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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 “consecutive $n$ strange group” (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 “consecutive 3 strange group”). 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 $b_{n+2} 、 b_{n+1} 、 b_{n}$;
(2) Find the unit digit of the integer part of $a_{2011}$.
|
-、(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
|
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
|
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
|
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
|
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
|
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
|
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
|
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:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The 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 “ $\checkmark$ ”), then $x_{k}$ points are obtained; if it is judged to be incorrect (i.e., marked with a “ $\times$ ”), 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. 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
|
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
|
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
|
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
|
7. (40 points) A cat caught 81 mice and arranged them in a circle, numbering them from $1 \sim 81$ in a clockwise direction. The cat starts counting from a certain mouse in a clockwise direction, continuously counting “$1, 2, 3$” and eating all the mice that are counted as 3. As the cat continues to count, the circle gets smaller and smaller until only two mice are left. It is known that one of the remaining mice with a higher number is 40. Question: From which numbered mouse did the cat start counting?
|
7. First, arrange the numbers of all the mice in 9 columns as follows.
\begin{tabular}{ccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\
19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 \\
28 & 29 & 30 & 31 & 32 & 33 & 34 & 35 & 36 \\
37 & 38 & 39 & 40 & 41 & 42 & 43 & 44 & 45 \\
46 & 47 & 48 & 49 & 50 & 51 & 52 & 53 & 54 \\
55 & 56 & 57 & 58 & 59 & 60 & 61 & 62 & 63 \\
64 & 65 & 66 & 67 & 68 & 69 & 70 & 71 & 72 \\
73 & 74 & 75 & 76 & 77 & 78 & 79 & 80 & 81
\end{tabular}
Assuming the cat starts counting from mouse number 1. In the first round, the mice in columns 3, 6, and 9 are eaten. In the second round, the mice in columns 4 and 8 are eaten. The remaining mice's numbers are rearranged into 9 columns as follows:
\begin{tabular}{ccccccccc}
1 & 2 & 5 & 7 & 10 & 11 & 14 & 16 & 19 \\
20 & 23 & 25 & 28 & 29 & 32 & 34 & 37 & 38 \\
41 & 43 & 46 & 47 & 50 & 52 & 55 & 56 & 59 \\
61 & 64 & 65 & 68 & 70 & 73 & 74 & 77 & 79
\end{tabular}
After the third and fourth rounds, the 3rd, 4th, 6th, 8th, and 9th columns in the rearranged 9 columns are eaten. At this point, the 16 surviving mice's numbers are:
\begin{tabular}{cccccccc}
1 & 2 & 10 & 14 & 20 & 23 & 29 & 34 \\
41 & 43 & 50 & 55 & 61 & 64 & 70 & 74
\end{tabular}
Continuing until only two mice remain, their numbers are 1 and 34. Given that the larger number is 40, the cat must have started counting from \(1 + 40 - 34 = 7\).
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (40 points) Given positive integers $a$, $b$, and $c$ to 甲, 乙, and 丙 respectively, each person only knows their own number. They are told that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, and each is asked the following two questions:
(1) Do you know the value of $a+b+c$?
(2) Do you know the values of $a$, $b$, and $c$ respectively?
甲 answers "no" to both questions, and 乙, after hearing 甲's answers, answers "yes" to the first question and "no" to the second question. After hearing both of their answers, how should 丙 answer these two questions?
|
10. First solve the equation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.
Assume $a \geqslant b \geqslant c$. If $c=3$, then $1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{3}{c}$.
Thus, $a=b=c=3$. Otherwise, $c=2$.
Then, $\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$.
Solving this, we get $a=b=4$ or $b=3, a=6$.
The sums corresponding to these three solutions are
$$
3+3+3=9,4+4+2=10,6+3+2=11 \text {. }
$$
Since Jia's answer to question (1) is negative, therefore, $a \neq 6$ or 4.
Similarly, $b=6$ or 4.
If $b=4$, since $a \neq 4$, then $a=2, c=4$.
Since Yi's answer to question (2) is negative, hence $b=6$.
Therefore, Bing knows $b=6$ and $a+b+c=11$.
Since Bing also knows the value of $c$, he must also know the value of $a$.
Thus, Bing answers "know" to both questions.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a, b$ be positive real numbers, and
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}, \\
(a-b)^{2}=4(a b)^{3} .
\end{array}
$$
Then $\log _{a} b=$ $\qquad$
|
3. -1 .
From $\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}$, we get $a+b \leqslant 2 \sqrt{2} a b$.
$$
\begin{array}{l}
\text { Also, }(a+b)^{2}=4 a b+(a-b)^{2} \\
=4 a b+4(a b)^{3} \\
\geqslant 4 \times 2 \sqrt{a b(a b)^{3}}=8(a b)^{2},
\end{array}
$$
which means $a+b \geqslant 2 \sqrt{2} a b$.
Thus, $a+b=2 \sqrt{2} a b$.
From the condition for equality in equation (1), we get $a b=1$.
Solving this together with equation (2), we have
Therefore, $\log _{a} b=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) As shown in Figure 4, given that $AB$ is the diameter of $\odot O$, chord $CD$ intersects $AB$ at point $E$. A tangent to $\odot O$ is drawn from point $A$ and intersects the extension of $CD$ at point $F$. Given $AC=8$, $CE:ED=6:5$, and $AE:EB=2:3$. Find the length of $AB$ and the value of $\tan \angle ECB$.
---
Translate the text into English, preserving the original text's line breaks and format. Directly output the translation result.
|
Let $C E=6 x, E D=5 x, A E=2 y$, $E B=3 y, D F=z$.
By the intersecting chords theorem, we have
$$
A E \cdot B E=C E \cdot E D \Rightarrow y=\sqrt{5} x \text {. }
$$
By the secant-tangent theorem, we have
$$
A F^{2}=D F \cdot C F=z(z+11 x) \text {. }
$$
By the Pythagorean theorem, we have
$$
\begin{array}{l}
A F^{2}=E F^{2}-A E^{2} \\
\Rightarrow z(z+11 x)=(5 x+z)^{2}-(2 y)^{2} .
\end{array}
$$
Substituting $y=\sqrt{5} x$ yields $z=5 x$, meaning $D$ is the midpoint of $E F$.
Therefore, $A D=D E=5 x$.
It is easy to see that $B C=B E=3 \sqrt{5} x, A B=5 \sqrt{5} x$.
$$
\begin{array}{l}
\text { Also, } A B^{2}-B C^{2}=A C^{2} \\
\Rightarrow(5 \sqrt{5} x)^{2}-(3 \sqrt{5} x)^{2}=8^{2} \\
\Rightarrow x=\frac{2 \sqrt{5}}{5} \Rightarrow A B=10 .
\end{array}
$$
Since $\angle E C B=\angle E A D=\angle D E A$, we have
$$
\tan \angle E C B=\tan \angle D E A=\frac{A F}{A E}=2 .
$$
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let the function
$$
f(x)=\frac{1+\ln (x+1)}{x} \text {, }
$$
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$.
untranslated part:
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$.
(Note: The last part was already in English, so it remains unchanged.)
|
11. Since $x>0$, we have
$$
k0)
$$
always holds.
$$
\text{Let } g(x)=\frac{(x+1)[1+\ln (x+1)]}{x}(x>0) \text{.}
$$
Then $g^{\prime}(x)=\frac{x-1-\ln (x+1)}{x^{2}}$.
Let $h(x)=x-1-\ln (x+1)$.
Obviously, $h(2)0$.
If $h\left(x_{0}\right)=0$, then $20$ (since $x>0$), then $h(x)$ is monotonically increasing in $(0,+\infty)$.
When $x \in\left(0, x_{0}\right)$, $h(x)0$, i.e., $g^{\prime}(x) >0$, then $g(x)$ is increasing in the interval $\left(x_{0},+\infty\right)$.
Thus, when $x>0$,
$$
\begin{array}{l}
g(x)_{\text{min}}=g\left(x_{0}\right) \\
=\frac{\left(x_{0}+1\right)\left[1+\ln \left(x_{0}+1\right)\right]}{x_{0}}.
\end{array}
$$
From equation (1) and the above equation, we get $g(x)_{\min }=x_{0}+1$.
Given that $k<g(x)_{\text{min}}$, i.e., $k<x_{0}+1$.
Since $3<x_{0}+1<4$ and $k$ is a positive integer, the maximum value of $k$ is 3.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a=\sqrt{3}-1$. Then $a^{2012}+2 a^{2011}-2 a^{2010}=$
|
2. 0 .
Notice that $a^{2}=(\sqrt{3}-1)^{2}=4-2 \sqrt{3}$. Then $a^{2}+2 a-2=0$.
Therefore, $a^{2012}+2 a^{2011}-2 a^{2010}$ $=a^{2010}\left(a^{2}+2 a-2\right)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $\triangle A B C$
the lengths of the three medians are 3,
4, 5. Then $S_{\triangle M B C}$ is $\qquad$
|
4. 8 .
As shown in Figure 4, extend GD to point $D^{\prime}$, making it twice as long.
Then the side lengths of $\triangle G D^{\prime} C$ are $\frac{2}{3}$ times the lengths of the three medians of $\triangle A B C$.
Therefore, it is a right triangle, and its area is $\frac{8}{3}$.
Additionally, the area of $\triangle G D^{\prime} C$ is equal to the area of $\triangle B G C$, and is $\frac{1}{3}$ of the area of $\triangle A B C$, hence
$$
S_{\triangle A B C}=8 .
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) There are $m$ regular $n$-sided polygons, and the sum of the interior angles of these $m$ regular polygons can be divided by 8. Find the minimum value of $m+n$.
|
Three, from the problem, we know that the total sum of the interior angles of these $m$ regular polygons is $m(n-2) \times 180$.
From $81[180 m(n-2)]$
$\Rightarrow 21 \mathrm{~m}(n-2) \Rightarrow 21 \mathrm{mn}$.
Thus, at least one of $m$ and $n$ is even.
Also, $m \geqslant 1, n \geqslant 3$, and both are integers.
To make $m+n$ the smallest, then
$(m, n)=(1,4),(2,3)$.
Therefore, the minimum value of $m+n$ is 5.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given an integer $n \geqslant 2$.
(1) Prove: The set $\{1,2, \cdots, n\}$ can be properly arranged into subsets $A_{1}, A_{2}, \cdots, A_{2^{n}}$, such that the number of elements in $A_{i}$ and $A_{i+1}\left(i=1,2, \cdots, 2^{n}\right.$, and $\left.A_{2^{n+1}}=A_{1}\right)$ differ by exactly 1.
(2) For subsets $A_{1}, A_{2}$,
$$
\begin{array}{l}
\cdots, A_{2^{n}}, \text { find all possible values of } \sum_{i=1}^{2 n}(-1)^{i} S\left(A_{i}\right) \text {, where } \\
\text { } S\left(A_{i}\right)=\sum_{x \in \lambda_{i}} x, S(\varnothing)=0 .
\end{array}
$$
(Supplied by Liang Yingde)
|
3. (1) Prove by mathematical induction: there exists a subsequence $A_{1}, A_{2}, \cdots, A_{2^{n}}$ that meets the requirements, and $A_{1}=\{1\}$, $A_{2^{n}}=\varnothing$.
When $n=2$, the sequence $\{1\}, \{1,2\}, \{2\}, \varnothing$ satisfies the requirements.
Assume that when $n=k$, there exists a subsequence $B_{1}, B_{2}, \cdots, B_{2^{k}}$ that meets the requirements.
For $n=k+1$, construct the sequence as follows:
$A_{1}=B_{1}=\{1\}$,
$A_{i}=B_{i-1} \cup\{k+1\}\left(i=2,3, \cdots, 2^{k}+1\right)$,
$A_{j}=B_{j-2^{k}}\left(j=2^{k}+2,2^{k}+3, \cdots, 2^{k+1}\right)$.
It is easy to verify that the sequence $A_{1}, A_{2}, \cdots, A_{2^{k+1}}$ meets the requirements.
In summary, for any positive integer $n(n \geqslant 2)$, there exists a subsequence that meets the requirements.
(2) Without loss of generality, let $A_{1}=\{1\}$. Since the number of elements in adjacent two subsets differs by 1, one must be odd and the other even. Therefore, the number of elements in a subset is the same as the parity of its index.
$$
\text { Hence } \sum_{i=1}^{2 n}(-1)^{i} S\left(A_{i}\right)=\sum_{i \in P} S(A)-\sum_{A \in Q} S(A) \text {, }
$$
where $P$ is the set of all even-element subsets of $\{1,2, \cdots, n\}$, and $Q$ is the set of all odd-element subsets of $\{1,2, \cdots, n\}$.
For any $x \in\{1,2, \cdots, n\}$, since it appears $\mathrm{C}_{n-1}^{k-1}$ times in all $k$-element subsets, its contribution to $\sum_{i \in P} S(A)-\sum_{i \in Q} S(A)$ is
$$
\sum_{k=1}^{n}(-1)^{k} \mathrm{C}_{n-1}^{k-1}=-(1-1)^{n-1}=0 .
$$
Therefore, $\sum_{i=1}^{2 n}(-1)^{i} S\left(A_{i}\right)=0$.
|
0
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
5. The function $f(x)$ is defined on $(0, \infty)$, and satisfies
$$
f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0 \text {. }
$$
Then the minimum value of $f(x)$ is $\qquad$
|
5.3.
From $f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0$, we get $f\left(\frac{1}{x}\right)-\frac{2}{x} f(x)+\frac{3}{x^{2}}=0$.
By solving the above two equations simultaneously, we obtain $f(x)=x^{2}+\frac{2}{x}$.
By the AM-GM inequality,
$$
x^{2}+\frac{2}{x}=x^{2}+\frac{1}{x}+\frac{1}{x} \geqslant 3\left(x^{2} \cdot \frac{1}{x} \cdot \frac{1}{x}\right)^{\frac{1}{3}}=3 \text {, }
$$
and when $x=1$, the equality holds.
Therefore, the minimum value of $f(x)$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The complex number $z$ satisfies
$$
|z|(3 z+2 \mathrm{i})=2(\mathrm{i} z-6) \text {. }
$$
Then $|z|$ equals $\qquad$ .
|
6. 2 .
Solution 1 Direct calculation shows that
$$
|3 z+2 \mathrm{i}|^{2}-|\mathrm{i} z-6|^{2}=8\left(|z|^{2}-4\right) \text {. }
$$
From this, if $|z|>2$, then
$$
\begin{array}{l}
|3 z+2 \mathrm{i}|>|\mathrm{i} z-6| \\
\Rightarrow|| z|(3 z+2 \mathrm{i})|>|2(\mathrm{i} z-6)|,
\end{array}
$$
which contradicts the given condition;
if $|z|<2$, then
$$
|| z \mid(3 z+2 \text { i) }|<| 2(z+6 \text { i) } \mid,
$$
which is also a contradiction.
Therefore, $|z|=2$.
Solution 2 Let $|z|=r$. Substituting into the condition, we get
$$
z=\frac{-(12+2 r \mathrm{i})}{3 r-2 \mathrm{i}} \text {. }
$$
Then $r^{2}=|z|^{2}=\frac{12^{2}+(2 r)^{2}}{(3 r)^{2}+(-2)^{2}}$.
Solving this, we get $r=2$, i.e., $|z|=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $\left(1+x-x^{2}\right)^{10}=a_{0}+a_{1} x+\cdots+a_{20} x^{20}$. Then $a_{0}+a_{1}+2 a_{2}+\cdots+20 a_{20}=$ $\qquad$
|
7. -9 .
Let $x=0$, we get $a_{0}=1$.
Differentiating both sides of the given equation, we get
$$
\begin{array}{l}
10\left(1+x-x^{2}\right)^{9}(1-2 x) \\
=a_{1}+2 a_{2} x+\cdots+20 a_{20} x^{19} . \\
\text { Let } x=1 \text {, we get. } \\
a_{1}+2 a_{2}+\cdots+20 a_{20}=-10 .
\end{array}
$$
Then $a_{0}+a_{1}+2 a_{2}+\cdots+20 a_{20}=-9$.
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 1, in the tetrahedron $D-ABC$, it is known that $DA \perp$ plane $ABC$, and $\triangle ABC$ is an equilateral triangle with a side length of 2. Then, when the tangent value of the dihedral angle $A-BD-C$ is 2, $V_{D-ABC}=$ $\qquad$
|
4.2.
Given $D A \perp$ plane $A B C$, we know
$D A \perp A B$, plane $D A B \cdot \perp$ plane $A B C$.
As shown in Figure 3, take the midpoint $O$ of $A B$.
Then, by the problem statement, $C O \perp A B, C O$
is on plane $D A B$, and
$$
C O=\sqrt{3} \text {. }
$$
Draw $O E \perp D B$ at point $E$, and connect $C E$. Then
$$
C E \perp D B \text {. }
$$
Thus, $\angle O E C$ is the plane angle of the dihedral angle $A-B D-C$.
Let $D A=x$. Then
$$
\begin{array}{l}
O E=\frac{1}{2} \times \frac{D A \cdot A B}{D B} \\
=\frac{1}{2} \times \frac{2 x}{\sqrt{x^{2}+4}}=\frac{x}{\sqrt{x^{2}+4}} .
\end{array}
$$
Given $\tan \angle O E C=\frac{O C}{O E}=2$, we have
$$
\sqrt{3} \times \sqrt{x^{2}+4}=2 x \Rightarrow x=2 \sqrt{3} \text {. }
$$
Therefore, $V_{D \rightarrow A B C}=\frac{1}{3} \times 2 \sqrt{3}\left(\frac{\sqrt{3}}{4} \times 2^{2}\right)=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given the equation in $x$
$$
x^{4}+2 x^{3}+(3+k) x^{2}+(2+k) x+2 k=0
$$
has real roots. If the product of all real roots is -2, then the sum of the squares of all real roots is $\qquad$ .
|
$$
\begin{array}{l}
\left(x^{4}+2 x^{3}+x^{2}\right)+\left[(2+k) x^{2}+(2+k) x\right]+2 k=0 \\
\Rightarrow\left(x^{2}+x\right)^{2}+(2+k)\left(x^{2}+x\right)+2 k=0 \\
\Rightarrow\left(x^{2}+x+2\right)\left(x^{2}+x+k\right)=0 .
\end{array}
$$
Since the original equation has real roots, and $x^{2}+x+2=0$ has no real roots, the equation $x^{2}+x+k=0$ has two real roots $x_{1}$ and $x_{2}$.
According to the problem, $x_{1} x_{2}=k=-2$.
Also, $x_{1}+x_{2}=-1$, so
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=5 \text{. }
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $A$ and $B$ be the common vertices of the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
and the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
Let $P$ and $M$ be two moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
$$
\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A M}+\overrightarrow{B M}),
$$
where $\lambda \in \mathbf{R},|\lambda|>1$. Let the slopes of the lines $A P, B P, A M, B M$ be $k_{1}, k_{2}, k_{3}, k_{4}$, respectively, and $k_{1}+k_{2}=5$. Then $k_{3}+k_{4}=$ $\qquad$
|
11. -5 .
Let $A(-a, 0), B(a, 0), P\left(x_{1}, y_{1}\right), M\left(x_{2}, y_{2}\right)$.
From $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A M}+\overrightarrow{B M})$, we know that points $O, P, M$ are collinear, and we can find that
$$
\begin{array}{l}
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a} \\
=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}} .
\end{array}
$$
Similarly, $k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}$.
Therefore, $k_{3}+k_{4}=-\left(k_{1}+k_{2}\right)=-5$.
|
-5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 5, given $\triangle A B C$ is inscribed in $\odot O$, chord $A F \perp B C$ at point $H, G$ is the midpoint of $B F$. Then $\frac{A C}{O G}=$
|
4. 2 .
Connect $F O$. Then
$$
\begin{array}{l}
\text { Rt } \triangle A B H \backsim \text { Rt } \triangle O \\
\Rightarrow \frac{A H}{B H}=\frac{O G}{F G} . \\
\text { Also } \triangle A C H \backsim \triangle B F H \\
\Rightarrow \frac{A H}{B H}=\frac{A C}{B F} .
\end{array}
$$
Therefore $A C=2 O G \Rightarrow \frac{A C}{O G}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The number of real numbers $a$ that make the equation
$$
x^{2}+a x+8 a=0
$$
have only integer solutions is $\qquad$ .
|
2. 8 .
Let the equation (1) have integer solutions $m, n (m \leqslant n)$. Then $m+n=-a, mn=8a$.
Thus, $(m+8)(n+8)=64$.
Solving for $(m, n)$
$$
\begin{array}{c}
=(-72,-9),(-40,-10),(-24,-12), \\
(-16,-16),(-7,56),(-6,24), \\
(-4,8),(0,0) .
\end{array}
$$
Correspondingly,
$$
\begin{array}{l}
a=-(m+n) \\
=81,50,36,32,-49,-18,-4,0,
\end{array}
$$
a total of 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. All prime numbers $p$ that make $2 p^{4}-p^{2}+36$ a perfect square are $\qquad$ .
|
5.2.
When $p=2$,
$$
2 p^{4}-p^{2}+36=64
$$
is a perfect square, thus, $p=2$ is the solution.
When $p=3$,
$$
2 p^{4}-p^{2}+36=189
$$
is not a perfect square.
When $p$ is an odd prime greater than 3, let
$2 p^{4}-p^{2}+36=k^{2}$ ( $k$ is an odd positive integer).
Then $p^{2}\left(2 p^{2}-1\right)=(k-6)(k+6)$
$\Rightarrow p \mid(k-6)$ or $p \mid(k+6)$.
However, $p$ cannot divide both $k-6$ and $k+6$. Therefore, one and only one of $k-6$ and $k+6$ is divisible by $p^{2}$.
If $p^{2} \mid(k-6)$, let $k-6=s p^{2}$ ( $s$ is an odd positive integer).
$$
\begin{array}{l}
\text { Then } p^{2}\left(2 p^{2}-1\right)=(k-6)(k+6) \\
=s p^{2}\left(s p^{2}+12\right) \\
\Rightarrow s\left(s p^{2}+12\right)=2 p^{2}-1 \\
\Rightarrow s^{2}<2 \Rightarrow s=1 \Rightarrow p^{2}=13,
\end{array}
$$
which is a contradiction.
If $p^{2} \mid(k+6)$, let $k+6=s p^{2}$ ( $s$ is an odd positive integer).
Similarly, there does not exist a prime $p$ greater than 3.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given a moving point $O$ inside $\triangle A B C$, rays $A O, B O, C O$ intersect the opposite sides at points $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. Let $B^{\prime} C^{\prime}$ intersect $A O$ at $D$, $C^{\prime} A^{\prime}$ intersect $B O$ at $E$, and $A^{\prime} B^{\prime}$ intersect $C O$ at $F$. Prove: Regardless of the position of point $O$, $\frac{O D}{A D}+\frac{O E}{B E}+\frac{O F}{C F}$ is always a constant.
|
Prove that, as shown in Figure 3, there exist three complete quadrilaterals.
From the complete quadrilateral \( A C^{\prime} O B^{\prime} B C \), the complete quadrilateral \( B A^{\prime} O C^{\prime} C A \), and the complete quadrilateral \( C B^{\prime} O A^{\prime} A B \), we get
\[
\begin{array}{l}
\frac{O D}{A D}=\frac{O A^{\prime}}{A A^{\prime}}=\frac{S_{\triangle O B C}}{S_{\triangle A B C}}, \\
\frac{O E}{B E}=\frac{O B^{\prime}}{B B^{\prime}}=\frac{S_{\triangle O C A}}{S_{\triangle A B C}}, \\
\frac{O F}{C F}=\frac{O C^{\prime}}{C C^{\prime}}=\frac{S_{\triangle O A B}}{S_{\triangle A B C}} . \\
\text { Therefore, } \frac{O D}{A D}+\frac{O E}{B E}+\frac{O F}{C F} \\
=\frac{S_{\triangle O B C}+S_{\triangle O C A}+S_{\triangle O A B}}{S_{\triangle A B C}} \\
=\frac{S_{\triangle A B C}}{S_{\triangle A B C}}=1 \text { (a constant). }
\end{array}
\]
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given real numbers $a, b$ satisfy
$$
6^{a}=2010,335^{b}=2010 \text {. }
$$
Then the value of $\frac{1}{a}+\frac{1}{b}$ is $\qquad$
|
12. 1 .
Notice, $6^{a b}=2010^{b}, 335^{a b}=2010^{a}$.
Then $(6 \times 335)^{a b}=2010^{a+b}$.
Thus $a b=a+b \Rightarrow \frac{1}{a}+\frac{1}{b}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. $[x]$ represents the greatest integer not exceeding the real number $x$. Then, in the Cartesian coordinate system $x O y$, the area of the figure formed by all points $(x, y)$ that satisfy $[x][y]=$ 2011 is $\qquad$
|
3.4.
Let $[x]=a,[y]=b$, meaning all such points $(x, y)$ form the region
$$
a \leqslant x<a+1, b \leqslant y<b+1
$$
bounded by these inequalities, with an area of 1.
Since 2011 is a prime number, the points $(x, y)$ that satisfy
$$
[x][y]=2011
$$
form 4 regions, each with an area of 1, making the total area 4.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If the two quadratic equations
$$
x^{2}+x+m=0 \text { and } m x^{2}+x+1=0
$$
each have two distinct real roots, but one of them is a common real root $\alpha$, then the range of the real root $\alpha$ is $\qquad$.
|
Given that the common real root of the two equations is $\alpha$, and $m \neq 1$. Then
$$
\begin{array}{l}
\alpha^{2}+\alpha+m=0, \\
m \alpha^{2}+\alpha+1=0 .
\end{array}
$$
From equation (1), we get $m=-\alpha^{2}-\alpha$.
Substituting into equation (2), we get
$$
\begin{array}{l}
\alpha^{4}+\alpha^{3}-\alpha-1=0 \\
\Rightarrow\left(\alpha^{2}-1\right)\left(\alpha^{2}+\alpha+1\right)=0 .
\end{array}
$$
Since $\alpha^{2}+\alpha+1=\left(\alpha+\frac{1}{2}\right)^{2}+\frac{3}{4}>0$, then
$$
\alpha^{2}-1=0 \Rightarrow \alpha= \pm 1 \text {. }
$$
When $\alpha=1$, from equation (1) we get $m=-2$, at this time, the given two equations become
$$
x^{2}+x-2=0 \text { and } 2 x^{2}-x-1=0,
$$
with distinct roots -2 and $\frac{1}{2}$.
When $\alpha=-1$, from equation (1) we get $m=0$, at this time, the equation $m x^{2}+x+1=0$ becomes a linear equation $x+1=0$, which contradicts the problem statement.
In summary, the value of the real root $\alpha$ is $\alpha=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
20. Given the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$, a line is drawn through its left focus $F_{1}$ intersecting the ellipse at points $A$ and $B$. Point $D(a, 0)$ is a point to the right of $F_{1}$. Connecting $A D$ and $B D$ intersects the left directrix of the ellipse at points $M$ and $N$. If the circle with diameter $M N$ passes exactly through point $F_{1}$, find the value of $a$.
|
20. It is known that $F_{1}(-3,0)$, the equation of the left directrix is $x=-\frac{25}{3}$, and $l_{A B}: y=k(x+3)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}y=k(x+3), \\ \frac{x^{2}}{25}+\frac{y^{2}}{16}=1\end{array}\right.$
$$
\Rightarrow\left(16+25 k^{2}\right) x^{2}+150 k^{2} x+225 k^{2}-400=0 .
$$
Then $x_{1}+x_{2}=-\frac{150 k^{2}}{16+25 k^{2}}, x_{1} x_{2}=\frac{225 k^{2}-400}{16+25 k^{2}}$
$$
\Rightarrow y_{1} y_{2}=k^{2}\left(x_{1}+3\right)\left(x_{2}+3\right)=-\frac{256 k^{2}}{16+25 k^{2}} \text {. }
$$
Let $M\left(-\frac{25}{3}, y_{3}\right), N\left(-\frac{25}{3}, y_{4}\right)$.
From the collinearity of $M, A, D$ we get
$$
y_{3}=\frac{(3 a+25) y_{1}}{3\left(a-x_{1}\right)} \text {. }
$$
Similarly, $y_{4}=\frac{(3 a+25) y_{2}}{3\left(a-x_{2}\right)}$.
$$
\text { Also, } \overrightarrow{F_{1} M}=\left(-\frac{16}{3}, y_{3}\right), \overrightarrow{F_{1} N}=\left(-\frac{16}{3}, y_{4}\right) \text {. }
$$
From the given information,
$$
\begin{array}{l}
\overrightarrow{F_{1} M} \perp \overrightarrow{F_{1} N} \Rightarrow \overrightarrow{F_{1} M} \cdot \overrightarrow{F_{1} N}=0 \\
\Rightarrow y_{3} y_{4}=-\frac{256}{9} \text {. } \\
\end{array}
$$
$$
\begin{array}{l}
\text { And } y_{3} y_{4}=\frac{(3 a+25)^{2} y_{1} y_{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)} \\
\Rightarrow-\frac{256 k^{2}}{16+25 k^{2}} \cdot \frac{(3 a+25)^{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)}=-\frac{256}{9} \\
\Rightarrow\left(1+k^{2}\right)\left(16 a^{2}-400\right)=0
\end{array}
$$
$\Rightarrow a= \pm 5$ (negative value is discarded).
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that $A B=4, A A_{1}=A D=2$, points $E, F, G$ are the midpoints of edges $A A_{1}, C_{1} D_{1}, B C$ respectively. Then the volume of the tetrahedron $B_{1}-E F G$ is $\qquad$
|
2.3.
Take point $H$ on the extension of $D_{1} A_{1}$ such that $A_{1} H=\frac{1}{2}$. Then $H E / / B_{1} G$. Therefore, $H E / /$ plane $B_{1} F G$.
Thus, $V_{B_{1}-E F G}=V_{E-B_{1} F C}=V_{H-B_{1} F G}=V_{G-B_{1} F H}$.
And $S_{\triangle B_{1} F H}=\frac{9}{2}$, the distance from point $G$ to plane $B_{1} F H$ is 2. Hence $V_{B_{1}-E F G}=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) In the geometric sequence $\left\{a_{n}\right\}$ where all terms are positive, what is the maximum number of terms that can be integers between $100 \sim 1000$?
|
9. Let the geometric sequence $\left\{a q^{n-1}\right\}$ satisfy
$$
100 \leqslant a q^{n-1} \leqslant 1000 \quad (a, q > 1) \text{ are integers.}
$$
Clearly, $q$ must be a rational number.
Let $q=\frac{t}{s}(t>s \geqslant 1, (t, s)=1)$.
Since $a q^{n-1}=a\left(\frac{t}{s}\right)^{n-1}$ is an integer, $a$ must be a multiple of $s^{n-1}$.
Let $t=s+1$. Thus, the sequence satisfies
$100 \leqslant a < a \cdot \frac{s+1}{s} < \cdots < a\left(\frac{s+1}{s}\right)^{n-1} \leqslant 1000$.
If $s \geqslant 3$, then
$1000 \geqslant a\left(\frac{s+1}{s}\right)^{n-1} \geqslant (s+1)^{n-1} \geqslant 4^{n-1}$.
Thus, $n \leqslant 5$.
If $s=1$, then
$1000 \geqslant a \cdot 2^{n-1} \geqslant 100 \times 2^{n-1}$.
Thus, $n \leqslant 4$.
If $s=2$, then
$1000 \geqslant a\left(\frac{3}{2}\right)^{n-1} \geqslant 100\left(\frac{3}{2}\right)^{n-1}$.
Thus, $n \leqslant 6$.
On the other hand, the sequence
$128, 192, 288, 432, 648, 972$
satisfies the given conditions.
Therefore, the maximum number of terms between 100 and 1000 is 6.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given 315, find
$$
S=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\cdots+\sqrt[2011]{2011}}}}
$$
the integer part of \( S \).
|
Let $n=2011$, and set
$$
\begin{array}{l}
a_{1}=\sqrt[n]{n}, \\
a_{k}=\sqrt[n-k+1]{n-k+1+a_{k-1}}(2 \leqslant k \leqslant n-1) .
\end{array}
$$
Then $S=a_{n-1}$.
We will prove by induction that:
$$
1<a_{m}<2(m=1,2, \cdots, n-1) \text {. }
$$
(1) When $m=1$, since $1<n<2^{n}$, we have
$$
1<\sqrt[n]{n}<2 \text {. }
$$
(2) Assume that when $m=k(1 \leqslant k \leqslant n-2)$, the conclusion holds. Then when $m=k+1$, since
$$
\begin{array}{l}
1<n-k+1<n-k+a_{k}<n-k+2 \\
\leqslant(1+1)^{n-k}=1+(n-k)+\cdots+1,
\end{array}
$$
we have $1<a_{k+1}=\sqrt[n-k]{n-k+a_{k}}<2$.
Therefore, when $m=k+1$, the conclusion holds.
In summary, $1<a_{m}<2(m=1,2, \cdots, n-1)$.
Thus, the integer part of $S$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given $a, b \in \mathbf{R}$, the equation about $x$
$$
x^{4}+a x^{3}+2 x^{2}+b x+1=0
$$
has one real root. Find the minimum value of $a^{2}+b^{2}$.
|
【Analysis】This is a quartic equation, and parameters $a, b$ are not easy to handle. We might as well regard $a, b$ as the main variables and $x$ as a parameter. Since $a^{2}+b^{2}$ represents the square of the distance from the moving point $P(a, b)$ to the origin, and $P(a, b)$ lies on the line
$$
x^{3} a + x b + x^{4} + 2 x^{2} + 1 = 0,
$$
then
$$
\begin{array}{l}
\sqrt{a^{2}+b^{2}} \geqslant \frac{x^{4}+2 x^{2}+1}{\sqrt{x^{6}+x^{2}}} = \frac{x^{4}+1+2 x^{2}}{|x| \sqrt{x^{4}+1}} \\
= \frac{\sqrt{x^{4}+1}}{|x|} + \frac{2|x|}{\sqrt{x^{4}+1}} \geqslant 2 \sqrt{2}.
\end{array}
$$
The equality holds if and only if $x = \pm 1$.
Therefore, $\left(a^{2}+b^{2}\right)_{\text {min }} = 8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The graph of the quadratic function $y=x^{2}-a x+2$ is symmetric about the line $x=1$. Then the minimum value of $y$ is $\qquad$ .
|
$$
=, 1.1 .
$$
From the condition, we know that $a=2$. Then $y=(x-1)^{2}+1$ has a minimum value of 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a=\sqrt{3}-1$. Then the value of $a^{2012}+2 a^{2011}-$ $2 a^{2010}$ is $\qquad$.
|
2. 0 .
From the condition we know
$$
\begin{array}{l}
(a+1)^{2}=3 \Rightarrow a^{2}+2 a-2=0 \text {. } \\
\text { Then } a^{2012}+2 a^{2011}-2 a^{2010} \\
=a^{2010}\left(a^{2}+2 a-2\right)=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In a chess tournament, there are $n$ female players and $9 n$ male players. Each player plays one game against each of the other $10 n-1$ players. The scoring system is as follows: the winner gets 2 points, the loser gets 0 points, and in the case of a draw, each player gets 1 point. After the tournament, it was found that the total score of all male players is 4 times the total score of all female players. Then, all possible values of $n$ are $\qquad$
|
4.1.
Let the total score of the girls be $m$. Then the total score of the boys is $4 m$. According to the problem,
$$
\begin{array}{l}
\frac{10 n(10 n-1)}{2} \times 2=5 m \\
\Rightarrow m=2 n(10 n-1) .
\end{array}
$$
Since each player competes in $10 n-1$ matches, the maximum score is $2(10 n-1)$, so the total score of the girls
$$
m \leqslant 2 n(10 n-1) \text {. }
$$
Therefore, each girl must win all her matches.
Thus, $n=1$.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the equations
$$
x^{2}+b x+1=0 \text { and } x^{2}-x-b=0
$$
have a common root, find the value of $b$.
|
Prompt: Example 1. Use the substitution method to find the common root \( x_{0}=-1 \).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The foci of the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ are $F_{1}$ and $F_{2}$. If a point $P$ on the ellipse makes $P F_{1} \perp P F_{2}$, then the area of $\triangle P F_{1} F_{2}$ is $\qquad$
|
2. 9 .
It is known that $F_{1} F_{2}=8, P F_{1}+P F_{2}=10$.
Then $\left(P F_{1}+P F_{2}\right)^{2}=10^{2}$.
In the right triangle $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}^{2}+P F_{2}^{2}=8^{2} \text{. }
$$
From equations (1) and (2), we get
$$
S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P F_{2}=9 .
$$
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given an arithmetic sequence $\left\{a_{n}\right\}$, the sum of the first 15 terms $S_{15}=30$. Then $a_{1}+a_{8}+a_{15}=$ $\qquad$
|
12.6.
From $S_{15}=30 \Rightarrow a_{1}+7 d=2$.
Therefore, $a_{1}+a_{8}+a_{15}=3\left(a_{1}+7 d\right)=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Let $x, y$ be real numbers. Then
$$
\max _{S x^{2}+4 y^{2}=10 x}\left(x^{2}+y^{2}\right)=
$$
$\qquad$
|
15. 4 .
$$
\begin{array}{l}
\text { Given } 5 x^{2}+4 y^{2}=10 x \\
\Rightarrow 4 y^{2}=10 x-5 x^{2} \geqslant 0 \\
\Rightarrow 0 \leqslant x \leqslant 2 .
\end{array}
$$
Then $4\left(x^{2}+y^{2}\right)=10 x-x^{2}$
$$
\begin{array}{l}
=25-(5-x)^{2} \leqslant 25-3^{2} \\
\Rightarrow x^{2}+y^{2} \leqslant 4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given that $x, y, z$ are real numbers, satisfying
$$
\begin{array}{c}
x=\sqrt{y^{2}-\frac{1}{16}}+\sqrt{z^{2}-\frac{1}{16}}, \\
y=\sqrt{z^{2}-\frac{1}{25}}+\sqrt{x^{2}-\frac{1}{25}}, \\
z=\sqrt{x^{2}-\frac{1}{36}}+\sqrt{y^{2}-\frac{1}{36}},
\end{array}
$$
and $x+y+z=\frac{m}{\sqrt{n}}\left(m, n \in \mathbf{Z}_{+}, \sqrt{n}\right.$ is the simplest radical form). Find $m+n$.
|
Solve as shown in Figure 11, construct an acute triangle $\triangle ABC$, such that
$$
\begin{array}{l}
BC=x, \\
CA=y, \\
AB=z .
\end{array}
$$
Draw perpendiculars from $A$, $B$, and $C$ to $BC$, $CA$, and $AB$ respectively, with the feet of the perpendiculars being $D$, $E$, and $F$. Let $AD=u$, $BE=v$, and $CF=w$. Then
$$
\begin{array}{l}
x=\sqrt{y^{2}-u^{2}}+\sqrt{z^{2}-u^{2}}, \\
y=\sqrt{z^{2}-v^{2}}+\sqrt{x^{2}-v^{2}}, \\
z=\sqrt{x^{2}-w^{2}}+\sqrt{y^{2}-w^{2}} .
\end{array}
$$
Consider the function
$$
f(u)=\sqrt{y^{2}-u^{2}}+\sqrt{z^{2}-u^{2}}.
$$
It is easy to see that $f(u)$ is strictly decreasing on $(0,+\infty)$. Hence $u=\frac{1}{4}$.
Similarly, $v=\frac{1}{5}$, $w=\frac{1}{6}$.
Let the area of $\triangle ABC$ be $S$. Then
$$
S=\frac{1}{2} AB \cdot CF=\frac{1}{2} BC \cdot AD=\frac{1}{2} CA \cdot BE,
$$
which means
$$
\begin{array}{l}
2 S=z w=x u=y v \\
\Rightarrow x=8 S, y=10 S, z=12 S \\
\Rightarrow x+y+z=30 S .
\end{array}
$$
By Heron's formula,
$$
\begin{array}{l}
S^{2}=15 S(15 S-8 S)(15 S-10 S)(15 S-12 S) \\
\Rightarrow S=\frac{1}{15 \sqrt{7}} \Rightarrow x+y+z=\frac{2}{\sqrt{7}} \\
\Rightarrow m+n=9 .
\end{array}
$$
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given two equations about $x$
$$
x^{2}-x+3 m=0, x^{2}+x+m=0(m \neq 0) \text {. }
$$
If one root of the first equation is three times a root of the second equation, then the value of the real number $m$ is $\qquad$
|
- 1. -2 .
Let one root of the latter equation be $\alpha$. And the former equation has a root $3 \alpha$, then
$$
\alpha^{2}+\alpha+m=0,
$$
and
$$
\begin{array}{l}
9 \alpha^{2}-3 \alpha+3 m=0 \\
\Rightarrow 3 \alpha^{2}-\alpha+m=0 .
\end{array}
$$
(2) - (1) gives
$$
2 \alpha^{2}-2 \alpha=0 \Rightarrow \alpha=0 \text { or } 1 \text {. }
$$
If $\alpha=0$, then $m=0$, which contradicts the assumption that $m \neq 0$. Therefore, $\alpha=1$. Consequently, $m=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the sequence $\left\{a_{n}\right\}$ satisfies the recurrence relation
$$
a_{n+1}=2 a_{n}+2^{n}-1\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
and $\left\{\frac{a_{n}+\lambda}{2^{n}}\right\}$ is an arithmetic sequence. Then the value of $\lambda$ is $\qquad$
|
6. -1 .
Notice,
$$
\begin{array}{l}
\frac{a_{n+1}+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2 a_{n}+2^{n}-1+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2^{n}-1-\lambda}{2^{n+1}} . \\
\text { From } \frac{2^{n}-1-\lambda}{2^{n+1}} \text { being a constant, we know } \lambda=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. There are 8 red, 8 white, and 8 yellow chopsticks. Without looking, how many chopsticks must be taken out to ensure that at least two pairs of chopsticks are of different colors? $\qquad$
|
10. 11.
Since among 11 chopsticks there must be a pair of the same color (let's say yellow), the number of black or white chopsticks must be at least 3, among which there must be a pair of the same color, i.e., both black or both white. Therefore, 11 chopsticks ensure success. However, if only 10 chopsticks are taken, it is possible to have 8 yellow, 1 black, and 1 white, which does not meet the requirement.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given real numbers $x, y$ satisfy
$$
3|x+1|+2|y-1| \leqslant 6 \text {. }
$$
Then the maximum value of $2 x-3 y$ is $\qquad$ (1)
|
Solve As shown in Figure 1, the figure determined by inequality (1) is the quadrilateral $\square A B C D$ and its interior, enclosed by four straight lines, where,
$$
\begin{array}{l}
A(-1,4), B(1,1), \\
C(-1,-2), D(-3,1) .
\end{array}
$$
Consider the family of lines $2 x-3 y=k$, i.e.,
$$
y=\frac{2}{3} x-\frac{k}{3} \text {, }
$$
which has the smallest intercept $-\frac{4}{3}$ at the lowest point $C$ of $\triangle A B C D$.
Therefore, when $x=-1, y=-2$, $2 x-3 y$ attains its maximum value of 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $x$ be a real number, and define $\lceil x\rceil$ as the smallest integer not less than the real number $x$ (for example, $\lceil 3.2 \rceil = 4, \lceil -\pi \rceil = -3$). Then, the sum of all real roots of the equation
$$
\lceil 3 x+1\rceil=2 x-\frac{1}{2}
$$
is equal to
|
10. -4 .
Let $2 x-\frac{1}{2}=k \in \mathbf{Z}$. Then
$$
x=\frac{2 k+1}{4}, 3 x+1=k+1+\frac{2 k+3}{4} \text {. }
$$
Thus, the original equation is equivalent to
$$
\begin{array}{l}
{\left[\frac{2 k+3}{4}\right]=-1 \Rightarrow-2<\frac{2 k+3}{4} \leqslant-1} \\
\Rightarrow-\frac{11}{2}<k \leqslant-\frac{7}{2} \\
\Rightarrow k=-5 \text { or }-4 .
\end{array}
$$
The corresponding $x$ values are $-\frac{9}{4},-\frac{7}{4}$.
Therefore, the sum of all real roots is -4.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$
has a maximum value of $g(m)$, find the minimum value of $g(m)$.
|
Three, 13. Let $t=\sqrt{100-m x}$. Then $x=\frac{100-t^{2}}{m}$.
Hence $y=\frac{100-t^{2}}{m}+t$
$$
=-\frac{1}{m}\left(t-\frac{m}{2}\right)^{2}+\frac{100}{m}+\frac{m}{4} \text {. }
$$
When $t=\frac{m}{2}$, $y$ has a maximum value $\frac{100}{m}+\frac{m}{4}$, that is,
$$
g(m)=\frac{100}{m}+\frac{m}{4} \geqslant 2 \sqrt{\frac{100}{m} \times \frac{m}{4}}=10 .
$$
Therefore, when and only when $m=20$, $g(m)$ has a minimum value of 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given the function
$$
f(x)=2\left(\sin ^{4} x+\cos ^{4} x\right)+m(\sin x+\cos x)
$$
has a maximum value of 5 for $x \in\left[0, \frac{\pi}{2}\right]$. Find the value of the real number $m$.
|
14. Notice,
$$
f(x)=2-(2 \sin x \cdot \cos x)^{2}+m(\sin x+\cos x)^{4} \text {. }
$$
Let $t=\sin x+\cos x$
$$
=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \in[1, \sqrt{2}] \text {. }
$$
Thus, $2 \sin x \cdot \cos x=t^{2}-1$.
Therefore, $f(x)=2-\left(t^{2}-1\right)^{2}+m t^{4}$
$=(m-1) t^{4}+2 t^{2}+1$.
Let $u=t^{2} \in[1,2]$.
By the problem, $g(u)=(m-1) u^{2}+2 u+1$ has a maximum value of 5 for $u \in[1,2]$.
When $m-1=0$,
$$
g(u)=2 u+1 \leqslant g(2)=5,
$$
Thus, $m=1$ satisfies the condition;
When $m-1>0$,
$$
g(u)_{\max } \geqslant g(2)>2 \times 2+1=5 \text {, }
$$
which is a contradiction;
When $m-1<0$,
$$
g(u)<2 u+1 \leqslant 5 \text {, }
$$
which is a contradiction.
In conclusion, the real number $m=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) If the two sides of a right triangle, $x, y$, are both prime numbers, and make the algebraic expressions $\frac{2 x-1}{y}$ and $\frac{2 y+3}{x}$ both positive integers, find the inradius $r$ of this right triangle.
|
(1) If $x>y$, then
$$
1 \leqslant \frac{2 y+3}{x}<\frac{2 x+3}{x}<4 \text {. }
$$
It is easy to see that $\frac{2 y+3}{x}=1$ or 2.
$$
\begin{array}{l}
\text { (i) From } \frac{2 y+3}{x}=1 \\
\Rightarrow x=2 y+3 \\
\Rightarrow \frac{2 x-1}{y}=\frac{2(2 y+3)-1}{y}=4+\frac{5}{y} \\
\Rightarrow y=5, x=13 ;
\end{array}
$$
(ii) From $\frac{2 y+3}{x}=2 \Rightarrow 2 x=2 y+3$, the equation is obviously not valid, leading to a contradiction.
(2) If $x=y$, then $\frac{2 x-1}{y}$ is not a positive integer, which contradicts the given condition.
(3) If $x<y$, then
$$
\begin{array}{l}
1 \leqslant \frac{2 x-1}{y}<2 \Rightarrow \frac{2 x-1}{y}=1 \Rightarrow y=2 x-1 \\
\Rightarrow \frac{2 y+3}{x}=\frac{2(2 x-1)+3}{x}=4+\frac{1}{x}
\end{array}
$$
It cannot be a positive integer, which contradicts the given condition.
In summary, only $y=5, x=13$ meets the conditions of the problem. The third side of the right triangle is 12 or $\sqrt{194}$.
$$
\text { Hence } r=\frac{5+12-13}{2}=2
$$
or $r=\frac{5+13-\sqrt{194}}{2}=\frac{18-\sqrt{194}}{2}$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the function
$$
f(x)=a x+\sin x
$$
has perpendicular tangents on its graph, then the real number $a$ is
$\qquad$ .
|
4.0.
Notice that, $f^{\prime}(x)=a+\cos x$.
If the function $f(x)$ has two perpendicular tangents, then there exist $x_{1}, x_{2} \in \mathbf{R}$, such that
$$
\begin{array}{l}
f^{\prime}\left(x_{1}\right) f^{\prime}\left(x_{2}\right)=-1 \\
\Leftrightarrow\left(a+\cos x_{1}\right)\left(a+\cos x_{2}\right)=-1 \\
\Leftrightarrow a^{2}+a\left(\cos x_{1}+\cos x_{2}\right)+\cos x_{1} \cdot \cos x_{2}+1=0 \\
\Leftrightarrow\left(a+\frac{\cos x_{1}+\cos x_{2}}{2}\right)^{2}+1- \\
\quad\left(\frac{\cos x_{1}-\cos x_{2}}{2}\right)^{2}=0 \\
\Leftrightarrow \cos x_{1}=-\cos x_{2}= \pm 1, a=0 .
\end{array}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 The equation $2 x^{2}+5 x y+2 y^{2}=2007$ has $\qquad$ different integer solutions.
(2007, National Junior High School Mathematics League Sichuan Preliminary Competition)
|
The original equation can be transformed into
$$
(2 x+y)(x+2 y)=2007 \text {. }
$$
Since $x$ and $y$ are integers, without loss of generality, assume $x \leqslant y$, so,
$$
2 x+y \leqslant x+2 y \text {. }
$$
Notice that, $31[2 x+y)+(x+2 y)]$.
Thus, from equation (1), we get the system of equations
$$
\left\{\begin{array} { l }
{ 2 x + y = 3 , } \\
{ x + 2 y = 6 6 9 ; }
\end{array} \left\{\begin{array}{l}
2 x+y=-669, \\
x+2 y=-3 .
\end{array}\right.\right.
$$
The integer solutions are
When $x \geqslant y$, the original equation has two more sets of integer solutions
$$
\left\{\begin{array} { l }
{ x = 4 4 5 , } \\
{ y = - 2 2 1 , }
\end{array} \left\{\begin{array}{l}
x=221, \\
y=-445 .
\end{array}\right.\right.
$$
In summary, the original equation has 4 sets of integer solutions.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Suppose the lengths of the two legs of a right triangle are $a$ and $b$, and the length of the hypotenuse is $c$. If $a$, $b$, and $c$ are all positive integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition.
(2010, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
|
Solve: By the Pythagorean theorem, we have $c^{2}=a^{2}+b^{2}$.
Also, $c=\frac{1}{3} a b-(a+b)$, so
$$
c^{2}=\left[\frac{1}{3} a b-(a+b)\right]^{2} \text {. }
$$
Rearranging gives $a b-6(a+b)+18=0$.
Thus, $(a-6)(b-6)=18$.
Since $a$ and $b$ are both positive integers, and without loss of generality, let $a<b$, then,
$$
\left\{\begin{array} { l }
{ a - 6 = 1 , } \\
{ b - 6 = 1 8 ; }
\end{array} \left\{\begin{array} { l }
{ a - 6 = 2 , } \\
{ b - 6 = 9 ; }
\end{array} \left\{\begin{array}{l}
a-6=3, \\
b-6=6 .
\end{array}\right.\right.\right.
$$
Solving gives $\left\{\begin{array}{l}a=7, \\ b=24 ;\end{array}\left\{\begin{array}{l}a=8, \\ b=15 ;\end{array}\left\{\begin{array}{l}a=9, \\ b=12 .\end{array}\right.\right.\right.$
The corresponding $c=25,17,15$.
Therefore, there are 3 right-angled triangles that satisfy the conditions.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 It is known that for any $x$,
$$
a \cos x + b \cos 2x \geqslant -1
$$
always holds. Find the minimum value of $a + b$.
(2009, Peking University Independent Admission Examination)
|
When $x=0$, $a+b \geqslant-1$. Taking $a=-\frac{4}{5}, b=-\frac{1}{5}$, then
$$
\begin{array}{l}
a \cos x+b \cos 2 x \\
=-\frac{2}{5}(\cos x+1)^{2}+\frac{3}{5} \\
\geqslant-\frac{2}{5}(1+1)^{2}+\frac{3}{5} \\
=-1 .
\end{array}
$$
Therefore, the minimum value of $a+b$ is -1.
|
-1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. If $a, b, c$ are positive numbers, and satisfy
$$
\begin{array}{c}
a+b+c=9, \\
\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{10}{9}, \\
\text { then } \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=
\end{array}
$$
|
B. 7 .
From the given information, we have
$$
\begin{array}{l}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \\
=\frac{9-b-c}{b+c}+\frac{9-c-a}{c+a}+\frac{9-a-b}{a+b} \\
=\frac{9}{b+c}+\frac{9}{c+a}+\frac{9}{a+b}-3 \\
=9 \times \frac{10}{9}-3=7 .
\end{array}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. A. As shown in Figure 4, the side length of square $ABCD$ is $2 \sqrt{15}, E, F$ are the midpoints of sides $AB, BC$ respectively, $AF$ intersects $DE, DB$ at points $M, N$. Then the area of $\triangle DMN$ is $\qquad$.
|
7. A. 8 .
Connect $D F$. Let the side length of the square $A B C D$ be $2 a$. From the problem, we easily know
$$
\begin{array}{l}
\triangle B F N \backsim \triangle D A N \\
\Rightarrow \frac{A D}{B F}=\frac{A N}{N F}=\frac{D N}{B N}=\frac{2}{1} \\
\Rightarrow A N=2 N F \Rightarrow A N=\frac{2}{3} A F .
\end{array}
$$
In the right triangle $\triangle A B F$, given $A B=2 a, B F=a$, we have
$$
A F=\sqrt{A B^{2}+B F^{2}}=\sqrt{5} a \text {. }
$$
Thus, $\cos \angle B A F=\frac{A B}{A F}=\frac{2 \sqrt{5}}{5}$.
From the problem, we know
$$
\begin{array}{c}
\triangle A D E \cong \triangle B A F \Rightarrow \angle A E D=\angle A F B \\
\Rightarrow \angle A M E=180^{\circ}-\angle B A F-\angle A E D \\
=180^{\circ}-\angle B A F-\angle A F B=90^{\circ} .
\end{array}
$$
Therefore, $A M=A E \cos \angle B A F=\frac{2 \sqrt{5}}{5} a$,
$$
M N=A N-A M=\frac{2}{3} A F-A M=\frac{4 \sqrt{5}}{15} a \text {. }
$$
Thus, $\frac{S_{\triangle M N D}}{S_{\triangle A F D}}=\frac{M N}{A F}=\frac{4}{15}$.
Since $S_{\triangle A F D}=\frac{1}{2}(2 a)(2 a)=2 a^{2}$, then
$$
S_{\triangle M N D}=\frac{4}{15} S_{\triangle A F D}=\frac{8}{15} a^{2} .
$$
Given $a=\sqrt{15}$, we have $S_{\triangle M N D}=8$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. A. 2 eighth-grade students and $m$ ninth-grade students participate in a single round-robin chess tournament, where each participant plays against every other participant exactly once. The scoring rule is: the winner of each match gets 3 points, the loser gets 0 points, and in the case of a draw, both players get 1 point. After the tournament, the total score of all students is 130 points, and the number of draws does not exceed half of the total number of matches. Then the value of $m$ is $\qquad$ .
|
9. A. 8 .
Let the number of draws be $a$, and the number of wins (losses) be $b$.
From the problem, we know
$$
2 a+3 b=130 \text {. }
$$
This gives $0 \leqslant b \leqslant 43$.
Also, $a+b=\frac{(m+1)(m+2)}{2}$
$$
\Rightarrow 2 a+2 b=(m+1)(m+2) \text {. }
$$
Thus, $0 \leqslant b=130-(m+1)(m+2) \leqslant 43$.
Therefore, $87 \leqslant(m+1)(m+2) \leqslant 130$.
From this, we get $m=8$ or 9 .
When $m=8$, $b=40, a=5$;
When $m=9$, $b=20, a=35, a>\frac{55}{2}$, which does not meet the problem's conditions.
Hence, $m=8$.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. Given that $n$ is even, and $1 \leqslant n \leqslant 100$. If there is a unique pair of positive integers $(a, b)$ such that $a^{2}=b^{2}+n$ holds, then the number of such $n$ is
|
B. 12.
From the given, we have $(a-b)(a+b)=n$, and $n$ is even, so $a-b$ and $a+b$ are both even.
Therefore, $n$ is a multiple of 4.
Let $n=4m$. Then $1 \leqslant m \leqslant 25$.
(1) If $m=1$, we get $b=0$, which contradicts that $b$ is a positive integer.
(2) If $m$ has at least two different prime factors, then there are at least two pairs of positive integers $(a, b)$ that satisfy
$$
\frac{a-b}{2} \cdot \frac{a+b}{2}=m.
$$
If $m$ is exactly a power of a prime, and this power is not less than 3, then there are at least two pairs of positive integers $(a, b)$ that satisfy equation (1).
(3) If $m$ is a prime number, or $m$ is exactly a power of a prime, and this power is 2, then there is a unique pair of positive integers $(a, b)$ that satisfies equation (1).
Since there is a unique pair of positive integers $(a, b)$, the possible values of $m$ are
$$
2,3,4,5,7,9,11,13,17,19,23,25,
$$
a total of 12.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. In a convex $n$-sided polygon, what is the maximum number of interior angles that can be $150^{\circ}$? Explain your reasoning.
|
B. Suppose in a convex $n$-sided polygon, there are $k$ interior angles equal to $150^{\circ}$. Then there are $n-k$ interior angles not equal to $150^{\circ}$.
(1) If $k=n$, then
$$
n \times 150^{\circ}=(n-2) \times 180^{\circ},
$$
we get $n=12$. Thus, in a regular dodecagon, all 12 interior angles are $150^{\circ}$.
(2) If $k(n-2) \times 180^{\circ}$,
$$
we get $k<12$, i.e., $k \leqslant 11$.
When $k=11$, there exists a convex $n$-sided polygon with 11 interior angles of $150^{\circ}$, and the remaining $n-k$ angles are
$$
\begin{aligned}
\alpha & =\frac{(n-2) \times 180^{\circ}-11 \times 150^{\circ}}{n-11} \\
& =\left(6-\frac{1}{n-11}\right) \times 30^{\circ} \in\left[165^{\circ}, 180^{\circ}\right) .
\end{aligned}
$$
(3) If $k<n$, and $8 \leqslant n \leqslant 11$, when $k=n-1$, there exists a convex $n$-sided polygon with $n-1$ interior angles of $150^{\circ}$, and the other angle is
$$
\begin{array}{l}
\alpha=(n-2) \times 180^{\circ}-(n-1) \times 150^{\circ} \\
=(n-7) \times 30^{\circ} \in\left[30^{\circ}, 120^{\circ}\right] .
\end{array}
$$
(4) If $k<n$, and $3 \leqslant n \leqslant 7$, from (3) we know $k \leqslant n-2$, when $k=n-2$, there exists a convex $n$-sided polygon with $n-2$ interior angles of $150^{\circ}$, and the other two angles are both $(n-2) \times 15^{\circ}$.
In summary, when $n=12$, the maximum value of $k$ is 12;
when $n \geqslant 13$, the maximum value of $k$ is 11;
when $8 \leqslant n \leqslant 11$, the maximum value of $k$ is $n-1$;
when $3 \leqslant n \leqslant 7$, the maximum value of $k$ is $n-2$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Equation
$$
|| \cdots|||x|-1|-2| \cdots|-2011|=2011
$$
has $\qquad$ solutions.
|
4. 4.
The solutions to the equation $||x|-1|=1$ are $x=0$ or $\pm 2$;
The solutions to the equation $|||x|-1|-2|=2$ are $x= \pm 1$ or $\pm 5$;
The solutions to the equation $||||x|-1|-2|-3|=3$ are $x= \pm 3$ or $\pm 9$;
In general, the solutions to the equation
$$
|1 \cdots||| x|-1|-2|\cdots|-n \mid=n(n \geqslant 2)
$$
are
$$
x= \pm \frac{n(n-1)}{2} \text { or } \pm \frac{n(n+3)}{2} \text {. }
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the parabola
$$
y=x^{2}+(k+1) x+1
$$
intersects the $x$-axis at two points $A$ and $B$, not both on the left side of the origin. The vertex of the parabola is $C$. To make $\triangle A B C$ an equilateral triangle, the value of $k$ is $\qquad$
|
$=1 .-5$.
From the problem, we know that points $A$ and $B$ are to the right of the origin, and
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\sqrt{(k+1)^{2}-4} \text {. } \\
\text { Then } \frac{\sqrt{3}}{2} \sqrt{(k+1)^{2}-4}=\left|1-\left(\frac{k+1}{2}\right)^{2}\right| \text {. }
\end{array}
$$
Solving this, we get $k=-5$.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $A M$ is the median of $\triangle A B C$ on side $B C$, $P$ is the centroid of $\triangle A B C$, and a line $E F$ through point $P$ intersects sides $A B$ and $A C$ at points $E$ and $F$ respectively. Then $\frac{B E}{A E}+\frac{C F}{A F}=$ $\qquad$
|
3. 1 .
Draw $B G$ and $C K$ parallel to $A M$ through points $B$ and $C$ respectively, intersecting line $E F$ at points $G$ and $K$. Then
$$
\frac{B E}{A E}=\frac{B G}{A P}, \frac{C F}{A F}=\frac{C K}{A P} \text {. }
$$
Adding the two equations gives
$$
\frac{B E}{A E}+\frac{C F}{A F}=\frac{B G+C K}{A P} .
$$
In trapezoid BCKG, we have
$$
P M=\frac{1}{2}(B G+C K) \text {. }
$$
Since $P$ is the centroid, $A P=2 P M$.
$$
\text { Therefore, } \frac{B E}{A E}+\frac{C F}{A F}=\frac{2 P M}{2 P M}=1 \text {. }
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the circumradius of $\triangle A B C$ is $1, \angle A$, $\angle B$, and $\angle C$'s angle bisectors intersect the circumcircle of $\triangle A B C$ at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Then
$$
\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\sin C}=
$$
$\qquad$
|
4. 2 .
Connect $B A_{1}$. By the Law of Sines, we have
$$
\begin{array}{l}
A A_{1} \cos \frac{A}{2}=2 \sin \left(B+\frac{A}{2}\right) \cdot \cos \frac{A}{2} \\
=\sin C+\sin B .
\end{array}
$$
Similarly, $B B_{1} \cos \frac{B}{2}=\sin C+\sin A$,
$$
C C_{1} \cos \frac{C}{2}=\sin A+\sin B \text {. }
$$
Therefore, the original expression $=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $f(x)=a \sin [(x+1) \pi]+b \sqrt[3]{x-1}+2$, where $a$ and $b$ are real constants. If $f(\lg 5)=5$, then $f(\lg 20)=$
|
5. -1 .
Obviously,
$$
f(x)=a \sin [(x-1) \pi]+b \sqrt[3]{x-1}+2 .
$$
Let $t=x-1$. Then
$$
f(t+1)=a \sin \pi t+b \sqrt[3]{t}+2=g(t)+2,
$$
where, $g(t)=a \sin \pi t+b \sqrt[3]{t}$ is an odd function.
According to the problem,
$$
\begin{aligned}
5 & =f(\lg 5)=f(1-\lg 2) \\
& =g(-\lg 2)+2=-g(\lg 2)+2 .
\end{aligned}
$$
Thus, $g(\lg 2)=-3$.
$$
\text { Therefore, } f(\lg 20)=f(1+\lg 2)=g(\lg 2)+2=-1 \text {. }
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the Cartesian coordinate system, $O$ is the origin, points $A(3, a) 、 B(3, b)$ make $\angle A O B=45^{\circ}$, where $a 、 b$ are integers, and $a>b$. Then the number of pairs $(a, b)$ that satisfy the condition is.
|
6.6.
Let $\angle A O X=\alpha, \angle B O X=\beta$. Then $\tan \alpha=\frac{a}{3}, \tan \beta=\frac{b}{3}$.
Given $a>b$, we have
$$
\begin{array}{l}
1=\tan 45^{\circ}=\tan (\alpha-\beta) \\
=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta}=\frac{3(a-b)}{9+a b} .
\end{array}
$$
Rearranging gives $(a+3)(b-3)=-18$.
Since $a>b$, and both are integers, we have
$$
\begin{array}{l}
(a+3, b-3) \\
=(18,-1),(9,-2),(6,-3), \\
\quad(3,-6),(2,-9),(1,-18) .
\end{array}
$$
Therefore, there are 6 pairs of integers $(a, b)$ that satisfy the condition.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$. Then the integer part of $\sum_{n=0}^{2011} \frac{1}{a_{n}+1}$ is
|
6.3.
From the problem, we have
$$
\begin{array}{l}
\frac{1}{a_{n+1}}=\frac{1}{a_{n}\left(a_{n}+1\right)}=\frac{1}{a_{n}}-\frac{1}{a_{n}+1} \\
\Rightarrow \frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \\
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\sum_{n=0}^{2011}\left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right) \\
\quad=\frac{1}{a_{0}}-\frac{1}{a_{2011}}=4-\frac{1}{a_{2011}} .
\end{array}
$$
Since $a_{2011}>1$, the integer part of $\sum_{n=0}^{2011} \frac{1}{a_{n}+1}$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. $[x]$ represents the greatest integer not exceeding the real number $x$. The area of the figure formed by points satisfying $[x]^{2}+[y]^{2}=50$ on the plane is $\qquad$ .
|
8. 12 .
First, consider the case in the first quadrant.
When $x>0, y>0$, from $[x]^{2}+[y]^{2}=50$, we get
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array} { l }
{ 7 \leqslant x < 8 , } \\
{ 1 \leqslant y < 2 }
\end{array} \left\{\begin{array} { l }
{ 5 \leqslant x < 6 , } \\
{ 5 \leqslant y < 6 ; }
\end{array} \left\{\begin{array}{l}
1 \leqslant x<2, \\
7 \leqslant y<8 .
\end{array}\right.\right.\right. \\
\end{array}
$$
Thus, the area of the figure formed in the first quadrant is 3.
Therefore, the area of the figure given in the problem is 12.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $\sqrt{2 \sqrt{3}-3}=\sqrt{\sqrt{3} x}-\sqrt{\sqrt{3} y}(x, y$ are rational numbers). Then $x-y=$ $\qquad$ .
|
3. 1 .
From the given, $\sqrt{2-\sqrt{3}}=\sqrt{x}-\sqrt{y}$.
Then $\sqrt{x}-\sqrt{y}=\sqrt{\frac{4-2 \sqrt{3}}{2}}=\frac{\sqrt{(\sqrt{3}-1)^{2}}}{\sqrt{2}}$ $=\frac{\sqrt{3}-1}{\sqrt{2}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$.
Thus, $x+y-2 \sqrt{x y}=\frac{3}{2}+\frac{1}{2}-2 \sqrt{\frac{3}{2} \times \frac{1}{2}}$.
Since $x, y$ are rational numbers, we have $x y=\frac{3}{2} \times \frac{1}{2}, x+y=\frac{3}{2}+\frac{1}{2}$.
Also, $x>y$, so $x=\frac{3}{2}, y=\frac{1}{2} \Rightarrow x-y=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { II. (25 points) (1) Given the parabola } \\
y=x^{2}-2 m x+4 m-8
\end{array}
$$
with vertex $A$, construct an inscribed equilateral $\triangle A M N$ (points $M, N$ are on the parabola). Find the area of $\triangle A M N$;
(2) If the parabola $y=x^{2}-2 m x+4 m-8$ intersects the $x$-axis at points with integer coordinates, find the integer value of $m$.
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|
(1) From the symmetry of the parabola and the equilateral triangle, we know that $M N \perp y$-axis.
As shown in Figure 8, let the axis of symmetry of the parabola intersect $M N$ at point $B$. Then
$$
A B=\sqrt{3} B M .
$$
Let $M(a, b)$ $(m<a)$. Then
$$
\begin{aligned}
& B M=a-m . \\
& \text { Also, } A B=y_{B}-y_{A} \\
= & \left(a^{2}-m^{2}\right)-2 m(a-m) \\
= & (a-m)^{2},
\end{aligned}
$$
Thus, $B M^{2}=A B=\sqrt{3} B M$, which means $B M=\sqrt{3}$.
Therefore, $S_{\triangle A M N}=\sqrt{3} B M^{2}=3 \sqrt{3}$.
(2) According to the problem, $\Delta=4\left[(m-2)^{2}+4\right]$ is a perfect square.
Let $(m-2)^{2}+4=n^{2}$, then
$$
(n+m-2)(n-m+2)=4 \text {. }
$$
Since $n+m-2$ and $n-m+2$ have the same parity, we have
$$
\left\{\begin{array} { l }
{ n + m - 2 = 2 , } \\
{ n - m + 2 = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
n+m-2=-2, \\
n-m+2=-2 .
\end{array}\right.\right.
$$
Solving these, we get $\left\{\begin{array}{l}m=2, \\ n=2\end{array}\right.$ or $\left\{\begin{array}{l}m=2, \\ n=-2 .\end{array}\right.$
In summary, $m=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given vectors
$$
a=(x-y+1, x-y), b=\left(x-y+1,10^{x}\right) \text {. }
$$
Then the number of all integer pairs $(x, y)$ that satisfy $a \cdot b=2012$ is $\qquad$
|
5.0.
From the problem, we have
$$
\begin{array}{l}
(x-y+1)^{2}+10^{x}(x-y)=2012 \\
\Rightarrow(x-y)\left(x-y+2+10^{x}\right)=2011 .
\end{array}
$$
Obviously, $x \neq 0$, otherwise, $y(y-3)=2011$, this equation has no integer solutions.
If $x>0$, then $10^{x}$ is a positive integer.
Thus, $x-y+2+10^{x}>x-y$.
We get the following two systems of equations:
$$
\begin{array}{l}
\left\{\begin{array}{l}
x-y+2+10^{x}=2011, \\
x-y=1 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x-y+2+10^{x}=-1, \\
x-y=-2011 .
\end{array}\right.
\end{array}
$$
Calculations show that neither system has integer solutions.
In conclusion, the number of integer pairs $(x, y)$ is 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\qquad$ 1. A six-digit number $\overline{a b c d e f}$, when multiplied by 4, becomes $\overline{f a b c d e}$. The number of six-digit numbers that satisfy this condition is $\qquad$.
|
$=1.6$
Let $\overline{a b c d e}=x$. Then
$$
\begin{array}{l}
4(10 x+f)=100000 f+x \\
\Rightarrow x=2564 f .
\end{array}
$$
Since $f$ is a single digit and $x$ is a five-digit number, it is easy to see that,
$$
f=4,5,6,7,8,9 \text {. }
$$
Therefore, the six-digit numbers $\overline{a b c d e f}$ that satisfy the condition are
$$
\begin{array}{l}
102564,128205,153846, \\
179487,205128,230769,
\end{array}
$$
a total of 6.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Given that $D$ is a point inside $\triangle A B C$, $E$ is the midpoint of side $A C$, $A B=6, B C=10, \angle B A D=$ $\angle B C D, \angle E D C=\angle A B D$. Find the length of $D E$.
---
The above text has been translated into English, preserving the original text's line breaks and format.
|
II. As shown in Figure 4, extend \( CD \) to point \( F \) such that \( DF = CD \), and connect \( AF \) and \( BF \).
Then \( AF \parallel DE \), and \( DE = \frac{1}{2} AF \).
Thus, \( \angle AFD = \angle EDC = \angle ABD \)
\(\Rightarrow A, F, B, D\) are concyclic
\(\Rightarrow \angle BFD = \angle BAD = \angle BCD \)
\(\Rightarrow BF = BC = 10 \).
Since \( D \) is the midpoint of \( FC \), it follows that \( BD \perp FC \).
Therefore, \( \angle FAB = \angle FDB = 90^\circ \).
In the right triangle \( \triangle FAB \),
\( FA = \sqrt{BF^2 - AB^2} = \sqrt{10^2 - 6^2} = 8 \).
Hence, \( DE = \frac{1}{2} FA = 4 \).
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Place 27 balls numbered $1 \sim 27$ into three bowls, Jia, Yi, and Bing, such that the average values of the ball numbers in bowls Jia, Yi, and Bing are $15$, $3$, and $18$, respectively, and each bowl must contain no fewer than 4 balls. Then the maximum value of the smallest ball number in bowl Jia is $\qquad$
|
4. 10 .
Let there be $a$, $b$, and $c$ balls in bowls 甲, 乙, and 丙, respectively. Then,
$$
\begin{aligned}
a+b+c & =27, \\
15 a+3 b+18 c & =\frac{27 \times 28}{2} .
\end{aligned}
$$
From equation (2), we get
$$
5 a+b+6 c=126 \text {. }
$$
If $b \geqslant 6$, then the average value of the ball numbers in bowl 乙 is not less than
$$
\frac{1+2+\cdots+b}{b}=\frac{b+1}{2}>3,
$$
which is a contradiction.
Therefore, $b \leqslant 5 \Rightarrow b=4$ or 5.
$6 \times$ (1) - (3) gives $a+5 b=36$.
Solving this, we get $(a, b)=(11,5),(16,4)$.
We discuss two cases.
(1) $(a, b, c)=(11,5,11)$.
In this case, bowl 乙 must contain balls numbered 1 to 5. When the ball numbers in bowl 甲 are $10 \sim 20$, the minimum ball number reaches its maximum value of 10.
(2) $(a, b, c)=(16,4,7)$.
In this case, bowl 乙 must contain balls numbered $(1,2,4,5)$ or $(1,2,3,6)$. When the ball numbers in bowl 甲 are $7 \sim 14,16 \sim 23$, the minimum ball number reaches its maximum value of 7.
In summary, the maximum value of the minimum ball number in bowl 甲 is 10.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and for all non-negative integers $m, n (m \geqslant n)$, we have
$$
a_{m+n}+a_{m-n}+m-n-1=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {. }
$$
Then the remainder when $a_{2012}$ is divided by 2012 is ـ. $\qquad$
|
5.1.
In equation (1), let $n=0$, we get
$$
a_{2 m}=4 a_{m}+2 m-2-a_{0} \text {. }
$$
In equation (2), let $m=0,1$, we get
$$
a_{0}=1, a_{2}=3 \text {. }
$$
In equation (1), let $n=1$, we get
$$
\begin{array}{l}
a_{m+1}+a_{m-1}+m-2=\frac{1}{2}\left(a_{2 m}+a_{2}\right) \\
=\frac{1}{2}\left(4 a_{m}+2 m-3+3\right)=2 a_{m}+m \\
\Rightarrow a_{m+1}-a_{m}=a_{m}-a_{m-1}+2 \\
\Rightarrow a_{m}-a_{m-1}=a_{1}-a_{0}+2(m-1)=2(m-1) \\
\Rightarrow a_{m}=a_{0}+\sum_{i=1}^{m}\left(a_{i}-a_{i-1}\right)=1+m(m-1) \\
\Rightarrow a_{m} \equiv 1(\bmod m) \\
\Rightarrow a_{2012} \equiv 1(\bmod 2012) .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given a cyclic quadrilateral $A_{1} A_{2} A_{3} A_{4}$ with an inscribed circle $\odot I$ that is tangent to the sides $A_{1} A_{2}$, $A_{2} A_{3}$, $A_{3} A_{4}$, and $A_{4} A_{1}$ at points $B_{1}$, $B_{2}$, $B_{3}$, and $B_{4}$ respectively, then
$$
\left(\frac{A_{1} A_{2}}{B_{1} B_{2}}\right)^{2}+\left(\frac{A_{2} A_{3}}{B_{2} B_{3}}\right)^{2}+\left(\frac{A_{3} A_{4}}{B_{3} B_{4}}\right)^{2}+\left(\frac{A_{4} A_{1}}{B_{4} B_{1}}\right)^{2}
$$
the minimum value is $\qquad$
|
8. 8 .
As shown in Figure 5, let the radius of $\odot I$ be $r$,
$$
\begin{array}{l}
A_{i} B_{i}=A_{i} B_{i-1} \\
=a_{i},
\end{array}
$$
where $i=1,2,3,4, B_{0}=B_{4}$.
Then $A_{i} A_{i+1}=a_{i}+a_{i+1}$,
$$
B_{i} B_{i+1}=\frac{2 A_{i+1} B_{i} \cdot B_{i} I}{\sqrt{A_{i+1} B_{i}^{2}+B_{i} I^{2}}}=2\left(a_{i+1}^{-2}+r^{-2}\right)^{-\frac{1}{2}} \text {. }
$$
From the right triangle $\triangle A_{i} B_{i} I \subset$ right triangle $\triangle I B_{i+2} A_{i+2}$
$$
\Rightarrow a_{i} a_{i+2}=r^{2}(i=1,2) \text {. }
$$
$$
\begin{array}{l}
\text { Then }\left(\frac{A_{1} A_{2}}{B_{1} B_{2}}\right)^{2}+\left(\frac{A_{2} A_{3}}{B_{2} B_{3}}\right)^{2}+\left(\frac{A_{3} A_{4}}{B_{3} B_{4}}\right)^{2}+\left(\frac{A_{4} A_{1}}{B_{4} B_{1}}\right)^{2} \\
=\frac{1}{4} \sum_{i=1}^{4}\left(a_{i}+a_{i+1}\right)^{2}\left(a_{i+1}^{-2}+r^{-2}\right) . \\
\text { Also } \sum_{i=1}^{4} \frac{\left(a_{i}+a_{i+1}\right)^{2}}{4 a_{i+1}^{2}} \\
=\frac{1}{4} \sum_{i=1}^{4}\left(1+\frac{2 a_{i}}{a_{i+1}}+\frac{a_{i}^{2}}{a_{i+1}^{2}}\right) \geqslant 4, \\
\sum_{i=1}^{4} \frac{\left(a_{i}+a_{i+1}\right)^{2}}{4 r^{2}} \geqslant \sum_{i=1}^{4} \frac{a_{i} a_{i+1}}{r^{2}} \\
\geqslant \frac{4}{r^{2}} \sqrt{\prod_{i=1}^{4} a_{i} a_{i+1}}=4, \\
\text { Therefore }\left(\frac{A_{1} A_{2}}{B_{1} B_{2}}\right)^{2}+\left(\frac{A_{2} A_{3}}{B_{2} B_{3}}\right)^{2}+\left(\frac{A_{3} A_{4}}{B_{3} B_{4}}\right)^{2}+\left(\frac{A_{4} A_{1}}{B_{4} B_{1}}\right)^{2} \geqslant 8 .
\end{array}
$$
Equality holds if and only if $a_{1}=a_{2}=a_{3}=a_{4}$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Write down all positive integers from 1 to 10000 from left to right, then remove those numbers that are divisible by 5 or 7, and concatenate the remaining numbers to form a new number. Try to find:
(1) The number of digits in the new number;
(2) The remainder when the new number is divided by 11.
|
Three, (1) Obviously, the remainder of the remaining numbers when divided by 35 is
$$
\begin{array}{l}
1,2,3,4,6,8,9,11,12,13,16,17,18, \\
19,22,23,24,26,27,29,31,32,33,34,
\end{array}
$$
denoted as $a_{i}(i=1,2, \cdots, 24)$ in sequence.
Among the numbers from $1 \sim 9$, 7 numbers remain, denoted as $b_{i}$ $(i=1,2, \cdots, 7)$ in sequence.
Among the numbers from $10 \sim 99$, since $99=35 \times 2+29$, there are $24 \times 2+20-7=61$ remaining numbers, denoted as $b_{i}(i=$ $8,9, \cdots, 68)$ in sequence.
Among the numbers from $100 \sim 999$, since $999=35 \times 28+19$, there are $24 \times 28+14-68=618$ remaining numbers, denoted as $b_{i}(i=69,70, \cdots, 686)$ in sequence.
Among the numbers from $1000 \sim 9999$, since
$$
9999=35 \times 285+24 \text {, }
$$
there are $24 \times 285+17-686=6171$ remaining numbers, denoted as $b_{i}(i=687,688, \cdots, 6857)$ in sequence.
In summary, the number of digits in the new number is
$$
7+61 \times 2+618 \times 3+6171 \times 4=26667 \text {. }
$$
(2) First, there are two conclusions:
(i) If $c_{i}(i=1,2, \cdots, k)$ are all $n$-digit positive integers, then
$$
\overline{c_{1} c_{2} \cdots c_{k}} \equiv \sum_{i=1}^{k} c_{i}(-1)^{(k-i) n}(\bmod 11) .
$$
(ii) The sum of any 11 consecutive terms in an arithmetic sequence composed of positive integers is divisible by 11.
Obviously, $b_{24 k+j}=35 k+a_{j}(j=1,2, \cdots, 24)$,
$$
\begin{array}{l}
\sum_{j=1}^{24} b_{24 k+j}=35 k \times 24+\sum_{j=1}^{24} a_{j} \\
\equiv 4 k+2(\bmod 11), \\
\sum_{j=1}^{24}(-1)^{j} b_{24 k+j}=\sum_{j=1}^{24}(-1)^{j} a_{j} \\
\equiv 5(\bmod 11) .
\end{array}
$$
Let the new number be $A=\overline{A_{1} A_{2} A_{3} A_{4}}$, where,
$$
\begin{array}{l}
A_{1}=\overline{1234689} \equiv 1-2+3-4+6-8+9 \\
\equiv 5(\bmod 11) \text {, } \\
A_{2}=\overline{b_{8} b_{9} \cdots b_{68}} \equiv \sum_{i=8}^{68} b_{i} \\
\equiv \sum_{k=0}^{2} \sum_{j=1}^{24} b_{24 k+j}-\sum_{j=1}^{7} b_{j}-\sum_{j=21}^{24} b_{48+j} \\
\equiv \sum_{k=0}^{2}(4 k+2)-0-\sum_{j=21}^{24}\left(70+a_{j}\right) \\
\equiv 18-4 \times 70-\sum_{j=11}^{24} a_{j} \\
\equiv 2-31 \equiv 4(\bmod 11) \text {, } \\
A_{3}=\overline{b_{69} b_{70} \cdots b_{686}} \equiv \sum_{i=69}^{686}(-1)^{i} b_{i} \\
\equiv \sum_{j=21}^{24}(-1)^{j} b_{48+j}+\sum_{k=3}^{27} \sum_{j=1}^{24}(-1)^{j} b_{2 k+j}+ \\
\sum_{j=1}^{14}(-1)^{j} b_{24 \times 28+j} \\
\equiv \sum_{j=21}^{24}(-1)^{j} a_{j}+5 \times 25+\sum_{j=1}^{14}(-1)^{j} a_{j} \\
\equiv 5 \times 26-\sum_{j=15}^{20}(-1)^{j} a_{j} \\
\equiv 5 \times 26-5 \equiv 4(\bmod 11) \text {, } \\
A_{4}=\overline{b_{687} b_{688} \cdots b_{6857}} \equiv \sum_{i=687}^{6857} b_{i} \equiv \sum_{i=687}^{785} b_{i} \\
\equiv \sum_{j=15}^{24} b_{24 \times 28+j}+\sum_{k=29}^{31} \sum_{j=1}^{24} b_{24 k+j}+\sum_{j=1}^{17} b_{24 \times 32+j} \\
=\sum_{j=15}^{24}\left(35 \times 28+a_{j}\right)+\sum_{k=2}^{31}(4 k+2)+\sum_{j=1}^{17}\left(35 \times 32+a_{j}\right) \\
=1+\sum_{j=1}^{24} a_{j}+\sum_{j=15}^{17} a_{j} \\
\equiv 1+2+3 \equiv 6(\bmod 11) \text {. } \\
\end{array}
$$
Therefore, $A \equiv \sum_{i=1}^{4} A_{i} \equiv 8(\bmod 11)$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Find all real roots of the equation
$$
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right)
$$
|
Solve:
$$
\begin{array}{l}
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right) \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}=x^{2}+2 x+4 \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}-3=x^{2}+2 x+1 \\
\Rightarrow \frac{-2\left(x^{3}+1\right)^{2}}{x^{6}+x^{3}+1}=(x+1)^{2} \\
\Rightarrow(x+1)^{2}\left[1+\frac{2\left(x^{2}-x+1\right)^{2}}{x^{6}+x^{3}+1}\right]=0 .
\end{array}
$$
Notice that, $x^{6}+x^{3}+1=\left(x^{3}+\frac{1}{2}\right)^{2}+\frac{3}{4}>0$.
Therefore, the original equation has a unique real root -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $\frac{1}{a}+\frac{1}{b}=\frac{5}{a+b}$, then $\frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}=$
|
2. 7 .
$$
\begin{array}{l}
\text { Given } \frac{1}{a}+\frac{1}{b}=\frac{5}{a+b} \Rightarrow \frac{b}{a}+\frac{a}{b}=3 \\
\Rightarrow \frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}=\left(\frac{b}{a}+\frac{a}{b}\right)^{2}-2=7 .
\end{array}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $x-1$ is a factor of $x^{3}+a x^{2}+1$, then the value of $a$ is $\qquad$ .
|
3. -2 .
Let $x^{3}+a x^{2}+1=(x-1)\left(x^{2}-m x-1\right)$, that is
$$
\begin{array}{l}
x^{3}+a x^{2}+1 \\
=x^{3}-(m+1) x^{2}+(m-1) x+1 .
\end{array}
$$
By comparing, we get $m-1=0, a=-(m+1)$.
Thus, $m=1, a=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (25 points) Let the two intersection points of the functions $y=2x$ and $y=\frac{4}{x}$ be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(x_{1}>x_{2}\right)$, and point $C(\sqrt{2},-2 \sqrt{2})$. Find the area of $\triangle ABC$.
|
From $\left\{\begin{array}{l}y=2 x, \\ y=\frac{4}{x},\end{array}\right.$ eliminating $y$ we get
$$
2 x=\frac{4}{x} \Rightarrow x^{2}=2 \Rightarrow x= \pm \sqrt{2} \text {. }
$$
Therefore, $A(\sqrt{2}, 2 \sqrt{2}), B(-\sqrt{2},-2 \sqrt{2})$.
Since point $C(\sqrt{2},-2 \sqrt{2})$, $\triangle A B C$ is a right triangle, and $\angle A C B=90^{\circ}$.
Obviously, $A C=4 \sqrt{2}, B C=2 \sqrt{2}$.
Thus, $S_{\triangle A C B}=\frac{1}{2} A C \cdot B C=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $a_{1}=1, a_{2}=3$,
$$
a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n} \text {, }
$$
when $m \geqslant n$, $a_{m}$ is divisible by 9. Then the minimum value of $n$ is $\qquad$ .
|
5.5.
Notice,
$$
\begin{array}{l}
a_{n+2}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right) \\
=\cdots=(n+2)(n+1) n \cdots \cdots \cdot 3\left(a_{2}-a_{1}\right) \\
=(n+2)!.
\end{array}
$$
Thus, $a_{n}=a_{1}+\sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)=\sum_{i=1}^{n} i!$.
Given $a_{1}=1, a_{2}=3$, we have
$$
a_{3}=9, a_{4}=33, a_{5}=153 \text {, }
$$
At this point, 153 is divisible by 9.
When $m>5$, $a_{m}=a_{5}+\sum_{k=6}^{m} k!$, and for $k \geqslant 6$, $k!$ is divisible by 9, so when $m \geqslant 5$, $a_{m}$ is divisible by 9.
Therefore, the smallest value of $n$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (12 points) Given
$$
\begin{array}{l}
f(x, y) \\
=x^{3}+y^{3}+x^{2} y+x y^{2}-3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{array}
$$
and $x, y \geqslant \frac{1}{2}$. Find the minimum value of $f(x, y)$.
|
12. When $x \neq y$, multiplying both sides of the function by $x-y$ gives
$$
\begin{array}{l}
(x-y) f(x, y) \\
=\left(x^{4}-y^{4}\right)-3\left(x^{3}-y^{3}\right)+3\left(x^{2}-y^{2}\right) . \\
\text { Let } g(x)=x^{4}-3 x^{3}+3 x^{2} .
\end{array}
$$
Then $f(x, y)=\frac{g(x)-g(y)}{x-y}$ is the slope of the line segment connecting two points on the graph of $g(x)$.
When $x=y$, $f(x, y)=4 x^{3}-9 x^{2}+6 x$.
Thus, we only need to find the minimum value of the derivative of $g(x)$ for $x \geqslant \frac{1}{2}$,
$$
h(x)=4 x^{3}-9 x^{2}+6 x
$$
It is easy to find that when $x \geqslant \frac{1}{2}$,
$$
h(x)=4 x^{3}-9 x^{2}+6 x
$$
the minimum value is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The ten positive integers from $1 \sim 10$ are written in a row in some order, denoted as $a_{1}, a_{2}, \cdots, a_{10}, S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\cdots, S_{10}=a_{1}+a_{2}+\cdots+a_{10}$. Then, among $S_{1}, S_{2}, \cdots, S_{10}$, the maximum number of primes that can occur is .
|
2. 7 .
Adding an odd number changes the sum to the opposite parity, and among even numbers, only 2 is a prime. Let $b_{i}$ be the $i$-th ($i=1,2,3,4,5$) odd number in this row. Then, when adding $b_{2}$ and $b_{4}$, the sums $S_{k}$ and $S_{n}$ are even numbers greater than 2. Therefore, $S_{k} \backslash S_{n}$ and $S_{10}=55$ must be composite numbers, meaning that in $S_{1}, S_{2}, \cdots, S_{10}$, the primes are no greater than 7. The example in Table 1 shows that there can be 7 primes.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$i$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$a_{i}$ & 7 & 6 & 4 & 2 & 9 & 1 & 3 & 5 & 10 & 8 \\
\hline$S_{i}$ & 7 & 13 & 17 & 19 & 28 & 29 & 32 & 37 & 47 & 55 \\
\hline
\end{tabular}
Therefore, in $S_{1}, S_{2}, \cdots, S_{10}$, there can be at most 7 primes.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (10 points) Prove: For any three distinct numbers $a, b, c$, we have
$$
\frac{(a+b-c)^{2}}{(a-c)(b-c)}+\frac{(b+c-a)^{2}}{(b-a)(c-a)}+\frac{(c+a-b)^{2}}{(c-b)(a-b)}
$$
is a constant.
|
$$
\begin{array}{l}
S=\frac{(a+b-c)^{2}}{(a-c)(b-c)}+\frac{(b+c-a)^{2}}{(b-a)(c-a)}+\frac{(c+a-b)^{2}}{(c-b)(a-b)}, \\
p=a^{2}+b^{2}+c^{2}-2 a b-2 b c-2 c a . \\
\text { Then } (a+b-c)^{2}=p+4 a b, \\
(b+c-a)^{2}=p+4 b c, \\
(c+a-b)^{2}=p+4 c a . \\
\text { Hence } S=\frac{p+4 a b}{(a-c)(b-c)}+\frac{p+4 b c}{(b-a)(c-a)}+ \\
\quad \frac{p+4 c a}{(c-b)(a-b)} \\
=p q+4 r,
\end{array}
$$
where,
$$
\begin{aligned}
q & =\frac{1}{(a-c)(b-c)}+\frac{1}{(b-a)(c-a)}+\frac{1}{(c-b)(a-b)} \\
& =\frac{(a-b)+(b-c)-(a-c)}{(a-b)(b-c)(a-c)}=0, \\
r & =\frac{a b}{(a-c)(b-c)}+\frac{b c}{(b-a)(c-a)}+\frac{c a}{(c-b)(a-b)} \\
& =\frac{a b(a-b)+b c(b-c)-a c(a-c)}{(a-b)(b-c)(a-c)} \\
& =\frac{\left(a^{2}-c^{2}\right) b+(c-a) b^{2}-a c(a-c)}{(a-b)(b-c)(a-c)} \\
& =\frac{(a+c) b-b^{2}-a c}{(a-b)(b-c)}=1 .
\end{aligned}
$$
Therefore, $S=p q+4 r=4$.
|
4
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
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