problem
stringlengths 15
4.7k
| solution
stringlengths 2
11.9k
| answer
stringclasses 51
values | problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
8. The constant term in the expansion of $\left(x^{2}+x-\frac{1}{x}\right)^{6}$ is $\qquad$ (answer with a specific number).
|
8. -5 .
From the conditions, the constant term is
$$
C_{6}^{2}(-1)^{4}+C_{6}^{3}(-1)^{3}=-5 .
$$
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let the sequence $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, satisfying $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$. Then the value of $S_{20}$ is $\qquad$.
|
9. 0 .
From the condition, we know $a_{1}=\frac{\left(a_{1}+1\right)^{2}}{4}$, solving this gives $a_{1}=1$.
When $n \geqslant 2$, from $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$, we know
$$
S_{n-1}=\frac{\left(a_{n-1}+1\right)^{2}}{4} \text {. }
$$
Then $a_{n}=\frac{1}{4}\left(a_{n}+1\right)^{2}-\frac{1}{4}\left(a_{n-1}+1\right)^{2}$
$$
\Rightarrow\left(a_{n}-1\right)^{2}=\left(a_{n-1}+1\right)^{2}
$$
$\Rightarrow a_{n}=a_{n-1}+2$ (discard) or $a_{n}=-a_{n-1}$.
Thus, $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and the common ratio -1.
$$
\text { Hence } S_{20}=\frac{1 \times\left[1-(-1)^{20}\right]}{1-(-1)}=0 \text {. }
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Let the function $f(x)=\sin x+\sqrt{3} \cos x+1$.
(1) Find the maximum and minimum values of the function $f(x)$ on $\left[0, \frac{\pi}{2}\right]$;
(2) If real numbers $a$, $b$, and $c$ satisfy
$$
a f(x)+b f(x-c)=1
$$
for any $x \in \mathbf{R}$, find the value of $\frac{b \cos c}{a}$.
|
Three, 13. (1) From the given conditions,
$$
f(x)=2 \sin \left(x+\frac{\pi}{3}\right)+1 \text {. }
$$
From $0 \leqslant x \leqslant \frac{\pi}{2} \Rightarrow \frac{\pi}{3} \leqslant x+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$.
Thus, $\frac{1}{2} \leqslant \sin \left(x+\frac{\pi}{3}\right) \leqslant 1$.
Therefore, when $x=\frac{\pi}{2}$, $f(x)$ has a minimum value of 2;
when $x=\frac{\pi}{6}$, $f(x)$ has a maximum value of 3.
(2) From the given conditions, for any $x \in \mathbf{R}$,
$$
\begin{array}{l}
2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}-c\right)+a+b=1 \\
\Rightarrow 2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}\right) \cdot \cos c- \\
2 b \cos \left(x+\frac{\pi}{3}\right) \cdot \sin c+(a+b-1)=0 \\
\Rightarrow 2(a+b \cos c) \cdot \sin \left(x+\frac{\pi}{3}\right)- \\
2 b \sin c \cdot \cos \left(x+\frac{\pi}{3}\right)+(a+b-1)=0 . \\
\text { Hence }\left\{\begin{array}{l}
a+b \cos c=0, \\
\sin c=0, \\
a+b-1=0 .
\end{array}\right.
\end{array}
$$
From $b \sin c=0$, we know $b=0$ or $\sin c=0$.
If $b=0$, then from $a+b \cos c=0$, we know $a=0$, which contradicts $a+b-1=0$.
If $\sin c=0$, then $\cos c=1$ (discard), $\cos c=-1$.
Solving gives $a=b=\frac{1}{2}, c=(2 k+1) \pi$.
Thus, $\frac{b \cos c}{a}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $\triangle A B C$ is an isosceles right triangle, $\angle A$ $=90^{\circ}$, and $\overrightarrow{A B}=a+b, \overrightarrow{A C}=a-b$.
If $a=(\cos \theta, \sin \theta)(\theta \in \mathbf{R})$, then $S_{\triangle A B C}$ $=$ . $\qquad$
|
4. 1.
From the problem, we know that $A B \perp A C,|A B|=|A C|$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
(a+b) \cdot(a-b)=0, \\
|a+b|=|a-b|
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
|a|=|b|, \\
a \cdot b=0 .
\end{array}\right.
\end{array}
$$
Since $|a|=1$, we have $|b|=1$.
According to the geometric meaning of vector addition and subtraction,
$$
|a+b|=|a-b|=\sqrt{2} \text {. }
$$
Therefore, $S_{\triangle A B C}=\frac{1}{2}|a+b| \cdot|a-b|=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In a regular tetrahedron $ABCD$, $AO \perp$ plane $BCD$, with the foot of the perpendicular being $O$. Let $M$ be a point on the line segment $AO$ such that $\angle BMC=90^{\circ}$. Then $\frac{AM}{MO}=$ $\qquad$
|
5. 1.
As shown in Figure 3, connect $O B$. Let the edge length of the regular tetrahedron $A B C D$ be $a$. Then
$$
\begin{array}{c}
O B=\frac{\sqrt{3}}{3} a, \\
M B=\frac{\sqrt{2}}{2} a . \\
\text { Therefore, } M O=\sqrt{M B^{2}-O B^{2}} \\
=\frac{\sqrt{6}}{6} a=\frac{1}{2} A O=A M . \\
\text { Hence, } \frac{A M}{M O}=1 .
\end{array}
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $p$ and $q$ be two different prime numbers. Then the remainder when $p^{q-1}+q^{p-1}$ is divided by $p q$ is $\qquad$
|
8. 1 .
Since $p$ and $q$ are different prime numbers, by Fermat's Little Theorem, we have
$$
p^{q-1} \equiv 1(\bmod q) \text {. }
$$
Also, $q^{p-1} \equiv 0(\bmod q)$, thus
$$
p^{q-1}+q^{p-1} \equiv(\bmod q) \text {. }
$$
Similarly, $p^{q-1}+q^{p-1} \equiv 1(\bmod p)$.
Therefore, $p^{q-1}+q^{p-1} \equiv 1(\bmod p q)$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 4, in the "dart-shaped" quadrilateral $ABCD$, $AB=4\sqrt{3}$, $BC=8$, $\angle A=\angle B=\angle C=30^{\circ}$. Then the distance from point $D$ to $AB$ is $\qquad$
|
3. 1.
As shown in Figure 9, extend $A D$ to intersect $B C$ at point $E$. Then $A E=B E$.
Draw $E F \perp A B$ at point $F$.
It is easy to see that $A F=B F=2 \sqrt{3}$,
$$
E F=2, A E=4 \text {. }
$$
Thus, $C E=4$.
Also, $\angle A D C=90^{\circ}$
$\Rightarrow D E=2$
$\Rightarrow D$ is the midpoint of $A E$
$\Rightarrow$ The distance from $D$ to $A B$ is equal to $\frac{1}{2} E F=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Given that $f(x)$ is a function defined on the set of real numbers $\mathbf{R}$, $f(0)=2$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(5+2 x)=f(-5-4 x), \\
f(3 x-2)=f(5-6 x) .
\end{array}
$$
Find the value of $f(2012)$.
|
In equation (1), let $x=\frac{1}{2} y-\frac{3}{2}(y \in \mathbf{R})$, we get $f(2+y)=f(1-2 y)$.
In equation (2), let $x=\frac{1}{3} y+\frac{2}{3}(y \in \mathbf{R})$, we get $f(y)=f(1-2 y)$.
Thus, $f(2+y)=f(y)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2012)=f(2 \times 1006+0)=f(0)=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let $x, y$ be real numbers. Then the minimum value of the algebraic expression
$$
2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5
$$
is $\qquad$ [1]
(2005, National Junior High School Mathematics League Wuhan CASIO Cup Selection Competition)
|
$$
\begin{array}{l}
\text { Original expression }= x^{2}+4 x y+4 y^{2}+x^{2}-4 x+4+ \\
y^{2}+2 y+1-10 \\
=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10 .
\end{array}
$$
Therefore, when $x=2, y=-1$, the minimum value of the required algebraic expression is -10.
|
-10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$
has a maximum value of $g(m)$, find the minimum value of $g(m)$.
(2011, National High School Mathematics League Sichuan Province Preliminary Contest)
|
First, use the discriminant method to find $g(m)$.
Notice that the original function is $y-x=\sqrt{100-m x}$. Squaring both sides and rearranging, we get
$$
x^{2}+(m-2 y) x+y^{2}-100=0 \text {. }
$$
By $\Delta \geqslant 0$, we have $y \leqslant \frac{m}{4}+\frac{100}{m}$.
Thus, $g(m)=\frac{m}{4}+\frac{100}{m}$
$$
\geqslant 2 \sqrt{\frac{m}{4} \times \frac{100}{m}}=10 \text {. }
$$
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given that $a$ and $b$ are real numbers, and
$$
a^{2}+a b+b^{2}=3 \text {. }
$$
If the maximum value of $a^{2}-a b+b^{2}$ is $m$, and the minimum value is $n$, find the value of $m+n$. ${ }^{[3]}$
$(2008$, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
|
Let $a^{2}-a b+b^{2}=k$.
From $\left\{\begin{array}{l}a^{2}+a b+b^{2}=3 \\ a^{2}-a b+b^{2}=k,\end{array}\right.$, we get $a b=\frac{3-k}{2}$.
Thus, $(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b$
$=3+\frac{3-k}{2}=\frac{9-k}{2}$.
Since $(a+b)^{2} \geqslant 0$, then $\frac{9-k}{2} \geqslant 0$, which means $k \leqslant 9$.
Therefore, $a+b= \pm \sqrt{\frac{9-k}{2}}, a b=\frac{3-k}{2}$.
Hence, the real numbers $a, b$ can be considered as the two roots of the quadratic equation
$$
x^{2} \mp \sqrt{\frac{9-k}{2}}+\frac{3-k}{2}=0
$$
Thus, $\Delta=\left(\sqrt{\frac{9-\bar{k}}{2}}\right)^{2}-4 \times \frac{3-k}{2}=\frac{3 k-3}{2} \geqslant 0$.
Solving this, we get $k \geqslant 1$.
Therefore, $1 \leqslant k \leqslant 9$.
So, the maximum value of $a^{2}-a b+b^{2}$ is $m=9$, and the minimum value is $n=1$.
Thus, $m+n=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-x y+y^{2}$ is $\qquad$ [4]
(2010, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
|
Solve: Completing the square for $x^{2}-x y+y^{2}$ yields
$$
\begin{array}{l}
x^{2}-x y+y^{2}=\frac{1}{4}(x+y)^{2}+\frac{3}{4}(x-y)^{2} . \\
\text { Also, }|x \pm y| \leqslant|x|+|y| \leqslant 1, \text { then } \\
x^{2}-x y+y^{2} \leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$
When $x$ and $y$ are such that one is 0 and the other is 1, the equality holds.
Therefore, the maximum value of $x^{2}-x y+y^{2}$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given that $x_{1}, x_{2}, \cdots, x_{6}$ are six different positive integers, taking values from $1,2, \cdots, 6$. Let
$$
\begin{aligned}
S= & \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\left|x_{3}-x_{4}\right|+ \\
& \left|x_{4}-x_{5}\right|+\left|x_{5}-x_{6}\right|+\left|x_{6}-x_{1}\right| .
\end{aligned}
$$
Then the minimum value of $S$ is $\qquad$ [5]
(2009, National Junior High School Mathematics League Sichuan Preliminary)
|
Since equation (1) is a cyclic expression about $x_{1}, x_{2}, \cdots, x_{6}$, we can assume $x_{1}=6, x_{j}=1(j \neq 1)$. Then
$$
\begin{aligned}
S \geqslant & \left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
& \left|\left(x_{j+1}-1\right)+\left(x_{j+2}-x_{j+1}\right)+\cdots+\left(6-x_{6}\right)\right| \\
= & 10 .
\end{aligned}
$$
When $x_{i}=7-i(i=1,2, \cdots, 6)$, $S$ achieves its minimum value of 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the function $S=|x-2|+|x-4|$.
(1) Find the minimum value of $S$;
(2) If for any real numbers $x, y$ the inequality
$$
S \geqslant m\left(-y^{2}+2 y\right)
$$
holds, find the maximum value of the real number $m$.
|
(1) Using the absolute value inequality, the answer is 2.
(2) From the problem, we know that for any real number $y$,
$$
m\left(-y^{2}+2 y\right) \leqslant 2
$$
holds. It is easy to find that the maximum value of $-y^{2}+2 y$ is 1. Therefore, $0 \leqslant m \leqslant 2$, and the maximum value of $m$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) A fly and $k$ spiders are placed at some intersections of a $2012 \times 2012$ grid. An operation consists of the following steps: first, the fly moves to an adjacent intersection or stays in place, then each spider moves to an adjacent intersection or stays in place (multiple spiders can occupy the same intersection). Assume that each spider and the fly always know the positions of the other spiders and the fly.
(1) Find the smallest positive integer $k$ such that the spiders can catch the fly in a finite number of operations, regardless of their initial positions;
(2) In a $2012 \times 2012 \times 2012$ three-dimensional grid, what is the conclusion for (1)?
[Note] In the problem, adjacent means that one intersection has only one coordinate different from another intersection's corresponding coordinate, and the difference is 1; catching means the spider and the fly are at the same intersection.
|
The minimum value of $k$ is 2.
(1) First, prove that a single spider cannot catch the fly.
Establish a Cartesian coordinate system, then the range of the spider and fly's movement is
$$
\{(x, y) \mid 0 \leqslant x, y \leqslant 2012, x, y \in \mathbf{N}\}.
$$
For each point in the above set, there are at least two points in the set that are adjacent to it. Therefore, if the spider is not adjacent to the fly after a move, the fly can remain still; if the spider is adjacent to the fly after a move, the fly can move to another adjacent point in the set. This way, the spider cannot catch the fly.
Second, if there are two spiders $S_{1}$ and $S_{2}$, let the distance between $S_{i}$ and the fly $F$ after $n$ moves be $a_{i, n} (i=1,2)$, where the distance between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is defined as
$$
d=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|.
$$
First, it can be achieved through several operations that
$$
S_{1}(0,0), S_{2}(2012,2012).
$$
Next, we prove:
(i) After each move, $S_{1}$ and $S_{2}$ can be appropriately moved so that the fly $F$ is inside or on the boundary of the rectangle with $S_{1}$ and $S_{2}$ as the diagonal (the line connecting $S_{1}$ and $S_{2}$ is not parallel to the coordinate axes);
(ii) Keep $a_{1, n}+a_{2, n}$ non-increasing, and it will strictly decrease after several moves.
In fact, let the unit vectors in the positive directions of the $x$ and $y$ axes be $\boldsymbol{i}$ and $\boldsymbol{j}$. The strategy is as follows:
If the fly's coordinate value is not inside or on the boundary of the rectangle with $S_{1}$ and $S_{2}$ as the diagonal after a move, then $S_{1}$ and $S_{2}$ move in that direction towards the grid where $\boldsymbol{F}$ is located; otherwise, $S_{1}$ can move $\boldsymbol{i}$ or $\boldsymbol{j}$, and $S_{2}$ can move $\boldsymbol{i}$ or $-\boldsymbol{j}$, appropriately chosen to ensure that the line connecting $S_{1}$ and $S_{2}$ is not parallel to the coordinate axes, and the fly's position is inside or on the boundary of the rectangle. By induction, (i) is easily proven.
Next, we prove (ii).
In fact, if the fly is strictly inside the rectangle with $S_{1}$ and $S_{2}$ as the diagonal before the $n$-th operation, then $a_{1, n}+a_{2, n}$ must decrease after that round; otherwise, the fly must be on one of the rectangle's sides and must move outside the rectangle, and by the rule, $a_{1, n}+a_{2, n}$ does not increase, but the grid is bounded, so after at most 4024 steps, the fly can no longer move outside the rectangle. At this point, $a_{1, n}+a_{2, n}$ strictly decreases.
Since the initial value of $a_{1,0}+a_{2,0}$ is 4024, it cannot decrease indefinitely, so after a finite number of operations, the spider can catch the fly.
(2) The conclusion remains unchanged, and the minimum value of $k$ is still 2.
Establish a spatial Cartesian coordinate system, then the range of $S_{i}$ and $F$ is $\{(x, y, z) \mid 0 \leqslant x, y, z \leqslant 2012, x, y, z \in \mathbf{N}\}$.
When $k=1$, the fly can adopt a similar strategy to ensure it is not caught;
When $k=2$, let $S_{1}$ and $S_{2}$ move to $(0,0,0)$ and $(2012,2012,2012)$ through several operations, and each operation keeps the fly inside or on the boundary (face, edge, vertex) of the cube with $S_{1}$ and $S_{2}$ as the body diagonal, and the line connecting $S_{1}$ and $S_{2}$ is not parallel to the planes $x O y$, $y O z$, and $z O x$. It can be completely similarly proven that after each operation, $a_{1, n}+a_{2, n}$ does not increase, and after at most $3 \times 2012$ steps, it will strictly decrease.
Since the initial value of $a_{1,0}+a_{2,0}$ is 6036, after a finite number of operations, the spider will catch the fly.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given the sets
$$
\begin{array}{l}
M=\{(x, y) \mid x(x-1) \leqslant y(1-y)\}, \\
N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant k\right\} .
\end{array}
$$
If $M \subset N$, then the minimum value of $k$ is $\qquad$ .
(2007, Shanghai Jiao Tong University Independent Admission Examination)
|
Notice that,
$$
M=\left\{(x, y) \left\lvert\,\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2}\right.\right\}
$$
represents a disk with center $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{\sqrt{2}}{2}$.
By $M \subset N \Rightarrow \sqrt{k} \geqslant \sqrt{2} \times \frac{1}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$
$$
\Rightarrow k_{\min }=2 \text {. }
$$
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Find the largest positive integer $k$ such that the set of positive integers can be partitioned into $k$ subsets $A_{1}, A_{2}, \cdots, A_{k}$, so that for all integers $n(n \geqslant 15)$ and all $i \in\{1,2, \cdots, k\}$, there exist two distinct elements in $A_{i}$ whose sum is $n$.
|
4. The largest positive integer $k$ is 3.
When $k=3$, let
$A_{1}=\{1,2,3\} \cup\{3 m \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{2}=\{4,5,6\} \cup\{3 m-1 \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{3}=\{7,8,9\} \cup\{3 m-21 m \in \mathbf{Z}$, and $m \geqslant 4\}$.
Then the sum of two different elements in $A_{1}$ can represent all
positive integers $n$ greater than or equal to $1+12=13$; the sum of two different elements in $A_{2}$ can represent all positive integers $n$ greater than or equal to $4+11=15$; the sum of two different elements in $A_{3}$ can represent all positive integers $n$ greater than or equal to $7+10=17$ and $7+8=15, 7+9=16$.
If $k \geqslant 4$, and there exist sets $A_{1}, A_{2}, \cdots, A_{k}$ that satisfy the conditions, then the sets $A_{1}, A_{2}, A_{3}, A_{4} \cup A_{5} \cup \cdots \cup A_{k}$ also satisfy the conditions.
Thus, we can assume $k=4$.
Let $B_{i}=A_{i} \cap\{1,2, \cdots, 23\}(i=1,2,3,4)$.
For each $i=1,2,3,4$, the integers $15,16, \cdots, 24$ can be written as the sum of two different elements in $B_{i}$. Therefore, $B_{i}$ must have at least five elements.
Since $\left|B_{1}\right|+\left|B_{2}\right|+\left|B_{3}\right|+\left|B_{4}\right|=23$, there exists $j \in\{1,2,3,4\}$ such that $\left|B_{j}\right|=5$.
Let $B_{j}=\left\{x_{1}, x_{2}, \cdots, x_{5}\right\}$. Then the sums of two different elements in $A_{j}$ that represent the integers $15,16, \cdots, 24$ are exactly the sums of all two different elements in $B_{j}$.
$$
\begin{array}{l}
\text { Hence } 4\left(x_{1}+x_{2}+\cdots+x_{5}\right) \\
=15+16+\cdots+24=195 .
\end{array}
$$
Since 195 is not divisible by 4, the above equation does not hold.
Thus, a contradiction arises.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. (50 points) Let the sequence $\left\{x_{n}\right\}$ satisfy
$$
x_{1}=1, x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \text {. }
$$
Find the units digit of $x_{2012}$.
|
7. Clearly, $x_{2}=7$, and for any positive integer $n, x_{n}$ is a positive integer.
By the property of the floor function, we have
$$
\begin{array}{l}
4 x_{n}+\sqrt{11} x_{n}>x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \\
>4 x_{n}+\sqrt{11} x_{n}-1 .
\end{array}
$$
Multiplying both sides of the above inequality by $4-\sqrt{11}$, we get
$$
\begin{array}{l}
5 x_{n}>(4-\sqrt{11}) x_{n+1}>5 x_{n}-(4-\sqrt{11}) \\
\Rightarrow 4 x_{n+1}-5 x_{n}<\sqrt{11} x_{n+1} \\
\quad<4 x_{n+1}-5 x_{n}+(4-\sqrt{11}) \\
\Rightarrow\left[\sqrt{11} x_{n+1}\right]=4 x_{n+1}-5 x_{n} \\
\Rightarrow x_{n+2}=4 x_{n+1}+\left[\sqrt{11} x_{n+1}\right] \\
\quad=4 x_{n+1}+\left(4 x_{n+1}-5 x_{n}\right) \\
\quad=8 x_{n+1}-5 x_{n} .
\end{array}
$$
Clearly, $x_{n+2} \equiv x_{n}(\bmod 2)$,
$$
x_{n+2} \equiv 3 x_{n+1}(\bmod 5) \text {. }
$$
By $x_{1}=1, x_{2}=7$, it is easy to prove using the second principle of mathematical induction that for any positive integer $n$,
$$
x_{n} \equiv 1(\bmod 2) \text {. }
$$
By $x_{2}=7$, it is easy to prove using the first principle of mathematical induction that for any positive integer $n \geq 2$,
$$
x_{n}=2 \times 3^{n-2}(\bmod 5) \text {. }
$$
Since $x_{2012} \equiv 1(\bmod 2)$,
$$
x_{2012} \equiv 2 \times 3^{2010} \equiv 2 \times 3^{2} \equiv 3(\bmod 5) \text {, }
$$
Therefore, the last digit of $x_{2012}$ is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A country has $n(n \geqslant 3)$ cities and two airlines. There is exactly one two-way flight between every pair of cities, and this two-way flight is operated exclusively by one of the airlines. A female mathematician wants to start from a city, pass through at least two other cities (each city is visited only once), and finally return to the starting city. She finds that no matter which starting city and intermediate cities she chooses, she cannot complete her journey using only one airline. Find the maximum value of $n$.
(Liang Yingde, problem contributor)
|
6. Solution 1 Consider each city as a vertex, each flight route as an edge, and each airline as a color. Then, the country's flight network can be seen as a complete graph with $n$ vertices whose edges are colored with two colors.
From the condition, we know that any cycle contains edges of both colors, meaning that the subgraph formed by edges of the same color has no cycles.
By a well-known conclusion, the number of edges in a simple graph without cycles is less than the number of vertices. Therefore, the number of edges of each color does not exceed $n-1$. This indicates that the total number of edges does not exceed $2(n-1)$.
On the other hand, the number of edges in a complete graph with $n$ vertices is $\frac{n(n-1)}{2}$.
Therefore, $\frac{n(n-1)}{2} \leqslant 2(n-1) \Rightarrow n \leqslant 4$.
When $n=4$, let the four cities be $A$, $B$, $C$, and $D$. If the routes $A B$, $B C$, and $C D$ are operated by the first airline, and the routes $A C$, $A D$, and $B D$ are operated by the second airline, it is easy to see that the routes operated by each airline form a chain and do not contain any cycles, i.e., this scenario satisfies the conditions of the problem.
In summary, the maximum value of $n$ is 4.
Solution 2 From the same reasoning as in Solution 1, we know that the subgraph formed by edges of the same color has no cycles.
By a well-known conclusion, a complete graph with six vertices, when colored with two colors, must contain a monochromatic triangle, i.e., a monochromatic cycle.
Therefore, $n \leqslant 5$.
We now prove: $n \neq 5$.
If not, let the five vertices be $A$, $B$, $C$, $D$, and $E$. Consider the four edges extending from vertex $A$. If at least three of these edges are the same color (e.g., $A B$, $A C$, $A D$ are the same color), then among $B C$, $B D$, and $C D$, there will be an edge of the same color as the aforementioned three edges, leading to a monochromatic triangle, which is a contradiction. If $B C$, $B D$, and $C D$ are all different colors from $A B$, $A C$, and $A D$, they will form a monochromatic triangle, which is a contradiction. If each vertex has exactly two edges of the same color, it is easy to see that the graph contains a Hamiltonian cycle, which is a contradiction.
Therefore, $n \leqslant 4$.
The rest follows the same reasoning as in Solution 1, showing that $n$ can be 4.
Hence, the maximum value of $n$ is 4.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. At the beginning, there are 111 pieces of clay of equal weight on the table. Perform the following operations on the clay: First, divide a part or all of the clay into several groups, with the same number of pieces in each group, then knead the clay in each group into one piece. It is known that after $m$ operations, there are exactly 11 pieces of clay with different weights on the table. Find the minimum value of $m$.
|
1. 2 .
Obviously, one operation can result in at most two different weights of clay blocks.
Below, we show that two operations can achieve the goal.
Assume without loss of generality that each block of clay initially weighs 1.
In the first operation, select 74 blocks and divide them into 37 groups, with two blocks in each group. After the first operation, there are 37 blocks weighing 1 and 37 blocks weighing 2 on the table.
In the second operation, select 36 blocks weighing 1 and 36 blocks weighing 2, and divide them into 9 groups, with 8 blocks in each group. Specifically, the $i$-th group ($1 \leqslant i \leqslant 9$) contains $i-1$ blocks weighing 2 and $9-i$ blocks weighing 1. After the second operation, there are 11 blocks of clay on the table, with weights
$$
1, 2, 9-i+2(i-1)=7+i \quad (1 \leqslant i \leqslant 9).
$$
This meets the requirement.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $f(x)$ represent a quartic polynomial in $x$. If $f(1)=f(2)=f(3)=0, f(4)=6$, $f(5)=72$, then the last digit of $f(2010)$ is $\qquad$. ${ }^{3}$
(2010, International Cities Mathematics Invitational for Youth)
|
【Analysis】It is given in the problem that $1,2,3$ are three zeros of the function $f(x)$, so we consider starting from the zero point form of the function.
Solution From the problem, we know
$$
f(1)=f(2)=f(3)=0 \text {. }
$$
Let the one-variable quartic polynomial function be
$$
f(x)=(x-1)(x-2)(x-3)(a x+b) \text {, }
$$
where $a, b \in \mathbf{R}$.
Therefore, $\left\{\begin{array}{l}f(4)=6(4 a+b)=6, \\ f(5)=24(5 a+b)=72 .\end{array}\right.$
Solving these, we get $a=2, b=-7$.
Thus, $f(x)=(x-1)(x-2)(x-3)(2 x-7)$.
Hence, $f(2010)$
$$
\begin{array}{l}
=(2010-1)(2010-2)(2010-3)(4020-7) \\
\equiv 2(\bmod 10) .
\end{array}
$$
Therefore, the last digit of $f(2010)$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Let $f(x)$ be a polynomial with integer coefficients, $f(0)=$
11. There exist $n$ distinct integers $x_{1}, x_{2}, \cdots, x_{n}$, such that $f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010$.
Then the maximum value of $n$ is $\qquad$ (6)
$(2010$, Xin Zhi Cup Shanghai High School Mathematics Competition)
|
Let $g(x)=f(x)-2010$.
$$
\begin{array}{l}
\text { By } f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010, \text { we have } \\
g\left(x_{1}\right)=g\left(x_{2}\right)=\cdots=g\left(x_{n}\right)=0,
\end{array}
$$
That is, $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ distinct integer roots of the integer-coefficient polynomial $g(x)$. Therefore,
$$
g(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right) q(x) \text {, }
$$
where $q(x)$ is an integer-coefficient polynomial.
Then $g(0)=f(0)-2010$
$$
=11-2010=-1999 \text {. }
$$
Note that 1999 is a prime number, and can be at most the product of three different integers, $-1999=(-1) \times 1 \times 1999$, so $n \leqslant 3$.
When $n=3$,
$$
\begin{array}{l}
g(x)=(x-1)(x+1)(x+1999) . \\
\text { Hence } f(x)=(x-1)(x+1)(x+1999)+2010 .
\end{array}
$$
Therefore, the maximum value of $n$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. On the edge $AS$ of the tetrahedron $S-ABC$, mark points $M, N$ such that $AM=MN=NS$. If the areas of $\triangle ABC$, $\triangle MBC$, and $\triangle SBC$ are $1$, $2$, and $\sqrt{37}$, respectively, find the area of $\triangle NBC$.
|
8. Let the areas of $\triangle A B C$, $\triangle M B C$, $\triangle N B C$, $\triangle S B C$ be $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$, and let $h_{1}$, $h_{2}$, $h_{3}$, $h_{4}$ be the heights from these triangles to the common base $B C$, as shown in Figure 2.
Points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $S^{\prime}$, $M^{\prime}$, $N^{\prime}$ represent the orthogonal projections of points $A$, $B$, $C$, $S$, $M$, $N$ onto a plane perpendicular to the edge $B C$, as shown in Figure 3. Points $B^{\prime}$ and $C^{\prime}$ coincide, and
$$
\begin{array}{l}
A^{\prime} M^{\prime}=M^{\prime} N^{\prime}=N^{\prime} S^{\prime}=a, A^{\prime} B^{\prime}=h_{1}, \\
M^{\prime} B^{\prime}=h_{2}, N^{\prime} B^{\prime}=h_{3}, S^{\prime} B^{\prime}=h_{4}.
\end{array}
$$
Considering that $M^{\prime} B^{\prime}$ and $N^{\prime} B^{\prime}$ are the medians of $\triangle A^{\prime} B^{\prime} N^{\prime}$ and $\triangle M^{\prime} B^{\prime} S^{\prime}$, respectively, we have
$$
\begin{array}{l}
h_{1}^{2}+h_{3}^{2}-2 h_{2}^{2}=2 a^{2}=h_{2}^{2}+h_{4}^{2}-2 h_{3}^{2} \\
\Leftrightarrow 3 h_{3}^{2}=3 h_{2}^{2}+h_{4}^{2}-h_{1}^{2} \\
\Rightarrow h_{3}=\sqrt{h_{2}^{2}+\frac{h_{4}^{2}-h_{1}^{2}}{3}} \\
\Rightarrow S_{3}=\sqrt{S_{2}^{2}+\frac{S_{4}^{2}-S_{1}^{2}}{3}}=4.
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given trapezoid $A B C D$ with bases $A D=3, B C=1$, the diagonals intersect at point $O$, two circles intersect base $B C$ at points $K, L$, these two circles are tangent at point $O$, and are tangent to line $A D$ at points $A, D$ respectively. Find $A K^{2}+D L^{2}$.
|
10. As shown in Figure 4, draw the common tangent of the two circles through point $O$, intersecting the lower base $AD$ at point $P$. By the properties of tangents, we have
$$
A P=O P=D P \text {. }
$$
This indicates that $\triangle A O D$ is a right triangle and is similar to $\triangle C D B$, with the similarity ratio $k=\frac{B C}{A D}=\frac{1}{3}$.
It is easy to see that $\triangle A K C \backsim \triangle A O K$.
Thus, $A K^{2}=A O \cdot A C$
$$
=A O \cdot \frac{A O+O C}{A O} \cdot A O=\frac{4}{3} A O^{2} \text {, }
$$
Similarly, $D L^{2}=\frac{4}{3} O D^{2}$.
Therefore, $A K^{2}+D L^{2}=\frac{4}{3} A O^{2}+\frac{4}{3} O D^{2}$
$$
=\frac{4}{3} A D^{2}=\frac{4}{3} \times 3^{2}=12 \text {. }
$$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Arrange all positive integers that satisfy the following conditions in descending order, denoted as $M$, and the $k$-th number as $b_{k}$: each number's any three consecutive digits form a non-zero perfect square. If $b_{16}-b_{20}=2^{n}$, find $n$.
Arrange all positive integers that satisfy the following conditions in descending order, denoted as $M$, and the $k$-th number as $b_{k}$: each number's any three consecutive digits form a non-zero perfect square. If $b_{16}-b_{20}=2^{n}$, find $n$.
|
II. Notice that, among three-digit numbers, the perfect squares are $100, 121, 144, 169, 196, 225, 256, 289$, $324, 361, 400, 441, 484, 529, 576, 625$, $676, 729, 784, 841, 900, 961$.
For each number in $M$, consider its leftmost three digits.
(1) The six-digit number with the leftmost three digits 100 is 100169, the five-digit numbers are $10016, 10049$, and the four-digit numbers are $1001, 1004, 1009$.
(2) The six-digit number with the leftmost three digits 400 is 400169, the five-digit numbers are 40016, 40049, and the four-digit numbers are $4001, 4004, 4009$.
(3) The six-digit number with the leftmost three digits 900 is 900169, the five-digit numbers are $90016, 90049$, and the four-digit numbers are $9001, 9004, 9009$.
(4) When the leftmost three digits are $144, 196, 225, 484, 625, 784$, the corresponding four-digit numbers are $1441, 1961, 2256, 4841, 6256, 7841$.
(5) The rest are all three-digit numbers. After sorting the above numbers in descending order, we find that $b_{16}=4009, b_{20}=1961$.
Thus, $b_{16}-b_{20}=2048=2^{11}$, which means $n=11$.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If five vertices of a regular nonagon are colored red, then there are at least $\qquad$ pairs of congruent triangles (each pair of triangles has different vertex sets) whose vertices are all red.
|
8. 4 .
A triangle with both vertices colored red is called a "red triangle". Thus, there are $\mathrm{C}_{5}^{3}=10$ red triangles. For a regular nonagon, the triangles formed by any three vertices are of only seven distinct types (the lengths of the minor arcs of the circumcircle of the regular nonagon corresponding to the sides of the triangles are
$$
\begin{array}{l}
(1,1,2),(1,2,3),(1,3,4),(1,4,4), \\
(2,2,4),(2,3,4),(3,3,3)
\end{array}
$$
assuming the circumference of the circumcircle of the regular nonagon is 9$)$.
If there are at most six distinct non-congruent red triangles, by Turán's theorem, there must be at least four pairs of congruent red triangles; if there are all seven distinct non-congruent red triangles, then particularly, there must be those corresponding to the minor arc lengths $(1,1,2)$ and $(3,3,3)$. Thus, essentially, it must be one of the two scenarios shown in Figure 5, but neither of them contains all seven distinct non-congruent red triangles, leading to a contradiction.
Therefore, there must be at least four pairs of such triangles.
Any of the scenarios in Figure 5 can serve as an example when there are four pairs.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (40 points) Find the largest positive real number $\lambda$ such that for all positive integers $n$ and positive real numbers $a_{i} (i=1,2, \cdots, n)$, we have
$$
1+\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant \lambda\left[\sum_{k=1}^{n} \frac{1}{\left(1+\sum_{s=1}^{k} a_{s}\right)^{2}}\right] .
$$
|
II. Define $S_{k}=\sum_{i=1}^{k} a_{i}+1$, and supplement the definition $S_{0}=1$. First, prove a lemma.
Lemma For any positive integer $k \geqslant 1$, we have $\frac{1}{a_{k}^{2}}+\frac{1}{S_{k-1}^{2}} \geqslant \frac{8}{S_{k}^{2}}$.
Proof Let $S_{k-1}=a_{k} t_{k}$. Then $S_{k}=a_{k}\left(t_{k}+1\right)$.
Thus, equation (1) $\Leftrightarrow\left(1+\frac{1}{t_{k}^{2}}\right)\left(t_{k}+1\right)^{2} \geqslant 8$ $\Leftrightarrow t_{k}^{2}+2 t_{k}+2+\frac{2}{t_{k}}+\frac{1}{t_{k}^{2}} \geqslant 8$.
The last inequality is easily proven by the AM-GM inequality.
Returning to the original problem.
By the lemma, we have $\frac{1}{a_{k}^{2}} \geqslant \frac{8}{S_{k}^{2}}-\frac{1}{S_{k-1}^{2}}$.
Thus, $\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant \sum_{k=1}^{n} \frac{7}{S_{k}^{2}}+\frac{1}{S_{n}^{2}}-1$.
Hence, $1+\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant 7 \sum_{k=1}^{n} \frac{1}{S_{k}^{2}}$.
Therefore, when $\lambda=7$, the inequality holds.
Next, we show that $\lambda=7$ is the best possible.
Let $a_{k}=2^{k-1}$, and let $n \rightarrow+\infty$, it is easy to see that $\lambda \leqslant 7$.
In conclusion, $\lambda_{\text {max }}=7$.
|
7
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: For $n$ consecutive positive integers, if each number is written in its standard prime factorization form, and each prime factor is raised to an odd power, such a sequence of $n$ consecutive positive integers is called a "consecutive $n$ odd group" (for example, when $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{1}$, $24=2^{3} \times 3^{1}$, then $22, 23, 24$ form a consecutive 3 odd group). The maximum possible value of $n$ in a consecutive $n$ odd group is $\qquad$ [1]
|
【Analysis】Notice that, in a connected $n$-singular group, if there exists a multiple of 4, then by the definition of a connected $n$-singular group, it must be a multiple of 8.
Let this number be $2^{k} A\left(k, A \in \mathbf{N}_{+}, k \geqslant 3, A\right.$ is an odd number). Then $2^{k} A+4$ and $2^{k} A-4$ are both numbers that are congruent to 4 modulo 8.
Therefore, the power of 2 in their prime factorizations is even.
Thus, a connected $n$-singular group cannot contain two multiples of 4 simultaneously. However, in any 8 consecutive positive integers, there must be a number that is congruent to 0 modulo 8 and a number that is congruent to 4 modulo 8. Therefore, in any 8 consecutive positive integers, there must be two multiples of 4.
Hence, $n \leqslant 7$.
Furthermore, $29,30, \cdots, 35$ is a connected 7-singular group, so $n_{\max }=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 The number of prime pairs \((a, b)\) that satisfy the equation
$$
a^{b} b^{a}=(2 a+b+1)(2 b+a+1)
$$
is \qquad (2]
(2011, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
|
【Analysis】If $a$ and $b$ are both odd, then
the left side of equation (1) $\equiv 1 \times 1 \equiv 1(\bmod 2)$,
the right side of equation (1) $\equiv(2 \times 1+1+1)(2 \times 1+1+1)$ $\equiv 0(\bmod 2)$.
Clearly, the left side is not congruent to the right side $(\bmod 2)$, a contradiction.
Therefore, at least one of $a$ and $b$ must be even.
Since $a$ and $b$ are both prime numbers, by symmetry, we can assume $a=2$ first.
Then $2^{b} b^{2}=(b+5)(2 b+3)$.
Clearly, $b \neq 2$, so $b \geqslant 3$.
If $b>3$, then $b \geqslant 5$. Thus,
$$
2^{b} b^{2} \geqslant 2^{5} b^{2}=4 b \times 8 b>(b+5)(2 b+3) \text {. }
$$
This contradicts equation (2).
When $b=3$, equation (2) holds.
Therefore, $a=2, b=3$.
By symmetry, $a=3, b=2$ also satisfies the condition.
In summary, the number of prime pairs is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Find the smallest positive integer $n$, such that there exist rational-coefficient polynomials $f_{1}, f_{2}, \cdots, f_{n}$, satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+\cdots+f_{n}^{2}(x) .
$$
(51st IMO Shortlist)
|
【Analysis】For the case $n=5$,
$$
x^{2}+7=x^{2}+2^{2}+1+1+1 \text {, }
$$
it meets the requirement.
Now we prove that $n \leqslant 4$ does not meet the requirement.
Assume there exist four rational coefficient polynomials $f_{1}(x)$, $f_{2}(x)$, $f_{3}(x)$, $f_{4}(x)$ (which may include the zero polynomial), such that $x^{2}+7=\sum_{i=1}^{4}\left(f_{i}(x)\right)^{2}$, and the degree of the polynomial $f_{i}(x)$ is at most 1.
Let $f_{i}(x)=a_{i} x+b_{i}$. Then
$$
\sum_{i=1}^{4} a_{i}^{2}=1, \sum_{i=1}^{4} a_{i} b_{i}=0, \sum_{i=1}^{4} b_{i}^{2}=7 .
$$
Let $p_{i}=a_{i}+b_{i}$, $q_{i}=a_{i}-b_{i}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{4} p_{i}^{2}=\sum_{i=1}^{4}\left(a_{i}^{2}+2 a_{i} b_{i}+b_{i}^{2}\right)=8, \\
\sum_{i=1}^{4} q_{i}^{2}=\sum_{i=1}^{4}\left(a_{i}^{2}-2 a_{i} b_{i}+b_{i}^{2}\right)=8, \\
\sum_{i=1}^{4} p_{i} q_{i}=\sum_{i=1}^{4}\left(a_{i}^{2}-b_{i}^{2}\right)=-6 .
\end{array}
$$
Let the least common multiple of the denominators of the rational numbers $p_{i}$, $q_{i}$ ($i=1,2,3,4$) be $m$. Then there exist integers $x_{i}$, $y_{i}$ (after converting $p_{i}$, $q_{i}$ to a common denominator $m$, the numerators of $p_{i}$, $q_{i}$), such that
$$
\left\{\begin{array}{l}
\sum_{i=1}^{4} x_{i}^{2}=8 m^{2}, \\
\sum_{i=1}^{4} y_{i}^{2}=8 m^{2}, \\
\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2} .
\end{array}\right.
$$
Assume $m$ is the smallest positive integer satisfying the above system of indeterminate equations.
$$
\text { From } \sum_{i=1}^{4} x_{i}^{2}=8 m^{2} \equiv 0(\bmod 8) \text {, we know } x_{1} \text {, } x_{2} \text {, } x_{3} \text {, }
$$
$x_{4}$ are all even.
$$
\text { From } \sum_{i=1}^{4} y_{i}^{2}=8 m^{2} \equiv 0(\bmod 8) \text {, we know } y_{1} \text {, } y_{2} \text {, } y_{3} \text {, }
$$
$y_{4}$ are all even.
From $\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2} \equiv 0(\bmod 4)$, we know $m$ is also even.
Thus, $\frac{x_{i}}{2}$, $\frac{y_{i}}{2}$, $\frac{m}{2}$ also satisfy the system of indeterminate equations.
This contradicts the minimality of $m$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Find all real numbers $x$ such that $4 x^{5}-7$ and $4 x^{13}-7$ are both perfect squares. ${ }^{[6]}$
(2008, German Mathematical Olympiad)
|
【Analysis】Let
$$
4 x^{5}-7=a^{2}, 4 x^{13}-7=b^{2}(a, b \in \mathbf{N}) \text {. }
$$
Then $x^{5}=\frac{a^{2}+7}{4}>1$ is a positive rational number, and $x^{13}=\frac{b^{2}+7}{4}$ is a positive rational number.
Therefore, $x=\frac{\left(x^{5}\right)^{8}}{\left(x^{13}\right)^{3}}$ is a positive rational number.
Let $x=\frac{p}{q}\left((p, q)=1, p, q \in \mathbf{Z}_{+}\right)$.
Then $\left(\frac{p}{q}\right)^{5}=\frac{a^{2}+7}{4}$ can only have $q=1$, i.e., $x$ is a positive:
integer.
Obviously, $x \geqslant 2$.
When $x=2$,
$$
\begin{array}{l}
4 x^{5}-7=121=11^{2}, \\
4 x^{13}-7=32761=181^{2}
\end{array}
$$
satisfies the conditions.
When $x$ is an odd number,
$$
a^{2}=4 x^{5}-7 \equiv 5(\bmod 8) \text {, }
$$
which is not valid.
Assume $x$ is a positive even number below.
When $x \geqslant 4$,
$$
\begin{array}{l}
(a b)^{2}=\left(4 x^{5}-7\right)\left(4 x^{13}-7\right) \\
=16 x^{18}-28 x^{13}-28 x^{7}+49 .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
16 x^{18}-28 x^{13}-28 x^{7}+49 \\
\left(4 x^{9}-\frac{7}{2} x^{4}-1\right)^{2},
\end{array}
$$
thus $x \geqslant 4$ and being even is not valid.
In summary, only $x=2$ satisfies the conditions of the problem.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leq 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at that time.
(Supplied by Li Shenghong)
|
4. Let $f(x)=a \cos x+b \cos 2 x+$ $c \cos 3 x+d \cos 4 x$.
From $f(0)=a+b+c+d$,
$f(\pi)=-a+b-c+d$,
$f\left(\frac{\pi}{3}\right)=\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2}$,
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.,
$$
a=1, b+d=1, c=-1
$$
In this case, 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
|
8. Let $m$ be a positive integer, $n=2^{m}-1$, and the set of $n$ points on the number line be $P_{n}=\{1,2, \cdots, n\}$.
A grasshopper jumps on these points, each step moving from one point to an adjacent point. Find the maximum value of $m$ such that for any $x, y \in P_{n}$, the number of ways to jump from point $x$ to point $y$ in 2012 steps (allowing intermediate visits to points $x, y$) is even.
|
8. When $m \geqslant 11$, $n=2^{m}-1>2013$.
Since there is only one way to jump from point 1 to point 2013 in 2012 steps, this is a contradiction, so $m \leqslant 10$.
We will now prove that $m=10$ satisfies the condition.
We use mathematical induction on $m$ to prove a stronger proposition: $\square$
For any $k \geqslant n=2^{m}-1$ and any $x, y \in P_{n}$, the number of ways to jump from point $x$ to point $y$ in $k$ steps is even.
When $m=1$, the number of ways to jump is necessarily 0, so the conclusion holds.
Assume that the conclusion holds for $m=l$.
For $k \geqslant n=2^{l+1}-1$, we divide the paths from point $x$ to point $y$ in $k$ steps into three categories, and we only need to prove that the number of paths in each case is even.
(1) The path never passes through point $2^{\prime}$.
In this case, points $x$ and $y$ are on the same side of point $2^{l}$, and by the induction hypothesis, the number of paths is even.
(2) The path passes through point $2^{l}$ exactly once.
Let the $i$-th step ($i \in\{0,1, \cdots, k\}$) be the step that reaches point $2^{l}$, where $i=0$ indicates that point $x$ is point $2^{l}$, and $i=k$ indicates that point $y$ is point $2^{l}$. We need to prove that for any $i$, the number of paths is even.
Let the path be
$x, a_{1}, \cdots, a_{i-1}, 2^{l}, a_{i+1}, \cdots, a_{k-1}, y$.
We divide this path into two sub-paths: from point $x$ to $a_{i-1}$, which is $i-1$ steps; and from point $a_{i+1}$ to $y$, which is $k-i-1$ steps (for $i=0$ or $k$, there is only one sub-path, which is $k-1$ steps).
Since $k \geqslant n=2^{l+1}-1$, if $i-1<2^{l}-1$ and $k-i-1<2^{l}-1$, then
$i-1 \leqslant 2^{l}-2$ and $k-i-1 \leqslant 2^{l}-2$.
Adding these, we get $k \leqslant 2^{l+1}-2$, which is a contradiction.
Therefore, either $i-1 \geqslant 2^{l}-1$ or $k-i-1 \geqslant 2^{l}-1$ must hold.
By the induction hypothesis, the number of paths for at least one of the sub-paths is even.
By the multiplication principle, the number of paths that reach point $2^{l}$ at the $i$-th step is even.
(3) The path passes through point $2^{l}$ at least twice.
In this case, we pair the paths between the first and second visits to point $2^{l}$ by reflecting them along point $2^{l}$, so the number of such paths is necessarily even.
By mathematical induction, the proposition is proved.
In summary, the maximum value of $m$ is 10.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, find the maximum value of $k$; if not, please briefly explain the reason.
|
提示: As the sum of the squares of the smallest 10 distinct prime numbers is
$$
\begin{array}{l}
4+9+25+49+121+169+289+361+529+841 \\
=2397>2010,
\end{array}
$$
thus, $k \leqslant 9$.
By analyzing the parity and the fact that the square of an odd number is congruent to 1 modulo 8, it is easy to prove that $k \neq 8, k \neq 9$.
Also, $2010=2^{2}+3^{2}+7^{2}+11^{2}+13^{2}+17^{2}+37^{2}$, hence $k_{\max }=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a$ is a constant, and real numbers $x, y, z$ satisfy
$$
(x-1)^{2}+(y-\sqrt{5})^{2}+(z+1)^{2}=a
$$
when, $-8 \leqslant 4 x-\sqrt{5} y+2 z \leqslant 2$. Then $a=$ $\qquad$
|
3. 1.
Let $4 x-\sqrt{5} y+2 z=k$. Then $-8 \leqslant k \leqslant 2$.
From the given equation, we have
$$
\frac{(4 x-4)^{2}}{16 a}+\frac{(-\sqrt{5} y+5)^{2}}{5 a}+\frac{(2 z+2)^{2}}{4 a}=1 \text {. }
$$
Using the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
(16 a+5 a+4 a)\left[\frac{(4 x-4)^{2}}{16 a}+\frac{(-\sqrt{5} y+5)^{2}}{5 a}+\frac{(2 z+2)^{2}}{4 a}\right] \\
\geqslant(4 x-4-\sqrt{5} y+5+2 z+2)^{2} \\
\Rightarrow 25 a \geqslant(k+3)^{2} \\
\Rightarrow|k+3| \leqslant 5 \sqrt{a} .
\end{array}
$$
Since $-8 \leqslant k \leqslant 2$, we have
$$
-5 \leqslant k+3 \leqslant 5 \Rightarrow|k+3| \leqslant 5 \text {. }
$$
Therefore, $5 \sqrt{a}=5 \Rightarrow a=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $x, y, z \in (0, \sqrt{2})$, and satisfying
$$
\left(2-x^{2}\right)\left(2-y^{2}\right)\left(2-z^{2}\right)=x^{2} y^{2} z^{2} \text{. }
$$
Then the maximum value of $x+y+z$ is
|
4.3.
$$
\begin{array}{l}
\text { Let } x=\sqrt{2} \cos \alpha, y=\sqrt{2} \cos \beta, \\
z=\sqrt{2} \cos \gamma\left(\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)\right) \text {. }
\end{array}
$$
Then the given equation transforms to
$$
\tan \alpha \cdot \tan \beta \cdot \tan \gamma=1 \text {. }
$$
Assume without loss of generality that $\tan \gamma \geqslant 1$. Then
$$
0<\tan \alpha \cdot \tan \beta \leqslant 1 \text {. }
$$
Since $\tan \alpha \cdot \tan \beta=\frac{1}{\tan \gamma}$, we have
$$
\begin{array}{l}
x+y+z=\sqrt{2}(\cos \alpha+\cos \beta+\cos \gamma) \\
=\sqrt{2}\left(\frac{1}{\sqrt{1+\tan ^{2} \alpha}}+\frac{1}{\sqrt{1+\tan ^{2} \beta}}+\frac{1}{\sqrt{1+\tan ^{2} \gamma}}\right) \\
=\sqrt{2}\left(\frac{\sqrt{1+\tan ^{2} \alpha}+\sqrt{1+\tan ^{2} \beta}}{\sqrt{1+\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \alpha \cdot \tan ^{2} \beta}}+\frac{1}{\sqrt{1+\tan ^{2} \gamma}}\right) .
\end{array}
$$
Since $\tan ^{2} \alpha+\tan ^{2} \beta \geqslant 2 \tan \alpha \cdot \tan \beta$,
$$
1+\tan ^{2} \gamma \geqslant \frac{1}{2}(1+\tan \gamma)^{2} \text {, }
$$
thus $x+y+z$
$$
\begin{array}{l}
\leqslant \sqrt{2}\left[\sqrt{1+\frac{2}{1+\tan \alpha \cdot \tan \beta}+\frac{1-\tan ^{2} \alpha \cdot \tan ^{2} \beta}{(1+\tan \alpha \cdot \tan \beta)^{2}}}+\frac{\sqrt{2}}{1+\tan \gamma}\right] \\
=\sqrt{2}\left(\sqrt{\frac{4}{1+\tan \alpha \cdot \tan \beta}}+\frac{\sqrt{2}}{1+\tan \gamma}\right) \\
=2\left(\sqrt{\frac{2}{1+\tan \alpha \cdot \tan \beta}}-\frac{1}{1+\tan \alpha \cdot \tan \beta}+1\right) \\
=-2\left(\frac{1}{\sqrt{1+\tan \alpha \cdot \tan \beta}}-\frac{\sqrt{2}}{2}\right)^{2}+3 \leqslant 3 .
\end{array}
$$
Equality holds if and only if $\alpha=\beta=\gamma=45^{\circ}$, i.e., $x=y=z=1$, in which case $(x+y+z)_{\text {max }}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $p$ and $5 p^{2}-2$ both be prime numbers: Find the value of $p$.
(2012, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
|
It is easy to prove that when $3 \times p$, $3 \mid \left(5 p^{2}-2\right)$.
Since $5 p^{2}-2>3$, thus $5 p^{2}-2$ is not a prime number, which contradicts the given condition.
Therefore, $3 \mid p$. Then $p=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-2 x-m=0$, and $2 x_{1}+x_{2}=0$. Then the value of $m$ is $\qquad$ .
|
By the relationship between roots and coefficients, we know $x_{1}+x_{2}=2$.
Also, $2 x_{1}+x_{2}=0$, then
$$
\begin{array}{l}
x_{1}+2=0 \\
\Rightarrow x_{1}=-2 \\
\Rightarrow(-2)^{2}-2 \times(-2)-m=0 \\
\Rightarrow m=8 .
\end{array}
$$
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a, b$ be positive real numbers, $m$ be a positive integer, and satisfy
$$
\left\{\begin{array}{l}
a+b \leqslant 14, \\
a b \geqslant 48+m .
\end{array}\right.
$$
Then the value of $m$ is
|
3.1.
Notice,
$$
\begin{array}{l}
14 \geqslant a+b \geqslant 2 \sqrt{a b} \geqslant 2 \sqrt{48+m} \\
\geqslant 2 \sqrt{48+1}=14 .
\end{array}
$$
Therefore, all equalities hold. Hence \( m=1 \).
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In a ball game competition, there are eight teams participating, and each pair of teams has to play a match. A team gets 2 points for a win, 1 point for a draw, and 0 points for a loss. If a team wants to ensure it enters the top four (i.e., its points must exceed those of at least four other teams), then the minimum points the team needs are $\qquad$
|
4. 11.
Since there are eight teams, there will be $\frac{8 \times 7}{2}=28$ matches, totaling $28 \times 2=56$ points.
If the top five teams draw with each other and each win against the bottom three teams, and the bottom three teams draw with each other, then there will be five teams each with 10 points, and the other three teams each with 2 points. Therefore, scoring 10 points does not guarantee a place in the top four.
If a team scores 11 points but is in fifth place, then the total points of the top five teams is at least 55. Thus, the total points of the bottom three teams is at most 1. However, the total points from the three matches among these three teams should be 6, leading to a contradiction.
Therefore, scoring 11 points definitely ensures a place in the top four.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $P$ be any point on the graph of the function $y=x+\frac{2}{x}(x>0)$, and draw perpendiculars from $P$ to the line $y=x$ and the $y$-axis, with the feet of the perpendiculars being $A$ and $B$ respectively. Then $\overrightarrow{P A} \cdot \overrightarrow{P B}=$ $\qquad$
|
- 1. -1 .
Solution 1 Let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)$.
Then $l_{P A}: y-\left(x_{0}+\frac{2}{x_{0}}\right)=-\left(x-x_{0}\right)$, which is $y=-x+2 x_{0}+\frac{2}{x_{0}}$.
Solving the above equation with $y=x$ yields point $A\left(x_{0}+\frac{1}{x_{0}}, x_{0}+\frac{1}{x_{0}}\right)$.
Also, point $B\left(0, x_{0}+\frac{2}{x_{0}}\right)$, then $\overrightarrow{P A}=\left(\frac{1}{x_{0}},-\frac{1}{x_{0}}\right), \overrightarrow{P B}=\left(-x_{0}, 0\right)$. Therefore, $\overrightarrow{P A} \cdot \overrightarrow{P B}=\frac{1}{x_{0}}\left(-x_{0}\right)=-1$.
Solution 2 As shown in Figure 3, let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)\left(x_{0}>0\right)$.
Figure 3
Then the distances from point $P$ to the line $x-y=0$ and the $y$-axis are respectively
$$
|P A|=\frac{\left|x_{0}-\left(x_{0}+\frac{2}{x_{0}}\right)\right|}{\sqrt{2}}=\frac{\sqrt{2}}{x_{0}},|P B|=x_{0} \text {. }
$$
Since $O, A, P, B$ are concyclic, we have
$$
\angle A P B=\pi-\angle A O B=\frac{3 \pi}{4} \text {. }
$$
Therefore, $\overrightarrow{P A} \cdot \overrightarrow{P B}=|\overrightarrow{P A}||\overrightarrow{P B}| \cos \frac{3 \pi}{4}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $\triangle A B C$ have interior angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively, and satisfy
$$
\begin{array}{l}
a \cos B-b \cos A=\frac{3}{5} c . \\
\text { Then } \frac{\tan A}{\tan B}=
\end{array}
$$
|
2.4.
Solution 1 From the given and the cosine rule, we have
$$
\begin{array}{l}
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c \\
\Rightarrow a^{2}-b^{2}=\frac{3}{5} c^{2} . \\
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\frac{a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}}{b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}} \\
=\frac{c^{2}+a^{2}-b^{2}}{c^{2}+b^{2}-a^{2}}=4 .
\end{array}
$$
Solution 2 Draw $C D \perp A B$ from point $C$, with the foot of the perpendicular at $D$. Then
$a \cos B=D B$,
$b \cos A=A D$.
From the given, we have
$D B-A D=\frac{3}{5} c$.
Figure 4
Also, $D B+D A=c$,
Solving these equations, we get
$$
\begin{array}{l}
A D=\frac{1}{5} c, D B=\frac{4}{5} c . \\
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\frac{C D}{A D}}{\frac{C D}{D B}}=\frac{D B}{A D}=4 .
\end{array}
$$
Solution 3 From the projection theorem, we have $a \cos B+b \cos A=c$.
Also, $a \cos B-b \cos A=\frac{3}{5} c$. Solving these equations, we get $a \cos B=\frac{4}{5} c, b \cos A=\frac{1}{5} c$.
Therefore, $\frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\frac{a \cos B}{b \cos A}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. For the parabola $y^{2}=2 p x(p>0)$, the focus is $F$, and the directrix is $l$. Points $A$ and $B$ are two moving points on the parabola, and they satisfy $\angle A F B=\frac{\pi}{3}$. Let $M$ be the midpoint of segment $A B$, and let $N$ be the projection of $M$ on $l$. Then the maximum value of $\frac{|M N|}{|A B|}$ is $\qquad$.
|
4. 1 .
Solution 1 Let $\angle A B F=\theta\left(0<\theta<\frac{2 \pi}{3}\right)$. Then by the Law of Sines, we have
$$
\frac{|A F|}{\sin \theta}=\frac{|B F|}{\sin \left(\frac{2 \pi}{3}-\theta\right)}=\frac{|A B|}{\sin \frac{\pi}{3}} .
$$
Thus, $\frac{|A F|+|B F|}{\sin \theta+\sin \left(\frac{2 \pi}{3}-\theta\right)}=\frac{|A B|}{\sin \frac{\pi}{3}}$, which means
$$
\begin{array}{l}
\frac{|A F|+|B F|}{|A B|}=\frac{\sin \theta+\sin \left(\frac{2 \pi}{3}-\theta\right)}{\sin \frac{\pi}{3}} \\
=2 \cos \left(\theta-\frac{\pi}{3}\right) .
\end{array}
$$
As shown in Figure 5, by the definition of a parabola and the midline theorem of a trapezoid, we get
$$
\begin{array}{l}
|M N| \\
=\frac{|A F|+|B F|}{2} .
\end{array}
$$
Thus, $\frac{|M N|}{|A B|}$
$$
=\cos \left(\theta-\frac{\pi}{3}\right) \text {. }
$$
Therefore, when $\theta=\frac{\pi}{3}$, $\frac{|M N|}{|A B|}$ reaches its maximum value of 1.
Solution 2 Similarly to Solution 1, we obtain equation (1).
In $\triangle A F B$, by the Law of Cosines, we have
$$
\begin{array}{l}
|A B|^{2}=|A F|^{2}+|B F|^{2}-2|A F||B F| \cos \frac{\pi}{3} \\
=(|A F|+|B F|)^{2}-3|A F||B F| \\
\geqslant(|A F|+|B F|)^{2}-3\left(\frac{|A F|+|B F|}{2}\right)^{2} \\
=\left(\frac{|A F|+|B F|}{2}\right)^{2}=|M N|^{2} .
\end{array}
$$
Equality holds if and only if $|A F|=|B F|$.
Therefore, the maximum value of $\frac{|M N|}{|A B|}$ is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let two regular tetrahedra $P-ABC$ and $Q-ABC$ be inscribed in the same sphere. If the dihedral angle between a lateral face and the base of the regular tetrahedron $P-ABC$ is $45^{\circ}$, then the tangent value of the dihedral angle between a lateral face and the base of the regular tetrahedron $Q-ABC$ is . $\qquad$
|
5.4.
As shown in Figure 6, connect $P Q$. Then $P Q \perp$ plane $A B C$, with the foot of the perpendicular $H$ being the center of the equilateral $\triangle A B C$, and $P Q$ passing through the center of the sphere $O$.
Connect $C H$ and extend it to intersect $A B$ at point $M$. Then $M$ is the midpoint of side $A B$, and $C M \perp A B$.
It is easy to see that $\angle P M H$ and $\angle Q M H$ are the plane angles of the dihedral angles formed by the lateral faces and the base of the regular tetrahedrons $P-A B C$ and $Q-A B C$, respectively.
$$
\begin{array}{l}
\text { Then } \angle P M H=45^{\circ} \Rightarrow P H=M H=\frac{1}{2} A H \text {. } \\
\text { By } \angle P A Q=90^{\circ}, A H \perp P Q \\
\Rightarrow A H^{2}=P H \cdot Q H \\
\Rightarrow A H^{2}=\frac{1}{2} A H \cdot Q H \\
\Rightarrow Q H=2 A H=4 M H \text {. } \\
\text { Therefore } \tan \angle Q M H=\frac{Q H}{M H}=4 .
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The function $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)+\arcsin x$. Then the solution set of $f(x)+f\left(2-x^{2}\right) \leqslant 0$ is $\qquad$
|
$-、 1 .\{-1\}$.
It is known that the function $f(x)$ is a monotonically increasing odd function defined on $[-1,1]$.
$$
\begin{array}{l}
\text { Given } f(x)+f\left(2-x^{2}\right) \leqslant 0 \\
\Rightarrow f(x) \leqslant f\left(x^{2}-2\right) \\
\Rightarrow-1 \leqslant x \leqslant x^{2}-2 \leqslant 1 \\
\Rightarrow x=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For the expression $\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{2013}$, when written as a decimal, find the digit before the decimal point.
|
Let $a_{n}=\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{n}-\frac{\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{n}$.
Then $a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$.
The last digits are
$$
\begin{array}{l}
1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7 \\
7,4,1,5,6,1,7,8,5,3,8,1,9,0,9,9 \\
8,7,5,2,7,9,6,5,1,6,7,3,0,3,3,6 \\
9,5,4,9,3,2,5,7,2,9,1,0,1,1, \cdots .
\end{array}
$$
It is not hard to see that the last digits form a periodic sequence with a period of 60.
Thus, the last digit of $a_{2013}$ is 8.
$$
\text { Also, }-1<\frac{\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{2013}<0 \text {, so } \frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{2013}
$$
has an integer part of $a_{2013}-1$.
Therefore, the digits before the decimal point are 7.
(Jin Lei, Xi'an Jiaotong University Affiliated High School,
710054)
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 The family of sets $\Omega$ consists of 11 five-element sets $A_{1}, A_{2}$, $\cdots, A_{11}$, where the intersection of any two sets is not empty. Let $A=\bigcup_{i=1}^{11} A_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, for any $x_{i} \in A$, the number of sets in $\Omega$ that contain the element $x_{i}$ is $k_{i}$, and let $m=\max \left\{k_{1}, k_{2}, \cdots, k_{n}\right\}$. Find the minimum value of $m$.
|
It is known that $\sum_{i=1}^{n} k_{i}=55$.
Notice that, the $k_{i}$ sets containing $x_{i}$ form
$$
\mathrm{C}_{k_{i}}^{2}=\frac{k_{i}\left(k_{i}-1\right)}{2}
$$
pairs of sets. Since the intersection of any two sets is not empty, the sum $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2}$ includes all pairs of sets, with some repetitions (because some pairs of sets may have more than one common element, and some elements may belong to multiple sets).
According to the condition, the 11 sets $A_{1}, A_{2}, \cdots, A_{11}$ in the family $\Omega$ have non-empty intersections, forming $\mathrm{C}_{11}^{2}=55$ unique pairs of sets.
Therefore, $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2} \geqslant 55$, which means
$$
\sum_{i=1}^{n} k_{i}\left(k_{i}-1\right) \geqslant 110 \text {. }
$$
Given $\max k_{i}=m$, from the above inequality we have
$$
\begin{array}{l}
110 \leqslant \sum_{i=1}^{n} k_{i}\left(k_{i}-1\right) \\
\leqslant \sum_{i=1}^{n} k_{i}(m-1)=55(m-1) .
\end{array}
$$
Thus, $m \geqslant 3$.
If $m=\max k_{i}=3$, then for any $i$, $k_{i} \leqslant 3$, which would imply that all $k_{i}=3$.
In fact, if some $k_{i} \leqslant 2$, meaning the number of sets containing element $x_{i}$ is no more than two, then at least nine sets in the family $\Omega$ do not contain $x_{i}$. Let these sets be $A_{1}, A_{2}, \cdots, A_{9}$, and let $A_{10}$ contain $x_{i}$, denoted as $A_{10}=\left\{x_{i}, a, b, c, d\right\}$. Since $A_{10} \cap A_{i} \neq \varnothing$, then
$$
A_{10} \cap A_{i} \subset\{a, b, c, d\}(i=1,2, \cdots, 9) .
$$
Therefore, among the four elements $a, b, c, d$, there must be one element (say $a$) that belongs to at least three sets in the family $\Omega$, i.e., it belongs to at least four sets among $A_{1}, A_{2}, \cdots, A_{9}, A_{10}$, which contradicts $\max k_{i}=3$.
Thus, all $k_{i}=3$.
However, this leads to $55=\sum_{i=1}^{n} k_{i}=3 n$, which is a contradiction.
Therefore, $m>3$, i.e., $m \geqslant 4$.
Furthermore, $m=4$ is achievable.
For this, a geometric construction method is used.
The "five-element set" is associated with the pentagram in Figure 3, which has 10 vertices and the midpoints of each side, totaling 15 points, labeled as 1, 2, ..., 15.
Figure 3
Construct 11 five-element sets of $A=\{1,2, \cdots, 15\}$ as follows: the 5 points on the circumference of the pentagram, the 5 points on each of the 5 sides, and the 5 points on the "cross" shapes like $(1,4,8,6,11)$, each forming a set, resulting in 11 sets.
Therefore, the minimum value of $m$ is 4.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 When $n$ is any real number and $k$ is a certain specific integer, the equation
$$
n(n+1)(n+2)(n+3)+1=\left(n^{2}+k n+1\right)^{2}
$$
holds. Then $k=$ $\qquad$ . [1]
(2010, Taiyuan Junior High School Mathematics Competition)
|
【Analysis】Since the left side of the given equation is a polynomial and the right side is in the form of a product, we only need to factorize the left side. The method of factorization is to use the overall idea and the complete square formula to handle it.
Solution Note that,
$$
\begin{array}{l}
n(n+1)(n+2)(n+3)+1 \\
=\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1 \\
=\left(n^{2}+3 n\right)^{2}+2\left(n^{2}+3 n\right)+1 \\
=\left(n^{2}+3 n+1\right)^{2} .
\end{array}
$$
Therefore, $k=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given $a, b, c \in \mathbf{R}$, and
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {. }
$$
Then there exists an integer $k$, such that the following equations hold for:
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}$;
(2) $\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}=\frac{1}{a^{2 k+1}+b^{2 k+1}+c^{2 k+1}}$;
(3) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k}=\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}$;
(4) $\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}=\frac{1}{a^{2 k}+b^{2 k}+c^{2 k}}$.
(2010, National High School Mathematics League Xinjiang Uygur Autonomous Region Preliminary)
|
【Analysis】The condition equation in this problem is a fractional equation, which is relatively complex. The key to solving this problem is to simplify the relationship between the letters $a, b, c$. First, eliminate the denominator and rearrange the condition equation into the form $f(a, b, c)=0$, then factorize $f(a, b, c)$. For polynomial factorization problems with multiple variables, the method of selecting the main variable is usually adopted, arranging the polynomial in descending order of the main variable $a$, which generally allows for a smooth factorization.
Solution From the given condition, we have
$$
\begin{array}{l}
(b+c) a^{2}+(b+c)^{2} a+b c(b+c)=0 \\
\Rightarrow(b+c)(c+a)(a+b)=0 \\
\Rightarrow b+c=0 \text { or } c+a=0 \text { or } a+b=0
\end{array}
$$
$\Rightarrow b+c=0$ or $c+a=0$ or $a+b=0$
$\Rightarrow a, b, c$ have at least two numbers that are opposites.
Assume $a+b=0$ without loss of generality.
$$
\text { Then }\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\left(\frac{1}{c}\right)^{2 k+1}=\frac{1}{c^{2 k+1}} \text {. }
$$
Therefore, equation (1) holds.
Similarly, equation (2) also holds, while equations (3) and (4) do not hold.
Thus, the number of equations that hold is 2.
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. The graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two points $A$ and $B$, with the vertex at $C$. If $\triangle A C B$ is a right triangle, then the value of the discriminant is $\qquad$.
|
3. 4 .
As shown in Figure 4.
From the problem, we know $\Delta=b^{2}-4 a c>0$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right), C\left(-\frac{b}{2 a}, \frac{4 a c-b^{2}}{4 a}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}$.
Then $\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}$
$=\left(-\frac{b}{a}\right)^{2}-\frac{4 c}{a}=\frac{b^{2}-4 a c}{a^{2}}$.
Assume $a>0$. Then
$A B=\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-4 a c}}{a}$.
Thus $\frac{\left|4 a c-b^{2}\right|}{4 a}=\frac{\sqrt{b^{2}-4 a c}}{2 a} \Rightarrow \sqrt{b^{2}-4 a c}=2$.
Therefore, $\Delta=b^{2}-4 a c=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. $a_{1}, a_{2}, a_{3}, \cdots$ is an arithmetic sequence, where $a_{1}>0, s_{n}$ represents the sum of the first $n$ terms. If $S_{3}=S_{11}$, in $S_{1}, S_{2}, S_{3}, \cdots$ the largest number is $S_{k}$, then $k=$ $\qquad$ .
|
-1.7 .
Let the common difference be $d$. Then
$$
\begin{array}{l}
a_{n}=a_{1}+(n-1) d . \\
\text { By } S_{3}=S_{11} \Rightarrow d=-\frac{2}{13} a_{1}<0 . \\
\text { Hence } a_{n}=a_{1}+(n-1)\left(-\frac{2}{13} a_{1}\right) \\
=\frac{a_{1}}{13}(15-2 n),
\end{array}
$$
and the largest positive integer $n$ for which $a_{n} \geqslant 0$ is $n=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Consider a tangent line to the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$, which intersects the two symmetry axes of the ellipse at points $A$ and $B$. Then the minimum length of segment $AB$ is $\qquad$ .
|
2. 8 .
Let the point of tangency be \( P(5 \cos \theta, 3 \sin \theta) \). Then the equation of the tangent line to the ellipse at point \( P \) is
$$
\frac{\cos \theta}{5} x + \frac{\sin \theta}{3} y = 1,
$$
which intersects the \( x \)-axis and \( y \)-axis at
$$
A\left(\frac{5}{\cos \theta}, 0\right), B\left(0, \frac{3}{\sin \theta}\right) \text{. }
$$
Therefore, \( A B^{2} = \frac{5^{2}}{\cos ^{2} \theta} + \frac{3^{2}}{\sin ^{2} \theta} \).
By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
& \frac{5^{2}}{\cos ^{2} \theta} + \frac{3^{2}}{\sin ^{2} \theta} \\
& = \left( \frac{5^{2}}{\cos ^{2} \theta} + \frac{3^{2}}{\sin ^{2} \theta} \right) \left( \cos ^{2} \theta + \sin ^{2} \theta \right) \\
\geqslant & (5 + 3)^{2} = 8^{2} .
\end{aligned}
$$
Thus, \( |A B| \geqslant 8 \).
The equality holds when \( \tan ^{2} \theta = \frac{3}{5} \).
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In rectangle $A B C D$, it is known that $A B=2, B C=3$, $E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. Rotate $\triangle F A B$ $90^{\circ}$ around $E F$ to $\triangle F A^{\prime} B^{\prime}$. Then the volume of the tetrahedron $A^{\prime} B^{\prime} C D$ is $\qquad$ .
|
3. 2 .
It is known that $E F=B C=3$, and the plane of $\triangle F A^{\prime} B^{\prime}$ divides the tetrahedron $A^{\prime} B^{\prime} C D$ into two tetrahedrons of equal volume, $C F A^{\prime} B^{\prime}$ and $D F A^{\prime} B^{\prime}$. Their heights are $C F=D F=1$, and $S_{\triangle F A^{\prime} B^{\prime}}=3$. Therefore,
$$
\begin{array}{l}
=2 \times \frac{1}{3} \times 1 \times 3=2 \text {. } \\
\end{array}
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find all real roots of the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right)
$$
All real roots. ${ }^{[4]}$
(2011, International Invitational Competition for Young Mathematicians in Cities)
|
【Analysis】The most basic method to solve higher-degree equations is to convert them into lower-degree equations for solving, that is, to handle them as linear or quadratic equations. Factorization is the most powerful tool to achieve such a transformation. The preferred method for factoring higher-degree polynomials is the trial root method (synthetic division). After rearranging the equation into $f(x)=0$, it is found that all coefficients are positive. If there is a rational root, it must be a negative root. The leading coefficient is 1, and the constant term is 3. If there is a rational root, it can only be -1 or -3.
Solution: The original equation can be transformed into
$$
x^{4}+3 x^{3}+6 x^{2}+7 x+3=0.
$$
It is easy to see that $x=-1$ is a solution of the equation.
$$
\begin{array}{l}
\text { Therefore, } x^{4}+3 x^{3}+6 x^{2}+7 x+3 \\
=(x+1)\left(x^{3}+2 x^{2}+4 x+3\right) \\
=(x+1)^{2}\left(x^{2}+x+3\right)=0.
\end{array}
$$
Notice that, $x^{2}+x+3=\left(x+\frac{1}{2}\right)^{2}+\frac{11}{4}>0$.
Therefore, the equation has a unique real root -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. For any $x, y \in [0,1]$, the function
$$
f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}
$$
has a maximum value of $\qquad$ .
|
7. 1.
Since $x, y \in [0,1]$, then $x \leqslant \sqrt{x}, y \leqslant \sqrt{y}$.
Let $x=\sin ^{2} \alpha, y=\sin ^{2} \beta\left(\alpha, \beta \in\left[0, \frac{\pi}{2}\right]\right)$.
Thus, $f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}$
$$
\begin{array}{l}
\leqslant \sqrt{x(1-y)}+\sqrt{y(1-x)} \\
=\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta \\
=\sin (\alpha+\beta) \leqslant 1,
\end{array}
$$
The equality holds if and only if $\alpha+\beta=\frac{\pi}{2}$, and
$$
x=\sqrt{x}, y=\sqrt{y} \text {. }
$$
At this point, $\{x, y\}=\{0,1\}$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. In the tetrahedron $ABCD$, it is known that $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$. The radius of the sphere that passes through $D$ and is tangent to the plane $ABC$ and internally tangent to the circumscribed sphere of the tetrahedron is 1, then the radius of the circumscribed sphere of the tetrahedron $ABCD$ is
|
12.3.
As shown in Figure 3, draw a perpendicular from point $D$ to plane $ABC$, with the foot of the perpendicular being $H$. Draw $DE \perp AB$ and $DF \perp AC$, with the feet of the perpendiculars being $E$ and $F$ respectively.
Then $HE \perp AB$, $HF \perp AC$, and
$AE = AF = AD \cos 45^{\circ} = \sqrt{6}$.
From $\triangle AEH \cong \triangle AFH \Rightarrow \angle HAE = 30^{\circ}$.
Thus, $AH = \frac{AE}{\cos 30^{\circ}} = 2\sqrt{2}$,
$$
DH = \sqrt{AD^2 - AH^2} = 2,
$$
Therefore, $DH$ is the diameter of a sphere with radius 1.
Thus, the center $O$ of the circumscribed sphere of tetrahedron $ABCD$ lies on the extension of $DH$.
Let the radius of the circumscribed sphere be $r$. Then
$$
r^2 = (r-2)^2 + (2\sqrt{2})^2 \Rightarrow r = 3.
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Let the lengths of the two legs of a right triangle be $a$ and $b$, and the length of the hypotenuse be $c$. If $a$, $b$, and $c$ are all integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition. ${ }^{(6)}$
(2010, National Junior High School Mathematics Competition, Tianjin Preliminary Contest)
|
【Analysis】In a right-angled triangle, the three sides satisfy the Pythagorean theorem. Given the condition $c=\frac{1}{3} a b-(a+b)$, one unknown can be eliminated to obtain a quadratic equation in two variables, which is generally difficult to solve. However, the problem of integer solutions to a quadratic equation in two variables can be handled through factorization.
By the Pythagorean theorem, we have $c^{2}=a^{2}+b^{2}$.
Substituting $c=\frac{1}{3} a b-(a+b)$ into the above equation, we get
$$
\begin{array}{l}
a^{2}+b^{2} \\
=\frac{1}{9}(a b)^{2}-\frac{2}{3} a b(a+b)+a^{2}+2 a b+b^{2} .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
a b-6(a+b)+18=0 \\
\Rightarrow(a-6)(b-6)=18=1 \times 18=2 \times 9=3 \times 6 .
\end{array}
$$
Since $a$ and $b$ are both positive integers, we can assume without loss of generality that $a<b$.
Solving, we get
$$
(a, b, c)=(7,24,25),(8,15,17),(9,12,15) \text {. }
$$
Thus, there are three right-angled triangles that satisfy the conditions.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the equation with respect to $x$
$$
x^{2}+2(m+3) x+m^{2}+3=0
$$
has two real roots $x_{1}$ and $x_{2}$, then the minimum value of $\left|x_{1}-1\right|+\left|x_{2}-1\right|$ is $\qquad$.
|
2.6 .
According to the problem, we have
$$
\begin{array}{l}
\Delta=[2(m+3)]^{2}-4\left(m^{2}+3\right) \geqslant 0 \\
\Rightarrow m \geqslant-1 .
\end{array}
$$
Then $x_{1}+x_{2}=-2(m+3)<0$.
When $x=1$, the left side of the equation is greater than 0, thus, $x_{1}$ and $x_{2}$ are on the same side of 1.
$$
\begin{array}{l}
\text { Hence }\left|x_{1}-1\right|+\left|x_{2}-1\right|=2-\left(x_{1}+x_{2}\right) \\
=2(m+4) \geqslant 6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) As shown in Figure 2, in the isosceles right triangle $\triangle ABC$, $\angle C=90^{\circ}$, points $D$ and $E$ are on side $BC$, and point $F$ is on the extension of $AC$, such that $BE=ED=CF$. Find the tangent value of $\angle CEF + \angle CAD$.
---
The translation preserves the original text's line breaks and formatting.
|
$\begin{array}{l}\text { I. Draw } D G \perp A B \text { at point } G . \\ \text { Let } A C=B C=x, \\ B E=E D=C F=y\left(0<y<\frac{x}{2}\right) . \\ \text { Then } G D=\sqrt{2} y \Rightarrow A G=\sqrt{2}(x-y)=\sqrt{2} C E \text {. } \\ \text { Therefore, } \mathrm{Rt} \triangle E C F \backsim \mathrm{Rt} \triangle A G D \\ \Rightarrow \angle C E F=\angle G A D \\ \Rightarrow \angle C E F+\angle C A D \\ =\angle G A D+\angle C A D=45^{\circ} \\ \Rightarrow \tan (\angle C E F+\angle C A D)=1 .\end{array}$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$ has a segment of digits $\overline{2012}$ in its decimal part, where $n$ is the smallest number satisfying the condition. Then $\left[\frac{m}{\sqrt{n}}\right]=$ $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
8. 2 .
Let $\frac{m}{n}=\overline{A . B 2012 C}$, where $A, B, C$ are digit strings, and the lengths of $A, B$ are $k, l (k, l \in \mathbf{N})$. Then
$$
\frac{10^{6}(m-n A)-n B}{n}=\overline{0.2012 C} \triangleq \frac{a}{b},
$$
where $(a, b)=1, 1 \leqslant b \leqslant n$.
Assume $\frac{m}{n}=\overline{0.2012 C}$.
$$
\begin{array}{l}
\text { Hence } \frac{2012}{10000} \leqslant \frac{m}{n}<\frac{2013}{10000} \\
\Rightarrow \frac{10000}{2013}<\frac{n}{m} \leqslant \frac{10000}{2012} .
\end{array}
$$
Let $n=5 m-r$. From
$$
\begin{array}{l}
\text { Equation (1) } \Rightarrow \frac{60}{2.012} \leqslant \frac{r}{m}<\frac{65}{2013} \\
\Rightarrow \frac{2013}{13}<\frac{5 m}{r}<\frac{503}{3} \\
\Rightarrow \frac{2000}{13} r<5 m-r \leqslant \frac{500}{3} r .
\end{array}
$$
Take $r=1, m=31$, then $n=5 m-r=154$. This is the minimum value.
Thus $\left[\frac{m}{\sqrt{n}}\right]=\left[\sqrt{\frac{961}{154}}\right]=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given that $a$ and $b$ are real numbers, and $a^{2} + ab + b^{2} = 3$. If the maximum value of $a^{2} - ab + b^{2}$ is $m$, and the minimum value is $n$, find the value of $m + n$. ${ }^{\text {[2] }}$
|
Let $a^{2}-a b+b^{2}=t$.
Combining this with the given equation, we get
$$
a b=\frac{3-t}{2}, a+b= \pm \sqrt{\frac{9-t}{2}} .
$$
Thus, $a$ and $b$ are the two real roots of the quadratic equation in $x$:
$$
x^{2} \pm \sqrt{\frac{9-t}{2}} x+\frac{3-t}{2}=0
$$
Therefore, $\Delta=\left( \pm \sqrt{\frac{9-t}{2}}\right)^{2}-4 \times \frac{3-t}{2} \geqslant 0$.
Solving this, we get $t \geqslant 1$.
Also, $\frac{9-t}{2} \geqslant 0$, which implies $t \leqslant 9$.
Thus, $1 \leqslant t \leqslant 9$.
Hence, $m=9, n=1$
$$
\Rightarrow m+n=10 \text {. }
$$
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initially 340 Given two points $A$ and $B$ on a straight line $l$, the distance between them is $10000 \mathrm{~cm}$. At points $A$ and $B$, there are two movable barriers, designated as Barrier 1 and Barrier 2, respectively. Assume there is a ping-pong ball between $A$ and $B$, moving along the straight line $l$ at a uniform speed. Initially, Barrier 1 and the ping-pong ball start from point $A$ and move along the line $l$ towards point $B$ at uniform speeds of $4 \mathrm{~cm} / \mathrm{s}$ and $5 \mathrm{~cm} / \mathrm{s}$, respectively, and assume the speed of Barrier 1 remains constant. When the ping-pong ball first hits Barrier 2, it reverses direction, and its speed decreases by $20 \%$, while Barrier 2 quickly moves $10 \mathrm{~cm}$ along the direction of $A B$, so the distance from Barrier 2 to point $A$ is now $10010 \mathrm{~cm}$. When the ping-pong ball first contacts Barrier 1 (referred to as the first contact between the ping-pong ball and Barrier 1) and reverses direction, its speed increases by $25 \%$. When the ping-pong ball hits Barrier 2 for the second time, it continues to move in the same manner, and Barrier 2 quickly moves $9 \mathrm{~cm}$ along the direction of $A B$. Each subsequent time the ping-pong ball hits Barrier 2, it continues to move in the same manner, and Barrier 2 moves $1 \mathrm{~cm}$ less than the previous time, until the number of times the ping-pong ball hits Barrier 2 exceeds 10, at which point Barrier 2 no longer moves. When the distance between Barrier 1 and Barrier 2 first equals $1 \mathrm{~cm}$, how many times has the ping-pong ball contacted Barrier 1?
|
Let's assume that when the ping-pong ball contacts the 2nd board for the $n$-th time, the position of the 2nd board after it moves quickly is $B_{n}$; when the ping-pong ball contacts the 1st board for the $n$-th time, its position is $A_{n}$, and we set
$$
\begin{array}{l}
B_{0}=B, A_{n} B_{n-1}=x_{n}, A_{n} B_{n}=y_{n} . \\
\text { Then } \frac{10000}{5}+\frac{x_{1}}{5(1-20 \%)}=\frac{10000-x_{1}}{4} \\
\Rightarrow x_{1}=\frac{10000(5-4)(1-20 \%)}{4+5(1-20 \%)}=1000 .
\end{array}
$$
Thus, $y_{1}=1010$.
Similarly, $x_{2}=101, y_{2}=110$;
$$
\begin{array}{l}
x_{3}=11, y_{3}=19 ; \\
x_{4}=1.9, y_{4}=8.9 .
\end{array}
$$
When the ping-pong ball hits the 2nd board for the 5th time, the distance between the 1st board and the 2nd board is
$8.9 \times\left(1-\frac{4}{5}\right)=1.78>1$.
Thus, $x_{5}=0.89, y_{5}=6.89$.
When the ping-pong ball hits the 2nd board for the 6th time, the distance between the 1st board and the 2nd board is
6. $89 \times\left(1-\frac{4}{5}\right)=1.378>1$.
Thus, $x_{6}=0.689, y_{6}=5.689$.
When the ping-pong ball hits the 2nd board for the 7th time, the distance between the 1st board and the 2nd board is
$5.689 \times\left(1-\frac{4}{5}\right)=1.1378>1$.
Thus, $x_{1}=0.5689, y_{7}=4.5689$.
When the ping-pong ball hits the 2nd board for the 8th time, the distance between the 1st board and the 2nd board is
4. $5689 \times\left(1-\frac{4}{5}\right)=0.91378<1$.
Therefore, when the distance between the 1st board and the 2nd board is first equal to $1 \mathrm{~cm}$, the 1st board has contacted the ping-pong ball 7 times.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This line is not translated as it seems to be an instruction and not part of the text to be translated. If you need this line translated as well, please let me know.
|
Notice,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1000} \text {. }
$$
Consider the integer-coefficient quadratic equation $x^{2}-10 x+1=0$ with $5+2 \sqrt{6}$ as one of its roots, the other root being $5-2 \sqrt{6}$.
$$
\begin{array}{l}
\text { Let } a=5+2 \sqrt{6}, b=5-2 \sqrt{6}, \\
u_{n}=a^{n}+b^{n}(n=1,2, \cdots) .
\end{array}
$$
It is easy to see that $a+b=10, a b=1$.
Since $a^{2}=10 a-1, b^{2}=10 b-1$, for all $n \in \mathbf{N}_{+}$, we have
$$
a^{n+2}=10 a^{n+1}-a^{n}, b^{n+2}=10 b^{n+1}-b^{n} \text {. }
$$
Adding the above two equations gives $u_{n+2}=10 u_{n+1}-u_{n}$, where,
$$
u_{1}=a+b=10, u_{2}=a^{2}+b^{2}=98 \text {. }
$$
Therefore, $u_{n}$ is a positive integer, and
$$
u_{n+4} \equiv-u_{n+2} \equiv u_{n}(\bmod 10)(n=1,2, \cdots) \text {. }
$$
Since $1006=4 \times 251+2$, we get
$$
u_{1006} \equiv u_{2} \equiv 8(\bmod 10) \text {. }
$$
Given $0<b<1$, we know
$$
\left[(\sqrt{2}+\sqrt{3})^{2012}\right]=u_{1000}-1=7(\bmod 10),
$$
which means 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 5 Let $x$ be a real number. Then
$$
|x-1|+|x+1|+|x+5|
$$
the minimum value is $\qquad$ (s)
|
Let $y=|x-1|+|x+1|+|x+5|$. When $x<-5$,
$$
y=1-x-x-1-x-5=-3 x-5 \text{; }
$$
When $-5 \leqslant x<-1$,
$$
y=1-x-x-1+x+5=-x+5 \text{; }
$$
When $-1 \leqslant x<1$,
$$
y=1-x+x+1+x+5=x+7 \text{; }
$$
When $x \geqslant 1$,
$$
y=x-1+x+1+x+5=3 x+5 \text{. }
$$
By the properties of linear functions, we know that when $x=-1$, $y_{\text {min }}=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the vectors $\overrightarrow{O A}=(1,0), \overrightarrow{O B}=(1,1)$, and $O$ be the origin. A moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O A} \leqslant 1, \\
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O B} \leqslant 2
\end{array}\right.
$$
Then the area of the figure formed by the point $Q(x+y, y)$ is
|
2. 2 .
From the problem, the point $P(x, y)$ satisfies
$$
\begin{array}{l}
\left\{\begin{array}{l}
0 \leqslant x \leqslant 1, \\
0 \leqslant x+y \leqslant 2 .
\end{array}\right. \\
\text { Let }\left\{\begin{array}{l}
x+y=u, \\
y=v .
\end{array}\right.
\end{array}
$$
Then the point $Q(u, v)$ satisfies
$$
\left\{\begin{array}{l}
0 \leqslant u-v \leqslant 1, \\
0 \leqslant u \leqslant 2 .
\end{array}\right.
$$
In the $u O v$ plane, draw
the plane region formed by
the point $Q(u, v)$ (the shaded part in Figure 2), its area equals 2.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $(a, b)$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-a-2 b$ is . $\qquad$
|
Hint: Use the method of completing the square. The original expression can be transformed into
$$
\left(a+\frac{1}{2} b-\frac{1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 \text {. }
$$
When $a+\frac{1}{2} b-\frac{1}{2}=0, b-1=0$, i.e., $a=0$, $b=1$, the original expression can achieve the minimum value of -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let positive real numbers $a, b, c$ satisfy $\frac{2}{a}+\frac{1}{b}=\frac{\sqrt{3}}{c}$. Then the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is $\qquad$ .
|
4. 9 .
From the given, we have $\frac{2 c}{a}+\frac{c}{b}=\sqrt{3}$.
By the Cauchy-Schwarz inequality and the AM-GM inequality, we get
$$
\begin{array}{l}
\frac{2 a^{2}+b^{2}}{c^{2}} \\
=\frac{1}{3}(2+1)\left[2\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}\right] \\
\geqslant \frac{1}{3}\left(\frac{2 a}{c}+\frac{b}{c}\right)^{2} \\
=\frac{1}{9}\left[\left(\frac{2 a}{c}+\frac{b}{c}\right)\left(\frac{2 c}{a}+\frac{c}{b}\right)\right]^{2} \\
=\frac{1}{9}\left(4+\frac{2 a}{b}+\frac{2 b}{a}+1\right)^{2} \\
\geqslant \frac{1}{9}(4+2 \times 2+1)^{2}=9,
\end{array}
$$
Equality holds if and only if $a=b=\sqrt{3} c$. Therefore, the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the equation in $x$
$$
x^{3}-4 x^{2}+5 x+a=0(a \in \mathbf{R})
$$
has three real roots $x_{1}, x_{2}, x_{3}$. Then the maximum value of $\max \left\{x_{1}, x_{2}, x_{3}\right\}$ is $\qquad$ .
|
6.2.
Assume $x_{3}=\max \left\{x_{1}, x_{2}, x_{3}\right\}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=4, \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=5
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x_{1}+x_{2}=4-x_{3}, \\
x_{1} x_{2}=5-x_{3}\left(x_{1}+x_{2}\right)=5-x_{3}\left(4-x_{3}\right) .
\end{array}\right.
\end{array}
$$
Thus, the quadratic equation with roots $x_{1} 、 x_{2}$ is
$$
\begin{array}{l}
x^{2}-\left(4-x_{3}\right) x-5+x_{3}\left(4-x_{3}\right)=0 \\
\Rightarrow \Delta=\left(4-x_{3}\right)^{2}-4\left[5-x_{3}\left(4-x_{3}\right)\right] \geqslant 0 \\
\Rightarrow 3 x_{3}^{2}-8 x_{3}+4 \leqslant 0 \\
\Rightarrow \frac{2}{3} \leqslant x_{3} \leqslant 2 .
\end{array}
$$
When $x_{3}=2$, $x_{1}=x_{2}=1, a=-2$.
Therefore, the maximum value of $\max \left\{x_{1}, x_{2}, x_{3}\right\}$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a>1$. Then when the graphs of $y=a^{x}$ and $y=\log _{a} x$ are tangent, $\ln \ln a=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3. -1.
Since the two functions are inverse functions of each other and are symmetric about the line $y=x$, the point of tangency lies on $y=x$.
Let the point of tangency be $\left(x_{0}, y_{0}\right)$. Then
$$
\begin{array}{l}
x_{0}=a^{x_{0}}, \\
a^{x_{0}} \ln a=1 .
\end{array}
$$
Substituting equation (1) into equation (2) gives $x_{0} \ln a=1$, i.e.,
$$
\ln a^{x_{0}}=1 \text{. }
$$
Substituting equation (1) into equation (3) gives
$$
\ln x_{0}=1 \Rightarrow x_{0}=e \text{. }
$$
Thus, $\ln a=\frac{1}{\mathrm{e}} \Rightarrow \ln \ln a=-1$.
|
-1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 3 Given $a b c=-1, a^{2} b+b^{2} c+c^{2} a=t$, $\frac{a^{2}}{c}+\frac{b}{c^{2}}=1$. Try to find the value of $a b^{5}+b c^{5}+c a^{5}$.
|
【Analysis】Using $a b c=-1$ can make common substitutions
$$
a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x} \text {. }
$$
This problem involves three letters, and can also be solved by the method of elimination.
Solution 1 Let $a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x}$.
$$
\begin{array}{l}
\text { Then } \frac{a^{2}}{c}+\frac{b}{c^{2}}=1 \text { becomes } \\
x^{2} y^{3}+y^{2} z^{3}+z^{2} x^{3}=0 . \\
\text { Hence } a b^{5}+b c^{5}+c a^{5}-3 \\
=\frac{x y^{4}}{z^{5}}+\frac{y z^{4}}{x^{5}}+\frac{z x^{4}}{y^{5}}-3 \\
=\frac{x^{6} y^{9}+y^{6} z^{9}+z^{6} x^{9}-3 x^{5} y^{5} z^{5}}{x^{5} y^{5} z^{5}} \\
=\frac{\left(x^{2} y^{3}+y^{2} z^{3}+z^{2} x^{3}\right)\left(x^{4} y^{6}+y^{4} z^{6}+z^{4} x^{6}-x^{2} y^{5} z^{3}-y^{2} z^{5} x^{3}-z^{2} x^{5} y^{3}\right)}{x^{5} y^{5}} \\
=0 \\
\Rightarrow a b^{5}+b c^{5}+c a^{5}=3 .
\end{array}
$$
Solution 2 From $a b c=-1 \Rightarrow b=-\frac{1}{a c}$.
Substitute into $\frac{a^{2}}{c}+\frac{b}{c^{2}}=1$, we get $a^{3} c^{2}=a c^{3}+1$.
$$
\begin{array}{l}
\text { Then } a b^{5}+b c^{5}+c a^{5} \\
=-\frac{1}{a^{4} c^{5}}-\frac{c^{4}}{a}+c a^{5}=\frac{a^{9} c^{6}-1-a^{3} c^{9}}{a^{4} c^{5}} \\
=\frac{\left(a c^{3}+1\right)^{3}-1-a^{3} c^{9}}{a^{4} c^{5}}=\frac{3\left(a^{2} c^{6}+a c^{3}\right)}{a^{4} c^{5}} \\
=\frac{3\left(a c^{3}+1\right)}{a^{3} c^{2}}=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. 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 $\tan (\alpha+\beta+\gamma)=$ $\qquad$
|
3. 11 .
Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then the given equations can be written as
$$
\begin{array}{l}
x+y+z=\frac{17}{6}, \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}, \\
\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5} . \\
\text { From (1) } \div \text { (3) we get } \\
x y z=-\frac{5}{6} .
\end{array}
$$
From (1) $\div$ (3) we get
From (2) $\times$ (4) we get $x y+y z+z x=\frac{2}{3}$.
Thus $\tan (\alpha+\beta+\gamma)=\frac{x+y+z-x y z}{1-(x y+y z+z x)}=11$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $a, b, c$ satisfy
$$
a+b+c=a^{2}+b^{2}+c^{2} \text {. }
$$
Then the maximum value of $a+b+c$ is $\qquad$
|
4.3.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
3(a+b+c) \\
=\left(1^{2}+1^{2}+1^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)
\end{array}
$$
$$
\begin{array}{l}
\geqslant(a+b+c)^{2} \\
\Rightarrow a+b+c \leqslant 3 .
\end{array}
$$
Equality holds if and only if \(a=b=c=1\).
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that the base edge length of a regular tetrahedron is 6, and the side edge is 4. Then the radius of the circumscribed sphere of this regular tetrahedron is $\qquad$
|
3. 4 .
From the problem, we know that the distance from point $A$ to the base $B C D$ is 2, and the radius of its circumscribed sphere is $R$. Then
$$
R^{2}-12=(R-2)^{2} \Rightarrow R=4 \text {. }
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $S_{4} \leqslant 4, S_{5} \geqslant 15$, then the minimum value of $a_{4}$ is $\qquad$ .
|
4. 7 .
Let the common difference be $d$. From the given conditions, we have
$$
\begin{array}{l}
2 a_{4} \leqslant 2+3 d, d \leqslant a_{4}-4 \\
\Rightarrow 2 a_{4} \leqslant 2+3 d \leqslant 2+3\left(a_{4}-3\right) \\
\Rightarrow a_{4} \geqslant 7 .
\end{array}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 As shown in Figure 3, in the acute triangle $\triangle ABC$, it is known that $BE \perp AC$ at point $E$, $CD \perp AB$ at point $D$, $BC=25$, $CE=7$, $BD=15$. If $BE$ and $CD$ intersect at point $H$, connect $DE$, and construct a circle with $DE$ as the diameter, which intersects $AC$ at another point $F$. Find the length of $AF$.
(2012, Joint Autonomous Admissions Examination of Tsinghua University and Other Universities)
|
Solve for DF. From the given conditions, we have
$$
\begin{array}{l}
\cos B=\frac{3}{5}, \cos C=\frac{7}{25} \\
\Rightarrow \sin B=\frac{4}{5}, \sin C=\frac{24}{25} \\
\Rightarrow \sin A=\sin B \cdot \cos C+\sin C \cdot \cos B=\frac{4}{5}=\sin B \\
\Rightarrow \angle A=\angle B \Rightarrow AC=BC=25 .
\end{array}
$$
From $CD \perp AB, DF \perp AC$, we know that $D$ and $F$ are the midpoints of $AB$ and $AE$, respectively.
Also, $AE=AC-CE=18$, so $AF=9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let two ellipses be
$$
\frac{x^{2}}{t^{2}+2 t-2}+\frac{y^{2}}{t^{2}+t+2}=1
$$
and $\frac{x^{2}}{2 t^{2}-3 t-5}+\frac{y^{2}}{t^{2}+t-7}=1$
have common foci. Then $t=$ $\qquad$ .
|
5.3.
Given that the two ellipses have a common focus, we have
$$
\begin{array}{l}
t^{2}+2 t-2-\left(t^{2}+t+2\right) \\
=2 t^{2}-3 t-5-\left(t^{2}+t-7\right) \\
\Rightarrow t=3 \text { or } 2 \text { (rejected). }
\end{array}
$$
Therefore, $t=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Calculate:
$$
\sum_{k=0}^{2013}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2014}^{k}}=
$$
$\qquad$
|
5. 0 .
Let $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2 n}^{k}}$.
And $\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n}-k-1}$, so $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}}$.
Perform a summation index transformation $l=2 n-k-1$, then
$$
\begin{array}{l}
a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}} \\
=\sum_{l=0}^{2 n-1}(-1)^{\prime} \frac{l+1}{\mathrm{C}_{2 n}^{l}}=-a_{n} . \\
\text { Hence } a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2 n}^{k}}=0 .
\end{array}
$$
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Let $a, b, c > 0$, and $a+b+c=3$. Find the maximum value of $a^{2} b+b^{2} c+c^{2} a+a b c$.
|
Let's assume $a \leqslant b \leqslant c$ or $c \leqslant b \leqslant a$.
Then $c(b-a)(b-c) \leqslant 0$
$$
\begin{aligned}
\Rightarrow & b^{2} c+c^{2} a \leqslant a b c+c^{2} b \\
\Rightarrow & a^{2} b+b^{2} c+c^{2} a+a b c \\
& \leqslant a^{2} b+c^{2} b+2 a b c \\
& =b(a+c)^{2}=b(3-b)^{2} \\
& \leqslant \frac{1}{2}\left(\frac{2 b+3-b+3-b}{3}\right)^{3}=4 .
\end{aligned}
$$
Equality holds if and only if $a=b=c=1$ or $\{a, b, c\}=\{0, 1, 2\}$.
Thus, the maximum value of $a^{2} b+b^{2} c+c^{2} a+a b c$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50) Find the maximum value of $n$ such that there are $n$ points in the plane, where among any three points, there must be two points whose distance is 1.
|
If there exist $n(n \geqslant 8)$ points satisfying the conditions of the problem, let $V=\left\{v, v_{1}, v_{2}, \cdots, v_{7}\right\}$ represent any eight of these points. When and only when the distance between two points is 1, connect an edge between these two points, forming a graph $G$.
If there exists a point (let's assume it is $v$) with a degree less than or equal to 3, then at least four points not connected to $v$ must be pairwise connected, which is clearly impossible.
Therefore, the degree of any point is at least 4.
In the point set $V=\left\{v, v_{1}, v_{2}, \cdots, v_{7}\right\}$, take a point (let's assume it is $v$) at the vertex of the convex hull $T$. Point $v$ is connected to $v_{1}, v_{2}, v_{3}, v_{4}$ (where $v_{1}, v_{2}, v_{3}, v_{4}$ are arranged in a counterclockwise direction). Then $\left|v_{1} v_{4}\right| \neq 1$ (otherwise, $v_{2}, v_{3}, v_{4}$ would have no two points connected, which is a contradiction). Therefore, $v_{2}, v_{3}$ must each be connected to at least one of $v_{1}, v_{4}$. Thus, $v_{2}, v_{3}$ are each connected to exactly one of $v_{1}, v_{4}$ (otherwise, assume $v_{2}$ is connected to both $v_{1}, v_{4}$, then $v_{1}, v_{3}, v_{4}$ would have no two points connected, which is a contradiction).
Moreover, $v_{2}, v_{3}$ cannot both be connected to the same point among $v_{1}, v_{4}$. Therefore, among the points $v_{1}, v_{2}, v_{3}, v_{4}$, only $\left|v_{1} v_{2}\right|=\left|v_{3} v_{4}\right|=1$ or $\left|v_{1} v_{3}\right|=\left|v_{2} v_{4}\right|=1$ (with $\left.\left|v_{2} v_{3}\right| \neq 1\right)$ can occur.
Thus, each of $v_{1}, v_{2}, v_{3}, v_{4}$ must be connected to at least two of $v_{5}, v_{6}, v_{7}$. Therefore, the points $v_{5}, v_{6}, v_{7}$ must draw at least eight edges to $v_{1}, v_{2}, v_{3}, v_{4}$. By the pigeonhole principle, at least one point (let's assume it is $v_{5}$) must draw at least three edges to $v_{1}, v_{2}, v_{3}, v_{4}$, meaning there exist three points among $v_{1}, v_{2}, v_{3}, v_{4}$ that are all 1 unit away from both $v$ and $v_{5}$, which is impossible.
Therefore, $n \leqslant 7$.
Take two rhombuses $A B C D$ and $A B_{1} C_{1} D_{1}$ with side lengths of 1, such that
$$
\angle B A D=\angle B_{1} A D_{1}=60^{\circ}, C C_{1}=1 .
$$
It is easy to prove that among $A, B, C, D, B_{1}, C_{1}, D_{1}$, any three points have at least two points whose distance is 1.
Therefore, the maximum value of $n$ is 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $x, y$ are positive real numbers, $n \in \mathrm{N}$, and $n \geqslant 2$. Prove:
$$
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \geqslant 4 \text {. }
$$
|
$$
\begin{array}{l}
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \\
\geqslant 2 \sqrt{\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}} \cdot \sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}}} \\
=2 \sqrt[2 n]{\frac{\left(2^{n}-1\right)\left(x^{2}+y^{2}\right)+\left(2^{2 n}-2^{n+1}+2\right) x y}{x y}} \\
\geqslant 2 \sqrt[2 n]{\frac{2\left(2^{n}-1\right) x y+\left(2^{2 n}-2^{n+1}+2\right) x y}{x y}} \\
=4 .
\end{array}
$$
The equality holds if and only if \( x = y \).
|
4
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let the functions $f(x)=\ln x, g(x)=\frac{1}{2} x^{2}$. If $x_{1}>x_{2}>0$, for what value of $m(m \in \mathbf{Z}, m \leqslant 1)$ is it always true that
$$
m\left(g\left(x_{1}\right)-g\left(x_{2}\right)\right)>x_{1} f\left(x_{1}\right)-x_{2} f\left(x_{2}\right)
$$
holds.
|
Introduce an auxiliary function
$$
t(x)=m g(x)-x f(x)=\frac{m}{2} x^{2}-x \ln x \quad (x>0) \text {. }
$$
By the problem, $x_{1}>x_{2}>0$.
Therefore, if the original inequality always holds for $x>0$, i.e., the function $t(x)$ is monotonically increasing, then
$$
t^{\prime}(x)=m x-\ln x-1 \geqslant 0
$$
always holds.
Thus, $m \geqslant \frac{\ln x+1}{x}$ always holds, which means
$$
m \geqslant\left(\frac{\ln x+1}{x}\right)_{\max } \text {. }
$$
Introduce another auxiliary function
$$
h(x)=\frac{\ln x+1}{x} \quad (x>0) \text {. }
$$
By $h^{\prime}(x)=\frac{-\ln x}{x^{2}}$, we know that the function $h(x)$ is monotonically increasing on $(0,1]$ and monotonically decreasing on $(1,+\infty)$, i.e.,
$$
h(x)_{\max }=h(1)=1 \quad (m \geqslant 1) .
$$
Since $m \in \mathbf{Z}, m \leqslant 1$, hence $m=1$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
3 x-3 \geqslant 6 x+a, \\
x \geqslant 1
\end{array}\right.
$$
the solution is $1 \leqslant x \leqslant 3$. Then $a=$
|
$=, 1 .-12$.
From the given, we have $1 \leqslant x \leqslant \frac{1}{3}(-a-3)$. Then $\frac{1}{3}(-a-3)=3 \Rightarrow a=-12$.
|
-12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $x$ and $y$ be two distinct non-negative integers, and satisfy $x y + 2x + y = 13$. Then the minimum value of $x + y$ is $\qquad$
|
3. 5 .
From the problem, we know that $(x+1)(y+2)=15$.
$$
\begin{array}{l}
\text { Then }(x+1, y+2) \\
=(15,1),(5,3),(3,5),(1,15) \\
\Rightarrow(x, y)=(14,-1),(4,1),(2,3),(0,13) .
\end{array}
$$
Therefore, the minimum value of $x+y$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. In the arithmetic sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{20} \simeq \frac{1}{a}, a_{201}=\frac{1}{b}, a_{2012}=\frac{1}{c} \text {. }
$$
Then $1992 a c-1811 b c-181 a b=$
|
12. 0 .
Let the common difference of the arithmetic sequence be $d$. Then, according to the problem, we have
$$
\begin{array}{l}
a_{201}-a_{20}=\frac{a-b}{a b}=181 d, \\
a_{2012}-a_{201}=\frac{b-c}{b c}=1811 d, \\
a_{2012}-a_{20}=\frac{a-c}{a c}=1992 d .
\end{array}
$$
Therefore, $1992 a c-1811 b c-181 a b$
$$
=\frac{a-c}{d}+\frac{c-b}{d}+\frac{b-a}{d}=0 .
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given constants $a, b$ satisfy $a, b>0, a \neq 1$, and points $P(a, b), Q(b, a)$ are both on the curve $y=\cos (x+c)$, where $c$ is a constant. Then $\log _{a} b=$ $\qquad$
|
13. 1 .
Given points $P(a, b), Q(b, a)$ are both on the curve
$$
y=\cos (x+c)
$$
we know
$$
\begin{array}{l}
a-b=\cos (b+c)-\cos (a+c) \\
=2 \sin \left(\frac{a+b+2 c}{2}\right) \cdot \sin \frac{a-b}{2} .
\end{array}
$$
Without loss of generality, assume $a \geqslant b$.
If $a>b$, then
$$
\left|\sin \frac{a+b+2 c}{2}\right| \leqslant 1,\left|\sin \frac{a-b}{2}\right|<\frac{a-b}{2} \text {. }
$$
Thus $|a-b|<2\left|\frac{a-b}{2}\right|=|a-b|$, which is a contradiction.
Since the graphs of $y=x$ and $y=\cos (x+c)$ must intersect, we know that $a=b$ holds.
Therefore, $\log _{a} b=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $m$ is an integer, the equation
$$
x^{2}-m x+3-n=0
$$
has two distinct real roots, the equation
$$
x^{2}+(6-m) x+7-n=0
$$
has two equal real roots, and the equation
$$
x^{2}+(4-m) x+5-n=0
$$
has no real roots.
$$
\text { Then }(m-n)^{2013}=
$$
|
3. According to the problem, we have
$$
\left\{\begin{array}{l}
m^{2}-4(3-n)>0, \\
(6-m)^{2}-4(7-n)=0, \\
(4-m)^{2}-4(5-n)\frac{5}{3}, m<3 \text {. }
\end{array}\right.
$$
Then $m=2$. Consequently, $n=3$.
Therefore, $(m-n)^{2013}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) In a dormitory of a school, there are several students. On New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator, who in turn gives a card back to each student. In this way, a total of 51 greeting cards were used. Question: How many students live in the dormitory?
|
Let there be $x$ students and $y$ administrators living in the dormitory $\left(x, y \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
x(x-1)+x y+y=51 \\
\Rightarrow y=\frac{51+x-x^{2}}{x+1}=\frac{49}{x+1}-x+2 .
\end{array}
$$
Since $x, y$ are positive integers, we know that $(x, y)=(6,3)$.
Therefore, there are 6 students living in the dormitory.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In $\triangle A B C$, it is known that $O$ is the circumcenter, the three altitudes $A D, B E, C F$ intersect at point $H$, line $E D$ intersects $A B$ at point $M$, and $F D$ intersects $A C$ at point $N$. Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=$ $\qquad$
|
3. 0 .
It is known that, $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{O H}$.
Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot \overrightarrow{M N}$
$$
\begin{aligned}
= & \overrightarrow{O A} \cdot(\overrightarrow{M F}+\overrightarrow{F E}+\overrightarrow{E N})+ \\
& \overrightarrow{O B} \cdot(\overrightarrow{M F}+\overrightarrow{F N})+\overrightarrow{O C} \cdot(\overrightarrow{M E}+\overrightarrow{E N}) \\
= & (\overrightarrow{O A}+\overrightarrow{O B}) \cdot \overrightarrow{M F}+(\overrightarrow{O A}+\overrightarrow{O C}) \cdot \overrightarrow{E N} .
\end{aligned}
$$
Since $O$ is the circumcenter of $\triangle A B C$, $\overrightarrow{O A}+\overrightarrow{O B}$ and $\overrightarrow{O A}+\overrightarrow{O C}$ pass through the midpoints of segments $A B$ and $A C$, respectively.
Therefore, the value of equation (1) is 0.
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the real number pair $(x, y)$ satisfies the equation $(x-2)^{2}+y^{2}=3$, let the minimum and maximum values of $\frac{y}{x}$ be $m$ and $n$ respectively. Then $m+n=$
|
$\begin{array}{l}\text { II. 1.0. } \\ \text { Let } y=t x \text {. Then }\left(1+t^{2}\right) x^{2}-4 x+1=0 \text {. } \\ \text { By } \Delta=(-4)^{2}-4\left(1+t^{2}\right) \geqslant 0 \\ \Rightarrow-\sqrt{3} \leqslant t \leqslant \sqrt{3} \\ \Rightarrow m=-\sqrt{3}, n=\sqrt{3} \\ \Rightarrow m+n=0 \text {. }\end{array}$
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 2, in
rhombus $A B C D$, it is
known that $\angle A B C=60^{\circ}$, line
$E F$ passes through point $D$, and
intersects the extensions of
$B A$ and $B C$ at points $E$ and $F$, respectively. $M$
is the intersection of $C E$ and $A F$. If $C M=4, E M=5$, then $C A=$ $\qquad$
|
3. 6 .
It is easy to prove $\triangle E A D \backsim \triangle D C F$.
$$
\begin{array}{l}
\text { Therefore, } \frac{E A}{A D}=\frac{D C}{C F} \Rightarrow \frac{E A}{A C}=\frac{A C}{C F} \\
\Rightarrow \triangle E A C \backsim \triangle A C F \\
\Rightarrow \angle A E C=\angle C A F \\
\Rightarrow C A^{2}=C E \cdot C M=9 \times 4 \\
\Rightarrow C A=6 .
\end{array}
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Let $x, y (x>y)$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying $x-y \geqslant \frac{xy}{31}$. Find the maximum value of the number of elements $n$ in this set of natural numbers.
|
Let's assume $a_{1}1 \\
\Rightarrow d_{6} \geqslant 2 \Rightarrow a_{7}=a_{6}+d_{6} \geqslant 8 \\
\Rightarrow d_{7} \geqslant \frac{8^{2}}{31-8}>2 \Rightarrow d_{7} \geqslant 3 . \\
\Rightarrow a_{8}=a_{7}+d_{7} \geqslant 11 \\
\Rightarrow d_{8} \geqslant \frac{11^{2}}{31-11}>6 \Rightarrow d_{8} \geqslant 7 \\
\Rightarrow a_{9}=a_{8}+d_{8} \geqslant 18 \\
\Rightarrow d_{9} \geqslant \frac{18^{2}}{31-18}>24 \Rightarrow d_{9} \geqslant 25 \\
\Rightarrow a_{10}=a_{9}+d_{9} \geqslant 43>31 .
\end{array}
$$
Therefore, $n-1 \leqslant 9 \Rightarrow n \leqslant 10$.
At the same time, the ten natural numbers $1,2,3,4,5,6,8, 11,18, 43$ meet the conditions of the problem.
Thus, the maximum value of $n$ is 10.
|
10
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $a, b, x \in \mathbf{N}_{+}$, and $a \leqslant b$. $A$ is the solution set of the inequality
$$
\lg b - \lg a < \lg x < \lg b + \lg a
$$
It is known that $|A|=50$. When $ab$ takes its maximum possible value,
$$
\sqrt{a+b}=
$$
|
4. 6 .
It is easy to know, $\frac{b}{a}<x<a b, a \neq 1$.
Therefore, $a \geqslant 2$,
$$
\begin{array}{l}
50 \geqslant a b-\frac{b}{a}-1=a b\left(1-\frac{1}{a^{2}}\right)-1 \geqslant \frac{3}{4} a b-1 \\
\Rightarrow a b \leqslant 68 .
\end{array}
$$
Upon inspection, when and only when $a=2, b=34$, the equality holds.
At this time, $\sqrt{a+b}=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The function defined on the domain $R$
$$
f(x)=|\lg | x-2||-1 \text{. }
$$
If $b<0$, then the equation concerning $x$
$$
f^{2}(x)+b f(x)=0
$$
has $\qquad$ distinct real roots.
|
5. 8 .
From the problem, we know that the graph of $y=\lg x$ is first symmetrical about the $y$-axis, then the part below the $x$-axis is flipped up, followed by a rightward shift of 2 units and a downward shift of 1 unit to form the graph of $f(x)$. The zeros of $g(x)=f^{2}(x)+b f(x)$ are the roots of the equation. It is easy to see that $g(x)$ has 8 zeros.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $z$ be a complex number with modulus 2. Then the sum of the maximum and minimum values of $\left|z-\frac{1}{z}\right|$ is $\qquad$ [2]
|
Given $|z|=2$, we know
$$
\begin{array}{l}
|z+1|^{2}=(z+1)(\bar{z}+1) \\
=z \bar{z}+z+\bar{z}+1=5+2 \operatorname{Re} z, \\
|z-1|^{2}=(z-1)(\bar{z}-1) \\
=z \bar{z}-z-\bar{z}+1=5-2 \operatorname{Re} z .
\end{array}
$$
Therefore, $\left|z-\frac{1}{z}\right|=\left|\frac{z^{2}-1}{z}\right|$
$$
=\frac{|z+1||z-1|}{|z|}=\frac{\sqrt{25-4 \operatorname{Re} z}}{2} \text {. }
$$
Its maximum value is $\frac{5}{2}$, and the minimum value is $\frac{3}{2}$.
Thus, the required result is 4.
1.3 Roots of Unity in Complex Numbers
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For a natural number $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and their sum is 17. If there exists a unique $n$ such that $S_{n}$ is also an integer, find $n .{ }^{(4]}$
|
Solve: Regarding $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the modulus of the complex number $(2 k-1)+a_{k} \mathrm{i}$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}} \\
=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} \mathrm{i}\right| \\
\geqslant\left|\sum_{k=1}^{n}\left[(2 k-1)+a_{k} \mathrm{i}\right]\right| \\
=\left|n^{2}+17 \mathrm{i}\right|=\sqrt{n^{4}+17^{2}}
\end{array}
$$
From the given condition, we have
$$
\begin{array}{l}
n^{4}+17^{2}=m^{2}\left(m \in \mathbf{N}_{+}, m=S_{n}\right) \\
\Rightarrow\left(m-n^{2}\right)\left(m+n^{2}\right)=289 \\
\Rightarrow m-n^{2}=1, m+n^{2}=289 \\
\Rightarrow n=12 .
\end{array}
$$
2.2 Application in Trigonometric Problems
The connection between complex numbers and trigonometric functions mainly relies on the exponential form (or trigonometric form) of complex numbers. By using the exponential form and operations (or trigonometric operations) of complex numbers, we can achieve the purpose of trigonometric evaluation and proof.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In rectangle $A B C D$, it is known that $A B=5, B C=9$, points $E, F, G, H$ are on sides $A B, B C, C D, D A$ respectively, such that $A E=C G=3, B F=D H=4, P$ is a point inside the rectangle. If the area of quadrilateral $A E P H$ is 15, then the area of quadrilateral $P F C G$ is $\qquad$
|
4. 11.
As shown in Figure 3, let the distances from $P$ to $AB$ and $AD$ be $a$ and $b$, respectively. Then the distances from $P$ to $BC$ and $CD$ are $5-b$ and $9-a$, respectively. Given that $S_{\text {quadrilateral } AEPH}=\frac{1}{2}(3a+5b)=15$, we have
$$
\begin{array}{l}
S_{\text {quadrilateral PFCG }}=\frac{1}{2}[5(5-b)+3(9-a)] \\
=\frac{1}{2}[52-(3a+5b)]=11 .
\end{array}
$$
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The number of integers $n$ that make $n^{4}-3 n^{2}+9$ a prime number is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
5.4.
If $n=0$, then $n^{4}-3 n^{2}+9=9$ is not a prime number.
If $n>0$, then
$$
\begin{array}{l}
n^{4}-3 n^{2}+9=\left(n^{2}+3\right)^{2}-(3 n)^{2} \\
=\left(n^{2}-3 n+3\right)\left(n^{2}+3 n+3\right) .
\end{array}
$$
Notice that, $n^{2}+3 n+3>3$.
Therefore, $n^{2}-3 n+3=1$.
Solving this gives $n=1$ or 2 (corresponding to $n^{4}-3 n^{2}+9=7$ or 13, both of which are prime numbers).
If $n<0$, then $n=-1$ or -2.
In summary, there are 4 integers $n$ that satisfy the condition.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given any positive integer $a$, define the integer sequence $x_{1}, x_{2}$, $\cdots$, satisfying
$$
x_{1}=a, x_{n}=2 x_{n-1}+1(n \geqslant 1) .
$$
If $y_{n}=2^{x_{n}}-1$, determine the maximum integer $k$ such that there exists a positive integer $a$ for which $y_{1}, y_{2}, \cdots, y_{k}$ are all prime numbers.
|
1. If $y_{i}$ is a prime number, then $x_{i}$ is also a prime number.
Otherwise, if $x_{i}=1$, then $y_{i}=1$ is not a prime number; if $x_{i}=m n($ integers $m, n>1)$, then
$\left(2^{m}-1\right) \mid\left(2^{x_{i}}-1\right)$, i.e., $x_{i}$ and $y_{i}$ are composite numbers.
Below, we prove by contradiction: For any odd prime $a$, at least one of $y_{1}, y_{2}, y_{3}$ is composite.
Otherwise, $x_{1}, x_{2}, x_{3}$ are all primes.
Since $x_{1} \geqslant 3$ is odd, we know
$x_{2}>3$, and $x_{2}=3(\bmod 4)$.
Therefore, $x_{3} \equiv 7(\bmod 8)$.
Thus, 2 is a quadratic residue of $x_{3}$, i.e., there exists $x \in \mathbf{N}_{+}$ such that
$$
x^{2}=2\left(\bmod x_{3}\right) \text {. }
$$
Therefore, $2^{x_{2}}=2^{\frac{x_{3}-1}{2}} \equiv x^{x_{3}-1} \equiv 1\left(\bmod x_{3}^{3}\right)$.
Thus, $x_{3} \mid y_{2}$.
Since $x_{2}>3$, then $2^{x_{2}}-1>2 x_{2}+1=x_{3}$.
Therefore, $y_{2}$ is composite.
Finally, if $a=2$, then
$$
y_{1}=3, y_{2}=31 \text {, and } 23 \mid y_{3}=\left(2^{11}-1\right) \text {. }
$$
Therefore, $k=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies
$$
\begin{array}{l}
f(x+1)=f(-x), \\
f(x)=\left\{\begin{array}{ll}
1, & -1<x \leqslant 0 \\
-1, & 0<x \leqslant 1 .
\end{array}\right.
\end{array}
$$
Then $f(f(3.5))=$ $\qquad$
|
2. -1 .
From $f(x+1)=-f(x)$, we know $f(x+2)=f(x)$.
Then $f(3.5)=f(-0.5)=1$.
Therefore, $f(f(3.5))=f(1)=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $P$ is a moving point on the line $l$:
$$
k x+y+4=0(k>0)
$$
$P A$ and $P B$ are the two tangents from $P$ to the circle $C$:
$$
x^{2}+y^{2}-2 y=0
$$
with points of tangency $A$ and $B$ respectively. If the minimum area of quadrilateral $P A C B$ is 2, then $k=$ $\qquad$
|
6.2.
$$
\begin{array}{l}
\text { Given } S_{\text {quadrilateral } P A C B}=P A \cdot A C=P A \\
=\sqrt{C P^{2}-C A^{2}}=\sqrt{C P^{2}-1},
\end{array}
$$
we know that the area is minimized when $|C P|$ is minimized, i.e., when $C P \perp l$. Also, $\sqrt{C P^{2}-1}=2$, then $C P=\sqrt{5}$. Using the point-to-line distance formula, we get
$$
C P=\sqrt{5}=\frac{5}{\sqrt{1+k^{2}}} \text {. }
$$
Since $k>0$, we have $k=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.