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stringlengths 2
5.64k
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stringlengths 2
13.5k
| answer
stringlengths 1
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| problem_type
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stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
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|---|---|---|---|---|---|---|---|---|
4. Let $D$, $E$, and $F$ be points on the sides $BC$, $AB$, and $AC$ of $\triangle ABC$, respectively, such that $AE = AF$, $BE = BD$, and $CF = CD$. Given that $AB \cdot AC = 2BD \cdot DC$, $AB = 12$, and $AC = 5$. Then the inradius $r$ of $\triangle ABC$ is $\qquad$
|
4.2.
As shown in Figure 6, let
$$
\begin{array}{l}
A E=A F=x, \\
B E=B D=y, \\
C D=C F=z .
\end{array}
$$
Then $A B=x+y$,
$$
\begin{array}{l}
A C=x+z, \\
B C=y+z .
\end{array}
$$
From $A B \cdot A C=2 B D \cdot D C$
$$
\begin{array}{l}
\Rightarrow(x+y)(x+z)=2 y z \\
\Rightarrow x^{2}+x y+x z=y z \\
\text { Also } A B^{2}+A C^{2}=(x+y)^{2}+(x+z)^{2} \\
=y^{2}+z^{2}+2\left(x^{2}+x y+x z\right) \\
=y^{2}+z^{2}+2 y z \\
=(y+z)^{2}=B C^{2},
\end{array}
$$
Then $\angle A=90^{\circ}$, and
$$
B C=\sqrt{A B^{2}+A C^{2}}=13 \text {. }
$$
Thus $r=\frac{A B \cdot A C}{A B+B C+C A}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) In the set of numbers $1,2, \cdots, 2009$, what is the maximum number of numbers that can be selected such that the sum of any two selected numbers is divisible by 100?
|
Three, let the $n$ numbers that meet the conditions be,
$$
a_{1}, a_{2}, \cdots, a_{n} \text {, }
$$
Take any three of these numbers (let them be $a_{k} \backslash a_{m} \backslash a_{f}$). Then
$$
\begin{array}{l}
a_{k}+a_{m}=100 k_{1}, \\
a_{k}+a_{4}=100 k_{2}, \\
a_{m}+a_{4}=100 k_{3},
\end{array}
$$
where $k_{1} γ k_{2} γ k_{3}$ are positive integers.
$$
\begin{array}{l}
\text { (1) }+ \text { (2)-(3) gives } \\
a_{k}=50\left(k_{1}+k_{2}-k_{3}\right) .
\end{array}
$$
Similarly, $a_{m}=\mathbf{5 0}\left(k_{1}+k_{3}-k_{2}\right)$,
$$
a_{1}=50\left(k_{2}+k_{3}-k_{1}\right) \text {. }
$$
Therefore, $a_{k} γ \grave{a}_{m} γ a$ are all multiples of 50.
However, the multiples can only be odd or even, and they must all be odd or all even to be divisible by 100, so the numbers that meet the conditions are $50,150, \cdots, 1950$ or $100,200, \cdots, 2000$, with 20 numbers in each set.
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given positive real numbers $a$, $b$, $c$ satisfy
$$
(1+a)(1+b)(1+c)=8 \text {. }
$$
Then the minimum value of $a b c+\frac{9}{a b c}$ is
|
-1.10 .
$$
\begin{array}{l}
\text { Given } 8=(1+a)(1+b)(1+c) \\
\geqslant 2 \sqrt{a} \cdot 2 \sqrt{b} \cdot 2 \sqrt{c}=8 \sqrt{a b c} \\
\Rightarrow a b c \leqslant 1 .
\end{array}
$$
Equality holds if and only if $a=b=c=1$.
It is easy to verify that the function
$$
f(x)=x+\frac{9}{x}
$$
is decreasing on $(0,3)$.
Since $0<a b c \leqslant 1<3$, when $a b c=1$,
$$
f(a b c)=a b c+\frac{9}{a b c}
$$
is minimized, and the minimum value is 10.
|
10
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given a finite arithmetic sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1$, and a common difference of 2, the arithmetic mean of all its terms is 2011. If one term is removed, the arithmetic mean of the remaining terms is an integer. Then the number of ways to remove a term is $\qquad$.
|
4.3.
According to the problem, we have
$$
\frac{1}{n}\left[n+\frac{n(n-1)}{2} \times 2\right]=2011 \text {. }
$$
Solving this, we get $n=2011$.
Thus, the sum of all terms in the sequence is $2011^{2}$.
Suppose that after removing the $k$-th term from the sequence, the arithmetic mean of the remaining terms is an integer, i.e.,
$$
\frac{2011^{2}-(2 k-1)}{2010} \in \mathbf{Z} \text {. }
$$
Noting that $2011^{2}=2010 \times 2012+1$, the above equation is equivalent to
$$
2010|2(1-k) \Rightarrow 1005|(1-k) \text {. }
$$
Since $1 \leqslant k \leqslant 2011$, the set of possible values for $k$ is $\{1,1006,2011\}$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. As shown in Figure 2, in plane $\alpha$, $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ are on opposite sides of line $l$, with no common points with $l$, and are symmetric about line $l$. Now, if plane $\alpha$ is folded along line $l$ to form a right dihedral angle, then the six points $A, B, C, A_{1}, B_{1}, C_{1}$ can determine $\qquad$ planes (answer with a number).
|
7.11.
Notice that after the fold, the three sets of four points
$$
\left(A, B, A_{1}, B_{1}\right) γ\left(B, C, B_{1}, C_{1}\right) γ\left(C, A, C_{1}, A_{1}\right)
$$
are all coplanar, therefore, these six points can determine
$$
C_{6}^{3}-3\left(C_{4}^{3}-1\right)=11 \text { (planes). }
$$
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (50 points) A company printed a batch of T-shirts, each T-shirt can have three different colors: red, yellow, and blue, and four different patterns. Now, this batch of T-shirts is to be distributed to $n$ new employees, with each employee receiving exactly 4 T-shirts with different patterns. Try to find the minimum value of $n$, such that there will always be two people who have received 4 T-shirts of the same color for two of the patterns.
---
Note: The translation maintains the original text's line breaks and formatting.
|
Four, the minimum value of $n$ is 19.
When $n=18$, the answer scenario shown in Table 1 does not meet the requirements.
[Note] In Table 1, (1), (2), (3), (4) are patterns, $A_{1}, A_{2}, \cdots, A_{18}$ are members, and $A$, $B$, $C$ represent red, yellow, and blue colors, respectively.
The proof below shows that when $n \geqslant 19$, there must exist two people who meet the requirements.
If the 4 cultural shirts of a case have the same color, then there must exist a sub-table in the form of a rectangle, where the letters (colors) in the four corner cells are the same.
If for a certain color (for example, red), let $A_{i}$ be assigned $x_{i}$ red cultural shirts. Then when $\sum_{i=1}^{n} \mathrm{C}_{x_{i}}^{2}>6$ (it is agreed that when $x_{i}<2$, $\mathrm{C}_{x_{i}}^{2}=0$), there must be two columns with the same pair; thus, there must be a rectangle with all four corners being $A$.
When $n \geqslant 19$, taking any 19 people, among all the colors of their cultural shirts, at least one color appears no less than
$$
\left[\frac{4 \times 19-1}{3}\right]+1=26 \text { times, }
$$
let's assume it is red.
Suppose $A_{i}(i=1,2, \cdots, 19)$ is assigned $x_{i}\left(x_{i}\right.$ is a non-negative integer) red cultural shirts. Then
$$
\sum_{i=1}^{19} x_{i} \geqslant 26 \text {. }
$$
By the adjustment method, it is easy to know that when $\sum_{i=1}^{19} \mathrm{C}_{x_{i}}^{2}$ takes the minimum value, for any $1 \leqslant j6 \text {. }
$$
This indicates that when $n \geqslant 19$, there must exist a sub-table in the form of a rectangle with all four corners being the same letter.
In summary, the minimum value of $n$ is 19.
In fact, if all the colors and patterns of the cultural shirts of all people are made into a table as above; if there exist two people with the same two patterns...
|
19
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 In $\triangle A B C$, it is known that $A B=A C=2$, and there are 100 different points $P_{1}, P_{2}, \cdots, P_{100}$ on side $B C$. Let $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C(i=1,2, \cdots, 100)$.
Find the value of $m_{1}+m_{2}+\cdots+m_{100}$.
|
Solve As shown in Figure 2, since $\triangle A B C$ is an isosceles triangle, applying the property we get
$$
\begin{aligned}
& A P_{i}^{2} \\
= & A B^{2}-B P_{i} \cdot P_{i} C .
\end{aligned}
$$
Therefore, $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C=A B^{2}=4$.
Thus, $m_{1}+m_{2}+\cdots+m_{100}=400$.
|
400
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given points $A, B, C, D$ lie on the same circle, and $BC = DC = 4, AC$ intersects $BD$ at point $E, AE = 6$. If the lengths of segments $BE$ and $DE$ are both integers, find the length of $BD$.
|
Apply the property to $\triangle B C D$ to get
$$
C E^{2}=C D^{2}-B E \cdot E D=16-6 \cdot E C \text {. }
$$
Solving, we get $E C=2$.
From $B E \cdot E D=A E \cdot E C=12$, and
$$
B D<B C+C D=8 \text {, }
$$
we solve to get $B D=4+3=7$.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A circle is filled with 12 positive integers, each taken from $\{1,2, \cdots, 9\}$ (each number can appear multiple times on the circle). Let $S$ represent the sum of all 12 numbers on the circle. If the sum of any three consecutive numbers on the circle is a multiple of 7, then the number of possible values for $S$ is $\qquad$ kinds.
|
4.9.
For any three consecutive numbers $a_{k}, a_{k+1}, a_{k+2}$ on a circle, $a_{k}+a_{k+1}+a_{k+2}$ can be 7, 14, or 21.
For any four consecutive numbers on the circle, if they are $a_{k}, a_{k+1}, a_{k+2}, a_{k+3}$, since
$$
a_{k}+a_{k+1}+a_{k+2} \text { and } a_{k+1}+a_{k+2}+a_{k+3}
$$
are both multiples of 7, it must be that
$7 \mid\left(a_{k+3}-a_{k}\right)$.
Thus, $a_{k}$ and $a_{k+3}$ are either equal or differ by 7.
Now, divide the circle into four segments, each with a sum of three consecutive numbers that can be 7, 14, or 21, so the total sum of the four segments can be any of the following:
$$
\{28,35,42,49,56,63,70,77,84\}
$$
There are a total of 9 possible cases.
One way to fill the circle is: first, sequentially fill 12 numbers on the circle:
$$
1,2,4,1,2,4,1,2,4,1,2,4,
$$
with a sum of 28; then, each time change one 1 to 8, or one 2 to 9, each operation increases the total sum by 7. Such operations can be performed 8 times.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, \text { and } \\
a_{n+2}=\left|a_{n+1}\right|-a_{n}
\end{array}\left(n \in \mathbf{N}_{+}\right) .
$$
Let $\left\{a_{n}\right\}$'s sum of the first $n$ terms be $S_{n}$. Then $S_{100}=$
|
- 1. 89.
From the given, $a_{k+9}=a_{k}$.
Then $S_{100}=a_{1}+11\left(a_{1}+a_{2}+\cdots+a_{9}\right)=89$.
|
89
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $n<100$. Then the largest integer $n$ such that the expansion of $(a+b)^{n}$ has three consecutive terms with coefficients in arithmetic progression is $\qquad$ . . .
|
3. 98.
Let the coefficients of three consecutive terms in the expansion of $(a+b)^{n}$ be $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1} (1 \leqslant k \leqslant n-1)$.
By the problem, we have $2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1}$.
Expanding and rearranging according to the definition of combination numbers, we get
$$
n^{2}-(4 k+1) n+4 k^{2}-2=0 \text{. }
$$
Thus, $n_{1,2}=\frac{4 k+1 \pm \sqrt{8 k^{2}+9}}{2}\left(n_{1,2} \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
8 k^{2}+9=(2 m+1)^{2} \\
\Rightarrow 2 k=m^{2}+m-2 .
\end{array}
$$
Substituting into equation (1), we get
$$
n_{1}=(m+1)^{2}-2, n_{2}=m^{2}-2 \text{. }
$$
From $(m+1)^{2}-2<100$, we know $n=98$.
|
98
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Among the positive integers less than 20, each time three numbers are taken without repetition, so that their sum is divisible by 3. Then the number of different ways to do this is $\qquad$ .
|
4.327.
Divide these 19 numbers into three categories based on the remainder when divided by 3:
$$
\begin{array}{l}
A_{1}: 3,6,9,12,15,18 ; \\
A_{2}: 2,5,8,11,14,17 ; \\
A_{3}: 1,4,7,10,13,16,19 .
\end{array}
$$
Thus, the number of ways to satisfy the conditions of the problem are only four scenarios.
(1) Choose any three numbers from $A_{1}$, there are $\mathrm{C}_{6}^{3}=20$ ways;
(2) Choose any three numbers from $A_{2}$, there are $\mathrm{C}_{6}^{3}=20$ ways;
(3) Choose any two numbers from $A_{3}$, there are $\mathrm{C}_{7}^{3}=35$ ways;
(4) Choose one number from each of $A_{1}$, $A_{2}$, and $A_{3}$, there are
$$
6 \times 6 \times 7=252
$$
ways.
Therefore, the total number of ways is
$$
20 \times 2+35+252=327 \text { (ways). }
$$
|
327
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Place 10 numbers on a given circle such that their total sum is 200, and the sum of any three consecutive numbers is not less than 58. Then the maximum value of the largest number among all sets of 10 numbers that satisfy the above requirements is $\qquad$
|
8. 26 .
Let the maximum number in all placements be $A$. Then
$$
A+3 \times 58 \leqslant 200 \Rightarrow A \leqslant 26 \text {. }
$$
In fact, $26,6,26,26,6,26,26,6,26,26$ satisfies.
|
26
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Let
$$
P=x^{4}+6 x^{3}+11 x^{2}+3 x+31 \text {. }
$$
Find the integer value(s) of $x$ that make $P$ a perfect square.
|
10. Since $P=\left(x^{2}+3 x+1\right)^{2}-3(x-10)$, therefore, when $x=10$, $P=131^{2}$ is a perfect square.
Next, we only need to prove: there are no other integer $x$ that satisfy the requirement.
(1) $x>10$.
If $P0$, then $P>\left(x^{2}+3 x\right)^{2}$.
Therefore, $\left(x^{2}+3 x\right)^{2}\left(x^{2}+3 x+1\right)^{2}$.
Let $P=y^{2}(y \in \mathbf{Z})$. Then
$$
|y|>\left|x^{2}+3 x+1\right| \text {, }
$$
which means $|y|-1 \geqslant \mid x^{2}+3 x+11$.
Thus, $y^{2}-2|y|+1 \geqslant\left(x^{2}+3 x+1\right)^{2}$, i.e., $-3(x-10)-2\left|x^{2}+3 x+1\right|+1 \geqslant 0$.
Solving this inequality yields the integer values of $x$ as
$$
\pm 2, \pm 1, 0, -3, -4, -5, -6 \text {. }
$$
However, the corresponding $P$ for these values are not perfect squares.
In conclusion, the integer value of $x$ that makes $P$ a perfect square is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n+1}-2 a_{n}=2^{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the smallest positive integer $n$ for which $a_{n}>10$ holds is $\qquad$ . .
|
3.3.
Given $a_{n+1}-2 a_{n}=2^{n+1}$, we know $\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=1$.
Thus, the sequence $\left\{\frac{a_{n}}{2^{n}}\right\}$ is an arithmetic sequence with a common difference of 1. Also, $a_{1}=2$, so $\frac{a_{n}}{2^{n}}=n \Rightarrow a_{n}=n \times 2^{n}$. Therefore, $\left\{a_{n}\right\}$ is an increasing sequence, and
$$
a_{2}=2 \times 2^{2}=8, a_{3}=3 \times 2^{3}=24 \text {. }
$$
Hence, the smallest positive integer $n$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$
where, $a_{i} \in\{0,1, \cdots, 6\}(i=0,1,2,3)$, and $a_{3} \neq 0$.
If positive integers $m γ n \in A$, and $m+n=2010(m>n)$, then the number of positive integers $m$ that satisfy the condition is $\qquad$.
|
6. 662 .
According to the problem, we know that $m$ and $n$ are four-digit numbers in base 7, and the largest four-digit number in base 7 is
$$
6 \times 7^{3}+6 \times 7^{2}+6 \times 7+6=2400,
$$
the smallest one is $1 \times 7^{3}=343$.
Since $m+n=2010(m>n)$, therefore,
$$
1006 \leqslant m \leqslant 1667 \text {. }
$$
Thus, the number of positive integers $m$ that meet the condition is
$$
1667-1006+1=662 \text { (numbers). }
$$
|
662
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Arrange the real solutions of the equation $x^{3}-3[x]=4$ in ascending order to get $x_{1}, x_{2}, \cdots, x_{k}$. Then the value of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}$ is $\qquad$ ( $[x]$ denotes the greatest integer less than or equal to the real number $x$).
|
8. 15 .
Notice that $x-1<[x] \leqslant x$.
Therefore, when $x \geqslant 3$,
$$
\begin{array}{l}
x^{3}-3[x] \geqslant x^{3}-3 x=x\left(x^{2}-3\right) \\
\geqslant 3 \times 6=18 ;
\end{array}
$$
When $x \leqslant-3$,
$$
\begin{array}{l}
x^{3}-3[x]<x^{3}-3(x-1)=x\left(x^{2}-3\right)+3 \\
\leqslant-3 \times 6+3=-15 .
\end{array}
$$
Therefore, the solution to the equation can only be within the interval $(-3,3)$, and $[x]$ can only take the values $-3, -2, -1, 0, 1, 2$.
If $[x]=-3$, then $x^{3}=-5$, which does not meet the requirement;
If $[x]=-2$, then $x^{3}=-2 \Rightarrow x=-\sqrt[3]{2}$, which meets the requirement;
If $[x]=-1$, then $x^{3}=1$, which does not meet the requirement;
If $[x]=0$, then $x^{3}=4$, which does not meet the requirement;
If $[x]=1$, then $x^{3}=7 \Rightarrow x=\sqrt[3]{7}$, which meets the requirement;
If $[x]=2$, then $x^{3}=10 \Rightarrow x=\sqrt[3]{10}$, which meets the requirement.
Thus, $x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}=-2+7+10=15$.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. If the positive integer $m$ makes it true that for any set of positive numbers $a_{1} γ a_{2} γ a_{3} γ a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$
then the minimum value of the positive integer $m$ is $\qquad$
|
9.3.
Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$
It is verified that $m=1, m=2$ do not meet the requirements.
Hence $m \geqslant 3$.
$$
\begin{array}{l}
\text { By } \frac{a_{1}^{3}+a_{2}^{3}+a_{3}^{3}}{3} \geqslant a_{1} a_{2} a_{3}, \\
\frac{a_{1}^{3}+a_{2}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{2} a_{4}, \\
\frac{a_{1}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{3} a_{4}, \\
\frac{a_{2}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{2} a_{3} a_{4}, \\
a_{1} a_{2} a_{3} a_{4}=1, \\
\text { then } a_{1}^{3}+a_{2}^{3}+a_{3}^{3}+a_{4}^{3} \\
\geqslant a_{1} a_{2} a_{3}+a_{1} a_{2} a_{4}+a_{1} a_{3} a_{4}+a_{2} a_{3} a_{4} \\
=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}} .
\end{array}
$$
Therefore, $m=3$ meets the requirements:
Thus, the smallest positive integer $m$ is 3.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. There are three numbers arranged in sequence: $3, 9, 8$. For any two adjacent numbers, the difference between the right number and the left number is written between these two numbers, resulting in a new sequence $3, 6, 9, -1, 8$, which is called the first operation; after the second similar operation, a new sequence $3, 3, 6, 3, 9, -10, -1, 9, 8$ is produced; continue this operation. Ask: Starting from the sequence 3, 9, 8, what is the sum of all numbers in the new sequence after the 100th operation?
|
For convenience, let the sequence of $n$ numbers be $a_{1}, a_{2}, \cdots, a_{n}$. According to the problem, the newly added numbers are $a_{2}-a_{1}, a_{3}-a_{2}, \cdots, a_{n}-a_{n-1}$. Therefore, the sum of the newly added numbers is
$$
\begin{array}{l}
\left(a_{2}-a_{1}\right) + \left(a_{3}-a_{2}\right) + \cdots + \left(a_{n}-a_{n-1}\right) \\
=a_{n}-a_{1} .
\end{array}
$$
The original sequence is three numbers $3, 9, 8$. After the first operation, the resulting sequence is $3, 6, 9, -1, 8$. By equation (1), the sum of the two newly added numbers is
$$
6 + (-1) = 5 = 8 - 3.
$$
After the second operation, by equation (1), the sum of the four newly added numbers is
$$
3 + 3 + (-10) + 9 = 5 = 8 - 3.
$$
Thus, after the 100th operation, the sum of all numbers in the new sequence is
$$
(3 + 9 + 8) + 100 \times (8 - 3) = 520.
$$
|
520
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
19. There are 10 red, 10 black, and 10 white balls. Now, all of them are to be placed into two bags, A and B, with the requirement that each bag must contain balls of all three colors, and the product of the number of balls of each color in bags A and B must be equal. How many ways are there to do this?
|
19. Let the number of red, black, and white balls in bag A be $x$, $y$, and $z$ respectively. Then $1 \leqslant x, y, z \leqslant 9$, and
$$
x y z=(10-x)(10-y)(10-z) \text {, }
$$
i.e., $x y z=500-50(x+y+z)+5(x y+y z+z x)$.
Thus, $5 \mid x y z$.
Therefore, one of $x, y, z$ must be 5.
Assume $x=5$, substituting into equation (1) gives $y+z=10$.
At this point, $y$ can take $1,2, \cdots, 9$ (correspondingly $z$ takes 9, 8, ..., 1), for a total of 9 ways.
Similarly, when $y=5$ or $z=5$, there are also 9 ways each.
However, when $x=y=z$, two ways are repeated, so there are a total of $9 \times 3-2=25$ ways.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
20. Let the ellipse be $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1), \operatorname{Rt} \triangle A B C$ with $A(0,1)$ as the right-angle vertex, and sides $A B, B C$ intersecting the ellipse at points $B, C$. If the maximum area of $\triangle A B C$ is $\frac{27}{8}$, find the value of $a$.
|
20. Let $l_{A B}: y=k x+1(k>0)$. Then $l_{A C}: y=-\frac{1}{k} x+1$.
From $\left\{\begin{array}{l}y=k x+1, \\ \frac{x^{2}}{a^{2}}+y^{2}=1,\end{array}\right.$,
$\left(1+a^{2} k^{2}\right) x^{2}+2 a^{2} k x=0$
$\Rightarrow x_{B}=\frac{-2 a^{2} k}{1+a^{2} k^{2}}$.
Thus, $|A B|=\sqrt{1+k^{2}} \cdot \frac{2 a^{2} k}{1+a^{2} k^{2}}$.
Similarly, $|A C|=\sqrt{1+\frac{1}{k^{2}}} \cdot \frac{2 a^{2} k}{a^{2}+k^{2}}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2}|A B||A C|$
$=2 a^{4} \frac{k\left(1+k^{2}\right)}{\left(1+a^{2} k^{2}\right)\left(a^{2}+k^{2}\right)}$
$=2 a^{4} \frac{k+\frac{1}{k}}{a^{2}\left(k^{2}+\frac{1}{k^{2}}\right)+a^{4}+1}$.
Let $t=k+\frac{1}{k} \geqslant 2$. Then
$S_{\triangle A B C}=\frac{2 a^{4} t}{a^{2} t^{2}+\left(a^{2}-1\right)^{2}}$
$=\frac{2 a^{4}}{a^{2} t+\frac{\left(a^{2}-1\right)^{2}}{t}}$.
And $a^{2} t+\frac{\left(a^{2}-1\right)^{2}}{t} \geqslant 2 a\left(a^{2}-1\right)$, when $t=\frac{a^{2}-1}{a}$, equality holds.
Thus, when $t=\frac{a^{2}-1}{a}$, $\left(S_{\triangle A B C}\right)_{\max }=\frac{a^{3}}{a^{2}-1}$.
From $\frac{a^{3}}{a^{2}-1}=\frac{27}{8}$
$\Rightarrow(a-3)\left(8 a^{2}-3 a-9\right)=0$
$\Rightarrow a=3, a=\frac{3+\sqrt{297}}{16}$.
From $\frac{a^{2}-1}{a}>2 \Rightarrow a>1+\sqrt{2}$.
Therefore, $a=\frac{3+\sqrt{297}}{16}$ is not valid, and is discarded.
Thus, $a=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the arithmetic mean of two positive numbers is $2 \sqrt{3}$, and the geometric mean is $\sqrt{3}$, what is the difference between these two numbers?
|
1. Let $x, y$ represent two numbers. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y=4 \sqrt{3}, \\
x y=3
\end{array}\right. \\
\Leftrightarrow|x-y|=\sqrt{(x-y)^{2}} \\
=\sqrt{(x+y)^{2}-4 x y}=6,
\end{array}
$$
which means the difference between the two numbers is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. It is known that fresh shiitake mushrooms contain $90 \% \sim 99 \%$ water, while dried shiitake mushrooms contain $30 \% \sim 45 \%$ water. Then, under the influence of drying, by what maximum factor can the weight of fresh shiitake mushrooms be reduced?
|
2. Let the weights of fresh mushrooms, baked mushrooms be $m_{1} \mathrm{~g}, m_{2} \mathrm{~g}$, and the weight of dried mushrooms be $x \mathrm{~g}$ (unknown). Then the range of the proportion of dried mushrooms in fresh mushrooms and baked mushrooms is
$$
\begin{array}{l}
0.01=1-0.99 \\
\leqslant \frac{x}{m_{1}} \leqslant 1-0.90=0.10, \\
0.55=1-0.45 \\
\leqslant \frac{x}{m_{2}} \leqslant 1-0.3=0.70 .
\end{array}
$$
Therefore, the maximum multiple of the weight of fresh mushrooms to the weight of baked mushrooms is
$$
\frac{m_{1}}{m_{2}}=\frac{\frac{x}{m_{2}}}{\frac{x}{m_{1}}}=\frac{0.70}{0.01}=70 .
$$
|
70
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given two circles $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent at point $A$, circle $\Gamma$ is externally tangent to $\Gamma_{1}$ and $\Gamma_{2}$ at points $B$ and $C$ respectively. Extend the chord $B A$ of circle $\Gamma_{1}$ to intersect circle $\Gamma_{2}$ at point $D$, extend the chord $C A$ of circle $\Gamma_{2}$ to intersect circle $\Gamma_{1}$ at point $E$, and extend the chords $E B$ and $D C$ to intersect circle $\Gamma$ at points $F$ and $G$ respectively: If $B C=5, B F=12$, find the length of $B G$.
|
5. First, prove that quadrilateral $B C G F$ is a rectangle:
In fact, let $K L, B M, C N$ be the common tangents of circles $\Gamma_{1}$ and $\Gamma_{2}$, $\Gamma_{1}$ and $\Gamma$, $\Gamma_{2}$ and $\Gamma$ (Figure 2). According to the inscribed angle and the angle between a tangent and a chord, we have
$$
\angle A B E=\angle E A K=\angle C A L=\angle A D C \text {. }
$$
Then $B E / / C D$
$$
\begin{array}{l}
\Rightarrow \angle C B E+\angle B C D=180^{\circ} . \\
\text { Also } \angle B C N=\angle C B M, \\
\angle A C N=\angle C A L=\angle A B E, \\
\angle A B M=\angle B A L=\angle A C D,
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore } \angle E B C=\angle D C B=90^{\circ} \\
\Rightarrow \angle B F G=\angle B C D=90^{\circ} \text {. }
\end{array}
$$
Thus, $B G=\sqrt{5^{2}+12^{2}}=13$.
|
13
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ be a function, and for any positive integers $m, n$, we have
$$
f(f(m)+f(n))=m+n .
$$
Find the value of $f(2011)$.
|
Since $f(x)$ is a function from the set of positive integers to the set of positive integers, let
$$
f(1)=p\left(p \in \mathbf{N}_{+}\right) \text {. }
$$
If $p>1$, then $p \geqslant 2$.
$$
\text { Let } p=1+b\left(b \in \mathbf{N}_{+}\right), f(b)=c\left(c \in \mathbf{N}_{+}\right) \text {. }
$$
On one hand,
$$
\begin{array}{l}
f(f(1)+f(1))=f(p+p) \\
=f(1+b+1+b)=f(2+2 b) \\
=f(f(f(1)+f(1))+f(f(b)+f(b))) \\
=f(f(2 f(1))+f(2 c))=2 f(1)+2 c \\
=2 p+2 c=2+2 b+2 c>2 .
\end{array}
$$
On the other hand, $f(f(1)+f(1))=2$. This is a contradiction.
Therefore, $p=1$.
Thus, $f(1)=1$,
$$
f(2)=f(1+1)=f(f(1)+f(1))=2 \text {. }
$$
Conjecture: $f(n)=n$.
We will prove this by mathematical induction.
When $n=1$, $f(1)=1$, so the conclusion holds.
Assume that when $n=k$, the conclusion holds.
When $n=k+1$,
$$
f(k+1)=f(f(k)+f(1))=k+1 \text {. }
$$
Thus, when $n=k+1$, the conclusion holds.
Therefore, for all $n \in \mathbf{N}_{+}, f(n)=n$.
Hence, $f(2011)=2011$.
(Pan Tie, Tianjin Experimental High School, 300074)
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For the quadratic function $y=x^{2}+b x+c$, the vertex of its graph is $D$, and it intersects the positive x-axis at points $A$ and $B$ from left to right, and the positive y-axis at point $C$. If $\triangle A B D$ and $\triangle O B C$ are both isosceles right triangles (where $O$ is the origin), then $b+2 c=$
|
2. 2 .
From the given information, we have
$$
\begin{array}{l}
C(0, c) γ A\left(\frac{-\dot{b}-\sqrt{b^{2}-4 c}}{2}, 0\right), \\
B\left(\frac{-b+\sqrt{b^{2}-4 c}}{2}, 0\right), D\left(-\frac{b}{2},-\frac{b^{2}-4 c}{4}\right) .
\end{array}
$$
Draw $D E \perp A B$ at point $E$. Then $2 D E=A B$,
which means $2 \times \frac{b^{2}-4 c}{4}=\sqrt{b^{2}-4 c}$.
Solving this, we get $\sqrt{b^{2}-4 c}=0$ (discard) or 2 .
$$
\begin{array}{l}
\text { Also, } O C=O B \\
\Rightarrow c=\frac{-b+\sqrt{b^{2}-4 c}}{2} \\
\Rightarrow b+2 c=\sqrt{b^{2}-4 c}=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The value of the positive integer $n$ that makes $2^{n}+256$ a perfect square is $\qquad$
|
3. 11.
When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$, if it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a natural number). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, then $k=1$, thus, $n=11$.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 2, given that $A B$ is the diameter of $\odot O$, chord $C D$ intersects $A B$ at point $E$, a tangent line through $A$ intersects the extension of $C D$ at point $F$, and $D$ is the midpoint of $E F$. If $D E=\frac{3}{4} C E, A C=$ $8 \sqrt{5}$, then $A B=$ . $\qquad$
|
4.24.
Let $C E=4 x, A E=y$. Then
$$
D F=D E=3 x, E F=6 x \text {. }
$$
Connect $A D, B C$.
Since $A B$ is the diameter of $\odot O$ and $A F$ is the tangent of $\odot O$, we have
$$
\angle E A F=90^{\circ}, \angle A C D=\angle D A F \text {. }
$$
Since $D$ is the midpoint of the hypotenuse $E F$ of the right triangle $\triangle A E F$,
$$
\begin{array}{l}
D A=D E=D F \\
\Rightarrow \angle A F D=\angle D A F=\angle A C F \\
\Rightarrow A F=A C=8 \sqrt{5} .
\end{array}
$$
In the right triangle $\triangle A E F$, by the Pythagorean theorem,
$$
E F^{2}=A E^{2}+A F^{2} \Rightarrow 36 x^{2}=y^{2}+320 \text {. }
$$
Let $B E=z$.
Since $\triangle A D E \backsim \triangle C B E$, we have
$$
\frac{A D}{B C}=\frac{D E}{B E}=\frac{A E}{C E} \text {. }
$$
Thus, $B C=C E$, and
$$
\begin{array}{l}
C E \cdot D E=A E \cdot B E \\
\Rightarrow y z=4 x \cdot 3 x=12 x^{2} .
\end{array}
$$
Therefore, $y^{2}+320=3 y z$.
In the right triangle $\triangle A C B$, by the Pythagorean theorem,
$$
A B^{2}=A C^{2}+B C^{2} \Rightarrow(y+z)^{2}=320+z^{2} \text {. }
$$
Thus, $y^{2}+2 y z=320$.
Solving equations (1) and (2) simultaneously, we get $y=8, z=16$.
Therefore, $A B=A E+B E=24$.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}
$$
The value equals $\qquad$ [5]
(2009, Beijing Middle School Mathematics Competition (Grade 8))
|
Solution 1
$$
p=\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \text {. }
$$
Then $p^{2}=2 x+2 \sqrt{x^{2}-(2 \sqrt{x-1})^{2}}$
$$
=2 x+2 \sqrt{(x-2)^{2}} \text {. }
$$
Since $1 \leqslant x \leqslant 2$, therefore,
$$
\begin{array}{l}
p^{2}=2 x+2 \sqrt{(x-2)^{2}} \\
=2 x+2(2-x)=4 .
\end{array}
$$
And $p \geqslant 0$, so $p=2$.
Solution 2 Let
$$
p=\sqrt{x+2 \sqrt{x-1}}, q=\sqrt{x-2 \sqrt{x-1}}
$$
Then $p^{2}+q^{2}=2 x, p q=2-x$.
Since $(p+q)^{2}=p^{2}+q^{2}+2 p q$
$$
=2 x+2(2-x)=4(p \geqslant 0, q \geqslant 0) \text {, }
$$
Therefore, $p+q=2$, that is, the original expression $=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $f(x)$ be a continuous even function, and when $x>0$, $f(x)$ is a strictly monotonic function. Then the sum of all $x$ that satisfy $f(x) = f\left(\frac{x+3}{x+4}\right)$ is ( ).
(A) -3
(B) -8
(C) 3
(D) 8
|
8. B.
When $f(x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $x=\frac{x+3}{x+4}$, we get $x^{2}+3 x-3=0$. At this point,
$$
x_{1}+x_{2}=-3 \text {. }
$$
Since $f(x)$ is a continuous even function, another scenario is $f(-x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $-x=\frac{x+3}{x+4}$, which gives
$$
x^{2}+5 x+3=0 \text {. }
$$
Therefore, $x_{3}+x_{4}=-5$.
Thus, the sum of all $x$ is
$$
-3+(-5)=-8 \text {. }
$$
|
-8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ [6] $(2008$, National Junior High School Mathematics Competition)
|
Solve: Given $a=\frac{\sqrt{5}-1}{2} \Rightarrow 2 a+1=\sqrt{5}$.
Square both sides and simplify to get
$$
a^{2}+a=1 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=\frac{a^{3}\left(a^{2}+a\right)-2 a^{3}-\left(a^{2}+a\right)+2}{a(a+1)(a-1)} \\
=\frac{1-a^{3}}{\left(a^{2}+a\right)(a-1)} \\
=-\left(a^{2}+a+1\right)=-2 .
\end{array}
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. If the constant term in the expansion of $\left(a \sqrt{x}-\frac{1}{\sqrt{x}}\right)^{6}$ is -160, then $\int\left(3 x^{2}-1\right) \mathrm{d} x=$. $\qquad$
|
$$
\begin{array}{l}
T_{r+1}=\mathrm{C}_{6}^{r}(a \sqrt{x})^{6-r}\left(-\frac{1}{\sqrt{x}}\right)^{r} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{\frac{6-r}{2}-\frac{r}{2}} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{3-r} .
\end{array}
$$
Let $3-r=0$. Then $r=3$.
The constant term is
$$
\begin{array}{l}
-\mathrm{C}_{6}^{3} a^{3}=-20 a^{3}=-160 \Rightarrow a=2 . \\
\text { Hence } \int_{0}^{3}\left(3 x^{2}-1\right) \mathrm{d} x=\left.\left(x^{3}-x\right)\right|_{0} ^{3}=18 .
\end{array}
$$
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given $a=\sqrt{5}-1$. Then the value of $2 a^{3}+7 a^{2}-2 a$ -12 is $\qquad$ [7]
(2010) "Mathematics Weekly" Cup National Junior High School Mathematics Competition)
|
Given $a=\sqrt{5}-1 \Rightarrow a+1=\sqrt{5}$. Squaring both sides, we get
$$
a^{2}+2 a=4 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=2 a\left(a^{2}+2 a\right)+3 a^{2}-2 a-12 \\
=3 a^{2}+6 a-12 \\
=3\left(a^{2}+2 a\right)-12=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $M=(5+\sqrt{24})^{2 n}\left(n \in \mathbf{N}_{+}\right), N$ be the fractional part of $M$. Then the value of $M(1-N)$ is $\qquad$
|
10.1.
Since $(5+\sqrt{24})^{2 n}+(5-\sqrt{24})^{2 n}$ is a positive integer, and $0<(5-\sqrt{24})^{2 n}<1$, therefore,
$$
\begin{array}{l}
N=1-(5-\sqrt{24})^{2 n} \\
M(1-N)=(5+\sqrt{24})^{2 n}(5-\sqrt{24})^{2 n}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. As shown in Figure $2, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$.
|
16. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), \( C(0, c) \), and assume \( b > c \). The line \( l_{P B} \) is given by \( y - b = \frac{y_{0} - b}{x_{0}} x \), which can be rewritten as
\[
\left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0.
\]
The distance from the circle's center \((1, 0)\) to \( P B \) is 1, so
\[
\frac{\left|y_{0} - b + x_{0} b\right|}{\sqrt{\left(y_{0} - b\right)^{2} + x_{0}^{2}}} = 1.
\]
Thus,
\[
\left(y_{0} - b\right)^{2} + x_{0}^{2} = \left(y_{0} - b\right)^{2} + 2 x_{0} b \left(y_{0} - b\right) + x_{0}^{2} b^{2}.
\]
Given \( x_{0} > 2 \), the equation simplifies to
\[
\left(x_{0} - 2\right) b^{2} + 2 y_{0} b - x_{0} = 0.
\]
Similarly,
\[
\left(x_{0} - 2\right) c^{2} + 2 y_{0} c - x_{0} = 0.
\]
Therefore,
\[
b + c = \frac{-2 y_{0}}{x_{0} - 2}, \quad b c = \frac{-x_{0}}{x_{0} - 2}.
\]
Thus,
\[
(b - c)^{2} = \frac{4 x_{0}^{2} + 4 y_{0}^{2} - 8 x_{0}}{\left(x_{0} - 2\right)^{2}}.
\]
Since \( P\left(x_{0}, y_{0}\right) \) lies on the parabola, \( y_{0}^{2} = 2 x_{0} \). Therefore,
\[
(b - c)^{2} = \frac{4 x_{0}^{2}}{\left(x_{0} - 2\right)^{2}} \Rightarrow b - c = \frac{2 x_{0}}{x_{0} - 2}.
\]
Hence,
\[
S_{\triangle P B C} = \frac{1}{2} (b - c) x_{0} = \frac{x_{0}}{x_{0} - 2} \cdot x_{0} = \left(x_{0} - 2\right) + \frac{4}{x_{0} - 2} + 4 \geqslant 2 \sqrt{4} + 4 = 8.
\]
The equality holds when \( \left(x_{0} - 2\right)^{2} = 4 \), which gives \( x_{0} = 4 \) and \( y_{0} = \pm 2 \sqrt{2} \).
Therefore, the minimum value of \( S_{\triangle P B C} \) is 8.
(Provided by Shi Dongyang)
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x$ be a positive integer, and $x<50$. Then the number of $x$ such that $x^{3}+11$ is divisible by 12 is $\qquad$.
|
$$
\begin{array}{l}
x^{3}+11=(x-1)\left(x^{2}+x+1\right)+12 \\
=(x-1) x(x+1)+(x-1)+12 .
\end{array}
$$
Notice that $6 \mid(x-1) x(x+1)$, so $6 \mid(x-1)$;
also, $2 \mid \left(x^{2}+x+1\right)$, hence $12 \mid (x-1)$.
Since $x<50$, we have
$$
x-1=0,12,24,36,48 \text {. }
$$
Thus, $x=1,13,25,37,49$, a total of 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $x_{1}$ and $x_{2}$ are the two real roots of the equation
$$
x^{2}-2009 x+2011=0
$$
real numbers $m$ and $n$ satisfy
$$
\begin{array}{l}
2009 m x_{1}+2009 n x_{2}=2009, \\
2010 m x_{1}+2010 n x_{2}=2010 .
\end{array}
$$
Then $2011 m x_{1}+2011 n x_{2}=$ $\qquad$
|
2. -2009 .
From the given, we know
$$
\begin{array}{l}
=2010 \times 2009-2011 \times 2009=-2009 . \\
\end{array}
$$
|
-2009
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 5, in $\triangle A B C$, $D$ is the midpoint of side $A B$, and $G$ is the midpoint of $C D$. A line through $G$ intersects $A C$ and $B C$ at points $P$ and $Q$, respectively. Then the value of $\frac{C A}{C P}+\frac{C B}{C Q}$ is $\qquad$.
|
4. 4 .
As shown in Figure 12, draw $A E / / P Q$ intersecting $C D$ at point $E$, and $B F / / P Q$ intersecting the extension of $C D$ at point $F$. Then $A E / / B F$.
Since $A D = B D$, therefore, $E D = F D$.
$$
\begin{array}{c}
\text { Also, } \frac{C A}{C P} = \frac{C E}{C G}, \\
\frac{C B}{C O} = \frac{C F}{C G}, \\
C E = C D - D E, C F = C D + F D, \\
\text { hence } \frac{C A}{C P} + \frac{C B}{C Q} = \frac{C E + C F}{C G} = \frac{2 C D}{C G} = 4 .
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that the set $M$ is a subset of $\{1,2, \cdots, 2011\}$, and the sum of any four elements in $M$ cannot be divisible by 3. Then $|M|_{\text {max }}=$ $\qquad$
|
- 1. 672.
Consider the set $A=\{3,6,9, \cdots, 2010\}$,
$$
\begin{array}{l}
B=\{1,4,7, \cdots, 2011\}, \\
C=\{2,5,8, \cdots, 2009\} .
\end{array}
$$
If $M \cap A \neq \varnothing$, then $|M \cap B|<3,|M \cap C|<3,|M \cap A|<4$. Therefore, $|M|<10$.
Now assume $M \cap A=\varnothing$.
If $|M \cap B| \geqslant 2$, then $|M \cap C|<2$.
$$
\begin{array}{l}
\text { Hence }|M|=|M \cap B|+|M \cap C| \\
\leqslant|B|+1=672 .
\end{array}
$$
If $|M \cap B| \leqslant 1$, then $|M| \leqslant|C|+1=671$. Therefore, $|M|_{\max }=672$.
For example, take $M=B \cup\{2\}$.
|
672
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $n \in \mathbf{N}, n \geqslant 2, a_{i} \in\{0,1, \cdots, 9\}$
$$
\begin{array}{l}
(i=1,2, \cdots, n), a_{1} a_{2} \neq 0 \text {, and } \\
\sqrt{a_{1} a_{2} \cdots a_{n}}-\sqrt{a_{2} a_{3} \cdots a_{n}}=a_{1}, \\
\end{array}
$$
then $n=$ $\qquad$, where $\overline{a_{1} a_{2} \cdots a_{n}}$ is the $n$-digit number formed by $a_{1}, a_{2}, \cdots$ $a_{n}$.
|
2. 2 .
Let $x=\overline{a_{2} a_{3} \cdots a_{n}} \in \mathbf{N}_{+}$. Then $\overline{a_{1} a_{2} \cdots a_{n}}=10^{n-1} a_{1}+x$.
By the given condition, $\sqrt{10^{n-1} a_{1}+x}-\sqrt{x}=a_{1}$. Therefore, $10^{n-1} a_{1}+x=x+a_{1}^{2}+2 a_{1} \sqrt{x}$.
Thus, $10^{n-1}-2 \sqrt{x}=a_{1}$.
If $n \geqslant 3$, then
$$
\begin{array}{l}
10^{n-1}-2 \sqrt{x} \geqslant 10^{n-1}-2 \sqrt{10^{n-1}} \\
=\left(\sqrt{10^{n-1}}-1\right)^{2}-1 \\
\geqslant\left(\sqrt{10^{2}}-1\right)^{2}-1=80,
\end{array}
$$
which contradicts $a_{1} \leqslant 9$.
Therefore, $n=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. For any positive integer $n$, let $a_{n}$ be the smallest positive integer such that $n \mid a_{n}$!. If $\frac{a_{n}}{n}=\frac{2}{5}$, then $n=$ $\qquad$ .
|
4.25.
From $\frac{a_{n}}{n}=\frac{2}{5} \Rightarrow a_{n}=\frac{2 n}{5} \Rightarrow 51 n$.
Let $n=5 k\left(k \in \mathbf{N}_{+}\right)$. If $k>5$, then $5 k \mid k!$.
Thus, $a_{\mathrm{n}} \leqslant k<\frac{2 n}{5}$, a contradiction.
Clearly, when $k=2,3,4$, $a_{n} \neq \frac{2 n}{5}$.
Also, $a_{25}=10=\frac{2}{5} \times 25$, so $n=25$.
|
25
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. On a straight line, three points $A$, $B$, and $C$ are arranged in sequence, and $A B=6, A C=24, D$ is a point outside the line, and $D A$ $\perp A B$. When $\angle B D C$ takes the maximum value, $A D=$ $\qquad$ .
|
5. 12 .
Let $\angle B D C=\theta\left(\theta<90^{\circ}\right), \triangle B C D$'s circumcircle $\odot O$ has a radius of $R$. Then $\sin \theta=\frac{B C}{2 R}$.
When $R$ decreases, $\theta$ increases.
Therefore, when $\odot O$ is tangent to $A D$ at point $D$, $\theta$ is maximized.
At this time, $A D^{2}=A B \cdot A C=12^{2} \Rightarrow A D=12$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that on the side $C A$ of $\angle A C B$ there are 2011 points $A_{1}, A_{2}, \cdots, A_{2011}$, and on the side $C B$ there are 2011 points $B_{1}, B_{2}, \cdots, B_{2011}$, satisfying
$$
\begin{array}{l}
A_{1} A_{2}=A_{2} A_{3}=\cdots=A_{2010} A_{2011}, \\
B_{1} B_{2}=B_{2} B_{3}=\cdots=B_{2010} B_{2011} .
\end{array}
$$
If $S_{\text {quadrilateral } A_{1} A_{3} B_{3} B_{1}}=2, S_{\text {quadrilateral } A_{2} A_{4} B_{4} B_{2}}=3$, then
$$
S_{\text {quadrilateral } A_{2} A_{2011} B_{2011} B_{2009}}=
$$
. $\qquad$
|
7.2010 .
$$
\begin{array}{l}
\text { Let } \angle A C B=\alpha, C A_{1}=a, C B_{1}=b, \\
A_{i} A_{i+1}=s, B_{i} B_{i+1}=t(i=1,2, \cdots, 2010), \\
S_{\text {trapezoid } A_{i} A_{i+1} B_{i+2} B_{i}}=S_{i}(i=1,2, \cdots, 2009) . \\
\text { Then } S_{i}=\frac{1}{2}\{[a+(i+1) s][b+(i+1) t]- \\
\quad[a+(i-1) s][b+(i-1) t]\} \sin \alpha \\
=(a t+b s+2 i s t) \sin \alpha .
\end{array}
$$
Therefore, $\left\{S_{i}\right\}$ is an arithmetic sequence.
Given $S_{1}=2, S_{2}=3$, we get $S_{2009}=S_{2010}$.
|
2010
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given positive integers $a, b, c$ satisfy
$$
(a!)(b!)=a!+b!+c! \text {. }
$$
then $c\left(a^{5}+b^{5}+c^{2}\right)+3=$ $\qquad$
|
8.2011 .
$$
\begin{array}{l}
\text { Given }(a!)(b!)=a!+b!+c! \\
\Rightarrow(a!-1)(b!-1)=c!+1 \text {. }
\end{array}
$$
Assume without loss of generality that $a \geqslant b$. Clearly, $c > a$. Then $(a!-1) \mid(c!+1)$.
Also, $c!+1=\frac{c!}{a!}(a!-1)+\frac{c!}{a!}+1$, so $(a!-1) \left\lvert\,\left(\frac{c!}{a!}+1\right)\right.$
$$
\text { Hence } \frac{c!}{a!}+1 \geqslant a!-1 \Rightarrow c!\geqslant a!(a!-2) \text {. }
$$
Also, $c!=(a!-1) \backslash a!\leqslant a!(a!-2)$, combining with equation (1) we get $a=b$, and $c!=a!(a!-2)$.
Clearly, $a \geqslant 3$.
If $c \geqslant a+3$, then $\frac{c!}{a!} \equiv 0(\bmod 3)$.
But $a!-2 \equiv 1(\bmod 3)$, a contradiction.
If $c=a+2$, then
$$
\begin{array}{l}
(a+2)(a+1)=a!-2 \\
\Rightarrow a^{2}+3 a+4=a! \\
\Rightarrow a \mid 4 \Rightarrow a=4,
\end{array}
$$
which does not satisfy equation (2), a contradiction.
If $c=a+1$, then
$$
a+1=a!-2 \Rightarrow a+3=a!\Rightarrow a=3 \text {. }
$$
Thus, $a=b=3, c=4$.
In this case, $c\left(a^{5}+b^{5}+c^{2}\right)+3=2011$.
|
2011
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. At a concert, there are 20 singers who will perform. For each singer, there is a set of other singers (possibly an empty set) such that he wishes to perform later than all the singers in this set. Question: Is there a way to have exactly 2,010 ways to order the singers so that all their wishes are satisfied?
|
1. Such examples exist.
A ranking of singers that satisfies everyone's wishes is called "good".
If for a set of wishes of $k$ singers there exist $N$ good rankings, then $N$ is called "achievable by $k$ singers" (or simply "$k$-achievable").
Next, we prove that 2010 is "20-achievable".
First, we prove a lemma.
Lemma Assume $n_{1}$ and $n_{2}$ are $k_{1}$-achievable and $k_{2}$-achievable, respectively. Then $n_{1} n_{2}$ is $\left(k_{1}+k_{2}\right)$-achievable.
Proof Suppose singers $A_{1}, A_{2}, \cdots, A_{k_{1}}$ (some of whom have wishes) can achieve $n_{1}$, and singers $B_{1}, B_{2}, \cdots, B_{k_{2}}$ (some of whom have wishes) can achieve $n_{2}$, and add the wish for each singer $B_{i}$: to perform later than all singers $A_{j}$. Then the good rankings of singers have the form
$$
\left(A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{k_{1}}}, B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{k_{2}}}\right),
$$
where, ( $\left.A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{k_{1}}}\right)$ is a good ranking for $A_{i}$, and $\left(B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{k_{2}}}\right)$ is a good ranking for $B_{j}$.
Conversely, every ranking of this form is clearly good, so the number of good rankings is $n_{1} n_{2}$.
Back to the original problem.
By the lemma, construct the number of achievable rankings for 4 singers, 3 singers, and 13 singers as $5, 6, 67$, respectively. Then
$$
2010=5 \times 6 \times 67
$$
is achievable by $4+3+13=20$ singers.
Examples of these three scenarios are shown in Figures 1 to 3 (the numbers in parentheses are the number of good rankings, and wishes are indicated by arrows).
In Figure 1, $c$ wishes to perform later than $a$ and $b$, and $d$ wishes to perform later than $b$. Then there are exactly 5 good rankings
$$
\begin{array}{l}
(a, b, c, d), (a, b, d, c), (b, a, c, d), \\
(b, a, d, c), (b, d, a, c).
\end{array}
$$
In Figure 2, every ranking is good, totaling 6, as they have no wishes.
In Figure 3, the order of $a_{1}, a_{2}, \cdots, a_{11}$ is fixed on this line, singer $x$ can be after each $a_{i}(i \leqslant 9)$, singer $y$ can be before each $a_{j}(j \geqslant 5)$, and when between $a_{i}$ and $a_{i+1}(5 \leqslant i \leqslant 8)$, the order of the two singers can be swapped.
Thus, the number of good rankings is
$$
9 \times 7+4=67.
$$
|
2010
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (50 points) Given that $p$ is a prime number, the fractional part of $\sqrt{p}$ is $x$, and the fractional part of $\frac{1}{x}$ is $\frac{\sqrt{p}-31}{75}$. Find all prime numbers $p$ that satisfy the condition.
|
3. Let $p=k^{2}+r$, where $k, r$ are integers, and satisfy $0 \leqslant r \leqslant 2 k$.
Since $\frac{\sqrt{p}-31}{75}$ is the fractional part of $\frac{1}{x}$, we have $0 \leqslant \frac{\sqrt{p}-31}{75} < 1$. Let
$\frac{1}{x}=\frac{1}{\sqrt{p}-k}=N+\frac{\sqrt{p}-31}{75}(N \geqslant 1)$.
Then $\frac{\sqrt{p}+k}{r}=N+\frac{\sqrt{p}-31}{75}$.
By equating the irrational and rational parts, we get
$\frac{1}{r}=\frac{1}{75}, \frac{k}{r}=\frac{75 N-31}{75}$
$\Rightarrow r=75, k=75 N-31$.
If $N \geqslant 2$, then $k \geqslant 75 \times 2-31=119$.
Thus, $[\sqrt{p}]=k \geqslant 119$, which contradicts $31 \leqslant \sqrt{p}<106$.
Therefore, $N=1, k=44$.
Hence, $p=44^{2}+75=2011$, and 2011 is a prime number.
|
2011
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. (50 points) A scientist stored the design blueprint of a time machine on a computer, setting the file opening password as a permutation of $\{1,2, \cdots, 64\}$. They also designed a program that, when eight positive integers between 1 and 64 are input each time, the computer will indicate the order (from left to right) of these eight numbers in the password. Please design an operation scheme such that the password can be determined with at most 45 inputs.
|
8. Prepare $n^{2}(n=8)$ cards, the front side of which are numbered $1,2, \cdots, n^{2}$, and the back side corresponds to the position of the number in the password (counting from the left). Of course, the operator does not know the numbers on the back side in advance.
First, divide the $n^{2}$ cards into $n$ groups (each group has $n$ cards). In the first $n$ operations, input the card numbers of each group once, so you can know the size order of the numbers on the back of each group (also recorded as the size order of the cards).
Arrange the cards in each group in ascending order and place them on the table, forming an $n \times n$ matrix $\left(a_{i j}\right)$, where $a_{i j}$ is the card number of the $i$-th card in the $j$-th group, and the corresponding number on the back is $b_{i j}$.
Second, input two more times, which are
$\left\{a_{1 j} \mid j=1,2, \cdots, n\right\},\left\{a_{n j} \mid j=1,2, \cdots, n\right\}$.
These $2 n$ cards are called βoriginal cardsβ (the former is βoriginal small cardβ, the latter is βoriginal large cardβ). Then
$$
\begin{array}{l}
\min _{1 \leqslant j \leqslant n} b_{1 j}=\min _{1 \leqslant j \leqslant n} \min _{1 \leqslant i \leqslant n} b_{i j}=\min _{1 \leqslant i, j \leqslant n} b_{i j}, \\
\max _{1 \leqslant j \leqslant n} b_{n j}=\max _{1 \leqslant j \leqslant n} \max _{1 \leqslant i \leqslant n} b_{i j}=\max _{1 \leqslant i, j \leqslant n} b_{i j} .
\end{array}
$$
Thus, we know the card numbers of the smallest and largest numbers in $\left\{b_{i j}\right\}$ (i.e., the first and last two numbers of the password). Remove these two cards and move the card below (or above) the front (or back) one to its position (called βnew small (or large) cardβ).
Then, input the card numbers of these two new cards and the $\frac{n}{2}-1$ smaller (or larger) cards among the original small (or large) cards, so you can know the card numbers of the smallest and largest numbers among the remaining $n^{2}-2$ $b_{i j}$ (i.e., the 2nd and $n^{2}-1$th numbers of the password).
Similarly, by following the two operation methods below, all cards can be arranged in ascending order of the numbers on their back, and the card numbers on the front form the password.
Operation A: When the number of new small and large cards is less than $\frac{n}{2}$, input $\frac{n}{2}$ smaller cards and $\frac{n}{2}$ larger cards (i.e., take $\frac{n}{2}$ cards from the 1st and $n$th rows on the table) after combining these new cards with some smaller (or larger) original cards. Then, you can know the card numbers of the smallest and largest cards on the table. Remove these two cards and move the card below (or above) the front (or back) one to its position (called new small (or large) card).
Operation B: When the number of new small (or large) cards is $\frac{n}{2}$, input the card numbers of all cards in the 1st (or $n$th) row (i.e., $\frac{n}{2}$ original small (or large) cards and $\frac{n}{2}$ new small (or large) cards, which are all renamed as original cards). Then, you can know the card number of the smallest (or largest) card on the table. Remove this card and move the card below (or above) it to its position (called new small (or large) card).
The feasibility of the operations is obvious, and except for the first $n$ operations, all subsequent operations can be considered as either operation A or B. The $n+1$th and $n+2$th operations can be considered as operation B, and the $n+3$th operation can be considered as operation A.
Assume that operations A and B are performed $x$ and $y$ times, respectively, until all cards are removed.
Since each operation A removes 2 cards, and each operation B removes 1 card, and the last operation can remove at most $n$ cards, we have
$$
2 x+y \leqslant n^{2}-n+2 \text {. }
$$
Note that each operation A can increase the number of new cards by at most 2, and each operation B can decrease the number of new cards by $\frac{n}{2}-1$.
If the last operation is operation B, then
$$
2 x+2 \geqslant\left(\frac{n}{2}-1\right)(y-2)+1 \text {; }
$$
If the last operation is operation A, then
$$
2(x-1)+2 \geqslant\left(\frac{n}{2}-1\right)(y-2) \text {. }
$$
From equations (1) and (2)
$$
\begin{array}{l}
\Rightarrow 2 x+y-1 \geqslant \frac{n}{2}(y-2) \\
\Rightarrow n^{2}-n+1 \geqslant \frac{n}{2}(y-2) \\
\Rightarrow \frac{y}{2} \leqslant n+\frac{1}{n} .
\end{array}
$$
Thus, the total number of operations is
$$
\begin{array}{l}
n+x+y=n+\frac{(2 x+y)+y}{2} \\
\leqslant n+\frac{n^{2}-n+2}{2}+n+\frac{1}{n} \\
=\frac{(n+1)(n+2)}{2}+\frac{1}{n}=45 \frac{1}{8} \\
\Rightarrow n+x+y \leqslant 45 .
\end{array}
$$
[Note] If a column is emptied ahead of time, a card can be taken from any row between the 2nd and $(n-1)$th rows to fill the gap.
(Proposed by Li Jianquan, Li Baoyi, Ding Yunlong, Pan Tie, Song Qiang)
|
45
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the set $M=\{2,0,11\}$. If $A \varsubsetneqq M$, and $A$ contains at least one even number, then the number of sets $A$ that satisfy the condition is $\qquad$ .
|
1. 5.
The sets $A$ that satisfy the conditions are $\{2\}, \{0\}, \{2,0\}, \{2,11\}, \{0,11\}$. There are 5 in total.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In the array of numbers shown in Figure 1, the three numbers in each row form an arithmetic sequence, and the three numbers in each column also form an arithmetic sequence. If $a_{22}=2$, then the sum of all nine numbers is equal to
δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζε¦δΈοΌ
5. In the array of numbers shown in Figure 1, the three numbers in each row form an arithmetic sequence, and the three numbers in each column also form an arithmetic sequence. If $a_{22}=2$, then the sum of all nine numbers is equal to
|
5.18.
From the problem, we have
$$
\begin{array}{l}
a_{11}+a_{13}=2 a_{12}, a_{21}+a_{23}=2 a_{22}, \\
a_{31}+a_{33}=2 a_{32}, a_{12}+a_{32}=2 a_{22} .
\end{array}
$$
Thus, the sum of all nine numbers is $9 a_{n}=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{1}{p}, & x=\frac{q}{p} ; \\
0, & x \neq \frac{q}{p},
\end{array}\right.
$$
where $p$ and $q$ are coprime, and $p \geqslant 2$. Then the number of $x$ values that satisfy $x \in[0,1]$ and $f(x)>\frac{1}{5}$ is $\qquad$ .
|
7.5.
Obviously, $x=\frac{q}{p}$ (otherwise, $f(x)=0$). At this point, from $f(x)=\frac{1}{p}>\frac{1}{5}$, we get $p<5$, i.e., $p=2,3,4$. When $p=2$, $x=\frac{1}{2}$; when $p=3$, $x=\frac{1}{3}, \frac{2}{3}$; when $p=4$, $x=\frac{1}{4}, \frac{3}{4}$.
Therefore, there are 5 values of $x$ that satisfy the condition.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $p$ and $q$ are both prime numbers, and $7p+q$, $2q+11$ are also prime numbers. Then $p^{q}+q^{p}=$ $\qquad$ .
|
8. 17 .
Since $7 p+q$ is a prime number, and $7 p+q>2$, $7 p+q$ must be an odd number.
Therefore, one of $p$ or $q$ must be even (which can only be 2).
Clearly, $q \neq 2$ (otherwise, $2 q+11=15$, which is not a prime number). Thus, $p=2$.
At this point, $14+q$ and $2 q+11$ are both prime numbers.
If $q=3 k+1\left(k \in \mathbf{N}_{+}\right)$, then
$$
14+q=3(k+5) \text {, }
$$
is not a prime number;
If $q=3 k+2\left(k \in \mathbf{N}_{+}\right)$, then
$$
2 q+11=3(2 k+5) \text {, }
$$
is not a prime number.
Therefore, $q=3 k\left(k \in \mathbf{N}_{+}\right)$, and the only solution is $q=3$.
Thus, $p^{q}+q^{p}=2^{3}+3^{2}=17$.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 As shown in Figure 7, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3$, the distance from point $P$ to line $A N$ is $P D=12, A D=$
30, and a line is drawn through $P$ intersecting
$A N$ and $A M$ at points
$B$ and $C$ respectively. Find the minimum
value of the area of $\triangle A B C$. ${ }^{[7]}$
|
Solve As shown in Figure 7, it can be proven: when the line moves to $B_{0} P=P C_{0}$, the area of $\triangle A B C$ is minimized.
Draw $C_{0} Q / / A N$, intersecting $C B$ at point $Q$.
Thus, $\triangle P C_{0} Q \cong \triangle P B_{0} B$.
Also, $S_{\triangle P C_{0} C} \geqslant S_{\triangle P C_{0} Q}=S_{\triangle P B_{0} B}$, then
$S_{\triangle \triangle B C} \geqslant S_{\triangle A B_{0} C_{0}}$.
Next, we find the area of $\triangle A B_{0} C_{0}$.
As shown in Figure 7, draw the altitude $C_{0} E$ from point $C_{0}$ to side $A B$, with $E$ as the foot of the perpendicular.
Since $C_{0} E / / P D$, and $C_{0} P=P B_{0}$, we get
$C_{0} E=2 P D=24$.
In the right triangle $\triangle C_{0} A E$,
$\tan \angle M A N=\frac{C_{0} E}{A E}=3$.
Therefore, $A E=8$.
Then $E D=A D-A E=B_{0} D$.
Hence $\left(S_{\triangle I B C}\right)_{\text {min }}=S_{\triangle M B_{0} C_{0}}=\frac{1}{2} A B_{0} \cdot C_{0} E$
$=\frac{1}{2} \times 52 \times 24=624$.
|
624
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. From the set $\{1,2, \cdots, 10\}$, any two non-adjacent numbers are taken and multiplied. Then the sum of all such products is equal to
|
5. 990 .
Take any two numbers, multiply them, and then find their sum:
$$
\begin{array}{l}
S_{1}=\frac{1}{2} \sum_{k=1}^{10} k\left(\sum_{i=1}^{10} i-k\right) \\
=\frac{1}{2} \sum_{k=1}^{10} k(55-k)=\frac{1}{2} \sum_{k=1}^{10}\left(55 k-k^{2}\right) \\
=\frac{1}{2}(55 \times 55-385)=1320,
\end{array}
$$
Among them, the sum that does not meet the condition is
$$
\begin{array}{l}
S_{2}=\sum_{k=1}^{9} k(k+1)=\sum_{k=1}^{2}\left(k^{2}+k\right) \\
=285+45=330 .
\end{array}
$$
Therefore, the required sum is
$$
S=S_{1}-S_{2}=1320-330=990 .
$$
|
990
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given positive integers $a_{1}, a_{2}, \cdots, a_{18}$ satisfying
$$
\begin{array}{l}
a_{1}<a_{2}<\cdots<a_{18}, \\
a_{1}+a_{2}+\cdots+a_{18}=2011 .
\end{array}
$$
Then the maximum value of $a_{9}$ is
|
7.193.
To maximize $a_{9}$, $a_{1}, a_{2}, \cdots, a_{8}$ should be as small as possible, and $a_{10}, a_{11}, \cdots, a_{18}$ should be as close to $a_{9}$ as possible. Therefore, we take $a_{1}, a_{2}, \cdots, a_{8}$ to be $1, 2, \cdots, 8$, respectively, with their sum being 36.
Let $a_{9}=n$. Then
$$
\begin{array}{l}
a_{10}=n+1, a_{11}=n+2, \cdots, a_{18}=n+9 \\
\Rightarrow n=193 .
\end{array}
$$
Thus, the maximum value of $a_{9}$ is 193.
|
193
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Define the sequence $\left\{a_{n}\right\}: a_{n}=n^{3}+4\left(n \in \mathbf{N}_{+}\right)$, let $d_{n}=\left(a_{n}, a_{n+1}\right)$. Then the maximum value of $d_{n}$ is $\qquad$
|
8.433.
Given $d_{n} \mid\left(n^{3}+4,(n+1)^{3}+4\right)$, we know $d_{n} \mid\left(n^{3}+4,3 n^{2}+3 n+1\right)$.
Then $d_{n} \mid\left[-3\left(n^{3}+4\right)+n\left(3 n^{2}+3 n+1\right)\right]$,
and $\square$
$$
\begin{array}{l}
d_{n} \mid\left(3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid\left(3 n^{2}+n-12,3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid\left(2 n+13,3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid(2 n+13,33 n-2) \\
\Rightarrow d_{n} \mid[-2(33 n-2)+33(2 n+13)] \\
\Rightarrow d_{n} \mid 433 .
\end{array}
$$
Therefore, $\left(d_{n}\right)_{\max } \leqslant 433$.
It is easy to see that, $\left(a_{210}, a_{211}\right)=433$.
Thus, $\left(d_{n}\right)_{\max }=433$.
|
433
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find all odd prime numbers $p$ such that
$$
p \mid \sum_{k=1}^{2011} k^{p-1} .
$$
|
If $p>2011$, then for $k(1 \leqslant k \leqslant 2011)$ by Fermat's Little Theorem we have
$k^{p-1} \equiv 1(\bmod p)$.
Thus $\sum_{k=1}^{2011} k^{p-1} \equiv 2011 \not\equiv 0(\bmod p)$, a contradiction.
Therefore, $p \leqslant 2011$.
Let $2011=p q+r(0 \leqslant r < q)$, then if $r>q$, we have $q=r$, so
$$
2011=p q+r=(p+1) r \text {. }
$$
This implies $p$ is even, which contradicts $p$ being a prime.
If $p \leqslant q$, then $2011=p q+r \geqslant p^{2}$.
Thus, the odd primes satisfying $p^{2} \leqslant 2011$ are
$$
3,5,7,11,13, 17, 19,23,29,31,37,41,43 \text {. }
$$
Checking, we find that only $p=3$ satisfies the condition.
In summary, $p=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The equation concerning $x, y$
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}
$$
has $\qquad$ groups of positive integer solutions $(x, y)$.
|
3. 12 .
From $\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}$, we get
$$
\begin{array}{l}
x y-2011 x-2011 y-2011=0 \\
\Rightarrow(x-2011)(y-2011) \\
\quad=2011 \times 2012 \\
=2^{2} \times 503 \times 2011 .
\end{array}
$$
Thus, the positive integer solutions of the original equation are
$$
(2+1)(1+1)(1+1)=12 \text { (sets). }
$$
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. For $n$ consecutive natural numbers, if each number is written in its standard prime factorization form, and each prime factor appears an odd number of times, such $n$ consecutive natural numbers are called a βconsecutive $n$ strange groupβ (for example, when $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{1}$, $24=2^{3} \times 3^{1}$, then $22, 23, 24$ form a βconsecutive 3 strange groupβ). The maximum possible value of $n$ in a consecutive $n$ strange group is . $\qquad$
|
5.7.
For example: $29, 30, 31, 32, 33, 34, 35$ is a consecutive 7 peculiar group.
Below is the proof: When $n \geqslant 8$, there does not exist a consecutive $n$ peculiar group.
Otherwise, one of them (denoted as $m$) is divisible by 8, so among these $n$ numbers, there must be $m-4$ or $m+4$, which is divisible by 4 but not by 8, contradicting the condition that "each prime factor appears an odd number of times."
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=(\sqrt{3}+\sqrt{2})^{2 n}\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
Let $b_{n}=a_{n}+\frac{1}{a_{n}}$.
(1) Find the recurrence relation between $b_{n+2} γ b_{n+1} γ b_{n}$;
(2) Find the unit digit of the integer part of $a_{2011}$.
|
-γ(1) From the given, we have
$$
\begin{array}{l}
b_{n}=(\sqrt{3}+\sqrt{2})^{2 n}+(\sqrt{3}-\sqrt{2})^{2 n} \\
\Rightarrow b_{n}=(5+2 \sqrt{6})^{n}+(5-2 \sqrt{6})^{n} .
\end{array}
$$
Then $b_{n+2}=[(5+2 \sqrt{6})+(5-2 \sqrt{6})]$.
$$
\begin{array}{l}
{\left[(5+2 \sqrt{6})^{n+1}+(5-2 \sqrt{6})^{n+1}\right]-} \\
(5+2 \sqrt{6})(5-2 \sqrt{6})\left[(5+2 \sqrt{6})^{n}+(5-2 \sqrt{6})^{n}\right] .
\end{array}
$$
Thus, $b_{n+2}=10 b_{n+1}-b_{n}$.
$$
\begin{array}{l}
\text { (2) Given } b_{1}=10, b_{2}=98, \\
b_{n+2}=10 b_{n+1}-b_{n},
\end{array}
$$
we know that $b_{n}$ is an integer.
If $b_{n}$ is divisible by 10, then $b_{n+2}$ is also divisible by 10.
Since $b_{1}=10$, it follows that $b_{2 k+1}(k \in \mathbf{N})$ is an integer, and the units digit of $b_{2011}$ is 0.
Also, $0<(5-2 \sqrt{6})^{2011}<1$, so the units digit of the integer part of $a_{2011}$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, let $n$ sets $A_{1}, A_{2}, \cdots, A_{n}$ be a partition of the set $A=\{1,2, \cdots, 29\}$, and the sum of any elements in $A_{i}(i=1,2$, $\cdots, n)$ does not equal 30. Find the minimum possible value of $n$.
[Note] If the non-empty subsets $A_{1}, A_{2}, \cdots, A_{n}$ $\left(n \in \mathbf{N}_{+}, n \geqslant 2\right)$ of set $A$ satisfy
$$
\begin{array}{l}
A_{i} \cap A_{j}=\varnothing(i \neq j), \\
A_{1} \cup A_{2} \cup \cdots \cup A_{n}=A,
\end{array}
$$
then $A_{1}, A_{2}, \cdots, A_{n}$ is called a partition of set $A$.
|
The minimum value of $n$ is 3.
First, we decompose
$$
\begin{array}{l}
A_{1}=\{1,2, \cdots, 7\}, \\
A_{2}=\{10,11, \cdots, 15,21,22\}, \\
A_{3}=\{8,9,16,17, \cdots, 20,23,24, \cdots, 29\}
\end{array}
$$
which satisfies the conditions.
Next, we prove that $n=2$ does not satisfy the conditions.
Assume, for contradiction, that $A$ can be decomposed into two sets $A_{1}$ and $A_{2}$, and without loss of generality, let $1 \in A_{1}$, and the smallest element in $A_{1}$ other than 1 be $k$. Clearly, $k$ exists.
(1) If $k \geqslant 9$, then $2,3,4,6,7,8 \in A_{2}$, and $2+3+4+6+7+8=30$, which is a contradiction.
(2) If $2 \leqslant k \leqslant 8$, then consider $k+1$.
If $k+1 \in A_{2}$, then $29-k > k+1$, and no matter which set the element $29-k$ belongs to, it leads to a contradiction. Therefore, $k+1 \in A_{1}$.
Similarly, $k+2 \in A_{1}, k+3 \in A_{1}, \cdots, 11 \in A_{1}$, then $9, 10, 11 \in A_{1}$. But $9+10+11=30$, which is a contradiction.
Thus, $n=2$ does not satisfy the conditions.
In conclusion, the minimum value of $n$ is 3.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 As shown in Figure 6, in the regular nonagon ABCDEFGHI, it is known that $A E=1$. Then the length of $A B + A C$ is
The translation preserves the original text's line breaks and format.
|
Solving: Since the sum of the interior angles of a regular nonagon is
$$
(9-2) \times 180^{\circ}=1260^{\circ} \text {, }
$$
and each interior angle is $140^{\circ}$, therefore,
$$
\angle C A B=\left(180^{\circ}-140^{\circ}\right) \div 2=20^{\circ} \text {. }
$$
Connecting $A H$, and drawing $H M$ and $G N$ perpendicular to $A E$ at points $M$ and $N$ respectively.
Then $\angle H A M=3 \times 20^{\circ}=60^{\circ}$
$$
\Rightarrow \angle A H M=30^{\circ}, A H=2 A M \text {. }
$$
Let $A M=E N=x, M N=y$.
Since quadrilateral $H G N M$ is a rectangle, $H G=y$, which is the side length of the regular nonagon.
Thus, $A B+A C=y+2 x$.
And $A E=x+y+x=1$, so $A B+A C=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.3. There is a convex 2011-gon on the blackboard. Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw?
|
9.3.4016.
Use induction to prove: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. We can sequentially draw $2n-6$ diagonals as follows: $A_{2} A_{4}, A_{3} A_{5}$, $A_{4} A_{6}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_{n-1}$.
We will use mathematical induction to prove that at most $2n-6$ diagonals can be drawn.
When $n=3$, the conclusion is obvious. Assume that the last diagonal drawn is $A_{1} A_{k}$. It can intersect at most one of the previously drawn diagonals (if it exists, let it be $d$) at an internal point. All the drawn diagonals, except for $A_{1} A_{k}$ and $d$, are entirely within the $k$-sided polygon $A_{1} A_{2} \cdots A_{k}$ and the $(n+2-k)$-sided polygon $A_{k} A_{k+1} \cdots A_{n} A_{1}$. By the induction hypothesis, we have at most
$$
\begin{array}{l}
(2 k-6)+[2(n+2-k)-6] \\
=2 n-8 \text { (diagonals). }
\end{array}
$$
|
4016
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let four distinct real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\left(a^{2011}-c^{2011}\right)\left(a^{2011}-d^{2011}\right)=2011, \\
\left(b^{2011}-c^{2011}\right)\left(b^{2011}-d^{2011}\right)=2011 . \\
\text { Then }(a b)^{2011}-(c d)^{2011}=(\quad) .
\end{array}
$$
Then $(a b)^{2011}-(c d)^{2011}=(\quad)$.
(A) -2012
(B) -2011
(C)2012
(D) 2011
|
4. B.
From $a \neq b$, we know $a^{2011} \neq b^{2011}$.
Thus, $a^{2011}$ and $b^{2011}$ are the two distinct real roots of the quadratic equation in $x$:
$$
\left(x-c^{2011}\right)\left(x-d^{2011}\right)=2011,
$$
which is
$$
x^{2}-\left(c^{2011} \div d^{2011}\right) x+(c d)^{2011}-2011=0.
$$
By Vieta's formulas, we get
$$
\begin{array}{l}
a^{2011} b^{2011}=(c d)^{2011}-2011 . \\
\text { Therefore, }(a b)^{2011}-(c d)^{2011}=-2011 .
\end{array}
$$
|
-2011
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. Figure 2 is a part of the HZ district map. A river runs through the district, with the two banks being the broken lines $A-B-C$ and $D-O-E$, and there are two locations $M$ and $N$. Two bridges perpendicular to the riverbanks and roads are to be built to connect $M$ and $N$ to both banks of the river, making the total bridge and road length between $M$ and $N$ the shortest. If
$$
\begin{array}{l}
A(3,15), B(3,3), C(15,3), D(0,15), \\
E(15,0), M(-16,12), N(11,-1),
\end{array}
$$
then the minimum total bridge and road length between $M$ and $N$ is ( ).
(A) 33
(B) 32
(C) 31
(D) 30
|
6. B.
Translate point $M(-16,12)$ 3 units to the left to get point $M^{\prime}(-13,12)$, and translate point $N(11,-1)$ 3 units upwards to get point $N^{\prime}(11,2)$. Connect $M^{\prime} N^{\prime}$, which intersects $A B$ and $B C$ at points $P$ and $Q$ respectively.
Then the minimum length of the road is
$$
M^{\prime} N^{\prime}=\sqrt{(-13-11)^{2}+(12-2)^{2}}=26 \text {. }
$$
Therefore, the minimum length of the bridge road between $M$ and $N$ is
$$
M^{\prime} N^{\prime}+3+3=26+6=32 \text {. }
$$
The minimum value is achieved when bridges are built at points $P$ and $Q$.
|
32
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given a prime number $p$ such that $p^{3}-6 p^{2}+9 p$ has exactly 30 positive divisors. Then the smallest value of $p$ is $\qquad$ .
|
4. 23 .
Obviously, when $p=2$ or 3, it does not meet the requirements of the problem.
Therefore, $p>3$.
Also, $p^{3}-6 p^{2}+9 p=p(p-3)^{2}$, at this point,
$(p, p-3)=(p, 3)=1$.
Since $p$ has two factors, $(p-3)^{2}$ has 15 factors.
And $15=5 \times 3$, to make $p$ the smallest, $p-3$ is also even, so it can only be
$(p-3)^{2}=2^{4} \times 3^{2}$ or $2^{4} \times 5^{2}$.
Thus, $p=15$ (discard) or 23 .
Therefore, the minimum value of $p$ is 23 .
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $x, y$ satisfy
$$
17\left(x^{2}+y^{2}\right)-30 x y-16=0 \text {. }
$$
then the maximum value of $f(x, y)=\sqrt{16 x^{2}+4 y^{2}-16 x y-12 x+6 y+9}$ is . $\qquad$
|
$-1.7$.
From $17\left(x^{2}+y^{2}\right)-30 x y-16=0$, we get
$$
\begin{array}{l}
(x+y)^{2}+16(x-y)^{2}=16 . \\
\text { Let }\left\{\begin{array}{l}
x+y=4 \cos \theta, \\
x-y=\sin \theta
\end{array}(\theta \in \mathbf{R})\right. \\
\Rightarrow\left\{\begin{array}{l}
x=2 \cos \theta+\frac{1}{2} \sin \theta, \\
y=2 \cos \theta-\frac{1}{2} \sin \theta .
\end{array}\right.
\end{array}
$$
Then $f(x, y)$
$$
=\sqrt{(3 \sin \theta+4 \cos \theta)^{2}-3(3 \sin \theta+4 \cos \theta)+9} \text {. }
$$
Therefore, when $3 \sin \theta+4 \cos \theta=-5$, $f(x, y)$ reaches its maximum value of 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $A$ be the set of values of $m$ for which the roots of the equation in $x$
$$
2(m+1) x^{2}-\left(m^{2}+m+16\right) x+8 m=0
$$
are both integers. Then $|A|=$
$\qquad$ .
|
2. 2 .
If $m=-1$, then $x=-\frac{1}{2}$, which does not meet the requirement. If $m \neq-1$, then we can get $x_{1}=\frac{m}{2}, x_{2}=\frac{8}{m+1}$. According to the problem, $m$ is an even number, and $(m+1) \mid 8$, so $m=0$ or -2.
Therefore, $|A|=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=a_{2}=3, a_{n+2}=3 a_{n+1}-2 a_{n}-1 \text {, }
$$
where, $n \in \mathbf{N}_{+}, S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$. Then the maximum value of $S_{n}$ is $\qquad$ .
|
4. 8 .
From the given information, we have
$$
\begin{array}{l}
a_{n+2}-a_{n+1}-1=2\left(a_{n+1}-a_{n}-1\right) \\
=2^{n}\left(a_{2}-a_{1}-1\right)=-2^{n} \\
\Rightarrow a_{n}-a_{n-1}=1-2^{n-2} \\
\Rightarrow a_{n}=n+3-2^{n-1}\left(n \in \mathbf{N}_{+}\right)
\end{array}
$$
Therefore, $a_{3}=2$.
Also, when $n \geqslant 4$,
$$
\begin{array}{l}
2^{n-1}-(n+3)=(1+1)^{n-1}-(n+3) \\
\geqslant 2\left(1+C_{n-1}^{1}\right)-(n+3)=n-3>0,
\end{array}
$$
which means $a_{n}<0$.
Hence, $\left(S_{n}\right)_{\max }=S_{3}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. From the set $\{1,2, \cdots, 2011\}$, any two different numbers $a, b$ are selected such that $a+b=n$ (where $n$ is some positive integer) with a probability of $\frac{1}{2011}$. Then the minimum value of $ab$ is
|
6.2010.
Let the number of ways such that $a+b=n$ be $k$. Then
$$
\frac{k}{\mathrm{C}_{2011}^{2}}=\frac{1}{2011} \Rightarrow k=1005 \text {. }
$$
Consider $a b$ to be as small as possible, and the number of ways such that $a+b=n$ is 1005.
Take $n=2011$. Then
$$
1+2010=2+2009=\cdots=1005+1006 \text {. }
$$
At this point, the number of ways such that $a+b=2011$ is exactly 1005.
Thus, the minimum value of $a b$ is $1 \times 2010=2010$.
|
2010
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (50 points) Each point in the plane is colored with one of $n$ colors, while satisfying:
(1) Each color has infinitely many points, and they are not all on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such that there exist four points of different colors that are concyclic.
|
Given $n \geqslant 4$.
If $n=4$, take a fixed circle $\odot O$ and three points $A, B, C$ on it. Color the arc $\overparen{A B}$ (including point $A$ but not $B$), the arc $\overparen{B C}$ (including point $B$ but not $C$), and the arc $\overparen{C A}$ (including point $C$ but not $A$) with colors 1, 2, and 3, respectively. Color all other points in the plane with color 4. This satisfies the condition and there do not exist four points of different colors on the same circle.
Therefore, $n \geqslant 5$.
When $n=5$, assume that there do not exist four points of different colors on the same circle. By condition (2), there exists a line $l$ with exactly two colors of points (assume $l$ has only points of colors 1 and 2). By condition (1), there exist points $A, B, C$ with colors 3, 4, and 5, respectively, that are not collinear. Let the circle passing through $A, B, C$ be $\odot O$ (as shown in Figure 3).
If $\odot O$ intersects $l$, then there exist four points of different colors on the same circle, which is a contradiction.
If $\odot O$ is disjoint from $l$, draw a perpendicular from point $O$ to $l$ intersecting $l$ at point $D$.
Assume $D$ is colored 1, and the perpendicular intersects $\odot O$ at points $E$ and $S$, as shown in Figure 3.
Assume $E$ is colored 3. Consider a point $F$ on $l$ with color 2, and let $FS$ intersect $\odot O$ at point $G$.
Since $E G \perp G F$, points $D, E, G, F$ are concyclic. Therefore, $G$ must be colored 3.
Also, one of $B, C$ must be different from $S$ (assume it is $B$), and $SB$ intersects $l$ at point $H$.
Since $E B \perp B H$, points $B, E, D, H$ are concyclic. Therefore, $H$ must be colored 1.
Thus, $S B \cdot S H = S E \cdot S D = S G \cdot S F$.
Hence, points $B, H, F, G$ are concyclic and have different colors, which is a contradiction.
Therefore, when $n=5$, there exist four points of different colors on the same circle.
Thus, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, $\frac{1}{x}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$. Try to find the minimum value of $x$.
---
The above text translated into English, preserving the original text's line breaks and format, is as follows:
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, $\frac{1}{x}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$. Try to find the minimum value of $x$.
|
Solve: It is easy to know that $x=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.
Let $d=(a, b), a=d a_{0}, b=d b_{0},\left(a_{0}, b_{0}\right)=1$.
Then $x=\frac{d^{2} a_{0}^{2} b_{0}^{2}}{a_{0}^{2}+b_{0}^{2}} \in \mathbf{N}_{+}$.
By $\left(a_{0}^{2}+b_{0}^{2}, a_{0}^{2} b_{0}^{2}\right)=1$, we know $\left(a_{0}^{2}+b_{0}^{2}\right) \mid d^{2}$.
To find the minimum value of $x$, we need $a_{0}, b_{0}, d$ to be as small as possible.
So, $a_{0}=1, b_{0}=2, d=5$.
At this point, $a=5, b=10$.
Therefore, $x_{\text {min }}=20$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given 10 points on a circle, color six of them black and the remaining four white. They divide the circumference into arcs that do not contain each other. Rule: arcs with both ends black are labeled with the number 2; arcs with both ends white are labeled with the number $\frac{1}{2}$; arcs with ends of different colors are labeled with the number 1. Multiply all these numbers together, and find their product.
|
Mark all the black points with $\sqrt{2}$, and all the white points with $\frac{1}{\sqrt{2}}$, then the number marked on each arc is exactly the product of the numbers at its two ends. Therefore, the product of the numbers marked on all these arcs is the square of the product of the numbers marked on all the points, i.e.,
$$
\left[(\sqrt{2})^{6}\left(\frac{1}{\sqrt{2}}\right)^{4}\right]^{2}=4 \text {. }
$$
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 4 Let $P$ be a polynomial of degree $3n$, such that
$$
\begin{array}{l}
P(0)=P(3)=\cdots=P(3 n)=2, \\
P(1)=P(4)=\cdots=P(3 n-2)=1, \\
P(2)=P(5)=\cdots=P(3 n-1)=0 .
\end{array}
$$
$$
\text { If } P(3 n+1)=730 \text {, find } n \text {. }
$$
|
Solving, we know
$$
\sum_{k=0}^{3 n+1}(-1)^{k} \dot{\mathrm{C}}_{3 n+1}^{k} P(3 n+1-k)=0,
$$
which means $729+2 \sum_{j=0}^{n}(-1)^{3 j+1} \mathrm{C}_{3 n+1}^{3 j+1}+\sum_{j=0}^{n}(-1)^{3 j} \mathrm{C}_{3 n+1}^{3 j}=0$
Using the multi-section formula, we can find
$$
\begin{array}{l}
\sum_{j=0}^{n}(-1)^{3 j+1} \mathrm{C}_{3 n+1}^{3 j+1} \\
=-2(\sqrt{3})^{3 n-1} \cos \frac{(2 n-1) \pi}{6}, \\
\sum_{j=0}^{n}(-1)^{3 j} \mathrm{C}_{3 n+1}^{3 j}=2(\sqrt{3})^{3 n-1} \cos \frac{(3 n+1) \pi}{6} .
\end{array}
$$
Thus, $729-4(\sqrt{3})^{3 n-1} \cos \frac{(3 n-1) \pi}{6}+$
$$
2(\sqrt{3})^{3 n-1} \cos \frac{(3 n+1) \pi}{6}=0 \text {. }
$$
By distinguishing the parity of $n$, it is easy to find that only $n=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Find the smallest positive integer $n$, such that there exist $n$ distinct positive integers $s_{1}, s_{2}, \cdots, s_{n}$, satisfying
$$
\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \cdots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010} .
$$
|
1. Suppose the positive integer $n$ satisfies the condition, and let $s_{1}=39$.
Thus, $n \geqslant 39$.
Below, we provide an example to show that $n=39$ satisfies the condition.
Take 39 different positive integers:
$$
2,3, \cdots, 33,35,36, \cdots, 40,67,
$$
which satisfy the given equation.
In conclusion, the minimum value of $n$ is 39.
|
39
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Find the smallest positive integer $n$, such that there exist rational coefficient polynomials $f_{1}, f_{2}, \cdots, f_{n}$, satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+\cdots+f_{n}^{2}(x) .
$$
|
3. Since $x^{2}+7=x^{2}+2^{2}+1^{2}+1^{2}+1^{2}$, then $n \leqslant 5$.
Thus, it only needs to be proven that $x^{2}+7$ is not equal to the sum of squares of no more than four rational coefficient polynomials.
Assume there exist four rational coefficient polynomials $f_{1}, f_{2}, f_{3}, f_{4}$ (some of which may be 0), satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+f_{3}^{2}(x)+f_{4}^{2}(x).
$$
Then the degrees of $f_{1}, f_{2}, f_{3}, f_{4}$ are at most 1.
Let $f_{i}(x)=a_{i} x+b_{i}\left(a_{i}, b_{i} \in \mathbf{Q}, i=1,2,3,4\right)$.
From $x^{2}+7=\sum_{i=1}^{4}\left(a_{i} x+b_{i}\right)^{2}$, we get
$$
\sum_{i=1}^{4} a_{i}^{2}=1, \sum_{i=1}^{4} a_{i} b_{i}=0, \sum_{i=1}^{4} b_{i}^{2}=7.
$$
Let $p_{i}=a_{i}+b_{i}, q_{i}=a_{i}-b_{i}(i=1,2,3,4)$.
Then $\sum_{i=1}^{4} p_{i}^{2}=\sum_{i=1}^{4} a_{i}^{2}+2 \sum_{i=1}^{4} a_{i} b_{i}+\sum_{i=1}^{4} b_{i}^{2}=8$,
$\sum_{i=1}^{4} q_{i}^{2}=\sum_{i=1}^{4} a_{i}^{2}-2 \sum_{i=1}^{4} a_{i} b_{i}+\sum_{i=1}^{4} b_{i}^{2}=8$,
$\sum_{i=1}^{4} p_{i} q_{i}=\sum_{i=1}^{4} a_{i}^{2}-\sum_{i=1}^{4} b_{i}^{2}=-6$.
This implies there exist integers $x_{i}, y_{i}(i=1,2,3,4)$, $m(m>0)$ satisfying the following equations
(1) $\sum_{i=1}^{4} x_{i}^{2}=8 m^{2}$,
(2) $\sum_{i=1}^{4} y_{i}^{2}=8 m^{2}$,
(3) $\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2}$.
Assume the above equations have a solution, consider a solution that makes $m$ the smallest.
Note that, when $x$ is odd, $x^{2} \equiv 1(\bmod 8)$; when $x$ is even,
$x^{2} \equiv 0(\bmod 8)$ or $x^{2} \equiv 4(\bmod 8)$.
From (1), we know that $x_{1}, x_{2}, x_{3}, x_{4}$ are all even, and from (2), we know that $y_{1}, y_{2}, y_{3}, y_{4}$ are also all even. Thus, the left side of (3) can be divided by 4. Therefore, $m$ is even.
Hence $\left(\frac{x_{1}}{2}, \frac{y_{1}}{2}, \frac{x_{2}}{2}, \frac{y_{2}}{2}, \frac{x_{3}}{2}, \frac{y_{3}}{2}, \frac{x_{4}}{2}, \frac{y_{4}}{2}, \frac{m}{2}\right)$ is also a solution to the above equations, which contradicts the minimal selection of $m$.
In conclusion, the minimum value of $n$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a_{1}, a_{2}, \cdots$ be integers, for any positive integer $n$ we have
$$
a_{n}=(n-1)\left[\left(\frac{a_{2}}{2}-1\right) n+2\right] \text {. }
$$
If $2001 a_{199}$, find the smallest positive integer $n(n>1)$, such that $200 \mathrm{l} a_{n}$.
|
3. From the given, we have
$$
a_{\mathrm{T99}}=198\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right] \text {. }
$$
From $2001 a_{199}$, we get
$$
100 \left\lvert\, 99\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right]\right. \text {. }
$$
Thus, $a_{2}$ is an even number.
Let $a_{2}=2 m$. Then
$$
1001[(m-1) \times 199+2] \text {, }
$$
which simplifies to $1001(m-3)$.
Hence, let $m-3=100 t$. Then
$$
\begin{array}{l}
a_{n}=(n-1)[(100 t+2) n+2] \\
=n(n-1) \times 100 t+2\left(n^{2}-1\right) .
\end{array}
$$
Since $200 \mid a_{n}$, i.e., $100 \mid\left(n^{2}-1\right)$, $n$ must be an odd number.
Let $n=2 k+1$. Then $251 k(k+1)$.
Since $25=5^{2},(k, k+1)=1$, we have $5^{2} \mid k$ or $5^{2} \mid(k+1)$.
Thus, $k_{\min }=24, n_{\text {min }}=49$.
It is easy to verify that the smallest positive integer $n$ that satisfies the condition is
49.
|
49
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $\triangle A B C$ with the three angles $\angle A, \angle B, \angle C$ opposite to the sides of lengths $a, b, c$, and $\angle A=2 \angle B$. Then $\frac{a^{2}}{b(b+c)}=$ $\qquad$ .
|
1. 1.
As shown in Figure 4, it is easy to know,
$$
\begin{array}{l}
\triangle A B C \backsim \triangle D A C \\
\Rightarrow \frac{A B}{D A}=\frac{A C}{D C}=\frac{B C}{A C} .
\end{array}
$$
Let $B D=x, D C=y$. Then
$$
\begin{array}{l}
\frac{c}{x}=\frac{b}{y}=\frac{a}{b} \Rightarrow \frac{b+c}{x+y}=\frac{a}{b} \\
\Rightarrow b(b+c)=a^{2} \Rightarrow \frac{a^{2}}{b(b+c)}=1 .
\end{array}
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $0<a<b<c<d<500$, and $a+d=b+c$. Also, $bc-ad=93$.
Then the number of ordered quadruples of integers ( $\dot{a}, b, c$,
$d)$ that satisfy the conditions is . $\qquad$
|
7.870.
Since $a+d=b+c$, we set
$$
(a, b, c, d)=(a, a+x, a+y, a+x+y) \text {, }
$$
where $x$ and $y$ are integers, and $0<x<y$.
Then $93=b c-a d$
$$
=(a+x)(a+y)-a(a+x+y)=x y \text {. }
$$
Therefore, $(x, y)=(1,93)$ or $(3,31)$.
First case
$$
(a, b, c, d)=(a, a+1, a+93, a+94) \text {, }
$$
where $a=1,2, \cdots, 405$;
Second case
$$
(a, b, c, d)=(a, a+3, a+31, a+34) \text {, }
$$
where $a=1,2, \cdots, 465$.
Thus, the total number of integer quadruples that satisfy the condition is
$$
405+465=870 \text { (sets). }
$$
|
870
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. The number of prime pairs $(a, b)$ that satisfy the equation
$$
a^{b} b^{a}=(2 a+b+1)(2 b+a+1)
$$
is $\qquad$.
|
9.2.
If $a=b$, then the equation is equivalent to $a^{2a}=(3a+1)^2$. But $a \mid a^{2a}, a \mid (3a+1)^2$, which is a contradiction, so $a \neq b$.
Assume without loss of generality that $a > b$. Since the difference between the two factors on the right side of the original equation is $a-b$, $a$ can only divide one of these factors, and $a^b$ also divides this factor.
Thus, $a^2 \leq a^b \leq 2a + b + 1 \leq 3a \Rightarrow a \leq 3$.
Therefore, $a=3, b=2$.
Verification shows that $(a, b)=(3,2)$ is a solution to the original equation.
By symmetry, $(a, b)=(2,3)$ is also a solution to the original equation.
In conclusion, the only solutions are $(3,2)$ and $(2,3)$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Each cell of a $2011 \times 2011$ grid is labeled with an integer from $1,2, \cdots, 2011^{2}$, such that each number is used exactly once. Now, the left and right boundaries, as well as the top and bottom boundaries of the grid, are considered the same, forming a torus (which can be viewed as the surface of a "doughnut"). Find the largest positive integer $M$ such that for any labeling method, there exist two adjacent cells (cells sharing a common edge) whose numbers differ (the larger minus the smaller) by at least $M$.
γNoteγUsing coordinates, a cell $(x, y)$ and $\left(x^{\prime}, y^{\prime}\right)$ are adjacent if:
$$
\begin{aligned}
x=x^{\prime}, y-y^{\prime} \equiv \pm 1(\bmod 2011) \\
\text { or } \quad y=y^{\prime}, x-x^{\prime} \equiv \pm 1(\bmod 2011) .
\end{aligned}
$$
|
6. Let $N=2011$.
Consider a general $N \times N$ table.
When $N=2$, the conclusion is obvious, and the required $M=2$. An example is shown in Table 1.
Table 1
\begin{tabular}{|l|l|}
\hline 1 & 2 \\
\hline 3 & 4 \\
\hline
\end{tabular}
When $N \geqslant 3$, first prove:
$M \geqslant 2 N-1$.
Starting from a state where each small square in the table is white, write the numbers $1,2, \cdots$ in the table while coloring the marked squares black. Stop the operation when the following condition is first met: every row or every column has at least two black squares. Let the last number written be $k$.
Before marking $k$, there must be one row and one column with at most one black square.
Assume, when marking $k$, every row has two black squares. At this point, there is at most one row with all black squares. This is because if there are two rows with all black squares, then if $k$ is marked in one of these two rows, each row already had two black squares before (using $N \geqslant 3$); if $k$ is marked in another row, then each column already had two black squares.
Color a black square red if it has an adjacent white square. Since, except for the possible all-black row, each row has two black squares and one white square, each of these rows has at least two red squares. Furthermore, the row adjacent to the possible all-black row must have at least one white square. Therefore, at least one black square in the all-black row is colored red. Thus, the number of red squares is at least $2(N-1)+1=2 N-1$.
Therefore, the smallest number in all red squares is at most $k+1-(2 N-1)$.
When the adjacent white square of this red square is marked (the number marked is at least $k+1$), the difference between these two adjacent squares is at least $2 N-1$.
Since $N=2011$, it is sufficient to construct an example for $N=2 n+1(\geqslant 2)$.
Table 2 provides an example where $M=2 N-1$.
Therefore, the required $M=4021$.
Table 2
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline$(2 n+1)^{2}-2$ & $(2 n+1)^{2}-9$ & $\cdots$ & $\cdots$ & $n(2 n-1)+1$ & $\cdots$ & $\cdots$ & $(2 n+1)^{2}-10$ & $(2 n+1)^{2}-3$ \\
\hline$(2 n+1)^{2}-8$ & $\cdots$ & $\cdots$ & $n(2 n-1)+2$ & $\cdots$ & $n(2 n-1)$ & $\cdots$ & $\cdots$ & $(2 n+1)^{2}-11$ \\
\hline$\vdots$ & $\vdots$ & $\ddots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\ddots$ & $\vdots$ & $\vdots$ \\
\hline$\cdots$ & $2 n^{2}$ & $\cdots$ & 8 & 2 & 6 & $\cdots$ & $2 n(n-1)+2$ & $2 n(n+1)+2$ \\
\hline $2 n^{2}+1$ & $\cdots$ & $\cdots$ & 3 & 1 & 5 & $\cdots$ & $\cdots$ & $2 n(n+1)+1$ \\
\hline$\cdots$ & $2 n^{2}+2$ & $\cdots$ & 10 & 4 & 12 & $\cdots$ & $2 n(n+1)$ & $\cdots$ \\
\hline$\vdots$ & $\vdots$ & $\ddots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\ddots$ & $\vdots$ & $\vdots$ \\
\hline$(2 n+1)^{2}-7$ & $\cdots$ & $\cdots$ & $n(2 n+1)$ & $\cdots$ & $n(2 n+1)+2$ & $\cdots$ & $\cdots$ & $(2 n+1)^{2}-4$ \\
\hline$(2 n+1)^{2}-1$ & $(2 n+1)^{2}-6$ & $\cdots$ & $\cdots$ & $n(2 n+1)+1$ & $\cdots$ & $\cdots$ & $(2 n+1)^{2}-5$ & $(2 n+1)^{2}$ \\
\hline
\end{tabular}
|
4021
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.1. A table of numbers consisting of $n$ rows and 10 columns, with each element being an integer from $0 \sim 9$, satisfies the following condition: for any row $A$ and any two columns $B$ and $C$, there exists another row $D$ such that $D$ differs from $A$ only in the numbers in columns $B$ and $C$. Prove: $n \geqslant 512$.
|
10.1. Let $R_{0}$ be the first row. Arbitrarily select $2 m$ columns $C_{1}$, $C_{2}, \cdots, C_{2 m}$. By the given condition, there exists a row $R_{1}$, which differs from $R_{0}$ only in $C_{1}$ and $C_{2}$; further, there exists a row $R_{2}$, which differs from $R_{1}$ only in $C_{3}$ and $C_{4}$; $\cdots \cdots$ there exists a row $R_{m}$, which differs from $R_{m-1}$ only in $C_{2 m-1}$ and $C_{2 m}$ (if $m=0$, then $R_{m}=R_{0}$).
Obviously, $R_{m}$ differs from $R_{0}$ only in $C_{1}, C_{2}, \cdots, C_{2 m}$. Therefore, the corresponding $R_{m}$ for different column selections are also different. Since the total number of different even column selections is $\mathrm{C}_{10}^{0}+\mathrm{C}_{10}^{2}+\mathrm{C}_{10}^{4}+\mathrm{C}_{10}^{6}+\mathrm{C}_{10}^{8}+\mathrm{C}_{10}^{10}=512$ (ways), thus, $n \geqslant 512$.
|
512
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
10.2. Nine real-coefficient quadratic polynomials
$$
x^{2}+a_{1} x+b_{1}, x^{2}+a_{2} x+b_{2}, \cdots, x^{2}+a_{9} x+b_{9}
$$
satisfy: The sequences $a_{1}, a_{2}, \cdots, a_{9}$ and $b_{1}, b_{2}, \cdots, b_{9}$ are both arithmetic sequences. It is known that the polynomial obtained by adding these nine polynomials has real roots. Question: At most how many of these nine polynomials do not have real roots?
|
$$
\begin{array}{l}
\text { 10. 2. Let } P_{i}(x)=x^{2}+a_{i} x+b_{i}(i=1,2, \cdots, 9), \\
P(x)=P_{1}(x)+P_{2}(x)+\cdots+P_{9}(x) .
\end{array}
$$
Notice that
$$
P_{i}(x)+P_{10-i}(x)=2 P_{5}(x)=\frac{2}{9} P(x)
$$
has real roots, denoted as $x_{0}$.
Then $P_{i}\left(x_{0}\right)+P_{10-i}\left(x_{0}\right)=0$.
Thus, $P_{i}\left(x_{0}\right) P_{10-i}\left(x_{0}\right) \leqslant 0$.
This indicates that at least one of $P_{i}$ and $P_{10-i}$ has real roots.
Therefore, at most four of $P_{i}$ do not have real roots, and among $x^{2}-4$, $x^{2}-3, \cdots, x^{2}+4$, exactly four do not have real roots.
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|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.7. For any positive integers $a, b (a>b>1)$, define the sequence $x_{n}=\frac{a^{n}-1}{b^{n}-1}(n=1,2, \cdots)$. It is known that the defined sequence does not have $d$ consecutive terms consisting of prime numbers. Find the minimum value of $d$.
untranslated part:
ε°δΈι’ηζζ¬ηΏ»θ―ζθ±ζοΌθ―·δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζγ
Note: The last part is a note and not part of the text to be translated. It is provided in Chinese and should not be included in the translated output.
|
10.7. The minimum value of $d$ is 3.
When $a=4, b=2$, $x_{1}=3, x_{2}=5$ are prime numbers.
Therefore, the minimum value of $d$ is greater than 2.
Next, we prove: there cannot be three consecutive terms all being prime numbers.
In fact, a stronger conclusion can be proven:
For $n \geqslant 2, x_{n}$ and $x_{n+1}$ cannot both be prime numbers.
Assume, for contradiction, that they are prime numbers $p$ and $q$. Then,
$$
\begin{array}{l}
(a-1)\left(a^{n-1}+a^{n-2}+\cdots+1\right) \\
=p(b-1)\left(b^{n-1}+b^{n-2}+\cdots+1\right), \\
(a-1)\left(a^{n}+a^{n-1}+\cdots+1\right) \\
=q(b-1)\left(b^{n}+b^{n-1}+\cdots+1\right) .
\end{array}
$$
Since $a^{n-1}+a^{n-2}+\cdots+1$ and $a^{n}+a^{n-1}+\cdots+1$ are coprime, we have
$$
(b-1) \mid (a-1).
$$
Let $a-1=k(b-1)$.
Substituting into the first equation, we get
$$
\begin{array}{l}
k\left(a^{n-1}+a^{n-2}+\cdots+1\right) \\
=p\left(b^{n-1}+b^{n-2}+\cdots+1\right).
\end{array}
$$
Since $a>b$, we have $1<k<p$.
Thus, $k \mid \left(b^{n-1}+b^{n-2}+\cdots+1\right)$.
Similarly, we have
$$
k<q, k \mid \left(b^{n}+b^{n-1}+\cdots+1\right),
$$
which contradicts the fact that $b^{n-1}+b^{n-2}+\cdots+1$ and $b^{n}+b^{n-1}+\cdots+1$ are coprime.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example $8 A, B, C, D, E$ five people participated in an exam, which has seven questions, all of which are true or false questions. The scoring criteria are: for each question, 1 point is awarded for a correct answer, 1 point is deducted for a wrong answer, and no points are awarded or deducted for unanswered questions. Table 1 records the answers given by $A, B, C, D, E$ five individuals. It is now known that $A, B, C, D$ each scored 2 points. Question: How many points should $E$ get? What is the correct answer for each question?
|
Let $x_{k}=\left\{\begin{array}{ll}1, & \text { if the conclusion of the } k \text {th question is correct; } \\ -1, & \text { if the conclusion of the } k \text {th question is incorrect, }\end{array}\right.$ where $k=1,2, \cdots, 7$.
At this point, if the conclusion is judged to be correct (i.e., marked with a β $\checkmark$ β), then $x_{k}$ points are obtained; if it is judged to be incorrect (i.e., marked with a β $\times$ β), then $-x_{k}$ points are obtained.
Since $A, B, C, D$ each scored 2 points, we can derive the system of equations:
$$
\left\{\begin{array}{l}
x_{1}+0 \cdot x_{2}-x_{3}+x_{4}-x_{5}+x_{6}+x_{7}=2, \\
x_{1}-x_{2}+x_{3}+x_{4}-x_{5}-x_{6}+0 \cdot x_{7}=2, \\
0 \cdot x_{1}+x_{2}-x_{3}-x_{4}+x_{5}-x_{6}+x_{7}=2, \\
-x_{1}-x_{2}-x_{3}+x_{4}+x_{5}+0 \cdot x_{6}-x_{7}=2 .
\end{array}\right.
$$
Adding these four equations yields
$$
x_{1}-x_{2}-2 x_{3}+2 x_{4}-x_{6}+x_{7}=8 \text {. }
$$
Noting that $x_{i}= \pm 1(i=1,2, \cdots, 7)$.
Therefore, the left side of the above equation is less than or equal to 8, while the right side is 8, hence
$$
\begin{array}{l}
x_{1}=1, x_{2}=-1, x_{3}=-1, \\
x_{4}=1, x_{6}=-1, x_{7}=1 .
\end{array}
$$
Substituting these results into the first equation of the system, we get $x_{5}=1$.
Thus, questions 1, 4, 5, and 7 are correct, while questions 2, 3, and 6 are incorrect.
Therefore, according to the problem, $E$ scores 4 points.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If the length, width, and height of a rectangular prism are all prime numbers, and the sum of the areas of two adjacent sides is 341, then the volume of this rectangular prism $V=$ $\qquad$ .
|
3. 638.
Let the length, width, and height of the rectangular prism be $x, y, z$. From the problem, we have
$$
\begin{array}{l}
x(y+z)=341=11 \times 31 \\
\Rightarrow(x, y+z)=(11,31),(31,11) .
\end{array}
$$
Since $y+z$ is odd, one of $y, z$ must be 2 (let's assume $z=2$).
Also, $11-2=9$ is not a prime number, so
$$
(x, y, z)=(11,29,2) \text {. }
$$
Therefore, $V=x y z=638$.
|
638
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let
$$
\begin{array}{l}
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \\
\text { then } f(1)+f(2)+\cdots+f(50)=
\end{array}
$$
|
$$
\begin{array}{l}
x^{2}-53 x+196=(x-4)(x-49) . \\
\text { Therefore, when } 4 \leqslant x \leqslant 49, f(x)=0 . \\
\text { Then } f(1)+f(2)+\cdots+f(50) \\
=f(1)+f(2)+f(3)+f(50)=660 \text {. }
\end{array}
$$
|
660
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $t=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}$. Then the sum of all real solutions of the equation $(t-1)(t-2)(t-3)=0$ with respect to $x$ is $\qquad$
|
3.4.
Notice that
$$
f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}
$$
is a monotonically decreasing function.
When $x=0,1,3$, its values are $3, 2, 1$ respectively, so the three roots of the equation are $x=0,1,3$, and their sum is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Two boxes are filled with black and white balls. The total number of balls in both boxes is 25. Each time, a ball is randomly drawn from each box. The probability that both balls are black is $\frac{27}{50}$, and the probability that both balls are white is $\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$. Then $m+n=$ $\qquad$ .
|
5.26.
Let the first box contain $x$ balls, of which $p$ are black, and the second box contain $25-x$ balls, of which $q$ are black. Then
$$
\begin{array}{l}
\frac{p}{x} \cdot \frac{q}{25-x}=\frac{27}{50}, \\
50 p q=27 x(25-x) .
\end{array}
$$
Thus, $x$ is a multiple of 5.
Substituting $x=5$ and $x=10$ into the two equations, we get
$$
\begin{array}{l}
x=5, p=3, q=18 ; \\
x=10, p=9, q=9 .
\end{array}
$$
For both cases, $\frac{m}{r_{0}}=\frac{1}{25}$. Therefore, $m+n=26$.
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The equation $x+y+z=2011$ satisfies $x<y<z$ to prove: the number of solutions $(x, y, z)$ is $\qquad$ groups.
|
5. 336005 .
The equation $x+y+z=2011$ has $\mathrm{C}_{2010}^{2}$ sets of positive integer solutions. In each solution, $x, y, z$ cannot all be equal, and there are $3 \times 1005$ sets of solutions where two of $x, y, z$ are equal. Therefore, the number of solutions that meet the requirements is
$$
\frac{C_{2010}^{2}-3 \times 1005}{3!}=336005 \text { (sets). }
$$
|
336005
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. For any real numbers $x, y, z$ not all zero, we have
$$
\begin{array}{l}
-6 x y + 18 z x + 36 y z . \\
\leqslant k\left(54 x^{2} + 41 y^{2} + 9 z^{2}\right) .
\end{array}
$$
Then the minimum value of the real number $k$ is
|
6.1.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{c}
{\left[(x+y)^{2}+(z+x)^{2}+(y+z)^{2}\right]\left(4^{2}+8^{2}+10^{2}\right)} \\
\geqslant[4(x+y)+8(z+x)+10(y+z)]^{2} .
\end{array}
$$
Simplifying, we get $-11r$.
$$
54 x^{2}+41 y^{2}+9 z^{2} \geqslant-6 x y+18 z x+36 y z \text {. }
$$
When $x: y: z=1: 3: 7$, the equality holds.
Thus, the minimum value of $k$ is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given real numbers $a, b, c, d$ satisfy $a^{4}+b^{4}=c^{4}+d^{4}=2011, a c+b d=0$. Find the value of $a b+c d$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Given $a c+b d=0 \Rightarrow a c=-b d$.
Raising both sides to the fourth power, we get $a^{4} c^{4}=b^{4} d^{4}$.
Since $c^{4}=2011-d^{4}, b^{4}=2011-a^{4}$, we have,
$$
\begin{array}{l}
a^{4}\left(2011-d^{4}\right)=\left(2011-a^{4}\right) d^{4} \\
\Rightarrow 2011 a^{4}-a^{4} d^{4}=2011 d^{4}-a^{4} d^{4} .
\end{array}
$$
Thus, $a^{4}=d^{4} \Rightarrow a= \pm d$.
Combining with the given conditions, we get
$$
b^{4}=c^{4} \Rightarrow b= \pm c \text {. }
$$
From equations (1) and (2), we have
$$
a=d, b=c
$$
or $a=-d, b=-c$
or $a=d, b=-c$
or $a=-d, b=c$.
In the first two cases,
$$
a b+c d= \pm(a c+b d)=0 ;
$$
In the last two cases,
$$
a b=-c d \Rightarrow a b+c d=0 \text {. }
$$
In summary, $a b+c d=0$.
(Chen Kuanhong, Land Reserve Center of Yueyang County, Hunan, 414100)
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given that $C D$ is the altitude on the hypotenuse $A B$ of the right triangle $\triangle A B C$, $\odot O$ is its circumcircle, $\odot O_{1}$ is internally tangent to arc $\overparen{A C}$ and tangent to $A B$ and $C D$, with $E$ being the tangency point on side $A B$; $\odot O_{2}$ is internally tangent to arc $\overparen{C B}$ and tangent to $A B$ and $C D$, with $F$ being the tangency point on side $A B$. Prove: $\frac{A E}{E D} \cdot \frac{D F}{F B} \cdot \frac{B C}{C A}=1$.
|
Prompt: From Example 8, we get $A F=A C, B E=B C$. Then prove that $C E$ bisects $\angle A C D$, and $C F$ bisects $\angle D C B$. Thus, $\frac{A E}{E D} \cdot \frac{D F}{F B} \cdot \frac{B C}{C A}=\frac{A C}{C D} \cdot \frac{C D}{C B} \cdot \frac{B C}{C A}=1$.
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 As shown in Figure 2, let $M$ be the centroid of $\triangle A B C$, and a line through $M$ intersects sides $A B$ and $A C$ at points $P$ and $Q$, respectively, and
$$
\frac{A P}{P B}=m, \frac{A Q}{Q C}=n \text {. }
$$
Then $\frac{1}{m}+\frac{1}{n}=$
|
Solution 1 As shown in Figure 2, draw lines through points $C$ and $B$ parallel to $PQ$ intersecting $AD$ or its extension at points $E$ and $F$. Then,
$$
\begin{array}{l}
\frac{1}{m}=\frac{P B}{A P}=\frac{F M}{A M}, \\
\frac{1}{n}=\frac{Q C}{A Q}=\frac{E M}{A M} .
\end{array}
$$
Since $D$ is the midpoint of side $BC$, and $BF \parallel CE$, it follows that $DE = DF$.
Thus, $FM + EM = 2DM$.
Furthermore, since $M$ is the centroid of $\triangle ABC$, we have $AM = 2DM$. Therefore, $\frac{1}{m} + \frac{1}{n} = \frac{FM + EM}{AM} = \frac{2MD}{AM} = 1$.
Solution 2 As shown in Figure 3, consider a special case where $PQ \parallel BC$.
Then,
$$
\begin{array}{l}
\frac{A P}{P B}=\frac{A Q}{Q C} \\
=\frac{A M}{M D}=2 .
\end{array}
$$
Thus,
$$
\ldots \frac{1}{m}+\frac{1}{n}=1 \text {. }
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given positive integers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a b+b c+c a+2(a+b+c)=8045, \\
a b c-a-b-c=-2 .
\end{array}\right.
$$
then $a+b+c=$ $\qquad$
|
$$
-1.2012 .
$$
Note that
$$
\begin{array}{l}
(a+1)(b+1)(c+1) \\
=a b c+a b+b c+c a+a+b+c+1 \\
=8045+(-2)+1=8044 .
\end{array}
$$
Since $a, b, c$ are positive integers, we have
$$
a+1 \geqslant 2, b+1 \geqslant 2, c+1 \geqslant 2 \text{. }
$$
Therefore, 8044 can only be factored as $2 \times 2 \times 2011$. Hence $a+b+c=2010+1+1=2012$.
|
2012
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a class, there are two types of students: one type always lies, and the other type never lies. Each student knows what type the other students are. During a gathering today, each student has to state what type the other students are, and all students together said "liar" 240 times. At a similar gathering yesterday, one student was absent, but all students together said "liar" 216 times. Then the total number of students who participated in today's gathering is.
|
2. 22 .
Consider four possible scenarios:
(1) If student $A$ is a liar and student $B$ is not a liar, then $A$ will say $B$ is a liar;
(2) If student $A$ is a liar and student $B$ is also a liar, then $A$ will say $B$ is not a liar;
(3) If student $A$ is not a liar and student $B$ is also not a liar, then $A$ will say $B$ is not a liar;
(4) If student $A$ is not a liar and student $B$ is a liar, then $A$ will say $B$ is a liar.
Thus, if there are $p$ liars and $q$ non-liars in the class, the number of times "liar" is mentioned is $2pq$.
Combining the given conditions, we have
$pq = 120$,
$(p-1)q = 108$ or $p(q-1) = 108$.
Solving these, we get $(p, q) = (10, 12)$ or $(12, 10)$.
Therefore, the total number of students attending today's gathering is 22.
|
22
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The positive integer $n$ has exactly 4 positive divisors (including 1 and $n$). It is known that $n+1$ is four times the sum of the other two divisors. Then $n=$
|
5.95.
Notice that a positive integer with exactly four positive divisors must be of the form $p^{3}$ or $p q$ (where $p$ and $q$ are primes, $p \neq q$).
In the first case, all positive divisors are $1, p, p^{2}, p^{3}$, then $1+p^{3}=4\left(p+p^{2}\right)$, but $p \nmid \left(1+p^{3}\right)$, which is a contradiction.
In the second case, all positive divisors are $1, p, q, p q$, then
$$
\begin{array}{l}
1+p q=4(p+q) \\
\Rightarrow(p-4)(q-4)=15 .
\end{array}
$$
Thus, $q$ is a prime and $q-4$ is a divisor of 15. Hence, $q \in\{5,7,19\}$.
Since $p$ is also a prime, the only solution is $\{p, q\}=\{5,19\}$. Therefore, $n=95$.
|
95
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given a positive integer $n$ that satisfies the following conditions:
(1) It is an eight-digit number, and all its digits are 0 or 1;
(2) Its first digit is 1;
(3) The sum of the digits in the even positions equals the sum of the digits in the odd positions.
How many such $n$ are there?
|
10.35.
From the fact that the sum of the digits in the even positions equals the sum of the digits in the odd positions, we know that the number of 1s in the even positions equals the number of 1s in the odd positions.
Since the first digit is fixed, there are only three positions in the odd positions that can be freely changed.
Therefore, the total is $\sum_{k=1}^{4} \mathrm{C}_{3}^{k-1} \mathrm{C}_{4}^{k}=4+18+12+1=35$ possible integers.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given real numbers $a, b, c$ satisfy
$$
\frac{a(b-c)}{b(c-a)}=\frac{b(c-a)}{c(b-a)}=k>0,
$$
where $k$ is some constant. Then the greatest integer not greater than $k$ is . $\qquad$
|
12. 0 .
Let the substitution be $x=a b, y=b c, z=c a$. Then
$$
x-z=k(y-x), y-x=k(y-z) \text {. }
$$
Thus, $x-z=k^{2}(y-z)$. Therefore,
$$
\begin{array}{l}
(y-x)+(x-z)+(z-y)=0 \\
\Leftrightarrow(y-z)\left(k^{2}+k-1\right)=0(x \neq y \neq z) \\
\Rightarrow k^{2}+k-1=0 .
\end{array}
$$
Also, $k>0 \Rightarrow k=\frac{-1+\sqrt{5}}{2} \Rightarrow[k]=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given that 15 rays share a common endpoint. Question: What is the maximum number of obtuse angles (considering the angle between any two rays to be the one not greater than $180^{\circ}$) that these 15 rays can form?
|
15. First, it is explained that constructing 75 obtuse angles is achievable.
The position of the rays is represented by their inclination angles, with 15 rays placed near the positions of $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$. Within each group, the five rays are sufficiently close to each other (as shown in Figure 10). Thus, the total number of obtuse angles formed is
$$
\mathrm{C}_{15}^{2}-3 \mathrm{C}_{5}^{2}=75
$$
Next, we prove that it is impossible to obtain more than 75 obtuse angles.
For any arrangement, if there do not exist three rays such that the angles between them are all obtuse, then the total number of obtuse angles does not exceed $\frac{\mathrm{C}_{15}^{2} \times 2}{3}=70$; if there exist three rays such that the angles between each pair are all obtuse, then any other ray can form at most two obtuse angles with these three rays. Removing these initial three rays, the total number of obtuse angles is reduced by at most
$$
3+2 \times 12=27 \text{.}
$$
In the remaining 12 rays, continue to search for such three rays (if such three rays do not exist, then the initial total number of obtuse angles does not exceed $27+44=71$), and remove these three rays. Thus, the number of obtuse angles is reduced by at most $3+2 \times 9=21$. Continue this process, and the number of obtuse angles is reduced by at most $3+2 \times 6=15$ and $3+2 \times 3=9$. In the end, only three rays remain.
Therefore, the initial number of obtuse angles is at most
$$
27+21+15+9+3=75
$$
|
75
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. (40 points) Find all real roots of the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right)
$$
|
1. The original equation can be transformed into
$$
x^{4}+3 x^{3}+6 x^{2}+7 x+3=0 \text {. }
$$
Let $x=-1$. Then $1-3+6-7+3=0$.
Therefore, $x+1$ is a factor.
$$
\begin{array}{l}
\text { Hence } x^{4}+3 x^{3}+6 x^{2}+7 x+3 \\
=(x+1)\left(x^{3}+2 x^{2}+4 x+3\right) \\
=(x+1)^{2}\left(x^{2}+x+3\right)=0 .
\end{array}
$$
Notice that $x^{2}+x+3=\left(x+\frac{1}{2}\right)^{2}+\frac{11}{4}>0$.
Thus, the only real root is -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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