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Example 3 As shown in Figure 2, the three altitudes of acute $\triangle A B C$ intersect at point $H$. How many triangles are there in Figure 2?
The text above is translated into English, preserving the original text's line breaks and format.
|
Solution 1: Let the set of triangles with vertex $A$ be $P_{1}$. Then, the triangles with vertices on $A B$ and $A D$ are 3, the triangles with vertices on $A D$ and $A C$ are 3, and the triangles with vertices on $A B$ and $A C$ are 3, giving $\left|P_{1}\right|=9$.
Similarly, the sets of triangles with vertices at $B, C$ are $P_{2}, P_{3}$, and we have $\left|P_{2}\right|=\left|P_{3}\right|=\left|P_{1}\right|=9$.
The set of triangles with vertices at $A, B$ is $P_{1} \cap P_{2}$, which has 4 elements ( $\triangle A B E, \triangle A B D, \triangle A B H, \triangle A B C$), so
$$
\left|P_{1} \cap P_{2}\right|=4.
$$
Similarly, $\left|P_{1} \cap P_{3}\right|=\left|P_{2} \cap P_{3}\right|=4$.
The set of triangles with vertices at $A, B, C$ is $P_{1} \cap P_{2} \cap P_{3}$, which has only 1 element $(\triangle A B C)$. By the principle of inclusion-exclusion, we get
$$
\begin{array}{l}
\left|P_{1} \cup P_{2} \cup P_{3}\right| \\
=\sum_{i=1}^{3}\left|P_{i}\right|-\sum_{1<i<j<3}\left|P_{i} \cap P_{j}\right|+\left|P_{1} \cap P_{2} \cap P_{3}\right| \\
=9 \times 3-4 \times 3+1=16 .
\end{array}
$$
Solution 2: From the given figure, we can see 6 lines intersecting pairwise, and from the required figure, we need to find three lines that intersect pairwise but not at the same point (forming a triangle). Thus, from $\mathrm{C}_{6}^{3}$, we subtract the 4 points where three lines intersect at $A, B, C, H$, giving $C_{8}^{3}-4=16$.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. There are four numbers, among which the sum of every three numbers is $24, 36, 28, 32$. Then the average of these four numbers is $\qquad$ .
|
10.10 .
The sum of these four numbers is $(24+36+28+32) \div 3=40$, so the average of these four numbers is 10 .
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. If $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, then $a^{2}+b^{2}=$ $\qquad$ .
|
11.3.
Given $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, we have $\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}+b^{2}\right)-6=0$.
Therefore, $a^{2}+b^{2}=3$ or -2.
Since $a^{2}+b^{2} \geqslant 0$, we have $a^{2}+b^{2}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. If real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x y+x+y+7=0, \\
3 x+3 y=9+2 x y,
\end{array}\right.
$$
then $x^{2} y+x y^{2}=$
|
13.6.
From $\left\{\begin{array}{l}x y+x+y+7=0, \\ 3 x+3 y=9+2 x y,\end{array}\right.$ we get $x y=-6, x+y=-1$.
Therefore, $x^{2} y+x y^{2}=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. A person's 5 trips to work (unit: $\mathrm{min}$) are $a, b, 8, 9, 10$. It is known that the average of this set of data is 9, and the variance is 2. Then the value of $|a-b|$ is $\qquad$.
|
15.4.
Since the average of $a, b, 8, 9, 10$ is 9 and the variance is 2, it follows that $a, b, 8, 9, 10$ are 5 consecutive integers, so $a=7, b=11$ or $a=11, b=7$. Therefore, the value of $|a-b|$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. If the integer $m$ makes the equation
$$
x^{2}-m x+m+2006=0
$$
have non-zero integer roots, then the number of such integers $m$ is
$\qquad$.
|
16.5.
Let the two integer roots of the equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=m, a \beta=m+2006, \\
\text { i.e., } \alpha \beta-(\alpha+\beta)+1=2006+1 \\
=2007=(\alpha-1)(\beta-1) .
\end{array}
$$
Thus, $\alpha-1= \pm 1, \pm 3, \pm 9$;
$$
\beta-1= \pm 2007, \pm 669, \pm 223 \text {. }
$$
Therefore, there are 5 such integers $m$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. In a math test, there are 20 questions. Each correct answer earns 5 points, no answer earns 0 points, and each wrong answer deducts 2 points. If Xiaoli's score in this test is a prime number, then the maximum number of questions Xiaoli answered correctly is $\qquad$.
|
17.17.
Let Xiaoli answer $a$ questions correctly and $b$ questions incorrectly, with the score being $5a-2b<100$.
Since the score is a prime number, we have $5a-2b=97, 91, 89, 83, \cdots$. Only when $a=17, b=1$, $5a-2b=83$ is the largest prime number.
Therefore, Xiaoli answered at most 17 questions correctly this time.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
18. Let the perimeter of a surface development of a $5 \mathrm{~cm} \times 4 \mathrm{~cm} \times 3 \mathrm{~cm}$ rectangular prism be $n \mathrm{~cm}$. Then the minimum value of $n$ is $\qquad$ .
|
18.50.
As shown in Figure 9, the perimeter of the unfolded rectangular prism is
$$
8 c+4 b+2 a \text {. }
$$
Therefore, the minimum value of the perimeter is
$$
\begin{array}{l}
8 \times 3+4 \times 4+2 \times 5 \\
=50 .
\end{array}
$$
|
50
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
20. There is a pentagon $A B C D E$. If the vertices $A$, $B$, $C$, $D$, $E$ are colored with one of the three colors: red, yellow, green, such that adjacent vertices are colored differently, then there are a total of different coloring methods.
|
20.30.
If point $A$ is colored red (as shown in Figure 11), there are 10 coloring methods.
Figure 11
Similarly, if point $A$ is colored yellow or green, there are also 10 coloring methods each. Therefore, there are a total of 30 different coloring methods.
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $x, y$ be real numbers, the algebraic expression
$$
5 x^{2}+4 y^{2}-8 x y+2 x+4
$$
has a minimum value of
|
9.3.
$$
\begin{array}{l}
\text { Since } 5 x^{2}+4 y^{2}-8 x y+2 x+4 \\
=5\left(x-\frac{4}{5} y+\frac{1}{5}\right)^{2}+\frac{4}{5}(y+1)^{2}+3,
\end{array}
$$
Therefore, when $y=-1, x=-1$, the original expression has a minimum value of 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. In a $3 \times 3$ grid filled with the numbers $1 \sim 9$, the largest number in each row is colored red, and the smallest number in each row is colored green. Let $M$ be the smallest number among the three red squares, and $m$ be the largest number among the three green squares. Then $M-m$ can have $\qquad$ different values.
|
14.8.
From the conditions, it is known that the values of $m$ and $M$ are positive integers from 3 to 7 (inclusive of 3 and 7).
Obviously, $M \neq m$, so the value of $M-m$ is
$$
1,2,3,4,-1,-2,-3,-4 \text {. }
$$
Therefore, $M-m$ can have 8 different values.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. In the $7 \times 7$ unit square grid shown in Figure 10, there are 64 grid points, and there are many squares with these grid points as vertices. How many different values of the areas of these squares are there?
|
17. Since the vertices of the square are grid points in the grid, as shown in Figure 13, the area of the shaded square is $a^{2}+b^{2}$.
Where, $0 \leqslant a+b \leqslant 7$.
Without loss of generality, let $a \geqslant b$. We can enumerate all possible values of $(a, b)$:
$$
\begin{array}{l}
(0,0),(1,0),(2,0),(3,0),(4,0),(5,0), \\
(6,0),(7,0) ; \\
(1,1),(2,1),(3,1),(4,1),(5,1),(6,1) ; \\
(2,2),(3,2),(4,2),(5,2) ; \\
(3,3),(4,3) .
\end{array}
$$
Among them, $(0,0)$ does not meet the requirements and is discarded.
In addition, $5^{2}=4^{2}+3^{2}$, that is, $(5,0)$ and $(4,3)$ give the same area.
Upon inspection, it is known that the other area values are all different, so there are 18 different values in total.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
18. Let $k$, $a$, $b$ be positive integers, and the quotients when $k$ is divided by $a^2$ and $b^2$ are $m$ and $m+116$ respectively.
(1) If $a$ and $b$ are coprime, prove that $a^2 - b^2$ is coprime with both $a^2$ and $b^2$;
(2) When $a$ and $b$ are coprime, find the value of $k$;
(3) If the greatest common divisor of $a$ and $b$ is 5, find the value of $k$.
|
18. (1) Let $s$ be the greatest common divisor of $a^{2}-b^{2}$ and $a^{2}$.
Then $a^{2}-b^{2}=s u, a^{2}=s v\left(s, v \in \mathbf{N}_{+}\right)$.
Thus, $a^{2}-\left(a^{2}-b^{2}\right)=b^{2}=s(v-u)$.
It is clear that $s$ is a divisor of $b^{2}$.
Since $a, b$ are coprime, $a^{2}, b^{2}$ are also coprime.
Therefore, $s=1$, meaning $a^{2}-b^{2}$ and $a^{2}$ are coprime.
Similarly, $a^{2}-b^{2}$ and $b^{2}$ are coprime.
(2) Since $k=m a^{2}=(m+116) b^{2}$, we have $m\left(a^{2}-b^{2}\right)=116 b^{2}(a>b)$.
Given that $a, b, m$ are all positive integers, $a^{2}-b^{2}$ divides $116 b^{2}$.
Since $a^{2}-b^{2}$ and $b^{2}$ are coprime, $a^{2}-b^{2}$ divides 116, i.e., $(a+b)(a-b)$ divides 116.
And $116=2^{2} \times 29, a+b$ and $a-b$ have the same parity, and $a+b>a-b>0$, so,
$$
\left\{\begin{array}{l}
a+b=29, \\
a-b=1
\end{array}\right.
$$
or $\left\{\begin{array}{l}a+b=2 \times 29, \\ a-b=2 .\end{array}\right.$
Solving these, we get $a=15, b=14$ or $a=30, b=28$.
Since $a, b$ are coprime, we have $a=15, b=14$.
Thus, $m=\frac{116 b^{2}}{a^{2}-b^{2}}=2^{4} \times 7^{2}$.
Therefore, $k=m a^{2}=2^{4} \times 7^{2} \times 15^{2}=176400$.
(3) If the greatest common divisor of $a, b$ is 5, let $a=5 a_{1}, b=$ $5 b_{1}$, then $a_{1}, b_{1}$ are coprime.
Similarly to (2), we have $m\left(a^{2}-b^{2}\right)=116 b^{2}$, i.e.,
$m\left(25 a_{1}^{2}-25 b_{1}^{2}\right)=116\left(25 b_{1}^{2}\right)$.
Thus, $m\left(a_{1}^{2}-b_{1}^{2}\right)=116 b_{1}^{2}$, and $a_{1}, b_{1}$ are coprime.
According to (2),
$$
m=2^{4} \times 7^{2}, a_{1}=15, b_{1}=14 .
$$
Therefore, $k=m a^{2}=m\left(5 a_{1}\right)^{2}=25 m a_{1}^{2}=4410000$.
|
4410000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that $a, b, c, d$ are even numbers, and $0<a<b<c<d, d-a=90, a, b, c$ form an arithmetic sequence, $b, c, d$ form a geometric sequence. Then $a+b+c+d=$ ( ).
(A) 384
(B) 324
(C) 284
(D) 194
|
9.D.
Let $a, b, c, d$ be $b-m, b, b+m, \frac{(b+m)^{2}}{b}$, respectively.
Also, $\frac{(b+m)^{2}}{b}-(b-m)=90$, which means
$$
b=\frac{m^{2}}{3(30-m)} \text {. }
$$
Since $a, b, c, d$ are even numbers, and $0<a<b<c<d$, we know that $m$ is a multiple of 6, and $m<30$.
Let $m=6k$, substituting into equation (1) yields
$$
b=\frac{2 k^{2}}{5-k}(k=1,2,3,4) \text {. }
$$
Substituting and checking, we find $k=4, b=32$.
Thus, $m=24, b=32, a, b, c, d$ are $8, 32, 56, 98$ respectively.
Therefore, $a+b+c+d=194$.
|
194
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given $\frac{\sin (\alpha+2 \beta)}{\sin \alpha}=3$, and $\beta \neq \frac{1}{2} k \pi$, $\alpha+\beta \neq n \pi+\frac{\pi}{2}(n, k \in \mathbf{Z})$. Then the value of $\frac{\tan (\alpha+\beta)}{\tan \beta}$ is $\qquad$
|
$\begin{array}{l}13.2 \text {. } \\ \frac{\tan (\alpha+\beta)}{\tan \beta}=\frac{\sin (\alpha+\beta) \cdot \cos \beta}{\cos (\alpha+\beta) \cdot \sin \beta} \\ =\frac{\frac{1}{2}[\sin (\alpha+2 \beta)+\sin \alpha]}{\frac{1}{2}[\sin (\alpha+2 \beta)-\sin \alpha]}=\frac{\frac{\sin (\alpha+2 \beta)}{\sin \alpha}+1}{\frac{\sin (\alpha+2 \beta)}{\sin \alpha}-1} \\ =\frac{3+1}{3-1}=2 .\end{array}$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Let $\left\{a_{n}\right\}$ be a sequence of positive numbers, and let the sum of the first $n$ terms be $b_{n}$. The product of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $c_{n}$, and $b_{n}+c_{n}=1$. Then the number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ that is closest to 2000 is $\qquad$.
|
14.1980.
According to the problem, we have $b_{n}=\frac{c_{n}}{c_{n-1}}(n \geqslant 2)$.
Also, $b_{n}+c_{n}=1$, so $\frac{c_{n}}{c_{n-1}}+c_{n}=1$, which means $\frac{1}{c_{n}}-\frac{1}{c_{n-1}}=1$. Given $c_{1}=b_{1}, c_{1}+b_{1}=1$, we can get $c_{1}=b_{1}=\frac{1}{2}$. Therefore, $c_{n}=\frac{1}{n+1}, b_{n}=\frac{n}{n+1}, \frac{1}{a_{n}}=n(n+1)$. Thus, the number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ closest to 2000 is $44 \times 45=1980$.
|
1980
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) Given a finite set of planar vectors $M$, for any three elements chosen from $M$, there always exist two elements $\boldsymbol{a}, \boldsymbol{b}$ such that $\boldsymbol{a}+\boldsymbol{b} \in M$. Try to find the maximum number of elements in $M$.
|
Three, the maximum number of elements in the set $M$ is 7.
Let points $A, B, C$ be any three points on a plane. Consider the 7-element set $M=\{\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C A}, \boldsymbol{B A}, \boldsymbol{C B}, \boldsymbol{A C}, \mathbf{0}\}$, which clearly satisfies the conditions. We will now prove that the number of elements in $M$ cannot exceed 7.
When all elements in $M$ are collinear, move all the starting points to the same point and draw a line $l$ parallel to all the elements. Project all elements of $M$ onto the line $l$. Thus, all vectors in $M$ correspond to coordinates on a number line with the common starting point of the elements in $M$ projected onto $l$ as the origin and an arbitrary direction on $l$ as the positive direction. Therefore, the problem can be transformed into a number set problem (from two-dimensional to one-dimensional).
Next, we prove that there are at most 7 elements in this number set.
First, we prove that there are at most 3 positive numbers in the number set. Assume that there are at least 4 positive elements, and the largest 4 numbers are $a_{1}, a_{2}, a_{3}, a_{4}$, with $a_{1} < a_{2} < a_{3} < a_{4}$. Then, $a_{1} + a_{2} + a_{4} > a_{1} + a_{4} > a_{4}$, so the sum $a_{i} + a_{4} \notin M (i=1,2,3)$. Since there is only one element greater than $a_{3}$, which is $a_{4}$, but $a_{2} + a_{3} > a_{1} + a_{3} > a_{3}$, in the set $\left\{a_{1}, a_{3}, a_{4}\right\}$ or $\left\{a_{2}, a_{3}, a_{4}\right\}$, at least one set has the property that the sum of any two elements is not in $M$. This contradicts the given conditions, so the number set has at most 3 positive numbers. Similarly, the number set has at most 3 negative numbers. Adding a 0, thus, the number set $M$ has at most 7 elements.
When the elements in $M$ are not all collinear, move all the starting points to the same point $O$. By the finiteness of $M$, we can draw a Cartesian coordinate system $x O y$ such that the elements in $M$ are not parallel to the coordinate axes.
Next, we prove that there are at most 3 elements in the upper half-plane.
First, we prove that all elements in the upper half-plane are not collinear. Assume there exist elements $a_{1}$ and $a_{2}$ in the upper half-plane that are collinear. Then, take the element $b$ with the smallest angle with $a_{1}$ and $a_{2}$. Consider the set $\left\{a_{1}, a_{2}, b\right\}$. By the choice of $b$, $a_{1} + b$ and $a_{2} + b$ are not in $M$ (the sum vector lies between the two vectors). Thus, there exists $a_{3}$ in $M$ such that $a_{3} = a_{1} + a_{2}$, so $a_{3}$ is collinear with $a_{1}$ and $a_{2}$. Consider the set $\left\{a_{1}, a_{3}, b\right\}$, similarly, there exists $a_{4}$ collinear with $a_{1}$ and $a_{2}$. Continuing this discussion, there are infinitely many elements in $M$ collinear with $a_{1}$ and $a_{2}$, which is a contradiction. Therefore, all elements in the upper half-plane are not collinear.
Next, we prove that there are at most 3 elements in the upper half-plane. Assume there are at least 4 elements in the upper half-plane. Take 4 adjacent elements $a, b, c, d$ in the upper half-plane in a counterclockwise direction. Consider the set $\{a, b, c\}$, then $b = a + c$; consider the set $\{b, c, d\}$, then $c = b + d$. Thus, $b = a + c = a + b + d$, which implies $a + d = 0$. This contradicts the fact that $a$ and $d$ are in the upper half-plane, so there are at most 3 elements in the upper half-plane. Similarly, there are at most 3 elements in the lower half-plane. Adding the zero vector, thus, the set $M$ has at most 7 elements.
In conclusion, the maximum number of elements in the set $M$ is 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given real numbers $a, b, c, d, e, f$ satisfy the system of equations:
$$
\left\{\begin{array}{l}
2 a+b+c+d+e+f=20, \\
a+2 b+c+d+e+f=40, \\
a+b+2 c+d+e+f=80, \\
a+b+c+2 d+e+f=160, \\
a+b+c+d+2 e+f=320, \\
a+b+c+d+e+2 f=640 .
\end{array}\right.
$$
Then the value of $f-e+d-c+b-a$ is $\qquad$ .
|
2. 420 .
$$
\begin{array}{l}
\text { (6) - (5) + (4) - (3) + (2)-(1) gives } \\
f-e+d-c+b-a \\
=640-320+160-80+40-20=420 .
\end{array}
$$
|
420
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The number of positive integers $n$ such that $n+1$ divides $n^{2006}+2006$ is $\qquad$.
Makes the positive integer $n$ such that $n+1$ can divide $n^{2006}+2006$ total $\qquad$.
Note: The second sentence seems to be a repetition or a different phrasing of the first. If it's meant to be a different statement, please clarify. Otherwise, I will assume the first translation is sufficient.
|
5. 5 .
From the problem, we have
$$
\begin{array}{l}
n^{2006}+2006 \equiv(-1)^{2006}+2006=2007 \\
\equiv 0(\bmod (n+1)) .
\end{array}
$$
Since $2007=3 \times 3 \times 223$, then
$$
n+1=3,9,223,669,2007 \text {. }
$$
Therefore, $n=2,8,222,668,2006$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given $n(n>1)$ integers (which can be the same) $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2007 .
$$
Then the minimum value of $n$ is $\qquad$
|
9. 5 .
Given $a_{1} a_{2} \cdots a_{n}=2007$, we know that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. Also, $a_{1}+a_{2}+\cdots+a_{n}=2007$ is an odd number, so $n$ is odd.
If $n=3$, i.e., $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=2007$, without loss of generality, assume $a_{1} \geqslant a_{2} \geqslant a_{3}$, then
$$
a_{1} \geqslant \frac{a_{1}+a_{2}+a_{3}}{3}=669, a_{2} a_{3} \leqslant \frac{2007}{a_{1}} \leqslant 3 \text {. }
$$
If $a_{1}=669$, then $a_{2} a_{3}=3$. Thus,
$a_{2}+a_{3} \leqslant 4, a_{1}+a_{2}+a_{3} \leqslant 673669$, only $a_{1}=2007, a_{2} a_{3}=1$ and $a_{2}+a_{3}$ $=0$, which is impossible.
Therefore, $n \geqslant 5$.
$$
\begin{array}{l}
\text { Also, } 2007+1+1+(-1)+(-1) \\
=2007 \times 1 \times 1 \times(-1) \times(-1)=2007,
\end{array}
$$
Thus, the minimum value of $n$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8.3. There are 19 cards. Can a non-zero digit be written on each card so that these 19 cards can be arranged to form a 19-digit number divisible by 11?
|
8.3. Answer: Yes.
Write a 2 on each of 10 cards, and a 1 on each of the remaining cards. It is well known that a positive decimal integer is divisible by 11 if and only if the difference $S$ between the sum of the digits in the odd positions and the sum of the digits in the even positions is a multiple of 11. Under all different arrangements of the 19 cards, we have
$$
-7 \leqslant S \leqslant 11,
$$
where only 11 is a multiple of 11, corresponding to placing the cards with 2s in all odd positions and the cards with 1s in all even positions.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $A_{1}, A_{2}, \cdots, A_{n}$ are 11 stations sequentially on a straight highway, and
$$
\begin{array}{l}
A_{i} A_{i+2} \leqslant 12(i=1,2, \cdots, 9), \\
A_{i} A_{i+3} \geqslant 17(i=1,2, \cdots, 8) .
\end{array}
$$
If $A_{1} A_{11}=56 \mathrm{~km}$, then $A_{2} A_{7}=$ $\qquad$ $\mathrm{km}$.
|
Ni, 1.29.
According to the problem, we have $A_{1} A_{10} \geqslant 3 \times 17=51(\mathrm{~km})$, then
$$
A_{10} A_{11} \leqslant 5 \mathrm{~km} \text {; }
$$
Also, $A_{8} A_{11} \geqslant 17 \mathrm{~km}, A_{8} A_{10} \leqslant 12 \mathrm{~km}$, then $A_{10} A_{11} \geqslant 5 \mathrm{~km}$.
Therefore, $A_{10} A_{11}=5 \mathrm{~km}$.
Similarly, $A_{1} A_{2}=5 \mathrm{~km}$.
And $A_{1} A_{7} \geqslant 34 \mathrm{~km}$, thus,
$$
A_{2} A_{7} \geqslant 29 \mathrm{~km} \text {. }
$$
Also, $A_{2} A_{10}=56-5 \times 2=46(\mathrm{~km}), A_{7} A_{10} \geqslant 17 \mathrm{~km}$, so $A_{2} A_{7} \leqslant 29 \mathrm{~km}$.
From equations (1) and (2), we get $A_{2} A_{7}=29 \mathrm{~km}$.
|
29
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of positive integer pairs $(x, y)$ that satisfy the equation
$$
\begin{array}{l}
x \sqrt{y}+y \sqrt{x}-\sqrt{2006 x}-\sqrt{2006 y}+\sqrt{2006 x y} \\
\quad=2006
\end{array}
$$
|
3.8 .
Given the equation can be transformed into
$$
(\sqrt{x}+\sqrt{y}+\sqrt{2006})(\sqrt{x y}-\sqrt{2006})=0 \text {. }
$$
Since $\sqrt{x}+\sqrt{y}+\sqrt{2006}>0$, we have
$$
\sqrt{x y}=\sqrt{2006} \text {, }
$$
which means $\square$
$$
\begin{array}{l}
x y=2006=1 \times 2006=2 \times 1003 \\
=17 \times 118=34 \times 59 .
\end{array}
$$
Thus, $(x, y)=(1,2006),(2006,1),(2,1003)$, $(1003,2),(17,118),(118,17),(34,59),(59,34)$. Therefore, there are 8 pairs of positive integers $(x, y)$ that satisfy the given equation.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The number of five-digit numbers where the sum of any two adjacent digits is divisible by 3 is $\qquad$ .
|
2.1254.
Classify the digits by their remainders, then the five-digit number $\overline{a b c d e}$ that meets the condition satisfies
$$
a \equiv c=e \equiv 2(\bmod 3), b \equiv d \equiv 1(\bmod 3),
$$
or $a=c=e=1(\bmod 3), b=d=2(\bmod 3)$,
or $a=b=c=d=e=0(\bmod 3)$.
If it is equation (1), then there are $3^{5}$ such five-digit numbers. If it is equation (2), then there are $3^{5}$ such five-digit numbers. If it is equation (3), then there are $3 \times 4^{4}$ such five-digit numbers.
Therefore, the total number of five-digit numbers that meet the condition is
$$
3^{5}+3^{5}+3 \times 4^{4}=1254 \text { (numbers). }
$$
|
1254
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A three-digit number $\overline{x y z}, 1 \leqslant x \leqslant 9,0 \leqslant y, z \leqslant 9$, and $x!+y!+z!=\overline{x y z}$. Then the value of $x+y+z$ is $\qquad$
|
5.10.
From $\overline{x y z}=x!+y!+z!$, we get
$$
100 x+10 y+z=x!+y!+z!.
$$
It is easy to see that $x, y, z \leqslant 6$. Otherwise,
$$
x!+y!+z!\geqslant 7!>1000.
$$
If $x=6$, then the left side of equation (1) is $700$, which is a contradiction.
Therefore, $x \leqslant 5$.
Similarly, $y \leqslant 5, z \leqslant 5$.
Thus, $x!+y!+z!\leqslant 3 \times 5!=360$. Hence, $x \leqslant 3$.
When $x=3$,
$$
294+10 y+z=y!+z!\leqslant 2 \times 5!=240,
$$
which is a contradiction.
When $x=2$,
$$
198+10 y+z=y!+z!.
$$
If $y \leqslant 4, z \leqslant 4$, then $y!+z!\leqslant 2 \times 4!<198$, which is a contradiction.
Therefore, one of $y, z$ must be 5.
(i) If $y=5,128+z=z$!, which is impossible;
(ii) If $z=5,83+10 y=y$!, then $y \mid 83$, which is also impossible.
When $x=1$, $99+10 y+z=y!+z!$.
If $y \leqslant 4, z \leqslant 4$, then $y!+z!\leqslant 48<99$, which is a contradiction.
Therefore, one of $y, z$ must be 5.
(i) If $y=5,29+z=z$!, which implies $z \mid 29$, which is impossible;
(ii) If $z=5,10 y=y!+16$, by calculation, $y$ can only be 4.
Thus, $x+y+z=10$.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Use several small cubes of the same size, with surfaces all white or all red, to form a large cube $A B C D-E F G H$ as shown in Figure 2. If the small cubes used on the diagonals $A G$, $B H$, $C E$, and $D F$ of the large cube are all red, and a total of 41 are used, and the rest of the large cube is made up of small cubes that are all white, then the number of small cubes with all white surfaces used is ( ) pieces.
(A) 688
(B) 959
(C) 1290
(D) 1687
|
5.C.
For a large cube, the number of small cubes used on each of the four space diagonals is the same, so the total number of small cubes used on the four space diagonals should be a multiple of 4. However, according to the given condition, the four space diagonals actually use 41 small cubes, which is an odd number. This indicates that for the given large cube, when calculating the number of small cubes used on each space diagonal, the small cube at the intersection of the four space diagonals is counted 4 times. Therefore, the number of red small cubes used on each space diagonal should be $(41+3) \div 4=11$.
At this point, there are also 11 small cubes on each edge of the large cube. Therefore, the number of white small cubes used is
$$
11 \times 11 \times 11-41=1290 \text {. }
$$
|
1290
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. If the graph of the inverse proportion function $y=\frac{k}{x}$ intersects the graph of the linear function $y=a x+b$ at points $A(-2, m)$ and $B(5, n)$, then the value of $3 a+b$ is
保留了源文本的换行和格式。
|
Ni.7.0.
Since points $A(-2, m)$ and $B(5, n)$ lie on the graph of an inverse proportion function, we have
$$
\left\{\begin{array}{l}
m=-\frac{k}{2}, \\
n=\frac{k}{5} .
\end{array}\right.
$$
Since points $A(-2, m)$ and $B(5, n)$ also lie on the line $y=a x+b$, we get
$$
\left\{\begin{array}{l}
m=-2 a+b \\
n=5 a+b
\end{array}\right.
$$
Thus,
$$
\left\{\begin{array}{l}
-2 a+b=-\frac{k}{2}, \\
5 a+b=\frac{k}{5} .
\end{array}\right.
$$
Then,
$$
\left\{\begin{array}{l}
a=\frac{k}{10}, \\
b=-\frac{3 k}{10}
\end{array}\right.
$$.
Solving, we get $b=-3 a$.
Therefore, $3 a+b=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $n$ be a positive integer, and $n^{2}+1085$ is a positive integer power of 3. Then the value of $n$ is $\qquad$
|
11.74.
Generally, the unit digit of $n^{2}$ (where $n$ is a positive integer) can only be 0, 1, 4, 5, 6, 9, so the unit digit of $n^{2}+1085$ can only be 5, 6, 9, 0, 1, 4; while the unit digit of $3^{m}$ (where $m$ is a positive integer) can only be 1, 3, 7, 9.
From the given, let $n^{2}+1085=3^{m}$ (where $n, m$ are positive integers), we can deduce that the unit digit of $3^{m}$ can only be 1 or 9, and $m$ is an even number.
Let $m=2k$ (where $k$ is a positive integer), then we have
$$
n^{2}+1085=3^{2k} \text{. }
$$
Rearranging gives $\left(3^{k}-n\right)\left(3^{k}+n\right)=1 \times 5 \times 7 \times 31$.
We get $\left\{\begin{array}{l}3^{k}-n=1, \\ 3^{k}+n=1085 ;\end{array}\left\{\begin{array}{l}3^{k}-n=5, \\ 3^{k}+n=217 ;\end{array}\right.\right.$
$$
\left\{\begin{array}{l}
3^{k}-n=7, \\
3^{k}+n=155 ;
\end{array} \left\{\begin{array}{l}
3^{k}-n=31, \\
3^{k}+n=35 .
\end{array}\right.\right.
$$
However, only the equation set $\left\{\begin{array}{l}3^{k}-n=7, \\ 3^{k}+n=155\end{array}\right.$ has a solution that meets the conditions: $\left\{\begin{array}{l}k=4, \\ n=74 .\end{array}\right.$
|
74
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given that $a$, $b$, and $c$ are positive integers, and the graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from points $A$ and $B$ to the origin $O$ are both less than 1, find the minimum value of $a+b+c$.
|
14. Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, where $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2}+b x+c=0$.
Given that $a$, $b$, and $c$ are positive integers, we have
$x_{1}+x_{2}=-\frac{b}{a} < 0$.
Thus, the equation $a x^{2}+b x+c=0$ has two negative real roots, i.e., $x_{1} < 0$.
Solving, we get $b>2 \sqrt{a c}$.
Since $|O A|=\left|x_{1}\right| < 0, a+c > b$.
$$
Given that $a$, $b$, and $c$ are positive integers, we have
$$
a+c \geqslant b+1 > 2 \sqrt{a c}+1.
$$
Thus, $a+c > 2 \sqrt{a c}+1$.
Then, $(\sqrt{a}-\sqrt{c})^{2} > 1, \sqrt{a}-\sqrt{c} > 1, \sqrt{a} > \sqrt{c}+1 \geqslant 2$.
Therefore, $a > 4$, i.e., $a \geqslant 5$.
Hence, $b > 2 \sqrt{a c} \geqslant 2 \sqrt{5 \times 1}=2 \sqrt{5}$, i.e., $b \geqslant 5$.
Thus, taking $a=5, b=5, c=1$, the equation $y=5 x^{2}+5 x+1$ satisfies the conditions.
Therefore, the minimum value of $a+b+c$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. At a party, 9 celebrities performed $n$ "trio dance" programs. If in these programs, any two people have collaborated exactly once, then $n=$ $\qquad$
|
12.12.
Consider 9 people as 9 points. If two people have performed in the same group, then a line segment is drawn between the corresponding two points. Thus, every pair of points is connected, resulting in a total of 36 line segments. Each trio dance corresponds to a triangle, which has three sides. When all sides appear and are not reused, the total number of triangles is \( n = \frac{36}{3} = 12 \), i.e., 12 performances. Such 12 trio dances can be arranged, for example:
$$
\begin{array}{l}
(1,2,3),(4,5,6),(7,8,9) ;(1,4,7),(2,5,8),(3,6,9) ; \\
(1,5,9),(2,6,7),(3,4,8) ;(1,6,8),(2,4,9),(3,5,7)
\end{array}
$$
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. (i) (Grade 11) In the arithmetic sequence $\left\{a_{n}\right\}: a_{n}=4 n -1\left(n \in \mathbf{N}_{+}\right)$, after deleting all numbers that can be divided by 3 or 5, the remaining numbers are arranged in ascending order to form a sequence $\left\{b_{n}\right\}$. Find the value of $b_{2006}$.
(ii) (Grade 12) Given $\odot P: x^{2}+y^{2}=2 x$ and the parabola $S: y^{2}=4 x$, a line $l$ is drawn through the center $P$ intersecting the two curves at four points, which are sequentially labeled as $A, B, C, D$ from top to bottom. If the lengths of segments $A B, B C, C D$ form an arithmetic sequence in this order, find the equation of the line $l$.
|
(i) Since $a_{n+15}-a_{n}=60$, $a_{n}$ is a multiple of 3 or 5 if and only if $a_{n+15}$ is a multiple of 3 or 5.
Now, divide the positive direction of the number line into a series of intervals of length 60:
$$
(0,+\infty)=(0,60] \cup(60,120] \cup(120,180] \cup \cdots \text {. }
$$
Note that the first interval contains 15 terms of $\left\{a_{n}\right\}$, namely 3, $7,11,15,19,23,27,31,35,39,43,47,51,55,59$, of which 8 terms belong to $\left\{b_{n}\right\}$: $b_{1}=7, b_{2}=11, b_{3}=19, b_{4}=23$, $b_{5}=31, b_{6}=43, b_{7}=47, b_{8}=59$. Therefore, each interval contains 15 terms of $\left\{a_{n}\right\}$ and 8 terms of $\left\{b_{n}\right\}$, and $b_{8 k+r}-b_{r}=60 k(k \in \mathbf{N}, 1 \leqslant r \leqslant 8)$.
Since $2006=8 \times 250+6$, and $b_{6}=43$, we have
$$
b_{2006}=60 \times 250+b_{6}=60 \times 250+43=15043 \text {. }
$$
(ii) The equation of $\odot P$ is $(x-1)^{2}+y^{2}=1$, with a diameter of $|B C|=2$ and the center at $P(1,0)$.
Let $l: k y=x-1$, i.e., $x=k y+1$, substituting into the parabola equation gives $y^{2}=4 k y+4$.
Let $A\left(x_{1}, y_{1}\right)$ and $D\left(x_{2}, y_{2}\right)$, then
$$
\left\{\begin{array}{l}
y_{1}+y_{2}=4 k, \\
y_{1} y_{2}=-4 .
\end{array}\right.
$$
Thus, $\left(y_{1}-y_{2}\right)^{2}=\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}$.
Hence, $|A D|^{2}=\left(y_{1}-y_{2}\right)^{2}+\left(x_{1}-x_{2}\right)^{2}$
$=\left(y_{1}-y_{2}\right)^{2}+\left(\frac{y_{1}^{2}-y_{2}^{2}}{4}\right)^{2}$
$=\left(y_{1}-y_{2}\right)^{2}\left[1+\left(\frac{y_{1}+y_{2}}{4}\right)^{2}\right]=16\left(k^{2}+1\right)^{2}$.
Therefore, $|A D|=4\left(k^{2}+1\right)$.
According to the problem,
$2|B C|=|A B|+|C D|=|A D|-|B C|$,
so, $|A D|=3|B C|=6$, i.e.,
$4\left(k^{2}+1\right)=6$.
Solving for $k$ gives $k= \pm \frac{\sqrt{2}}{2}$.
Thus, the equation of $l$ is $x=\frac{\sqrt{2}}{2} y+1$ or $x=-\frac{\sqrt{2}}{2} y+1$.
|
15043
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Find the smallest positive real number $k$ such that for any 4 distinct real numbers $a, b, c, d$ not less than $k$, there exists a permutation $p, q, r, s$ of $a, b, c, d$ such that the equation $\left(x^{2}+p x+q\right)\left(x^{2}+r x+s\right)=0$ has 4 distinct real roots. (Feng Zhigang)
|
2. On one hand, if $k4(d-a)>0, c^{2}-4 b>4(c-b)$ $>0$, therefore, equations (1) and (2) both have two distinct real roots.
Secondly, if equations (1) and (2) have a common real root $\beta$, then
$$
\left\{\begin{array}{l}
\beta^{2}+d \beta+a=0, \\
\beta^{2}+\phi \beta+b=0 .
\end{array}\right.
$$
Subtracting the two equations yields $\beta=\frac{b-a}{d-c}>0$, at this time, $\beta^{2}+d \beta+a>$ 0, which is a contradiction.
Therefore, equations (1) and (2) do not have a common real root.
Thus, $k=4$ meets the requirement.
In summary, $k=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Arrange the numbers $1,2, \cdots, 13$ in a row $a_{1}, a_{2}$, $\cdots, a_{13}$, where $a_{1}=13, a_{2}=1$, and ensure that $a_{1}+a_{2}+$ $\cdots+a_{k}$ is divisible by $a_{k+1}(k=1,2, \cdots, 12)$. Then the value of $a_{4}$ $+a_{5}+\cdots+a_{12}$ is $\qquad$ .
|
II. 1.68.
Since $a_{1}+a_{2}+\cdots+a_{12}$ is divisible by $a_{13}$, then $a_{1}+a_{2}+\cdots+a_{12}+a_{13}$ is also divisible by $a_{13}$, meaning $a_{13}$ is a factor of $a_{1}+a_{2}+\cdots+a_{12}+a_{13}=13 \times 7$. Given that $a_{1}=13$, $a_{2}=1$, we have $a_{13}=7$.
Also, since $a_{1}+a_{2}$ is divisible by $a_{3}$, then $a_{3}$ is a factor of $a_{1}+a_{2}=2 \times 7$. Therefore, $a_{3}=2$. We can sequentially obtain a series of numbers that satisfy the conditions: $13,1,2,8,3,9,4,10,5,11,6,12,7$. Thus,
$$
a_{4}+a_{5}+\cdots+a_{12}=13 \times 7-13-1-2-7=68 .
$$
|
68
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $a$, $b$, $c$ be positive integers, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-b c-c a=19 \text {. }
$$
Then the minimum value of $a+b+c$ is $\qquad$
|
4. 10 .
By the cyclic symmetry of $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant c$.
Let $a-b=m, b-c=n$, then $a-c=m+n$, and
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[m^{2}+n^{2}+(m+n)^{2}\right],
\end{array}
$$
which is $m^{2}+m n+n^{2}-19=0$.
For the equation in $m$,
$$
\Delta=n^{2}-4\left(n^{2}-19\right) \geqslant 0 \text {. }
$$
Thus, $0 \leqslant n \leqslant \frac{2 \sqrt{57}}{3}0)$ has no solution.
Therefore, when $a=6, b=3, c=1$, $a+b+c$ achieves its minimum value, which is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) In a population of 1000 individuals numbered $0,1,2, \cdots, 999$, they are sequentially divided into 10 groups: Group 0 includes numbers $0,1, \cdots, 99$, Group 1 includes numbers $100,101, \cdots, 199$, and so on. Now, one number is to be selected from each group to form a sample of size 10, with the following rule: If a number $x$ is arbitrarily chosen from Group 0, then the numbers in the subsequent groups are obtained by shifting positions, i.e., the last two digits of the number drawn from Group $k$ are the last two digits of $x + a k$.
(1) When $x=24, a=33$, write down the 10 numbers in the sample;
(2) If $a=33$ and one of the 10 numbers in the sample has the last two digits 87, find the sum of all possible values of $x$ that satisfy the condition;
(3) Does there exist a two-digit positive integer $a$ such that at least two of the 10 numbers in the sample have the same last two digits? If it exists, find the value of $a$; if not, explain the reason.
|
(1) When $x=24, a=33$, the values of $x+a k$ corresponding to $k$ taking the values $0,1, \cdots, 9$ are
$$
24,57,90,123,156,189,222,255,288,321 \text {. }
$$
Therefore, the 10 numbers of the sample drawn are
$$
24,157,290,323,456,589,622,755,888,921 \text {. }
$$
(2) Since a number with the last two digits 87 can appear in each group, and when $k$ takes the values $0,1, \cdots, 9$, the values of $33 k$ are
$$
0,33,66,99,132,165,198,231,264,297 \text {, }
$$
Thus, all possible values of $x$ are
$$
87,54,21,88,55,22,89,56,23,90 \text {. }
$$
Their sum is 585.
(3) Suppose the last two digits of the numbers drawn from the $i$-th group and the $j$-th group are the same, i.e., the last two digits of $x+a i$ and $x+a j$ are the same. Therefore, the last two digits of $a|i-j|$ must be 0, and $1 \leqslant|i-j| \leqslant 9$.
Since $a$ is a two-digit positive integer, all values of $a$ that satisfy the condition are $20,25,40,50,60,75,80$.
|
585
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 1, label the six vertices of a regular hexagon with the numbers $0,1,2,3,4,5$ in sequence. An ant starts from vertex 0 and crawls counterclockwise, moving 1 edge on the first move, 2 edges on the second move, $\cdots \cdots$ and $2^{n-1}$ edges on the $n$-th move. After 2005 moves, the value at the vertex where the ant is located is
|
Ni.1.1.
According to the problem, after 2005 movements, the ant has crawled over
$$
1+2+\cdots+2^{2004}=2^{2005}-1
$$
edges.
Let $x=2^{2005}-1$. Consider the remainder of $x$ modulo 6.
Since $x \equiv 1(\bmod 2), x \equiv 1(\bmod 3)$, we have
$$
3 x \equiv 3(\bmod 6), 2 x \equiv 2(\bmod 6), 5 x \equiv 5(\bmod 6) \text {. }
$$
Also, since $(5,6)=1$, it follows that $x \equiv 1(\bmod 6)$.
Since the value at the vertex where the ant is located is equivalent to the remainder of $x$ modulo 6, the value at the vertex where the ant is located is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $x \in\left(0, \frac{\pi}{2}\right), \sin ^{2} x, \sin x \cdot \cos x$, and $\cos ^{2} x$ cannot form a triangle, and the length of the interval of $x$ that satisfies the condition is $\arctan k$. Then $k$ equals $\qquad$ (Note: If $x \in(a, b), b>a$, then the length of such an interval is $b-a$).
|
2.2.
First, find the range of $x$ that can form a triangle.
(1) If $x=\frac{\pi}{4}$, then $\sin ^{2} \frac{\pi}{4} 、 \cos ^{2} \frac{\pi}{4} 、 \sin \frac{\pi}{4} \cdot \cos \frac{\pi}{4}$ can form a triangle.
(2) If $x \in\left(0, \frac{\pi}{4}\right)$, at this time,
$$
\sin ^{2} x\cos ^{2} x$, that is,
$$
\tan ^{2} x+\tan x-1>0 \text {. }
$$
Solving this, we get $\arctan \frac{\sqrt{5}-1}{2}<x<\frac{\pi}{4}$.
(3) If $x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, similarly, we get
$$
\frac{\pi}{4}<x<\arctan \frac{\sqrt{5}+1}{2} \text {. }
$$
Thus, the range of $x$ for which these three positive numbers can form the side lengths of a triangle is $\left(\arctan \frac{\sqrt{5}-1}{2}, \arctan \frac{\sqrt{5}+1}{2}\right)$. The length of this interval is $\arctan \frac{\sqrt{5}+1}{2}-\arctan \frac{\sqrt{5}-1}{2}=\arctan \frac{1}{2}$. Therefore, the length of the interval of $x$ that satisfies the condition is $\frac{\pi}{2}-\arctan \frac{1}{2}=\arctan k$. It is easy to find that $k=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Divide each side of a square into 8 equal parts, and take the division points as vertices (excluding the vertices of the square), the number of different triangles that can be formed is ( ).
(A) 1372
(B) 2024
(C) 3136
(D) 4495
|
6. C.
Solution 1: First, note that the three vertices of the triangle are not on the same side of the square. Choose any three sides of the square so that each of the three vertices lies on one of them, which can be done in 4 ways.
Next, select a point on each of the chosen three sides, which can be done in $7^{3}$ ways.
The number of such triangles is $4 \times 7^{3}=1372$.
Additionally, if two vertices of the triangle are on one side of the square and the third vertex is on another side, first choose one side to have two vertices of the triangle, which can be done in 4 ways; then choose any two points on this side, which can be done in 21 ways; finally, choose one point from the remaining 21 points as the third vertex.
The number of such triangles is $4 \times 21 \times 21=1764$.
In summary, the number of different triangles is
$$
1372+1764=3136 \text {. }
$$
Solution 2: $C_{28}^{3}-4 C_{7}^{3}=3136$.
|
3136
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. The sum of the first $m$ terms of the arithmetic sequence $\left\{a_{n}\right\}$ is 90, and the sum of the first $2 m$ terms is 360. Then the sum of the first $4 m$ terms is $\qquad$ .
|
$=\mathbf{7 . 1 4 4 0}$.
Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k}$, it is easy to know that $S_{m} 、 S_{2 m}-S_{m}$ 、 $S_{3 m}-S_{2 m}$ form an arithmetic sequence. Therefore, $S_{3 m}=810$.
It is also easy to know that $S_{2 m}-S_{m} 、 S_{3 m}-S_{2 m} 、 S_{4 m}-S_{3 m}$ form an arithmetic sequence, then $S_{4 m}=3 S_{3 m}-3 S_{2 m}+S_{m}=1440$.
|
1440
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\frac{1}{2} \sin 2 y+a=0 .
\end{array}\right.
$$
then the value of $\cos (x+2 y)$ is
|
8.1.
Let $f(t)=t^{3}+\sin t$. Then $f(t)$ is monotonically increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
From the original system of equations, we get $f(x)=f(-2 y)=2 a$, and since $x$ and $-2 y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, it follows that $x=-2 y$, hence $x+2 y=0$. Therefore, $\cos (x+2 y)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If the six edges of a tetrahedron are $2,3,4,5$, 6,7, then there are $\qquad$ different shapes (if two tetrahedra can be made to coincide by appropriate placement, they are considered the same shape).
|
6.10.
Let the edge of length $k$ be denoted as $l_{k}(k \in\{2,3,4,5,6,7\})$. Consider $l_{2}$ and $l_{3}$.
(1) If $l_{2}$ and $l_{3}$ are coplanar, then the other side of the plane must be $l_{4}$.
(i) If $l_{2}$, $l_{3}$, and $l_{4}$ form a triangle in a clockwise direction (all referring to the direction of the three sides when viewed from inside the shape, the same below), as shown in Figure 6, then side $D A$ cannot be $l_{6}$ (otherwise, it would make the three sides of $\triangle B C D$ be $2,5,7$, which is a contradiction).
If $D A=l_{5},\{D B, D C\}=\left\{l_{6}, l_{7}\right\}$, there are 2 cases;
If $D A=l_{7},\{D B, D C\}=\left\{l_{5}, l_{6}\right\}$, there are also 2 cases.
A total of 4 cases are obtained.
(ii) If $l_{2}$, $l_{3}$, and $l_{4}$ form a triangle in a counterclockwise direction, similarly, 4 cases are obtained.
(2) If $l_{2}$ and $l_{3}$ are skew, let $A B=l_{2}, C D=l_{3}$. Then the remaining four edges, each one must be adjacent to $l_{2}$ and $l_{3}$. Therefore, the other side of the plane containing $l_{2}$ and $l_{7}$ must be $l_{6}$.
(i) If $l_{2}$, $l_{6}$, and $l_{7}$ form a triangle in a clockwise direction, without loss of generality, let $A C=l_{6}, B C=l_{7}$, as shown in Figure 7. The remaining two edges, $B D$ cannot be $l_{4}$, so only $B D=l_{5}, A D=l_{4}$, yielding 1 case.
(ii) If $l_{2}$, $l_{6}$, and $l_{7}$ form a triangle in a counterclockwise direction, similarly, 1 case is obtained.
Therefore, in this problem, there are 10 different shapes.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.100 Arrange chairs in a circle, with $n$ people sitting on the chairs, such that when one more person sits down, they will always be sitting next to one of the original $n$ people. The minimum value of $n$ is $\qquad$ .
|
9.34.
From the problem, we know that after $n$ people sit down, there are at most two empty chairs between any two people. If we can arrange it so that there are exactly two empty chairs between every two people, then $n$ is minimized. In this case, if we number the chairs where people are sitting, we get an arithmetic sequence: $1, 4, 7, \cdots, 100$.
Thus, $100 = 1 + 3(n-1)$. Solving for $n$ gives $n = 34$.
|
34
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. In $\triangle A B C$, $A B=\sqrt{30}, A C=\sqrt{6}, B C$ $=\sqrt{15}$, there is a point $D$ such that $A D$ bisects $B C$ and $\angle A D B$ is a right angle, the ratio $\frac{S_{\triangle A D B}}{S_{\triangle A B C}}$ can be written as $\frac{m}{n}$. ($m, n$ are coprime positive integers). Then $m+n=$
|
10.65.
Let the midpoint of $BC$ be $E$, and $AD=\frac{x}{2}$.
By the median formula, we get $AE=\frac{\sqrt{57}}{2}$.
By the Pythagorean theorem, we get $120-15+57=2 \sqrt{57} x$, solving for $x$ gives $x=\frac{81}{\sqrt{57}}$.
Thus, $\frac{m}{n}=\frac{AD}{2AE}=\frac{x}{2\sqrt{57}}=\frac{27}{38}$.
Therefore, $m+n=27+38=65$.
|
65
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. (16 points) Does there exist a smallest positive integer $t$, such that the inequality
$$
(n+t)^{n+t}>(1+n)^{3} n^{n} t^{t}
$$
holds for any positive integer $n$? Prove your conclusion.
|
$$
\text { Three, 13. Take }(t, n)=(1,1),(2,2),(3,3) \text {. }
$$
It is easy to verify that when $t=1,2,3$, none of them meet the requirement.
When $t=4$, if $n=1$, equation (1) obviously holds.
If $n \geqslant 2$, we have
$$
\begin{array}{l}
4^{4} n^{n}(n+1)^{3}=n^{n-2}(2 n)^{2}(2 n+2)^{3} \times 2^{3} \\
\leqslant\left[\frac{(n-2) n+2 \times 2 n+3(2 n+2)+2^{3}}{n+4}\right]^{n+4} \\
=\left(\frac{n^{2}+8 n+14}{n+4}\right)^{n+4}<\left(\frac{n^{2}+8 n+16}{n+4}\right)^{n+4} \\
=(n+4)^{n+4} .
\end{array}
$$
Therefore, equation (1) holds.
Thus, $t=4$ satisfies that for any positive integer $n$, equation (1) always holds.
|
4
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
15. (22 points) Let $A$ and $B$ be the common left and right vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Let $P$ and $Q$ be moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
$$
A P+B P=\lambda(A Q+B Q)(\lambda \in \mathbf{R},|\lambda|>1) \text {. }
$$
Let the slopes of the lines $A P$, $B P$, $A Q$, and $B Q$ be $k_{1}$, $k_{2}$, $k_{3}$, and $k_{4}$, respectively.
(1) Prove that $k_{1}+k_{2}+k_{3}+k_{4}=0$;
(2) Let $F_{1}$ and $F_{2}$ be the right foci of the ellipse and the hyperbola, respectively. If $P F_{2} \parallel Q F_{1}$, find the value of $k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{4}^{2}$.
|
15. (1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}}.
$$
Similarly,
$$
k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}.
$$
Let $O$ be the origin, then
$$
2 O P=A P+B P=\lambda(A Q+B Q)=2 \lambda O Q .
$$
Therefore, $O P=\lambda O Q$.
Hence, $O, P, Q$ are collinear.
Thus, $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$.
From equations (1) and (2), we get $k_{1}+k_{2}+k_{3}+k_{4}=0$.
(2) Since point $Q$ is on the ellipse, we have $\frac{x_{2}^{2}}{a^{2}}+\frac{y_{2}^{2}}{b^{2}}=1$.
From $O P=\lambda O Q$, we get $\left(x_{1}, y_{1}\right)=\lambda\left(x_{2}, y_{2}\right)$.
Therefore, $x_{2}=\frac{1}{\lambda} x_{1}, y_{2}=\frac{1}{\lambda} y_{1}$.
Thus, $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=\lambda^{2}$.
Since point $P$ is on the hyperbola, we have $\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}=1$.
From equations (3) and (4), we get $x_{1}^{2}=\frac{\lambda^{2}+1}{2} a^{2}, y_{1}^{2}=\frac{\lambda^{2}-1}{2} b^{2}$.
Since $P F_{2} \parallel Q F_{1}$, we have $\left|O F_{2}\right|=\lambda\left|O F_{1}\right|$.
Therefore, $\lambda^{2}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}, \frac{x_{1}^{2}}{y_{1}^{2}}=\frac{\left(\lambda^{2}+1\right) a^{2}}{\left(\lambda^{2}-1\right) b^{2}}=\frac{a^{4}}{b^{4}}$.
From equation (1), we get $\left(k_{1}+k_{2}\right)^{2}=\frac{4 b^{4}}{a^{4}} \cdot \frac{x_{1}^{2}}{y_{1}^{2}}=4$.
Similarly, $\left(k_{3}+k_{4}\right)^{2}=4$.
On the other hand, $k_{1} k_{2}=\frac{y_{1}}{x_{1}+a} \cdot \frac{y_{1}}{x_{1}-a}=\frac{b^{2}}{a^{2}}$.
Similarly, $k_{3} k_{4}=-\frac{b^{2}}{a^{2}}$.
Therefore,
$$
k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{4}^{2}
=\left(k_{1}+k_{2}\right)^{2}+\left(k_{3}+k_{4}\right)^{2}-2\left(k_{1} k_{2}+k_{3} k_{4}\right)=8.
$$
|
8
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. Points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola $y=2 x^{2}$ are symmetric with respect to the line $y=x+m$. If $2 x_{1} x_{2}=-1$, then the value of $2 m$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
|
3. A.
$$
\begin{array}{l}
\text { Given } \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-1, \\
\frac{y_{1}+y_{2}}{2}=\frac{x_{1}+x_{2}}{2}+m, \\
2 x_{1} x_{2}=-1 \text { and } y_{1}=2 x_{1}^{2}, y_{2}=2 x_{2}^{2},
\end{array}
$$
we get
$$
\begin{array}{l}
x_{2}-x_{1}=y_{1}-y_{2}=2\left(x_{1}^{2}-x_{2}^{2}\right) \Rightarrow x_{1}+x_{2}=-\frac{1}{2}, \\
2 m=\left(y_{1}+y_{2}\right)-\left(x_{1}+x_{2}\right)=2\left(x_{1}^{2}+x_{2}^{2}\right)+\frac{1}{2} \\
=2\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}+\frac{1}{2}=3 .
\end{array}
$$
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. If $\frac{1-\cos \theta}{4+\sin ^{2} \theta}=\frac{1}{2}$, then
$$
\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=
$$
|
ニ、7.9.
From the condition, we get $2-2 \cos \theta=4+\sin ^{2} \theta$, then
$$
\begin{array}{l}
(\cos \theta-1)^{2}=4 \Rightarrow \cos \theta=-1 . \\
\text { Therefore, }\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=9 .
\end{array}
$$
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The sequence $\left\{x_{n}\right\}: 1,3,3,3,5,5,5,5,5, \cdots$ is formed by arranging all positive odd numbers in ascending order, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$ (where $[x]$ denotes the greatest integer not greater than $x$), then $a+b+c+d=$ . $\qquad$
|
8.3.
Given $x_{k^{2}+1}=x_{k^{2}+2}=\cdots=x_{(k+1)^{2}}=2 k+1$, that is, when $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, we have $x_{n}=2 k+1(k=[\sqrt{n-1}])$.
Therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Hence, $(a, b, c, d)=(2,1,-1,1), a+b+c+d=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. If each element in set $A$ can be expressed as the product of two different numbers from 1, $2, \cdots, 9$, then the maximum number of elements in set $A$ is $\qquad$.
|
10.31.
From $1,2, \cdots, 9$ each time take a pair of numbers to form a product, a total of $C_{G}^{2}=$ 36 values are obtained. However, there are repetitions, the repeated cases are
$$
\begin{array}{l}
1 \times 6=2 \times 3,1 \times 8=2 \times 4,2 \times 9=3 \times 6, \\
2 \times 6=3 \times 4,3 \times 8=4 \times 6,
\end{array}
$$
a total of 5.
Therefore, the maximum number of elements in set $A$ is $C_{9}^{2}-5=31$.
|
31
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Use five different colors to color the five vertices of the "pentagram" in Figure 1 (each vertex is colored with one color, and some colors may not be used), so that the two vertices on each line segment are of different colors. Then the number of different coloring methods is .
|
12.1020.
Convert it into a coloring problem of a disk with 5 sectors, each to be colored with one of 5 colors, such that no two adjacent sectors have the same color (as shown in Figure 4).
Let the number of ways to color a disk with $k$ sectors using 5 colors be $x_{k}$. Then, we have $x_{k}+x_{k-1}=5 \times 4^{k-1}$. Therefore,
$$
\begin{array}{l}
x_{5}=\left(x_{5}+x_{4}\right)-\left(x_{4}+x_{3}\right)+ \\
\left(x_{3}+x_{2}\right)-x_{2} \\
=5\left(4^{4}-4^{3}+4^{2}-4\right)=1020 .
\end{array}
$$
|
1020
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. The sequence $\left\{a_{n}\right\}$ satisfies: $a_{0}=1, a_{n}=\left[\sqrt{S_{n-1}}\right]$ $(n=1,2, \cdots,[x]$ represents the greatest integer not greater than $x$, $\left.S_{k}=\sum_{i=0}^{k} a_{i}(k=0,1, \cdots)\right)$. Find the value of $a_{2006}$.
|
15. Observe the initial terms of the sequence (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$a_{n}$ & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 4 & 4 \\
\hline$S_{n}$ & 1 & 2 & 3 & 4 & 6 & 8 & 10 & 13 & 16 & 20 & 24 \\
\hline \hline$n$ & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & \\
\hline$a_{n}$ & 4 & 5 & 5 & 6 & 6 & 7 & 7 & 8 & 8 & 8 & \\
\hline$S_{n}$ & 28 & 33 & 38 & 44 & 50 & 57 & 64 & 72 & 80 & 88 & \\
\cline { 1 - 2 } & &
\end{tabular}
Notice that the sequence $\left\{a_{n}\right\}$ is strictly increasing, and each positive integer 1, 2, $\cdots$ appears in the sequence $\left\{a_{n}\right\}$ in order. Except for the first term $a_{0}=1$, each number of the form $2^{k} (k=0,1, \cdots)$ appears three times consecutively, and other numbers appear twice consecutively.
In general, the following properties of the sequence $\left\{a_{n}\right\}$ can be proven:
(1) For any $k \in \mathbf{N}$, if we denote $m=2^{k+1}+k+1$, then $a_{m-2}=a_{m-1}=a_{m}=2^{k}$.
(2) For any $k \in \mathbf{N}_{+}$, if we denote $m_{0}=2^{k}+k$, then when $1 \leqslant r \leqslant 2^{k-1}$, we have
$$
a_{m_{0}+2 r-1}=a_{m_{0}+2 r}=2^{k-1}+r \text {. }
$$
We use induction on $k$.
From the terms listed above, we know that the conclusion holds for $k \leqslant 2$. Assume that properties (1) and (2) hold for $k \leqslant n$, i.e., when $m=2^{n+1}+n+1$, $a_{m-2}=a_{m-1}=a_{m}=2^{n}$, then
$$
\begin{array}{l}
S_{m}=a_{0}+2\left(1+2+\cdots+2^{n}\right)+\left(2^{0}+2^{1}+\cdots+2^{n}\right) \\
=2^{2 n}+3 \times 2^{n} .
\end{array}
$$
We then use induction on $r$ for $1 \leqslant r \leqslant 2^{n}$.
When $r=1$, since $\left(2^{n}+1\right)^{2}0 \text {, }
$$
i.e., $\left(2^{n}+p+1\right)^{2} \leqslant S_{m+2 p}<\left(2^{n}+p+2\right)^{2}$.
Therefore, $a_{m+2 p+1}=\left[\sqrt{S_{m+2 p}}\right]=2^{n}+p+1$.
Since $S_{m+2 p}<S_{m+2 p+1}$
$$
\begin{array}{l}
=S_{m+2 p}+a_{m+2 p+1} \\
=2^{2 n}+2(p+2) \times 2^{n}+p^{2}+2 p+1 \\
<\left(2^{n}+p+2\right)^{2},
\end{array}
$$
Therefore, $a_{m+2 p+2}=\left[\sqrt{S_{m+2 p+1}}\right]=2^{n}+p+1$.
Hence, by induction, when $m=2^{n+1}+n+1, 1 \leqslant r \leqslant 2^{n}$, we have
$$
a_{m+2 r-1}=a_{m+2 r}=2^{n}+r \text {. }
$$
In particular, when $r=2^{n}$, the above formula becomes
$$
a_{2^{n+2}+n}=a_{2^{n+2}+n+1}=2^{n+1} \text {. }
$$
Also, from (1), we get
$$
\begin{array}{l}
S_{m+2 r}=2^{2 n}+(2 r+3) \times 2^{n}+r(r+1) . \\
\text { When } r=2^{n}, m=2^{n+1}+n+1, \text { we have } \\
S_{2^{n+2}+n+1}=2^{2 n}+\left(2^{n+1}+3\right) \times 2^{n}+2^{n}\left(2^{n}+1\right) \\
=2^{2 n+2}+2 \times 2^{n+1}<\left(2^{n+1}+1\right)^{2} .
\end{array}
$$
Therefore, $a_{2^{n+2}{ }_{n+2}}=\left[\sqrt{S_{2^{n+2}+n+1}}\right]=2^{n+1}$.
From (2) and (3), we know that for $m=2^{k+1}+k+1$, when $k=n+1$, we also have $a_{m-2}=a_{m-1}=a_{m}=2^{k}$.
Thus, properties (1) and (2) hold.
Since $2^{10}+10<2006<2^{11}+11$, let $m_{0}=2^{10}+10$, then $k-1=9, r=\frac{2006-m_{0}}{2}=486$.
Therefore, $a_{2006}=a_{m_{0}+2 r}=2^{9}+r=512+486=998$.
|
998
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. 5 people participate in 4 groups, each group has two people, and each person must participate in at least one group. Then, the number of different groupings is ( ).
(A) 135
(B) 130
(C) 125
(D) 120
|
10.A.
5 people can form $C_{5}^{2}=10$ pairs. Every 4 pairs are called a grouping method. A valid grouping method should include all 5 people. Since fewer than 4 people cannot form 4 pairs, all invalid grouping methods, i.e., those containing only 4 people, total $\mathrm{C}_{5}^{4} \mathrm{C}_{6}^{4}=75$ (each 4 people can form $\mathrm{C}_{4}^{2}=6$ pairs). Therefore, the number of valid grouping methods is $C_{10}^{4}-C_{5}^{4} C_{6}^{4}=135$.
|
135
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $a_{n}$ be the coefficient of $x$ in the binomial expansion of $(3-\sqrt{x})^{n}(n=2,3, \cdots)$. Then the value of $\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{18}}{a_{18}}$ is . $\qquad$
|
$$
\begin{array}{l}
\text { Since } a_{n}=\mathrm{C}_{n}^{2} \cdot 3^{n-2}, \text { then, } \\
\frac{3^{n}}{a_{n}}=\frac{3^{n}}{\mathrm{C}_{n}^{2} \cdot 3^{n-2}}=3^{2} \times \frac{2}{n(n-1)}=\frac{18}{n(n-1)} . \\
\text { Therefore, } \frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{18}}{a_{18}} \\
=18\left(\frac{1}{2 \times 1}+\frac{1}{3 \times 2}+\cdots+\frac{1}{18 \times 17}\right) \\
=18\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{17}-\frac{1}{18}\right)=17 .
\end{array}
$$
|
17
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. (12 points) Find the area of the figure formed by the set of points on the right-angle coordinate plane $b O a$
$$
S=\left\{(b, a) \mid f(x)=a x^{3}+b x^{2}-3 x\right.
$$
is a monotonic function on $\mathbf{R}$, and $a \geqslant-1\}$.
|
Three, 15. When $a=0$, by $f(x)$ being monotonic on $\mathbf{R}$, we know $b=0$. $f(x)$ being monotonic on $\mathbf{R}$ $\Leftrightarrow f^{\prime}(x)$ does not change sign on $\mathbf{R}$.
Since $f^{\prime}(x)=3 a x^{2}+2 b x-3$, therefore, by $\Delta=4 b^{2}+36 a \leqslant 0$,
we get, $a \leqslant-\frac{1}{9} b^{2}$.
Thus, the points $(b, a)$ that satisfy this condition form a closed figure in the $b O a$ coordinate plane, bounded by the curve $y=-\frac{1}{9} x^{2}$ and the line $y=-1$, with an area of
$$
6-2 \int_{0}^{3} \frac{1}{9} x^{2} \mathrm{~d} x=4
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any $n \in \mathbf{N}_{+}$, $a_{n}+a_{n+1}=2007$, and $31\left(a_{n}^{2}+a_{n} a_{n+1}+a_{n+1}^{2}\right)$. Then the sum of all different values of the sum of the first 2007 terms of the sequence $\left\{a_{n}\right\}$ is $\qquad$.
|
3. 1345368366 .
From $31\left(a_{n}^{2}+a_{n} a_{n+1}+a_{n+1}^{2}\right)$, we get
$$
31\left[a_{n}^{2}+a_{n+1}\left(a_{n}+a_{n+1}\right)\right] \text {. }
$$
Also, $31\left(a_{n}+a_{n+1}\right)$, so, $31 a_{n}$.
Since $a_{1}=2007-a_{2}<2006$, we have
$$
\begin{array}{l}
a_{1} \leqslant 2004, a_{1}=3 k(k=1,2, \cdots, 668) . \\
\text { Hence } a_{1}+a_{2}+\cdots+a_{2007} \\
=a_{1}+\left(a_{2}+a_{3}\right)+\left(a_{4}+a_{5}\right)+\cdots+\left(a_{200}+a_{2000}\right) \\
=a_{1}+2007 \times 1003 \\
=3 k+2007 \times 1003 .
\end{array}
$$
Therefore, $a_{1}+a_{2}+\cdots+a_{2007}$
Then $S=\sum_{k=1}^{668}(3 k+2007 \times 1003)=1345368366$.
|
1345368366
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The number of prime pairs $(p, q)$ that satisfy $\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=q$ is $\qquad$ .
|
5.2.
(1) When $p=2$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1$, which is not a prime number, contradiction.
(2) When $p=3$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1+1=2$ is a prime number.
(3) When $p=5$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=2+1=3$ is a prime number.
(4) When $p>5$, since $p$ is a prime number, $p=6k+1$ or $6k+5$.
If $p=6k+1$, then $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=3k+2k+k=6k$, which is not a prime number, contradiction;
If $p=6k+5$, then $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=(3k+2)+(2k+1)+k=6k+3$, which is not a prime number, contradiction.
Therefore, the number of prime pairs $(p, q)$ is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $f(x)$ is a function defined on $\mathbf{R}$, $f\left(\frac{\pi}{4}\right)=0$, and for any $x, y \in \mathbf{R}$, we have
$$
f(x)+f(y)=2 f\left(\frac{x+y}{2}\right) f\left(\frac{x-y}{2}\right) .
$$
Then $f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left(\frac{2007 \pi}{4}\right)$ $=$ . $\qquad$
|
6.0 .
Let $\frac{x-y}{2}=\frac{\pi}{4}$, then
$$
\begin{array}{l}
f(x)+f\left(x-\frac{\pi}{2}\right)=2 f\left(x-\frac{\pi}{4}\right) f\left(\frac{\pi}{4}\right)=0 . \\
\text { Therefore } f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left(\frac{2007 \pi}{4}\right)=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points) Find the maximum value of a prime number $p$ with the following property: there exist two permutations (which can be the same) of $1, 2, \cdots, p$, $a_{1}, a_{2}, \cdots, a_{p}$ and $b_{1}, b_{2}, \cdots, b_{p}$, such that the remainders of $a_{1} b_{1}$, $a_{2} b_{2}, \cdots, a_{p} b_{p}$ when divided by $p$ are all distinct.
|
Lemma (Wilson's Theorem): For any prime $p$, we have $(p-1)! \equiv -1 \pmod{p}$.
Proof of the Lemma: When $p=2,3$, equation (1) is obviously true.
When $p>3$, we prove: For any $2 \leqslant k \leqslant p-2$, there must exist $k^{\prime}\left(2 \leqslant k^{\prime} \leqslant p-2, k^{\prime} \neq k\right)$, such that $k k^{\prime} \equiv 1 \pmod{p}$.
In fact, for $2 \leqslant k \leqslant p-2$, we have $(k, p)=1$.
Thus, $k, 2k, \cdots, pk$ form a complete residue system modulo $p$.
Therefore, there must exist $k^{\prime}\left(1 \leqslant k^{\prime} \leqslant p\right)$, such that $k k^{\prime} \equiv 1 \pmod{p}$.
Clearly, $k^{\prime} \neq p$, otherwise $k k^{\prime} \equiv 0 \pmod{p}$, which is a contradiction.
Moreover, if $k^{\prime}=1$, then by $k k^{\prime} \equiv 1 \pmod{p}$, we get $k \equiv 1 \pmod{p}$, which contradicts $2 \leqslant k \leqslant p-2$.
If $k^{\prime}=p-1$, then by $k(p-1) \equiv 1 \pmod{p}$, we get $-k \equiv 1 \pmod{p}$, which contradicts $2 \leqslant k \leqslant p-2$.
If $k^{\prime}=k$, by $k k^{\prime} \equiv 1 \pmod{p}$, we get $k^2 \equiv 1 \pmod{p}$. Therefore, $p \mid (k+1)(k-1)$, but $1 \leqslant k-1 \leqslant p-3$, which contradicts equation (2), so $p \leqslant 2$.
When $p=2$, let
$$
a_{1}=1, a_{2}=2, b_{1}=1, b_{2}=2.
$$
At this point, $a_{1} b_{1}=1=1 \pmod{2}$,
$$
a_{2} b_{2}=4 \equiv 0 \pmod{2}.
$$
Therefore, $p=2$ satisfies the condition. Hence, $p_{\max}=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) $n$ people exchange greetings by phone during a holiday. It is known that each person makes calls to at most three friends, any two people make at most one call to each other, and among any three people, at least two of them, one person has called the other. Find the maximum value of $n$.
|
Three, first prove the lemma.
Lemma In a simple graph $G$ of order $n$ without $K_{3}$, then
$$
f(G) \leqslant\left[\frac{n^{2}}{4}\right],
$$
where $f(G)$ represents the number of edges in $G$.
Proof of the lemma: Let $A$ be the vertex with the maximum degree, and let the set of vertices adjacent to $A$ be $M=\left\{A_{1}, A_{2}, \cdots, A_{r}\right\}$, and the set of vertices not adjacent to $A$ be $N=\left\{B_{1}, B_{2}, \cdots, B_{s}\right\}(r+s+1=n)$. Since $G$ contains no triangles, there are no edges in $M$. Thus, the other edges of $G$ are either in $N$ or between $M$ and $N$. These edges are all induced by the vertices $B_{1}, B_{2}, \cdots, B_{s}$. Therefore,
$$
\begin{array}{l}
f(G) \leqslant d(A)+d\left(B_{1}\right)+d\left(B_{2}\right)+\cdots+d\left(B_{s}\right) \\
\leqslant r+\underbrace{r+\cdots+r}_{s}=(s+1) r \leqslant\left(\frac{s+r+1}{2}\right)^{2}=\frac{n^{2}}{4} .
\end{array}
$$
Since $f(G) \in \mathbf{Z}$, we have
$$
f(G) \leqslant\left[\frac{n^{2}}{4}\right] \text {. }
$$
Now, prove the original problem.
Use $n$ points to represent $n$ people. If a person $A$ calls another person $B$'s home phone, then draw a directed edge from $A$ to $B$, resulting in a simple directed graph $G$.
On one hand, $\bar{G}$ contains no triangles. By the lemma,
$f(\bar{G}) \leqslant\left[\frac{n^{2}}{4}\right]$.
Thus, $f(G)=\mathrm{C}_{n}^{2}-f(\bar{G})$
$\geqslant C_{n}^{2}-\left[\frac{n^{2}}{4}\right]=\left[\frac{(n-1)^{2}}{4}\right]$.
On the other hand, $f(G)=\sum_{i=1}^{n} d^{+}\left(x_{i}\right) \leqslant \sum_{i=1}^{n} 3=3 n$.
Therefore, $\left[\frac{(n-1)^{2}}{4}\right] \leqslant 3 n$.
When $n$ is odd, equation (1) becomes $\frac{(n-1)^{2}}{4} \leqslant 3 n$, solving for $n \leqslant 13$;
When $n$ is even, equation (1) becomes $\frac{n^{2}-2 n}{4} \leqslant 3 n$, solving for $n \leqslant 14$.
In summary, $n \leqslant 14$.
Finally, $n=14$ is possible. Construct two $K_{7}$, for each heptagon $A_{1} A_{2} \cdots A_{7}$, let $A_{i}$ point to $A_{i+1}, A_{i+2}, A_{i+3}(i=1,2, \cdots, 7, A_{i+7}=A_{i})$, then the graph meets the conditions.
First, each vertex as a starting point exactly induces 3 directed edges, thus, each person calls at most 3 friends' home phones.
Second, for any three vertices, by the pigeonhole principle, there must be two vertices $A_{i} 、 A_{j}(i<j)$ such that $A_{j}$ calls $A_{i}$'s home phone.
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M$ and $N$ are points on the diagonals $A C$ and $B F$ respectively, and $A M=F N$. If $A B=1$, then the maximum value of $M N$ is $\qquad$
|
5.1.
As shown in Figure 2, draw $M P \perp A B$ at $P$, and connect $P N$. It can be proven that $P N \perp A B$. Thus, $\angle M P N=120^{\circ}$.
Let $A M=F N=x$.
Therefore, $\frac{M P}{1}=\frac{A M}{\sqrt{2}}$,
which means $M P=\frac{\sqrt{2}}{2} x$.
Hence, $P N=\frac{\sqrt{2}-x}{\sqrt{2}}\left(\frac{P N}{1}=\frac{B N}{\sqrt{2}}=\frac{B F-N F}{\sqrt{2}}=\frac{\sqrt{2}-x}{\sqrt{2}}\right)$.
Then, $M N^{2}=M P^{2}+P N^{2}-2 P M \cdot P N \cos 120^{\circ}$
$$
\begin{array}{l}
=\frac{x^{2}}{2}+\left(\frac{\sqrt{2}-x}{\sqrt{2}}\right)^{2}-2 \frac{\sqrt{2}}{2} x \cdot \frac{\sqrt{2}-x}{\sqrt{2}} \cos 120^{\circ} \\
=\frac{1}{2}\left(x^{2}-\sqrt{2} x+2\right) \\
=\frac{1}{2}\left(x-\frac{\sqrt{2}}{2}\right)^{2}+\frac{3}{4}(0 \leqslant x \leqslant \sqrt{2}) .
\end{array}
$$
When $x=0$ or $x=\sqrt{2}$, $M N_{\text {max }}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $\cos \beta+\sin \beta \cdot \cot \theta=\tan 54^{\circ}$, $\cos \beta-\sin \beta \cdot \cot \theta=\tan 18^{\circ}$.
Then the value of $\tan ^{2} \theta$ is $\qquad$
|
6.1
From the given conditions, we have
$\cos \beta=\frac{1}{2}\left(\tan 54^{\circ}+\tan 18^{\circ}\right)=\frac{1}{2 \cos 54^{\circ}}$,
$\sin \beta=\frac{\tan \theta}{2}\left(\tan 54^{\circ}-\tan 18^{\circ}\right)=\frac{\tan \theta}{2 \cos 18^{\circ}}$.
Since $\sin ^{2} \beta+\cos ^{2} \beta=1$, we have
$\frac{1}{\cos ^{2} 54^{\circ}}+\frac{\tan ^{2} \theta}{\cos ^{2} 18^{\circ}}=4$.
Thus, $\tan ^{2} \theta=\left(4-\frac{1}{\cos ^{2} 54^{\circ}}\right) \cos ^{2} 18^{\circ}$
$$
\begin{array}{l}
=4 \cos ^{2} 18^{\circ}-\frac{\sin ^{2} 72^{\circ}}{\sin ^{2} 36}=4 \cos ^{2} 18^{\circ}-4 \cos ^{2} 36^{\circ} \\
=4\left(\cos 18^{\circ}+\cos 36^{\circ}\right)\left(\cos 18^{\circ}-\cos 36^{\circ}\right)
\end{array}
$$
$$
\begin{array}{l}
=16 \cos 27^{\circ} \cdot \cos 9^{\circ} \cdot \sin 27^{\circ} \cdot \sin 9^{\circ} \\
=\frac{4 \sin 18^{\circ} \cdot \cos 18^{\circ} \cdot \sin 54^{\circ}}{\cos 18^{\circ}}=\frac{2 \sin 36^{\circ} \cdot \cos 36^{\circ}}{\cos 18^{\circ}} \\
=\frac{\sin 72^{\circ}}{\cos 18^{\circ}}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { 2. Let } M=\frac{2 \cos 34^{\circ}-\cos 22^{\circ}}{\cos 14^{\circ}} \text {, } \\
N=\sin 56^{\circ} \cdot \sin 28^{\circ} \cdot \sin 14^{\circ} \text {. } \\
\text { Then } \frac{M}{N}=
\end{array}
$$
|
2.8.
$$
\begin{array}{l}
M=\frac{2 \cos 34^{\circ}-\sin 68^{\circ}}{\cos 14^{\circ}}=\frac{2 \cos 34^{\circ}\left(1-\sin 34^{\circ}\right)}{\cos 14^{\circ}} \\
=\frac{2 \cos 34^{\circ}\left(1-\cos 56^{\circ}\right)}{\cos 14^{\circ}}=\frac{4 \cos 34^{\circ} \cdot \sin ^{2} 28^{\circ}}{\cos 14^{\circ}} \\
=8 \cos 34^{\circ} \cdot \sin 28^{\circ} \cdot \sin 14^{\circ}=8 N .
\end{array}
$$
Therefore, $\frac{M}{N}=8$.
Note: A more general conclusion of this problem is
$$
\frac{2 \sin 4 \alpha-\sin 8 \alpha}{\cos \alpha}=8 \sin \alpha \cdot \sin 2 \alpha \cdot \sin 4 \alpha \text{. }
$$
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. From the 6 face diagonals of the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, two tetrahedra $A-B_{1} C D_{1}$ and $A_{1}-B C_{1} D$ can be formed. If the combined volume of the two tetrahedra is 1 (overlapping parts are counted only once), then the volume of the rectangular prism is $\qquad$ .
|
3.2.
Let the length, width, and height of the rectangular prism be $a$, $b$, and $c$ respectively, then the volume of the rectangular prism $V_{0}=a b c$. The volumes of the two tetrahedra (as shown in Figure 4) are both
$$
\begin{array}{l}
V_{1}=V_{0}-V_{A_{1}-A B D}-V_{A_{1}-B B C_{1}}-V_{A_{1}-D_{1} D C_{1}}-V_{C_{1}-B C D} \\
=V_{0}-4 \times \frac{1}{6} a b c=\frac{1}{3} V_{0} .
\end{array}
$$
The common part of the two tetrahedra is an octahedron with vertices at the centers of the 6 faces (as shown in Figure 5), and its volume is
$$
V_{2}=\frac{1}{3} \times \frac{1}{2} S_{\text {quadrilateral } \triangle B C D} \cdot h=\frac{1}{6} V_{0} .
$$
Therefore, the total volume occupied by the two tetrahedra is
$$
1=2 V_{1}-V_{2}=\frac{2}{3} V_{0}-\frac{1}{6} V_{0}=\frac{1}{2} V_{0} \Rightarrow V_{0}=2 .
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Arrange all positive divisors of 8128 in ascending order as $a_{1}, a_{2}, \cdots, a_{n}$, then $\sum_{k=1}^{n} k a_{k}=$ $\qquad$ .
|
5.211335 .
Given $8128=2^{6}\left(2^{7}-1\right)$, we have
$$
\begin{array}{l}
a_{n}=\left\{\begin{array}{ll}
2^{n-1}, & 1 \leqslant n \leqslant 7 ; \\
2^{n-8}\left(2^{7}-1\right), & 8 \leqslant n \leqslant 14 .
\end{array}\right. \\
\text { Then } \sum_{k=1}^{n} k a_{k}=\sum_{k=1}^{14} k \cdot 2^{k-1}-\sum_{k=1}^{7}(k+7) 2^{k-1} . \\
\text { Also } \sum_{k=1}^{14} k \cdot 2^{k-1}=\sum_{k=1}^{14}\left(k \cdot 2^{k}-k \cdot 2^{k-1}\right) \\
=14 \times 2^{14}+\sum_{k=1}^{13}\left[k \cdot 2^{k}-(k+1) 2^{k}\right]-1 \\
=14 \times 2^{14}-\sum_{k=0}^{13} 2^{k}=14 \times 2^{14}-\left(2^{14}-1\right) \\
=13 \times 2^{14}+1, \\
\sum_{k=1}^{7}(k+7) 2^{k-1}=13 \times 2^{7}-6, \\
\text { Therefore } \sum_{k=1}^{14} k a_{k}=\left(13 \times 2^{14}+1\right)-\left(13 \times 2^{7}-6\right) \\
=13 \times 2^{7}\left(2^{7}-1\right)+7=211335 .
\end{array}
$$
|
211335
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Among the 8 vertices, 6 centers of the faces (bottom faces), and the center of the body, a total of 15 points, if a plane formed by any 3 different and non-collinear points is perpendicular to a line formed by another 2 different points, then these 5 points are called an "orthogonal 5-point group". Therefore, the total number of orthogonal 5-point groups formed by these 15 points is ( ).
(A)366
(B) 516
(C) 1428
(D) 1824
|
6.D.
As shown in Figure 6, let $O$ be the center of the body, and $O_{i} (i = 1,2, \cdots, 6)$ be the centers of the respective faces.
Classify and discuss according to the plane formed by the 3 points (referred to as the "3-point plane").
Observation shows that there are only 4 possible 3-point planes: side faces (such as $A_{1} B_{1} C_{1} D_{1}$), diagonal faces (such as $A_{1} C_{1} C A$), equilateral triangles formed by three vertices (such as $\triangle A B_{1} C$), and central faces (such as $O_{3} O_{5} O_{4} O_{6}$).
(1) When the side face is $A_{1} B_{1} C_{1} D_{1}$, the points $A_{1}, B_{1}, C_{1}, D_{1}$ can form 4 non-collinear 3-point groups (3-point planes, hereinafter the same), and the points $O_{2}, O, O_{1}$ can form 3 2-point groups. Thus, the 3-point groups and 2-point groups can combine to form $4 \times 3 = 12$ orthogonal 5-point groups. Additionally, by taking point $O_{1}$ and any two points from $\{A_{1}, B_{1}, C_{1}, D_{1}\}$, 4 non-collinear 3-point groups can be formed. For example, $\{O_{1}, A_{1}, B_{1}\}$, and the 2-point groups perpendicular to this 3-point group are $\{O_{2}, O\}$, $\{D, D_{1}\}$, $\{C, C_{1}\}$. Therefore, $4 \times 3 = 12$ orthogonal 5-point groups can be formed.
Noting that there are 6 side faces, the total number of orthogonal 5-point groups formed by 3-point groups from side faces is $6 \times (12 + 12) = 144$.
(2) When the diagonal face is $A_{1} C_{1} C A$, the 3-point groups formed by $\{A, A_{1}, C_{1}, C, O_{1}, O, O_{2}\}$ can form $\left(C_{7}^{3} - 5\right) \times 4 + \left(C_{6}^{3} - 3\right) \times 4 = 188$ orthogonal 5-point groups.
Since there are 6 diagonal faces, a total of $6 \times 188 = 1128$ orthogonal 5-point groups of this type are formed.
(3) When the equilateral triangle is $\triangle A B_{1} C$, the 3-point groups are formed by $\{A, B_{1}, C, O_{2}, O_{3}, O_{5}\}$, and the 2-point groups are formed by $\{D_{1}, O, B\}$. Therefore, there are $\left(C_{6}^{3} - 3\right) \times C_{3}^{2} = 51$ orthogonal 5-point groups.
Since there are 8 equilateral triangles, a total of $8 \times 51 = 408$ orthogonal 5-point groups of this type are formed.
(4) When the central face is $O_{3} O_{5} O_{4} O_{6}$, the 3-point groups are formed by $\{O, O_{3}, O_{5}, O_{4}, O_{6}\}$, and there are $\left(C_{5}^{3} - 2\right) \times 5 + C_{4}^{3} \times 2 = 48$ orthogonal 5-point groups.
Since there are 3 central faces, a total of $48 \times 3 = 144$ orthogonal 5-point groups of this type are formed.
In summary, there are a total of $144 + 1128 + 408 + 144 = 1824$ orthogonal 5-point groups.
|
1824
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. Among all the eight-digit numbers formed by the digits $1,2, \cdots, 8$ without repetition, the number of those divisible by 11 is. $\qquad$
|
3.4608 .
Since there are 4 odd numbers in $1,2, \cdots, 8$, the algebraic sum after arbitrarily adding “+” and “-” signs is always even. Because the difference between the sum of the largest four numbers and the sum of the smallest four numbers in $1,2, \cdots, 8$ is no more than 16, for each eight-digit number that meets the conditions, the sum of the four digits in the odd positions must equal the sum of the four digits in the even positions. Since $1+2+\cdots+8=36$, divide $1,2, \cdots, 8$ into two groups of four numbers each, with the sum of each group being 18, and consider the group containing 8, the sum of the other three numbers in the group is 10, there are only four cases:
$$
(1,2,7,8),(1,3,6,8),(1,4,5,8),(2,3,5,8) \text {. }
$$
For each case, the group containing 8 can be placed in the odd positions or the even positions, resulting in $2 \times 4!\times 4!$ numbers.
In the four cases, a total of $8 \times 4!\times 4!=4608$ eight-digit numbers that meet the conditions are obtained.
|
4608
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points) Define a "Hope Set" (Hope Set) abbreviated as HS as follows: HS is a non-empty set that satisfies the condition "if $x \in \mathrm{HS}$, then $2 x \notin \mathrm{HS}$". How many "Hope Subsets" are there in the set $\{1,2, \cdots, 30\}$? Please explain your reasoning.
|
Sure, here is the translation:
```
II. Below, “ $a \longmapsto 2 a$ ” represents the doubling relationship between $a$ and $2 a$. Notice:
$\left\{\begin{array}{ll}\text { (1) } & 15 \longmapsto 30, \\ \text { (2) } & 13 \longmapsto 26, \\ \text { (3) } & 11 \longmapsto 22, \\ \text { (4) } & 9 \longmapsto 18,\end{array}\right.$
$\left\{\begin{array}{ll}(5) & 7 \longmapsto 14,14 \longmapsto 28, \\ \text { (6) } & 5 \longmapsto 10,10 \longmapsto 20,\end{array}\right.$
(7) $3 \longmapsto 6,6 \longmapsto 12,12 \longrightarrow 24$,
(8) $1 \longmapsto 2,2 \longmapsto 4,4 \longmapsto 8,8 \longmapsto 16$.
Clearly, $17,19,21,23,25,27,29$ do not affect whether HS is a hopeful subset (since these numbers are not divisible by 2, and each number's double is greater than 30), so there are $2^{7}$ ways to assign these 7 numbers.
In (1), 15 and 30 cannot both be taken, so there are $2^{2}-1=3$ ways.
Similarly, in (2), (3), and (4), there are 3 ways each.
Next, we use a recursive algorithm.
In (5), if 7 is taken, then 14 cannot be taken, and 28 can be taken or not, giving 2 ways; if 7 is not taken, then by (1), there are 3 ways for 14 and 28 (the situation for 14 and 28 is the same as in (1)). Therefore, there are $2+3=5$ ways in (5).
Similarly, there are 5 ways in (6).
In (7), if 3 is taken, then 6 cannot be taken, and by (1), there are 3 ways for 12 and 24; if 3 is not taken, then by (5), there are 5 ways for $6,12,24$. Therefore, there are $3+5=8$ ways in (7).
In (8), if 1 is taken, then 2 cannot be taken, and by (5), there are 5 ways for 4,8,16; if 1 is not taken, then by (7), there are 8 ways for $2,4,8,16$. Therefore, there are $5+8=13$ ways in (8).
Considering the empty set $\varnothing$ (i.e., none of $1,2, \cdots, 30$ are taken), the number of hopeful subsets of $\{1,2, \cdots, 30\}$ is
$$
2^{7} \times 3^{4} \times 5^{2} \times 8 \times 13-1=26956799 \text {. }
$$
```
|
26956799
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the sum of the squares of the first 101 positive integers starting from a positive integer $k$ be equal to the sum of the squares of the next 100 positive integers. Then the value of $k$ is $\qquad$ .
|
2.20100.
From the problem, we get
$$
k^{2}+\sum_{i=1}^{100}(k+i)^{2}=\sum_{i=1}^{100}(k+100+i)^{2} \text {. }
$$
Then $k^{2}=\sum_{i=1}^{100}\left[(k+100+i)^{2}-(k+i)^{2}\right]$
$$
=200 \sum_{i=1}^{100}(k+50+i)=10000(2 k+201) \text {. }
$$
Thus, $k^{2}-20000 k-2010000=0$.
Taking the positive root, we get $k=20100$.
|
20100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. For any real numbers $a, b$, the inequality
$$
\max \{|a+b|,|a-b|,|2006-b|\} \geqslant c
$$
always holds, then the maximum value of the constant $c$ is $\qquad$ (where, $\max \{x, y, z\}$ denotes the maximum of $x, y, z$).
|
5.1003 .
Let $\max \{|a+b|,|a-b|,|2006-b|\}=M$, then $M \geqslant|a+b|, M \geqslant|a-b|, M \geqslant|2006-b|$.
Therefore, $4 M \geqslant|a+b|+|b-a|+2|2006-b|$
$$
\geqslant|(a+b)+(b-a)+2(2006-b)|=4012,
$$
which means $M \geqslant 1003$.
When $a=0, b=1003$, $M=1003$.
Thus, the minimum value of $M$ is 1003.
Therefore, the maximum value of $c$ is 1003.
|
1003
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (14 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=1$, and for $n \geqslant 2$,
$a_{n}=\left\{\begin{array}{ll}a_{\frac{n}{2}}+1, & \text { when } n \text { is even; } \\ \frac{1}{a_{n-1}}, & \text { when } n \text { is odd. }\end{array}\right.$
It is known that $a_{n}=\frac{30}{19}$. Find the positive integer $n$.
|
10. From the given conditions, it is easy to know that $a_{n}>0(n=1,2, \cdots)$.
From $a_{1}=1$, we know that:
when $n$ is even, $a_{n}>1$;
when $n(n>1)$ is odd, $a_{n}=\frac{1}{a_{n-1}}1$, so $n$ is even.
Thus, $a_{\frac{n}{2}}=\frac{30}{19}-1=\frac{11}{19}1, \frac{n}{2}-1$ is even;
$a_{\frac{n-2}{4}}=\frac{19}{11}-1=\frac{8}{11}1, \frac{n-6}{4}$ is even;
$a_{\frac{n-6}{8}}=\frac{11}{8}-1=\frac{3}{8}1, \frac{n-14}{8}$ is even;
$a_{\frac{n-14}{16}}=\frac{8}{3}-1=\frac{5}{3}>1, \frac{n-14}{16}$ is even;
$a_{\frac{n-14}{32}}=\frac{5}{3}-1=\frac{2}{3}1, \frac{n-46}{32}$ is even;
$a_{\frac{n-46}{64}}=\frac{3}{2}-1=\frac{1}{2}1, \frac{n-110}{64}$ is even;
$a_{\frac{n-110}{128}}=2-1=1$.
Therefore, $\frac{n-110}{128}=1$. Solving for $n$ gives $n=238$.
|
238
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 There is a sequence of numbers, the 1st number is 105, the 2nd number is 85, starting from the 3rd number, each number is the average of the two numbers before it. What is the integer part of the 19th number?
|
Notice that "the average of any two numbers $a$ and $b$, $\frac{a+b}{2}$, corresponds to the midpoint of the segment between the points representing $a$ and $b$ on the number line," we can consider placing these 19 numbers on the number line, as shown in Figure 2.
From the property mentioned above, starting from the 6th number, the numbers represented on the number line are all between 91 and 92. Therefore, the integer part of the 19th number is 91.
|
91
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x=\frac{1}{2-\sqrt{5}}$. Then $x^{3}+3 x^{2}-5 x+1=$
|
From $x=\frac{1}{2-\sqrt{5}}=-2-\sqrt{5}$, we get $x+2=-\sqrt{5}$. Therefore, $(x+2)^{2}=5$, which means $x^{2}+4 x=1$.
Then $x^{3}+3 x^{2}-5 x+1=\left(x^{2}+4 x-1\right)(x-1)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The quadratic equation in $x$
$$
6 x^{2}-(2 m-1) x-(m+1)=0
$$
has a root $\alpha$, given that $\alpha$ satisfies $|\alpha| \leqslant 2000$, and makes $\frac{3}{5} \alpha$ an integer. Then the number of possible values for $m$ is $\qquad$.
|
3.2 401 .
From the quadratic formula, we get
$$
x_{1,2}=\frac{(2 m-1) \pm(2 m+5)}{12},
$$
which gives the roots as $-\frac{1}{2}$ and $\frac{m+1}{3}$.
If $\alpha=-\frac{1}{2}$, then $\frac{3}{5} \alpha=-\frac{3}{10}$ is not an integer. Therefore,
$$
\alpha=\frac{m+1}{3} \text {. }
$$
Given $|\alpha| \leqslant 2000$, we have $-2000 \leqslant \frac{m+1}{3} \leqslant 2000$, which implies $-6000 \leqslant m+1 \leqslant 6000$, and $\alpha$ must make $\frac{3}{5} \alpha=\frac{m+1}{5}$ an integer. Thus,
$$
\begin{array}{l}
m=-1+5 n(n=0, \pm 1, \pm 2, \cdots) . \\
\text { Hence }-6000 \leqslant(-1+5 n)+1 \leqslant 6000 .
\end{array}
$$
Solving this, we get $-1200 \leqslant n \leqslant 1200$.
Thus, $n=0, \pm 1, \pm 2, \cdots, \pm 1200$.
Therefore, the number of possible values for $m$ is $1200 \times 2+1=2401$.
|
2401
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Sustainable development has become a common concern for human development, and China has made it a basic national policy. To protect the environment and maintain ecological balance, a policy of zoned grazing has been implemented in the northern pastoral areas of China. Mr. Li, a herdsman, has a grassland that can be used for raising cattle and sheep. However, once the grass on the grassland is eaten up, the grassland will lose its ability to regenerate and will no longer be able to support cattle and sheep. If the grassland is used to raise 200 cattle and 200 sheep, the grass will be eaten up in 10 days; if it is used to raise 100 cattle and 200 sheep, the grass will be eaten up in 20 days; if it is used to raise 50 cattle and 300 sheep, the grass will be eaten up in 30 days. Cattle and sheep need to be raised for a full year before they can be sold (otherwise, the herdsman can only produce and sell them himself), and when sold, each cow is worth 2500 RMB, and each sheep is worth 300 RMB. If Mr. Li wants to maximize his income from raising cattle and sheep in one year, please design a feasible breeding plan for Mr. Li.
(Note: The original amount of grass on the grassland should not decrease after one year.)
|
Let the amount of grass consumed by each cow and each sheep per day be $x$ and $y$, respectively, the amount of grass that grows on the pasture each day be $z$, and the original amount of grass on the pasture be $A$. According to the problem, we have
$$
\left\{\begin{array}{l}
200 \times 10 x + 200 \times 10 y = 10 z + A, \\
100 \times 20 x + 200 \times 20 y = 20 z + A, \\
50 \times 30 x + 300 \times 30 y = 30 z + A.
\end{array}\right.
$$
Solving these equations, we get $x = 6y$, $z = 200y$, and $A = 12000y$.
To maximize the profit after one year of breeding, it is necessary to ensure that the total amount of grass consumed by the cows and sheep each day does not exceed the amount of grass that the pasture can regenerate. Otherwise, the pasture will lose its ability to regenerate, and the cows and sheep will have no grass to eat. To maximize the profit, let Mr. Li raise $m$ cows and $n$ sheep, then $mx + ny \leq z$. At this point, Mr. Li's income is $w = 2500m + 300n$.
Therefore, $6my + ny \leq 200y \Rightarrow 6m + n \leq 200$.
Clearly, since $x = 6y$, the amount of grass consumed by one cow per day is equivalent to the amount consumed by 6 sheep per day, but the value of one cow is 2500 RMB, while 6 sheep are worth $300 \times 6 = 1800$ RMB. Therefore, to maximize the profit after one year, Mr. Li should raise as many cows as possible. Since $6m + n \leq 200$, when $m = 33$ and $n = 2$, Mr. Li's profit is maximized, at which point the profit is
$$
w = 2500 \times 33 + 300 \times 2 = 83100 \text{ RMB.}
$$
Therefore, to maximize his profit after one year, Mr. Li can raise 33 cows and 2 sheep, at which point the maximum profit is 83100 RMB.
|
83100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $a, b, x, y$ satisfy
$$
\begin{array}{l}
a x+b y=3, a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, a x^{4}+b y^{4}=42 .
\end{array}
$$
Then $a x^{5}+b y^{5}=$ $\qquad$
|
4.20 .
$$
\begin{array}{l}
\text { Given } a x^{3}+b y^{3}=16 \\
\Rightarrow\left(a x^{3}+b y^{3}\right)(x+y)=16(x+y) \\
\Rightarrow\left(a x^{4}+b y^{4}\right)+x y\left(a x^{2}+b y^{2}\right)=16(x+y) \\
\Rightarrow 42+7 x y=16(x+y), \\
a x^{2}+b y^{2}=7 \\
\Rightarrow\left(a x^{2}+b y^{2}\right)(x+y)=7(x+y) \\
\Rightarrow\left(a x^{3}+b y^{3}\right)+x y(a x+b y)=7(x+y) \\
\Rightarrow 16+3 x y=7(x+y),
\end{array}
$$
Solving equations (1) and (2) simultaneously, we get $x+y=-14, x y=-38$.
$$
\begin{array}{l}
\text { Then } a x^{4}+b y^{4}=42 \\
\Rightarrow\left(a x^{4}+b y^{4}\right)(x+y)=42(x+y) \\
\Rightarrow\left(a x^{5}+b y^{5}\right)+x y\left(a x^{3}+b y^{3}\right)=42(x+y) \\
\Rightarrow a x^{5}+b y^{5}=42(x+y)-16 x y=20 .
\end{array}
$$
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let for any natural numbers $m$, $n$ satisfying $\frac{m}{n}<\sqrt{7}$, the inequality $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\lambda}{n^{2}}$ always holds. Then the maximum value of $\lambda$ is $\qquad$.
|
5.3.
The original inequality $\Leftrightarrow 7 n^{2}-m^{2} \geqslant \lambda$.
Since $7 n^{2} \equiv 0(\bmod 7), m^{2} \equiv 0,1,2,4(\bmod 7)$, then $\left(7 n^{2}-m^{2}\right)_{\text {min }}=3$, so, $\lambda_{\text {max }}=3$.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (20 points) Given the semi-ellipse $\frac{x^{2}}{4}+y^{2}=1(y>0)$, two perpendicular lines are drawn through a fixed point $C(1,0)$ intersecting the ellipse at points $P$ and $Q$, respectively. Here, $O$ is the origin, and $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse.
(1) Find the minimum value of $\left|P F_{1}+P F_{2}\right|$;
(2) Does there exist a line such that the sum of the x-coordinates of $P F_{1}$ and $P F_{2}$ is minimized? If not, explain why.
|
(1) Since $P O$ is the median of $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}+P F_{2}=2 P O, \left|P F_{1}+P F_{2}\right|=2|P O| \text {. }
$$
Therefore, when $P$ is at the vertex of the minor axis, $\left|P F_{1}+P F_{2}\right|$ achieves its minimum value of 2.
(2) From the problem analysis, the slopes of the lines $C P$ and $C Q$ both exist. Let the slopes of the lines $C P$ and $C Q$ be
$$
k_{C P}=k, k_{C Q}=-\frac{1}{k} \text {. }
$$
Then the equations of these two lines are
$$
l_{C P}: y=k(x-1), l_{C Q}: y=-\frac{1}{k}(x-1) \text {. }
$$
Since $P F_{1}+P F_{2}=2 P O, Q F_{1}+Q F_{2}=2 Q O$, it is sufficient to satisfy $O P \perp O Q$.
Then $k_{O P} k_{O Q}=\frac{k\left(x_{P}-1\right)}{x_{P}} \cdot \frac{-\frac{1}{k}\left(x_{Q}-1\right)}{x_{Q}}=-1$
$$
\Rightarrow x_{P}+x_{Q}=1 \text {. }
$$
From the ellipse equation, we get $y_{P}=\frac{\sqrt{4-x_{P}^{2}}}{2}, y_{Q}=\frac{\sqrt{4-x_{Q}^{2}}}{2}$, then
$$
\begin{array}{l}
k_{O P} k_{Q Q}=\frac{\sqrt{4-x_{P}^{2}}}{2 x_{P}} \cdot \frac{\sqrt{4-x_{Q}^{2}}}{2 x_{Q}}=-1 \\
\Rightarrow \sqrt{16-4\left(x_{P}^{2}+x_{Q}^{2}\right)+x_{P}^{2} x_{Q}^{2}}=-4 x_{P} x_{Q} . \\
\text { Let } x_{P} x_{Q}=t \leqslant 0, \text { and } x_{P}+x_{Q}=1 .
\end{array}
$$
Let $x_{P} x_{Q}=t \leqslant 0$, and $x_{P}+x_{Q}=1$.
Thus $\sqrt{16-4(1-2 t)+t^{2}}=-4 t$
$$
\Rightarrow 15 t^{2}-8 t-12=0 \text {. }
$$
Solving, we get $t=-\frac{2}{3}$ or $t=\frac{6}{5}$ (discard).
Then $\left\{\begin{array}{l}x_{p}+x_{Q}=1, \\ x_{p} x_{Q}=-\frac{2}{3}\end{array}\right.$.
Solving, we get $\left\{\begin{array}{l}x_{P}=\frac{3+\sqrt{33}}{6}, \\ x_{Q}=\frac{3-\sqrt{33}}{6} ;\end{array}\left\{\begin{array}{l}x_{P}=\frac{3-\sqrt{33}}{6}, \\ x_{Q}=\frac{3+\sqrt{33}}{6}\end{array}\right.\right.$.
Therefore, there exist lines such that $\boldsymbol{P F} F_{1}+P F_{2}$ and $\boldsymbol{Q} F_{1}+Q F_{2}$ are perpendicular to each other.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
204 Given that $a, b, c$ are positive numbers satisfying $a+b+c=1$. Prove:
$$
\frac{a^{3}+b}{a(b+c)}+\frac{b^{3}+c}{b(c+a)}+\frac{c^{3}+a}{c(a+b)} \geqslant 5 .
$$
|
Prove: $\frac{a^{3}+b}{a(b+c)}=\frac{a^{2}(1-b-c)+b}{a(b+c)}$
$$
\begin{array}{l}
=\frac{a^{2}+b}{a(b+c)}-a=\frac{a(1-b-c)+b}{a(b+c)}-a \\
=\frac{a+b}{a(b+c)}-1-a .
\end{array}
$$
Similarly, $\frac{b^{3}+c}{b(c+a)}=\frac{b+c}{b(c+a)}-1-b$,
$$
\frac{c^{3}+a}{c(a+b)}=\frac{c+a}{c(a+b)}-1-c \text {. }
$$
Adding the above three equations, we get
$$
\begin{array}{l}
\frac{a^{3}+b}{a(b+c)}+\frac{b^{3}+c}{b(c+a)}+\frac{c^{3}+a}{c(a+b)} \\
=\frac{a+b}{a(b+c)}+\frac{b+c}{b(c+a)}+\frac{c+a}{c(a+b)}-4 \\
\geqslant 3 \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{a(b+c) b(c+a) c(a+b)}}-4 \\
=\frac{3}{\sqrt[3]{a b c}}-4 \geqslant \frac{9}{a+b+c}-4=5 .
\end{array}
$$
Therefore, the original inequality holds.
|
5
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 For all $a, b, c \in \mathbf{R}_{+}$, find
$$
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}}
$$
the minimum value.
|
Explanation: Make the substitution
$$
\begin{array}{l}
x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 a c}}, \\
z=\frac{c}{\sqrt{c^{2}+8 a b}} .
\end{array}
$$
Then $x, y, z \in (0, +\infty)$.
Thus, $x^{2}=\frac{a^{2}}{a^{2}+8 b c}$, which means $\frac{1}{x^{2}}-1=\frac{8 b c}{a^{2}}$.
Similarly, $\frac{1}{y^{2}}-1=\frac{8 a c}{b^{2}}, \frac{1}{z^{2}}-1=\frac{8 a b}{c^{2}}$.
Multiplying the above three equations yields
$$
\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right)\left(\frac{1}{z^{2}}-1\right)=512 \text{. }
$$
If $x+y+z\frac{\prod\left[\left(\sum x\right)^{2}-x^{2}\right]}{x^{2} y^{2} z^{2}} \\
=\frac{\prod[(y+z)(2 x+y+z)]}{x^{2} y^{2} z^{2}} \\
\geqslant \frac{\prod\left(2 \sqrt{y z} \cdot 4 \sqrt[4]{x^{2} y z}\right)}{x^{2} y^{2} z^{2}} \\
=\frac{\prod\left(8 \sqrt{x^{2} y^{3} z^{3}}\right)}{x^{2} y^{2} z^{2}}=512 .
\end{array}
$$
This leads to a contradiction.
Therefore, $x+y+z \geqslant 1$.
Thus, when $a=b=c$, the minimum value is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Find
$$
\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}
$$
the minimum value.
|
Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}(x, y, z \in \mathbf{R}_{+})$, then
$$
\begin{array}{l}
\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \\
=\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{[y(y+2 x)+z(z+2 y)+x(x+2 z)] \cdot} \\
\quad\left(\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}\right) \\
\geqslant(x+y+z)^{2} .
\end{array}
$$
Thus, $\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}$
$$
\begin{array}{l}
\geqslant \frac{(x+y+z)^{2}}{[y(y+2 x)+z(z+2 y)+x(x+2 z)]} \\
=1,
\end{array}
$$
which means $\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \geqslant 1$.
The equality holds if and only if $a=b=c=1$. Therefore, the minimum value is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given $x>y>0$, and $x y=1$. Find the minimum value of $\frac{3 x^{3}+125 y^{3}}{x-y}$.
|
Let the minimum value (target) be a positive number $t$, then
$$
\frac{3 x^{3}+125 y^{3}}{x-y} \geqslant t \text {, }
$$
which means $3 x^{3}+125 y^{3}+y t \geqslant x t$.
By $x y=1$ and the AM-GM inequality, we have
$3 x^{3}+125 y^{3}+y t$
$=x^{3}+x^{3}+x^{3}+125 y^{3}+y t$
$\geqslant 5 \sqrt[5]{x^{9} \cdot 125 y^{4} t}=5 \sqrt[5]{125 x^{5} t}$
$=5 x \sqrt[5]{125 t}$.
Let $5 x \sqrt[5]{125 t}=x t$, solving for $t$ gives $t=25$.
Note that the equality in the above inequality holds if and only if $x^{3}=125 y^{3}=y t$.
Therefore, when $x=\sqrt{5}, y=\frac{\sqrt{5}}{5}$, $\frac{3 x^{3}+125 y^{3}}{x-y}$ achieves its minimum value of 25.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a, b, c \in \mathbf{R}_{+}$, and $a+b+c=1$. Find
$$
\frac{3 a^{2}-a}{1+a^{2}}+\frac{3 b^{2}-b}{1+b^{2}}+\frac{3 c^{2}-c}{1+c^{2}}
$$
the minimum value.
|
( Hint: First prove $\frac{3 a^{2}-a}{1+a^{2}} \geqslant \frac{9}{10}\left(a-\frac{1}{3}\right)$. Similarly, obtain the other two inequalities. Then add the three inequalities, to get the minimum value of 0. )
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 The inequality $x^{2}+|2 x-4| \geqslant p$ holds for all real numbers $x$. Find the maximum value of the real number $p$.
保留了原文的换行和格式，如上所示。
|
Explanation 1: Transform the original inequality into
$|x-2| \geqslant-\frac{x^{2}}{2}+\frac{p}{2}$.
Let $y_{1}=|x-2|, y_{2}=-\frac{x^{2}}{2}+\frac{p}{2}$.
Thus, solving inequality (1) is equivalent to determining the range of $x$ values for which the graph of the function $y_{1}=|x-2|$ is above the graph of the function $y_{2}=-\frac{x^{2}}{2}+\frac{p}{2}$. In this problem, we are given that this range is all real numbers, and we need to find the maximum value of the parameter $p$.
As shown in Figure 5,
draw the graphs of
the functions $y_{1}=|x-2|$
and $y_{2}=-\frac{x^{2}}{2}$
$+\frac{p}{2}$.
It is easy to see that the critical case where the graph of $y_{1}=|x-2|$ is above the graph of $y_{2}=-\frac{x^{2}}{2}+\frac{p}{2}$ is when $p=3$. Therefore, the maximum value of $p$ is 3.
Explanation 2: Let $y=x^{2}+|2 x-4|$. We only need to draw the graph of the function $y=x^{2}+|2 x-4|$.
When $x \geqslant 2$,
$$
y=x^{2}+2 x-4 \text {; }
$$
When $x<2$,
$$
y=x^{2}-2 x+4 \text {. }
$$
As shown in Figure 6, it is clear that the maximum value of $p$ is 3.
Note: Using function graphs to combine numerical and geometric insights is another effective approach to discussing parameters, such as the number of real roots of an equation or the solution set of an inequality. It is important to note that although the graphs are rough sketches, the key conditions, key points, and trends must be accurately represented, otherwise, there will be misjudgments.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Color each vertex of a square pyramid with one color, and make the endpoints of the same edge different colors. If only 5 colors are available, then the total number of different coloring methods is $\qquad$
|
Solution: According to the problem, for the quadrilateral pyramid $S-ABCD$, the vertices $S$, $A$, and $B$ are colored with different colors, which gives a total of $5 \times 4 \times 3=60$ coloring methods.
When $S$, $A$, and $B$ are already colored, let's assume their colors are $1$, $2$, and $3$ respectively. If $C$ is colored with color 2, then $D$ can be colored with color 3, 4, or 5, giving 3 coloring methods; if $C$ is colored with color 4, then $D$ can be colored with color 3 or 5, giving 2 coloring methods; if $C$ is colored with color 5, then $D$ can be colored with color 3 or 4, also giving 2 coloring methods. Therefore, when $S$, $A$, and $B$ are already colored, $C$ and $D$ still have 7 coloring methods.
Thus, the total number of coloring methods is $60 \times 7=420$.
These two approaches highlight the special nature of the problem. When the quadrilateral pyramid is generalized to an $n$-sided pyramid, the limitations of these methods become immediately apparent.
|
420
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Algebraic expression
$$
\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}
$$
The minimum value is $\qquad$
|
Notice:
$$
\begin{array}{l}
\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}-9} \\
=\sqrt{(0-x)^{2}+[(-2)-0]^{2}}+ \\
\sqrt{(12-x)^{2}+(3-0)^{2}} .
\end{array}
$$
Establish a rectangular coordinate system as shown in Figure 8.
Let $A(0,-2)$, $B(12,3)$, and $P(x, 0)$. Then,
$$
\begin{array}{l}
|P A|=\sqrt{x^{2}+4}, \\
|P B|=\sqrt{(12-x)^{2}+9} .
\end{array}
$$
Thus, $|P A|+|P B| \geqslant|A B|=\sqrt{12^{2}+5^{2}}=13$, with equality holding if and only if point $P$ coincides with $P'$. Therefore, the minimum value of the given algebraic expression is 13.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $CA=4$, $P$ is the midpoint of the semicircular arc $\overparen{AC}$, connect $BP$, the line segment $BP$ divides the figure $APCB$ into two parts. The absolute value of the difference in the areas of these two parts is $\qquad$.
|
Ni.6.4.
As shown in Figure 8, let $A C$ and $B P$ intersect at point $D$, and the point symmetric to $D$ with respect to the circle center $O$ is denoted as $E$. The line segment $B P$ divides the figure $A P C B$ into two parts, and the absolute value of the difference in the areas of these two parts is the area of $\triangle B E P$, which is twice the area of $\triangle B O P$. Thus,
$$
\begin{array}{l}
S_{\triangle B P O}=\frac{1}{2} P O \cdot C O \\
=\frac{1}{2} \times 2 \times 2=2 .
\end{array}
$$
Therefore, the absolute value of the difference in the areas of these two parts is 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12.B. The real numbers $a, b, c$ satisfy $a \leqslant b \leqslant c$, and $ab + bc + ca = 0, abc = 1$. Find the largest real number $k$ such that the inequality $|a+b| \geqslant k|c|$ always holds.
|
12. B. When $a=b=-\sqrt[3]{2}, c=\frac{\sqrt[3]{2}}{2}$, the real numbers $a, b, c$ satisfy the given conditions, at this time, $k \leqslant 4$.
Below is the proof: The inequality $|a+b| \geqslant 4|c|$ holds for all real numbers $a, b, c$ that satisfy the given conditions.
From the given conditions, we know that $a, b, c$ are all not equal to 0, and $c>0$.
Since $ab=\frac{1}{c}>0, a+b=-\frac{1}{c^{2}}<0$, it follows that $a \leqslant b<0$.
By the relationship between the roots and coefficients of a quadratic equation, $a, b$ are the two real roots of the quadratic equation $x^{2}+\frac{1}{c^{2}} x+\frac{1}{c}=0$.
Then $\Delta=\frac{1}{c^{4}}-\frac{4}{c} \geqslant 0 \Rightarrow c^{3} \leqslant \frac{1}{4}$.
Therefore, $|a+b|=-(a+b)=\frac{1}{c^{2}} \geqslant 4 c=4|c|$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11.B. Given the parabolas $C_{1}: y=-x^{2}-3 x+4$ and $C_{2}: y=x^{2}-3 x-4$ intersect at points $A$ and $B$, point $P$ is on parabola $C_{1}$ and lies between points $A$ and $B$; point $Q$ is on parabola $C_{2}$, also lying between points $A$ and $B$.
(1) Find the length of segment $A B$;
(2) When $P Q \parallel y$-axis, find the maximum length of $P Q$.
|
11. B. (1) Solve the system of equations $\left\{\begin{array}{l}y=-x^{2}-3 x+4, \\ y=x^{2}-3 x-4,\end{array}\right.$ to get
$$
\left\{\begin{array} { l }
{ x _ { 1 } = - 2 , } \\
{ y _ { 1 } = 6 , }
\end{array} \left\{\begin{array}{l}
x_{2}=2, \\
y_{2}=-6 .
\end{array}\right.\right.
$$
Therefore, $A(-2,6)$ and $B(2,-6)$.
Thus, $A B=\sqrt{(2+2)^{2}+(-6-6)^{2}}=4 \sqrt{10}$.
(2) As shown in Figure 11, when
$P Q / / y$ axis, set
$$
\begin{array}{l}
P\left(t,-t^{2}-3 t+4\right), \\
Q\left(t, t^{2}-3 t-4\right) \\
(-2<t<2) .
\end{array}
$$
Then $P Q=2\left(4-t^{2}\right)$
$\leqslant 8$.
Equality holds if and only if $t=0$.
Therefore, the maximum length of $P Q$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $a b=1$, and $\frac{1}{1-2^{x} a}+\frac{1}{1-2^{y+1} b}=1$, then the value of $x+y$ is $\qquad$.
|
8. -1 .
Given that both sides of the equation are multiplied by $\left(1-2^{x} a\right)\left(1-2^{y+1} b\right)$, we get
$$
2-2^{x} a-2^{y+1} b=1-2^{x} a-2^{y+1} b+2^{x+y+1} a b \text {, }
$$
which simplifies to $1=2 \times 2^{x+y}$.
Therefore, $x+y=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let the sum of 10 consecutive positive integers not greater than 2006 form the set $S$, and the sum of 11 consecutive positive integers not greater than 2006 form the set $T$, then the number of elements in $S \cap T$ is $\qquad$ .
|
11.181.
$S$ is the set of all integers ending in 5, starting from 55 to
$$
20015=\sum_{i=1}^{10}(1996+i)=10 \times 1996+55
$$
Similarly, $T$ is the set of integers starting from 66 and increasing by 11 each time, with the largest number being
$$
22011=\sum_{i=1}^{11}(1995+i)=11 \times 1995+66 \text {. }
$$
In $T$, one out of every 10 elements has a units digit of 5, so there are $199=\left[\frac{1996}{10}\right]$ elements in $T$ with a units digit of 5, which are
$$
t_{k}=55+110 k(k=1,2, \cdots, 199) .
$$
From $55+110 k \leqslant 20015$, we solve to get $k \leqslant 181$.
Thus, $|S \cap T|=181$.
|
181
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Let $a>1, b>1$. Prove:
$$
\frac{a^{4}}{(b-1)^{2}}+\frac{b^{4}}{(a-1)^{2}} \geqslant 32 .
$$
|
15. Let $\left\{\begin{array}{l}x=a-1, \\ y=b-1,\end{array}\right.$ then $\left\{\begin{array}{l}a=x+1, \\ b=y+1 .\end{array}\right.$
The left side of the inequality
$$
\geqslant \frac{(2 \sqrt{x})^{4}}{y^{2}}+\frac{(2 \sqrt{y})^{4}}{x^{2}}=16\left(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}\right) \geqslant 32 \text {. }
$$
|
32
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
10. As shown in Figure 2, in Pascal's Triangle, the numbers above the diagonal form the sequence: $1,3,6,10, \cdots$, let the sum of the first $n$ terms of this sequence be $S_{n}$. Then, as $n \rightarrow +\infty$, the limit of $\frac{n^{3}}{S(n)}$ is $\qquad$
|
10.6 .
It is known that $S_{n}=\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\cdots+\mathrm{C}_{n+1}^{2}=\mathrm{C}_{\mathrm{n}+2}^{3}$. Therefore, $\lim _{n \rightarrow+\infty} \frac{n^{3}}{S(n)}=\lim _{n \rightarrow+\infty} \frac{n^{3} \times 6}{(n+2)(n+1) n}=6$.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The coefficient of $x^{150}$ in the expansion of $\left(1+x+x^{2}+\cdots+x^{100}\right)^{3}$, after combining like terms, is $\qquad$
|
5.7651 .
By the polynomial multiplication rule, the problem can be transformed into finding the number of natural number solutions of the equation
$$
s+t+r=150
$$
that do not exceed 100.
Obviously, the number of natural number solutions of equation (1) is $\mathrm{C}_{152}^{2}$.
Next, we find the number of natural number solutions of equation (1) that exceed 100.
Since their sum is 150, there can only be one number greater than 100.
Assume without loss of generality that $s>100$, and transform equation (1) into
$$
(s-101)+t+r=49 \text {. }
$$
Let $s^{\prime}=s-101$.
Then the number of natural number solutions of the equation $s^{\prime}+t+r=49$ is $\mathrm{C}_{51}^{2}$.
Therefore, the coefficient of $x^{150}$ is $\mathrm{C}_{52}^{2}-\mathrm{C}_{3}^{1} \mathrm{C}_{51}^{2}=7651$.
|
7651
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) Given sets $A_{1}, A_{2}, \cdots, A_{n}$ are different subsets of the set $\{1,2, \cdots, n\}$, satisfying the following conditions:
(i) $i \notin A_{i}$ and $\operatorname{Card}\left(A_{i}\right) \geqslant 3, i=1,2, \cdots, n$;
(ii) $i \in A_{j}$ if and only if $j \notin A_{i}(i \neq j$, $i, j=1,2, \cdots, n)$.
Try to answer the following questions:
(1) Find $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)$;
(2) Find the minimum value of $n$.
|
(1) Let $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ sets, and let $r_{i}$ be the number of sets that contain element $i$ for $i=1,2, \cdots, n$. Then,
$$
\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n} r_{i} \text {. }
$$
Assume the $r_{1}$ sets that contain element 1 are
$$
A_{2}, A_{3}, \cdots, A_{r_{1}+1} \text {. }
$$
By condition (ii), we know that $2 \notin A_{1}, 3 \notin A_{1}, \cdots \cdots r_{1}+1 \notin A_{1}, r_{1}+2 \in A_{1}, r_{1}+3 \in A_{1}, \cdots \cdots n \in A_{1}$.
Also, by condition (i), we know that $1 \notin A_{1}$.
Thus, $\operatorname{Card}\left(A_{1}\right)=n-\left(r_{1}+2\right)+1=n-r_{1}-1$.
Similarly, $\operatorname{Card}\left(A_{i}\right)=n-r_{i}-1$ for $i=1,2, \cdots, n$.
Therefore, $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n}\left(n-r_{i}-1\right)$ $=\sum_{i=1}^{n}(n-1)-\sum_{i=1}^{n} r_{i}=\sum_{i=1}^{n} r_{i}$.
Thus, $\sum_{i=1}^{n} r_{i}=\frac{1}{2} \sum_{i=1}^{n}(n-1)=\frac{1}{2} n(n-1)$.
(2) Since $\operatorname{Card}\left(A_{i}\right) \geqslant 3$, we have $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right) \geqslant 3 n$.
And since $A_{i} \subseteq\{1,2, \cdots, n\}$ for $i=1,2, \cdots, n$, at least one of the elements $1,2, \cdots, n$ must be in at least 3 sets. Assume this element is 1, i.e., $r_{1} \geqslant 3$.
$$
\begin{array}{l}
\text { Also, } \operatorname{Card}\left(A_{1}\right)=n-r_{1}-1 \geqslant 3, \text { so } n \geqslant r_{1}+4 \geqslant 7 \text {. And } \\
A_{1}=\{2,3,4\}, A_{2}=\{3,4,5\}, A_{3}=\{4,5,6\}, \\
A_{4}=\{5,6,7\}, A_{5}=\{6,7,1\}, A_{6}=\{7,1,2\}, \\
A_{7}=\{1,2,3\}
\end{array}
$$
satisfy the conditions.
Therefore, the minimum value of $n$ is 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Take any 21 points on a circle. Prove: Among all the arcs with these points as endpoints, there are no fewer than 100 arcs that do not exceed $120^{\circ}$.
|
(Assuming among all points, the chord with $A_{1}$ as an endpoint is the least, and denoting the chords with $A_{1}$ as an endpoint as $A_{1} A_{2}, A_{1} A_{3}, \cdots$, $A_{1} A_{n}$, a total of $n-1$ chords, and the chords with $A_{2}, A_{3}, \cdots, A_{n}$ as endpoints are no less than $n-1$ each. Therefore, these $n$ points have at least $\frac{n(n-1)}{2}$ chords. Among the remaining $21-n$ points, take any 2 points $A_{i} 、 A_{j}(i \neq j, i 、 j=$ $n+1, n+2, \cdots, 21)$. In this group of three points, there must be a chord. According to the selection of $A_{1}$, this chord cannot be $A_{1} A_{i} 、 A_{1} A_{j}$, but can only be $A_{i} A_{j}$. Therefore, between any two points among these $21-n$ points, there is a chord, a total of $\frac{(21-n)(21-n-1)}{2}$ chords. In summary, the total number of chords is at least $y=\frac{n(n-1)}{2}+\frac{(21-n)(21-n-1)}{2}=\left(n-\frac{21}{2}\right)^{2}+$ $\frac{399}{4}$. Therefore, when $n=10$ or 11, $y$ reaches its minimum value of 100.)
|
100
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $S$ be a subset of the set $\{1,2, \cdots, 50\}$ with the following property: the sum of any two distinct elements of $S$ cannot be divisible by 7. Then, what is the maximum number of elements that $S$ can have?
(43rd American High School Mathematics Examination)
|
For two different natural numbers $a$ and $b$, if $7 \times (a+b)$, then the sum of their remainders when divided by 7 is not 0. Therefore, the set $\{1,2, \cdots, 50\}$ can be divided into 7 subsets based on the remainders when divided by 7. Among them, each element in $K_{i}$ has a remainder of $i$ when divided by 7 ($i=1,2, \cdots, 6, 0$), then
$$
\begin{array}{l}
K_{1}=\{1,8,15,22,29,36,43,50\}, \\
K_{2}=\{2,9,16,23,30,37,44\}, \\
K_{3}=\{3,10,17,24,31,38,45\}, \\
K_{4}=\{4,11,18,25,32,39,46\}, \\
K_{5}=\{5,12,19,26,33,40,47\}, \\
K_{6}=\{6,13,20,27,34,41,48\}, \\
K_{0}=\{7,14,21,28,35,42,49\} .
\end{array}
$$
From the problem, we get
(1) If $S$ contains one element from $K_{i} (i=1,2, \cdots, 6)$, then it can contain all elements of this set, but it cannot contain elements from $K_{7-i}$ at the same time;
(2) $S$ can contain at most one element from $K_{0}$;
(3) The largest subset $S$ must contain all elements of $K_{1}$.
Thus, $S=K_{1} \cup K_{2} \cup K_{3} \cup \{7\}$ is the largest set that meets the requirements, and the maximum number of elements in $S$ is
$$
8+7+7+1=23 \text{. }
$$
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let $X=\{1,2, \cdots, 2001\}$. Find the smallest positive integer $m$ that satisfies the following condition: for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ (where $u$ and $v$ can be the same), such that $u+v$ is a power of 2.
(2001, China Mathematical Olympiad)
|
Explanation: When $u$ and $v$ are $2^{r} + a$ and $2^{r} - a$ respectively, $u + v = 2 \times 2^{r} = 2^{r+1}$ is a power of 2. Based on this, $X$ is divided into the following 5 subsets:
$$
\begin{array}{l}
2001 = 2^{10} + 977 \geqslant x \geqslant 2^{10} - 977 = 47, \\
46 = 2^{5} + 14 \geqslant x \geqslant 2^{5} - 14 = 18, \\
17 = 2^{4} + 1 \geqslant x \geqslant 2^{4} - 1 = 15, \\
14 = 2^{3} + 6 \geqslant x \geqslant 2^{3} - 6 = 2, \\
x = 1. \\
\text{Let } Y = \{2001, 2000, \cdots, 1025\} \cup \\
\{46, 45, \cdots, 33\} \cup \{17\} \cup \\
\{14, 13, \cdots, 9\}. \\
\end{array}
$$
Thus, $|Y| = 998$, and for any $u, v \in Y$, $u + v$ is not a power of 2.
Proof: When $u, v \in Y$, assume $u \geqslant v$ and $2^{r} < u \leqslant 2^{r} + a < 2^{r+1}$, where, when $r$ takes the values $10, 5, 4, 3$, the corresponding $a$ values are $977, 14, 1, 6$ respectively.
(1) If $2^{r} < v \leqslant u$, then $2^{r+1} < u + v < 2^{r+2}$, so $u + v$ cannot be a power of 2.
(2) If $1 \leqslant v < 2^{r}$, then when $2^{r} < u \leqslant 2^{r} + a$ ($1 \leqslant a < 2^{r}$), $1 \leqslant v < 2^{r} - a$. Thus, $2^{r} < u + v < 2^{r+1}$, so $u + v$ cannot be a power of 2.
Therefore, the sum of any two numbers in the subset $Y$ is not a power of 2.
Hence, the smallest positive integer $m \geqslant 999$.
Divide $X$ into 999 mutually disjoint subsets:
$$
A_{i} = \{1024 - i, 1024 + i\} (i = 1, 2, \cdots,
$$
977),
$$
\begin{array}{l}
B_{j} = \{32 - j, 32 + j\} (j = 1, 2, \cdots, 14), \\
C = \{15, 17\}, \\
D_{k} = \{8 - k, 8 + k\} (k = 1, 2, \cdots, 6), \\
E = \{1, 8, 16, 32, 1024\}.
\end{array}
$$
For any 999-element subset $W$ of $X$, if $W \cap E \neq \varnothing$, then the double of any element taken from it is a power of 2; if $W \cap E = \varnothing$, then the 999 elements of $W$ belong to the preceding 998 2-element subsets. By the pigeonhole principle, there must be different $u, v$ in $W$ that belong to the same subset. Clearly, $u + v$ is a power of 2.
In summary, the smallest positive integer $m = 999$.
Note: This solution divides the set $X$ twice. The first division divides $X$ into 5 subsets, constructing a set $Y$ that does not meet the requirements of the problem and contains the most elements; the second division divides the set into 999 subsets, and then applies the pigeonhole principle.
|
999
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let $M=\{1,2, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: when $x \in A$, $15 x \notin A$. Then the maximum number of elements in $A$ is $\qquad$
(1995, National High School Mathematics Competition)
|
Construct the subset $A$ as follows:
Notice that $1995 \div 15=133$, and let
$$
A_{1}=\{134,135, \cdots, 1995\} \text {, }
$$
then $\left|A_{1}\right|=1862$.
Also, $133=15 \times 8+13$, and let
$A_{2}=\{9,10, \cdots, 133\}$,
then $\left|A_{2}\right|=125$.
Let $A_{3}=\{1,2, \cdots, 8\}$, then $\left|A_{3}\right|=8$.
Thus, $M=A_{1} \cup A_{2} \cup A_{3}, A_{i} \cap A_{j}=\varnothing(i \neq j)$.
When $a \in A_{1}$, $15 a \notin A$; when $a \in A_{2}$, at least one of $a$ and $15 a$ does not belong to $A$; when $a \in A_{3}$, $15 a \notin A_{1}$. Therefore, let $A=A_{1} \cup A_{3}$, then $A$ is a subset that meets the requirements, and
$$
|A|=\left|A_{1}\right|+\left|A_{3}\right|=1862+8=1870 .
$$
|
1870
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $S=\{1,2, \cdots, 2005\}$. If any set of $n$ pairwise coprime numbers in $S$ contains at least one prime number, find the minimum value of $n$.
(2005, China Western Mathematical Olympiad)
|
Explanation: First, take a subset of $S$
$$
A_{0}=\left\{1,2^{2}, 3^{2}, 5^{2}, \cdots, 41^{2}, 43^{2}\right\},
$$
then $\left|A_{0}\right|=15, A_{0}$ contains any two numbers that are coprime, but there are no primes in it. This indicates that $n \geqslant 16$.
Second, it can be proven: For any $A \subseteq S, n=|A|=16, A$ contains any two numbers that are coprime, then $A$ must contain a prime number.
Proof by contradiction.
Assume $A$ contains no primes. Let
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{16}\right\}.
$$
Consider two cases.
(1) If $1 \notin A$, then $a_{1}, a_{2}, \cdots, a_{16}$ are all composite numbers. Also, $\left(a_{i}, a_{j}\right)=1(1 \leqslant i < j \leqslant 16)$. Let the smallest prime factor of $a_{i}$ be $p_{i}$, then
$$
a_{1} \geqslant p_{1}^{2} \geqslant 2^{2}, a_{2} \geqslant p_{2}^{2} \geqslant 3^{2}, \cdots \cdots
$$
$$
a_{16} \geqslant p_{16}^{2} \geqslant 47^{2}>2005,
$$
which is a contradiction.
(2) If $1 \in A$, without loss of generality, let $a_{16}=1, a_{1}, a_{2}, \cdots, a_{15}$ be composite numbers.
Similarly, we have
$$
\begin{array}{l}
a_{1} \geqslant p_{1}^{2} \geqslant 2^{2}, a_{2} \geqslant p_{2}^{2} \geqslant 3^{2}, \cdots \cdots \\
a_{15} \geqslant p_{15}^{2} \geqslant 47^{2}>2005,
\end{array}
$$
which is a contradiction.
Thus, $A$ must contain a prime number, i.e., when $n=|A|=16$, the conclusion holds.
In summary, the minimum value of $n$ is 16.
Note: The solution to this problem uses the strategy of exploring special cases (taking the subset $A_{0}$) and verifying general cases.
|
16
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Let sets $A$ and $B$ both consist of positive numbers, with $|A|=10, |B|=9$, and set $A$ satisfies the following condition: If $x, y, u, v \in A, x+y=u+v$, then \begin{aligned}\{x, y\} & =\{u, v\} . \text{ Let } \\ A & +B=\{a+b \mid a \in A, b \in B\} .\end{aligned}
Prove: $|A+B| \geqslant 50$ ($|X|$ denotes the number of elements in set $X$).
(2005, National High School Mathematics League, Fujian Province Preliminary Contest)
|
Explanation: Consider the general case.
Let $|A|=m,|B|=n$,
$A+B=\left\{s_{1}, s_{2}, \cdots, s_{k}\right\}$.
Suppose $s_{i}(1 \leqslant i \leqslant k)$ can be expressed in $f(i)$ ways as $a+b$, where $a \in A, b \in B$, i.e.,
$$
s_{i}=a_{i 1}+b_{i 1}=a_{i 2}+b_{i 2}=\cdots=a_{i(i)}+b_{i f(i)}.
$$
Then $f(1)+f(2)+\cdots+f(k)=\mathrm{C}_{m}^{1} \mathrm{C}_{n}^{1}=m n$.
For any $1 \leqslant r<t \leqslant f(i)$, consider the set $\left\{b_{i r}, b_{i t}\right\}$, then there are $\mathrm{C}_{f(i)}^{2}$ such sets. For $i=1,2, \cdots, k$, there are $\mathrm{C}_{f(1)}^{2}+\mathrm{C}_{f(2)}^{2}+\cdots+\mathrm{C}_{f(k)}^{2}$ sets in total.
We will prove that these sets are all distinct.
If not, then there exist $1 \leqslant i<j$ and $b_{r}, b_{i} \in B$ $(r \neq t)$ such that
$$
s_{i}=x+b_{r}=u+b_{t}, s_{j}=v+b_{r}=y+b_{t},
$$
where $x, y, u, v \in A$.
Thus, $x+y=u+v$.
By the problem's condition, $\{x, y\}=\{u, v\}$.
If $x=u, y=v$, then $b_{r}=b_{t}$, which is impossible;
If $x=v, y=u$, then $s_{i}=s_{j}$, which is impossible.
Therefore, $\mathrm{C}_{f(1)}^{2}+\mathrm{C}_{f(2)}^{2}+\cdots+\mathrm{C}_{f(k)}^{2} \leqslant \mathrm{C}_{n}^{2}$.
Simplifying, we get
$$
\begin{array}{l}
\left(f^{2}(1)+f^{2}(2)+\cdots+f^{2}(k)-\right. \\
\quad(f(1)+f(2)+\cdots+f(k)) \\
\leqslant n^{2}-n .
\end{array}
$$
By the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
f^{2}(1)+f^{2}(2)+\cdots+f^{2}(k) \\
\geqslant \frac{1}{k}(f(1)+f(2)+\cdots+f(k))^{2}=\frac{1}{k} m^{2} n^{2} .
\end{array}
$$
Thus, $\frac{1}{k} m^{2} n^{2}-m n \leqslant n^{2}-n$, i.e.,
$$
k \geqslant \frac{m^{2} n}{m+n-1} .
$$
When $m=10, n=9$, $k \geqslant 50$.
Comment: The key to solving this problem is to generalize the problem and derive the inequality $\mathrm{C}_{f(1)}^{2}+\mathrm{C}_{f(2)}^{2}+\cdots+\mathrm{C}_{f(k)}^{2} \leqslant \mathrm{C}_{n}^{2}$.
|
50
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
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