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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 th... | 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... | 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 ... | -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(... | 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
$$
(2, p)=1,(3, p)=1,(6,... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. There are two people, A and B. A, on a car, notices that B is walking in the opposite direction. $1 \mathrm{~min}$ later, A gets off the car to chase B. If A's speed is twice that of B, but slower than the car's speed by $\frac{4}{5}$, then the time it takes for A to catch up with B after getting off the car is $\ma... | 8.11.
Let the speed of A be $x \mathrm{~m} / \mathrm{min}$, then according to the problem, the speed of B is $\frac{x}{2} \mathrm{~m} / \mathrm{min}$, and the speed of the car is $5 x \mathrm{~m} / \mathrm{min}$.
Let the time it takes for A to catch up with B be $t \mathrm{~min}$. According to the problem, we have $\f... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. If a convex $n$-sided polygon has exactly 4 obtuse interior angles, then the maximum number of sides $n$ of this polygon is $\qquad$ . | 10.7.
Since a convex $n$-sided polygon has exactly 4 obtuse interior angles, the sum of these 4 angles is greater than $360^{\circ}$ and less than $720^{\circ}$. The other $n-4$ angles are right angles or acute angles, so the sum of these $n-4$ angles is no more than $(n-4) \times 90^{\circ}$ and greater than $0^{\cir... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. If in a $4 \times 4$ grid of 16 cells, each cell is filled with a number such that the sum of the numbers in all adjacent cells of each cell is 1, then the sum of the 16 numbers in the grid is $\qquad$
(Note: Adjacent cells refer to cells that share exactly one edge) | 11.6.
First, fill the 16 squares with the English letters $A, B, C$, $\cdots, O, P$ (as shown in Figure 8). Given that for each square, the sum of the numbers in all adjacent squares is 1, so,
$$
\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... | 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. Sup... | 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 $... | 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... | -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)
Analysi... | 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,
$$
\begin{array}{l}
\left|m_{i}-m_{j}\r... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given $a 、 b 、 c$ are real numbers, and
$$
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+\f... | -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 ... | 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 $\Delt... | 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
$$
\begin{array}{l}
\left(\frac{a}{b+c}-\frac{a}{a+b+c}\right)+\left(\frac{b}{a+c}-\frac{b}{a+b+c}\right)+ \\
\left(\frac{c}{a+b}-\frac{c}{a+b+c}\right)=0 .
\en... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 2, two diameters $A C$ and $B D$ of the large circle intersect perpendicularly at point $O$. Four semicircles are drawn outside the large circle with $A B$, $B C$, $C D$, and $D A$ as diameters, respectively. The total area of the four "crescent" shaded regions in the figure is $2 \mathrm{~cm}^{2}... | $=.1 .1$.
By the Pythagorean theorem, we know $A D^{2}+C D^{2}=A C^{2}$, so the area of the upper half of the large circle is equal to the sum of the areas of the two semicircles with diameters $A D$ and $C D$. Similarly, the area of the lower half of the large circle is equal to the sum of the areas of the two semicir... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (15 points) Given non-zero real numbers $a, b, c$ satisfy $a+b+c=0$. Prove:
(1) $a^{3}+b^{3}+c^{3}=3 a b c$;
(2) $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9$. | Three, (1) From $a+b+c=0$, we get $a+b=-c$. Therefore, $(a+b)^{3}=-c^{3}$.
Thus, $a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=-c^{3}$.
Hence $a^{3}+b^{3}+c^{3}=-3 a b(a+b)=-3 a b(-c)=3 a b c$.
$$
\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... | 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_{... | 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... | 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, $a+d=41, b+c=101$.... | 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 \l... | 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 $a \sqrt{2-b^{2}}+b \sqrt{2-a^{2}}$
$$
\begin{array}{l}
=2(\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \alp... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a regular $n$-sided polygon inscribed in a circle. If the line connecting a vertex to the center of the circle is exactly on the perpendicular bisector of a certain side, then $n \equiv$ $\qquad$ $(\bmod 2)$ | 2.1.
As shown in Figure 8, if the line connecting vertex $A_{1}$ and the center $O$ perpendicularly bisects side $A_{i} A_{j}$, then $\triangle A_{1} A_{i} A_{j}$ is an isosceles triangle. Since equal chords subtend equal arcs, the number of vertices in $\overparen{A_{1} A_{i}}$ is equal, and their sum is even. Adding... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 As shown in Figure 1, quadrilateral $ABCD$ is a rectangle. Two people, A and B, start from points $A$ and $B$ respectively at the same time, and move counterclockwise along the rectangle. In which circle does B possibly catch up with A for the first time? In which circle does B certainly catch up with A at th... | Explanation: Let $A D=B C=a \mathrm{~m}, A B=C D=b . \mathrm{m}$, and suppose the first time Yi catches up with Jia is after
$$
\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 {. }
$$
At this point, the number... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Question 4 Let the function $f(x)=\frac{1}{k-x}$, and denote
$$
\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}
$$
Multiplyin... | -1 | Algebra | proof | Yes | Yes | cn_contest | false |
4. For a positive integer $n$, if there exist positive integers $a$ and $b$ such that $n=a+b+a b$, then $n$ is called a "good number". For example, $3=1+1+1 \times 1$, so 3 is a good number. Then, among the 20 positive integers from $1 \sim 20$, the number of good numbers is $\qquad$ . | 4.12.
$n+1=a+b+a b+1=(a+1)(b+1)$ is a composite number, so the required $n$ is the number obtained by subtracting 1 from the composite numbers between 2 and 21. The composite numbers between 2 and 21 are $4,6,8,9,10,12,14,15,16,18,20,21$, a total of 12. Therefore, the required $n$ has 12 values: $3,5,7,8,9,11,13,14,15,... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the set $M=\left\{a \left\lvert\, a=\frac{x+y}{t}\right., 2^{x}+2^{y}=\right.$ $2^{t}$, where $x, y, t, a$ are all integers $\}$. Then the sum of all elements in the set $M$ is ( ).
(A) 1
(B) 4
(C) 7
(D) 8 | 6. D.
Assume $x \leqslant y$, then $2^{x}=2^{x}+2^{y} \leqslant 2^{y}+2^{y}=2^{y+1}$.
Thus, $t \leqslant y+1$.
From $2^{x}>0$, we get $2^{t}=2^{x}+2^{y}>2^{y}$. Therefore, $t>y$.
So, $y<t \leqslant y+1$.
Given that $x, y, t$ are all integers, then $t=y+1$. Hence, $2^{y+1}=$ $2^{x}+2^{y}$, which means $2^{x}=2^{y}$. Th... | 8 | Algebra | MCQ | Yes | Yes | cn_contest | false |
7. Given a fixed point $A(4, \sqrt{7})$. If a moving point $P$ is on the parabola $y^{2}=4 x$, and the projection of point $P$ on the $y$-axis is point $M$, then the maximum value of $|P A|-|P M|$ is $\qquad$. | $$
\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 {. }
$$
If $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, then the value of $S_{2003}-2 S_{2004}+S_{2000}$ is $\qquad$ | 9.3.
According to the problem, when $n$ is even, we have $a_{1}+a_{2}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n}{2}$ pairs, so $S_{n}=\frac{n}{2}$. Therefore, $S_{2004}=1002$.
When $n$ is odd, we have $a_{2}+a_{3}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n-1}{2}$ pairs, so
$$... | 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
$$
\boldsymbol{x}=\boldsymbol{a}+(3-t) \boldsymbol{b} \text { and } \boldsymbol{y... | 13.1.
Given $\boldsymbol{a} \cdot \boldsymbol{b}=0$.
$$
\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}^... | 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 & ... | 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} \fra... | 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 ... | 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 $x$. The... | 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
$$
\begin{array}{l}
1+3+5+\cdots+2 k-3+1 \\
\leqslant n0$, sol... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $x$, $y$, $z$ are positive real numbers, and $x y z(x+y+z)=1$. Then the minimum value of $(x+y)(y+z)$ is $\qquad$ | \begin{array}{l}=1.2. \\ (x+y)(y+z)=y^{2}+(x+z) y+x z \\ =y(x+y+z)+x z=y \cdot \frac{1}{x y z}+x z \\ =\frac{1}{x z}+x z \geqslant 2 .\end{array} | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given the quadratic trinomial $a x^{2}+b x+c$ $(a>0)$.
(1) When $c<0$, find the maximum value of the function
$$
y=-2\left|a x^{2}+b x+c\right|-1
$$
(2) For any real number $k$, the line $y=k(x-$
1) $-\frac{k^{2}}{4}$ intersects the parabola $y=a x^{2}+b x+c$ at exactly one point, find the value of ... | (1) Given $a>0, c<0$, we know that $y^{\prime}=a x^{2}+b x+c$ intersects the $x$-axis, and $y_{\text {nuin }}^{\prime}<0$. Therefore,
$$
\left|y^{\prime}\right|=\left|a x^{2}+b x+c\right| \geqslant 0 \text {. }
$$
Thus, the minimum value of $\left|y^{\prime}\right|$ is 0.
At this point, $y_{\text {man }}=-2 \times 0-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The number of positive integers $n$ that make $1^{2 \times 15}+2^{2005}+\cdots+n^{205}$ divisible by $n+2$ is | $=1.0$.
Let $S_{n}=1^{2008}+2^{2000}+\cdots+n^{2005}$, then
$$
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}
$$
Fro... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $B C$ respectively. Find the distance from point $D$ to the plane $B_{1} E F$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result dire... | Solution: As shown in Figure 10, in the rectangular coordinate system, we have
$$
\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 | 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... | 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 o... | 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 p... | 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
$$
\left\{\begin{array} { l }
{ m - 1 0 - y = 1 , } \\
{ m - 1 0 + y = 3 }
\end{array} \text { or... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. It is known that $\alpha^{2005}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$. Find the sum of the coefficients of this polynomial.
（Zhu Huawei provided the problem） | 1. Solution 1: In the expansion of $\alpha^{k}+\beta^{k}$, let $\alpha+\beta=1$, $\alpha \beta=1$, the sum of the coefficients we are looking for is $S_{k}=\alpha^{k}+\beta^{k}$. From
$$
\begin{array}{l}
(\alpha+\beta)\left(\alpha^{k-1}+\beta^{k-1}\right) \\
=\left(\alpha^{k}+\beta^{k}\right)+\alpha \beta\left(\alpha^{... | 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 w... | 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 $f(x)+f(4-x)=\frac{2}{2^{x-2}+1}+\frac{2}{2^{2-x}+1}$
$$
=\frac{2\left(2^{x-2}+1\right)}{2^{x-2}+1}=2 \text {, ... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) Given that $a$, $b$, and $c$ are all prime numbers greater than 3, and $2a + 5b = c$.
(1) Prove that there exists a positive integer $n > 1$ such that the sum $a + b + c$ of all three prime numbers $a$, $b$, and $c$ that satisfy the given condition is divisible by $n$;
(2) Find the maximum value of $... | (1) Since $c=2a+5b$, we have
$$
a+b+c=3a+6b=3(a+2b).
$$
Also, since $a, b, c$ are all prime numbers greater than 3, it follows that $3 \mid (a+b+c)$, i.e., there exists a positive integer $n>1$ (for example, $n=3$) such that $n! \mid (a+b+c)$.
(2) Since $a, b, c$ are all prime numbers greater than 3, $a, b, c$ are not... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
12. The positive integer solutions $(x, y)$ of the equation $2 x^{2}-x y-3 x+y+2006=0$ are $\qquad$ pairs.
| 12.4.
From $2 x^{2}-x y-3 x+y+2006=0$, we get $y=\frac{2 x^{2}-3 x+2006}{x-1}=2 x-1+\frac{2005}{x-1}$.
Therefore, $x-1$ can take the values $1,5,401,2005$. Hence, the equation has 4 pairs of positive integer solutions. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. A and B take turns tossing a fair coin, and whoever tosses heads first wins, at which point the game ends, and the loser gets to toss first in the next game.
(1) Find the probability that the first person to toss wins in any given game;
(2) Suppose they play a total of 10 games, and A tosses first in the first game... | 15. (1) In any given match, the probability of the first person to throw winning is $\frac{1}{2}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{5}+\cdots=\frac{2}{3}$.
(2) From (1), the probability of the second person to throw winning is $1-\frac{2}{3}=\frac{1}{3}$.
Given $P_{1}=\frac{2}{3}$, for $2 \leqslant... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the polynomial
$$
\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}
$$
can be divided by $x^{2}+3 x-2$, and $\alpha^{2}+3 \alpha... | 3.0 .
Since $\alpha^{2}+3 \alpha-2=0$, $\alpha$ is a root of the equation $x^{2}+3 x-2=0$. Therefore,
$$
\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 \... | 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 ... | -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 $... | 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
$$
\begin{array}{l}
b_{n}=\frac{1}{1+a_{n}}=\frac{a_{n}^{2}}{a_{n} a_{n+1}}=\frac{a_{n+1}-a_{n}}{a_{n} a_{n+1}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}, \\
P_{n}=b_{1} b_{2} \cdots b_{n}=\frac... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
17. (12 points) Draw a line through point $P(3+2 \sqrt{2}, 4)$ that intersects the $x$-axis and $y$-axis at points $M$ and $N$, respectively. Find the maximum value of $OM + ON - MN$ (where $O$ is the origin). | 17. A circle is drawn through the point $P(3+2 \sqrt{2}, 4)$, tangent to the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and such that point $P$ lies on the major arc $\overparen{A B}$. The equation of the circle is $(x-3)^{2}+(y-3)^{2}=9$. Thus, the tangent line through point $P$ intersects the $x$-axis ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. Given the inequality $|a x-3| \leqslant b$ has the solution set $\left[-\frac{1}{2}, \frac{7}{2}\right]$. Then $a+b=$ $\qquad$ . | 13.6.
From $|a x-3| \leqslant b$, we get $3-b \leqslant a x \leqslant 3+b$. It is easy to see that $a \neq 0$, so $\frac{3-b}{a}+\frac{3+b}{a}=-\frac{1}{2}+\frac{7}{2}$. Solving this, we get $a=2$.
Therefore, $\frac{3-b}{a}=-\frac{1}{2}$. Solving this, we get $b=4$. Hence, $a+b=6$. | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
18. The inequality $x+2 \sqrt{2 x y} \leqslant a(x+y)$ holds for all positive numbers $x, y$. Then the minimum value of the real number $a$ is $\qquad$ | 18.2.
Since $x+2 \sqrt{2 x y} \leqslant x+(x+2 y)=2(x+y)$, then the minimum value of $a$ is 2. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
25. The coordinates of point $P(x, y)$ satisfy the following relations:
$$
\left\{\begin{array}{l}
2 x+y \geqslant 15, \\
x+3 y \geqslant 27, \\
x \geqslant 2, \\
y \geqslant 3,
\end{array}\right.
$$
and $x, y$ are both integers. Then the minimum value of $x+y$ is
$\qquad$, and at this time, the coordinates of point $... | $25.12,(3,9)$ or $(4,8)$.
Adding equations (1), (2), and (3), we get $x+y \geqslant 11$.
If $x+y=11$, then the equalities in equations (1), (2), and (3) hold, and the system of equations has no integer solutions.
If $x+y=12$, from $2 x+y \geqslant 15$, we get $x \geqslant 3$;
from $x+3 y \geqslant 27$, we get $2 y \geq... | 12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. For an integer $m$, its unit digit is denoted by $f(m)$, and let $a_{n}=f\left(2^{n+1}-1\right)(n=1,2, \cdots)$. Then $a_{2006}$ $=$ . $\qquad$ | 3.7.
It is known that $f\left(2^{m}\right)=f\left(2^{m+4}\right)$. Therefore,
$$
a_{2 \omega 0}=f\left(2^{2 \omega 1}-1\right)=f\left(2^{2 \omega 07}\right)-1=f(8)-1=7 .
$$ | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $A$ be a finite set, for any $x, y \in A$, if $x \neq y$, then $x+y \in A$. Then, the maximum number of elements in $A$ is $\qquad$ . | Let the number of elements in $A$ be $n$.
If $n>3$, then there must be two elements with the same sign. Without loss of generality, assume there are two positive numbers. Let the largest two positive numbers be $a, b$ (with $a < b$). Then, $a+b \notin A$, which is a contradiction.
Therefore, $n \leqslant 3$.
Also, $A=\... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the solution set of the equation $x^{2}-x-1=\left(x^{2}-1\right) \pi^{x}-$ $x \pi^{x^{2}-1}$ be $M$. Then the sum of the cubes of all elements in $M$ is ( ).
(A) 0
(B) 2
(C) 4
(D) 5 | 4.C.
Obviously, $x=0, \pm 1$ are all solutions to the original equation.
When $x \neq 0, \pm 1$, from
$$
\left(x^{2}-1\right)\left(\pi^{x}-1\right)=x\left(\pi^{x^{2}-1}-1\right)
$$
and $\pi$ being a transcendental number, we know $\pi^{x}-1=\pi^{x^{2}-1}-1$, which means $x=x^{2}-1$. At this point, there are two real ... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 7: Xiao Li and his brother attended a gathering, along with two other pairs of brothers. After meeting, some people greeted each other with handshakes, but no one shook hands with their own brother, and no one shook hands with the same person twice. At this point, Xiao Li noticed that, apart from himself, every... | Explanation: From the problem, it is easy to see that the maximum number of handshakes for each person is 4, and the minimum is 0. Also, except for Xiao Li, each person has a different number of handshakes. Therefore, the 5 people must have handshakes 0 times, 1 time, 2 times, 3 times, and 4 times respectively.
Using ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
13. Let $f(x)$ be an odd function with a period of 2, and $f\left(-\frac{2}{5}\right)=3$. If $\sin \alpha=\frac{\sqrt{5}}{5}$, then the value of $f(4 \cos 2 \alpha)$ is | 13. -3.
Since $\sin \alpha=\frac{\sqrt{5}}{5}$, therefore,
$$
\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 | 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 \leqsla... | 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
$$
\left\{\begin{array}{l}
2^{2}+x^{... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $m$ be a real number not less than -1, such that the equation in $x$
$$
x^{2}+2(m-2) x+m^{2}-3 m+3=0
$$
has two distinct real roots $x_{1}$ and $x_{2}$.
(1) If $x_{1}^{2}+x_{2}^{2}=6$, find the value of $m$;
(2) Find the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$.
(2000, N... | Explanation: Since the equation has two distinct real roots, we have
$$
\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... | 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... | 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., ... | 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. I... | 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 maxi... | 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 latu... | 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)... | 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... | 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=\co... | 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 ... | 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
$$
\begin{array}{l}
=2^{2006}-2^{2005}-2^{2004}-\cdots-2^{3}-2^{2}+2 \\
=2^{2006}(2-1)-2^{2005}-\cdots-2^{3}-2^{2}+2 \\
=2^{2005}(2-1)-2^{2004}-\cdots-2^{3... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Place the numbers 1, 2, $3, \cdots, 9$ into the 9 circles in Figure 4, such that the sum of the numbers in the three circles on each side of $\triangle ABC$ and $\triangle DEF$ is 18.
(1) Provide one valid arrangement;
(2) How many different arrangements are there? Prove your conclusion. | 15. (1) Figure 9 gives one arrangement that meets the requirements.
(2) There are 6 different ways to fill in the numbers.
Let the sum of the three numbers in circles $A, B, C$ be $x$; the sum of the three numbers in circles $D, E, F$ be $y$; and the sum of the remaining three circles be $z$. Clearly, we have
$$
x+y+z... | 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 $\quad(21,22,23, \cdots, 52)$
G... | 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 progressio... | 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 exce... | 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}... | 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 b... | 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,... | 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
$$
\begin{array}{l}
x y \leqslant \frac{1}{4}(x+y)^{2} \leqslant \frac{1}{4} \text {. } \\
\text { Hence } f(x, y) \\
=6\left(x^{3}+y^{3}\right)+6 x y(x+y)-4 x y- \\
4\left(x^{2}+y^{2}\right)-3(x+y)+5 \\
=6\left(x y-\frac{1}{4}\right)(x+y-1)+(6 x+1) \text {. } \\
\left(x-\frac{1}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let $0 \leqslant x \leqslant \pi, 0 \leqslant y \leqslant 1$. Try to find the minimum value of the function
$$
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 {. }
$$
Thus, when $\frac{1}{2} \leqslant y \leqslant 1$, $f(x, y) \geqslant 0$, and the equality holds when $x=0$.
When $0 \leqslant yx>\sin x$,
and $\sin (x+\delta)=\sin x \cdot \cos \delta+\cos x \... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. The number of real solutions to the equation $\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+\right.$ $\left.x^{2004}\right)=2006 x^{2005}$ is $\qquad$ | 11.1.
$$
\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}=... | 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
$$
\left\{\alpha_{1}^{2}, \alpha_{1} \alpha_{2}, \cdots, \alpha_{1} \alpha_{n-1}\right\}=\left\{\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Find
$$
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}+... | 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+... | 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 n... | 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,... | 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_{... | 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 | 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... | 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:
$$
\begin{array}{l}
A_{1}=\{3,7,11,15,19,23,27,4,8,16,24\}, \\
A_{2}=\{1,5,9,13,17,... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. If 5 consecutive natural numbers are all composite, then this group of numbers is called a "twin 5 composite". So, among the natural numbers not exceeding 100, there are $\qquad$ groups of twin 5 composite. | Ni, 6.10.
It is easy to know that the prime numbers not exceeding 100 are
$$
\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 ; 54,55,56,5... | 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
$$
\Delta=a^{2... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given real numbers $a, b, c$ satisfy
$$
\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
$$
Thus, we have $\Delta=(-8)^{2}-4\left(c^{2}+16\right) \geqslant 0$.
Hence, $c^{2} \leqslant 0$.
I... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For the quadratic equation in $x$
$$
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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