right) \\
& =144,
\end{aligned}
$$
and $\left(1+\alpha_{i}\right)\left(1+\alpha_{2}\right)\left(1+\alpha_{3}\right)\left(1+\alpha_{4}\right)$
$$
\geq(i+3)(1+2)(1+1)(1+1)=48 \text {, }
$$
there must be another $\alpha_{j}>0(j \geqslant 5)$, and $\alpha_{j} \leqslant$
2. To make $n$ as small as possible, it is advisable to take $0 \leqslant \alpha_{5} \leqslant 2$. By
$$
\begin{array}{l}
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right)\left(\alpha_{5}+1\right) \\
=144\left(\alpha_{5} \neq 0 \text { when }\right)
\end{array}
$$
or $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right)\left(\alpha_{5}+1\right)$
$$
=144 \text { (when } \alpha_{5}=0 \text {). }
$$
Considering the possible factorizations of $144=2^{4} \times 3^{2}$, and comparing the corresponding sizes of $n$, we find that the smallest $n$ is
$$
n=2^{5} \times 3^{2} \times 5 \times 7 \times 11=110880 \text {. }
$$
|
110880
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 1, if
the three sides of $\triangle A B C$ are
$n+x, n+2 x, n$
$+3 x$, and the height $A D$ from $B C$ is $n$, where $n$
is a positive integer, and $0<x \leqslant$
1. Then the number of triangles that satisfy the above conditions is $\qquad$.
|
6.12 .
Let $a=n+x, b=n+2 x, c=n+3 x, s=\frac{1}{2}(a+b+c)$. By Heron's formula and the general area formula, we have
$$
\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2} n(n+2 x) .
$$
Simplifying, we get $12 x=n$.
And $0<x=\frac{n}{12} \leqslant 1$,
so $0<n \leqslant 12$,
which means $n$ has exactly 12 possibilities (correspondingly giving $x$ values).
Therefore, the number of triangles sought is 12 .
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the sum of positive integers $a$ and $b$ is $n$, then $n$ can be transformed into $a b$. Can this method be used several times to change 22 into 2001?
|
1. Reverse calculation, $2001=3 \times 667$ from $3+667=670$; $670=10 \times 67$ from $10+67=77$; $77=7 \times 11$ from $7+11=18$. From any $n=1+(n-1)$, we can get $n-1=1 \times(n-1)$. Therefore, starting from 22, we can sequentially get 21, $20,19,18,77,670$ and 2001.
|
2001
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Three-digit number $\overline{a b c}=a^{2}+1+(\overline{b c})^{2}$. Then $\overline{a b c}=$
|
3.726 .
From $100 a+\overline{b c}=a^{2}+(\overline{b c})^{2}+1$, we have $(\overline{b c}-1) \overline{b c}=100 a-a^{2}-1$.
Obviously, the left side is even,
so the right side $a$ must be odd to make both sides equal.
When $a=1$, $(\overline{b c}-1) \overline{b c}=98=2 \times 7^{2}$;
When $a=3$, $(\overline{b c}-1) \overline{b c}=290=2 \times 5 \times 29$;
When $a=5$, $(\overline{b c}-1) \overline{b c}=474=2 \times 3 \times 79$;
When $a=7$, $(\overline{b c}-1) \overline{b c}=650=26 \times(26-1)$;
When $a=9$, $(\overline{b c}-1) \overline{b c}=818=2 \times 409$.
Only when $a=7$, $\overline{b c}=26$ satisfies.
Therefore, $\overline{a b c}=726$.
|
726
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $x, y \in \mathbf{R}$. If $2 x, 1, y-1$ form an arithmetic sequence, and $y+3,|x+1|+|x-1|, \cos (\arccos x)$ form a geometric sequence, then the value of $(x+1)(y+1)$ is $\qquad$ .
|
2.4.
Given that $2 x, 1, y-1$ form an arithmetic sequence, we have
$$
y=3-2 x \text{. }
$$
From $\cos (\arccos x)=x$ and $-1 \leqslant x \leqslant 1$, it follows that
$$
|x+1|+|x-1|=2 \text{. }
$$
Thus, $y+3, 2, x$ form a geometric sequence, which gives
$$
x(y+3)=4 \text{. }
$$
Substituting (1) into (2), we get $2 x^{2}-6 x+4=0$.
Solving this, we find $x_{1}=2$ (discard) and $x_{2}=1$. Hence, $y=1$.
Therefore, $(x+1)(y+1)=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $a, b$ be two positive numbers, and $a>b$. Points $P, Q$ are on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the line connecting point $A(-a, 0)$ and $Q$ is parallel to the line $O P$, and intersects the $y$-axis at point $R$, where $O$ is the origin, then $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=$ $\qquad$ .
|
5.2.
Let $A Q:\left\{\begin{array}{l}x=-a+t \cos \theta \\ y=t \sin \theta\end{array}\right.$ ( $t$ is a parameter ).
Substitute (1) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$t=\frac{2 a b^{2} \cos \theta}{b^{2} \cos ^{2} \theta + a^{2} \sin ^{2} \theta}$.
Then $|A Q|=\frac{2 a b^{2}|\cos \theta|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$.
In (1), let $x=0$, we get $t=\frac{a}{\cos \theta}$.
Then $|A R|=\frac{a}{|\cos \theta|}$.
Also, let $O P:\left\{\begin{array}{l}x=t^{\prime} \cos \theta \\ y=t^{\prime} \sin \theta\end{array}\right.$ ( $t^{\prime}$ is a parameter).
Substitute (2) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$t^{\prime 2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$.
Then $|O P|^{2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$.
Therefore, $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=\ldots=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Draw 63 lines on the coordinate plane: $y=b, y=\sqrt{3} x + 2b, y=-\sqrt{3} x+2b$, where $b=-10,-9,-8, \ldots, 8,9,10$. These lines divide the plane into several equilateral triangles. The number of equilateral triangles with side length $\frac{2}{\sqrt{3}}$ is $\qquad$ $-$
|
6.660.
The six outermost straight lines determine a regular hexagon with a side length of $\frac{20}{\sqrt{3}}$. The three straight lines passing through the origin $O$ divide this hexagon into six equilateral triangles with a side length of $\frac{20}{\sqrt{3}}$. Since the side length of each large triangle is 10 times that of the small triangles, and each large triangle is divided into $10^{2}$ small triangles, the number of small triangles with a side length of $\frac{2}{\sqrt{3}}$ inside the hexagon is $6 \times 10^{2}=600$ (units). Additionally, there are 10 small triangles adjacent to each side of the hexagon (as shown in Figure 3). Therefore, the total number of equilateral triangles with a side length of $\frac{2}{\sqrt{3}}$ is $6 \times 10^{2}+6 \times 10=660$.
|
660
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given the equation about $x$
Figure 3 $\left(a^{2}-1\right)\left(\frac{x}{x-1}\right)^{2}-(2 a+7)\left(\frac{x}{x-1}\right)+1=0$ has real roots.
(1) Find the range of values for $a$;
(2) If the two real roots of the original equation are $x_{1}$ and $x_{2}$, and $\frac{x_{1}}{x_{1}-1} + \frac{x_{2}}{x_{2}-1}=\frac{3}{11}$, find the value of $a$.
|
15.1) Let $\frac{x}{x-1}=t$, then $t \neq 1$. The original equation can be transformed into $\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0$.
When $a^{2}-1=0$, i.e., $a= \pm 1$, the equation becomes
$-9 t+1=0$ or $-5 t+1=0$,
which means $\frac{x}{x-1}=\frac{1}{9}$ or $\frac{x}{x-1}=\frac{1}{5}$.
Thus, $x=-\frac{1}{8}$ or $x=\frac{1}{4}$.
Therefore, when $a= \pm 1$, the original equation has real roots.
When $a \neq \pm 1$, the original equation has real roots if $\Delta \geqslant 0$.
By $\Delta=[-(2 a+7)]^{2}-4\left(a^{2}-1\right) \geqslant 0$,
we get $a \geqslant-\frac{53}{28}$.
In summary, when $a \geqslant-\frac{53}{28}$, the original equation has real solutions.
(2) From the given, $\frac{x_{1}}{x_{1}-1}$ and $\frac{x_{2}}{x_{2}-1}$ are the two roots of the equation
$$
\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0
$$
Using the relationship between roots and coefficients, we get $\frac{2 a+7}{a^{2}-1}=\frac{3}{11}$. Thus,
$$
3 a^{2}-22 a-80=0,
$$
Solving this, we get $a_{1}=10, a_{2}=-\frac{8}{3}$.
From (1), we know that $a \geqslant-\frac{53}{28}$.
Since $a_{2}=-\frac{8}{3}<-\frac{53}{28}$, $a_{2}$ should be discarded.
Therefore, $a=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given 997 points in the plane, the midpoints of the line segments connecting every pair of points are colored red. Prove that there are at least 1991 red points. Can you find 997 points that result in exactly 1991 red points?
|
(遈示: Let $A B$ be the longest line connecting two points among 997 points. Point $A$ is connected to the other 996 points, and the midpoints of these lines are all within the circle centered at $A$ with a radius of $\frac{A B}{2}$. Point $B$ is connected to the other 996 points, and the midpoints of these lines are all within the circle centered at $B$ with a radius of $\frac{A B}{2}$. The two circles are externally tangent at the midpoint of $A B$, so there are at least $996 \times 2-1=1991$ red midpoints. On the number line, take 997 points corresponding to $1,2, \cdots, 997$, and exactly 1991 red midpoints are obtained.)
|
1991
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (20 points) Given that $x$ and $y$ are real numbers, and satisfy
$$
\begin{array}{l}
x y + x + y = 17, \\
x^{2} y + x y^{2} = 66 .
\end{array}
$$
Find the value of $x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4}$.
|
Given the conditions $x y + x + y = 17$ and $x y (x + y) = 66$, we know that $x y$ and $x + y$ are the two real roots of the equation
$$
t^{2} - 17 t + 66 = 0
$$
Solving this equation, we get
$$
t_{1} = 6, \quad t_{2} = 11.
$$
Thus, $x y = 6$ and $x + y = 11$; or $x y = 11$ and $x + y = 6$.
When $x y = 6$ and $x + y = 11$, $x$ and $y$ are the roots of the equation
$$
u^{2} - 11 u + 6 = 0
$$
Since
$$
\Delta_{1} = (-11)^{2} - 4 \times 6 = 121 - 24 > 0,
$$
the equation (2) has real roots.
$$
\begin{array}{l}
\text{At this time, } x^{2} + y^{2} = (x + y)^{2} - 2 x y \\
= 11^{2} - 2 \times 6 = 109. \\
x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4} \\
= x^{4} + y^{4} + x^{2} y^{2} + x y (x^{2} + y^{2}) \\
= (x^{2} + y^{2})^{2} - x^{2} y^{2} + x y (x^{2} + y^{2}) \\
= 109^{2} - 6^{2} + 6 \times 109 = 12499.
\end{array}
$$
When $x y = 11$ and $x + y = 6$, $x$ and $y$ are the roots of the equation
$$
v^{2} - 6 v + 11 = 0
$$
Since
$$
\Delta_{2} = (-6)^{2} - 4 \times 11 = 36 - 44 < 0,
$$
the equation (3) has no real roots.
Therefore, the value of $x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4}$ is 12499.
|
12499
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (16 points) Find all four-digit numbers that satisfy the following conditions: they are divisible by 111, and the quotient obtained is equal to the sum of the digits of the four-digit number.
|
$$
\overline{a b c d}=a \times 10^{3}+b \times 10^{2}+c \times 10+d
$$
satisfies the condition. Then
$$
\begin{array}{l}
\frac{a \times 10^{3}+b \times 10^{2}+c \times 10+d}{11 i} \\
=9 a+b+\frac{a-11 b+10 c+d}{11 i} . \\
\because-98 \leqslant a-11 b+10 c+d \leqslant 108, \text { and } \overline{a b c d} \text { is divisible by }
\end{array}
$$
111,
$$
\therefore a-11 b+c+d=0 \text {, }
$$
i.e., $11 b=a+10 c+d$.
According to the problem, $9 a+b=a+b+c+d$,
i.e., $8 a=c+d$.
Substituting into (1) yields
$$
11 b=9(a+c) \text {. }
$$
And from $c+d \leqslant 18$, we know $a=1$ or 2.
Thus, from equation (2) we get
$$
b=9, a=2, c=9 \text {. }
$$
Then, from $8 a=c+d$, we get $d=7$.
Therefore, the required four-digit number is 2997.
|
2997
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (16 points) (1) In a $4 \times 4$ grid paper, some small squares are painted red, and then 2 rows and 2 columns are crossed out. If no matter how they are crossed out, at least one red small square remains uncrossed, how many small squares must be painted red at least? Prove your conclusion.
(2) If the “$4 \times 4$ grid paper” in the previous problem is changed to “$n \times n$ grid paper $(n \geqslant 5)$,” with other conditions unchanged, how many small squares must be painted red at least? Prove your conclusion.
|
Three, (1) At least 7 cells need to be colored.
If the number of colored cells $\leqslant 4$, then by appropriately crossing out 2 rows and 2 columns, all the colored cells can be crossed out.
If the number of colored cells is 5, then at least one row has 2 cells colored. By crossing out this row, the remaining colored cells do not exceed 3. Then, by crossing out 1 more row and 2 columns, all the colored cells can be crossed out.
If the number of colored cells is 6, then at least one row has 3 cells colored, or at least two rows each have 2 cells colored. Therefore, by crossing out 2 rows, at least 4 colored cells can be crossed out, leaving no more than 2 colored cells. By crossing out 2 more columns, they can all be crossed out.
Color 7 cells as shown in Figure 4, then by crossing out 2 rows, at most 4 colored cells can be crossed out, and the remaining colored cells are in different 3 columns. By crossing out 2 more columns, they cannot all be crossed out.
(2) At least 5 cells need to be colored.
This is because, if the number of colored cells $\leqslant 4$, then by crossing out 2 rows and 2 columns, they can all be crossed out.
Color 5 cells as shown in Figure 5, then by arbitrarily crossing out 2 rows and 2 columns, there will be colored cells that are not crossed out.
|
5
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: 20 teams participate in the national football championship. To ensure that in any three teams that have already played, there are two teams that have already played against each other, what is the minimum number of matches that need to be played?
(3rd All-Union Mathematical Olympiad)
|
Solution: Suppose that in any three teams that have already played, there are two teams that have played against each other. Let team $A$ be the team with the fewest matches played, having played $k$ matches. Thus, the $k$ teams that have played against team $A$, as well as team $A$ itself, have all played no fewer than $k$ matches.
Among the $19-k$ teams that have not played against team $A$, each team has played against the other $18-k$ teams. Thus, in any three of these teams, they have all played against each other.
It is evident that twice the total number of matches is no less than
$$
\begin{array}{l}
k(k+1)+(19-k)(18-k) \\
=2 k^{2}-36 k+342 \\
=2(k-9)^{2}+180 \geqslant 180,
\end{array}
$$
which means the number of matches is no less than 90.
Now, let's construct an example: divide the 20 teams into two groups, with each group of 10 teams playing against each other, but teams from different groups do not play against each other. This schedule meets the requirements of the problem, meaning at least $2 \mathrm{C}_{0}^{2}=90$ matches are needed.
|
90
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In a convex $n$-sided polygon, the difference between any two adjacent interior angles is $18^{\circ}$. Try to find the maximum value of $n$.
The difference between any two adjacent interior angles of a convex $n$-sided polygon is $18^{\circ}$. Try to find the maximum value of $n$.
|
Solution: Complete in three steps.
(1) Prove that $n$ is even. Let the convex $n$-gon be $A_{1} A_{2} A_{3} \cdots$
$A_{n}$, then $A_{2}=A_{1} \pm 18^{\circ}$, (where “ + ” or “-” is chosen)
$A_{3}=A_{2} \pm 18^{\circ}=A_{1} \pm 18^{\circ} \pm 18^{\circ}$, (choose “ + ” or “ - ” as before) $\qquad$
$$
\begin{array}{l}
A_{k}=A_{1} \pm \underbrace{18^{\circ} \pm 18^{\circ} \pm \cdots \pm 18^{\circ}}_{\text {total }} \\
=A_{1}+p_{k} \cdot 18^{\circ}-q_{k} \cdot 18^{\circ} .
\end{array}
$$
where $p_{k}$ is the number of “ + ” signs, and $q_{k}$ is the number of “ - ” signs, so $p_{k}+q_{k}=k-1$.
In the above formula, let $k=n+1$, then we know
$$
\begin{aligned}
A_{n+1} & =A_{1}+p_{n+1} \cdot 18^{\circ}-q_{n+1} \cdot 18^{\circ} \\
& =A_{1}+p_{n+1} \cdot 18^{\circ}-\left(n+1-1-p_{n+1}\right) \cdot 18^{\circ} \\
& =A_{1} .
\end{aligned}
$$
(Actually, vertex $A_{n+1}$ is the same as $A_{1}$)
Thus, $p_{n+1}=n-p_{n+1}$,
which means $n=2 p_{n+1}$, proving that $n$ is even.
(2) Prove that $nn x \geqslant(n-2) \cdot 180^{\circ}+n \cdot 9^{\circ} \text {. }$
$$
Then $n0$,
indicating that there exists a convex 38-gon that meets the conditions.
In summary, the maximum value of $n$ is 38.
(Yang Yongliang and Xun Yangtao, No. 5 Middle School of Bozhou City, Hunan Province, 414411)
|
38
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 In a football invitational tournament, 16 cities participate, each city sending Team A and Team B. According to the competition rules, each pair of teams plays at most one match, and the two teams from the same city do not play against each other. After several days of the tournament, it was found that except for Team A of City $A$, all other teams have played a different number of matches. How many matches has Team B of City $A$ played? Please prove your conclusion.
$(1985$, National High School Mathematics Competition)
|
Solution: Consider the general case: Suppose there are $n$ cities participating, and let the number of matches played by Team B from City A be $a_{n}$. Clearly, $a_{1}=$ 0.
Since there are $n$ cities participating, according to the competition rules, each team can play at most $2(n-1)$ matches. Given the conditions, the number of matches played by the $2 n-1$ teams, excluding Team A from City A, covers $0,1,2, \cdots, 2(n-1)$. Assume Team A from City B has played $2(n-1)$ matches, meaning that except for Team B from City B, all other teams have played at least one match. Therefore, the number of matches played by Team B from City B is 0.
Now, remove the two teams from City B, and for the remaining $n-1$ cities, the number of matches played by the teams, excluding Team A from City A, are respectively:
$$
\begin{array}{l}
1-1=0, \\
2-1=1, \\
\cdots \cdots, \\
(2 n-3)-1=2(n-2) .
\end{array}
$$
The number of matches played by Team B from City A is $a_{n}-1$,
thus $a_{n-1}=a_{n}-1(n \geqslant 2)$.
It is evident that the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 0 and common difference 1, hence,
$$
a_{n}=0+(n-1) \cdot 1=n-1 \text {. }
$$
In this problem, $n=16$, so, $a_{16}=15$, which means Team B from City A has played 15 matches.
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 There is a type of sports competition with $M$ events, and athletes $A$, $B$, and $C$ participate. In each event, the first, second, and third places receive $p_{1}$, $p_{2}$, and $p_{3}$ points respectively, where $p_{1}$, $p_{2}$, $p_{3} \in \mathbf{Z}^{+}$, and $p_{1}>p_{2}>p_{3}$. In the end, $A$ scores 22 points, $B$ and $C$ both score 9 points. It is also known that $B$ came first in the 100-meter dash. Find the value of $M$.
(18th Canadian High School Mathematics Competition)
|
Solution: Consider the total score of three people, we have
$$
\begin{array}{l}
M\left(p_{1}+p_{2}+p_{3}\right)=22+9+9=40 . \\
\because p_{1}, p_{2}, p_{3} \in \mathbf{Z}^{+}, \text { and } p_{1}>p_{2}>p_{3}, \\
\therefore p_{1}+p_{2}+p_{3} \geqslant 3+2+1=6 .
\end{array}
$$
Thus, $6 M \leqslant 40, M \leqslant 6$.
Since $p_{1}>p_{2}>p_{3}$, and $B$ and $C$ have the same score, then $M \neq 1$. In addition, $M$ must divide 40, so $M$ can be $2, 4, 5$.
(1) If $M=2$, i.e., there is one more competition besides the 100-meter dash. Since $B$ is the first in the 100-meter dash and has a total score of only 9, it must be that $9 \geqslant p_{1}+p_{3}$. Therefore, $p_{1} \leqslant 8$. This way, $A$ cannot score 22 points in two competitions.
(2) If $M=4$, then for $B$'s score, we have $9 \geqslant p_{1} + 3 p_{3}$. So $p_{1} \leqslant 6$.
If $p_{1} \leqslant 5$, then $A$ can score at most 19 points in 4 competitions (since $B$ is the first in the 100-meter dash), but $A$ actually scored 22 points, so $p_{1}=6, p_{3}=1$.
$$
\begin{array}{l}
\because 4\left(p_{1}+p_{2}+p_{3}\right)=40, \\
\therefore p_{2}+p_{3}=4 . \\
\text { Thus, } p_{2}=3 .
\end{array}
$$
But $A$ can win at most 3 first places (since $B$ is already the first in the 100-meter dash) and 1 second place, scoring at most $3 \times 6 + 3 = 21$. This contradicts the given conditions.
(3) If $M=5$, from $5\left(p_{1}+p_{2}+p_{3}\right)=40$, we get $p_{1}+p_{2}+p_{3}=8$.
If $p_{3} \geqslant 2$, then $p_{1}+p_{2}+p_{3} \geqslant 4+3+2=9$, which is a contradiction.
So, $p_{3}=1$.
Also, $p_{1}$ must be at least 5, otherwise $A$ can score at most $4 \times 4 + 3 = 19$ points in 5 competitions, which is a contradiction.
So, $p_{1} \geqslant 5$.
If $p_{1} \geqslant 6$, then $p_{2}+p_{3} \leqslant 2$, which is also impossible. Thus, only $p_{1}=5$, and hence, $p_{2}=2, p_{3}=1$.
In conclusion, $M=5$.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 In a certain competition, each player plays exactly one game against each of the other players. A win earns the player 1 point, a loss 0 points, and a draw 0.5 points for each player. After the competition, it is found that exactly half of each player's points were earned in games against the 10 lowest-scoring players (the 10 lowest-scoring players earned half of their points in games against each other). Find the total number of players in the competition.
(3rd American Invitational Mathematics Examination)
|
Solution: Let there be a total of $n$ players, then they collectively score $\mathrm{C}_{n}^{2}$ points. The 10 players with the lowest scores collectively score $\mathrm{C}_{10}^{2}=45$ points through their matches against each other, which is half of their total score, so these 10 players collectively score 90 points. The remaining $n-10$ players collectively score $\mathrm{C}_{n-10}^{2}$ points through their matches against each other, which is also half of their total score, so these $n-10$ players collectively score $2 \mathrm{C}_{n-10}^{2}$ points. Therefore, we have
$$
\mathrm{C}_{n}^{2}=90+2 \mathrm{C}_{n-10}^{2} \text {, }
$$
which simplifies to $n^{2}-41 n+400=0$.
Solving this, we get $n=16$ or 25.
However, if there are only 16 players, then there would be 6 high-scoring players, who collectively score 30 points, averaging 5 points per person. Meanwhile, the 10 low-scoring players average $90 \div 10=9$ points per person, which is a contradiction.
Therefore, $n=25$, meaning there are a total of 25 participants.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 After 20 figure skaters have performed, 9 judges each assign them ranks from 1 to 20. It is known that for each skater, the difference between any two ranks does not exceed 3. If the sum of the ranks each skater receives is arranged in an increasing sequence: $C_{1} \leqslant C_{2} \leqslant \cdots \leqslant C_{20}$, find the maximum value of $C_{1}$.
(2nd All-Union Mathematical Olympiad)
|
If 9 judges all give a certain athlete the first place, then $C_{1}=9$. If two athletes are both judged as the first place, then one of them gets no less than 5 first places, and the other 4 ranks no higher than fourth. Therefore, $C_{1}=5 \times 1+$ $4 \times 4=21$.
If 3 athletes all get the first place, then their other ranks do not exceed fourth, and their total ranks do not exceed $1 \times 9+3 \times 9+4 \times 9=72$, so $C_{1} \leqslant 24$.
If 4 athletes all get the first place, then their total ranks do not exceed $1 \times 9+2 \times 9+3 \times 9+4 \times$ $9=90$, so the rank of one of them is no more than 24. The case where 5 or more athletes all get the first place is impossible. Therefore, $C_{1} \leqslant 24$.
Now, an example where $C_{1}=24$ is given:
The judges give the top 3 athletes the ranks
$$
1,1,1,3,3,3,4,4,4 \text {; }
$$
The ranks given to the last three athletes are
$$
2,2,2,5,5,5,6,6,6 \text {; }
$$
The ranks of the remaining athletes are given arbitrarily between the two. This way, $\left(C_{1}\right)_{\text {max }}=24$.
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In the final stage of a professional bowling tournament, the top five players compete as follows: First, the fifth-place player competes with the fourth-place player, the loser gets fifth place, the winner competes with the third-place player; the loser gets third place, the winner competes with the first-place player, the loser gets second place, the winner gets first place. How many different possible orders of finish are there?
(39th American High School Mathematics Examination)
|
(Due to determining the top five rankings, 4 races were held, and each race has two possible outcomes. By the multiplication principle, there are $2^{4}=16$ possible ranking orders for the top five winners.)
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In a cross-country race, there are two teams participating, each with 5 members. When the athletes reach the finish line, their ranks are recorded, and the team of the $n$-th arriving athlete scores $n$ points, with the team having the lower total score winning. Now, assuming no two team members arrive at the finish line simultaneously, how many possible scores can the winning team have?
(40th American High School Mathematics Examination)
|
(The hint: The sum of the scores of 10 athletes is $1+2+\cdots+10$ $=55$, thus the total score of the winning team does not exceed $\left[\frac{55}{2}\right]=27$. Also, the total score of the winning team is no less than $1+2+\cdots+5=15$, it is easy to see that the score of the winning team can take all values from $15,16, \cdots, 27$. Therefore, the winning team's score has 27 $-15+1=13$ possible values.)
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example $1\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}=$ $\qquad$
(1999, National Junior High School Mathematics League Wuhan Selection Competition)
|
Let $3^{999}=a, 5^{099}=b, 7^{999}=c$, then
$$
\begin{array}{l}
\text { Original expression }=\frac{c}{a} \sqrt{\frac{a^{2}+(a b)^{2}}{c^{2}+(b c)^{2}}} \\
=\frac{c}{a} \sqrt{\frac{a^{2}\left(1+b^{2}\right)}{c^{2}\left(1+b^{2}\right)}} . \\
=\frac{c}{a} \cdot \frac{a}{c}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let the parabola
$$
y=x^{2}+(2 a+1) x+2 a+\frac{5}{4}
$$
intersect the $x$-axis at only one point.
(1) Find the value of $a$;
(2) Find the value of $a^{18}+323 a^{-6}$.
(1998, National Junior High School Mathematics Competition)
|
Solution: (1) From the given condition, the equation
$$
x^{2}+(2 a+1) x+2 a+\frac{5}{4}=0
$$
has two equal real roots, hence
$$
\Delta=(2 a+1)^{2}-4\left(2 a+\frac{5}{4}\right)=0,
$$
which simplifies to $a^{2}-a-1=0$.
Solving for $a$, we get $a=\frac{1 \pm \sqrt{5}}{2}$.
(2) From (1), we know $a^{2}-a-1=0$, thus
$$
\begin{array}{l}
a-a^{-1}=1, \\
a^{2}+a^{-2}=3, \\
\begin{aligned}
a^{6}+a^{-6} & =\left(a^{2}+a^{-2}\right)^{3}-3\left(a^{2}+a^{-2}\right) \\
& =3^{3}-3 \times 3=18, \\
a^{12}+a^{-12} & =\left(a^{6}+a^{-6}\right)^{2}-2 \\
& =18^{2}-2=322 .
\end{aligned}
\end{array}
$$
Substituting $a^{12}+a^{-12}$ with 322, we get
$$
\begin{aligned}
323 & =a^{12}+a^{-12}+1 \\
\therefore & a^{18}+323 a^{-6} \\
= & a^{18}+\left(a^{12}+a^{-12}+1\right) a^{-6} \\
= & \left(a^{18}+a^{-18}\right)+\left(a^{6}+a^{-6}\right) \\
= & \left(a^{6}+a^{-6}\right)\left(a^{12}-1+a^{-12}\right) \\
& +\left(a^{6}+a^{-6}\right) \\
= & \left(a^{6}+a^{-6}\right)\left(a^{12}+a^{-12}\right) \\
= & 18 \times 322=5796 .
\end{aligned}
$$
|
5796
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given $a^{2}+a+1=0$. Then, $a^{1992}+$ $a^{322}+a^{2}=$ $\qquad$
(Harbin 15th Junior High School Mathematics Competition)
|
Given the condition $a \neq 1$, from $(a-1)(a^2 + a + 1) = 0$, we get $a^3 - 1 = 0$, hence $a^3 = 1$.
Substituting 1 for $a^3$ and 0 for $a^2 + a + 1$, we get
$$
\begin{aligned}
\text { Original expression } & =\left(a^{3}\right)^{664}+\left(a^{3}\right)^{107} a+a^{2} \\
& =1^{664}+1^{107} a+a^{2} \\
& =1+a+a^{2}=0 .
\end{aligned}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Given that $x$ and $y$ are positive integers, and satisfy the conditions $x y + x + y = 71, x^{2} y + x y^{2} = 880$. Find the value of $x^{2} + y^{2}$.
(Jiangsu Province 14th Junior High School Mathematics Competition)
|
Solution: From the conditions, we have
$$
\begin{array}{l}
x y+(x+y)=71, \\
x y(x+y)=880,
\end{array}
$$
Therefore, $x y$ and $x+y$ are the two roots of the equation
$$
z^{2}-71 z+880=0
$$
Solving this equation, we get
$$
x+y=16, x y=55,
$$
or $x+y=55, x y=16$.
When $x+y=55, x y=16$, $x$ and $y$ are the two roots of the equation $z^{2}-55 z+16=0$. Solving this equation, we find that its roots are not positive integers, so this solution should be discarded. Similarly, we can verify that $x$ and $y$ satisfying $x+y=16, x y=55$ meet the conditions. Therefore, substituting 16 and 55 for $x+y$ and $x y$ respectively, we get
$$
\begin{array}{l}
x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=16^{2}-2 \times 55=256-110=146 .
\end{array}
$$
|
146
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $f(x)=\sum_{0}^{5}\left[2^{i} x\right]$. Find $f(3.14)$.
|
Given: $\because x=3.14, \frac{4}{2^{5}} \leqslant 0.14<\frac{5}{2^{5}}$, so take $p=3, k=4$.
$$
\begin{aligned}
\therefore f(3.14) & =\sum_{0}^{5} 2^{i} \times 3+\sum_{0}^{5}\left[\frac{4}{2^{i}}\right] \\
& =3 \times 63+7=196 .
\end{aligned}
$$
|
196
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In a $3 \times 3$ square grid, fill in the numbers as shown in the table below. The operation on the table is as follows: each operation involves adding a number to two adjacent numbers in the grid (adjacent means two small squares that share a common edge).
\begin{tabular}{|l|l|l|}
\hline 0 & 3 & 2 \\
\hline 6 & 7 & 0 \\
\hline 4 & 9 & 5 \\
\hline
\end{tabular}
Can it be done after several operations such that
(1) all the numbers in the grid are 0;
(2) the numbers in the four corners are all 1, and the rest are 0.
|
(Tip: (1) After 5 operations, all cells can be 0.
(2) As shown in the table below, the invariant $S=a-$ $b+c-d+e-f+g-h+k$.
\begin{tabular}{|l|l|l|}
\hline$a$ & $b$ & $c$ \\
\hline$d$ & $e$ & $f$ \\
\hline$g$ & $h$ & $k$ \\
\hline
\end{tabular}
The initial state is
$$
S=0-3+2-6+7-0+4-9+5=0,
$$
The target state is
$$
S=1-0+1-0+0-0+1-0+1=4 \text {. }
$$
Therefore, it is impossible.)
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A bus starts a 100-kilometer journey at 12:20 PM. There is a computer on the bus that at 1:00 PM, 2:00 PM, 3:00 PM, 4:00 PM, 5:00 PM, and 6:00 PM, says: “If the average speed in the future is the same as the average speed in the past, then it will take one more hour to reach the destination.” Is this possible? If so, how far has the bus traveled by 6:00 PM?
|
1. The computer is correct.
At $k: 00$ PM, where $1 \leqslant k \leqslant 6$, the distance the car has traveled is $\frac{60 k-20}{60 k+40}$ of the total distance, and the average speed of the car up to this point is $\frac{1}{60 k+40}$. If it continues to travel the remaining $\frac{60}{60 k+40}$ at the same speed, it will take exactly one hour to reach the destination. Note that between $(k-1): 00$ PM and $k: 00$ PM, the distance traveled is
$$
\frac{60 k-20}{60 k+40}-\frac{60 k-80}{60 k-20}=\frac{3600}{(60 k+40)(60 k-20)},
$$
and these numbers are all determined. At $6: 00$ PM, the distance traveled is $\frac{340}{400}=\frac{17}{20}$, which means the car has traveled 85 kilometers.
|
85
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The cube of an $n$-digit number is an $m$-digit number, can $n+m=$ 2001?
|
2. Notice that $10^{500}$ has 501 digits, and $\left(10^{500}\right)^{3}$ has 1501 digits, totaling 2002 digits.
For any positive number $a>10^{500}, a$ and $a^{3}$ have at least 2002 digits combined.
If $a<10^{500}$, it has at most 500 digits, and $a^{3}<10^{1500}$ has at most 1500 digits. Therefore, the sum of the number of digits being 2001 is impossible.
|
2001
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. According to the "Personal Income Tax Law of the People's Republic of China," citizens do not need to pay tax on the portion of their monthly salary and wages that does not exceed 800 yuan. The portion exceeding 800 yuan is the monthly taxable income, which is taxed according to the following table in segments:
\begin{tabular}{|c|c|}
\hline Monthly taxable income & Tax rate \\
\hline The portion not exceeding 500 yuan & $5 \%$ \\
\hline The portion exceeding 500 yuan to 2000 yuan & $10 \%$ \\
\hline The portion exceeding 2000 yuan to 5000 yuan & $15 \%$ \\
\hline$\ldots .$. & $\ldots .$. \\
\hline
\end{tabular}
A person's tax for January is 30 yuan. Then his monthly salary and wages for that month are $\qquad$ yuan.
|
$=、 11.1350$
|
1350
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The solution to the equation $\sqrt{13-\sqrt{13+x}}$ $=x$ is $\qquad$
|
3.3.
Let $\sqrt{13+x}=y$ then
$$
\left\{\begin{array}{l}
\sqrt{13-y}=x, \\
\sqrt{13+x}=y .
\end{array}\right.
$$
(2) $)^{2}-(1)^{2}$ gives $(x+y)(x-y+1)=0$.
From the original equation, we know $x>0$, and from (2), $y>0$, so $x+y \neq 0$. Thus, $x-y+1=0$, which means $y=x+1$. Therefore, it is easy to get $x=3$ (discard $x=-4$).
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) A chemical plant, starting from January this year, if it does not improve its production environment and continues to produce as it is, will earn 700,000 yuan per month. At the same time, it will receive penalties from the environmental protection department, with the first month's penalty being 30,000 yuan, and increasing by 20,000 yuan each subsequent month. If the plant invests 5,000,000 yuan to add recycling and purification equipment (assuming no time for equipment renovation), it can not only improve the environment but also significantly reduce raw material costs. According to estimates, the cumulative net income for the first 5 months after production is a quadratic function of production time \( n \) (in months). The cumulative income for the first 1, first 2, and first 3 months can reach 1,010,000 yuan, 2,040,000 yuan, and 3,090,000 yuan, respectively, and will stabilize at the level of the fifth month thereafter. At the same time, the plant will not only avoid penalties but also receive a one-time reward of 1,000,000 yuan from the environmental protection department. How many months will it take for the investment to start paying off, i.e., for the net income after the investment to exceed the net income without the investment?
|
Three, let the cumulative income for $n$ months without modifying the equipment, under the original conditions, be $a(n)$, and let the cumulative income for $n$ months after modifying the equipment be $b(n)$. From the given conditions, we have $a(n)=70 n$, and we can assume
$$
b(n)=a n^{2}+b n+c, \quad n \leqslant 5.
$$
Thus,
$$
\left\{\begin{array}{l}
b(1)=a+b+c=101, \\
b(2)=4 a+2 b+c=204, \\
b(3)=9 a+3 b+c=309.
\end{array}\right.
$$
Solving these, we get $a=1, b=100, c=0$.
$$
\therefore b(n)=\left\{\begin{array}{ll}
n^{2}+100 n, & (n \leqslant 5) \\
b(5)+(n-5 \llbracket b(5)-b(4)], & (n>5)
\end{array}\right.
$$
That is $\square$
$$
\left.\begin{array}{l}
b(n)=\left\{\begin{array}{l}
n^{2}+100 n, \quad(n \leqslant 5) \\
109 n-20. \quad(n>5)
\end{array}\right. \\
\because b(5)-500+100=125a(n)-\left[3 n+\frac{2 n(n-1)}{2}\right](n>5)
\end{array}
\right.
$$
That is $n^{2}+41 n-420>0$.
When $n \geqslant 9$, the above inequality holds.
Therefore, the investment will start to pay off after 9 months.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ has $n$ different points $P_{1}, P_{2}, \ldots, P_{n}, F$ is the right focus, $\left\{\left|P_{i} F\right|\right.$ \} forms an arithmetic sequence with a common difference $d>\frac{1}{100}$. Then the maximum value of $n$ is ( ).
(A ) Fang data $) 200$
( C) 99
( D ) 100
|
5.( B ).
$$
\begin{array}{l}
\because\left|P_{n} F\right|=\left|P_{1} F\right|+(n-1) d, \\
\therefore n=1+\frac{\left|P_{n} F\right|-\left|P_{1} F\right|}{d} .
\end{array}
$$
In an ellipse, the maximum value of the focal radius $\left|P_{i} F\right|$ is $a+c$, and the minimum value is $a-c$, so
$$
\begin{array}{l}
\left|P_{n} F\right|-\left|P_{1} F\right| \leqslant a+c-(a-c)=2 c=2, \\
\text { hence } n \leqslant 1+\frac{2}{d}<1+2 \times 100=201 .
\end{array}
$$
Therefore, the maximum value of $n$ is 200.
|
200
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $x, y, z \in \mathbf{R}, x y+y z+z x=-1$. Then the minimum value of $x^{2}+5 y^{2}+8 z^{2}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3.4 .
$$
\begin{array}{l}
\text { Given }(x+2 y+2 z)^{2}+(y-2 z)^{2} \geqslant 0 , \\
x^{2}+5 y^{2}+8 z^{2} \geqslant-4(x y+y z+z x)=4 ,
\end{array}
$$
we have $x=\frac{3}{2} , y=-\frac{1}{2} , z=-\frac{1}{4}$ ,
or $x=-\frac{3}{2}, y=\frac{1}{2}, z=\frac{1}{4}$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $S_{n}$ be the sum of the elements of all 3-element subsets of the set $A=\left\{1, \frac{1}{2}, \cdots, \frac{1}{2^{n-1}}\right\}$. Then $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=$ $\qquad$ .
|
6.1 .
For any element in set $A$, it appears in a subset containing 3 elements $\mathrm{C}_{n-1}^{2}$ times, then
$$
\begin{array}{l}
S_{n}=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^{n-1}}\right) \mathrm{C}_{n-1}^{2} . \\
\text { Therefore, } \lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=\lim _{n \rightarrow \infty} \frac{\frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}} \cdot \frac{(n-1)(n-2)}{2}}{n^{2}}=1 .
\end{array}
$$
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (20 points) The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=$ $b_{0}=1, a_{n}=a_{n-1}+2 b_{n-1}, b_{n}=a_{n-1}+b_{n-1}$, $(n=1,2 \ldots)$. Find the value of $a_{2001}^{2}-2 b_{2001}^{2}$.
|
$$
\begin{array}{l}
a_{0}^{2}-2 b_{0}^{2}=-1, \\
a_{1}^{2}-2 b_{1}^{2}=1, \\
\ldots \ldots
\end{array}
$$
Conjecture $a_{n}^{2}-2 b_{n}^{2}=(-1)^{n+1}(n=0,1 \ldots)$.
We will prove this by mathematical induction.
Assume the proposition holds for $n=k$, that is,
$$
a_{k}^{2}-2 b_{k}^{2}=(-1)^{k+1} \text {. }
$$
Then for $n=k+1$,
$$
\begin{array}{l}
a_{k+1}^{2}-2 b_{k+1}^{2}=\left(a_{k}+2 b_{k}\right)^{2}-2\left(a_{k}+b_{k}\right)^{2} \\
=-\left(a_{k}^{2}-2 b_{k}^{2}\right)=(-1)^{k+2} .
\end{array}
$$
Therefore, by mathematical induction, the conjecture is true.
Thus, $a_{2001}^{2}-2 b_{2001}^{2}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, (20 points) For the parabola $y^{2}=2 p x(p>0)$ with focus $F$, does there exist an inscribed isosceles right triangle such that one of its legs passes through $F$? If it exists, how many are there? If not, explain the reason.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
As shown in Figure 4, let \( A\left(x_{A}, y_{A}\right) \), \( B\left(x_{B}, y_{B}\right) \), \( C\left(x_{C}, y_{C}\right) \), and the inclination angle of \( AC \) be \( \theta \). Clearly, \( \theta \neq 0 \), \( \frac{\pi}{2} \). Suppose \( \theta \in \left(0, \frac{\pi}{2}\right) \). By the focal chord length formula, we have
\[
\begin{array}{l}
|A C|=\frac{2 p}{\sin ^{2} \theta} . \\
\because\left\{\begin{array}{l}
y_{A}-y_{C}=|A C| \sin \theta=\frac{2 p}{\sin \theta}, \\
y_{A}+y_{C}=2 p \frac{x_{A}-x_{C}}{y_{A}-y_{C}}=2 p \cot \theta,
\end{array}\right.
\end{array}
\]
\[
\left\{\begin{array}{l}
y_{C}-y_{B}=|C B| \cos \theta=|A C| \cos \theta=\frac{2 p \cos \theta}{\sin ^{2} \theta}, \\
y_{C}+y_{B}=2 p \frac{x_{C}-x_{B}}{y_{C}-y_{B}}=-2 p \tan \theta .
\end{array}\right.
\]
From (1), we get \( y_{C}=p \frac{\cos \theta-1}{\sin \theta} \),
From (2), we get \( y_{C}=p \frac{\cos ^{2} \theta-\sin ^{3} \theta}{\sin ^{2} \theta \cos \theta} \),
\(\therefore \cos\) squared minus \(\sin ^{3} \theta\),
which means \(\cos \theta+\sin \theta=\tan \theta\),
or equivalently, \(\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)=\tan \theta\).
The problem reduces to whether (3) has a solution in \(\left(0, \frac{\pi}{2}\right)\). From the graphs of trigonometric functions, (3) has exactly one intersection point in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), indicating that there exists an isosceles right triangle that satisfies the given conditions. By symmetry, when \(\theta \in \left(\frac{\pi}{2}, \pi\right)\), there is also one intersection point.
In summary, there exist exactly two isosceles right triangles that satisfy the given conditions.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 A class participated in a math competition, with a total of $a$, $b$, and $c$ three questions. Each question either scores full marks or 0 points, where question $a$ is worth 20 points, and questions $b$ and $c$ are worth 25 points each. After the competition, every student answered at least one question correctly, and those who answered all questions correctly are $\mathrm{i}$ people, and 15 people answered two questions correctly. The sum of the number of people who answered question $a$ correctly and the number of people who answered question $b$ correctly is 29; the sum of the number of people who answered question $a$ correctly and the number of people who answered question $c$ correctly is 25; the sum of the number of people who answered question $b$ correctly and the number of people who answered question $c$ correctly is 20. What is the average score of the class?
(1909, National Junior High School Mathematics Competition)
|
Solution: Let $x, y, z$ represent the number of correct answers for questions $a$, $b$, and $c$, respectively. Then we have
$$
\left\{\begin{array} { l }
{ x + y = 2 9 , } \\
{ x + z = 2 5 , } \\
{ y + z = 2 0 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=17, \\
y=12, \\
z=8 .
\end{array}\right.\right.
$$
The number of people who answered 1 question correctly is
$$
37-1 \times 3-2 \times 15=4 \text { (people). }
$$
The total number of people in the class is $1+4+15=20$ (people).
Thus, the average score is
$$
\frac{17 \times 20+(12+8) \times 25}{20}=42 \text { (points). }
$$
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $a$ is a natural number, there exists a linear polynomial with integer coefficients and $a$ as the leading coefficient, which has two distinct positive roots less than 1. Then, the minimum value of $a$ is $\qquad$ .
|
2.5.
Let $f(x)=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, $00$. Then from $f(0)$ and $f(1)$ being positive integers, we get $f(0) f(1) \geqslant 1$, that is, $a^{2} x_{1} x_{2}\left(1-x_{1}\right)\left(1-x_{2}\right) \geqslant 1$.
Also, $x(1-x) \leqslant \frac{1}{4}$, with equality when $x=\frac{1}{2}$. Since $x_{1} \neq x_{2}$, we have $x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right)4$.
When $a=5$, the equation $5 x^{2}-5 x+1=0$ has two distinct positive roots in the interval $(0,1)$. Therefore, the minimum value of $a$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (20 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=3, a_{n}=$ $3^{a_{n-1}}(n \geqslant 2)$. Find the last digit of $a_{n}(n \geqslant 2)$.
|
Let's prove, when $n \geqslant 2$, we have
$$
a_{n}=4 m+3, m \in \mathbf{N} \text {. }
$$
Using mathematical induction:
(i) When $n=2$, $a_{2}=3^{3}=4 \times 6+3$, so equation (1) holds.
(ii) Assume when $n=k(k \geqslant 2)$, equation (1) holds, i.e.,
$$
a_{k}=4 m+3, m \in \mathbf{N} \text {. }
$$
Then $a_{k+1}=3^{a_{k}}=3^{4 m+3}=(4-1)^{4 m+3}$
$$
\begin{aligned}
= & \mathrm{C}_{4 m+3}^{0} \cdot 4^{4 m+3}(-1)^{0}+\mathrm{C}_{4 m+3}^{1} 4^{4 m+2}(-1)^{1} \\
& +\cdots+\mathrm{C}_{4 m+3}^{4 m+2} 4^{1}(-1)^{4 m+2} \\
& +\mathrm{C}_{4 m+3}^{4 m+3} 4^{0}(-1)^{4 m+3} \\
= & 4 t-1=4(t-1)+3 .
\end{aligned}
$$
Here, $t=\mathrm{C}_{4 m+3}^{0} 4^{4 m+2}+\mathrm{C}_{4 m+3}^{1} 4^{4 m+1}(-1)^{1}+\cdots+$ $\mathrm{C}_{4 m+3}^{4 m+2}(-1)^{4 m+2}$ is an integer, and since $\left\{a_{n}\right\}$ is an increasing sequence, we have $a_{k+1} \geqslant a_{k} \geqslant \cdots \geqslant a_{2}=27$, i.e., $4 t-1 \geqslant 27$. Thus, $t \geqslant 7, t-1 \geqslant 6$, i.e., $t-1 \in \mathbf{N}$.
Therefore, when $n=k+1$, equation (1) holds.
Thus, equation (1) holds, and we have
$$
a_{n}=3^{a_{n-1}}=3^{4 m+3}=(81)^{m} \cdot 27 \text {, }
$$
and its last digit is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 On a circle with a circumference of $300 \mathrm{~cm}$, there are two balls, A and B, moving in uniform circular motion at different speeds. Ball A starts from point A and moves in a clockwise direction, while ball B starts from point B at the same time and moves in a counterclockwise direction. The two balls meet at point C. After meeting, the two balls reverse their directions and continue moving in uniform circular motion, but ball A's speed is doubled, and ball B's speed is halved. They meet for the second time at point D (as shown in Figure 1). It is known that $\overparen{A M C}=40 \mathrm{~cm}$ and $\overparen{B N D}=20 \mathrm{~cm}$. Find the length of $\overparen{A C B}$.
|
Solution: Let the length of $\overparen{A C P}$ be $x \mathrm{~cm}$, and the original speeds of A and B be $v_{1}$ and $v_{2}\left(v_{1} \neq v_{2}\right)$, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{40}{v_{1}}=\frac{x-40}{v_{2}}, \\
\frac{300-20-(x-40)}{2 v_{1}}=\frac{x-40+20}{\frac{1}{2} v_{2}} .
\end{array}\right.
$$
(1) $\div$ (2) gives $\frac{2 \times 40}{320-x}=\frac{x-40}{2(x-20)}$.
Solving, we get $x_{1}=120, x_{2}=80$.
When $x=80$, substituting into (1) gives $v_{1}=v_{2}$, which contradicts the problem statement, so it is discarded. Therefore, the length of $\overparen{A C B}$ is $120 \mathrm{~cm}$.
|
120
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 A mall installs an escalator between the first and second floors, which travels upwards at a uniform speed. A boy and a girl start walking up the escalator to the second floor (the escalator itself is also moving). If the boy and the girl both move at a uniform speed, and the boy walks twice as many steps per minute as the girl, it is known that the boy walked 27 steps to reach the top of the escalator, while the girl walked 18 steps to reach the top (assuming the boy and girl each step on 1 step at a time).
(1) How many steps are visible on the escalator?
(2) If there is a staircase from the second floor to the first floor, with the same number of steps as the escalator, and the two children each walk down the stairs at their original speeds, then reach the bottom 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?
|
Solution: (1) Let the girl's speed be $x$ steps/min, the escalator's speed be $y$ steps/min, and the stairs have $s$ steps, then the boy's speed is $2x$ steps/min. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y} .
\end{array}\right.
$$
Dividing the two equations yields $\frac{3}{4}=\frac{s-27}{s-18}$.
Solving for $s$ gives $s=54$, meaning the escalator has 54 steps exposed.
(2) Let the boy pass the escalator $m$ times and the stairs $n$ times when he catches up with the girl for the first time. Among them, $0 \leqslant m-n \leqslant 1$, and at least one of $m$ and $n$ is an integer, at this time the girl has passed the escalator $(m-1)$ times and the stairs $(n-1)$ times.
From the system of equations in (1), we can solve for $y=2x$, then the boy ascends the escalator at $4x$ steps/min, and the girl ascends the escalator at $3x$ steps/min. Therefore,
$$
\frac{54 m}{4 x}+\frac{54 n}{2 x}=\frac{54(m-1)}{3 x}+\frac{54(n-1)}{x} .
$$
Simplifying gives $m+6 n=16$. (One of $m$ and $n$ must be a positive integer, and $0 \leqslant m-n \leqslant 1$).
Thus, we can use a table method to solve.
From the table, it is evident that only $m=3, n=2 \frac{1}{6}$ meets the requirements. Therefore, the number of steps the boy has walked is: $3 \times 27+\frac{13}{6} \times 54=198$ (steps).
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. In a regular hexagon divided into six areas for观赏 plants, 加园: $\mathrm{i}$, it is required that the same type of plant be in the same area, and adjacent areas must have different plants. There are 4 different types of plants available, resulting in $\qquad$ planting schemes.
Note: The term "观赏 plants" is directly translated as "ornamental plants" in English, but I kept it in its original form to maintain the source text's integrity. If you prefer, I can translate it to "ornamental plants." Please let me know.
|
12.732.
Consider planting the same type of plant at $A$, $C$, and $E$, there are $4 \times 3 \times 3 \times 3=108$ methods.
Consider planting two types of plants at $A$, $C$, and $E$, there are $3 \times 4 \times 3 \times 3 \times 2 \times 2=432$ methods.
Consider planting three types of plants at $A$, $C$, and $E$, there are $\mathrm{P}_{4}^{3} \times 2 \times 2 \times 2=192$ methods.
Therefore, the total number of methods is $108+432+192=732$.
|
732
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given an integer $n > 3$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}$, $x_{n+1}, x_{n+2}$ satisfy the condition
$$
0 < x_{1} < x_{2} < \cdots < x_{n} < x_{n+1} < x_{n+2}.
$$
Find the minimum value of
$$
\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{\left(\sum_{k=1}^{n} \frac{x_{k+1} x_{k+2}}{x_{k+1}^{2} + x_{k} x_{k+2}}\right)\left(\sum_{l=1}^{n} \frac{x_{l+1}^{2} + x_{l} x_{l+2}}{x_{l} x_{l+1}}\right)}
$$
and determine all sets of real numbers $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ that achieve this minimum value.
(Zhang Zhusheng provided)
|
Solution: (I) Let $t_{i}=\frac{x_{i+1}}{x_{i}}(>1), 1 \leqslant i \leqslant n+1$. The expression in the problem can be written as
$$
\frac{\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} i_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i=1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)} \text {. }
$$
We observe that
$$
\begin{aligned}
( & \left(\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
= & \left(\sum_{i=1}^{n} t_{i}-\sum_{i=1}^{n} \frac{t_{i,}^{2}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
= & \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
& -\left(\sum_{i=1}^{n} \frac{t_{i}^{2}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
\leqslant & \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
& \quad-\left(\sum_{i=1}^{n} \frac{t_{i}}{\sqrt{t_{i}+t_{i+1}}} \sqrt{t_{i}+t_{i+1}}\right)^{2} \\
= & \left(\sum_{i=1}^{n} t_{i}\right)^{2}+\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right)-\left(\sum_{i=1}^{n} t_{i}\right)^{2} \\
= & \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right) .
\end{aligned}
$$
Therefore, for the real number sequence $00$, we have
$$
x_{k}=t_{k-1} t_{k-2} \cdots t_{1} a=a b^{k-1} c^{\frac{(k-1)(k-2)}{2}}, 2 \leqslant k \leqslant n
$$
+2 .
$$
\begin{array}{l}
\because x_{2}>x_{1}, \therefore b=\frac{x_{2}}{x_{1}}>1 . \\
\text{Also, since } t_{j}=b c^{j-1}>1,1 \leqslant j \leqslant n+1, \\
\therefore c>\sqrt[n]{\frac{1}{b}}\left(\geqslant \sqrt[j]{\frac{1}{b}}, 1 \leqslant j \leqslant n+1\right) .
\end{array}
$$
(III) Conclusion:
(i) For the real number sequence $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}$, $x_{n+2}$, the minimum value of the expression in the problem is 1.
(ii) The conditions for the expression to achieve its minimum value are $00, b>1, c>\sqrt[\pi]{\frac{1}{b}}$.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A university has no more than 5000 registered students, of which $\frac{1}{3}$ are freshmen, $\frac{2}{7}$ are sophomores, $\frac{1}{5}$ are juniors, and the remainder are seniors. In the mathematics department's student list, freshmen account for $\frac{1}{40}$ of all freshmen in the university, sophomores account for $\frac{1}{16}$ of all sophomores in the university, juniors account for $\frac{1}{9}$ of all juniors in the university, and seniors account for $\frac{1}{3}$ of the mathematics department's students. Therefore, the mathematics department has ) students.
$\begin{array}{llll}\text { (A) } 181 & \text { (B) } 182 & \text { (C) } 183 & \text { (D) } 184\end{array}$
|
4. (C).
Let the total number of students in the school be $n$, and the number of students in the mathematics department be $c$. Then $\frac{1}{3} n$ are first-year students, $\frac{2}{7} n$ are second-year students, and $\frac{1}{5} n$ are third-year students. The number of first-year students in the mathematics department is $\frac{1}{40} \cdot \frac{1}{3} n$, the number of second-year students is $\frac{1}{16} \cdot \frac{2}{7} n$, and the number of third-year students is $\frac{1}{9} \cdot \frac{1}{5} n$. We have
$$
\begin{array}{l}
\frac{1}{40} \cdot \frac{1}{3} n+\frac{1}{16} \cdot \frac{2}{7} n+\frac{1}{9} \cdot \frac{1}{5} n=c-\frac{1}{3} c, \\
\frac{1}{120} n+\frac{1}{56} n+\frac{1}{45} n=\frac{2}{3} c .
\end{array}
$$
The least common multiple of $120$, $56$, and $45$ is $2520$, and $2 \times$ $2520>5000$, so $n=2520$. It is easy to find that $c=183$.
|
183
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $16^{9 m}=a, 4^{37 n}=\frac{1}{a}$, then $(36 m+74 n-1)^{2000}$
|
Ni.1.1.
Since $a=16^{9 m}=\left(2^{4}\right)^{9 m}=2^{36 m}$, and $\frac{1}{a}=4^{37 n}=2^{74 n}$, then $1=a \times \frac{1}{a}=2^{36 m} \cdot 2^{74 n}=2^{36 m+74 n}$,
we have $36 m+74 n=0$.
Therefore, $(36 m+74 n-1)^{2000}=(-1)^{20000}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 2, in
Rt $\triangle A B C$,
$\angle C=90^{\circ}$, point $M$
is the intersection of the three
medians of the triangle. Perpendiculars are drawn from $M$
to $A B$, $B C$, and $A C$,
with the feet of the perpendiculars being $D$, $E$, and $F$, respectively. If $A C=3$, $B C=12$, then the area of $\triangle D E F$ is $\qquad$
|
4.4.
$$
\begin{array}{l}
AB = \sqrt{3^2 + 12^2} = \sqrt{153} = 3 \sqrt{17}, \\
S_{\triangle ABC} = \frac{1}{2} \times 3 \times 12 = 18.
\end{array}
$$
Draw a perpendicular from $C$ to $AB$, with the foot of the perpendicular at $H$, and let it intersect $EF$ at $G$. We have
$$
\begin{array}{l}
S_{\triangle ABC} = \frac{1}{2} \cdot AB \cdot CH = 18, \\
CH = \frac{18 \times 2}{3 \sqrt{17}} = \frac{12}{\sqrt{17}}. \\
\because ME = FC = \frac{1}{3} AC = 1, CE = \frac{1}{3} BC = 4. \\
\therefore EF = \sqrt{CF^2 + CE^2} = \sqrt{1 + 4^2} = \sqrt{17}. \\
\because EF \cdot CG = CE \cdot CF,
\end{array}
$$
i.e., $CG = \frac{CE \cdot CF}{EF} = \frac{4 \times 1}{\sqrt{17}} = \frac{4}{\sqrt{17}}$.
$$
\begin{array}{l}
\therefore GH = CH - CG = \frac{12}{\sqrt{17}} - \frac{4}{\sqrt{17}} = \frac{8}{\sqrt{17}}. \\
\text{Therefore, } S_{\triangle DEF} = \frac{1}{2} \cdot EF \cdot GH = \frac{1}{2} \times \sqrt{17} \times \frac{8}{\sqrt{17}} = 4.
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) In a square array of 16 rows and 16 columns composed of red and blue dots, adjacent dots of the same color are connected by a line segment of the same color, and adjacent dots of different colors are connected by a yellow line segment. It is known that there are 133 red dots, of which 32 are on the boundary of the array, and 2 are at the corners of the array. If there are 196 yellow line segments, how many blue line segments are there?
|
Three, each row has 15 line segments, 16 rows have a total of $15 \times 16=240$ horizontal line segments; 16 columns similarly have 240 vertical line segments; in total, there are 480 line segments. It is known that 195 of these line segments are yellow, so the remaining 284 line segments are red and blue.
There are $m$ red line segments. Below, we will count the number of times red points appear as endpoints of line segments using two methods.
Each yellow line segment has one red endpoint, and each red line segment has two red endpoints, so the total number of red endpoints is $195+2 m$. On the other hand, the 32 red points on the edges are each endpoints of 3 line segments, the 2 red points at the corners are each endpoints of 2 line segments, and the 99 red points inside the grid are each endpoints of 4 line segments, so the total number of red endpoints is $32 \times 3+2 \times 2+99 \times 4=496$. Therefore, $195+2 m=496$. Hence, $m=150$. Thus, the number of blue line segments is $284-150=134$.
（Yang Jin, No. 13 Middle School, Wuhu City, Anhui Province, 241002）
|
134
|
Combinatorics
|
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.