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10. (14 points) As shown in Figure 1, in the triangular prism $A B C-A_{1} B_{1} C_{1}$, it is known that the side edge $A A_{1} \perp$ plane $A B C$, and $\triangle A B C$ is an equilateral triangle with a side length of 2. $M$ is a point on $A A_{1}$, $A A_{1}=4, A_{1} M=1, P$ is a point on the edge $B C$, and the shortest distance from $P$ along the side surface of the prism through the edge $C C_{1}$ to point $M$ is $3 \sqrt{2}$. Let this shortest distance intersect $C C_{1}$ at point $N$.
(1) Prove: $A_{1} B / /$ plane $M N P$;
(2) Find the tangent value of the dihedral angle (acute angle) formed by plane $M N P$ and plane $A B C$.
|
10. (1) From $A A_{1} \perp$ plane $A B C$ and $\triangle A B C$ being an equilateral triangle, we know that all the side faces are congruent rectangles.
As shown in Figure 3, rotate the side face $B C_{1}$ by $120^{\circ}$ so that it lies in the same plane as the side face $A C_{1}$. Point $P$ moves to the position of $P_{1}$, and connect $M P_{1}$. Then $M P_{1}$ is the shortest path from point $P$ along the lateral surface of the prism through $C C_{1}$ to point $M$.
Let $P C=x$. Then $P_{1} C=x$.
In the right triangle $\triangle M A P_{1}$, note that,
$$
\begin{array}{l}
(2+x)^{2}+3^{2}=18 \\
\Rightarrow x=1 .
\end{array}
$$
Thus, $P$ is the midpoint of $B C$.
Therefore, $N C=1$.
Let $A_{1} C$ intersect $M N$ at point $Q$. Then $Q$ is the midpoint of $A_{1} C$.
So, $A_{1} B / / P Q$.
Hence, $A_{1} B / /$ plane $M N P$.
(2) As shown in Figure 3, connect $C H$ and $P P_{1}$. Then $P P_{1}$ is the intersection line of plane $M N P$ and plane $A B C$.
Draw $N H \perp P P_{1}$ at point $H$.
Since $C C_{1} \perp$ plane $A B C$, it follows that $C H \perp P P_{1}$.
Therefore, $\angle N H C$ is the plane angle of the dihedral angle formed by plane $N M P$ and plane $A B C$.
In the right triangle $\triangle P H C$, we have
$$
\angle P C H=\frac{1}{2} \angle P C P_{1}=60^{\circ} \text {, }
$$
Then $C H=\frac{1}{2}$.
In the right triangle $\triangle N C H$,
$$
\tan \angle N H C=\frac{N C}{C H}=2 .
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In $\triangle A B C$, it is known that $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of side $A B$, and $\triangle A C M$ is folded along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Then the distance from point $M$ to plane $A B C$ is $\qquad$
|
3. 1 .
From the problem, we know that $M A=M B=M C=2$.
Therefore, the projection $O$ of $M$ on the plane $A B C$ is the circumcenter of $\triangle A B C$.
Given $A B=2 \sqrt{2}, A C=2, B C=2 \sqrt{3}$, we know that $A B^{2}+A C^{2}=B C^{2}$.
Thus, $O$ is the midpoint of the hypotenuse $B C$ of the right triangle $\triangle A B C$.
Hence, $M O=\sqrt{M C^{2}-O C^{2}}=1$ is the desired result.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 In $\triangle A B C$, it is known that $\angle A: \angle B: \angle C = 4: 2: 1, \angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively.
(1) Prove: $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$;
(2) Find the value of $\frac{(a+b-c)^{2}}{a^{2}+b^{2}+c^{2}}$.
|
(1) Proof As shown in Figure 3, construct the angle bisector of $\angle ABC$, intersecting the circumcircle of $\triangle ABC$ at point $M$, and construct the angle bisector of $\angle BAC$, intersecting the circumcircle of $\triangle ABC$ at point $N$.
Then $AM=MC=AB=c, AM \parallel BC$;
$$
CN=NB=AC=b, AB \parallel CN \text{. }
$$
Thus, quadrilaterals $ABCM$ and $ABNC$ are both isosceles trapezoids, and we have
$$
BM=AC=b, AN=BC=a.
$$
Note that $M$ and $N$ are the midpoints of arcs $\overparen{AC}$ and $\overparen{BC}$ (excluding $B$ or $A$), respectively. By Corollary 2(2), we have
$$
\begin{array}{l}
MB^2 - MA^2 = AB \cdot BC, \\
NA^2 - NB^2 = AB \cdot AC,
\end{array}
$$
which gives $b^2 - c^2 = ac, a^2 - b^2 = bc$.
Adding these two equations, we get
$$
\begin{array}{l}
a^2 - ac - c(b+c) \\
\Rightarrow \frac{b+c}{a} = \frac{a-c}{c} = \frac{a}{c} - 1.
\end{array}
$$
Also, from $b^2 + bc = a^2$, we have
$$
\frac{b+c}{a} = \frac{a}{b}.
$$
From equations (1) and (2), we get
$$
\frac{a}{c} - \frac{a}{b} = 1 \Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{c}.
$$
(2) Solution From equation (3), we have $ab = ac + bc$.
Thus, $\frac{(a+b-c)^2}{a^2 + b^2 + c^2}$
$$
= \frac{a^2 + b^2 + c^2 + 2(ab - bc - ac)}{a^2 + b^2 + c^2} = 1.
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the equation in $x$
$$
(a-1) x^{2}+2 x-a-1=0
$$
has roots that are all integers. Then the number of integer values of $a$ that satisfy this condition is
$\qquad$.
|
4.5.
When $a=1$, $x=1$.
When $a \neq 1$, it is easy to see that $x=1$ is an integer root of the equation.
Furthermore, from $1+x=\frac{2}{1-a}$ and $x$ being an integer, we know
$$
1-a= \pm 1, \pm 2 \text {. }
$$
Therefore, $a=-1,0,2,3$.
In summary, there are 5 integer values of $a$ that satisfy the condition.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let the quadratic function
$$
f(x)=a x^{2}+b x+c
$$
satisfy the following conditions:
(i) When $x$ is a real number, its minimum value is 0, and
$$
f(x-1)=f(-x-1)
$$
holds;
(ii) There exists a real number $m (m>1)$, such that there exists a real number $t$, as long as $1 \leqslant x \leqslant m$, then $f(x+t) \leqslant x$ holds.
|
(1) In condition (ii), let $x=1$, we get $f(1)=1$.
From condition (i), we know that the quadratic function opens upwards and is symmetric about $x=-1$, so we can assume the quadratic function to be
$$
f(x)=a(x+1)^{2}(a>0) .
$$
Substituting $f(1)=1$ into the above equation, we get $a=\frac{1}{4}$.
Therefore, $f(x)=\frac{1}{4}(x+1)^{2}$.
(2) Suppose there exists $t \in \mathbf{R}$, such that for $1 \leqslant x \leqslant m$, we have $f(x+t) \leqslant x$.
Taking $x=1$, we have
$$
(t+2)^{2} \leqslant 4 \Rightarrow-4 \leqslant t \leqslant 0 \text {. }
$$
Taking $x=m$, we have
$$
\begin{array}{l}
(m+t+1)^{2} \leqslant 4 m \\
\Rightarrow m^{2}+2(t-1) m+\left(t^{2}+2 t+1\right) \leqslant 0 \\
\Rightarrow 1-t-\sqrt{-4 t} \leqslant m \leqslant 1-t+\sqrt{-4 t} .
\end{array}
$$
Thus, $m \leqslant 1-t+\sqrt{-4 t}$
$$
\leqslant 1-(-4)+\sqrt{-4 \times(-4)}=9 \text {. }
$$
When $t=-4$, for any $x(1 \leqslant x \leqslant 9)$, we always have
$$
\begin{array}{l}
f(x-4)-x=\frac{1}{4}\left(x^{2}-10 x+9\right) . \\
=\frac{1}{4}(x-1)(x-9) \leqslant 0 .
\end{array}
$$
Therefore, the maximum value of $m$ is 9.
(Sun Yan, Anhui Provincial Anqing City Educational Research Office, 246003, Huang Quanfu, Anhui Provincial Huaining County Jiangzhen Middle School, 246142)
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. From the arithmetic sequence $2,5,8, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$ Ω.
|
7.8.
First, let's take $x_{1}, x_{2}, \cdots, x_{k}$ from the known sequence such that
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1 .
$$
Let $y_{i}=\frac{x_{1} x_{2} \cdots x_{k}}{x_{i}}$. Then
$$
y_{1}+y_{2}+\cdots+y_{k}=x_{1} x_{2} \cdots x_{k} \text {. }
$$
It is easy to see that for any $n$, $x_{n}=2(\bmod 3)$.
Thus, taking modulo 3 on both sides of equation (1) gives
$$
2^{k-1} k \equiv 2^{k}(\bmod 3) \Rightarrow k \equiv 2(\bmod 3) \text {. }
$$
When $k=2$,
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}} \leqslant \frac{1}{2}+\frac{1}{5}<1 \text {; }
$$
When $k=5$,
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{5}} \leqslant \frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}<1 \text {; }
$$
When $k=8$, taking $2, 5, 8, 11, 20, 41, 110, 1640$ satisfies the condition.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Can 2010 be written as the sum of squares of $k$ distinct positive integers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason. ${ }^{[2]}$
(2010, Beijing Middle School Mathematics Competition (Grade 8))
|
Estimate the approximate range of $k$ first, then discuss by classification.
Let $p_{i}$ be a prime number.
If 2010 can be written as the sum of squares of $k$ prime numbers, then by
$$
\begin{array}{l}
2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}+17^{2}+19^{2}+23^{2}+29^{2} \\
=2397>2010,
\end{array}
$$
we know $k \leqslant 9$.
(1) When $k=9$, if
$$
2010=p_{1}^{2}+p_{2}^{2}+\cdots+p_{9}^{2} \text {, }
$$
among them, there must be one square of an even number and eight squares of odd prime numbers.
The left side of the equation is congruent to 2 modulo 8, while the right side is congruent to 4 modulo 8, so the equation cannot hold.
(2) When $k=8$, if
$$
2010=p_{1}^{2}+p_{2}^{2} \cdots+p_{8}^{2} \text {, }
$$
these eight addends are all squares of odd prime numbers.
The left side of the equation is congruent to 2 modulo 8, while the right side is congruent to 0 modulo 8, so the equation cannot hold.
(3) When $k=7$, after trial calculation, we find
$$
2^{2}+3^{2}+7^{2}+11^{2}+13^{2}+17^{2}+37^{2}=2010 \text {. }
$$
In summary, the maximum value of $k$ is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A. Let $a=\sqrt[3]{3}, b$ be the fractional part of $a^{2}$. Then $(b+2)^{3}=$ $\qquad$
|
γγ6. A.9.
From $2<a^{2}<3$, we know $b=a^{2}-2=\sqrt[3]{9}-2$. Therefore, $(b+2)^{3}=(\sqrt[3]{9})^{3}=9$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. In the Cartesian coordinate system $x O y$, it is known that $O$ is the origin, point $A(10,100), B\left(x_{0}, y_{0}\right)$, where $x_{0} γ y_{0}$ are integers, and points $O γ A γ B$ are not collinear. For all points $B$ that satisfy the above conditions, find the minimum area of $\triangle O A B$.
|
B. Draw a line parallel to the $x$-axis through point $B$, intersecting line $O A$ at point $C\left(\frac{y_{0}}{10}, y_{0}\right)$.
Thus, $B C=\left|x_{0}-\frac{y_{0}}{10}\right|$.
Since points $O$, $A$, and $B$ are not collinear, we have
$$
\begin{array}{l}
S_{\triangle O A B}=\frac{1}{2} B C \times 100=50\left|x_{0}-\frac{y_{0}}{10}\right| \\
=5\left|10 x_{0}-y_{0}\right| \neq 0 .
\end{array}
$$
Given that $\left|10 x_{0}-y_{0}\right|$ is an integer, then $\left|10 x_{0}-y_{0}\right| \geqslant 1$.
Therefore, $S_{\triangle O A B}=5\left|10 x_{0}-y_{0}\right| \geqslant 5$.
When $x_{0}=1, y_{0}=9$, $S_{\triangle O A B}=5$.
Thus, the minimum area of $\triangle O A B$ is 5.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. A. If placing the positive integer $M$ to the left of the positive integer $m$ results in a new number that is divisible by 7, then $M$ is called the "magic number" of $m$ (for example, placing 86 to the left of 415 results in the number 86415, which is divisible by 7, so 86 is called the magic number of 415). Find the smallest positive integer $n$ such that there exist distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying that for any positive integer $m$, at least one of $a_{1}, a_{2}, \cdots, a_{n}$ is a magic number of $m$.
|
14. A. If $n \leqslant 6$, take $m=1,2, \cdots, 7$. By the pigeonhole principle, there must be a positive integer $M$ among $a_{1}, a_{2}, \cdots, a_{n}$ that is a common magic number of $i$ and $j (1 \leqslant i<j \leqslant 7)$, i.e.,
$$
7 \mid(10 M+i), 7 \mid(10 M+j) \text {. }
$$
Then $7 \mid (j-i)$. But $0<j-i \leqslant 6$, which is a contradiction. Therefore, $n \geqslant 7$.
Furthermore, when $a_{1}, a_{2}, \cdots, a_{n}$ are $1,2, \cdots, 7$, for any positive integer $m$, let it be a $k$-digit number ($k$ is a positive integer).
Then the remainders of $10^{k} i+m (i=1,2, \cdots, 7)$ when divided by 7 are all different. If not, there exist positive integers $i, j (1 \leqslant i<j \leqslant 7)$, such that
$$
\begin{array}{l}
7 \mid\left[\left(10^{k} j+m\right)-\left(10^{k} i+m\right)\right] \\
\Rightarrow 7 \mid 10^{k}(j-i) \Rightarrow 7 \mid (j-i),
\end{array}
$$
which is a contradiction.
Therefore, there must exist a positive integer $i (1 \leqslant i \leqslant 7)$ such that $7 \mid \left(10^{k} i+m\right)$, i.e., $i$ is a magic number of $m$. Thus, the minimum value of $n$ is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If real numbers $x, y, z$ satisfy the equation
$$
\sqrt{x+9+\sqrt{x-7}}+\frac{|x+y-z|}{4}=4 \text {, }
$$
then the units digit of $(5 x+3 y-3 z)^{2013}$ is $\qquad$
|
3. 4 .
It is known that $x \geqslant 7$, then
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}} \geqslant 4, \\
\frac{|x+y-z|}{4} \geqslant 0 .
\end{array}\right.
$$
Combining the given equations, we have
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}}=4, \\
\frac{|x+y-z|}{4}=0 .
\end{array}\right.
$$
Therefore, $x=7, x+y-z=0$.
Thus, $(5 x+3 y-3 z)^{2013}=14^{2013}$, and its unit digit is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given real numbers $m, n$ satisfy $m-n=\sqrt{10}$, $m^{2}-3 n^{2}$ is a prime number. If the maximum value of $m^{2}-3 n^{2}$ is $a$, and the minimum value is $b$, then $a-b=$ $\qquad$
|
5.11 .
Let $m^{2}-3 n^{2}=p$ (where $p$ is a prime number).
From $m-n=\sqrt{10}$, we get
$$
m=\sqrt{10}+n \text {. }
$$
Substituting equation (2) into equation (1) and simplifying, we get
$$
\begin{array}{l}
2 n^{2}-2 \sqrt{10} n+p-10=0 \\
\Rightarrow \Delta=40-8 p+80 \geqslant 0 \\
\Rightarrow p \leqslant 15 \\
\Rightarrow a=13, b=2 \\
\Rightarrow a-b=11
\end{array}
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the Cartesian coordinate system $x O y$, it is known that there are three points $A(a, 1), B(2, b), C(3,4)$.
If the projections of $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are the same, then $3 a-4 b=$ $\qquad$
|
3. 2 .
Solution 1 The projections of vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are $\frac{\overrightarrow{O A} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}, \frac{\overrightarrow{O B} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}$, respectively.
According to the problem, we have
$$
\overrightarrow{O A} \cdot \overrightarrow{O C}=\overrightarrow{O B} \cdot \overrightarrow{O C} \text {, }
$$
which means $3 a+4=6+4 b$.
Thus, $3 a-4 b=2$.
Solution 2 Since the projections of vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are the same, $A B \perp O C$, that is,
$$
\overrightarrow{A B} \cdot \overrightarrow{O C}=0 \text {. }
$$
Therefore, $3(2-a)+4(b-1)=0$, which simplifies to $3 a-4 b=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, $8 n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$. [4]
(2009, "Mathematics Weekly Cup" National Junior High School Mathematics Competition
|
Let $a_{1}, a_{2}, \cdots, a_{n}$ be such that removing $a_{i} (i=1, 2, \cdots, n)$ leaves the arithmetic mean of the remaining $n-1$ numbers as a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Thus, for any $1 \leqslant i<j \leqslant n$, we have
$$
b_{i}-b_{j}=\frac{a_{j}-a_{i}}{n-1} \text {. }
$$
Therefore, $(n-1) \mid\left(a_{j}-a_{i}\right)$.
Since $b_{1}-b_{n}=\frac{a_{n}-a_{1}}{n-1}=\frac{2008}{n-1}$ is a positive integer, we have
$$
(n-1) \mid\left(2^{3} \times 251\right) \text {. }
$$
Also, $a_{n}-1=\sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)$
$$
\geqslant \sum_{i=1}^{n-1}(n-1)=(n-1)^{2} \text {, }
$$
so $(n-1)^{2} \leqslant 2008$, which implies $n \leqslant 45$.
Combining with equation (1), we get $n \leqslant 9$.
On the other hand, let
$$
a_{i+1}=8 i+1(i=0,1, \cdots, 7), a_{9}=8 \times 251+1 \text {. }
$$
Then these nine numbers satisfy the given conditions.
Hence, the maximum value of $n$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. If for any $x \in\left(-\frac{1}{2}, 1\right)$, we have
$$
\frac{x}{1+x-2 x^{2}}=\sum_{k=0}^{\infty} a_{k} x^{k},
$$
then $a_{3}+a_{4}=$ . $\qquad$
|
9. -2 .
In equation (1), let $x=0$, we get $a_{0}=0$. Then $\frac{1}{1+x-2 x^{2}}=\sum_{k=1}^{\infty} a_{k} x^{k-1}$. Substituting $x=0$ into the above equation, we get $a_{1}=1$. Then $\frac{-1+2 x}{1+x-2 x^{2}}=\sum_{k=2}^{\infty} a_{k} x^{k-2}$. Substituting $x=0$ into the above equation, we get $a_{2}=-1$. Similarly, $a_{3}=3, a_{4}=-5$. Therefore, $a_{3}+a_{4}=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. If $0 \leqslant x_{i} \leqslant 1(i=1,2, \cdots, 5)$, then
$$
M=x_{1}-x_{2}^{3}+x_{2}-x_{3}^{3}+x_{3}-x_{4}^{3}+x_{4}-x_{5}^{3}+x_{5}-x_{1}^{3}
$$
the maximum value is $\qquad$
|
10.4.
If there exists a positive integer $j$, such that
$$
x_{j}=x_{j+1}\left(j=1,2, \cdots, 5, x_{6}=x_{1}\right) \text {, }
$$
then $M \leqslant 4$.
When $x_{1}=0, x_{2}=1, x_{3}=0, x_{4}=1, x_{5}=0$, the equality holds.
If for any positive integer $i$, we have $x_{i} \neq x_{i+1}$ $(i=1,2, \cdots 5)$, then either $x_{i-1}x_{i+1}$, or $x_{i-1}>x_{i}$ and $x_{i}<x_{i+1}$.
For both cases, we have
$$
\left|x_{i-1}-x_{i}\right|^{3}+\left|x_{i}-x_{i+1}\right|^{3}<\left|x_{i-1}-x_{i+1}\right|^{3} \leqslant 1 \text {. }
$$
Thus, $M<1+3=4$.
In summary, the maximum value of $M$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given a moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
2 x+y \leqslant 2, \\
x \geqslant 0, \\
\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1 .
\end{array}\right.
$$
Then the area of the figure formed by the moving point $P(x, y)$ is
|
8. 2 .
From equation (1), we have
$$
\begin{array}{l}
x+\sqrt{x^{2}+1} \geqslant \sqrt{y^{2}+1}-y \\
\Rightarrow \ln \left(x+\sqrt{x^{2}+1}\right) \geqslant \ln \left(\sqrt{y^{2}+1}-y\right) .
\end{array}
$$
It is easy to see that $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)$ is a strictly increasing function.
$$
\begin{array}{l}
\text { Therefore, equation (1) } \Leftrightarrow f(x) \geqslant f(-y) \\
\Leftrightarrow x \geqslant -y \Leftrightarrow x+y \geqslant 0 .
\end{array}
$$
From $\left\{\begin{array}{l}2 x+y \leqslant 2 \\ x+y \geqslant 0\end{array}\right.$, we get the intersection point coordinates as $(2,-2)$.
Thus, the area of the region formed by the moving point $P$ is
$$
\frac{1}{2} \times 2 \times 2=2
$$
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The positive integer $n=$ $\qquad$ that makes $2^{n}+256$ a perfect square.
|
When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$.
If it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a positive integer). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, hence $k=1$.
Thus, $n=11$.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 (1) Find the minimum value of the sum of the digits of integers of the form $3 n^{2}+n+1\left(n \in \mathbf{Z}_{+}\right)$;
(2) Does there exist a number of this form whose sum of digits is 1999?
|
(1) Since
$$
3 n^{2}+n+1=n(3 n+1)+1
$$
is an odd number, the last digit must be odd.
If the sum of the digits is 1, then the number is 1, but $n$ is a positive integer, $3 n^{2}+n+1 \geqslant 5$, which is a contradiction.
If the sum of the digits is 2 and it is an odd number, then the first and last digits must both be 1, with all the middle digits being 0. Let's assume it is $10^{a}+1$ $\left(a \in \mathbf{Z}_{+}\right)$.
$$
\begin{array}{l}
\text { Hence } 3 n^{2}+n+1=10^{a}+1 \\
\Rightarrow n(3 n+1)=10^{a} . \\
\text { But }(3 n+1, n)=(1, n)=1, \text { and } 3 n+1>n,
\end{array}
$$
Thus, $n=2^{a}, 3 n+1=5^{a}$.
So, $3 \times 2^{a}+1=5^{a}$.
But when $a>1$,
$$
\left(\frac{5}{2}\right)^{a}>3+1>3+\frac{1}{2^{a}},
$$
which is a contradiction.
Therefore, $a$ can only be 1.
Thus, $n=2$, and in this case, $3 n+1=7 \neq 5$, which is a contradiction.
Therefore, the sum of the digits must be at least 3.
When $n=8$, $3 n^{2}+n+1=201$, and the sum of the digits is exactly 3, so 3 is the minimum value.
$$
\begin{array}{l}
\text { (2) Let } n=10^{m}-1 \text {. Then } \\
3 n^{2}+n+1=3 \times 10^{2 m}-6 \times 10^{m}+3+10^{m} \\
=2 \underbrace{99 \cdots 95}_{m-1 \uparrow} \underbrace{00 \cdots 03}_{m-1 \uparrow} . \\
\text { Let } 2+9(m-1)+5+3=1999 .
\end{array}
$$
So, we only need to take $m=222$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given that $p$ is a prime number greater than 3. Find
$$
\prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)
$$
the value.
|
Let $\omega=\mathrm{e}^{\frac{2 \pi i}{p}}$. Then
$$
\begin{array}{l}
\omega^{p}=1, \omega^{-\frac{p}{2}}=-1, 2 \cos \frac{2 k \pi}{p}=\omega^{k}+\omega^{-k} . \\
\text { Hence } \prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)=\prod_{k=1}^{p-1}\left(1+\omega^{k}+\omega^{-k}\right) \\
=\prod_{k=1}^{p-1} \omega^{-k}\left(\omega^{2 k}+\omega^{k}+1\right) \\
=\omega^{-\left(\frac{p(p-1)}{2}\right)} \prod_{k=1}^{p-1}\left(\omega^{2 k}+\omega^{k}+1\right) \\
=(-1)^{p-1} \prod_{k=1}^{p-1} \frac{1-\omega^{3 k}}{1-\omega^{k}} .
\end{array}
$$
Since $p$ is a prime number greater than 3, $(-1)^{p-1}=1$; also, $p$ and 3 are coprime, $\{1,2, \cdots, p-1\}$ and $\{3 \times 1$, $3 \times 2, \cdots, 3 \times(p-1)\}$ both represent the reduced residue system modulo $p$.
Thus, $\prod_{k=1}^{p-1}\left(1-\omega^{3 k}\right)=\prod_{k=1}^{p-1}\left(1-\omega^{k}\right)$.
In conclusion, $\prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The sum of all real numbers $m$ that satisfy $(2-m)^{m^{2}-m-2}=1$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
|
2. A.
When $2-m=1$, i.e., $m=1$, it satisfies the condition.
When $2-m=-1$, i.e., $m=3$,
$$
(2-m)^{m^{2}-m-2}=(-1)^{4}=1 \text{, }
$$
it satisfies the condition.
When $2-m \neq \pm 1$, i.e., $m \neq 1$ and $m \neq 3$, by the condition, $m^{2}-m-2=0$, and $2-m \neq 0$.
Solving gives $m=-1$.
Therefore, the sum is $1+3+(-1)=3$.
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $x, y, z$ satisfy
$$
x+y=4,|z+1|=x y+2 y-9 \text {. }
$$
then $x+2 y+3 z=$ $\qquad$
|
$=, 1.4$.
From $x+y=4$, we get $x=4-y$. Then
$$
\begin{array}{l}
|z+1|=x y+2 y-9 \\
=6 y-y^{2}-9=-(y-3)^{2} \\
\Rightarrow z=-1, y=3 \Rightarrow x=1 \\
\Rightarrow x+2 y+3 z=4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. A cube is painted red on all its faces, then cut into $n^{3}(n>2)$ identical smaller cubes. If the number of smaller cubes with only one face painted red is the same as the number of smaller cubes with no faces painted red, then $n=$ $\qquad$ .
|
2.8.
The total number of small cubes with only one face painted red is $6(n-2)^{2}$, and the total number of small cubes with no faces painted red is $(n-2)^{3}$. According to the problem,
$$
6(n-2)^{2}=(n-2)^{3} \Rightarrow n=8 \text {. }
$$
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Given real numbers $a, b, c, d$ satisfy $2a^2 + 3c^2 = 2b^2 + 3d^2 = (ad - bc)^2 = 6$. Find the value of $\left(a^2 + \dot{b}^2\right)\left(c^2 + d^2\right)$.
|
Let $m=a^{2}+b^{2}, n=c^{2}+d^{2}$. Then
$$
\begin{array}{l}
2 m+3 n=2 a^{2}+2 b^{2}+3 c^{2}+3 d^{2}=12 . \\
\text { By }(2 m+3 n)^{2}=(2 m-3 n)^{2}+24 m n \geqslant 24 m n \\
\Rightarrow 12^{2} \geqslant 24 m n \\
\Rightarrow m n \leqslant 6 . \\
\text { Also } m n=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\
=a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2} \\
=(a c+b d)^{2}+(a d-b c)^{2} \\
\geqslant(a d-b c)^{2}=6,
\end{array}
$$
From equations (1) and (2), we get $m n=6$, that is
$$
\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=6 \text {. }
$$
[Note] Real numbers $a, b, c, d$ that satisfy the conditions exist and are not unique, for example,
$$
(a, b, c, d)=\left(\sqrt{2}, 1, \frac{\sqrt{6}}{3},-\frac{2 \sqrt{3}}{3}\right)
$$
is one of the solutions.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) If positive numbers $a, b, c$ satisfy
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=3 \text {, }
$$
find the value of the algebraic expression
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$
|
Three, since $a, b, c$ have cyclic symmetry, without loss of generality, assume $0 < a < b < c$, then
$$
c-a>b>0, c-b>a>0 \text {. }
$$
Thus $\frac{b^{2}+c^{2}-a^{2}}{2 b c}=1+\frac{(c-b)^{2}-a^{2}}{2 b c}>1$,
$$
\begin{array}{l}
\frac{c^{2}+a^{2}-b^{2}}{2 c a}=1+\frac{(c-a)^{2}-b^{2}}{2 c a}>1, \\
\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{(a+b)^{2}-c^{2}}{2 a b}-13,
$$
Contradiction.
(2) If $c-1, \\
\frac{a^{2}+b^{2}-c^{2}}{2 a b}=1+\frac{(a-\dot{b})^{2}-c^{2}}{2 a b}<1 .
\end{array}
$$
From the above three equations, we get
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}<3,
$$
Contradiction.
Combining (1) and (2), we know $c=a+b$.
Therefore, $\frac{b^{2}+c^{2}-a^{2}}{2 b c}=1, \frac{c^{2}+a^{2}-b^{2}}{2 c a}=1$,
$$
\begin{array}{l}
\frac{a^{2}+b^{2}-c^{2}}{2 a b}=-1 . \\
\text { Hence } \frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $\mathrm{i}$ is the imaginary unit. If
$$
z=1+\mathrm{i}+\cdots+\mathrm{i}^{2013},
$$
denote the complex conjugate of $z$ as $\bar{z}$, then $z \cdot \bar{z}=$ $\qquad$
|
8. 2 .
Let $a_{n}=\mathrm{i}^{n}$. Then $a_{n+4}=a_{n}$, and
$$
1+\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}=0 \text {. }
$$
Thus $z=1+\mathrm{i} \Rightarrow \bar{z}=1-\mathrm{i}$
$$
\Rightarrow z \cdot \bar{z}=(1+\mathrm{i})(1-\mathrm{i})=1-\mathrm{i}^{2}=2 .
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Given a positive integer $n$. Find $\sum_{k=1}^{n}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
---
Please note that the format and line breaks have been preserved as requested.
|
$$
\text { Three, let } n=2^{m} a_{m}+2^{m-1} a_{m-1}+\cdots+2^{1} a_{1}+a_{0}
$$
where, $a_{m} \neq 0$.
At this point, $2^{m} \leqslant n<2^{m+1}$, so, $\left[\log _{2} n\right]=m$.
If $k \geqslant m+2$, then
$$
\frac{n}{2^{k}}-\frac{1}{2}<\frac{2^{m+1}}{2^{m+2}}-\frac{1}{2}=0 \text {, }
$$
at this time, $\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=-1$.
If $k=m+1$, then
$$
\frac{n}{2^{k}}-\frac{1}{2}=\frac{n}{2^{m+1}}-\frac{1}{2} \in\left[0, \frac{1}{2}\right) \text {, }
$$
at this time, $\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=0$.
If $k=m$, then
$$
\begin{array}{l}
{\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=\left[\sum_{i=0}^{m} \frac{a_{m-t}}{2^{t}}-\frac{1}{2}\right]} \\
=\left[a_{m}+\frac{a_{m-1}}{2}-\frac{1}{2}\right]=a_{m-1} .
\end{array}
$$
If $k \leqslant m-1$, then,
$$
\begin{array}{l}
{\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=\left[\sum_{t=0}^{m} 2^{m-k-t} a_{m-t}-\frac{1}{2}\right]} \\
=\sum_{t=k}^{m} 2^{t-k} a_{t}+a_{k-1}-1 .
\end{array}
$$
Then $\sum_{k=1}^{\left[\log _{2} n\right]}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=\sum_{k=1}^{m}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]$
$$
\begin{array}{l}
=\sum_{k=1}^{m-1}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]+a_{m-1} \\
=a_{m-1}+\sum_{k=1}^{m-1}\left(\sum_{t=k}^{m} 2^{t-k} a_{t}+a_{k-1}-1\right) \\
=a_{m-1}-a_{m}+\sum_{t=1}^{m} \sum_{k=1}^{t} 2^{t-k} a_{t}+\sum_{k=1}^{m-1} a_{k-1}-\sum_{k=1}^{m-1} 1 \\
=a_{m}\left(2^{m}-2\right)+\sum_{t=1}^{m-1}\left(2^{t}-1\right) a_{t}+\sum_{k=1}^{m} a_{k-1}-(m-1) \\
=\sum_{t=0}^{m} 2^{t} a_{t}-m-1=n-m-1 .
\end{array}
$$
Therefore, $\sum_{k=1}^{n}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]$
$$
\begin{array}{l}
=\sum_{k=1}^{m}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]+\left[\frac{n}{2^{m+1}}-\frac{1}{2}\right]+\sum_{k=m+2}^{n}\left[\frac{n}{2^{k}}-\frac{1}{2}\right] \\
=(n-m-1)+0+\sum_{k=m+2}^{n}(-1) \\
=(n-m-1)-(n-m-1)=0 .
\end{array}
$$
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $a, b, c > 0$, and $a^{2} + b^{2} + c^{2} + abc = 4$. Prove:
$$
\begin{array}{l}
\sqrt{\frac{(2-a)(2-b)}{(2+a)(2+b)}} + \sqrt{\frac{(2-b)(2-c)}{(2+b)(2+c)}} + \\
\sqrt{\frac{(2-c)(2-a)}{(2+c)(2+a)}} = 1 .
\end{array}
$$
|
$$
\begin{array}{l}
16=4 a^{2}+4 b^{2}+4 c^{2}+4 a b c \text {. } \\
\text { Then } \sqrt{\frac{(2-a)(2-b)}{(2+a)(2+b)}}+\sqrt{\frac{(2-b)(2-c)}{(2+b)(2+c)}}+ \\
\sqrt{\frac{(2-c)(2-a)}{(2+c)(2+a)}} \\
=\frac{\sqrt{(4-a)^{2}\left(4-b^{2}\right)}}{(2+a)(2+b)}+ \\
\frac{\sqrt{\left(4-b^{2}\right)\left(4-c^{2}\right)}}{(2+b)(2+c)}+ \\
\frac{\sqrt{\left(4-c^{2}\right)\left(4-a^{2}\right)}}{(2+c)(2+a)} \\
=\frac{\sqrt{16-4 a^{2}-4 b^{2}+a^{2} b^{2}}}{(2+a)(2+b)}+ \\
\frac{\sqrt{16-4 b^{2}-4 c^{2}+b^{2} c^{2}}}{(2+b)(2+c)}+ \\
\frac{\sqrt{16-4 c^{2}-4 a^{2}+c^{2} a^{2}}}{(2+c)(2+a)} \\
=\frac{\sqrt{4 c^{2}+4 a b c+a^{2} b^{2}}}{(2+a)(2+b)}+ \\
\frac{\sqrt{4 a^{2}+4 a b c+b^{2} c^{2}}}{(2+b)(2+c)}+ \\
\frac{\sqrt{4 b^{2}+4 a b c+c^{2} a^{2}}}{(2+c)(2+a)} \\
=\frac{2 c+a b}{(2+a)(2+b)}+\frac{2 a+b c}{(2+b)(2+c)}+ \\
\frac{2 b+c a}{(2+c)(2+a)} \\
=\frac{(2 c+a b)(2+c)+(2 a+b c)(2+a)+(2 b+c a)(2+b)}{(2+a)(2+b)(2+c)} \\
=\frac{2\left(a^{2}+b^{2}+c^{2}\right)+2(a b+b c+c a)+4(a+b+c)+3 a b c}{(2+a)(2+b)(2+c)} \\
=\frac{a b c+2(a b+b c+c a)+4(a+b+c)+8}{(2+a)(2+b)(2+c)} \\
=\frac{(2+a)(2+b)(2+c)}{(2+a)(2+b)(2+c)}=1 . \\
\end{array}
$$
|
1
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Given
$$
f(x)=\frac{1+\ln (x+1)}{x}, g(x)=\frac{k}{x+1} .
$$
Find the largest positive integer $k$, such that for any positive number $c$, there exist real numbers $a$ and $b$ satisfying $-1<a<b<c$, and
$$
f(c)=f(a)=g(b) .
$$
|
11. For positive integer $k$, it is clear that $g(x)=\frac{k}{x+1}$ is a decreasing function on the interval $(-1,+\infty)$.
Thus, for any positive number $c$,
$$
f(c)=g(b)>g(c) \text {. }
$$
When $x>0$, the inequality
$$
\begin{array}{l}
f(x)>g(x) \\
\Leftrightarrow k0) \text {. }
$$
Then $h^{\prime}(x)=\frac{x-1-\ln (x+1)}{x^{2}}$.
$$
\text { Let } \varphi(x)=x-1-\ln (x+1)(x>0) \text {. }
$$
Then $\varphi^{\prime}(x)=\frac{x}{x+1}>0$.
Hence, $\varphi(x)$ is an increasing function when $x>0$.
Also, $\varphi(2)=1-\ln 30 \text {, }
$$
Therefore, there exists a unique positive real number $x_{0}$ such that
$$
\varphi\left(x_{0}\right)=x_{0}-1-\ln \left(x_{0}+1\right)=0 \text {. }
$$
Thus, $h^{\prime}\left(x_{0}\right)=0$, and $x_{0} \in(2,3)$.
Hence, when $x \in\left(0, x_{0}\right)$, $h^{\prime}(x)0$, $h(x)$ is an increasing function.
Therefore, when $x>0$, combining with equation (2), the minimum value of $h(x)$ is $h\left(x_{0}\right)=x_{0}+1 \in(3,4)$.
Combining with equation (1), we have the positive integer
$k \leqslant 3$.
We now prove that when $k=3$, for $-10 . \\
\text { Let } \tau(x)=1-2 x+(x+1) \ln (x+1) \text {, where } \\
\text { }-1\tau(0)>0$.
Therefore, equation (4) holds.
Note that,
$g(x)=\frac{3}{x+1}(x \in(-1,+\infty))$ has a range of $(0,+\infty)$,
$f(x)=\frac{1+\ln (x+1)}{x}(x \in(0,+\infty))$ also has a range of $(0,+\infty)$,
$$
f(x)=\frac{1+\ln (x+1)}{x}(x \in(-1,0)) \text { has a range }
$$
of $\mathbf{R}$.
Combining the graphs of the functions, we know that for any positive number $c$, there exist real numbers $a γ b$ satisfying $-1<a<b<c$, and
$$
f(c)=f(a)=g(b) \text {. }
$$
In summary, the maximum value of the positive integer $k$ is 3.
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
354 Divide the sides of the equilateral $\triangle A B C$ into four equal parts, and draw lines parallel to the other two sides through each division point. The 15 points formed by the intersections of the sides of $\triangle A B C$ and these parallel lines are called lattice points. Among these 15 lattice points, if $n$ points are chosen, there will definitely be three points that can form an isosceles triangle (including equilateral triangles). Find the minimum value of $n$.
---
The problem is to find the minimum value of $n$ such that, when $n$ points are chosen from the 15 lattice points, there will always be three points that can form an isosceles triangle (including equilateral triangles).
|
Solve for the minimum value of $n$ being 6.
Let the three equal division points from point $A$ to $B$ on side $AB$ be $L, F, W$; the three equal division points from point $B$ to $C$ on side $BC$ be $V, D, U$; and the three equal division points from point $C$ to $A$ on side $CA$ be $N, E, M$. Denote the intersection point of lines $MV, LU, EF$ as $P$; the intersection point of lines $MV, WN, DF$ as $Q$; and the intersection point of lines $WN, LU, DE$ as $R$.
If the minimum value of $n$ is 5, and we take five grid points on side $AB$, then there do not exist three grid points that can form an isosceles triangle. Therefore, $n \geqslant 6$.
Below, we prove: If among these 15 grid points, we take $n$ points such that no three points can form an isosceles triangle, then $n \leqslant 5$.
Assume the $n$ points taken are red points, and the other points are blue points.
Since $P, Q, R$ cannot all be red points, we discuss the following three scenarios:
(1) If only two of $P, Q, R$ are red points, assume $Q, R$ are red points. Thus, $P, A, D$ are all blue points, and $U, V$ and $E, F$ are also all blue points.
Since at most one of $M, C$ can be a red point (otherwise, $M, C$ and $R$ form an isosceles triangle), at most one of $N, C$ can be a red point (otherwise, $N, C$ and $R$ form an isosceles triangle), and at most one of $M, N$ can be a red point (otherwise, $M, N$ and $Q$ form an isosceles triangle), at most one of $M, N, C$ can be a red point.
Similarly, at most one of $L, W, B$ can be a red point.
Therefore, $n \leqslant 5$.
(2) If only one of $P, Q, R$ is a red point, assume $P$ is a red point.
(i) If $D$ is a red point, then $W, N$ are both blue points, and at most one of $F, L, M, E$ can be a red point.
If $L$ (or $M$) is a red point, then $A$ is a blue point, and at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
If $F$ (or $E$) is a red point, then $A$ is a blue point, and at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
(ii) If $D$ is a blue point and $A$ is a red point, then $L, M$ and $W, N$ are all blue points, at most one of $E, F$ can be a red point; at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
(iii) If $D, A$ are both blue points, since at most one of $L, M$ can be a red point, assume $L$ is a red point, then $E, F, M$ are all blue points, at most one of $W, N$ can be a red point; at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
If $L, M$ are both blue points, when $E, F$ are both red points, $W, N$ and $B, C$ are all blue points; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
When $E, F$ have at most one red point, at most one of $W, N$ can be a red point; at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
(3) If $P, Q, R$ are all blue points, when each of the three-point groups $A, L, M$, $B, V, W$, $C, N, U$, $D, E, F$ have at most one red point, then there are at most four red points, otherwise, among $A, L, M$, $B, V, W$, $C, N, U$, $D, E, F$, there must be a three-point group that contains at least two red points.
(i) If $D, E, F$ contain at least two red points, assume $E, F$ are red points, then $D, A$ and $B, C$ are all blue points, at most one of $L, M$ can be a red point; at most one of $W, V$ can be a red point; at most one of $N, U$ can be a red point. Therefore, $n \leqslant 5$.
(ii) If among $A, L, M$, $B, V, W$, $C, N, U$ there is a three-point group that contains at least two red points, assume $B, V, W$ contain at least two red points.
If $V, W$ are red points, then $B, E$ and $F, D$ are all blue points. If one of $L, U$ is a red point, assume $L$ is a red point, then $C, N$ are all blue points, at most one of $A, M$ can be a red point, even if $U$ is a red point, $n \leqslant 5$; if $L, U$ are both blue points, then at most one of $M, C$ can be a red point, at most one of $A, N$ can be a red point, therefore, $n \leqslant 5$.
If $B, V$ are red points, then $W$ is a blue point.
1) If $F$ is a red point, then $D, M, N, E$ are all blue points, so at most one of $A, C$ can be a red point, at most one of $L, U$ can be a red point. Therefore, $n \leqslant 5$.
2) If $F$ is a blue point, $D$ is a red point, then $E, L$ are both blue points, at most one of $A, C$ can be a red point; at most one of $M, N, U$ can be a red point. Therefore, $n \leqslant 5$.
3) If $F, D$ are blue points, $L$ is a red point, then $U, N$ are both blue points, at most one of $A, C$ can be a red point; at most one of $M, E$ can be a red point. Therefore, $n \leqslant 5$.
4) If $F, D, L$ are blue points, $U$ is a red point, then $A$ is a blue point, at most one of $M, C$ can be a red point; at most one of $N, E$ can be a red point. Therefore, $n \leqslant 5$.
5) If $F, D, L, U$ are blue points, then at most one of $M, N$ can be a red point; at most one of $A, C$ can be a red point. Even if $E$ is a red point, $n \leqslant 5$.
Using the contrapositive, when $n \geqslant 6$, we know that there must exist
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 What is the maximum number of rational points (points with both coordinates being rational numbers) that can lie on a circle in the plane, given that the center of the circle is not a rational point.
|
γAnalysisγIf $A, B, C$ are three rational points on a circle,
then the midpoint $D$ of $AB$ is a rational point, the slope of $AB$ is a rational number or infinite, so the equation of the perpendicular bisector of $AB$ is a linear equation with rational coefficients. Similarly, the equation of the perpendicular bisector of $BC$ is also a linear equation with rational coefficients. Therefore, the intersection of the perpendicular bisector of $AB$ and the perpendicular bisector of $BC$ is a rational point, i.e., the center of the circle is a rational point, which does not meet the condition.
Two examples of rational points: $(0,0)$ and $(1,0)$, the center of the circle is $\left(\frac{1}{2}, \sqrt{2}\right)$, which satisfies the condition.
Therefore, at most two rational points can exist on a circle such that the center of the circle is not a rational point.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let $A, B, C$ be three non-collinear lattice points on a plane, and the side lengths of $\triangle ABC$ are all positive integers. Find the minimum value of $AB$ and the minimum perimeter.
|
γAnalysisγIf $A B=1$, we might as well set $A(0,0), B(1,0)$. Then $|A C-B C|<A B=1$, which can only be $A C=B C$.
Thus, point $C$ lies on the perpendicular bisector of $A B$.
Therefore, the x-coordinate of point $C$ is $\frac{1}{2}$, meaning $C$ cannot be a lattice point.
Hence, $A B$ cannot be 1.
If $A B=2$, we might as well set $A(0,0), B(2,0)$.
Then $|A C-B C|<2$, which can only be $|A C-B C|=0,1$.
If $A C=B C \Rightarrow x_{c}-1 \Rightarrow A C^{2}=1+y_{c}^{2}$, thus, the difference of two perfect squares cannot be 1.
If $B C-A C=1$
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{l}
A C^{2}=x_{c}^{2}+y_{c}^{2}=k^{2}, \\
B C^{2}=\left(x_{C}-2\right)^{2}+y_{C}^{2}=(k+1)^{2}
\end{array}\right. \\
\Rightarrow 2 k+1=4-4 x_{C} .
\end{array}
$$
The parity of both sides of the above equation is different, so the equation does not hold, i.e., $B C-A C$ is not equal to 1.
Therefore, $A B$ cannot be 2.
If $A B=3$, we might as well set $A(0,0), B(3,0)$. Then $C(3,4)$ satisfies the conditions of the problem.
Therefore, the minimum value of $A B$ is 3.
Noting that the three vertices of $\triangle A B C$ are lattice points, using the area formula of analytic geometry, we know
$$
2 S_{\triangle A B C}=\left(\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right)
$$
The absolute value is a positive integer.
Also, the three sides of $\triangle A B C$ are all integers, with side lengths greater than or equal to 3, thus, using Heron's formula, the side lengths of 3, 3, 3; 3, 3, 4; 3, 4, 4; and 3, 3, 5 are all impossible (the area is irrational); 3, 4, 5 satisfies the conditions.
Therefore, the minimum perimeter is 12.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 On a plane, there exist $n$ points, no three of which are collinear, and when these $n$ points are arbitrarily labeled as $A_{1}, A_{2}, \cdots, A_{n}$, the broken line $A_{1} A_{2} \cdots A_{n}$ does not intersect itself. Find the maximum value of $n$.
|
γAnalysisγWhen $n=2,3$, it is obviously true.
When $n=4$, if the convex hull of the four points is a triangle, then it satisfies the condition.
Next, we show that when $n \geqslant 5$, it is impossible to satisfy the condition.
Consider only five points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ among these $n$ points.
If the convex hull of these five points is a pentagon $A B C D E$ or a quadrilateral $A B C D$, then the broken line $A C B D$ does not satisfy the condition.
Assume the convex hull of points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ is $\triangle P_{1} P_{2} P_{3}$, i.e., points $P_{4}, P_{5}$ are inside $\triangle P_{1} P_{2} P_{3}$. The line connecting $P_{4}, P_{5}$ does not pass through $P_{1}, P_{2}, P_{3}$, so it must intersect two sides of $\triangle P_{1} P_{2} P_{3}$. Assume without loss of generality that $P_{4} P_{5}$ intersects $P_{1} P_{3}$ and $P_{2} P_{3}$. Then, quadrilateral $P_{1} P_{2} P_{4} P_{5}$ is a convex quadrilateral, which does not satisfy the condition.
Therefore, $n \geqslant 5$ does not satisfy the condition, i.e., the maximum value of $n$ is 4.
Using the convex hull to analyze problems involving finite point sets is a common method in combinatorial geometry.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Find the smallest $n \in \mathbf{N}_{+}$, such that for any finite point set $M$ in the plane, if any $n$ points in $M$ can be covered by two lines, then there must exist two lines that can cover the point set $M$.
|
The required minimum value is 6.
When $|M| \leqslant 6$, the proposition is obviously true.
If $|M|>6$, for any six points $A_{1}, A_{2}, \cdots, A_{6}$ in $M$, they can be covered by two lines $l_{1}$ and $l_{2}$. Without loss of generality, assume that $l_{1}$ contains at least three points $A_{1}, A_{2}, A_{3}$. Let $M_{1}=M \cap l_{1}$ and $M_{2}=M \backslash l_{1}$.
If $\left|M_{2}\right| \leqslant 2$, the proposition is obviously true.
Without loss of generality, assume $\left|M_{2}\right| \geqslant 3$, and take $A_{1}, A_{2}, A_{3} \in M_{1}$, $B_{1}, B_{2}, B_{3} \in M_{2}$.
Then there exist two lines $l_{1}^{\prime}$ and $l_{2}^{\prime}$ that cover these six points. Among $A_{1}, A_{2}, A_{3}$, there must be two points on one of the lines, say $A_{1}, A_{2}$ are on $l_{1}^{\prime}$, then $l_{1}=l_{1}^{\prime}$. Points $B_{1}, B_{2}, B_{3}$ are not on $l_{1}^{\prime}$, so $B_{1}, B_{2}, B_{3}$ are on $l_{2}^{\prime}$. Replacing point $B_{3}$ with another point $B_{k}$ in $M_{2}$, then $B_{1}, B_{2}, B_{k}$ are collinear. Therefore, there exist two lines that can cover the point set $M$, and the proposition is true.
When $n=5$, there exists a counterexample. Take the three vertices and the midpoints of the three sides of a regular $\triangle ABC$. Any five of these points can be covered by two lines, but these six points cannot be covered by two lines.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. (50 points) Given that $a$, $b$, and $c$ are three distinct real numbers. If any two of the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$
have exactly one common root, find the value of $a^{2}+b^{2}+$ $c^{2}$.
|
2. From equations (1) and (2), we know their common root is $p=\frac{b-c}{b-a}$.
Similarly, the common roots of equations (2) and (3), and equations (1) and (3) are $q=\frac{c-a}{c-b}$ and $r=\frac{a-b}{a-c}$, respectively.
Thus, $p q r=-1$.
If any two of $p, q, r$ are equal, assume $p=q$, then the three equations have a common root $p$. Therefore,
$$
p=r \Rightarrow p=r=q=-1 \Rightarrow a=b=c,
$$
which is a contradiction.
Assume $p, q, r$ are distinct. Then the three equations have the form
$$
\begin{array}{l}
(x-p)(x-r)=0, \\
(x-p)(x-q)=0, \\
(x-q)(x-r)=0 .
\end{array}
$$
Thus, $a=-p-r=q r, b=-q-p=r p$,
$$
\begin{array}{l}
c=-r-q=p q . \\
\text { Hence }-2(p+q+r)=p q+q r+r p, \\
-1=(q+1)(p+1)(r+1) \\
=p q+q r+p+p+q+r .
\end{array}
$$
Therefore, $p+q+r=1, p q+q r+r p=-2$.
Then $a^{2}+b^{2}+c^{2}$
$$
=2\left(p^{2}+q^{2}+r^{2}+p q+q r+r p\right)=6 .
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4.12 The acrobats are numbered $1, 2, \cdots$, 12. They are to be arranged in two circles, $A$ and $B$, each with six people. In circle $B$, each acrobat stands on the shoulders of two adjacent acrobats in circle $A$. If the number of each acrobat in circle $B$ is equal to the sum of the numbers of the two acrobats below him, then such an arrangement is called a "tower". How many structurally different towers can be formed?
[Note] Towers that are the same after rotation or reflection are considered the same structure. For example, with 8 people, draw a circle and fill in the numbers of the acrobats in the bottom layer inside the circle, and the numbers of the acrobats in the top layer outside the circle. In Figure 2, the three diagrams are all towers, but the last two can be obtained from the first one by rotation or reflection, so they belong to the same structure.
|
4. Let the sums of the elements in circles $A$ and $B$ be $x$ and $y$ respectively. Then $y=2x$. Therefore,
$$
3x = x + y = 1 + 2 + \cdots + 12 = 78.
$$
Solving for $x$ gives $x = 26$.
Clearly, $1, 2 \in A$ and $11, 12 \in B$.
Let $A = \{1, 2, a, b, c, d\} (a < b < c < d)$. Then $a + b + c + d = 23$, and $a \geq 3, 8 \leq d \leq 10$ (if $d \leq 7$, then $a + b + c + d \leq 4 + 5 + 6 + 7 = 22$, a contradiction).
(1) If $d = 8$, then
$$
A = \{1, 2, a, b, c, 8\}, c \leq 7, a + b + c = 15.
$$
Thus, $(a, b, c) = (3, 5, 7)$ or $(4, 5, 6)$, i.e.,
$$
A = \{1, 2, 3, 5, 7, 8\} \text{ or } \{1, 2, 4, 5, 6, 8\}.
$$
If $A = \{1, 2, 3, 5, 7, 8\}$, then
$$
B = \{4, 6, 9, 10, 11, 12\}.
$$
Since $B$ contains $4, 6, 11, 12$, in the $A$ circle, 1 must be adjacent to 3, 1 must be adjacent to 5, 5 must be adjacent to 7, and 8 must be adjacent to 3. In this case, there is only one arrangement, leading to one tower, as shown in Figure 6(a).
$$
\begin{array}{l}
\text{If } A = \{1, 2, 4, 5, 6, 8\}, \text{ then} \\
B = \{3, 7, 9, 10, 11, 12\}.
\end{array}
$$
Similarly, in the $A$ circle, 1 must be adjacent to 2, 5 must be adjacent to 6, and 4 must be adjacent to 8. In this case, there are two arrangements, leading to two towers, as shown in Figures 6(b) and (c).
(2) If $d = 9$, then
$$
A = \{1, 2, a, b, c, 9\}, c \leq 8, a + b + c = 14.
$$
Thus, $(a, b, c) = (3, 5, 6)$ or $(3, 4, 7)$, i.e.,
$$
A = \{1, 2, 3, 5, 6, 9\} \text{ or } \{1, 2, 3, 4, 7, 9\}.
$$
If $A = \{1, 2, 3, 5, 6, 9\}$, then
$$
B = \{4, 7, 8, 10, 11, 12\}.
$$
To get $4, 10, 12$ in the $B$ circle, in the $A$ circle, 1, 3, and 9 must be pairwise adjacent, which is impossible;
If $A = \{1, 2, 3, 4, 7, 9\}$, then
$$
B = \{5, 6, 8, 10, 11, 12\}.
$$
To get $6, 8, 12$ in the $B$ circle, in the $A$ circle, 2 must be adjacent to 4, 1 must be adjacent to 7, and 9 must be adjacent to 3. In this case, there are two arrangements, leading to two towers, as shown in Figure 7.
(3) If $d = 10$, then
$$
A = \{1, 2, a, b, c, 10\}, c \leq 9, a + b + c = 13.
$$
Thus, $(a, b, c) = (3, 4, 6)$, i.e.,
$$
A = \{1, 2, 3, 4, 6, 10\},
$$
$$
B = \{5, 7, 8, 9, 11, 12\}.
$$
To get $8, 9, 11, 12$ in the $B$ circle, in the $A$ circle, 6 must be adjacent to 2, 6 must be adjacent to 3, 10 must be adjacent to 1, and 10 must be adjacent to 2. In this case, there is only one arrangement, leading to one tower, as shown in Figure 8.
Therefore, there are 6 structurally distinct towers.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Remove any $2 \times 2$ small square from the corner of a $3 \times 3$ grid, and the remaining shape is called a "corner shape" (Figure 3 is an example of a corner shape). Now, place some non-overlapping corner shapes in a $10 \times 10$ grid (Figure 4). The boundaries of the corner shapes must coincide with the boundaries or grid lines of the grid. Find the maximum value of the positive integer $k$, such that no matter how $k$ corner shapes are placed, it is always possible to place one more complete corner shape in the grid.
|
7. First, $k_{\max }$
$<8$. This is because, if eight corner shapes are placed in the manner shown in Figure 9, it is impossible to place another corner shape in the grid.
Next, we prove that after placing seven corner shapes arbitrarily, it is still possible to place another complete corner shape.
Cover the 5th and 6th rows and the 5th and 6th columns of the $10 \times 10$ grid, leaving four $4 \times 4$ smaller grids. After placing seven corner shapes, since each corner shape cannot intersect with two of the aforementioned $4 \times 4$ smaller grids, according to the pigeonhole principle, there must exist a $4 \times 4$ smaller grid $S$ such that at most one corner shape intersects with $S$. Since a corner shape can be contained within a $3 \times 3$ smaller grid, the part of the smaller grid $S$ occupied by the corner shape must be contained within a $3 \times 3$ smaller grid at one of its corners.
As shown in Figure 10, a new corner shape can be placed in the remaining part of the smaller grid $S$ after removing a $3 \times 3$ smaller grid from one of its corners. Therefore, $k=7$ satisfies the condition.
In conclusion, $k_{\max }=7$.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 4, in the Cartesian coordinate system, $O$ is the origin, the diagonals of $\square A B O C$ intersect at point $M$, and the hyperbola $y=\frac{k}{x}(x<0)$ passes through points $B$ and $M$. If the area of $\square A B O C$ is 24, then $k=$ . $\qquad$
|
2, 1. -8.
Let $M\left(\frac{k}{y}, y\right)$. Then $B\left(\frac{k}{2 y}, 2 y\right), C\left(\frac{3 k}{2 y}, 0\right)$.
From $S_{\text {OABOC }}=4 S_{\triangle O C M}=2\left|x_{c} y_{M}\right|$
$\Rightarrow 24=2\left|\frac{3 k}{2}\right| \Rightarrow|k|=8 \Rightarrow k=-8$.
|
-8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find all odd prime numbers $p$ such that $p \mid \sum_{k=1}^{103} k^{p-1}$.
untranslated text remains the same as requested.
|
If $p>103$, then for $1 \leqslant k \leqslant 103$, we have
$$
\begin{array}{l}
k^{p-1} \equiv 1(\bmod p), \\
\sum_{k=1}^{103} k^{p-1} \equiv 103(\bmod p) .
\end{array}
$$
Therefore, $p \leqslant 103$.
Let $103=p q+r(0 \leqslant r < q)$, then $r=q, 103=p q+r=(p+1) r$. Since 103 is a prime number, we get $p=102, r=1$, which is a contradiction.
If $p \leqslant q$, then
$$
103=p q+r \geqslant p^{2} \Rightarrow p=3,5,7 \text {. }
$$
When $p=3$, $q=34 \equiv 1=r(\bmod 3)$;
When $p=5$, $q=20 \equiv 2(\bmod 3), 2 \neq r=3$;
When $p=7$, $q=14 \equiv 2(\bmod 3), 2 \neq r=5$. Therefore, $p=3$ is the solution.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Find all positive integers that are coprime with all terms of the sequence $\left\{a_{n}\right\}$ satisfying
$$
a_{n}=2^{n}+3^{n}+6^{n}-1\left(n \in Z_{+}\right)
$$
|
Solution: Clearly, $\left(1, a_{n}\right)=1$.
Let $m(m>1)$ be a positive integer that is coprime with all terms in $\left\{a_{n}\right\}$, and let $p$ be a prime factor of $m$.
If $p>3$, then by Fermat's Little Theorem,
$$
\begin{array}{l}
2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), \\
6^{p-1} \equiv 1(\bmod p) .
\end{array}
$$
Let $2^{p-1}=m p+1,3^{p-1}=n p+1$,
$$
6^{p-1}=t p+1\left(m, n, t \in \mathbf{Z}_{+}\right) \text {. }
$$
Then $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$
$$
\begin{array}{l}
=\frac{m p+1}{2}+\frac{n p+1}{3}+\frac{t p+1}{6}-1 \\
=\frac{3 m p+2 n p+t p}{6}=p \cdot \frac{3 m+2 n+t}{6} .
\end{array}
$$
Since $a_{p-2}$ is an integer, $(p, 6)=1$, thus,
$$
6|(3 m+2 n+t), p| a_{p-2} \text {. }
$$
This contradicts $\left(m, a_{p-2}\right)=1$.
If $p=2$ or 3, then $a_{2}=48=2^{4} \times 3 \Rightarrow p \mid a_{2}$, which is also a contradiction.
Therefore, the only positive integer that is coprime with all terms in the sequence $\left\{a_{n}\right\}$ is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Prove: For any positive integer $n, 3^{n}+2 \times 17^{n}$ is not a multiple of 5, and find the smallest positive integer $n$, such that
$$
11 \mid\left(3^{n}+2 \times 17^{n}\right) .
$$
|
$$
\begin{array}{l}
3^{2 k}+2 \times 17^{2 k} \equiv(-1)^{k}+2 \times 2^{2 k} \\
\equiv 3(-1)^{k}(\bmod 5) \\
3^{2 k+1}+2 \times 17^{2 k+1} \equiv 3(-1)^{k}+4(-1)^{k} \\
\equiv 2(-1)^{k}(\bmod 5)
\end{array}
$$
Therefore, $3^{n}+2 \times 17^{n}$ is not a multiple of 5.
$$
\begin{array}{l}
\text { Also, } 3^{n}+2 \times 17^{n} \equiv 3^{n}+2 \times 6^{n} \\
\equiv 3^{n}\left(1+2^{n+1}\right) \equiv 0(\bmod 11) \\
\Leftrightarrow 111\left(1+2^{n+1}\right), \\
\text { so } n_{\min }=4 .
\end{array}
$$
|
4
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 What is the minimum degree of the highest term of a polynomial with rational coefficients that has $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots?
(2013, Joint Autonomous Admission Examination of Peking University and Other Universities)
|
Notice that the polynomial
$$
f(x)=\left(x^{2}-2\right)\left[(x-1)^{3}-2\right]
$$
has roots $\sqrt{2}$ and $1-\sqrt[3]{2}$, and its degree is 5. Therefore, the degree of the highest term of a rational-coefficient polynomial with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots cannot be less than 5.
If there exists a rational-coefficient polynomial of degree no more than 4
$$
g(x)=a x^{4}+b x^{3}+c x^{2}+d x+e
$$
with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots, where $a, b, c, d, e$ are not all zero, then we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
g(\sqrt{2})=(4 a+2 c+e)+(2 b+d) \sqrt{2}=0, \\
g(1-\sqrt[3]{2}) \\
=-(7 a+b-c-d-e)-(2 a+3 b+2 c+d) \sqrt[3]{2}+ \\
(6 a+3 b+c) \sqrt[3]{4}=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
4 a+2 c+e=0, \\
2 b+d=0, \\
7 a+b-c-d-e=0, \\
2 a+3 b+2 c+d=0, \\
6 a+3 b+c=0 .
\end{array}\right.
\end{array}
$$
(1) + (3) gives
$$
11 a+b+c-d=0.
$$
(6) + (2), (6) + (4) respectively give
$$
\begin{array}{l}
11 a+3 b+c=0, \\
13 a+4 b+3 c=0 .
\end{array}
$$
(7) - (5) gives $a=0$.
Substituting into equations (7), (8) gives $b=c=0$,
Substituting into equations (1), (2) gives $d=e=0$.
Thus, $a=b=c=d=e=0$, which contradicts the assumption that $a, b, c, d, e$ are not all zero.
In conclusion, the minimum degree of the highest term of a rational-coefficient polynomial with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $x$ is an integer, and satisfies the inequality system
$$
\left\{\begin{array}{l}
x-1>0, \\
2 x-1<4,
\end{array}\right.
$$
then $x=$ $\qquad$
|
$$
=, 1.2 \text {. }
$$
From the given, we know that $1<x<\frac{5}{2}$. Therefore, the integer $x=2$.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let real numbers $x, y, z$ satisfy
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=27 \text {. }
$$
Then the maximum value of $|y-z|$ is $\qquad$
|
3. 6 .
The original equation is equivalent to a quadratic equation in $x$
$$
\begin{array}{l}
x^{2}-(y+z) x+y^{2}+z^{2}-y z-27=0 . \\
\text { And } \Delta=(y+z)^{2}-4\left(y^{2}+z^{2}-y z-27\right) \geqslant 0 \\
\Rightarrow(y-z)^{2} \leqslant 36 \Rightarrow|y-z| \leqslant 6 .
\end{array}
$$
When $|y-z|=6$, and $x=\frac{y+z}{2}$, the equality holds.
Therefore, the maximum value of $|y-z|$ is 6 .
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $x_{1}, x_{2}, \cdots, x_{15}$ take values of 1 or -1. Let
$$
S=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{15} x_{1} \text {. }
$$
Then the smallest positive integer that $S$ can take is $\qquad$
|
4.3.
Let $y_{i}=x_{i} x_{i+1}(i=1,2, \cdots, 15)$, with the convention that $x_{16}=$ $x_{1}$. Then $y_{i}=1$ or -1.
In $y_{1}, y_{2}, \cdots, y_{15}$, let there be $a$ values that are 1 and $b$ values that are -1. Clearly, $a+b=15$.
Also, $1^{a}(-1)^{b}=y_{1} y_{2} \cdots y_{15}=\left(x_{1} x_{2} \cdots x_{15}\right)^{2}$ $=1$, so $b$ must be even.
Since $S$ is a positive integer, $S=a-b=15-2 b \geqslant 3$, at this point, $b=6$.
On the other hand, in $x_{1}, x_{2}, \cdots, x_{15}$,
$$
x_{2}=x_{5}=x_{8}=-1 \text {, }
$$
the rest are all 1, then $S=3$.
Therefore, the smallest positive integer that $S$ can take is 3.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. (25 points) As shown in Figure 1, given that $E$ is a point on side $A B$ of square $A B C D$, and the symmetric point of $A$ with respect to $D E$ is $F, \angle B F C = 90^{\circ}$. Find the value of $\frac{A B}{A E}$.
|
2. As shown in Figure 4, extend $E F$ to intersect $B C$ at point $M$, connect $D M$, and let it intersect $C F$ at point $G$.
Then, Rt $\triangle D F M \cong$ Rt $\triangle D C M$. Therefore, $\angle F D M = \angle M D C$, and $F M = C M$.
Thus, $M$ is the midpoint of $B C$.
$$
\begin{array}{l}
\text { Also, } \angle E D M = \angle E D F + \angle F D M \\
= \frac{1}{2} \angle A D F + \frac{1}{2} \angle F D C = 45^{\circ} .
\end{array}
$$
Rotate $\triangle M D C$ around point $D$ by $90^{\circ}$ counterclockwise to get $\triangle H D A$.
Then, $\triangle M D E \cong \triangle H D E$.
Therefore, $E M = H E = A E + M C$.
Let the side length of the square be $1$, and $A E = x$.
By $E B^{2} + B M^{2} = E M^{2}$, we have
$$
\begin{array}{l}
(1-x)^{2} + \left(\frac{1}{2}\right)^{2} = \left(x + \frac{1}{2}\right)^{2} \\
\Rightarrow x = \frac{1}{3} \Rightarrow \frac{A B}{A E} = 3 .
\end{array}
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (25 points) On a circle, there are $n$ different positive integers $a_{1}$, $a_{2}, \cdots, a_{n}$ placed in a clockwise direction. If for any number $b$ among the ten positive integers $1, 2, \cdots, 10$, there exists a positive integer $i$ such that $a_{i}=b$ or $a_{i}+a_{i+1}=b$, with the convention that $a_{n+1}=a_{1}$, find the minimum value of the positive integer $n$.
|
3. From the conditions, we know that the $2n$ numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n}+a_{1}$ should include the ten positive integers $1,2, \cdots, 10$. Therefore, $2 n \geqslant 10 \Rightarrow n \geqslant 5$.
When $n=5$, the ten numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n}+a_{1}$ are all distinct and take values from $1 \sim 10$.
$$
\begin{array}{l}
\text { Hence } a_{1}+a_{2}+\cdots+a_{n}+\left(a_{1}+a_{2}\right)+ \\
\left(a_{2}+a_{3}\right)+\cdots+\left(a_{n}+a_{1}\right) \\
=1+2+\cdots+10 \\
\Rightarrow 3\left(a_{1}+a_{2}+\cdots+a_{n}\right)=55,
\end{array}
$$
which is a contradiction.
Thus, when $n=5$, there do not exist five positive integers $a_{1}, a_{2}, \cdots, a_{5}$ that satisfy the conditions.
When $n=6$, construct $a_{1}=1, a_{2}=10, a_{3}=2, a_{4}=6, a_{5}=3, a_{6}=4$, and verify that these six numbers satisfy the conditions.
In conclusion, the minimum value of the positive integer $n$ is 6.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Find the largest positive integer $n(n \geqslant 3)$, such that there exists a convex $n$-gon, where the tangent values of all its interior angles are integers.
(Proposed by the Problem Committee)
|
On the one hand, since each interior angle of a regular octagon is $135^{\circ}$, and its tangent value is -1, $n=8$ satisfies the condition.
On the other hand, if $n \geqslant 9$, let the exterior angles of the $n$-sided polygon be $\angle A_{1}, \angle A_{2}, \cdots, \angle A_{n}\left(0<\angle A_{1} \leqslant \angle A_{2}\right.$ $\left.\leqslant \cdots \leqslant \angle A_{n}\right)$. Then
$$
\begin{array}{l}
\angle A_{1}+\angle A_{2}+\cdots+\angle A_{n}=2 \pi \\
\Rightarrow 0<\angle A_{1} \leqslant \frac{2 \pi}{n}<\frac{\pi}{4} \Rightarrow 0<\tan A_{1}<1 .
\end{array}
$$
Thus, the tangent value of the corresponding interior angle of $\angle A_{1}$ is not an integer. Therefore, when $n \geqslant 9$, there is no polygon that satisfies the condition. In summary, the maximum value of $n$ is 8.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Find the remainder when $10^{10}(100$ ones$)$ is divided by 7.
|
\begin{array}{l}\text { Sol } 10^{10} \equiv(7+3)^{10^{10}} \equiv 3^{10} \\ \equiv(7+2)^{50 \cdots 0} \equiv 2^{5 \times 10^{9}} \equiv 2^{3 \times 106 \cdots 6+2} \\ \equiv 4(7+1)^{166 \cdots 6} \equiv 4(\bmod 7) .\end{array}
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Seven, let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1}=\left(1+\frac{k}{n}\right) a_{n}+1(n=1,2, \cdots) .
\end{array}
$$
Find all positive integers $k$ such that every term in the sequence $\left\{a_{n}\right\}$ is an integer.
(Zhang Lei)
|
When $k=1$, $a_{2}=3, a_{3}=\frac{11}{2}$, which does not satisfy the condition.
When $k=2$, by the given condition we have
$$
\frac{a_{n+1}}{(n+1)(n+2)}=\frac{a_{n}}{n(n+1)}+\frac{1}{(n+1)(n+2)} \text {. }
$$
Thus, $\frac{a_{n}}{n(n+1)}=\frac{a_{1}}{1 \times 2}+\sum_{i=2}^{n} \frac{1}{i(i+1)}=1-\frac{1}{n+1}$.
Therefore, $a_{n}=n^{2}$ is a positive integer, satisfying the condition.
When $k \geqslant 3$, by the given condition we have
$$
\begin{array}{l}
\frac{a_{n+1}}{(n+1)(n+2) \cdots(n+k)} \\
=\frac{a_{n}}{n(n+1) \cdots(n+k-1)}+\frac{1}{(n+1)(n+2) \cdots(n+k)} .
\end{array}
$$
Thus, $\frac{a_{n}}{n(n+1) \cdots(n+k-1)}$
$$
\begin{array}{l}
=\frac{a_{1}}{1 \times 2 \times \cdots \times k}+\sum_{i=2}^{n} \frac{1}{i(i+1) \cdots(i+k-1)} \\
=\frac{1}{k-1}\left[\frac{1}{1 \times 2 \times \cdots \times(k-1)}-\frac{1}{(n+1)(n+2) \cdots(n+k-1)}\right] \\
\Rightarrow a_{n}=\frac{1}{k-1} \cdot \frac{n(n+1) \cdots(n+k-1)}{1 \times 2 \times \cdots \times(k-1)}-\frac{n}{k-1} \\
=\frac{(n+k-1)(n+k-2)}{(k-1)^{2}} \mathrm{C}_{n+k-3}^{k-2}-\frac{n}{k-1} .
\end{array}
$$
When $n+k-2=(k-1)^{2}, n=(k-1)^{2}-k+2$, by $\frac{(n+k-1)(n+k-2)}{(k-1)^{2}} \mathrm{C}_{n+k-3}^{k-2} \in \mathbf{N}, \frac{n}{k-1}$ $\notin \mathbf{N}$, we know
$$
a_{n}=\frac{(n+k-1)(n+k-2)}{(k-1)^{2}} \mathrm{C}_{n+k-3}^{k-2}-\frac{n}{k-1} \notin \mathbf{N} \text {. }
$$
In summary, the value of the positive integer $k$ that satisfies the condition is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $S$ be a set of $n(n \geqslant 5)$ points in the plane. If any four points chosen from $S$ have at least one point connected to the other three, then which of the following conclusions is correct? $\qquad$
(1) There is no point in $S$ that is connected to all other points;
(2) There is at least one point in $S$ that is connected to all other points;
(3) There are at most two points in $S$ that are not connected to all other points;
(4) There are at most two points in $S$ that are connected to all other points.
|
-γ1. (2).
In the point set $S$, all points are connected to each other, which clearly satisfies the problem. Therefore, conclusions (1) and (4) are incorrect.
Suppose $A$, $B$, and $C$ are three points in the point set $S$ that are not connected to each other, but the remaining $n-3$ points are all connected to each other. This also clearly satisfies the problem. Hence, conclusion (3) is also incorrect.
If all points in the point set $S$ are connected to each other, then conclusion (2) is obviously true; otherwise, assume points $P$ and $Q$ are not connected. Take any two other points in the point set $S$, say $R$ and $T$. Among the four points $P$, $Q$, $R$, and $T$, there must be one point that is connected to the other three. This point cannot be $P$ or $Q$, so it must be one of $R$ or $T$ (let's say $R$). By the arbitrariness of points $R$ and $T$, we know that there is at least one point in the point set $S$ that is connected to all other points. Therefore, conclusion (2) is correct.
|
2
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that the three vertices of $\triangle A B C$ are all on the parabola $y^{2}=2 p x(p>0)$, and the centroid of $\triangle A B C$ is exactly the focus of the parabola. If the equation of the line on which side $B C$ lies is $4 x+y$ $-20=0$, then $p=$ $\qquad$ .
|
6. $p=8$.
Let $A\left(\frac{y_{1}^{2}}{2 p}, y_{1}\right), B\left(\frac{y_{2}^{2}}{2 p}, y_{2}\right), C\left(\frac{y_{3}^{2}}{2 p}, y_{3}\right)$.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
y^{2}=2 p x, \\
4 x+y-20=0
\end{array}\right. \\
\Rightarrow 2 y^{2}+p y-20 p=0 .
\end{array}
$$
From the given information,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{y_{1}^{2}}{2 p}+\frac{y_{2}^{2}}{2 p}+\frac{y_{3}^{2}}{2 p}=\frac{3 p}{2}, \\
y_{1}+y_{2}+y_{3}=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
y_{2}+y_{3}=-\frac{p}{2}, \\
y_{2} y_{3}=-10 p, \\
y_{1}+y_{2}+y_{3}=0, \\
y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=3 p^{2} .
\end{array}\right.
\end{array}
$$
From equations (1) and (3), we get $y_{1}=\frac{p}{2}$.
Substituting into equation (4) and solving with (1) and (2) yields $p=8$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given $\tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$, $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}$, $\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha=-\frac{17}{5}$.
Then $\tan (\alpha+\beta+\gamma)=$ $\qquad$ [5]
(2012, Xin Zhi Cup Shanghai High School Mathematics Competition)
|
Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then $x+y+z=\frac{17}{6}$,
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}$,
$\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5}$.
(1) $\div$ (3) gives
$x y z=-\frac{5}{6}$.
(2) $\times$ (4) gives $x y+y z+z x=\frac{2}{3}$.
Therefore, $\tan (\alpha+\beta+\gamma)=\frac{x+y+z-x y z}{1-(x y+y z+z x)}=11$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.
[2]
|
Notice that,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1006} \text {. }
$$
Clearly, $0<(5-2 \sqrt{6})^{1006}<1$,
$$
\begin{array}{l}
(5+2 \sqrt{6})^{1006}+(5-2 \sqrt{6})^{1006} \\
=2\left(C_{1006}^{0} 5^{1006}+C_{1006}^{2} 5^{1004} \times 24+\cdots+\right. \\
\left.\quad C_{1006}^{1006} 24^{503}\right) \in \mathbf{Z}_{+} .
\end{array}
$$
Therefore, the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$ is
$$
2\left(\mathrm{C}_{1006}^{0} 5^{1006}+\mathrm{C}_{1006}^{2} 5^{1004} \times 24+\cdots+\mathrm{C}_{1006}^{100} 24^{503}\right)-1 \text {. }
$$
And $2\left(\mathrm{C}_{1006}^{0} 5^{1006}+\mathrm{C}_{1006}^{2} 5^{1004} \times 24+\cdots+\right.$
$$
\begin{array}{l}
\left.C_{1006}^{1006} 24^{503}\right)-1 \\
\equiv 2 \times 24^{503}-1 \equiv 2 \times(25-1)^{503}-1 \\
\equiv 2 \times(-1)^{503}-1 \equiv 7(\bmod 10),
\end{array}
$$
That is, the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$ is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Let $x=(15+\sqrt{220})^{19}+(15+\sqrt{200})^{22}$. Find the unit digit of the number $x$.
|
Solve the conjugate expression
$$
y=(15-\sqrt{220})^{19}+(15-\sqrt{220})^{82} \text {. }
$$
Then $x+y$
$$
\begin{aligned}
= & (15+\sqrt{220})^{19}+(15-\sqrt{220})^{19}+ \\
& (15+\sqrt{220})^{82}+(15-\sqrt{220})^{82} .
\end{aligned}
$$
By the binomial theorem, for any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
(15+\sqrt{220})^{n}+(15-\sqrt{220})^{n} \\
=2\left(\mathrm{C}_{n}^{0} 15^{n}+\mathrm{C}_{n}^{2} 15^{n-2} \times 220+\cdots\right) \in \mathbf{Z}_{+},
\end{array}
$$
and the last digit is zero.
Therefore, $x+y$ is a positive integer with the last digit being zero. Also, $0<15-\sqrt{220}=\frac{5}{15+\sqrt{220}}<\frac{5}{25}=\frac{1}{5}$, and $(15-\sqrt{220})^{82}<(15-\sqrt{220})^{19}$, so $0<y<2(15-\sqrt{220})^{19}<2 \times 0.2^{19}<0.4$. Therefore, the last digit of $x$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
A=\{2,0,1,3\}, \\
B=\left\{x \mid -x \in A, 2-x^{2} \notin A\right\} .
\end{array}
$$
Then the sum of all elements in set $B$ is
|
,$- 1 .-5$.
It is easy to know that $B \subseteq\{-2,0,-1,-3\}$.
When $x=-2,-3$, $2-x^{2}=-2,-7 \notin A$; when $x=0,-1$, $2-x^{2}=2,1 \in A$.
Therefore, the set $B=\{-2,-3\}$.
Thus, the sum of all elements in set $B$ is -5.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In the Cartesian coordinate system $x O y$, it is known that points $A$ and $B$ lie on the parabola $y^{2}=4 x$, and satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B}=-4, F$ is the focus of the parabola. Then $S_{\triangle O F A} \cdot S_{\triangle O F B}=$ $\qquad$ .
|
2. 2 .
From the problem, we know the point $F(1,0)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Thus, $x_{1}=\frac{y_{1}^{2}}{4}, x_{2}=\frac{y_{2}^{2}}{4}$.
Then $-4=\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2}$
$$
\begin{array}{l}
=\frac{1}{16}\left(y_{1} y_{2}\right)^{2}+y_{1} y_{2} \\
\Rightarrow \frac{1}{16}\left(y_{1} y_{2}+8\right)^{2}=0 \\
\Rightarrow y_{1} y_{2}=-8 .
\end{array}
$$
Therefore, $S_{\triangle O F A} \cdot S_{\triangle O F B}$
$$
\begin{array}{l}
=\left(\frac{1}{2}|O F|\left|y_{1}\right|\right)\left(\frac{1}{2}|O F|\left|y_{2}\right|\right) \\
=\frac{1}{4}|O F|^{2}\left|y_{1} y_{2}\right|=2 .
\end{array}
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In $\triangle A B C$, it is known that
$$
\sin A=10 \sin B \cdot \sin C, \cos A=10 \cos B \cdot \cos C \text {. }
$$
Then $\tan A=$ $\qquad$
|
3. 11 .
From $\sin A-\cos A$
$$
\begin{array}{l}
=10(\sin B \cdot \sin C-\cos B \cdot \cos C) \\
=-10 \cos (B+C)=10 \cos A \\
\Rightarrow \sin A=11 \cos A \\
\Rightarrow \tan A=11 .
\end{array}
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$
$(2009$, National High School Mathematics League Jilin Province Preliminary)
|
Hint: The fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is $1-(\sqrt{2}-\sqrt{3})^{2010}$.
Since $0.9<1-(\sqrt{2}-\sqrt{3})^{2010}<1$, the first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The number of positive integers $n$ such that $n+1$ divides $n^{2012}+2012$ is $\qquad$ .
|
From the problem, we know
$$
\begin{array}{l}
n^{2012}+2012 \equiv(-1)^{2012}+2012 \\
=2013 \equiv 0(\bmod n+1) .
\end{array}
$$
Since $2013=3 \times 11 \times 61$, $n$ has $2^{3}-1=7$ solutions.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) As shown in Figure 5, $P$ is a point outside circle $\odot O$, $PA$ is tangent to $\odot O$ at point $A$, and $PBC$ is a secant of $\odot O$. $AD \perp PO$ at point $D$. If $PB=4$, $CD=5$, $BC=6$, find the length of $BD$.
|
Connect $O A, O B, O C$. Then $O B=O C$. It is easy to see that
$$
\begin{array}{l}
P A \perp O A, \\
P A^{2}=P B \cdot P C=P D \cdot P O \\
\Rightarrow \frac{P O}{P C}=\frac{P B}{P D} .
\end{array}
$$
Since $\angle B P D$ is a common angle, we have
$$
\begin{array}{l}
\triangle P O B \sim \triangle P C D . \\
\text { Then } \frac{P C}{C D}=\frac{P O}{O B}=\frac{P O}{O C} .
\end{array}
$$
At the same time, $\triangle P O C \sim \triangle P B D \Rightarrow \frac{P O}{O C}=\frac{P B}{B D}$.
$$
\text { Hence } \frac{P C}{C D}=\frac{P B}{B D} \Rightarrow B D=\frac{P B \cdot C D}{P C}=\frac{4 \times 5}{4+6}=2 \text {. }
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For each positive integer $n$, let the tangent line to the curve $y=x^{n+1}$ at the point $(1,1)$ intersect the $x$-axis at a point with abscissa $x_{n}$. Let $a_{n}=\lg x_{n}$. Then
$$
a_{1}+a_{2}+\cdots+a_{99}=
$$
$\qquad$
|
3. -2 .
Given $y=x^{n+1}$, we know $y^{\prime}=(n+1) x^{n}$.
Then according to the condition, we have
$$
\begin{array}{l}
\frac{0-1}{x_{n}-1}=y^{\prime}(1)=n+1 \Rightarrow x_{n}=\frac{n}{n+1} . \\
\text { Therefore, } a_{1}+a_{2}+\cdots+a_{99}=\lg \left(x_{1} x_{2} \cdots x_{99}\right) \\
=\lg \left(\frac{1}{2} \times \frac{2}{3} \times \cdots \times \frac{99}{100}\right)=-2 .
\end{array}
$$
|
-2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let the ellipse $C$ be:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{20}=1(a>2 \sqrt{5})
$$
with its left focus at $F$, and point $P(1,1)$. It is known that there exists a line $l$ passing through point $P$ intersecting the ellipse at points $A$ and $B$, with $M$ being the midpoint of $A B$, such that $|F M|$ is the geometric mean of $|F A|$ and $|F B|$. Find the smallest positive integer value of $a$, and the equation of $l$ at this value of $a$.
|
11. By the median length formula, we have
$|F M|^{2}=\frac{1}{2}\left(|F A|^{2}+|F B|^{2}\right)-\frac{1}{4}|A B|^{2}$.
Also, $|F M|^{2}=|F A||F B|$, so
$|A B|^{2}=2(|F A|-|F B|)^{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and
$c=\sqrt{a^{2}-20}$.
Then $|F A|-|F B|=\frac{c}{a} x_{1}-\frac{c}{a} x_{2}$.
Thus, $\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=\frac{2 c^{2}}{a^{2}}\left(x_{1}-x_{2}\right)^{2}$, which means
$\left(y_{1}-y_{2}\right)^{2}=\frac{2 c^{2}-a^{2}}{a^{2}}\left(x_{1}-x_{2}\right)^{2}$
$=\frac{a^{2}-40}{a^{2}}\left(x_{1}-x_{2}\right)^{2}$.
Therefore, $a^{2} \geqslant 40$.
When $a \in \mathbf{N}_{+}$, $a \geqslant 7$.
When $a=7$, $\left(y_{1}-y_{2}\right)^{2}=\frac{9}{49}\left(x_{1}-x_{2}\right)^{2}$.
Thus, the slope of line $l$ is $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}= \pm \frac{3}{7}$.
Since point $P(1,1)$ is clearly inside the ellipse, both lines $y-1= \pm \frac{3}{7}(x-1)$ satisfy the condition.
In summary, the smallest positive integer value of $a$ is 7, and at this time, the equations of line $l$ are
$3 x-7 y+4=0$ or $3 x+7 y-10=0$.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) Remove a $2 \times 2$ small square from any corner of a $3 \times 3$ grid, and the remaining shape is called a "corner shape". Now, place some non-overlapping corner shapes in a $9 \times 9$ grid, with the requirement that the boundaries of the corner shapes coincide with the boundaries or grid lines of the grid. Find the maximum positive integer $k$ such that, no matter how $k$ corner shapes are placed, it is always possible to place one more complete corner shape in the grid.
|
Three, first, $k_{\max }<6$. This is because, if 6 corner shapes are placed in the manner shown in Figure 4, it is impossible to place another complete corner shape on this grid.
Now, place 5 corner shapes in any manner.
Next, we will prove that it is still possible to place another complete corner shape.
Consider the five shaded areas $A, B, C, D, E$ in Figure 5. If there exists a region that does not intersect with any of the 5 placed corner shapes, then it is clearly possible to place a complete corner shape in that region.
Assume that all five regions contain a part of a corner shape. Note that a corner shape cannot intersect with two of the regions $A, B, C, D, E$ simultaneously. Therefore, each of the five regions $A, B, C, D, E$ corresponds to a unique corner shape that intersects with it, which we denote as $f(A), f(B), f(C), f(D), f(E)$.
According to Figure 5, the four $4 \times 4$ sub-grids in the corners of the $9 \times 9$ grid are labeled as $a, b, c, d$.
Note that the corner shape $f(E)$ cannot intersect with all four regions $a, b, c, d$ simultaneously. Without loss of generality, assume $f(E)$ does not intersect with $a$. Considering that the corner shapes $f(B), f(C), f(D)$ also do not intersect with $a$, the only corner shape that intersects with $a$ is $f(A)$. However, the part of region $a$ occupied by the corner shape $f(A)$ must be contained within a $3 \times 3$ sub-grid in one of the corners of $a$. Without loss of generality, assume it is contained in the shaded area shown in Figure 6, then it is possible to place a new corner shape in the remaining part of $a$ as shown in Figure 6.
The above demonstrates that $k=5$ satisfies the problem's requirements.
In conclusion, $k_{\max }=5$.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leqslant 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at this time.
(Ninth China Southeast Mathematical Olympiad)
|
Let
$$
\begin{array}{l}
f(x)=a \cos x+b \cos 2 x+c \cos 3 x+d \cos 4 x . \\
\text { By } f(0)=a+b+c+d, \\
f(\pi)=-a+b-c+d, \\
f\left(\frac{\pi}{3}\right)=\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2},
\end{array}
$$
then \(a+b-c+d\)
$$
=f(0)+\frac{2}{3} f(\pi)+\frac{4}{3} f\left(\frac{\pi}{3}\right) \leqslant 3 .
$$
Equality holds if and only if \(f(0)=f(\pi)=f\left(\frac{\pi}{3}\right)=1\), i.e., \(a=1, b+d=1, c=-1\).
At this point, let \(t=\cos x(-1 \leqslant t \leqslant 1)\). Then
$$
\begin{array}{l}
f(x)-1 \\
=\cos x+b \cos 2 x-\cos 3 x+d \cos 4 x-1 \\
=t+(1-d)\left(2 t^{2}-1\right)-\left(4 t^{3}-3 t\right)+ \\
d\left(8 t^{4}-8 t^{2}+1\right)-1 \\
=2(t-1)(t+1)(2 t-1)[2 d t-(1-d)] \\
\leqslant 0 \\
\end{array}
$$
for any real number \(t \in[-1,1]\).
Thus, \(d>0\), and \(\frac{2 d}{2}=\frac{1-d}{1}\), i.e., \(d=\frac{1}{2}\).
Therefore, the maximum value of \(a+b-c+d\) is 3, and at this time,
$$
(a, b, c, d)=\left(1, \frac{1}{2},-1, \frac{1}{2}\right) \text {. }
$$
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. A moving point moves on the integer points in the first quadrant of the Cartesian coordinate system (including the integer points on the $x$-axis and $y$-axis of the first quadrant), with the movement rules being $(m, n) \rightarrow(m+1, n+1)$ or $(m, n) \rightarrow$ $(m+1, n-1)$. If the moving point starts from the origin and reaches the point $(6,2)$ after 6 steps, then there are $\qquad$ different movement paths.
|
$\begin{array}{l}\text { 17. } 9 \\ C_{6}^{2}-C_{6}^{1}=9\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 17. } 9 \\ C_{6}^{2}-C_{6}^{1}=9\end{array}$
Note: The original text is already in a mathematical format, which is universal and does not require translation. However, if you intended to have the problem statement or any surrounding text translated, please provide that context.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure $4, \triangle A B C$ has an incircle $\odot O_{1}$ that touches side $B C$ at point $D, \odot O_{2}$ is the excircle of $\triangle A B C$ inside $\angle A$. If $O_{1} B=6, O_{1} C=3, O_{1} D=2$, then $O_{1} O_{2}=$ $\qquad$
|
$$
\begin{aligned}
& \angle O_{1} B O_{2}=\angle O_{1} C O_{2}=90^{\circ} \\
\Rightarrow & O_{1} γ C γ O_{2} γ B \text { are concyclic } \\
\Rightarrow & \angle O_{1} C D=\angle O_{1} O_{2} B \\
\Rightarrow & \triangle O_{1} D C \backsim \triangle O_{1} B O_{2} \\
\Rightarrow & O_{1} O_{2}=\frac{O_{1} B \cdot O_{1} C}{O_{1} D}=\frac{3 \times 6}{2}=9 .
\end{aligned}
$$
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Among $m$ students, it is known that in any group of three, two of them know each other, and in any group of four, two of them do not know each other. Then the maximum value of $m$ is $\qquad$
|
3. 8 .
When $m=8$, the requirement is satisfied.
It only needs to prove: $m \leqslant 8$.
First, prove that the following two scenarios are impossible.
(1) If a student $A$ knows at least 6 people, by Ramsey's theorem, among these 6 people, there exist 3 people who either all know each other or all do not know each other. If it is the former, then $A$ and these 3 people form a group of 4 people who all know each other, which contradicts the given condition; if it is the latter, this contradicts the given condition that among any 3 people, there are 2 who know each other.
(2) If a student $A$ knows at most $m-5$ people, then at least 4 people do not know $A$, thus, these 4 people all know each other, which contradicts the given condition.
Secondly, when $m \geqslant 10$, either (1) or (2) must occur, which is impossible.
When $m=9$, to make (1) and (2) not occur, each student must know exactly 5 other people, thus, the number of friend pairs (pairs of people who know each other) generated by these 9 people is $\frac{9 \times 5}{2} \notin Z$, which is a contradiction.
In summary, the maximum value of $m$ is 8.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If $n$ is a positive integer greater than 1, then
$$
\begin{array}{l}
\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cos \frac{6 \pi}{n}+\cdots+\cos \frac{2 n \pi}{n} \\
=
\end{array}
$$
|
5. 0 .
$$
\sum_{k=1}^{n} \cos \frac{2 k \pi}{n}=\operatorname{Re} \sum_{k=1}^{n} \mathrm{e}^{\frac{2 k \pi i}{n}}=0 .
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 A philanthropist recruits members for his club in the following way: each member can introduce two others to join, where these two are not introduced by anyone else; and each new member can also introduce two others to join. For a member $A$, the members introduced by $A$ and the members introduced by those introduced by $A$ are collectively referred to as $A$'s "downlines." If each of the two members $B_{1}, B_{2}$ introduced by a member $A$ has at least 200 downlines, then member $A$ will receive a dinner voucher from the philanthropist as a reward at the end of the year.
Determine: If a total of 2012 people have become members, then at most how many people can enjoy the philanthropist's dinner at the end of the year?
|
Solve using reverse thinking.
For this, let $200=k$, and denote the minimum number of members when exactly $r$ people receive coupons as $f(r)$. Members who join without being introduced (or without a superior) are called "bosses", and the members they introduce are called "followers". Each boss along with all their followers and subordinates form a "gang"; each gang forms a tree diagram.
To minimize the number of members, in any gang, a boss who cannot win a prize can actually be removed. Thus, some of the followers under their control may become bosses, forming new gangs along with the subtrees they manage.
When $r=1$, the minimum number of members is $2k+3$ (at this time, only $A$ wins, and $A, B_1, B_2$ are all members, and $B_1, B_2$ each have at least $k$ subordinates), i.e., $f(1)=2k+3$.
When $r=2$, the winners are $A$ and one of $B_1, B_2$ (let it be $B_1$), and according to the above discussion, the branch containing $B_1$ (including $B_1$) has at least $2k+3$ people. Since $A$ is a winner, $B_2$ must have at least $k$ subordinates. Therefore, the number of members is at least
$$
1+(2k+3)+(k+1)=3k+5,
$$
i.e., $f(2)=3k+5$.
When $r=3$, there are two scenarios for the winners.
(1) $A, B_1, B_2$;
(2) $A, B_1$, and one of $B_1$'s subordinates (let it be $C_1$).
For (1), according to the above discussion, the branch containing $B_1$ (including $B_1$) has at least $2k+3$ people; the branch containing $B_2$ (including $B_2$) also has at least $2k+3$ people. Therefore, including $A$, the number of members is at least $4k+7$.
For (2), $B_1$'s branch has two winners, and this branch has at least $3k+5$ people; since $A$ is a winner, $B_2$ must have at least $k$ subordinates. Therefore, the number of members is at least
$$
1+(3k+5)+(k+1)=4k+7.
$$
Thus, both scenarios result in $f(3)=4k+7$.
$$
\text{By } f(2)-f(1)=k+2, f(3)-f(2)=k+2,
$$
we can guess that for each positive integer $n$,
$$
\begin{array}{l}
f(r+1)-f(r)=k+2. \\
\text{Thus, } f(r)=r(k+2)+(k+1).
\end{array}
$$
It is sufficient to prove equation (1) for the case of only one gang.
In fact, if there are $s$ gangs, each gang has $n_1, n_2, \cdots, n_s$ people, and the number of winners are $r_1, r_2, \cdots, r_s$, then
$$
\begin{array}{l}
\text{Total number of members } n=\sum_{i=1}^{1} n_{i}, \\
\text{Total number of winners } r=\sum_{i=1}^{1} r_{i}. \\
\text{Thus, } n=\sum_{i=1}^{s} n_{i} \geqslant \sum_{i=1}^{\prime} f\left(r_{i}\right) \\
=\sum_{i=1}^{s}\left[r_{i}(k+2)+(k+1)\right] \\
=(k+2) \sum_{i=1}^{s} r_{i}+s(k+1) \\
=r(k+2)+s(k+1) \\
\geqslant r(k+2)+(k+1).
\end{array}
$$
Next, we prove equation (1) using mathematical induction.
When $r=1,2,3$, it has already been established.
Assume equation (1) holds for fewer than $r$ winners.
When there are $r$ winners and the number of members in this gang has reached the minimum, the boss $A$ must be a winner. Let the two direct followers of boss $A$ be $B, C$. If the number of people in $B$'s sub-gang (including $B$ and all their followers) is $n_B$, and the number of winners among them is $r_B$, similarly define $n_C, r_c$.
According to the induction hypothesis, if $r_B>0$ (i.e., $B$'s sub-gang has winners), then the number of members in this sub-gang is
$$
n_B \geqslant r_B(k+2)+(k+1).
$$
If $r_B=0$, since $A$ is a winner, $B$ must have introduced at least $k$ people, including $B$ themselves, $B$'s sub-gang has at least $k+1$ members. Thus,
$$
n_B \geqslant k+1=r_B(k+2)+(k+1).
$$
Therefore, in any case,
$$
n_B \geqslant r_B(k+2)+(k+1).
$$
Similarly, $n_C \geqslant r_c(k+2)+(k+1)$.
Since the total number of members in $A$'s gang is $n=n_B+n_C+1$, and the total number of winners is $r=r_B+r_C+1$, we have
$$
\begin{array}{l}
n=n_B+n_C+1 \\
\geqslant\left(r_B+r_C\right)(k+2)+2(k+1)+1 \\
=r(k+2)+(k+1), \\
f(r)=r(k+2)+(k+1).
\end{array}
$$
Thus, $f(r)=r(k+2)+(k+1)$.
According to equation (1), when there are $r$ winners, the number of members $n$ satisfies $n \geqslant f(r)=r(k+2)+(k+1)$.
When $n=2012, k=200$, the above equation becomes
$$
202 r \leqslant 1811 \Rightarrow r \leqslant\left[\frac{1811}{202}\right]=8.
$$
Therefore, at most 8 people can receive the philanthropist's banquet at the end of the year.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 If the positive integer $m$ makes it true that for any set of positive numbers $a_{1}, a_{2}, a_{3}, a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$
then the minimum value of the positive integer $m$ is $\qquad$ [2]
|
Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$
Verification shows that $m=1, m=2$ do not meet the requirements.
Hence $m \geqslant 3$.
$$
\begin{array}{l}
\text { By } \frac{a_{1}^{3}+a_{2}^{3}+a_{3}^{3}}{3} \geqslant a_{1} a_{2} a_{3}, \frac{a_{1}^{3}+a_{2}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{2} a_{4}, \\
\frac{a_{1}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{3} a_{4}, \frac{a_{2}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{2} a_{3} a_{4}, \\
a_{1} a_{2} a_{3} a_{4}=1,
\end{array}
$$
we get
$$
\begin{array}{l}
a_{1}^{3}+a_{2}^{3}+a_{3}^{3}+a_{4}^{3} \\
\geqslant a_{1} a_{2} a_{3}+a_{1} a_{2} a_{4}+a_{1} a_{3} a_{4}+a_{2} a_{3} a_{4} \\
=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}} .
\end{array}
$$
Therefore, $m=3$ meets the requirements.
Thus, the smallest positive integer $m$ is 3.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { Example } 6 \text { Let } u=1+\frac{x^{3}}{3!}+\frac{x^{6}}{6!}+\cdots, \\
v=\frac{x}{1!}+\frac{x^{4}}{4!}+\frac{x^{7}}{7!}+\cdots, w=\frac{x^{2}}{2!}+\frac{x^{5}}{5!}+\frac{x^{8}}{8!}+\cdots
\end{array}
$$
Prove: $u^{3}+v^{3}+w^{3}-3 u v w=1$.
|
Let $\lambda=\mathrm{e}^{\frac{2 \pi}{3}}$ be a unit cube root.
Then $1+\lambda+\lambda^{2}=0$.
Thus $u^{3}+v^{3}+w^{3}-3 u v w$
$$
=(u+v+w)\left(u+\lambda v+\lambda^{2} w\right)\left(u+\lambda^{2} v+\lambda w\right)
$$
(By Corollary 3)
$$
=\mathrm{e}^{x} \mathrm{e}^{\lambda x} \mathrm{e}^{\lambda^{2} x}=\mathrm{e}^{0}=1 .
$$
|
1
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $x, y, z \in \mathbf{R}_{+}$, satisfying $x^{2}+y^{2}+z^{2}=1$. Then $\min \left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)=$ $\qquad$ .
|
Notice,
$$
\begin{array}{l}
\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)\left(x^{2}+y^{2}+z^{2}\right) \\
\geqslant 3 \sqrt[3]{\frac{1}{x^{2}} \cdot \frac{1}{y^{2}} \cdot \frac{1}{z^{2}}} \times 3 \sqrt[3]{x^{2} y^{2} z^{2}}=9,
\end{array}
$$
when and only when $x=y=z=\frac{\sqrt{3}}{3}$,
$$
\min \left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)=9 .
$$
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Calculate:
$$
\frac{2013^{2}+2011}{2011^{2}-2013} \times \frac{4020^{2}-8040}{2011 \times 2014-4}=
$$
$\qquad$
|
$=1.4$.
Let $a=2$ 011. Then
$$
\begin{array}{l}
\text { Original expression }=\frac{(a+2)^{2}+a}{a^{2}-a-2} \cdot \frac{(2 a-2)^{2}-4(a-1)}{a(a+3)-4} \\
=\frac{(a+1)(a+4)}{(a-2)(a+1)} \cdot \frac{4(a-1)(a-2)}{(a+4)(a-1)}=4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. A sequence of numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is constructed according to the following rule: $a_{1}=7, a_{k}=a_{k-1}^{2}$'s digit sum $+1(k=2,3, \cdots)$. For example, $a_{2}=14, a_{3}=17$, and so on.
Then $a_{2013}=$ $\qquad$ .
|
2.8.
From the problem, we know
$$
\begin{array}{l}
a_{1}=7, a_{2}=14, a_{3}=17, a_{4}=0, \\
a_{5}=5, a_{6}=8, a_{7}=11, a_{8}=5 .
\end{array}
$$
Thus, $a_{8}=a_{5}$, meaning from $a_{5}$ onwards, the sequence repeats with a period of 3.
Therefore, $a_{2013}=a_{6+3 \times 669}=8$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1$, and $a_{2 n}=a_{n}, a_{2 n+1}=a_{n}+1\left(n \in \mathbf{Z}_{+}\right)$.
Then $a_{2013}=$ . $\qquad$
|
2.9.
From the problem, we know
$$
\begin{array}{l}
a_{2013}=a_{1006}+1=a_{503}+1=a_{251}+2 \\
=a_{125}+3=a_{62}+4=a_{31}+4=a_{15}+5 \\
=a_{7}+6=a_{3}+7=a_{1}+8=9 .
\end{array}
$$
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $M$ be a moving point on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. Given points $F(1,0)$ and $P(3,1)$. Then the maximum value of $2|M F|-|M P|$ is $\qquad$.
|
2.1.
Notice that $F$ is the right focus of the ellipse, and the right directrix of the ellipse is $l: x=4$. Then $2|M F|$ is the distance from point $M$ to $l$.
Draw a perpendicular line $M A$ from point $M$ to $l$, and draw a perpendicular line $P B$ from point $P$ to $M A$, where $A$ and $B$ are the feet of the perpendiculars, respectively. Therefore,
$$
\begin{array}{l}
2|M F|-|M P|=|M A|-|M P| \\
\leqslant|M A|-|M B|=|A B|,
\end{array}
$$
where $|A B|$ is the distance from point $P$ to the right directrix, which equals 1.
Thus, when the y-coordinate of point $M$ is 1, the maximum value of $2|M F|-|M P|$ is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $x, y \in \mathbf{R}$, and $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $x+y-x y$ is $\qquad$ .
|
3. 1 .
Notice that, $x+y-x y=x(1-y)+y$.
When $y$ is fixed, the expression can only achieve its maximum value when $x$ is as large as possible; similarly, when $x$ is fixed, $y$ should also be as large as possible. Therefore, we might as well assume that $x$ and $y$ are non-negative, and $x^{2}+y^{2}=1$.
Let $x+y=t(t \in[1, \sqrt{2}])$.
From $t^{2}=x^{2}+y^{2}+2 x y=1+2 x y \Rightarrow x y=\frac{t^{2}-1}{2}$.
Thus, $x+y-x y=t-\frac{t^{2}-1}{2}=1-\frac{(t-1)^{2}}{2} \leqslant 1$, and
when $x=1, y=0$, the equality holds.
Therefore, the maximum value sought is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the function
$$
f(x)=A \cos \left(\omega x+\frac{\pi}{4} \omega\right)(A>0)
$$
is decreasing on $\left(0, \frac{\pi}{8}\right)$. Then the maximum value of $\omega$ is
|
Ni,6.8.
Assume $\omega>0$.
To make $f(x)$ a decreasing function in $\left(0, \frac{\pi}{8}\right)$, combining the image of the cosine-type function, we must have
$$
\begin{array}{l}
\frac{T}{2} \geqslant \frac{\pi}{8} \Rightarrow \frac{\pi}{\omega} \geqslant \frac{\pi}{8} \Rightarrow \omega \leqslant 8 . \\
\text { When } \omega=8 \text {, } \\
f(x)=A \cos (8 x+2 \pi)(A>0),
\end{array}
$$
Obviously, in $\left(0, \frac{\pi}{8}\right)$, it is a decreasing function, which fits.
Therefore, the maximum value of $\omega$ is 8.
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. For the real number $x$, the functions are
$$
f(x)=\sqrt{3 x^{2}+7}, g(x)=x^{2}+\frac{16}{x^{2}+1}-1,
$$
then the minimum value of the function $g(f(x))$ is . $\qquad$
|
$-, 1.8$.
From the problem, we have
$$
g(f(x))=3 x^{2}+7+\frac{16}{3 x^{2}+8}-1.
$$
Let $t=3 x^{2}+8(t \geqslant 8)$. Then
$$
h(t)=g(f(x))=t+\frac{16}{t}-2 \text{. }
$$
It is easy to see that $h(t)$ is a monotonically increasing function on the interval $[8,+\infty)$. Therefore, $h(t) \geqslant h(8)=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $a_{1}, a_{2}, \cdots, a_{10}$ and $b_{1}, b_{2}, \cdots, b_{10}$ are 20 distinct real numbers. If the equation
$$
\begin{array}{l}
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right| \\
=\left|x-b_{1}\right|+\left|x-b_{2}\right|+\cdots+\left|x-b_{10}\right|
\end{array}
$$
has a finite number of solutions, then this equation has at most $\qquad$ solutions.
|
7.9.
$$
\text { Let } \begin{aligned}
f(x)= & \left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right|- \\
& \left|x-b_{1}\right|-\left|x-b_{2}\right|-\cdots-\left|x-b_{10}\right| .
\end{aligned}
$$
Thus, by the problem statement, $f(x)=0$.
Let $c_{1}<c_{2}<\cdots<c_{20}$ be the elements of the set
$$
\left\{a_{1}, a_{2}, \cdots, a_{10}, b_{1}, b_{2}, \cdots, b_{10}\right\}
$$
arranged in increasing order, and in
$$
\left(-\infty, c_{1}\right],\left[c_{1}, c_{2}\right], \cdots,\left[c_{19}, c_{20}\right],\left[c_{20},+\infty\right)
$$
these 21 intervals, the function $f(x)$ is linear in each.
Notice that, in the interval $\left(-\infty, c_{1}\right]$,
$$
f(x)=a_{1}+a_{2}+\cdots+a_{10}-b_{1}-b_{2}-\cdots-b_{10}=m,
$$
and in the interval $\left[c_{20},+\infty\right)$, $f(x)=-m$.
Since the number of roots of the equation is finite, we have $m \neq 0$.
Moving along the number line from left to right. Initially, the coefficient of $x$ in $f(x)$ is 0. Each time we pass a $c_{i}\left(1 \leqslant i \leqslant 20, i \in \mathbf{Z}_{+}\right)$, the way one of the absolute values is removed changes, causing the coefficient of $x$ to change by $\pm 2$ (increase by 2 or decrease by 2). This indicates that the coefficient of $x$ is always even and does not change sign before it becomes 0. Therefore, the coefficient of $x$ in any two adjacent intervals is either both non-negative or both non-positive. Thus, $f(x)$ is either non-increasing or non-decreasing in such an interval union. As a result, if $f(x)=0$ has only a finite number of roots, then it has at most one root in each of the intervals $\left[c_{1}, c_{3}\right], \cdots,\left[c_{17}, c_{19}\right],\left[c_{19}, c_{20}\right]$. Furthermore, since $f\left(c_{1}\right)$ and $f\left(c_{20}\right)$ have different signs, and $f(x)$ changes sign at each root, the equation $f(x)=0$ has an odd number of roots. Therefore, it has at most nine roots.
On the other hand, it is not difficult to verify that if
$$
\begin{array}{l}
a_{1}=1, a_{2}=4, a_{3}=5, a_{4}=8, a_{5}=9, \\
a_{6}=12, a_{7}=13, a_{8}=16, a_{9}=17, a_{10}=19.5, \\
b_{1}=2, b_{2}=3, b_{3}=6, b_{4}=7, b_{5}=10, \\
b_{6}=11, b_{7}=14, b_{8}=15, b_{9}=18, b_{10}=19,
\end{array}
$$
then the equation $f(x)=0$ has exactly nine roots.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (16 points) Given the sequence $\left\{F_{n}\right\}$ satisfies
$$
\begin{array}{l}
F_{1}=F_{2}=1, \\
F_{n+2}=F_{n+1}+F_{n}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
If $F_{a} γ F_{b} γ F_{c} γ F_{d}(a<b<c<d)$ are the side lengths of a convex quadrilateral, find the value of $d-b$.
|
From the given, we know that $F_{a}+F_{b}+F_{c}>F_{d}$.
If $c \leqslant d-2$, then
$$
F_{a}+\left(F_{b}+F_{c}\right) \leqslant F_{a}+F_{d-1} \leqslant F_{d},
$$
which is a contradiction.
Therefore, $c=d-1$.
Thus, the side lengths of the quadrilateral are $F_{a} γ F_{b} γ F_{d-1} γ F_{d}$.
If $b \leqslant d-3$, then
$$
\left(F_{a}+F_{b}\right)+F_{d-1} \leqslant F_{d-2}+F_{d-1}=F_{d},
$$
which is a contradiction.
Therefore, $b=d-2$, at this point,
$$
F_{a}+\left(F_{d-2}+F_{d-1}\right)=F_{a}+F_{d}>F_{d} \text {. }
$$
Thus, the side lengths of the quadrilateral are $F_{a} γ F_{d-2} γ F_{d-1} γ F_{d}$.
Hence, $d-b=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Let the equation $x^{2}-m x-1=0$ have two real roots $\alpha, \beta (\alpha<\beta)$, and the function $f(x)=\frac{2 x-m}{x^{2}+1}$.
(1) Find the value of $\alpha f(\alpha)+\beta f(\beta)$;
(2) Determine the monotonicity of $f(x)$ in the interval $(\alpha, \beta)$, and provide a proof;
(3) If $\lambda, \mu$ are both positive real numbers, prove:
$$
\left|f\left(\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}\right)-f\left(\frac{\mu \alpha+\lambda \beta}{\lambda+\mu}\right)\right|<|\alpha-\beta| .
$$
|
Three, 13. (1) Given that $\alpha, \beta$ are the roots of the equation $x^{2}-m x-1=0$, we know
$$
\begin{array}{l}
\alpha+\beta=m, \alpha \beta=-1 . \\
\text { Then } f(\alpha)=\frac{2 \alpha-m}{\alpha^{2}+1}=\frac{2 \alpha-(\alpha+\beta)}{\alpha^{2}-\alpha \beta} \\
=\frac{\alpha-\beta}{\alpha(\alpha-\beta)}=\frac{1}{\alpha} \\
\Rightarrow \alpha f(\alpha)=1 .
\end{array}
$$
Similarly, $\beta f(\beta)=1$.
Thus, $\alpha f(\alpha)+\beta f(\beta)=2$.
(2) Note that,
$$
f^{\prime}(x)=-\frac{2\left(x^{2}-m x-1\right)}{\left(x^{2}+1\right)^{2}}=-\frac{2(x-\alpha)(x-\beta)}{\left(x^{2}+1\right)^{2}} \text {. }
$$
Therefore, when $x \in(\alpha, \beta)$, $f^{\prime}(x)>0$.
Thus, $f(x)$ is monotonically increasing on $(\alpha, \beta)$.
(3) By $\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}-\alpha=\frac{\mu(\beta-\alpha)}{\lambda+\mu}>0$,
$$
\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}-\beta=\frac{\lambda(\alpha-\beta)}{\lambda+\mu}<0,
$$
we know $\alpha<\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}<\beta$.
Hence, by (2), $f(\alpha)<f\left(\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}\right)<f(\beta)$.
Similarly, $f(\alpha)<f\left(\frac{\mu \alpha+\lambda \beta}{\lambda+\mu}\right)<f(\beta)$.
Thus, $\left|f\left(\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}\right)-f\left(\frac{\mu \alpha+\lambda \beta}{\lambda+\mu}\right)\right|<|f(\alpha)-f(\beta)|$.
By (1), we know $f(\alpha)=\frac{1}{\alpha}, f(\beta)=\frac{1}{\beta}, \alpha \beta=-1$.
Then $|f(\alpha)-f(\beta)|=\left|\frac{1}{\alpha}-\frac{1}{\beta}\right|$
$$
=\left|\frac{\beta-\alpha}{\alpha \beta}\right|=|\alpha-\beta| \text {. }
$$
Thus, $\left|f\left(\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}\right)-f\left(\frac{\mu \alpha+\lambda \beta}{\lambda+\mu}\right)\right|<|\alpha-\beta|$.
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. For the convex pentagon $A B C D E$, the side lengths are sequentially $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. It is known that a quadratic trinomial in $x$ satisfies:
When $x=a_{1}$ and $x=a_{2}+a_{3}+a_{4}+a_{5}$, the value of the quadratic trinomial is 5;
When $x=a_{1}+a_{2}$, the value of the quadratic trinomial is $p$;
When $x=a_{3}+a_{4}+a_{5}$, the value of the quadratic trinomial is $q$.
Then $p-q=$ $\qquad$
|
4. 0 .
Let the quadratic trinomial be
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {, }
$$
and let the axis of symmetry of its graph be $x=x_{0}$.
By the problem, we know
$$
f\left(a_{1}\right)=f\left(a_{2}+a_{3}+a_{4}+a_{5}\right)=5,
$$
and $\square$
$$
\begin{array}{l}
a_{1} \neq a_{2}+a_{3}+a_{4}+a_{5} . \\
\text { Hence } a_{1}+\left(a_{2}+a_{3}+a_{4}+a_{5}\right)=2 x_{0} \\
\Rightarrow\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}+a_{5}\right)=2 x_{0} \\
\Rightarrow f\left(a_{1}+a_{2}\right)=f\left(a_{3}+a_{4}+a_{5}\right) \\
\Rightarrow p=q \Rightarrow p-q=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Find the smallest integer \( n (n > 1) \), such that there exist \( n \) integers \( a_{1}, a_{2}, \cdots, a_{n} \) (allowing repetition) satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2013 .
$$
12. (20 points) Let positive integers \( a, b, c, d \) satisfy
$$
a^{2}=c(d+13), b^{2}=c(d-13) \text {. }
$$
Find all possible values of \( d \).
|
11. Since $a_{1} a_{2} \cdots a_{n}=2013$, it follows that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. From $a_{1}+a_{2}+\cdots+a_{n}=2013$ being odd, we know that $n$ is odd (otherwise, the sum of an even number of odd numbers should be even, which is a contradiction).
If $n=3$, then $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=2013$.
Assume without loss of generality that $a_{1} \geqslant a_{2} \geqslant a_{3}$. Then
$a_{1} \geqslant \frac{a_{1}+a_{2}+a_{3}}{3}=\frac{2013}{3}=671$
$\Rightarrow a_{2} a_{3}=\frac{2013}{a_{1}} \leqslant \frac{2013}{671}=3$.
If $a_{1}=671$, then $a_{2} a_{3}=3$. Thus, $a_{2}+a_{3} \leqslant 4$.
Hence $a_{1}+a_{2}+a_{3} \leqslant 671+4=675671$, at this point, $a_{2} a_{3}<3$.
Therefore, $a_{2} a_{3}=1, a_{1}=2013$, which contradicts $a_{1}+a_{2}+a_{3}=2013$.
When $n=5$, there exist $-1, -1, 1, 1, 2013$ satisfying
$(-1)+(-1)+1+1+2013$
$=(-1) \times(-1) \times 1 \times 1 \times 2013=2013$.
Thus, the minimum value of $n$ is 5.
12. From the given, we have $\frac{a^{2}}{b^{2}}=\frac{c(d+13)}{c(d-13)}=\frac{d+13}{d-13}$.
Let $(a, b)=t, a_{1}=\frac{a}{t}, b_{1}=\frac{b}{t}$. Then
$\frac{a_{1}^{2}}{b_{1}^{2}}=\frac{a^{2}}{b^{2}}=\frac{d+13}{d-13}$.
Since $\left(a_{1}, b_{1}\right)=1$, we know $\left(a_{1}^{2}, b_{1}^{2}\right)=1$, meaning the left side of equation (1) is a reduced fraction. Assume without loss of generality that
$d+13=k a_{1}^{2}, d-13=k b_{1}^{2}\left(k \in \mathbf{Z}_{+}\right)$.
Thus, $k\left(a_{1}^{2}-b_{1}^{2}\right)=26$, i.e., $a_{1}^{2}-b_{1}^{2}=\frac{26}{k}$ is a positive divisor of 26 and can be decomposed into the product of two unequal positive integers $a_{1}+b_{1}$ and $a_{1}-b_{1}$ with the same parity, hence $\frac{26}{k}=13$.
Then $a_{1}+b_{1}=13, a_{1}-b_{1}=1, k=2$. Thus, $a_{1}=7$.
Furthermore, $d=k a_{1}^{2}-13=2 \times 7^{2}-13=85$.
Upon verification, when $(a, b, c, d)=(14,12,2,85)$, it satisfies the conditions. Therefore, the value of $d$ is 85.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Write all positive integers in ascending order in a row. Then the 2013th digit from left to right is
|
$-1.7$
All single-digit numbers occupy 9 positions, all two-digit numbers occupy $2 \times 90=180$ positions, and next come the three-digit numbers in sequence. Since $2013-9-180=1824$, and $\frac{1824}{3}=608$, because $608+99=707$, the 2013th digit is the last digit of the three-digit number 707, which is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Equation
$$
\sin \pi x=\left[\frac{x}{2}-\left[\frac{x}{2}\right]+\frac{1}{2}\right]
$$
The sum of all real roots of the equation in the interval $[0,2 \pi]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
7.12. Let $\left\{\frac{x}{2}\right\}=\frac{x}{2}-\left[\frac{x}{2}\right]$. Then for any real number $x$, we have $0 \leqslant\left\{\frac{x}{2}\right\}<1$.
Thus, the original equation becomes
$$
\sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right] \text {. }
$$
(1) If $0 \leqslant\left\{\frac{x}{2}\right\}<\frac{1}{2}$, then
$$
\begin{array}{l}
\Rightarrow \sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right]=0 \\
\Rightarrow \pi x=k \pi(k \in \mathbf{Z}) \\
\Rightarrow x=k(k \in \mathbf{Z}) .
\end{array}
$$
Combining $x \in[0,2 \pi]$, we know $x=0,1, \cdots, 6$.
Upon inspection, $x=0,2,4,6$ meet the requirements of the problem.
$$
\begin{array}{l}
\text { (2) If } \frac{1}{2} \leqslant\left\{\frac{x}{2}\right\}<1 \\
\Rightarrow \sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right]=1 \\
\Rightarrow \pi x=2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z}) \\
\Rightarrow x=2 k+\frac{1}{2}(k \in \mathbf{Z}) .
\end{array}
$$
In this case, $\left\{\frac{x}{2}\right\}=0$, which does not meet the requirements.
Therefore, the values of $x$ that satisfy the conditions are $0,2,4,6$, and their sum is 12.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $f(x)$ is an increasing function on $\mathbf{R}$, and for any $x \in$ $\mathbf{R}$, we have
$$
f\left(f(x)-3^{x}\right)=4 .
$$
Then $f(2)=$
|
8. 10 .
From the problem, we know that $f(x)-3^{x}$ is a constant. Let's assume $f(x)-3^{x}=m$.
Then $f(m)=4, f(x)=3^{x}+m$.
Therefore, $3^{m}+m=4 \Rightarrow 3^{m}+m-4=0$.
It is easy to see that the equation $3^{m}+m-4=0$ has a unique solution $m=1$.
Thus, $f(x)=3^{x}+1$,
and hence, $f(2)=3^{2}+1=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that the elements of set $A$ are all integers, the smallest is 1, and the largest is 200, and except for 1, every number in $A$ is equal to the sum of two numbers (which may be the same) in $A$. Then the minimum value of $|A|$ is $\qquad$ ( $|A|$ represents the number of elements in set $A$).
|
9. 10 .
It is easy to know that the set
$$
A=\{1,2,3,5,10,20,40,80,160,200\}
$$
meets the requirements, at this time, $|A|=10$.
Next, we will show that $|A|=9$ does not meet the requirements.
Assume the set
$$
A=\left\{1, x_{1}, x_{2}, \cdots, x_{7}, 200\right\},
$$
where $x_{1}<x_{2}<\cdots<x_{7}$ meets the requirements.
$$
\begin{array}{l}
\text { Then } x_{1}=1+1=2, x_{2} \leqslant 2+2=4, x_{3} \leqslant 8, \\
x_{4} \leqslant 16, x_{5} \leqslant 32, x_{6} \leqslant 64, x_{7} \leqslant 128 . \\
\text { By } x_{6}+x_{7} \leqslant 64+128=192<200 \text {, we know that } \\
200=x_{7}+x_{7} \Rightarrow x_{7}=100 ;
\end{array}
$$
Similarly, by $x_{5}+x_{6} \leqslant 32+64=96<100$, we know that
$$
x_{7}=100=x_{6}+x_{6} \Rightarrow x_{6}=50 \text {; }
$$
By $x_{4}+x_{5} \leqslant 16+32=48<50$, we know that
$$
x_{6}=50=x_{5}+x_{5} \Rightarrow x_{5}=25 \text {; }
$$
By $x_{3}+x_{4} \leqslant 8+16=24<25$, we know that
$$
x_{5}=25=x_{4}+x_{4} \Rightarrow x_{4}=\frac{25}{2} \text {, }
$$
which contradicts the fact that $x_{4}$ is an integer.
Thus, $|A|=9$ does not meet the requirements, i.e., $|A| \neq 9$.
Similarly, $|A| \leqslant 8$ does not meet the requirements.
In summary, the minimum value of $|A|$ is 10.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given the set
$$
P=\left\{x \mid x=7^{3}+a \times 7^{2}+b \times 7+c, a γ b γ c\right. \text { are positive integers not }
$$
exceeding 6 $\}$.
If $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ elements in set $P$ that form an arithmetic sequence, find the maximum value of $n$.
|
15. (1) Clearly,
$$
\begin{array}{l}
7^{3}+7^{2}+7+1, 7^{3}+7^{2}+7+2, \\
7^{3}+7^{2}+7+3, 7^{3}+7^{2}+7+4, \\
7^{3}+7^{2}+7+5, 7^{3}+7^{2}+7+6
\end{array}
$$
These six numbers are in the set $P$ and form an arithmetic sequence.
(2) Prove by contradiction: Any seven different numbers in the set $P$ cannot form an arithmetic sequence.
Let $x_{1}, x_{2}, \cdots, x_{7}$ be seven different elements in the set $P$ that form an arithmetic sequence, with a common difference of $d (d>0)$.
By the properties of the elements in the set $P$, we know that no element in the set $P$ is a multiple of 7.
Thus, by the pigeonhole principle, among the seven numbers $x_{1}, x_{2}, \cdots, x_{7}$, there exist two numbers that have the same remainder when divided by 7, and their difference is divisible by 7.
Without loss of generality, assume $x_{i}-x_{j} (i, j \in \{1,2, \cdots, 7\}, i<j)$ is divisible by 7. Then
$$
7 \mid (j-i) d \Rightarrow 7 \mid d \text{. }
$$
Let $d=7 m (m \in \mathbf{Z}_{+})$, and suppose
$$
x_{1}=7^{3}+a_{1} \times 7^{2}+a_{2} \times 7+a_{3},
$$
where $a_{1}, a_{2}, a_{3}$ are positive integers not exceeding 6.
Then $x_{i}=7^{3}+a_{1} \times 7^{2}+a_{2} \times 7+a_{3}+7(i-1) m$, where $i=2,3, \cdots, 7$.
$$
\begin{array}{l}
\text{By } x_{7} \leqslant 7^{3}+6 \times 7^{2}+6 \times 7+6, \\
x_{7} \geqslant 7^{3}+1 \times 7^{2}+1 \times 7+1+7 \times(7-1) m,
\end{array}
$$
we know $1 \leqslant m \leqslant 6$, i.e., the common difference $d$ can only be
$$
7 \times 1, 7 \times 2, \cdots, 7 \times 6 \text{. }
$$
Since $1 \leqslant m \leqslant 6$, and $(7, m)=1$, the remainders of $m, 2 m, \cdots, 6 m$ when divided by 7 are all different, and are one of 1, 2, $\cdots, 6$.
Therefore, there exists $k \in \{1,2, \cdots, 6\}$ such that $a_{2}+k m$ is divisible by 7.
Let $a_{2}+k m=7 t (t \in \mathbf{Z}_{+})$. Then
$$
\begin{array}{l}
x_{k+1}=7^{3}+a_{1} \times 7^{2}+a_{2} \times 7+a_{3}+7 k m \\
=7^{3}+a_{1} \times 7^{2}+(a_{2}+k m) \times 7+a_{3} \\
=7^{3}+(a_{1}+t) \times 7^{2}+a_{3} .
\end{array}
$$
Thus, the coefficient of 7 (i.e., the second digit from the left) in the base-7 representation of $x_{k+1}$ is 0, which contradicts $x_{k+1} \in P$.
Therefore, any seven different numbers in the set $P$ cannot form an arithmetic sequence.
Hence, the maximum value of $n$ is 6.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $A=\{2,4, \cdots, 2014\}, B$ be any non-empty subset of $A$, and $a_{i} γ a_{j}$ be any two elements in set $B$. There is exactly one isosceles triangle with $a_{i} γ a_{j}$ as side lengths. Then the maximum number of elements in set $B$ is $\qquad$
|
7. 10 .
By symmetry, without loss of generality, assume $a_{i}<a_{j}$. Then there must exist an isosceles triangle with $a_{j}$ as the waist and $a_{i}$ as the base, and there is only one isosceles triangle with $a_{i}$ and $a_{j}$ as side lengths.
Thus, $a_{i}+a_{i} \leqslant a_{j} \Rightarrow a_{j} \geqslant 2 a_{i}$.
This indicates that, when the elements of set $B$ are arranged in ascending order, each subsequent element is at least twice the previous one.
Therefore, when the number of elements is maximized, the set
$$
B=\{2,4,8,16,32,64,128,256,512,1024\} \text {. }
$$
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 2, given that $D$ is the intersection of the tangents to the circumcircle $\odot O$ of $\triangle A B C$ at points $A$ and $B$, the circumcircle of $\triangle A B D$ intersects line $A C$ and segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. If $\frac{B C}{B F}=2$, find $\frac{B G}{G E}$.
(2010-2011, Hungarian Mathematical Olympiad)
|
Let $C D$ intersect $A B$ at point $M$.
If the center $O$ is not on the line segment $B C$, from $\frac{B C}{B F}=2$, we know that $F$ is the midpoint of $B C$.
Then, by the perpendicular diameter theorem,
$$
F O \perp B C \Rightarrow \angle O F B=90^{\circ} \text {. }
$$
Since $D A$ and $D B$ are tangent to $\odot O$, we have
$$
\angle O A D=\angle O B D=90^{\circ} \text {. }
$$
Thus, $O, A, B, D$ are concyclic.
Since $F, A, B, D$ are concyclic, then $F, O, A, B, D$ are concyclic. Therefore,
$$
\angle O A B=\angle O F B=90^{\circ} \text {. }
$$
This contradicts $\angle O A D=90^{\circ}$.
Therefore, point $F$ coincides with $O$.
Next, consider the ratio of some segments
$$
\begin{aligned}
\frac{A M}{M B} & =\frac{S_{\triangle A C D}}{S_{\triangle B C D}}=\frac{\frac{1}{2} A C \cdot A D \sin \angle C A D}{\frac{1}{2} B C \cdot B D \sin \angle C B D} \\
= & \frac{A C}{B C} \cdot \frac{A D}{B D} \cdot \frac{\sin \left(90^{\circ}+\angle C A F\right)}{1} \\
= & \frac{A C}{B C} \cdot \frac{A D}{B D} \cos \angle C A F .
\end{aligned}
$$
Since $\angle C A F=\angle A C F=\angle B C A$, by the tangent length theorem, we know $A D=B D$.
Thus, $\frac{A M}{B M}=\frac{A C}{B C} \cos \angle B C A=\left(\frac{A C}{B C}\right)^{2}$.
Applying Menelaus' theorem to $\triangle B A E$ and the transversal $G M C$, we get
$$
\frac{B G}{G E} \cdot \frac{E C}{C A} \cdot \frac{A M}{M B}=1 \text {. }
$$
Therefore, $\frac{B G}{G E}=\frac{C A}{C E} \cdot \frac{M B}{M A}=\left(\frac{B C}{A C}\right)^{2} \frac{A C}{C E}=\frac{B C^{2}}{A C \cdot C E}$.
By the secant theorem and $F$ being the midpoint of $B C$, we have
$$
A C \cdot C E=C F \cdot C B=\frac{1}{2} B C^{2} \text {. }
$$
Therefore, $\frac{B G}{G E}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In $\triangle A B C$, it is known that $a, b, c$ are the sides opposite to $\angle A$, $\angle B$, and $\angle C$ respectively, and $a c+c^{2}=b^{2}-a^{2}$. If the longest side of $\triangle A B C$ is $\sqrt{7}$, and $\sin C=2 \sin A$, then the length of the shortest side of $\triangle A B C$ is $\qquad$.
|
6. 1 .
From $a c+c^{2}=b^{2}-a^{2}$
$\Rightarrow \cos B=-\frac{1}{2} \Rightarrow \angle B=\frac{2 \pi}{3}$.
Thus, the longest side is $b$.
Also, $\sin C=2 \sin A \Rightarrow c=2 a$.
Therefore, $a$ is the shortest side.
By the cosine rule,
$$
(\sqrt{7})^{2}=a^{2}+4 a^{2}-2 a \times 2 a \times\left(-\frac{1}{2}\right) \text {. }
$$
Solving, we get $a=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has one focus at $F_{1}(-\sqrt{3}, 0)$, and passes through the point $H\left(\sqrt{3}, \frac{1}{2}\right)$. Let the upper and lower vertices of the ellipse $E$ be $A_{1}$ and $A_{2}$, respectively, and let $P$ be any point on the ellipse different from $A_{1}$ and $A_{2}$. The lines $P A_{1}$ and $P A_{2}$ intersect the $x$-axis at points $M$ and $N$, respectively. If the line $O T$ is tangent to the circle $\odot G$ passing through points $M$ and $N$, with the point of tangency being $T$, prove that the length of segment $O T$ is a constant, and find this constant value.
|
12. From the problem, we have
$$
a^{2}-b^{2}=3, \frac{3}{a^{2}}+\frac{1}{4 b^{2}}=1 \text {. }
$$
Solving, we get $a^{2}=4, b^{2}=1$.
Thus, the equation of the ellipse $E$ is $\frac{x^{2}}{4}+y^{2}=1$.
From this, we know the points $A_{1}(0,1), A_{2}(0,-1)$. Let point $P\left(x_{0}, y_{0}\right)$.
Then $l_{P A_{1}}: y-1=\frac{y_{0}-1}{x_{0}} x$,
$$
l_{P A_{1}}: y+1=\frac{y_{0}+1}{x_{0}} x \text {. }
$$
Let $y=0$, we get
$$
x_{N}=\frac{-x_{0}}{y_{0}-1}, x_{M}=\frac{x_{0}}{y_{0}+1} \text {. }
$$
Thus, $|O M||O N|=\left|\frac{x_{0}}{y_{0}+1} \cdot \frac{-x_{0}}{y_{0}-1}\right|=\left|\frac{x_{0}^{2}}{y_{0}^{2}-1}\right|$.
Also, $\frac{x_{0}^{2}}{4}+y_{0}^{2}=1$, so, $x_{0}^{2}=4\left(1-y_{0}^{2}\right)$.
Therefore, $|O M||O N|=\left|\frac{x_{0}^{2}}{y_{0}^{2}-1}\right|=4$.
By the secant-tangent theorem, we have
$$
O T^{2}=|O M||O N|=4 \Rightarrow|O T|=2 .
$$
|
2
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given that $a, b, c$ are positive real numbers. Prove:
$$
\frac{\sqrt{a^{2}+3 b c}}{a}+\frac{\sqrt{b^{2}+3 a c}}{b}+\frac{\sqrt{c^{2}+3 a b}}{c} \geqslant 6 \text {. }
$$
|
15. Notice,
$$
\begin{array}{l}
\frac{\sqrt{a^{2}+3 b c}}{a}=\frac{\sqrt{a^{2}+b c+b c+b c}}{a} \\
\geqslant \frac{\sqrt{4 \sqrt[4]{a^{2} b^{3} c^{3}}}}{a}=\frac{2 \sqrt[8]{a^{2} b^{3} c^{3}}}{a} .
\end{array}
$$
Similarly, $\frac{\sqrt{b^{2}+3 a c}}{b} \geqslant \frac{2 \sqrt[8]{a^{3} b^{2} c^{3}}}{b}$,
$$
\begin{array}{l}
\frac{\sqrt{c^{2}+3 a b}}{c} \geqslant \frac{2 \sqrt[8]{a^{3} b^{3} c^{2}}}{c} . \\
\text { Therefore, } \frac{\sqrt{a^{2}+3 b c}}{a}+\frac{\sqrt{b^{2}+3 a c}}{b}+\frac{\sqrt{c^{2}+3 a b}}{c} \\
\geqslant \frac{2 \sqrt[8]{a^{2} b^{3} c^{3}}}{a}+\frac{2 \sqrt[8]{a^{3} b^{2} c^{3}}}{b}+\frac{2 \sqrt[8]{a^{3} b^{3} c^{2}}}{c} \\
\geqslant 2 \times 3 \sqrt[3]{\frac{\sqrt[8]{a^{8} b^{8} c^{8}}}{a b c}}=6 .
\end{array}
$$
|
6
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $a, b, c \in \mathbf{R}_{+}$, and
$$
a+b+c=12, a b+b c+c a=45 \text{. }
$$
Then $\min \max \{a, b, c\}=$ $\qquad$
|
6. 5 .
Let $a=\max \{a, b, c\}$.
From $a+b+c=12$, we get $a \geqslant 4$.
$$
\begin{array}{l}
\text { By }(a-b)(a-c) \geqslant 0 \\
\Rightarrow a^{2}-a(12-a)+b c \geqslant 0 \\
\Rightarrow b c \geqslant 12 a-2 a^{2} . \\
\text { Also } 45=a b+b c+c a=b c+a(12-a) \\
\geqslant 12 a-2 a^{2}+a(12-a),
\end{array}
$$
Then $(a-5)(a-3) \geqslant 0$.
Thus, $a \geqslant 5$.
When $a=b=5, c=2$, the equality holds.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The unit digit of $\left[\frac{10^{10000}}{10^{100}+9}\right]$ is
|
8. 1 .
Notice that,
$$
\frac{10^{10000}}{10^{100}+9}=\frac{\left(10^{100}\right)^{100}-3^{200}}{10^{100}+9}+\frac{3^{200}}{10^{100}+9} \text {. }
$$
And $\left(10^{100}\right)^{100}-3^{200}=\left[\left(10^{100}\right)^{2}\right]^{50}-\left(9^{2}\right)^{50}$, so $\left[\left(10^{100}\right)^{2}-9^{2}\right] \mid\left(10^{10000}-9^{200}\right)$.
Since $\left(10^{100}+9\right) \mid \left[\left(10^{100}\right)^{2}-9^{2}\right]$, therefore, $\frac{10^{10000}-3^{200}}{10^{100}+9}$ is an integer.
By $\frac{3^{200}}{10^{100}+9}=\frac{9^{100}}{10^{100}+9}<1$, we know
$$
\begin{array}{l}
{\left[\frac{10^{10000}}{10^{100}+9}\right]=\frac{10^{10000}-3^{200}}{10^{100}+9}} \\
=\frac{10^{10000}-9^{100}}{10^{100}+9}=\frac{10^{10000}-81^{50}}{10^{100}+9},
\end{array}
$$
where the unit digit of the denominator is 9, and the unit digit of the numerator is 9, so the unit digit of the quotient is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Given $\odot O: x^{2}+y^{2}=4$, circle $M$ :
$$
(x-5 \cos \theta)^{2}+(y-5 \sin \theta)^{2}=1(\theta \in \mathbf{R}) \text {, }
$$
Through any point $P$ on circle $M$, draw two tangents $P E$ and $P F$ to $\odot O$, with the points of tangency being $E$ and $F$. Try to find the minimum value of $\overrightarrow{P E} \cdot \overrightarrow{P F}$.
|
9. The center of circle $M$ is on the circle $x^{2}+y^{2}=25$.
Let $|P E|=|P F|=d$.
In the right triangle $\triangle P E O$, it is easy to see that
$4 \leqslant|P O| \leqslant 6,|O E|=2$.
Thus, $2 \sqrt{3} \leqslant d \leqslant 4 \sqrt{2}$.
Also, $\overrightarrow{P E} \cdot \overrightarrow{P F}=|\overrightarrow{P E}||\overrightarrow{P F}| \cos \angle E P F$
$=d^{2} \cos \angle E P F$.
Let $\angle O P E=\theta$. Then,
$\tan \theta=\frac{|O E|}{|P E|}=\frac{2}{d}$.
Thus, $\cos \angle E P F=\cos 2 \theta$
$=\cos ^{2} \theta-\sin ^{2} \theta$
$=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{d^{2}-4}{d^{2}+4}$.
Therefore, $\overrightarrow{P E} \cdot \overrightarrow{P F}=d^{2} \cdot \frac{d^{2}-4}{d^{2}+4}$
$=\left(d^{2}+4\right)+\frac{32}{d^{2}+4}-12$.
Let $d^{2}+4=x$. Then $x \in[16,36]$.
Using the function $f(x)=x+\frac{32}{x}$ which is monotonically increasing in the interval $[16,36]$, we know that when $d^{2}+4=16$,
$$
(\overrightarrow{P E} \cdot \overrightarrow{P F})_{\min }=6
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let
$$
\begin{array}{l}
f(x)=a_{1} x^{2013}+a_{2} x^{2012}+\cdots+a_{2013} x+a_{2014} \\
=x^{13}\left(x^{10}+x^{2}+x\right)^{2000}, \\
b_{0}=1, b_{1}=2, b_{n+2}+b_{n}=b_{n+1}(n \in \mathbf{N}) .
\end{array}
$$
Find the value of $\sum_{i=1}^{2013} a_{i} b_{i}$.
|
11. From the recurrence relation of $\left\{b_{n}\right\}$, we have
$$
\begin{array}{l}
b_{2}=1, b_{3}=-1, b_{4}=-2, b_{5}=-1, \\
b_{6}=1=b_{0}, b_{7}=2=b_{1} .
\end{array}
$$
Thus, $b_{6 k}=1, b_{6 k+1}=2, b_{6 k+2}=1, b_{6 k+3}=-1$, $b_{6 k+4}=-2, b_{6 k+5}=-1(k \in \mathbf{N})$.
Let $\lambda=\frac{1}{2}+\frac{\sqrt{3}}{2}i$. Then
$$
\lambda^{6}=1, \lambda^{10}=-\lambda \text{. }
$$
Therefore, $f(\lambda)=\lambda^{13}\left(\lambda^{10}+\lambda^{2}+\lambda\right)^{200}=\lambda^{13}\left(\lambda^{2}\right)^{200}$
$$
=\lambda^{413}=\frac{1}{2}-\frac{\sqrt{3}}{2}i.
$$
Thus, $\operatorname{Re} f(\lambda)=\frac{1}{2}$.
From $f(\lambda)=\sum_{i=1}^{2014} a_{i} \lambda^{2014-i}$, we know
$\operatorname{Re} f(\lambda)$
$$
=-a_{1}-\frac{1}{2} a_{2}+\frac{1}{2} a_{3}+a_{4}+\frac{1}{2} a_{5}-\frac{1}{2} a_{6} \cdots+\frac{1}{2} a_{2013}+a_{2014}.
$$
Since $a_{2014}=f(0)=0$, we have
$\operatorname{Re} f(\lambda)=-\frac{1}{2} \sum_{i=1}^{2013} a_{i} b_{i}=\frac{1}{2}$.
Thus, $\sum_{i=1}^{2013} a_{i} b_{i}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If the complex number $x$ satisfies $x+\frac{1}{x}=-1$, then $x^{2013}+\frac{1}{x^{2013}}=$ $\qquad$ .
|
8. 2 .
Given that $x^{2}+x+1=0$.
Since the discriminant $\Delta=-3<0$, $x$ is a complex number.
Also, $x^{3}-1=(x-1)\left(x^{2}+x+1\right)=0$
$\Rightarrow x^{3}=1$.
Therefore, $x^{2013}+\frac{1}{x^{2013}}=\left(x^{3}\right)^{671}+\frac{1}{\left(x^{3}\right)^{671}}=1+1=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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