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13. If $n^{2}+100$ can be divided by $n+10$, then, the maximum positive integer value of $n$ that satisfies the condition is $\qquad$
|
13. 190.
Since $n^{2}+100$ and $n^{2}-100$ are both divisible by $n+10$, therefore, 200 is divisible by $n+10$.
Hence, the largest positive integer that satisfies the condition is 190.
|
190
|
Number Theory
|
math-word-problem
|
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
|
Six. (20 points) Let two positive real numbers $x, y, z$. Try to find the maximum value of the algebraic expression $\frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z}$.
|
$$
\begin{array}{l}
\text { Six, because } \frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z} \\
=\frac{16 x+\frac{9 \sqrt{x \cdot 18 y}}{3}+\frac{3 \sqrt[3]{x \cdot 18 y \cdot 36 z}}{2}}{x+2 y+z} \\
\leqslant \frac{16 x+\frac{3(x+18 y)}{2}+\frac{x+18 y+36 z}{2}}{x+2 y+z}=18,
\end{array}
$$
Therefore, the equality holds if and only if $x: y: z=36: 2: 1$.
(Provided by Yang Yinghui)
$$
|
18
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Let $y=\frac{x}{1+x}=f(x)$, and $f(1)$ represents the value of $y$ when $x=1$, i.e., $f(1)=\frac{1}{1+1}=\frac{1}{2}$; $f\left(\frac{1}{2}\right)$ represents the value of $y$ when $x=\frac{1}{2}$, i.e., $f\left(\frac{1}{2}\right)=$ $\frac{\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3} ; \cdots \cdots$. Try to find
$$
\begin{array}{l}
f\left(\frac{1}{2006}\right)+f\left(\frac{1}{2005}\right)+\cdots+f\left(\frac{1}{2}\right)+ \\
f(1)+f(0)+f(1)+f(2)+\cdots+ \\
f(2005)+f(2006)
\end{array}
$$
the value of.
|
Explanation: Given $y=\frac{x}{1+x}=f(x)$, we have
$$
\begin{array}{l}
f\left(\frac{1}{n}\right)=\frac{\frac{1}{n}}{1+\frac{1}{n}}=\frac{1}{1+n} \\
f(n)=\frac{n}{1+n} \\
f(0)=\frac{0}{1+0}=0 .
\end{array}
$$
Thus, $f(n)+f\left(\frac{1}{n}\right)=\frac{1}{n+1}+\frac{n}{n+1}=1$.
Therefore, the original expression is
$$
\begin{aligned}
= & f\left(\frac{1}{2006}\right)+f(2006)+f\left(\frac{1}{2005}\right)+ \\
& f(2005)+\cdots+f(1)+f(1)+f(0) \\
= & 1+1+\cdots+1+1+0=2006
\end{aligned}
$$
|
2006
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $A=\{1,2,3,4,5,6\}, f: A \rightarrow A$, the number of mappings $f$ that satisfy $f(f(x)) \neq x$ is $\qquad$
|
2.7360 .
First, there should be $f(x) \neq x$, and the number of mappings satisfying $f(x) \neq x$ is $5^{6}$.
If $x \rightarrow a, a \rightarrow x(a \neq x)$, then it is called a pair of cyclic correspondences.
If there is one pair of cyclic correspondences and $f(x) \neq x$, the number of such mappings is $C_{6}^{2} \cdot 5^{4}$; if there are two pairs of cyclic correspondences and $f(x) \neq x$, the number of such mappings is $\frac{\mathrm{C}_{6}^{2} \mathrm{C}_{4}^{2}}{\mathrm{~A}_{2}^{2}} \cdot 5^{2}$; if there are three pairs of cyclic correspondences and $f(x) \neq x$, the number of such mappings is $\frac{C_{6}^{2} C_{4}^{2} C_{2}^{2}}{A_{3}^{3}}$.
Therefore, the number of mappings that meet the conditions is
$$
5^{6}-C_{6}^{2} \cdot 5^{4}+\frac{C_{6}^{2} C_{4}^{2}}{A_{2}^{2}} \cdot 5^{2}-\frac{C_{6}^{2} C_{4}^{2} C_{2}^{2}}{A_{3}^{3}}=7360
$$
|
7360
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\frac{2}{2007}\left[\left(1^{2}+3^{2}+5^{2}+\cdots+2005^{2}\right)-\right. \\
\left.\left(2^{2}+4^{2}+6^{2}+\cdots+2006^{2}\right)\right] .
\end{array}
$$
|
Answer: -2006 .
|
-2006
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $P$ be a regular 2006-gon. If an end of a diagonal of $P$ divides the boundary of $P$ into two parts, each containing an odd number of sides of $P$, then the diagonal is called a "good edge". It is stipulated that each side of $P$ is a good edge.
Given 2003 non-intersecting diagonals inside $P$ that partition $P$ into several triangles, how many isosceles triangles with two good edges can there be in this partition?
|
2. If an isosceles triangle has two good sides, it is briefly referred to as a "good triangle". Let $\triangle A B C$ be a good triangle, and $A B, B C$ be the good sides. Then, there are an odd number of edges between points $A$ and $B$; the same applies to $B$ and $C$. We say these edges belong to the good $\triangle A B C$.
Thus, in each of these two groups, at least one edge does not belong to any other good triangle. This is because the good triangles with three vertices between $A$ and $B$ have two equal-length sides, thus, there are an even number of edges belonging to it. Apart from all the edges belonging to any other good triangle, there must be one edge left that does not belong to any other good triangle. We designate such two edges (one in each group) to correspond to $\triangle A B C$.
For each good triangle, designate a pair of edges, with no two triangles sharing the designated edges. Thus, it follows that under this partition, there are at most 1003 good triangles, and it is easy to draw a partition that achieves this value.
|
1003
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the value of $2-2^{2}-2^{3}-\cdots-2^{2005}+2^{2000}$.
|
Explanation: Utilizing the characteristic $2^{n+1}-2^{n}=2^{n}$, rearrange the terms of the original expression in reverse order.
Therefore, the original expression
$$
\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}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. A construction company
has contracted two projects, each to be constructed by two different teams. According to the progress of the projects, the construction company can adjust the number of people in the two teams at any time. If 70 people are transferred from Team A to Team B, then the number of people in Team B will be twice that of Team A; if a certain number of people are transferred from Team B to Team A, then the number of people in Team A will be three times that of Team B. How many people are there in Team A at least?
|
14. Let team A have $x$ people, then team B has $[2(x-70)-70]$ people, which means team B has $(2 x-210)$ people. Suppose $y$ people are transferred from team B to team A, making the number of people in team A three times that of team B. Then
$$
3(2 x-210-y)=x+y,
$$
which simplifies to $x=126+\frac{4}{5} y$.
Given $y>0$, $y$ must be at least 5, meaning $x \geqslant 126+4=130$.
Therefore, team A has at least 130 people.
|
130
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Place the numbers 1, 2, $3, \cdots, 9$ into the 9 circles in Figure 4, such that the sum of the numbers in the three circles on each side of $\triangle ABC$ and $\triangle DEF$ is 18.
(1) Provide one valid arrangement;
(2) How many different arrangements are there? Prove your conclusion.
|
15. (1) Figure 9 gives one arrangement that meets the requirements.
(2) There are 6 different ways to fill in the numbers.
Let the sum of the three numbers in circles $A, B, C$ be $x$; the sum of the three numbers in circles $D, E, F$ be $y$; and the sum of the remaining three circles be $z$. Clearly, we have
$$
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
|
7. In the quadratic equation $x^{2}+m x+n=0$, the coefficients $m, n$ can take values from $1,2,3,4,5,6$. Then, among the different equations obtained, the number of equations with real roots is ( ).
(A) 20
(B) 19
(C) 16
(D) 10
|
7.B.
$\Delta=m^{2}-4 n \geqslant 0$, then $m^{2} \geqslant 4 n$. When $n=1$, $m=2,3,4,5,6$, a total of 5; when $n=2$, $m=3,4,5,6$, a total of 4; when $n=3$, $m=4,5,6$, a total of 3; when $n=4$, $m=4,5,6$, a total of 3; when $n=5$, $m=5,6$, a total of 2; when $n=6$, $m=5,6$, a total of 2. In total, 19.
|
19
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
13. As shown in Figure 4, the area of isosceles right $\triangle ABC$ is $98$, and $D$ is a point on the hypotenuse $BC$ such that $BD: DC = 2: 5$. Then the area of the square $ADEF$ with side $AD$ is $\qquad$
|
13.116.
Draw $D G \perp A B, D H \perp A C$, with the feet of the perpendiculars at $G$ and $H$ respectively. It is easy to see that $A B=A C=14$.
Also, $B D: D C=G D: H D=2: 5$, so $D G=4, D H=10$. Therefore, $A D=\sqrt{4^{2}+10^{2}}$. Hence, $S_{\text {square ADEF }}=116$.
|
116
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. (16 points) In an activity class, the teacher asked each student to make a rectangular box with a lid as shown in Figure 6. The length, width, and height of the rectangular box are $x \mathrm{~cm}, y \mathrm{~cm}$, and $z \mathrm{~cm}$, respectively. When Xiao Yang presented his box, he told the class: “The length, width, and height of the box I made are all positive integers, and upon measurement, they satisfy $x y=x z+3, y z=x y+x z-7$." Please calculate how many square centimeters of cardboard are needed to make such a box (ignoring the seams)?
|
Three, 15. Since $x y=x z+3$, therefore, $x(y-z)=3$.
Also, $x, y, z$ are all positive integers, then
$$
\left\{\begin{array}{l}
x = 3, \\
y - z = 1
\end{array} \text { or } \left\{\begin{array}{l}
x=1, \\
y-z=3 .
\end{array}\right.\right.
$$
(1) When $\left\{\begin{array}{l}x=3, \\ y=z+1\end{array}\right.$, by $y z=x y+x z-7$, that is
$$
(z+1) z=3(z+1)+3 z-7,
$$
rearranging gives
$$
z^{2}-5 z+4=0 .
$$
Thus, the surface area of the rectangular box is
$$
\begin{array}{l}
2(x y+x z+y z) \\
=2(3 \times 2+3 \times 1+2 \times 1)=22\left(\mathrm{~cm}^{2}\right),
\end{array}
$$
or $2(x y+x z+y z)$
$$
=2(3 \times 5+3 \times 4+5 \times 4)=94\left(\mathrm{~cm}^{2}\right) \text { (discard). }
$$
(2) When $\left\{\begin{array}{l}x=1, \\ y=z+3\end{array}\right.$, by $y z=x y+x z-7$, that is
$$
(z+3) z=z+3+z-7,
$$
rearranging gives
$$
z^{2}+z+4=0 .
$$
This equation has no real solutions.
Therefore, to make such a box, at least $22 \mathrm{~cm}^{2}$ of cardboard is needed.
|
22
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. (18 points) Among 200 small balls numbered $1, 2, \cdots, 200$, any $k$ balls are drawn such that there must be two balls with numbers $m$ and $n$ satisfying
$$
\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} .
$$
Therefore, the minimum value of $k$ is 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If a natural number $n$ allows the vertical addition
$$
n+(n+1)+(n+2)
$$
to not produce any carry, then $n$ is called a "continuous number". For example, 12 is a continuous number, because $12+13+14$ does not produce a carry; but 13 is not a continuous number. Then, the number of continuous numbers less than 1000 is ( ).
(A) 27
(B) 47
(C) 48
(D) 60
|
5.C.
There are $0,1,2$, a total of 3 single-digit numbers;
There are $\overline{a 0}, \overline{a 1}, \overline{a 2}, a=1,2,3$, a total of $3 \times 3=9$ two-digit numbers;
There are $\overline{a b 0}, \overline{a b 1}, \overline{a b 2}, a=1,2,3, b=0,1,2,3$, a total of $3 \times 4 \times 3=36$ three-digit numbers.
Therefore, there are $3+9+36=48$ continuous numbers less than 1000.
|
48
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let natural numbers $x, y$ satisfy
$$
x<y, x^{3}+19 y=y^{3}+19 x
$$
Then $x+y=$ $\qquad$
|
2.5.
From $x^{3}-y^{3}=19(x-y), x<y$, we get $x^{2}+x y+y^{2}=19$.
Thus, $3 x^{2}<x^{2}+x y+y^{2}=19$.
Since $x$ is a natural number, we have $x=2$.
Therefore, $y=3$, so $x+y=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The 2008 Olympic Games will be held in Beijing. Figure 3 is the Olympic five-ring logo, where $a, b, c, d, e, g, h, i$ are filled with different numbers from $1 \sim 9$. If the sum of the numbers in each circle is equal, then the maximum value of this sum is $\qquad$
|
3.14 .
Let the sum of the numbers in each circle be $N$, then
$$
(1+2+\cdots+9)+(b+d+f+h)=5 N .
$$
Therefore, $N=\frac{b+d+f+h}{5}+9$.
To maximize $N$, $b+d+f+h$ must be the largest possible value and a multiple of 5.
Since $b+d+f+h=6+7+8+9=30$ is the maximum, and $N=15$, at this point, no matter how you fill in the numbers, there will be repeated or missing numbers. Therefore, $N$ cannot exceed 14.
After testing, when $a=5, b=9, c=2, d=3, e=4, f=7$, $g=1, h=6, i=8$, the conditions of the problem are satisfied.
Thus, the maximum sum of the numbers is 14.
|
14
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(20 points) Read the material and answer the questions:
As shown in Figure 5, the shapes of rhombuses, rectangles, and squares differ. We refer to the degree of similarity between these shapes as "closeness."
When studying closeness,
the closeness of similar figures
should be equal.
(1) Let the two adjacent interior angles of a rhombus be $m^{\circ}$ and $n^{\circ}$. Define the closeness of the rhombus as $|m-n|$. Thus, the smaller the $|m-n|$, the closer the rhombus is to a square.
(i) If one interior angle of the rhombus is $80^{\circ}$, then the closeness of the rhombus is $\qquad$ ;
(ii) When the closeness of the rhombus equals $\qquad$, the rhombus is a square.
(2) Let the lengths of the two sides of a rectangle be $a$ and $b$. Define the closeness of the rectangle as $|a-b|$. Thus, the smaller the $|a-b|$, the closer the rectangle is to a square.
Do you think this statement is reasonable? If not, explain why and provide a reasonable definition for the closeness of a rectangle.
|
(1)(i)20 (ii)0 (2)Unreasonable
In rectangles with side lengths of $a, b$ and $2a, 2b$, it is clear that these two rectangles are similar, but $|2a-2b| \neq |a-b|$.
The proximity of a rectangle can be defined as $\frac{a}{b}(a \leqslant b)$, then the closer $\frac{a}{b}$ is to 1, the closer the rectangle is to a square.
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the sequence $\left\{a_{n}\right\}$ consists of positive integers and is an increasing sequence, and satisfies $a_{n+2}=a_{n+1}+2 a_{n}\left(n \in \mathbf{N}_{+}\right)$. If $a_{5}$ $=52$, then $a_{7}=(\quad)$.
(A) 102
(B) 152
(C) 212
(D) Insufficient conditions, $a_{7}$ cannot be uniquely determined
|
2.C.
$$
\begin{array}{l}
a_{3}=a_{2}+2 a_{1}, \\
a_{4}=a_{3}+2 a_{2}=3 a_{2}+2 a_{1}, \\
a_{5}=a_{4}+2 a_{3}=5 a_{2}+6 a_{1} .
\end{array}
$$
The positive integer solutions to the equation $5 a_{2}+6 a_{1}=52$ are $a_{1}=2, a_{2}=8$ or $a_{1}=7, a_{2}=2$.
Since $a_{2}>a_{1}$, we have $a_{1}=2, a_{2}=8$.
Thus, $a_{7}=a_{6}+2 a_{5}=21 a_{2}+22 a_{1}=212$.
|
212
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 "Tossing a coin continuously until two consecutive heads appear" occurs in all possible ways on the $n(n \geqslant 2)$-th toss.
How many such ways are there?
|
Explanation: Let the number of methods that satisfy the conditions be $a_{n}$.
When $n=2$, there is only 1 case of "positive positive", $a_{2}=1$;
When $n=3$, there is only 1 case of "negative positive positive", $a_{3}=1$.
For $n \geqslant 4$, the number of methods $a_{n}$ where "positive positive" appears exactly at the $n$-th toss includes two scenarios:
(1) The 1st toss is negative, and "positive positive" appears in the subsequent $n-1$ tosses, with $a_{n-1}$ methods;
(2) The 1st toss is positive, then the 2nd toss must be negative, and "positive positive" appears in the subsequent $n-2$ tosses, with $a_{n-2}$ methods.
Thus, we have $\left\{\begin{array}{l}a_{n}=a_{n-1}+a_{n-2}, \\ a_{2}=1, a_{3}=1 .\end{array}\right.$
Therefore, the problem of finding the number of methods is transformed into finding the general term of the Fibonacci sequence, also using the recursive method.
1.2 Understanding of the Recursive Method
From the above example, we can make an initial definition of recursive relations, the recursive method, and how to use the recursive method.
1.2.1 Recursive Relation
Let $f(1), f(2), \cdots, f(n), \cdots$ represent the various states of a changing process. If state $f(n)$ is dependent on previous states $f(n-1), f(n-2)$, etc., and can be expressed as an equation or inequality (see Example 7)
$f(n)=F(f(n-1), f(n-2), \cdots, f(1))$
or $f(n) \geqslant F(f(n-1), f(n-2), \cdots, f(1))$
or $f(n) \leqslant F(f(n-1), f(n-2), \cdots, f(1))$.
Then such a relationship is called a recursive relation, and when it is an equality, it is commonly known as a recursive equation or recursive formula.
In this case, one can start from some initial conditions and gradually deduce the result at any point in time. That is, the initial conditions and the recursive relation (both are indispensable) can completely determine a process (for example, in Example 1, from $a_{1}=1, a_{2}=1$ and the recursive relation (1), the Fibonacci sequence can be written as: $1,1,2,3,5,8,13,21,34,55,89,144,233$, $377, \cdots$. The answer is $\left.a_{13}=233\right)$.
|
233
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) Let real numbers $p, q, r$ satisfy: there exists $a$ which is one of $p, q, r$, and the other two are exactly the two real roots of the equation
$$
x^{2}+(a-3) x+a^{2}-3 a=0
$$
Find the minimum possible value of $p^{3}+q^{3}+r^{3}$.
|
Three, let $a=p$, then $q, r$ are exactly the two roots of the equation. By Vieta's formulas, we know $p+q+r=3$ and $qr=p^2-3p$.
From $\Delta=(p-3)^2-4(p^2-3p) \geqslant 0$, we know $-1 \leqslant p \leqslant 3$.
$$
\begin{array}{l}
\text { Then } p^2+q^2+r^2=p^2+(q+r)^2-2qr \\
=p^2+(3-p)^2-2(p^2-3p)=9 \text {. }
\end{array}
$$
Thus, $p^3+q^3+r^3$
$$
\begin{array}{l}
=3pqr+(p+q+r)[p^2+q^2+r^2-p(q+r)-qr] \\
=3p(p^2-3p)+3[9-p(3-p)-(p^2-3p)] \\
=3p^3-9p^2+27,
\end{array}
$$
where $-1 \leqslant p \leqslant 3$.
For $f(p)=3p^3-9p^2+27$, taking the derivative gives
$$
f'(p)=9p^2-18p=9p(p-2) \text {. }
$$
Thus, $f(p)$ is increasing on $[2,3]$, decreasing on $[0,2]$, and increasing on $[-1,0]$.
Therefore, the minimum value of $f(p)$ on $[-1,3]$ is
$$
\min \{f(-1), f(2)\}=15 \text {. }
$$
Hence, the minimum value of $p^3+q^3+r^3$ is 15.
When $p=2, q=2, r=-1$, the minimum value can be achieved and the conditions are satisfied.
|
15
|
Algebra
|
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
|
15. A football team participates in the World Cup, and in a certain match, 11 players start the game, using a 4-4-2 formation (4 defenders, 4 midfielders, 2 forwards). There are 7 substitute players (2 defenders, 2 midfielders, 2 forwards, and 1 goalkeeper). The rules allow for a maximum of three substitutions. If the team maintains the 4-4-2 formation after substitutions, and substituted players do not return to the field, and no players are sent off during the entire match, it is found that the two main forwards and the goalkeeper play the entire match, and substitutions only occur between defenders and defenders or midfielders and midfielders. How many:
(1) different possible lineups of the 11 players on the field at the end of the match are there?
(2) different possible match processes are there (if the 11 players on the field at any time are different, the matches are considered different processes)?
|
15. (1) Let the total number of substitutions in the match be $t$, where $t=0,1,2,3$.
When $t=0$, 11 players play the entire match, which is only 1 way.
When $t=1$, the substituted player can be a midfielder or a defender, resulting in $\mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1} + \mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1} = 16$ ways.
When $t=2$, the substitutions can be two midfielders, two defenders, or one midfielder and one defender, resulting in $\mathrm{C}_{4}^{2} + \mathrm{C}_{4}^{2} + \left(\mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1}\right)\left(\mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1}\right) = 76$ ways.
When $t=3$, the substitutions can be two midfielders and one defender, or two defenders and one midfielder, resulting in $\mathrm{C}_{4}^{2} \mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1} + \mathrm{C}_{4}^{2} \mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1} = 96$ ways.
In total, there are $1 + 16 + 76 + 96 = 189$ ways.
(2) When $t \geqslant 2$, each final lineup corresponds to $t!$ different match processes. Therefore, the total number of different possible match processes is $1 + 16 + 76 \times 2! + 96 \times 3! = 745$ ways.
|
189
|
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
|
3. As shown in Figure $9, A B$ is the diameter of a semicircle, and $C$ is a point on the semicircular arc. One side $D G$ of the square $D E F G$ lies on the diameter $A B$,
and the other side $D E$ passes through the incenter $I$ of $\triangle A B C$, with point
$E$ on the semicircular arc. If the area of the square $D E F G$ is
100, find the area of $\triangle A B C$.
|
(Given: $A D=\frac{b+c-a}{2}, B D=\frac{c+a-b}{2}$. From $D E^{2}$ $=A D \cdot D B$, we get $100=\frac{b+c-a}{2} \cdot \frac{c+a-b}{2}=\frac{a b}{2}$, hence $\left.S_{\triangle B B C}=100.\right)$
|
100
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $c(X)$ denote the number of subsets of set $X$. If $n$ sets $A_{1}, A_{2}, \cdots, A_{n}$ with different numbers of elements satisfy:
$$
\left(\left|A_{1}\right|+1\right)\left(\left|A_{2}\right|+1\right) \cdots\left(\left|A_{n}\right|+1\right)=2006
$$
and
$$
\begin{array}{l}
c\left(A_{1}\right)+c\left(A_{2}\right)+\cdots+c\left(A_{n}\right) \\
=c\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)+ \\
\quad c\left(A_{1} \cap A_{2} \cap \cdots \cap A_{n}\right)(n \geqslant 2),
\end{array}
$$
then the minimum value of $\max \left\{\left|A_{1}\right|,\left|A_{2}\right|, \cdots,\left|A_{n}\right|\right\}$ is
$\qquad$
|
2.58.
Assume $\left|A_{1}\right|>\left|A_{2}\right|>\cdots>\left|A_{n}\right| \geqslant 0$.
When $n \geqslant 3$, by $c(X)=2^{|X|}$ and the condition, we know that $2^{\left|A_{1}\right|}+2^{\left|A_{2}\right|}+\cdots+2^{\left|A_{n}\right|}=2^{\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|}+2^{\left|A_{1} \cap A_{2} \cap \cdots \cap A_{n}\right|}$.
The left side of the above equation is greater than $2^{\left|A_{1}\right|}+2^{\left|A_{n}\right|}$; the right side is less than or equal to $2^{\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|}+2^{\left|A_{n}\right|}$, hence $\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right| \geqslant\left|A_{1}\right|+1$.
Also, the left side of the above equation is less than $2^{\left|A_{1}\right|+1}$, so $\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|b)$, then $(a+1)(b+1)=2006=2 \times 17 \times 59=34 \times 59$. Therefore, $a \geqslant 59-1=58$.
When $a=58, b=33$, the minimum value is achieved.
|
58
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that the ellipse $C_{1}$ and the hyperbola $C_{2}$ share foci $F_{1}(3,0), F_{2}(-3,0)$, and have coincident minor axes. Then the number of lattice points inside the region enclosed by the intersection points of $C_{1}$ and $C_{2}$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3.25.
From the problem, we can let
$$
C_{1}: \frac{x^{2}}{m+9}+\frac{y^{2}}{m}=1, C_{2}: \frac{x^{2}}{9-m}-\frac{y^{2}}{m}=1,
$$
where $0<m<9$.
Let the intersection point be $P\left(x_{0}, y_{0}\right)$. Then, since point $P$ lies on $C_{1}$ and $C_{2}$, we have $\frac{x_{0}^{2}}{m+9}+\frac{x_{0}^{2}}{9-m}=2$, which simplifies to $x_{0}^{2}=9-\frac{m^{2}}{9}$.
Thus, $y_{0}^{2}=\frac{m^{2}}{9}$, where $x_{0}^{2}, y_{0}^{2} \neq 0,9$.
Therefore, $x_{0}^{2}+y_{0}^{2}=9$.
From the sketch (omitted), it is known that there are 25 lattice points within the enclosed region.
|
25
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given that $a, b$ are positive real numbers, the equations
$$
x^{2}+a x+2 b=0 \text{ and } x^{2}+2 b x+a=0
$$
both have real roots. Find the minimum value of $a^{2}+b^{2}$.
|
Explanation: Since the given equations all have real roots, we obtain the constraint conditions
$$
\left\{\begin{array}{l}
a^{2}-8 b \geqslant 0 \\
b^{2}-a \geqslant 0 \\
a>0, b>0
\end{array}\right.
$$
To find the minimum value of $a^{2}+b^{2}$, using algebraic methods is clearly challenging, so we consider a numerical and geometric transformation.
As shown in Figure 1, we draw the right half of the parabola $a^{2}=8 b(a>0)$, and then draw the upper half of the parabola $b^{2}=a(b>0)$. The intersection point of the two curves is $M(4,2)$. Thus, the point $P(a, b)$ that satisfies equations (1), (2), and (3) lies in the shaded region $G$ (including the boundary) shown in Figure 1.
Since $a^{2}+b^{2}$ represents the square of the distance from point $P(a, b)$ to the origin $O$, it is easy to see that when $a=4, b=2$,
$$
\left(a^{2}+b^{2}\right)_{\min }=|O M|^{2}=20 .
$$
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let the real number $a$ be such that the quadratic equation $5 x^{2}-5 a x+66 a-1715=0$ has two integer roots. Then all such $a$ are $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3.870.
Let the two integer roots be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$. By the relationship between roots and coefficients, we have $a=x_{1}+x_{2}$, thus, $a$ is an integer. From the original equation, we get
$$
\begin{array}{l}
a=\frac{5 x^{2}-1715}{5 x-66}=x+13+\frac{x-857}{5 x-66} \in \mathbf{Z} \\
\Leftrightarrow \frac{x-857}{5 x-66} \in \mathbf{Z} \\
\Leftrightarrow \frac{5(x-857)}{5 x-66} \in \mathbf{Z} \text { (since } 5 \text { and } 5 x-66 \text { are coprime) } \\
\Leftrightarrow \frac{-5 \times 857+66}{5 x-66} \in \mathbf{Z} \Leftrightarrow \frac{4219}{5 x-66} \in \mathbf{Z} \\
\Leftrightarrow 5 x-66= \pm 1 \text { or } \pm 4219 \text { (since } 4219 \text { is a prime number) } \\
\Leftrightarrow x=13 \text { or } 857 .
\end{array}
$$
Therefore, $a=13+857$ or $13+13$ or $857+857$, which means $a=870$ or 26 or 1714.
Since the equation has integer roots, $5 \mid a$, which contradicts $a=26,1714$. Hence, $a=870$.
|
870
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If from the set $S=\{1,2, \cdots, 20\}$, we take a three-element subset $A=\left\{a_{1}, a_{2}, a_{3}\right\}$, such that it simultaneously satisfies: $a_{2}-a_{1} \geqslant 5,4 \leqslant a_{3}-a_{2} \leqslant 9$, then the number of all such subsets $A$ is $\qquad$ (answer with a specific number).
|
6.251.
$a_{2}-a_{1} \geqslant 5 \Leftrightarrow\left(a_{2}-4\right)-a_{1} \geqslant 1$,
$4 \leqslant a_{3}-a_{2} \leqslant 9 \Leftrightarrow 1 \leqslant\left(a_{3}-7\right)-\left(a_{2}-4\right) \leqslant 6$.
Let $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-4, a_{3}^{\prime}=a_{3}-7$, then $a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime} \in \mathbf{Z}$, satisfying $1 \leqslant a_{1}^{\prime}<a_{2}^{\prime}<a_{3}^{\prime} \leqslant 13$, and
$$
a_{3}^{\prime}-a_{2}^{\prime} \leqslant 6 .
$$
Obviously, the sets of $\left(a_{1}, a_{2}, a_{3}\right)$ and the new sets of $\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}\right)$ form a one-to-one correspondence. Below, we count the number of $\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}\right)$ that satisfy (1), for which we categorize and count.
(1) If $a_{2}^{\prime} \leqslant 7$, then $1 \leqslant a_{1}^{\prime}<a_{2}^{\prime} \leqslant 7,1 \leqslant a_{3}^{\prime}-a_{2}^{\prime} \leqslant 6$. When $a_{1}^{\prime}, a_{2}^{\prime}$ are determined, $a_{3}^{\prime}-a_{2}^{\prime}$ has 6 possible values $\left(a_{3}^{\prime}-a_{2}^{\prime}=1,2,3,4,5,6\right)$, so, $\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}\right)$ has
$$
6 \times C_{7}^{2}=6 \times \frac{7 \times 6}{2}=126 \text { (sets). }
$$
(2) If $8 \leqslant a_{2}^{\prime} \leqslant 12$, then $1 \leqslant a_{3}^{\prime}-a_{2}^{\prime} \leqslant 13-a_{2}^{\prime}(\leqslant 5<6)$. When $a_{2}^{\prime}\left(8 \leqslant a_{2}^{\prime} \leqslant 12\right)$ is determined, $a_{1}^{\prime}$ has $a_{2}^{\prime}-1$ possible values, and $a_{3}^{\prime}-a_{2}^{\prime}$ has $13-a_{2}^{\prime}$ possible values $\left(a_{3}^{\prime}-a_{2}^{\prime}=1,2, \cdots, 13-\right.$ $\left.a_{2}^{\prime}\right)$, so, $\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}\right)$ has
$$
\begin{array}{l}
\sum_{a_{2}^{\prime}=8}^{12}\left(a_{2}^{\prime}-1\right)\left(13-a_{2}^{\prime}\right) \\
=7 \times 5+8 \times 4+9 \times 3+10 \times 2+11 \times 1 \\
=125 \text { (sets). }
\end{array}
$$
In summary, the number of $\left(a_{1}, a_{2}, a_{3}\right)$ is 251.
|
251
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Try to find the minimum value of the function $f(x, y)=6\left(x^{2}+y^{2}\right) \cdot (x+y)-4\left(x^{2}+x y+y^{2}\right)-3(x+y)+5$ in the region $A=\{(x, y) \mid x>0, y>0\}$.
|
Explanation: If $x+y \leqslant 1$, then
$$
\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
$$
$>2$.
In summary, for any $(x, y) \in A$, we have $f(x, y) \geqslant 2$.
When and only when $x=y=\frac{1}{2}$, $f(x, y)_{\min }=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points) Try to find the smallest positive integer $m$, such that the following conditions are satisfied simultaneously:
(1) $\left[\frac{2}{1977} m^{2}\right] \geqslant m+2006$ ( $[x]$ denotes the greatest integer not exceeding $x$);
(2) $99^{m}$ leaves a remainder of 11 when divided by 190.
|
From condition (1), we know that $m > 1977$.
Otherwise, if $m \leqslant 1977$, then we have
$$
\begin{array}{l}
{\left[\frac{2}{1977} m^{2}\right] \leqslant \frac{2}{1977} m^{2} \leqslant \frac{2}{1977} \times 1977 m} \\
= 2m \cdot 1977.
$$
Let $m = 1977 + k$ ($k$ is a positive integer), and substitute it into the inequality in condition (1):
$$
2 \times 1977 + 4k + \left[\frac{2}{1977} k^{2}\right] \geqslant 1977 + k + 2006,
$$
which simplifies to
$$
\left[\frac{2}{1977} k^{2}\right] + 3k \geqslant 29.
$$
Since $f(k) = \left[\frac{2}{1977} k^{2}\right] + 3k$ is a strictly increasing function, and $k = 10$ satisfies inequality (1), while $k = 9$ does not satisfy inequality (1), the solution to inequality (1) is $k \geqslant 10$. Therefore, the solution to the inequality in condition (1) is
$$
m = 1977 + k \geqslant 1987.
$$
Since 99 and 190 are coprime, by Euler's theorem, we have
$$
99^{1500} \equiv 1 \pmod{190},
$$
which implies $99^{72} \equiv 1 \pmod{190}$.
Lemma: If $99^a \equiv 1 \pmod{190}$, where $a$ is a positive integer, and $t$ is the smallest positive integer such that $99^t \equiv 1 \pmod{190}$, then $t \mid a$.
Proof of the lemma: Clearly, $a \geqslant t$ (by the minimality of $t$).
Let $a = t q + r$ ($q \in \mathbf{N}_{+}$, $r \in \mathbf{N}$, and $0 \leqslant r < t$).
From $99^a \equiv 1 \pmod{190}$ and $99^t \equiv 1 \pmod{190}$, we get
$$
(99^t)^q 99^r \equiv 1 \pmod{190},
$$
which simplifies to
$$
99^r \equiv 1 \pmod{190}.
$$
By the minimality of $t$ and $0 \leqslant r < t$, we must have $r = 0$. Thus, $a = t q$, which means $t \mid a$.
Returning to the original problem.
From equation (2) and the lemma, we know $t \mid 72 = 2^3 \times 3^2$.
Therefore, $t \in \{2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72\}$.
Now, we need to find the value of $t$.
Notice that
$$
99^2 \equiv 9801 \equiv 111 \pmod{190},
$$
$$
99^3 \equiv 111 \times 99 \equiv -31 \pmod{190},
$$
$$
99^4 \equiv -31 \times 99 \equiv -29 \pmod{190},
$$
$$
99^6 \equiv (-31)^2 \equiv 11 \pmod{190},
$$
$$
99^8 \equiv (-29)^2 \equiv 81 \pmod{190},
$$
$$
99^9 \equiv 11 \times (-31) \equiv 39 \pmod{190},
$$
$$
99^{12} \equiv 11^2 \equiv -69 \pmod{190},
$$
$$
99^{18} \equiv 39^2 \equiv 1 \pmod{190},
$$
so $t = 18$.
Given $99^m \equiv 11 \equiv 99^6 \pmod{190}$, and $m \geqslant 1987$, and 99 and 190 are coprime, we have $99^{m-6} \equiv 1 \pmod{190}$.
By $t = 18$ and the above lemma, we get $18 \mid (m-6)$, which means $m = 6 + 18n$ ($n \in \mathbf{N}$).
From $m \geqslant 1987$, we get $18n + 6 \geqslant 1987$. Thus,
$$
n \geqslant 111, \quad m \geqslant 18 \times 111 + 6 = 2004.
$$
Therefore, the smallest positive integer $m$ is $2004$.
|
2004
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let $0 \leqslant x \leqslant \pi, 0 \leqslant y \leqslant 1$. Try to find the minimum value of the function
$$
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
|
Example 9 Let $a, b, c$ be positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$.
(2005, National High School Mathematics League, Fujian Province Preliminary
|
Let the two real roots of the equation be $x_{1}$ and $x_{2}$. By Vieta's formulas, we know $x_{1}x_{2} = \frac{c}{a} = 9$. Therefore,
\[ b^{2} \geqslant 4ac = 4 \times \frac{a}{c} \times c^{2} > 4 \times 9 \times 1^{2} = 36. \]
Thus, $b \geqslant 7$.
Also, $\frac{b}{a} = (-x_{1}) + (-x_{2}) \Rightarrow \frac{3}{2} b \geqslant \frac{21}{2}$, which implies $a \geqslant 11$.
(1) When $b = 7$, from $4a \leqslant 4ac \leqslant b^{2}$ and $a \geqslant 11$, we know $a = 11$ or $12$, and $c = 1$. However, the equation $11x^{2} + 7x + 1 = 0$ has roots $-\frac{7 + \sqrt{5}}{22} - \frac{1}{3}$, which gives
\[ 4 + \sqrt{16 - a} > 0. \]
Since $f(15) = 0$, $a$ can only be 16.
At this point, $a + b + c = 25$, and the equation $16x^{2} + 8x + 1 = 0$ has roots $x_{1} = x_{2} = -\frac{1}{4}$, which satisfies the problem's conditions.
(3) When $b \geqslant 9$, $a > \frac{3}{2}b \geqslant \frac{27}{2}$, so $a \geqslant 14$. Therefore, $a + b + c \geqslant 14 + 9 + 1 = 24$.
If $a + b + c < 25$, then it can only be $a = 14$, $b = 9$, and $c = 1$. In this case, the equation $14x^{2} + 9x + 1 = 0$ has roots $-\frac{1}{2}$ and $-\frac{1}{7}$, which does not meet the problem's conditions.
Thus, $a + b + c \geqslant 25$.
In summary, the minimum value of $a + b + c$ is 25.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let sets $A$ and $B$ be sets composed of positive integers, $|A|=10, |B|=9$, and set $A$ satisfies the following condition:
If $x, y, u, v \in A, x+y=u+v$, then $\{x, y\}=\{u, v\}$.
Let $A+B=\{a+b \mid a \in A, b \in B\}$.
Prove: $|A+B| \geqslant 50$ ( $|X|$ represents the number of elements in set $X$).
(2005, National High School Mathematics League Fujian Province Preliminary)
This problem is strengthened as follows:
Let sets $A$ and $B$ be sets composed of positive integers, $|A|=10, |B|=9$, and set $A$ satisfies the following condition:
If $x, y, u, v \in A, x+y=u+v$, then $\{x, y\}=\{u, v\}$.
Let $A+B=\{a+b \mid a \in A, b \in B\}$.
Prove: $|A+B| \geqslant 54$ ( $|X|$ represents the number of elements in set $X$).
|
Prove: Let $A=\left\{a_{1}, a_{2}, \cdots, a_{10}\right\}$,
$$
\begin{array}{l}
B=\left\{b_{1}, b_{2}, \cdots, b_{9}\right\}, \\
C_{i}=\left\{a_{1}+b_{i}, a_{2}+b_{i}, \cdots, a_{10}+b_{i}\right\}
\end{array}
$$
where $a_{i}(i=1,2, \cdots, 10)$ are distinct.
Prove by contradiction.
$$
\left|C_{i} \cap C_{j}\right| \leqslant 1(i \neq j \text { and } i, j \in\{1,2, \cdots, 9\}) \text {. }
$$
If $C_{i} \cap C_{j} \geqslant 2$, then without loss of generality, assume
$$
\begin{array}{l}
a_{1}+b_{1}=a_{2}+b_{2}, \\
a_{3}+b_{1}=a_{4}+b_{2} .
\end{array}
$$
From equation (1),
$$
a_{1}-a_{2}=b_{2}-b_{1} \text {, }
$$
From equation (2),
$$
a_{3}-a_{4}=b_{2}-b_{1} \text {. }
$$
From equations (3) and (4),
$$
\begin{array}{l}
a_{1}-a_{2}=a_{3}-a_{4} \Rightarrow a_{1}+a_{4}=a_{2}+a_{3} \\
\Rightarrow\left\{a_{1}, a_{4}\right\}=\left\{a_{2}, a_{3}\right\},
\end{array}
$$
which contradicts the fact that $a_{i}(i=1,2, \cdots, 10)$ are distinct.
Therefore, $\left|C_{i} \cap C_{j}\right| \leqslant 1$.
$$
\begin{array}{l}
\text { Hence }|A+B|=\left|\bigcup_{i=1}^{9} C_{i}\right| \\
\geqslant \sum_{i=1}^{9}\left|C_{i}\right|-\sum_{1 \leqslant i<j \leqslant 9}\left|C_{i} \cap C_{j}\right| \\
\geqslant 90-C_{9}^{2}=54 .
\end{array}
$$
|
54
|
Combinatorics
|
proof
|
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
|
12. Calculate:
\[
\begin{array}{l}
\sqrt{2005 \times 2006 \times 2007 \times 2008+1}-2006^{2} \\
=
\end{array}
\]
|
12.2005 .
Let $n=2005$. Then
$$
\begin{array}{l}
\text { Original expression }=\sqrt{n(n+1)(n+2)(n+3)+1}-(n+1)^{2} \\
=\sqrt{\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1}-(n+1)^{2} \\
=\sqrt{\left(n^{2}+3 n+1\right)^{2}}-(n+1)^{2}=n=2005 .
\end{array}
$$
|
2005
|
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
|
18. $A 、 n$ are natural numbers, and
$$
A=n^{2}+15 n+26
$$
is a perfect square. Then $n$ equals $\qquad$
|
18.23.
Let $A=n^{2}+15 n+26=(n+2)(n+13)$ be a perfect square, we can set $n+2=a^{2}, n+13=b^{2}(a, b$ are positive integers and $b$ $>a)$. Therefore, $11=b^{2}-a^{2}=(b+a)(b-a)$. Then we have $\left\{\begin{array}{l}b+a=11, \\ b-a=1 .\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}a=5, \\ b=6 .\end{array}\right.$
Thus, $n=a^{2}-2=23$.
|
23
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
19. A rectangular prism has integer length, width, and height, and its volume is exactly $2006 \mathrm{~cm}^{3}$. After painting its surface red, it is then cut into small cubes with edge lengths of $1 \mathrm{~cm}$. If there are 178 small cubes with three red faces, then the number of small cubes with exactly two red faces is _.
|
19.1824 .
From the problem, we know that one of the edges of the cuboid must be $1 \mathrm{~cm}$, otherwise, only 8 small cubes would have three faces painted red. Let's assume the length, width, and height of the cuboid are $x \mathrm{~cm}, y \mathrm{~cm}, 1 \mathrm{~cm}$, respectively. Then we have
$$
\left\{\begin{array}{l}
2(x-2)+2(y-2)=178, \\
x y=2006 .
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=59, \\
y=34 .
\end{array}\right.\right.
$$
Therefore, the number of small cubes with exactly two faces painted red is
$$
(x-2)(y-2)=57 \times 32=1824 \text { (cubes). }
$$
|
1824
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let the first term and common difference of an arithmetic sequence be positive integers, the number of terms be a prime number no less than 3, and the sum of all terms be 2006. Then the number of such sequences is $\qquad$.
|
Ni, 7.15.
Let the first term and common difference of an arithmetic sequence be $a$ and $d$, and the number of terms be $n$, then
$$
n a+\frac{1}{2} n(n-1) d=2006,
$$
i.e., $n[2 a+(n-1) d]=2 \times 2006=2 \times 2 \times 17 \times 59$.
Since $2 a+(n-1) d \geqslant 2+n-1=n+1>n$, hence $n=17,59$.
(1) If $n=17$, then $a+8 d=118$, there are 14 sets of positive integer solutions;
(2) If $n=59$, then $a+29 d=34$, there is 1 set of positive integer solutions. Therefore, there are a total of 15 sets of positive integer solutions.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
(x-11)^{5}+15(x-11)=5, \\
(y-4)^{5}+15(y-4)=-5 .
\end{array}\right.
$$
Then $x+y$ equals $\qquad$ .
|
8. 15 .
Since the function $f(t)=t^{5}+15 t$ is strictly monotonically increasing in the real number domain, and $f(x-11)=f(4-y)$, then $x-11=4-y$. Therefore, $x+y=15$.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. The 8 vertices of a cube can form
$\qquad$ non-equilateral triangles.
|
11.48.
Each vertex leads to three edges whose endpoints can form an equilateral triangle, making a total of 8 equilateral triangles. Therefore, the answer is $\mathrm{C}_{8}^{3}-8=48$.
|
48
|
Geometry
|
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
|
6. In a mathematics competition, there are three types of questions: multiple-choice questions (6 questions), fill-in-the-blank questions (4 questions), and problem-solving questions (3 questions). Among them, each multiple-choice question and each fill-in-the-blank question is worth 7 points, and the problem-solving questions are worth 20 points for the 1st question and 25 points each for the 2nd and 3rd questions, with a total full score of 140 points. The scoring criteria are: 7 points for a correct answer to a multiple-choice or fill-in-the-blank question, and 0 points for an unanswered or incorrect answer; for problem-solving questions, there are 6 levels of scores: 0 points, 5 points, 10 points, 15 points, 20 points, and 25 points. Therefore, the maximum number of different scores that can be obtained in this exam is ( . ) types.
(A) 141
(B) 129
(C) 105
(D) 117
|
G.D.
(1) The possible scores for multiple-choice and fill-in-the-blank questions are $0,7,14, \cdots, 70$, a total of 11 possibilities. The possible scores for problem-solving questions are $0,5,10, \cdots, 70$, a total of 15 possibilities. Therefore, there can be $11 \times 15=165$ results.
(2) The following 23 scores $0,5,7,10,12,14,15,17,19$, $20,21,22,24,25,26,27,28,29,30,31,32,33,34$ can be obtained and each has only one way to achieve; also, $35=7 \times 5=5 \times 7$, meaning 35 points can be represented by getting 5 questions right worth 7 points each or 7 questions right worth 5 points each. Therefore, the 23 scores mentioned earlier, when each increased by 35, have two ways to achieve. Thus, these new 23 scores each have two ways to achieve and are all less than 70. By symmetry, subtracting these new 23 scores from 140 also results in numbers that have two ways to represent, and these numbers are precisely the scores between 71 and 140 that can be achieved. In total, there are $2 \times 23=46$ such numbers.
(3) Since $70=7 \times 10=5 \times 14=7 \times 5+5 \times 7$, there are three ways to achieve 70 points.
Therefore, the number of different scores that can be achieved without repetition is
$$
165-46-2=117 \text { (ways). }
$$
|
117
|
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
|
Given that $k$ and $a$ are positive integers, and $2004k + a$, $2004(k+1) + a$ are both perfect squares.
(1) How many ordered pairs of positive integers $(k, a)$ are there?
(2) Identify the minimum value of $a$ and explain your reasoning.
|
II. (1) Let $2004k + a = m^2$,
(1)
$$
2004(k+1) + a = n^2 \text{, }
$$
where $m, n$ are positive integers, then
$$
n^2 - m^2 = 2004 \text{. }
$$
Thus, $(n+m)(n-m) = 2004 = 2 \times 2 \times 3 \times 167$.
Noting that $m+n$ and $n-m$ have the same parity, we have
$$
\left\{\begin{array}{l}
n + m = 1002, \\
n - m = 2
\end{array} \text{ or } \left\{\begin{array}{l}
n + m = 334, \\
n - m = 6
\end{array}\right.\right.
$$
Solving these, we get $\left\{\begin{array}{l}n=502, \\ m=500\end{array}\right.$ or $\left\{\begin{array}{l}n=170, \\ m=164\end{array}\right.$
When $n=502, m=500$, from equation (1) we have
$$
2004k + a = 250000 \text{. }
$$
Thus, $a = 2004(124 - k) + 1504$.
Since $k$ and $a$ are positive integers, $k$ can take values $1, 2, \cdots, 124$, resulting in 124 pairs of positive integer solutions $(k, a)$.
When $n=170, m=164$, from equation (1) we have
$2004k + a = 26896$.
Thus, $a = 2004(13 - k) + 844$.
Since $k$ and $a$ are positive integers, $k$ can take values $1, 2, \cdots, 13$, resulting in 13 pairs of positive integer solutions $(k, a)$.
Therefore, the total number of positive integer solutions $(k, a)$ is $124 + 13 = 137$ (pairs).
(2) The smallest positive integer $a$ satisfying equation (3) is 1504. The smallest positive integer $a$ satisfying equation (4) is 844.
Thus, the smallest value of $a$ is 844.
|
137
|
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
|
8. Remove all positive integers that are divisible by 4 and those that leave a remainder of 1 when divided by 4, and arrange the remaining numbers in ascending order to form a sequence $\left\{a_{n}\right\}$:
$$
2,3,6,7,10,11, \cdots \text {. }
$$
The sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is denoted as $S_{n}$, where $n=$ $1,2, \cdots$. Find
$$
S=\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2000}}\right]
$$
where $[x]$ denotes the greatest integer not exceeding $x$.
|
8. It is known that $a_{2 n-1}=4 n-2, a_{2 n}=4 n-1(n=1,2$, $\cdots$ ). Therefore,
$$
\begin{aligned}
& S_{2 n}=\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}\right)+\cdots+\left(a_{2 n-1}+a_{2 n}\right) \\
= & 5+13+21+\cdots+(8 n-3) \\
= & \frac{(5+8 n-3) n}{2}=(2 n)^{2}+n, \\
& S_{2 n-1}=S_{2 n}-a_{2 n}=4 n^{2}+n-(4 n-1) \\
= & (2 n-1)^{2}+n .
\end{aligned}
$$
Thus, $(2 n)^{2}<S_{2 n}<(2 n+1)^{2}$,
$$
\begin{array}{l}
(2 n-1)^{2}<S_{2 n-1}<(2 n)^{2} . \\
\text { Hence }\left[\sqrt{S_{2 n}}\right]=2 n,\left[\sqrt{S_{2 n-1}}\right]=2 n-1 .
\end{array}
$$
Therefore, $\left[\sqrt{S_{n}}\right]=n$. Thus,
$$
\begin{array}{l}
S=\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2000}}\right] \\
=1+2+\cdots+2006 \\
=\frac{2006 \times 2007}{2}=2013021 .
\end{array}
$$
|
2013021
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given a sequence of positive integers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, where $a_{1}=2^{2006}$, and for each positive integer $i, a_{i+1}$ is equal to the square of the sum of the digits of $a_{i}$. Find the value of $a_{2006}$.
|
3. Since the remainders when $2, 2^{2}, 2^{3}, 2^{4}, 2^{5}, 2^{6}, \cdots$ are divided by 9 are $2, 4, 8, 7, 5, 1, \cdots$, therefore, the remainders when $2^{m+6}$ and $2^{m}$ are divided by 9 are equal.
But $2006 = 334 \times 6 + 2$, so the remainder when $a_{1}$ is divided by 9 is 4. Thus, the remainder when the sum of the digits of $a_{1}$ is divided by 9 is 4.
Since the remainders when $a_{2}$ and $4^{2}$ are divided by 9 are equal, the remainder when $a_{2}$ is divided by 9 is 7.
Since the remainders when $a_{3}$ and $7^{2}$ are divided by 9 are equal, the remainder when $a_{3}$ is divided by 9 is 4, and the remainder when the sum of the digits of $a_{3}$ is divided by 9 is 4.
On the other hand, $a_{1} = 2^{2006} < 2^{3 \times 666} < 10^{666}$;
the sum of the digits of $a_{1}$ does not exceed $9 \times 669 = 6021$, so $a_{2} \leqslant 6021^{2} < 37 \times 10^{6}$;
the sum of the digits of $a_{2}$ does not exceed $9 \times 7 + 2 = 65$, so $a_{3} \leqslant 65^{2} = 4225$;
the sum of the digits of $a_{3}$ does not exceed $9 \times 3 + 3 = 30$, so $a_{4} \leqslant 30^{2}$.
Since $a_{4}$ is the square of the sum of the digits of $a_{3}$, $a_{4}$ is the square of a number that leaves a remainder of 4 when divided by 9.
Since $a_{4} \leqslant 30^{2}$, $a_{4}$ is one of $4^{2}, 13^{2}, 22^{2}$, i.e., one of $16, 169, 484$.
Since $a_{5}$ is one of $49, 256$, then
$$
a_{6} = 169, a_{7} = 256, a_{8} = 169, \cdots \text{.}
$$
By analogy, we get $a_{20006} = 169$.
|
169
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $\odot O_{1}$ and $\odot O_{2}$ are externally tangent, their radii are $112$ and $63$, respectively. The segment $A B$ is intercepted by their two external common tangents on their internal common tangent. Then, the length of $A B$ is $\qquad$ .
|
4.168.
From the tangent theorem, we easily know that
$$
\begin{array}{l}
A B=\text { length of the external common tangent }=\sqrt{(112+63)^{2}-(112-63)^{2}} \\
=2 \sqrt{112 \times 63}=168 .
\end{array}
$$
|
168
|
Geometry
|
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
|
7. In $\triangle A B C$, $A C=B C, \angle A C B=90^{\circ}$, $D 、 E$ are points on side $A B$, $A D=3, B E=4$, $\angle D C E=45^{\circ}$. Then the area of $\triangle A B C$ is $\qquad$
|
7:36.
As shown in Figure 5, rotate $\triangle C E B$ $90^{\circ}$ clockwise to get $\triangle C E^{\prime} A$. Connect $E^{\prime} D$.
It is easy to see that $A E^{\prime}=B E=4$, $\angle E^{\prime} A D=90^{\circ}$, so
$$
\begin{array}{l}
E^{\prime} D=\sqrt{A E^{\prime 2}+A D^{2}} \\
=\sqrt{4^{2}+3^{2}}=5 .
\end{array}
$$
Since $\angle D C E=45^{\circ}=\angle D C E^{\prime}$, we know $\triangle D C E \cong \triangle D C E^{\prime}$.
Therefore, $D E=E^{\prime} D=5, A B=12$.
Thus, $S_{\triangle A B C}=\frac{1}{4} A B^{2}=36$.
|
36
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. There are two sets of square paper pieces of the same size and the same number, one set black and one set white. Xiao Zhang first uses the white paper pieces to form a rectangle without any gaps in the middle, then uses the black paper pieces to surround the already formed white rectangle to create a larger rectangle, and then continues with white paper pieces. This process is repeated, and after Xiao Zhang has used the black paper pieces 5 times, the black and white paper pieces are exactly used up. Therefore, the black paper pieces are at least $\qquad$ pieces.
|
10.350.
Let the first white rectangle be $a \times b$, then the first black rectangle is $(a+2)(b+2)$, the second white rectangle is $(a+4) \times (b+4), \cdots \cdots$, and the fifth black rectangle is $(a+18)(b+18)$.
Obviously, the difference in perimeter between each black rectangle and its inner white rectangle (number of paper pieces) is 8.
Given that the number of black and white square paper pieces is equal, we have
$$
5 \times 8=(a-2)(b-2) \text {. }
$$
Solving this, we get $(a, b)=(3,42),(4,22),(6,12),(7,10)$.
Therefore, the corresponding number of black square paper pieces is
$$
\frac{1}{2}(a+18)(b+18)=630,440,360,350 \text {. }
$$
Thus, the black square paper pieces are at least 350.
Note: This problem must emphasize that "only one layer of paper pieces can be added each time (except for the first white rectangle)."
Otherwise, starting from the outermost layer, they can be taken as (black, white, black, white, white, black, white, black, white, black, black, white), with the innermost white rectangle being $1 \times 2$ (as shown in Figure 6).
Therefore, the difference in the number of black and white square paper pieces is
$$
8+8-8-8-8+8=0 .
$$
In this case, the number of black square paper pieces is $\frac{24 \times 23}{2}=276$.
|
350
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $x$ and $y$ are real numbers, and satisfy
$$
\left(x+\sqrt{x^{2}+2008}\right)\left(y+\sqrt{y^{2}+2008}\right)=2008 \text {. }
$$
Then the value of $x^{2}-3 x y-4 y^{2}-6 x-6 y+2008$
is $\qquad$
|
II, 1.2008.
From the given, we have
$$
x+\sqrt{x^{2}+2008}=\frac{2008}{y+\sqrt{y^{2}+2008}} .
$$
Rationalizing the denominator, we get
$$
x+\sqrt{x^{2}+2008}=\sqrt{y^{2}+2008}-y .
$$
Similarly, $y+\sqrt{y^{2}+2008}=\sqrt{x^{2}+2008}-x$.
$$
\begin{array}{l}
\text { (1) }+ \text { (2) gives } x+y=0 \text {. } \\
\text { Therefore, } x^{2}-3 x y-4 y^{2}-6 x-6 y+2008 \\
=(x+y)(x-4 y)-6(x+y)+2008 \\
=2008 \text {. }
\end{array}
$$
|
2008
|
Algebra
|
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
|
One. (20 points) There are three cylindrical containers $M$, $N$, and $P$, whose axial cross-sections are shown in figures 6(a), (b), and (c), respectively. The internal base areas are $S_{1} \mathrm{~cm}^{2}$, $S_{2} \mathrm{~cm}^{2}$, and $S_{3} \mathrm{~cm}^{2}$, and the internal heights are $h_{1} \mathrm{~cm}$, $h_{2} \mathrm{~cm}$, and $h_{3} \mathrm{~cm}$. The volumes are $V_{1} \mathrm{~cm}^{3}$, $V_{2} \mathrm{~cm}^{3}$, and $V_{3} \mathrm{~cm}^{3}$, respectively, with $V_{1} > V_{2} > V_{3}$. These three cylindrical containers $M$, $N$, and $P$ can be combined to form six different shapes of containers (by stacking the three containers from top to bottom in the order of $P N M$, $N P M$, $P M N$, $M P N$, $N M P$, and $M N P$), and their volumes are the sum of the volumes of $M$, $N$, and $P$.
Water is uniformly injected into these 6 containers at the same rate until they are full. The water depth $h(\mathrm{~cm})$ versus injection time $t(\mathrm{~s})$ for three of these containers is shown in figures 7, 8, and 9.
(1) Which of the six containers do the relationships in figures 7, 8, and 9 represent?
(2) Find the values of $h_{1}$, $h_{2}$, and $h_{3}$, and the ratio $S_{1}: S_{2}: S_{3}$.
(3) If $V_{3}=\frac{9}{m^{2}+4 m+4.125} \mathrm{~cm}^{3}$ and the injection rate is $\frac{n^{2}+6 n+45}{n+3} \mathrm{~cm}^{3} / \mathrm{s}$, where $m$ and $n$ are constants, find the sum of the volumes of containers $M$, $P$, and $N$.
|
(1) From Figure 7, we know that filling containers $M$, $N$, and $P$ takes $60 \mathrm{~s}$. From Figure 8, we know that filling two of the three containers $M$, $N$, and $P$ takes $54 \mathrm{~s}$. Therefore, filling the top container in Figure 8 takes $60-54=6(\mathrm{~s})$.
Similarly, filling the top container in Figure 9 takes $60-24=36(\mathrm{~s})$. Thus, filling the third container takes $60-6-36=18$ (s).
Since the water flow rate is constant and $V_{1}>V_{2}>V_{3}$, it follows that filling containers $M$, $N$, and $P$ takes $36 \mathrm{~s}$, $18 \mathrm{~s}$, and $6 \mathrm{~s}$, respectively.
Therefore, the patterns reflected in Figures 7, 8, and 9 are $N P M$, $P M N$, and $M N P$ respectively.
(2) Let the water flow rate be $V \mathrm{~cm}^{3} / \mathrm{s}$. From Figures 7, 8, 9, and the results of (1), we have:
$$
\left\{\begin{array}{l}
h_{1} + h_{3} = 24, \\
h_{2} + h_{1} = 30, \\
h_{2} + h_{3} = 18;
\end{array} \quad \left\{\begin{array}{l}
V_{1} + V_{2} + V_{3} = 60 V, \\
V_{2} + V_{1} = 54 V, \\
V_{3} + V_{2} = 24 V.
\end{array}\right.\right.
$$
Solving these equations, we get $h_{1}=18, h_{2}=12, h_{3}=6; V_{1}=36 \mathrm{~V}, V_{2}=18 \mathrm{~V}, V_{3}=6 \mathrm{~V}$.
Thus, $S_{1}=\frac{V_{1}}{h_{1}}=\frac{36 V}{18}=2 V$,
$$
\begin{array}{l}
S_{2}=\frac{V_{2}}{h_{2}}=\frac{18 V}{12}=\frac{3}{2} V, \\
S_{3}=\frac{V_{3}}{h_{3}}=\frac{6 V}{6}=V.
\end{array}
$$
Therefore, $S_{1}: S_{2}: S_{3}=2 V: \frac{3}{2} V: V=4: 3: 2$.
(3) Since $m^{2}+4 m+4.125=(m+2)^{2}+0.125$, the minimum value of $m^{2}+4 m+4.125$ is 0.125 when $m=-2$. Therefore, $V_{3} \leqslant 72$.
Given the water flow rate $V=\frac{n^{2}+6 n+45}{n+3}$, we have $n^{2}+(6-V) n+(45-3 V)=0$. Thus, $\Delta=(6-V)^{2}-4 \times 1 \times(45-3 V) \geqslant 0$.
Hence, $V \geqslant 12$ or $V \leqslant-12$ (discarded). When $n=3$, the minimum value of $V$ is 12.
From (2), we know $V_{3}=6 V$.
Since $V_{3} \leqslant 72$ and $6 V \geqslant 72$, we have $V_{3}=6 V=72$.
At this point, $m=-2, n=3$. Therefore, $V=12$.
Thus, $V_{1}+V_{2}+V_{3}=60 \mathrm{~V}=60 \times 12=720\left(\mathrm{~cm}^{3}\right)$.
Therefore, the total volume is $720 \mathrm{~cm}^{3}$.
|
720
|
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
|
3. In a certain basketball tournament, Xiao Ming played 10 games. In the 6th, 7th, 8th, and 9th games, he scored 23 points, 14 points, 11 points, and 20 points, respectively. His average score in the first 9 games was higher than his average score in the first 5 games. If the average score of the 10 games he played exceeds 18 points, then the minimum score he could have in the 10th game is $\qquad$ .
|
3.29.
Let the average score of the first 5 games be $x$, then the average score of the first 9 games is
$$
\frac{5 x+23+14+11+20}{9}=\frac{5 x+68}{9} .
$$
According to the problem, $\frac{5 x+68}{9}>x$.
Solving this, we get $x<17$.
Therefore, the total score of the first 5 games is at most $5 \times 17-1=84$ points.
Let the score of the 10th game be $y$, then the total score of the 10 games he played is at least $18 \times 10+1=181$ points.
Thus, we have $y+84+68 \geqslant 181$, which means $y \geqslant 29$.
Therefore, the minimum score of the 10th game is 29.
|
29
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\left\{\begin{array}{l}
2006(x-y)+2000(y-z)+2008(z-x)=0, \\
2000^{2}(x-y)+2000^{2}(y-z)+2000^{2}(z-x)=2008 .
\end{array}\right.
$$
Then the value of $z-y$ is $\qquad$ .
|
$=、 1.2008$
Let $z-y=t$.
From the first equation, we get $z-2 x+y=0$.
Thus, $z-x=\frac{t}{2}$, and consequently, $x-y=\frac{t}{2}$.
Therefore, from the second equation, we can find that
$$
t=z-y=2008 .
$$
|
2008
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the quadratic equation with integer coefficients
$$
x^{2}+(a+3) x+2 a+3=0
$$
has one positive root $x_{1}$ and one negative root $x_{2}$, and $\left|x_{1}\right|<\left|x_{2}\right|$, then
$$
a=
$$
$\qquad$
|
$=、 1 .-2$.
Since the two roots of the equation are not equal, we have $\Delta>0$, that is
$$
(a+3)^{2}>4(2 a+3) \text {. }
$$
Solving this, we get $a>3$ or $a-3, a<-\frac{3}{2}$, which means $-3<a<-\frac{3}{2}$.
Since $a$ is an integer, then $a=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A three-digit number $x y z$ (where $x$, $y$, and $z$ are distinct), rearranging its digits to form the largest and smallest possible three-digit numbers. If the difference between the largest and smallest three-digit numbers is equal to the original three-digit number, then this three-digit number is $\qquad$
|
3.495.
Let the three-digit number $\overline{y z}$, after rearrangement, form the largest three-digit number $\overline{a b c}(a>b>c)$, then the smallest three-digit number is $\overline{c b a}$.
Since $1 \leqslant a \leqslant 9,1 \leqslant b \leqslant 9,1 \leqslant c \leqslant 9$, and
$$
\begin{array}{l}
\overline{a b c}-\overline{c b a}=(100 a+10 b+c)-(100 c+10 b+a) \\
=99(a-c)
\end{array}
$$
is a multiple of 99, hence the three-digit number $\overline{x y z}$ sought is also a multiple of 99.
The three-digit multiples of 99 are only 8: $198,297,396$, $495,594,693,792,891$.
Upon verification, only 495 meets the requirement.
|
495
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) It is known that the unit digit of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{8}^{3}-x_{9}^{3}$ is 1, where $x_{1}, x_{2}, \cdots, x_{9}$ are nine different numbers from 2001, 2002, ..., 2009, and $8 x_{9}>$ $x_{1}+x_{2}+\cdots+x_{8}$. Find the value of $x_{9}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note at the end is not part of the translation and is provided for context. The translation itself is above.
|
Three, because $x_{1}, x_{2}, \cdots, x_{9}$ are nine different numbers from $2001, 2002, \cdots, 2009$, and the unit digits of $2001, 2002, \cdots, 2009$ are $1, 2, 3, 4, 5, 6, 7, 8, 9$. After cubing these unit digits, the resulting unit digits are $1, 8, 7, 4, 5, 6, 3, 2, 9$. Therefore, the unit digit of $x_{1}^{3} + x_{2}^{3} + \cdots + x_{9}^{3}$ must be 5.
Furthermore, $x_{1}^{3} + x_{2}^{3} + \cdots + x_{8}^{2} - x_{9}^{3}$
$= x_{1}^{3} + x_{2}^{3} + \cdots + x_{8}^{3} + x_{9}^{3} - 2 x_{9}^{3}$.
Assume the unit digit of $x_{9}^{3}$ is $a_{9}$.
Thus, the unit digit of $x_{1}^{3} + x_{2}^{3} + \cdots + x_{9}^{3}$ has two cases:
(1) When $5 - 2 a_{9} > 0$, it is $5 - 2 a_{9}$;
(2) When $5 - 2 a_{9} < 0$, it is $15 - 2 a_{9}$.
Since $x_{1} + x_{2} + \cdots + x_{8} < x_{1} + x_{2} + \cdots + x_{8}$, therefore, $x_{9} = 2008$.
(From the 40th Middle School of Shusheng City, Shandong Province, 277200)
|
2008
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A person rolls a die, adding up the numbers rolled each time, and stops if the total exceeds 20. Then, when he stops, the number he is most likely to have rolled is $\qquad$
|
4.21.
Consider the value obtained after the dice roll just before the one that exceeds 20 is $x$.
If $x=15$, then only rolling a 6 can result in 21;
If $x=16$, then rolling a 5 or 6 can result in 21 or 22, with each number having a probability of $\frac{1}{2}$;
If $x=17$, then rolling a 4, 5, or 6 can result in 21, 22, or 23, with each number having a probability of $\frac{1}{3}$;
If $x=18$, then rolling a 3, 4, 5, or 6 can result in 21, 22, 23, or 24, with each number having a probability of $\frac{1}{4}$;
If $x=19$, then rolling a 2, 3, 4, 5, or 6 can result in 21, 22, 23, 24, or 25, with each number having a probability of $\frac{1}{5}$;
If $x=20$, then rolling a 1, 2, 3, 4, 5, or 6 can result in 21, 22, 23, 24, 25, or 26, with each number having a probability of $\frac{1}{6}$.
Therefore, the probability of getting 21 is greater than the probability of getting any other number. Hence, 21 is the most likely total sum to be obtained.
|
21
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Given the quadratic function
$$
y=x^{2}+2 m x-n^{2} \text {. }
$$
(1) If the graph of this quadratic function passes through the point $(1,1)$, and let the larger of the two numbers $m, n+4$ be $P$, find the minimum value of $P$;
(2) If $m, n$ vary, these functions represent different parabolas. If each parabola intersects the coordinate axes at three distinct points, prove that the circles passing through these two intersection points all pass through the same fixed point, and find the coordinates of this fixed point.
|
(1) From the quadratic function passing through the point $(1,1)$, we get $m=\frac{n^{2}}{2}$.
Notice that
$$
\begin{array}{l}
m-(n+4)=\frac{n^{2}}{2}-(n+4) \\
=\frac{1}{2}\left(n^{2}-2 n-8\right)=\frac{1}{2}(n-4)(n+2),
\end{array}
$$
Therefore, $P=\left\{\begin{array}{ll}\frac{n^{2}}{2}, & n \leqslant-2 \text { or } n \geqslant 4 ; \\ n+4, & -2<n<4 .\end{array}\right.$
Using the function graph, we can see that when $n=-2$, $P_{\min }=2$.
(2) The graph intersects the coordinate axes at three different points, which can be set as $A\left(x_{1}, 0\right) 、 B\left(x_{2}, 0\right) 、 C\left(0,-n^{2}\right)$.
Also, $x_{1} x_{2}=-n^{2}$. If $n=0$, it would contradict the condition of three intersection points, so $x_{1} x_{2}=-n^{2}<0$. Therefore, $x_{1} 、 x_{2}$ are on opposite sides of the origin.
Also, $\left|x_{1} x_{2}\right|=n^{2} \times 1$, so there exists a point $P_{0}(0,1)$ such that $|O A| \cdot|O B|=\left|O P_{0}\right| \cdot|O C|$.
Thus, points $A 、 B 、 C 、 P_{0}$ are concyclic, meaning these circles must pass through the fixed point $P_{0}(0,1)$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given two sets of numbers, set $A$ is: $1,2, \cdots, 100$; set $B$ is: $1^{2}, 2^{2}, \cdots, 100^{2}$. For a number $x$ in set $A$, if there is a number $y$ in set $B$ such that $x+y$ is also a number in set $B$, then $x$ is called an "associated number". Therefore, the number of such associated numbers in set $A$ is $\qquad$.
|
3.73.
Let $x+y=a^{2}, y=b^{2}$, then $1 \leqslant b<a \leqslant 100$.
And $x=a^{2}-b^{2}=(a+b)(a-b) \leqslant 100$, since $a+b$ and $a-b$ have the same parity, hence $a+b \geqslant(a-b)+2$.
(1) If $a-b=1$, then $a+b$ is odd, and $3 \leqslant a+b \leqslant$ 99. Thus, $a+b$ can take the values $3,5,7, \cdots, 99$, a total of 49 values, at this time, the corresponding $x$ can also take these 49 values.
(2) If $a-b=2$, then $a+b$ is even, and $4 \leqslant a+b \leqslant$ 50. Thus, $a+b$ can take the values $4,6,8, \cdots, 50$, a total of 24 values, at this time, the corresponding $x$ can take the values $8,12,16, \cdots, 100$ for 24 values.
In other cases, the values of $x$ obtained all fall into the above scenarios.
If $a-b=$ is odd, then $a+b=$ is odd.
And $x=a^{2}-b^{2} \geqslant a+b \geqslant 3$, falling into (1).
If $a-b=$ is even, then $a+b=$ is even.
And $x=(a-b)(a+b)$ is a multiple of 4, and $a-b \geqslant 2$, $a+b \geqslant 4$, hence $x \geqslant 8$, falling into (2).
Therefore, there are a total of $49+24=73$ such $x$.
|
73
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In pentagon $A B C D E$, $\angle A=\angle C=90^{\circ}$, $A B=B C=D E=A E+C D=3$. Then the area of this pentagon is $(\quad)$.
(A) 9
(B) 10.5
(C) 12
(D) 13.5
|
3. A.
As shown in Figure 3, extend $DC$ to point $F$ such that $CF = AE$, and connect $BE$, $BD$, and $BF$. Then,
$$
DF = DE = 3.
$$
Also, $\triangle BCF \cong \triangle BAE$, so
$$
BE = BF.
$$
Since $BD = BD$, we have
$$
\triangle BED \cong \triangle BFD.
$$
Therefore, $S_{\triangle BRD} = S_{\triangle BFD}$. Hence, $S_{\text{pentagon } 1BCDE} = S_{\text{quadrilateral } BRDF} = 2S_{\triangle BFD}$
$$
= 2 \times \frac{1}{2} DF \cdot BC = 9.
$$
|
9
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Find the positive integer solutions of the indeterminate equation
$$
29 a+30 b+31 c=2196
$$
|
One, transform the original equation into
$$
\left\{\begin{array}{l}
29(a+b+c)+(b+2 c)=2196 \\
31(a+b+c)-(2 a+b)=2196 .
\end{array}\right.
$$
Since \(a, b, c\) are positive integers, from equation (1) we get
$$
\begin{array}{l}
29(a+b+c)=2196-(b+2 c) \\
\leqslant 2196-(1+2 \times 1)=2193 .
\end{array}
$$
Therefore, \(a+b+c \leqslant 75 \frac{18}{29}\).
From equation (2) we get
$$
\begin{array}{l}
31(a+b+c)=2196+(2 a+b) \\
\geqslant 2196+(2 \times 1+1)=2199 .
\end{array}
$$
Therefore, \(a+b+c \geqslant 70 \frac{29}{31}\).
From equations (3) and (4) we get
$$
a+b+c=71,72,73,74 \text { or } 75 \text {. }
$$
When \(a+b+c=71\), combining with the original equation, we solve to get
$$
b=5-2 a, c=a+66 \text {. }
$$
From \(b \geqslant 1\), we get \(5-2 a \geqslant 1\). Solving this, we get \(a \leqslant 2\), so \(a=1\) or 2. In this case, the original equation has 2 sets of positive integer solutions.
Similarly, when \(a+b+c=72,73,74,75\), we can respectively find that the original equation has \(17, 33, 24, 10\) sets of positive integer solutions.
Therefore, the total number of positive integer solutions to the original equation is
$$
2+17+33+24+10=86 \text { (sets). }
$$
|
86
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the three sides of $\triangle A B C$ are $A B=$ $2 \sqrt{a^{2}+576}, B C=\sqrt{a^{2}+14 a+625}, A C=$ $\sqrt{a^{2}-14 a+625}$, where $a>7$. Then the area of $\triangle A B C$ is $\qquad$
|
4.168.
Notice
$$
\begin{array}{l}
A B^{2}=(2 a)^{2}+48^{2}, \\
B C^{2}=(a+7)^{2}+24^{2}, \\
A C^{2}=(a-7)^{2}+24^{2} .
\end{array}
$$
As shown in Figure 6, with $A B$ as the hypotenuse, construct a right triangle $\triangle A B D$ on one side of $\triangle A B C$, such that
$$
\begin{array}{l}
B D=2 a, A D=48, \\
\angle A D B=90^{\circ} .
\end{array}
$$
Take a point $E$ on $B D$ such that $B E=a+7, E D=a-7$, and take the midpoint $F$ of $A D$, construct rectangle $E D F C_{1}$.
Since $B C_{1}^{2}=B E^{2}+E C_{1}^{2}=(a+7)^{2}+24^{2}=B C^{2}$,
$$
A C_{1}^{2}=C_{1} F^{2}+A F^{2}=(a-7)^{2}+24^{2}=A C^{2} \text {, }
$$
thus point $C$ coincides with point $C_{1}$.
And $S_{\triangle A B D}=48 a, S_{\triangle C B D}=24 a, S_{\triangle A C D}=24(a-7)$,
then $S_{\triangle B B C}=S_{\triangle A B D}-S_{\triangle C B D}-S_{\triangle A C D}=168$.
|
168
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $b$ take even numbers from 1 to 11, and $c$ take any positive integer. Then the number of quadratic equations $x^{2}-b x+c=0$ that can be formed with two distinct real roots is ( ).
(A) 50
(C) 55
(C) 57
(D) 58
|
5.A.
When $b=2$, from $b^{2}-4 c=4-4 c>0$, we get $c<1$, so, $c=0$.
When $b=4$, from $b^{2}-4 c=16-4 c>0$, we get $c<4$, so, $c=0,1,2,3$.
When $b=6$, from $b^{2}-4 c=36-4 c>0$, we get $c<9$, so, $c=0,1, \cdots, 8$.
When $b=8$, from $b^{2}-4 c=64-4 c>0$, we get $c<16$, so, $c=0,1, \cdots, 15$.
When $b=10$, from $b^{2}-4 c=100-4 c>0$, we get $c<25$, so, $c=1,2, \cdots, 24$.
Therefore, the number of equations that satisfy the conditions is
$$
3+8+15+24=50 \text { (equations). }
$$
|
50
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Calculate: $\frac{1^{2}}{1^{2}-100+5000}+\frac{3^{2}}{3^{2}-300+5000}+$ $\frac{5^{2}}{5^{2}-500+5000}+\cdots+\frac{99^{2}}{99^{2}-9900+5000}=$ $\qquad$
|
2.50.
Notice that
$$
\begin{array}{l}
\frac{k^{2}}{k^{2}-100 k+5000}+\frac{\left(100-k^{2}\right)}{(100-k)^{2}-100(100-k)+5000} \\
=\frac{k^{2}}{k^{2}-100 k+5000}+\frac{10000-200 k+k^{2}}{k^{2}-100 k+5000} \\
=\frac{2\left(k^{2}-100 k+5000\right)}{k^{2}-100 k+5000}=2 .
\end{array}
$$
Let $k=1,3,5, \cdots, 49$, and add them up, we know the original expression $=2 \times 25=50$.
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Let $S=\{1,2, \cdots, 98\}$. Find the smallest natural number $n$, such that in any $n$-element subset of $S$, one can always select 10 numbers, and no matter how these 10 numbers are divided into two groups, there is always one group in which one number is coprime with the other 4 numbers, and in the other group, there is always one number that is not coprime with the other 4 numbers.
(1998, China Mathematical Olympiad)
|
Explanation: First, there are 49 even numbers in $S$, and the 49-element subset formed by them obviously does not meet the requirement. Therefore, the smallest natural number $n \geqslant 50$.
To prove that in any 50-element subset $T$ of $S$, there are 10 numbers that meet the requirement, we first prove a strengthened proposition:
In any 50-element subset $T$ of $S$, it is possible to select 10 numbers such that one of them is coprime with the other 9, and the other 9 numbers are pairwise not coprime.
$$
\begin{array}{l}
\text { Let } M_{1}=\{2 k \mid k=1,2, \cdots, 49\}, \\
M_{2}=\{6 k-31 k=1,2, \cdots, 16\}, \\
M_{3}=\{5,25,35,55,65,85,95,77,91\}, \\
M_{4}=\{7,49,11,13,17,19,23,29,31,37, \\
41,43,47\} . \\
M_{5}=\{1,53,59,61,67,71,73,79,83,89,97\} .
\end{array}
$$
Clearly, $M_{1}, M_{2}, M_{3}, M_{4}, M_{5}$ are pairwise disjoint,
$$
\begin{aligned}
\bigcup_{i=1}^{5} M_{i}=S & \text { and } \\
\left|M_{1}\right| & =49,\left|M_{2}\right|=16,\left|M_{3}\right|=9, \\
\left|M_{4}\right| & =13,\left|M_{5}\right|=11 .
\end{aligned}
$$
Assume there exists $T \subset S$, such that $|T|=50$ and the conclusion of the strengthened proposition does not hold.
(1) If $\left|T \cap M_{2}\right| \geqslant 9$, by the contradiction assumption, the following 15 odd numbers
$$
\{37,41,43,47\} \cup M_{5}
$$
cannot be in $T$, so $T$ must contain at least 16 even numbers.
(2) When $\left|T \cap M_{2}\right| \leqslant 8$, at least 8 of the 16 odd numbers in $M_{2}$ are not in $T$, so $T$ must contain at least 9 even numbers. By the contradiction assumption, $T \cap M_{5}=\varnothing$, so $T$ must contain at least 20 even numbers.
Combining (1) and (2), it is clear that, in any case, $T$ contains at least 16 even numbers.
(3) When $\left|T \cap M_{1}\right| \geqslant 16$, because for each odd number in $M_{4} \cup$ $M_{5}$, the number of even numbers in $S$ that are not coprime with it does not exceed 7, these 24 odd numbers are not in $T$. Therefore, $T$ contains at least 25 even numbers.
(4) When $\left|T \cap M_{1}\right| \geqslant 25$, because for each odd number in $M_{3} \cup$ $M_{4} \cup M_{5}$, the number of even numbers that are not coprime with it does not exceed 15, by the contradiction assumption, they are not in $T$, so $T$ contains at least 34 even numbers.
(5) When $\left|T \cap M_{1}\right| \geqslant 34$, because for each odd number in $M_{2}$, the number of even numbers that are not coprime with it does not exceed 23, by the contradiction assumption, they are not in $T$, thus, $T$ contains no odd numbers, which is impossible.
In summary, the smallest natural number $n=50$.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given $a+b+c=0, a^{2}+b^{2}+c^{2}=4$. Then, the value of $a^{4}+b^{4}+c^{4}$ is $\qquad$ .
|
11.8.
From the known equations, we get $a+b=-c, a^{2}+b^{2}=4-c^{2}$.
$$
\begin{array}{l}
\text { Also, } a b=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] \\
=\frac{1}{2}\left[(-c)^{2}-\left(4-c^{2}\right)\right]=c^{2}-2 .
\end{array}
$$
Therefore, $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}$
$$
=\left(4-c^{2}\right)^{2}-2\left(c^{2}-2\right)^{2}=8-c^{4} \text {. }
$$
Thus, $a^{4}+b^{4}+c^{4}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. In the park, there are two rivers $O M$ and $O N$ converging at point $O$ (as shown in Figure 6, $\angle M O N=60^{\circ}$. On the peninsula formed by the two rivers, there is an ancient site $P$. It is planned to build a small bridge $Q$ and $R$ on each of the two rivers, and to construct three small roads to connect the two bridges $Q$, $R$, and the ancient site $P$. If the distance from the ancient site $P$ to the two rivers is $50 \sqrt{3} \mathrm{~m}$ each, then the minimum value of the sum of the lengths of the three small roads is $\qquad$ m.
|
12.300.
As shown in Figure 10, let the symmetric points of point $P$ with respect to $OM$ and $ON$ be $P_{1}$ and $P_{2}$, respectively. Connect $P_{1}P_{2}$, intersecting $OM$ and $ON$ at points $Q$ and $R$. By symmetry, we have
$$
\begin{array}{l}
PQ = P_{1}Q, \\
PR = P_{2}R, \\
PQ + QR + RP \\
= P_{1}Q + QR + RP_{2} = P_{1}P_{2}.
\end{array}
$$
Now we prove that $PQ + QR + RP$ is the minimum value sought.
Take any point $Q'$ on $OM$ (not coinciding with point $Q$), and any point $R'$ on $ON$ (not coinciding with point $R$). Connect $P_{1}Q'$, $Q'R'$, and $R'P_{2}$. By symmetry, we have
$$
PQ' = P_{1}Q', \quad PR' = P_{2}R'.
$$
Then $PQ' + Q'R' + R'P = P_{1}Q' + Q'R' + R'P_{2}$
$$
> P_{1}P_{2} = PQ + QR + RP.
$$
Therefore, $PQ + QR + RP$ is the minimum value sought.
Given in $\triangle PP_{1}P_{2}$,
$$
\begin{array}{l}
PP_{1} = PP_{2} = 100\sqrt{3}, \\
\angle P_{1}PP_{2} = 120^{\circ}, \\
\angle P_{1} = \angle P_{2} = 30^{\circ},
\end{array}
$$
Construct $PK \perp P_{1}P_{2}$, with the foot of the perpendicular at $K$. Then in the right triangle $\triangle PP_{1}K$,
$$
\begin{array}{l}
PK = \frac{1}{2} PP_{1} = 50\sqrt{3}, \\
P_{1}K = \sqrt{PP_{1}^2 - PK^2} \\
= \sqrt{(100\sqrt{3})^2 - (50\sqrt{3})^2} = 150.
\end{array}
$$
Thus, $P_{1}P_{2} = 2P_{1}K = 300$.
Therefore, the minimum value of the sum of the lengths of the three paths is
$$
PQ + QR + RP = 300 \text{ m}.
$$
|
300
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) Given $n$ four-element sets $A_{1}, A_{2}, \cdots, A_{n}$, any two of which have exactly one common element, and
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=n .
$$
Find the maximum value of $n$. Here $\operatorname{Card} A$ is the number of elements in set $A$.
|
Consider any element $a \in A_{1} \cup A_{2} \cup \cdots \cup A_{n}$.
If each $A_{i}$ contains $a$, then by the given condition, the other elements in each $A_{i}$ are all different. Hence,
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=3 n+1>n,
$$
which contradicts the given condition.
Therefore, there must be an $A_{i}$ that does not contain $a$.
Without loss of generality, assume $a \notin A_{1}$. If the number of sets containing $a$ is greater than or equal to 5, then by the given condition, $A_{1}$ shares a common element with each of these 5 sets (this element is certainly not $a$), and these 5 elements are all different (if they were the same, then this common element would be a common element of 2 sets containing $a$, which would mean these two sets have 2 common elements, contradicting the given condition), thus, Card $A_{1} \geqslant 5$, which is a contradiction. Therefore, the number of sets containing $a$ is less than or equal to 4.
On the other hand, since
Card $A_{1}+$ Card $A_{2}+\cdots+\operatorname{Card} A_{n}=4 n$, each element belongs to exactly 4 sets.
Without loss of generality, assume the sets containing element $b$ are $A_{1}, A_{2}, A_{3}, A_{4}$. By the above conclusion,
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup A_{3} \cup A_{4}\right)=3 \times 4+1=13.
$$
If $n>13$, then there exists an element $c \notin A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$. Let the set containing $c$ be $A_{5}$, then $A_{5}$ is not one of $A_{1}, A_{2}, A_{3}, A_{4}$. Therefore, $A_{5}$ does not contain $b$. And $A_{5}$ shares a common element with each of $A_{1}, A_{2}, A_{3}, A_{4}$ (certainly not $b$), these 4 common elements are all different (for the same reason as above), and none of them is $c$, thus, $\operatorname{Card} A_{5} \geqslant 5$, which is a contradiction.
Therefore, $n \leqslant 13$.
$n \leqslant 13$ is possible. For example, it is not hard to verify that the following 13 sets meet the requirements.
$$
\begin{array}{l}
\{0,1,2,3\},\{0,4,5,6\},\{0,7,8,9\},\{0,10,11,12\}, \\
\{10,1,4,7\},\{10,2,5,8\},\{10,3,6,9\},\{11,1,5,9\}, \\
\{11,2,6,7\},\{11,3,4,8\},\{12,1,6,8\},\{12,2,4,9\}, \\
\{12,3,5,7\}.
\end{array}
$$
Thus, the maximum value of $n$ is 13.
(Wang Cuixi provided)
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Choose several colors from 6 different given colors to color a cube, with each face colored with one color and ensuring that any two adjacent faces sharing a common edge are of different colors. How many different coloring schemes are there (if one of the two colored cubes can be rotated to match the coloring state of the other, then the original coloring schemes of the two are considered the same)?
(1996, National High School Mathematics Competition)
|
Explanation: Clearly, the maximum number of colors used is 6, and the minimum is 3. Below, we use the grouping method to count.
(1) When using 6 colors, each color is used to paint one face. According to the problem, the first color can be placed on top by flipping. At this point, the bottom is one of the remaining 5 colors, giving 5 different coloring methods. The other 4 colors are used to paint the front, back, left, and right faces. We can choose one of these 4 colors to paint the front face. Then, the back face is one of the remaining 3 colors, giving 3 different coloring methods. The remaining 2 colors are used to paint the left and right faces, giving 2 different coloring methods. Therefore, the total number of different coloring methods is $5 \times 3 \times 2 = 30$.
(2) When using 5 colors, we select 5 colors from the 6 available colors, and then choose one of these 5 colors to paint two opposite faces, which can be considered as the top and bottom faces, giving $6 \times 5 = 30$ different coloring methods. The remaining 4 colors are used to paint the 4 side faces. We choose one of these 4 colors to paint the front face. Then, the back face is one of the remaining 3 colors, giving 3 different coloring methods. The remaining 2 colors are used to paint the left and right faces. Unlike in (1), at this point, the cube can be rotated 180 degrees around the axis through the centers of the front and back faces, so there is only 1 coloring method. Therefore, the total number of different coloring methods is 90.
(3) When using 4 colors, we select 4 colors from the 6 available colors, and then choose 2 of these 4 colors to paint two opposite faces. The number of different coloring methods is $\mathrm{C}_{6}^{4} \mathrm{C}_{4}^{2} = 90$.
(4) When using 3 colors, the number of different coloring methods is $\mathrm{C}_{6}^{3} = 20$.
In summary, the total number of different coloring methods is 230.
Comment: Although this problem is just a fill-in-the-blank question, it is not easy to solve correctly. The standard answer given at the time was 320, which is incorrect. The reason for the error lies in (2), where the last 2 colors are used to paint the two faces. Since the cube can be rotated, there is only 1 method, but the standard answer incorrectly considered it as 2 methods.
|
230
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six, find the smallest real number $m$, such that for any positive real numbers $a, b, c$ satisfying $a + b + c = 1$, we have
$$
m\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1 \text {. }
$$
(Xiong Bin)
|
Six, Solution 1: When $a=b=c=\frac{1}{3}$, we have $m \geqslant 27$.
Next, we prove the inequality
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1
$$
for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.
Since for $0<x<1$, we have
$$
\begin{array}{l}
27 x^{3} \geqslant 6 x^{2}+5 x-\frac{4}{3} \\
\Leftrightarrow 81 x^{3}-18 x^{2}-15 x+4 \geqslant 0 \\
\Leftrightarrow(3 x-1)^{2}(9 x+4) \geqslant 0,
\end{array}
$$
then $27 x^{3} \geqslant 6 x^{2}+5 x-\frac{4}{3}(0<x<1)$.
Thus, $27 a^{3} \geqslant 6 a^{2}+5 a-\frac{4}{3}$,
$27 b^{3} \geqslant 6 b^{2}+5 b-\frac{4}{3}$,
$27 c^{3} \geqslant 6 c^{2}+5 c-\frac{4}{3}$.
Adding these three inequalities, we get
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1 \text {. }
$$
Therefore, the minimum value of $m$ is 27.
Solution 2: When $a=b=c=\frac{1}{3}$, we have $m \geqslant 27$.
Next, we prove the inequality
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1
$$
for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.
Since $(a-b)^{2}(a+b) \geqslant 0$, we have
$$
a^{3}+b^{3} \geqslant a^{2} b+a b^{2} \text {. }
$$
Similarly, $b^{3}+c^{3} \geqslant b^{2} c+b c^{2}$, $c^{3}+a^{3} \geqslant c^{2} a+c a^{2}$.
Thus, $2\left(a^{3}+b^{3}+c^{3}\right)$
$$
\geqslant a^{2} b+b^{2} c+c^{2} a+a b^{2}+b c^{2}+c a^{2} \text {. }
$$
Hence, $3\left(a^{3}+b^{3}+c^{3}\right)$
$$
\begin{array}{l}
\geqslant a^{3}+b^{3}+c^{3}+a^{2} b+b^{2} c+c^{2} a+a b^{2}+b c^{2}+c a^{2} \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)=a^{2}+b^{2}+c^{2} .
\end{array}
$$
Then, $6\left(a^{2}+b^{2}+c^{2}\right)+1$
$$
\begin{array}{l}
=6\left(a^{2}+b^{2}+c^{2}\right)+(a+b+c)^{2} \\
\leqslant 6\left(a^{2}+b^{2}+c^{2}\right)+3\left(a^{2}+b^{2}+c^{2}\right) \\
=9\left(a^{2}+b^{2}+c^{2}\right) \\
\leqslant 27\left(a^{3}+b^{3}+c^{3}\right) .
\end{array}
$$
Therefore, the minimum value of $m$ is 27.
|
27
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$ and $b$ are non-negative real numbers, and
$$
a^{2005}+b^{2005}=1, a^{2006}+b^{2006}=1 \text {. }
$$
then $a^{2007}+b^{2007}=$ $\qquad$
|
Ni, 1.1.
If $a$ or $b$ is greater than 1, then $a^{2005} + b^{2005} > 1$; if $0 < a, b < 1$, then $a^{2005} > a^{2006}$, $b^{2005} > b^{2006}$. Therefore, $a^{2000} + b^{2000} > a^{2006} + b^{2000}$, which means $1 > 1$, a contradiction. Hence, one of $a$ or $b$ is 0, and the other is 1. Thus, $a^{200} + b^{20 n} = 1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Zhang Hua wrote a five-digit number, which can be divided by 9 and 11. If the first, third, and fifth digits are removed, the resulting number is 35; if the first three digits are removed, the resulting number can be divided by 9; if the last three digits are removed, the resulting number can also be divided by 9. Then, this number is $\qquad$
|
4.63954.
Let this five-digit number be $\overline{a 3 b 5 c}$, then $9 \mid \overline{5 c}$.
So, $9 \mid(5+c)$.
Since $5 \leqslant 5+c \leqslant 14$, thus $5+c=9, c=4$.
Similarly, $9 \mid \overline{a 3}$, then $9 \mid(a+3)$.
Since $4 \leqslant a+3 \leqslant 12$, so $a=6$.
Also, $91 \overline{a 3 b 5 c}$, then
$9 \mid(a+3+b+5+c)$, i.e., $9 \mid(18+b)$.
So, $b=0$ or 9.
Also, $11 \mid \overline{a 3 b 5 c}$, then $11 \mid(a+b+c-5-3)$.
Thus, $11 \mid(2+b)$.
Therefore, $b=9$.
So, this five-digit number is 63954.
|
63954
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let the function $f(x)=x^{2}+a x+b(a, b \in$
$\mathbf{R})$. If there exists a real number $m$, such that
$$
|f(m)| \leqslant \frac{1}{4} \text {, and }|f(m+1)| \leqslant \frac{1}{4},
$$
find the maximum and minimum values of $\Delta=a^{2}-4 b$.
|
Solution 1: If $\Delta=a^{2}-4 b$
$$
\frac{-a-\sqrt{\Delta+1}}{2} \leqslant x \leqslant \frac{-a-\sqrt{\Delta-1}}{2}
$$
or $\frac{-a+\sqrt{\Delta-1}}{2} \leqslant x \leqslant \frac{-a+\sqrt{\Delta+1}}{2}$.
If $|f(m)| \leqslant \frac{1}{4}$, and $|f(m+1)| \leqslant \frac{1}{4}$, then it must be true that
$$
\frac{-a+\sqrt{\Delta-1}}{2}-\frac{-a-\sqrt{\Delta-1}}{2} \leqslant 1
$$
or $\frac{-a-\sqrt{\Delta-1}}{2}-\frac{-a-\sqrt{\Delta+1}}{2} \geqslant 1$
or $\frac{-a+\sqrt{\Delta+1}}{2}-\frac{-a+\sqrt{\Delta-1}}{2} \geqslant 1$,
which means $\sqrt{\Delta-1} \leqslant 1$ or $\sqrt{\Delta+1}-\sqrt{\Delta-1} \geqslant 1$.
Solving this, we get $\Delta \leqslant 2$.
When $a=1, b=-\frac{1}{4}$,
$|f(0)|=|f(-1)|=\frac{1}{4}$, and $\Delta=2$,
Therefore, the maximum value of $\Delta$ is 2.
Solution 2: Since $1=\left|\left(m+1-x_{0}\right)-\left(m-x_{0}\right)\right|$
$$
\leqslant\left|m+1-x_{0}\right|+\left|m-x_{0}\right|,
$$
thus, $\left|m+1-x_{0}\right|$ and $\left|m-x_{0}\right|$ must have one that is not less than $\frac{1}{2}$.
If $\Delta=a^{2}-4 b<0$, then
$$
\begin{array}{l}
f(m)=\left(m+\frac{a}{2}\right)^{2}-\frac{\Delta}{4}, \\
f(m+1)=\left(m+1+\frac{a}{2}\right)^{2}-\frac{\Delta}{4}
\end{array}
$$
one of them must be greater than $\frac{1}{4}$.
This contradicts the given condition, so, $\Delta \geqslant 0$.
The rest is the same as Solution 1.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given positive numbers $a, b, c$ satisfying $a+b+c=3$.
Prove:
$$
\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+ \\
\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5 .
\end{array}
$$
|
$$
\begin{array}{l}
\text { 5. } \frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}=\frac{a^{2}+9}{2 a^{2}+(3-a)^{2}} \\
=\frac{1}{3}\left(1+\frac{2 a+6}{a^{2}-2 a+3}\right)=\frac{1}{3}\left[1+\frac{2 a+6}{(a-1)^{2}+2}\right] \\
\leqslant \frac{1}{3}\left(1+\frac{2 a+6}{2}\right)=\frac{1}{3}(4+a) .
\end{array}
$$
Similarly, $\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}} \leqslant \frac{1}{3}(4+b)$,
$$
\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant \frac{1}{3}(4+c) \text {. }
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \\
\leqslant \frac{1}{3}(12+a+b+c)=5 .
\end{array}
$$
|
5
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. The product of all integers $m$ that make $m^{2}+m+7$ a perfect square is ( ).
(A) 84
(B) 86
(C) 88
(D) 90
|
3. A.
Let $m^{2}+m+7=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
Then $m^{2}+m+7-k^{2}=0$.
Solving for $m$, we get $m=\frac{-1 \pm \sqrt{4 k^{2}-27}}{2}$.
Since $m$ is an integer, we should have $4 k^{2}-27=n^{2}\left(n \in \mathbf{N}_{+}\right)$, which means $(2 k+n)(2 k-n)=27$.
Solving, we get $\left\{\begin{array}{l}n=13, \\ k=7\end{array}\right.$ or $\left\{\begin{array}{l}n=3, \\ k=3 .\end{array}\right.$
Thus, $m_{1}=-7, m_{2}=6, m_{3}=-2, m_{4}=1$.
Therefore, $m_{1} m_{2} m_{3} m_{4}=84$.
|
84
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Let $x, y$ be non-negative integers, $x+2y$ is a multiple of 5, $x+y$ is a multiple of 3, and $2x+y \geqslant 99$. Try to find the minimum value of $S=7x+5y$.
|
Let $x+2 y=5 A, x+y=3 B, A, B$ be integers.
Since $x, y \geqslant 0$, it follows that $A, B \geqslant 0$.
It is also easy to see that $x=6 B-5 A, y=5 A-3 B$, thus,
$$
\begin{array}{l}
2 x+y=9 B-5 A \geqslant 99, \\
S=7 x+5 y=27 B-10 A .
\end{array}
$$
Therefore, the problem becomes: For integers $A, B \geqslant 0, 6 B \geqslant 5 A \geqslant 3 B, 9 B \geqslant 5 A+99$, find the minimum value of $S$.
Since $15 A \geqslant 9 B \geqslant 99+5 A$, it follows that $10 A \geqslant 99, A \geqslant 10$, then $9 B \geqslant 50+99=149, B \geqslant 17$.
Thus, $5 A \geqslant 51, A \geqslant 11$.
Furthermore, $9 B \geqslant 55+99=154, B \geqslant 18$.
If $B \geqslant 19$, then
$$
S=9 B+2(9 B-5 A) \geqslant 9 \times 19+2 \times 99=369.
$$
If $B=18$, then $5 A \leqslant 9 B-99=63, A \leqslant 12, A$ can only be 12 or 11. Among these, $A=12$ makes $S=27 \times 18-10 \times 12=366$ the minimum value, and this is the only case (at this point, $x=48, y=6$).
|
366
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) As shown in Figure 4, in $\triangle ABC$, $\angle BAC=90^{\circ}$, $AB=AC$, points $D_{1}$ and $D_{2}$ are on $AC$, and satisfy $AD_{1}=CD_{2}$, $AE_{1} \perp BD_{1}$, $AE_{2} \perp BD_{2}$, intersecting $BC$ at points $E_{1}$ and $E_{2}$, respectively. Prove that: $\frac{CE_{2}}{BE_{2}}+\frac{CE_{1}}{BE_{1}}=1$.
---
The translation retains the original text's line breaks and formatting.
|
$$
\begin{array}{l}
\frac{C E_{1}}{B E_{1}}=\frac{A F}{A B}, \\
\angle E_{1} A C=\angle F C A . \\
\text { Also, } \angle A B D_{1}=90^{\circ}-\angle B A E_{1}=\angle E_{1} A C, \text { then } \\
\angle A B D_{1}=\angle A C F .
\end{array}
$$
Since $\angle B A D_{1}=\angle C A F=90^{\circ}, A B=A C$, therefore, $\triangle A B D_{1} \cong \triangle A C F$, which means $A D_{1}=A F$.
Thus, $\frac{C E_{1}}{B E_{1}}=\frac{A D_{1}}{A B}$.
Similarly, $\frac{C E_{2}}{B E_{2}}=\frac{A D_{2}}{A B}$.
$$
\begin{array}{l}
\text { Therefore, } \frac{C E_{1}}{B E_{1}}+\frac{C E_{2}}{B E_{2}}=\frac{A D_{1}+A D_{2}}{A B} \\
=\frac{A D_{1}+A C-C D_{2}}{A B}=\frac{A C}{A B}=1 .
\end{array}
$$
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 2, the three edges of the cube are $AB$, $BC$, and $CD$, and $AD$ is the body diagonal. Points $P$, $Q$, and $R$ are on $AB$, $BC$, and $CD$ respectively, with $AP=5$, $PB=15$, $BQ=15$, and $CR=10$. What is the area of the polygon formed by the intersection of the plane $PQR$ extended in all directions with the cube?
|
Explanation: Since $B P = B Q$, therefore, $P Q \parallel A C$. Thus, the line through point $R$ and parallel to $P Q$ intersects $A F$ at point $U$, and $A U = C R$.
Since the line through point $R$ and parallel to $P Q$ lies in the plane $P Q R$, $U$ is a vertex of the polygon formed by the intersection. Also, the midpoint of $U R$ is the center of the cube, so the points on the polygon formed by the intersection are symmetric about the center of the cube. Therefore, its area is twice the area of the trapezoid $P Q R U$.
It is easy to know that $U R = 20 \sqrt{2}, P Q = 15 \sqrt{2}, P U = 5 \sqrt{5}$.
Thus, the required area is
$$
\begin{array}{l}
(U R + P Q) \sqrt{P U^{2} - \left(\frac{U R - P Q}{2}\right)^{2}} \\
= 35 \times \sqrt{2} \times \sqrt{\frac{225}{2}} = 525 .
\end{array}
$$
|
525
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that the length, width, and height of a rectangular prism are all integers, and the volume equals the surface area. Then the maximum value of its volume is
|
2.882.
Lemma When the sum of the reciprocals of two integers is a constant, the larger their difference, the greater their product.
Let the length, width, and height be $a$, $b$, and $c$, respectively, then
$$
a b c=2(b c+c a+a b) \Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2} .
$$
Assume without loss of generality that $a \geqslant b \geqslant c$, then $c \geqslant 3$, and when $c=3$, we have $\frac{1}{a}+\frac{1}{b}=\frac{1}{6} \Rightarrow b \geqslant 7$.
When $b=7$, there is a solution $(a, b, c)=(42,7,3)$.
Therefore, the maximum value of $a b c$ is 882.
Note: All solutions under $a \geqslant b \geqslant c$ can be found to be $(42,7,3)$, $(24,8,3)$, $(18,9,3)$, $(15,10,3)$, $(12,12,3)$, $(20,5,4)$, $(12,6,4)$, $(8,8,4)$, $(10,5,5)$, and $(6,6,6)$, but it is more complicated.
|
882
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $[x]$ denote the greatest integer not exceeding $x$. Then the last two digits of $\left[\frac{10^{2006}}{10^{99}-3}\right]$ are $\qquad$ .
|
3.23.
From $10^{2000}=10^{59 \times 34}-3^{34}+3^{34}$, we know
$$
\left[\frac{10^{2006}}{10^{99}-3}\right]=\frac{10^{99 \times 34}-3^{34}}{10^{59}-3}=\sum_{k=0}^{33} 10^{59(33-k)} \times 3^{k} \text {. }
$$
Therefore, the last two digits we are looking for are the remainder of $3^{33}$ modulo 100.
Also, $3^{32}=(10-1)^{16}=-\mathrm{C}_{16}^{15} \times 10+1(\bmod 100)$ $=41(\bmod 100)$,
so $3 \times 3^{32}=3 \times 41(\bmod 100) \equiv 23(\bmod 100)$.
Thus, the last two digits we are looking for are 23.
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5: Through the vertex $A$ of the regular tetrahedron $ABCD$, make a section in the shape of an isosceles triangle, and let the angle between the section and the base $BCD$ be $75^{\circ}$: The number of such sections that can be made is $\qquad$.
|
Explanation: Let the edge length of a regular tetrahedron be 1. Draw $AO \perp$ plane $BCD$ at point $O$, then $AO=\frac{\sqrt{6}}{3}$. With $O$ as the center and $\frac{\sqrt{6}}{3} \cot 75^{\circ}$ as the radius, draw a circle on plane $BCD$. It is easy to see that this circle is inside $\triangle BCD$, and the intersection line of the desired section with plane $BCD$ is a tangent to this circle.
When the tangent is parallel to one side of $\triangle BCD$, the corresponding section $\triangle AMN$ is an isosceles triangle, and there are 6 such sections.
When $CB_1 = DC_1$, and $B_1C_1$ is tangent to the circle (as shown in Figure 6), the corresponding section $\triangle AB_1C_1$ is an isosceles triangle, and there are 6 such sections.
Draw $BE$ tangent to the circle, intersecting $CD$ at point $E$. By $\triangle BCE \cong \triangle ACE$, we get $BE = AE$, and the corresponding section $\triangle ABE$ is also an isosceles triangle, and there are 6 such sections.
In summary, there are 18 sections that meet the conditions.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If the quadratic function $f(x)=a x^{2}+b x+c$ has values whose absolute values do not exceed 1 on $[0,1]$, what is the maximum value of $|a|+$ $|b|+|c|$?
|
Solution 1: For $x \in [0,1]$, we have $|f(x)| \leqslant 1$, thus
$$
\begin{array}{l}
|f(0)| \leqslant 1, \left|f\left(\frac{1}{2}\right)\right| \leqslant 1, |f(1)| \leqslant 1. \\
\text { Also } f(0)=c, \\
f\left(\frac{1}{2}\right)=\frac{1}{4} a+\frac{1}{2} b+c, \\
f(1)=a+b+c,
\end{array}
$$
Solving simultaneously, we get
$$
\begin{array}{l}
a=2 f(0)-4 f\left(\frac{1}{2}\right)+2 f(1), \\
b=-3 f(0)+4 f\left(\frac{1}{2}\right)-f(1), \\
c=f(0) .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence }|a|=\left|2 f(0)-4 f\left(\frac{1}{2}\right)+2 f(1)\right| \\
\leqslant 2|f(0)|+4\left|f\left(\frac{1}{2}\right)\right|+2|f(1)| \\
\leqslant 2+4+2=8 .
\end{array}
$$
Similarly, $|b| \leqslant 3+4+1=8, |c| \leqslant 1$.
Therefore, $|a|+|b|+|c| \leqslant 17$.
Moreover, when $f(0)=f(1)=-f\left(\frac{1}{2}\right)=1$, i.e.,
$$
a=8, b=-8, c=1
$$
or $f(0)=f(1)=-f\left(\frac{1}{2}\right)=-1$,
i.e., $a=-8, b=8, c=-1$, the equality holds.
Thus, the maximum value of $|a|+|b|+|c|$ is 17.
Solution 2: For $x \in (0,1)$, we have
$$
\left\{\begin{array}{l}
f(0)=c, \\
f(1)=a+b+c, \\
f(x)=a x^{2}+b x+c .
\end{array}\right.
$$
Substituting (1) into (2) and (3), we get
$$
\begin{array}{l}
a=\frac{1}{x(1-x)}[-f(x)+f(1) x+f(0)(1-x)], \\
b=\frac{1}{x(1-x)}\left[f(x)-f(1) x^{2}-f(0)\left(1-x^{2}\right)\right] .
\end{array}
$$
From (4), for all $x \in (0,1)$, we have
$$
\begin{array}{l}
|a| \leqslant \frac{1}{x(1-x)}[|f(x)|+|f(1)| x+|f(0)|(1-x)] \\
\leqslant \frac{1}{x(1-x)}[1+x+(1-x)] \\
=\frac{2}{x(1-x)} .
\end{array}
$$
But $x(1-x) \leqslant\left[\frac{x+(1-x)}{2}\right]^{2}=\frac{1}{4}$, with equality holding only when $x=\frac{1}{2}$, hence
$$
|a| \leqslant \min _{x \in(0,1)} \frac{2}{x(1-x)}=8,
$$
with equality holding only when $x=\frac{1}{2}$.
Substituting into (6), the conditions for equality are
$$
\left\{\begin{array}{l}
x=\frac{1}{2}, \\
f(1)=1, \\
f(0)=1, \\
f\left(\frac{1}{2}\right)=-1
\end{array} \text { or } \left\{\begin{array}{l}
x=\frac{1}{2}, \\
f(1)=-1, \\
f(0)=-1, \\
f\left(\frac{1}{2}\right)=1 .
\end{array}\right.\right.
$$
This gives $\left\{\begin{array}{l}a=8, \\ c=1\end{array}\right.$ or $\left\{\begin{array}{l}a=-8, \\ c=-1 .\end{array}\right.$
Similarly, from (5) we have
$$
\begin{array}{l}
|b| \leqslant \frac{1}{x(1-x)}\left[1+x^{2}+\left(1-x^{2}\right)\right]=\frac{2}{x(1-x)}, \\
|b| \leqslant \min _{x \in(0,1)} \frac{2}{x(1-x)}=8 .
\end{array}
$$
The conditions for equality are
$$
\left\{\begin{array}{l}
x=\frac{1}{2}, \\
f(1)=1, \\
f(0)=1, \\
f\left(\frac{1}{2}\right)=-1
\end{array} \text { or } \left\{\begin{array}{l}
x=\frac{1}{2}, \\
f(1)=-1, \\
f(0)=-1, \\
f\left(\frac{1}{2}\right)=1 .
\end{array}\right.\right.
$$
This gives $\left\{\begin{array}{l}b=-8, \\ c=1\end{array}\right.$ or $\left\{\begin{array}{l}b=8, \\ c=-1 .\end{array}\right.$
From (1), (7), and (8), we get
$|a|+|b|+|c| \leqslant 8+8+1=17$.
Furthermore, when $a=8, b=-8, c=1$, for $x \in [0,1]$, we have
$$
\begin{array}{l}
f(x)=8 x^{2}-8 x+1=-1+2(2 x-1)^{2} \geqslant-1, \\
f(x)=8 x^{2}-8 x+1=1-8 x(1-x) \leqslant 1 .
\end{array}
$$
Therefore, the maximum value of $|a|+|b|+|c|$ is 17.
|
17
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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