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Three. (20 points) A tetrahedron has 6 edges and 4 triangular faces. Given that the 6 edge lengths are exactly the positive integers $n, n+$ $1, n+2, n+3, n+4, n+5$. If the perimeter of a certain triangular face is a multiple of 3, then this triangle is painted red; otherwise, it is painted yellow. Question: What is the maximum number of yellow triangles? Prove your conclusion.
|
Three, at most 3 yellow triangles.
Since 6 consecutive positive integers divided by 3 must have remainders of 2 zeros, 2 ones, and 2 twos, denoted as $a, a, b, b, c, c$. If there are 4 yellow triangles, take any yellow $\triangle ABC$, then it must have two sides with the same remainder when divided by 3, denoted as $a, a$ for sides $AB \equiv AC \equiv a(\bmod 3)$. Since the two $a$s are already used, $AD$ must not have a remainder of $a$ when divided by 3, denoted as $AD \equiv b(\bmod 3)$, then among $BD, CD$,
at most one can have a remainder of $b$ when divided by 3 (as shown in Figure 10).
When $BD \equiv c(\bmod 3)$
$\triangle ABD$ is a red triangle;
When $CD \equiv c(\bmod 3)$
$\triangle ACD$ is a red triangle.
Therefore, there cannot be 4 yellow triangles.
Furthermore, when $AB=3, BC=4, CA=5, AD=6, CD=8, BD=7$, except for $\triangle ABC$, there are 3 yellow triangles.
Thus, the maximum number of yellow triangles is 3.
(Luo Zengru, Department of Mathematics, Shaanxi Normal University, 710062)
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$.
|
Explanation: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, with $x_{1}<x_{2}$, and let $C(0, c)$, where $c \neq 0$. Since $\triangle A B C$ is a right triangle, it follows that $x_{1}<0<x_{2}$, and it must be that $\angle A C B=90^{\circ}$. The graph can only be as shown in Figure 2 or Figure 3. In either case, we have $O C^{2}=O A \cdot O B$, which means
$$
c^{2}=-x_{1} x_{2}=-\frac{c}{a} .
$$
Therefore, $a c=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 7 Let $a, b, c$ be positive real numbers. Prove:
$$
\begin{array}{l}
\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+a+c)^{2}}{2 b^{2}+(c+a)^{2}}+ \\
\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8^{[3]} .
\end{array}
$$
|
Prove: Let $s=a+b+c$,
$$
f(t)=\frac{(t+s)^{2}}{2 t^{2}+(s-t)^{2}}, t \in[0, s) \text {. }
$$
Since $f(t)=\frac{1}{3}+\frac{2}{3} \times \frac{4 s t+s^{2}}{3\left(t-\frac{s}{3}\right)^{2}+\frac{2}{3} s^{2}}$
$$
\leqslant \frac{1}{3}+\frac{4 s t+s^{2}}{s^{2}}=4\left(\frac{1}{3}+\frac{t}{s}\right),
$$
thus $f(a)+f(b)+f(c)$
$$
\leqslant 4\left(\frac{1}{3} \times 3+\frac{a+b+c}{s}\right)=8 \text {. }
$$
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. The sequence $a_{1}, a_{2}, \cdots$ is defined as follows:
$$
a_{n}=2^{n}+3^{n}+6^{n}-1(n=1,2,3, \cdots) \text {. }
$$
Find all positive integers that are coprime with every term of this sequence.
(Poland provided)
|
4. The only positive integer that satisfies the condition is 1.
The following is the proof: For any prime number $p$, it must be a divisor of some term in the sequence $\left\{a_{n}\right\}$.
For $p=2$ and $p=3$, they are divisors of $a_{2}=48$.
For every prime number $p$ greater than 3, since
$$
(2, p)=1,(3, p)=1,(6, p)=1 \text {, }
$$
by Fermat's Little Theorem, we have
$$
\begin{array}{l}
2^{p-1} \equiv 1(\bmod p), \\
3^{p-1} \equiv 1(\bmod p), \\
6^{p-1} \equiv 1(\bmod p) .
\end{array}
$$
Then $3 \times 2^{p-1}+2 \times 3^{p-1}+6^{p-1}$
$$
\equiv 3+2+1=6(\bmod p) \text {, }
$$
i.e., $6 \times 2^{p-2}+6 \times 3^{p-2}+6 \times 6^{p-2} \equiv 6(\bmod p)$.
Thus, $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0(\bmod p)$.
That is, $p \mid a_{p-2}$.
For any positive integer $n$ greater than 1, it must have a prime factor $p$.
If $p \in\{2,3\}$, then $\left(n, a_{2}\right)>1$;
If $p \geqslant 5$, then $\left(n, a_{p-2}\right)>1$.
Therefore, positive integers greater than 1 do not meet the requirement.
Since 1 is coprime with all positive integers, the only positive integer that satisfies the condition is 1.
Note: The average score for this problem is 3.78.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. There are two people, A and B. A, on a car, notices that B is walking in the opposite direction. $1 \mathrm{~min}$ later, A gets off the car to chase B. If A's speed is twice that of B, but slower than the car's speed by $\frac{4}{5}$, then the time it takes for A to catch up with B after getting off the car is $\mathrm{min}$.
|
8.11.
Let the speed of A be $x \mathrm{~m} / \mathrm{min}$, then according to the problem, the speed of B is $\frac{x}{2} \mathrm{~m} / \mathrm{min}$, and the speed of the car is $5 x \mathrm{~m} / \mathrm{min}$.
Let the time it takes for A to catch up with B be $t \mathrm{~min}$. According to the problem, we have $\frac{x}{2}+5 x=x t-\frac{x}{2} t$.
Solving for $t$ yields $t=11$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. If a convex $n$-sided polygon has exactly 4 obtuse interior angles, then the maximum number of sides $n$ of this polygon is $\qquad$ .
|
10.7.
Since a convex $n$-sided polygon has exactly 4 obtuse interior angles, the sum of these 4 angles is greater than $360^{\circ}$ and less than $720^{\circ}$. The other $n-4$ angles are right angles or acute angles, so the sum of these $n-4$ angles is no more than $(n-4) \times 90^{\circ}$ and greater than $0^{\circ}$. Therefore, we have the inequality
$$
360^{\circ}+0^{\circ}<(n-2) \times 180^{\circ}<720^{\circ}+(n-4) \times 90^{\circ} \text {. }
$$
Solving this, we get $4<n<8$. Therefore, the number of sides $n$ is at most 7.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. If in a $4 \times 4$ grid of 16 cells, each cell is filled with a number such that the sum of the numbers in all adjacent cells of each cell is 1, then the sum of the 16 numbers in the grid is $\qquad$
(Note: Adjacent cells refer to cells that share exactly one edge)
|
11.6.
First, fill the 16 squares with the English letters $A, B, C$, $\cdots, O, P$ (as shown in Figure 8). Given that for each square, the sum of the numbers in all adjacent squares is 1, so,
$$
\begin{aligned}
16= & 2(A+D+M+P)+3(B+C+E+H+ \\
& I+L+N+O)+4(F+G+J+K) \\
= & 2(A+B+C+\cdots+O+P)+(B+C+E+ \\
& H+I+L+N+O)+2(F+G+J+K) \\
= & 2(A+B+C+\cdots+O+P)+(B+E+J+
\end{aligned}
$$
$$
\begin{array}{l}
G)+(C+F+K+H)+(F+I+N+K)+ \\
(G+J+O+L) .
\end{array}
$$
$$
\begin{array}{l}
\text { and } B+E+J+G=C+F+K+H \\
=F+I+N+K=G+J+O+L=1,
\end{array}
$$
Therefore, 16 $=2(A+B+C+\cdots+O+P)+4$. Hence, the sum of the 16 numbers in the grid is
$$
A+B+C+\cdots+O+P=6
$$
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given that $p$, $q$, $\frac{2 q-1}{p}$, and $\frac{2 p-1}{q}$ are all integers, and $p>1$, $q>1$. Find the value of $p+q$.
|
13. If $\frac{2 q-1}{p} \geqslant 2, \frac{2 p-1}{q} \geqslant 2$, then
$$
2 q-1 \geqslant 2 p, 2 p-1 \geqslant 2 q \text {. }
$$
Adding the two inequalities gives $2 p+2 q-2 \geqslant 2 p+2 q$. This is clearly a contradiction.
Therefore, at least one of $\frac{2 q-1}{p}$ and $\frac{2 p-1}{q}$ must be less than 2. Suppose $\frac{2 q-1}{p}<2, q>1$, then $\frac{2 q-1}{p}=1$, i.e., $2 q-1=p$.
Also, $\frac{2 p-1}{q}=\frac{4 q-3}{q}$ is an integer, i.e., $4-\frac{3}{q}$ is an integer, so $q=1$ or $q=3$.
Since $q>1$, then $q=3$, and simultaneously $p=2 q-1=5$, hence $p+q=8$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $a, b, x, y$ are non-negative real numbers, and $a+b=27$. Try to find the maximum value of $\lambda$, such that the inequality
$$
\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant \lambda\left(a x^{2} y+b x y^{2}\right)^{2}
$$
always holds, and find the conditions for equality.
|
Let $a=0, b=27, x=27, y=2$, then the original inequality is $\lambda \leqslant 4$.
We will now prove that $\lambda=4$ makes the given inequality always true.
It is only necessary to prove under the original conditions that
$$
\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant 4(a x+b y)^{2} x^{2} y^{2}.
$$
When $x$ or $y$ is 0, inequality (1) is obviously true.
When $x$ and $y$ are both greater than 0, divide both sides of inequality (1) by $x^{3} y^{3}$ and let $t=\frac{x}{y}$, then inequality (1) becomes
$$
\left(a t+\frac{b}{t}+4\right)^{3} \geqslant 4\left(a^{2} t+\frac{b^{2}}{t}+2 a b\right).
$$
Here $t \in \mathbf{R}_{+}$. Since:
$$
\begin{array}{l}
\left(a t+\frac{b}{t}\right)(a+b) \\
=a^{2} t+\frac{b^{2}}{t}+a b\left(t+\frac{1}{t}\right) \\
\geqslant a^{2} t+\frac{b^{2}}{t}+2 a b,
\end{array}
$$
it is only necessary to prove
$$
\left(a t+\frac{b}{t}+4\right)^{3} \geqslant 4\left(a t+\frac{b}{t}\right) \times 27.
$$
Since $a t+\frac{b}{t}+4=a t+\frac{b}{t}+2+2$
$$
\geqslant 3 \sqrt[3]{4\left(a t+\frac{b}{t}\right)},
$$
we have $\left(a t+\frac{b}{t}+4\right)^{3} \geqslant 27 \times 4\left(a t+\frac{b}{t}\right)$.
Therefore, $\lambda_{\text {max }}=4$.
The equality in inequality (1) holds if and only if $a t+\frac{b}{t}=2, t=1$ or $x=0, b=0$ or $x=0, y=0$ or $y=0, a=0$.
Since $a+b=27>2$, the equality in inequality (1) holds if and only if $x=0, a=27, b=0$ or $y=0, a=0, b=27$. Or $x=0, y=0$.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $x=\frac{13}{4+\sqrt{3}}$, then $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+3}{x^{3}-7 x^{2}+5 x+15}=$ $\qquad$
|
$$
\text { II.1. }-5 \text {. }
$$
Since $x=\frac{13}{4+\sqrt{3}}=4-\sqrt{3}$, then $x^{2}-8 x+13=0$.
$$
\begin{array}{l}
\text { Hence } x^{4}-6 x^{3}-2 x^{2}+18 x+3 \\
=\left(x^{2}+2 x+1\right)\left(x^{2}-8 x+13\right)-10=-10, \\
x^{3}-7 x^{2}+5 x+15 \\
=(x+1)\left(x^{2}-8 x+13\right)+2=2 .
\end{array}
$$
Therefore, the original expression $=\frac{-10}{2}=-5$.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3: There are $n$ people, and it is known that any two of them make at most one phone call. The total number of phone calls made among any $n-2$ people is equal, and it is $3^{k}$ times, where $k$ is a positive integer. Find all possible values of $n$.
(2000, National High School Mathematics Competition)
Analysis: Carefully savor the quantitative relationships given in the problem, and you can obtain its quantitative characteristics.
Assume the $n$ people are $A_{1}, A_{2}, \cdots, A_{n}$, and the number of calls made by $A_{i}$ is $m_{i}$. The number of calls between $A_{i}$ and $A_{j}$ is $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1. Thus, the quantitative characteristic is
$$
m_{i}+m_{j}-\lambda_{i j}=\frac{1}{2} \sum_{s=1}^{n} m_{s}-3^{k}=c,
$$
where $c$ is a constant. Consequently, we conjecture that $m_{i}(i=1,2, \cdots, n)$ is a constant.
|
Let $n$ people be denoted as $A_{1}, A_{2}, \cdots, A_{n}$. Let the number of calls made by $A_{i}$ be $m_{i}$, and the number of calls between $A_{i}$ and $A_{j}$ be $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1.
Clearly, $n \geqslant 5$. Therefore,
$$
\begin{array}{l}
\left|m_{i}-m_{j}\right|=\left|\left(m_{i}+m_{s}\right)-\left(m_{j}+m_{s}\right)\right| \\
=\left|\lambda_{i s}-\lambda_{j}\right| \leqslant 1,1 \leqslant i, j, s \leqslant n .
\end{array}
$$
Let $m_{i}=\max \left\{m_{s}, 1 \leqslant s \leqslant n\right\}$,
$$
m_{j}=\min \left\{m_{s}, 1 \leqslant s \leqslant n\right\} .
$$
Thus, $m_{i}-m_{j} \leqslant 1$.
If $m_{i}-m_{j}=1$, then for any $s \neq i, j, 1 \leqslant s \leqslant n$, we have
$$
\begin{array}{l}
\left(m_{i}+m_{s}-\lambda_{i s}\right)-\left(m_{j}+m_{s}-\lambda_{j s}\right) \\
=1-\left(\lambda_{i s}-\lambda_{j s}\right)=0,
\end{array}
$$
which means $\lambda_{i s}-\lambda_{j s} \equiv 1,1 \leqslant i, j, s \leqslant n$.
Therefore, $\lambda_{i s} \equiv 1, \lambda_{j i} \equiv 0, s \neq i, j$, and $1 \leqslant s \leqslant n$.
Thus, $m_{i} \geqslant n-2, m_{j} \leqslant 1$.
Hence, $m_{i}-m_{j} \geqslant n-3 \geqslant 2$, which is a contradiction.
Therefore, $m_{i}-m_{j}=0$.
Then $\lambda_{i j} \equiv 0$ or $\lambda_{i j} \equiv 1$.
If $\lambda_{i j} \equiv 0$, then $m_{s} \equiv 0,1 \leqslant s \leqslant n$, which is a contradiction.
If $\lambda_{i j} \equiv 1$, then $m_{s}=n-1,1 \leqslant s \leqslant n$.
Thus, $(n-2)(n-3)=3^{k} \times 2$.
Let $n-2=2 \times 3^{k_{1}}, n-3=3^{k_{2}}, k_{1} \geqslant k_{2}$.
From $2 \times 3^{k_{1}}-3^{k_{2}}=1$, we get
$$
3^{k_{2}}=1 \text{. }
$$
Therefore, $k_{1}=k_{2}=0$. This contradicts $k \geqslant 1$.
Let $n-2=3^{k_{1}}, n-3=2 \times 3^{k_{2}}, k_{1} \geqslant k_{2}+1$.
From $3^{k_{1}}-2 \times 3^{k_{2}}=1$, we get
$$
k_{2}=0, k_{1}=1 \text{. }
$$
Therefore, $n=5$ is the solution.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $a 、 b 、 c$ are real numbers, and
$$
a^{2}+b^{2}+c^{2}+2 a b=1, a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{8} \text {, }
$$
The roots of the quadratic equation $(a+b) x^{2}-(2 a+c) x-(a+b)=0$ are $\alpha 、 \beta$. Find the value of $2 \alpha^{3}+\beta^{-5}-\beta^{-1}$.
|
Solution: From the given, we have
$$
\left\{\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)+2 a b=1, \\
2 a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{4} .
\end{array}\right.
$$
Thus, \(a^{2}+b^{2}+c^{2}\) and \(2 a b\) are the roots of the equation \(t^{2}-t+\frac{1}{4}=0\).
Since the roots of the equation \(t^{2}-t+\frac{1}{4}=0\) are \(t_{1}=t_{2}=\frac{1}{2}\), we have
$$
a^{2}+b^{2}+c^{2}=2 a b=\frac{1}{2} \text {. }
$$
Solving this, we get \(a=b= \pm \frac{1}{2}, c=0\).
Thus, the given equation can be transformed into
$$
x^{2}-x-1=0 \text {. }
$$
Since \(\alpha, \beta\) are the roots of equation (1), then \(\alpha+\beta=1\), and
$$
\left\{\begin{array}{l}
\alpha^{2}-\alpha-1=0, \\
\beta^{2}-\beta-1=0 .
\end{array}\right.
$$
From equation (2), we get \(\alpha^{2}=\alpha+1\). Therefore,
$$
\alpha^{3}=\alpha \cdot \alpha^{2}=\alpha(\alpha+1)=\alpha^{2}+\alpha=2 \alpha+1 \text {. }
$$
Clearly, \(\beta \neq 0\).
Dividing both sides of equation (3) by \(\beta\) and \(\beta^{2}\), and rearranging, we get
$$
\frac{1}{\beta}=\beta-1, \frac{1}{\beta^{2}}=1-\frac{1}{\beta}=2-\beta \text {. }
$$
$$
\begin{array}{l}
\text { and } \beta^{-3}=\beta^{-1} \cdot \beta^{-2}=(\beta-1)(2-\beta) \\
=3 \beta-\beta^{2}-2=2 \beta-3, \\
\beta^{-5}=\beta^{-2} \cdot \beta^{-3} \\
=(2-\beta)(2 \beta-3)=7 \beta-2 \beta^{2}-6 \\
=7 \beta-2(\beta+1)-6=5 \beta-8,
\end{array}
$$
Thus, \(2 \alpha^{3}+\beta^{-5}-\beta^{-1}=4(\alpha+\beta)-5=-1\).
(Zhang Zhecai, Shengzhou Middle School, Jiaojiang District, Taizhou City, Zhejiang Province, 318000)
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 When $m=$ $\qquad$, the polynomial
$$
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m
$$
can be factored into the product of two linear factors.
(1992, Zhengzhou City Junior High School Mathematics Competition)
|
Solution: First, factorize the quadratic term, we have
$$
12 x^{2}-10 x y+2 y^{2}=(3 x-y)(4 x-2 y) \text {. }
$$
Therefore, we can set
$$
\begin{array}{l}
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m \\
=(3 x-y+a)(4 x-2 y+b) \\
=(3 x-y)(4 x-2 y)+(4 a+3 b) x- \\
(2 a+b) y+a b .
\end{array}
$$
Comparing the coefficients of the linear terms, we get
$$
\left\{\begin{array}{l}
4 a+3 b=11, \\
2 a+b=5 .
\end{array}\right.
$$
Solving these, we find $a=2, b=1$.
Comparing the constant term, we get $m=a b=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The axial cross-section of a wine glass is part of a parabola, whose equation is $x^{2}=2 y(0 \leqslant y<15)$. If a glass ball with a radius of 3 is placed inside the cup, then the distance from the highest point of the ball to the bottom of the cup is $\qquad$
|
5.8 .
As shown in the figure, let the center of the sphere be at $(0, b)$.
Then we have
$$
x^{2}+(y-6)^{2}=9 \text {. }
$$
Therefore, the system of equations
$$
\left\{\begin{array}{l}
x^{2}=2 y \\
x^{2}+(y-b)^{2}=9
\end{array}\right.
$$
has two solutions. For $y$,
$$
y^{2}+2(1-b) y+b^{2}-9=0 \text {. }
$$
By $\Delta=0$, we solve to get $b=5$.
Thus, the distance from the highest point of the sphere to the bottom of the cup is
$$
d=b+r=5+3=8 .
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 4 If real numbers $a, b, c$ satisfy
$$
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1 \text{, }
$$
find the value of $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}$.
(1999, Changsha Junior High School Mathematics Competition)
|
Solution: Construct the identity
$$
\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1 \text {. }
$$
Subtract the above identity from the given equation to get
$$
\begin{array}{l}
\left(\frac{a}{b+c}-\frac{a}{a+b+c}\right)+\left(\frac{b}{a+c}-\frac{b}{a+b+c}\right)+ \\
\left(\frac{c}{a+b}-\frac{c}{a+b+c}\right)=0 .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
\frac{a^{2}}{(b+c)(a+b+c)}+\frac{b^{2}}{(a+c)(a+b+c)}+ \\
\frac{c^{2}}{(a+b)(a+c+b)}=0,
\end{array}
$$
which means $\square$
$$
\begin{array}{l}
\frac{1}{a+b+c}\left(\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}\right)=0 . \\
\text { Hence } \frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 2, two diameters $A C$ and $B D$ of the large circle intersect perpendicularly at point $O$. Four semicircles are drawn outside the large circle with $A B$, $B C$, $C D$, and $D A$ as diameters, respectively. The total area of the four "crescent" shaded regions in the figure is $2 \mathrm{~cm}^{2}$. Then the radius of the large circle is $\qquad$ $\mathrm{cm}$.
|
$=.1 .1$.
By the Pythagorean theorem, we know $A D^{2}+C D^{2}=A C^{2}$, so the area of the upper half of the large circle is equal to the sum of the areas of the two semicircles with diameters $A D$ and $C D$. Similarly, the area of the lower half of the large circle is equal to the sum of the areas of the two semicircles with diameters $A B$ and $B C$. Therefore, the area of the square $A B C D$ is equal to the total area of the four "lunar" shapes. It is easy to calculate that the radius of the large circle $O D$ is $1 \mathrm{~cm}$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (15 points) Given non-zero real numbers $a, b, c$ satisfy $a+b+c=0$. Prove:
(1) $a^{3}+b^{3}+c^{3}=3 a b c$;
(2) $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9$.
|
Three, (1) From $a+b+c=0$, we get $a+b=-c$. Therefore, $(a+b)^{3}=-c^{3}$.
Thus, $a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=-c^{3}$.
Hence $a^{3}+b^{3}+c^{3}=-3 a b(a+b)=-3 a b(-c)=3 a b c$.
$$
\begin{array}{l}
\text { (2) }\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{c}{a-b} \\
=1+\left(\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{c}{a-b}=1+\frac{2 c^{2}}{a b} \text {. }
\end{array}
$$
Similarly, $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{a}{b-c}=1+\frac{2 a^{2}}{b c}$,
$$
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{b}{c-a}=1+\frac{2 b^{2}}{a c} \text {. }
$$
Therefore, $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)$ $=1+\frac{2 c^{2}}{a b}+1+\frac{2 a^{2}}{b c}+1+\frac{2 b^{2}}{a c}$
$$
=3+\frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}=3+\frac{2 \times 3 a b c}{a b c}=9 \text {. }
$$
|
9
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $n(n \geqslant 3)$ be a positive integer. If there are $n$ lattice points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane such that: when $\left|P_{i} P_{j}\right|$ is a rational number, there exists $P_{k}$ such that $\left|P_{i} P_{k}\right|$ and $\left|P_{j} P_{k}\right|$ are both irrational; when $\left|P_{i} P_{j}\right|$ is an irrational number, there exists $P_{k}$ such that $\left|P_{i} P_{k}\right|$ and $\left|P_{j} P_{k}\right|$ are both rational, then $n$ is called a "good number".
(1) Find the smallest good number;
(2) Is 2005 a good number?
|
6. We assert that the smallest good number is 5, and 2005 is a good number.
In a triplet $\left(P_{i}, P_{j}, P_{k}\right)$, if $\left|P_{i} P_{j}\right|$ is a rational number (or irrational number), and $\left|P_{i} P_{k}\right|, \left|P_{j} P_{k}\right|$ are irrational numbers (or rational numbers), then $\left(P_{i}, P_{j}, P_{k}\right)$ is called a good triplet.
(1) $n=3$ is clearly not a good number.
$n=4$ is also not a good number.
If otherwise, assume $P_{1}, P_{2}, P_{3}, P_{4}$ satisfy the conditions. As shown in Figure 4, without loss of generality, assume $\left|P_{1} P_{2}\right|$ is a rational number and $\left(P_{1}, P_{2}, P_{3}\right)$ is a good triplet, then $\left(P_{2}, P_{3}, P_{4}\right)$ is a good triplet. Clearly, $\left(P_{2}, P_{4}, P_{1}\right)$ and $\left(P_{2}, P_{4}, P_{3}\right)$ are not good triplets. Therefore, $P_{1}, P_{2}, P_{3}, P_{4}$ do not satisfy the conditions. Contradiction.
$n=5$ is a good number.
The following five lattice points satisfy the conditions:
$$
A_{5}=\{(0,0),(1,0),(5,3),(8,7),(0,7)\} \text {. }
$$
As shown in Figure 5.
(2) Let $A=\{(1,0),(2,0), \cdots,(669,0)\}$,
$$
\begin{array}{l}
B=\{(1,1),(2,1), \cdots,(668,1)\}, \\
C=\{(1,2),(2,2), \cdots,(668,2)\}, \\
S_{2005}=A \cup B \cup C .
\end{array}
$$
For any positive integer $n$, it is easy to prove that $n^{2}+1$ and $n^{2}+4$ are not perfect squares. It is not difficult to prove that for any two points $P_{i}$ and $P_{j}$ in the set $S_{2005}, \left|P_{i} P_{j}\right|$ is a rational number if and only if $P_{i} P_{j}$ is parallel to one of the coordinate axes. Therefore, 2005 is a good number.
Note: When $n=6$, we have $A_{6}=A_{5} \cup\{(-24,0)\}$;
When $n=7$, we have $A_{7}=A_{6} \cup\{(-24,7)\}$.
It can be verified that $n=6,7$ are both good numbers.
When $n \geqslant 8$, they can be arranged in three rows like $n=2005$, indicating that for $n \geqslant 8$, all $n$ are good numbers.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given positive integers $a, b, c, d$ satisfying $b<a<d<c$, and the sums of each pair are $26, 27, 41, 101, 115, 116$. Find the value of $(100a + b) - (100d - c)$.
|
Three, from $b<a<d<c$, we can get
$$
\begin{array}{l}
a+b<b+d<b+c<a+c<d+c, \\
b+d<a+d<a+c .
\end{array}
$$
Therefore, $a+d$ and $b+c$ are both between $b+d$ and $a+c$.
It is easy to know that $a+b=26, b+d=27, a+c=115, d+c=116$.
If $a+d=101$, then $b+c=41$.
Solving this gives $b=-24$ (discard).
Thus, $a+d=41, b+c=101$.
Solving this gives $a=20, b=6, c=95, d=21$.
Therefore, $(100 a+b)-(100 d-c)=2006-2005=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. If $2^{6}+2^{9}+2^{n}$ is a perfect square, then the positive integer $n=$ $\qquad$ .
|
11. 10 .
$$
2^{6}+2^{9}=2^{6}\left(1+2^{3}\right)=2^{6} \times 3^{2}=24^{2} \text {. }
$$
Let $24^{2}+2^{n}=a^{2}$, then
$$
(a+24)(a-24)=2^{n} \text {. }
$$
Thus, $a+24=2^{r}, a-24=2^{t}$,
$$
2^{r}-2^{t}=48=2^{4} \times 3,2^{t}\left(2^{r-t}-1\right)=2^{4} \times 3 \text {. }
$$
Then $t=4, r-t=2$.
So $r=6, n=t+r=10$.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x=\frac{3-\sqrt{5}}{2}$. Then $x^{3}-3 x^{2}+3 x+$ $\frac{6}{x^{2}+1}=$ . $\qquad$
|
$$
=, 1.6 \text {. }
$$
Given $x=\frac{3-\sqrt{5}}{2}$, we know that
$$
\begin{array}{l}
x^{2}-3 x+1=0, x+\frac{1}{x}=3 . \\
\text { Therefore, } x^{3}-3 x^{2}+3 x+\frac{6}{x^{2}+1} \\
=x\left(x^{2}-3 x\right)+3 x+\frac{6}{x^{2}+1}=2 x+\frac{6}{3 x} \\
=2 x+\frac{2}{x}=2\left(x+\frac{1}{x}\right)=6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a$ and $b$ are real numbers, and $a \geqslant 1$. If the equation $x^{2}-2 b x-\left(a-2 b^{2}\right)=0$ has real solutions, and satisfies $2 a^{2}-a b^{2}-5 a+b^{2}+4=0$, then $a^{2}+b^{2}=$
|
3.6 .
From the equation $x^{2}-2 b x-\left(a-2 b^{2}\right)=0$ having real solutions, we get $\Delta=4 b^{2}+4\left(a-2 b^{2}\right) \geqslant 0$, which means $a \geqslant b^{2}$.
From $2 a^{2}-a b^{2}-5 a+b^{2}+4=0$, we can derive $2 a^{2}-5 a+4=a b^{2}-b^{2}=b^{2}(a-1) \leqslant a(a-1)$. Therefore, $2 a^{2}-5 a+4 \leqslant a^{2}-a$, which simplifies to $(a-2)^{2} \leqslant 0$.
Solving this, we get $a=2, b^{2}=2$.
Thus, $a^{2}+b^{2}=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The maximum value of the algebraic expression $a \sqrt{2-b^{2}}+b \sqrt{2-a^{2}}$ is $\qquad$ .
|
5.2.
It is known that $|a| \leqslant \sqrt{2},|b| \leqslant \sqrt{2}$. Let
$$
a=\sqrt{2} \sin \alpha, b=\sqrt{2} \sin \beta, \alpha, \beta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text {. }
$$
Then $a \sqrt{2-b^{2}}+b \sqrt{2-a^{2}}$
$$
\begin{array}{l}
=2(\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \alpha) \\
=2 \sin (\alpha+\beta) \leqslant 2 .
\end{array}
$$
Equality holds if and only if $\alpha+\beta=\frac{\pi}{2}$, i.e., $a^{2}+b^{2}=2, a, b \geqslant 0$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a regular $n$-sided polygon inscribed in a circle. If the line connecting a vertex to the center of the circle is exactly on the perpendicular bisector of a certain side, then $n \equiv$ $\qquad$ $(\bmod 2)$
|
2.1.
As shown in Figure 8, if the line connecting vertex $A_{1}$ and the center $O$ perpendicularly bisects side $A_{i} A_{j}$, then $\triangle A_{1} A_{i} A_{j}$ is an isosceles triangle. Since equal chords subtend equal arcs, the number of vertices in $\overparen{A_{1} A_{i}}$ is equal, and their sum is even. Adding the points $A_{1} 、 A_{i} 、 A_{j}$, $n$ must be odd. Therefore, $n \equiv 1(\bmod 2)$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 As shown in Figure 1, quadrilateral $ABCD$ is a rectangle. Two people, A and B, start from points $A$ and $B$ respectively at the same time, and move counterclockwise along the rectangle. In which circle does B possibly catch up with A for the first time? In which circle does B certainly catch up with A at the latest? Please explain your reasoning.
|
Explanation: Let $A D=B C=a \mathrm{~m}, A B=C D=b . \mathrm{m}$, and suppose the first time Yi catches up with Jia is after
$$
\frac{2 a+b}{74-65}=\frac{2 a+b}{9}
$$
minutes. The distance Yi has run when he first catches up with Jia is
$$
\frac{2 a+b}{9} \times 74(\mathrm{~m}) \text {. }
$$
At this point, the number of laps Yi has run is $p$, then
$$
\begin{array}{l}
p=\frac{74(2 a+b)}{9} \div(2 a+2 b) \\
=\frac{37(2 a+b)}{9(a+b)}=4+\frac{38 a+b}{9(a+b)} \\
=9-\frac{7 a+44 b}{9(a+b)} .
\end{array}
$$
Thus, when $4\frac{35}{2} b$, we have $\frac{7 a+44 b}{9(a+b)}<1$, so Yi will certainly catch up with Jia by the time he completes the 9th lap at most.
Note: The key to solving this problem is the skillful use of partial fraction decomposition.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 4 Let the function $f(x)=\frac{1}{k-x}$, and denote
$$
\underbrace{f(f(\cdots)}_{\text {j times }}
$$
as $f_{j}(x)$. Given that $f_{n}(x)=x$, and
$$
f_{i}(x) \neq f_{j}(x)(i \neq j, 1 \leqslant i, j \leqslant n) \text {. }
$$
Prove: $\prod_{i=1}^{n} f_{i}(x)=-1$.
|
Proof: Consider the recursive sequence $g_{j}=k g_{j-1}-g_{j-2}$, $g_{0}=0, g_{1}=1$. We have
$$
\begin{array}{l}
f_{1}(x)=\frac{1}{k-x}=\frac{x g_{0}-g_{1}}{x g_{1}-g_{2}}, \\
f_{2}(x)=\frac{x g_{1}-g_{2}}{x g_{2}-g_{3}}, \\
\cdots \cdots \\
f_{n}(x)=\frac{x g_{n-1}-g_{n}}{x g_{n}-g_{n+1}} .
\end{array}
$$
Multiplying the above equations, we get
$$
\begin{array}{l}
\prod_{i=1}^{n} f_{i}(x)=\frac{x g_{0}-g_{1}}{x g_{n}-g_{n+1}} . \\
\text { Also, } f_{n}(x)=x=\frac{x g_{n-1}-g_{n}}{x g_{n}-g_{n+1}} \\
=\frac{x g_{n-1}-g_{n}}{x g_{n}-k g_{n}+g_{n-1}},
\end{array}
$$
Then, $g_{n} x^{2}-g_{n} k x+g_{n}=0$ always holds, i.e.,
$$
\left\{\begin{array}{l}
g_{n}=0, \\
-g_{n} k=0, \\
g_{n}=0 .
\end{array}\right.
$$
Thus, $g_{n}=0$.
So, $g_{n}=0=k g_{n-1}-g_{n-2}$.
This gives $g_{n-1}=\frac{g_{n-2}}{g_{2}}=k g_{n-2}-g_{n-3}$.
Therefore, $g_{n-2}=\frac{g_{n-3}}{\frac{g_{3}}{g_{2}}}$, i.e., $\frac{g_{n-2}}{g_{2}}=\frac{g_{n-3}}{g_{3}}$.
Thus, $g_{n-1}=\frac{g_{n-2}}{g_{2}}=\frac{g_{n-3}}{g_{3}}=\cdots$.
If $n$ is even,
$$
g_{n-1}=\frac{g_{n-2}}{g_{2}}=\frac{g_{n-3}}{g_{3}}=\cdots=\frac{g_{\frac{n}{2}}}{g_{\frac{n}{2}}}=1 ;
$$
If $n$ is odd,
$$
\begin{array}{l}
g_{n-1}=\frac{g_{n-2}}{g_{2}}=\frac{g_{n-3}}{g_{3}}=\cdots=\frac{g_{\left[\frac{n}{2}\right]+1}}{g_{\left[\frac{n}{2}\right]}} \\
=\frac{g_{\left[\frac{n}{2}\right]}}{g_{\left[\frac{n}{2}\right]+1}}=\frac{g_{\left[\frac{n}{2}\right]}+g_{\left[\frac{n}{2}\right]+1}}{g_{\left[\frac{n}{2}\right]}+g_{\left[\frac{n}{2}\right]+1}}=1 .
\end{array}
$$
Hence, $g_{n+1}=k g_{n}-g_{n-1}=-1$.
Therefore, $\prod_{i=1}^{n} f_{i}(x)=\frac{x g_{0}-g_{1}}{x g_{n}-g_{n+1}}=-1$.
|
-1
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. For a positive integer $n$, if there exist positive integers $a$ and $b$ such that $n=a+b+a b$, then $n$ is called a "good number". For example, $3=1+1+1 \times 1$, so 3 is a good number. Then, among the 20 positive integers from $1 \sim 20$, the number of good numbers is $\qquad$ .
|
4.12.
$n+1=a+b+a b+1=(a+1)(b+1)$ is a composite number, so the required $n$ is the number obtained by subtracting 1 from the composite numbers between 2 and 21. The composite numbers between 2 and 21 are $4,6,8,9,10,12,14,15,16,18,20,21$, a total of 12. Therefore, the required $n$ has 12 values: $3,5,7,8,9,11,13,14,15,17,19,20$.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let the set $M=\left\{a \left\lvert\, a=\frac{x+y}{t}\right., 2^{x}+2^{y}=\right.$ $2^{t}$, where $x, y, t, a$ are all integers $\}$. Then the sum of all elements in the set $M$ is ( ).
(A) 1
(B) 4
(C) 7
(D) 8
|
6. D.
Assume $x \leqslant y$, then $2^{x}=2^{x}+2^{y} \leqslant 2^{y}+2^{y}=2^{y+1}$.
Thus, $t \leqslant y+1$.
From $2^{x}>0$, we get $2^{t}=2^{x}+2^{y}>2^{y}$. Therefore, $t>y$.
So, $y<t \leqslant y+1$.
Given that $x, y, t$ are all integers, then $t=y+1$. Hence, $2^{y+1}=$ $2^{x}+2^{y}$, which means $2^{x}=2^{y}$. Therefore, $x=y=t-1$.
Thus, $a=\frac{x+y}{t}=2-\frac{2}{t}$.
Here $a, t \in \mathbf{Z}$, so $t= \pm 1, \pm 2$.
Then $a=0,1,3,4$.
Therefore, the sum of all elements in set $M$ is $0+1+3+4=8$.
|
8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given a fixed point $A(4, \sqrt{7})$. If a moving point $P$ is on the parabola $y^{2}=4 x$, and the projection of point $P$ on the $y$-axis is point $M$, then the maximum value of $|P A|-|P M|$ is $\qquad$.
|
$$
\begin{array}{l}
|P M|=|P N|-|M N|=|P F|-1 \downarrow \\
\text { Then }|P A|-|P M|=|P A|-(|P F|-1) \\
=(|P A|-|P F|)+1 \leqslant|A F|+1=4+1=5 .
\end{array}
$$
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n}+a_{n+1}=1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
If $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, then the value of $S_{2003}-2 S_{2004}+S_{2000}$ is $\qquad$
|
9.3.
According to the problem, when $n$ is even, we have $a_{1}+a_{2}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n}{2}$ pairs, so $S_{n}=\frac{n}{2}$. Therefore, $S_{2004}=1002$.
When $n$ is odd, we have $a_{2}+a_{3}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n-1}{2}$ pairs, so
$$
S_{n}=a_{1}+\frac{n-1}{2}=2+\frac{n-1}{2}=\frac{n+3}{2} \text {. }
$$
Therefore, $S_{2003}=1003, S_{2005}=1004$.
Thus, $S_{2003}-2 S_{2004}+S_{2005}$
$$
=1003-2 \times 1002+1004=3 .
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Let two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ in a plane be perpendicular to each other, and $|\boldsymbol{a}|=2,|\boldsymbol{b}|=1$. Also, $k$ and $t(t \geqslant 0)$ are two real numbers that are not both zero. If the vectors
$$
\boldsymbol{x}=\boldsymbol{a}+(3-t) \boldsymbol{b} \text { and } \boldsymbol{y}=-k \boldsymbol{a}+t^{2} \boldsymbol{b}
$$
are perpendicular to each other, then the maximum value of $k$ is $\qquad$
|
13.1.
Given $\boldsymbol{a} \cdot \boldsymbol{b}=0$.
$$
\begin{array}{l}
\boldsymbol{x} \cdot \boldsymbol{y}=[\boldsymbol{a}+(3-t) \boldsymbol{b}] \cdot\left(-k \boldsymbol{a}+t^{2} \boldsymbol{b}\right) \\
=-k \boldsymbol{a}^{2}+\left[-k(3-t)+t^{2}\right] \boldsymbol{a} \cdot \boldsymbol{b}+t^{2}(3-t) \boldsymbol{b}^{2} \\
=-k|\boldsymbol{a}|^{2}+t^{2}(3-t)|\boldsymbol{b}|^{2} \\
=-4 k+t^{2}(3-t) .
\end{array}
$$
According to the problem, $-4 k+t^{2}(3-t)=0$.
Therefore, $k=-\frac{1}{4} t^{3}+\frac{3}{4} t^{2}$.
Let $k=f(t)$, then
$$
f^{\prime}(t)=-\frac{3}{4} t^{2}+\frac{3}{2} t=-\frac{3}{4} t(t-2) .
$$
Since $t \neq 0$ (otherwise $k=0$), when $t=2$, $f^{\prime}(t)=0$, and $00, t>2$ when $f^{\prime}(t)<0$. Therefore, when $t=2$, $k$ reaches its maximum value of 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If there exists a permutation $a_{1}$, $a_{2}, \cdots, a_{n}$ of $1,2, \cdots, n$, such that $k+a_{k}(k=1,2, \cdots, n)$ are all perfect squares, then $n$ is called a "middle number". Then, in the set $\{15,17,2000\}$, the number of elements that are middle numbers is $\qquad$ .
|
5.3.
(1) 15 is a median number. Because in the arrangement of Table 1, $k+a_{k}$ $(k=1,2, \cdots, 15)$ are all perfect squares:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$k$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
\hline$a_{k}$ & 15 & 14 & 13 & 12 & 11 & 10 & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
\hline$k+a_{k}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ \\
\hline
\end{tabular}
(2) 17 is a median number. Because in the arrangement of Table 2, $k+a_{k}$ $(k=1,2, \cdots, 17)$ are all perfect squares:
Table 2
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$k$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\
\hline$a_{k}$ & 3 & 7 & 6 & 5 & 4 & 10 & 2 & 17 & 16 & 15 & 14 & 13 & 12 & 11 & 1 & 9 & 8 \\
\hline$k+a_{k}$ & $2^{2}$ & $3^{2}$ & $3^{2}$ & $3^{2}$ & $3^{2}$ & $4^{2}$ & $3^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $4^{2}$ & $5^{2}$ & $5^{2}$ \\
\hline
\end{tabular}
(3) 2006 is a median number. Because it can be arranged in such a way that $k+a_{k}(k=1,2, \cdots, 2006)$ are all perfect squares:
For $k$ taking $1,2, \cdots, 17, a_{k}$ can correspond as in (2); for $k$ taking 18, $a_{k}=18$ can correspond; for $k$ taking $19,20, \cdots, 2006$, $a_{k}$ can sequentially take $2006,2005, \cdots, 19$ to correspond.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given the set $M=\left\{x \left\lvert\, x=\lim _{n \rightarrow \infty} \frac{2^{n+1}-2}{\lambda^{n}+2^{n}}\right.\right.$, $\lambda$ is a constant, and $\lambda+2 \neq 0\}$. Then the sum of all elements of $M$ is $\qquad$ .
|
10.3.
When $|\lambda|>2$, $x=\lim _{n \rightarrow \infty} \frac{2\left(\frac{2}{\lambda}\right)^{n}-2\left(\frac{1}{\lambda}\right)^{n}}{1+\left(\frac{2}{\lambda}\right)^{n}}=0$;
When $\lambda=2$, $x=\lim _{n \rightarrow \infty}\left(1-\frac{1}{2^{n}}\right)=1$;
When $|\lambda|<2$, $x=\lim _{n \rightarrow \infty} \frac{2-2\left(\frac{1}{2}\right)^{n}}{\left(\frac{\lambda}{2}\right)^{n}+1}=2$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=2, a_{n}=\frac{1+a_{n-1}}{1-a_{n-1}}(n \geqslant 2) \text {. }
$$
Then the value of $a_{2006}$ is $\qquad$ .
|
3.2.
Since $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}, a_{5}=$ 2, we conjecture that $a_{n}$ is a periodic sequence with a period of 4. And
$$
a_{n+4}=\frac{1+a_{n+3}}{1-a_{n+3}}=\frac{1+\frac{1+a_{n+2}}{1-a_{n+2}}}{1-\frac{1+a_{n+2}}{1-a_{n+2}}}=-\frac{1}{a_{n+2}}=a_{n} .
$$
Therefore, $a_{2008}=a_{4 \times 501+1}=a_{1}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5, 5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$, such that for all integers $n$, $a_{n}=$ $b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$. Then $b+c+d$ equals
|
5.2.
Divide the known sequence into groups as follows:
$$
\begin{array}{l}
(1),(3,3,3),(5,5,5,5,5), \cdots, \\
(\underbrace{2 k-1,2 k-1, \cdots, 2 k-1}_{2 k-1 \uparrow}),
\end{array}
$$
Let $a_{n}$ be in the $k$-th group, where $a_{n}=2 k-1$. Then we have
$$
\begin{array}{l}
1+3+5+\cdots+2 k-3+1 \\
\leqslant n0$, solving this gives $\sqrt{n-1}<k \leqslant \sqrt{n-1}+1$.
Therefore, $k=[\sqrt{n-1}+1]=[\sqrt{n-1}]+1$.
Thus, $a_{n}=2[\sqrt{n-1}]+1$.
Hence, $b+c+d=2+(-1)+1=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $x$, $y$, $z$ are positive real numbers, and $x y z(x+y+z)=1$. Then the minimum value of $(x+y)(y+z)$ is $\qquad$
|
\begin{array}{l}=1.2. \\ (x+y)(y+z)=y^{2}+(x+z) y+x z \\ =y(x+y+z)+x z=y \cdot \frac{1}{x y z}+x z \\ =\frac{1}{x z}+x z \geqslant 2 .\end{array}
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given the quadratic trinomial $a x^{2}+b x+c$ $(a>0)$.
(1) When $c<0$, find the maximum value of the function
$$
y=-2\left|a x^{2}+b x+c\right|-1
$$
(2) For any real number $k$, the line $y=k(x-$
1) $-\frac{k^{2}}{4}$ intersects the parabola $y=a x^{2}+b x+c$ at exactly one point, find the value of $a+b+c$.
|
(1) Given $a>0, c<0$, we know that $y^{\prime}=a x^{2}+b x+c$ intersects the $x$-axis, and $y_{\text {nuin }}^{\prime}<0$. Therefore,
$$
\left|y^{\prime}\right|=\left|a x^{2}+b x+c\right| \geqslant 0 \text {. }
$$
Thus, the minimum value of $\left|y^{\prime}\right|$ is 0.
At this point, $y_{\text {man }}=-2 \times 0-1=-1$.
(2) From $\left\{\begin{array}{l}y=a x^{2}+b x+c, \\ y=k(x-1)-\frac{k^{2}}{4} \text {, we get }\end{array}\right.$ $a x^{2}+(b-k) x+c+\frac{k^{2}}{4}+k=0$.
Let $\Delta=(b-k)^{2}-4 a\left(c+\frac{k^{2}}{4}+k\right)=0$.
Rearranging gives
$$
(1-a) k^{2}-(2 b+4 a) k+b^{2}-4 a c=0 \text {. }
$$
According to the problem,
$$
1-a=0,2 b+4 a=0, b^{2}-4 a c=0 \text {. }
$$
Solving these equations, we get $a=1, b=-2, c=1$.
Thus, $a+b+c=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The number of positive integers $n$ that make $1^{2 \times 15}+2^{2005}+\cdots+n^{205}$ divisible by $n+2$ is
|
$=1.0$.
Let $S_{n}=1^{2008}+2^{2000}+\cdots+n^{2005}$, then
$$
S_{n}=n^{2008}+(n-1)^{2005}+\cdots+1^{2008} \text {. }
$$
By misalignment addition, we get
$$
\begin{aligned}
2 S_{n}= & 2+\left(2^{2005}+n^{2005}\right)+\left[3^{2005}+(n-1)^{2000}\right]+ \\
& \cdots+\left(n^{2005}+2^{2005}\right) .
\end{aligned}
$$
From $k^{20 \infty}+(n+2-k)^{2005} \equiv 0(\bmod (n+2))(k \geqslant 2)$, we know $2 S_{n} \equiv 2(\bmod (n+2))$, which means $S_{n}(n \geqslant 2)$ cannot be divisible by $n+2$.
Also, $S_{1}$ cannot be divisible by $n+2$, so, $S_{n}$ cannot be divisible by $n+2$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $B C$ respectively. Find the distance from point $D$ to the plane $B_{1} E F$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Solution: As shown in Figure 10, in the rectangular coordinate system, we have
$$
\begin{array}{l}
D(0,0,0), \\
B_{1}(1,1,1), \\
E\left(1, \frac{1}{2}, 0\right), \\
E\left(\frac{1}{2}, 1,0\right) .
\end{array}
$$
Thus, $B_{1} E=\left(0,-\frac{1}{2},-1\right)$,
$$
\boldsymbol{B}_{1} \boldsymbol{F}=\left(-\frac{1}{2}, 0,-1\right) \text {. }
$$
Let the normal vector of plane $B_{1} E F$ be $\boldsymbol{n}=(x, y, z)$.
Since $\boldsymbol{n} \perp B_{1} E, \boldsymbol{n} \perp B_{1} F$, we get
$-\frac{1}{2} y-z=0, -\frac{1}{2} x-z=0$.
Let $z=1$, then $x=-2, y=-2$.
Therefore, $\boldsymbol{n}=(-2,-2,1)$.
Also, $D E=\left(1, \frac{1}{2}, 0\right)$, so the distance from point $D$ to plane $B_{1} E F$ is
$$
d=\frac{|n \cdot D E|}{|n|}=\frac{\left|(-2,-2,1) \cdot\left(1, \frac{1}{2}, 0\right)\right|}{3}=1 \text {. }
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In a $9 \times 9$ grid, there are 81 small squares. In each small square, write a number. If in every row and every column, there are at most three different numbers, it can be guaranteed that there is a number in the grid that appears at least $n$ times in some row and at least $n$ times in some column. What is the maximum value of $n$? Prove your conclusion.
In a $9 \times 9$ grid, there are 81 small squares. In each small square, write a number. If in every row and every column, there are at most three different numbers, it can be guaranteed that there is a number in the grid that appears at least $n$ times in some row and at least $n$ times in some column. What is the maximum value of $n$? Prove your conclusion.
|
3. If a $9 \times 9$ grid is divided into 9 $3 \times 3$ grids, and each small cell in the same $3 \times 3$ grid is filled with the same number, and the numbers in any two different $3 \times 3$ grids are different, then each row and each column will have exactly three different numbers. Therefore, the maximum value of $n$ is no more than 3.
The following proof shows that as long as each row and each column have at most three different numbers, it can be guaranteed that there is a number in the grid that appears at least 3 times in a row and at least 3 times in a column.
When a number appears no less than 3 times in a row, mark the cells containing this number in that row with a symbol (1). Since each row has at most three different numbers, there are at most four cells in the same row that are not marked with a symbol (1), and at least five cells are marked with a symbol (1). Therefore, at least $5 \times 9$ cells in the entire grid are marked with a symbol (1).
Similarly, when a number appears no less than 3 times in a column, mark the cells containing this number in that column with a symbol (2).
By the same reasoning, at least $5 \times 9$ cells in the entire grid are marked with a symbol (2).
Since $5 \times 9 + 5 \times 9 > 9 \times 9$, there is at least one cell that is marked with both symbol (1) and symbol (2).
Clearly, the number in this cell appears at least 3 times in the row it is in and at least 3 times in the column it is in.
In conclusion, the maximum value of $n$ is 3.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $\frac{(2 x+z)^{2}}{(x+y)(-2 y+z)}=8$. Then $2 x+$ $4 y-z+6=$ $\qquad$
|
1. Hint: $(2 x+4 y-z)^{2}=0, 2 x+4 y-z+6=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A computer user plans to purchase single-piece software and boxed disks, priced at 60 yuan and 70 yuan each, respectively, with a budget of no more than 500 yuan. According to the needs, the user must buy at least 3 pieces of software and at least 2 boxes of disks. How many different purchasing options are there?
|
4. First buy 3 pieces of software and 2 boxes of disks. With the remaining 180 yuan, if you do not buy software, you can buy 0, 1, or 2 more boxes of disks; if you buy 1 more piece of software, you can buy 0 or 1 more box of disks; if you buy 2 more pieces of software, you cannot buy any more disks; if you buy 3 more pieces of software, you also cannot buy any more disks. There are 7 ways to make the purchase.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given the equation $6 x^{2}+2(m-13) x+12-m$ $=0$ has exactly one positive integer solution. Then the value of the integer $m$ is
|
5. Since $\Delta=4(m-13)^{2}-24(12-m)$ is a perfect square, there exists a non-negative integer $y$ such that
$$
(m-13)^{2}-6(12-m)=y^{2} \text {, }
$$
i.e., $(m-10-y)(m-10+y)=3$.
Since $m-10-y \leqslant m-10+y$, we have
$$
\left\{\begin{array} { l }
{ m - 1 0 - y = 1 , } \\
{ m - 1 0 + y = 3 }
\end{array} \text { or } \left\{\begin{array}{l}
m-10-y=-3, \\
m-10+y=-1 .
\end{array}\right.\right.
$$
Solving these, we get $m=12$ or $m=8$.
If $m=12$, then the original equation has no positive integer solutions, which is a contradiction.
If $m=8$, then the original equation yields $x=1$ or $x=\frac{2}{3}$. This satisfies the conditions.
Therefore, the required $m=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. It is known that $\alpha^{2005}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$. Find the sum of the coefficients of this polynomial.
（Zhu Huawei provided the problem）
|
1. Solution 1: In the expansion of $\alpha^{k}+\beta^{k}$, let $\alpha+\beta=1$, $\alpha \beta=1$, the sum of the coefficients we are looking for is $S_{k}=\alpha^{k}+\beta^{k}$. From
$$
\begin{array}{l}
(\alpha+\beta)\left(\alpha^{k-1}+\beta^{k-1}\right) \\
=\left(\alpha^{k}+\beta^{k}\right)+\alpha \beta\left(\alpha^{k-2}+\beta^{k-2}\right),
\end{array}
$$
we have $S_{k}=S_{k-1}-S_{k-2}$.
Thus, $S_{k}=\left(S_{k-2}-S_{k-3}\right)-S_{k-2}=-S_{k-3}$.
Similarly, $S_{k-3}=-S_{k-6}$.
Therefore, $S_{k}=S_{k-6}$.
Hence, the sequence $\left\{S_{k}\right\}$ is a periodic sequence with a period of 6.
So, $S_{2000}=S_{1}=1$.
Solution 2: In the expansion of $\alpha^{k}+\beta^{k}$, let $\alpha+\beta=1$, $\alpha \beta=1$, the sum of the coefficients we are looking for is $S_{k}=\alpha^{k}+\beta^{k}$. Then $\alpha, \beta$ are the roots of the equation $x^{2}-x+1=0$. Solving this, we get
$$
\alpha=\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}, \beta=\cos \frac{\pi}{3}-\mathrm{i} \sin \frac{\pi}{3} .
$$
Thus, $\alpha^{k}+\beta^{k}$
$$
\begin{aligned}
= & \left(\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}\right)^{k}+\left(\cos \frac{\pi}{3}-\mathrm{i} \sin \frac{\pi}{3}\right)^{k} \\
& =\left(\cos \frac{k \pi}{3}+\mathrm{i} \sin \frac{k \pi}{3}\right)+\left(\cos \frac{k \pi}{3}-\mathrm{i} \sin \frac{k \pi}{3}\right) \\
= & 2 \cos \frac{k \pi}{3} .
\end{aligned}
$$
Taking $k=2005$, we get $S_{k}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $n$ students be such that among any 3 of them, 2 know each other, and among any 4 of them, 2 do not know each other. Find the maximum value of $n$.
(Tang Lihua
|
8. The maximum value of $n$ is 8.
When $n=8$, the example shown in Figure 7 satisfies the requirements, where $A_{1}, A_{2}, \cdots, A_{8}$ represent 8 students, and the line between $A_{i}$ and $A_{j}$ indicates that $A_{i}$ and $A_{j}$ know each other.
Suppose $n$ students meet the requirements of the problem, we will prove that $n \leqslant 8$. For this, we first prove that the following two cases are impossible.
(1) If a person $A$ knows at least 6 people, denoted as $B_{1}, B_{2}, \cdots, B_{6}$, by Ramsey's theorem, among these 6 people, either there exist 3 people who do not know each other, which contradicts the given condition that any 3 people have 2 who know each other; or there exist 3 people who all know each other, in which case, $A$ and these 3 people form a group of 4 people who all know each other, which also contradicts the given condition.
(2) If a person $A$ knows at most $n-5$ people, then there are at least 4 people who do not know $A$, thus, these 4 people must all know each other, which is a contradiction.
Secondly, when $n \geqslant 10$, either (1) or (2) must occur, so $n$ does not meet the requirements at this time.
When $n=9$, to make (1) and (2) not occur, each person must know exactly 5 other people. Thus, the number of friend pairs (pairs of people who know each other) generated by these 9 people is $\frac{9 \times 5}{2} \notin \mathbf{N}_{+}$, which is a contradiction.
From the above, we know that the $n$ that meets the requirements is $\leqslant 8$.
In summary, the maximum value of $n$ is 8.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (20 points) (1) Given $a+\log _{2}(2 a+6)=11$ and $b+2^{b-1}=14$. Find the value of $a+b$.
(2) Given $f(x)=\frac{2}{2^{x-2}+1}$. Find
$$
f(-1)+f(0)+f(1)+f(2)+f(3)+f(4)+f(5)
$$
the value.
|
(1) Let $t=\log _{2}(2 a+6)$, then we have $t+2^{t-1}=14$. Since the function $f(x)=x+2^{x-1}$ is an increasing function on $\mathbf{R}$, we have $b$ $=t$. Therefore, $a+t=11$, which means $a+b=11$.
(2) Since $f(x)+f(4-x)=\frac{2}{2^{x-2}+1}+\frac{2}{2^{2-x}+1}$
$$
=\frac{2\left(2^{x-2}+1\right)}{2^{x-2}+1}=2 \text {, }
$$
then $f(-1)+f(0)+f(1)+f(2)+f(3)+f(4)+f(5)=7$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (20 points) Given that $a$, $b$, and $c$ are all prime numbers greater than 3, and $2a + 5b = c$.
(1) Prove that there exists a positive integer $n > 1$ such that the sum $a + b + c$ of all three prime numbers $a$, $b$, and $c$ that satisfy the given condition is divisible by $n$;
(2) Find the maximum value of $n$ from the previous question.
|
(1) Since $c=2a+5b$, we have
$$
a+b+c=3a+6b=3(a+2b).
$$
Also, since $a, b, c$ are all prime numbers greater than 3, it follows that $3 \mid (a+b+c)$, i.e., there exists a positive integer $n>1$ (for example, $n=3$) such that $n! \mid (a+b+c)$.
(2) Since $a, b, c$ are all prime numbers greater than 3, $a, b, c$ are not multiples of 3.
If $a \equiv 1 \pmod{3}$ and $b \equiv 2 \pmod{3}$, then
$$
c=2a+5b=2+10=0 \pmod{3}.
$$
This contradicts the fact that $c$ is not a multiple of 3.
Similarly, if $a \equiv 2 \pmod{3}$ and $b \equiv 1 \pmod{3}$, it will also lead to a contradiction.
Therefore, it must be that $a \equiv b \equiv 1 \pmod{3}$ or $a \equiv b \equiv 2 \pmod{3}$.
Thus, $a+2b=3a \equiv 0 \pmod{3}$.
Hence, $9 \mid (a+b+c)$.
When $a=7$ and $b=13$, $c=2 \times 7 + 5 \times 13 = 79$ is a prime number, and $a+b+c=99=9 \times 11$;
When $a=7$ and $b=19$, $c=2 \times 7 + 5 \times 19 = 109$ is a prime number, and $a+b+c=135=9 \times 15$.
Therefore, among all $n \mid (a+b+c)$, the largest is 9.
|
9
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
12. The positive integer solutions $(x, y)$ of the equation $2 x^{2}-x y-3 x+y+2006=0$ are $\qquad$ pairs.
|
12.4.
From $2 x^{2}-x y-3 x+y+2006=0$, we get $y=\frac{2 x^{2}-3 x+2006}{x-1}=2 x-1+\frac{2005}{x-1}$.
Therefore, $x-1$ can take the values $1,5,401,2005$. Hence, the equation has 4 pairs of positive integer solutions.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. A and B take turns tossing a fair coin, and whoever tosses heads first wins, at which point the game ends, and the loser gets to toss first in the next game.
(1) Find the probability that the first person to toss wins in any given game;
(2) Suppose they play a total of 10 games, and A tosses first in the first game. Let the probability that A wins the $k$-th game be $P_{k}$. If $1<3^{9}\left(2 P_{k}-1\right)<81$, find the value of $k$.
|
15. (1) In any given match, the probability of the first person to throw winning is $\frac{1}{2}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{5}+\cdots=\frac{2}{3}$.
(2) From (1), the probability of the second person to throw winning is $1-\frac{2}{3}=\frac{1}{3}$.
Given $P_{1}=\frac{2}{3}$, for $2 \leqslant k \leqslant 10$, we have $P_{k}=\frac{1}{3} P_{k-1}+\frac{2}{3}\left(1-P_{k-1}\right)=\frac{2}{3}-\frac{1}{3} P_{k-1}$. Therefore, $P_{k}-\frac{1}{2}=-\frac{1}{3}\left(P_{k-1}-\frac{1}{2}\right)$. Hence, $P_{k}=\frac{1}{2}+\frac{(-1)^{k-1}}{3^{k-1}}\left(P_{1}-\frac{1}{2}\right)=\frac{1}{2}+\frac{(-1)^{k-1}}{2 \times 3^{k}}$. Since $1<3^{9}\left(2 P_{k}-1\right)<81$, we have $\frac{1}{2}+\frac{1}{2 \times 3^{9}}<P_{k}<\frac{1}{2}+\frac{1}{2 \times 3^{3}}$. Substituting $P_{k}$ and simplifying, we get $\frac{1}{3^{9}}<\frac{(-1)^{k-1}}{3^{k}}<\frac{1}{3^{5}}$. Therefore, $k$ is odd, and $5<k<9$. Hence, $k=7$.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the polynomial
$$
\begin{aligned}
a_{0}+ & \left(a_{1}+4\right) x+ \\
& \left(a_{2}-10\right) x^{2}+\left(a_{3}+6\right) x^{3}+\left(a_{4}-1\right) x^{4}+ \\
& \left(a_{5}-1\right) x^{5}+a_{6} x^{6}+\cdots+a_{2 \alpha 5} x^{2 \omega 5}
\end{aligned}
$$
can be divided by $x^{2}+3 x-2$, and $\alpha^{2}+3 \alpha-2=0$. Then
$$
a_{0}+a_{1} \alpha+a_{2} \alpha^{2}+\cdots+a_{2 \alpha 6} \alpha^{20 \% 5}
$$
has the value $\qquad$ .
|
3.0 .
Since $\alpha^{2}+3 \alpha-2=0$, $\alpha$ is a root of the equation $x^{2}+3 x-2=0$. Therefore,
$$
\begin{array}{l}
a_{0}+\left(a_{1}+4\right) \alpha+\left(a_{2}-10\right) \alpha^{2}+\left(a_{3}+6\right) \alpha^{3}+ \\
\left(a_{4}-1\right) \alpha^{4}+\left(a_{5}-1\right) \alpha^{5}+a_{6} \alpha^{6}+\cdots+a_{2 \cos } \alpha^{2005} \\
=M\left(\alpha^{2}+3 \alpha-2\right)=0,
\end{array}
$$
where $M$ is the quotient. Therefore,
$$
\begin{array}{l}
a_{0}+a_{1} \alpha+a_{2} \alpha^{2}+\cdots+a_{2} \cos \alpha^{2008} \\
=\alpha^{5}+\alpha^{4}-6 \alpha^{3}+10 \alpha^{2}-4 \alpha \\
=\left(\alpha^{3}-2 \alpha^{2}+2 \alpha\right)\left(\alpha^{2}+3 \alpha-2\right)=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the quadratic function $y=a x^{2}+13 x$ passes through two lattice points (points with integer coordinates) in the first quadrant, and their y-coordinates are both prime numbers. Then $a=$ $\qquad$ -
|
4. -6 .
Let the lattice point satisfying the given conditions be $(m, n)$.
Then $n=a m^{2}+13 m=m(a m+13)$.
Since $n$ is a prime number, we have $m_{1}=1$ and $a m_{1}+13$ is a prime number, or $a m_{2}+13=1$ and $m_{2}$ is a prime number.
Therefore, $m_{1}=1$ and $a+13$ is a prime number, and $a m_{2}=-12=-2 \times 2 \times 3$ and $m_{2}$ is a prime number.
Thus, $a=-6$ satisfies the conditions.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The smallest natural number $n$ that satisfies $n \sin 1 > 1 + 5 \cos 1$ is $\qquad$ .
|
6.5 .
Since $\frac{\pi}{4}n \sin 1>1+5 \cos 1>1+5 \cos \frac{\pi}{3}=\frac{7}{2} \text {. }$
Thus, $n>\frac{7}{\sqrt{3}}>4$.
When $n=5$, it is easy to prove that $5 \sin 1>1+5 \cos 1$, which means
$$
5(\sin 1-\cos 1)>1 \text {, }
$$
Squaring it gives $\sin 2<\frac{24}{25}$, which is obviously true.
In fact, because $\frac{3 \pi}{10}<1<\frac{\pi}{3}$, so,
$$
\begin{array}{l}
\sin \frac{2 \pi}{3}<\sin 2<\sin \frac{3 \pi}{5}=\frac{\sqrt{10+2 \sqrt{5}}}{4} \\
<\frac{\sqrt{14.5}}{4}<\frac{24}{25} .
\end{array}
$$
Therefore, the smallest natural number $n=5$.
|
5
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. (12 points) The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}(n \in \mathbf{N}), \\
b_{n}=\frac{1}{1+a_{n}}, S_{n}=b_{1}+b_{2}+\cdots+b_{n}, \\
P_{n}=b_{1} b_{2} \cdots b_{n} .
\end{array}
$$
Try to find the value of $2 P_{n}+S_{n}$.
|
Three, 15. Since $a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}, n \in \mathbf{N}$, therefore, $a_{n+1}=a_{n}\left(a_{n}+1\right)$. Then
$$
\begin{array}{l}
b_{n}=\frac{1}{1+a_{n}}=\frac{a_{n}^{2}}{a_{n} a_{n+1}}=\frac{a_{n+1}-a_{n}}{a_{n} a_{n+1}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}, \\
P_{n}=b_{1} b_{2} \cdots b_{n}=\frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdots \cdots \frac{a_{n}}{a_{n+1}}=\frac{1}{2 a_{n+1}}, \\
S_{n}=b_{1}+b_{2}+\cdots+b_{n} \\
=\left(\frac{1}{a_{1}}-\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{2}}-\frac{1}{a_{3}}\right)+\cdots+\left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right) \\
=2-\frac{1}{a_{n+1}} .
\end{array}
$$
Therefore, $2 P_{n}+S_{n}=\frac{1}{a_{n+1}}+\left(2-\frac{1}{a_{n+1}}\right)=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. (12 points) Draw a line through point $P(3+2 \sqrt{2}, 4)$ that intersects the $x$-axis and $y$-axis at points $M$ and $N$, respectively. Find the maximum value of $OM + ON - MN$ (where $O$ is the origin).
|
17. A circle is drawn through the point $P(3+2 \sqrt{2}, 4)$, tangent to the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and such that point $P$ lies on the major arc $\overparen{A B}$. The equation of the circle is $(x-3)^{2}+(y-3)^{2}=9$. Thus, the tangent line through point $P$ intersects the $x$-axis and $y$-axis at points $M_{1}$ and $N_{1}$, respectively, making the circle the incircle of $\mathrm{Rt} \triangle O M_{1} N_{1}$. Therefore, $O M_{1}+O N_{1}-M_{1} N_{1}=6$.
If the line $M N$ through point $P$ does not touch the circle, then a tangent line parallel to $M N$ is drawn, intersecting the $x$-axis and $y$-axis at points $M_{0}$ and $N_{0}$, respectively. Thus, $O M_{0}+O N_{0}-M_{0} N_{0}=6$.
By the length of the broken line $M_{0} M N N_{0}$ being greater than the length of $M_{0} N_{0}$ and the tangent segment theorem, we have
$$
\begin{array}{l}
O M+O N-M N \\
=\left(O M_{0}-M M_{0}\right)+\left(O N_{0}-N N_{0}\right)-M N \\
=\left(O M_{0}+O N_{0}-M_{0} N_{0}\right)+\left[M_{0} N_{0}-\left(M_{0} M+M N+N N_{0}\right)\right] \\
<O M_{0}+O N_{0}-M_{0} N_{0}=6 .
\end{array}
$$
Therefore, the maximum value of $O M+O N-M N$ is 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given the inequality $|a x-3| \leqslant b$ has the solution set $\left[-\frac{1}{2}, \frac{7}{2}\right]$. Then $a+b=$ $\qquad$ .
|
13.6.
From $|a x-3| \leqslant b$, we get $3-b \leqslant a x \leqslant 3+b$. It is easy to see that $a \neq 0$, so $\frac{3-b}{a}+\frac{3+b}{a}=-\frac{1}{2}+\frac{7}{2}$. Solving this, we get $a=2$.
Therefore, $\frac{3-b}{a}=-\frac{1}{2}$. Solving this, we get $b=4$. Hence, $a+b=6$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
18. The inequality $x+2 \sqrt{2 x y} \leqslant a(x+y)$ holds for all positive numbers $x, y$. Then the minimum value of the real number $a$ is $\qquad$
|
18.2.
Since $x+2 \sqrt{2 x y} \leqslant x+(x+2 y)=2(x+y)$, then the minimum value of $a$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
25. The coordinates of point $P(x, y)$ satisfy the following relations:
$$
\left\{\begin{array}{l}
2 x+y \geqslant 15, \\
x+3 y \geqslant 27, \\
x \geqslant 2, \\
y \geqslant 3,
\end{array}\right.
$$
and $x, y$ are both integers. Then the minimum value of $x+y$ is
$\qquad$, and at this time, the coordinates of point $P$ are $\qquad$.
|
$25.12,(3,9)$ or $(4,8)$.
Adding equations (1), (2), and (3), we get $x+y \geqslant 11$.
If $x+y=11$, then the equalities in equations (1), (2), and (3) hold, and the system of equations has no integer solutions.
If $x+y=12$, from $2 x+y \geqslant 15$, we get $x \geqslant 3$;
from $x+3 y \geqslant 27$, we get $2 y \geqslant 15$, hence $x \geqslant 3, y \geqslant 8$.
In this case, the coordinates of point $P$ are $(3,9)$ or $(4,8)$.
(Hong Wangbao provided)
|
12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For an integer $m$, its unit digit is denoted by $f(m)$, and let $a_{n}=f\left(2^{n+1}-1\right)(n=1,2, \cdots)$. Then $a_{2006}$ $=$ . $\qquad$
|
3.7.
It is known that $f\left(2^{m}\right)=f\left(2^{m+4}\right)$. Therefore,
$$
a_{2 \omega 0}=f\left(2^{2 \omega 1}-1\right)=f\left(2^{2 \omega 07}\right)-1=f(8)-1=7 .
$$
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $A$ be a finite set, for any $x, y \in A$, if $x \neq y$, then $x+y \in A$. Then, the maximum number of elements in $A$ is $\qquad$ .
|
Let the number of elements in $A$ be $n$.
If $n>3$, then there must be two elements with the same sign. Without loss of generality, assume there are two positive numbers. Let the largest two positive numbers be $a, b$ (with $a < b$). Then, $a+b \notin A$, which is a contradiction.
Therefore, $n \leqslant 3$.
Also, $A=\{0,1,-1\}$ satisfies the condition, so the maximum value of $n$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let the solution set of the equation $x^{2}-x-1=\left(x^{2}-1\right) \pi^{x}-$ $x \pi^{x^{2}-1}$ be $M$. Then the sum of the cubes of all elements in $M$ is ( ).
(A) 0
(B) 2
(C) 4
(D) 5
|
4.C.
Obviously, $x=0, \pm 1$ are all solutions to the original equation.
When $x \neq 0, \pm 1$, from
$$
\left(x^{2}-1\right)\left(\pi^{x}-1\right)=x\left(\pi^{x^{2}-1}-1\right)
$$
and $\pi$ being a transcendental number, we know $\pi^{x}-1=\pi^{x^{2}-1}-1$, which means $x=x^{2}-1$. At this point, there are two real solutions $x_{1}, x_{2}$.
From $x^{3}=x^{2}+x=2 x+1$, we know
$$
x_{1}^{3}+x_{2}^{3}=2\left(x_{1}+x_{2}\right)+2=4 \text {. }
$$
Therefore, the sum of the cubes of all elements in $M$ is
$$
0^{3}+1^{3}+(-1)^{3}+x_{1}^{3}+x_{2}^{3}=4 .
$$
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7: Xiao Li and his brother attended a gathering, along with two other pairs of brothers. After meeting, some people greeted each other with handshakes, but no one shook hands with their own brother, and no one shook hands with the same person twice. At this point, Xiao Li noticed that, apart from himself, everyone had a different number of handshakes. How many times did Xiao Li shake hands? How many times did Xiao Li's brother shake hands?
|
Explanation: From the problem, it is easy to see that the maximum number of handshakes for each person is 4, and the minimum is 0. Also, except for Xiao Li, each person has a different number of handshakes. Therefore, the 5 people must have handshakes 0 times, 1 time, 2 times, 3 times, and 4 times respectively.
Using points to represent people and lines between points to represent handshakes. Let $F$ and $E$ represent Xiao Li and his partner, and $A$, $B$, $C$, $D$ represent the other two pairs of brothers, where $A$ and $B$, $C$ and $D$ are brothers.
First, consider the point with 4 lines. First, exclude point $E$, because if point $E$ connects 4 lines, these 4 lines must connect to $A$, $B$, $C$, and $D$ respectively. Thus, the point without any line connected would be a contradiction. Therefore, the point connecting 4 lines must be among $A$, $B$, $C$, and $D$. Assume it is point $C$, and these 4 lines are $C A$, $C B$, $C E$, and $C F$. Then, the point without any line connected is $D$ (as shown in Figure 2).
Next, consider the point that only connects 1 line. First, exclude point $E$, if point $E$ only connects 1 line, then $A$ and $B$ can each connect at most 1 line to point $F$. This way, there would be no point connecting 3 lines, which is a contradiction. Therefore, the point that only connects 1 line can only be $A$ or $B$. Assume it is $A$. At this point, $E$ can only connect 1 line to $E B$, so there must be 1 line between $B$ and $F$ (as shown in Figure 3).
In summary, Xiao Li shakes hands twice, and Xiao Li's partner shakes hands twice.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Let $f(x)$ be an odd function with a period of 2, and $f\left(-\frac{2}{5}\right)=3$. If $\sin \alpha=\frac{\sqrt{5}}{5}$, then the value of $f(4 \cos 2 \alpha)$ is
|
13. -3.
Since $\sin \alpha=\frac{\sqrt{5}}{5}$, therefore,
$$
\cos 2 \alpha=1-2 \sin ^{2} \alpha=\frac{3}{5} \text{. }
$$
Also, $f(x)$ is an odd function with a period of 2, and $f\left(-\frac{2}{5}\right)=$
3, so,
$$
f(4 \cos 2 \alpha)=f\left(\frac{12}{5}\right)=f\left(\frac{2}{5}\right)=-f\left(-\frac{2}{5}\right)=-3 .
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Let $x>1, y>1, S=\min \left\{\log _{x} 2, \log _{2} y\right.$ , $\left.\log _{y}\left(8 x^{2}\right)\right\}$. Then the maximum value of $S$ is $\qquad$ .
|
15.2.
From the problem, we have $\log _{x} 2 \geqslant S, \log _{2} y \geqslant S, \log _{r}\left(8 x^{2}\right) \geqslant S$, then $S \leqslant \log ,\left(8 x^{2}\right)=\frac{3+2 \log _{2} x}{\log _{2} y}=\frac{3+\frac{2}{\log _{x} 2}}{\log _{2} y} \leqslant \frac{3+\frac{2}{S}}{S}$. Therefore, $S^{3}-3 S-2 \leqslant 0$, which is $(S-2)(S+1)^{2} \leqslant 0$, yielding $S \leqslant 2$.
When $x=\sqrt{2}, y=4$, equality holds.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let $a, b, c > 0$. Prove:
$$
f=\sum \frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}} \leqslant 8 \text {. }
$$
(32nd United States of America Mathematical Olympiad)
|
Explanation: By homogeneity, we may assume $a+b+c=3$. Then
$$
\begin{aligned}
f & =\sum \frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}}=\sum \frac{a^{2}+6 a+9}{3\left(a^{2}-2 a+3\right)} \\
& =\frac{1}{3} \sum\left(1+\frac{8 a+6}{(a-1)^{2}+2}\right) \\
& \leqslant \frac{1}{3} \sum(4 a+4)=8 .
\end{aligned}
$$
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 3, in $\triangle A B C$, $M$ is the midpoint of side $B C$, and $M D \perp A B, M E \perp A C$, with $D$ and $E$ being the feet of the perpendiculars. If $B D=2, C E=1$, and $D E \parallel B C$, then $D M^{2}$ equals $\qquad$
|
4.I.
Let $D M=x, M E=y, A D=z, A E=w$.
Since $D E / / B C$, we have $\frac{z}{2}=\frac{w}{1}$, i.e., $z=2 w$.
Because $B M=M C$, we have $S_{\triangle A K M}=S_{\triangle A M}$, i.e., $\frac{1}{2} x(z+2)=\frac{1}{2} y(w+1)$.
Thus, $2 x+x z=y+y w$.
By the Pythagorean theorem, we get
$$
\left\{\begin{array}{l}
2^{2}+x^{2}=B M^{2}=M C^{2}=1^{2}+y^{2}, \\
x^{2}+z^{2}=A M^{2}=y^{2}+w^{2},
\end{array}\right.
$$
i.e., $\left\{\begin{array}{l}x^{2}+3=y^{2}, \\ x^{2}+z^{2}=y^{2}+w^{2}\end{array}\right.$
Subtracting (2) from (3) gives $z^{2}-3=w^{2}$.
Substituting $z=2 w$ into the above equation yields $4 w^{2}=3+w^{2}$.
Thus, $w=1, z=2$.
Substituting them into equation (1) gives $y=2 x$.
Then, from equation (2), we get $x^{2}=1$, i.e., $D M^{2}=1$.
Note: We find $A B=4, A C=2, B C=2 \sqrt{5}, B C$ is the longest side, so $\angle B, \angle C$ are acute angles, and points $D, E$ are indeed on the sides.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $m$ be a real number not less than -1, such that the equation in $x$
$$
x^{2}+2(m-2) x+m^{2}-3 m+3=0
$$
has two distinct real roots $x_{1}$ and $x_{2}$.
(1) If $x_{1}^{2}+x_{2}^{2}=6$, find the value of $m$;
(2) Find the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$.
(2000, National Junior High School Mathematics Competition)
|
Explanation: Since the equation has two distinct real roots, we have
$$
\begin{array}{l}
\Delta=4(m-2)^{2}-4\left(m^{2}-3 m+3\right) \\
=-4 m+4>0 .
\end{array}
$$
Thus, $m<1$.
Also, $m \geqslant-1$, so, $-1 \leqslant m<1$.
By the relationship between roots and coefficients, we get
$$
x_{1}+x_{2}=-2(m-2), x_{1} x_{2}=m^{2}-3 m+3 \text {. }
$$
(1) Since $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$
$$
=2 m^{2}-10 m+10=6 \text {, }
$$
Then we have $2 m^{2}-10 m+4=0$.
Solving the equation and considering $-1 \leqslant m<1$, we get
$$
m=\frac{5-\sqrt{17}}{2} \text {. }
$$
$$
\begin{array}{l}
\text { (2) } \frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}} \\
=\frac{m x_{1}^{2}\left(1-x_{2}\right)+m x_{2}^{2}\left(1-x_{1}\right)}{\left(1-x_{1}\right)\left(1-x_{2}\right)} \\
=\frac{m\left(x_{1}^{2}+x_{2}^{2}\right)-m x_{1} x_{2}\left(x_{1}+x_{2}\right)}{1-\left(x_{1}+x_{2}\right)+x_{1} x_{2}} \\
=2\left(m^{2}-3 m+1\right) . \\
\text { Let } y=2\left(m^{2}-3 m+1\right) \\
=2\left(m-\frac{3}{2}\right)^{2}-\frac{5}{2}(-1 \leqslant m<1) .
\end{array}
$$
According to the graph of the function $y$, $y$ is monotonically decreasing on $-1 \leqslant m<1$.
Thus, when $m=-1$, $y$ reaches its maximum value of 10, i.e., the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$ is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. If $\sin \alpha \cdot \cos \beta=1$, then $\cos \alpha \cdot \sin \beta=$
|
12.0.
From $\sin \alpha \cdot \cos \beta=1$ and the boundedness of sine and cosine functions, we know that $\sin \alpha=\cos \beta=1$ or $\sin \alpha=\cos \beta=-1$. Therefore, $\cos \alpha=\sin \beta=0$, which means $\cos \alpha \cdot \sin \beta=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Simplify $\log _{\sqrt{2}} \sin \frac{7 \pi}{8}+\log _{\sqrt{2}} \sin \frac{3 \pi}{8}$, the result is $\qquad$ .
|
13. -3 .
$$
\begin{array}{l}
\text { Original expression }=\log _{\sqrt{2}}\left(\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}\right)=\log _{\sqrt{2}}\left(\frac{1}{2} \sin \frac{\pi}{4}\right) \\
=\log _{\sqrt{2}} \frac{\sqrt{2}}{4}=\log _{2} \frac{1}{2} 2^{-\frac{3}{2}}=-3 .
\end{array}
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the complex numbers $z_{1}, z_{2}$ satisfy
$$
\left|z_{1}-z_{2}\right|=\sqrt{3}, z_{1}^{2}+z_{1}+q=0, z_{2}^{2}+z_{2}+q=0 \text {. }
$$
then the real number $q=$ $\qquad$
|
2.1.
From $\left|z_{1}-z_{2}\right|=\sqrt{3}$, we know $z_{1} \neq z_{2}$.
By the definition of the roots of the equation, $z_{1}$ and $z_{2}$ are the two imaginary roots of the quadratic equation $x^{2}+x+q=0$, then we have
$$
\Delta=1-4 q<0 \text{. }
$$
Solving the equation, we get $z_{1,2}=\frac{-1 \pm \sqrt{4 q-1} \mathrm{i}}{2}$.
Thus, $\left|z_{1}-z_{2}\right|=\sqrt{4 q-1}=\sqrt{3}$.
Solving for $q$ yields $q=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. $a$ and $b$ are rational numbers, and satisfy the equation
$$
a+b \sqrt{3}=\sqrt{6} \times \sqrt{1+\sqrt{4+2 \sqrt{3}}} \text {. }
$$
Then the value of $a+b$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8
|
5. B.
Since $\sqrt{6} \times \sqrt{1+\sqrt{4+2 \sqrt{3}}}=\sqrt{6} \times \sqrt{1+(1+\sqrt{3})}$ $=\sqrt{12+6 \sqrt{3}}=3+\sqrt{3}$,
thus, $a+b \sqrt{3}=3+\sqrt{3}$, which means $(a-3)+(b-1) \sqrt{3}=0$.
Since $a, b$ are rational numbers, then $a=3, b=1$, hence $a+b=4$.
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. The sum of the x-coordinates of the x-axis intercepts of the graph of the function $y=x^{2}-2006|x|+2008$ is $\qquad$
|
$=.1 .0$.
The original problem is transformed into solving the equation
$$
x^{2}-2006|x|+2008=0
$$
We are to find the sum of all real roots.
If a number $x_{0}$ is a root of equation (1), then its opposite number $-x_{0}$ is also a root of equation (1). Therefore, the sum of all real roots of the equation is 0, i.e., the sum of the x-coordinates of the points where it intersects the x-axis is 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If $(x+1)(y+1)=x^{2} y+x y^{2}=6$, then, $x^{2}+y^{2}$ equals $(\quad)$.
(A) 6
(B) 5
(C) 4
(D) 3
|
5.B.
From $\left\{\begin{array}{l}x y+(x+y)=5, \\ x y(x+y)=6,\end{array}\right.$ we get $\left\{\begin{array}{l}x+y=3, \\ x y=2\end{array}\right.$ or $\left\{\begin{array}{l}x+y=2, \\ x y=3\end{array}\right.$ (discard). Therefore, $x^{2}+y^{2}=(x+y)^{2}-2 x y=9-4=5$.
|
5
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
15. The school offers four extracurricular interest classes in Chinese, Math, Foreign Language, and Natural Science for students to voluntarily sign up for. The number of students who want to participate in the Chinese, Math, Foreign Language, and Natural Science interest classes are 18, 20, 21, and 19, respectively. If the total number of students in the class is 25, how many students at least have signed up for all four interest classes?
|
Three, 15. The number of people not attending the Chinese interest class is 7, the number of people not attending the Math interest class is 5, the number of people not attending the Foreign Language interest class is 4, and the number of people not attending the Natural Science interest class is 6. Therefore, the maximum number of people who do not attend at least one interest class is 22, which means the minimum number of people who attend all four interest classes is 3.
Construct a scenario where exactly 3 people attend all four interest classes as follows:
3 people attend all four interest classes, 7 people attend the three interest classes except for Chinese, 5 people attend the three interest classes except for Math, 4 people attend the three interest classes except for Foreign Language, and 6 people attend the three interest classes except for Natural Science.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The line $l$ passes through the focus of the parabola $y^{2}=a(x+1)(a>0)$ and is perpendicular to the $x$-axis. If the segment cut off by $l$ on the parabola is 4, then $a=$ $\qquad$ .
|
$=.1 .4$.
Since the parabolas $y^{2}=a(x+1)$ and $y^{2}=a x$ have the same length of the focal chord with respect to their directrix, the general equation $y^{2}=a(x+1)$ can be replaced by the standard equation $y^{2}=a x$ for solving, and the value of $a$ remains unchanged. Using the formula for the length of the latus rectum, we get $a=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The number of right-angled triangles with integer side lengths and area (numerically) equal to the perimeter is $\qquad$
|
4.2.
Let the legs of a right-angled triangle be $a, b$, and $c = \sqrt{a^{2}+b^{2}} (a \leqslant b)$, then we have
$$
\frac{1}{2} a b = a + b + \sqrt{a^{2} + b^{2}}.
$$
Thus, $\frac{1}{2} a b - a - b = \sqrt{a^{2} + b^{2}}$.
Squaring both sides and simplifying, we get
$$
a b - 4 a - 4 b + 8 = 0.
$$
Then, $(a-4)(b-4) = 8$.
Since $a, b$ are positive integers, when $a = 5$, $b = 12$; when $a = 6$, $b = 8$.
Therefore, there are 2 triangles that satisfy the problem.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A chord $AB$ is drawn through a focus $F$ of the ellipse $\frac{x^{2}}{6^{2}}+\frac{y^{2}}{2^{2}}=1$. If $|A F|=m,|B F|=n$, then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$
|
5.3.
As shown in Figure 2, draw $A A_{1}$ perpendicular to the directrix at point $A_{1} . A E \perp x$-axis at point $E$.
$$
\begin{array}{l}
\text { Since } \frac{c}{a}=e=\frac{|F A|}{\left|A A_{1}\right|} \\
=\frac{|F A|}{|D F|+|F E|} \\
=\frac{m}{m \cos \alpha+\frac{b^{2}}{c}} .
\end{array}
$$
Therefore, $\frac{1}{m}=\frac{a-c \cos \alpha}{b^{2}}$.
Similarly, $\frac{1}{n}=\frac{a+c \cos \alpha}{b^{2}}$. Hence $\frac{1}{m}+\frac{1}{n}=\frac{2 a}{b^{2}}=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (20 points) Given the ellipse $C: \frac{x^{2}}{4}+y^{2}=1$ and a fixed point $P(t, 0)(t>0)$, a line $l$ with a slope of $\frac{1}{2}$ passes through point $P$ and intersects the ellipse $C$ at two distinct points $A$ and $B$. For any point $M$ on the ellipse, there exists $\theta \in[0,2 \pi]$, such that $O M=\cos \theta \cdot O A+\sin \theta \cdot O B$ holds. Try to find the value of the real number $t$ that satisfies the condition.
|
Let $M(x, y)$ and points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. The equation of line $AB$ is $x-2 y-t=0$.
From the cubic equation, we get
$2 x^{2}-2 t x+t^{2}-4=0$ and $8 y^{2}+4 t y+t^{2}-4=0$.
Thus, $x_{1} x_{2}=\frac{t^{2}-4}{2}$ and $y_{1} y_{2}=\frac{t^{2}-4}{8}$.
Given $O M=\cos \theta \cdot O A+\sin \theta \cdot O B$, we have
$$
\begin{array}{l}
x=x_{1} \cos \theta+x_{2} \sin \theta, \\
y=y_{1} \cos \theta+y_{2} \sin \theta .
\end{array}
$$
Using $4=x^{2}+4 y^{2}$
$$
\begin{aligned}
= & \left(x_{1} \cos \theta+x_{2} \sin \theta\right)^{2}+4\left(y_{1} \cos \theta+y_{2} \sin \theta\right)^{2} \\
= & \left(x_{1}^{2}+4 y_{1}^{2}\right) \cos ^{2} \theta+\left(x_{2}^{2}+4 y_{2}^{2}\right) \sin ^{2} \theta+ \\
& 2 \sin \theta \cdot \cos \theta \cdot\left(x_{1} x_{2}+4 y_{1} y_{2}\right) \\
= & 4\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+2 \sin \theta \cdot \cos \theta . \\
& \left(x_{1} x_{2}+4 y_{1} y_{2}\right),
\end{aligned}
$$
We get $2 \sin \theta \cdot \cos \theta \cdot\left(x_{1} x_{2}+4 y_{1} y_{2}\right)=0$.
Since $\theta \in[0,2 \pi]$ is arbitrary, we know
$$
x_{1} x_{2}+4 y_{1} y_{2}=\frac{t^{2}-4}{2}+4 \times \frac{t^{2}-4}{8}=t^{2}-4=0 \text {. }
$$
Solving for $t$, we get $t=2$.
Substituting back, the condition is satisfied.
Therefore, the value of $t$ that satisfies the condition is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the value of $2-2^{2}-2^{3}-\cdots-2^{2005}+2^{2000}$.
|
Explanation: Utilizing the characteristic $2^{n+1}-2^{n}=2^{n}$, rearrange the terms of the original expression in reverse order.
Therefore, the original expression
$$
\begin{array}{l}
=2^{2006}-2^{2005}-2^{2004}-\cdots-2^{3}-2^{2}+2 \\
=2^{2006}(2-1)-2^{2005}-\cdots-2^{3}-2^{2}+2 \\
=2^{2005}(2-1)-2^{2004}-\cdots-2^{3}-2^{2}+2 \\
=\cdots \\
=2^{2}(2-1)+2=6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Place the numbers 1, 2, $3, \cdots, 9$ into the 9 circles in Figure 4, such that the sum of the numbers in the three circles on each side of $\triangle ABC$ and $\triangle DEF$ is 18.
(1) Provide one valid arrangement;
(2) How many different arrangements are there? Prove your conclusion.
|
15. (1) Figure 9 gives one arrangement that meets the requirements.
(2) There are 6 different ways to fill in the numbers.
Let the sum of the three numbers in circles $A, B, C$ be $x$; the sum of the three numbers in circles $D, E, F$ be $y$; and the sum of the remaining three circles be $z$. Clearly, we have
$$
x+y+z=1+2+\cdots+9=45.
$$
In Figure 9, the sum of the numbers in the three circles on each of the six edges is 18, so we have
$$
z+3 y+2 x=6 \times 18=108.
$$
Subtracting (1) from (2) gives
$$
x+2 y=108-45=63.
$$
Adding the sums of the numbers in the three circles on each of the edges $AB, BC, CA$, we get
$$
2 x+y=3 \times 18=54.
$$
Solving equations (3) and (4) simultaneously, we get $x=15, y=24$, and thus $z=6$.
Among the numbers $1,2, \cdots, 9$, the only three numbers that sum to 24 are $7, 8, 9$, so the circles $D, E, F$ can only be filled with the numbers $7, 8, 9$, and there are 6 different ways to do this.
Clearly, once the numbers in these three circles are determined, the numbers in the remaining six circles are also determined according to the problem's requirements.
Therefore, the conclusion is that there are 6 different ways to fill in the numbers.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. (18 points) Among 200 small balls numbered $1, 2, \cdots, 200$, any $k$ balls are drawn such that there must be two balls with numbers $m$ and $n$ satisfying
$$
\frac{2}{5} \leqslant \frac{n}{m} \leqslant \frac{5}{2} \text {. }
$$
Determine the minimum value of $k$ and explain the reasoning.
|
17. Divide 200 balls numbered $1 \sim 200$ into 6 groups, such that the ratio of the numbers of any two balls in each group is no less than $\frac{2}{5}$ and no more than $\frac{5}{2}$.
Grouping as follows:
Group 1 $(1,2)$
Group 2 $\quad(3,4,5,6,7)$
Group 3 $(8,9,10, \cdots, 20)$
Group 4 $\quad(21,22,23, \cdots, 52)$
Group 5 $\quad(53,54,55, \cdots, 132)$
Group 6 $\quad(133,134,135, \cdots, 200)$
When $k=7$, among the 200 balls in these 6 groups, if 7 balls are randomly drawn, according to the pigeonhole principle, there must be two balls in the same group, and the numbers of these two balls are $m$ and $n$, which must satisfy
$$
\frac{2}{5} \leqslant \frac{n}{m} \leqslant \frac{5}{2} .
$$
Therefore, the minimum value of $k$ is 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let natural numbers $x, y$ satisfy
$$
x<y, x^{3}+19 y=y^{3}+19 x
$$
Then $x+y=$ $\qquad$
|
2.5.
From $x^{3}-y^{3}=19(x-y), x<y$, we get $x^{2}+x y+y^{2}=19$.
Thus, $3 x^{2}<x^{2}+x y+y^{2}=19$.
Since $x$ is a natural number, we have $x=2$.
Therefore, $y=3$, so $x+y=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, color the numbers in $S=\{0,1,2, \cdots, n\}$ with two colors arbitrarily. Find the smallest positive integer $n$, such that there must exist $x, y, z \in S$ of the same color, satisfying $x+y=2 z$.
|
Let $A=\{0,2,5,7\}$ and $B=\{1,3,4,6\}$. Then neither $A$ nor $B$ contains a 3-term arithmetic progression. By coloring the elements of $A$ red and the elements of $B$ yellow, we can see that $n>7$.
When $n \geqslant 8$, we will prove that there must be a color for which the numbers form a 3-term arithmetic progression.
In fact, we only need to consider the first 9 numbers $0,1, \cdots, 8$. Suppose these numbers are divided into two subsets $A$ and $B$ (i.e., they are colored with two different colors), and without loss of generality, assume $0 \in A$.
Assume the conclusion is false.
(1) Suppose $1 \in A$, then $2 \in B$.
(i) If $3 \in A$, then by $5+1=2 \times 3$, we have $5 \in B$. Similarly, $\Rightarrow$ $8 \in A \Rightarrow 4 \in B \Rightarrow 6 \in A$, and by $0, 3, 6 \in A$ we reach a contradiction.
(ii) If $3 \in B$, then by $2+4=2 \times 3$, we have $4 \in A$. Similarly, $\Rightarrow$ $7 \in B \Rightarrow 5 \in A \Rightarrow 6 \in B \Rightarrow 8 \in A$, and by $0, 4, 8 \in A$ we reach a contradiction.
(2) Suppose $1 \in B$.
(i) If $2 \in B$, then by $1+3=2 \times 2$, we have $3 \in A$. Similarly, $\Rightarrow$ $6 \in B \Rightarrow 4 \in A \Rightarrow 5 \in B, 8 \in B$, and by $2, 5, 8 \in B$ we reach a contradiction.
(ii) If $2 \in A$, then by $0+4=2 \times 2$, we have $4 \in B$. Similarly, $\Rightarrow$ $7 \in A$.
If $3 \in B$, then $5 \in A \Rightarrow 6 \in B \Rightarrow 8 \in A$, and by $2, 5, 8 \in A$ we reach a contradiction.
If $3 \in A$, then $5 \in B \Rightarrow 6 \in A$, and by $0, 3, 6 \in A$ we reach a contradiction.
Therefore, $n_{\min }=8$.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The set of integer points on the plane
$$
S=\{(a, b) \mid 1 \leqslant a, b \leqslant 5(a, b \in \mathbf{Z})\},
$$
$T$ is a set of integer points on the plane, such that for any point $P$ in $S$, there exists a point $Q$ in $T$ different from $P$, such that the line segment $P Q$ contains no other integer points except $P$ and $Q$. What is the minimum number of elements in $T$?
(Supplied by Chen Yonggao)
|
5. The minimum number is 2.
First, we prove that $T$ cannot consist of only one point.
Otherwise, suppose
$$
T=\left\{Q\left(x_{0}, y_{0}\right)\right\}.
$$
In $S$, take a point $P\left(x_{1}, y_{1}\right)$ such that $\left(x_{1}, y_{1}\right) \neq \left(x_{0}, y_{0}\right)$, and $x_{1}$ has the same parity as $x_{0}$, and $y_{1}$ has the same parity as $y_{0}$.
Then the midpoint of segment $PQ$ is an integer point. This is a contradiction.
The case where $T$ contains two points is shown in Figure 3.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let the set
$$
\begin{array}{l}
M=\{1,2, \cdots, 19\}, \\
A=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\} \subseteq M .
\end{array}
$$
Find the smallest $k$, such that for any $b \in M$, there exist $a_{i}, a_{j} \in A$, satisfying $b=a_{i}$ or $b=a_{i} \pm a_{j}\left(a_{i}, a_{j}\right.$ can be the same).
(Supplied by Li Shenghong)
|
6. By the hypothesis, in $A$, there are $k(k+1)$ possible combinations. Thus, $k(k+1) \geqslant 19$, i.e., $k \geqslant 4$.
When $k=4$, we have $k(k+1)=20$. Suppose $a_{1}<a_{2}<a_{3}<a_{4}$. Then $a_{4} \geqslant 10$.
(1) When $a_{4}=10$, we have $a_{3}=9$. At this time, $a_{2}=8$ or 7.
If $a_{2}=8$, then $20,10-9=1,9-8=1$, so it is impossible for all 19 numbers in $A$ to be generated. If $a_{2}=7$, then $a_{1}=6$ or $a_{1}=5$. Since $20,10-9=1,7-6=1$ or $20,9-7=2,7-5=2$, this is impossible.
(2) When $a_{4}=11$, we have $a_{3}=8$. At this time, $a_{2}=7$ and $a_{1}=6$, which is impossible.
(3) When $a_{4}=12$, we have $a_{3}=7$. At this time, $a_{2}=6, a_{1}=5$, which is impossible.
(4) When $a_{4}=13$, we have $a_{3}=6, a_{2}=5, a_{1}=4$, which is impossible.
(5) When $a_{4}=14$, we have $a_{3}=5, a_{2}=4, a_{1}=3$, which is impossible.
(6) When $a_{4}=15$, we have $a_{3}=4, a_{2}=3, a_{1}=2$, which is impossible.
(7) When $a_{4}=16$, we have $a_{3}=3, a_{2}=2, a_{1}=1$, which is impossible.
(8) When $a_{4} \geqslant 17$, it is impossible.
Therefore, $k \geqslant 5$.
If we take $A=\{1,3,5,9,16\}$, then $A$ satisfies the condition.
Thus, the smallest $k=5$.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Try to find the minimum value of the function $f(x, y)=6\left(x^{2}+y^{2}\right) \cdot (x+y)-4\left(x^{2}+x y+y^{2}\right)-3(x+y)+5$ in the region $A=\{(x, y) \mid x>0, y>0\}$.
|
Explanation: If $x+y \leqslant 1$, then
$$
\begin{array}{l}
x y \leqslant \frac{1}{4}(x+y)^{2} \leqslant \frac{1}{4} \text {. } \\
\text { Hence } f(x, y) \\
=6\left(x^{3}+y^{3}\right)+6 x y(x+y)-4 x y- \\
4\left(x^{2}+y^{2}\right)-3(x+y)+5 \\
=6\left(x y-\frac{1}{4}\right)(x+y-1)+(6 x+1) \text {. } \\
\left(x-\frac{1}{2}\right)^{2}+(6 y+1)\left(y-\frac{1}{2}\right)^{2}+ \\
(x+y-1)^{2}+2 \\
\geqslant 2 \text {. } \\
\end{array}
$$
When and only when $x=y=\frac{1}{2}$, the equality holds.
If $x+y>1$, then
$$
x^{2}+y^{2} \geqslant \frac{1}{2}(x+y)^{2}>\frac{1}{2} \text {. }
$$
Hence $f(x, y)$
$$
=6\left(x^{2}+y^{2}-\frac{1}{2}\right)(x+y-1)+2(x-y)^{2}+2
$$
$>2$.
In summary, for any $(x, y) \in A$, we have $f(x, y) \geqslant 2$.
When and only when $x=y=\frac{1}{2}$, $f(x, y)_{\min }=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let $0 \leqslant x \leqslant \pi, 0 \leqslant y \leqslant 1$. Try to find the minimum value of the function
$$
f(x, y)=(2 y-1) \sin x+(1-y) \sin (1-y) x
$$
|
It is known that for all $0 \leqslant x \leqslant \pi$, we have
$$
\sin x \geqslant 0, \sin (1-y) x \geqslant 0 \text {. }
$$
Thus, when $\frac{1}{2} \leqslant y \leqslant 1$, $f(x, y) \geqslant 0$, and the equality holds when $x=0$.
When $0 \leqslant yx>\sin x$,
and $\sin (x+\delta)=\sin x \cdot \cos \delta+\cos x \cdot \sin \delta$
$\leqslant \sin x+\delta \cos x$.
Therefore, $\frac{\sin x}{x}-\frac{\sin (x+\delta)}{x+\delta}$
$=\frac{(x+\delta) \sin x-x \sin (x+\delta)}{x(x+\delta)}$
$\geqslant \frac{(x+\delta) \sin x-x(\sin x+\delta \cos x)}{x(x+\delta)}$
$=\frac{\delta \sin x-\delta x \cos x}{x(x+\delta)}=\frac{\delta \cos x(\tan x-x)}{x(x+\delta)}$
$>0$.
Hence, the function $f(x)=\frac{\sin x}{x}$ is decreasing on $\left(0, \frac{\pi}{2}\right)$.
Since $g(x)=\sin x$ is decreasing on $\left[\frac{\pi}{2}, \pi\right]$, it follows that $f(x)=\frac{\sin x}{x}$ is also decreasing on $\left[\frac{\pi}{2}, \pi\right]$.
Moreover, $00$,
then $\frac{\sin (1-y) x}{1-y} \leqslant \frac{1-y}{1-2 y} \sin (1-y) x$,
$\sin x \leqslant \frac{1-y}{1-2 y} \sin (1-y) x$.
Thus, $(2 y-1) \sin x+(1-y) \sin (1-y) x \geqslant 0$,
which means $f(x, y) \geqslant 0$.
When $y=0$, the equality holds.
In summary, when $x=0$ or $y=0$,
$$
f(x, y)_{\min }=0 \text {. }
$$
Note: The monotonicity of the function $f(x)=\frac{\sin x}{x}$ on $\left(0, \frac{\pi}{2}\right)$ can also be determined using the derivative method combined with $\tan x>x$, $x \in\left(0, \frac{\pi}{2}\right)$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. The number of real solutions to the equation $\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+\right.$ $\left.x^{2004}\right)=2006 x^{2005}$ is $\qquad$
|
11.1.
$$
\begin{array}{l}
\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=2006 x^{2005} \\
\Leftrightarrow\left(x+\frac{1}{x^{2005}}\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=2006 \\
\Leftrightarrow x+x^{3}+x^{5}+\cdots+x^{2005}+\frac{1}{x^{2008}}+\frac{1}{x^{2003}} \\
\quad+\cdots+\frac{1}{x}=2006 \\
\Leftrightarrow x+\frac{1}{x}+x^{3}+\frac{1}{x^{3}}+\cdots+x^{2005}+\frac{1}{x^{2005}} \\
\geqslant 2 \times 1003=2006 .
\end{array}
$$
For the equality to hold, it must be that $x=\frac{1}{x}, x^{3}=\frac{1}{x^{3}}, \cdots, x^{2005}=\frac{1}{x^{2005}}$, i.e., $x= \pm 1$.
However, when $x \leqslant 0$, the original equation is not satisfied.
Therefore, $x=1$ is the only solution to the original equation.
Thus, the number of real solutions to the original equation is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given $x=2 \sqrt{2}+1$. Then the fraction
$$
\frac{x^{2}-2 x-9}{x^{3}-11 x-15}=
$$
$\qquad$
|
13.2
Given $x=2 \sqrt{2}+1$, we know $x-1=2 \sqrt{2}$, then $x^{2}-2 x-7=0$. Therefore, $\frac{x^{2}-2 x-9}{x^{3}-11 x-15}=\frac{\left(x^{2}-2 x-7\right)-2}{(x+2)\left(x^{2}-2 x-7\right)-1}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. There are 2006 distinct complex numbers, such that the product of any two of them (including self-multiplication) is one of these 2006 numbers. Find the sum of these 2006 numbers.
|
Three, 13. Let there be $n$ distinct complex numbers $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$.
If one of them is 0, without loss of generality, let $\alpha_{n}=0$, then the set
$$
\left\{\alpha_{1}^{2}, \alpha_{1} \alpha_{2}, \cdots, \alpha_{1} \alpha_{n-1}\right\}=\left\{\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n-1}\right\} .
$$
Thus, by $\alpha_{1}^{2} \cdot \alpha_{1} \alpha_{2} \cdots \cdot \alpha_{1} \alpha_{n-1}=\alpha_{1} \cdot \alpha_{2} \cdots \cdot \alpha_{n-1}$, we get $\alpha_{1}^{n-1}=1$.
Similarly, $a_{k}^{n-1}=1(k=1,2, \cdots, n-1)$.
Since $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n-1}$ are distinct complex numbers, we have $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n-1}=0$, which means
$$
\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=0 .
$$
If these $n$ complex numbers are all non-zero, similarly, we get $\alpha_{k}^{n}=1$. Since $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ are distinct complex numbers, we have $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=0$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Find
$$
2 \sum_{k=1}^{n} k^{3} \mathrm{C}_{n}^{k}-3 n \sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}+n^{2} \sum_{k=1}^{n} k \mathrm{C}_{n}^{k}
$$
the value.
|
14. Solution 1: Since $k \mathrm{C}_{n}^{k}=n \mathrm{C}_{n-1}^{k-1}$, we have:
$$
\begin{array}{l}
\sum_{k=1}^{n} k \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} n \mathrm{C}_{n-1}^{k-1}=n 2^{n-1} \\
\sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} k n \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=1}^{n}(k-1) \mathrm{C}_{n-1}^{k-1}+n \sum_{k=1}^{n} \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=2}^{n}(n-1) \mathrm{C}_{n-2}^{k-2}+n 2^{n-1} \\
=n(n-1) 2^{n-2}+n 2^{n-1}=n(n+1) 2^{n-2} \\
\sum_{k=1}^{n} k^{3} \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} k^{2} n \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=1}^{n}(k-1)^{2} \mathrm{C}_{n-1}^{k-1}+n \sum_{k=1}^{n}(2 k-1) \mathrm{C}_{n-1}^{k-1} \\
=n(n-1) n 2^{n-3}+2 n(n+1) 2^{n-2}-n 2^{n-1} \\
=n^{2}(n+3) 2^{n-3}
\end{array}
$$
Thus, $2 \sum_{k=1}^{n} k^{3} \mathrm{C}_{n}^{k}-3 n \sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}+n^{2} \sum_{k=1}^{n} k \mathrm{C}_{n}^{k}$
$$
\begin{array}{l}
=\left[2 n^{2}(n+3)-3 n \times 2 n(n+1)+n^{2} \times 4 n\right] 2^{n-3} \\
=0
\end{array}
$$
Solution 2: The original expression $=\sum_{k=1}^{n} k \mathrm{C}_{n}^{k}\left(2 k^{2}-3 n k+n^{2}\right)$
$$
=\sum_{k=1}^{n} k(n-k)(n-2 k) \mathrm{C}_{n}^{k} \text {. }
$$
Let $M_{k}=k(n-k)(n-2 k) \mathrm{C}_{n}^{k}$, then $M_{n-k}=-M_{k}$.
When $n$ is even, the $\frac{n}{2}$-th term and the $n$-th term are 0; when $n$ is odd, the $n$-th term is 0.
In summary, the original expression equals 0.
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Arrange $n$ squares of different sizes without overlapping, so that the total area of the resulting figure is exactly 2,006. The minimum value of $n$ is $\qquad$ $\therefore$.
|
3.3.
Let the side lengths of $n$ squares be $x_{1}, x_{2}, \cdots, x_{n}$, then $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=2006$.
Since $x_{i}^{2} \equiv 0$ or $1(\bmod 4)$, and $2006 \equiv 2(\bmod 4)$, there must be at least two odd numbers among $x_{i}$.
If $n=2$, then $x_{1}$ and $x_{2}$ are both odd, let them be $2 p+1$ and $2 q+1$, then $(2 p+1)^{2}+(2 q+1)^{2}=2006$.
Thus, $p^{2}+p+q^{2}+q=501$.
However, $p^{2}+p$ and $q^{2}+q$ are both even, which is a contradiction.
If $n=3$, we can set
$$
x_{1}=2 p+1, x_{2}=2 k, x_{3}=2 q+1 \text {, }
$$
then $(2 p+1)^{2}+(2 k)^{2}+(2 q+1)^{2}=2006$,
which means $p^{2}+p+k^{2}+q^{2}+q=501$.
Clearly, $k$ is odd and $k \leqslant \sqrt{501}$. Hence $k \leqslant 21$.
When $k=1$, $p^{2}+p+q^{2}+q=500$ has no positive integer solutions;
When $k=3$, $p^{2}+p+q^{2}+q=492$ has solutions $p=8, q=20$.
Thus, $17^{2}+6^{2}+41^{2}=2006$. Therefore, $n_{\min }=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4.5 soccer teams are conducting a round-robin tournament (each pair of teams plays one match). It is known that Team A has played 3 matches, Team B has played more matches than Team A, Team C has played fewer matches than Team A, and Team D and Team E have played the same number of matches, but Team D and Team E have not played against each other. Therefore, the total number of matches played is $\qquad$
|
4.6.
Team B has played 4 matches. If Team C has only played 1 match, then Team C hasn't played against Team A, and Team A must have played against Teams D and E, so Teams D and E have each played 2 matches. Therefore, the total number of matches is
$$
(3+4+1+2+2) \div 2=6 \text {. }
$$
If Team C has played 2 matches, and Teams D and E have each played $x$ matches, then $3+4+2+x+x=9+2x$ is not an even number, leading to a contradiction.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) As shown in Figure 3, $\triangle A B C$ is divided into six smaller triangles by three concurrent lines $A D$, $B E$, and $C F$. If the areas of $\triangle B P F$, $\triangle C P D$, and $\triangle A P E$ are all 1, find the areas of $\triangle A P F$, $\triangle D P B$, and $\triangle E P C$.
|
Three, let the areas of $\triangle A P F$, $\triangle D P B$, and $\triangle E P C$ be $x$, $y$, and $z$ respectively.
Since $\frac{S_{\triangle D P B}}{S_{\triangle D P C}}=\frac{B D}{D C}, \frac{S_{\triangle D A B}}{S_{\triangle A K C}}=\frac{B D}{D C}$,
then $\frac{S_{\triangle D P B}}{S_{\triangle D P C}}=\frac{S_{\triangle D X B}}{S_{\triangle A M C}}$, i.e., $\frac{y}{1}=\frac{x+y+1}{z+2}$.
Thus, $x-y=y z-1$.
Similarly, $y-z=z x-1, z-x=x y-1$.
Assume $x \geqslant y \geqslant z$, then
$$
y z-1=x-y \geqslant 0, x y-1=z-x \leqslant 0 \text {. }
$$
Therefore, $x y \leqslant y z$, i.e., $x \leqslant z$.
Thus, $x=y=z$.
From $0=x-y=y z-1=z^{2}-1$, we get $z=1$.
Therefore, $x=y=z=1$.
Hence, the areas of the three triangles are all 1.
(Respectfully, Nanjing University Affiliated High School, 210008)
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A building has 4 elevators, each of which can stop at three floors (not necessarily consecutive floors, and not necessarily the lowest floor). For any two floors in the building, there is at least one elevator that can stop at both. How many floors can this building have at most?
|
2. Let the building have $n$ floors, then the number of floor pairs is $\frac{n(n-1)}{2}$. Each elevator stops at 3 floors, which gives $\frac{3 \times 2}{2}=3$ floor pairs, so, $4 \times 3 \geqslant$ $\frac{n(n-1)}{2}$. Therefore, $n \leqslant 5$.
When $n=5$, the floors served by the four elevators are $(1,4,5),(2,4,5),(3,4,5),(1,2,3)$
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Divide the numbers $1,2, \cdots, 30$ into $k$ groups (each number can only appear in one group) such that the sum of any two different numbers in each group is not a perfect square. Find the minimum value of $k$.
Put the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
3. First consider the numbers $6,19,30$.
Since $6+19=5^{2}, 6+30=6^{2}, 19+30=7^{2}$, these 3 numbers must be in 3 different groups. Therefore, $k \geqslant 3$.
Next, divide the 30 numbers $1,2, \cdots, 30$ into the following 3 groups:
$$
\begin{array}{l}
A_{1}=\{3,7,11,15,19,23,27,4,8,16,24\}, \\
A_{2}=\{1,5,9,13,17,21,25,29,6,14,18,26\}, \\
A_{3}=\{2,10,12,20,22,28,30\},
\end{array}
$$
such that they satisfy the conditions of the problem.
Since the remainder of a perfect square when divided by 4 can only be 0 or 1, it is easy to verify that $A_{1}, A_{2}, A_{3}$ satisfy the conditions of the problem.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If 5 consecutive natural numbers are all composite, then this group of numbers is called a "twin 5 composite". So, among the natural numbers not exceeding 100, there are $\qquad$ groups of twin 5 composite.
|
Ni, 6.10.
It is easy to know that the prime numbers not exceeding 100 are
$$
\begin{array}{l}
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47, \\
53,59,61,67,71,73,79,83,89,97 .
\end{array}
$$
There are 10 groups of twin 5 composites, namely
$$
\begin{array}{l}
24,25,26,27,28 ; 32,33,34,35,36 ; \\
48,49,50,51,52 ; 54,55,56,57,58 ; \\
62,63,64,65,66 ; 74,75,76,77,78 ; \\
84,85,86,87,88 ; 90,91,92,93,94 ; \\
91,92,93,94,95 ; 92,93,94,95,96 .
\end{array}
$$
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For real numbers $a, b, c$ satisfying
$$
a+b+c=0, \, abc=2 \text{.}
$$
then $u=|a|^{3}+|b|^{3}+|c|^{3}$ has the minimum value of
$\qquad$.
|
2.10.
From the problem, we know that $a$, $b$, and $c$ must be one positive and two negative.
Without loss of generality, let $a>0$, $b<0$, and $c<0$.
Since $b+c=-a$ and $bc=\frac{2}{a}$, it follows that $b$ and $c$ are the two negative roots of the equation $x^{2}+a x+\frac{2}{a}=0$. Therefore, we have
$$
\Delta=a^{2}-\frac{8}{a} \geqslant 0 \text {. }
$$
Solving this, we get $a^{3} \geqslant 8$.
Thus, $u=a^{3}-b^{3}-c^{3}$
$$
\begin{array}{l}
=a^{3}-(b+c)\left[(b+c)^{2}-3 b c\right] \\
=a^{3}+a\left(a^{2}-\frac{6}{a}\right)=2 a^{3}-6 \\
\geqslant 2 \times 8-6=10 .
\end{array}
$$
Equality holds if and only if $a^{3}=8$, i.e., $a=2$.
In this case, $b=c=-1$.
Therefore, the minimum value of $u$ is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given real numbers $a, b, c$ satisfy
$$
\left(-\frac{1}{a}+b\right)\left(\frac{1}{a}+c\right)+\frac{1}{4}(b-c)^{2}=0 \text {. }
$$
Then the value of the algebraic expression $a b+a c$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2
|
3. A.
The given equation can be rewritten as
$$
4(a b+1)(a c+1)+(a b-a c)^{2}=0,
$$
which simplifies to $(a b+a c)^{2}+4(a b+a c)+4=0$, or equivalently $[(a b+a c)+2]^{2}=0$.
Thus, $a b+a c=-2$.
|
-2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a-b+c=7, \\
a b+b c+b+c^{2}+16=0 .
\end{array}
$$
Then the value of $\left(a^{-1}-b^{-1}\right)^{a k c}(a+b+c)^{a+b+c}$ is $\qquad$ .
|
2. -1 .
From equation (1), we get
$$
(-b)+(a+c+1)=8 \text {. }
$$
From equation (2), we get
$$
(a+c+1)(-b)=c^{2}+16 \text {. }
$$
Therefore, $a+c+1$ and $-b$ are the two roots of the equation
$$
x^{2}-8 x+c^{2}+16=0
$$
Thus, we have $\Delta=(-8)^{2}-4\left(c^{2}+16\right) \geqslant 0$.
Hence, $c^{2} \leqslant 0$.
It is easy to see that $c=0$. Consequently, $x_{1}=x_{2}=4$, which means $a+c+1=4$, $-b=4$, or equivalently, $a=3, b=-4$.
$$
\begin{array}{l}
\text { Therefore, }\left(a^{-1}-b^{-1}\right)^{2 k}(a+b+c)^{a+b+c} \\
=\left(3^{-1}+4^{-1}\right)^{0}[3+(-4)+0]^{3-4+0}=-1 \text {. }
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For the quadratic equation in $x$
$$
m^{2} x^{2}+(2 m+3) x+1=0
$$
there are two real roots whose product is 1; for the quadratic equation in $x$
$$
x^{2}+(2 a+m) x+2 a+1-m^{2}=0
$$
there is a real root that is greater than 0 and less than 4. Then the integer value of $a$ is . $\qquad$
|
2, -1.
According to the problem, we have $m^{2}=1, m= \pm 1$, and
$$
\Delta=(2 m+3)^{2}-4 m^{2}=12 m+9>0 \text {. }
$$
Thus, $m=1$.
Let $f(x)=x^{2}+(2 a+1) x+2 a$, then we should have
$$
f(0) f(4)=2 a(20+10 a)<0,
$$
which means $a(a+2)<0$.
Therefore, $-2<a<0$. Hence, $a=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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